Let \(\lambda \in \varLambda \). By definition of \(\varLambda \), there exists \(T_{\lambda }>100\) such that (1.6) is true. Let \(T_{2}\geqslant e^{900}(1+T_{\lambda })\) be such that
$$\begin{aligned} t^{\frac{2}{3}} \lambda (t) < \frac{t}{100}, \quad t \geqslant T_{2} \end{aligned}$$
(4.1)
Let \(T_{0}\geqslant T_{2}\) be otherwise arbitrary. For the whole paper, we work in the region \(t \geqslant T_{0}\), and C denotes a constant, independent of \(T_{0}\), unless otherwise specified. We define g by
$$\begin{aligned} g(t) := t^{\alpha } \end{aligned}$$
where \(\alpha \) is any number satisfying
$$\begin{aligned} \alpha > \frac{1}{2}+\frac{C_{l}}{2}, \quad \alpha < \frac{2}{3}-\frac{5}{6}C_{u}-\frac{C_{l}}{6} \end{aligned}$$
(4.2)
(Note that (1.7) implies that \(\frac{2}{3} - \frac{5}{6} C_{u}-\frac{C_{l}}{6} > \frac{1}{2}+\frac{C_{l}}{2}\), so such \(\alpha \) as above exists. We also remind the reader that some intuition behind this choice of g is given in Sect. 3.) We note that the definition of \(\varLambda \) implies that, for \(x \geqslant t \geqslant T_{\lambda }\),
$$\begin{aligned}{} & {} \left( \frac{x}{t}\right) ^{-C_{l}} \leqslant \frac{\lambda (x)}{\lambda (t)} \leqslant \left( \frac{x}{t}\right) ^{C_{u}}, \quad t \mapsto \frac{\lambda (t)}{t^{\frac{1}{30}}} \text { is decreasing,} \nonumber \\{} & {} \quad \quad \frac{\sup _{y \in [100,x]}(\lambda (y)\log (y))}{\sup _{y \in [100,t]}(\lambda (y)\log (y))} \leqslant C \left( \frac{x}{t}\right) ^{C_{u}} \frac{\log (x)}{\log (t)} \end{aligned}$$
(4.3)
All of the properties of our ansatz which will be used in the later sections are listed in the following proposition, whose proof is completed in this section.
Proposition 1
[Approximate solution to (1.1)]. There exists \(T_{3}>0\) so that, for all \(T_{0}>T_{3}\), there exists \(u_{ansatz} \in C^{2}([T_{0},\infty ) \times (0,\infty ))\), satisfying Lemma 55 such that, if \(u(t,r)=Q_{\frac{1}{\lambda (t)}}(r)+u_{ansatz}(t,r)\), then,
$$\begin{aligned} -\partial _{t}^{2}u+\partial _{r}^{2}u +\frac{1}{r}\partial _{r}u-\frac{\sin (2u)}{2r^{2}}=-F_{5}(t,r) \end{aligned}$$
where, for
$$\begin{aligned} \delta := \frac{1}{2}\text {min}\{4\alpha -C_{l}-2,1-2C_{u}-\alpha ,4-5C_{u}-6\alpha ,\frac{1}{2}-2C_{l}-15C_{u}\} \end{aligned}$$
and \(C_{5},C \geqslant 0\),
$$\begin{aligned} \begin{aligned}{}&{} \frac{||F_{5}(t,r)||_{L^{2}(r dr)}}{\lambda (t)^{2}} \leqslant \frac{C_{5} \log ^{30}(t)}{t^{4+2\delta }},\\{}&{} ||L_{\frac{1}{\lambda (t)}}F_{5}(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \log ^{2}(t)}{t^{2+5\alpha }}+ \frac{C \log ^{6}(t)}{t^{4+2\alpha -3C_{u}}}+\frac{C \log ^{30}(t)}{t^{9/2-C_{l}-15C_{u}}}, \quad t \geqslant T_{0} \end{aligned} \end{aligned}$$
Moreover, there exists \(v_{rad}\) satisfying
$$\begin{aligned}{} & {} -\partial _{t}^{2}v_{rad}+\partial _{r}^{2}v_{rad}+\frac{1}{r}\partial _{r}v_{rad}-\frac{v_{rad}}{r^{2}}=0, \quad E(v_{rad},\partial _{t}v_{rad})<\infty \nonumber \\{} & {} ||\partial _{t}(Q_{\frac{1}{\lambda (t)}}+u_{ansatz}-v_{rad})||_{L^{2}(r dr)} \leqslant \frac{C \log (t)}{t^{1-C_{u}}}, \quad ||u_{ansatz}-v_{rad}||_{\dot{H}^{1}_{e}} \leqslant \frac{C \log (t)}{t^{1-C_{u}}} \nonumber \\ \end{aligned}$$
(4.4)
Remark
The function \(u_{ansatz}\) is explicitly given in (4.209), where we refer the reader to Sect. 2.1 for the definitions of the constituent functions. \(F_{5}\) is explicitly given in (4.210) and \(v_{rad}\) is given in (4.216), in terms of \(v_{2}, v_{2,2}\), and \(v_{2,4}\), which are defined in (4.37) (with \(v_{2,0}\) given in (4.43)), (4.50) (with \(v_{2,3}\) given in (4.134)), and (4.189) (with \(v_{2,5}\) given in (4.187)), respectively.
We begin by constructing an approximate solution to
$$\begin{aligned} -\partial _{tt}u_{1}+\partial _{rr}u_{1}+\frac{1}{r}\partial _{r}u_{1} - \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}}u_{1} = \partial _{t}^{2}Q_{1}(\frac{r}{\lambda (t)}) \end{aligned}$$
(4.5)
starting by matching explicit solutions to certain approximations of the operator on the left-hand side.
4.1 Small r corrections
4.1.1 First iteration
We first consider the ODE
$$\begin{aligned} \partial _{rr}u_{ell}+\frac{1}{r}\partial _{r}u_{ell}-\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}}u_{ell} = \partial _{t}^{2}Q_{1}(\frac{r}{\lambda (t)}), \quad u_{ell}(t,0)=0 \end{aligned}$$
(4.6)
We get
$$\begin{aligned} u_{ell}(t,r) = v_{ell}(t,\frac{r}{\lambda (t)}) \end{aligned}$$
where
$$\begin{aligned} v_{ell}(t,R) = f_{1}(R) \lambda '(t)^{2} + \lambda (t) \lambda ''(t) f_{2}(R) + \frac{c_{1}(t) R}{1+R^{2}} \end{aligned}$$
(4.7)
and
$$\begin{aligned} f_{1}(R) = \frac{R (-2+R^{2})}{2(1+R^{2})}, \quad f_{2}(R) = \frac{R^{2}(-1+2R^{2}) - (-1+R^{4})\log (1+R^{2}) + 2 R^{2} Li_{2}(-R^{2})}{2R(1+R^{2})} \end{aligned}$$
and \(c_{1}\) will be chosen later.
4.1.2 Second iteration.
We define a second order correction, \(u_{ell,2}(t,r)\), which is useful in the region of small r, by the following solution to
$$\begin{aligned}{} & {} \partial _{rr}u_{ell,2}(t,r)+\frac{1}{r}\partial _{r}u_{ell,2}(t,r)-\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}} u_{ell,2}(t,r) = \partial _{t}^{2}\left( v_{ell}(t,\frac{r}{\lambda (t)})\right) \end{aligned}$$
(4.8)
$$\begin{aligned}{} & {} u_{ell,2}(t,r) = v_{ell,2}(t,\frac{r}{\lambda (t)}) \end{aligned}$$
(4.9)
where \(v_{ell,2}\) is defined by
$$\begin{aligned} v_{ell,2}(t,R)= & {} - \frac{\phi _{0}(R)}{2} \int _{0}^{R} \lambda (t)^{2} \partial _{t}^{2}\left( v_{ell}(t,\frac{r}{\lambda (t)})\right) \Bigr |_{r=s\lambda (t)}s e_{2}(s) ds \nonumber \\{} & {} + e_{2}(R) \int _{0}^{R} \lambda (t)^{2} \partial _{t}^{2}\left( v_{ell}(t,\frac{r}{\lambda (t)})\right) \Bigr |_{r=s\lambda (t)} \frac{s \phi _{0}(s)}{2} ds \end{aligned}$$
(4.10)
In order to compute explicit terms in an expansion of \(v_{ell,2}(t,R)\) for large R, we let
$$\begin{aligned} v_{ell,lgR}(t,R) = \frac{1}{2} R \lambda '(t)^2+R \lambda (t) (1-\log (R)) \lambda ''(t) \end{aligned}$$
(4.11)
be the leading behavior of \(v_{ell}(t,R)\) for large R (recall (4.7)), and define
$$\begin{aligned} err_{0}(t,r) = \lambda (t)^{2} \partial _{t}^{2}\left( v_{ell,lgR}(t,\frac{r}{\lambda (t)})\right) = r \lambda (t)^{2} \partial _{t}^{2}\left( \frac{\lambda '(t)^{2}}{2 \lambda (t)}+\lambda ''(t)-\lambda ''(t) \log (\frac{r}{\lambda (t)})\right) \end{aligned}$$
To understand \(v_{ell,2}(t,R)\) for large R, we let
$$\begin{aligned}\begin{aligned} v_{ell,2,0}(t,R)&= - \frac{\phi _{0}(R)}{2} \int _{0}^{R} err_{0}(t,s\lambda (t))s e_{2}(s) ds + e_{2}(R) \int _{0}^{R} err_{0}(t,s\lambda (t)) \frac{s \phi _{0}(s)}{2} ds\end{aligned} \end{aligned}$$
We get
$$\begin{aligned} \begin{aligned} v_{ell,2,0}(t,R)&= f_{3}(R) \left( 2 \lambda '(t)^{4} - 7 \lambda (t) \lambda '(t)^{2} \lambda ''(t) + 4 \lambda (t)^{2} \lambda ''(t)^{2}\right. \\&\quad \qquad \qquad \left. + 6 \lambda (t)^{2} \lambda '(t)\lambda '''(t)\right) +f_{4}(R) \lambda (t)^{3} \lambda ''''(t) \end{aligned} \end{aligned}$$
(4.12)
where
$$\begin{aligned} f_{3}(R) = \frac{2\left( R^{2}\left( -2+6R^{2}+R^{4}\right) -2(-1+R^{4})\log (1+R^{2})+4R^{2}Li_{2}(-R^{2})\right) }{32(R+R^{3})} \\ \begin{aligned} f_{4}(R)= \frac{1}{32(R+R^{3})}&\left( R^{2}\left( -12+44R^{2}+7R^{4}-4(-2+6R^{2}+R^{4})\log (R)\right) \right. \\&\left. +8\left( -1+R^{4}\right) (-1+\log (R))\log (1+R^{2})\right. \\&\left. +4\left( -1+4R^{2}+R^{4}-4R^{2}\log (R)\right) Li_{2}(-R^{2})+16R^{2}Li_{3}(-R^{2})\right) \end{aligned} \end{aligned}$$
For later convenience, we split
$$\begin{aligned} v_{ell,2,0}(t,\frac{r}{\lambda (t)})=v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})+soln_{1}(t,\frac{r}{\lambda (t)}) \end{aligned}$$
and write
$$\begin{aligned} v_{ell,2}(t,R) = v_{ell,2,0,main}(t,R)+soln_{1}(t,R)+soln_{2}(t,R) \end{aligned}$$
(4.13)
where
$$\begin{aligned}{} & {} v_{ell,2,0,main}(t,s) = \frac{s^{3} \lambda (t)^{3}}{8}\left( \partial _{t}^{2}\left( \frac{\lambda '(t)^{2}}{2\lambda (t)}+\lambda ''(t)+\lambda ''(t)\log (\lambda (t))\right) \right. \nonumber \\{} & {} \qquad \qquad \quad \qquad \qquad \quad \left. -\lambda ''''(t)\log (s\lambda (t)) + \frac{3}{4} \lambda ''''(t)\right) \end{aligned}$$
(4.14)
$$\begin{aligned}{} & {} \begin{aligned} soln_{1}(t,s)&= \left( f_{3}(s)-f_{3,0}(s)\right) \left( 2\lambda '(t)^{4}-7 \lambda (t) \lambda '(t)^{2}\lambda ''(t) + 4 \lambda (t)^{2}\lambda ''(t)^{2}\right. \\&\quad \left. +6 \lambda (t)^{2}\lambda '(t)\lambda '''(t)\right) + \left( f_{4}(s)-f_{4,0}(s)\right) \lambda (t)^{3} \lambda ''''(t)\end{aligned}\nonumber \\ \end{aligned}$$
(4.15)
and
$$\begin{aligned} f_{3,0}(x) = \frac{x^{3}}{16}, \quad f_{4,0}(x) = \frac{1}{32} x^3 (7-4 \log (x)) \end{aligned}$$
Finally,
$$\begin{aligned}{} & {} soln_{2}(t,\frac{s}{\lambda (t)}) = u_{ell,2}(t,s) - v_{ell,2,0,main}(t,\frac{s}{\lambda (t)})-soln_{1}(t,\frac{s}{\lambda (t)})\nonumber \\{} & {} \quad =-\frac{\phi _{0}(\frac{s}{\lambda (t)})}{2}\int _{0}^{\frac{s}{\lambda (t)}} err_{1}(t,q\lambda (t)) q e_{2}(q)dq+\frac{e_{2}(\frac{s}{\lambda (t)})}{2}\int _{0}^{\frac{s}{\lambda (t)}} err_{1}(t,q\lambda (t)) q \phi _{0}(q)dq \end{aligned}$$
(4.16)
where
$$\begin{aligned} err_{1}(t,r)=\lambda (t)^{2}\partial _{t}^{2}\left( v_{ell}(t,\frac{r}{\lambda (t)})-v_{ell,lgR}(t,\frac{r}{\lambda (t)})\right) \end{aligned}$$
Note in particular, that
$$\begin{aligned} v_{ell,2,0}(t,\frac{r}{\lambda (t)}){} & {} = \frac{r^{3}}{8} \partial _{t}^{2}\left( \frac{\lambda '(t)^{2}}{2\lambda (t)}+\lambda ''(t)-\lambda ''(t) \log (\frac{r}{\lambda (t)})\right) + \frac{3}{32} r^{3} \lambda ''''(t)\nonumber \\{} & {} \quad +O\left( r \log ^{2}(r)\right) , \quad r \rightarrow \infty \end{aligned}$$
(4.17)
This observation will be important when we match the small r corrections to the large r corrections. Also, the leading behavior of \(err_{1}(t,r)\) in the region \(r \geqslant \lambda (t)\) is \(err_{1,0}\) given by
$$\begin{aligned} err_{1,0}(t,r){} & {} = \lambda (t)^{2} \partial _{t}^{2}\left( \lambda (t)\left( \frac{-3}{2r}\lambda '(t)^{2}\right. \right. \nonumber \\{} & {} \quad \left. \left. +\frac{\lambda (t)\lambda ''(t)}{r}\left( -(2+\frac{\pi ^{2}}{6})+ \log (\frac{r}{\lambda (t)}) -2 \log ^{2}(\frac{r}{\lambda (t)}) \right) +\frac{c_{1}(t)}{r}\right) \right) \nonumber \\ \end{aligned}$$
(4.18)
4.2 First wave iteration
We consider the PDE
$$\begin{aligned} -\partial _{tt}u_{w}+\partial _{rr}u_{w}+\frac{1}{r}\partial _{r}u_{w}-\frac{u_{w}}{r^{2}} = \partial _{t}^{2} Q_{1}(\frac{r}{\lambda (t)}) \end{aligned}$$
(4.19)
We consider the solution \(u_{w}\), that can be written as
$$\begin{aligned} u_{w}(t,r) = v_{1}(t,r) + v_{2}(t,r) \end{aligned}$$
(4.20)
where \(v_{1}\) solves
$$\begin{aligned} -\partial _{tt}v_{1}+\partial _{rr}v_{1}+\frac{1}{r}\partial _{r}v_{1}-\frac{v_{1}}{r^{2}} = \partial _{t}^{2} Q_{1}(\frac{r}{\lambda (t)}) \end{aligned}$$
with 0 Cauchy data at infinity, and \(v_{2}\) solves
$$\begin{aligned} {\left\{ \begin{array}{ll}-\partial _{tt}v_{2}+\partial _{rr}v_{2}+\frac{1}{r}\partial _{r}v_{2}-\frac{v_{2}}{r^{2}} = 0\\ v_{2}(0)=0\\ \partial _{t}v_{2}(0,r)=v_{2,0}(r)\end{array}\right. } \end{aligned}$$
where \(v_{2,0}\in L^{2}(r dr)\) will be chosen later. To describe \(v_{1}(t,r)\), we decompose it as
$$\begin{aligned} v_{1}(t,r) = w_{1}(t,r) + v_{ex}(t,r) \end{aligned}$$
(4.21)
where \(w_{1}\) solves
$$\begin{aligned} -\partial _{tt}w_{1}+\partial _{rr}w_{1}+\frac{1}{r}\partial _{r}w_{1}-\frac{w_{1}}{r^{2}} = \frac{-2 \lambda ''(t)}{r} \end{aligned}$$
(4.22)
with 0 Cauchy data at infinity. We use the same procedure used in Section 4.2 of [25] to obtain (4.25) of [25], to derive the following expression which can be directly checked to be the solution to (4.22) with 0 Cauchy data at infinity.
$$\begin{aligned} \begin{aligned}w_{1}(t,r)&= \frac{-1}{2 \pi }\int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{2\pi } \left( -2\lambda ''(s)\right) \frac{\left( r+\rho \cos (\theta )\right) }{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )} d\theta \\&= \int _{t}^{\infty } ds \frac{\lambda ''(s)}{r} \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}}\left( 1+\text {sgn}(r^{2}-\rho ^{2})\right) \\&= \frac{2}{r} \int _{t}^{t+r} ds \lambda ''(s) (s-t) + \frac{2}{r} \int _{t+r}^{\infty } ds \lambda ''(s) \left( (s-t) - \sqrt{(s-t)^{2}-r^{2}}\right) \\&=\frac{-2}{r}\left( \lambda (t+r)-\lambda (t)-r\lambda '(t+r)\right) + \frac{2}{r} \int _{t+r}^{\infty } ds \lambda ''(s) \left( (s-t) - \sqrt{(s-t)^{2}-r^{2}}\right) \end{aligned}\nonumber \\ \end{aligned}$$
(4.23)
(For \(r \ne \rho \), we can evaluate the following integral using Cauchy’s residue theorem).
$$\begin{aligned} \int _0^{2 \pi } \frac{\rho \cos (\theta )+r}{\rho ^2+2 \rho r \cos (\theta )+r^2} \, d\theta = \frac{\pi \left( \text {sgn}\left( r^2-\rho ^2\right) +1\right) }{r} \end{aligned}$$
First, we note that \(w_{1}(t,\cdot ) \in C^{\infty }((0,\infty ))\). This follows from
$$\begin{aligned} w_{1}(t,r) = \frac{-2}{r}\left( \lambda (t+r)-\lambda (t)-r\lambda '(t+r)\right) + 2 r \int _{1}^{\infty } \lambda ''(t+r y)\left( y-\sqrt{y^{2}-1}\right) dy \qquad \end{aligned}$$
(4.24)
and the smoothness and symbol type estimates on \(\lambda \). Next, we introduce some notation. Let
$$\begin{aligned}{} & {} \begin{aligned} w_{1,main}(t,r)&= r\left( \log (2)+\frac{1}{2}\right) \lambda ''(t) + r \int _{t}^{2t} \frac{(\lambda ''(s) - \lambda ''(t))}{s-t} ds + r \lambda ''(t) \log (\frac{t}{r})\\&\quad + r \int _{2t}^{\infty } \frac{\lambda ''(s) ds}{(s-t)} \end{aligned} \end{aligned}$$
(4.25)
$$\begin{aligned}{} & {} w_{1,sub}(t,r) = w_{1}(t,r) -w_{1,main}(t,r)\nonumber \\{} & {} \begin{aligned}&w_{1,cubic,main}(t,r) \\&\quad =\frac{3}{32} r^{3} \lambda ''''(t)+\frac{r^{3}}{8}\left( \lambda ''''(t) \left( \log (2)+\frac{1}{2}\right) -\log (r) \lambda ''''(t) + \log (t) \lambda ''''(t)\right. \\&\qquad \left. + \int _{t}^{2t} \frac{\lambda ''''(s)-\lambda ''''(t)}{s-t} ds + \int _{2t}^{\infty } \frac{\lambda ''''(s) ds}{s-t}\right) \end{aligned} \end{aligned}$$
(4.26)
The following lemma will be useful later on when we match the small r and large r corrections.
Lemma 1
For \(0 \leqslant j \leqslant 8\), \(0 \leqslant k \leqslant 3\) and \(r \leqslant t\),
$$\begin{aligned}\begin{aligned}&\Bigr |\partial _{t}^{j}\partial _{r}^{k}\left( w_{1,sub}(t,r)-w_{1,cubic,main}(t,r)-\frac{r^{5} \lambda ^{(6)}(t)}{576}\left( 5+\log (8)\right) \right. \\&\left. -\frac{r^{5}}{192} \left( \int _{t}^{2t} \frac{\left( \lambda ^{(6)}(s)-\lambda ^{(6)}(t)\right) ds}{s-t} + \lambda ^{(6)}(t) \log (\frac{t}{r})+\int _{2t}^{\infty } \frac{\lambda ^{(6)}(s) ds}{s-t}\right) \right) \Bigr | \\&\leqslant \frac{C \lambda (t) r^{7-k}}{t^{8+j}} \left( \log (t)+|\log (r)|\right) \end{aligned} \end{aligned}$$
Proof
As a first step, we start with (4.24), and subtract and add the Taylor polynomial centered at 0, of degree 3, of \(\lambda (t+r)-\lambda (t)-r \lambda '(t+r) + \frac{r^{2}}{2} \lambda ''(t+r)\), regarded as a function of r for each fixed t. For the integral term of (4.24), we first note that
$$\begin{aligned} y-\sqrt{y^{2}-1} = \frac{1}{2y} + O\left( \frac{1}{y^{3}}\right) \end{aligned}$$
Therefore, we have
$$\begin{aligned}{} & {} 2 r \int _{1}^{\infty } \lambda ''(t+ry) (y-\sqrt{y^{2}-1}) dy = 2 r \int _{1}^{\infty } \frac{\lambda ''(t+r y)}{2y} dy \nonumber \\{} & {} \quad + 2 r \int _{1}^{\infty } \lambda ''(t+r y) \left( y-\sqrt{y^{2}-1}-\frac{1}{2y}\right) dy \end{aligned}$$
(4.27)
After integrating by parts twice in the second integral on the right-hand side, we get the following expression for \(w_{1}\), whose second line is equal to \(w_{1,main}(t,r)\), and the other terms vanish faster than \(O(r \log (r))\) as r approaches zero.
$$\begin{aligned} w_{1}(t,r)= & {} \frac{-2}{r} \left( \lambda (t+r) - \lambda (t) -r \lambda '(t+r) +\frac{r^{2}}{2} \lambda ''(t+r) - \frac{r^{3}}{6} \lambda '''(t)\right) \nonumber \\{} & {} +r\left( \log (2)+\frac{1}{2}\right) \lambda ''(t) \nonumber \\{} & {} + r \int _{t}^{2t} \frac{(\lambda ''(s) - \lambda ''(t))}{s-t} ds + r \lambda ''(t) \log (\frac{t}{r}) + r \int _{2t}^{\infty } \frac{\lambda ''(s) ds}{(s-t)}\nonumber \\{} & {} +r (\log (2) + \frac{1}{2}) \left( \lambda ''(t+r) - \lambda ''(t) - r \lambda '''(t)\right) \nonumber \\{} & {} -\left( \frac{r^{2}}{6} - r^{2} (1+\log (\frac{1}{2}))\right) \left( \lambda '''(t+r) - \lambda '''(t)\right) \nonumber \\{} & {} - r \int _{t}^{t+r} \frac{\lambda ''(s) - \lambda ''(t) - (s-t) \lambda '''(t)}{(s-t)} ds\nonumber \\{} & {} +\frac{1}{r} \int _{t+r}^{\infty } \lambda ''''(s) \left( \frac{(s-t)^{3}}{3} \left( 1-\left( 1-\frac{r^{2}}{(s-t)^{2}}\right) ^{3/2} - \frac{3 r^{2}}{2 (s-t)^{2}}\right) \right. \nonumber \\{} & {} \left. + r^{2}(s-t) \left( 1-\sqrt{1-\frac{r^{2}}{(s-t)^{2}}}+\log \left( 1+\frac{\sqrt{1-\frac{r^{2}}{(s-t)^{2}}}-1}{2}\right) \right) \right) ds\nonumber \\ \end{aligned}$$
(4.28)
Repeating this process, we make manifest the terms composing \(w_{1,cubic,main}(t,r)\), and recall the notation for \(P_{k,a}(f)\) given just above (2.2).
$$\begin{aligned}{} & {} \begin{aligned} w_{1}(t,r)&= r \left( \log (2)+\frac{1}{2}\right) \lambda ''(t) + r \int _{t}^{2t} \frac{\left( \lambda ''(s)-\lambda ''(t)\right) }{s-t} ds + r \lambda ''(t) \log (\frac{t}{r}) \\&\quad + r \int _{2t}^{\infty } \frac{\lambda ''(s) ds}{s-t} -\frac{2}{r}\left( \lambda (t+r)-\lambda (t)-r \lambda '(t+r)+\frac{r^{2}}{2} \lambda ''(t+r)-\frac{r^{3}}{6} \lambda '''(t) \right. \\&\quad \left. - \frac{r^{4}}{8} \lambda ''''(t) - \frac{r^{5}}{20} \lambda '''''(t)\right) \\&+\frac{r^{3}}{8} \lambda ''''(t) \left( \log (2)+\frac{1}{2}\right) +\frac{3}{32} r^{3} \lambda ''''(t)\\&+r\left( \log (2)+\frac{1}{2}\right) \left( \lambda ''(t+r)-P_{3,0}(r \mapsto \lambda ''(t+r))\right) \\&-r^{2}\left( \frac{-5}{6}+\log (2)\right) \left( \lambda '''(t+r)-P_{2,0}(r\mapsto \lambda '''(t+r))\right) \\&-r \int _{t}^{t+r} \frac{\left( \lambda ''(s)-P_{3,t}(\lambda '')(s)\right) }{s-t}ds\\&-\frac{r^{3}}{8} \log (r) \lambda ''''(t) + \frac{r^{3}}{8} \log (t)\lambda ''''(t) + \frac{r^{3}}{8} \int _{t}^{2t} \frac{\lambda ''''(s)-\lambda ''''(t)}{s-t} ds\\&+\frac{r^{3}}{8} \int _{2t}^{\infty } \frac{\lambda ''''(s)}{s-t} ds - \frac{r^{3}}{8} \int _{t}^{t+r} \frac{\lambda ''''(s)-\lambda ''''(t)-(s-t)\lambda '''''(t)}{s-t} ds\\&-\frac{r^{3}}{96} \left( 41-60\log (2)\right) \left( \lambda ''''(t+r)-\lambda ''''(t)-r\lambda '''''(t)\right) \\&+ \left( \lambda '''''(t+r)-\lambda '''''(t)\right) r^{4} \frac{(299-420\log (2))}{1440}+\frac{1}{r}\int _{t+r}^{\infty }\lambda ''''''(s) K(r,s-t) ds\end{aligned} \end{aligned}$$
(4.29)
$$\begin{aligned}{} & {} \begin{aligned}K(r,w)&=\frac{1}{1440} \left( 24 w^{5} \left( 1-\left( \sqrt{1-\frac{r^{2}}{w^{2}}}+\frac{r^{2}}{2w^{2}}+\frac{r^{4}}{8w^{4}}\right) \right) \right. \\&\quad +332 r^{2}w^{3} \left( 1-\left( \sqrt{1-\frac{r^{2}}{w^{2}}}+\frac{r^{2}}{2w^{2}}\right) \right) \\&\left. -64 r^{4} w\left( \sqrt{1-\frac{r^{2}}{w^{2}}}-1\right) + 240 r^{2} w^{3} \left( \log \left( \frac{1}{2}\left( 1+\sqrt{1-\frac{r^{2}}{w^{2}}}\right) \right) +\frac{r^{2}}{4w^{2}}\right) \right. \\&\left. +180 r^{4} w \log \left( \frac{1}{2}\left( 1+\sqrt{1-\frac{r^{2}}{w^{2}}}\right) \right) \right) \end{aligned}\nonumber \\ \end{aligned}$$
(4.30)
Finally, to prove the lemma, we treat (4.29) line by line. We show in detail how we obtain the leading part of terms of different forms. In the following computations, we work in the region \(r \leqslant t\). By Taylor’s theorem, we have
$$\begin{aligned}{} & {} |\frac{-2}{r}\left( \lambda (t+r)-\lambda (t)-r\lambda '(t+r)+\frac{r^{2}}{2}\lambda ''(t+r)-\frac{r^{3}}{6} \lambda '''(t)\right. \\{} & {} \qquad \left. -\frac{r^{4}}{8} \lambda ''''(t)-\frac{r^{5}}{20} \lambda '''''(t)\right) -\left( \frac{-\lambda ^{(6)}(t) r^{5}}{36} - \frac{\lambda ^{(7)}(t) r^{6}}{168}\right) |\\{} & {} \quad \leqslant C \frac{r^{7}\lambda (t)}{t^{8}} \end{aligned}$$
where we used (4.3), and the fact that \(r \leqslant t\). Other terms in (4.29) of the same form are treated with the same argument. Next, we have
$$\begin{aligned} -r \int _{t}^{t+r} \frac{ds}{(s-t)} \left( \frac{(s-t)^{4}}{24}\lambda ^{(6)}(t)+\frac{(s-t)^{5}}{120} \lambda ^{(7)}(t)\right) =-r \left( \frac{r^{4}}{4 \cdot 24} \lambda ^{(6)}(t) + \frac{r^{5}}{600} \lambda ^{(7)}(t)\right) \end{aligned}$$
and
$$\begin{aligned}\begin{aligned}&\Bigl |-r \int _{t}^{t+r} \frac{\left( \lambda ''(s)-P_{3,t}(\lambda '')(s)\right) }{s-t}ds-\left( -r \int _{t}^{t+r} \frac{ds}{(s-t)} \left( \frac{(s-t)^{4}}{24}\lambda ^{(6)}(t)+\frac{(s-t)^{5}}{120} \lambda ^{(7)}(t)\right) \right) \Bigr |\\&= |-r \int _{t}^{t+r} \frac{\left( \lambda ''(s)-P_{5,t}(\lambda '')(s)\right) ds}{s-t}|\leqslant C \frac{\lambda (t) r^{7}}{t^{8}}\end{aligned} \end{aligned}$$
where we again used Taylor’s theorem and (4.3). Again, other terms in (4.29) of the same form are treated with the same argument. Finally, we recall (4.30), and note that
$$\begin{aligned} K(r,w) - \frac{r^6}{192 w} = O\left( \frac{r^{8}}{w^{3}}\right) , \quad w \rightarrow \infty \end{aligned}$$
Therefore, with the same procedure used in (4.27),
$$\begin{aligned} \begin{aligned}&\frac{1}{r}\int _{t+r}^{\infty } \lambda ^{(6)}(s) K(r,s-t) ds\\&\quad =\frac{r^{5}}{192} \left( \int _{t}^{2t} \frac{\left( \lambda ^{(6)}(s)-\lambda ^{(6)}(t)\right) }{(s-t)} ds + \lambda ^{(6)}(t) \log (\frac{t}{r}) + \int _{2t}^{\infty } \frac{\lambda ^{(6)}(s)}{(s-t)} ds\right) \\&\qquad -\frac{r^{5}}{192} \int _{t}^{t+r} \frac{\left( \lambda ^{(6)}(s)-\lambda ^{(6)}(t)-(s-t)\lambda ^{(7)}(t)\right) }{(s-t)} ds - \frac{r^{6}}{192}\lambda ^{(7)}(t) - \lambda ^{(6)}(t) \frac{r^{5} (218-315 \log (2))}{2880} \\&\qquad +\lambda ^{(7)}(t) r^{6} \frac{(2413-3465 \log (2))}{100800} + \frac{\left( \lambda ^{(7)}(t+r)-\lambda ^{(7)}(t)\right) }{100800}\cdot r^{6}\left( 2413-3465 \log (2)\right) \\&\qquad -\left( \lambda ^{(6)}(t+r)-\lambda ^{(6)}(t)-r \lambda ^{(7)}(t)\right) r^{5} \frac{(218-315 \log (2))}{2880}-r^{6} \lambda ^{(7)}(t)\frac{(218-315 \log (2))}{2880}+\text {Err}\end{aligned} \end{aligned}$$
where
$$\begin{aligned} |\text {Err}| \leqslant \frac{1}{r} \int _{t+r}^{\infty } |W(s) \lambda ^{(8)}(s)| ds \end{aligned}$$
and, for \(y=\frac{s-t}{r}\)
$$\begin{aligned} W(s)= & {} \frac{r^7}{2880} \left( -\frac{128}{35} \left( \sqrt{1-\frac{1}{y^2}}-1\right) y+15 y \log \left( \frac{1}{2} \sqrt{1-\frac{1}{y^2}}+\frac{1}{2}\right) \right. \\{} & {} -\frac{1779}{35} \left( \sqrt{1-\frac{1}{y^2}}+\frac{1}{2 y^2}-1\right) y^3\\{} & {} -\frac{1518}{35} \left( \sqrt{1-\frac{1}{y^2}}-\left( -\frac{1}{8 y^4}-\frac{1}{2 y^2}+1\right) \right) y^5\\{} & {} -\frac{8}{7} \left( \sqrt{1-\frac{1}{y^2}}+\left( \frac{1}{16 y^6}+\frac{1}{8 y^4}+\frac{1}{2 y^2}\right) -1\right) y^7\\{} & {} \left. +12 \left( 2 y^5+5 y^3\right) \left( \frac{3}{32 y^4}+\frac{1}{4 y^2}+\log \left( \frac{1}{2} \sqrt{1-\frac{1}{y^2}}+\frac{1}{2}\right) \right) -\frac{45}{8 y}\right) \end{aligned}$$
We finish the proof by noting that
$$\begin{aligned} W(s) = O\left( \frac{r^{8}}{(s-t)}\right) , \quad s \rightarrow \infty \end{aligned}$$
and estimating \(\text {Err}\). The higher derivatives are treated similarly. \(\square \)
We remark that \(w_{1,main}(t,r)\), along with the expression of \(u_{ell}\) will end up determining the data for the free wave \(v_{2}\), as \(w_{1,main}(t,r)\) will turn out to be the leading contribution of \(w_{1}\) in the matching region.
We also record some pointwise estimates on \(w_{1}\) and some of its derivatives.
Lemma 2
We have the following estimates. For \(0 \leqslant k \leqslant 8\) and \(0 \leqslant j \leqslant 3\),
$$\begin{aligned} |\partial _{t}^{k}\partial _{r}^{j}w_{1}(t,r)| \leqslant {\left\{ \begin{array}{ll} \frac{C r^{1-j} \lambda (t) \left( \log (t)+|\log (r)|\right) }{t^{2+k}}, \quad r \leqslant t\\ \frac{C}{r^{1+j} t^{k}} \sup _{x \in [t,t+r]}\left( \lambda (x)\right) , \quad r \geqslant t\end{array}\right. } \end{aligned}$$
(4.31)
Proof
By Lemma 1, it suffices to consider (4.24) in the region \(r \geqslant t\):
$$\begin{aligned}\begin{aligned}&|\frac{2}{r}\left( \lambda '(t+r) r -(\lambda (t+r)-\lambda (t))\right) | \\&\quad \leqslant \frac{C}{r}\left( \lambda (t+r)+\lambda (t)+ r \frac{\lambda (t+r)}{t+r}\right) \leqslant \frac{C}{r}\left( \lambda (t+r)+\lambda (t)\right) \end{aligned} \end{aligned}$$
Then, we have
$$\begin{aligned}\begin{aligned}&|2r \int _{1}^{\infty } \lambda ''(t+r y) \left( y-\sqrt{y^{2}-1}\right) dy| \leqslant \frac{C r \lambda (t+r)}{t+r} \int _{1}^{\infty } \frac{dy}{y(t+r y)}\\ {}&\quad \leqslant \frac{C r \lambda (t+r)}{r (t+r)} \frac{\log (1+\frac{t}{r})}{\frac{t}{r}} \leqslant \frac{C \lambda (t+r)}{t+r}, \quad r \geqslant t\end{aligned} \end{aligned}$$
where we used (4.3). Then, using the symbol type estimates on \(\lambda \), we get (4.31) for \(r \geqslant t\). \(\square \)
Next, we study \(v_{ex}\) ((4.21)) which solves the following equation with 0 Cauchy data at infinity.
$$\begin{aligned} -\partial _{t}^{2}v_{ex}+\partial _{r}^{2}v_{ex}+\frac{1}{r}\partial _{r}v_{ex}-\frac{v_{ex}}{r^{2}}= \partial _{t}^{2} Q_{1}(\frac{r}{\lambda (t)}) + \frac{2 \lambda ''(t)}{r}:=RHS(t,r) \end{aligned}$$
(4.32)
From integral identities of [9], we have
$$\begin{aligned} \widehat{RHS}(t,\xi ) = 2 \xi K_{0}(\xi \lambda (t))\lambda (t) \lambda '(t)^{2} + \frac{2\lambda ''(t)}{\xi } \left( 1-\xi \lambda (t) K_{1}(\xi \lambda (t))\right) \end{aligned}$$
(4.33)
We have the following representation formula for \(v_{ex}\)
Lemma 3
For \(r >0\), \(v_{ex}\) is given by
$$\begin{aligned} v_{ex}(t,r)= & {} \int _{t}^{\infty } ds \int _{0}^{\infty } d\xi J_{1}(r\xi ) \sin ((t-s)\xi ) \widehat{RHS}(s,\xi )\nonumber \\= & {} - \int _{0}^{\infty } \frac{J_{1}(r\xi )}{\xi } \widehat{RHS}(t,\xi ) d\xi -\int _{0}^{\infty } d\xi J_{1}(r\xi ) \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi ^{2}} \partial _{s}^{2}\widehat{RHS}(s,\xi )\nonumber \\ \end{aligned}$$
(4.34)
Proof
From asymptotics of \(K_{1}, K_{0}\) (see [9]), we get, for \(k=0,1,2\),
$$\begin{aligned} |\partial _{s}^{k}\widehat{RHS}(s,\xi )| \leqslant C\frac{\lambda (s)^{2}}{s^{2+k}} \frac{\xi \lambda (s)}{1+\xi ^{2}\lambda (s)^{2}} \langle \log (\xi \lambda (s))\rangle \end{aligned}$$
(4.35)
Using
$$\begin{aligned} |J_{1}(r\xi )| \leqslant \frac{C}{\sqrt{r \xi }}, \quad r >0 \end{aligned}$$
the first integral on the right-hand side of (4.34) converges absolutely, for \(r >0\). Therefore, by the Fubini theorem, and (4.35), we get
$$\begin{aligned}\begin{aligned}&\int _{t}^{\infty } ds \int _{0}^{\infty } d\xi J_{1}(r\xi ) \sin ((t-s)\xi ) \widehat{RHS}(s,\xi ) \\&\quad = \int _{0}^{\infty } d\xi \int _{t}^{\infty }ds J_{1}(r\xi ) \partial _{s}\left( \frac{\cos ((t-s)\xi )}{\xi }\right) \widehat{RHS}(s,\xi )\\&\quad =- \int _{0}^{\infty } \frac{J_{1}(r\xi )}{\xi } \widehat{RHS}(t,\xi ) d\xi -\int _{0}^{\infty } d\xi J_{1}(r\xi ) \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi ^{2}} \partial _{s}^{2}\widehat{RHS}(s,\xi )\end{aligned} \end{aligned}$$
Moreover, using
$$\begin{aligned} |\sin ((t-s)\xi )|\leqslant 1, \quad |J_{1}'(x)| \leqslant C, \quad |J_{1}''(x)| \leqslant \frac{C}{\sqrt{x}}, \quad x >0, \text { and } (4.35) \end{aligned}$$
the dominated convergence theorem shows that, for all \(r_{0}>0\), and \(k=1,2\),
$$\begin{aligned}{} & {} \partial _{r}^{k}\left( \int _{0}^{\infty } d\xi J_{1}(r\xi ) \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi ^{2}} \partial _{s}^{2}\widehat{RHS}(s,\xi )\right) \Bigr |_{r=r_{0}} \\{} & {} \quad = \int _{0}^{\infty } d\xi \partial _{r}^{k}(J_{1}(r\xi ))|_{r=r_{0}} \cdot \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi ^{2}} \partial _{s}^{2}\widehat{RHS}(s,\xi ) \end{aligned}$$
A similar argument shows that we can differentiate (up to) two times under the integral sign in t. Then, by the fact that the first term on the second line of (4.34) is a solution to the following ODE (which follows, for instance, by direct computation of the integral, see (4.36))
$$\begin{aligned} \partial _{r}^{2}f + \frac{1}{r} \partial _{r}f - \frac{f}{r^{2}} = RHS(t,r) \end{aligned}$$
we see that \(v_{ex}\), the solution to (4.32) with zero Cauchy data at infinity, is given by (4.34). \(\square \)
Note that we could not have done the integration by parts in the s variable to get from the first to the second line of (4.34) if RHS was simply equal to \(\partial _{t}^{2}Q_{1}(\frac{r}{\lambda (t)})\), because there would be too large of a singularity of the resulting integrand at low frequencies. Also, we have
$$\begin{aligned}{} & {} v_{ex,ell}(t,r):=-\int _{0}^{\infty } \frac{J_{1}(r\xi )}{\xi } \widehat{RHS}(t,\xi ) d\xi \nonumber \\{} & {} \quad = \frac{-\left( 2 \lambda (t) \log (1+\frac{r^{2}}{\lambda (t)^{2}})\lambda '(t)^{2} + \left( \lambda (t)^{2} \log (1+\frac{r^{2}}{\lambda (t)^{2}})+r^{2} \log (1+\frac{\lambda (t)^{2}}{r^{2}})\right) \lambda ''(t)\right) }{2 r}\nonumber \\ \end{aligned}$$
(4.36)
Finally, we consider \(v_{2}\) solving
$$\begin{aligned} {\left\{ \begin{array}{ll} -\partial _{tt}v_{2}+\partial _{rr}v_{2}+\frac{1}{r}\partial _{r}v_{2}-\frac{v_{2}}{r^{2}} = 0\\ v_{2}(0)=0\\ \partial _{t}v_{2}(0)=v_{2,0}\end{array}\right. } \end{aligned}$$
with \(v_{2,0}\) not yet chosen. For \(\widehat{v_{2,0}}\in C^{0}((0,\infty ))\) such that \(|\widehat{v_{2,0}}(\xi )| \leqslant C \left( \sqrt{\xi }(1+\xi )^{5/2}\right) ^{-1}\), we have
$$\begin{aligned} \begin{aligned}v_{2}(t,r)&= \int _{0}^{\infty } J_{1}(r\xi ) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi \\&= -\frac{r}{4 \pi } \int _{0}^{\pi } \sin ^{2}(\theta ) \left( F(t+r\cos (\theta ))+F(t-r\cos (\theta ))\right) d\theta \\&=\frac{-r}{4} F(t) -\frac{r^{3}}{32}F''(t) \\&-\frac{r}{2\pi } \int _{0}^{\pi } \sin ^{2}(\theta ) \left( F(t+r \cos (\theta ))-F(t)-\frac{r^{2}}{2} \cos ^{2}(\theta ) F''(t)\right) d\theta \end{aligned} \end{aligned}$$
(4.37)
where
$$\begin{aligned} F(t) = -2 \int _{0}^{\infty } \xi \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi \end{aligned}$$
(4.38)
and we used
$$\begin{aligned} J_{1}(x) = \frac{x}{\pi } \int _{0}^{\pi } \cos (x\cos (\theta ))\sin ^{2}(\theta ) d\theta \end{aligned}$$
4.3 First order matching
The error of \(u_{ell}\) in solving (4.5) is \(\partial _{t}^{2} u_{ell}(t,r)\), which is large for large values of r, because of the growth of \(u_{ell}(t,r)\) for large r. On the other hand, the error of \(u_{w}\) in solving (4.5) is \(\left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) u_{w}(t,r)\). The largest contributions to this error term arise from substituting the second line of (4.28) plus (4.37), expanded for \(r \ll t\) into \(u_{w}\) in the error term. Therefore, we choose the data for \(v_{2}\) in order to match the largest contributions to the error terms of \(u_{ell}\) and \(u_{w}\). In particular, \(v_{ell,lgR}\), defined in (4.11) and recalled here makes the largest contribution to the error term of \(u_{ell}\):
$$\begin{aligned} v_{ell,lgR}(t,R) = R \left( \frac{1}{2}\lambda '(t)^{2} + \lambda (t) \lambda ''(t) - \lambda (t) \lambda ''(t) \log (R)\right) \end{aligned}$$
(4.39)
Also, as per Lemma 1, the main contribution from \(w_{1}(t,r)\) is
$$\begin{aligned} w_{1,main}(t,r)= & {} r\left( \log (2)+\frac{1}{2}\right) \lambda ''(t) + r \int _{t}^{2t} \frac{(\lambda ''(s) - \lambda ''(t))}{s-t} ds \\{} & {} + r \lambda ''(t) \log (\frac{t}{r}) + r \int _{2t}^{\infty } \frac{\lambda ''(s) ds}{(s-t)} \end{aligned}$$
and, by (4.37), the main contribution from \(v_{2}\) is
$$\begin{aligned} v_{2,main}(t,r) = \frac{-r}{4} \left( -2 \int _{0}^{\infty } \xi \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi \right) \end{aligned}$$
(4.40)
Note that the \(r \log (r)\) terms from \(w_{1,main}\) and (4.39) (where \(R = \frac{r}{\lambda (t)}\)) are the same. Therefore, it is possible to choose \(v_{2,0}\) so that (for example, for all \(t \geqslant 2 T_{\lambda }\))
$$\begin{aligned} \frac{r}{\lambda (t)} \left( \frac{1}{2}\lambda '(t)^{2} + \lambda (t) \lambda ''(t) - \lambda (t) \lambda ''(t) \log (\frac{r}{\lambda (t)})\right) =v_{2,main}(t,r)+w_{1,main}(t,r) \end{aligned}$$
In other words, we choose \(v_{2,0}\) to satisfy, for all \(t \geqslant 2 T_{\lambda }\),
$$\begin{aligned}{} & {} -2 \int _{0}^{\infty } \xi \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi \nonumber \\{} & {} \quad = 4\left( \left( \log (2)-\frac{1}{2}\right) \lambda ''(t) + \int _{t}^{2t} \left( \frac{\lambda ''(s)-\lambda ''(t)}{s-t}\right) ds+\lambda ''(t) \log (\frac{t}{\lambda (t)}) \right. \nonumber \\{} & {} \qquad \left. + \int _{2t}^{\infty } \frac{\lambda ''(s) ds}{s-t} - \frac{\lambda '(t)^{2}}{2 \lambda (t)}\right) \end{aligned}$$
(4.41)
We remind the reader of the discussion following (3.4), which compares (4.41) to the relation between the radiation and \(\lambda (t)\) from [25].
Since we will only need (4.41) to be true for all t sufficiently large, and since we only assume \(\lambda (t)\) to be defined for \(t > 50\), we use a (relatively unimportant) cutoff, \(\psi \in C^{\infty }([0,\infty ))\), such that
$$\begin{aligned} \psi (x) = {\left\{ \begin{array}{ll} 0, \quad x \leqslant T_{\lambda }\\ 1, \quad x \geqslant 2 T_{\lambda }\end{array}\right. }, \quad 0 \leqslant \psi (x) \leqslant 1 \end{aligned}$$
(4.42)
where we recall that \(T_{\lambda }>100\) is part of the definition of \(\varLambda \), see (1.6), and then get
$$\begin{aligned} \widehat{v_{2,0}}(\xi ) = \frac{-1}{\pi \xi } \int _{0}^{\infty } H(t) \sin (t\xi ) dt \end{aligned}$$
(4.43)
with
$$\begin{aligned} H(t){} & {} = 4\left( \left( \log (2)-\frac{1}{2}\right) \lambda ''(t) + \int _{t}^{2t} \left( \frac{\lambda ''(s)-\lambda ''(t)}{s-t}\right) ds+\lambda ''(t) \log (\frac{t}{\lambda (t)}) \right. \nonumber \\{} & {} \quad \left. + \int _{2t}^{\infty } \frac{\lambda ''(s) ds}{s-t} - \frac{\lambda '(t)^{2}}{2 \lambda (t)}\right) \psi (t) \end{aligned}$$
(4.44)
We can now record some basic estimates on \(\widehat{v_{2,0}}\) and \(v_{2}\). We remark that the estimates of Lemma 4 below imply that the conditions on \(\widehat{v_{2,0}}\) stated just above (4.37) are satisfied.
Lemma 4
For \(0 \leqslant k\), and \(N \geqslant 1\), there exist \(C_{k},C_{k,N}\) such that
$$\begin{aligned} \xi ^{k}|\widehat{v_{2,0}}^{(k)}(\xi )| \leqslant {\left\{ \begin{array}{ll} C_{k} \int _{100}^{\frac{1}{\xi }} \frac{\lambda (\sigma ) \log (\sigma ) d\sigma }{\sigma } + C_{k} \lambda (\frac{1}{\xi }) \log (\frac{1}{\xi }), \quad \xi \leqslant \frac{1}{100}\\ \frac{C_{k,N}}{\xi ^{N}}, \quad \xi > \frac{1}{100}\end{array}\right. } \end{aligned}$$
Proof
We start with (4.44). By the mean value theorem, there exists \(x \in [t,s]\) such that
$$\begin{aligned}\begin{aligned} |H(t)|&\leqslant C\left( \frac{\lambda (t)}{t^{2} } + \int _{t}^{2t} |\lambda '''(x)|ds + \log (t) |\lambda ''(t)| + \frac{1}{t} \int _{2t}^{\infty } \frac{\lambda (s) ds}{s^{2}} + \frac{ \lambda (t)}{t^{2}}\right) |\psi (t)|\\&\quad \leqslant \frac{C \lambda (t) \log (t)}{t^{2}} \mathbbm {1}_{\{t \geqslant T_{\lambda }\}}\end{aligned} \end{aligned}$$
where we used (4.3). Using the symbol-type nature of the estimates on \(\lambda \), we get, for \(k \geqslant 0\),
$$\begin{aligned} |H^{(k)}(t)| \leqslant C_{k} \mathbbm {1}_{\{t \geqslant T_{\lambda }\}} \frac{\lambda (t) \log (t)}{t^{2+k}} \end{aligned}$$
(4.45)
Then, for \(\xi \geqslant \frac{1}{100}\), we integrate by parts in the formula
$$\begin{aligned} \widehat{v_{2,0}}(\xi ) = \frac{-1}{\pi \xi } \int _{0}^{\infty } H(\sigma ) \sin (\sigma \xi ) d\sigma \end{aligned}$$
noting that no boundary terms arise, to get that, for each \(k \in \mathbb {N}\), there exists \(C_{k}>0\) such that
$$\begin{aligned} |\widehat{v_{2,0}}(\xi )| \leqslant \frac{C_{k}}{\xi ^{k}}, \quad \xi \geqslant \frac{1}{100} \end{aligned}$$
For \(\xi \leqslant \frac{1}{100}\), we have
$$\begin{aligned} |\widehat{v_{2,0}}(\xi )| \leqslant {\left\{ \begin{array}{ll} C \int _{T_{\lambda }}^{\frac{1}{\xi }} \frac{\lambda (\sigma ) \log (\sigma ) d\sigma }{\sigma } + \frac{C}{\xi } \int _{\frac{1}{\xi }}^{\infty } \frac{\lambda (\sigma ) \log (\sigma )}{\sigma ^{2} } d\sigma \leqslant C \int _{100}^{\frac{1}{\xi }} \frac{\lambda (\sigma ) \log (\sigma )}{\sigma } d\sigma + C \lambda (\frac{1}{\xi }) \log (\frac{1}{\xi }),\text { } \xi \leqslant \text {min}\{\frac{1}{100},\frac{1}{T_{\lambda }}\}\\ \frac{C}{\xi } \int _{\frac{1}{\xi }}^{\infty } \frac{\lambda (\sigma ) \log (\sigma )}{\sigma ^{2}} d\sigma \leqslant C \int _{100}^{\infty } \frac{\log (\sigma )}{\sigma ^{2-C_{u}}} d\sigma \leqslant C \lambda (\frac{1}{\xi }) \log (\frac{1}{\xi }), \quad \frac{1}{T_{\lambda }}\leqslant \xi \leqslant \frac{1}{100}\end{array}\right. } \end{aligned}$$
where we used \(T_{\lambda }\geqslant 100\) and (4.3) for the first case, and the fact that \(\text {min}_{x\in [100,T_{\lambda }]}(\lambda (x)) \log (100) >0\) for the second. In particular, the constants C depend on (the fixed function) \(\lambda \), but not on \(\xi \). Finally, Lemma 4 for \(k >0\) follows from (4.45) and
$$\begin{aligned} \widehat{v_{2,0}}(\xi ) = \frac{-1}{\pi \xi ^{2}} \int _{0}^{\infty } H(\frac{\omega }{\xi }) \sin (\omega ) d\omega \end{aligned}$$
\(\square \)
Lemma 5
For \(0 \leqslant k\leqslant 7\), \(j=0,1\), we have
$$\begin{aligned} |\partial _{t}^{k}\partial _{r}^{j}v_{2}(t,r)| \leqslant \frac{C r^{1-j} \lambda (t) \log (t)}{t^{2+k}}, \quad r \leqslant \frac{t}{2} \end{aligned}$$
For all \(j+k \leqslant 7\), we have the following two estimates
$$\begin{aligned} |\partial _{t}^{k}\partial _{r}^{j}v_{2}(t,r)| \leqslant \frac{C \log (r)}{\sqrt{r}\langle |t-r|\rangle ^{\frac{1}{2}+k+j}} \sup _{x \in [100,r]}\left( \lambda (x) \log (x)\right) , \quad r \geqslant \frac{t}{2} \end{aligned}$$
and
$$\begin{aligned} |\partial _{t}^{j}\partial _{r}^{k}v_{2}(t,r)| \leqslant \frac{C}{\sqrt{r}}, \quad r \geqslant \frac{t}{2} \\ |\left( \partial _{t}+\partial _{r}\right) v_{2}(t,r)| \leqslant \frac{C \log (r) \sup _{x \in [100,r]}\left( \lambda (x)\log (x)\right) }{r^{3/2}\sqrt{\langle t-r \rangle }}, \quad r >\frac{t}{2}\\ |\partial _{r}^{2} v_{2}(t,r)| \leqslant \frac{C r \lambda (t) \log (t)}{t^{4}}, \quad r \leqslant \frac{t}{2} \end{aligned}$$
Proof
To estimate \(v_{2}\), we start with the region \(r \leqslant \frac{t}{2}\). Here, we use (4.37) and (4.45), to get
$$\begin{aligned}\begin{aligned} |v_{2}(t,r)|&\leqslant C r \int _{0}^{\pi } |F(t+r \cos (\theta ))|d\theta \leqslant C r \int _{0}^{\pi } \frac{\lambda (t+r \cos (\theta )) \log (t+r \cos (\theta ))}{(t+r \cos (\theta ))^{2}} d\theta \\&\leqslant \frac{C r \lambda (t) \log (t)}{t^{2}}\end{aligned} \end{aligned}$$
where we used (4.3). For the region \(r \geqslant \frac{t}{2}\), we start with the first term on the right-hand side of (4.37), namely
$$\begin{aligned} v_{2}(t,r) = \int _{0}^{\infty } J_{1}(r\xi ) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi \end{aligned}$$
One estimate which will be useful in the region \(r \geqslant \frac{t}{2}\) (in particular, it is useful in the region \(|t-r| \leqslant 100\)) is the following. Using \(|J_{1}(x)| \leqslant \frac{C}{\sqrt{x}}, x>0\) (though a much better estimate is true for small \(x>0\)) we get
$$\begin{aligned} |v_{2}(t,r)|\leqslant & {} \frac{C}{\sqrt{r}}\int _{0}^{\frac{1}{100}} \frac{1}{\sqrt{\xi }}\left( \frac{\log (\frac{1}{\xi })}{\xi ^{C_{u}}}+\int _{100}^{\frac{1}{\xi }} \frac{\log (\sigma ) d\sigma }{\sigma ^{1-C_{u}}}\right) d\xi + \frac{C}{\sqrt{r}}\int _{\frac{1}{100}}^{\infty } \frac{d\xi }{\xi ^{5}}\nonumber \\\leqslant & {} \frac{C}{\sqrt{r}} \int _{0}^{\frac{1}{100}} \frac{\log (\frac{1}{\xi }) d\xi }{\xi ^{C_{u}+1/2}}+\frac{C}{\sqrt{r}} \leqslant \frac{C}{\sqrt{r}} \end{aligned}$$
(4.46)
where we also used (4.3). The higher derivatives of \(v_{2}\) are treated similarly. To treat the region where \(|t-r| > 100\) and \(r \geqslant \frac{t}{2}\), we use
$$\begin{aligned} v_{2}(t,r) = \int _{0}^{\frac{1}{r}} J_{1}(r\xi ) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi + \int _{\frac{1}{r}}^{\infty } J_{1}(r\xi ) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi \end{aligned}$$
(4.47)
We start with
$$\begin{aligned}\begin{aligned} |\int _{0}^{\frac{1}{r}} J_{1}(r\xi ) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi |&\leqslant C \int _{r}^{\infty } d\sigma \int _{0}^{\frac{1}{\sigma }} d\xi \frac{r \xi \lambda (\sigma ) \log (\sigma )}{\sigma } \\&\quad + C \int _{100}^{r} d\sigma \int _{0}^{\frac{1}{r}} d\xi r \xi \frac{\lambda (\sigma )\log (\sigma )}{\sigma }\\&\quad + C \int _{0}^{\frac{1}{r}} d\xi r \xi \lambda (\frac{1}{\xi }) \log (\frac{1}{\xi })\\&\leqslant \frac{C r \lambda (r) \log (r)}{r} \int _{r}^{\infty } \frac{d\sigma }{\sigma ^{2}} + \frac{C}{r} \int _{100}^{r} \frac{\lambda (\sigma ) \log (\sigma ) d\sigma }{\sigma }\\&\quad + \frac{C r \lambda (r) \log (r)}{r^{2}} \\&\leqslant \frac{C}{r} \sup _{\sigma \in [100,r]}\left( \lambda (\sigma ) \log (\sigma )\right) \log (r)\end{aligned} \end{aligned}$$
Here, we used (4.3), and Lemma 4 in the region \(\xi \leqslant \frac{1}{100}\), since \(\frac{1}{r} \ll \frac{1}{100}\), along with Fubini’s theorem to switch the order of the \(\xi \) and \(\sigma \) integrals. It remains to treat the following integral.
$$\begin{aligned}\begin{aligned}&\int _{\frac{1}{r}}^{\infty } J_{1}(r\xi ) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi = -\int _{\frac{1}{r}}^{\infty } \frac{1}{\sqrt{\pi r \xi }} \left( \cos (r \xi )-\sin (r\xi )\right) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi \\&\quad +\int _{\frac{1}{r}}^{\infty } \left( J_{1}(r\xi )+\left( \frac{\cos (r\xi )}{\sqrt{\pi r \xi }} -\frac{\sin (r\xi )}{\sqrt{\pi r \xi }}\right) \right) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi \end{aligned} \end{aligned}$$
We start with
$$\begin{aligned} \begin{aligned}&\int _{\frac{1}{r}}^{\infty } \frac{1}{\sqrt{\pi r \xi }} \left( \cos (r \xi )-\sin (r\xi )\right) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi \\&\quad =\frac{1}{2\sqrt{\pi r}} \int _{\frac{1}{r}}^{\infty }\\&\frac{1}{\sqrt{\xi }} \left( \sin ((t+r)\xi )+\sin ((t-r)\xi ) + \cos ((t+r)\xi )-\cos ((t-r)\xi )\right) \widehat{v_{2,0}}(\xi ) d\xi \\&\quad =\frac{1}{2\sqrt{\pi r}} \int _{\frac{1}{\sqrt{r}}}^{\infty } \left( \sin ((t+r)\omega ^{2})+\cos ((t+r)\omega ^{2})+\sin ((t-r)\omega ^{2})\right. \\&\qquad \left. -\cos ((t-r)\omega ^{2})\right) \widehat{v_{2,0}}(\omega ^{2}) \cdot 2 d\omega \end{aligned} \end{aligned}$$
We will show in detail how to treat the term involving \(\sin ((t-r)\omega ^{2})\). The other terms can be treated with a similar argument.
$$\begin{aligned}\begin{aligned} \frac{1}{\sqrt{\pi r}} \int _{\frac{1}{\sqrt{r}}}^{\infty } \sin ((t-r)\omega ^{2}) \widehat{v_{2,0}}(\omega ^{2}) d\omega&= \frac{1}{\sqrt{\pi r}} \int _{\frac{1}{\sqrt{r}}}^{\frac{1}{\sqrt{|t-r|}}} \sin ((t-r)\omega ^{2}) \widehat{v_{2,0}}(\omega ^{2}) d\omega \\&\quad + \frac{1}{\sqrt{\pi r}} \int _{\frac{1}{\sqrt{|t-r|}}}^{\infty } \sin ((t-r)\omega ^{2}) \widehat{v_{2,0}}(\omega ^{2}) d\omega \end{aligned} \end{aligned}$$
For the first term, we get
$$\begin{aligned} \begin{aligned}&|\frac{1}{\sqrt{\pi r}} \int _{\frac{1}{\sqrt{r}}}^{\frac{1}{\sqrt{|t-r|}}} \sin ((t-r)\omega ^{2}) \widehat{v_{2,0}}(\omega ^{2}) d\omega | \leqslant \frac{C}{\sqrt{r}} \int _{\frac{1}{\sqrt{r}}}^{\frac{1}{\sqrt{|t-r|}}} \sup _{x \in [100,\frac{1}{\omega ^{2}}]}\left( \lambda (x) \log (x)\right) \log (\frac{1}{\omega ^{2}}) d\omega \\&\quad \leqslant \frac{C \sup _{x \in [100,r]}\left( \lambda (x) \log (x)\right) \log (r)}{\sqrt{r}\sqrt{|t-r|}}\end{aligned} \end{aligned}$$
On the other hand, we integrate by parts to get
$$\begin{aligned}\begin{aligned}&|\frac{1}{\sqrt{\pi r}} \int _{\frac{1}{\sqrt{|t-r|}}}^{\infty } 2 \omega \sin ((t-r)\omega ^{2}) \frac{\widehat{v_{2,0}}(\omega ^{2})}{2 \omega } d\omega |\leqslant \frac{C}{\sqrt{r}} \left( \frac{|\widehat{v_{2,0}}(\frac{1}{|t-r|})|}{\sqrt{|t-r|}} + \int _{\frac{1}{\sqrt{|t-r|}}}^{\infty } \frac{|\widehat{v_{2,0}}'(\omega ^{2})| + \frac{|\widehat{v_{2,0}}(\omega ^{2})|}{\omega ^{2}}}{|t-r|} d\omega \right) \end{aligned} \end{aligned}$$
This gives
$$\begin{aligned}\begin{aligned}&|\frac{1}{\sqrt{\pi r}} \int _{\frac{1}{\sqrt{|t-r|}}}^{\infty } 2 \omega \sin ((t-r)\omega ^{2}) \frac{\widehat{v_{2,0}}(\omega ^{2})}{2 \omega } d\omega |\\&\quad \leqslant \frac{C}{\sqrt{r}} \frac{|\widehat{v_{2,0}}(\frac{1}{|t-r|})|}{\sqrt{|t-r|}} + \frac{C}{\sqrt{r} |t-r|} \int _{\frac{1}{\sqrt{|t-r|}}}^{\frac{1}{10}} \frac{1}{\omega ^{2}} \sup _{x \in [100,|t-r|]}\left( \lambda (x) \log (x)\right) \log (\frac{1}{\omega ^{2}})d\omega \\&\qquad +\frac{C}{\sqrt{r}|t-r|} \int _{\frac{1}{10}}^{\infty } \frac{d\omega }{\omega ^{50}}\\&\quad \leqslant \frac{C \log (|t-r|)}{\sqrt{r}\sqrt{|t-r|}} \sup _{x\in [100,|t-r|]}\left( \lambda (x) \log (x)\right) \end{aligned} \end{aligned}$$
Finally
$$\begin{aligned}\begin{aligned}&|\int _{\frac{1}{r}}^{\infty } \left( J_{1}(r\xi )+\left( \frac{\cos (r\xi )}{\sqrt{\pi r \xi }} -\frac{\sin (r\xi )}{\sqrt{\pi r \xi }}\right) \right) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) d\xi |\\&\quad \leqslant C \int _{\frac{1}{r}}^{\infty } \frac{1}{(r\xi )^{3/2}} |\widehat{v_{2,0}}(\xi )| d\xi \\&\quad \leqslant C \int _{\frac{1}{r}}^{\frac{1}{100}} \frac{d\xi }{(r\xi )^{3/2}} \sup _{x \in [100,\frac{1}{\xi }]}\left( \lambda (x) \log (x)\right) \log (\frac{1}{\xi })+C \int _{\frac{1}{100}}^{\infty } \frac{d\xi }{(r\xi )^{3/2}} \frac{1}{\xi ^{101}}\\&\quad \leqslant \frac{C}{r} \sup _{x \in [100,r]}\left( \lambda (x) \log (x)\right) \log (r)\end{aligned} \end{aligned}$$
Combining this with (4.46) finishes the estimation of \(v_{2}(t,r)\) in the region \(r \geqslant \frac{t}{2}\). The derivatives of \(v_{2}\) are treated similarly. The only important difference in the procedure used to estimate \(\partial _{t}v_{2}(t,r)+\partial _{r}v_{2}(t,r)\) is that we exploit the fact that \(\left( \partial _{t}+\partial _{r}\right) \sin ((t-r)\xi ) =0\), and similarly with \(\cos ((t-r)\xi )\). \(\square \)
Finally, we define \(v_{2,sub}\) by
$$\begin{aligned} \begin{aligned} v_{2,sub}(t,r)&= \frac{-r}{2\pi } \int _{0}^{\pi } \sin ^{2}(\theta ) \left( F(t+r\cos (\theta ))-F(t)\right) d\theta =v_{2}(t,r)-v_{2,main}(t,r)\quad \end{aligned} \end{aligned}$$
(4.48)
(where we recall that F is defined in (4.38)). We also define
$$\begin{aligned} v_{2,cubic,main}(t,r) = \frac{-r^{3}}{32} F''(t) \end{aligned}$$
Then, the \(v_{2}\) analog of Lemma 1 is
Lemma 6
We have the following estimates. For \(0 \leqslant j \leqslant 8\), \(0 \leqslant k \leqslant 2\) and \(r \leqslant \frac{t}{2}\),
$$\begin{aligned} |\partial _{t}^{j}\partial _{r}^{k}\left( v_{2,sub}(t,r)-v_{2,cubic,main}(t,r)+\frac{r^{5}}{768} F^{(4)}(t)\right) | \leqslant \frac{C r^{7-k} \lambda (t)\log (t)}{t^{8+j}} \end{aligned}$$
Proof
We expand (4.37) and directly estimate, as in Lemma 5\(\square \)
4.4 Second wave iteration
The second wave correction, \(u_{w,2}\) is defined as the solution to
$$\begin{aligned} -\partial _{tt}u_{w,2}+\partial _{rr}u_{w,2}+\frac{1}{r}\partial _{r}u_{w,2}-\frac{u_{w,2}}{r^{2}}=\left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) \left( w_{1}+v_{2}\right) :=RHS_{2}(t,r)\qquad \end{aligned}$$
(4.49)
with 0 Cauchy data at infinity. (We carried out the first order matching before defining \(u_{w,2}\) so that we could choose \(v_{2,0}\) before having to consider the equation defining \(u_{w,2}\)). Later on, we will add, to \(u_{w,2}\), a free wave, \(v_{2,2}\), solving
$$\begin{aligned} {\left\{ \begin{array}{ll}-\partial _{tt}v_{2,2}+\partial _{rr}v_{2,2}+\frac{1}{r}\partial _{r}v_{2,2}-\frac{v_{2,2}}{r^{2}}=0\\ v_{2,2}(0,r)=0\\ \partial _{t}v_{2,2}(0,r) = v_{2,3}(r)\end{array}\right. } \end{aligned}$$
(4.50)
with \(v_{2,3}\) chosen so as to satisfy a third order matching condition which we will describe later on. We start by proving estimates on \(\widehat{RHS_{2}}\), which will allow us to justify a representation formula for \(u_{w,2}\). For this, we start with the following definitions.
$$\begin{aligned} RHS_{2,1}(t,r) = \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) w_{1}, \quad RHS_{2,2}(t,r) = \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) v_{2}\qquad \end{aligned}$$
(4.51)
Then, we have the following.
Lemma 7
Recalling that \(w_{1}\) is defined in (4.23), we have, for \(2 \leqslant k \leqslant 4\) and \(0 \leqslant j \leqslant 1\),
$$\begin{aligned} \begin{aligned}&\xi ^{j}t^{k}|\partial _{\xi }^{j}\partial _{t}^{k} \widehat{RHS_{2,1}}(t,\xi )|+ \xi ^{j} |\partial _{\xi }^{j}\widehat{RHS_{2,1}}(t,\xi )| \leqslant {\left\{ \begin{array}{ll} \frac{C \xi \lambda (t)^{3} \log ^{2}(t)}{t^{2} }, \quad \xi \leqslant \frac{1}{\lambda (t)}\\ \frac{C (\log (t)+|\log (\xi )|)}{\sqrt{\lambda (t)}\xi ^{5/2} t^{2} }, \quad \frac{1}{\lambda (t)} < \xi \end{array}\right. }\end{aligned} \end{aligned}$$
Proof
In the regions where \(\xi \leqslant \frac{1}{\lambda (t)}\), we use
$$\begin{aligned} \widehat{RHS_{2,1}}(t,\xi ) = \int _{0}^{\infty } J_{1}(r\xi ) \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) w_{1}(t,r) r dr \end{aligned}$$
and Lemma 2. When \(\xi \geqslant \frac{1}{\lambda (t)}\), we use
$$\begin{aligned} \widehat{RHS_{2,1}}(t,\xi ) = \frac{1}{\xi ^{2}}\int _{0}^{\infty } J_{1}(r\xi ) H_{1}\left( \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) w_{1}(t,r)\right) r dr \end{aligned}$$
where
$$\begin{aligned} H_{1}(g) = -\partial _{r}^{2}g-\frac{1}{r}\partial _{r}g+\frac{g}{r^{2}} \end{aligned}$$
and then use the estimates on \(w_{1}\) from Lemma 2. We use the same procedure to estimate \(\partial _{t}^{k} \widehat{RHS_{2,1}}(t,\xi )\). To estimate \(\partial _{\xi }\partial _{t}^{k}\widehat{RHS_{2,1}}(t,\xi )\), we start with
$$\begin{aligned} \widehat{RHS_{2,1}}(t,\xi ) = \int _{0}^{\infty } RHS_{2,1}(t,\frac{x}{\xi }) J_{1}(x) \frac{x dx}{\xi ^{2}} \end{aligned}$$
differentiate under the integral, and use the symbol-type nature of the estimates in Lemma 2. \(\square \)
Note that some of the estimates in the following lemma can be combined into a single estimate with multiple cases of numbers of derivatives, but is presented as is for convenience.
Lemma 8
Let
$$\begin{aligned} f(x) = \left( \frac{\cos (2Q_{1}(x))-1}{x^{2}}\right) =\frac{-8}{(1+x^{2})^{2}} \end{aligned}$$
and
$$\begin{aligned} K(y,z) = \int _{0}^{\infty } J_{1}(xy) f(x) J_{1}(x z) x dx \end{aligned}$$
Then,
$$\begin{aligned} K(y,z)= & {} 4 \cdot {\left\{ \begin{array}{ll} z I_{0}(z) K_{1}(y)-y I_{1}(z)K_{2}(y), \quad 0< z< y\\ y I_{0}(y) K_{1}(z) - z I_{1}(y) K_{2}(z), \quad 0< y< z\end{array}\right. }\nonumber \\:= & {} 4 \cdot {\left\{ \begin{array}{ll} K_{y>z}(y,z), \quad 0< z< y\\ K_{z>y}(y,z), \quad 0< y < z\end{array}\right. } \end{aligned}$$
(4.52)
and, we have the following estimates, for \(z \ne y\), \(0 \leqslant j \leqslant 1\), and \(0 \leqslant k \leqslant 4\):
$$\begin{aligned} |\partial _{y}^{j}\partial _{z}^{k} K(y,z)| \leqslant \left( \text {max}\{1,\frac{1}{y}\}\right) ^{j} {\left\{ \begin{array}{ll} C e^{-|y-z|} \left( \frac{|z-y|}{\sqrt{zy}} + \frac{\sqrt{\text {max}\{y,z\}}}{\text {min}\{y,z\}^{3/2}}\right) , \quad y,z>1\\ C \sqrt{y} e^{-y} z^{\frac{1}{2}\left( (-1)^{k}+1\right) }, \quad y>1>z\\ C \sqrt{z} e^{-z} y, \quad z>1>y\end{array}\right. } \end{aligned}$$
For \(1>y,z\), \(y \ne z\), and \(0 \leqslant j, k \leqslant 1\),
$$\begin{aligned}\begin{aligned}&|\partial _{y}^{j}\partial _{z}^{k}K(y,z)| \leqslant C y^{1-j} z^{1-k} (1+|\log (\text {max}\{y,z\})|), \quad |\partial _{y}^{j}\partial _{z}^{2+k}K(y,z)| \leqslant \frac{C y^{-j} \text {min}\{y,z\}}{z^{k}\text {max}\{y,z\}}\\&|\partial _{y}^{j}\partial _{z}^{4+k} K(y,z)| \leqslant {\left\{ \begin{array}{ll} \frac{C z^{1-k}}{y^{1+j}}, \quad 1>y>z\\ \frac{C y^{1-j}}{z^{3+k}}, \quad 1>z>y \end{array}\right. }\end{aligned} \end{aligned}$$
If \(1 \leqslant j \leqslant 2\), \(0 \leqslant n \leqslant 1\), and \(0 \leqslant k \leqslant 4-j\), then, for \(0 < \omega , \xi \) and \(\omega \ne \xi \), we have
$$\begin{aligned}\begin{aligned}&|\partial _{\xi }^{n}\partial _{\omega }^{k}\partial _{t}^{j}\left( K(\xi \lambda (t),\omega \lambda (t))\right) | \\&\quad \leqslant \frac{C \lambda (t)^{k}}{t^{j}} \left( \text {max}\{\lambda (t),\frac{1}{\xi }\}\right) ^{n}{\left\{ \begin{array}{ll} e^{-\lambda (t)|\xi -\omega |} \left( \frac{\lambda (t)^{j} |\xi -\omega |^{j+1}}{\sqrt{\omega \xi }}+\frac{\sqrt{\text {max}\{\xi ,\omega \}}}{\text {min}\{\xi ,\omega \}^{3/2}\lambda (t)}\right) , \quad \omega , \xi> \frac{1}{\lambda (t)}\\ e^{-\lambda (t) \xi } \left( \xi \lambda (t)\right) ^{1/2+j} \left( \omega \lambda (t)\right) ^{\frac{1}{2}\left( (-1)^{k}+1\right) }, \quad \xi> \frac{1}{\lambda (t)}> \omega \\ e^{-\omega \lambda (t)} \xi \lambda (t) \left( \omega \lambda (t)\right) ^{\frac{1}{2}+j}, \quad \omega> \frac{1}{\lambda (t)} > \xi \end{array}\right. }\end{aligned} \end{aligned}$$
For \(1 \leqslant j \leqslant 2\) , \(0 \leqslant n,k \leqslant 1\), and \(\frac{1}{\lambda (t)} \geqslant \omega , \xi \), \(\omega \ne \xi \), we have
$$\begin{aligned} |\partial _{\xi }^{n}\partial _{t}^{j} \partial _{\omega }^{k}\left( K(\xi \lambda (t),\omega \lambda (t))\right) | \leqslant \frac{C \lambda (t)^{2}}{t^{j}} \omega ^{1-k}\xi ^{1-n}\left( 1+|\log (\text {max}\{\xi ,\omega \}\lambda (t))|\right) \end{aligned}$$
For \(0 \leqslant n,k \leqslant 1\), \(1 \leqslant j \leqslant 2\), and \(\frac{1}{\lambda (t)} > \omega , \xi \), \(\omega \ne \xi \), we have
$$\begin{aligned} |\partial _{\xi }^{n}\partial _{\omega }^{2+k}\partial _{t}^{j}\left( K(\xi \lambda (t),\omega \lambda (t))\right) | \leqslant \frac{C \lambda (t)^{2}}{t^{j} } \frac{\text {min}\{\omega ,\xi \}}{\text {max}\{\omega ,\xi \} \omega ^{k}\xi ^{n}} \end{aligned}$$
Proof
We start with the table of Gradshteyn and Ryzhik, [9], entry 6.541, 1. A special case of this identity says that, for \(c>0\),
$$\begin{aligned} \int _{0}^{\infty } J_{1}(x y) J_{1}(x z) \frac{x dx}{x^{2}+c^{2}} = {\left\{ \begin{array}{ll} I_{1}(z c) K_{1}(y c), \quad 0< z< y\\ I_{1}(y c) K_{1}(z c), \quad 0< y < z\end{array}\right. } \end{aligned}$$
Upon differentiating in c (it is possible to differentiate under the integral, by the dominated convergence theorem), and setting \(c=1\), we get
$$\begin{aligned} K(y,z) = -8 \int _{0}^{\infty } \frac{J_{1}(x y) J_{1}(x z) x dx}{(1+x^{2})^{2}} = 4 \cdot {\left\{ \begin{array}{ll} z I_{0}(z) K_{1}(y)-y I_{1}(z)K_{2}(y), \quad 0< z< y\\ y I_{0}(y) K_{1}(z) - z I_{1}(y) K_{2}(z), \quad 0< y < z\end{array}\right. } \end{aligned}$$
At this stage, the estimates on K(y, z) and its derivatives follow from a straightforward application of asymptotics of Bessel functions, which can be found from numerous sources, for example, [9]. To estimate the time derivatives of \(K(\xi \lambda (t), \omega \lambda (t))\), we let \(V=y\partial _{y}+z\partial _{z}\), and note that
$$\begin{aligned} \partial _{t}\left( K(\xi \lambda (t),\omega \lambda (t))\right) = \frac{\lambda '(t)}{\lambda (t)} V(K)\Bigr |_{\begin{array}{c} y=\xi \lambda (t)\\ z=\omega \lambda (t) \end{array}}. \end{aligned}$$
For \(1 \leqslant j \leqslant 4\), we can iterate this to obtain expressions for \(\partial _{t}^{j}\left( K(\xi \lambda (t),\omega \lambda (t))\right) \) in terms of \(V^{n}(K)\) for \(1 \leqslant n \leqslant j\). Then, we note that, for all \(1 \leqslant j \leqslant 4\), there exist constants \(c_{k,j,w}\) such that
$$\begin{aligned} V^{j}(K)(y,z) = \sum _{w=0}^{j} z^{w} \sum _{k=0}^{j} c_{k,j,w} (y-z)^{k} \partial _{y}^{k} \left( \partial _{y}+\partial _{z}\right) ^{w} K(y,z). \end{aligned}$$
This observation (and a decomposition of the above form, except with y and z switched (note that V is symmetric in y and z)), combined with the same procedure used above, allows us to estimate the time derivatives of \(K(\xi \lambda (t), \omega \lambda (t))\). \(\square \)
We can now estimate \(\widehat{RHS_{2,2}}(t,\xi )\).
Lemma 9
With \(RHS_{2,2}\) given in (4.51), and \(m_{\geqslant 1}(x) = 1-m_{\leqslant 1}(x)\), where \(m_{\leqslant 1}\) is defined in (2.1), we have
$$\begin{aligned} \widehat{RHS_{2,2}}(t,\xi ) = \int _{0}^{\infty } \widehat{v_{2,0}}(\omega ) \sin (t\omega ) K(\xi \lambda (t), \omega \lambda (t)) d\omega \end{aligned}$$
(4.53)
and, for any \(N>10\), there exists \(C_{N}>0\) such that we have the following estimates for \(0 \leqslant k \leqslant 1\):
$$\begin{aligned}{} & {} |\partial _{\xi }^{k}\widehat{RHS_{2,2}}(t,\xi )| \leqslant \frac{C \lambda (t) \log ^{2}(t)}{t^{2}} \sup _{x \in [100,\frac{t}{2}]}\left( \lambda (x) \log (x)\right) {\left\{ \begin{array}{ll} \lambda (t)\xi ^{1-k} \log (t), \quad \xi \leqslant \frac{1}{\lambda (t)}\\ \lambda (t)^{k}\sqrt{\xi \lambda } e^{-\xi \lambda (t)} + \frac{C_{N}\lambda (t)^{k}}{\langle \xi \rangle ^{N}}, \quad \xi> \frac{1}{\lambda (t)}\end{array}\right. }\nonumber \\{} & {} \begin{aligned}&|\partial _{\xi }^{k}\left( \partial _{t}^{2} \widehat{RHS_{2,2}}(t,\xi )-\frac{8 \widehat{v_{2,0}}(\xi ) \xi \lambda (t)^{2} m_{\geqslant 1}(\xi t) \sin (\xi t)}{t^{4}}\right) | \\&\leqslant \frac{C \lambda (t) \log ^{2}(t)}{t^{4}} \sup _{x \in [100,\frac{t}{2}]}\left( \lambda (x) \log (x)\right) {\left\{ \begin{array}{ll} \xi ^{1-k} \lambda (t) \log (t), \quad \xi \leqslant \frac{1}{\lambda (t)}\\ \left( \xi \lambda (t)\right) ^{5/2} \lambda (t)^{k} e^{-\xi \lambda (t)} + \frac{C_{N} \lambda (t)^{k} (1+\lambda (t))}{\langle \xi \rangle ^{N}}, \quad \xi > \frac{1}{\lambda (t)} \end{array}\right. }\end{aligned} \end{aligned}$$
(4.54)
Proof
Equation (4.53) follows from insertion of (4.37) into (4.51), and Fubini’s theorem. To prove the stated estimates, we start with the decomposition
$$\begin{aligned} \begin{aligned} \partial _{t}^{2}\widehat{RHS_{2,2}}(t,\xi )&= - \int _{0}^{\infty } d\omega \widehat{v_{2,0}}(\omega ) \omega ^{2} \sin (t\omega ) K(\xi \lambda (t),\omega \lambda (t))\\&\quad +2 \int _{0}^{\infty } d\omega \widehat{v_{2,0}}(\omega ) \omega \cos (\omega t) \partial _{t}\left( K(\xi \lambda (t),\omega \lambda (t))\right) \\&\quad +\int _{0}^{\infty } d\omega \widehat{v_{2,0}}(\omega ) \sin (\omega t) \partial _{t}^{2}\left( K(\xi \lambda (t),\omega \lambda (t))\right) \end{aligned} \end{aligned}$$
(Strictly speaking, we first split the integral in (4.53) over the regions \(\omega \leqslant \xi \) and \(\omega \geqslant \xi \), differentiate under the integral sign, and then combine the resulting integrals). Define \(int_{i}(t,\xi )\) to be the ith line on the right-hand side of the expression above. Starting with \(int_{1}\), we make the following decomposition (where \(m_{\leqslant 1}\) is defined in (2.1))
$$\begin{aligned}{} & {} int_{1}(t,\xi ) = int_{1,a}(t,\xi ) +int_{1,b}(t,\xi ), \\{} & {} int_{1,a}(t,\xi )= -\int _{0}^{\infty } d\omega \widehat{v_{2,0}}(\omega ) \omega ^{2} \sin (\omega t) m_{\leqslant 1}(\omega t) K(\xi \lambda (t), \omega \lambda (t)) \end{aligned}$$
For \(int_{1,a}\), we separately treat the cases \(\xi < \frac{1}{t}\), \(\frac{1}{\lambda (t)}> \xi > \frac{1}{t}\), and \(\xi > \frac{1}{\lambda (t)}\). In each case, we directly estimate \(int_{1,a}\) using Lemmas 4 and 8. To estimate \(int_{1,b}\), we decompose the integral into the regions \(\omega \leqslant \xi \) and \(\omega \geqslant \xi \), and integrate by parts four times in the \(\omega \) variable. We recall (4.52) and note that
$$\begin{aligned} 4\left( \partial _{z}^{3} K_{y>z}(y,z)-\partial _{z}^{3} K_{z>y}(y,z)\right) \Bigr |_{z=y}=\frac{8}{y} \end{aligned}$$
which explains the form of (4.54). Then, we consider the two cases \(\frac{1}{\lambda (t)} < \frac{1}{100}\), and \(\frac{1}{\lambda (t)} > \frac{1}{100}\). Within each of these two cases, we then consider various regions of \(\xi \). For example, in the case \(\frac{1}{\lambda (t)} < \frac{1}{100}\), we consider the regions
$$\begin{aligned} \xi< \frac{1}{2 t}, \quad \frac{1}{2t}< \xi< \frac{1}{\lambda (t)}, \quad \frac{1}{\lambda (t)}< \xi< \frac{1}{100}, \quad \frac{1}{100} < \xi \end{aligned}$$
Then, we directly estimate the resulting integrals using Lemmas 4 and 8. After this, we combine the estimates in the separate cases \(\frac{1}{\lambda (t)} < \frac{1}{100}\), and \(\frac{1}{\lambda (t)} > \frac{1}{100}\). We then use the same procedure to estimate \(int_{2}, int_{3}\) (where we integrate by parts \(4-i+1\) times to treat \(int_{i,b}\)) and combine everything to get (4.54) for \(k=0\).
To estimate \(\partial _{\xi }\partial _{t}^{2}\widehat{RHS_{2,2}}(t,\xi )\), we start with
$$\begin{aligned} \begin{aligned} \partial _{t}^{2} \widehat{RHS}_{2,2}(t,\xi )&= \int _{0}^{\infty } \widehat{v_{2,0}}(\omega ) \left( -\omega ^{2}\sin (t\omega ) K(\xi \lambda (t),\omega \lambda (t)) \right. \\&\quad + 2 \omega \cos (\omega t) \partial _{t}\left( K(\xi \lambda (t),\omega \lambda (t))\right) \\&\quad \left. + \sin (\omega t) \partial _{t}^{2}\left( K(\xi \lambda (t),\omega \lambda (t))\right) \right) m_{\leqslant 1}(\omega t) d\omega \\&\quad +\frac{8 \widehat{v_{2,0}}(\xi ) \xi \lambda (t)^{2} m_{\geqslant 1}(\xi t) \sin (\xi t)}{t^{4}}\\&\quad -\int _{0}^{\infty } \frac{\sin (\omega t)}{t^{4}}\partial _{\omega }^{4}\left( \widehat{v_{2,0}}(\omega ) \omega ^{2} K(\xi \lambda (t),\omega \lambda (t)) m_{\geqslant 1}(\omega t)\right) d\omega \\&\quad +2 \int _{0}^{\infty } \frac{\sin (\omega t)}{t^{3}} \partial _{\omega }^{3}\left( \widehat{v_{2,0}}(\omega ) \omega \partial _{t}\left( K(\xi \lambda (t),\omega \lambda (t))\right) m_{\geqslant 1}(\omega t)\right) d\omega \\&\quad -\int _{0}^{\infty } \frac{\sin (\omega t)}{t^{2}} \partial _{\omega }^{2}\left( \widehat{v_{2,0}}(\omega )m_{\geqslant 1}(\omega t) \partial _{t}^{2}\left( K(\xi \lambda (t),\omega \lambda (t))\right) \right) d\omega \end{aligned}\nonumber \\ \end{aligned}$$
(4.55)
where \(m_{\geqslant 1}(x) = 1-m_{\leqslant 1}(x)\). We split each integral of (4.55) into the regions \(\omega \leqslant \xi \) and \(\omega \geqslant \xi \), and differentiate with respect to \(\xi \). Finally, we use the estimates of Lemmas 8, 4 to finish the proof of (4.54). The estimation of \(\partial _{\xi }^{k}\widehat{RHS}_{2,2}(t,\xi )\) for \(k=0,1\) is done similarly. \(\square \)
With the same procedure used to establish Lemma 3, we get that \(u_{w,2}\), the solution to (4.49) with zero Cauchy data at infinity, is given by the following.
$$\begin{aligned} u_{w,2}(t,r)= & {} \int _{t}^{\infty } ds \int _{0}^{\infty } d\xi \sin ((t-s)\xi ) J_{1}(r\xi ) \widehat{RHS_{2}}(s,\xi ) \nonumber \\= & {} - \int _{0}^{\infty } J_{1}(r\xi ) \frac{\widehat{RHS_{2}}(t,\xi )}{\xi } d\xi \nonumber \\{} & {} - \int _{0}^{\infty } d\xi \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi ^{2}} \partial _{s}^{2}\widehat{RHS_{2}}(s,\xi ) J_{1}(r\xi ) \end{aligned}$$
(4.56)
Using Lemmas 2 and 5, we can use Fubini’s theorem to get that the first integral in (4.56) is
$$\begin{aligned} u_{w,2,ell}(t,r){} & {} :=- \int _{0}^{\infty } J_{1}(r\xi ) \frac{\widehat{RHS_{2}}(t,\xi )}{\xi } d\xi \nonumber \\{} & {} = -\frac{1}{2}\left( \frac{1}{r}\int _{0}^{r} s^{2} RHS_{2}(t,s) ds + r \int _{r}^{\infty } RHS_{2}(t,s) ds\right) \end{aligned}$$
(4.57)
It will be useful to compute the following function, which will turn out to contain the leading behavior of \(u_{w,2,ell}\) in the matching region.
$$\begin{aligned} u_{w,2,ell,0}(t,r):=-\frac{1}{2}\left( \frac{1}{r}\int _{0}^{r} s^{2} RHS_{2,0}(t,s) ds + r \int _{r}^{\infty } RHS_{2,0}(t,s) ds\right) \end{aligned}$$
(4.58)
with
$$\begin{aligned} RHS_{2,0}(t,s):= & {} \left( \frac{\cos (2Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) s \left( f_{1}(t) - \lambda ''(t)\log (s)\right) \nonumber \\= & {} \left( \frac{\cos (2Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) \left( w_{1,main}+v_{2,main}\right) (t,s) \end{aligned}$$
(4.59)
and
$$\begin{aligned} f_{1}(t) = \lambda ''(t) + \frac{\lambda '(t)^{2}}{2 \lambda (t)} + \lambda ''(t) \log (\lambda (t)) \end{aligned}$$
(4.60)
where we recall the definitions of \(w_{1,main}\) and \(v_{2,main}\) in (4.25) and (4.40), which are the main parts of \(w_{1}\) and \(v_{2}\), respectively, in the matching region. We have
$$\begin{aligned} u_{w,2,ell,0}(t,r)= & {} \frac{2 f_{1}(t) \lambda (t)^2 \log \left( \frac{r^2}{\lambda (t)^2}+1\right) }{r}\nonumber \\{} & {} -\frac{\lambda (t)^2 \lambda ''(t) \left( \text {Li}_2\left( -\frac{r^2}{\lambda (t)^2}\right) +2 \log (r) \left( \log \left( \frac{r^2}{\lambda (t)^2}+1\right) -\frac{r^2}{\lambda (t)^2+r^2}\right) +\log \left( \frac{r^2}{\lambda (t)^2}+1\right) \right) }{r}\nonumber \\{} & {} -r \lambda ''(t) \left( \frac{2 \lambda (t)^2 \log (r)}{\lambda (t)^2+r^2}+\log \left( \frac{\lambda (t)^2}{r^2}+1\right) \right) \end{aligned}$$
(4.61)
The leading behavior of \(u_{w,2,ell,0}\) in the matching region is \(u_{w,2,ell,0,cont}\), which is defined by
$$\begin{aligned} \begin{aligned}&u_{w,2,ell,0,cont}(t,r):= \frac{-2}{r}\left( \log (r)\left( -\lambda (t)\lambda '(t)^{2}-\lambda (t)^{2}\lambda ''(t)-2 \lambda (t)^{2} \log (\lambda (t))\lambda ''(t)\right) \right. \\&\quad \left. +\lambda (t) \log (\lambda (t))\lambda '(t)^{2}+\lambda (t)^{2}\log (\lambda (t))\lambda ''(t) + \lambda (t)^{2}\log ^{2}(\lambda (t))\lambda ''(t)\right. \\&\quad +\left. \lambda (t)^{2}\log ^{2}(r)\lambda ''(t)+\frac{1}{2} \lambda (t)^2 \lambda ''(t)-\frac{1}{12} \pi ^2 \lambda (t)^2 \lambda ''(t)\right) \end{aligned} \end{aligned}$$
(4.62)
In the above computation, we have used asymptotics of \(\text {Li}_{2}\), for example, from [9]. In the course of proving Proposition 2, which will occur when we do a third order matching, we will record refined estimates on the difference between \(u_{w,2}\) and its leading parts in the matching region. On the other hand, we will also need estimates which are global in the spatial coordinate, of the difference between the full correction \(u_{w,2}\), and the piece \(u_{w,2,ell,0,cont}\). Now that we have established Lemmas 7 and 9, we can estimate \(u_{w,2}-u_{w,2,ell,0,cont}\).
Lemma 10
For \(0 \leqslant k \leqslant 1\), the following two estimates are true:
$$\begin{aligned}{} & {} |\partial _{r}^{k}\left( u_{w,2}(t,r)-u_{w,2,ell}\right) (t,r)|\nonumber \\{} & {} \quad \leqslant Cr \frac{\lambda (t)^{2-k}(1+\lambda (t))\log ^{5}(t+r)}{t^{4}} \sup _{x \in [100,\frac{t}{2}]}\left( \lambda (x) \log (x)\right) , \quad r \geqslant g(t) \lambda (t)\nonumber \\ \end{aligned}$$
(4.63)
and
$$\begin{aligned} |\partial _{r}^{k}u_{w,2}(t,r)| \leqslant C \lambda (t)^{\frac{1}{2}-k}(1+\lambda (t)^{2+k}) \log ^{2}(t)\log ^{2}(r) \frac{\sup _{x \in [100,t]} \left( \lambda (x) \log (x)\right) }{\sqrt{r}t^{2}}, \quad r \geqslant \frac{t}{2}\qquad \end{aligned}$$
(4.64)
For \(0 \leqslant j \leqslant 8\) and \(0 \leqslant k \leqslant 1\), the following two estimates are true.
$$\begin{aligned} |\partial _{r}^{k}\partial _{t}^{j}\left( u_{w,2,ell}-u_{w,2,ell,0}\right) (t,r)| \leqslant \frac{C r^{1-k} \lambda (t)^{2}\log (t)}{t^{4+j}} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) , \quad r \leqslant \frac{t}{2}\qquad \end{aligned}$$
(4.65)
and
$$\begin{aligned} |\partial _{r}^{k}\left( u_{w,2,ell}- u_{w,2,ell,0}\right) (t,r)| \leqslant \frac{C \lambda (t)^{2} \log (r)}{r^{1+k} t^{2}} \sup _{x \in [100,t+r]} \left( \lambda (x) \log (x)\right) , \quad r > \frac{t}{2} \end{aligned}$$
For \(0 \leqslant j,k \leqslant 5\),
$$\begin{aligned} |\partial _{t}^{j}\partial _{r}^{k}\left( u_{w,2,ell,0}(t,r)-u_{w,2,ell,0,cont}(t,r)\right) | \leqslant {\left\{ \begin{array}{ll} \frac{C \lambda (t)^{3} \left( \log ^{2}(r) + \log ^{2}(t)\right) }{r^{1+k} t^{2+j}}, \quad r \leqslant \lambda (t)\\ \frac{C \lambda (t)^{5} \left( |\log (r)|+\log (t)\right) }{r^{3+k} t^{2+j} }, \quad r \geqslant \lambda (t)\end{array}\right. }\qquad \end{aligned}$$
(4.66)
For \(0 \leqslant k \leqslant 1\),
$$\begin{aligned}{} & {} r|\partial _{r} u_{w,2,ell}(t,r)| + t^{k}|\partial _{t}^{k}u_{w,2,ell}(t,r)| \nonumber \\{} & {} \quad \leqslant {\left\{ \begin{array}{ll} \frac{C r \lambda (t)(\log (t)+|\log (r)|)}{t^{2}}, \quad r \leqslant \lambda (t)\\ \frac{C \lambda (t)^{2} \log (t)}{r t^{2}} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) , \quad \lambda (t)< r < \frac{t}{2}\\ \frac{C \lambda (t)^{2} \log (r) \sup _{x \in [100,t+r]}\left( \lambda (x) \log (x)\right) }{\sqrt{r} t^{5/2}}, \quad r > \frac{t}{2}\end{array}\right. } \end{aligned}$$
(4.67)
For \(0 \leqslant k \leqslant 2\),
$$\begin{aligned} |\partial _{t}^{2+k}u_{w,2,ell}(t,r)| \leqslant {\left\{ \begin{array}{ll} \frac{C r \lambda (t)(\log (t)+|\log (r)|)}{t^{4+k}}, \quad r \leqslant \lambda (t)\\ \frac{C \lambda (t)^{2} \log (t)}{r t^{4+k}} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) , \quad \lambda (t)< r < \frac{t}{2}\\ \frac{C \lambda (t)^{2} \log (r) \sup _{x \in [100,t+r]}\left( \lambda (x) \log (x)\right) }{\sqrt{r}}\left( \frac{1}{r^{4}\langle t-r \rangle ^{\frac{1}{2}+k}}+\frac{1}{t^{9/2+k}}\right) , \quad r > \frac{t}{2}\end{array}\right. } \end{aligned}$$
(4.68)
Finally, we have the following estimate, for all \(r >0\). (Recall the definition of E in (1.3)).
$$\begin{aligned} \sqrt{E(u_{w,2}(t),\partial _{t}u_{w,2})} +|u_{w,2}(t,r)| \leqslant \frac{C \log (t)}{t^{1-C_{u}}} \end{aligned}$$
(4.69)
Proof
We have
$$\begin{aligned} \begin{aligned}&u_{w,2,ell}(t,r)-u_{w,2,ell,0}(t,r) \\&= -\frac{1}{2}\left( \frac{1}{r}\int _{0}^{r} s^{2}\left( RHS_{2}(t,s)-RHS_{2,0}(t,s)\right) ds \right. \\&\quad \left. + r \int _{r}^{\infty } \left( RHS_{2}(t,s)-RHS_{2,0}(t,s)\right) ds\right) \end{aligned} \end{aligned}$$
(4.70)
where we recall the definitions of \(RHS_{2,0}\) and \(RHS_{2}\) from (4.59), and (4.49), respectively. Inside the integrals, in the region \(s \leqslant \frac{t}{2}\), we use
$$\begin{aligned} RHS_{2}(t,s) - RHS_{2,0}(t,s) = \left( \frac{\cos (2Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) \left( v_{2,sub}(t,s) + w_{1,sub}(t,s)\right) \end{aligned}$$
and the estimates on \(v_{2,sub}\) and \(w_{1,sub}\) from Lemmas 1 and 6. (Recall the definitions of \(w_{1,sub}\) and \(v_{2,sub}\) in (4.26) and (4.48), respectively). On the other hand, in the region \(s \geqslant \frac{t}{2}\), we use
$$\begin{aligned} \left( RHS_{2}- RHS_{2,0}\right) (t,s) = \left( \frac{\cos (2Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) \left( v_{2}(t,s) + w_{1}(t,s) - s \left( f_{1}(t)-\lambda ''(t) \log (s)\right) \right) \end{aligned}$$
where \(f_{1}\) is given by (4.60), and the estimates from Lemmas 2 and 5. The only detail to note is that, when establishing (4.65) for \(j \geqslant 1\), we will have terms of the following form, for \(1 \leqslant k \leqslant j\).
$$\begin{aligned} -\frac{r}{2} \int _{\frac{t}{2}}^{\infty } g(s,t)\partial _{t}^{k} v_{2}(t,s) ds \end{aligned}$$
For these, since the estimates on \(\partial _{t}^{k}v_{2}(t,s)\) in the region \(s \geqslant \frac{t}{2}\) from Lemma 5 are not a factor of \(t^{-k}\) better than the corresponding estimates on \(v_{2}(t,s)\), we write the following (iterating as needed if \(j >2\))
$$\begin{aligned} \partial _{t}^{k}v_{2}(t,s) = {\left\{ \begin{array}{ll} \left( \partial _{t}+\partial _{s}\right) v_{2}(t,s) -\partial _{s}v_{2}(t,s), \quad k=1\\ \partial _{s}^{2}v_{2}+\frac{1}{s}\partial _{s}v_{2}-\frac{v_{2}}{s^{2}}, \quad k=2\end{array}\right. } \end{aligned}$$
(4.71)
and integrate by parts for the terms involving s derivatives, which is why (4.65) has a factor of \(\frac{1}{t^{j}}\).
To estimate \(u_{w,2}-u_{w,2,ell}\), we first note that
$$\begin{aligned} u_{w,2}(t,r)-u_{w,2,ell}(t,r) = \int _{t}^{\infty } ds \int _{0}^{\infty } d\xi \frac{\sin ((s-t)\xi )}{\xi ^{2}} \partial _{s}^{2}\widehat{RHS_{2}}(s,\xi ) J_{1}(r\xi )\qquad \end{aligned}$$
(4.72)
Then, we recall (4.54), and write
$$\begin{aligned} \partial _{s}^{2}\widehat{RHS_{2}}(s,\xi ) = \partial _{s}^{2}\widehat{RHS_{2,3}}(s,\xi )+ \frac{8 \widehat{v_{2,0}}(\xi ) \xi \lambda (t)^{2} m_{\geqslant 1}(\xi t) \sin (\xi t)}{t^{4}} \end{aligned}$$
(4.73)
We split the s integral in (4.72) involving \(\partial _{s}^{2}\widehat{RHS_{2,3}}(s,\xi )\) into the regions \(s-t \leqslant r\) and \(s-t \geqslant r\). Then, when \(s-t \leqslant r\), we split the \(\xi \) integral in (4.72) over the three regions \(\xi \leqslant \frac{1}{r}\), \(\frac{1}{r} \leqslant \xi \leqslant \frac{1}{s-t}\) and \(\frac{1}{s-t} < \xi \). We make a similar splitting in the region \(s-t \geqslant r\), with the following detail. In the region \(\xi \geqslant \frac{1}{s-t}\), and \(s-t \geqslant r\), we integrate by parts in the \(\xi \) variable, integrating \(\sin ((s-t)\xi )\), and differentiating the rest of the integrand. Then, we use
$$\begin{aligned} |J_{1}(x)| \leqslant {\left\{ \begin{array}{ll} C x , \quad x \leqslant 1\\ \frac{C}{\sqrt{x}}, \quad x \geqslant 1\end{array}\right. } \end{aligned}$$
along with (4.3) and Lemmas 7 and 9 to estimate each resulting integral. Recalling (4.73), we now need to consider the \(\frac{8 \widehat{v_{2,0}}(\xi ) \xi \lambda (t)^{2} m_{\geqslant 1}(\xi t) \sin (\xi t)}{t^{4}}\) contribution to (4.72). For this, we first use
$$\begin{aligned} \sin ((s-t)\xi ) \sin (\xi s) = \frac{1}{2}\left( \cos (t\xi )-\cos ((2s-t)\xi )\right) \end{aligned}$$
and then split the \(\xi \) integral in (4.72) over the regions \(\xi \leqslant \frac{1}{t}\) and \(\xi \geqslant \frac{1}{t}\), integrating by parts in \(\xi \) in the second integral (where we integrate \(\cos (t\xi )-\cos ((2s-t)\xi )\) and differentiate the rest of the integrand). This establishes (4.63) for \(k=0\). To establish (4.63) for \(k=1\), we start by differentiating (4.72) in r, and then use the same procedure as for \(u_{w,2}-u_{w,2,ell}\).
To estimate \(u_{w,2}(t,r)\) for \(r \geqslant \frac{t}{2}\), we start with
$$\begin{aligned} u_{w,2}(t,r) = \int _{t}^{\infty } ds \int _{0}^{\infty } d\xi \sin ((t-s)\xi )\widehat{RHS_{2}}(s,\xi ) J_{1}(r\xi ) \end{aligned}$$
(4.74)
Then, we use the same procedure as above, except that, whenever \(|s-t-r|\geqslant 1\), \(\xi \geqslant \frac{1}{s-t}\) and \(\xi \geqslant \frac{1}{r}\), we write
$$\begin{aligned} J_{1}(x) = \frac{\sin (x)}{\sqrt{\pi } \sqrt{x}}-\frac{\cos (x)}{\sqrt{\pi } \sqrt{x}} + J_{1}(x) -\left( \frac{\sin (x)}{\sqrt{\pi } \sqrt{x}}-\frac{\cos (x)}{\sqrt{\pi } \sqrt{x}}\right) \end{aligned}$$
and use
$$\begin{aligned} \sin (y) \left( \frac{\sin (x)}{\sqrt{\pi x }}-\frac{\cos (x)}{\sqrt{\pi x }}\right) =\frac{-\sin (y-x)-\sin (x+y)+\cos (y-x)-\cos (x+y)}{2\sqrt{ \pi x }} \end{aligned}$$
with \(y=(s-t)\xi \) and \(x=r\xi \). Then, we integrate by parts in \(\xi \), and estimate all of the resulting terms directly. We use the same procedure for \(\partial _{r}u_{w,2}(t,r)\) in the region \(r \geqslant \frac{t}{2}\). This yields an extra factor of \(1+\frac{1}{\lambda (t)}\) relative to the estimates for \(u_{w,2}(t,r)\) in the region \(r \geqslant \frac{t}{2}\). (We can differentiate under the integral sign in (4.74) by the dominated convergence theorem, and the fact that \(\xi \widehat{RHS}_{2}(s,\xi ) \in L^{1}_{s,\xi }((t,\infty ) \times (0,\infty ))\)).
Finally, to obtain (4.67), we recall the definition of \(u_{w,2,ell}\), in (4.57).
$$\begin{aligned} u_{w,2,ell}(t,r) = -\frac{1}{2}\left( \frac{1}{r}\int _{0}^{r} s^{2} RHS_{2}(t,s) ds + r \int _{r}^{\infty } RHS_{2}(t,s) ds\right) \end{aligned}$$
If \(r \leqslant \frac{t}{2}\), we write
$$\begin{aligned} \partial _{t}^{k} u_{w,2,ell}(t,r){} & {} = \frac{-1}{2r} \int _{0}^{r} s^{2}\partial _{t}^{k}RHS_{2}(t,s) ds - \frac{r}{2}\int _{\frac{t}{2}}^{\infty } \partial _{t}^{k}RHS_{2}(t,s) ds\nonumber \\{} & {} \quad -\frac{r}{2}\int _{r}^{\frac{t}{2}} \partial _{t}^{k}RHS_{2}(t,s) ds \end{aligned}$$
(4.75)
If \(r \geqslant \frac{t}{2}\), we have
$$\begin{aligned} \partial _{t}^{k} u_{w,2,ell}(t,r){} & {} = \frac{-1}{2r} \int _{0}^{\frac{t}{2}} s^{2}\partial _{t}^{k}RHS_{2}(t,s) ds - \frac{1}{2r}\int _{\frac{t}{2}}^{r} s^{2}\partial _{t}^{k}RHS_{2}(t,s) ds \nonumber \\{} & {} \quad -\frac{r}{2}\int _{r}^{\infty } \partial _{t}^{k}RHS_{2}(t,s) ds \end{aligned}$$
(4.76)
The point of this decomposition is the following. The function \(\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\) is a symbol in t and r. The estimates on \(w_{1}(t,r)\) and \(v_{2}(t,r)\) from Lemmas 2 and 5 improve by a factor of \(\frac{1}{t}\) with each time derivative taken (for example, for up to 4 derivatives), in the region \(r \leqslant \frac{t}{2}\). On the other hand, \(v_{2}(t,r)\) is a free wave, and our estimates on, for example, \(\partial _{t}v_{2}(t,r)\) from Lemma 5 are not a power of t better than the estimates on \(v_{2}(t,r)\) in the region \(r \geqslant \frac{t}{2}\). Therefore, when \(s \geqslant \frac{t}{2}\), we rewrite \(\partial _{t}^{k} v_{2}(t,s)\) as in (4.71) (iterating these as needed, to treat \(k=3,4\)). Then, we insert this re-writing into (4.75) and (4.76), and integrate by parts in s to remove all s derivatives from \(v_{2}\) (or \(\left( \partial _{t}+\partial _{s}\right) v_{2}\), as appropriate). A direct estimation of all of the integral and boundary terms then gives rise to (4.67) and (4.68). The form of the estimates (4.67) and (4.68) are different in the region \(r \geqslant \frac{t}{2}\) because, for sufficiently regular f,
$$\begin{aligned}\begin{aligned}&-\frac{1}{2r} \int _{\frac{t}{2}}^{r} s^{2}f(t,s)\partial _{s}v_{2}(t,s) ds - \frac{r}{2}\int _{r}^{\infty } f(t,s)\partial _{s}v_{2}(t,s) ds\\&\quad =\frac{1}{2r} \left( \frac{t}{2}\right) ^{2} f(t,\frac{t}{2}) v_{2}(t,\frac{t}{2}) + \frac{1}{2r} \int _{\frac{t}{2}}^{r} v_{2}(t,s) \partial _{s}\left( s^{2}f(t,s)\right) ds + \frac{r}{2}\int _{r}^{\infty } v_{2}(t,s) \partial _{s}f(t,s) ds\end{aligned} \end{aligned}$$
In other words, the boundary terms at r arising from one integration by parts exactly cancel. This does not happen if \(\partial _{s}v_{2}(t,s)\) in the first line is replaced by \(\partial _{s}^{k} v_{2}(t,s)\), and we integrate by parts k times, for \(k \geqslant 2\). The estimate (4.66) follows from the explicit formulae (4.61) and (4.62). Finally, we prove the energy estimate, (4.69). By the dominated convergence theorem, and fact that \(\xi \widehat{RHS}_{2}(s,\xi ) \in L^{1}_{s,\xi }((t,\infty ) \times (0,\infty ))\), we can differentiate once in either t or r under the integral in (4.74). The estimate on \(||\partial _{t}u_{w,2}||_{L^{2}(r dr)}\) implied by (4.69) then follows directly from Minkowski’s inequality, the \(L^{2}\) isometry property of the Hankel transform of order 1, and the estimates from Lemmas 2 and 5. Next, we have
$$\begin{aligned} \partial _{r}u_{w,2}(t,r)+\frac{u_{w,2}(t,r)}{r} = \int _{t}^{\infty } ds \int _{0}^{\infty } d\xi \sin ((t-s)\xi ) \xi J_{0}(r\xi ) \widehat{RHS}_{2}(s,\xi ) \end{aligned}$$
Using again Minkowski’s inequality, and the \(L^{2}\) isometry property of the Hankel transforms of orders 0 and 1, we get
$$\begin{aligned} ||\partial _{r}u_{w,2}(t,r)+\frac{u_{w,2}(t,r)}{r}||_{L^{2}(r dr)} \leqslant C \int _{t}^{\infty } ds ||RHS_{2}(s,r)||_{L^{2}(r dr)}<\infty \end{aligned}$$
where the last inequality is true by the estimates from Lemmas 2 and 5. Therefore, by the dominated convergence theorem,
$$\begin{aligned} \int _{0}^{\infty } \left( \partial _{r}u_{w,2}(t,r)+\frac{u_{w,2}(t,r)}{r}\right) ^{2} rdr = \lim _{M \rightarrow \infty } \int _{\frac{1}{M}}^{M} \left( \partial _{r}u_{w,2}(t,r)+\frac{u_{w,2}(t,r)}{r}\right) ^{2} rdr \end{aligned}$$
Then, we note that
$$\begin{aligned} \left( \partial _{r}u_{w,2}(t,r)+\frac{u_{w,2}(t,r)}{r}\right) ^{2} = (\partial _{r}u_{w,2}(t,r))^{2}+\frac{u_{w,2}(t,r)^{2}}{r^{2}} + \frac{\partial _{r}(u_{w,2}^{2})}{r} \end{aligned}$$
For each \(M>0\),
$$\begin{aligned} \int _{\frac{1}{M}}^{M} \frac{\partial _{r}(u_{w,2}^{2})(t,r)}{r} rdr = u_{w,2}^{2}(t,M)-u_{w,2}^{2}(t,\frac{1}{M}) \end{aligned}$$
Again because of \(\xi \widehat{RHS}_{2}(s,\xi ) \in L^{1}_{s,\xi }((t,\infty ) \times (0,\infty ))\) and the dominated convergence theorem,
$$\begin{aligned} \lim _{r \rightarrow 0} \left( \int _{t}^{\infty } ds \int _{0}^{\infty } d\xi \sin ((s-t)\xi )\widehat{RHS_{2}}(s,\xi ) J_{1}(r\xi )\right) =0 \end{aligned}$$
Then, by (4.64), we get
$$\begin{aligned} \lim _{M \rightarrow \infty } \int _{\frac{1}{M}}^{M} \frac{\partial _{r}(u_{w,2}^{2})(t,r)}{r} rdr = 0 \end{aligned}$$
Therefore,
$$\begin{aligned} \infty > \int _{0}^{\infty } \left( \partial _{r}u_{w,2}(t,r)+\frac{u_{w,2}(t,r)}{r}\right) ^{2} rdr = \lim _{M \rightarrow \infty } \int _{\frac{1}{M}}^{M} \left( (\partial _{r}u_{w,2}(t,r))^{2}+\frac{u_{w,2}(t,r)^{2}}{r^{2}}\right) rdr \end{aligned}$$
By the monotone convergence theorem, we thus get
$$\begin{aligned} \int _{0}^{\infty } \left( (\partial _{r}u_{w,2}(t,r))^{2}+\frac{u_{w,2}(t,r)^{2}}{r^{2}}\right) rdr \leqslant C \left( \int _{t}^{\infty } ds ||RHS_{2}(s,r)||_{L^{2}(r dr)}\right) ^{2} \end{aligned}$$
Finally, (4.69) follows from
$$\begin{aligned} \begin{aligned}|u_{w,2}(t,r)|&\leqslant \int _{t}^{\infty } \int _{0}^{\infty } |\widehat{RHS_{2}}(s,\xi )| \sqrt{\xi } \frac{|J_{1}(r\xi )|}{\sqrt{\xi }} d\xi ds\\&\quad \leqslant \int _{t}^{\infty } \left( \int _{0}^{\infty } |\widehat{RHS_{2}(s,\xi )}|^{2} \xi d\xi \right) ^{1/2} \left( \int _{0}^{\infty } \frac{|J_{1}(x)|^{2}}{x} dx\right) ^{1/2}ds\\&\leqslant C \int _{t}^{\infty }||RHS_{2}(s,r)||_{L^{2}(r dr)}ds\end{aligned}\nonumber \\ \end{aligned}$$
(4.77)
\(\square \)
4.5 Second order matching, part 1
The first question is whether the choice of \(v_{2,0}\) allows for matching between the next order ( roughly on the order of \(\frac{r^{3} \lambda (t)}{t^{4}}\)) terms in \(w_{1}(t,r)\) and \(v_{ell,2}(t,\frac{r}{\lambda (t)})\). For clarity, we recall the fourth, eighth, and ninth lines of (4.29), and the expression (4.17). The main contribution of \(v_{ell,2,0}(t,\frac{r}{\lambda (t)})\) to the matching is
$$\begin{aligned} v_{ell,2,0,main}(t,\frac{r}{\lambda (t)}) = \frac{r^{3}}{8} \partial _{t}^{2}\left( \frac{\lambda '(t)^{2}}{2\lambda (t)} + \lambda ''(t) - \lambda ''(t) \log (\frac{r}{\lambda (t)})\right) + \frac{3}{32} r^{3} \lambda ''''(t) \end{aligned}$$
The \(r^{3}\) contribution of \(w_{1}(t,r)\) is (recall Lemma (1))
$$\begin{aligned}\begin{aligned}&w_{1,cubic,main}(t,r) \\&=\frac{3}{32} r^{3} \lambda ''''(t)+\frac{r^{3}}{8}\left( \lambda ''''(t) \left( \log (2)+\frac{1}{2}\right) -\log (r) \lambda ''''(t) + \log (t) \lambda ''''(t)\right. \\&\quad \left. + \int _{t}^{2t} \frac{\lambda ''''(s)-\lambda ''''(t)}{s-t} ds + \int _{2t}^{\infty } \frac{\lambda ''''(s) ds}{s-t}\right) \end{aligned}\end{aligned}$$
Finally, the \(v_{2}\) contribution at this order is
$$\begin{aligned} v_{2,cubic,main}(t,r) = \frac{-r^{3}}{32} F''(t) \end{aligned}$$
Recall that we already chose \(v_{2,0}\) to allow for matching of terms of the form rf(t) coming from \(v_{ell}(t,\frac{r}{\lambda (t)})\) and \(w_{1}(t,r)+v_{2}(t,r)\). Note that the coefficient of \(r^{3}\) given in the expression above for \(v_{2,cubic,main}\) is precisely one-eighth of two time derivatives of the r coefficient of \(v_{2}\) (see (4.37)). Although the \(r^{3}\) coefficient of \(v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})\) and that of \(w_{1,cubic,main}(t,r)\) are not precisely one-eighth of two time derivatives of the main r coefficients in the matching region of \(v_{ell}(t,\frac{r}{\lambda (t)})\) and \(w_{1}(t,r)\), respectively, the \(r^{3}\) coefficient of the difference of these two functions is precisely one-eighth of two time derivatives of the main r coefficient in the matching region of \(v_{ell}(t,\frac{r}{\lambda (t)})-w_{1}(t,r)\). Note the important cancellation between the \(\frac{3}{32} r^{3} \lambda ''''(t)\) terms when \(v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})\) and \(w_{1,cubic,main}(t,r)\) are subtracted. Therefore, the matching of the \(r^{3}\) terms is already accomplished with our previous choice of \(v_{2,0}\). In other words, by the choice of \(v_{2,0}\), we have
$$\begin{aligned} v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})-\left( w_{1,cubic,main}(t,r)+v_{2,cubic,main}(t,r)\right) =0 \end{aligned}$$
4.6 Second order matching, part 2
Next, we choose the coefficient \(c_{1}(t)\) in (4.7). For convenience, we define \(v_{ell,sub}\) by
$$\begin{aligned} v_{ell,sub}(t,R)= & {} v_{ell}(t,R) - R \left( \frac{1}{2}\lambda '(t)^{2} + \lambda (t) \lambda ''(t) - \lambda (t) \lambda ''(t) \log (R)\right) \nonumber \\= & {} \left( 2c_{1}(t)- 3\lambda '(t)^{2}\right) f_{4}(R) + \lambda (t) \lambda ''(t) f_{3}(R)\end{aligned}$$
(4.78)
with
$$\begin{aligned}{} & {} f_{4}(R) = \frac{R}{2(1+R^{2})} \\{} & {} f_{3}(R) = \frac{R^{2} \left( -3+2\left( 1+R^{2}\right) \log (R)\right) -\left( R^{4}-1\right) \log (1+R^{2})+2R^{2} Li_{2}(-R^{2})}{2R(1+R^{2})} \end{aligned}$$
The function \(c_{1}(t)\) appears as part of the coefficient of the \(\frac{1}{r}\) term in a large r expansion of \(u_{ell,sub}(t,r)=v_{ell,sub}(t,\frac{r}{\lambda (t)})\). On the other hand, there are also higher order terms, of size \(\frac{\log (r)}{r}\) and \(\frac{\log ^{2}(r)}{r}\) appearing in the expansion of \(u_{ell,sub}(t,r)\) for large r, and \(c_{1}(t)\) does not appear in the coefficients of these terms. However, it turns out that these terms happen to exactly match corresponding \(\frac{\log (r)}{r}\) and \(\frac{\log ^{2}(r)}{r}\) terms coming from expansions of \(v_{ex}\) and \(u_{w,2,ell,0}\), which means that we only need to match terms of size \(\frac{1}{r}\), which can be done by choosing an appropriate \(c_{1}(t)\). To show this, we note the following.
The main contribution of \(v_{ell,sub}(t,\frac{r}{\lambda (t)})\) in the matching region is
$$\begin{aligned}{} & {} v_{ell,sub,cont}(t,r)\nonumber \\{} & {} \quad = \frac{c_{1}(t) \lambda (t)-\frac{3}{2} \lambda (t) \lambda '(t)^2}{r} \nonumber \\{} & {} \qquad -\frac{\lambda (t)^2 \lambda ''(t) \left( -6 \log (r) (4 \log (\lambda (t))+1)+6 \log (\lambda (t)) (2 \log (\lambda (t))+1)+12 \log ^2(r)+\pi ^2+12\right) }{6 r}\nonumber \\ \end{aligned}$$
(4.79)
On the other hand, the main contribution of \(v_{ex}(t,r)\) in the matching region is (recall (4.36))
$$\begin{aligned} v_{ex,cont}(t,r)= & {} \frac{1}{2r} \left( \log (r)\left( -4 \lambda (t)\lambda '(t)^{2}-2 \lambda (t)^{2}\lambda ''(t)\right) \right. \nonumber \\{} & {} \left. + 4\lambda (t)\log (\lambda (t))\lambda '(t)^{2} -\lambda (t)^{2}\lambda ''(t) + 2 \lambda (t)^{2}\lambda ''(t)\log (\lambda (t))\right) \nonumber \\ \end{aligned}$$
(4.80)
Finally, the main contribution of \(u_{w,2,ell,0}(t,r)\) in the matching region is (recall (4.62))
$$\begin{aligned}\begin{aligned}&u_{w,2,ell,0,cont}(t,r) = \frac{-2}{r}\left( \log (r)\left( -\lambda (t)\lambda '(t)^{2}-\lambda (t)^{2}\lambda ''(t)-2 \lambda (t)^{2} \log (\lambda (t))\lambda ''(t)\right) \right. \\&\left. +\lambda (t) \log (\lambda (t))\lambda '(t)^{2}+\lambda (t)^{2}\log (\lambda (t))\lambda ''(t) + \lambda (t)^{2}\log ^{2}(\lambda (t))\lambda ''(t)\right. \\&+\left. \lambda (t)^{2}\log ^{2}(r)\lambda ''(t)+\frac{1}{2} \lambda (t)^2 \lambda ''(t)-\frac{1}{12} \pi ^2 \lambda (t)^2 \lambda ''(t)\right) \end{aligned} \end{aligned}$$
These expressions give
$$\begin{aligned}{} & {} v_{ell,sub,cont}(t,r)-\left( u_{w,2,ell,0,cont}(t,r)+v_{ex,cont}(t,r)\right) \\{} & {} \quad = -\frac{\lambda (t) \left( -6 c_{1}(t)+\left( 3+2 \pi ^2\right) \lambda (t) \lambda ''(t)+9 \lambda '(t)^2\right) }{6 r} \end{aligned}$$
(As previously mentioned, the \(\frac{\log (r)}{r}\) and \(\frac{\log ^{2}(r)}{r}\) terms from \(v_{ell,sub,cont}\), \(u_{w,2,ell,0,cont}\) and \(v_{ex,cont}\) cancel in the above expression). We therefore choose \(c_{1}(t)\) so as to make
$$\begin{aligned} v_{ell,sub,cont}(t,r)-\left( u_{w,2,ell,0,cont}(t,r)+v_{ex,cont}(t,r)\right) =0 \end{aligned}$$
by defining
$$\begin{aligned} c_{1}(t) = \frac{3}{2}\lambda '(t)^{2}+\frac{\lambda (t) \lambda ''(t)}{2}+\frac{\pi ^{2}}{3} \lambda (t)\lambda ''(t) \end{aligned}$$
(4.81)
Note that the matching condition used to determine the initial velocity of \(v_{2}\) is precisely
$$\begin{aligned} u_{ell}(t,r) - v_{ell,sub}(t,\frac{r}{\lambda (t)}) = v_{2}(t,r) - v_{2,sub}(t,r) + w_{1}(t,r)-w_{1,sub}(t,r) \nonumber \\ \end{aligned}$$
(4.82)
where we recall that \(w_{1,sub}\) and \(v_{2,sub}\) were defined in (4.26) and (4.48), respectively.
4.7 Third order matching
It will be convenient for us to define
$$\begin{aligned}{} & {} v_{ex,sub}(t,r) := v_{ex}(t,r)-v_{ex,ell}(t,r) \nonumber \\{} & {} \quad = -\int _{0}^{\infty } d\xi J_{1}(r\xi ) \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi ^{2}} \partial _{s}^{2}\widehat{RHS}(s,\xi ) \end{aligned}$$
(4.83)
We will now choose \(v_{2,2}\) (recall (4.50)) so as to match the principal terms (which are of the form \(r \log ^{k}(r) g_{k}(t)\)) in \(u_{ell,2}(t,r)-v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})\) and \(v_{ex,sub}(t,r)+u_{w,2}(t,r) - u_{w,2,ell,0,cont}(t,r)\). The main proposition of this section is the following.
Proposition 2
(Third order matching). For \(v_{2,2}\) defined in (4.50), where \(v_{2,3}\) is defined by (4.134), and \(F_{3}\) and \(G_{3}\) are given in (4.131) and (4.133), respectively, we have the following estimate. For \(0 \leqslant j,k \leqslant 1\) or \(k=2,j=0\), and \(g(t)\lambda (t) \leqslant r \leqslant 2 g(t)\lambda (t)\),
$$\begin{aligned}\begin{aligned} |&\partial _{t}^{j}\partial _{r}^{k}\left( u_{ell,2}(t,r)-v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})-\left( u_{w,2}(t,r)-u_{w,2,ell,0,cont}(t,r)\right. \right. \\&\quad \left. \left. +v_{2,2}(t,r)+v_{ex}-v_{ex,cont}-q_{4,2}\right) \right) | \\&\leqslant \frac{C \lambda (t)^{2-k} \log (t)}{t^{2+j} g(t)^{3+k}} + \frac{C g(t)^{2-k} \lambda (t)^{4-k} \log ^{4}(t) \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) }{t^{5+j}}\end{aligned} \end{aligned}$$
where \(q_{4,2}\) is defined in (4.116) and satisfies the following inequality for \(k=0,1\), and \(r >0\).
$$\begin{aligned}\begin{aligned}&|\partial _{t}^{k}q_{4,2}(t,r)| + \sqrt{E(\partial _{t}^{k}q_{4,2},\partial _{t}^{k+1}q_{4,2})} +||\partial _{r}^{2} q_{4,2}||_{L^{2}((\lambda (t)g(t), 2 \lambda (t)g(t)),r dr)}\leqslant \frac{C \lambda (t)^{5} \log ^{3}(t)}{t^{5} }\end{aligned} \end{aligned}$$
Proof
We start by computing the leading parts of \(u_{ell,2}(t,r)-v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})\) and \(v_{ex,sub}(t,r)+u_{w,2}(t,r) - u_{w,2,ell,0,cont}(t,r)\) in the matching region, \(r \sim g(t)\lambda (t)\).
We recall our expressions for \(v_{ex,sub}\) and \(u_{w,2}-u_{w,2,ell}\):
$$\begin{aligned} v_{ex,sub}(t,r)= & {} -\int _{0}^{\infty } d\xi J_{1}(r\xi ) \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi ^{2}} \partial _{s}^{2}\widehat{RHS}(s,\xi )\nonumber \\ u_{w,2}(t,r)-u_{w,2,ell}(t,r)= & {} \int _{t}^{\infty } ds \int _{0}^{\infty } d\xi \frac{\sin ((s-t)\xi )}{\xi ^{2}} \partial _{s}^{2} \widehat{RHS_{2}}(s,\xi ) J_{1}(r\xi ) \nonumber \\ \end{aligned}$$
(4.84)
In particular, these functions solve the following equations with 0 Cauchy data at infinity
$$\begin{aligned} \left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) \left( u_{w,2}-u_{w,2,ell}\right) = \partial _{t}^{2} u_{w,2,ell} \\ \left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) v_{ex,sub} = \partial _{t}^{2}v_{ex,ell} \end{aligned}$$
In order to compute their leading behavior in the matching region, \(r \sim g(t)\lambda (t)\), we define \(u_{w,2,sub,0}(t,r)\) and \(v_{ex,sub,0}(t,r)\) to be the solutions to the following equations with 0 Cauchy data at infinity
$$\begin{aligned}{} & {} \left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) u_{w,2,sub,0} = \partial _{t}^{2} u_{w,2,ell,0,cont} \nonumber \\{} & {} \left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) v_{ex,sub,0}= \partial _{t}^{2} v_{ex,cont} \end{aligned}$$
(4.85)
Each of these right-hand sides are of the following form (recall Sect. 4.6)
$$\begin{aligned} \frac{1}{r}\sum _{k=0}^{2}f_{k}(t)\log ^{k}(r) \end{aligned}$$
In particular, the right-hand sides of the equations for \(u_{w,2,sub,0}\) and \(v_{ex,sub,0}\) are technically singular at \(r=0\), even though the right-hand sides of the equations for \(u_{w,2}-u_{w,2,ell}\) and \(v_{ex,sub}\) are not. This will not cause any problems for us, and is done so that the principal parts of \(u_{w,2,sub,0}\) and \(v_{ex,sub,0}\) can be exactly calculated.
We will first prove the following lemma which will lead to fairly explicit formulae for \(u_{w,2,sub,0}\) and \(v_{ex,sub,0}\). After the lemma we will provide some information about how the formulae arise.
Lemma 11
Let \(f_{k}\) be the functions arising from expressing either \(\partial _{t}^{2}v_{ex,cont}\) or \(\partial _{t}^{2}u_{w,2,ell,0,cont}\) in the form \(\frac{1}{r}\sum _{k=0}^{2}f_{k}(t)\log ^{k}(r)\). If u is given by
$$\begin{aligned} \begin{aligned} u(t,r) = -\int _{t}^{\infty } f_{1}(s) K_{0}\left( \frac{r}{s-t},r\right) ds \end{aligned} \end{aligned}$$
(4.86)
where
$$\begin{aligned}\begin{aligned} K_{0}(x,r) = \frac{1}{x} {\left\{ \begin{array}{ll} \left( \left( 1-\sqrt{1-x^2}\right) \left( \log (r)-\log (x)\right) +\sqrt{1-x^2}-1+\sqrt{1-x^2} \log \left( \sqrt{1-x^2}+1\right) \right. \\ \left. +\log \left( \sqrt{1-x^2}+1\right) -2 \sqrt{1-x^2} \log \left( 2 \sqrt{1-x^2}\right) \right) , \quad 0<x<1\\ \left( \sqrt{x^{2}-1} \sin ^{-1}\left( \frac{1}{x}\right) +\log (r)-1\right) , \quad x>1\end{array}\right. } \end{aligned} \end{aligned}$$
then, u solves
$$\begin{aligned} -\partial _{tt}u+\partial _{rr}u+\frac{1}{r}\partial _{r}u-\frac{u}{r^{2}} = \frac{f_{1}(t) \log (r)}{r}, \quad r>0 \end{aligned}$$
with 0 Cauchy data at infinity. Similarly, if u is defined by
$$\begin{aligned} u(t,r) = \int _{t}^{\infty } f_{2}(s) K_{2}(\frac{s-t}{r},r)ds \end{aligned}$$
(4.87)
with
$$\begin{aligned} K_{2}(a,r)=\sum _{k=0}^{2} F_{k}(a) \log ^{k}(r) \end{aligned}$$
where, for \(a>1\), we have
$$\begin{aligned}&F_{0}(a)=a \left( \text {Li}_2\left( \frac{\sqrt{a^2-1}}{a}\right) +\text {Li}_2\left( -\frac{a}{\sqrt{a^2-1}}\right) +\text {Li}_2\left( 1-2 a \left( a+\sqrt{a^2-1}\right) \right) \right. \\&\quad \left. \left. -\text {Li}_2\left( 2-2 a \left( a+\sqrt{a^2-1}\right) \right) +2\right) +\frac{1}{2}\left( 4 \sqrt{a^2-1} \text {Li}_2\left( 2 a \left( a-\sqrt{a^2-1}\right) -1\right) \right. \right. \\&\quad +\frac{1}{4} a \log ^2\left( 1-\frac{1}{a^2}\right) \\&\quad \left. +2 \left( \sqrt{a^2-1}+a\right) \log ^2\left( \sqrt{a^2-1}+a\right) +4 \left( -a \log \left( 2 \left( a^2-1\right) \right) \right. \right. \\&\quad -\sqrt{a^2-1} \left( \log \left( a^2-1\right) -1+\log (4)\right) \\&\quad \left. \left. +a+a \log (a)\right) \log \left( \sqrt{a^2-1}+a\right) +\frac{1}{6} \sqrt{a^2-1} \left( 12 \log \left( a^2-1\right) \left( \log \left( a^2-1\right) -2+\log (16)\right) \right. \right. \\&\quad \left. \left. -\pi ^2+24+24 (\log (2)-1) \log (4)\right) -8 a\right) , \quad a>1\\&F_{1}(a) = 2 \left( -\sqrt{a^2-1}-\left( \sqrt{a^2-1}+a\right) \log \left( \sqrt{a^2-1}+a\right) +\sqrt{a^2-1} \log \left( 4 \left( a^2-1\right) \right) +a\right) ,\\&\quad \quad a>1\\&F_{2}(a) = \sqrt{a^2-1}-a, \quad a>1 \end{aligned}$$
and, for \(0<a<1\), we have
then, u solves
$$\begin{aligned} -\partial _{tt}u+\partial _{rr}u+\frac{1}{r}\partial _{r}u-\frac{u}{r^{2}} = \frac{f_{2}(t) \log ^{2}(r)}{r}, \quad r>0 \end{aligned}$$
with 0 Cauchy data at infinity.
Proof
To verify that (4.86) is the claimed solution to the Cauchy problem, we start with
$$\begin{aligned} u(t,r) = -\int _{0}^{1} K_{0}(\frac{1}{y},r) f_{1}(t+r y) r dy -\int _{1}^{\infty } K_{0}(\frac{1}{y},r) f_{1}(t+r y) r dy \end{aligned}$$
and note that the explicit formulae for \(K_{0}(x,r)\) in the regions \(x<1\) and \(x>1\), combined with the definitions of \(f_{k}\) and the Dominated convergence theorem allow us to differentiate up to two times in either t or r under the integral signs. An integration by parts then establishes the lemma. (The same procedure is used to verify the stated property of (4.87)). \(\square \)
Remark
To see how the formulae from the lemma can be formally derived, we could first use the same procedure used for \(w_{1}\) to obtain, for example, for \(u_{w,2,sub,0}\), that
$$\begin{aligned} u_{w,2,sub,0}(t,r) = \int _{t}^{\infty } u_{w,2,sub,0,s}(t,r)ds \end{aligned}$$
where
$$\begin{aligned}{} & {} u_{w,2,sub,0,s}(t,r) \nonumber \\{} & {} \quad = \frac{-1}{2\pi } \int _{0}^{s-t}\frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{2\pi } d\theta \frac{\left( r+\rho \cos (\theta )\right) }{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )} \nonumber \\{} & {} \qquad \sum _{k=0}^{2} f_{k}(s) \log ^{k}\left( \sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}\right) \end{aligned}$$
(4.88)
The integral of the \(k=0\) term in the sum above has been evaluated when computing \(w_{1}\), via Cauchy’s residue theorem. This same procedure can not directly be applied to the integrals of the \(k \ne 0\) terms. We can still explicitly compute these integrals, with a procedure involving introducing a parameter into the integrals, differentiating in this parameter, and then using Cauchy’s residue theorem. This gives the following. For \(r \ne \rho \), we have
$$\begin{aligned}{} & {} \begin{aligned}&\int _{0}^{2\pi } 4 \log (r^{2}+\rho ^{2}+2 r \rho \cos (\theta ))\frac{\left( r+\rho \cos (\theta )\right) }{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )} d\theta \\&\quad ={\left\{ \begin{array}{ll} \frac{8 \pi }{r} \left( \log (r^{2})+\log (1-\frac{\rho ^{2}}{r^{2}})\right) , \quad r>\rho \\ \frac{-8 \pi }{r} \log (1-\frac{r^{2}}{\rho ^{2}}), \quad \rho>r\end{array}\right. } \end{aligned}\\{} & {} \begin{aligned}&\int _{0}^{2\pi } \frac{\log ^{2}(r^{2}+\rho ^{2}+2 r \rho \cos (\theta ))}{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}\left( r+\rho \cos (\theta )\right) d\theta \\&\quad = {\left\{ \begin{array}{ll} \frac{4 \pi }{r} \left( \log ^{2}(1-\frac{\rho ^{2}}{r^{2}})+2 \log (r)\log (r-\frac{\rho ^{2}}{r})+\text {Li}_{2}(\frac{\rho ^{2}}{r^{2}})\right) , \quad r> \rho \\ \frac{-4\pi }{r} \log (\rho ^{2}-r^{2})\log (1-\frac{r^{2}}{\rho ^{2}}), \quad \rho >r\end{array}\right. } \end{aligned} \end{aligned}$$
Carrying out the \(\rho \) integral in (4.88) then leads to the formulae in the Lemma statement.
Using the information from Lemma 11, we can calculate the principal parts of \(u_{w,2,sub,0}(t,r)\) and \(v_{ex,sub,0}(t,r)\) in the matching region, \(r \sim g(t)\lambda (t)\). We describe in detail how this is done for \(u_{w,2,sub,0}\), since \(v_{ex,sub,0}\) can be treated in the same way. First, we introduce some notation. Let
$$\begin{aligned} \begin{aligned} K_{2,0}(\frac{s-t}{r},r)&=\frac{r \left( -12 \log (s-t) \log (4 (s-t))+\pi ^2-12 \log ^2(2)\right) }{24 (s-t)}\end{aligned} \end{aligned}$$
(4.89)
We define \(F_{k,0}\) by
$$\begin{aligned} K_{2,0}(a,r) = \sum _{k=0}^{2} F_{k,0}(a) \log ^{k}(r) \end{aligned}$$
(4.90)
We also recall the definition of \(u_{w,2,ell,0,cont}\), (4.62), and write
$$\begin{aligned} \partial _{t}^{2}u_{w,2,ell,0,cont}(t,r) = \frac{g_{0}''(t)}{r}+\frac{g_{1}''(t)\log (r)}{r}+\frac{g_{2}''(t) \log ^{2}(r)}{r} \end{aligned}$$
(4.91)
for
$$\begin{aligned}{} & {} \begin{aligned} g_{0}(t) = -2&\left( \frac{1}{2} \lambda (t)^2 \lambda ''(t)-\frac{1}{12} \pi ^2 \lambda (t)^2 \lambda ''(t)+\lambda (t)^2 \lambda ''(t) \log ^2(\lambda (t))\right. \\&\quad \left. +\lambda (t)^2 \lambda ''(t) \log (\lambda (t))+\lambda (t) \lambda '(t)^2 \log (\lambda (t))\right) \end{aligned} \\{} & {} g_{1}(t) = -2 \left( -\lambda (t)^2 \lambda ''(t)-2 \lambda (t)^2 \lambda ''(t) \log (\lambda (t))-\lambda (t) \lambda '(t)^2\right) \\{} & {} g_{2}(t) = -2 \lambda (t)^2 \lambda ''(t) \end{aligned}$$
Finally, we let
$$\begin{aligned} c_{0}=\int _{1}^{\infty } \left( F_{0}(a)-F_{0,0}(a)\right) da \end{aligned}$$
Note that the exact value of \(c_{0}\) is not needed for our purposes, though the precise value of many other constants appearing in the following lemma are needed.
Lemma 12
[Leading part of \(u_{w,2,sub,0}\)]. Let \(u_{w,2,sub,0,cont}\) be defined by
$$\begin{aligned}\begin{aligned}&u_{w,2,sub,0,cont}(t,r)\\&\quad =-\frac{1}{2}g_{0}''(t) r -\frac{r}{2} g_{0}''(t) \log (\frac{t}{r}) - \frac{r}{2} \int _{t}^{2t}\frac{(g_{0}''(s)-g_{0}''(t))}{(s-t)} ds \\&\qquad - \frac{r}{2} \int _{2t}^{\infty } \frac{g_{0}''(s)}{s-t} ds - g_{0}''(t) \frac{r}{4} \left( \log (4)-1\right) \\&\qquad -\frac{r}{16} g_{1}''(t) \left( -12+\pi ^{2}+8 \log (r)\right) +-g_{1}''(t)r \left( \frac{\log (\frac{t}{r})\log (2)}{2}+\frac{1}{4}\left( \log ^{2}(t)-\log ^{2}(r)\right) \right) \\&\qquad -\frac{r}{2} \int _{2t}^{\infty } \frac{g_{1}''(s)}{(s-t)} \log (2(s-t)) ds-\frac{r}{2} \int _{t}^{2t}\left( g_{1}''(s)-g_{1}''(t)\right) \frac{\log (2(s-t))}{(s-t)} ds\\&\qquad + \frac{-r g_{1}''(t)}{24}\left( 6 (\log (4)-1) \log (r)-2 \pi ^2+15+6 \log ^2(2)\right) \\&\qquad -\frac{r g_{2}''(t)}{16} \left( 2 \log (r) \left( 4 \log (r)+\pi ^2-12\right) -28 \zeta (3)+\pi ^2+28\right) \\&\qquad +g_{2}''(t) \frac{r}{24} \log \left( \frac{r}{t}\right) \left( 4 \log (t) \log (8 r t)+4 \log (r) \log (8 r)-\pi ^2+12 \log ^2(2)\right) \\&\qquad +\int _{t}^{2t} \left( g_{2}''(s)-g_{2}''(t)\right) K_{2,0}(\frac{s-t}{r},r) ds+\int _{2t}^{\infty } g_{2}''(s)K_{2,0}(\frac{s-t}{r},r) ds\\&\qquad + g_{2}''(t) r \left( c_{0}+\frac{1}{12} \left( 2 \pi ^2-15-6 \log ^2(2)\right) \log (r) +\frac{1}{4}\left( 1-\log (4)\right) \log ^{2}(r)\right) \end{aligned} \end{aligned}$$
Then, for \(0 \leqslant j+ k \leqslant 2\)
$$\begin{aligned}{} & {} |r^{k}t^{j}\partial _{r}^{k}\partial _{t}^{j}\left( u_{w,2,sub,0}(t,r)-u_{w,2,sub,0,cont}(t,r)\right) | \leqslant \frac{C r^{3} \lambda (t)^{3} \log ^{5}(t)}{t^{6} }, \\{} & {} \quad \quad \lambda (t)g(t) \leqslant r \leqslant 2 g(t)\lambda (t) \end{aligned}$$
Proof
Using (4.91) and our calculations for \(w_{1}\), along with Lemma 11, we have
$$\begin{aligned} \begin{aligned}u_{w,2,sub,0}(t,r)&= \frac{-1}{r} \int _{t}^{t+r} g_{0}''(s)(s-t) ds -\frac{1}{r} \int _{t+r}^{\infty } g_{0}''(s)\left( (s-t)-\sqrt{(s-t)^{2}-r^{2}}\right) ds\\&\quad -\int _{t}^{\infty } g_{1}''(s) K_{0}\left( \frac{r}{s-t},r\right) ds + \int _{t}^{\infty } g_{2}''(s)K_{2}\left( \frac{s-t}{r},r\right) ds\end{aligned}\nonumber \\ \end{aligned}$$
(4.92)
The first two terms on the right-hand side of the above expression are treated with the identical procedure used to study \(w_{1}\), simply replacing \(\lambda \) in the \(w_{1}\) expressions with \(\frac{-g_{0}}{2}\). The leading part of the sum of these two terms in the matching region \(r \sim g(t)\lambda (t)\) is (note that we will prove concrete estimates on the terms we claim are subleading)
$$\begin{aligned}{} & {} -\frac{1}{2}g_{0}''(t) r -\frac{r}{2} g_{0}''(t) \log (\frac{t}{r}) - \frac{r}{2} \int _{t}^{2t}\frac{(g_{0}''(s)-g_{0}''(t))}{(s-t)} ds\\{} & {} \quad - \frac{r}{2} \int _{2t}^{\infty } \frac{g_{0}''(s)}{s-t} ds - g_{0}''(t) \frac{r}{4} \left( \log (4)-1\right) \end{aligned}$$
In fact, by inspecting (4.28), we have
$$\begin{aligned}\begin{aligned}&|\frac{-1}{r} \int _{t}^{t+r} g_{0}''(s)(s-t) ds -\frac{1}{r} \int _{t+r}^{\infty } g_{0}''(s)\left( (s-t)-\sqrt{(s-t)^{2}-r^{2}}\right) ds \\&\qquad - \left( -\frac{1}{2}g_{0}''(t) r -\frac{r}{2} g_{0}''(t) \log (\frac{t}{r}) - \frac{r}{2} \int _{t}^{2t}\frac{(g_{0}''(s)-g_{0}''(t))}{(s-t)} ds \right. \\&\qquad \left. - \frac{r}{2} \int _{2t}^{\infty } \frac{g_{0}''(s)}{s-t} ds - g_{0}''(t) \frac{r}{4} \left( \log (4)-1\right) \right) | \\&\quad \leqslant \frac{C r^{3} \lambda (t)^{3} \log ^{2}(t) \left( 1+\log (\frac{t}{r})\right) }{t^{6} }, \quad r \leqslant t\end{aligned} \end{aligned}$$
Estimates on the t and r derivatives are done similarly, as in the proof of Lemma 1. The third and fourth terms of (4.92) are treated with the same argument, which is essentially the same argument we used for \(w_{1}\), but with a few differences in the details. For clarity, we start with the third term. We have
$$\begin{aligned}{} & {} -\int _{t}^{\infty } g_{1}''(s) K_{0}\left( \frac{r}{s-t},r\right) ds\nonumber \\{} & {} \quad = -\int _{t}^{t+r} g_{1}''(s)K_{0}\left( \frac{r}{s-t},r\right) ds - \int _{t+r}^{\infty } g_{1}''(s) K_{0}\left( \frac{r}{s-t},r\right) ds \end{aligned}$$
(4.93)
With \(y=\frac{s-t}{r}\), we get
$$\begin{aligned}\begin{aligned} -\int _{t}^{t+r} g_{1}''(s)K_{0}\left( \frac{r}{s-t},r\right) ds&= -r\int _{0}^{1} g_{1}''(t+r y) K_{0}\left( \frac{1}{y},r\right) dy \\&= -r\int _{0}^{1} g_{1}''(t+ry) \left( K_{0,0}(\frac{1}{y})+K_{0,1}(\frac{1}{y})\log (r)\right) dy\end{aligned} \end{aligned}$$
where we note that, for \(x >1\),
$$\begin{aligned}{} & {} K_{0}(x,r) = K_{0,0}(x) + K_{0,1}(x) \log (r), \\{} & {} \quad K_{0,0}(x) = \frac{\sqrt{x^{2}-1} \sin ^{-1}\left( \frac{1}{x}\right) -1}{x}, \quad K_{0,1}(x) = \frac{1}{x} \end{aligned}$$
Since \(g(t)\lambda (t) \ll t\), the first term in the following decomposition is lower order relative to the second term, for \(r \sim g(t)\lambda (t)\)
$$\begin{aligned}\begin{aligned}&-r\int _{0}^{1} g_{1}''(t+ry) \left( K_{0,0}(\frac{1}{y})+K_{0,1}(\frac{1}{y})\log (r)\right) dy\\&\quad =-rg_{1}''(t)\int _{0}^{1} \left( K_{0,0}(\frac{1}{y})+K_{0,1}(\frac{1}{y})\log (r)\right) dy \\&\qquad -r\int _{0}^{1} \left( g_{1}''(t+ry)-g_{1}''(t)\right) \left( K_{0,0}(\frac{1}{y})+K_{0,1}(\frac{1}{y})\log (r)\right) dy\end{aligned} \end{aligned}$$
We thus get that the leading contribution from \(-r\int _{0}^{1} g_{1}''(t+ry) \left( K_{0,0}(\frac{1}{y})+K_{0,1}(\frac{1}{y})\log (r)\right) dy\) is
$$\begin{aligned} -\frac{r}{16} g_{1}''(t) \left( -12+\pi ^{2}+8 \log (r)\right) \end{aligned}$$
(4.94)
We treat the next term in (4.93), namely \(-\int _{t+r}^{\infty } g_{1}''(s) K_{0}\left( \frac{r}{s-t},r\right) ds.\) For this term, we start by decomposing \(K_{0}(x,r)\) into the leading piece for small x plus the remainder.
$$\begin{aligned}{} & {} K_{0}(x,r) = K_{0,sm}(x,r) + O\left( x^{3}\left( 1+|\log (\frac{r}{x})|\right) \right) , \quad x \rightarrow 0, \\{} & {} \quad K_{0,sm}(x,r) = \frac{1}{2} x (\log (r)-\log (x)+\log (2)) \end{aligned}$$
We therefore get
$$\begin{aligned} -\int _{t+r}^{\infty } g_{1}''(s) K_{0}\left( \frac{r}{s-t},r\right) ds= & {} -g_{1}''(t) \int _{t+r}^{2t} K_{0,sm}(\frac{r}{s-t},r) ds \nonumber \\{} & {} - \int _{t}^{2t} \left( g_{1}''(s)-g_{1}''(t)\right) K_{0,sm}(\frac{r}{s-t},r) ds \nonumber \\{} & {} +\int _{t}^{t+r}\left( g_{1}''(s)-g_{1}''(t)\right) K_{0,sm}(\frac{r}{s-t},r) ds \nonumber \\{} & {} - \int _{2t}^{\infty } g_{1}''(s) K_{0,sm}(\frac{r}{s-t},r)ds\nonumber \\{} & {} -\int _{t+r}^{\infty } g_{1}''(s) \left( K_{0}-K_{0,sm}\right) (\frac{r}{s-t},r) ds \end{aligned}$$
(4.95)
This decomposition also appears in the \(w_{1}\) expression (4.28). The leading contribution from the first four terms of the right-hand side of (4.95) is
$$\begin{aligned} \begin{aligned}&-g_{1}''(t)r \left( \frac{\log (\frac{t}{r})\log (2)}{2}+\frac{1}{4}\left( \log ^{2}(t)-\log ^{2}(r)\right) \right) \\&-\frac{r}{2} \int _{2t}^{\infty } \frac{g_{1}''(s)}{(s-t)} \log (2(s-t)) ds-\frac{r}{2} \int _{t}^{2t}\left( g_{1}''(s)-g_{1}''(t)\right) \frac{\log (2(s-t))}{(s-t)} ds\end{aligned} \end{aligned}$$
(4.96)
Moreover, the difference between \(-\int _{t}^{\infty } g_{1}''(s) K_{0}\left( \frac{r}{s-t},r\right) ds\) and the leading parts of the terms considered thus far is equal to
$$\begin{aligned}{} & {} -\int _{t}^{t+r} g_{1}''(s) K_{0}(\frac{r}{s-t},r) ds - \left( \frac{-r g_{1}''(t)}{16} \left( -12+\pi ^{2} + 8 \log (r)\right) \right) \\{} & {} \quad -\int _{t+r}^{\infty } g_{1}''(s) K_{0}\left( \frac{r}{s-t},r\right) ds-\left( -g_{1}''(t)r \left( \frac{\log (\frac{t}{r})\log (2)}{2}+\frac{1}{4}\left( \log ^{2}(t)-\log ^{2}(r)\right) \right) \right. \\{} & {} \quad \left. -\frac{r}{2} \int _{2t}^{\infty } \frac{g_{1}''(s)}{(s-t)} \log (2(s-t)) ds-\frac{r}{2} \int _{t}^{2t}\left( g_{1}''(s)-g_{1}''(t)\right) \frac{\log (2(s-t))}{(s-t)} ds\right) \\{} & {} \quad = \int _{t}^{t+r} \left( g_{1}''(s)-g_{1}''(t)\right) K_{0,sm}(\frac{r}{s-t},r) ds - \int _{t+r}^{\infty } g_{1}''(s) \left( K_{0}-K_{0,sm}\right) (\frac{r}{s-t},r) ds \\{} & {} \quad -r \int _{0}^{1} \left( g_{1}''(t+r y)-g_{1}''(t)\right) K_{0}(\frac{1}{y},r) dy \end{aligned}$$
We have
$$\begin{aligned}\begin{aligned}&-r \int _{0}^{1} \left( g_{1}''(t+r y)-g_{1}''(t)\right) K_{0}(\frac{1}{y},r) dy + \int _{t}^{t+r} \left( g_{1}''(s)-g_{1}''(t)\right) K_{0,sm}(\frac{r}{s-t},r) ds\\&\quad = -r \int _{0}^{1} \left( g_{1}''(t+r y)-g_{1}''(t)-r y g_{1}'''(t)\right) K_{0}(\frac{1}{y},r) dy \\&\qquad + \int _{t}^{t+r} \left( g_{1}''(s)-g_{1}''(t)-(s-t) g_{1}'''(t)\right) K_{0,sm}(\frac{r}{s-t},r) ds\\&\qquad -r^{2} \int _{0}^{1} y g_{1}'''(t) K_{0}(\frac{1}{y},r) dy + \int _{t}^{t+r} g_{1}'''(t) (s-t) K_{0,sm}(\frac{r}{s-t},r) ds\end{aligned} \end{aligned}$$
which gives
$$\begin{aligned}\begin{aligned}&|-r \int _{0}^{1} \left( g_{1}''(t+r y)-g_{1}''(t)\right) K_{0}(\frac{1}{y},r) dy + \int _{t}^{t+r} \left( g_{1}''(s)-g_{1}''(t)\right) K_{0,sm}(\frac{r}{s-t},r) ds\\&-\left( \frac{r^{2} g_{1}'''(t)}{18}\left( -7+\log (512)+3 \log (r)\right) \right) |\\&\leqslant \frac{C r^{3} \left( 1+|\log (r)|\right) \lambda (t)^{3} \log (t)}{t^{6} }\end{aligned} \end{aligned}$$
The higher derivatives are estimated similarly, for example, by writing
$$\begin{aligned}{} & {} \int _{t}^{t+r} \left( g_{1}''(s)-g_{1}''(t)-(s-t) g_{1}'''(t)\right) K_{0,sm}(\frac{r}{s-t},r) ds \\{} & {} \quad = \frac{r}{2}\int _{0}^{1}\left( g_{1}''(t+r y)-g_{1}''(t)-r y g_{1}'''(t)\right) \log (2 r y) \frac{dy}{y} \end{aligned}$$
and using the symbol-type estimates on \(\lambda \) to justify the differentiation under the integral, exactly as in the proof of Lemma 1. We now turn attention to the fifth term on the right-hand side of (4.95). The point here is that \((K_{0}-K_{0,sm})(\frac{r}{s-t},r)\) decays faster in \(s-t\) than does \(K_{0,sm}(\frac{r}{s-t},r)\). To exploit this, we integrate by parts in s in the following integral, and the symbol type estimates on \(g_{1}''(s)\) show that the non-boundary terms obtained are subleading relative to the boundary terms (this will be proven once we estimate the terms that we claim are subleading)
$$\begin{aligned} \begin{aligned}&-\int _{t+r}^{\infty } g_{1}''(s) \left( K_{0}-K_{0,sm}\right) (\frac{r}{s-t},r) ds\\&=-g_{1}''(t+r) \int _{t+r}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy \\&\quad - \int _{t+r}^{\infty } \left( \int _{s}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy\right) g_{1}'''(s) ds\end{aligned} \end{aligned}$$
Recalling that \(g(t)\lambda (t) \ll t\), the leading behavior of \(-\int _{t+r}^{\infty } g_{1}''(s) \left( K_{0}-K_{0,sm}\right) (\frac{r}{s-t},r) ds\) in the matching region is
$$\begin{aligned}{} & {} -g_{1}''(t) \int _{1}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{1}{q},r) r dq = \frac{-r g_{1}''(t)}{24}\left( 6 (\log (4)-1) \log (r) -2 \pi ^2+15+6 \log ^2(2)\right) \nonumber \\ \end{aligned}$$
(4.97)
The subleading part of \(-\int _{t+r}^{\infty } g_{1}''(s) \left( K_{0}-K_{0,sm}\right) (\frac{r}{s-t},r) ds\) is therefore given by
$$\begin{aligned}\begin{aligned}&-\int _{t+r}^{\infty } g_{1}''(s) \left( K_{0}-K_{0,sm}\right) (\frac{r}{s-t},r) ds-\left( \frac{-r g_{1}''(t)}{24}\left( 6 (\log (4)-1) \log (r)\right. \right. \\&\quad \left. \left. -2 \pi ^2+15+6 \log ^2(2)\right) \right) \\&= -\left( g_{1}''(t+r)-g_{1}''(t)\right) \int _{t+r}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy \\&\quad - \int _{t+r}^{\infty } \int _{s}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy g_{1}'''(s) ds\end{aligned} \end{aligned}$$
We have
$$\begin{aligned}\begin{aligned}&-\left( g_{1}''(t+r)-g_{1}''(t)\right) \int _{t+r}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy\\&=-r \left( g_{1}''(t+r)-g_{1}''(t)-r g_{1}'''(t)\right) \int _{1}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{1}{z},r) dz\\&\quad -r^{2} g_{1}'''(t) \int _{1}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{1}{z},r) dz\end{aligned} \end{aligned}$$
which gives
$$\begin{aligned} \begin{aligned}&|-\left( g_{1}''(t+r)-g_{1}''(t)\right) \int _{t+r}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy\\&\quad -\left( -r^{2} g_{1}'''(t) \int _{1}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{1}{z},r) dz\right) |\\&\leqslant \frac{C r^{3} \log (t) \lambda (t)^{3} \left( 1+|\log (r)|\right) }{t^{6} }\end{aligned} \end{aligned}$$
and the higher derivatives are estimated similarly. On the other hand, we have
$$\begin{aligned}\begin{aligned}&-\int _{t+r}^{\infty } \int _{s}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy g_{1}'''(s) ds\\&= - g_{1}'''(t+r) \int _{t+r}^{\infty } \int _{q}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy dq\\&\quad -\int _{t+r}^{\infty } \int _{s}^{\infty } \int _{q}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) g_{1}''''(s)dy dq ds\end{aligned} \end{aligned}$$
Therefore, for \(r \sim g(t)\lambda (t)\),
$$\begin{aligned}\begin{aligned}&|-\int _{t+r}^{\infty } \int _{s}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy g_{1}'''(s) ds\\&\quad -\left( -r^{2} g_{1}'''(t) \int _{1}^{\infty } (x-1) \left( K_{0}-K_{0,sm}\right) (\frac{1}{x},r) dx\right) |\\&\leqslant \frac{C r^{3} \log (t) \lambda (t)^{3}}{t^{6} } \left( 1+|\log (r)|\right) \left( \log (1+\frac{t}{r})+1\right) \end{aligned} \end{aligned}$$
As with the previous terms, the higher derivatives of \(-\int _{t+r}^{\infty } \int _{s}^{\infty } \int _{q}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) g_{1}''''(s)dy dq ds\) can be estimated by letting \(x=\frac{y-t}{r}\), then, \(w=\frac{q-t}{r}\), and \(z=\frac{s-t}{r}\), and differentiating under the integral. Finally, we get
$$\begin{aligned}\begin{aligned}&|-\left( g_{1}''(t+r)-g_{1}''(t)\right) \int _{t+r}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy\\&\quad -\int _{t+r}^{\infty } \int _{s}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy g_{1}'''(s) ds\\&-\left( -r^{2} g_{1}'''(t) \int _{1}^{\infty } x \left( K_{0}-K_{0,sm}\right) (\frac{1}{x},r) dx\right) |\\&\leqslant \frac{C r^{3} \log (t) \lambda (t)^{3} (1+|\log (r)|)(\log (1+\frac{t}{r})+1)}{t^{6} }\end{aligned} \end{aligned}$$
Using
$$\begin{aligned} -r^{2} g_{1}'''(t) \int _{1}^{\infty } x \left( K_{0}-K_{0,sm}\right) (\frac{1}{x},r) dx = \frac{-r^{2} g_{1}'''(t)}{18} \left( -7+\log (512)+3 \log (r)\right) \end{aligned}$$
we finally get the following expression which identifies the leading part of \(-\int _{t}^{\infty } g_{1}''(s) K_{0}(\frac{r}{s-t},r) ds\).
$$\begin{aligned}{} & {} |-\int _{t}^{\infty } g_{1}''(s) K_{0}(\frac{r}{s-t},r) ds - \left( \frac{-r g_{1}''(t)}{16} \left( -12+\pi ^{2} + 8 \log (r)\right) \right) \\{} & {} \quad -\left( -g_{1}''(t)r \left( \frac{\log (\frac{t}{r})\log (2)}{2}+\frac{1}{4}\left( \log ^{2}(t)-\log ^{2}(r)\right) \right) \right. \\{} & {} \quad \left. -\frac{r}{2} \int _{2t}^{\infty } \frac{g_{1}''(s)}{(s-t)} \log (2(s-t)) ds-\frac{r}{2} \int _{t}^{2t}\left( g_{1}''(s)-g_{1}''(t)\right) \frac{\log (2(s-t))}{(s-t)} ds\right) \\{} & {} \quad -\left( \frac{-r g_{1}''(t)}{24} \left( 6 \left( \log (4)-1\right) \log (r) -2 \pi ^{2} + 15 + 6 \log ^{2}(2)\right) \right) |\\{} & {} \quad \leqslant \frac{C r^{3} \log (t) \lambda (t)^{3} \left( 1+|\log (r)|\right) \left( \log (1+\frac{t}{r})+1\right) }{t^{6} }, \quad \lambda (t)g(t) \leqslant r \leqslant 2 g(t)\lambda (t) \end{aligned}$$
Notice that the \(r^{2}\) terms from
$$\begin{aligned}{} & {} -\left( g_{1}''(t+r)-g_{1}''(t)\right) \int _{t+r}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy\\{} & {} \quad -\int _{t+r}^{\infty } \int _{s}^{\infty } \left( K_{0}-K_{0,sm}\right) (\frac{r}{y-t},r) dy g_{1}'''(s) ds \end{aligned}$$
and
$$\begin{aligned} -r \int _{0}^{1} \left( g_{1}''(t+r y)-g_{1}''(t)\right) K_{0}(\frac{1}{y},r) dy + \int _{t}^{t+r} \left( g_{1}''(s)-g_{1}''(t)\right) K_{0,sm}(\frac{r}{s-t},r) ds \end{aligned}$$
canceled to give the above expression.
The same procedure is carried out for the \(K_{2}\) term in (4.92). The details are as follows. The analog of (4.94) is
$$\begin{aligned}\begin{aligned}&r g_{2}''(t) \int _{0}^{1} \left( F_{0}(a)+F_{1}(a)\log (r)+F_{2}(a)\log ^{2}(r)\right) da\\&=-\frac{r g_{2}''(t)}{16} \left( 2 \log (r) \left( 4 \log (r)+\pi ^2-12\right) -28 \zeta (3)+\pi ^2+28\right) \end{aligned} \end{aligned}$$
where \(\zeta \) denotes the Riemann zeta function. We remark that one part of this computation involves the following.
$$\begin{aligned} \text {Im}\left( \text {Li}_{2}\left( \frac{1}{2}-\frac{i a}{2 \sqrt{1-a^{2}}}\right) \right) = \int _{0}^{a} \left( \frac{x \sin ^{-1}(x)}{x^{2}-1}-\frac{\log (4(1-x^{2}))}{2\sqrt{1-x^{2}}}\right) dx \end{aligned}$$
and
$$\begin{aligned}{} & {} \int _0^1 \frac{1}{2} \left( \cos ^{-1}(x)-x \sqrt{1-x^2}\right) \left( \frac{x \sin ^{-1}(x)}{x^2-1}-\frac{\log \left( 4 \left( 1-x^2\right) \right) }{2 \sqrt{1-x^2}}\right) \, dx \\{} & {} \quad = \frac{1}{32} \left( -21 \zeta (3)+\pi ^2+\log (256)\right) \end{aligned}$$
The analog of (4.96) is
$$\begin{aligned}\begin{aligned}&g_{2}''(t) \frac{r}{24} \log \left( \frac{r}{t}\right) \left( 4 \log (t) \log (8 r t)+4 \log (r) \log (8 r)-\pi ^2+12 \log ^2(2)\right) \\&+\int _{t}^{2t} \left( g_{2}''(s)-g_{2}''(t)\right) K_{2,0}(\frac{s-t}{r},r) ds+\int _{2t}^{\infty } g_{2}''(s)K_{2,0}(\frac{s-t}{r},r) ds\end{aligned} \end{aligned}$$
where we recall that \(K_{2,0}\) is defined in (4.89). The analog of (4.97) is
$$\begin{aligned} g_{2}''(t) r \int _{1}^{\infty } \left( K_{2}-K_{2,0}\right) (a,r) da \end{aligned}$$
We have
$$\begin{aligned} \int _{1}^{\infty } \left( K_{2}-K_{2,0}\right) (a,r) da = c_{0}+d_{1}\log (r)+d_{2}\log ^{2}(r) \end{aligned}$$
The exact value of \(c_{0}\) is not important for our purposes, but we compute \(d_{1}\) and \(d_{2}\) explicitly. From direct integration,
$$\begin{aligned} d_{2} = \int _1^{\infty } \left( \sqrt{a^2-1}-a+\frac{1}{2 a}\right) \, da = \frac{1}{4}\left( 1-\log (4)\right) \end{aligned}$$
To compute \(d_{1}\), we recall that \(K_{2}(a,r)\) satisfies the wave equation:
$$\begin{aligned} -K_{2}+r \partial _{r}K_{2} + r^{2}\partial _{r}^{2} K_{2}+a \partial _{a}K_{2}-2 a r \partial _{ar}K_{2}+(a^{2}-1)\partial _{a}^{2}K_{2}=0, \quad a >1 \end{aligned}$$
and \(K_{2} = \sum _{k=0}^{2} F_{k}(a) \log ^{k}(r).\) Integrating the equation solved by \(K_{2}\) in the a variable, one relation we get is (for \(M>1\))
$$\begin{aligned} \int _{1}^{M} F_{1}(a) da = -\int _{1}^{M} F_{2}(a) da + \left( a F_{1}(a) + \frac{(1-a^{2})}{2} F_{0}'(a) + \frac{a F_{0}(a)}{2}\right) \Bigr |_{a=1}^{M} \end{aligned}$$
Recalling the definitions of \(F_{k,0}\) in (4.90), we have
$$\begin{aligned}{} & {} d_{1}=\int _{1}^{\infty }\left( F_{1}(a)-F_{1,0}(a)\right) da = \lim _{M \rightarrow \infty } \int _{1}^{M}\left( F_{1}(a)-F_{1,0}(a)\right) da\nonumber \\{} & {} \quad = \frac{1}{12} \left( 2 \pi ^2-15-6 \log ^2(2)\right) \end{aligned}$$
(4.98)
Therefore,
$$\begin{aligned} g_{2}''(t) r \int _{1}^{\infty } \left( K_{2}-K_{2,0}\right) (a,r) da= & {} g_{2}''(t) r \left( c_{0}+\frac{1}{12} \left( 2 \pi ^2-15-6 \log ^2(2)\right) \log (r) \right. \\{} & {} \left. +\frac{1}{4}\left( 1-\log (4)\right) \log ^{2}(r)\right) \end{aligned}$$
For estimating the subleading terms, we use the same procedure used for the \(K_{0}\) integral term previously. In particular, we have
$$\begin{aligned} \begin{aligned}&|\int _{t}^{\infty } g_{2}''(s) K_{2}(\frac{s-t}{r},r) ds {-} r g_{2}''(t) \int _{0}^{1} K_{2}(y,r) dy - r g_{2}''(t) \int _{1}^{\infty } (K_{2}-K_{2,0})(a,r) da\\&-g_{2}''(t) \int _{t+r}^{2t} K_{2,0}(\frac{s-t}{r},r) ds - \int _{t}^{2t}\left( g_{2}''(s)-g_{2}''(t)\right) K_{2,0}(\frac{s-t}{r},r) ds\\&\quad -\int _{2t}^{\infty } g_{2}''(s) K_{2,0}(\frac{s-t}{r},r) ds\\&-r^{2} g_{2}'''(t) \int _{0}^{1} y K_{2}(y,r) dy + r^{2} g_{2}'''(t) \int _{0}^{1} y K_{2,0}(y,r) dy \\&\quad - r^{2} g_{2}'''(t) \int _{1}^{\infty } a \left( K_{2}-K_{2,0}\right) (a,r) da|\\&\leqslant \frac{C r^{3} \lambda (t)^{3} \left( 1+\log ^{2}(r)\right) \left( 1+|\log (\frac{r}{t})|^{3}\right) }{t^{6} }, \quad r \leqslant t\end{aligned} \end{aligned}$$
(4.99)
The higher derivatives are estimated similarly, with the same procedure used for the \(g_{1}''(t)\) terms just studied. Finally, we will show that the last three terms on the left-hand side of the above inequality exactly cancel, just as was the case for the analogous terms arising from the \(K_{0}\) integral previously. By direct computation, we have
$$\begin{aligned} \int _{0}^{1} y K_{2,0}(y,r) dy= & {} -\frac{\log ^2(r)}{2}-\log (2) \log (r)\\{} & {} +\log (2 r)+\frac{\pi ^2}{24}-1-\frac{1}{2} \log ^2(2) \\ \int _{0}^{1} y K_{2}(y,r) dy= & {} -\frac{1}{3} \log ^2(r)+\frac{2 \log (r)}{9}+\frac{1}{18} (16 \log (2)-4)\\{} & {} +\frac{1}{108} \left( -9 \pi ^2+88-96 \log (2)\right) \end{aligned}$$
Finally, using a similar procedure as in (4.98), we get
$$\begin{aligned}{} & {} \int _{1}^{\infty } \left( F_{0}(a)-F_{0,0}(a)\right) a da \\{} & {} \quad = \lim _{M\rightarrow \infty }\left( -\frac{2}{3} \int _{1}^{M} a F_{2}(a) da -\frac{a^{2} F_{0}(a)}{3} \Bigr |_{a=1}^{M} + \frac{2}{3} \int _{1}^{M} a^{2} F_{1}'(a) da\right. \\{} & {} \qquad \left. -\frac{1}{3} \left( (a^{3}-a) F_{0}'(a) -(3 a^{2}-1) F_{0}(a)\right) \Bigr |_{a=1}^{M} - \int _{1}^{M} F_{0,0}(a) a da\right) \\{} & {} \quad =\frac{\pi ^2}{8}-\frac{43}{27}-\frac{1}{2} \log ^2(2)+\log (2) \end{aligned}$$
A straightforward computation then gives
$$\begin{aligned} \int _{1}^{\infty } a \left( K_{2}-K_{2,0}\right) (a,r) da= & {} -\frac{\log ^2(r)}{6}+\left( \frac{7}{9}-\log (2)\right) \log (r)\\{} & {} +\frac{\pi ^2}{8}-\frac{43}{27}-\frac{1}{2} \log ^2(2)+\log (2) \end{aligned}$$
In total, we then note that
$$\begin{aligned}\begin{aligned}&-\frac{\log ^2(r)}{2}-\log (2) \log (r)+\log (2 r)+\frac{\pi ^2}{24}-1-\frac{1}{2} \log ^2(2)\\&-\left( -\frac{1}{3} \log ^2(r)+\frac{2 \log (r)}{9}+\frac{1}{18} (16 \log (2)-4)+\frac{1}{108} \left( -9 \pi ^2+88-96 \log (2)\right) \right) \\&-\left( -\frac{\log ^2(r)}{6}+\left( \frac{7}{9}-\log (2)\right) \log (r)+\frac{\pi ^2}{8}-\frac{43}{27}-\frac{1}{2} \log ^2(2)+\log (2)\right) =0\end{aligned} \end{aligned}$$
which verifies that the last three terms on the left-hand side of (4.99) exactly cancel. Combining our computations and estimates above finishes the proof of the lemma. \(\square \)
Next, we note that
$$\begin{aligned} \partial _{t}^{2}v_{ex,cont}(t,r) = \frac{1}{r}\left( h_{0}''(t)+\log (r)h_{1}''(t)\right) \end{aligned}$$
where
$$\begin{aligned} h_{0}(t) = -2 \log (\lambda (t))f_{1}(t)+\lambda (t)^{2}f_{2}(t), \quad h_{1}(t) = 2 f_{1}(t) \end{aligned}$$
and
$$\begin{aligned} f_{1}(t) = \frac{-1}{2}\left( 2\lambda (t)\lambda '(t)^{2}+\lambda (t)^{2}\lambda ''(t)\right) , \quad f_{2}(t) = \frac{-\lambda ''(t)}{2} \end{aligned}$$
Using the same procedure as for \(u_{w,2,sub,0}\) (in fact, the same computations, except with \(g_{j}\) replaced with \(h_{j}\), for \(j=0,1\)) we get the following lemma.
Lemma 13
[Leading part of \(v_{ex,sub,0}\)]. Let
$$\begin{aligned}\begin{aligned}&v_{ex,sub,0,cont}(t,r)\\&=\frac{-r}{2} h_{0}''(t) -\frac{r}{2}h_{0}''(t) \log (\frac{t}{r}) - \frac{r}{2} \int _{t}^{2t} \frac{\left( h_{0}''(s)-h_{0}''(t)\right) }{(s-t)} ds \\&\quad - \frac{r}{2}\int _{2t}^{\infty } \frac{h_{0}''(s) ds}{s-t}-r h_{0}''(t)\frac{\left( \log (4)-1\right) }{4}-\frac{r}{8}f_{1}''(t)\left( -12+\pi ^{2}+8 \log (r)\right) \\&\quad -2 f_{1}''(t)\left( \frac{r \log (2) \log (\frac{t}{r})}{2}+\frac{r}{4}\left( \log ^{2}(t)-\log ^{2}(r)\right) \right) \\&\quad -r \int _{t}^{2t} \frac{\left( f_{1}''(s)-f_{1}''(t)\right) }{(s-t)} \log (2(s-t)) ds\\&\quad -r \int _{2t}^{\infty } f_{1}''(s) \left( \frac{ \log (2(s-t))}{(s-t)}\right) ds\\&\quad -\frac{r f_{1}''(t)}{12}\left( 6 \left( \log (4)-1\right) \log (r)-2 \pi ^{2}+15+6 \log ^{2}(2)\right) \end{aligned} \end{aligned}$$
Then, for \(0 \leqslant j+k \leqslant 2\),
$$\begin{aligned} |r^{k}t^{j}\partial _{r}^{k}\partial _{t}^{j}\left( v_{ex,sub,0}-v_{ex,sub,0,cont}\right) |(t,r) \leqslant \frac{C r^{3} \lambda (t)^{3} \log ^{3}(t)}{t^{6} }, \quad g(t)\lambda (t) \leqslant r \leqslant 2g(t)\lambda (t) \end{aligned}$$
Next, we consider
$$\begin{aligned}{} & {} u_{w,2,ell}(t,r)-u_{w,2,ell,0}(t,r)\nonumber \\{} & {} \quad =-\frac{1}{2}\left( \frac{1}{r}\int _{0}^{r} s^{2}\left( RHS_{2}(t,s)-RHS_{2,0}(t,s)\right) ds + r \int _{r}^{\infty } \left( RHS_{2}(t,s)-RHS_{2,0}(t,s)\right) ds\right) \nonumber \\{} & {} \quad =\frac{-1}{2}\left( \frac{1}{r}\int _{0}^{r} s^{2}\left( RHS_{2}(t,s)-RHS_{2,0}(t,s)\right) ds + r\int _{0}^{\infty } \left( RHS_{2}(t,s)-RHS_{2,0}(t,s)\right) ds\right. \nonumber \\{} & {} \qquad \left. - r \int _{0}^{r} \left( RHS_{2}(t,s)-RHS_{2,0}(t,s)\right) ds\right) \end{aligned}$$
(4.100)
We proceed to study each integral term on the right-hand side. The integral over \((0,\infty )\) requires the longest argument:
Lemma 14
We have
$$\begin{aligned}{} & {} \int _{0}^{\infty }\left( RHS_{2}(t,s)-RHS_{2,0}(t,s)\right) ds= \lambda (t) \int _{t}^{\infty } W_{3}(\frac{s-t}{\lambda (t)}) \lambda ''''(s) ds\nonumber \\{} & {} \quad + \int _{0}^{\infty } d\xi \left( \frac{-8}{\xi \lambda (t)^{2}}+4 \xi K_{2}(\xi \lambda (t)) + 2\xi \right) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) \end{aligned}$$
(4.101)
where
$$\begin{aligned}\begin{aligned} W_{3}(x)&= \frac{2}{3} \left( x \log \left( 64 x^6\right) +4 x^3 \log (2 x)+2 \left( x^2+1\right) ^{3/2} \log \left( 2 x \left( x-\sqrt{x^2+1}\right) +1\right) +x\right) \\&=O\left( \frac{\log (x)}{x}\right) , \quad x \rightarrow \infty \end{aligned} \end{aligned}$$
Proof
We first note
$$\begin{aligned}{} & {} r\int _{0}^{\infty } \left( RHS_{2}(t,s)-RHS_{2,0}(t,s)\right) ds \\{} & {} \quad = r\int _{0}^{\infty } \left( \left( \frac{\cos (2 Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) \left( v_{2}+w_{1}\right) - RHS_{2,0}(t,s)\right) ds \end{aligned}$$
Using the representation formula (4.37) for \(v_{2}\), and Fubini’s theorem, we have
$$\begin{aligned}{} & {} \int _{0}^{\infty }\left( \frac{\cos (2Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) v_{2}(t,s) ds\nonumber \\{} & {} \quad = F(t) + \int _{0}^{\infty } d\xi \left( \frac{-8}{\xi \lambda (t)^{2}}+4 \xi K_{2}(\xi \lambda (t)) + 2\xi \right) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) \end{aligned}$$
(4.102)
where we used
$$\begin{aligned}{} & {} \int _{0}^{\infty } dr \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) J_{1}(r\xi ) \nonumber \\{} & {} \quad = -\frac{8}{\xi \lambda (t)^{2}} + 4 \xi K_{2}(\xi \lambda (t)) = -2\xi + O\left( \xi ^{3}\log (\xi )\right) , \quad \xi \rightarrow 0 \end{aligned}$$
(4.103)
and we recall that F is defined in (4.38). Notice that the first term on the right-hand side of (4.102) is the leading part, given the smallness at low frequencies of the O term in (4.103). Next, using
$$\begin{aligned} w_{1}(t,r) =\frac{2}{r} \int _{t}^{t+r} \lambda ''(s)(s-t)ds + 2 r \int _{1}^{\infty } \lambda ''(t+r y) \left( y-\sqrt{y^{2}-1}\right) dy \end{aligned}$$
we get
$$\begin{aligned}{} & {} \int _{0}^{\infty } \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) w_{1}(t,r) dr\nonumber \\{} & {} \quad =\int _{0}^{\infty } dw \lambda ''(t+w) w \frac{8}{\lambda (t)^{2}} \left( \frac{\lambda (t)^{2}}{\lambda (t)^{2}+w^{2}}-\log \left( 1+\frac{\lambda (t)^{2}}{w^{2}}\right) \right) \nonumber \\{} & {} \qquad +\int _{0}^{\infty } dw \lambda ''(t+w)\frac{4}{\lambda (t)^{2}} \left( \frac{-2 \lambda (t)^{2} w}{\lambda (t)^{2}+w^{2}} + w \log (16) + 2 w \log (1+\frac{w^{2}}{\lambda (t)^{2}}) \right. \nonumber \\{} & {} \qquad \left. + \frac{\left( \lambda (t)^{2}+2w^{2}\right) \log (1+\frac{2 w (w-\sqrt{\lambda (t)^{2}+w^{2}})}{\lambda (t)^{2}})}{\sqrt{\lambda (t)^{2}+w^{2}}}\right) \end{aligned}$$
(4.104)
Using the definitions of \(v_{2,sub}\), \(w_{1,sub}\), and the first order matching (see, e.g. (4.39)), we get
$$\begin{aligned} RHS_{2,0}(t,r) =\frac{r}{\lambda (t)} \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) \left( \frac{1}{2}\lambda '(t)^{2}+\lambda (t)\lambda ''(t) -\lambda (t)\lambda ''(t) \log (\frac{r}{\lambda (t)})\right) \end{aligned}$$
which gives
$$\begin{aligned} -\int _{0}^{\infty } RHS_{2,0}(t,s) ds = \frac{4 }{\lambda (t)} \left( \frac{\lambda '(t)^{2}}{2}+\lambda (t)\lambda ''(t)\right) \end{aligned}$$
(4.105)
Given that \(RHS_{2,0}\) was the leading part of \(RHS_{2}\) in the matching region, the integrals (4.102), (4.105), and (4.104) have cancellation when added together. We show this in detail now. Recalling the definition of F(t), we have (for all \(t \geqslant T_{0}\))
$$\begin{aligned} \begin{aligned}&F(t)\\&= 4 \left( \left( \log (2)-\frac{1}{2}\right) \lambda ''(t)+\int _{t}^{2t} \left( \frac{\lambda ''(s)-\lambda ''(t)}{s-t}\right) ds + \lambda ''(t) \log (\frac{t}{\lambda (t)})\right. \\&\quad \left. +\int _{2t}^{\infty } \frac{\lambda ''(s) ds}{s-t}-\frac{(\lambda '(t))^{2}}{2\lambda (t)}\right) \qquad \quad \end{aligned}\nonumber \\ \end{aligned}$$
(4.106)
Therefore, the \(\lambda '(t)^{2}\) terms from F(t) and (4.105) cancel. Next, we determine the leading terms of the first term on the right-hand side of (4.104). The integral is
$$\begin{aligned} \frac{8}{\lambda (t)^{2}}\int _{0}^{\infty } \lambda ''(t+w) w \left( \frac{\lambda (t)^{2}}{\lambda (t)^{2}+w^{2}}-\log (1+\frac{\lambda (t)^{2}}{w^{2}})\right) dw \end{aligned}$$
Given the decay (for large w) of the part of the integrand multiplying \(\lambda ''(t+w)\), we integrate by parts in w, differentiating the symbol \(\lambda ''(t+w)\), and integrating (backwards from infinity) the rest of the integrand. This gives
$$\begin{aligned} \begin{aligned}&\frac{8}{\lambda (t)^{2}}\int _{0}^{\infty } \lambda ''(t+w) w \left( \frac{\lambda (t)^{2}}{\lambda (t)^{2}+w^{2}}-\log (1+\frac{\lambda (t)^{2}}{w^{2}})\right) dw \\&= -4 \lambda ''(t)-\frac{4 \pi }{3}\lambda (t)\lambda '''(t) \\&+\frac{4}{3\lambda (t)^{2}} \int _{0}^{\infty } \!\left( w \lambda (t)^{2}\!-\!2 \tan ^{-1}\left( \frac{\lambda (t)}{w}\right) \lambda (t)^{3} \!-\! w^{3} \log (1\!+\!\frac{\lambda (t)^{2}}{w^{2}})\right) \lambda ''''(t+w) dw\end{aligned}\nonumber \\ \end{aligned}$$
(4.107)
Notice the cancellation between the \(\lambda ''(t)\) terms in the expression above and (4.105). Next, we treat the second term on the right-hand side of (4.104). We start with
$$\begin{aligned}\begin{aligned}&\frac{4}{\lambda (t)^{2}} \int _{0}^{t} dw \lambda ''(t+w) G_{2}(w,\lambda (t))\end{aligned} \end{aligned}$$
where
$$\begin{aligned}\begin{aligned} G_{2}(w,\lambda (t))&= \left( \frac{-2 \lambda (t)^{2}w}{\lambda (t)^{2}+w^{2}}+w \log (16) + 2 w \log (1+\frac{w^{2}}{\lambda (t)^{2}})\right. \\&\quad \left. +\frac{\left( \lambda (t)^{2}+2w^{2}\right) }{\sqrt{\lambda (t)^{2}+w^{2}}}\log (1+\frac{2 w (w-\sqrt{\lambda (t)^{2}+w^{2}})}{\lambda (t)^{2}})\right) =\lambda (t) G_{3}(\frac{w}{\lambda (t)})\end{aligned} \end{aligned}$$
Since there will be some cancellation between the integral under consideration and the second term in (4.106), we start by writing \(\lambda ''(t+w)=\lambda ''(t)+\lambda ''(t+w)-\lambda ''(t)\), we have
$$\begin{aligned}{} & {} \frac{4}{\lambda (t)^{2}} \int _{0}^{t} dw \lambda ''(t+w) G_{2}(w,\lambda (t))\nonumber \\{} & {} \quad =-2 \lambda ''(t) \left( -1+2 \log (2)-2 \log (\frac{\lambda (t)}{t})\right) + \frac{4 \lambda ''(t)}{\lambda (t)^{2}} \lambda (t)^{2} G_{4}(\frac{t}{\lambda (t)}) \nonumber \\{} & {} \qquad +\frac{4}{\lambda (t)^{2}} \int _{0}^{t} dw \left( \lambda ''(t+w)-\lambda ''(t)\right) G_{2}(w,\lambda (t)) \end{aligned}$$
(4.108)
where
$$\begin{aligned} \begin{aligned}G_{4}(x)&= \frac{1}{2} \left( 2 x^2 \log \left( 4 \left( x^2+1\right) \right) +2x^{2} \sqrt{\frac{1}{x^2}+1} \log \left( 1-2 \left( \sqrt{\frac{1}{x^2}+1}-1\right) x^2\right) \right. \\&\quad \left. +2 \log (2 x)-1\right) \\&=O\left( \frac{\log (x)}{x^{2}}\right) , \quad x \rightarrow \infty \end{aligned} \end{aligned}$$
(4.109)
Note that the first term on the right-hand side of (4.108) cancels with all of the \(\lambda ''(t)\) terms outside the integral operators in (4.106). Next, we note that
$$\begin{aligned} G_{3}(x) = \frac{-1}{x} + O\left( \frac{\log (x)}{x^{3}}\right) , \quad x \rightarrow \infty \end{aligned}$$
(4.110)
Recalling that \(G_{2}(w,\lambda (t))=\lambda (t)G_{3}(\frac{w}{\lambda (t)})\), we have
$$\begin{aligned}\begin{aligned}&\frac{4}{\lambda (t)^{2}} \int _{0}^{t} dw \left( \lambda ''(t+w)-\lambda ''(t)\right) G_{2}(w,\lambda (t))\\&=\frac{-4}{\lambda (t)^{2}} \int _{0}^{t} dw \left( \lambda ''(t+w)-\lambda ''(t)\right) \frac{\lambda (t)^{2}}{w} \\&+\frac{4}{\lambda (t)^{2}}\int _{0}^{t} dw \left( \lambda ''(t+w) -\lambda ''(t)\right) \left( G_{2}(w,\lambda (t))+\frac{\lambda (t)^{2}}{w}\right) \end{aligned} \end{aligned}$$
The first term on the right-hand side of the above expression cancels with the second term on the right-hand side of (4.106), and \(G_{2}(w,\lambda (t))+\frac{\lambda (t)^{2}}{w}\) decays much more quickly for large w than \(G_{2}(w,\lambda (t))\) does, given (4.110). Using the same procedure as in (4.107), we get
$$\begin{aligned} \begin{aligned}&\frac{4}{\lambda (t)^{2}}\int _{0}^{t} dw \left( \lambda ''(t+w)-\lambda ''(t)\right) \left( G_{2}(w,\lambda (t))+\frac{\lambda (t)^{2}}{w}\right) \\&=\frac{4}{\lambda (t)^{2}} \left( \left( \lambda ''(2t)-\lambda ''(t)\right) \lambda (t)^2 G_{4}(\frac{t}{\lambda (t)})\right. \\&\left. +\lambda (t)^{3}\left( - \lambda '''(2t) W_{2}(\frac{t}{\lambda (t)}) +\frac{\lambda '''(t)}{3}\pi + \int _{t}^{2t} W_{2}(\frac{s-t}{\lambda (t)}) \lambda ''''(s) ds\right) \right) \end{aligned}\nonumber \\ \end{aligned}$$
(4.111)
where we recall that \(G_{4}\) was defined in (4.109), and
$$\begin{aligned}\begin{aligned} W_{2}(x) = \frac{1}{6}&\left( x^3 \log (16)+2 \left( x^2+1\right) ^{3/2} \log \left( 2 x^2-2 x \sqrt{x^2+1} +1\right) +2 x^3 \log \left( x^2+1\right) \right. \\&\left. -x+x \log (64)+6 x \log (x)+4 \tan ^{-1}(\frac{1}{x})\right) \end{aligned} \end{aligned}$$
Note that the second term on the right-hand side of our previous computation (4.108) cancels with the term \(\frac{4}{\lambda (t)^{2}}\left( -\lambda ''(t)\lambda (t)^{2}G_{4}(\frac{t}{\lambda (t)})\right) \) from (4.111).
Lastly, we consider
$$\begin{aligned}{} & {} \frac{4}{\lambda (t)^{2}}\int _{t}^{\infty } dw \lambda ''(t+w) G_{2}(w,\lambda (t)) = -4 \int _{2t}^{\infty } ds \frac{\lambda ''(s)}{(s-t)}\\{} & {} \quad + \frac{4}{\lambda (t)^{2}} \int _{t}^{\infty } dw \lambda ''(t+w)\left( G_{2}(w,\lambda (t))+\frac{\lambda (t)^{2}}{w}\right) \end{aligned}$$
The first term on the right-hand side of the above expression cancels with the fourth term on the right-hand side of (4.106). Then, we treat the following integral with the same procedure used in (4.107)
$$\begin{aligned}\begin{aligned}&\frac{4}{\lambda (t)^{2}} \int _{t}^{\infty } dw \lambda ''(t+w) \left( G_{2}(w,\lambda (t))+\frac{\lambda (t)^{2}}{w}\right) \\&=\frac{4}{\lambda (t)^{2}} \left( -\lambda ''(2t)\lambda (t)^{2} G_{4}(\frac{t}{\lambda (t)}) + \lambda '''(2t) \lambda (t)^{3} W_{2}(\frac{t}{\lambda (t)}) \right. \\&\quad \left. +\lambda (t)^{3} \int _{t}^{\infty } dw \lambda ''''(t+w) W_{2}(\frac{w}{\lambda (t)}) dw\right) \end{aligned} \end{aligned}$$
After combining all of our computations, we end up with (4.101), completing the proof of the lemma. \(\square \)
Finally, to compute the principal part (in the matching region \(r \sim \lambda (t)g(t)\)) of the other two integrals in (4.100), we start with
$$\begin{aligned} RHS_{2}(t,r) - RHS_{2,0}(t,r) = \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) \left( v_{2,sub}(t,r) + w_{1,sub}(t,r)\right) \end{aligned}$$
Next, we replace \(v_{2,sub}\) and \(w_{1,sub}\) by their principal parts, which are \(v_{2,cubic,main}\) and \(w_{1,cubic,main}\), respectively. Then, we use part 1 of second order matching, which says that
$$\begin{aligned} v_{2,cubic,main}(t,r) + w_{1,cubic,main}(t,r) = v_{ell,2,0,main}(t,\frac{r}{\lambda (t)}) \end{aligned}$$
Whence, we get
$$\begin{aligned}\begin{aligned}&\text {leading part of } \frac{1}{r}\int _{0}^{r} s^{2}\left( \frac{\cos (2Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) v_{ell,2,0,main}(t,\frac{s}{\lambda (t)}) ds\\&:= (u_{w,2,ell}-u_{w,2,ell,0})_{princ,1}(t,r)\\&=-\left( 2 r \left( 2 j_{1}(t) \lambda (t)^2+2 j_{2}(t) \lambda (t)^2 \log (r)-j_{2}(t) \lambda (t)^2\right) \right) \end{aligned}\\ \begin{aligned}&\text {leading part of }-r \int _{0}^{r} \left( \frac{\cos (2Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) v_{ell,2,0,main}(t,\frac{s}{\lambda (t)})ds\\&:=(u_{w,2,ell}-u_{w,2,ell,0})_{princ,2}(t,r)\\&=\frac{1}{3} r \lambda (t)^2 \left( -24 j_{1}(t) \log (\lambda (t))+24 j_{1}(t) \log (r)-12 j_{1}(t)-12 j_{2}(t) \log ^2(\lambda (t))\right. \\&\left. -12 j_{2}(t) \log (\lambda (t))+12 j_{2}(t) \log ^2(r)-\pi ^2 j_{2}(t)\right) \end{aligned} \end{aligned}$$
where
$$\begin{aligned} j_{1}(t) = \frac{3}{32} \lambda ''''(t)+\frac{1}{8} \partial _{t}^{2}\left( \lambda ''(t)+\lambda ''(t) \log (\lambda (t))+\frac{\lambda '(t)^2}{2 \lambda (t)}\right) \\ j_{2}(t) = \frac{-1}{8}\lambda ''''(t) \end{aligned}$$
and we used
$$\begin{aligned} v_{ell,2,0,main}(t,\frac{r}{\lambda (t)}) = r^{3} j_{1}(t)+r^{3}\log (r) j_{2}(t) \end{aligned}$$
We now estimate the difference between \(u_{w,2,ell}-u_{w,2,ell,0}\) and its principal part in the matching region, which will help us study other contributions to the third order matching.
Lemma 15
Let
$$\begin{aligned}\begin{aligned}&(u_{w,2,ell}(t,r)-u_{w,2,ell,0}(t,r))_{princ}\\&:= -\frac{1}{2}\left( (u_{w,2,ell}-u_{w,2,ell,0})_{princ,1}(t,r)+(u_{w,2,ell}-u_{w,2,ell,0})_{princ,2}(t,r)\right) \\&-\frac{r}{2}\lambda (t)\int _{t}^{\infty } W_{3}(\frac{s-t}{\lambda (t)}) \lambda ''''(s) ds - \frac{r}{2} \int _{0}^{\infty } \left( \frac{-8}{\xi \lambda (t)^{2}}+4 \xi K_{2}(\xi \lambda (t))+2 \xi \right) \sin (t\xi ) \widehat{v_{2,0}}(\xi )d\xi \end{aligned} \end{aligned}$$
Then, for \(0 \leqslant k \leqslant 2\), and \(0 \leqslant j \leqslant 8\) we have
$$\begin{aligned}\begin{aligned}&|\partial _{t}^{j}\partial _{r}^{k}\left( u_{w,2,ell}(t,r)-u_{w,2,ell,0}(t,r)-(u_{w,2,ell}(t,r)-u_{w,2,ell,0}(t,r))_{princ}\right) | \\&\leqslant \frac{C}{r^{1+k}} \frac{\lambda (t)^{3} \log ^{3}(t)}{t^{4+j}}\left( \frac{r^{4}}{t^{2}}+\lambda (t)^{2}\right) , \quad \lambda (t) \leqslant r \leqslant \frac{t}{2}\end{aligned} \end{aligned}$$
Moreover, for \(0 \leqslant m+k \leqslant 1\), \(m,k\geqslant 0\),
$$\begin{aligned}{} & {} |\partial _{t}^{2+k}\partial _{r}^{m}\left( u_{w,2,ell}(t,r)-u_{w,2,ell,0}(t,r)\right) | \leqslant \frac{C \lambda (t)^{2}\log (t) \sup _{y \in [100,t]}\left( \lambda (y)\log (y)\right) }{t^{9/2}\langle t-r\rangle ^{1/2+k+m}},\nonumber \\{} & {} \quad \quad \frac{t}{2} \leqslant r \leqslant 2t \end{aligned}$$
(4.112)
In addition, for \(s \geqslant t\), and \(0 \leqslant k \leqslant 2\),
$$\begin{aligned}{} & {} |\partial _{s}^{4+k} \left( u_{w,2,ell}-u_{w,2,ell,0}\right) (s,y)| \nonumber \\{} & {} \quad \leqslant C \lambda (s)^{2} \log ^{3}(s) \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) {\left\{ \begin{array}{ll} \frac{y}{s^{8+k}}, \quad y \leqslant \frac{s}{2}\\ \frac{1}{s^{9/2}t^{5/2+k}}, \quad \frac{s}{2} \leqslant y \leqslant s-t+2g(t)\lambda (t)\end{array}\right. }\nonumber \\ \end{aligned}$$
(4.113)
Proof
We first note that
$$\begin{aligned} \begin{aligned}&u_{w,2,ell}(t,r)-u_{w,2,ell,0}(t,r)-(u_{w,2,ell}(t,r)-u_{w,2,ell,0}(t,r))_{princ}\\&=-\frac{1}{2r}\left( \int _{0}^{r} s^{2} \left( \frac{\cos (2Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) \left( w_{1,sub}(t,s)+v_{2,sub}(t,s)\right. \right. \\&\quad \left. -\left( v_{2,cubic,main}(t,s)+w_{1,cubic,main}(t,s)\right) \right) ds\\&\left. +\int _{0}^{r} s^{2} \left( \frac{\cos (2Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) v_{ell,2,0,main}(t,\frac{s}{\lambda (t)})ds-r (u_{w,2,ell}-u_{w,2,ell,0})_{princ,1}(t,r)\right) \\&+\frac{r}{2}\left( \int _{0}^{r} \left( \frac{\cos (2Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) \left( w_{1,sub}(t,s)+v_{2,sub}(t,s)\right. \right. \\&\quad \left. -\left( v_{2,cubic,main}(t,s)+w_{1,cubic,main}(t,s)\right) \right) ds\\&+\left. \int _{0}^{r} \left( \frac{\cos (2Q_{1}(\frac{s}{\lambda (t)}))-1}{s^{2}}\right) v_{ell,2,0,main}(t,\frac{s}{\lambda (t)}) ds +\frac{1}{r}(u_{w,2,ell}-u_{w,2,ell,0})_{princ,2}(t,r) \right) \end{aligned}\nonumber \\ \end{aligned}$$
(4.114)
where we used
$$\begin{aligned} v_{2,cubic,main}(t,s)+w_{1,cubic,main}(t,s) = v_{ell,2,0,main}(t,\frac{s}{\lambda (t)}) \end{aligned}$$
which follows from the second order matching. Next, we use Lemmas 1 and 6, noting that \( s \leqslant r \leqslant \frac{t}{2}\) in the integrals which compose the first and third terms of (4.114).
The estimate (4.112) follows directly from Lemma 10 and (4.61), and (4.113) is proven with the same procedure as in Lemma 10. \(\square \)
Next, we need to understand \(w_{3}(t,r):=u_{w,2}(t,r)-u_{w,2,ell}(t,r)-u_{w,2,sub,0}\), since \(w_{3}\) will turn out to contribute terms which are logarithmically smaller than the largest contributions of \(u_{w,2,sub,0}(t,r)\) in the matching region, but not quite perturbative. Recalling the equations that these functions solve, (4.84) and (4.85), we see that \(w_{3}\) solves the following equation with 0 Cauchy data at infinity.
$$\begin{aligned} -\partial _{t}^{2}w_{3}+\partial _{r}^{2}w_{3}+\frac{1}{r}\partial _{r}w_{3}-\frac{w_{3}}{r^{2}} = \partial _{t}^{2}\left( u_{w,2,ell}-u_{w,2,ell,0,cont}\right) \end{aligned}$$
Therefore,
$$\begin{aligned} w_{3}(t,r) = \int _{t}^{\infty }w_{3,s}(t,r) ds, \quad w_{3,s}\text { solves } {\left\{ \begin{array}{ll} -\partial _{t}^{2}w_{3,s}+\partial _{r}^{2}w_{3,s}+\frac{1}{r}\partial _{r}w_{3,s}-\frac{w_{3,s}}{r^{2}} =0\\ w_{3,s}(s,r)=0\\ \partial _{1}w_{3,s}(s,r) = \partial _{1}^{2}(u_{w,2,ell}-u_{w,2,ell,0,cont})(s,r)\end{array}\right. } \end{aligned}$$
By the finite speed of propagation, in the region \(r \leqslant t\), we have
$$\begin{aligned} w_{3,s}(t,r) = v_{3,s}(t,r), \quad v_{3,s} \text { solves } {\left\{ \begin{array}{ll} -\partial _{t}^{2}v_{3,s}+\partial _{r}^{2}v_{3,s}+\frac{1}{r}\partial _{r}v_{3,s}-\frac{v_{3,s}}{r^{2}} =0\\ v_{3,s}(s,r)=0\\ \partial _{1}v_{3,s}(s,r) = \psi _{\leqslant 1}(r-s) \partial _{1}^{2}(u_{w,2,ell}-u_{w,2,ell,0,cont})(s,r)\end{array}\right. } \end{aligned}$$
where
$$\begin{aligned} \psi _{\leqslant 1}\in C^{\infty }(\mathbb {R}), \text { and }\psi _{\leqslant 1}(x) = {\left\{ \begin{array}{ll} 1, \quad x \leqslant 0\\ 0, \quad x \geqslant 1\end{array}\right. } \end{aligned}$$
Since we will only be interested in estimating \(w_{3}(t,r)\) in the matching region \(g(t)\lambda (t) \leqslant r \leqslant 2 g(t)\lambda (t)\), it will suffice to estimate \(v_{3}\) given below
$$\begin{aligned} v_{3}(t,r) :=\int _{t}^{\infty } v_{3,s}(t,r) ds \end{aligned}$$
I.e., \(v_{3}\) solves
$$\begin{aligned} -\partial _{t}^{2}v_{3}+\partial _{r}^{2}v_{3}+\frac{1}{r}\partial _{r}v_{3}-\frac{v_{3}}{r^{2}} = \psi _{\leqslant 1}(r-t) \partial _{t}^{2}(u_{w,2,ell}-u_{w,2,ell,0,cont})(t,r):=RHS_{3}(t,r)\qquad \nonumber \\ \end{aligned}$$
(4.115)
with 0 Cauchy data at infinity. Let
$$\begin{aligned} ellsoln=\frac{-1}{2}\left( \frac{1}{r}\int _{0}^{r} s^{2} RHS_{3}(t,s)ds+r \int _{r}^{\infty } RHS_{3}(t,s) ds\right) \end{aligned}$$
and
$$\begin{aligned} Q(t) = \frac{-1}{2} \int _{0}^{\infty } \psi _{\leqslant 1}(s-t) \partial _{t}^{2}\left( u_{w,2,ell}-u_{w,2,ell,0}\right) (t,s) ds \end{aligned}$$
Note that ellsoln satisfies
$$\begin{aligned} \partial _{r}^{2}ellsoln+\frac{1}{r}\partial _{r}ellsoln-\frac{ellsoln}{r^{2}} = RHS_{3}(t,r) \end{aligned}$$
Lemma 16
We have the following estimates. For \(0 \leqslant j,k \leqslant 1\),
$$\begin{aligned}{} & {} |r^{k}\partial _{r}^{k}\partial _{t}^{j}\left( ellsoln-rQ(t)\right) | \\{} & {} \quad \leqslant {\left\{ \begin{array}{ll} \frac{C r \lambda (t)^{3} \left( |\log ^{3}(r)|+\log ^{3}(t)\right) }{t^{4+j}}, \quad r \leqslant \lambda (t)\\ \frac{C r^{3} \lambda (t)^{2} \log ^{3}(t) \sup _{x \in [100,t]} \left( \lambda (x) \log (x)\right) }{t^{6+j}}+\frac{C}{r} \frac{\lambda (t)^{5} \log ^{2}(t)}{t^{4+j}}, \quad \lambda (t)< r < \frac{t}{2}\\ \frac{C \lambda (t)^{2} \log ^{3}(t) \sup _{x \in [100,t]} \left( \lambda (x) \log (x)\right) }{t^{3+\frac{j}{2}}}, \quad \frac{t}{2} \leqslant r \leqslant 2t\end{array}\right. } \\{} & {} |\partial _{r}^{2}\left( ellsoln-r Q(t)\right) | \leqslant \frac{C \lambda (t)^{5} \log ^{2}(t)}{r^{3} t^{4} } + \frac{C r \lambda (t)^{2} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{3}(t)}{t^{6}} ,\\{} & {} \quad \quad g(t)\lambda (t)\leqslant r \leqslant 2g(t)\lambda (t) \end{aligned}$$
Proof
We have
$$\begin{aligned}\begin{aligned}&ellsoln-r Q(t) \\&\quad =\frac{-1}{2r} \int _{0}^{r} s^{2}\psi _{\leqslant 1}(s-t) \partial _{t}^{2}\left( u_{w,2,ell}-u_{w,2,ell,0,cont}\right) (t,s) ds \\&\qquad + \frac{r}{2}\int _{0}^{r} \psi _{\leqslant 1}(s-t) \partial _{t}^{2}\left( u_{w,2,ell}-u_{w,2,ell,0}\right) (t,s) ds\\&\qquad -\frac{r}{2}\int _{r}^{\infty } \psi _{\leqslant 1}(s-t) \partial _{t}^{2}\left( u_{w,2,ell,0}-u_{w,2,ell,0,cont}\right) (t,s) ds\end{aligned} \end{aligned}$$
We then estimate directly, using Lemmas 15 and 10 and the definitions of \(u_{w,2,ell,0}\) and \(u_{w,2,ell,0,cont}\) from (4.58), (4.61), and (4.62). We use the same procedure to estimate the higher derivatives, except for the last estimate in the lemma statement, which is obtained by using
$$\begin{aligned} \left( \partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) \left( ellsoln-rQ(t)\right) = \psi _{\leqslant 1}(r-t) \partial _{t}^{2}\left( u_{w,2,ell}-u_{w,2,ell,0,cont}\right) \end{aligned}$$
and Lemmas 15 and 10. \(\square \)
We return to \(v_{3}\), which we recall solves (4.115), with 0 Cauchy data at infinity. If we define
$$\begin{aligned} q_{3}(t,r) := v_{3}(t,r)-\left( ellsoln-rQ(t)\right) \psi _{\leqslant 1}(r-t) \end{aligned}$$
then, \(q_{3}\) solves the following equation. Moreover, our estimates from Lemmas 16 and 10 show that \(q_{3}(t,r)\) also has 0 Cauchy data at infinity.
$$\begin{aligned}\begin{aligned} -\partial _{t}^{2} q_{3}+\partial _{r}^{2}q_{3}+\frac{1}{r}\partial _{r}q_{3}-\frac{q_{3}}{r^{2}}&= \psi _{\leqslant 1}(r-t) \left( \partial _{t}^{2}\left( ellsoln-rQ(t)\right) \right. \\&\quad \left. +\partial _{t}^{2}\left( u_{w,2,ell}-u_{w,2,ell,0,cont}\right) \cdot (1-\psi _{\leqslant 1}(r-t))\right) \\&\quad +\psi _{\leqslant 1}'(r-t)\left( -2 \left( \partial _{t}+\partial _{r}\right) \left( ellsoln-rQ(t)\right) \right. \\&\quad \left. -\frac{(ellsoln-rQ(t))}{r}\right) \\&:=RHS_{4}(t,r)\end{aligned} \end{aligned}$$
In other words, we have
$$\begin{aligned} q_{3}(t,r) = \int _{t}^{\infty } q_{3,s}(t,r)ds, \quad q_{3,s} \text { solves } {\left\{ \begin{array}{ll}\left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) q_{3,s}=0\\ q_{3,s}(s,r) =0\\ \partial _{1}q_{3,s}(s,r) =RHS_{4}(s,r)\end{array}\right. } \end{aligned}$$
We recall that we only need to estimate \(q_{3}(t,r)\) for \(r \leqslant 2 g(t)\lambda (t) < t-2\), and
$$\begin{aligned} \psi _{\leqslant 1}(x) =1, \quad \psi _{\leqslant 1}'(x) =0, \quad x\leqslant 0 \end{aligned}$$
Therefore, by the finite speed of propagation, for \(r \leqslant 2g(t)\lambda (t)\), we have
$$\begin{aligned} q_{3,s}(t,r) = q_{4,s}(t,r), \quad q_{4,s} \text { solves } {\left\{ \begin{array}{ll}\left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) q_{4,s}=0\\ q_{4,s}(s,r) =0\\ \partial _{1}q_{4,s}(s,r) =\psi _{\leqslant 1}(r-s) \partial _{s}^{2}\left( ellsoln-rQ(s)\right) \end{array}\right. } \end{aligned}$$
So, it suffices to estimate
$$\begin{aligned} q_{4}(t,r):=\int _{t}^{\infty } q_{4,s}(t,r) ds \end{aligned}$$
We recall that
$$\begin{aligned}\begin{aligned}&\psi _{\leqslant 1}(r-t) \partial _{t}^{2}\left( ellsoln(t,r)-r Q(t)\right) \\&\quad =\psi _{\leqslant 1}(r-t) \partial _{t}^{2}\left( \frac{-1}{2r} \int _{0}^{r} s^{2}\partial _{t}^{2}\left( u_{w,2,ell}-u_{w,2,ell,0}\right) (t,s) \psi _{\leqslant 1}(s-t) ds\right. \\&\qquad \left. -\frac{1}{2r} \int _{0}^{r} s^{2} \partial _{t}^{2}\left( u_{w,2,ell,0}-u_{w,2,ell,0,cont}\right) (t,s) \psi _{\leqslant 1}(s-t) ds\right. \\&\qquad +\frac{r}{2}\int _{0}^{r} \partial _{t}^{2}\left( u_{w,2,ell}-u_{w,2,ell,0}\right) (t,s) \psi _{\leqslant 1}(s-t) ds\\&\qquad \left. -\frac{r}{2} \int _{r}^{\infty } \partial _{t}^{2}\left( u_{w,2,ell,0}-u_{w,2,ell,0,cont}\right) (t,s) \psi _{\leqslant 1}(s-t) ds\right) \end{aligned} \end{aligned}$$
We correspondingly decompose \(q_{4}\) into
$$\begin{aligned} q_{4}(t,r) :=q_{4,1}(t,r)+q_{4,2}(t,r) \end{aligned}$$
where \(q_{4,2}\) solves the following equation with 0 Cauchy data at infinity (and \(q_{4,1}=q_{4}-q_{4,2}\))
$$\begin{aligned} \begin{aligned}&\left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) q_{4,2}(t,r) \\&= \psi _{\leqslant 1}(r-t) \partial _{t}^{2}\left( \frac{-1}{2r} \int _{0}^{r} s^{2} \partial _{t}^{2}\left( u_{w,2,ell,0}-u_{w,2,ell,0,cont}\right) (t,s) \psi _{\leqslant 1}(s-t) ds\right) \\&-\psi _{\leqslant 1}(r-t) \partial _{t}^{2}\left( \frac{r}{2} \int _{r}^{\infty } \partial _{t}^{2}\left( u_{w,2,ell,0}-u_{w,2,ell,0,cont}\right) (t,s) \psi _{\leqslant 1}(s-t) ds\right) \\&:=RHS_{4,2}(t,r)\end{aligned} \end{aligned}$$
(4.116)
To estimate \(q_{4,2}\), it will suffice to use energy estimates.
Lemma 17
We have the following estimates, for \(k=0,1\), and \(r>0\):
$$\begin{aligned}\begin{aligned}&|\partial _{t}^{k}q_{4,2}(t,r)| + \sqrt{E(\partial _{t}^{k}q_{4,2},\partial _{t}^{k+1}q_{4,2})} +||\partial _{r}^{2} q_{4,2}||_{L^{2}((\lambda (t)g(t), 2 \lambda (t)g(t)),r dr)}\leqslant \frac{C \lambda (t)^{5} \log ^{3}(t)}{t^{5} }\end{aligned} \end{aligned}$$
Proof
We first note that the right-hand side of (4.116) includes the term
$$\begin{aligned}{} & {} \psi _{\leqslant 1}(r-t) \left( \frac{-1}{2r} \int _{0}^{r} s^{2}\psi _{\leqslant 1}''(s-t)\partial _{t}^{2}(u_{w,2,ell,0}-u_{w,2,ell,0,cont})ds \right. \\{} & {} \left. \quad - \frac{r}{2}\int _{r}^{\infty } \psi _{\leqslant 1}''(s-t)\partial _{t}^{2}(u_{w,2,ell,0}-u_{w,2,ell,0,cont}) ds\right) \end{aligned}$$
We integrate by parts in each integral, integrating \(\psi _{\leqslant 1}''(s-t)\), and note that the boundary contributions from each integral at \(s=r\) cancel. Then, we directly insert the estimates from Lemma 10 into the other terms in (4.116), and use the same procedure used for (4.69). Finally, the symbol type estimates on \(u_{w,2,ell,0}-u_{w,2,ell,0,cont}\) from Lemma 10 show that \(\partial _{t}q_{4,2}\) solves
$$\begin{aligned} \left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) \partial _{t}q_{4,2}(t,r) =\partial _{t}RHS_{4,2}(t,r) \end{aligned}$$
also with 0 Cauchy data at infinity. Then, we use the same procedure used to estimate \(q_{4,2}\). Finally, we estimate
$$\begin{aligned} ||\partial _{r}^{2} q_{4,2}||_{L^{2}((\lambda (t)g(t), 2 \lambda (t)g(t)),r dr)} \end{aligned}$$
using the equation solved by \(q_{4,2}\) and our earlier estimates from the proof of this lemma. \(\square \)
Now, we study \(q_{4,1}\). Let
$$\begin{aligned} RHS_{4,1}(t,r)= & {} \psi _{\leqslant 1}(r-t) \partial _{t}^{2} \left( \frac{-1}{2r} \int _{0}^{r} s^{2}\partial _{t}^{2}\left( u_{w,2,ell}-u_{w,2,ell,0}\right) (t,s) \psi _{\leqslant 1}(s-t) ds\right. \nonumber \\{} & {} \left. +\frac{r}{2}\int _{0}^{r} \partial _{t}^{2}\left( u_{w,2,ell}-u_{w,2,ell,0}\right) (t,s) \psi _{\leqslant 1}(s-t) ds\right) \end{aligned}$$
(4.117)
so that we have
$$\begin{aligned} q_{4,1}(t,r) = \int _{t}^{\infty } q_{4,1,s}(t,r) ds, \quad q_{4,1,s} \text { solves } {\left\{ \begin{array}{ll}\left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) q_{4,1,s}=0\\ q_{4,1,s}(s,r) =0\\ \partial _{1}q_{4,1,s}(s,r) =RHS_{4,1}(s,r)\end{array}\right. } \end{aligned}$$
Since we only need to estimate \(q_{4,1}(t,r)\) (and hence \(q_{4,1,s}(t,r)\)) in the region \(r \leqslant 2 g(t)\lambda (t) <\frac{t}{2}\), the finite speed of propagation shows that \(q_{4,1,s}(t,r)\) only depends on \(RHS_{4,1}(s,y)\) for \(y \leqslant s-t+2 g(t)\lambda (t) < s-\frac{t}{2}\). Given the limits of the integrals in the terms defining \(RHS_{4,1}\), this means that we may replace \(\psi _{\leqslant 1}(s-t)\) and \(\psi _{\leqslant 1}(r-t)\) by 1 when considering \(q_{4,1}(t,r)\) in the region \(r \leqslant 2g(t)\lambda (t)\). Now, we estimate \(RHS_{4,1}\).
Lemma 18
For \(0 \leqslant k,j \leqslant 2\), \(j=3, 0 \leqslant k \leqslant 1\), and \(s \geqslant t\), we have
$$\begin{aligned}{} & {} y^{k} |\partial _{y}^{k}\partial _{s}^{j}RHS_{4,1}(s,y)| \\{} & {} \quad \leqslant C \lambda (s)^{2} \log ^{3}(s) \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) {\left\{ \begin{array}{ll} \frac{y^{3}}{s^{8+j}}, \quad y \leqslant \frac{s}{2}\\ \frac{1}{s^{5/2} t^{5/2+j}}, \quad \frac{s}{2}< y < s-t+2 g(t)\lambda (t)\end{array}\right. } \end{aligned}$$
Proof
We re-write (4.117) as
$$\begin{aligned} RHS_{4,1}(s,y){} & {} = \int _{0}^{y} F(w,y) \partial _{s}^{4}\left( RHS_{2}-RHS_{2,0}\right) (s,w) dw \nonumber \\{} & {} \quad - \frac{y^{3}}{16} \int _{0}^{\infty } \partial _{s}^{4}\left( RHS_{2}-RHS_{2,0}\right) (s,w) dw \end{aligned}$$
(4.118)
where
$$\begin{aligned} F(w,y) = \frac{-\left( w^{4}-y^{4}+4 w^{2}y^{2}\log (y/w)\right) }{16 y} \end{aligned}$$
This is of the same form as (4.70), and is treated in the same way, noting that \(\partial _{1}^{j}F(y,y)=0, \quad j=0,1,2\). In addition, we use \(s-t+2g(t)\lambda (t) < s-\frac{t}{2}\), which implies that
$$\begin{aligned} \frac{1}{\langle s-y\rangle } \leqslant \frac{C}{t}, \quad \frac{s}{2}< y < s-t+2 g(t)\lambda (t) \end{aligned}$$
We then directly differentiate (4.118) to treat higher derivatives of \(RHS_{4,1}\). \(\square \)
Let
$$\begin{aligned} q_{4,1,0}(t,r):=\frac{-r}{2} \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \left( \partial _{2}RHS_{4,1}(s,\rho )+\frac{RHS_{4,1}(s,\rho )}{\rho }\right) \qquad \end{aligned}$$
(4.119)
Now, we can estimate \(q_{4,1}\).
Lemma 19
For \(0 \leqslant k \leqslant 1\), and \(0 \leqslant j \leqslant 2\) or \(k=2,j=0\), we have
$$\begin{aligned}{} & {} |r^{k}\partial _{r}^{k}\partial _{t}^{j}\left( q_{4,1}(t,r)-q_{4,1,0}(t,r)\right) | \leqslant \frac{C r^{2} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \lambda (t)^{2}\log ^{3}(t)}{t^{5+j}},\nonumber \\{} & {} \quad \quad \lambda (t) g(t) \leqslant r \leqslant 2 \lambda (t)g(t) \end{aligned}$$
(4.120)
Also, for \(0 \leqslant j \leqslant 3\),
$$\begin{aligned} |\partial _{t}^{j} q_{4,1,0}(t,r)| \leqslant \frac{C r \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \lambda (t)^{2} \log ^{3}(t)}{t^{4+j}} \end{aligned}$$
Proof
The spherical means representation formula gives (see also (4.99), pg. 24 of [25])
$$\begin{aligned} \begin{aligned} q_{4,1}(t,r)&= -\frac{1}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t}\frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}}\\&\quad \int _{0}^{2\pi } \frac{RHS_{4,1}(s,\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )})}{\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}} \left( r+\rho \cos (\theta )\right) d\theta \end{aligned} \end{aligned}$$
(4.121)
If
$$\begin{aligned} G(s,r,\rho ) = \int _{0}^{2\pi } d\theta \frac{\left( r+\rho \cos (\theta )\right) }{\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}} RHS_{4,1}(s,\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}), \quad \rho >0 \end{aligned}$$
Then, by the dominated convergence theorem and Lemma 18, \(\partial _{r}G(s,r,\rho )\) can be computed by differentiation under the integral sign, and
$$\begin{aligned} G(s,0,\rho )=0 \implies G(s,r,\rho ) = r \int _{0}^{1} \partial _{2}G(s,\beta r,\rho ) d\beta \end{aligned}$$
Therefore, we have
$$\begin{aligned} q_{4,1}(t,r) = \frac{-r}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1} d\beta \int _{0}^{2\pi } I_{RHS_{4,1}}(s,r\beta ,\rho ,\theta ) d\theta \end{aligned}$$
(4.122)
where we recall the notation (2.2). Recalling (4.119), we get
$$\begin{aligned} \begin{aligned}&q_{4,1}(t,r)-q_{4,1,0}(t,r) = \frac{-r}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1}d\beta \\&\quad \int _{0}^{2\pi } d\theta \text {integrand}_{4,1,2}(s,r\beta ,\rho ,\theta )\end{aligned} \end{aligned}$$
(4.123)
where
$$\begin{aligned}\begin{aligned} \text {integrand}_{4,1,2}(s,y,\rho ,\theta )&= I_{RHS_{4,1}}(s,y,\rho ,\theta ) \\&\quad -\left( \partial _{2}RHS_{4,1}(s,\rho ) \cos ^{2}(\theta ) + \frac{RHS_{4,1}(s,\rho )}{\rho }\sin ^{2}(\theta )\right) \end{aligned} \end{aligned}$$
We note that
$$\begin{aligned} \text {integrand}_{4,1,2}(s,y,\rho ,\theta )= I_{RHS_{4,1}}(s,y,\rho ,\theta )- I_{RHS_{4,1}}(s,0,\rho ,\theta ) \end{aligned}$$
Also, for all \(r\geqslant 0, \rho >3r, s \geqslant T_{0}, \theta \in [0,2\pi ],\) and \(\beta \in [0,1]\),
$$\begin{aligned}{} & {} y \mapsto I_{RHS_{4,1}}(s,y,\rho ,\theta ) \in C^{1}([0,r\beta ]),\\{} & {} \quad \quad \text {(note that, in this setting, }\sqrt{y^{2}+\rho ^{2}+2 y \rho \cos (\theta )} \geqslant C \rho >0) \end{aligned}$$
Therefore, when \(s-t \geqslant 3r\), and \(\rho >3r\) in the integral in (4.123), we use
$$\begin{aligned} \text {integrand}_{4,1,2}(s,r\beta ,\rho ,\theta ) = \int _{0}^{r\beta } \partial _{2}I_{RHS_{4,1}}(s,y,\rho ,\theta ) dy \end{aligned}$$
and we directly substitute our estimates from Lemma 18 into (4.123) for the other regions, to get (4.120) for \(k=0, j=0\). Next, we use (4.123) to get
$$\begin{aligned} \partial _{r} \left( q_{4,1}-q_{4,1,0}\right) (t,r) = -\frac{1}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{2\pi } d\theta \text {integrand}_{4,1,2}(s,r,\rho ,\theta ) \end{aligned}$$
which is treated with the same argument used for (4.123). This gives (4.120) for \(k=1, j=0\). To estimate \(\partial _{t}^{j} \left( q_{4,1}-q_{4,1,0}\right) (t,r)\), we return to (4.123), and let \(w=s-t\). Then, the dominated convergence theorem, along with the estimates of Lemma 18 allow us to differentiate under the resulting integral signs, and we get, for \(j=1,2\),
$$\begin{aligned}\begin{aligned}&\partial _{t}^{j}\left( q_{4,1}-q_{4,1,0}\right) (t,r) = \frac{-r}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1}d\beta \\&\quad \int _{0}^{2\pi } d\theta \partial _{s}^{j}\text {integrand}_{4,1,2}(s,r\beta ,\rho ,\theta )\end{aligned} \end{aligned}$$
We then repeat the same procedure used for \(q_{4,1}-q_{4,1,0}\). To estimate \(q_{4,1,0}\), we use
$$\begin{aligned} q_{4,1,0}(t,r) = \frac{-r}{2} \int _{0}^{\infty } dw \int _{0}^{w} \frac{\rho d\rho }{\sqrt{w^{2}-\rho ^{2}}} \left( \partial _{2}RHS_{4,1}(t+w,\rho )+\frac{RHS_{4,1}(t+w,\rho )}{\rho }\right) \end{aligned}$$
differentiate under the integral sign, then substitute the estimates from Lemma 18. We prove (4.120) for \(j=0,k=2\) by using the equation solved by \(q_{4,1}\) and our previous estimates from this lemma. \(\square \)
Next, we study the analogous quantities related to \(v_{ex,sub}\) which we recall is defined in (4.83). In particular, we start by studying
$$\begin{aligned} w_{5}(t,r):= v_{ex,sub}(t,r)-v_{ex,sub,0}(t,r) \end{aligned}$$
The function \(w_{5}\) satisfies the following equation with 0 Cauchy data at infinity
$$\begin{aligned} \left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) w_{5}(t,r) = \partial _{t}^{2} \left( v_{ex,ell}-v_{ex,cont}\right) = \partial _{t}^{2}v_{ex,ell,1}(t,r) \end{aligned}$$
where we define
$$\begin{aligned} v_{ex,ell,1}(t,r) :=v_{ex}(t,r)-v_{ex,sub}(t,r)-v_{ex,cont}(t,r)=v_{ex,ell}(t,r)-v_{ex,cont}(t,r) \end{aligned}$$
and recall that \(v_{ex,ell}\) and \(v_{ex,cont}\) are explicitly given in (4.36) and (4.80), respectively. We have the following lemma
Lemma 20
For \(0 \leqslant k,j \leqslant 2\), we have
$$\begin{aligned} |\partial _{t}^{j}\partial _{r}^{k} v_{ex,ell,1}(t,r)| \leqslant \frac{C \lambda (t)^{2-k}}{t^{2+j}g(t)^{3+k}}, \quad g(t)\lambda (t) \leqslant r \leqslant 2g(t)\lambda (t) \end{aligned}$$
Proof
We directly estimate \(v_{ex,ell,1}\) from the formulae (4.36) and (4.80). \(\square \)
Next, we define \(ellsoln_{2}\) by
$$\begin{aligned} ellsoln_{2}(t,r):= \frac{-1}{2}\left( \frac{1}{r}\int _{0}^{r} s^{2}\partial _{t}^{2}v_{ex,ell,1}(t,s) ds + r \int _{r}^{\infty } \partial _{t}^{2}v_{ex,ell,1}(t,s) ds\right) \end{aligned}$$
Then, we have the following lemma.
Lemma 21
For all \(j,k \geqslant 0\), there exists \(C_{j,k}>0\) such that
$$\begin{aligned} r^{k}t^{j}|\partial _{r}^{k}\partial _{t}^{j}ellsoln_{2}(t,r)| \leqslant C_{j,k}\cdot {\left\{ \begin{array}{ll} \frac{1}{r} \frac{\lambda (t)^{5}(1+|\log (\frac{r}{\lambda (t)})|)}{t^{4} }, \quad r \geqslant \lambda (t)\\ r \frac{\lambda (t)^{3}(1+\log ^{2}(\frac{r}{\lambda (t)}))}{t^{4} }, \quad r \leqslant \lambda (t)\end{array}\right. } \end{aligned}$$
(4.124)
In addition, \(\partial _{t}^{2}ellsoln_{2}(t,r) = F_{ellsoln_{2}}(t,r)+\log ^{2}(r) r g_{1}''(t)+r \log (r) g_{2}''(t)\), where, for each \(t \geqslant T_{0}\),
$$\begin{aligned} r \mapsto F_{ellsoln_{2}}(t,r) \text { admits a }C^{2}\text { extension to }[0,\infty ) \end{aligned}$$
For \(0 \leqslant j+k \leqslant 2\),
$$\begin{aligned} |\partial _{r}^{k}\partial _{t}^{j} F_{ellsoln_{2}}(t,r)| \leqslant C {\left\{ \begin{array}{ll} \frac{r^{1-k}\lambda (t)^{3} \log ^{2}(t)}{t^{6+j}}, \quad 0 \leqslant k \leqslant 1\\ \frac{r \lambda (t)(1+|\log (\frac{r}{\lambda (t)})|)}{t^{6+j}}, \quad k=2\end{array}\right. }, \quad r \leqslant \lambda (t) \\ |g_{1}^{(k)}(t)| + |g_{2}^{(k)}(t)| \leqslant \frac{C_{k} \log (t)\lambda (t)^{3}}{t^{4+k}}, \quad k \geqslant 0 \end{aligned}$$
Proof
We have
$$\begin{aligned}\begin{aligned} ellsoln_{2}(t,r)&= \frac{-1}{2}\left( \frac{1}{r}\int _{0}^{r} s^{2}\partial _{t}^{2}v_{ex,ell,1}(t,s) ds + r \int _{r}^{\infty } \partial _{t}^{2}v_{ex,ell,1}(t,s) ds\right) \\&=-\frac{1}{2}\left( \frac{1}{r}f_{7}(t,\frac{r}{\lambda (t)})+r f_{8}(t,\frac{r}{\lambda (t)})\right) \end{aligned} \end{aligned}$$
where
$$\begin{aligned} \begin{aligned}f_{7}(t,R)&= \frac{\lambda (t)}{8} \left( \frac{-8(-2 R^{2}+3(1+R^{2}) \log (1+R^{2})) \lambda '(t)^{4}}{1+R^{2}}\right. \\&\quad \left. -24 \lambda (t) \left( R^{2} \log (1+\frac{1}{R^{2}})+3 \log (R^{2}+1)\right) \lambda '(t)^{2}\lambda ''(t)\right. \\&\quad \left. -4 \lambda (t)^{2} \left( R^{2}\log (1+\frac{1}{R^{2}}) + \log (1+R^{2})\right) \left( 3\lambda ''(t)^{2}+4 \lambda '(t)\lambda '''(t)\right) \right. \\&\quad \left. - \lambda (t)^{3} \left( -R^{2}+R^{2}(2+R^{2}) \log (1+\frac{1}{R^{2}}) + \log (1+R^{2})\right) \lambda ''''(t)\right) \end{aligned} \end{aligned}$$
and
$$\begin{aligned}\begin{aligned} f_{8}(t,R)&= \frac{1}{4} \left( \frac{-4(2+(1+R^{2})\log (1+\frac{1}{R^{2}})) \lambda '(t)^{4}}{(1+R^{2})\lambda (t)} \right. \\&\quad +12 \left( -2 \log (1+\frac{1}{R^{2}}) + \text {Li}_{2}(-\frac{1}{R^{2}})\right) \lambda '(t)^{2}\lambda ''(t)\\&\quad +2 \lambda (t) \text {Li}_{2}(\frac{-1}{R^{2}}) \left( 3\lambda ''(t)^{2}+4 \lambda '(t)\lambda '''(t)\right) \\&\quad \left. +\lambda (t)^{2}(-1+(1+R^{2})\log (1+\frac{1}{R^{2}}) + \text {Li}_{2}(\frac{-1}{R^{2}})) \lambda ''''(t)\right) \end{aligned} \end{aligned}$$
The lemma statement now follows from inspection. \(\square \)
We define
$$\begin{aligned} q_{5}(t,r):= w_{5}(t,r)-ellsoln_{2}(t,r)\psi _{\leqslant 1}(r-t) \end{aligned}$$
so that \(q_{5}\) solves the following equation with 0 Cauchy data at infinity.
$$\begin{aligned} \begin{aligned}&-\partial _{t}^{2}q_{5}+\partial _{r}^{2}q_{5}(t,r) + \frac{1}{r}\partial _{r}q_{5}-\frac{q_{5}}{r^{2}}\\&\quad = \psi _{\leqslant 1}(r-t)\partial _{t}^{2} ellsoln_{2} + \left( 1-\psi _{\leqslant 1}(r-t)\right) \partial _{t}^{2} v_{ex,ell,1}(t,r)\\&\qquad -\psi _{\leqslant 1}'(r-t) \left( 2\left( \partial _{t}+\partial _{r}\right) ellsoln_{2}+\frac{1}{r}ellsoln_{2}\right) \end{aligned} \end{aligned}$$
(4.125)
We define \(q_{5,0}\) by
$$\begin{aligned} q_{5,0}(t,r)= \frac{-r}{2} \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \left( \partial _{112}ellsoln_{2}(s,\rho ) + \frac{\partial _{11}ellsoln_{2}(s,\rho )}{\rho }\right) \qquad \quad \end{aligned}$$
(4.126)
Then, we have the following lemma.
Lemma 22
For \(0 \leqslant k \leqslant 1\) and \(0 \leqslant j \leqslant 2\),
$$\begin{aligned} |t^{j}r^{k}\partial _{t}^{j}\partial _{r}^{k} \left( q_{5}-q_{5,0}\right) (t,r)| \leqslant \frac{C r^{2}\lambda (t)^{3} \left( \log ^{3}(t)+|\log (r)|^{3}\right) }{t^{5} }, \quad r \leqslant 2 g(t)\lambda (t) \end{aligned}$$
For all \(j \geqslant 0\), there exists \(C_{j}>0\) such that
$$\begin{aligned}{} & {} |\partial _{t}^{j} q_{5,0}(t,r)| \leqslant \frac{C_{j} r \log ^{2}(t)\lambda (t)^{3}}{t^{4+j}}\\{} & {} |\partial _{r}^{2}\left( q_{5}-q_{5,0}\right) (t,r)| \leqslant \frac{C \lambda (t)^{3} \log ^{3}(t)}{t^{5}}, \quad g(t)\lambda (t) \leqslant r \leqslant 2 g(t)\lambda (t) \end{aligned}$$
Proof
First, we claim that in the region \(r \leqslant 2 g(t)\lambda (t)\),
$$\begin{aligned} \begin{aligned} q_{5}(t,r)&=\frac{-1}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t}\frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \\&\quad \int _{0}^{2\pi }d\theta \partial _{1}^{2}ellsoln_{2}(s,\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}) \frac{(r+\rho \cos (\theta ))}{\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}}\end{aligned} \end{aligned}$$
(4.127)
To verify this, we recall (4.125), and use the finite speed of propagation. The only item remaining is to show that the integral on the right -hand side of (4.127) solves (4.125) with \(\partial _{t}^{2}ellsoln_{2}\) on the right-hand side, and zero Cauchy data at infinity. For this purpose, we note that
$$\begin{aligned}\begin{aligned}&\frac{-1}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t}\frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \\&\quad \int _{0}^{2\pi }d\theta \partial _{1}^{2}ellsoln_{2}(s,\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}) \frac{(r+\rho \cos (\theta ))}{\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}}\\&\quad = \frac{-1}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t}\frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \\&\quad \int _{0}^{2\pi }d\theta F_{ellsoln_{2}}(s,\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}) \frac{(r+\rho \cos (\theta ))}{\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}}\\&\qquad +\frac{-r^{3}}{2\pi } \int _{0}^{\infty } dy \int _{0}^{y}\frac{w dw}{\sqrt{y^{2}-w^{2}}} \int _{0}^{2\pi }d\theta (1+w \cos (\theta )) \left( \log ^{2}(r) g_{1}''(t+r y) \right. \\&\qquad \left. + \log (r) \left( \log (1+w^{2}+2 w \cos (\theta ))g_{1}''(t+r y)+g_{2}''(t+r y)\right) \right. \\&\qquad \left. +\log ^{2}(\sqrt{1+w^{2}+2 w \cos (\theta )})g_{1}''(t+r y)\right. \\&\qquad \left. +\log (\sqrt{1+w^{2}+2 w \cos (\theta )})g_{2}''(t+r y))\right) \end{aligned} \end{aligned}$$
The point of this splitting is that, by Lemma 21 and the dominated convergence theorem, we can differentiate up to two times in (t, r) under the integral sign. Then, we can proceed as in any standard verification of the spherical means formula. Then, we return to (4.127) and use (4.124) and the same procedure used to establish Lemma 19. \(\square \)
Now, we compute the leading behavior of \(u_{ell,2}(t,r)-v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})\) in the matching region. We first recall that \(v_{ell,2,0}\) and \(v_{ell,2,0,main}\) are explicitly given in (4.12) and (4.14), respectively. So, it suffices to explain how to compute the leading behavior of \(u_{ell,2}(t,r)-v_{ell,2,0}(t,\frac{r}{\lambda (t)})\). From (4.16), we have
$$\begin{aligned} \begin{aligned}u_{ell,2}(t,r)-v_{ell,2,0}(t,\frac{r}{\lambda (t)})&= \frac{-\phi _{0}(\frac{r}{\lambda (t)})}{2}\int _{0}^{\frac{r}{\lambda (t)}} err_{1}(t,s\lambda (t)) s e_{2}(s) ds\\&\quad + e_{2}(\frac{r}{\lambda (t)}) \int _{0}^{\frac{r}{\lambda (t)}} err_{1}(t,s\lambda (t)) \frac{s \phi _{0}(s)}{2} ds\\&= \frac{-\phi _{0}(\frac{r}{\lambda (t)})}{2}\int _{0}^{\frac{r}{\lambda (t)}} err_{1,0}(t,s\lambda (t)) s e_{2}(s) ds \\&\quad + \frac{-\phi _{0}(\frac{r}{\lambda (t)})}{2}\int _{0}^{\frac{r}{\lambda (t)}} \left( err_{1}-err_{1,0}\right) (t,s\lambda (t)) s e_{2}(s) ds\\&\quad +\frac{r}{2\lambda (t)} \int _{0}^{\frac{r}{\lambda (t)}} err_{1}(t,s\lambda (t)) \frac{s \phi _{0}(s)}{2}ds\\&\quad + \left( e_{2}(\frac{r}{\lambda (t)})-\frac{r}{2\lambda (t)}\right) \int _{0}^{\frac{r}{\lambda (t)}} err_{1}(t,s\lambda (t)) \frac{s \phi _{0}(s)}{2}ds\end{aligned} \end{aligned}$$
where we recall that \(err_{1,0}(t,r)\) is defined in (4.18). Using the definition of \(v_{ell,2,0,main}\) given in (4.14), we can write down the leading behavior of \(u_{ell,2}(t,r) - v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})\) in the region \(r \sim g(t)\lambda (t)\), which we denote by \(u_{e,3}(t,r)\).
$$\begin{aligned} u_{e,3}(t,r)= & {} \frac{-2 \lambda (t)}{2 r}\int _{0}^{\frac{r}{\lambda (t)}} err_{1,0}(t,s\lambda (t)) \frac{s}{2} s ds \nonumber \\{} & {} +\frac{r}{2\lambda (t)} \int _{0}^{\infty }\left( err_{1}(t,s\lambda (t))-err_{1,0}(t,s\lambda (t))\right) s \frac{\phi _{0}(s)}{2} ds \nonumber \\{} & {} +\frac{r}{2\lambda (t)} \int _{0}^{\frac{r}{\lambda (t)}} err_{1,0}(t,s\lambda (t)) \frac{\phi _{0}(s) s ds}{2} \nonumber \\{} & {} +\frac{r \lambda (t)^{2}\lambda ''''(t)}{96}\left( 111-2\pi ^{2}-108 \log (\frac{r}{\lambda (t)}) + 24 \log ^{2}(\frac{r}{\lambda (t)})\right) \nonumber \\{} & {} + \frac{r}{ 16} (5-4 \log (\frac{r}{\lambda (t)})) \left( \partial _{t}^{2}\left( \frac{\lambda '(t)^{2}}{2 \lambda (t)}+\lambda ''(t) (1+\log (\lambda (t)))\right) 2 \lambda (t)^{2} \right. \nonumber \\{} & {} \left. -2\lambda (t)^{2}\lambda ''''(t)(1+\log (\lambda (t)))\right) \end{aligned}$$
(4.128)
By computing the integrals, we get
$$\begin{aligned}{} & {} u_{e,3}(t,R \lambda (t)) \nonumber \\{} & {} \quad =\frac{\left( 2\lambda '(t)^{4}-7 \lambda (t)\lambda '(t)^{2}\lambda ''(t)+4 \lambda (t)^{2}\lambda ''(t)^{2} + 6 \lambda (t)^{2}\lambda '(t)\lambda '''(t)\right) R\left( 5-4 \log (R)\right) }{16} \nonumber \\{} & {} \qquad +\frac{1}{96}\left( 111-2\pi ^{2}-108 \log (R)+24 \log ^{2}(R)\right) R \lambda (t)^{3}\lambda ''''(t) \nonumber \\{} & {} \qquad +\frac{3}{4} R \lambda '(t)^{4}+\frac{3}{8}R \left( -2 \lambda '(t)^{4}+5 \lambda (t)\lambda '(t)^{2}\lambda ''(t)\right) +\frac{3}{4} R \lambda (t)^{2}\partial _{t}\left( \lambda '(t)\lambda ''(t)\right) \nonumber \\{} & {} \qquad +\frac{1}{24} R \lambda (t)^{2}\left( 21+\pi ^{2}-18\log (R)+12 \log ^{2}(R)\right) \partial _{t}^{2}\left( \lambda (t)\lambda ''(t)\right) \nonumber \\{} & {} \qquad +\frac{1}{24} R \lambda (t)^{2}\left( 39+\pi ^{2}-42 \log (R) + 12 \log ^{2}(R)\right) \left( \partial _{t}^{2}\left( \lambda (t)\lambda ''(t)\right) -\lambda (t)\lambda ''''(t)\right) \nonumber \\{} & {} \qquad +\frac{1}{12}R \lambda (t)\left( 60+\pi ^{2}-54 \log (R)+12 \log ^{2}(R)\right) \lambda '(t)^{2}\lambda ''(t) \nonumber \\{} & {} \qquad -\frac{1}{4}R \lambda (t)^{2}c_{1}''(t)-\frac{R}{2}\left( \lambda (t)c_{1}'(t)\lambda '(t)+\frac{c_{1}(t)}{2} \left( -2 \lambda '(t)^{2}+\lambda (t)\lambda ''(t)\right) \right) -\frac{R}{2}c_{1}(t)\lambda '(t)^{2}\nonumber \\{} & {} \qquad +\frac{R}{8} \lambda '(t)^{4}\left( 17-12 \log (R)\right) -\frac{R}{4}\left( 5 \lambda (t)\lambda '(t)^{2} \lambda ''(t)-2\lambda '(t)^{4}\right) \left( -3+2\log (R)\right) \nonumber \\{} & {} \qquad +\frac{R}{4} \left( 3-8 \log (R)\right) \lambda (t)^{2}\partial _{t}\left( \lambda ''(t)\lambda '(t)\right) \nonumber \\{} & {} \qquad +\frac{R \lambda (t)^{2}\partial _{t}^{2}\left( \lambda (t)\lambda ''(t)\right) }{24}\left( 6+\pi ^{2}+12 \zeta (3) -\left( 36+2 \pi ^{2}\right) \log (R)+12 \log ^{2}(R) - 8 \log ^{3}(R)\right) \nonumber \\{} & {} \qquad -\frac{R\lambda (t)^{2}}{24}\left( \partial _{t}^{2}\left( \lambda (t)\lambda ''(t)\right) -\lambda (t)\lambda ''''(t)\right) \left( -27 -2 \pi ^{2}-12 \zeta (3)+(36+2\pi ^{2})\log (R) \right. \nonumber \\{} & {} \qquad \left. -24 \log ^{2}(R)+8 \log ^{3}(R)\right) \nonumber \\{} & {} \qquad + R\frac{\lambda (t)\lambda ''(t)\lambda '(t)^{2}}{72} \left( 255+17 \pi ^{2}+72 \zeta (3)-(360+12 \pi ^{2})\log (R) \right. \nonumber \\{} & {} \qquad \quad \left. + 252 \log ^{2}(R)-48 \log ^{3}(R)\right) \nonumber \\{} & {} \qquad +R \frac{c_{1}''(t)\lambda (t)^{2}}{4} \left( -1+2 \log (R)\right) +R\left( \lambda (t)\lambda '(t)c_{1}'(t)\right. \nonumber \\{} & {} \qquad \quad \left. +\frac{c_{1}(t)}{2}\left( \lambda (t)\lambda ''(t)-2\lambda '(t)^{2}\right) \right) (-1+\log (R))\nonumber \\{} & {} \qquad + R\frac{c_{1}(t)}{12} \lambda '(t)^{2}\left( -17+12 \log (R)\right) +R\frac{\lambda (t)^{3}\lambda ''''(t)}{48} \left( \pi ^{2}-12 \log ^{2}(R)\right) \nonumber \\{} & {} \qquad + \frac{R}{2}\log (R) \lambda (t)^{3} \left( \partial _{t}^{2}\left( \frac{\lambda '(t)^{2}}{2 \lambda (t)} + \lambda ''(t)(1+\log (\lambda (t)))\right) -\log (\lambda (t))\lambda ''''(t)\right) \end{aligned}$$
(4.129)
Now, we estimate the difference between \(u_{ell,2}(t,r)-v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})\) and \(u_{e,3}(t,r)\).
Lemma 23
For \(u_{ell,2}\) defined in (4.9), \(v_{ell,2,0,main}\) defined in (4.14), and \(u_{e,3}\) defined in (4.128) we have, for \(0 \leqslant j+k \leqslant 2\),
$$\begin{aligned} |\partial _{t}^{j}\partial _{r}^{k}\left( u_{ell,2}(t,r)-v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})-u_{e,3}(t,r)\right) | \leqslant \frac{C \lambda (t)^{5} (1+\log ^{4}(\frac{r}{\lambda (t)}))}{r^{1+k}t^{4+j}}, \quad r > \lambda (t) \end{aligned}$$
Proof
We estimate each term in the difference between (4.15) plus (4.16) and (4.128) directly, using the symbol type estimates on \(\lambda (t)\), and the fact that \(err_{1}(t,r)\) and \(err_{1,0}(t,r)\) are symbols in (t, r). \(\square \)
Combining our work, we get
$$\begin{aligned} \begin{aligned}&\text {leading part of }\left( u_{w,2}-u_{w,2,ell,0}+v_{ex,sub}\right) (t,r):=u_{w,3}(t,r)\\&\quad =u_{w,2,sub,0,cont}(t,r) + v_{ex,sub,0,cont}(t,r)\\&\qquad -\frac{1}{2}\left( (u_{w,2,ell}-u_{w,2,ell,0})_{princ,1}(t,r)+(u_{w,2,ell}-u_{w,2,ell,0})_{princ,2}(t,r)\right) \\&\qquad -\frac{r}{2} \lambda (t) \int _{t}^{\infty } W_{3}(\frac{s-t}{\lambda (t)}) \lambda ''''(s) ds-\frac{r}{2} \int _{0}^{\infty } d\xi \left( \frac{-8}{\xi \lambda (t)^{2}}+4 \xi K_{2}(\xi \lambda (t)) + 2\xi \right) \sin (t\xi ) \widehat{v_{2,0}}(\xi )\\&\qquad -\frac{r}{2} \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \left( \partial _{112}ellsoln_{2}(s,\rho ) + \frac{\partial _{11}ellsoln_{2}(s,\rho )}{\rho }\right) \\&\qquad -\frac{r}{2} \int _{0}^{\infty } dw \int _{0}^{w} \frac{\rho d\rho }{\sqrt{w^{2}-\rho ^{2}}} \left( \partial _{2}RHS_{4,1}(t+w,\rho )+\frac{RHS_{4,1}(t+w,\rho )}{\rho }\right) \end{aligned}\nonumber \\ \end{aligned}$$
(4.130)
A careful inspection of all of the terms shows that the \(r \log ^{k}(r)\) terms from \(u_{e,3}\) and \(u_{w,3}\) all exactly match, for \(k=1,2,3\). In other words, we have
$$\begin{aligned} \begin{aligned}&u_{e,3}(t,r)-u_{w,3}(t,r) \\&\quad = r F_{2}(t)+\frac{r\lambda (t)}{2}\int _{t}^{\infty } W_{3}(\frac{s-t}{\lambda (t)}) \lambda ''''(s) ds\\&\qquad +\frac{r}{2}\int _{t}^{2t} \left( \frac{(g_{0}+h_{0})''(s)-(g_{0}+h_{0})''(t)}{(s-t)} + \frac{\left( (g_{1}+2 f_{1})''(s)-(g_{1}+2f_{1})''(t)\right) \log (2(s-t))}{(s-t)}\right. \\&\qquad \left. -\frac{\left( g_{2}''(s)-g_{2}''(t)\right) \left( -12 \log (s-t)\log (4(s-t))+\pi ^{2}-12 \log ^{2}(2)\right) }{12(s-t)}\right) ds\\&\qquad +\frac{r}{2} \int _{2t}^{\infty } \left( (g_{0}+h_{0})''(s)+(g_{1}+2f_{1})''(s)\log (2(s-t))\right. \\&\qquad \left. -\frac{g_{2}''(s) \left( -12 \log (s-t)\log (4(s-t))+\pi ^{2}-12 \log ^{2}(2)\right) }{12}\right) \frac{ds}{(s-t)}\\&\qquad + \frac{r}{2} \int _{0}^{\infty } d\xi \left( \frac{-8}{\xi \lambda (t)^{2}}+4 \xi K_{2}(\xi \lambda (t))+2 \xi \right) \sin (t\xi ) \widehat{v_{2,0}}(\xi )\\&\qquad +\frac{r}{2} \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \left( \partial _{112}ellsoln_{2}(s,\rho ) + \frac{\partial _{11}ellsoln_{2}(s,\rho )}{\rho }\right) \\&\qquad +\frac{r}{2} \int _{0}^{\infty } dw \int _{0}^{w} \frac{\rho d\rho }{\sqrt{w^{2}-\rho ^{2}}} \left( \partial _{2}RHS_{4,1}(t+w,\rho )+\frac{RHS_{4,1}(t+w,\rho )}{\rho }\right) \\&\quad :=r F_{3}(t)\end{aligned} \end{aligned}$$
(4.131)
where \(F_{2}(t) = F_{2,0}(t)+F_{2,1}(t)\), and
$$\begin{aligned} F_{2,0}(t)= & {} \frac{3 \lambda '(t)^4}{4 \lambda (t)}+\frac{ (12 \log (\lambda (t))+17) \lambda '(t)^4}{8 \lambda (t)} -\frac{ c_{1}(t) \lambda '(t)^2}{2 \lambda (t)}\\{} & {} \quad +\frac{c_{1}(t) (-12 \log (\lambda (t))-17) \lambda '(t)^2}{12 \lambda (t)}\\{} & {} \quad +\frac{1}{12} \left( 12 \log ^2(\lambda (t))+54 \log (\lambda (t))+\pi ^2+60\right) \lambda ''(t) \lambda '(t)^2+\frac{3}{4} \partial _{t} \left( \lambda '(t) \lambda ''(t)\right) \lambda (t)\\{} & {} \quad +\frac{1}{4} \partial _{t} \left( \lambda ''(t) \lambda '(t)\right) \lambda (t) (8 \log (\lambda (t))+3)\\{} & {} \quad -\frac{ \left( 5 \lambda (t) \lambda '(t)^2 \lambda ''(t)-2 \lambda '(t)^4\right) (-2 \log (\lambda (t))-3)}{4 \lambda (t)}\\{} & {} \quad +\frac{1}{24} \partial _{t}^{2} \left( \lambda (t) \lambda ''(t)\right) \lambda (t) \left( 12 \log ^2(\lambda (t))+18 \log (\lambda (t))+\pi ^2+21\right) \\{} & {} \quad +\frac{1}{24} \partial _{t}^{2} \left( \lambda (t) \lambda ''(t)\right) \lambda (t) \left( 8 \log ^3(\lambda (t))+12 \log ^2(\lambda (t))\right. \\{} & {} \quad \left. +\left( 36+2 \pi ^2\right) \log (\lambda (t))+\pi ^2+12 \zeta (3)+6\right) \\{} & {} \quad -\frac{1}{24} \lambda (t) \left( \partial _{t}^{2} \left( \lambda (t) \lambda ''(t)\right) -\lambda (t) \lambda ''''(t)\right) \left( -8 \log ^3(\lambda (t))\right. \\{} & {} \quad \quad -24 \log ^2(\lambda (t))-\left( 36+2 \pi ^2\right) \log (\lambda (t))\\{} & {} \quad \left. -2 \pi ^2-12 \zeta (3)-27\right) \\{} & {} \quad +\frac{ (-\log (\lambda (t))-1) \left( \lambda (t) c_{1}'(t) \lambda '(t)+\frac{1}{2} c_{1}(t) \left( \lambda (t) \lambda ''(t)-2 \lambda '(t)^2\right) \right) }{\lambda (t)}-\frac{1}{4} \lambda (t) c_{1}''(t)\\{} & {} \quad +\frac{1}{4} \lambda (t) (-2 \log (\lambda (t))-1) c_{1}''(t)\\{} & {} \quad +\frac{\lambda ''(t)}{72} \lambda '(t)^2 \left( 48 \log ^3(\lambda (t))+252 \log ^2(\lambda (t))+\left( 360+12 \pi ^2\right) \log (\lambda (t))\right. \\{} & {} \quad \quad \left. +17 \pi ^2+72 \zeta (3)+255\right) \\{} & {} \quad +\frac{3 \left( 5 \lambda (t) \lambda '(t)^2 \lambda ''(t)-2 \lambda '(t)^4\right) }{8 \lambda (t)}-\frac{ \left( \lambda (t) c_{1}'(t) \lambda '(t)+\frac{1}{2} c_{1}(t) \left( \lambda (t) \lambda ''(t)-2 \lambda '(t)^2\right) \right) }{2 \lambda (t)}\\{} & {} \quad +\frac{(4 \log (\lambda (t))+5) \left( 2 \lambda '(t)^4-7 \lambda (t) \lambda ''(t) \lambda '(t)^2+6 \lambda (t)^2 \lambda '''(t) \lambda '(t)+4 \lambda (t)^2 \lambda ''(t)^2\right) }{16 \lambda (t)}\\{} & {} \quad +\frac{1}{96} \lambda (t)^2 \left( 24 \log ^2(\lambda (t))+108 \log (\lambda (t))-2 \pi ^2+111\right) \lambda ''''(t)\\{} & {} \quad \quad +\frac{1}{48} \lambda (t)^2 \left( \pi ^2-12 \log ^2(\lambda (t))\right) \lambda ''''(t)\\{} & {} \quad +\frac{1}{24} \lambda (t) \left( 12 \log ^2(\lambda (t))+42 \log (\lambda (t))+\pi ^2+39\right) \left( \partial _{t}^{2} \left( \lambda (t) \lambda ''(t)\right) -\lambda (t) \lambda ''''(t)\right) \end{aligned}$$
$$\begin{aligned}{} & {} F_{2,1}(t) =-\frac{1}{2} \log (\lambda (t)) \lambda (t)^2 \left( \partial _{t}^{2} \left( \frac{\lambda '(t)^2}{2 \lambda (t)}+(\log (\lambda (t))+1) \lambda ''(t)\right) -\log (\lambda (t)) \lambda ''''(t)\right) \\{} & {} \quad -\left( c_{0} g_{2}''(t)-\frac{1}{2} g_{0}''(t)-\frac{1}{4} (\log (4)-1) g_{0}''(t)-\frac{1}{2} \log (t) g_{0}''(t)-\frac{1}{16} \left( \pi ^2-12\right) g_{1}''(t)\right) \\{} & {} \quad -\left( -\left( \frac{\log ^2(t)}{4}+\frac{1}{2} \log (2) \log (t)\right) g_{1}''(t)-\frac{1}{24} \left( -2 \pi ^2+15+6 \log ^2(2)\right) g_{1}''(t)\right) \\{} & {} \quad -\left( -\frac{1}{16} \left( -28 \zeta (3)+\pi ^2+28\right) g_{2}''(t)\right. \\{} & {} \quad \quad \left. -\frac{1}{24} \log (t) \left( 4 \log (t) (\log (t)+\log (8))-\pi ^2+12 \log ^2(2)\right) g_{2}''(t)\right) \\{} & {} \quad -\left( -\frac{1}{8} \left( \pi ^2-12\right) f_{1}''(t)-2 f_{1}''(t) \left( \frac{1}{4} \log ^2(t)+\frac{1}{2} \log (2) \log (t)\right) -\frac{1}{2} h_{0}''(t)\right) \\{} & {} \quad -\left( -\frac{1}{4} (\log (4)-1) h_{0}''(t)-\frac{1}{2} \log (t) h_{0}''(t)-\frac{1}{12} \left( -2 \pi ^2+15+6 \log ^2(2)\right) f_{1}''(t)\right) \\{} & {} \quad +\frac{1}{2}\left( \frac{1}{3} \lambda (t)^2 \left( -24 j_{1} \log (\lambda (t))-12 j_{1}-12 j_{2} \log ^2(\lambda (t))-12 j_{2} \log (\lambda (t))-\pi ^2 j_{2}\right) \right) \\{} & {} \quad +\frac{1}{2}\left( -2 \left( 2 j_{1} \lambda (t)^2-j_{2} \lambda (t)^2\right) \right) \end{aligned}$$
We quickly record some symbol-type estimates on \(F_{3}(t)\).
Lemma 24
There exists \(C>0\) such that, for \(0 \leqslant k \leqslant 3 \),
$$\begin{aligned} t^{k}|F_{3}^{(k)}(t)| \leqslant \frac{C \lambda (t)^{2}}{t^{4}} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{3}(t) \end{aligned}$$
(4.132)
Proof
Using the symbol type estimates on \(\lambda \), Lemmas 19 and 22, and integration by parts in the region \(\xi t \geqslant 1\) to treat the fifth term in (4.131), we get
$$\begin{aligned} |F_{3}(t)| \leqslant \frac{C \lambda (t)^{2}}{t^{4}} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{3}(t) \end{aligned}$$
We prove (4.132) for \(k>0\) as follows. Using the symbol-type estimates on \(\lambda \), we see that \(c_{1},f_{n},h_{n}, g_{n}\), and \(j_{n}\) are all symbols, with estimates implying that their contributions to \(F_{3}^{(k)}(t)\) are bounded above by the right-hand side of (4.132). Moreover,
$$\begin{aligned} \int _{t}^{\infty } W_{3}\left( \frac{s-t}{\lambda (t)}\right) \lambda ''''(s) ds = \int _{0}^{\infty } W_{3}(w) \lambda ''''(\lambda (t)w+t) \lambda (t) dw \end{aligned}$$
is a symbol in t because \(\lambda (t)\) is. We similarly note that the sum of the third and fourth terms on the right-hand side of (4.131) is a symbol in t. Finally, we study the following integral
$$\begin{aligned} I_{7}(t)=\int _{0}^{\infty } d\xi \left( \frac{-8}{\xi \lambda (t)^{2}}+4 \xi K_{2}(\xi \lambda (t))+2 \xi \right) \sin (t\xi ) \widehat{v_{2,0}}(\xi ) \end{aligned}$$
Letting
$$\begin{aligned} H(x) = \frac{-8}{x} + 4 x K_{2}(x) + 2x, \quad x>0 \end{aligned}$$
we get, for \(0 \leqslant k \leqslant 7\),
$$\begin{aligned} |H^{(k)}(x)| \leqslant C {\left\{ \begin{array}{ll} x^{3-k} (1+|\log (x)|), \quad x \leqslant \frac{1}{2}\\ {\left\{ \begin{array}{ll} x^{1-k}, \quad 0 \leqslant k \leqslant 1\\ \frac{1}{x^{1+k}}, \quad 2 \leqslant k \end{array}\right. }, \quad x> \frac{1}{2}\end{array}\right. } \end{aligned}$$
Then, by Lemma (4) and inspection of the following formula, \(I_{7}(t)\) is a symbol in t with \(|t^{k}I_{7}^{(k)}(t)|\) bounded above by the right-hand side of (4.132) (for \(0 \leqslant k \leqslant 3\)).
$$\begin{aligned} I_{7}(t) = \frac{1}{\lambda (t)} \int _{0}^{\infty } \frac{d \omega }{t} H\left( \frac{\omega \lambda (t)}{t}\right) \sin (\omega ) \widehat{v_{2,0}}(\frac{\omega }{t}) \end{aligned}$$
\(\square \)
Now, we can choose the free wave, \(v_{2,2}\) from (4.50), so that the leading part of \(v_{2,2}(t,r)\) in the region \(r \sim g(t)\lambda (t)\) is exactly equal to \(u_{e,3}-u_{w,3}\). As in (4.37), the leading part of \(v_{2,2}(t,r)\) in the matching region is given by
$$\begin{aligned} \frac{-r}{4}\left( -2 \int _{0}^{\infty } \xi \sin (t\xi ) \widehat{v_{2,3}}(\xi ) d\xi \right) :=\frac{-r}{4} G_{3}(t) \end{aligned}$$
(4.133)
and we choose the initial velocity, \(v_{2,3}\) by the following, which is the analog of (4.41).
$$\begin{aligned} \widehat{v_{2,3}}(\xi ) = \frac{-1}{\pi \xi } \int _{0}^{\infty } (-4F_{3}(t)) \psi _{2}(t) \sin (t\xi ) dt, \quad \psi _{2}(x) = {\left\{ \begin{array}{ll} 0, \quad x \leqslant T_{2}\\ 1, \quad x \geqslant 2 T_{2}\end{array}\right. }, \quad 0 \leqslant \psi _{2}(x) \leqslant 1\qquad \nonumber \\ \end{aligned}$$
(4.134)
(We recall that \(T_{2}\) is defined in (4.1)). Note that the sine transform inversion formula implies that
$$\begin{aligned} G_{3}(t) = -4 F_{3}(t)\psi _{2}(t),\quad t>0 \end{aligned}$$
(4.135)
Up to this point, all of our computations and estimates were valid for all \(t \geqslant T_{0}\) for any \(T_{0} \geqslant T_{2}\). In particular, they are valid for all \(t \geqslant T_{2}\). At this stage, we restrict \(T_{0}\) so that \(T_{0} \geqslant 2T_{2}\) but is otherwise arbitrary. Thus, we have \(G_{3}(t) = -4 F_{3}(t)\) for \(t \geqslant T_{0}\). We start with some estimates on \(\widehat{v_{2,3}}(\xi )\):
Lemma 25
For \(0 \leqslant k \leqslant 3\),
$$\begin{aligned} |\xi ^{k}\partial _{\xi }^{k}\widehat{v_{2,3}}(\xi )| \leqslant {\left\{ \begin{array}{ll} C , \quad \xi < \frac{1}{100}\\ \frac{C \xi ^{k}}{\xi ^{4}}, \quad \xi \geqslant \frac{1}{100}\end{array}\right. } \end{aligned}$$
Proof
We use the same procedure as in Lemma 4. For instance, if \(\xi < \frac{1}{100}\), then,
$$\begin{aligned} \widehat{v_{2,3}}(\xi ) = \frac{-1}{\pi \xi } \int _{0}^{\frac{1}{\xi }} (-4F_{3}(t)) \psi _{2}(t) \sin (t\xi ) dt - \frac{1}{\pi \xi } \int _{\frac{1}{\xi }}^{\infty } (-4F_{3}(t)) \psi _{2}(t) \sin (t\xi ) dt \end{aligned}$$
So,
$$\begin{aligned}\begin{aligned}&|\widehat{v_{2,3}}(\xi )|\\&\leqslant C \int _{0}^{\frac{1}{\xi }} \frac{\lambda (t)^{2}\sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{3}(t) \mathbbm {1}_{\{t \geqslant T_{2}\}} dt}{t^{3}}\\&\quad + \frac{C}{\xi } \int _{\frac{1}{\xi }}^{\infty } \frac{\lambda (t)^{2} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{3}(t) \mathbbm {1}_{\{t \geqslant T_{2}\}} dt}{t^{4}}\\&\leqslant C\end{aligned} \end{aligned}$$
where we used (4.3). Also, the estimates established in this lemma are not rapidly decaying because we only estimated up to 3 derivatives of \(F_{3}\) in Lemma 24. This was due to the fact that the seventh term of (4.131) was estimated using Lemma 19. \(\square \)
Now, we can estimate \(v_{2,2}\)
Lemma 26
We have the following estimates. For \(0 \leqslant j \leqslant 2, \quad 0 \leqslant k \leqslant 1\),
$$\begin{aligned} |\partial _{t}^{j}\partial _{r}^{k}v_{2,2}(t,r)| \leqslant {\left\{ \begin{array}{ll}C r^{1-k} \frac{\lambda (t)^{2}}{t^{4+j}} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{3}(t), \quad r \leqslant \frac{t}{2}\\ \frac{C}{\langle t-r\rangle ^{5/2+j+k} \sqrt{t}} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{3}(t), \quad t>r>\frac{t}{2}\\ \frac{C}{\sqrt{r}\langle t-r \rangle ^{\frac{1}{2}+j+k}}, \quad r>t, \quad j+k \leqslant 2\end{array}\right. }\\ |\partial _{r}^{2}v_{2,2}(t,r)| \leqslant {\left\{ \begin{array}{ll}\frac{C r \lambda (t)^{2}}{t^{6}} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{3}(t), \quad r \leqslant \frac{t}{2}\\ \frac{C}{\langle t-r\rangle ^{9/2} \sqrt{t}} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{3}(t), \quad t>r>\frac{t}{2}\\ \frac{C}{\sqrt{r}\langle t-r \rangle ^{\frac{5}{2}}} , \quad r>t\end{array}\right. } \end{aligned}$$
For all \(0 \leqslant j+k \leqslant 3\),
$$\begin{aligned} |\partial _{t}^{j}\partial _{r}^{k}v_{2,2}(t,r)| \leqslant \frac{C}{\sqrt{r}}, \quad r \geqslant \frac{t}{2} \end{aligned}$$
Finally, for \(0 \leqslant j,k \leqslant 1\) or \(k=2,j=0\),
$$\begin{aligned} |\partial _{t}^{j}\partial _{r}^{k}\left( v_{2,2}(t,r)-r F_{3}(t)\right) | \leqslant \frac{C r^{3-k} \lambda (t)^{2} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{3}(t)}{t^{6+j}}, \quad r \leqslant \frac{t}{2}\qquad \end{aligned}$$
(4.136)
Proof
In the region \(r \leqslant \frac{t}{2}\), we use (4.3), Lemma 24, and the same procedure used to estimate \(v_{2}\). Next, we treat the region \(t> r >\frac{t}{2}\). Here, we use
$$\begin{aligned} v_{2,2}(t,r) = \frac{-r}{4\pi } \int _{0}^{\pi } \sin ^{2}(\theta )\left( G_{3}(t+r\cos (\theta ))+G_{3}(t-r\cos (\theta ))\right) d\theta \end{aligned}$$
(4.137)
Note that \(t\pm r\cos (\theta ) >0\) since \(t>r\), so \(G_{3}(t\pm r\cos (\theta )) = -4 \left( F_{3}\cdot \psi _{2}\right) (t\pm r \cos (\theta ))\), by (4.135). It suffices to treat the following integral
$$\begin{aligned}\begin{aligned}&|\frac{-r}{4\pi } \int _{0}^{\pi }\sin ^{2}(\theta ) G_{3}(t+r\cos (\theta )) d\theta | \\&\quad \leqslant C r \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{3}(t) \int _{0}^{\pi } \frac{\sin ^{2}(\theta ) d\theta }{(t+r\cos (\theta ))^{4}}\end{aligned} \end{aligned}$$
where we use the fact that \(\psi _{2}(x) =0, \quad x < 100\). We then use Cauchy’s residue theorem, recalling that \(t>r\), to get
$$\begin{aligned} \int _{0}^{\pi } \frac{\sin ^{2}(\theta ) d\theta }{(t+r\cos (\theta ))^{4}} = \frac{1}{2}\int _{0}^{2\pi } \frac{\sin ^{2}(\theta ) d\theta }{(t+r\cos (\theta ))^{4}} = \frac{\pi t}{2 \left( t^2-r^2\right) ^{5/2}} \end{aligned}$$
(4.138)
and this gives
$$\begin{aligned} |v_{2,2}(t,r)| \leqslant \frac{C}{(t-r)^{5/2} \sqrt{t}} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{3}(t), \quad t>r>\frac{t}{2} \end{aligned}$$
The higher derivatives are treated similarly. Note that \(G_{3}(t) = -2 \int _{0}^{\infty } \xi \sin (t\xi ) \widehat{v_{2,3}}(\xi ) d\xi \) is an odd function of t, and
$$\begin{aligned} G_{3}(t) = -4 F_{3}(t)\psi _{2}(t),\quad t>0 \end{aligned}$$
Also, if \(r >t\), then, the argument of \(G_{3}\) is negative in a region of the \(\theta \) integral in (4.137). So, the procedure of estimating \(v_{2,2}(t,r)\) in the region \(r\geqslant t\) using (4.137) is more involved than in the region \(r <t\). Instead, we simply use
$$\begin{aligned} v_{2,2}(t,r) = \int _{0}^{\infty } \widehat{v_{2,3}}(\xi ) J_{1}(r\xi ) \sin (t\xi ) d\xi \end{aligned}$$
and proceed exactly as in the proof of Lemma 5. We also do the same procedure as in (4.46) to finish the proof of all of the estimates in the lemma statement except for (4.136). The estimate (4.136) follows from
$$\begin{aligned}\begin{aligned} v_{2,2}(t,r) -r F_{3}(t)&= \frac{-r}{4 \pi } \int _{0}^{\pi } \sin ^{2}(\theta )\left( G_{3}(t_{+})-G_{3}(t)-r \cos (\theta ) G_{3}'(t)\right. \\&\quad \left. +G_{3}(t_{-})-G_{3}(t)+r \cos (\theta ) G_{3}'(t)\right) d\theta \end{aligned} \end{aligned}$$
where
$$\begin{aligned} t_{\pm } = t \pm r \cos (\theta ) \end{aligned}$$
\(\square \)
Finally, we are ready to obtain the main result of this section, Proposition 2. We start with the decomposition
$$\begin{aligned}\begin{aligned}&u_{ell,2}(t,r)-v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})-\left( v_{ex}(t,r)-v_{ex,cont}(t,r)+u_{w,2}(t,r)\right. \\&\qquad \left. -u_{w,2,ell,0,cont}(t,r)+v_{2,2}(t,r)\right) \\&\quad = u_{ell,2}(t,r)-v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})-u_{e,3}(t,r)\\&\qquad -\left( v_{ex,sub,0}-v_{ex,sub,0,cont}+q_{5}-q_{5,0}+v_{ex,ell}-v_{ex,cont}+u_{w,2,sub,0}-u_{w,2,sub,0,cont}\right) \\&\qquad -\left( q_{4,2}+(ellsoln-r Q(t))\psi _{\leqslant 1}(r-t)+u_{w,2,ell}-u_{w,2,ell,0}\right. \\&\qquad \left. -(u_{w,2,ell}(t,r)-u_{w,2,ell,0}(t,r))_{princ}\right) \\&\qquad -\left( q_{4,1}-q_{4,1,0}+u_{w,2,ell,0}-u_{w,2,ell,0,cont}+v_{2,2}-r F_{3}(t)+ellsoln_{2}(t,r)\psi _{\leqslant 1}(r-t)\right) \end{aligned} \end{aligned}$$
and use Lemmas 10, 12, 13, 15–17, 19–23, 26. \(\square \)
4.8 Joining the small r and large r solutions
Define
$$\begin{aligned} u_{e}(t,r) = u_{ell}(t,r)+u_{ell,2}(t,r), \quad u_{wave}(t,r) = u_{w}(t,r)+u_{w,2}(t,r)+v_{2,2}(t,r)\nonumber \\ \end{aligned}$$
(4.139)
where we recall that \(u_{w}\) is defined in (4.20). The point of the matching done in the previous sections is that we can transition between \(u_{e}(t,r)\), which is accurate for small r and \(u_{wave}(t,r)\), which is accurate for large r with an expression of the form
$$\begin{aligned} \chi _{\leqslant 1}\left( \frac{r}{g(t) \lambda (t)}\right) u_{e}(t,r) + \left( 1-\chi _{\leqslant 1}\left( \frac{r}{g(t)\lambda (t)}\right) \right) u_{wave}(t,r) \end{aligned}$$
for an appropriate choice of \(\chi _{\leqslant 1}\) (which will be defined later), and not incur large error terms when derivatives act on \(\chi _{\leqslant 1}\), upon substitution of the above expression into the left-hand side of (4.5). As mentioned previously, the basic idea of this procedure is inspired by matched asymptotic expansions. (The books [3, 23] have more information about matched asymptotic expansions for ODEs). In addition, the idea is motivated by the fact that the correction denoted by \(v_{1}\), defined in (4.12), pg. 11 of [25], has a leading order cancellation with \(v_{2}\), defined in (4.63), pg. 18 of [25] near the origin, and simultaneously a leading order cancellation with \(Q_{1}(\frac{r}{\lambda (t)})-\pi \) for large r, reminiscent of procedures used to match asymptotic expansions in various regions.
We now define a cutoff \(\chi _{\geqslant 1}\in C^{\infty }([0,\infty ))\), with the following properties.
Lemma 27
[Properties of \(\chi _{\geqslant 1}\)] There exists a function \(\chi _{\geqslant 1}\in C^{\infty }([0,\infty ))\) satisfying
$$\begin{aligned} \chi _{\geqslant 1}(x)= & {} {\left\{ \begin{array}{ll} 0, \quad x \leqslant 1\\ 1, \quad x \geqslant 2\end{array}\right. }, \nonumber \\ \text { } \int _{0}^{\infty } \frac{\chi _{\geqslant 1}(x)}{x^{3}}dx= & {} \int _{0}^{\infty } x^{3} \left( 1-\chi _{\geqslant 1}(x)\right) dx = \int _{0}^{\infty } x^{3} \log (x) \left( 1-\chi _{\geqslant 1}(x)\right) dx=0\nonumber \\ \end{aligned}$$
(4.140)
In particular, this implies that, for all \(k \geqslant 0\), there exists \(C_{k}>0\) such that
$$\begin{aligned} |\widehat{\frac{\chi _{\geqslant 1}(\cdot )}{\left( \cdot \right) ^{5}}}(\eta )| \leqslant {\left\{ \begin{array}{ll} C \eta ^{3} \langle \log (\eta )\rangle , \quad \eta < 1\\ \frac{C_{k}}{\eta ^{k}}, \quad \eta \geqslant 1\end{array}\right. } \end{aligned}$$
Proof
A direct computation shows that
$$\begin{aligned} \chi _{\geqslant 1}(x) = \left( \int _{-\infty }^{\infty } \alpha (y) dy\right) ^{-1} \int _{-\infty }^{x}\alpha (s) ds \end{aligned}$$
where
$$\begin{aligned} \alpha (s) = \frac{1}{s^{4}} \frac{d}{ds}\left( s \frac{d}{ds}\left( s^{7}\phi '(s)\right) \right) , \quad \phi (x) = {\left\{ \begin{array}{ll} e^{-\frac{1}{1-(2x-3)^{2}}}, \quad |x-\frac{3}{2}| < \frac{1}{2}\\ 0, \quad |x-\frac{3}{2}| \geqslant \frac{1}{2}\end{array}\right. } \end{aligned}$$
satisfies (4.140), given that
$$\begin{aligned} \alpha \in C^{\infty }_{c}([1,2]), \quad \int _{0}^{\infty } \alpha (s)\frac{ds}{s^{2}}= \int _{0}^{\infty } \alpha (s) s^{4} ds = \int _{0}^{\infty } \alpha (s) s^{4} \log (s) ds =0, \quad \int _{-\infty }^{\infty }\alpha (s)ds\ne 0 \end{aligned}$$
To verify the stated estimates on \(\widehat{\frac{\chi _{\geqslant 1}(\cdot )}{\left( \cdot \right) ^{5}}}(\eta )\), we start with the case of \(\eta >1\). We have
$$\begin{aligned} \widehat{\frac{\chi _{\geqslant 1}(\cdot )}{\left( \cdot \right) ^{5}}}(\eta ) = \int _{0}^{\infty } \frac{\chi _{\geqslant 1}(x)}{x^{5}} J_{1}(\eta x) x dx = \int _{0}^{\infty } \frac{\chi _{\geqslant 1}(x)}{x^{5}} \frac{H_{1}\left( J_{1}(\eta x)\right) }{\eta ^{2}} x dx \end{aligned}$$
where
$$\begin{aligned} H_{1}(f)(x) = -f''(x) - \frac{1}{x}f'(x) + \frac{f(x)}{x^{2}} \end{aligned}$$
Then,
$$\begin{aligned} |\widehat{\frac{\chi _{\geqslant 1}(\cdot )}{\left( \cdot \right) ^{5}}}(\eta )| \leqslant \frac{C_{k}}{\eta ^{k}}, \quad \eta > 1 \end{aligned}$$
follows from repeated integration by parts, noting that there are no non-zero boundary terms obtained in the process. For \(\eta <1\), we use the integral condition on \(\chi _{\geqslant 1}\). In particular, we have
$$\begin{aligned} \widehat{\frac{\chi _{\geqslant 1}(\cdot )}{\left( \cdot \right) ^{5}}}(\eta ) = \frac{\eta }{2} \int _{0}^{\frac{1}{\eta }} \frac{\chi _{\geqslant 1}(x)}{x^{3}} dx + O\left( \eta ^{3} \int _{0}^{\frac{1}{\eta }} \frac{|\chi _{\geqslant 1}(x)|}{x} dx\right) + O\left( \int _{\frac{1}{\eta }}^{\infty } \frac{|\chi _{\geqslant 1}(x)|}{x^{4}} dx\right) \end{aligned}$$
which implies the estimate in the lemma statement. \(\square \)
We recall that \(u_{e}\) and \(u_{wave}\) are defined in (4.139), and define \(\chi _{\leqslant 1}(x) = 1-\chi _{\geqslant 1}(x)\), and
$$\begin{aligned} u_{c}(t,r) = \chi _{\leqslant 1}(\frac{r}{\lambda (t) g(t)}) u_{e}(t,r) + \left( 1-\chi _{\leqslant 1}(\frac{r}{\lambda (t) g(t)})\right) u_{wave}(t,r) \end{aligned}$$
(4.141)
Let \(h(t) = \lambda (t) g(t)\). The error term of \(u_{c}\) in solving the linear PDE (4.5) is
$$\begin{aligned} \begin{aligned}&-\left( \left( -\partial _{tt}+\partial _{rr}+\frac{1}{r}\partial _{r}-\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}}\right) u_{c} -\partial _{t}^{2}Q_{1}(\frac{r}{\lambda (t)})\right) \\&= \chi _{\leqslant 1}(\frac{r}{h(t)}) \partial _{tt}u_{ell,2}(t,r) \\&+ \left( 1-\chi _{\leqslant 1}(\frac{r}{h(t)})\right) \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) \left( v_{ex}(t,r)+u_{w,2}(t,r)+v_{2,2}(t,r)\right) \\&-\left( u_{e}-u_{wave}\right) \left( -\chi _{\leqslant 1}''(\frac{r}{h(t)}) \frac{r^{2} h'(t)^{2}}{h(t)^{4}} - \chi _{\leqslant 1}'(\frac{r}{h(t)})\left( -\frac{r h''(t)}{(h(t))^{2}} + \frac{2 r h'(t)^{2}}{h(t)^{3}}\right) \right. \\&\quad \qquad \qquad \qquad \qquad \left. + \frac{1}{r} \frac{\chi _{\leqslant 1}'(\frac{r}{h(t)})}{h(t)} + \frac{\chi _{\leqslant 1}''(\frac{r}{h(t)})}{h(t)^{2}}\right) \\&- 2 \chi _{\leqslant 1}'(\frac{r}{h(t)}) \frac{r h'(t)}{h(t)^{2}} \partial _{t}\left( u_{e}-u_{wave}\right) - \frac{2}{h(t)} \chi _{\leqslant 1}'(\frac{r}{h(t)}) \partial _{r}(u_{e}-u_{wave}) \end{aligned}\nonumber \\ \end{aligned}$$
(4.142)
Some of the error terms in (4.142) are already perturbative, while some will need additional corrections. We first record estimates on those terms which are already perturbative. We recall the definitions of \(v_{ex,ell}\) in (4.36), and \(u_{w,2}\) in (4.56), and start with the following lemma.
Lemma 28
If
$$\begin{aligned} e_{ex,ell}(t,r) = \left( 1-\chi _{\leqslant 1}(\frac{r}{\lambda (t)g(t)})\right) \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) v_{ex,ell}(t,r) \end{aligned}$$
and
$$\begin{aligned} e_{w,2}(t,r) = \left( 1-\chi _{\leqslant 1}(\frac{r}{\lambda (t)g(t)})\right) \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) u_{w,2}(t,r) \end{aligned}$$
then, for \(k=0,1\),
$$\begin{aligned} ||L_{\frac{1}{\lambda (t)}}^{k}(e_{ex,ell})(t,r)||_{L^{2}(r dr)}\leqslant & {} \frac{C \lambda (t)^{1-k} \log (t)}{t^{2} g(t)^{4+k}} \end{aligned}$$
(4.143)
$$\begin{aligned} ||L_{\frac{1}{\lambda (t)}}^{k}(e_{w,2})(t,r)||_{L^{2}(r dr)}\leqslant & {} \frac{C \lambda (t)^{2-k}}{t^{4}} \frac{\left( 1+\lambda (t)\right) \log ^{5}(t) \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) }{g(t)^{2}} \nonumber \\{} & {} + \frac{C \lambda (t)^{1-k} \log ^{2}(t)}{g(t)^{4+k} t^{2}} \end{aligned}$$
(4.144)
Proof
The estimate in (4.143) follows from straightforward estimation of the expression (4.36) for \(v_{ex,ell}\). To establish (4.144), we write \(u_{w,2} = u_{w,2}-u_{w,2,ell}+u_{w,2,ell}-u_{w,2,ell,0}+u_{w,2,ell,0}\), and use Lemma 10 as well as (4.61). \(\square \)
We recall the definitions of \(v_{ell,sub}\) from (4.78) and \(v_{ell,sub,cont}\) from (4.79), and define
$$\begin{aligned} v_{ell,sub,1}(t,R) = v_{ell,sub}(t,R)-v_{ell,sub,cont}(t,R\lambda (t)) \end{aligned}$$
(4.145)
We will now describe the difference \(u_{e}-u_{wave}\). We have
$$\begin{aligned}{} & {} u_{e}(t,r)=u_{ell}(t,r)+u_{ell,2}(t,r) \\{} & {} u_{wave}(t,r)=w_{1}(t,r)+v_{ex}(t,r)+v_{2}(t,r)+u_{w,2}(t,r)+v_{2,2}(t,r) \end{aligned}$$
We recall the first order matching (4.82), which says that
$$\begin{aligned} u_{ell}(t,r) - v_{ell,sub}(t,\frac{r}{\lambda (t)}) = v_{2}(t,r)-v_{2,sub}(t,r)+w_{1}(t,r)-w_{1,sub}(t,r) \end{aligned}$$
From the second order matching, we have
$$\begin{aligned} v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})=w_{1,cubic,main}(t,r)+v_{2,cubic,main}(t,r) \end{aligned}$$
and
$$\begin{aligned} v_{ell,sub,cont}(t,r)=v_{ex,cont}(t,r)+u_{w,2,ell,0,cont}(t,r) \end{aligned}$$
From the third order matching, we have the estimates from Proposition 2. These give
$$\begin{aligned}\begin{aligned}&u_{e}(t,r)-u_{wave}(t,r)\\&=u_{ell,2}(t,r)-v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})\\&-\left( u_{w,2}(t,r)-u_{w,2,ell,0,cont}(t,r) +v_{2,2}(t,r)+v_{ex}-v_{ex,cont}-q_{4,2}\right) \\&-q_{4,2}(t,r)+v_{ell,sub,1}(t,\frac{r}{\lambda (t)})\\&-\left( w_{1,sub}(t,r)-w_{1,cubic,main}(t,r)-\frac{r^{5} \lambda ^{(6)}(t)}{576}\left( 5+\log (8)\right) \right. \\&\left. -\frac{r^{5}}{192} \left( \int _{t}^{2t} \frac{\left( \lambda ^{(6)}(s)-\lambda ^{(6)}(t)\right) ds}{s-t} + \lambda ^{(6)}(t) \log (\frac{t}{r})+\int _{2t}^{\infty } \frac{\lambda ^{(6)}(s) ds}{s-t}\right) \right) \\&- \frac{r^{5}}{192} \left( \int _{t}^{2t} \frac{\left( \lambda ^{(6)}(s)-\lambda ^{(6)}(t)\right) ds}{s-t} + \lambda ^{(6)}(t) \log (\frac{t}{r})+\int _{2t}^{\infty } \frac{\lambda ^{(6)}(s) ds}{s-t}\right) \\&-\frac{r^{5} \lambda ^{(6)}(t)}{576}\left( 5+\log (8)\right) -\left( v_{2,sub}(t,r)-v_{2,cubic,main}(t,r)+\frac{r^{5}}{768} F^{(4)}(t)\right) +\frac{r^{5}}{768} F^{(4)}(t)\end{aligned} \end{aligned}$$
where any function appearing without arguments is evaluated at the point (t, r). We also recall that F is defined in (4.38) (see also (4.41)). Using
$$\begin{aligned}\begin{aligned} F^{(4)}(t)&= 4\left( \left( \log (2)-\frac{1}{2}\right) \lambda ^{(6)}(t) + \int _{t}^{2t} \frac{\left( \lambda ^{(6)}(s)-\lambda ^{(6)}(t)\right) }{s-t} ds + \lambda ^{(6)}(t) \log (t)\right. \\&\quad \left. - \partial _{t}^{4}\left( \lambda ''(t) \log (\lambda (t))\right) -\partial _{t}^{4}\left( \frac{\lambda '(t)^{2}}{2\lambda (t)}\right) + \int _{2t}^{\infty } \frac{\lambda ^{(6)}(s)}{s-t} ds\right) \end{aligned} \end{aligned}$$
we get
$$\begin{aligned}\begin{aligned}&u_{e}(t,r)-u_{wave}(t,r)\\&=u_{ell,2}(t,r)-v_{ell,2,0,main}(t,\frac{r}{\lambda (t)}) \\&-\left( u_{w,2}(t,r)-u_{w,2,ell,0,cont}(t,r)+v_{2,2}(t,r)+v_{ex}-v_{ex,cont}-q_{4,2}\right) \\&-q_{4,2}(t,r)+v_{ell,sub,1}(t,\frac{r}{\lambda (t)})\\&-\left( w_{1,sub}(t,r)-w_{1,cubic,main}(t,r)-\frac{r^{5} \lambda ^{(6)}(t)}{576}\left( 5+\log (8)\right) \right. \\&\left. -\frac{r^{5}}{192} \left( \int _{t}^{2t} \frac{\left( \lambda ^{(6)}(s)-\lambda ^{(6)}(t)\right) ds}{s-t} + \lambda ^{(6)}(t) \log (\frac{t}{r})+\int _{2t}^{\infty } \frac{\lambda ^{(6)}(s) ds}{s-t}\right) \right) \\&-\left( v_{2,sub}(t,r)-v_{2,cubic,main}(t,r)+\frac{r^{5}}{768} F^{(4)}(t)\right) \\&-\frac{r^{5}}{1152}\left( 6\left( \partial _{t}^{4}\left( \lambda ''(t)\log (\lambda (t))\right) +\partial _{t}^{4}\left( \frac{\lambda '(t)^{2}}{2\lambda (t)}\right) \right) +(13-6 \log (r))\lambda ^{(6)}(t)\right) \end{aligned} \end{aligned}$$
All of the terms (and sufficiently many of their derivatives) appearing in this expression have been estimated already, except for the last term (which will be eliminated with another correction, utilizing the properties of \(\chi _{\geqslant 1}\)) and \(v_{ell,sub,1}(t,\frac{r}{\lambda (t)})\). We now record estimates on \(v_{ell,sub,1}(t,\frac{r}{\lambda (t)})\), and another estimate on \(v_{ex,sub}\) which will be useful later on.
Lemma 29
For \(j=0,1\) and \(k=0,1,2\), the following estimate is true.
$$\begin{aligned} |\partial _{t}^{j}\partial _{r}^{k}\left( v_{ell,sub,1}(t,\frac{r}{\lambda (t)})\right) | \leqslant C \frac{\lambda (t)^{2-k} \log ^{2}(t)}{t^{2+j} g(t)^{3+k}}, \quad g(t)\lambda (t) \leqslant r \leqslant 2g(t)\lambda (t) \end{aligned}$$
In addition, we have
$$\begin{aligned}{} & {} |v_{ex}(t,r)| \leqslant \frac{C \log ^{2}(t+r) \lambda (t)^{3}}{\sqrt{r}t^{5/2}}, \quad r \geqslant \frac{t}{2} \end{aligned}$$
(4.146)
$$\begin{aligned}{} & {} |\partial _{t}^{k}v_{ex,sub}(t,r)| \leqslant \frac{C r \lambda (t)^{3} \log ^{3}(t+r)}{t^{4+k} }, \quad g(t) \lambda (t) \leqslant r, \quad k=0,1,2 \\{} & {} |\partial _{r}\partial _{t}^{k}v_{ex,sub}(t,r)| \leqslant \frac{C \lambda (t)^{3} \log ^{3}(t+r)}{t^{4+k} }, \quad g(t) \lambda (t) \leqslant r, \quad k=0,1\nonumber \\{} & {} |\partial _{r}^{2} v_{ex,sub}(t,r)| \leqslant \frac{C \lambda (t)^{3} \log ^{3}(t)}{r t^{4} }, \quad g(t) \lambda (t) \leqslant r \leqslant t\nonumber \end{aligned}$$
(4.147)
Finally, we also have the following (non-sharp, but sufficient) estimates
$$\begin{aligned} |\partial _{r}v_{ex}(t,r)| \leqslant \frac{C \lambda (t)^{\frac{5}{2}}}{r^{\frac{3}{2}}t^{2}}, \quad |\partial _{t}v_{ex}(t,r)| \leqslant \frac{C \lambda (t)^{3/2}}{t r^{3/2}}, \quad r \geqslant g(t)\lambda (t) \end{aligned}$$
(4.148)
Proof
We directly estimate \(v_{ell,sub,1}\) using its definition, (4.145), and the explicit formulae for \(v_{ell,sub}\) and \(v_{ell,sub,cont}\) from (4.78) and (4.79), respectively. To obtain (4.147), we start with the definition (4.83), use Fubini’s theorem to switch the order of s and \(\xi \) integrals, and (for \(r \geqslant g(t) \lambda (t)\)) divide the s integral into three regions.
$$\begin{aligned} {\textbf {a}}: t \leqslant s \leqslant t+\lambda (t), \quad {\textbf {b}}: t+\lambda (t) \leqslant s \leqslant t+r, \quad {\textbf {c}}: t+r \leqslant s < \infty \end{aligned}$$
(4.149)
In each above region of s integration, we then divide the \(\xi \) region of integration into four subintervals, based on the scales \(\frac{1}{r}, \frac{1}{\lambda (t)}, \frac{1}{s-t}\). For example, in region a of the s integration, we have the four regions of the \(\xi \) variable
$$\begin{aligned} 0< \xi \leqslant \frac{1}{r}, \quad \frac{1}{r}< \xi \leqslant \frac{1}{\lambda (t)}, \quad \frac{1}{\lambda (t)}< \xi \leqslant \frac{1}{s-t}, \quad \frac{1}{s-t}< \xi <\infty \end{aligned}$$
In the regions where \(\xi \leqslant \frac{1}{s-t}\), we use
$$\begin{aligned} |\sin ((t-s)\xi )| \leqslant C (s-t) \xi \end{aligned}$$
On the other hand, when \(\xi >\frac{1}{s-t}\), we integrate by parts in the \(\xi \) variable. To be clear, we show how the s integral over the region c is treated.
$$\begin{aligned}\begin{aligned}&-\int _{t+r}^{\infty } ds \int _{0}^{\infty } d\xi \frac{\sin ((t-s)\xi )}{\xi ^{2}} J_{1}(r\xi ) \partial _{s}^{2} \widehat{RHS}(s,\xi )\\&\quad = - \int _{t+r}^{\infty } ds \int _{0}^{\frac{1}{s-t}} d\xi \frac{\sin ((t-s)\xi )}{\xi ^{2}} J_{1}(r\xi ) \partial _{s}^{2} \widehat{RHS}(s,\xi )\\&\qquad -\int _{t+r}^{\infty } ds \int _{\frac{1}{s-t}}^{\infty } d\xi \frac{\sin ((t-s)\xi )}{\xi ^{2}} J_{1}(r\xi ) \partial _{s}^{2} \widehat{RHS}(s,\xi )\end{aligned} \end{aligned}$$
The integral over the region \(\xi \geqslant \frac{1}{s-t}\) is then further treated as
$$\begin{aligned}\begin{aligned}&-\int _{t+r}^{\infty } ds \int _{\frac{1}{s-t}}^{\infty } d\xi \frac{\sin ((t-s)\xi )}{\xi ^{2}} J_{1}(r\xi ) \partial _{s}^{2} \widehat{RHS}(s,\xi )\\&\quad =-\int _{t+r}^{\infty } ds J_{1}(\frac{r}{s-t}) \partial _{1}^{2}\widehat{RHS}(s,\frac{1}{s-t}) (s-t)^{2} \frac{\cos (1)}{(t-s)}\\&\qquad -\int _{t+r}^{\infty } ds \int _{\frac{1}{s-t}}^{\frac{1}{r}} \frac{\cos ((t-s)\xi )}{(t-s)} \partial _{\xi } \left( \frac{J_{1}(r\xi ) \partial _{s}^{2}\widehat{RHS}(s,\xi )}{\xi ^{2}}\right) d\xi \\&\qquad -\int _{t+r}^{\infty } ds \int _{\frac{1}{r}}^{\frac{1}{\lambda (t)}} \frac{\cos ((t-s)\xi )}{(t-s)} \partial _{\xi } \left( \frac{J_{1}(r\xi ) \partial _{s}^{2}\widehat{RHS}(s,\xi )}{\xi ^{2}}\right) d\xi \\&\qquad -\int _{t+r}^{\infty } ds \int _{\frac{1}{\lambda (t)}}^{\infty } \frac{\cos ((t-s)\xi )}{(t-s)} \partial _{\xi } \left( \frac{J_{1}(r\xi ) \partial _{s}^{2}\widehat{RHS}(s,\xi )}{\xi ^{2}}\right) d\xi \end{aligned} \end{aligned}$$
Finally, we use the formula for \(\widehat{RHS}(s,\xi )\), namely (4.33), to get, for \(0 \leqslant k \leqslant 5\)
$$\begin{aligned} \xi s^{k}|\partial _{\xi }\partial _{s}^{k}\widehat{RHS}(s,\xi )|+s^{k} |\partial _{s}^{k} \widehat{RHS}(s,\xi )| \leqslant {\left\{ \begin{array}{ll}\frac{C\xi \lambda (s)^{3} (1+|\log (\xi \lambda (s))|)}{s^{2} }, \quad \xi \lambda (s) < \frac{1}{2}\\ \frac{C \lambda (s)^{2}}{s^{2} \xi \lambda (s)}, \quad \xi \lambda (s) > \frac{1}{2}\end{array}\right. }\nonumber \\ \end{aligned}$$
(4.150)
Then, we use
$$\begin{aligned} |J_{1}(x)| \leqslant C {\left\{ \begin{array}{ll} x, \quad x \leqslant 1\\ \frac{1}{\sqrt{x}}, \quad x > 1\end{array}\right. } \end{aligned}$$
as appropriate to estimate each integral above. We get (4.146) by using the same procedure as above, except that we don’t integrate by parts when \(\xi \geqslant \frac{1}{s-t}\) and \(s \leqslant t+r\), combined with (4.36). For (4.147), we use
$$\begin{aligned} \partial _{t}^{k}v_{ex,sub}(t,r) =-\int _{0}^{\infty } d\xi J_{1}(r\xi ) \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi ^{2}} \partial _{s}^{2+k}\widehat{RHS}(s,\xi ), \quad k=1,2 \end{aligned}$$
and the same procedure used for \(k=0\). Finally, to estimate \(\partial _{r}v_{ex,sub}(t,r)\), we start with
$$\begin{aligned} \partial _{r}v_{ex,sub}(t,r) = -\int _{0}^{\infty } d\xi J_{1}'(r\xi ) \xi \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi ^{2}} \partial _{s}^{2}\widehat{RHS}(s,\xi ) \end{aligned}$$
We make the same decomposition as described in (4.149). This time, we integrate by parts in \(\xi \) (integrating \(J_{1}'(r\xi )\)) when \(t \leqslant s \leqslant t+\lambda (t)\), and \(\frac{1}{r} \leqslant \xi \leqslant \frac{1}{s-t}\). We also integrate by parts in s (integrating \(\sin ((t-s)\xi )\)) when \(\xi \geqslant \frac{1}{s-t}\) and \(t+\lambda (t) \leqslant s \). Next, using
$$\begin{aligned} \partial _{tr}v_{ex,sub}(t,r) = -\int _{0}^{\infty } d\xi J_{1}'(r\xi ) \xi \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi ^{2}} \partial _{s}^{3}\widehat{RHS}(s,\xi ) \end{aligned}$$
and the same argument above, we obtain the estimates on \(\partial _{tr}v_{ex,sub}(t,r)\) in the lemma statement. Finally, we use the fact that \(v_{ex,sub}\) satisfies
$$\begin{aligned} -\partial _{tt}v_{ex,sub}+\partial _{rr}v_{ex,sub}+\frac{1}{r}\partial _{r}v_{ex,sub}-\frac{v_{ex,sub}}{r^{2}} = \partial _{t}^{2} v_{ex,ell}(t,r) \end{aligned}$$
to estimate \(\partial _{r}^{2}v_{ex,sub}(t,r)\). To obtain the estimate (4.148), we start with
$$\begin{aligned} \begin{aligned} \partial _{r}v_{ex,sub}(t,r)&=-\int _{0}^{\infty } d\xi m_{\leqslant 1}(r\xi ) J_{1}'(r\xi ) \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi } \partial _{s}^{2} \widehat{RHS}(s,\xi )\\&-\int _{0}^{\infty } d\xi (1-m_{\leqslant 1}(r\xi )) J_{1}'(r\xi ) \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi } \partial _{s}^{2} \widehat{RHS}(s,\xi )\end{aligned} \end{aligned}$$
(4.151)
where \(m_{\leqslant 1}\) is defined in (2.1). For the first term on the right-hand side of (4.151), we simply directly insert (4.150) into the integral, and estimate. For the second term, we integrate by parts, to get
$$\begin{aligned}\begin{aligned}&-\int _{0}^{\infty } d\xi (1-m_{\leqslant 1}(r\xi )) J_{1}'(r\xi ) \int _{t}^{\infty } ds \frac{\sin ((t-s)\xi )}{\xi } \partial _{s}^{2} \widehat{RHS}(s,\xi )\\&= \frac{1}{r} \int _{t}^{\infty } ds \int _{0}^{\infty } d\xi J_{1}(r\xi ) \partial _{\xi }\left( \left( 1-m_{\leqslant 1}(r\xi )\right) \frac{\sin ((t-s)\xi )}{\xi } \partial _{s}^{2} \widehat{RHS}(s,\xi )\right) d\xi \end{aligned} \end{aligned}$$
and then, we directly estimate using (4.150). Next, using (4.36) and \(v_{ex}(t,r)=v_{ex,ell}(t,r)+v_{ex,sub}(t,r)\), we get
$$\begin{aligned} |\partial _{r}v_{ex}(t,r)| \leqslant \frac{C \lambda (t)^{\frac{5}{2}}}{r^{\frac{3}{2}}t^{2}}, \quad r \geqslant g(t)\lambda (t) \end{aligned}$$
Finally, the dominated convergence theorem and the formula for \(v_{ex}\), namely (4.34), give
$$\begin{aligned} \partial _{t}v_{ex}(t,r) = -\int _{0}^{\infty } dw \int _{0}^{\infty } d\xi J_{1}(r\xi ) \sin (w\xi ) \partial _{1}\widehat{RHS}(w+t,\xi ) \end{aligned}$$
Then, we use the same procedure used in (4.151) to finish the proof of (4.148). \(\square \)
Finally, we let
$$\begin{aligned}\begin{aligned} (u_{e}-u_{wave})_{0}(t,r)&=u_{e}-u_{wave}+\frac{r^{5}}{1152}\left( 6\left( \partial _{t}^{4}\left( \lambda ''(t)\log (\lambda (t))\right) +\partial _{t}^{4}\left( \frac{\lambda '(t)^{2}}{2\lambda (t)}\right) \right) \right. \\&\quad \left. +(13-6 \log (r))\lambda ^{(6)}(t)\right) \end{aligned} \end{aligned}$$
and we note the following lemma.
Lemma 30
We have the following estimate, for \(0 \leqslant j,k \leqslant 1\) or \(k=2,j=0\), and \(g(t) \lambda (t) \leqslant r \leqslant 2 g(t)\lambda (t)\)
$$\begin{aligned}\begin{aligned}&|\partial _{t}^{j}\partial _{r}^{k} \left( \left( u_{e}-u_{wave}\right) _{0}+q_{4,2}\right) |\\&\leqslant \frac{C \lambda (t)^{2-k} \log ^{2}(t)}{t^{2+j} g(t)^{3+k}}+\frac{C g(t)^{2-k} \lambda (t)^{4-k} \log ^{4}(t) \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) }{t^{5+j}}\\&\quad +\frac{C \lambda (t)^{8-k} g(t)^{7-k} \log (t)}{t^{8+j} }\end{aligned} \end{aligned}$$
Proof
We directly combine the results of Proposition 2, Lemma 1, Lemma 6, and Lemma 29\(\square \)
We now can estimate the \((u_{e}-u_{wave})_{0}\) contribution to the error terms of our ansatz (4.141) which involve at least one derivative of \(\chi _{\leqslant 1}\). Let
$$\begin{aligned}\begin{aligned}&e_{match,0}(t,r)\\&=-\left( u_{e}-u_{wave}\right) _{0}\left( -\chi _{\leqslant 1}''(\frac{r}{h(t)}) \frac{r^{2} h'(t)^{2}}{h(t)^{4}} - \chi _{\leqslant 1}'(\frac{r}{h(t)})\left( -\frac{r h''(t)}{(h(t))^{2}} + \frac{2 r h'(t)^{2}}{h(t)^{3}}\right) \right. \\&\quad \left. + \frac{1}{r} \frac{\chi _{\leqslant 1}'(\frac{r}{h(t)})}{h(t)} + \frac{\chi _{\leqslant 1}''(\frac{r}{h(t)})}{h(t)^{2}}\right) \\&- 2 \chi _{\leqslant 1}'(\frac{r}{h(t)}) \frac{r h'(t)}{h(t)^{2}} \partial _{t}\left( u_{e}-u_{wave}\right) _{0} - \frac{2}{h(t)} \chi _{\leqslant 1}'(\frac{r}{h(t)}) \partial _{r}(u_{e}-u_{wave})_{0}\end{aligned} \end{aligned}$$
Then, we have the following estimates.
Lemma 31
[Estimates on the matching-induced error terms] For \(0 \leqslant k \leqslant 1\),
$$\begin{aligned}\begin{aligned}&||L_{\frac{1}{\lambda (t)}}^{k}(e_{match,0})(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \lambda (t)^{5} \log ^{2}(t)}{t^{2} h(t)^{4+k}}\\&\quad +\frac{C h(t)^{1-k} \lambda (t)^{2} \log ^{4}(t) \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) }{t^{5}}+\frac{C \lambda (t) h(t)^{6-k} \log (t)}{t^{8} } \end{aligned} \end{aligned}$$
Proof
We directly use Lemma 30 and the estimates on \(q_{4,2}\) from Proposition 2. \(\square \)
Next, we consider the error terms of \(u_{c}\) which involve derivatives of \(\chi _{\leqslant 1}\), and which result from replacing \(u_{e}-u_{wave}\) with \((u_{e}-u_{wave})_{0}\). In particular, let
$$\begin{aligned} \begin{aligned} e_{5}(t,r)&=-f_{5}(t,r)\left( -\chi _{\leqslant 1}''(\frac{r}{h(t)}) \frac{r^{2} h'(t)^{2}}{h(t)^{4}} - \chi _{\leqslant 1}'(\frac{r}{h(t)})\left( -\frac{r h''(t)}{(h(t))^{2}} + \frac{2 r h'(t)^{2}}{h(t)^{3}}\right) \right. \\&\quad \left. + \frac{1}{r} \frac{\chi _{\leqslant 1}'(\frac{r}{h(t)})}{h(t)} + \frac{\chi _{\leqslant 1}''(\frac{r}{h(t)})}{h(t)^{2}}\right) \\&\qquad - 2 \chi _{\leqslant 1}'(\frac{r}{h(t)}) \frac{r h'(t)}{h(t)^{2}} \partial _{t}f_{5}(t,r) - \frac{2}{h(t)} \chi _{\leqslant 1}'(\frac{r}{h(t)}) \partial _{r}f_{5}(t,r)\end{aligned} \end{aligned}$$
where
$$\begin{aligned} f_{5}(t,r) = -\frac{r^{5}}{1152}\left( 6\left( \partial _{t}^{4}\left( \lambda ''(t)\log (\lambda (t))\right) +\partial _{t}^{4}\left( \frac{\lambda '(t)^{2}}{2\lambda (t)}\right) \right) +(13-6 \log (r))\lambda ^{(6)}(t)\right) \end{aligned}$$
The following piece of \(e_{5}\) turns out to be perturbative. Let
$$\begin{aligned}\begin{aligned} e_{5,1}(t,r)&=-f_{5}(t,r)\left( -\chi _{\leqslant 1}''(\frac{r}{h(t)}) \frac{r^{2} h'(t)^{2}}{h(t)^{4}} - \chi _{\leqslant 1}'(\frac{r}{h(t)})\left( -\frac{r h''(t)}{(h(t))^{2}} + \frac{2 r h'(t)^{2}}{h(t)^{3}}\right) \right) \\&\quad - 2 \chi _{\leqslant 1}'(\frac{r}{h(t)}) \frac{r h'(t)}{h(t)^{2}} \partial _{t}f_{5}(t,r) \end{aligned} \end{aligned}$$
Lemma 32
We have the following estimates for \(k=0,1\).
$$\begin{aligned} ||L_{\frac{1}{\lambda (t)}}^{k}(e_{5,1})(t,r)||_{L^{2}(r dr)} \leqslant \frac{C h(t)^{6-k} \lambda (t) \log (t)}{t^{8}} \end{aligned}$$
Proof
This follows from a straightforward and direct computation \(\square \)
Next, we need to consider \(e_{5,0}\), which is given by
$$\begin{aligned} \begin{aligned} e_{5,0}(t,r)&:=e_{5}(t,r)-e_{5,1}(t,r)=-f_{5}(t,r)\left( \frac{1}{r} \frac{\chi _{\leqslant 1}'(\frac{r}{h(t)})}{h(t)} + \frac{\chi _{\leqslant 1}''(\frac{r}{h(t)})}{h(t)^{2}}\right) \\&\quad - \frac{2}{h(t)} \chi _{\leqslant 1}'(\frac{r}{h(t)}) \partial _{r}f_{5}(t,r)\end{aligned}\nonumber \\ \end{aligned}$$
(4.152)
The point is that, although \(e_{5,0}(t,r)\) does not decay fast enough (in \(L^{2}\), for example) to be perturbative, it is orthogonal to \(\phi _{0}(\frac{r}{\lambda (t)})\) to leading order. This is because \(e_{5,0}(t,r)\) is supported in the region \(\lambda (t) \ll h(t) \leqslant r \leqslant 2h(t)\). So, the leading order behavior of \((2 \lambda (t))^{-1}\langle \phi _{0}(\frac{\cdot }{\lambda (t)}),e_{5,0}(t,\cdot )\rangle _{L^{2}(r dr)}\) is
$$\begin{aligned} \begin{aligned} \int _{0}^{\infty } e_{5,0}(t,r) dr&= \int _{0}^{\infty } \chi _{\leqslant 1}(x) \left( \left( 24 p_{0}(t)+10 p_{1}(t)\right) h(t)^{3} x^{3}\right. \\&\quad \qquad \quad \left. + 24 p_{1}(t) h(t)^{3} x^{3} \left( \log (h(t))+\log (x)\right) \right) h(t) dx\\&=0 \end{aligned} \end{aligned}$$
(4.153)
where we integrated by parts, defined \(p_{j}(t)\) by
$$\begin{aligned} f_{5}(t,r) = (p_{0}(t)+p_{1}(t) \log (r))r^{5} \end{aligned}$$
and used Lemma 27. Therefore, we will add a term to \(u_{c}\) which will be an appropriate truncation of a solution to the ODE
$$\begin{aligned} \partial _{RR}w(t,R) + \frac{1}{R} \partial _{R}w(t,R)-\frac{\cos (2Q_{1}(R))}{R^{2}} w(t,R) = F(t,R) \end{aligned}$$
(4.154)
for an appropriate choice of F. (We leave F general here, since we will use a correction of this form to eliminate error terms of \(u_{c}\) other than just \(e_{5,0}\)). For the class of F which we will need to consider, we will use the following particular solution to (4.154) (recall the notation (2.3))
$$\begin{aligned} \begin{aligned} w(t,R)&= - \frac{R}{1+R^{2}} \int _{0}^{R} F(t,s) s e_{2}(s) ds + e_{2}(R) \int _{0}^{R} \frac{F(t,s) s^{2}}{1+s^{2}} ds\end{aligned} \end{aligned}$$
(4.155)
Returning to the correction associated to \(e_{5,0}\), we establish the following lemma
Lemma 33
Let \(w_{5,0}\) denote the function defined in (4.155) where
$$\begin{aligned} F(t,R) = \lambda (t)^{2}e_{5,0}(t,R\lambda (t)) \end{aligned}$$
Then, we have the following estimate, for \(0 \leqslant j+k \leqslant 2\), \(j=2, k=1\), \(j=0,k=3\), and \(j=1,k=2\).
$$\begin{aligned} |R^{k}t^{j}\partial _{t}^{j}\partial _{R}^{k} w_{5,0}(t,R)| \leqslant {\left\{ \begin{array}{ll} 0, \quad R \leqslant \frac{h(t)}{\lambda (t)}\\ C \frac{\lambda (t) h(t)^{5} \log (t)}{t^{6} } , \quad \frac{h(t)}{\lambda (t)} \leqslant R \leqslant \frac{2 h(t)}{\lambda (t)}\\ \frac{C}{R} \frac{h(t)^{6} \log (t)}{t^{6} } + \frac{C R h(t)^{2} \lambda (t)^{4} \log (t)}{t^{6} }, \quad R \geqslant \frac{2 h(t)}{\lambda (t)} \end{array}\right. } \end{aligned}$$
Proof
By a straightforward insertion of the definition of \(e_{5,0}\), (4.152), into the following integral, we have
$$\begin{aligned} |\int _{0}^{R} \lambda (t)^{2} e_{5,0}(t,s\lambda (t)) s e_{2}(s) ds| \leqslant {\left\{ \begin{array}{ll} 0, \quad R \leqslant \frac{h(t)}{\lambda (t)}\\ \frac{C h(t)^{6} \log (t)}{t^{6} }, \quad R \geqslant \frac{h(t)}{\lambda (t)}\end{array}\right. } \end{aligned}$$
On the other hand, we use (4.153) to get
$$\begin{aligned} \lambda (t)^{2} \int _{0}^{R} e_{5,0}(t,s\lambda (t)) \frac{s^{2} ds}{1+s^{2}}= & {} -\lambda (t)^{2} \int _{R}^{\infty } e_{5,0}(t,s\lambda (t)) ds\\+ & {} \lambda (t)^{2} \int _{0}^{R} e_{5,0}(t,s\lambda (t))\left( \frac{-1}{1+s^{2}}\right) ds \end{aligned}$$
(Note that the support properties of \(\chi _{\leqslant 1}\) imply that the left-hand side of the above equation vanishes when \(R \leqslant \frac{h(t)}{\lambda (t)}\)). Then, we get
$$\begin{aligned} |\lambda (t)^{2} \int _{0}^{R} e_{5,0}(t,s\lambda (t)) \frac{s^{2} ds}{1+s^{2}}| \leqslant {\left\{ \begin{array}{ll} 0, \quad R \leqslant \frac{h(t)}{\lambda (t)}\\ \frac{C h(t)^{4} \lambda (t)^{2} \log (t)}{t^{6} }, \quad \frac{h(t)}{\lambda (t)} \leqslant R \leqslant \frac{2 h(t)}{\lambda (t)}\\ \frac{C h(t)^{2} \lambda (t)^{4} \log (t)}{t^{6} }, \quad R \geqslant \frac{2h(t)}{\lambda (t)} \end{array}\right. } \end{aligned}$$
This gives the estimate of the lemma statement for \(j=k=0\). Next, we have
$$\begin{aligned}\begin{aligned} \partial _{R}w_{5,0}(t,R)&= - \partial _{R}\left( \frac{R}{1+R^{2}}\right) \int _{0}^{R} \lambda (t)^{2} e_{5,0}(t,s\lambda (t)) s e_{2}(s) ds\\&+ e_{2}'(R) \int _{0}^{R} \frac{\lambda (t)^{2} e_{5,0}(t,s\lambda (t)) s^{2}}{1+s^{2}} ds\end{aligned} \end{aligned}$$
So, \(\partial _{R}w_{5,0}(t,R)\) is the same expression as \(w_{5,0}(t,R)\), except with an extra derivative on the coefficients of each integral term. Our proof of the lemma for \(j=0,k=0\) therefore immediately implies the lemma statement is true for \(j=0,k=1\). We prove the \(j=0,k=2\) case of the lemma statement by noting that
$$\begin{aligned} \partial _{R}^{2} w_{5,0}(t,R) = \lambda (t)^{2} e_{5,0}(t,R\lambda (t)) -\frac{1}{R}\partial _{R}w_{5,0}(t,R)+\frac{\cos (2Q_{1}(R))}{R^{2}} w_{5,0}(t,R) \end{aligned}$$
Finally, the symbol-type estimates on \(\lambda (t)\), definition of \(e_{5,0}\), and the fact that
$$\begin{aligned} \int _{0}^{\infty } \lambda (t)^{2} e_{5,0}(t,s\lambda (t)) ds =0 \text { for all t implies } \partial _{t}^{j} \left( \int _{0}^{\infty } \lambda (t)^{2} e_{5,0}(t,s\lambda (t)) ds\right) =0 \end{aligned}$$
finishes the proof of the lemma. \(\square \)
We define \(f_{5,0}(t,r)\), the function to be added to our ansatz, by
$$\begin{aligned} f_{5,0}(t,r) = m_{\leqslant 1}(\frac{r}{t}) w_{5,0}(t,\frac{r}{\lambda (t)}) \end{aligned}$$
(4.156)
where we recall the definition of \(m_{\leqslant 1}\) in (2.1). Then, we define the error term of \(f_{5,0}\) by
$$\begin{aligned}\begin{aligned} e_{f_{5,0}}(t,r)&:= -\left( -\partial _{tt}f_{5,0} + \partial _{rr}f_{5,0}+\frac{1}{r}\partial _{r}f_{5,0}-\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}} f_{5,0}\right) +e_{5,0}(t,r)\\&=-m_{\leqslant 1}''(\frac{r}{t}) \left( \frac{-r^{2}}{t^{4}} + \frac{1}{t^{2}}\right) w_{5,0}(t,\frac{r}{\lambda (t)})\\&\quad -m_{\leqslant 1}'(\frac{r}{t}) \left( \left( \frac{-2r}{t^{3}} + \frac{1}{r t}\right) w_{5,0}(t,\frac{r}{\lambda (t)})+\frac{2r}{t^{2}} \partial _{t}\left( w_{5,0}(t,\frac{r}{\lambda (t)})\right) \right. \\&\quad \left. +\frac{2}{t}\partial _{r}\left( w_{5,0}(t,\frac{r}{\lambda (t)})\right) \right) \\&\quad +m_{\leqslant 1}(\frac{r}{t})\partial _{t}^{2}\left( w_{5,0}(t,\frac{r}{\lambda (t)})\right) +e_{5,0}(t,r) \left( 1-m_{\leqslant 1}(\frac{r}{t})\right) \end{aligned} \end{aligned}$$
Then, an insertion of our estimates from Lemma 33 into the above expression for \(e_{f_{5,0}}\) gives the following lemma.
Lemma 34
For \(k=0,1\),
$$\begin{aligned} ||L_{\frac{1}{\lambda (t)}}^{k}(e_{f_{5,0}})(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \lambda (t)}{t^{2}} \left( \frac{h(t)^{6-k} \log ^{2}(t)}{t^{6}} + \frac{h(t)^{2-k}\lambda (t)^{2} \log ^{2}(t)}{t^{4}}\right) \end{aligned}$$
Now, we will consider the error term in (4.142) involving \(\left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) v_{ex,sub}(t,r)\).
Lemma 35
Let \(w_{ex,sub}(t,R)\) be defined by the expression (4.155) for
$$\begin{aligned} F(t,s) = \left( \frac{\cos (2Q_{1}(s))-1}{s^{2}}\right) v_{ex,sub}(t,s\lambda (t)) \chi _{\geqslant 1}(\frac{s}{g(t)}) \end{aligned}$$
Then, we have
$$\begin{aligned} w_{ex,sub}(t,\frac{r}{\lambda (t)})=0, \quad r \leqslant \lambda (t) g(t) \end{aligned}$$
For \(0 \leqslant k+j \leqslant 2\), \(j=1,k=2\), and \(j=2,k=1\),
$$\begin{aligned} |\partial _{t}^{k}\partial _{r}^{j}\left( w_{ex,sub}(t,\frac{r}{\lambda (t)})\right) | \leqslant \frac{C \lambda (t)^{5} \log ^{3}(t)}{r^{1+j} t^{4+k} } + \frac{C r^{1-j} \lambda (t)^{3} \log ^{2}(t)}{t^{4+k} g(t)^{2}}, \quad g(t) \lambda (t) \leqslant r \leqslant t \end{aligned}$$
Proof
We define \(int_{ex,sub}\) by the integral
$$\begin{aligned} int_{ex,sub}(t,R) = \int _{0}^{R} \frac{F(t,s) s^{2}}{1+s^{2}} ds = -\int _{0}^{\infty } d\xi \int _{t}^{\infty } dx \frac{\sin ((t-x)\xi )}{\xi ^{2}} \partial _{x}^{2}\widehat{RHS}(x,\xi ) I_{1}(R,\xi ,t) \end{aligned}$$
where
$$\begin{aligned} I_{1}(R,\xi ,t) = \int _{0}^{R} \left( \frac{\cos (2Q_{1}(s))-1}{s^{2}}\right) J_{1}(s\lambda (t)\xi ) \frac{\chi _{\geqslant 1}(\frac{s}{g(t)}) s^{2}}{1+s^{2}} ds \end{aligned}$$
and we used Fubini’s theorem. Note that \(I_{1}(R,\xi ,t) = 0\) for \(R \leqslant g(t)\). For \(R \geqslant g(t)\), we decompose \(I_{1}\) as
$$\begin{aligned} I_{1}(R,\xi ,t) = I_{100}(\xi ,t)+I_{101}(\xi ,t) + I_{11}(R,\xi ,t) \end{aligned}$$
where
$$\begin{aligned} I_{100}(\xi ,t)= & {} \frac{-8}{g(t)^{3}} \widehat{\frac{\chi _{\geqslant 1}(\cdot )}{\left( \cdot \right) ^{5}}}(\xi g(t) \lambda (t)) \\ I_{101}(\xi ,t)= & {} \int _{0}^{\infty } \left( \frac{(\cos (2Q_{1}(s))-1)}{s^{2}}\frac{s^{2}}{1+s^{2}}+\frac{8}{s^{4}}\right) J_{1}(s\lambda (t)\xi ) \chi _{\geqslant 1}(\frac{s}{g(t)}) ds \end{aligned}$$
and
$$\begin{aligned} I_{11}(R,\xi ,t)= - \int _{R}^{\infty } \left( \frac{\cos (2Q_{1}(s))-1}{s^{2}}\right) J_{1}(s\lambda (t)\xi ) \chi _{\geqslant 1}(\frac{s}{g(t)}) \frac{s^{2} ds}{1+s^{2}} \end{aligned}$$
We claim that, for \(0 \leqslant j \leqslant 2\),
$$\begin{aligned} |\partial _{t}^{j}I_{100}(\xi ,t)|\leqslant & {} \frac{C}{g(t)^{3}t^{j}} {\left\{ \begin{array}{ll} \xi ^{3} g(t)^{3}\lambda (t)^{3} \langle \log (\xi g(t) \lambda (t))\rangle , \quad \xi \leqslant \frac{1}{g(t)\lambda (t)}\\ \frac{C_{k}}{(\xi g(t)\lambda (t))^{k}}, \quad \xi \geqslant \frac{1}{g(t)\lambda (t)}\end{array}\right. } \\ |\partial _{t}^{j}I_{101}(\xi ,t)|\leqslant & {} \frac{C}{t^{j}} {\left\{ \begin{array}{ll} \frac{ \xi \lambda (t)}{g(t)^{4}}, \quad \xi \lambda (t) \leqslant \frac{1}{g(t)}\\ \frac{C_{k}}{g(t)^{5} (\xi \lambda (t) g(t))^{2k}}, \quad k \geqslant 5, \quad \xi \lambda (t) \geqslant \frac{1}{g(t)}\end{array}\right. } \\ |\partial _{t}^{j}\left( I_{11}(\frac{r}{\lambda (t)},\xi ,t)\right) |\leqslant & {} \frac{C}{t^{j}}{\left\{ \begin{array}{ll} \frac{ \lambda (t)^{3}}{ \xi ^{3/2} r^{9/2}}, \quad \xi \geqslant \frac{1}{\lambda (t) g(t)}\\ \frac{ \xi \lambda (t)^{3}}{r^{2} }, \quad \xi \leqslant \frac{1}{r}\\ \frac{ \lambda (t)^{3}}{r^{7/2} \sqrt{\xi }}, \quad \frac{1}{r} \leqslant \xi \leqslant \frac{1}{g(t)\lambda (t)}\end{array}\right. } \end{aligned}$$
For \(j=0\), \(I_{100}\) is estimated by using Lemma 27, while, for all \(0 \leqslant j \leqslant 2\), \(I_{101}\) and \(I_{11}\) are estimated by integrating by parts when \(\xi \lambda (t) \geqslant \frac{1}{g(t)}\), and directly estimating otherwise (as in Lemma 27). For \(j >0\), the most delicate estimate is on \(\partial _{t}^{j}I_{100}(\xi ,t)\) in the region \(\xi \leqslant \frac{1}{\lambda (t) g(t)}\). We write
$$\begin{aligned} I_{100}(\xi ,t) = -8 \int _{0}^{\infty } \frac{J_{1}(g(t)\lambda (t) y \xi )}{g(t)^{3} y^{4}} \chi _{\geqslant 1}(y) dy \end{aligned}$$
which gives
$$\begin{aligned} \partial _{t}I_{100}(\xi ,t) = \frac{-3 g'(t)}{g(t)} I_{100}(\xi ,t) - 8 \int _{0}^{\infty } \frac{J_{1}'(g(t)\lambda (t) y \xi )}{g(t)^{3} y^{4}} (g(t)\lambda (t))' y \xi \chi _{\geqslant 1}(y) dy \end{aligned}$$
Then,
$$\begin{aligned}\begin{aligned}&- 8 \int _{0}^{\infty } \frac{J_{1}'(g(t)\lambda (t) y \xi )}{g(t)^{3} y^{4}} (g(t)\lambda (t))' y \xi \chi _{\geqslant 1}(y) dy\\&\quad = -8 \int _{0}^{\frac{1}{g(t)\lambda (t) \xi }} \frac{1}{2} \frac{(g(t)\lambda (t))'}{g(t)^{3} y^{4}} y \xi \chi _{\geqslant 1}(y) dy\\&\qquad -8\int _{0}^{\frac{1}{g(t)\lambda (t)\xi }} \frac{\left( J_{1}'(g(t)\lambda (t)y\xi )-\frac{1}{2}\right) \left( g(t)\lambda (t)\right) '}{g(t)^{3}y^{4}} y \xi \chi _{\geqslant 1}(y) dy\\&\quad -8 \int _{\frac{1}{g(t)\lambda (t)\xi }}^{\infty } \frac{J_{1}'(g(t)\lambda (t) y \xi )}{g(t)^{3} y^{4}} (g(t)\lambda (t))' y \xi \chi _{\geqslant 1}(y) dy\end{aligned} \end{aligned}$$
and to estimate the first term on the right hand side of the above expression, we use Lemma 27. We use a similar procedure for \(\partial _{t}^{2} I_{100}(\xi ,t)\).
$$\begin{aligned}\begin{aligned}&\partial _{t}^{2}\left( -int_{ex,sub}(t,\frac{r}{\lambda (t)})\right) = \partial _{t}^{2}\left( \int _{0}^{\infty } d\xi \int _{t}^{\infty } dx \frac{\sin ((t-x)\xi )}{\xi ^{2}} \partial _{x}^{2}\widehat{RHS}(x,\xi ) I_{1}(\frac{r}{\lambda (t)},\xi ,t)\right) \\&=\partial _{t}^{2}\left( -\int _{0}^{\infty } d\xi \int _{0}^{\infty } dw \frac{\sin (w\xi )}{\xi ^{2}} \partial _{1}^{2}\widehat{RHS}(t+w,\xi ) \left( I_{100}(\xi ,t)+I_{101}(\xi ,t)+I_{11}(\frac{r}{\lambda (t)},\xi ,t)\right) \right) \end{aligned} \end{aligned}$$
After differentiating under the integral sign in the last integral, we let \(x=w+t\), and split the x integration into two regions: \(t \leqslant x \leqslant t+\frac{1}{\xi }\) and \(x > t+\frac{1}{\xi }\). In the latter region, we integrate by parts in the x variable, integrating \(\sin ((t-x)\xi )\). Then, we make a similar decomposition of the \(\xi \) integral as was made while proving Lemma 29. This gives, for \(k=0,1,2\),
$$\begin{aligned}\begin{aligned} |\partial _{t}^{k}\left( int_{ex,sub}(t,\frac{r}{\lambda (t)})\right) |&\leqslant \frac{C \lambda (t)^{4} \log ^{2}(t)}{t^{4+k} g(t)^{2}}, \quad t \geqslant r \geqslant g(t) \lambda (t)\end{aligned} \end{aligned}$$
We recall the definition of \(e_{2}\) in (2.3). The next integral to consider is
$$\begin{aligned}\begin{aligned} int_{ex,sub,2}(t,R)&= \int _{0}^{R} \left( \frac{\cos (2Q_{1}(s))-1}{s^{2}}\right) v_{ex,sub}(t,s\lambda (t)) \chi _{\geqslant 1}(\frac{s}{g(t)}) s e_{2}(s) ds\\&=-\int _{0}^{\infty } d\xi \int _{t}^{\infty } dx \frac{\sin ((t-x)\xi )}{\xi ^{2}} \partial _{x}^{2} \widehat{RHS}(x,\xi ) I_{2}(R,\xi ,t)\end{aligned} \end{aligned}$$
where
$$\begin{aligned} I_{2}(R,\xi ,t) = \int _{0}^{R} \left( \frac{\cos (2Q_{1}(s))-1}{s^{2}}\right) J_{1}(s\lambda (t) \xi ) \chi _{\geqslant 1}(\frac{s}{g(t)}) s e_{2}(s) ds \end{aligned}$$
As with \(I_{1}\) above, \(I_{2}(R,\xi ,t) =0\) for \(R \leqslant g(t)\). A direct estimation gives, for \(0 \leqslant k \leqslant 2\),
$$\begin{aligned} |\partial _{t}^{k}\left( I_{2}(\frac{r}{\lambda (t)},\xi ,t)\right) | \leqslant \frac{C}{t^{k}} {\left\{ \begin{array}{ll} \log (\frac{r}{\lambda (t) g(t)}) \lambda (t) \xi , \quad \xi \leqslant \frac{1}{r}\\ \xi \lambda (t) \langle \log (\xi \lambda (t) g(t))\rangle , \quad \frac{1}{r} \leqslant \xi \leqslant \frac{1}{\lambda (t) g(t)}\\ \frac{C}{g(t)^{3/2} \sqrt{\lambda (t) \xi }}, \quad \xi \geqslant \frac{1}{\lambda (t) g(t)}\end{array}\right. } \end{aligned}$$
Then, as above, we get, for \(k=0,1,2\),
$$\begin{aligned} |\partial _{t}^{k}\left( int_{ex,sub,2}(t,\frac{r}{\lambda (t)})\right) | \leqslant \frac{C \lambda (t)^{4} \log ^{3}(t)}{t^{4+k} }, \quad g(t) \leqslant \frac{r}{\lambda (t)} \leqslant \frac{t}{\lambda (t)} \end{aligned}$$
Recalling (4.155), we have
$$\begin{aligned} w_{ex,sub}(t,\frac{r}{\lambda (t)}) = -\frac{\phi _{0}(\frac{r}{\lambda (t)})}{2} int_{ex,sub,2}(t,\frac{r}{\lambda (t)}) + e_{2}(\frac{r}{\lambda (t)}) int_{ex,sub}(t,\frac{r}{\lambda (t)}) \end{aligned}$$
which gives, for \(0 \leqslant k \leqslant 2\),
$$\begin{aligned} |\partial _{t}^{k}\left( w_{ex,sub}(t,\frac{r}{\lambda (t)})\right) | \leqslant \frac{C \lambda (t)^{5} \log ^{3}(t)}{r t^{4+k} } + \frac{C r \lambda (t)^{3} \log ^{2}(t)}{t^{4+k} g(t)^{2}}, \quad g(t) \lambda (t) \leqslant r \leqslant t \end{aligned}$$
Next, we note that
$$\begin{aligned} \partial _{r} \left( w_{ex,sub}(t,\frac{r}{\lambda (t)})\right) = \frac{-\phi _{0}'(\frac{r}{\lambda (t)})}{2\lambda (t)} int_{ex,sub,2}(t,\frac{r}{\lambda (t)}) + e_{2}'(\frac{r}{\lambda (t)}) \frac{1}{\lambda (t)} int_{ex,sub}(t,\frac{r}{\lambda (t)}) \end{aligned}$$
Finally, we estimate \(\partial _{r}^{2}\left( w_{ex,sub}(t,\frac{r}{\lambda (t)})\right) \) using the equation solved by \(w_{ex,sub}\), Lemma 29, and our previous estimates from the proof of this lemma. \(\square \)
The truncation of \(w_{ex,sub}\) which we will add to our ansatz is
$$\begin{aligned} f_{ex,sub}(t,r) = m_{\leqslant 1}(\frac{r}{t}) w_{ex,sub}(t,\frac{r}{\lambda (t)}) \end{aligned}$$
(4.157)
where we recall that \(m_{\leqslant 1}\) was defined in (2.1). We define the error term of \(f_{ex,sub}(t,r)\) as
$$\begin{aligned}\begin{aligned} e_{ex,sub}(t,r)&= -\left( -\partial _{tt}f_{ex,sub} + \partial _{rr}f_{ex,sub}+\frac{1}{r}\partial _{r}f_{ex,sub}-\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}} f_{ex,sub}\right) \\&\quad +\left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) v_{ex,sub}(t,r) \chi _{\geqslant 1}(\frac{r}{\lambda (t)g(t)})\end{aligned} \end{aligned}$$
We get
$$\begin{aligned}\begin{aligned} e_{ex,sub}(t,r)&= -m_{\leqslant 1}''(\frac{r}{t}) \left( \frac{-r^{2}}{t^{4}} + \frac{1}{t^{2}}\right) w_{ex,sub}(t,\frac{r}{\lambda (t)})\\&\quad -m_{\leqslant 1}'(\frac{r}{t}) \left( \left( \frac{-2r}{t^{3}} + \frac{1}{r t}\right) w_{ex,sub}(t,\frac{r}{\lambda (t)})+\frac{2r}{t^{2}} \partial _{t}\left( w_{ex,sub}(t,\frac{r}{\lambda (t)})\right) \right. \\&\quad \left. +\frac{2}{t}\partial _{r}\left( w_{ex,sub}(t,\frac{r}{\lambda (t)})\right) \right) \\&\quad +m_{\leqslant 1}(\frac{r}{t})\partial _{t}^{2}\left( w_{ex,sub}(t,\frac{r}{\lambda (t)})\right) \\&\quad +\left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) v_{ex,sub}(t,r) \chi _{\geqslant 1}(\frac{r}{\lambda (t)g(t)}) \left( 1-m_{\leqslant 1}(\frac{r}{t})\right) \end{aligned} \end{aligned}$$
Lemma 35 directly gives
Lemma 36
For \(k=0,1\),
$$\begin{aligned} ||L_{\frac{1}{\lambda (t)}}^{k}(e_{ex,sub})(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \lambda (t)^{3-k} \log ^{2}(t)}{t^{4} g(t)^{2+k}} \end{aligned}$$
Lemma 37
Let \(g_{ell,2}(t,R)\) be defined by the expression (4.155) for the choice
$$\begin{aligned} F(t,s) = \lambda (t)^{2} \chi _{\leqslant 1}(\frac{s}{g(t)}) \partial _{1}^{2} u_{ell,2}(t,s\lambda (t)) \end{aligned}$$
and let
$$\begin{aligned} f_{ell,2}(t,r) = g_{ell,2}(t,\frac{r}{\lambda (t)}) m_{\leqslant 1}(\frac{r}{t}) \end{aligned}$$
Then, for \(0 \leqslant j \leqslant 2, \quad 0 \leqslant k \leqslant 3\),
$$\begin{aligned} |R^{k}t^{j}\partial _{t}^{j}\partial _{R}^{k}g_{ell,2}(t,R)| \leqslant \frac{C \lambda (t)^{6}}{t^{6}} {\left\{ \begin{array}{ll} R^{5}\left( 1+\log ^{2}(R)\right) , \quad R \leqslant 1\\ R^{5} \log (t), \quad 1 \leqslant R \leqslant 2 g(t)\\ R g(t)^{2} \log ^{3}(t) + \frac{g(t)^{6} \log (t)}{R}, \quad R > 2 g(t)\end{array}\right. } \end{aligned}$$
Also, letting \(e_{ell,2}\) denote the error term of \(f_{ell,2}\):
$$\begin{aligned}\begin{aligned} e_{ell,2}(t,r)&= -\left( -\partial _{tt}f_{ell,2} + \partial _{rr}f_{ell,2}+\frac{1}{r}\partial _{r}f_{ell,2}-\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}}f_{ell,2}\right) \\&\quad +\chi _{\leqslant 1}(\frac{r}{g(t)\lambda (t)}) \partial _{1}^{2}u_{ell,2}(t,r)\end{aligned} \end{aligned}$$
we have, for \(k=0,1\),
$$\begin{aligned} ||L_{\frac{1}{\lambda (t)}}^{k}(e_{ell,2})(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \lambda (t)^{7-k} \log ^{3}(t) g(t)^{6-k}}{t^{8}} \end{aligned}$$
Proof
We will take advantage of two of the orthogonality conditions in Lemma 27 when studying \(g_{ell,2}\). We recall that \(u_{ell,2}\) is defined in (4.9), and we use (4.13) to get
$$\begin{aligned}\begin{aligned} g_{ell,2}(t,R)&=\frac{e_{2}(R)}{2} \lambda (t)^{2} \int _{0}^{R} \chi _{\leqslant 1}(\frac{s}{g(t)}) \left( \partial _{t}^{2}\left( v_{ell,2,0,main}(t,\frac{r}{\lambda (t)})\right) \Bigr |_{r=s\lambda (t)} \right. \\&\quad + \partial _{t}^{2}\left( soln_{1}(t,\frac{r}{\lambda (t)})\right) \Bigr |_{r=s\lambda (t)} \\&\quad +\left. \partial _{t}^{2}\left( soln_{2}(t,\frac{r}{\lambda (t)})\right) \Bigr |_{r=s\lambda (t)} \right) s \phi _{0}(s) ds\\&\quad - \frac{\phi _{0}(R)}{2} \lambda (t)^{2} \int _{0}^{R} \chi _{\leqslant 1}(\frac{s}{g(t)}) \partial _{1}^{2} u_{ell,2}(t,s\lambda (t)) s e_{2}(s) ds\end{aligned} \end{aligned}$$
where, for the reader’s convenience, we recall that
$$\begin{aligned} v_{ell,2,0,main}(t,s)= & {} \frac{s^{3} \lambda (t)^{3}}{8}\left( \partial _{t}^{2}\left( \frac{\lambda '(t)^{2}}{2\lambda (t)}+\lambda ''(t)+\lambda ''(t)\log (\lambda (t))\right) \right. \\{} & {} \left. -\lambda ''''(t)\log (s\lambda (t)) + \frac{3}{4} \lambda ''''(t)\right) \end{aligned}$$
We then estimate \(g_{ell,2}\) by using the orthogonality conditions of Lemma 27 to treat the term in the above expression for \(g_{ell,2}\) which involves \(v_{ell,2,0,main}\), and estimate the rest of \(g_{ell,2}\) directly. We have the following two estimates for \(0 \leqslant k \leqslant 1\) and \(0 \leqslant j \leqslant 2\),
$$\begin{aligned}{} & {} |\partial _{t}^{j}\partial _{s}^{k}\left( \partial _{t}^{2}\left( soln_{1}(t,\frac{r}{\lambda (t)})\right) \Bigr |_{r=s\lambda (t)}\right) | \leqslant \frac{C \lambda (t)^{4}}{t^{6+j}} {\left\{ \begin{array}{ll} s^{5-k}(1+|\log (s)|), \quad s \leqslant 1\\ s^{1-k} (1+\log ^{2}(s)), \quad s \geqslant 1\end{array}\right. } \end{aligned}$$
(4.158)
$$\begin{aligned}{} & {} |\partial _{t}^{j}\partial _{s}^{k}\left( \partial _{t}^{2}\left( soln_{2}(t,\frac{r}{\lambda (t)})\right) \Bigr |_{r=s\lambda (t)}\right) | \leqslant \frac{C \lambda (t)^{4}}{t^{6+j}} {\left\{ \begin{array}{ll} s^{3-k} (1+\log ^{2}(s)), \quad s \leqslant 1\\ s^{1-k} (1+\log ^{3}(s)), \quad s \geqslant 1\end{array}\right. } \end{aligned}$$
(4.159)
This results in the estimates on \(g_{ell,2}\) and \(e_{ell,2}\) in the lemma statement. \(\square \)
Finally, we treat the linear error term associated to \(v_{2,2}\) which we recall is a free wave added to \(u_{w,2}\), and is chosen so as to allow for the third order matching. This error term is
$$\begin{aligned} e_{2,2}(t,r) = \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) v_{2,2}(t,r) \left( 1-\chi _{\leqslant 1}(\frac{r}{h(t)})\right) \end{aligned}$$
We define \(u_{2,2}\) to be the solution to the following equation, with 0 Cauchy data at infinity.
$$\begin{aligned} -\partial _{t}^{2}u_{2,2}+\partial _{r}^{2}u_{2,2}+\frac{1}{r}\partial _{r}u_{2,2}-\frac{u_{2,2}}{r^{2}}=e_{2,2}(t,r) \end{aligned}$$
Then, we have the following lemma.
Lemma 38
For \(\frac{h(t)}{2} \leqslant r \leqslant \frac{t}{2}\), and \(0 \leqslant k \leqslant 1\), \(0 \leqslant j \leqslant 2\),
$$\begin{aligned} \begin{aligned}&t^{j}r^{k}|\partial _{t}^{j}\partial _{r}^{k}u_{2,2}(t,r)| \leqslant \frac{C r \lambda (t)^{4} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{5}(t)}{t^{4} h(t)^{2}} \\&\quad + \frac{C r\lambda (t)^{2} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{4}(t)}{t^{6}}\end{aligned} \end{aligned}$$
(4.160)
Also, for \(\frac{h(t)}{2} \leqslant r \leqslant \frac{t}{2}\),
$$\begin{aligned}\begin{aligned} |\partial _{r}^{2}u_{2,2}(t,r)|&\leqslant \frac{C \lambda (t)^{4} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{5}(t)}{t^{4} r h(t)^{2}} \\&\quad + \frac{C \lambda (t)^{2} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{4}(t)}{r t^{6}}\end{aligned} \end{aligned}$$
In addition, for all \(r>0\),
$$\begin{aligned} \sqrt{E(u_{2,2},\partial _{t}u_{2,2})} + |u_{2,2}(t,r)| \leqslant \frac{C \lambda (t)^{2} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{3}(t)}{t^{3}}\nonumber \\ \end{aligned}$$
(4.161)
Finally, for \(0 \leqslant j \leqslant 2\),
$$\begin{aligned}{} & {} |\partial _{t}^{j}\partial _{r}u_{2,2}(t,r)| \leqslant \frac{C \lambda (t)^{4} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{3}(t)}{\langle t-r \rangle ^{1+j} h(t)^{2} t^{3}} \nonumber \\{} & {} + \frac{C \lambda (t)^{2} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{3}(t)}{ t^{5/2} \langle t-r \rangle ^{7/2+j}}, \quad \frac{t}{2}< r < t \nonumber \\ \end{aligned}$$
(4.162)
$$\begin{aligned}{} & {} |\partial _{r}u_{2,2}(t,r)| \leqslant \frac{C \log ^{4}(t)}{t^{\frac{5}{2}-5C_{u}}}, \quad r>0 \end{aligned}$$
(4.163)
Proof
As in (4.122), we have
$$\begin{aligned} u_{2,2}(t,r) = \frac{-r}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1}d\beta \int _{0}^{2\pi } d\theta I_{e_{2,2}}(s,r\beta ,\rho ,\theta ) \end{aligned}$$
where we recall the notation (2.2). We start with the estimate of the lemma statement in the region \(h(t) \leqslant r \leqslant 2h(t) \leqslant \frac{t}{2}\). For all \(\rho \leqslant s-t\), \(r \leqslant \frac{t}{2}\), and \(0 \leqslant \beta \leqslant 1\), we have
$$\begin{aligned} \sqrt{\beta ^{2}r^{2}+\rho ^{2}+2\beta r \rho \cos (\theta )} \leqslant \beta r + \rho \leqslant s-t + \frac{t}{2}<s \end{aligned}$$
From Lemma 26, we get
$$\begin{aligned}\begin{aligned} |I_{e_{2,2}}(s,r \beta ,\rho ,\theta )| \leqslant C&\left( \frac{\lambda (s)^{2}}{s^{4}} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \frac{\log ^{3}(s) \lambda (s)^{2}}{(h(s)^{2}+\beta ^{2}r^{2}+\rho ^{2}+2\beta r \rho \cos (\theta ))^{2}}\right. \\&\left. +\frac{\lambda (s)^{2} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,s]}\left( \lambda (x)^{2}\right) \log ^{3}(s)}{\langle s-\sqrt{r^{2}\beta ^{2}+\rho ^{2}+2 \beta r \rho \cos (\theta )}\rangle ^{7/2} s^{9/2}}\right) \end{aligned} \end{aligned}$$
We then make the analogous decomposition as in (4.123), and use
$$\begin{aligned} s- \sqrt{\beta ^{2}r^{2}+\rho ^{2}+2 \beta r \rho \cos (\theta )} \geqslant s-(r+\rho ) \geqslant t-r \geqslant C t \end{aligned}$$
which is true for \(\rho \leqslant s-t\) and all \(r \leqslant \frac{t}{2}\). This gives (4.160) for \(k =0,j=0\), and \(\frac{h(t)}{2} \leqslant r \leqslant 2h(t)\). For \(k=1,j=0\), we note that
$$\begin{aligned} \partial _{r}u_{2,2}(t,r) = \frac{-1}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{2\pi } d\theta I_{e_{2,2}}(s,r,\rho ,\theta ) \end{aligned}$$
which, when combined with the same procedure used for \(u_{2,2}\), completes the proof of (4.160) for \(j=0\) and \(0 \leqslant k \leqslant 1\), in the region \(\frac{h(t)}{2} \leqslant r \leqslant 2h(t)\). To treat higher j, we simply note that
$$\begin{aligned} \partial _{t}^{j}u_{2,2}(t,r) = \frac{-r}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1}d\beta \int _{0}^{2\pi } d\theta \partial _{1}^{j}I_{e_{2,2}}(s,r\beta ,\rho ,\theta ) \end{aligned}$$
and use the same procedure used for \(j=0\).
In the region \(2h(t) \leqslant r \leqslant \frac{t}{2}\), a slightly more complicated argument is needed because factors of \(\frac{r}{h(t)}\) are no longer controlled by a constant in this region. We have
$$\begin{aligned} \begin{aligned}&|u_{2,2}(t,r)| \\&\leqslant C r \int _{t}^{\infty } ds \int _{0}^{s-t}\frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}}\int _{0}^{1}d\beta \int _{0}^{2\pi } d\theta \left( \frac{\lambda (s)^{4} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \log ^{3}(s)}{s^{4}(r^{2}\beta ^{2}+\rho ^{2}+2 r \beta \rho \cos (\theta )+h(s)^{2})^{2}}\right. \\&\left. +\frac{\lambda (s)^{2} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,s]}\left( \lambda (x)^{2}\right) \log ^{3}(s)}{s^{9/2} t^{7/2}} \right) \end{aligned}\nonumber \\ \end{aligned}$$
(4.164)
The second term of the integrand of the expression (4.164) is estimated with the following simple procedure.
$$\begin{aligned}\begin{aligned}&|r \int _{t}^{\infty } ds \int _{0}^{s-t}\frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}}\int _{0}^{1}d\beta \\&\quad \int _{0}^{2\pi } d\theta \frac{\lambda (s)^{2} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,s]}\left( \lambda (x)^{2}\right) \log ^{3}(s)}{s^{9/2} t^{7/2}}|\\&\leqslant C r \int _{t}^{\infty } ds \frac{\lambda (s)^{2} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,s]}\left( \lambda (x)^{2}\right) \log ^{3}(s)}{s^{7/2} t^{7/2}}\\&\leqslant \frac{C r \lambda (t)^{2} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{3}(t)}{t^{6}}, \quad r \leqslant \frac{t}{2}\end{aligned} \end{aligned}$$
Using Cauchy’s residue theorem appropriately, we get the following, for \(a,\rho ,y>0\).
$$\begin{aligned} \int _{0}^{2\pi } \frac{d\theta }{(y^{2}+\rho ^{2}+2 \rho y \cos (\theta )+a^{2})^{2}} = \frac{2\pi (a^{2}+\rho ^{2}+y^{2})}{((a^{2}+(\rho -y)^{2})(a^{2}+(\rho +y)^{2}))^{3/2}}\nonumber \\ \end{aligned}$$
(4.165)
So, for the first term of the integrand in (4.164), we get
$$\begin{aligned}\begin{aligned}&|r \int _{t}^{\infty } ds \int _{0}^{s-t}\frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}}\int _{0}^{1}d\beta \int _{0}^{2\pi } d\theta \frac{\lambda (s)^{4} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \log ^{3}(s)}{s^{4}(r^{2}\beta ^{2}+\rho ^{2}+2 r \beta \rho \cos (\theta )+h(s)^{2})^{2}}|\\&\quad \leqslant C r \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1}d\beta \frac{\lambda (s)^{4} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \log ^{3}(s)}{s^{4} \sqrt{h(s)^{2}+\rho ^{2}+r^{2}\beta ^{2}} (h(s)^{2}+(\rho -r\beta )^{2})^{3/2}}\\&\quad \leqslant C r \int _{t}^{t+3h(t)} ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \frac{\lambda (s)^{4} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \log ^{3}(s)}{s^{4} h(s)^{4}}\\&\qquad +C r \int _{t+3h(t)}^{t+3r} ds \int _{0}^{3h(t)} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \frac{\lambda (s)^{4} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \log ^{3}(s)}{s^{4}h(s)^{4}}\\&\qquad +C r \int _{t+3h(t)}^{t+3r} ds \int _{3h(t)}^{s-t} \frac{d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \frac{\lambda (s)^{4} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \log ^{3}(s)}{s^{4}} \\&\quad \int _{0}^{1}\frac{d\beta }{(h(s)^{2}+(\rho -r\beta )^{2})^{3/2}}\\&\qquad +C r \int _{t+3r}^{\infty } ds \int _{0}^{3h(t)} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \frac{\lambda (s)^{4} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \log ^{3}(s)}{s^{4} h(s)^{4}}\\&\qquad +C r \int _{t+3r}^{\infty } ds \int _{3h(t)}^{3r} \frac{d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \frac{\lambda (s)^{4} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \log ^{3}(s)}{s^{4}} \\&\quad \int _{0}^{1}\frac{d\beta }{(h(s)^{2}+(\rho -r \beta )^{2})^{3/2}}\\&\qquad +C r \int _{t+3r}^{\infty } ds \int _{3r}^{s-t} \frac{d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \frac{\lambda (s)^{4} \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \log ^{3}(s)}{s^{4}(r+\rho )^{3}}\\&\quad \leqslant \frac{C r \lambda (t)^{4} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{5}(t)}{t^{4}h(t)^{2}}, \quad 2h(t) \leqslant r \leqslant \frac{t}{2}\end{aligned} \end{aligned}$$
The derivatives of \(u_{2,2}\) in the region \(2h(t) \leqslant r \leqslant \frac{t}{2}\) are treated in a similar way. A similar procedure establishes (4.162). Finally, the estimate on \(\partial _{r}^{2} u_{2,2}\) is obtained by noting that
$$\begin{aligned} \partial _{r}^{2} u_{2,2}(t,r) = e_{2,2}(t,r) + \frac{u_{2,2}(t,r)}{r^{2}} - \frac{1}{r}\partial _{r}u_{2,2}(t,r) + \partial _{t}^{2}u_{2,2}(t,r) \end{aligned}$$
For (4.161), we again use Lemma 26 to get that
$$\begin{aligned} |e_{2,2}(t,r)| \leqslant \mathbbm {1}_{\{r \geqslant h(t)\}} {\left\{ \begin{array}{ll} \frac{C \lambda (t)^{2}}{r^{4}} \frac{r \lambda (t)^{2}}{t^{4}} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{3}(t), \quad h(t) \leqslant r \leqslant \frac{t}{2}\\ \frac{C \lambda (t)^{2}}{r^{4}} \frac{\sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{3}(t)}{\langle t-r \rangle ^{5/2}\sqrt{t}}, \quad \frac{t}{2} \leqslant r \leqslant t\\ \frac{C \lambda (t)^{2}}{r^{9/2} \sqrt{\langle t-r \rangle }}, \quad r \geqslant t\end{array}\right. } \end{aligned}$$
which gives
$$\begin{aligned} ||e_{2,2}(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \lambda (t)^{2} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{3}(t)}{t^{4}} \end{aligned}$$
Then, the same procedure used in (4.69) (energy estimate) finishes the proof of (4.161). Finally, we prove (4.163) by using Lemma 26 to get
$$\begin{aligned} |I_{e_{2,2}}(s,r,\rho ,\theta )| \leqslant \frac{C \log ^{4}(s)}{s^{\frac{9}{2}-5C_{u}}} \end{aligned}$$
which implies
$$\begin{aligned} |\partial _{r}u_{2,2}(t,r)| \leqslant C \int _{t}^{\infty } ds \frac{(s-t) \log ^{4}(s)}{s^{\frac{9}{2}-5C_{u}}} \leqslant \frac{C \log ^{4}(t)}{t^{\frac{5}{2}-5C_{u}}} \end{aligned}$$
\(\square \)
Note that
$$\begin{aligned} (1-\chi _{\leqslant 1}(\frac{2r}{h(t)}))(1-\chi _{\leqslant 1}(\frac{r}{h(t)}))=(1-\chi _{\leqslant 1}(\frac{r}{h(t)})) \end{aligned}$$
So, we will add the following truncation of \(u_{2,2}\) into our ansatz:
$$\begin{aligned} f_{2,2}(t,r) := u_{2,2}(t,r) \left( 1-\chi _{\leqslant 1}(\frac{2r}{h(t)})\right) \end{aligned}$$
(4.166)
We define the error term associated to \(f_{2,2}\) by
$$\begin{aligned}\begin{aligned} e_{f_{2,2}}(t,r)&: = \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) v_{2,2}(t,r) \left( 1-\chi _{\leqslant 1}(\frac{r}{h(t)})\right) \\&\quad -\left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{1}{r^{2}}\right) f_{2,2}(t,r) \\&\quad + \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) f_{2,2}\\&=u_{2,2}(t,r) \left( \frac{2}{rh(t)} \chi _{\leqslant 1}'(\frac{2r}{h(t)}) - \frac{4r}{h(t)^{3}} \chi _{\leqslant 1}'(\frac{2r}{h(t)}) h'(t)^{2} - \frac{4r^{2} h'(t)^{2}}{h(t)^{4}} \chi _{\leqslant 1}''(\frac{2r}{h(t)})\right. \\&\quad \left. + \frac{4 \chi _{\leqslant 1}''(\frac{2r}{h(t)})}{h(t)^{2}}+\frac{2r h''(t)}{h(t)^{2}} \chi _{\leqslant 1}'(\frac{2r}{h(t)})\right) \\&\quad +\frac{4 \chi _{\leqslant 1}'(\frac{2 r}{h(t)})}{h(t)^{2}} \left( h(t)\partial _{r}u_{2,2}+r h'(t) \partial _{t}u_{2,2}\right) \\&\quad + \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) u_{2,2}(t,r) \left( 1-\chi _{\leqslant 1}(\frac{2r}{h(t)})\right) \end{aligned} \end{aligned}$$
Then, a direct estimation gives the following lemma.
Lemma 39
For \(k=0,1\),
$$\begin{aligned} ||L_{\frac{1}{\lambda (t)}}^{k}(e_{f_{2,2}})(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \lambda (t)^{4} \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \log ^{5}(t)}{t^{4} h(t)^{2+k}} \end{aligned}$$
Now, we let
$$\begin{aligned}{} & {} u_{a}(t,r) = u_{c}(t,r)+f_{5,0}(t,r)+f_{ex,sub}(t,r)+f_{ell,2}(t,r)+f_{2,2}(t,r) \nonumber \\{} & {} u_{corr}(t,r) = u_{a}(t,r)+u_{1}(t,r) \end{aligned}$$
(4.167)
where we recall that \(u_{c}\) is defined by (4.141). The equation for \(u_{1}\) which results from substituting \(Q_{\frac{1}{\lambda (t)}}(r)+u_{corr}\) into (1.1) is
$$\begin{aligned}\begin{aligned}&-\partial _{t}^{2} u_{1}+\partial _{r}^{2}u_{1}+\frac{1}{r}\partial _{r}u_{1}-\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}} u_{1}\\&\quad =e_{a}(t,r)+\left( \frac{\cos (2u_{corr})-1}{2r^{2}}\right) \sin (2Q_{1}(\frac{r}{\lambda (t)})) + \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}}\left( \sin (2u_{corr})-2u_{corr}\right) \end{aligned} \end{aligned}$$
where
$$\begin{aligned} e_{a}(t,r) = \left( e_{ex,ell}+e_{w,2}+e_{match,0}+e_{5,1}+e_{f_{5,0}}+e_{ex,sub}+e_{ell,2}+e_{f_{2,2}}\right) (t,r) \end{aligned}$$
Combining Lemmas 39, 37, 36, 34, 32, 31, 28 and using (1.7) and (4.2) gives the following lemma.
Lemma 40
$$\begin{aligned}\begin{aligned}&\frac{||e_{a}||_{L^{2}(r dr)}}{\lambda (t)^{2}} \leqslant \frac{C \log ^{6}(t)}{t^{4+\text {min}\{4\alpha -C_{l}-2,1-\alpha -2C_{u},4-5C_{u}-6\alpha \}}}\end{aligned} \\ \begin{aligned}&||L_{\frac{1}{\lambda (t)}}\left( e_{a}\right) ||_{L^{2}(r dr)} \leqslant \frac{C \log ^{6}(t)}{t^{4+2\alpha -3C_{u}}}+\frac{C \log ^{2}(t)}{t^{2+5\alpha }}+\frac{C \log ^{5}(t)}{t^{5-3C_{u}}}+\frac{C \log ^{3}(t)}{t^{8-6C_{u}-5\alpha }}\end{aligned} \end{aligned}$$
4.9 First set of nonlinear interactions
It only remains to treat the terms which are nonlinear in \(u_{corr}=u_{a}+u_{1}\). For this, we define
$$\begin{aligned}\begin{aligned} NL(t,r)&= \left( \frac{\cos (2u_{corr})-1}{2r^{2}}\right) \sin (2Q_{1}(\frac{r}{\lambda (t)})) + \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}}\left( \sin (2u_{corr})-2u_{corr}\right) \\&=\left( \frac{\cos (2u_{1})-1}{2r^{2}}\right) \sin (2Q_{1}(\frac{r}{\lambda (t)})+2u_{a})\\&+\left( \frac{\sin (2u_{1})-2u_{1}}{2r^{2}}\right) \cos (2Q_{1}(\frac{r}{\lambda (t)})+2u_{a}) + \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \sin (2u_{a})-2u_{a}\right) \\&+u_{1}\left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)})+2u_{a})-\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}}\right) +\frac{\sin (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \cos (2u_{a})-1\right) \end{aligned} \end{aligned}$$
We focus on the \(u_{1}\)-independent terms of the expression above. In particular, we define
$$\begin{aligned} N_{a}(t,r) = \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \sin (2u_{a})-2u_{a}\right) + \frac{\sin (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \cos (2u_{a})-1\right) \end{aligned}$$
We recall that
$$\begin{aligned} u_{a} = \chi _{\leqslant 1}(\frac{r}{h(t)}) u_{e}+\left( 1-\chi _{\leqslant 1}(\frac{r}{h(t)})\right) u_{wave} + f_{5,0}+f_{ex,sub}+f_{ell,2}+f_{2,2} \end{aligned}$$
We write
$$\begin{aligned} u_{a} = u_{a,0}+u_{a,1} \end{aligned}$$
(4.168)
with
$$\begin{aligned} u_{a,0}(t,r)= & {} \chi _{\leqslant 1}(\frac{r}{h(t)}) u_{ell}(t,r) + \left( 1-\chi _{\leqslant 1}(\frac{r}{h(t)})\right) \left( w_{1}(t,r)+v_{2}(t,r)+v_{2,2}(t,r)\right) \\ u_{a,1}= & {} \chi _{\leqslant 1}(\frac{r}{h(t)}) \left( u_{e}-u_{ell}\right) + \left( 1-\chi _{\leqslant 1}(\frac{r}{h(t)})\right) \left( v_{ex}+u_{w,2}\right) + f_{5,0}+f_{ex,sub}+f_{ell,2}+f_{2,2} \end{aligned}$$
Then, we get
$$\begin{aligned}\begin{aligned} N_{a}(t,r)&= \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \sin (2u_{a,0})-2u_{a,0}\right) +\frac{\sin (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \cos (2u_{a,0})-1\right) \\&\quad +\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \sin (2u_{a,0})\left( \cos (2u_{a,1})-1\right) + \left( \sin (2u_{a,1})-2u_{a,1}\right) \cos (2u_{a,0})\right. \\&\quad \qquad \qquad \qquad \qquad \qquad \left. + \left( \cos (2u_{a,0})-1\right) 2u_{a,1}\right) \\&\quad +\frac{\sin (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \left( \cos (2u_{a,1})-1\right) \cos (2u_{a,0}) - 2 \sin (2u_{a,0}) u_{a,1}\right. \\&\quad \qquad \qquad \qquad \qquad \qquad \left. -\sin (2u_{a,0})\left( \sin (2u_{a,1})-2u_{a,1}\right) \right) \\&:=N_{0}+N_{1}\end{aligned} \end{aligned}$$
where
$$\begin{aligned} \begin{aligned}&N_{1}(t,r)\\&\quad =\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \sin (2u_{a,0})\left( \cos (2u_{a,1})-1\right) + \left( \sin (2u_{a,1})-2u_{a,1}\right) \cos (2u_{a,0})\right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \left. + \left( \cos (2u_{a,0})-1\right) 2u_{a,1}\right) \\&\qquad +\frac{\sin (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \left( \cos (2u_{a,1})-1\right) \cos (2u_{a,0}) - 2 \sin (2u_{a,0}) u_{a,1}\right. \\&\qquad \qquad \qquad \qquad \qquad \qquad -\sin (2u_{a,0})\left( \sin (2u_{a,1})\right. \left. \left. -2u_{a,1}\right) \right) \end{aligned}\nonumber \\ \end{aligned}$$
(4.169)
We proceed to estimate the term \(N_{1}\), which turns out to be perturbative.
Lemma 41
We have the following estimates on \(N_{1}\).
$$\begin{aligned}{} & {} \frac{||N_{1}(t,r)||_{L^{2}(r dr)}}{\lambda (t)^{2}} \leqslant \frac{C \log ^{3}(t)}{t^{6-3C_{u}-2\alpha }} + \frac{C \log ^{13}(t)}{t^{5-\frac{3}{2}C_{l}-9C_{u}}},\\{} & {} ||L_{\frac{1}{\lambda (t)}}(N_{1})(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \log ^{3}(t)}{t^{6-5C_{u}-2\alpha -C_{l}}} + \frac{C \log ^{13}(t)}{t^{5-\frac{19}{2}C_{u}-C_{l}}} \end{aligned}$$
Proof
We estimate \(N_{1}\) by combining the following estimates on various terms of \(u_{a,0}\) and \(u_{a,1}\). We use the explicit formulae following (4.7) and (4.81) to estimate \(u_{ell}\). We use the decomposition (4.13), along with (4.14) and the analogous estimates to (4.158) and (4.159) to estimate \(u_{ell,2}\). We use (4.146) in the region \(r \geqslant \frac{t}{2}\), and (4.147) and the expression for \(v_{ex,ell}\), namely (4.36) in the region \(g(t)\lambda (t) \leqslant r \leqslant \frac{t}{2}\), to estimate \(v_{ex}\). We use Lemma 10 to estimate \(u_{w,2}\). (In particular, we use (4.63) and (4.67) in the region \(g(t)\lambda (t) \leqslant r \leqslant \frac{t}{2}\), and (4.64) in the region \(r \geqslant \frac{t}{2}\)). Next, we use Lemma (2), Lemma (5), and Lemma 26 to estimate \(w_{1}\), \(v_{2}\), and \(v_{2,2}\), respectively. Finally, we use Lemma 38, and (4.166) (for \(f_{2,2}\)), Lemma 37 (for \(f_{ell,2}\)), Lemma 33 and (4.156) (for \(f_{5,0}\)), and Lemma 35 and (4.157) (for \(f_{ex,sub}\)). This gives rise to
$$\begin{aligned} \begin{aligned}&|u_{a,0}(t,r)| \\&\leqslant C {\left\{ \begin{array}{ll} \frac{r \lambda (t) \log (t)}{t^{2}}, \quad r \leqslant \frac{t}{2}\\ \frac{\sup _{x \in [100,r]}\left( \lambda (x) \log (x)\right) \log (r)}{\sqrt{r} \sqrt{\langle t-r \rangle }} + \frac{ \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \log ^{3}(t) \mathbbm {1}_{\{t> r> \frac{t}{2}\}}}{\langle t-r \rangle ^{5/2} \sqrt{t}}, \quad r > \frac{t}{2}\end{array}\right. }\end{aligned} \end{aligned}$$
(4.170)
and
$$\begin{aligned} |u_{a,1}(t,r)| \leqslant C {\left\{ \begin{array}{ll} \frac{r^{3}(\log ^{2}(r)+\log ^{2}(t))}{t^{4-C_{u}}}, \quad r \leqslant 2 h(t)\\ \frac{\lambda (t) \log ^{3}(t)}{t^{2-2C_{u}}r} + \frac{C r \lambda (t) \log ^{6}(t)}{t^{4-3C_{u}}}, \quad 2 h(t) \leqslant r \leqslant \frac{t}{2}\\ \frac{\log ^{5}(t)}{t^{\frac{5}{2}-\frac{7}{2}C_{u}}} \cdot \text {min}\{1,\frac{\lambda (t)}{t^{\frac{C_{u}-C_{l}}{2}}}\}, \quad r \geqslant \frac{t}{2}\end{array}\right. } \end{aligned}$$
(4.171)
Then, a straightforward, but slightly long computation gives the estimate on \(||N_{1}(t,r)||_{L^{2}(r dr)}\) from the lemma statement. Recall that
$$\begin{aligned} L_{\frac{1}{\lambda (t)}}N_{1}(t,r) = \partial _{r}N_{1}(t,r)-\frac{\cos (Q_{1}(\frac{r}{\lambda (t)}))}{r} N_{1}(t,r) \end{aligned}$$
To estimate \(||L_{\frac{1}{\lambda (t)}}N_{1}(t,r)||_{L^{2}(r dr)}\), we use the procedure outlined at the beginning of the proof to estimate \(\partial _{r}u_{a,0}\) and \(\partial _{r}u_{a,1}\). The only difference here is the following. After differentiating \(N_{1}(t,r)\) with respect to r, we use the pointwise estimates on \(\partial _{r}u_{2,2}(t,r)\) given in (4.160) for the region \(r \leqslant \frac{t}{2}\). We use Holder’s inequality and (4.161) to estimate the terms involving \(\partial _{r}u_{2,2}(t,r)\) in the region \(r \geqslant \frac{t}{2}\). \(\square \)
It remains to treat \(N_{0}\), which we recall is defined by
$$\begin{aligned} N_{0}(t,r) = \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \sin (2u_{a,0})-2u_{a,0}\right) + \frac{\sin (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \cos (2u_{a,0})-1\right) \nonumber \\ \end{aligned}$$
(4.172)
where
$$\begin{aligned} u_{a,0}(t,r) = \chi _{\leqslant 1}(\frac{r}{h(t)}) u_{ell}(t,r) + \left( 1-\chi _{\leqslant 1}(\frac{r}{h(t)})\right) \left( w_{1}(t,r)+v_{2}(t,r)+v_{2,2}(t,r)\right) \end{aligned}$$
We define \(u_{N_{0}}\) to be the solution to the following equation with 0 Cauchy data at infinity.
$$\begin{aligned} -\partial _{t}^{2}u_{N_{0}}+\partial _{r}^{2}u_{N_{0}}+\frac{1}{r}\partial _{r}u_{N_{0}}-\frac{u_{N_{0}}}{r^{2}} = N_{0}(t,r) \end{aligned}$$
(4.173)
Then, we have
Lemma 42
We have the following estimates. For \(0 \leqslant k \leqslant 2\) and \(0 \leqslant j \leqslant 1\),
$$\begin{aligned}{} & {} |\partial _{t}^{k}\partial _{r}^{j}u_{N_{0}}(t,r)| \leqslant \frac{C r^{1-j} \left( \sup _{x\in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \log ^{3}(t)}{t^{4+k}}, \quad r \leqslant \frac{t}{2} \qquad \end{aligned}$$
(4.174)
$$\begin{aligned}{} & {} \begin{aligned}&|u_{N_{0}}(t,r)| + \sqrt{E(u_{N_{0}},\partial _{t}u_{N_{0}})}\\&\quad \leqslant \frac{C \left( \sup _{x\in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \left( \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \right) ^{3} \log ^{9}(t)}{t^{2}} \end{aligned} \end{aligned}$$
(4.175)
Finally,
$$\begin{aligned}{} & {} \begin{aligned}&|\partial _{r}u_{N_{0}}(t,r)|\\&\leqslant \frac{C \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \log ^{3}(t)}{t^{3/2} \langle t-r \rangle ^{5/2}} \\&\quad + \frac{C \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \left( \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \right) ^{3} \log ^{9}(t)}{t^{3/2} \langle t-r \rangle ^{17/2}} \\&\quad + \frac{C \lambda (t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{2} \left( \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \right) ^{2} \log ^{6}(t)}{t^{2} \langle t-r \rangle ^{6}}, \quad \frac{t}{2}< r < t\end{aligned} \end{aligned}$$
(4.176)
$$\begin{aligned}{} & {} ||\partial _{r}u_{N_{0}}(t,r)||_{L^{\infty }_{r}} \leqslant \frac{C \log ^{12}(t)}{t^{\frac{3}{2}-9C_{u}}} \end{aligned}$$
(4.177)
Proof
As in (4.121), we have
$$\begin{aligned} u_{N_{0}}(t,r) = -\frac{r}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1}d\beta \int _{0}^{2\pi } d\theta I_{N_{0}}(s,r\beta ,\rho ,\theta ) \end{aligned}$$
where we use the notation defined in (2.2). We then estimate \(u_{a,0}\) and its derivatives using the procedure described in the proof of Lemma 41. This results in the following estimate for \(j=0,1,2\) and \(k=0,1\).
$$\begin{aligned}{} & {} |\partial _{t}^{j}\partial _{r}^{k} N_{0}(t,r)| \nonumber \\{} & {} \quad \leqslant C\left( \frac{1}{r}+\frac{1}{\langle t-r \rangle }\right) ^{k} \left( \frac{1}{t}+\frac{1}{\langle t-r \rangle }\right) ^{j} \nonumber \\{} & {} \qquad {\left\{ \begin{array}{ll} \frac{r \lambda (t)^{3} \log ^{3}(t)}{t^{4} (r^{2}+\lambda (t)^{2})}, \quad r \leqslant \frac{t}{2}\\ \frac{\lambda (t)}{t^{4}} \frac{\left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{2}}{\langle t-r \rangle ^{5}} \left( \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \right) ^{2} \log ^{6}(t) \\ + \frac{\left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \log ^{3}(t)}{r^{7/2} \langle t-r \rangle ^{3/2}}\\ +\frac{\left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \left( \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \right) ^{3} \log ^{9}(t)}{\langle t-r \rangle ^{15/2} t^{7/2}}, \quad \frac{t}{2}< r < t\end{array}\right. } \end{aligned}$$
(4.178)
We start with the region \(r \leqslant \frac{t}{2}\). Using an analog of (4.165), we get
$$\begin{aligned}{} & {} |u_{N_{0}}(t,r)| \nonumber \\{} & {} \quad \leqslant C r \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1}d\beta \left( \frac{\lambda (s)^{3} \log ^{3}(s)}{s^{4} \sqrt{(\lambda (s)^{2}+(\rho +r\beta )^{2})(\lambda (s)^{2} +(\rho -r \beta )^{2})}}\right. \nonumber \\{} & {} \quad \left. +\frac{\lambda (s) \left( \sup _{x\in [100,s]}\left( \lambda (x) \log (x)\right) \right) ^{2} \left( \sup _{x \in [100,s]}\left( \lambda (x)^{2}\right) \right) ^{2} \log ^{6}(s)}{s^{4} (s-(\rho +r))^{6}}\right. \nonumber \\{} & {} +\left. \frac{\left( \sup _{x\in [100,s]}\left( \lambda (x) \log (x)\right) \right) ^{3} \log ^{3}(s)}{s^{7/2} (s-(\rho +r))^{5/2}}\right. \nonumber \\{} & {} \left. +\frac{\left( \sup _{x\in [100,s]}\left( \lambda (x) \log (x)\right) \right) ^{3} \left( \sup _{x \in [100,s]}\left( \lambda (x)^{2}\right) \right) ^{3} \log ^{9}(s)}{s^{7/2} (s-(\rho +r))^{17/2}}\right) \end{aligned}$$
(4.179)
where we used the fact that, if \(0\leqslant \beta \leqslant 1\), then,
$$\begin{aligned} \langle s-\sqrt{\rho ^{2}+r^{2}\beta ^{2}+2 r \beta \rho \cos (\theta )}\rangle \geqslant C\left( s-(\rho +r)\right) . \end{aligned}$$
By a direct estimation, we get
$$\begin{aligned} |\int _{0}^{1}\frac{d\beta }{\sqrt{\lambda (s)^{2}+(r\beta -\rho )^{2}}} \frac{1}{(\lambda (s)+r\beta +\rho )}| \leqslant \frac{C \log (s)}{(r+\rho )(\lambda (s)+\rho )} \end{aligned}$$
and therefore,
$$\begin{aligned}{} & {} r \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1}d\beta \frac{\lambda (s)^{3} \log ^{3}(s)}{s^{4} \log ^{2b}(s) \sqrt{(\lambda (s)^{2}+(\rho +r \beta )^{2})(\lambda (s)^{2}+(\rho -r\beta )^{2})}}\\{} & {} \quad \leqslant \frac{C r \lambda (t)^{3} \log ^{6}(t)}{t^{4}} \end{aligned}$$
The other terms of (4.179) are treated with a similar argument, using \(s-(r+\rho ) \geqslant s-r-(s-t) \geqslant C t\) (since \(r \leqslant \frac{t}{2}\)). This gives (4.174), for \(k=0,j=0\). To obtain (4.174) for \(k=1,2\) and \(j=0\), we first note that
$$\begin{aligned} \partial _{t}^{k} u_{N_{0}}(t,r) = -\frac{r}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1}d\beta \int _{0}^{2\pi } d\theta \partial _{1}^{k}I_{N_{0}}(s,r\beta ,\rho ,\theta ) \end{aligned}$$
Then, after applying (4.178), we obtain an extra factor of
$$\begin{aligned} \left( \frac{1}{s}+\frac{1}{\langle s-\sqrt{r^{2}\beta ^{2}+\rho ^{2}+2 r \beta \rho \cos (\theta )}\rangle }\right) ^{k}\leqslant \left( \frac{C}{s} + \frac{C}{s-(r+s-t)}\right) ^{k} \leqslant \frac{C}{t^{k}} \end{aligned}$$
in the integrand of the expression for \(\partial _{t}^{k} u_{N_{0}}(t,r)\), relative to that for \(u_{N_{0}}\), which we just estimated. This immediately gives (4.174) for \(k=1,2\). Next, we use the same procedure as in (4.77), to get (4.175). Finally, we use
$$\begin{aligned} \partial _{r}u_{N_{0}}(t,r) = \frac{-1}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{2\pi } d\theta I_{N_{0}}(s,r,\rho ,\theta ) \end{aligned}$$
and a similar argument used to estimate \(u_{N_{0}}\), to get (4.174) for \(j=1,k=0\). For completeness, we show how to estimate the most delicate integral, which is
$$\begin{aligned} \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \frac{\lambda (s)^{3} \log ^{3}(s)}{s^{4} \sqrt{\lambda (s)^{2}+(\rho +r)^{2}}\sqrt{\lambda (s)^{2}+(\rho -r)^{2}}} \end{aligned}$$
Here, we use
$$\begin{aligned} \frac{d}{d\rho }\left( \text {arctanh}(\frac{\rho -r}{\sqrt{a^{2}+(\rho -r)^{2}}})\right) = \frac{1}{\sqrt{a^{2}+(r-\rho )^{2}}} \end{aligned}$$
and estimate the integral directly. A similar argument establishes (4.176). Finally, to obtain (4.177), we again estimate \(u_{a,0}\) and \(\partial _{r}u_{a,0}\) using the procedure described in the proof of Lemma 41 to get, for all \(r>0, \theta \in [0,2\pi ], \rho \leqslant s-t\),
$$\begin{aligned} |I_{N_{0}}(s,r,\rho ,\theta )| \leqslant \frac{C \log ^{12}(s)}{s^{\frac{7}{2}-9C_{u}}} \end{aligned}$$
This gives
$$\begin{aligned} |\partial _{r}u_{N_{0}}(t,r)| \leqslant C \int _{t}^{\infty } ds (s-t) \frac{C \log ^{12}(s)}{s^{\frac{7}{2}-9C_{u}}} \leqslant \frac{C \log ^{12}(t)}{t^{\frac{3}{2}-9C_{u}}} \end{aligned}$$
\(\square \)
The linear error term of \(u_{N_{0}}\) is \(e_{N_{0}}\) defined by
$$\begin{aligned} e_{N_{0}}(t,r):= \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) u_{N_{0}}(t,r) \end{aligned}$$
(4.180)
Then, a straightforward estimation, using Lemma 42 gives the following (recall (2.1)).
Lemma 43
We have the following estimates for \(k=0,1\).
$$\begin{aligned} \begin{aligned}&||L_{\frac{1}{\lambda (t)}}^{k}(e_{N_{0}}(t,r) (1-m_{\leqslant 1}(\frac{r}{2h(t)})))||_{L^{2}(r dr)} \\&\leqslant \frac{C \lambda (t)^{2} \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \log ^{3}(t)}{t^{4}}\left( \frac{1}{h(t)^{2+k}}+\frac{\left( \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \right) ^{3} \log ^{6}(t)}{t^{1+k}}\right) \end{aligned} \end{aligned}$$
It remains to treat \(m_{\leqslant 1}(\frac{r}{2h(t)}) e_{N_{0}}(t,r)\). For the reader’s convenience, we repeat the outline (that was given in Section 3, just below (3.5)) of the procedure to be used. We start with solving the equation
$$\begin{aligned} \partial _{rr}u_{N_{0},ell}+\frac{1}{r}\partial _{r}u_{N_{0},ell}-\frac{\cos (2Q_{\frac{1}{\lambda (t)}}(r))}{r^{2}}u_{N_{0},ell} = m_{\leqslant 1}(\frac{r}{2h(t)}) e_{N_{0}}(t,r)\nonumber \\ \end{aligned}$$
(4.181)
and then inserting the following function into the ansatz
$$\begin{aligned} u_{N_{0},corr}(t,r):=\left( u_{N_{0},ell}(t,r) - \frac{r \lambda (t)}{4} \langle m_{\leqslant 1}(\frac{R \lambda (t)}{2h(t)}) e_{N_{0}}(t,R \lambda (t)),\phi _{0}(R)\rangle _{L^{2}(R dR)}\right) m_{\leqslant 1}(\frac{2 r}{t}) + v_{2,4}(t,r)\nonumber \\ \end{aligned}$$
(4.182)
where \(v_{2,4}\) solves
$$\begin{aligned} -\partial _{tt}v_{2,4}+\partial _{rr}v_{2,4}+\frac{1}{r}\partial _{r}v_{2,4}-\frac{v_{2,4}}{r^{2}}=0 \end{aligned}$$
(4.183)
and \(v_{2,4}(t,r)\) matches \(\frac{r \lambda (t)}{4} \langle m_{\leqslant 1}(\frac{R \lambda (t)}{2h(t)}) e_{N_{0}}(t,R \lambda (t)),\phi _{0}(R)\rangle _{L^{2}(R dR)}\) for small r. (Recall that \(m_{\leqslant 1}\) was defined in (2.1)). In particular, we choose the initial velocity, say \(v_{2,5}\) of \(v_{2,4}\) by requiring, for all t sufficiently large,
$$\begin{aligned} \frac{-r}{4}\cdot -2 \int _{0}^{\infty } \xi \sin (t\xi ) \widehat{v_{2,5}}(\xi ) d\xi =\frac{r \lambda (t)}{4} \langle m_{\leqslant 1}(\frac{R \lambda (t)}{2h(t)}) e_{N_{0}}(t,R \lambda (t)),\phi _{0}(R)\rangle _{L^{2}(R dR)} \end{aligned}$$
We refer the reader to the discussion in Section 3, just below (3.5), for some intuition regarding this procedure.
We start by estimating \(u_{N_{0},ell}(t,r)- \frac{r \lambda (t)}{4} \langle m_{\leqslant 1}(\frac{R \lambda (t)}{2h(t)}) e_{N_{0}}(t,R \lambda (t)),\phi _{0}(R)\rangle \).
Lemma 44
We have the following estimates. For \(0 \leqslant k \leqslant 1\) and \(0 \leqslant j \leqslant 2\),
$$\begin{aligned} \begin{aligned}&t^{j}r^{k} |\partial _{t}^{j}\partial _{r}^{k}\left( u_{N_{0},ell}(t,r)-\frac{r \lambda (t)}{4} \langle m_{\leqslant 1}(\frac{R \lambda (t)}{2h(t)}) e_{N_{0}}(t,R \lambda (t)),\phi _{0}(R)\rangle \right) |\\&\leqslant {\left\{ \begin{array}{ll} \frac{C r \log ^{3}(t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{t^{4}} , \quad r \leqslant \lambda (t)\\ \frac{C (\log (t)+\log (\frac{r}{\lambda (t)})) \lambda (t)^{2}\left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \log ^{3}(t)}{r t^{4}} , \quad \lambda (t) < r\end{array}\right. }\end{aligned} \end{aligned}$$
(4.184)
In addition, we have
$$\begin{aligned}{} & {} |\partial _{r}^{2} \left( u_{N_{0},ell}(t,r)-\frac{r \lambda (t)}{4} \langle m_{\leqslant 1}(\frac{R \lambda (t)}{2h(t)}) e_{N_{0}}(t,R \lambda (t)),\phi _{0}(R)\rangle \right) | \nonumber \\{} & {} \quad \leqslant C {\left\{ \begin{array}{ll} \frac{r}{\lambda (t)^{2}} \frac{\left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \log ^{3}(t)}{t^{4}}, \quad r \leqslant \lambda (t)\\ \frac{ \lambda (t)^{2}(\log (t)+\log (\frac{r}{\lambda (t)}))\left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \log ^{3}(t)}{r^{3}t^{4}} , \quad r \geqslant \lambda (t)\end{array}\right. } \end{aligned}$$
(4.185)
Finally, for \(0 \leqslant k \leqslant 2\),
$$\begin{aligned} |\partial _{t}^{k}\langle e_{N_{0}}(t,R\lambda (t)) m_{\leqslant 1}(\frac{R \lambda (t)}{2 h(t)}),\phi _{0}(R)\rangle | \leqslant \frac{C \log ^{3}(t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{t^{4+k}\lambda (t)}\nonumber \\ \end{aligned}$$
(4.186)
Proof
We consider the particular solution given by
$$\begin{aligned} u_{N_{0},ell}(t,r) = v_{N_{0},ell}(t,\frac{r}{\lambda (t)}) \end{aligned}$$
where
$$\begin{aligned}\begin{aligned} v_{N_{0},ell}(t,R)&= e_{2}(R) \int _{0}^{R} \lambda (t)^{2} m_{\leqslant 1}(\frac{s}{2g(t)}) e_{N_{0}}(t,s\lambda (t)) \phi _{0}(s) \frac{s ds}{2}\\&\quad -\phi _{0}(R) \int _{0}^{R} \lambda (t)^{2} m_{\leqslant 1}(\frac{s}{2g(t)}) e_{N_{0}}(t,s\lambda (t)) e_{2}(s) \frac{s ds}{2}\end{aligned} \end{aligned}$$
We remind the reader that \(e_{N_{0}}\) is defined in (4.180). Let \(int_{N_{0},i}(t,r)\) denote the ith term on the right-hand side of the above equation, evaluated at \(R=\frac{r}{\lambda (t)}\). In the region \(r \leqslant \lambda (t)\), we separately estimate \(int_{N_{0},1}\), \(int_{N_{0},2}\), and \(\frac{r \lambda (t)}{4} \langle m_{\leqslant 1}(\frac{R \lambda (t)}{2h(t)}) e_{N_{0}}(t,R \lambda (t)),\phi _{0}(R)\rangle \), using Lemma 42. This gives (4.184), in the region \(r \leqslant \lambda (t)\), and for \(k=j=0\). When \(r \geqslant \lambda (t)\), we take advantage of the fact that
$$\begin{aligned} |e_{2}(R)-\frac{R}{2}| \leqslant C\frac{(1+\log (R))}{R}, \quad R \geqslant 1 \end{aligned}$$
In particular, we estimate \(int_{N_{0},2}(t,r)\) directly, using Lemma 42 and write
$$\begin{aligned}\begin{aligned}&int_{N_{0},1}(t,r) - \frac{r \lambda (t)}{4} \langle m_{\leqslant 1}(\frac{R \lambda (t)}{2h(t)}) e_{N_{0}}(t,R \lambda (t)),\phi _{0}(R)\rangle \\&\quad =\left( \frac{e_{2}(\frac{r}{\lambda (t)})}{2}-\frac{r}{4 \lambda (t)}\right) \int _{0}^{\frac{r}{\lambda (t)}} \lambda (t)^{2} m_{\leqslant 1}(\frac{s}{2g(t)}) e_{N_{0}}(t,s\lambda (t)) \phi _{0}(s) s ds\\&\qquad -\frac{r \lambda (t)}{4} \int _{\frac{r}{\lambda (t)}}^{\infty } m_{\leqslant 1}(\frac{s}{2 g(t)}) e_{N_{0}}(t,s\lambda (t)) \phi _{0}(s) s ds\end{aligned} \end{aligned}$$
Directly estimating the integrals using Lemma 42, we obtain (4.184), in the region \(r \geqslant \lambda (t)\), and \(k=j=0\). To obtain (4.184) for \(k=1\), we use
$$\begin{aligned}\begin{aligned} \partial _{r}u_{N_{0},ell}(t,r)&=\frac{-\phi _{0}'(\frac{r}{\lambda (t)})}{\lambda (t)} \int _{0}^{\frac{r}{\lambda (t)}} \lambda (t)^{2} m_{\leqslant 1}(\frac{s}{2g(t)}) e_{N_{0}}(t,s\lambda (t)) e_{2}(s) \frac{s ds}{2} \\&\quad + \frac{e_{2}'(\frac{r}{\lambda (t)})}{\lambda (t)} \int _{0}^{\frac{r}{\lambda (t)}} \lambda (t)^{2} m_{\leqslant 1}(\frac{s}{2g(t)}) e_{N_{0}}(t,s\lambda (t)) \phi _{0}(s) \frac{s ds}{2}\end{aligned} \end{aligned}$$
and then repeat the same procedure used to obtain (4.184) for \(k=0\). For the estimate (4.184) for \(j>0\), we first re-write \(u_{N_{0},ell}\) as follows, and then differentiate in t directly.
$$\begin{aligned}\begin{aligned} u_{N_{0},ell}(t,r)&=-\phi _{0}(\frac{r}{\lambda (t)}) \int _{0}^{r} m_{\leqslant 1}(\frac{x}{2h(t)}) e_{N_{0}}(t,x) e_{2}(\frac{x}{\lambda (t)})\frac{x dx}{2}\\&\quad +e_{2}(\frac{r}{\lambda (t)}) \int _{0}^{r} m_{\leqslant 1}(\frac{x}{2h(t)}) e_{N_{0}}(t,x) \phi _{0}(\frac{x}{\lambda (t)}) \frac{x dx}{2}\end{aligned} \end{aligned}$$
Using the same procedure as for (4.184) with \(j=0\), and noting the symbol-type nature of the estimates in Lemma 42, we finish the proof of (4.184) for \(j>0\). Finally, to get (4.185), we use the equation solved by \(u_{N_{0},ell}\). The estimates in (4.186) follow directly from Lemma 42. \(\square \)
Just after (4.135), we restricted \(T_{0}\) to satisfy \(T_{0} \geqslant 2T_{2}\), and all of our computations and estimates are valid for all \(t \geqslant T_{0}\) for any \(T_{0} \geqslant 2 T_{2}\). At this stage, we restrict \(T_{0}\) so that \(T_{0} \geqslant 4T_{2}\) but is otherwise arbitrary. Then, recalling the cutoff \(\psi _{2}\) defined in (4.134), we define a function \(v_{2,5}\) by
$$\begin{aligned} \widehat{v_{2,5}}(\xi ) = \frac{1}{\pi \xi } \int _{0}^{\infty }\lambda (t) \langle m_{\leqslant 1}(\frac{R}{2 g(t)}) e_{N_{0}}(t,R\lambda (t)),\phi _{0}(R)\rangle _{L^{2}(R dR)} \psi _{2}(\frac{t}{2}) \sin (t\xi ) dt\nonumber \\ \end{aligned}$$
(4.187)
Letting
$$\begin{aligned} F_{2,4}(t) = -2 \int _{0}^{\infty } \xi \sin (t \xi ) \widehat{v_{2,5}}(\xi ) d\xi \end{aligned}$$
the inversion of the sine transform gives
$$\begin{aligned} F_{2,4}(t)= -\lambda (t) \psi _{2}(\frac{t}{2})\langle m_{\leqslant 1}(\frac{R}{2 g(t)}) e_{N_{0}}(t,R\lambda (t)),\phi _{0}(R)\rangle _{L^{2}(R dR)}, \quad t >0 \end{aligned}$$
Therefore, (4.186) gives the following estimate for \(0 \leqslant k \leqslant 2\) and \(t >0\):
$$\begin{aligned} |\partial _{t}^{k}F_{2,4}(t)| \leqslant C \mathbbm {1}_{\{t>2T_{2}\}} \frac{\left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}\log ^{3}(t)}{t^{4+k}} \end{aligned}$$
We have the following estimates on \(\widehat{v_{2,5}}\).
Lemma 45
For \(0 \leqslant k \leqslant 2\),
$$\begin{aligned} |\xi ^{k}\partial _{\xi }^{k} \widehat{v_{2,5}}(\xi )|\leqslant {\left\{ \begin{array}{ll} C, \quad \xi \leqslant \frac{1}{100}\\ \frac{C}{\xi ^{3-k}}, \quad \xi > \frac{1}{100}\end{array}\right. } \end{aligned}$$
(4.188)
Proof
We recall
$$\begin{aligned} \widehat{v_{2,5}}(\xi ) = \frac{-1}{\pi \xi } \int _{0}^{\infty } F_{2,4}(t) \sin (t\xi ) dt \end{aligned}$$
In the region \(\xi \leqslant \frac{1}{100}\), we simply directly estimate as follows
$$\begin{aligned} |\widehat{v_{2,5}}(\xi )| \leqslant \frac{C}{\xi } \int _{0}^{\frac{1}{\xi }} |F_{2,4}(t)| t \xi dt + \frac{C}{\xi } \int _{\frac{1}{\xi }}^{\infty } |F_{2,4}(t)| dt \end{aligned}$$
If \(\xi > \frac{1}{100}\), then, we have
$$\begin{aligned} \widehat{v_{2,5}}(\xi ) = \frac{1}{\pi \xi ^{3}} \int _{0}^{\infty } \sin (t\xi ) F_{2,4}''(t) dt \end{aligned}$$
where we note that we can integrate by parts only twice (simply because of the estimates we have on \(u_{N_{0}}\), which gave rise to estimates on up to two derivatives of \(F_{2,4}\)) which is why the decay for large \(\xi \) in the estimate (4.188) is not as strong as the analogous estimate for \(\widehat{v_{2,0}}\) in Lemma 4. To estimate the derivatives of \(\widehat{v_{2,5}}(\xi )\), we first re-write its formula as
$$\begin{aligned} \widehat{v_{2,5}}(\xi ) = \frac{-1}{\pi \xi ^{2}} \int _{0}^{\infty } F_{2,4}(\frac{\sigma }{\xi })\sin (\sigma ) d\sigma \end{aligned}$$
and then differentiate in \(\xi \). Then, we estimate as above. \(\square \)
We define \(v_{2,4}\) to be the solution to the following Cauchy problem.
$$\begin{aligned} {\left\{ \begin{array}{ll}-\partial _{t}^{2}v_{2,4}+\partial _{r}^{2}v_{2,4}+\frac{1}{r}\partial _{r}v_{2,4}-\frac{v_{2,4}}{r^{2}}=0\\ v_{2,4}(0,r)=0\\ \partial _{t}v_{2,4}(0,r) = v_{2,5}(r)\end{array}\right. } \end{aligned}$$
(4.189)
Now, we obtain estimates on \(v_{2,4}(t,r)-\left( \frac{-r}{4} F_{2,4}(t)\right) \)
Lemma 46
For \(0 \leqslant k \leqslant 2\),
$$\begin{aligned} |\partial _{r}^{k}\left( v_{2,4}(t,r)+\frac{r}{4}F_{2,4}(t)\right) | \leqslant \frac{C r^{3-k}\log ^{3}(t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{t^{6}} , \quad r \leqslant \frac{t}{2}\nonumber \\ \end{aligned}$$
(4.190)
For \(0 \leqslant k \leqslant 2\), \(0 \leqslant j \leqslant 1\) and \(j+k \leqslant 2\).
$$\begin{aligned} |\partial _{t}^{k}\partial _{r}^{j} v_{2,4}(t,r)| \leqslant \frac{C r^{1-j} \log ^{3}(t)\left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{t^{4+k}} , \quad r \leqslant \frac{t}{2} \end{aligned}$$
For \(0 \leqslant k+j \leqslant 2\), we have the following estimate.
$$\begin{aligned} |\partial _{r}^{k}\partial _{t}^{j}v_{2,4}(t,r)| \leqslant \frac{C \log ^{3}(t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{\sqrt{t} \langle t-r\rangle ^{5/2+k+j}}, \quad \frac{t}{2}\leqslant r < t \end{aligned}$$
(4.191)
Finally, for \(0 \leqslant k \leqslant 1\),
$$\begin{aligned} |\partial _{r}^{k}v_{2,4}(t,r)| \leqslant \frac{C}{\sqrt{r} \langle t-r \rangle ^{\frac{1}{2}+k}}, \quad r>t \end{aligned}$$
(4.192)
Proof
The proof of this lemma uses the same procedure as the proof of Lemma 26. In particular, we have
$$\begin{aligned} v_{2,4}(t,r) = \frac{-r}{2\pi } \int _{0}^{\pi } \sin ^{2}(\theta ) F_{2,4}(t+r\cos (\theta )) d\theta \end{aligned}$$
(4.193)
In the region \(r \leqslant \frac{t}{2}\), we have
$$\begin{aligned} v_{2,4}(t,r) +\frac{r}{4} F_{2,4}(t) = \frac{-r}{2\pi } \int _{0}^{\pi } \sin ^{2}(\theta ) \left( F_{2,4}(t+r\cos (\theta ))-F_{2,4}(t)\right) d\theta \end{aligned}$$
and this immediately gives rise to (4.190).
Next, we note that
$$\begin{aligned} v_{2,4}(t,r)= \int _{0}^{\infty } d\xi J_{1}(r\xi ) \sin (t\xi ) \widehat{v_{2,5}}(\xi ) \end{aligned}$$
which leads to the following estimate, for all \(r \geqslant \frac{t}{2}\).
$$\begin{aligned} \begin{aligned}|v_{2,4}(t,r)|&\leqslant C \int _{0}^{\frac{2}{r}} d\xi r \xi + C \int _{\frac{1}{r}}^{\frac{1}{100}} \frac{d\xi }{\sqrt{r\xi }} + C \int _{\frac{1}{100}}^{\infty } \frac{d\xi }{\sqrt{r\xi }} \frac{1}{\xi ^{3}}\leqslant \frac{C}{\sqrt{r}}\end{aligned} \end{aligned}$$
(4.194)
The same computation leads to
$$\begin{aligned} |\partial _{r}^{k}\partial _{t}^{j}v_{2,4}(t,r)| \leqslant \frac{C}{\sqrt{r}}, \quad r \geqslant \frac{t}{2}, \quad 0 \leqslant j+k \leqslant 2 \end{aligned}$$
On the other hand, if \(t> r > \frac{t}{2}\), we have (using (4.193))
$$\begin{aligned} |v_{2,4}(t,r)| \leqslant C r \log ^{3}(t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}\int _{0}^{\pi } \frac{\sin ^{2}(\theta )d\theta }{(t+r\cos (\theta ))^{4}} \end{aligned}$$
Recalling (4.138), the above estimate, combined with (4.194), gives (4.191). (The same procedure is used to estimate derivatives of \(v_{2,4}(t,r)\) in the region \(\frac{t}{2}< r < t\)). Finally, to establish (4.192), we use the identical procedure used to establish the analogous estimates in Lemma 5. In particular, we do the same procedure which starts with (4.47). (The amount of high frequency decay in the estimate (4.188) is sufficient for this). \(\square \)
We can now define and estimate the linear error term associated to \(u_{N_{0},corr}\). For ease of notation, let
$$\begin{aligned} h_{2}(t):=\lambda (t)\langle m_{\leqslant 1}(\frac{R \lambda (t)}{2h(t)}) e_{N_{0}}(t,R \lambda (t)),\phi _{0}(R)\rangle _{L^{2}(R dR)} \end{aligned}$$
We first recall the definition of \(u_{N_{0},corr}\) in (4.182):
$$\begin{aligned} u_{N_{0},corr}(t,r) = \left( u_{N_{0},ell}(t,r) - \frac{r h_{2}(t)}{4} \right) m_{\leqslant 1}(\frac{2 r}{t}) + v_{2,4}(t,r) \end{aligned}$$
Then, we define the linear error term associated to \(u_{N_{0},corr}\) by
$$\begin{aligned}\begin{aligned} e_{N_{0},corr}(t,r)&= -\left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}}\right) u_{N_{0},corr}(t,r)+m_{\leqslant 1}(\frac{r}{2h(t)}) e_{N_{0}}(t,r)\end{aligned} \end{aligned}$$
We therefore get
$$\begin{aligned}\begin{aligned} e_{N_{0},corr}(t,r)&= \partial _{t}^{2}\left( u_{N_{0},ell}(t,r)-\frac{r}{4}h_{2}(t)\right) m_{\leqslant 1}(\frac{2r}{t})\\&\quad - 2\partial _{t}\left( u_{N_{0,ell}}(t,r)-\frac{r}{4}h_{2}(t)\right) m_{\leqslant 1}'(\frac{2r}{t})\cdot \frac{2r}{t^{2}}\\&\quad +\left( u_{N_{0},ell}(t,r)-\frac{r}{4}h_{2}(t)\right) \left( m_{\leqslant 1}''(\frac{2r}{t})\frac{4r^{2}}{t^{4}}+\frac{4r}{t^{3}} m_{\leqslant 1}'(\frac{2r}{t})-\frac{4 m_{\leqslant 1}''(\frac{2r}{t})}{t^{2}}\right) \\&\quad -4\partial _{r}\left( u_{N_{0},ell}(t,r)-\frac{r}{4}h_{2}(t)\right) \frac{m_{\leqslant 1}'(\frac{2r}{t})}{t}-\frac{1}{r} m_{\leqslant 1}'(\frac{2r}{t}) \frac{2}{t} \left( u_{N_{0},ell}-\frac{r}{4}h_{2}(t)\right) \\&\quad +\left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) \left( v_{2,4}(t,r)-\frac{r}{4}h_{2}(t)\right) m_{\leqslant 1}(\frac{2r}{t})\\&\quad +\left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) v_{2,4}(t,r) \left( 1-m_{\leqslant 1}(\frac{2r}{t})\right) \end{aligned} \end{aligned}$$
where we used the fact that
$$\begin{aligned} (m_{\leqslant 1}(\frac{2r}{t})-1)\cdot m_{\leqslant 1}(\frac{r}{2h(t)})=0 \end{aligned}$$
We write
$$\begin{aligned} e_{N_{0},corr}(t,r) = e_{N_{0},corr,1}(t,r)+e_{N_{0},corr,2}(t,r) \end{aligned}$$
where
$$\begin{aligned} e_{N_{0},corr,2}(t,r) = \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) v_{2,4}(t,r) \left( 1-m_{\leqslant 1}(\frac{2r}{t})\right) \end{aligned}$$
(4.195)
Then, it turns out that \(e_{N_{0},corr,1}\) is perturbative, as per the following lemma.
Lemma 47
For \(k=0,1\),
$$\begin{aligned} ||L_{\frac{1}{\lambda (t)}}^{k}(e_{N_{0},corr,1})(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \lambda (t)^{2-k} \log ^{9/2}(t)}{t^{6}} \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \end{aligned}$$
Proof
This is a direct consequence of Lemma 44 and 46\(\square \)
We need to add one more term to our ansatz in order to eliminate \(e_{N_{0},corr,2}\), given in (4.195). In particular, we define \(u_{N_{0},corr,2}\) to be the solution to the following equation with 0 Cauchy data at infinity.
$$\begin{aligned} -\partial _{t}^{2}u_{N_{0},corr,2}+\partial _{r}^{2}u_{N_{0},corr,2}+\frac{1}{r}\partial _{r}u_{N_{0},corr,2}-\frac{1}{r^{2}}u_{N_{0},corr,2}=e_{N_{0},corr,2}(t,r)\nonumber \\ \end{aligned}$$
(4.196)
Then, we prove the following lemma.
Lemma 48
We have the following estimates on \(u_{N_{0},corr,2}\). For \(0 \leqslant k \leqslant 1\) and \(0 \leqslant j \leqslant 1\),
$$\begin{aligned} |\partial _{t}^{j}\partial _{r}^{k}u_{N_{0},corr,2}(t,r)| \leqslant \frac{C r^{1-k}\lambda (t)^{2} \log ^{3}(t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{ t^{5/2} \langle t-r \rangle ^{7/2+j}}, \quad r \leqslant t \nonumber \\ \end{aligned}$$
(4.197)
In addition,
$$\begin{aligned} |\partial _{t}^{2}u_{N_{0},corr,2}(t,r)|+|\partial _{r}^{2}u_{N_{0},corr,2}(t,r)| \leqslant \frac{C \lambda (t)^{2}\log ^{3}(t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{\langle t-r\rangle ^{9/2}t^{5/2}}, \quad r \leqslant t \nonumber \\ \end{aligned}$$
(4.198)
For all \(r>0\),
$$\begin{aligned} |u_{N_{0},corr,2}(t,r)| + \sqrt{E(u_{N_{0},corr,2},\partial _{t}u_{N_{0},corr,2})} \leqslant \frac{C \lambda (t)^{2} \log ^{3}(t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{t^{3}}\nonumber \\ \end{aligned}$$
(4.199)
Finally,
$$\begin{aligned} ||\partial _{r}u_{N_{0},corr,2}(t,r)||_{L^{\infty }_{r}} \leqslant \frac{C \log ^{6}(t)}{t^{\frac{5}{2}-5C_{u}}} \end{aligned}$$
(4.200)
Proof
As in Lemma 42, we have
$$\begin{aligned} u_{N_{0},corr,2}(t,r) = \frac{-r}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t}\frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1}d\beta \int _{0}^{2\pi } I_{e_{N_{0},corr,2}}(s,r\beta ,\rho ,\theta )d\theta \end{aligned}$$
where we recall the notation (2.2). From Lemma 46, we get the following estimate.
$$\begin{aligned} |I_{e_{N_{0},corr,2}}(s,r\beta ,\rho ,\theta )|\leqslant & {} C \frac{\lambda (s)^{2} \log ^{3}(s) \left( \sup _{x \in [100,s]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{s^{9/2} \langle t-r \rangle ^{7/2}}, \\{} & {} s \geqslant t \geqslant r,\quad \rho \leqslant s-t, \quad 0 \leqslant \beta \leqslant 1 \end{aligned}$$
Using (4.3), we get (4.197) for \(k=j=0\). The formulae
$$\begin{aligned} \partial _{r}u_{N_{0},corr,2}(t,r) = \frac{-1}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t} \frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{2\pi } d\theta I_{e_{N_{0},corr,2}}(s,r,\rho ,\theta )\nonumber \\ \end{aligned}$$
(4.201)
and
$$\begin{aligned} \partial _{t}u_{N_{0},corr,2}(t,r) = \frac{-r}{2\pi } \int _{t}^{\infty } ds \int _{0}^{s-t}\frac{\rho d\rho }{\sqrt{(s-t)^{2}-\rho ^{2}}} \int _{0}^{1}d\beta \int _{0}^{2\pi } \partial _{1}I_{e_{N_{0},corr,2}}(s,r\beta ,\rho ,\theta )d\theta , \end{aligned}$$
along with the same procedure used to establish (4.197) for \(k=j=0\), implies (4.197) for all larger k, j in the lemma statement. Next, we note that
$$\begin{aligned} u_{N_{0},corr,2}(t,r)= & {} \frac{-1}{2\pi } \int _{0}^{\infty } dw \int _{0}^{w} \frac{\rho d\rho }{\sqrt{w^{2}-\rho ^{2}}}\\{} & {} \int _{0}^{2\pi } \frac{e_{N_{0},corr,2}(t+w,\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )})}{\sqrt{r^{2}+\rho ^{2}+2 r \rho \cos (\theta )}} \left( r+\rho \cos (\theta )\right) d\theta . \end{aligned}$$
Differentiation under the integral sign, combined with the procedure used to prove (4.197) gives the estimate on \(\partial _{t}^{2}u_{N_{0},corr,2}\) in (4.198). We estimate \(\partial _{r}^{2}u_{N_{0},corr,2}\) by directly differentiating (4.201). Next,
$$\begin{aligned} ||e_{N_{0},corr,2}(t,r)||_{L^{2}(rdr)} \leqslant \frac{C \lambda (t)^{2} \log ^{3}(t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{t^{4}} \end{aligned}$$
and this implies (4.199). Finally, using Lemma 46, we get
$$\begin{aligned} |I_{e_{N_{0},corr,2}}(s,r,\rho ,\theta )| \leqslant \frac{C \lambda (s)^{2} \log ^{6}(s)}{s^{\frac{9}{2}-3C_{u}}}, \quad s \geqslant t, \quad r >0, \rho \leqslant s-t, \theta \in [0,2\pi ] \end{aligned}$$
and this gives
$$\begin{aligned} |\partial _{r}u_{N_{0},corr,2}(t,r)| \leqslant C \int _{t}^{\infty } ds (s-t) \frac{\log ^{6}(s)}{s^{\frac{9}{2}-5 C_{u}}} \leqslant \frac{C \log ^{6}(t)}{t^{\frac{5}{2}-5C_{u}}} \end{aligned}$$
\(\square \)
Let \(m_{\geqslant 1}:\mathbb {R}\rightarrow \mathbb {R}\) be any function satisfying
$$\begin{aligned} m_{\geqslant 1}(x) = {\left\{ \begin{array}{ll} 1, \quad x \geqslant \frac{1}{4}\\ 0, \quad x \leqslant \frac{1}{8}\end{array}\right. }, \quad m_{\geqslant 1} \in C^{\infty }(\mathbb {R}) \end{aligned}$$
(4.202)
Then, we will add \(m_{\geqslant 1}(\frac{r}{t}) u_{N_{0},corr,2}(t,r)\) to our ansatz in order to eliminate the error term \(e_{N_{0},corr,2}\). Accordingly, we define the error term, \(err_{N_{0},corr,2}\), associated to \(m_{\geqslant 1}(\frac{r}{t}) u_{N_{0},corr,2}(t,r)\), by
$$\begin{aligned}\begin{aligned} err_{N_{0},corr,2}(t,r)&=-\left( -\partial _{t}^{2}+\partial _{r}^{2}+\frac{1}{r}\partial _{r}-\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}}\right) \left( m_{\geqslant 1}(\frac{r}{t}) u_{N_{0},corr,2}(t,r)\right) \\&\quad +e_{N_{0},corr,2}(t,r) \\&=-\left( \frac{m_{\geqslant 1}''(\frac{r}{t})}{t^{2}}\left( 1-\frac{r^{2}}{t^{2}}\right) + \frac{m_{\geqslant 1}'(\frac{r}{t})}{r t}\left( 1-\frac{2r^{2}}{t^{2}}\right) \right) u_{N_{0},corr,2}(t,r)\\&-\frac{2 m_{\geqslant 1}'(\frac{r}{t})}{t}\left( \partial _{r}u_{N_{0},corr,2}(t,r)+\frac{r}{t}\partial _{t}u_{N_{0},corr,2}(t,r)\right) \\&-\left( \frac{1-\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}}\right) m_{\geqslant 1}(\frac{r}{t}) u_{N_{0},corr,2}(t,r)\end{aligned} \end{aligned}$$
where we used the fact that
$$\begin{aligned} (1-m_{\geqslant 1}(\frac{r}{t})) e_{N_{0},corr,2}(t,r) =0 \end{aligned}$$
which follows from the definition of \(m_{\geqslant 1}\), and \(e_{N_{0},corr,2}\), which we recall is given in (4.195). The following lemma shows that \(err_{N_{0},corr,2}\) is small enough to be treated with our final, perturbative argument.
Lemma 49
We have the following estimates for \(k=0,1\).
$$\begin{aligned} ||L_{\frac{1}{\lambda (t)}}^{k}(err_{N_{0},corr,2})(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \lambda (t)^{2}(1+\lambda (t)^{2}) \log ^{3}(t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{t^{6+k}} \end{aligned}$$
Proof
We directly apply Lemma 48. The only point to note is that, to estimate \(\left( \frac{1-\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{r^{2}}\right) m_{\geqslant 1}(\frac{r}{t}) u_{N_{0},corr,2}(t,r)\), we use (4.197) in the region \(r \leqslant t-t^{3/7}\) and (4.199) for the region \(r \geqslant t-t^{3/7}\). \(\square \)
4.10 Second set of nonlinear interactions
The next step is to treat the nonlinear interactions between the newest addition to our ansatz, namely
$$\begin{aligned} u_{n}(t,r):=u_{N_{0}}(t,r)+u_{N_{0},corr}(t,r)+m_{\geqslant 1}(\frac{r}{t}) u_{N_{0},corr,2}(t,r), \end{aligned}$$
(4.203)
and \(u_{a}\). The interactions between \(v_{2,4}\) (part of \(u_{N_{0},corr}\)) and itself, along with the interactions between \(v_{2,4}\) and \(u_{a,0}\) are not quite perturbative, but the rest of the nonlinear interactions are. In order to show this, we note that
$$\begin{aligned}\begin{aligned} \sin (2Q_{1}(\frac{r}{\lambda (t)})+2u_{a}+2u_{n})&= 2r^{2}\left( N_{0}+N_{1}+N_{2}+N_{3}\right) \\&+\cos (2 Q_{1}(\frac{r}{\lambda (t)}))\cdot 2 (u_{a}+u_{n})+\sin (2 Q_{1}(\frac{r}{\lambda (t)}))\end{aligned} \end{aligned}$$
where \(N_{0}\) is defined in (4.172), \(N_{1}\) is defined in (4.169), and we define
$$\begin{aligned} \begin{aligned}N_{2}(t,r):&=\frac{\cos (2 Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} (-2 (u_{a0}+v_{2,4})+\sin (2 (u_{a,0}+v_{2,4}))-(\sin (2 u_{a,0})-2 u_{a,0}))\\&\quad +\frac{\sin (2 Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} (\cos (2 u_{a,0}+2 v_{2,4})-\cos (2 u_{a,0}))\\ \end{aligned}\nonumber \\ \end{aligned}$$
(4.204)
We therefore have
$$\begin{aligned} \begin{aligned}N_{3}(t,r)&= \frac{\sin (2 Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \cos (2 u_{a}+2 u_{n})-\cos (2 u_{a})-\left( \cos (2 u_{a,0}+2 v_{2,4})-\cos (2 u_{a,0})\right) \right) \\&\quad +\frac{\cos (2 Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \sin (2 u_{a}+2 u_{n})-2 (u_{a}+u_{n})-(\sin (2 u_{a})-2 u_{a})\right. \\&\quad \qquad \qquad \qquad \qquad \left. -(\sin (2 u_{a,0}+2 v_{2,4})-2 (u_{a,0}+v_{2,4}))+\sin (2u_{a,0})-2 u_{a,0}\right) \end{aligned} \end{aligned}$$
where we recall the definition of \(u_{a}\), from (4.168). We define \(u_{new}\) by
$$\begin{aligned}\begin{aligned} u_{new}(t,r):&=u_{n}(t,r)-v_{2,4}(t,r)\end{aligned} \end{aligned}$$
We remark that \(N_{2}\) contains the interactions between \(v_{2,4}\) and \(u_{a,0}\), except for the \(u_{a,0}\) self-interactions, which were already contained in \(N_{0}\). To give the reader an idea of how we estimate \(N_{3}\), we re-write its expression as follows.
$$\begin{aligned}{} & {} N_{3}(t,r)\nonumber \\{} & {} \quad =\frac{\cos (2 Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( (\cos (2 u_{a})-1) ((\cos (2 u_{new})-1) \sin (2 v_{2,4}))\right. \nonumber \\{} & {} \qquad \qquad \qquad \qquad \qquad \left. +(\cos (2 u_{new}+2 v_{2,4})-1) (\sin (2 u_{a,0}) (\cos (2 u_{a,1})-1))\right. \nonumber \\{} & {} \qquad \qquad \qquad \qquad \qquad \left. +\sin (2 v_{2,4}) (\cos (2 u_{a,0}) (\cos (2 u_{a,1})-1))\right. \nonumber \\{} & {} \qquad \qquad \qquad \qquad \qquad \left. +\sin (2 u_{a,0}) ((\cos (2 u_{new})-1) \cos (2 v_{2,4}))\right. \nonumber \\{} & {} \qquad \qquad \qquad \qquad \qquad \left. +(\cos (2 u_{new})-1) \sin (2 v_{2,4})-2 u_{new}+\sin (2 u_{new})\right) \nonumber \\{} & {} \qquad +\frac{\sin (2 Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( \cos (2 (u_{a}+ u_{n}))-\cos (2 u_{a})-\cos (2 u_{a,0}+2 v_{2,4})+\cos (2 u_{a,0})\right) \nonumber \\{} & {} \qquad +\frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))}{2r^{2}} \left( (\cos (2 u_{a})-1) \sin (2 u_{new}) \cos (2 v_{2,4})\right. \nonumber \\{} & {} \qquad \qquad \qquad \qquad \qquad +(\cos (2 (u_{new}+ v_{2,4}))-1) \cos (2 u_{a,0}) \sin (2 u_{a,1})\nonumber \\{} & {} \qquad \qquad \qquad \qquad \qquad \left. -\sin (2 v_{2,4}) \sin (2 u_{a,0}) \sin (2 u_{a,1})-\sin (2 u_{a,0}) \sin (2 u_{new}) \sin (2 v_{2,4})\right. \nonumber \\{} & {} \qquad \qquad \qquad \qquad \qquad \left. +\sin (2 u_{new}) (\cos (2 v_{2,4})-1)\right) \end{aligned}$$
(4.205)
Lemma 50
We have the following estimates.
$$\begin{aligned}\begin{aligned}||\frac{N_{3}(t,r)}{\lambda (t)^{2}}||_{L^{2}(r dr)} \leqslant \frac{C \log ^{24}(t)}{t^{9/2-15C_{u}-2C_{l}}}, \quad ||L_{\frac{1}{\lambda (t)}}(N_{3}(t,r))||_{L^{2}(r dr)} \leqslant \frac{C \log ^{24}(t)}{t^{9/2-15C_{u}-C_{l}}}\end{aligned} \end{aligned}$$
Proof
We note that, as per (4.202),
$$\begin{aligned} 1-m_{\geqslant 1}(\frac{r}{2t})=0, \quad r \geqslant \frac{t}{2} \end{aligned}$$
Using Lemma 42, Lemma 44, and Lemma 48, we get
$$\begin{aligned} |u_{new}(t,r)| \leqslant {\left\{ \begin{array}{ll} \frac{C r \log ^{4}(t) \left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3}}{t^{4}}, \quad r \leqslant \frac{t}{2}\\ \frac{\left( \sup _{x \in [100,t]}\left( \lambda (x) \log (x)\right) \right) ^{3} \left( \sup _{x \in [100,t]}\left( \lambda (x)^{2}\right) \right) ^{3} \log ^{9}(t)}{t^{2}}, \quad \frac{t}{2} \leqslant r\end{array}\right. }\qquad \end{aligned}$$
(4.206)
We then straightforwardly estimate the expression (4.205), for \(N_{3}\) (as well as its r derivative), using (4.171), (4.170), and Lemma 46 (to estimate \(u_{a,1}\), \(u_{a,0}\), and \(v_{2,4}\), respectively). \(\square \)
Next, we add a correction, \(u_{N_{2}}\) to improve the error term \(N_{2}\), defined in (4.204). In particular, we define \(u_{N_{2}}\) to be the solution to the following equation with 0 Cauchy data at infinity.
$$\begin{aligned} -\partial _{t}^{2}u_{N_{2}}+\partial _{r}^{2}u_{N_{2}}+\frac{1}{r}\partial _{r}u_{N_{2}}-\frac{u_{N_{2}}}{r^{2}}=N_{2}(t,r) \end{aligned}$$
(4.207)
Then, we have the following lemma.
Lemma 51
For \(0 \leqslant k \leqslant 1\), and \(r \leqslant t\)
$$\begin{aligned}\begin{aligned}&r^{k}|\partial _{r}^{k} u_{N_{2}}(t,r)|\\&\leqslant C r (r+\lambda (t)) \left( \frac{\log ^{7}(t)}{t^{5-3C_{u}}}\right) + \frac{C r \lambda (t)}{t^{2}\langle t-r \rangle ^{4}}\left( t^{4C_{u}}\log ^{8}(t)+\frac{t^{6C_{u}} \log ^{12}(t)}{\langle t-r \rangle ^{2}}\right) \\&\quad +\frac{C r \log ^{10}(t) t^{5C_{u}}}{\langle t-r \rangle ^{9/2} t^{3/2}} \left( 1+\frac{t^{4C_{u}} \log ^{8}(t)}{\langle t-r \rangle ^{4}}\right) \end{aligned} \end{aligned}$$
In addition, for all \(r>0\),
$$\begin{aligned}\begin{aligned}&|u_{N_{2}}(t,r)| + \sqrt{E(u_{N_{2}},\partial _{t}u_{N_{2}})}\leqslant \frac{C \log ^{18}(t)}{t^{2-9C_{u}}}\end{aligned} \end{aligned}$$
Finally,
$$\begin{aligned} ||\partial _{r}u_{N_{2}}(t,r)||_{L^{\infty }_{r}} \leqslant \frac{C \log ^{18}(t)}{t^{\frac{3}{2}-9C_{u}}} \end{aligned}$$
(4.208)
Proof
The lemma is proven with the same procedure as in Lemma 42, using the estimates from Lemma 46 and (4.170). \(\square \)
It only remains to estimate the linear error term associated to \(u_{N_{2}}(t,r)\) as well as its nonlinear interactions with all of the previous terms added into our ansatz. We start with the linear error term, which we denote by \(e_{N_{2}}\):
$$\begin{aligned} e_{N_{2}}(t,r):= \left( \frac{\cos (2Q_{1}(\frac{r}{\lambda (t)}))-1}{r^{2}}\right) u_{N_{2}}(t,r) \end{aligned}$$
By directly estimating \(e_{N_{2}}\) using the estimates from Lemmas 51, we get the following lemma.
Lemma 52
For \(k=0,1\),
$$\begin{aligned}\begin{aligned}&||L_{\frac{1}{\lambda (t)}}^{k}(e_{N_{2}})(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \log ^{18}(t)}{t^{5-11C_{u}-kC_{l}}}\end{aligned} \end{aligned}$$
4.11 Third set of nonlinear interactions
Finally, it remains to treat the nonlinear interactions between \(u_{N_{2}}\) and all of the previous terms of our ansatz (the sum of which is equal to \(u_{a}+u_{n}\)). I.e., it suffices to estimate \(N_{4}\) given by
$$\begin{aligned}\begin{aligned}&\sin (2Q_{1}(\frac{r}{\lambda (t)})+2u_{a}+2u_{n}+2u_{N_{2}})\\&\quad = 2r^{2}\left( N_{0}+N_{1}+N_{2}+N_{3}\right) +\cos (2Q_{1}(\frac{r}{\lambda (t)}))\cdot 2(u_{a}+u_{n}+u_{N_{2}}) \\&\qquad + \sin (2Q_{1}(\frac{r}{\lambda (t)}))+2r^{2}N_{4}(t,r)\end{aligned} \end{aligned}$$
So,
$$\begin{aligned}\begin{aligned} N_{4}(t,r)&=\frac{1}{2r^{2}} \left( \sin (2Q_{1}(\frac{r}{\lambda (t)})+2u_{a}+2u_{n})(\cos (2u_{N_{2}})-1)\right. \\&\quad +\cos (2Q_{1}(\frac{r}{\lambda (t)}))(\sin (2u_{N_{2}})-2u_{N_{2}})\\&\quad +\left. \left( \cos (2Q_{1}(\frac{r}{\lambda (t)})+2u_{a}+2u_{n})-\cos (2Q_{1}(\frac{r}{\lambda (t)}))\right) \sin (2u_{N_{2}})\right) \end{aligned} \end{aligned}$$
Lemma 53
$$\begin{aligned}\begin{aligned}&\lambda (t)^{-2} ||N_{4}(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \log ^{30}(t)}{t^{\frac{9}{2}-2C_{l}-15C_{u}}}, \quad ||L_{\frac{1}{\lambda (t)}}(N_{4})(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \log ^{30}(t)}{t^{\frac{9}{2}-15C_{u}-C_{l}}}\end{aligned} \end{aligned}$$
Proof
We have
$$\begin{aligned} |N_{4}(t,r)| \leqslant \frac{C}{r^{2}} \left( \left( \frac{r \lambda (t)}{(r^{2}+\lambda (t)^{2})} |u_{N_{2}}|+u_{N_{2}}^{2}\right) (|u_{N_{2}}|+|u_{a}|+|u_{n}|)+(u_{a}^{2}+u_{n}^{2}) |u_{N_{2}}|\right) \end{aligned}$$
where we recall the definitions of \(u_{a}\) and \(u_{n}\) in (4.168), and (4.203), respectively. We then use Lemmas 51 and 46, and estimates (4.170), (4.171), and (4.206), to estimate \(u_{N_{2}}\) and the various terms in \(u_{a}\) and \(u_{n}\), and then straightforwardly estimate \(N_{4}\), and \(L_{\frac{1}{\lambda (t)}}N_{4}\). \(\square \)
We recall that \(u_{a}\) is defined in (4.167), \(u_{n}\) is defined in (4.203), and \(u_{N_{2}}\) is defined in (4.207), and define our final ansatz by the following.
$$\begin{aligned} u_{ansatz}(t,r):=u_{a}(t,r)+u_{n}(t,r)+u_{N_{2}}(t,r) \end{aligned}$$
(4.209)
The error term of \(u_{ansatz}\), denoted by \(F_{5}\), is equal to the following.
$$\begin{aligned} F_{5}(t,r) = e_{N_{0}}\left( 1-m_{\leqslant 1}(\frac{r}{2h(t)})\right) +e_{N_{0},corr,1}+err_{N_{0},corr,2}+N_{1}+N_{3}+e_{N_{2}}+N_{4}+e_{a}\nonumber \\ \end{aligned}$$
(4.210)
Lemma 54
[Estimates on the error of \(u_{ansatz}\)] There exists \(C_{5}>0\) such that, if
$$\begin{aligned} \delta := \frac{1}{2}\text {min}\{4\alpha -C_{l}-2,1-2C_{u}-\alpha ,4-5C_{u}-6\alpha ,\frac{1}{2}-2C_{l}-15C_{u}\} \end{aligned}$$
then, \(2\delta >C_{l}\), and
$$\begin{aligned}{} & {} \frac{||F_{5}(t,r)||_{L^{2}(r dr)}}{\lambda (t)^{2}} \leqslant \frac{C_{5} \log ^{30}(t)}{t^{4+2\delta }}, \\{} & {} ||L_{\frac{1}{\lambda (t)}}F_{5}(t,r)||_{L^{2}(r dr)} \leqslant \frac{C \log ^{2}(t)}{t^{2+5\alpha }}+ \frac{C \log ^{6}(t)}{t^{4+2\alpha -3C_{u}}}+\frac{C \log ^{30}(t)}{t^{9/2-C_{l}-15C_{u}}} \end{aligned}$$
Proof
The inequality \(2\delta > C_{l}\) follows from (1.7) and (4.2). The ith term in (4.210) is estimated in the ith Lemma in the following list: Lemma 43, 47, 49, 41, 50, 52, 53, 40. \(\square \)
We finish this section with some estimates on various quantities involving \(u_{ansatz}\).
Lemma 55
We have the following estimates.
$$\begin{aligned}{} & {} \begin{aligned} ||\frac{8 \lambda (t) u_{ansatz}(t,r)}{r(r^{2}+\lambda (t)^{2})}||_{L^{\infty }_{r}}&\leqslant \frac{4}{t^{2}} \left( M^{2}+\frac{C_{2}}{3} \left( 3+2\pi ^{2}\right) \right) + \frac{C \log (t)}{t^{2+2\alpha }} + \frac{C \log ^{6}(t)}{t^{\frac{7}{2}-4C_{u}}} \end{aligned}\nonumber \\ \qquad \end{aligned}$$
(4.211)
$$\begin{aligned}{} & {} ||\frac{u_{ansatz}(t,r)}{r}||_{L^{\infty }_{r}} \leqslant \frac{C \log ^{6}(t)}{t^{\frac{3}{2}-3C_{u}}} \nonumber \\{} & {} ||\partial _{r}u_{ansatz}(t,r)||_{L^{\infty }_{r}} \leqslant \frac{C \log ^{6}(t)}{t^{\frac{1}{2}-3C_{u}}} \end{aligned}$$
(4.212)
$$\begin{aligned}{} & {} ||\frac{u_{ansatz}(t,r) \partial _{r}u_{ansatz}(t,r)}{r}||_{L^{\infty }_{r}(\{r \leqslant \lambda (t)\})} \leqslant \frac{C \log ^{7}(t)}{t^{\frac{5}{2}-4C_{u}}} \end{aligned}$$
(4.213)
$$\begin{aligned}{} & {} ||\frac{u_{ansatz}(t,r)\partial _{r}u_{ansatz}(t,r)}{r^{2}}||_{L^{\infty }_{r}(\{r \geqslant \lambda (t)\})} \leqslant \frac{C \log ^{7}(t)}{t^{\frac{5}{2}-3C_{u}}}\\{} & {} \lambda (t) ||\frac{\partial _{r}u_{ansatz}(t,r)}{r^{2}+\lambda (t)^{2}}||_{L^{\infty }_{r}} \leqslant \frac{1}{2t^{2}} \left( M^{2}+\frac{C_{2}}{3}(3+2\pi ^{2})\right) +\frac{C \log ^{6}(t)}{t^{\frac{5}{2}-4C_{u}}} \nonumber \end{aligned}$$
(4.214)
Proof
We recall (4.7), and note that
$$\begin{aligned} \frac{8 \lambda (t)u_{ell}(t,R\lambda (t))}{R\lambda (t)^{3}(R^{2}+1)} = \frac{4}{(1+R^{2})} \left( \frac{\lambda '(t)}{\lambda (t)}\right) ^{2} + \frac{\lambda ''(t)}{\lambda (t)} m_{1}(R) \end{aligned}$$
where
$$\begin{aligned} m_{1}(R) = \frac{8 \text {Li}_2\left( -R^2\right) +\frac{8}{3} \left( 3 R^2+\pi ^2\right) -\frac{4 \left( R^4-1\right) \log \left( R^2+1\right) }{R^2}}{(1+R^{2})^{2}}:=\frac{num_{1}(R)}{(1+R^{2})^{2}} \end{aligned}$$
We start with
$$\begin{aligned} num_{1}'(R) = \frac{8(1+R^{2})(R^{2}-(1+R^{2})\log (1+R^{2}))}{R^{3}} \end{aligned}$$
Then, since
$$\begin{aligned} \partial _{R}\left( R^{2}-(1+R^{2})\log (1+R^{2})\right) = -2 R \log (1+R^{2}) \leqslant 0 \end{aligned}$$
and
$$\begin{aligned} \left( R^{2}-(1+R^{2})\log (1+R^{2})\right) \Bigr |_{R=0}=0 \end{aligned}$$
we have
$$\begin{aligned} R^{2}-(1+R^{2})\log (1+R^{2}) \leqslant 0, \quad R \geqslant 0 \end{aligned}$$
Therefore,
$$\begin{aligned} num_{1}'(R) \leqslant 0, \quad R >0 \end{aligned}$$
So,
$$\begin{aligned} num_{1}(R) \geqslant num_{1}(1) = 2\pi ^{2}+8 > 0, \quad 0 < R \leqslant 1 \end{aligned}$$
Therefore,
$$\begin{aligned} R \mapsto \frac{num_{1}(R)}{(1+R^{2})^{2}}=m_{1}(R) \text { is non-increasing on }(0,1] \end{aligned}$$
Hence, for \(0 \leqslant R \leqslant 1\),
$$\begin{aligned} |m_{1}(R)| \leqslant \lim _{R \rightarrow 0^{+}} m_{1}(R) = \frac{4(3+2\pi ^{2})}{3} \end{aligned}$$
If \(R \geqslant 1\), then,
$$\begin{aligned}\begin{aligned} |num_{1}(R)|&\leqslant |num_{1}(1)| + 8 \int _{1}^{R} \frac{(1+s^{2})(s^{2}+(1+s^{2})\log (1+s^{2}))}{s^{3}} ds\\&\leqslant 2 \left( 2+\pi ^{2}+2R^{2}+2\log (1+R^{2})+2(R^{2}+2 \log (1+R^{2}))\log (1+R^{2})\right) \\ {}&:=m_{2}(R)\end{aligned} \end{aligned}$$
If \(1 \leqslant R \leqslant 2\), then,
$$\begin{aligned} \frac{d}{dR}\left( \frac{m_{2}(R)}{(1+R^{2})^{2}}\right){} & {} = \frac{-8R (\pi ^{2}+\log (1+R^{2})(-3+R^{2}+4 \log (1+R^{2})))}{(1+R^{2})^{3}}\\{} & {} \quad \leqslant \frac{R}{(1+R^{2})^{3}} \left( -8\pi ^{2}+24\log (5)\right) <0 \end{aligned}$$
Therefore, for \(1 \leqslant R \leqslant 2\),
$$\begin{aligned} |m_{1}(R)| \leqslant \frac{|num_{1}(R)|}{(1+R^{2})^{2}} \leqslant \frac{m_{2}(R)}{(1+R^{2})^{2}} \leqslant \frac{m_{2}(1)}{2^{2}} = 2+\frac{\pi ^{2}}{2}+2 \log ^{2}(2) +\log (4) \end{aligned}$$
Finally, if \(R \geqslant 2\), then,
$$\begin{aligned} |m_{1}(R)| \leqslant \frac{m_{2}(R)}{(1+R^{2})^{2}} \leqslant \frac{4}{5}+\frac{2}{25} \left( 2+\pi ^2\right) +\frac{8 \log ^2(5)}{25}+\frac{4 \log (5)}{5}+\frac{4 \log (5)}{25}\qquad \end{aligned}$$
(4.215)
This gives (4.211). To prove (4.212), we estimate \(\partial _{r}u_{a}(t,r)\) using the procedure detailed in the proof of Lemma 41. We then estimate \(\partial _{r}u_{n}(t,r)\) using (4.177), (4.200), (4.206) and Lemma 46. Finally, we estimate \(\partial _{r}u_{N_{2}}\) using (4.208). With the same procedure, we also get (4.213) and (4.214). Next,
$$\begin{aligned} \frac{\partial _{r}u_{ell}(t,r)}{r^{2}+\lambda (t)^{2}}\Bigr |_{r=R\lambda (t)} = \frac{\lambda '(t)^{2}}{2(1+R^{2})\lambda (t)^{3}}+\frac{\lambda ''(t)}{\lambda (t)^{2}} m_{3}(R) \end{aligned}$$
for
$$\begin{aligned} m_{3}(R)= & {} \frac{1}{(1+R^{2})^{3}}\left( 1+3R^{2}+\frac{\pi ^{2}}{3}(1-R^{2})\right. \\{} & {} \left. -\frac{(1+7R^{2}+7R^{4}+R^{6})\log (1+R^{2})}{2R^{2}} +(1-R^{2})\text {Li}_{2}(-R^{2})\right) \end{aligned}$$
With a similar procedure as that used to obtain (4.215), and its analogs for other regions of R, we get
$$\begin{aligned} |m_{3}(R)| \leqslant \frac{3+2\pi ^{2}}{6}, \quad R> 0 \end{aligned}$$
\(\square \)
The proof of our main proposition of this section, Proposition 1, is finished, once we establish (4.4). Let
$$\begin{aligned} v_{rad}(t,r)=v_{2}(t,r)+v_{2,2}(t,r)+v_{2,4}(t,r) \end{aligned}$$
(4.216)
We first note that
$$\begin{aligned} \begin{aligned}u_{ansatz}(t,r) - v_{rad}&= \chi _{\leqslant 1}(\frac{r}{\lambda (t) g(t)})(u_{ell}+u_{ell,2}) + (1-\chi _{\leqslant 1}(\frac{r}{\lambda (t) g(t)}))(w_{1}+v_{ex}+u_{w,2})\\&\quad -\chi _{\leqslant 1}(\frac{r}{\lambda (t)g(t)}) (v_{2}+v_{2,2}) + f_{5,0}+f_{ex,sub}+f_{ell,2}+f_{2,2}+u_{N_{0}}\\&\quad +(u_{N_{0},ell}-\frac{r\lambda (t)}{4} \langle m_{\leqslant 1}(\frac{R \lambda (t)}{2h(t)}) e_{N_{0}}(t,R\lambda (t)),\phi _{0}(R)\rangle _{L^{2}(R dR)})m_{\leqslant 1}(\frac{2r}{t})\\&\quad +m_{\geqslant 1}(\frac{r}{t}) u_{N_{0},corr,2}+u_{N_{2}}\end{aligned} \end{aligned}$$
Since, for non-constant \(\lambda \), \(\partial _{t}Q_{\frac{1}{\lambda (t)}}(r), \partial _{t}\left( \chi _{\geqslant 1}(\frac{r}{g(t)\lambda (t)})w_{1}(t,r)\right) \not \in L^{2}(r dr)\), the most delicate term to estimate will be the following, which is finite.
$$\begin{aligned} ||\partial _{t}\left( Q_{\frac{1}{\lambda (t)}}(r)+\chi _{\geqslant 1}(\frac{r}{g(t)\lambda (t)})w_{1}(t,r)\right) ||_{L^{2}(r dr)} \end{aligned}$$
(4.217)
We note that
$$\begin{aligned} \chi _{\geqslant 1}(\frac{r}{\lambda (t)g(t)}) =1-\chi _{\leqslant 1}(\frac{r}{\lambda (t)g(t)}) =1, \quad r \geqslant 2\lambda (t) g(t) \end{aligned}$$
Therefore, we will show in detail how to estimate (4.217), when the \(L^{2}\) norm is taken over the region \(r \geqslant 2g(t)\lambda (t)\), since the norm in the region \(r \leqslant 2 g(t)\lambda (t)\) is a direct estimation, using Lemma 2. From (4.24),
$$\begin{aligned} w_{1}(t,r) = -\frac{2}{r} \left( \lambda (t+r)-\lambda (t)-r \lambda '(t+r)\right) +2 r \int _{1}^{\infty } \lambda ''(t+r y) (y-\sqrt{y^{2}-1}) dy \nonumber \\ \end{aligned}$$
(4.218)
and
$$\begin{aligned} Q_{1}(\frac{r}{\lambda (t)}) = \pi -2\arctan (\frac{\lambda (t)}{r}) \end{aligned}$$
Therefore,
$$\begin{aligned}\begin{aligned} Q_{1}(\frac{r}{\lambda (t)})+w_{1}(t,r)&= \pi -2\arctan (\frac{\lambda (t)}{r})+2\frac{\lambda (t)}{r}\\&\quad -\frac{2}{r}\left( \lambda (t+r)\!-\!r\lambda '(t\!+\!r)\right) \!+\!2 r \int _{1}^{\infty } \lambda ''(t\!+\!ry) (y-\sqrt{y^{2}-1}) dy\end{aligned} \end{aligned}$$
Therefore,
$$\begin{aligned}\begin{aligned}\partial _{t}\left( Q_{1}(\frac{r}{\lambda (t)})+w_{1}(t,r)\right)&=\frac{2 \lambda '(t)}{r} \left( \frac{\lambda (t)^{2}}{r^{2}+\lambda (t)^{2}}\right) - \frac{2}{r}\left( \lambda '(t+r)-r\lambda ''(t+r)\right) \\&\quad +2 r \int _{1}^{\infty } \lambda '''(t+r y) (y-\sqrt{y^{2}-1}) dy\end{aligned} \end{aligned}$$
and this gives
$$\begin{aligned} ||\partial _{t}\left( Q_{1}(\frac{r}{\lambda (t)})+\chi _{\geqslant 1}(\frac{r}{\lambda (t)g(t)}) w_{1}(t,r)\right) ||_{L^{2}(r \geqslant 2 \lambda (t) g(t), r dr)}^{2} \leqslant \frac{C \log ^{2}(t)}{t^{2-2C_{u}}} \end{aligned}$$
We complete the proof of (4.4) by using (4.7) and (4.81) (for \(u_{ell}\)), (4.9), and the formulae following it, for \(u_{ell,2}\), Lemma 2 (for \(w_{1}\)), Lemma 29, for \(v_{ex}\), Lemma 10, for \(u_{w,2}\), Lemma 5 for \(v_{2}\), Lemma 26 for \(v_{2,2}\), Lemma 33, for \(f_{5,0}\), Lemma 35, for \(f_{ex,sub}\), Lemma 37, for \(f_{ell,2}\), Lemma 38, for \(f_{2,2}\), Lemma 42, for \(u_{N_{0}}\), Lemma 44, Lemma 48 for \(u_{N_{0},corr,2}\), Lemma 51 for \(u_{N_{2}}\).
