Time-Asymptotic Expansion with Pointwise Remainder Estimates for 1D Viscous Compressible Flow

Abstract

We construct a time-asymptotic expansion with pointwise remainder estimates for solutions to 1D compressible Navier–Stokes equations. The leading-order term is the well-known diffusion wave and the higher-order terms are a newly introduced family of waves which we call higher-order diffusion waves. In particular, these provide an accurate description of the power-law asymptotics of the solution around the origin $x=0$, where the diffusion wave decays exponentially. The expansion is valid locally and also globally in the $L^p({\mathbb {R}})$-norm for all $1\le p\le \infty$. The proof is based on pointwise estimates of Green’s function.

1 Introduction

The equations

$$\begin{aligned} \left\{ \begin{array}{ll} v_t-u_x=0, &{} x\in {\mathbb {R}},\, t>0, \\ u_t+p(v)_x=\nu (u_x/v)_x, &{} x\in {\mathbb {R}},\, t>0, \\ v(x,0)=v_0(x),\, u(x,0)=u_0(x), &{} x\in {\mathbb {R}} \end{array}\right. \end{aligned}$$

(1)

describe the motion of a 1D viscous compressible flow. Here v(x, t) is the specific volume (the reciprocal of the density $\rho$) and u(x, t) is the flow velocity; t is the time and x is the Lagrangian mass coordinate related to the Eulerian coordinate X by $x=\int _{X_0(t)}^{X}\rho (X',t)\, dX'$, where $X_0(t)$ is the trajectory of a particle moving with the fluid and initially placed at $X_0(0)=0$. The system above models barotropic flow, that is, the pressure p(v) does not depend on the temperature. We assume that $p'(v)<0$ and $p''(v)\ne 0$ for $v>0$ and that the viscous coefficient $\nu$ is a positive constant. The system is often called the p-system and is a typical example of quasilinear hyperbolic–parabolic viscous conservation laws.

The purpose of this paper is to construct a time-asymptotic expansion of the solution to (1) together with pointwise estimates for the remainder. We shall consider solutions close to the steady state $(v_S,u_S)\equiv (1,0)$. To study the long-time asymptotics of such solutions, it is convenient to consider

$$\begin{aligned} u_1=\frac{p''(1)}{4c}[-(v-1)+u/c], \quad u_2=\frac{p''(1)}{4c}[(v-1)+u/c] \end{aligned}$$

(2)

instead of (v, u). Here $c=\sqrt{-p'(1)}$ is the speed of sound for the state $(v_S,u_S)$.

It is well-known that $u_i$ has the diffusion wave $\theta _i$ as its asymptotic profile. Here $\theta _i$ ($i=1,2$) is the self-similar solution to the convective viscous Burgers equation

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _t \theta _i+\lambda _i \partial _x \theta _i+\partial _x(\theta _{i}^{2}/2)=\frac{\nu }{2}\partial _{x}^{2}\theta _i, &{} x\in {\mathbb {R}},t>0, \\ \lim _{t\searrow -1}\theta _i(x,t)=M_i \delta (x), \quad x\in {\mathbb {R}}, \end{array}\right. \end{aligned}$$

(3)

where $\lambda _i=(-1)^{i-1}c$, $M_i=\int _{-\infty }^{\infty }u_i(x,0)\, dx$, and $\delta (x)$ is the Dirac delta function. An explicit formula for $\theta _i$ is available through the use of Cole–Hopf transformation:

$$\begin{aligned} \theta _i(x,t)=\frac{\sqrt{\nu }\left( e^{\frac{M_i}{\nu }}-1 \right) }{\sqrt{2(t+1)}}e^{-\frac{(x-\lambda _i(t+1))^2}{2\nu (t+1)}}\left[ \sqrt{\pi }+\left( e^{\frac{M_i}{\nu }}-1 \right) \int _{\frac{x-\lambda _i(t+1)}{\sqrt{2\nu (t+1)}}}^{\infty }e^{-y^2}\, dy \right] ^{-1}.\nonumber \\ \end{aligned}$$

(4)

The diffusion wave $\theta _i$ describes the leading-order asymptotics in the $L^p({\mathbb {R}})$-norm. In fact, we have the following optimal decay estimates [14]:

$$\begin{aligned} \Vert \theta _i(\cdot ,t) \Vert _{L^p}\lesssim t^{-(1-1/p)/2} \quad \text {and} \quad \Vert (u_i-\theta _i)(\cdot ,t) \Vert _{L^p}\lesssim t^{-(3/2-1/p)/2} \quad (1\le p\le \infty ). \end{aligned}$$

The key to proving the $L^p$-decay estimates above—especially for $p=1$—is the pointwise estimates for Green’s function of the linearization of (1) around $(v_S,u_S)$. These, in fact, allow us to obtain pointwise estimates for the solution itself [10]:

$$\begin{aligned} |(u_i-\theta _i)(x,t)|{} & {} \lesssim [(x-\lambda _i(t+1))^2+(t+1)]^{-3/4}\nonumber \\{} & {} \quad +[|x+\lambda _i(t+1)|^3+(t+1)^2]^{-1/2}. \end{aligned}$$

(5)

The $L^p$-decay estimates are obtained by integrating this.

Pointwise estimates (5) allow us to deduce not just global $L^p$-estimates but also local ones. In particular, we have $|(u_i-\theta _i)(x,t)|\lesssim t^{-3/4}$ for $x=\lambda _i t+O(1)$. Since $\theta _i(x,t)\lesssim t^{-1/2}$ for $x=\lambda _i t+O(1)$, the diffusion wave $\theta _i$ also describes the leading-order asymptotics locally around the characteristic line $x=\lambda _i t$. However, the situation is different around the origin $x=0$. As can be seen from (4), the diffusion wave $\theta _i$ decays exponentially fast around the origin $x=0$ but (5) implies $|(u_i-\theta _i)(x,t)|\lesssim t^{-3/2}$ for $x=O(1)$. Thus the diffusion wave $\theta _i$ provides almost no information about the long-time asymptotics around $x=0$; we need new waves to capture the asymptotic behavior there.

In [12], van Baalen, Popović, and Wayne constructed a time-asymptotic expansion of $u_i$ in an $L^2$-framework. The leading-order term of the expansion is the diffusion wave $\theta _i$ but the first higher-order term beyond $\theta _i$ turns out to be a wave decaying algebraically as $t^{-3/2}$ around the origin. It is then natural to expect that this new wave captures the leading-order asymptotics of the flow around $x=0$. However, the decay estimate for the remainder of the expansion is given in the $H^1({\mathbb {R}})$-norm. This implies only a far from optimal decay estimate around $x=0$. For this reason, we cannot conclude that the higher-order term describes the leading-order asymptotics of the flow around $x=0$.

To overcome this issue, we construct a time-asymptotic expansion of $u_i$ with pointwise estimates for the remainder. The leading-order term is the diffusion wave $\theta _i$ and the higher-order terms are higher-order diffusion waves $\xi _{i;n}$ ($n\ge 1$) defined in the next section. It turns out that $|\xi _{i;n}(x,t)|\lesssim t^{-(2-1/2^n)}$ for $x=O(1)$ as $t\rightarrow \infty$. Setting $n=1$, we see that $|\xi _{i;1}(x,t)|\lesssim t^{-3/2}$ for $x=O(1)$. The pointwise estimates for the remainder imply $|(u_i-\xi _{i;1})(x,t)|\lesssim t^{-7/4}$ for $x=O(1)$, thus it is rigorously proved that $\xi _{i;1}$ describes the leading-order asymptotics of $u_i$ for $x=O(1)$. In addition, thanks to the pointwise estimates, our asymptotic expansion is valid not only in the $L^2({\mathbb {R}})$-norm but also in the $L^1({\mathbb {R}})$-norm.

The proof is based on pointwise estimates of Green’s function, and the basic strategy follows that of [10]. The most non-trivial part of the proof is perhaps the definition of the higher-order diffusion waves $\xi _{i;n}$ ($n\ge 1$); see (7). Although the differential equation defining $\xi _{i;n}$ does not seem to have a simple solution formula such as (4),¹ we use its structure (by the help of Lemma A.1) to analyze cancellation effects which are crucial in nonlinear estimates; see the proof of Lemma 3.6.

Before concluding the introduction, we briefly comment on related works. Diffusion wave approximations and pointwise estimates of solutions has been extensively studied for hyperbolic–parabolic systems [10], hyperbolic–elliptic systems [3], hyperbolic balance laws [13, 15], the Boltzmann equation [8], and so on. In these works, nonlinear diffusion waves similar to $\theta _i$ were constructed and pointwise estimates of solutions were obtained. However, to the best of our knowledge, time-asymptotic expansions with pointwise estimates have not been obtained previously. We mention that the author already analyzed the second-order term $\xi _{i;1}$ in connection with a fluid–structure interaction problem in [7]; the complete asymptotic expansion, however, was not given. We also comment that for multidimensional incompressible Navier–Stokes equations, time-asymptotic expansions were studied for example in [1, 2]. Because the nonlinearity is weaker compared to the 1D case, nonlinear waves similar to $\xi _{i;n}$ do not appear in these works.

In the next section, we state our main results. These are proven in Sect. 3.

2 Main Results

To state our main results (Theorem 2.1) we start by defining and discussing the properties of the higher-order diffusion waves $\xi _{i;n}$ ($n\ge 1$) mentioned in the introduction.

2.1 Higher-Order Diffusion Waves

Let (v, u) be the solution to (1). Then define $u_i$ by (2) and set

$$\begin{aligned} M_i=\int _{-\infty }^{\infty }u_i(x,0)\, dx. \end{aligned}$$

(6)

Let $\xi _{i;0}=\theta _i/2$ with $\theta _i$ defined by (3). We then define the higher-order diffusion waves $\xi _{i;n}$ ($n\ge 1$) inductively by the equations

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _t \xi _{i;n}+\lambda _i \partial _x \xi _{i;n}+\partial _x(\theta _i \xi _{i;n})+\partial _x(\theta _{i'}\xi _{i';n-1})=\frac{\nu }{2}\partial _{x}^{2}\xi _{i;n}, &{} x\in {\mathbb {R}},t>0, \\ \xi _{i;n}(x,0)=0, &{} x\in {\mathbb {R}}. \end{array}\right. \end{aligned}$$

(7)

Here $\lambda _i=(-1)^{i-1}c$ and $i'=3-i$, that is, $1'=2$ and $2'=1$. We remind the reader that $c=\sqrt{-p'(1)}>0$.

Although we do not have a simple explicit formula for $\xi _{i;n}$ like (4), we can still understand its asymptotic behavior quite well. To explain this, we introduce

$$\begin{aligned} \alpha _n=2-\frac{1}{2^{n+1}}, \quad \beta _n=\frac{3}{2}-\frac{1}{2^{n+1}} \quad (n\ge -1) \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \psi _n(x,t;\lambda )&=[(x-\lambda (t+1))^2+(t+1)]^{-\alpha _n/2}, \\ {\tilde{\psi }}_{n}(x,t;\lambda )&=[|x-\lambda (t+1)|^{\alpha _n}+(t+1)^{\beta _n}]^{-1}, \\ \Psi _{i;n}(x,t)&=\psi _n(x,t;\lambda _i)+{\tilde{\psi }}_n(x,t;\lambda _{i'}). \end{aligned} \end{aligned}$$

(8)

Then we have the following decay estimates for $\xi _{i;n}$ (we postpone the proof until we later prove a finer version in Lemma 3.1):

Proposition 2.1

Let $n\ge 1$ and $\varepsilon =\max (M_1,M_2)$. For $k\ge 0$, if $\varepsilon$ is sufficiently small, we have

$$\begin{aligned} |\partial _{x}^{k}\xi _{i;n}(x,t)|\le C_{n,k}\varepsilon ^{n+1}(t+1)^{-k/2}\Psi _{i;n-1}(x,t) \end{aligned}$$

for some positive constant $C_{n,k}$. In particular, when $|x|\le K$ for some fixed $K>0$, we have

$$\begin{aligned} |\xi _{i;n}(x,t)|\le C_{n,K}(t+1)^{-\alpha _{n-1}} \end{aligned}$$

for some $C_{n,K}>0$. Moreover, for any $1\le p\le \infty$, there exists $C_{n,p}>0$ such that

$$\begin{aligned} \Vert \xi _{i;n}(\cdot ,t) \Vert _{L^p}\le C_{n,p}(t+1)^{-(\alpha _{n-1}-1/p)/2}. \end{aligned}$$

We can also prove more detailed estimates if we focus on x with $(-1)^{i-1}x\ge 0$. Let

$$\begin{aligned} g(z)= & {} \partial _x e^{-\frac{x^2}{2\nu }}=-(x/\nu )e^{-\frac{x^2}{2\nu }}, \end{aligned}$$

(9)

$$\begin{aligned} f_{i;0}(z)= & {} \frac{\sqrt{\nu }}{\sqrt{2}}\left( e^{\frac{M_i}{\nu }}-1 \right) e^{-\frac{z^2}{2\nu }}\left[ \sqrt{\pi }+\left( e^{\frac{M_i}{\nu }}-1 \right) \int _{z}^{\infty }e^{-\xi ^2}\, d\xi \right] ^{-1}, \end{aligned}$$

(10)

and

$$\begin{aligned} f_{i;n}(z)=\int _{(-1)^{i-1}z}^{\infty }[\xi -(-1)^{i-1}z]^{-(1-1/2^n)}\xi e^{-\frac{\xi ^2}{2\nu }}\, d\xi . \end{aligned}$$

(11)

We then have the following asymptotic formula. This is obtained from Lemma 3.2 proved in the next section.

Proposition 2.2

Let $n\ge 1$ and $\varepsilon =\max (M_1,M_2)$. For any $K>0$, if $\varepsilon$ is sufficiently small, there exist $A_{i;n}$, $B_{i;n}$, and $C_n>0$ such that

$$\begin{aligned}&\left| \left\{ \xi _{i;n}(x,t)-\frac{A_{i;n}}{(t+1)^{\alpha _{n-1}/2}}f_{i;n}\left( \frac{x-\lambda _i(t+1)}{\sqrt{t+1}} \right) -\frac{B_{i;n}}{(t+1)^{\alpha _{n-1}/2}}g\left( \frac{x-\lambda _i(t+1)}{\sqrt{t+1}} \right) \right\} \right| \\&\quad \le C_n \varepsilon ^{n+1}\psi _n(x,t;\lambda _i) \end{aligned}$$

for x with $-K\le (-1)^{i-1}x$. The constants $A_{i;n}$ and $B_{i;n}$ are determined from $(M_1,M_2)$ defined by (6).

Remark 2.1

The function $f_{i;n}$ appears in [12, Section 4]. It is shown that $f_{i;n}(z)$ decays exponentially as $(-1)^{i-1}z\rightarrow \infty$ but decays algebraically as $f_{i;n}(z)\sim z^{-\alpha _{n-1}}$ in the limit $(-1)^{i-1}z\rightarrow -\infty$. In particular, if $|x|\le K$ for some fixed $K>0$, we have

$$\begin{aligned} (t+1)^{-\alpha _{n-1}/2}f_{i;n}\left( \frac{x-\lambda _i(t+1)}{\sqrt{t+1}} \right) \sim t^{-\alpha _{n-1}} \end{aligned}$$

Remark 2.2

With some additional effort, we can show that, for $-K\le (-1)^{i-1}x$, the higher-order diffusion waves $\xi _{i;n}$ ($n\ge 1$) are asymptotically equivalent to the higher-order terms of the asymptotic expansion constructed in [12].

2.2 Time-Asymptotic Expansion with Pointwise Remainder Estimates

Let $\varvec{u}_0=(v_0-1,u_0)$ and denote its anti-derivatives by $\varvec{u}_{0}^{\pm }$, that is,

$$\begin{aligned} \varvec{u}_{0}^{-}(x)=\int _{-\infty }^{x}\varvec{u}_0(y)\, dy, \quad \varvec{u}_{0}^{+}(x)=\int _{x}^{\infty }\varvec{u}_0(y)\, dy. \end{aligned}$$

Our main theorem is the following:

Theorem 2.1

For $\varvec{u}_0=(v_0-1,u_0)\in H^6({\mathbb {R}})\times H^6({\mathbb {R}})$, let (v, u) be the solution to (1). Define $u_i$, $\theta _i$, and $\xi _{i;n}$ by (2), (3), and (7), respectively. Set

$$\begin{aligned} u_{i;1}=\xi _{i;1}+\gamma _{i'}\partial _x \theta _{i'}, \quad u_{i;n}=\xi _{i;n}+\gamma _{i'}\partial _x \xi _{i';n-1} \quad (n\ge 2), \end{aligned}$$

where $i'=3-i$ and $\gamma _i=(-1)^i \nu /(4c)$. Then for $n\ge 1$, there exist positive constants $\delta _n$ and $C_n$ such that if

$$\begin{aligned} \begin{aligned} \delta&{:}{=}\Vert \varvec{u}_0 \Vert _6+\sup _{x\in {\mathbb {R}}}[(|x|+1)^{\alpha _n}|\varvec{u}_0(x)|+(|x|+1)^{5/4}|\varvec{u}_{0}'(x)|] \\&\quad +\sup _{x>0}[(|x|+1)^{\beta _n}(|\varvec{u}_{0}^{-}(-x)|+|\varvec{u}_{0}^{+}(x)|)]\le \delta _n, \end{aligned} \end{aligned}$$

(12)

the solution (v, u) satisfies the pointwise estimates

$$\begin{aligned} \biggl | \biggl ( u_i-\theta _i-\sum _{k=1}^{n}u_{i;k} \biggr )(x,t) \biggr | \le C_n\delta \Psi _{i;n}(x,t) \end{aligned}$$

for all $x\in {\mathbb {R}}$ and $t\ge 0$. Here $\Psi _{i;n}$ is defined by (8).

As a corollary, we obtain the following $L^p$-decay estimates. Combining this with Proposition 2.1, it follows that $u_i \sim \theta _i+\sum _{n=1}^{\infty }\xi _{i;n}$ is a time-asymptotic expansion in the $L^p({\mathbb {R}})$-norm for all $1\le p\le \infty$.

Corollary 2.1

Under the assumptions of Theorem 2.1, we have the optimal $L^p$-decay estimate

$$\begin{aligned} \biggl \Vert \biggl ( u_i-\theta _i-\sum _{k=1}^{n}\xi _{i;k} \biggr )(\cdot ,t) \biggr \Vert _{L^p}\le C_n \delta (t+1)^{-(\alpha _n-1/p)/2} \quad (1\le p\le \infty ). \end{aligned}$$

Proof

The same bound for $u_i-\theta _i-\sum _{k=1}^{n}u_{i;k}$ easily follows from Theorem 2.1. We can replace $u_{i;k}$ by $\xi _{i;k}$ thanks to (4) and Proposition 2.1. $\quad \square$

We also obtain the following local-in-space decay estimates:

Corollary 2.2

Under the assumptions of Theorem 2.1, when $|x|\le K$ for some fixed $K>0$, we have

$$\begin{aligned} \biggl | \biggl ( u_i-\sum _{k=1}^{n}\xi _{i;k} \biggr )(x,t) \biggr |\le C_{n,K}\delta (t+1)^{-\alpha _n}. \end{aligned}$$

Moreover, there exist constants $\{ A_{i;k} \}_{k=1}^{n}$ determined from $(M_1,M_2)$ such that

$$\begin{aligned} \biggl | u_i(x,t)-\sum _{k=1}^{n}\frac{A_{i;k}}{(t+1)^{\alpha _{k-1}/2}}f_{i;k}\left( \frac{x-\lambda _i(t+1)}{\sqrt{t+1}} \right) \biggr | \le C_{n,K}\delta (t+1)^{-\alpha _n}. \end{aligned}$$

Here $M_i$ and $f_{i;k}$ are defined by (6) and (11), respectively.

Proof

Again, the same bound for $u_i-\theta _i-\sum _{k=1}^{n}u_{i;k}$ easily follows from Theorem 2.1. We can then replace $u_{i;k}$ by $\xi _{i;k}$ thanks to (4) and Proposition 2.1. The second inequality follows from Proposition 2.2. $\quad \square$

By Corollary 2.2, and also Remark 2.1, we now have a detailed picture of the power-law asymptotics of the solution around $x=0$ where the diffusion waves decay exponentially.

Remark 2.3

The term $\gamma _{i'}\partial _x \theta _i$ and $\gamma _{i'}\partial _x \xi _{i;n}$ are both neglected in the two corollaries above. These are negligible in the $L^p({\mathbb {R}})$-norm and locally around $x=0$ but are important in the neighborhood of the other characteristic line $x=-\lambda _i t$. For this reason, these terms are required in the statement of Theorem 2.1.

Remark 2.4

The rather strong $H^6$-regularity is required to invoke pointwise estimates of $\partial _x(u_i-\theta _i)$ provided by [10, Theorem 2.6 and Remark 2.8]. The proof involves energy estimates up to the $H^6({\mathbb {R}})$-norm. These also imply a unique global-in-time existence theorem in appropriate Sobolev spaces. Off course, global-in-time existence of solutions can be proved with much lower regularity [4, 9], but proving detailed pointwise estimates for such data seems to be difficult at this point.

Remark 2.5

We add a comment on taking the limit $n\rightarrow \infty$ in Theorem 2.1 and also on a possible route to expand the solution to even higher order. A careful examination of the proof shows that the constant $C_n$ in Theorem 2.1 grows as $2^n$. So we cannot simply take the limit. However, as pointed out in [12, p. 1955], it might be possible to take the limit by adding a logarithmic weight:

$$\begin{aligned} \biggl | \biggl ( u_i-\theta _i-\sum _{k=1}^{\infty }u_{i;k} \biggr )(x,t) \biggr | \le C_{\infty }\delta \log (t+2)\Psi _{i;\infty }(x,t). \end{aligned}$$

Here, $C_{\infty }$ is a constant independent of n and $\Psi _{i;\infty }(x,t)=\lim _{n\rightarrow \infty }\Psi _{i;n}(x,t)$. Since $\Vert \Psi _{i;\infty }(\cdot ,t) \Vert _{L^{\infty }}\lesssim t^{-1}$, to study an asymptotic expansion beyond the order $O(t^{-1}\log t)$, it seems that we need to identify waves describing this order. Such waves are identified for example in [5] for generalized Burgers equations. Analogous results for hyperbolic–parabolic systems are, as far as I know, not known. If such waves are identified, we might be able to expand the solution beyond the order $O(t^{-1}\log t)$. And drawing an analogy between the heat equation, terms beyond this order should also depend on higher-order moments $\int _{-\infty }^{\infty }x^k u_i(x,0)\, dx$ and not just on $M_i=\int _{-\infty }^{\infty }u_i(x,0)\, dx$.

3 Proof

The following function appears frequently in the subsequent part of the paper:

$$\begin{aligned} \Theta _{\alpha }(x,t;\lambda ,\mu )=(t+1)^{-\alpha /2}e^{-\frac{(x-\lambda (t+1))^2}{\mu (t+1)}}. \end{aligned}$$

(13)

Here $\lambda \in {\mathbb {R}}$ and $\alpha ,\mu >0$. Note that

$$\begin{aligned} |\theta _i(x,t)|\le A_0|M_i|\Theta _1(x,t;\lambda _i,2\nu ), \quad \Theta _{\alpha _n}(x,t;\lambda ,\mu )\le B_0 \psi _n(x,t;\lambda ) \end{aligned}$$

(14)

for some positive constants $A_0$ and $B_0$. In what follows, the symbols C and $\nu ^*$ denote sufficiently large constants.

3.1 Pointwise Estimates of the Higher-Order Diffusion Waves

We start with the proofs of Propositions 2.1 and 2.2.

Proposition 2.1 follows from the following finer version:

Lemma 3.1

Let $n\ge 1$ and $\varepsilon =\max (M_1,M_2)$. If $\varepsilon$ is sufficiently small, we have

$$\begin{aligned} \begin{aligned}&|\partial _{x}^{k}\xi _{i;n}(x,t)-(-1)^i (2c)^{-1}\partial _{x}^{k}(\theta _{i'}\xi _{i';n-1})(x,t)| \\&\quad \le C_{n,k}\varepsilon ^{n+1}(t+1)^{-k/2}\psi _{n-1}(x,t;\lambda _i) \end{aligned} \end{aligned}$$

(15)

for any integer $k\ge 0$. In particular, we have

$$\begin{aligned} |\partial _{x}^{k}\xi _{i;n}(x,t)|&\le C_{n,k}\varepsilon ^{n+1}(t+1)^{-k/2}[\psi _{n-1}(x,t;\lambda _i)+\Theta _{2\beta _{n-1}}(x,t;\lambda _{i'},\nu ^*)] \\&\le C_{n,k}\varepsilon ^{n+1}(t+1)^{-k/2}\Psi _{i;n-1}(x,t). \end{aligned}$$

Proof

We assume $t\ge 4$ in what follows (the lemma is otherwise easier to prove). The lemma is trivial for $n=0$ if we set $\xi _{i;0}=\theta _i/2$ and $\xi _{i';-1}=0$. So it suffices to prove the lemma for n assuming that it holds for $n-1\ge 0$. In what follows, we only prove the case of $i=1$ and $k=0$ since the other cases are similar. Note first that, by (7) and Duhamel’s principle, we have $\xi _{i;n}(x,t)=\zeta _{1;n}(x,t)+\eta _{1;n}(x,t)$, where

$$\begin{aligned} \zeta _{1;n}(x,t)= -(2\pi \nu )^{-1/2}\int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}\partial _x (\theta _2 \xi _{2;n-1})(y,s)\, dyds\nonumber \\ \end{aligned}$$

(16)

and

$$\begin{aligned} \eta _{1;n}(x,t)= -(2\pi \nu )^{-1/2}\int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}\partial _x (\theta _1 \xi _{1;n})(y,s)\, dyds.\nonumber \\ \end{aligned}$$

(17)

We first consider $\zeta _{1;n}(x,t)$. Set $I(x,t)=-\sqrt{2\pi \nu }\zeta _{1;n}(x,t)$ and $f=\theta _2 \xi _{2;n-1}$. By Lemma A.1, we have

$$\begin{aligned} I(x,t)=(2c)^{-1}\sqrt{2\pi \nu }f(x,t)+I_1(x,t)+I_2(x,t), \end{aligned}$$

where

$$\begin{aligned} I_1(x,t)=\int _{0}^{t^{1/2}}\int _{-\infty }^{\infty }\partial _x \left\{ (t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}} \right\} f(y,s)\, dyds \end{aligned}$$

and

$$\begin{aligned} I_2(x,t)&=-(2c)^{-1}\int _{-\infty }^{\infty }(t-t^{1/2})^{-1/2}e^{-\frac{(x-y-c(t-\sqrt{t}))^2}{2\nu (t-\sqrt{t})}}f(y,t^{1/2})\, dy \\&\quad -(2c)^{-1}\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}L_2 f(y,s)\, dyds \\&{=}{:}I_{21}(x,t)+I_{22}(x,t). \end{aligned}$$

Here $L_2=\partial _t-c\partial _x-(\nu /2)\partial _{x}^{2}$. By the induction hypothesis, we have

$$\begin{aligned} |f(x,t)|\le C\varepsilon ^{n+1}\Theta _{\alpha _{n-2}+1}(x,t;-c,\nu ^*). \end{aligned}$$

By Lemmas A.2 and A.3, we obtain

$$\begin{aligned} |I_1(x,t)|+|I_{21}(x,t)|\le C\varepsilon ^{n+1}\Theta _{\alpha _{n-1}}(x,t;c,\nu ^*)\le C\varepsilon ^{n+1}\psi _{n-1}(x,t;c). \end{aligned}$$

Next, note that (3) and (7) imply

$$\begin{aligned} |L_2 f(x,t)|\le C\varepsilon ^{n+1}\Theta _{\alpha _{n-2}+3}(x,t;-c,\nu ^*). \end{aligned}$$

Then by Lemma A.4, we obtain

$$\begin{aligned} |I_{22}(x,t)|\le C\varepsilon ^{n+1}\psi _{n-1}(x,t;c). \end{aligned}$$

We have thus proved that

$$\begin{aligned} |\zeta _{1;n}(x,t)+(2c)^{-1}(\theta _2 \xi _{2;n-1})(x,t)|\le C\varepsilon ^{n+1}\psi _{n-1}(x,t;c). \end{aligned}$$

(18)

We next consider $\eta _{1;n}(x,t)$. Note that it is the solution to

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _t \eta _{1;n}+c\partial _x \eta _{1;n}+\partial _x(\theta _1 \eta _{1;n})=\frac{\nu }{2}\partial _{x}^{2}\eta _{1;n}-\partial _x(\theta _1 \zeta _{1;n}), &{} x\in {\mathbb {R}},t>0, \\ \eta _{1;n}(x,0)=0, &{} x\in {\mathbb {R}}. \end{array}\right. \end{aligned}$$

This variable coefficient equation can be solved by an iteration scheme. Let $\eta _{1;n}^{(1)}$ be the solution to

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _t \eta _{1;n}^{(1)}+c\partial _x \eta _{1;n}^{(1)}=\frac{\nu }{2}\partial _{x}^{2}\eta _{1;n}^{(1)}-\partial _x (\theta _1 \zeta _{1;n}), &{} x\in {\mathbb {R}},t>0, \\ \eta _{1;n}^{(1)}(x,0)=0, &{} x\in {\mathbb {R}} \end{array}\right. \end{aligned}$$

and $\eta _{1;n}^{(k)}$ ($k\ge 2$) be the solution to

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _t \eta _{1;n}^{(k)}+c\partial _x \eta _{1;n}^{(k)}=\frac{\nu }{2}\partial _{x}^{2}\eta _{1;n}^{(k)}-\partial _x(\theta _1 \eta _{1;n}^{(k-1)}) &{} x\in {\mathbb {R}},t>0, \\ \eta _{1;n}^{(k)}(x,0)=0, &{} x\in {\mathbb {R}}. \end{array}\right. \end{aligned}$$

Then we can write $\eta _{1;n}$ as

$$\begin{aligned} \eta _{1;n}(x,t)=\sum _{k=1}^{\infty }\eta _{1;n}^{(k)}(x,t). \end{aligned}$$

(19)

We now give bounds for $\eta _{1;n}^{(k)}$ ($k\ge 1$) inductively. Note first that

$$\begin{aligned} \eta _{1;n}^{(1)}(x,t)=-(2\pi \nu )^{-1/2}\int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}\partial _x (\theta _1 \zeta _{1;n})(y,s)\, dyds\nonumber \\ \end{aligned}$$

(20)

and that (18) implies

$$\begin{aligned} |(\theta _1 \zeta _{1;n})(x,t)|\le A_1 \varepsilon ^{n+2}\Theta _{\alpha _{n-1}+1}(x,t;c,\nu ') \end{aligned}$$

for some positive constants $A_1$ and $\nu '$. Then by [10, Lemma 3.2], we obtain

$$\begin{aligned} |\eta _{1;n}^{(1)}(x,t)|\le M A_1 \varepsilon ^{n+2}\Theta _{\alpha _{n-1}}(x,t;c,\nu ') \end{aligned}$$

for some $M>0$. This means that the inequality

$$\begin{aligned} |\eta _{1;n}^{(l)}(x,t)|\le MA_1 (MA_0)^{l-1}\varepsilon ^{n+l+1}\Theta _{\alpha _{n-1}}(x,t;c,\nu ') \end{aligned}$$

(21)

holds for $l=1$. We then show that (21) holds for $l=k+1$ assuming that it holds for $l=k$. By the induction hypothesis and (14), we have

$$\begin{aligned} |(\theta _1 \eta _{1;n}^{(k)})(x,t)|\le A_1 (MA_0)^{k}\varepsilon ^{n+k+2}\Theta _{\alpha _{n-1}+1}(x,t;c,\nu '). \end{aligned}$$

Applying [10, Lemma 3.2] again, this time to the integral representation

$$\begin{aligned} \eta _{1;n}^{(k+1)}(x,t)=-(2\pi \nu )^{-1/2}\int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}\partial _x (\theta _1 \eta _{1;n}^{(k)})(y,s)\, dyds, \end{aligned}$$

we obtain

$$\begin{aligned} |\eta _{1;n}^{(k+1)}(x,t)|\le MA_1 (MA_0)^k \varepsilon ^{n+k+2}\Theta _{\alpha _{n-1}}(x,t;c,\nu '). \end{aligned}$$

Therefore, (21) holds for any $l\ge 1$, and by taking $\varepsilon$ sufficiently small, we get

$$\begin{aligned} |\eta _{1;n}(x,t)|=\biggl | \sum _{k=1}^{\infty }\eta _{1;n}^{(k)}(x,t) \biggr |\le C\varepsilon ^{n+2}\Theta _{\alpha _{n-1}}(x,t;c,\nu ')\le C\varepsilon ^{n+2}\psi _{n-1}(x,t;c). \end{aligned}$$

Combining this with (18), we obtain (15). $\quad \square$

Remark 3.1

The proof above can be modified to show that

$$\begin{aligned} \begin{aligned}&|\partial _{x}^{k}\xi _{i;m}(x,t)-(-1)^i (2c)^{-1}\partial _{x}^{k}(\theta _{i'}\xi _{i';m-1})(x,t)| \\&\quad \le C_{n,k}\varepsilon ^{m+1}(t+1)^{-k/2}\psi _{n-1}(x,t;\lambda _i) \end{aligned} \end{aligned}$$

holds for all $m\ge n$ with the smallness of $\varepsilon$ depending only on n and k.

We next prove Proposition 2.2. (The proof is rather lengthy and may be skipped; the rest of the paper can be read independently.) Define $\zeta _{i;n}$ and $\eta _{i;n}$ by (16) and (17), respectively. Then Proposition 2.2 is a direct consequence of the following lemma.

Lemma 3.2

Let $n\ge 1$ and $\varepsilon =\max (M_1,M_2)$. Fix $k\ge 0$. For any $K>0$, if $\varepsilon$ is sufficiently small, there exist $A_{i;n}$, $B_{i;n}$, and $C_{n,k}>0$ such that

$$\begin{aligned}{} & {} \left| \partial _{x}^{k}\left\{ \zeta _{i;n}(x,t)-\frac{A_{i;n}}{(t+1)^{\alpha _{n-1}/2}}f_{i;n}\left( \frac{x-\lambda _i(t+1)}{\sqrt{t+1}} \right) \right\} \right| \nonumber \\{} & {} \quad \le C_{n,k}\varepsilon ^{n+1}(t+1)^{-k/2}\psi _n(x,t;\lambda _i) \end{aligned}$$

(22)

and

$$\begin{aligned}{} & {} \left| \partial _{x}^{k}\left\{ \eta _{i;n}(x,t)-\frac{B_{i;n}}{(t+1)^{\alpha _{n-1}/2}}g\left( \frac{x-\lambda _i(t+1)}{\sqrt{t+1}} \right) \right\} \right| \nonumber \\{} & {} \quad \le C_{n,k}\varepsilon ^{n+2}(t+1)^{-k/2}\psi _n(x,t;\lambda _i) \end{aligned}$$

(23)

when $-K\le (-1)^{i-1}x$. Here g and $f_{i;n}$ are defined by (9) and (11), respectively.

Proof

The lemma is proved by induction in n.

We first consider the case of $n=1$. Let $i=1$ and $k=0$ (the other cases are similar). We start with the proof of (22). Note that (16) implies

$$\begin{aligned} \zeta _{1;1}(x,t)=-\frac{1}{2\sqrt{2\pi \nu }}\partial _x \int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}\theta _{2}^{2}(y,s)\, dyds. \end{aligned}$$

In addition, by (4) and (10), we have

$$\begin{aligned} \theta _i(x,t)=(t+1)^{-1/2}f_{i;0}\left( \frac{x-\lambda _i (t+1)}{\sqrt{t+1}} \right) . \end{aligned}$$

Hence we may write

$$\begin{aligned} \theta _{2}^{2}(x,t)=\frac{a_{1;1}}{(t+1)\sqrt{2\pi \nu }}e^{-\frac{(x+c(t+1))^2}{2\nu (t+1)}}+\partial _x r(x,t), \end{aligned}$$

where

$$\begin{aligned} a_{1;1}=\int _{-\infty }^{\infty }f_{2;0}^{2}(z)\, dz \end{aligned}$$

and

$$\begin{aligned} r(x,t)=\int _{-\infty }^{x}\left( \theta _{2}^{2}(z,t)-\frac{a_{1;1}}{(t+1)\sqrt{2\pi \nu }}e^{-\frac{(z+c(t+1))^2}{2\nu (t+1)}} \right) \, dz. \end{aligned}$$

Noting that $\lim _{x\rightarrow \infty }r(x,t)=0$, we can show that

$$\begin{aligned} |r(x,t)|\le C\varepsilon ^2 \Theta _2(x,t;-c,\nu ^*), \quad |L_2 r(x,t)|\le C\varepsilon ^2 \Theta _4(x,t;-c,\nu ^*), \end{aligned}$$

where $L_2=\partial _t-c\partial _x-(\nu /2)\partial _{x}^{2}$. We then have

$$\begin{aligned} \zeta _{1;1}(x,t)&=-\frac{a_{1;1}}{4\pi \nu }\partial _x \int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}(s+1)^{-1}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}e^{-\frac{(y+c(s+1))^2}{2\nu (s+1)}}\, dyds \\&\quad -\frac{1}{2\sqrt{2\pi \nu }}\partial _x \int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}\partial _y r(y,s)\, dyds \\&=-\frac{a_{1;1}}{2\sqrt{2\pi \nu }}\partial _x \int _{0}^{t}(t+1)^{-1/2}(s+1)^{-1/2}e^{-\frac{(x-c(t-s)+c(s+1))^2}{2\nu (t+1)}}\, ds \\&\quad -\frac{1}{2\sqrt{2\pi \nu }}\partial _x \int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}\partial _y r(y,s)\, dyds. \end{aligned}$$

Concerning the second term, similar calculations leading to the bound of $\zeta _{1;n}(x,t)$ in Lemma 3.1 imply

$$\begin{aligned} \left| \partial _x \int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}\partial _y r(y,s)\, dyds \right| \le C\varepsilon ^2 \psi _n(x,t;c) \end{aligned}$$

for $-K\le x$. For the first term, note that a simple change of variable yields

$$\begin{aligned}{} & {} -\frac{(t+1)^{-3/4}}{\nu \sqrt{2c}}f_{1;1}\left( \frac{x-c(t+1)}{\sqrt{t+1}} \right) \\{} & {} \quad =\partial _x \int _{-1}^{\infty }(t+1)^{-1/2}(s+1)^{-1/2}e^{-\frac{(x-c(t-s)+c(s+1))^2}{2\nu (t+1)}}\, ds. \end{aligned}$$

Therefore,

$$\begin{aligned}&\zeta _{1;1}(x,t)-\frac{a_{1;1}}{2\nu \sqrt{4\pi \nu c}}(t+1)^{-3/4}f_{1;1}\left( \frac{x-c(t+1)}{\sqrt{t+1}} \right) \\&\quad =O(\varepsilon ^2)\partial _x \int _{s\in (-1,0)\cup (t,\infty )}(t+1)^{-1/2}(s+1)^{-1/2}e^{-\frac{(x-c(t-s)+c(s+1))^2}{2\nu (t+1)}}\, ds \\&\qquad +O(\varepsilon ^2)\psi _n(x,t;c). \end{aligned}$$

We then set

$$\begin{aligned} I(x,t)=\partial _x \int _{s\in (-1,0)\cup (t,\infty )}(t+1)^{-1/2}(s+1)^{-1/2}e^{-\frac{(x-c(t-s)+c(s+1))^2}{2\nu (t+1)}}\, ds \end{aligned}$$

and show that $|I(x,t)|\le C\Theta _2(x,t;c,\nu ^*)$ for $-K\le x$. We first consider Case (i) $|x-c(t+1)|\le (t+1)^{1/2}$. In this case, we simply have

$$\begin{aligned} |I(x,t)|\le C(t+1)^{-1}\le C\Theta _2(x,t;c,\nu ^*). \end{aligned}$$

We next consider Case (ii) $-K\le x\le c(t+1)-(t+1)^{1/2}$. The integral over $(-1,0)$ is easy to handle. For $s\in (t,\infty )$ on the other hand, when t is large (the case when t is not large is easier), we have

$$\begin{aligned} 0\le & {} c(t+1)-x-2K\le x-c(t-s)+c(s+1), \\ 0\le & {} c(s+1)-K\le x-c(t-s)+c(s+1). \end{aligned}$$

Hence

$$\begin{aligned}&\left| \partial _x \int _{t}^{\infty }(t+1)^{-1/2}(s+1)^{-1/2}e^{-\frac{(x-c(t-s)+c(s+1))^2}{2\nu (t+1)}}\, ds \right| \\&\quad \le C\int _{0}^{\infty }(s+1)^{-1/2}e^{-\frac{s^2}{C(t+1)}}\, ds\cdot \Theta _2(x,t;c,\nu ^*)\le C\Theta _2(x,t;c,\nu ^*). \end{aligned}$$

We end the analysis of $\zeta _{1;1}$ by considering Case (iii) $x\ge c(t+1)+(t+1)^{1/2}$. When $s>-1$, we have

$$\begin{aligned}{} & {} 0\le x-c(t+1)\le x-c(t-s)+c(s+1)=x-c(t+1)+2c(s+1), \\{} & {} 0\le 2c(s+1)\le x-c(t-s)+c(s+1). \end{aligned}$$

From these, it follows that $|I(x,t)|\le C\Theta _2(x,t;c,\nu ^*)$ as in Case (ii). These prove (22) for $n=1$ by setting

$$\begin{aligned} A_{1;1}=\frac{a_{1;1}}{2\nu \sqrt{4\pi \nu c}}. \end{aligned}$$

We next prove (23) by using the series representation (19). We first consider

$$\begin{aligned} \eta _{1;1}^{(1)}(x,t)=-(2\pi \nu )^{-1/2}\int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}\partial _x (\theta _1 \zeta _{1;1})(y,s)\, dyds. \end{aligned}$$

The bound (22) for $\zeta _{1;1}(x,t)$ implies

$$\begin{aligned} (\theta _1 \zeta _{1;1})(x,t)=\theta _1(x,t)\frac{A_{1;1}}{(t+1)^{3/4}}f_{1;1}\left( \frac{x-c(t+1)}{\sqrt{t+1}} \right) +O(\varepsilon ^3)\Theta _{\alpha _n+1}(x,t;c,\nu ^*). \end{aligned}$$

This holds for all $x\in {\mathbb {R}}$ since $\theta _1(x,t)$ decays exponentially for $x\le -K$. Plugging this into (20) and arguing similarly to the analysis of $\zeta _{1;1}$, we get

$$\begin{aligned} \eta _{1;1}^{(1)}(x,t)&=-\frac{A_{1;1}b_{1;1}}{2\pi \nu }\partial _x \int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}(s+1)^{-5/4}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}e^{-\frac{(y-c(s+1))^2}{2\nu (s+1)}}\, dyds \\&\quad +O(\varepsilon ^3)\psi _n(x,t;c) \\&=-\frac{A_{1;1}b_{1;1}}{\sqrt{2\pi \nu }}\partial _x \int _{0}^{t}(t+1)^{-1/2}(s+1)^{-3/4}e^{-\frac{(x-c(t+1))^2}{2\nu (t+1)}}\, ds+O(\varepsilon ^3)\psi _n(x,t;c) \\&=-\frac{4A_{1;1}b_{1;1}}{\sqrt{2\pi \nu }}(t+1)^{-3/4}g\left( \frac{x-c(t+1)}{\sqrt{t+1}} \right) +O(\varepsilon ^3)\psi _n(x,t;c), \end{aligned}$$

where

$$\begin{aligned} b_{1;1}=\int _{-\infty }^{\infty }(f_{1;0}f_1)(z)\, dz. \end{aligned}$$

Similar analysis for $\eta _{1;1}^{(k)}(x,t)$ ($k\ge 2$) shows that

$$\begin{aligned} \eta _{1;1}^{(k)}(x,t)=O(\varepsilon ^{k+2})(t+1)^{-3/4}g\left( \frac{x-c(t+1)}{\sqrt{t+1}} \right) +O(\varepsilon ^{k+2})\psi _n(x,t;c). \end{aligned}$$

Taking the sum $\sum _{k=1}^{\infty }$, it follows that (23) holds for $n=1$ with

$$\begin{aligned} B_{1;1}=-\frac{4A_{1;1}b_{1;1}}{\sqrt{2\pi \nu }}+O(\varepsilon ^4). \end{aligned}$$

We next prove the lemma for n assuming that it holds for $n-1$. Let $i=1$ and $k=0$ (the other cases are similar). The induction hypothesis and Lemma 3.1 imply

$$\begin{aligned} h(x,t)&{:}{=}(\theta _2 \xi _{2;n-1})(x,t)-\frac{A_{2;n-1}}{(t+1)^{\alpha _{n-2}/2}}\theta _2(x,t)f_{2;n-1}\left( \frac{x+c(t+1)}{\sqrt{t+1}} \right) \\&\quad -\frac{B_{2;n-1}}{(t+1)^{\alpha _{n-2}/2}}\theta _2(x,t)g\left( \frac{x+c(t+1)}{\sqrt{t+1}} \right) =O(\varepsilon ^{n+1})\Theta _{\alpha _{n-1}+1}(x,t;-c,\nu ^*) \end{aligned}$$

for all $x\in {\mathbb {R}}$ (not just for $x\le K$). We also have $\partial _x h(x,t)=O(\varepsilon ^{n+1})\Theta _{\alpha _{n-1}+2}(x,t;-c,\nu ^*)$. Using these, we can show that

$$\begin{aligned} L_2 h(x,t)=O(\varepsilon ^{n+1})\Theta _{\alpha _{n-1}+3}(x,t;-c,\nu ^*), \end{aligned}$$

where $L_2=\partial _t-c\partial _x-(\nu /2)\partial _{x}^{2}$. Then similar calculations leading to the bound of $\zeta _{1;1}(x,t)$ above imply (22) with

$$\begin{aligned} A_{1;n}=\frac{1}{\nu \sqrt{4\pi \nu c}}(A_{2;n-1}a_{1;n}+B_{2;n-1}b_{1;n}), \end{aligned}$$

where

$$\begin{aligned} a_{1;n}=\int _{-\infty }^{\infty }(f_{2;0}f_{2;n-1})(-z)\, dz, \quad b_{1;n}=\int _{-\infty }^{\infty }(gf_{2;0})(z)\, dz. \end{aligned}$$

The bound (23) for $\eta _{1;n}(x,t)$ is proved in a way similar to that for $n=1$. This ends the proof of the lemma. $\quad \square$

For the proof of Theorem 2.1, it is convenient to unify $(\xi _{i;n})_{n=1}^{\infty }$ into a single function

$$\begin{aligned} \Xi _i(x,t)=\sum _{n=1}^{\infty }\xi _{i;n}(x,t). \end{aligned}$$

(24)

Taking the infinite sum of (7), we see that $(\Xi _1,\Xi _2)$ is the solution to the system

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _t \Xi _1+c\partial _x \Xi _1+\partial _x(\theta _{2}^{2}/2+\theta _1 \Xi _1+\theta _2 \Xi _2)=\frac{\nu }{2}\partial _{x}^{2}\Xi _1, &{} x\in {\mathbb {R}},t>0, \\ \partial _t \Xi _2-c\partial _x \Xi _2+\partial _x(\theta _{1}^{2}/2+\theta _1 \Xi _1+\theta _2 \Xi _2)=\frac{\nu }{2}\partial _{x}^{2}\Xi _2, &{} x\in {\mathbb {R}},t>0, \\ \Xi _1(x,0)=\Xi _2(x,0)=0, &{} x\in {\mathbb {R}}. \end{array}\right. \end{aligned}$$

(25)

Then Lemma 3.1 and Remark 3.1 imply the following:

Lemma 3.3

Let

$$\begin{aligned} \Xi _{i;n}(x,t)=\sum _{m=n+1}^{\infty }\xi _{i;m}(x,t) \quad (n\ge -1). \end{aligned}$$

Here $\xi _{i;0}=\theta _i/2$. Then for $n\ge 0$, if $\varepsilon =\max (M_1,M_2)$ is sufficiently small, we have

$$\begin{aligned} |\partial _{x}^{k}\Xi _{i;n}(x,t)-(-1)^i (2c)^{-1}\partial _{x}^{k}(\theta _{i'}\Xi _{i';n-1})(x,t)|\le C_{n,k}\varepsilon ^{n+2}(t+1)^{-k/2}\psi _n(x,t;\lambda _i) \end{aligned}$$

for any integer $k\ge 0$. In particular, we have

$$\begin{aligned} |\partial _{x}^{k}\Xi _{i;n}(x,t)|\le C_{n,k}\varepsilon ^{n+2}(t+1)^{-k/2}\Psi _{i;n}(x,t). \end{aligned}$$

3.2 Proof of Theorem 2.1

Let us explain the strategy to prove Theorem 2.1. Note first that by Lemma 3.3, it suffices to prove the following:

Theorem 3.1

For $\varvec{u}_0=(v_0-1,u_0)\in H^6({\mathbb {R}})\times H^6({\mathbb {R}})$, let (v, u) be the solution to (1). Define $u_i$, $\theta _i$, and $\Xi _i$ by (2), (3), and (25), respectively. Then for $n\ge 1$, there exist positive constants $\delta _n$ and $C_n$ such that if $\delta \le \delta _n$, where $\delta$ is defined by (12), the solution (v, u) satisfies the pointwise estimates

$$\begin{aligned} |(u_i-\theta _i-\Xi _i-\gamma _{i'}\partial _x \theta _{i'}-\gamma _{i'}\partial _x \Xi _{i'})(x,t)|\le C_n\delta \Psi _{i;n}(x,t) \end{aligned}$$

for all $x\in {\mathbb {R}}$ and $t\ge 0$. Here $\gamma _i=(-1)^i \nu /(4c)$ and $i'=3-i$.

To prove Theorem 3.1, we set

$$\begin{aligned} v_i=u_i-\theta _i-\Xi _i-\gamma _{i'}\partial _x \theta _{i'}-\gamma _{i'}\partial _x \Xi _{i'} \end{aligned}$$

(26)

and define P(t) by

$$\begin{aligned} P(t){:}{=}\sum _{i=1}^{2}\sup _{0\le s\le t}\left| v_i(\cdot ,s)\Psi _{i;n}(\cdot ,s)^{-1} \right| _{L^{\infty }}. \end{aligned}$$

(27)

Our goal is then to prove the inequality

$$\begin{aligned} P(t)\le C\delta +C(\delta +P(t))^2 \quad (t\ge 0). \end{aligned}$$

(28)

From this inequality, taking $\delta$ sufficiently small, we can conclude that $P(t)\le C\delta$ for all $t\ge 0$ by a standard argument (see Sect. 3.2.4).

Remark 3.2

For the argument above to work, we first need to show that P(t) is finite. This can be proved, for example, by examining the iterative scheme in [6, Section 2.1] for the construction of the local-in-time solution to (1). The key step of the scheme consists of solving a variable coefficient parabolic equation, and by the Levi parametrix method, we can prove a gaussian upper bound for the fundamental solution. This bound allows us to control the spatial decay of each approximate solution, and by taking the limit, we can check that P(t) is finite at least for a short period of time. By the calculations below, it follows that (28) and hence $P(t)\le C\delta$ hold for this short duration. Then a standard continuity argument shows that $P(t)\le C\delta$ actually holds for all $t\ge 0$.

The proof of (28) is based on pointwise estimates of Greens’ function [10, 11] which we shall explain in the next section. We also give an integral formulation of (1). In the remaining sections, we prove bounds for the terms appearing in the integral equations which yield (28).

3.2.1 Pointwise Estimates of Green’s Function and Integral Equations

Our equations (1) can be written in the form

$$\begin{aligned} \varvec{u}_t+A\varvec{u}_x= \begin{pmatrix} 0 &{} 0 \\ 0 &{} \nu \end{pmatrix} \varvec{u}_{xx}+ \begin{pmatrix} 0 \\ N_x \end{pmatrix} \end{aligned}$$

(29)

with

$$\begin{aligned} \varvec{u}= \begin{pmatrix} v-1 \\ u \end{pmatrix}, \quad A= \begin{pmatrix} 0 &{} -1 \\ -c^2 &{} 0 \end{pmatrix}, \quad N=-p(v)+p(1)-c^2(v-1)-\nu \frac{v-1}{v}u_x.\nonumber \\ \end{aligned}$$

(30)

The matrix A has right and left eigenvectors $r_i$ and $l_i$ ($i=1,2$), corresponding to the eigenvalue $\lambda _i=(-1)^{i-1}c$, given by

$$\begin{aligned} r_i=\frac{2c}{p''(1)} \begin{bmatrix} (-1)^i \\ c \end{bmatrix}, \quad l_i=\frac{p''(1)}{4c} \begin{bmatrix} (-1)^i&1/c \end{bmatrix}. \end{aligned}$$

We note that (2) can be written as $u_i=l_i(v-1 \, u)^{T}$.

We define Green’s function $G=G(x,t)\in {\mathbb {R}}^{2\times 2}$ for the linearization of (29) as the solution to

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _t G+A\partial _x G= \begin{pmatrix} 0 &{} 0 \\ 0 &{} \nu \end{pmatrix} \partial _{x}^{2}G, &{} x\in {\mathbb {R}},\, t>0, \\ G(x,0)=\delta (x)I_2, &{} x\in {\mathbb {R}}, \end{array}\right. \end{aligned}$$

where $\delta (x)$ is the Dirac delta function and $I_2$ is the $2\times 2$ identity matrix. In addition, define $G^*=G^*(x,t)\in {\mathbb {R}}^{2\times 2}$ by

$$\begin{aligned} G^*(x,t)=\frac{1}{2(2\pi \nu t)^{1/2}}e^{-\frac{(x-ct)^2}{2\nu t}} \begin{pmatrix} 1 &{} -1/c \\ -c &{} 1 \end{pmatrix} +\frac{1}{2(2\pi \nu t)^{1/2}}e^{-\frac{(x+ct)^2}{2\nu t}} \begin{pmatrix} 1 &{} 1/c \\ c &{} 1 \end{pmatrix}. \end{aligned}$$

The next theorem is of fundamental importance in our analysis.

Theorem 3.2

([10, Theorem 5.8] and [11, Theorem 1.3]) For any $k\ge 0$, we have

$$\begin{aligned}{} & {} \biggl | \partial _{x}^{k}G(x,t)-\partial _{x}^{k}G^*(x,t)-e^{-\frac{c^2}{\nu }t}\sum _{j=0}^{k}\delta ^{(k-j)}(x)Q_j(t) \biggr |\\{} & {} \quad \le C(t+1)^{-\frac{1}{2}}t^{-\frac{k+1}{2}}\sum _{i=1}^{2}e^{-\frac{(x-\lambda _i t)^2}{Ct}}, \end{aligned}$$

where $\delta ^{(k)}(x)$ is the k-th derivative of the Dirac delta function and $Q_j=Q_j(t)$ is a $2\times 2$ polynomial matrix. In particular,

$$\begin{aligned} Q_0= \begin{pmatrix} 1 &{} 0 \\ 0 &{} 0 \end{pmatrix}, \quad Q_1= \begin{pmatrix} 0 &{} -1/\nu \\ -c^2/\nu &{} 0 \end{pmatrix}. \end{aligned}$$

Moreover, with $\gamma _i=(-1)^i \nu /(4c)$, we have

$$\begin{aligned} \begin{aligned}&\partial _{x}^{k}G(x,t)-\partial _{x}^{k}G^*(x,t)-\partial _{x}^{k+1}\sum _{i=1}^{2}\gamma _i \frac{e^{-\frac{(x-\lambda _i t)^2}{2\nu t}}}{(2\pi \nu t)^{1/2}} \begin{pmatrix} -1 &{} 0 \\ 0 &{} 1 \end{pmatrix}\\&\qquad -e^{-\frac{c^2}{\nu }t}\sum _{j=0}^{k}\delta ^{(k-j)}(x)Q_j(t) \\&\quad =O(1)(t+1)^{-\frac{1}{2}}t^{-\frac{k+1}{2}}e^{-\frac{(x-ct)^2}{Ct}} \begin{pmatrix} 1 &{} -1/c \\ -c &{} 1 \end{pmatrix}\\&\qquad +O(1)(t+1)^{-\frac{1}{2}}t^{-\frac{k+1}{2}}e^{-\frac{(x+ct)^2}{Ct}} \begin{pmatrix} 1 &{} 1/c \\ c &{} 1 \end{pmatrix} \\&\qquad +O(1)(t+1)^{-\frac{1}{2}}t^{-\frac{k+2}{2}}\sum _{i=1}^{2}e^{-\frac{(x-\lambda _i t)^2}{Ct}}. \end{aligned} \end{aligned}$$

Here O(1) is a bounded scalar function.

For the analysis of $u_i$, we need pointwise estimates for

$$\begin{aligned} g_i= \begin{pmatrix} g_{i1}&g_{i2} \end{pmatrix} {:}{=}l_i G \begin{pmatrix} r_1&r_2 \end{pmatrix}, \quad g_{i}^{*}= \begin{pmatrix} g_{i1}^{*}&g_{i2}^{*} \end{pmatrix} {:}{=}l_i G^* \begin{pmatrix} r_1&r_2 \end{pmatrix}. \end{aligned}$$

We note that

$$\begin{aligned} g_{ij}^{*}=(2\pi \nu t)^{-1/2}e^{-\frac{(x-\lambda _i t)^2}{2\nu t}}\delta _{ij}, \end{aligned}$$

(31)

where $\delta _{ij}$ is the Kronecker delta. Then Theorem 3.2 implies

$$\begin{aligned}{} & {} \biggl | \partial _{x}^{k}g_i(x,t)-\partial _{x}^{k}g_{i}^{*}(x,t)-e^{-\frac{c^2}{\nu }t}\sum _{j=0}^{k}\delta ^{(k-j)}(x)q_{ik}(t) \biggr |\nonumber \\{} & {} \quad \le C(t+1)^{-\frac{1}{2}}t^{-\frac{k+1}{2}}\sum _{i=1}^{2}e^{-\frac{(x-\lambda _i t)^2}{Ct}}, \end{aligned}$$

(32)

where

$$\begin{aligned} q_{ik}(t)=l_i Q_k(t) \begin{pmatrix} r_1&r_2 \end{pmatrix}. \end{aligned}$$

Moreover, we have

$$\begin{aligned} \begin{aligned}&\partial _{x}^{k}g_i(x,t)-\partial _{x}^{k}g_{i}^{*}(x,t)-\gamma _{i'}\partial _{x}^{k+1}g_{i'}^{*}(x,t)-e^{-\frac{c^2}{\nu }t}\sum _{j=0}^{k}\delta ^{(k-j)}(x)q_{ik}(t) \\&\quad =O(1)(t+1)^{-\frac{1}{2}}t^{-\frac{k+1}{2}}e^{-\frac{(x-ct)^2}{Ct}}+O(1)(t+1)^{-\frac{1}{2}}t^{-\frac{k+2}{2}}e^{-\frac{(x+ct)^2}{Ct}}. \end{aligned} \end{aligned}$$

(33)

We next write down an integral equation for $v_i$ defined by (26). Let

$$\begin{aligned} n=l_i \begin{pmatrix} 0 \\ N \end{pmatrix} =\frac{p''(1)}{4c^2}N, \quad n^*=-\theta _{1}^{2}/2-\theta _{2}^{2}/2-\theta _1 \Xi _1-\theta _2 \Xi _2, \end{aligned}$$

where $\Xi _i$ and N are defined by (24) and (30), respectively. Then by Duhamel’s principle, we obtain

Lemma 3.4

The function $v_i$ defined by (26) satisfies the integral equation

$$\begin{aligned} \begin{aligned} v_i(x,t)&=\int _{-\infty }^{\infty }g_i(x-y,t) \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} (y,0)\, dy-\int _{-\infty }^{\infty }g_{i}^{*}(x-y,t) \begin{pmatrix} \theta _1 \\ \theta _2 \end{pmatrix} (y,0)\, dy \\&\quad -\gamma _{i'}\int _{-\infty }^{\infty }\partial _x g_{i'}^{*}(x-y,t) \begin{pmatrix} \theta _1 \\ \theta _2 \end{pmatrix} (y,0)\, dy \\&\quad +\int _{0}^{t}\int _{-\infty }^{\infty }g_{ii}^{*}(x-y,t-s)\partial _x (n-n^*)(y,s)(y,s)\, dyds \\&\quad +\sum _{j=1}^{2}\int _{0}^{t}\int _{-\infty }^{\infty }(g_{ij}-g_{ij}^{*})(x-y,t-s)\partial _x n(y,s)\, dyds \\&\quad -\gamma _{i'}\int _{0}^{t}\int _{-\infty }^{\infty }\partial _x g_{i'i'}^{*}(x-y,t-s)\partial _x n^*(y,s)(y,s)\, dyds. \end{aligned} \end{aligned}$$

Here $\gamma _i=(-1)^i \nu /(4c)$ and $i'=3-i$.

We set

$$\begin{aligned} {\mathcal {I}}_i(x,t)&=\int _{-\infty }^{\infty }g_i(x-y,t) \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} (y,0)\, dy-\int _{-\infty }^{\infty }g_{i}^{*}(x-y,t) \begin{pmatrix} \theta _1 \\ \theta _2 \end{pmatrix} (y,0)\, dy\nonumber \\&\quad -\gamma _{i'}\int _{-\infty }^{\infty }\partial _x g_{i'}^{*}(x-y,t) \begin{pmatrix} \theta _1 \\ \theta _2 \end{pmatrix} (y,0)\, dy \end{aligned}$$

(34)

and

$$\begin{aligned} \begin{aligned} {\mathcal {N}}_i(x,t)&=\int _{0}^{t}\int _{-\infty }^{\infty }g_{ii}^{*}(x-y,t-s)\partial _x (n-n^*)(y,s)(y,s)\, dyds \\&\quad +\sum _{j=1}^{2}\int _{0}^{t}\int _{-\infty }^{\infty }(g_{ij}-g_{ij}^{*})(x-y,t-s)\partial _x n(y,s)\, dyds \\&\quad -\gamma _{i'}\int _{0}^{t}\int _{-\infty }^{\infty }\partial _x g_{i'i'}^{*}(x-y,t-s)\partial _x n^*(y,s)(y,s)\, dyds. \end{aligned} \end{aligned}$$

(35)

Lemma 3.4 may then be written as

$$\begin{aligned} v_i(x,t)={\mathcal {I}}_i(x,t)+{\mathcal {N}}_i(x,t). \end{aligned}$$

In the next two sections, we prove pointwise estimates for ${\mathcal {I}}_i(x,t)$ and ${\mathcal {N}}_i(x,t)$.

3.2.2 Contribution from the Initial Data

Our goal in this section is to prove the following pointwise estimates for ${\mathcal {I}}_i(x,t)$ defined by (34):

Lemma 3.5

For any $n\ge 1$, there exist positive constants $\delta _n$ and $C_n$ such that if (12) holds, then we have

$$\begin{aligned} |{\mathcal {I}}_i(x,t)|\le C\delta \Psi _{i;n}(x,t) \end{aligned}$$

for all $x\in {\mathbb {R}}$ and $t\ge 0$.

Proof

We assume $t\ge 1$ below (the case when $t<1$ is easier to handle). Let

$$\begin{aligned} {\mathcal {I}}_{i,1}(x,t)=\int _{-\infty }^{\infty }g_{i}^{*}(x-y,t) \begin{pmatrix} u_1-\theta _1 \\ u_2-\theta _2 \end{pmatrix} (y,0)\, dy \end{aligned}$$

and

$$\begin{aligned} {\mathcal {I}}_{i,2}(x,t)&=\int _{-\infty }^{\infty }(g_i-g_{i}^{*})(x-y,t) \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} (y,0)\, dy \\&\quad -\gamma _{i'}\int _{-\infty }^{\infty }\partial _x g_{i'}^{*}(x-y,t) \begin{pmatrix} \theta _1 \\ \theta _2 \end{pmatrix} (y,0)\, dy. \end{aligned}$$

Then of course ${\mathcal {I}}_i(x,t)={\mathcal {I}}_{i,1}(x,t)+{\mathcal {I}}_{i,2}(x,t)$.

We first show

$$\begin{aligned} |{\mathcal {I}}_{i,1}(x,t)|\le C\delta \Psi _{i;n}(x,t). \end{aligned}$$

For this purpose, set

$$\begin{aligned} \eta _j(x)=\int _{-\infty }^{x}(u_j-\theta _j)(y,0)\, dy \end{aligned}$$

and $\eta =(\eta _1 \, \eta _2)^{T}$. We then have

$$\begin{aligned} {\mathcal {I}}_{i,1}(x,t)=\int _{-\infty }^{\infty }g_{i}^{*}(x-y,t)\partial _x \eta (y)\, dy. \end{aligned}$$

By the definition of $M_i$, see (6), we have

$$\begin{aligned} \eta _j(x){:}{=}\int _{-\infty }^{x}(u_j-\theta _j)(y,0)\, dy=-\int _{x}^{\infty }(u_j-\theta _j)(y,0)\, dy. \end{aligned}$$

This and (12) imply

$$\begin{aligned} |\eta _j(x)|\le C\delta (|x|+1)^{-\beta _n}. \end{aligned}$$

We first consider Case (i) $|x-\lambda _i t|\le (t+1)^{1/2}$. In this case, integration by parts and (31) yield

$$\begin{aligned} |{\mathcal {I}}_{i,1}(x,t)|&=\left| \int _{-\infty }^{\infty }\partial _x g_{i}^{*}(x-y,t)\eta (y)\, dy \right| \\&\le C(t+1)^{-1}\int _{-\infty }^{\infty }|\eta _i(x)|\, dx\le C\delta (t+1)^{-1}\le C\delta \Psi _{i;n}(x,t). \end{aligned}$$

We next consider Case (ii) $(t+1)^{1/2}<|x-\lambda _i t|<t+1$ with $x-\lambda _i t>0$ (the case when $x-\lambda _i t<0$ is similar). Again, by integration by parts,

$$\begin{aligned} |{\mathcal {I}}_{i,1}(x,t)|&\le C(t+1)^{-1}e^{-\frac{(x-\lambda _i t)^2}{Ct}}\int _{-\infty }^{(x-\lambda _i t)/2}|\eta _i(y)|\, dy \\&\quad +C\delta (t+1)^{-1}\int _{(x-\lambda _i t)/2}^{\infty }e^{-\frac{(x-y-\lambda _i t)^2}{Ct}}(y+1)^{-\beta _n}\, dy \\&\le C\delta (t+1)^{-1}e^{-\frac{(x-\lambda _i t)^2}{Ct}}+C\delta (t+1)^{-1/2}(|x-\lambda _i t|+1)^{-\beta _n}\le C\delta \Psi _{i;n}(x,t). \end{aligned}$$

We finally consider Case (iii) $|x-\lambda _i t|\ge t+1$. For brevity, we assume $x-\lambda _i t>0$. In this case, by (12), we have

$$\begin{aligned} |{\mathcal {I}}_{i,1}(x,t)|&\le C(t+1)^{-1/2}e^{-\frac{(x-\lambda _i t)^2}{Ct}}\int _{-\infty }^{(x-\lambda _i t)/2}|(u_i-\theta _i)(y,0)|\, dy \\&\quad +C\delta (t+1)^{-1/2}\int _{(x-\lambda _i t)/2}^{\infty }e^{-\frac{(x-y-\lambda _i t)^2}{2\nu t}}(y+1)^{-\alpha _n}\, dy \\&\le C\delta e^{-\frac{t}{C}}e^{-\frac{(x-\lambda _i t)^2}{Ct}}+C\delta (|x-\lambda _i t|+1)^{-\alpha _n}\le C\delta \Psi _{i;n}(x,t). \end{aligned}$$

We next show

$$\begin{aligned} |{\mathcal {I}}_{i,2}(x,t)|\le C\delta \Psi _{i;n}(x,t). \end{aligned}$$

Writing

$$\begin{aligned} {\mathcal {I}}_{i,2}(x,t)&=\int _{-\infty }^{\infty }(g_i-g_{i}^{*}-\gamma _{i'}\partial _x g_{i'}^{*})(x-y,t) \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} (y,0)\, dy \\&\quad +\gamma _{i'}\int _{-\infty }^{\infty }\partial _x g_{i'}^{*}(x-y,t) \begin{pmatrix} u_1-\theta _1 \\ u_2-\theta _2 \end{pmatrix} (y,0)\, dy \end{aligned}$$

and applying (33), we see that it suffices to show that

$$\begin{aligned} A(x,t)&=\gamma _{i'}\partial _x \int _{-\infty }^{\infty }g_{i'}^{*}(x-y,t) \begin{pmatrix} u_1-\theta _1 \\ u_2-\theta _2 \end{pmatrix} (y,0)\, dy, \\ B(x,t)&=(t+1)^{-1}\int _{-\infty }^{\infty }e^{-\frac{(x-y-\lambda _i t)^2}{Ct}}\left| \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} \right| (y,0)\, dy, \\ C(x,t)&=(t+1)^{-3/2}\int _{-\infty }^{\infty }e^{-\frac{(x-y-\lambda _{i'}t)^2}{Ct}}\left| \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} \right| (y,0)\, dy \\ D(x,t)&=e^{-\frac{c^2}{\nu }t}\left| \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} \right| (x,0) \end{aligned}$$

are all bounded by $C\delta \Psi _{i;n}(x,t)$. First, this is trivial for D(x, t). Next, since $A(x,t)=\gamma _{i'}\partial _x {\mathcal {I}}_{i',1}(x,t)$, modifying the calculations above for ${\mathcal {I}}_{i,1}(x,t)$ yield the bound for A(x, t). The bound for B(x, t) is also obtained in a way similar to that for ${\mathcal {I}}_{i,1}(x,t)$ (except that we don’t need $\eta _i$ in the analysis).

Let us finally consider C(x, t). First, Case (i) $|x-\lambda _{i'}t|\le (t+1)^{1/2}$ is easy:

$$\begin{aligned} |C(x,t)|\le C\delta (t+1)^{-3/2}\le C\delta \Psi _{i;n}(x,t). \end{aligned}$$

Case (ii) $|x-\lambda _{i'}t|>(t+1)^{1/2}$ with $x-\lambda _{i'}t>0$ is as follows:

$$\begin{aligned} |C(x,t)|&\le C(t+1)^{-3/2}e^{-\frac{(x-\lambda _{i'}t)^2}{Ct}}\int _{-\infty }^{(x-\lambda _{i'}t)/2}\left| \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} \right| (y,0)\, dy \\&\quad +C\delta (t+1)^{-3/2}\int _{(x-\lambda _{i'}t)/2}^{\infty }e^{-\frac{(x-y-\lambda _{i'}t)^2}{Ct}}(y+1)^{-\alpha _n}\, dy \\&\le C\delta (t+1)^{-3/2}e^{-\frac{(x-\lambda _{i'}t)^2}{Ct}}+C\delta (t+1)^{-1}(|x-\lambda _{i'}t|+1)^{-\alpha _n}\le C\delta \Psi _{i;n}(x,t). \end{aligned}$$

The case when $x-\lambda _{i'}t<0$ is similar. This ends the proof of the lemma. $\quad \square$

3.2.3 Contribution from the Nonlinear Terms

Our goal in this section is to prove the following pointwise estimates for ${\mathcal {N}}_i(x,t)$ defined by (35):

Lemma 3.6

For any $n\ge 1$, there exist positive constants $\delta _n$ and $C_n$ such that if (12) holds, then we have

$$\begin{aligned} |{\mathcal {N}}_i(x,t)|\le C(\delta +P(t))^2 \Psi _{i;n}(x,t) \end{aligned}$$

for all $x\in {\mathbb {R}}$ and $t\ge 0$.

To prove this lemma, we first prove some preparatory lemmas. To state these, we introduce the notation

$$\begin{aligned} {\mathcal {I}}_i[f](x,t){:}{=}\int _{0}^{t}\int _{-\infty }^{\infty }\partial _x \left\{ (t-s)^{-1/2}e^{-\frac{(x-y-\lambda _i(t-s))^2}{2\nu (t-s)}} \right\} f(y,s)\, dyds \end{aligned}$$

for a function $f=f(x,t)$.

Lemma 3.7

Let $n\ge 1$ and $\varepsilon =\max (M_1,M_2)$. If $\varepsilon$ is sufficiently small, we have

$$\begin{aligned} |{\mathcal {I}}_i[\theta _{i'}\Xi _i](x,t)|\le C\varepsilon ^3 \Psi _{i;n}(x,t). \end{aligned}$$

Proof

We only prove the lemma for $i=1$ (the other case is similar). By Lemma 3.3, we have

$$\begin{aligned} |(\theta _2 \Xi _1)(x,t)+(2c)^{-1}(\theta _{2}^{3}/2+\theta _{2}^{2}\Xi _2)(x,t)|\le C\varepsilon ^3 \Theta _{4}(x,t;-c,\nu ^*). \end{aligned}$$

Since

$$\begin{aligned} \Theta _4(x,t;-c,\nu ^*)\le C(t+1)^{-(2+1/2^{n+1})/2}\psi _n(x,t;-c), \end{aligned}$$

Lemma A.7 implies

$$\begin{aligned} |{\mathcal {I}}_1[\Theta _4(\cdot ,\cdot ;-c,\nu ^*)](x,t)|\le C\Psi _{1;n}(x,t). \end{aligned}$$

Next, note that (3) and (25) imply

$$\begin{aligned} L_2 \theta _2=-\partial _x(\theta _{2}^{2}/2), \quad L_2 \Xi _2=-\partial _x(\theta _{1}^{2}/2+\theta _1 \Xi _1+\theta _2 \Xi _2), \end{aligned}$$

where $L_2=\partial _t-c\partial _x-(\nu /2)\partial _{x}^{2}$. Using these, similar to the bound for $\zeta _{1;n}$ in the proof of Lemma 3.1, we can show that

$$\begin{aligned} |{\mathcal {I}}_1[\theta _{2}^{3}/2+\theta _{2}^{2}\Xi _2](x,t)|\le C\varepsilon ^3 \Psi _{1;n}(x,t). \end{aligned}$$

Combining these, we obtain the lemma. $\quad \square$

We next show the following:

Lemma 3.8

Let $n\ge 1$ and $\varepsilon =\max (M_1,M_2)$. If $\varepsilon$ is sufficiently small, we have

$$\begin{aligned} |{\mathcal {I}}_i[\partial _x \theta _{i'}^{2}](x,t)|\le C\varepsilon ^2 \Psi _{i;n}(x,t). \end{aligned}$$

Proof

We only prove the lemma for $i=1$ (the other case is similar). Applying Lemma A.1 yields

$$\begin{aligned} {\mathcal {I}}_1[\partial _x \theta _{2}^{2}](x,t)=(2c)^{-1}\sqrt{2\pi \nu }\partial _x \theta _{2}^{2}(x,t)+I_1(x,t)+I_2(x,t), \end{aligned}$$

where

$$\begin{aligned} I_1(x,t)=\int _{0}^{t^{1/2}}\int _{-\infty }^{\infty }\partial _x \left\{ (t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}} \right\} \partial _x \theta _{2}^{2}(y,s)\, dyds \end{aligned}$$

and

$$\begin{aligned} I_2(x,t)&=-(2c)^{-1}\int _{-\infty }^{\infty }(t-t^{1/2})^{-1/2}e^{-\frac{(x-y-c(t-\sqrt{t}))^2}{2\nu (t-\sqrt{t})}}\partial _x \theta _{2}^{2}(y,t^{1/2})\, dy \\&\quad -(2c)^{-1}\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}\partial _x L_2 \theta _{2}^{2}(y,s)\, dyds \\&{=}{:}I_{21}(x,t)+I_{22}(x,t). \end{aligned}$$

By Lemmas A.2 and A.3, we obtain

$$\begin{aligned} |I_1(x,t)|+|I_{21}(x,t)|\le C\varepsilon ^2 \log (t+2)\Theta _2(x,t;c,\nu ^*)\le C\varepsilon ^2 \Psi _{1;n}(x,t). \end{aligned}$$

For $I_{22}(x,t)$, we apply Lemma A.1 (without the integral on $[0,t^{1/2}]$), which yields

$$\begin{aligned} {\mathcal {I}}_{22}(x,t)=-(2c)^{-2}\sqrt{2\pi \nu }L_2 \theta _{2}^{2}(x,t)+J_2(x,t), \end{aligned}$$

where

$$\begin{aligned} J_2(x,t)&=(2c)^{-2}\int _{-\infty }^{\infty }(t-t^{1/2})^{-1/2}e^{-\frac{(x-y-c(t-\sqrt{t}))^2}{2\nu (t-\sqrt{t})}}L_2 \theta _{2}^{2}(y,t^{1/2})\, dy \\&\quad +(2c)^{-2}\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}L_{2}^{2}\theta _{2}^{2}(y,s)\, dyds \\&{=}{:}J_{21}(x,t)+J_{22}(x,t). \end{aligned}$$

By some tedious calculations, we obtain

$$\begin{aligned} L_2 \theta _{2}^{2}=-2\partial _x(\theta _{2}^{3}/3)-\nu (\partial _x \theta _2)^2 \end{aligned}$$

and

$$\begin{aligned} L_{2}^{2}\theta _{2}^{2}=\partial _{x}^{2}(\theta _{2}^{4}/2)+\nu \partial _x[(\partial _x \theta _2)\partial _x \theta _{2}^{2}]+\nu (\partial _x \theta _2)\partial _{x}^{2}\theta _{2}^{2}+\nu ^2(\partial _{x}^{2}\theta _2)^2. \end{aligned}$$

Since $|L_2 \theta _{2}^{2}(x,t)|\le C\varepsilon ^2 \Theta _4(x,t;-c,\nu ^*)$, Lemma A.3 yields

$$\begin{aligned} |J_{21}(x,t)|\le C\varepsilon ^2 \Theta _{5/2}(x,t;c,\nu ^*)\le C\varepsilon ^2 \Psi _{1;n}(x,t). \end{aligned}$$

And since $|L_{2}^{2}\theta _{2}^{2}(x,t)|\le C\varepsilon ^2 \Theta _6(x,t;-c,\nu ^*)$, Lemma A.4 implies

$$\begin{aligned} |J_{22}(x,t)|\le C\varepsilon ^2 [(x-c(t+1))^2+(t+1)]^{-5/4}\le C\varepsilon ^2 \Psi _{1;n}(x,t;c). \end{aligned}$$

This proves the lemma. $\quad \square$

Similarly, we can now show the following:

Lemma 3.9

Let $n\ge 1$ and $\varepsilon =\max (M_1,M_2)$. If $\varepsilon$ is sufficiently small, we have

$$\begin{aligned} |{\mathcal {I}}_i[\partial _x(\theta _{i'}\Xi _{i'})](x,t)|\le C\varepsilon ^3 \Psi _{i;n}(x,t). \end{aligned}$$

We move on to prove the following:

Lemma 3.10

Let $n\ge 1$. If $\delta$ defined by (12) is sufficiently small, we have

$$\begin{aligned} |{\mathcal {I}}_i[\theta _{i'}v_{i'}](x,t)|\le C(\delta +P(t))^2 \Psi _{i;n}(x,t). \end{aligned}$$

Here P(t) is defined by (27).

Proof

We only prove the lemma for $i=1$ (the other case is similar). Set $f=\theta _2 v_2$. Then Lemma A.1 implies

$$\begin{aligned} {\mathcal {I}}_1[\theta _2 v_2](x,t)=(2c)^{-1}\sqrt{2\pi \nu }f(x,t)+I_1(x,t)+I_2(x,t), \end{aligned}$$

where

$$\begin{aligned} I_1(x,t)=\int _{0}^{t^{1/2}}\int _{-\infty }^{\infty }\partial _x \left\{ (t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}} \right\} f(y,s)\, dyds \end{aligned}$$

and

$$\begin{aligned} I_2(x,t)&=-(2c)^{-1}\int _{-\infty }^{\infty }(t-t^{1/2})^{-1/2}e^{-\frac{(x-y-c(t-\sqrt{t}))^2}{2\nu (t-\sqrt{t})}}f(y,t^{1/2})\, dy \\&\quad -(2c)^{-1}\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{2\nu (t-s)}}L_2 f(y,s)\, dyds \\&{=}{:}I_{21}(x,t)+I_{22}(x,t). \end{aligned}$$

By Lemmas A.2 and A.3, we obtain

$$\begin{aligned} |I_1(x,t)|+|I_{21}(x,t)|\le C\delta P(t)\Theta _{\alpha _{n+1}}(x,t;c,\nu ^*)\le C\delta P(t)\Psi _{1;n}(x,t). \end{aligned}$$

To bound $I_{22}(x,t)$, we first note that

$$\begin{aligned} L_2 v_2&=L_2(u_2-\theta _2-\Xi _2-\gamma _1 \partial _x \theta _1-\gamma _1 \partial _x \Xi _1) \\&=\frac{\nu }{2}\partial _{x}^{2}u_1+\partial _x n+\partial _x(\theta _{2}^{2}/2)+\partial _x(\theta _{1}^{2}/2+\theta _1 \Xi _1+\theta _2 \Xi _2) \\&\quad +\gamma _1 \partial _{x}^{2}(\theta _{1}^{2}/2+2c\theta _1)+\gamma _1 \partial _{x}^{2}(\theta _{2}^{2}/2+\theta _1 \Xi _1+\theta _2 \Xi _2+2c\Xi _1) \\&=\frac{\nu }{2}\partial _{x}^{2}(u_1-\theta _1-\Xi _1)+\partial _x(n-n_*)-\gamma _1 \partial _{x}^{2}n_*. \end{aligned}$$

Then set

$$\begin{aligned} F=\frac{\nu }{2}\partial _x(u_1-\theta _1-\Xi _1)+n-n_*-\gamma _1 \partial _x n_* \end{aligned}$$

and

$$\begin{aligned} G=\theta _2 F-\nu v_2 \partial _x \theta _2. \end{aligned}$$

We note that

$$\begin{aligned} L_2 f=-v_2 \partial _x(\theta _{2}^{2}/2)+\partial _x G-(\partial _x \theta _2)F+\nu v_2 \partial _{x}^{2}\theta _2. \end{aligned}$$

By [10, Theorem 2.6 and Remark 2.8], we have

$$\begin{aligned} |\partial _x(u_1-\theta _1)(x,t)|\le C\delta (t+1)^{-1/2}\Psi _{1;0}(x,t), \quad |\partial _{x}^{2}u_1(x,t)|\le C\delta (t+1)^{-3/2}.^2 \end{aligned}$$

² In addition, applying Taylor’s theorem, we see that

$$\begin{aligned} |(n-n_*)(x,t)|\le C(\delta +P(t))^2(t+1)^{-1/2}[\psi _n(x,t;c)+\psi _n(x,t;-c)]. \end{aligned}$$

These imply

$$\begin{aligned} |G(x,t)|\le C(\delta +P(t))^2 (t+1)^{-1}\Theta _{\alpha _n}(x,t;-c,\nu ^*) \end{aligned}$$

and

$$\begin{aligned} |L_2 f(x,t)-\partial _x G(x,t)|\le C(\delta +P(t))^2 (t+1)^{-3/2}\Theta _{\alpha _n}(x,t;-c,\nu ^*). \end{aligned}$$

Using these and integration by parts, we get

$$\begin{aligned} |I_{22}(x,t)|&\le C(\delta +P(t))^2 \int _{t^{1/2}}^{t/2}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{C(t-s)}}(s+1)^{-3/2}\psi _n(y,s;-c)\, dyds \\&\quad +C(\delta +P(t))^2 \int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }(t-s)^{-1}e^{-\frac{(x-y-c(t-s))^2}{C(t-s)}}(s+1)^{-1}\psi _n(y,s;-c)\, dyds. \end{aligned}$$

Applying Lemmas A.7 and A.9, we obtain

$$\begin{aligned} |I_{22}(x,t)|\le C(\delta +P(t))^2 \Psi _{1;n}(x,t). \end{aligned}$$

This proves the lemma. $\quad \square$

The lemma below can be shown in a similar manner.

Lemma 3.11

Let $n\ge 1$. If $\delta$ defined by (12) is sufficiently small, we have

$$\begin{aligned} |{\mathcal {I}}_i[\Xi _{i'}^{2}](x,t)|+|{\mathcal {I}}_i[\Xi _{i'}v_{i'}](x,t)|\le C(\delta +P(t))^3 \Psi _{i;n}(x,t). \end{aligned}$$

Set

$$\begin{aligned} n_a=-\frac{p''(1)^2}{8c^2}(v-1)^2, \quad n_b=-\frac{\nu p''(1)}{4c^2}(v-1)u_x, \quad n_c=n-n_a-n_b.\nonumber \\ \end{aligned}$$

(36)

Of course $n=n_a+n_n+n_c$. Correspondingly, set

$$\begin{aligned} \begin{aligned} {\mathcal {N}}_{i,a}(x,t)&=\int _{0}^{t}\int _{-\infty }^{\infty }g_{ii}^{*}(x-y,t-s)\partial _x (n_a-n^*)(y,s)(y,s)\, dyds \\&\quad +\sum _{j=1}^{2}\int _{0}^{t}\int _{-\infty }^{\infty }(g_{ij}-g_{ij}^{*})(x-y,t-s)\partial _x n_a(y,s)\, dyds \\&\quad -\gamma _{i'}\int _{0}^{t}\int _{-\infty }^{\infty }\partial _x g_{i'i'}^{*}(x-y,t-s)\partial _x n^*(y,s)(y,s)\, dyds, \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} {\mathcal {N}}_{i,b}(x,t)&=\int _{0}^{t}\int _{-\infty }^{\infty }g_{ii}^{*}(x-y,t-s)\partial _x n_b(y,s)(y,s)\, dyds \\&\quad +\sum _{j=1}^{2}\int _{0}^{t}\int _{-\infty }^{\infty }(g_{ij}-g_{ij}^{*})(x-y,t-s)\partial _x n_b(y,s)\, dyds, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} {\mathcal {N}}_{i,c}(x,t)&=\int _{0}^{t}\int _{-\infty }^{\infty }g_{ii}^{*}(x-y,t-s)\partial _x n_c(y,s)(y,s)\, dyds \\&\quad +\sum _{j=1}^{2}\int _{0}^{t}\int _{-\infty }^{\infty }(g_{ij}-g_{ij}^{*})(x-y,t-s)\partial _x n_c(y,s)\, dyds. \end{aligned} \end{aligned}$$

Then ${\mathcal {N}}_i(x,t)={\mathcal {N}}_{i,a}(x,t)+{\mathcal {N}}_{i,b}(x,t)+{\mathcal {N}}_{i,c}(x,t)$; see (35).

We next prove the following:

Lemma 3.12

Let $n\ge 1$. If $\delta$ defined by (12) is sufficiently small, we have

$$\begin{aligned} |{\mathcal {N}}_{i,a}(x,t)|\le C(\delta +P(t))^2 \Psi _{i;n}(x,t). \end{aligned}$$

Proof

Let $i=1$ (the case of $i=2$ is similar). By integration by parts, we have

$$\begin{aligned} {\mathcal {N}}_{1,a}(x,t)&=\int _{0}^{t}\int _{-\infty }^{\infty }\partial _x g_{11}^{*}(x-y,t-s)(n_a-n^*)(y,s)\, dyds \\&\quad +\sum _{j=1}^{2}\int _{0}^{t}\int _{-\infty }^{\infty }\partial _x (g_{1j}-g_{1j}^{*})(x-y,t-s)n_a(y,s)\, dyds \\&\quad -\gamma _2 \int _{0}^{t}\int _{-\infty }^{\infty }\partial _{x}^{2}g_{22}^{*}(x-y,t-s)n^*(y,s)\, dyds. \end{aligned}$$

By some tedious calculations, we can show that

$$\begin{aligned} \begin{aligned}&|(n_a-n^*)(x,t)-[\theta _2 \Xi _1+\gamma _2 \partial _x(\theta _{2}^{2}/2)+\gamma _2 \partial _x(\theta _2 \Xi _2)-\theta _2 v_2-\Xi _{2}^{2}/2-\Xi _2 v_2](x,t)| \\&\le C(\delta +P(t))^2[(t+1)^{-1/2}\psi _n(x,t;c)+(t+1)^{-\alpha _n/2}\psi _n(x,t;-c)]. \end{aligned} \end{aligned}$$

Then Lemmas 3.7–3.11, A.6, and A.7 yield

$$\begin{aligned} \left| \int _{0}^{t}\int _{-\infty }^{\infty }\partial _x g_{11}^{*}(x-y,t-s)(n_a-n^*)(y,s)\, dyds \right| \le C(\delta +P(t))^2 \Psi _{i;n}(x,t). \end{aligned}$$

It remains to show that

$$\begin{aligned} |I(x,t)|\le C(\delta +P(t))^2 \Psi _{1;n}(x,t), \end{aligned}$$

where

$$\begin{aligned} I(x,t)&=\sum _{j=1}^{2}\int _{0}^{t}\int _{-\infty }^{\infty }\partial _x (g_{1j}-g_{1j}^{*})(x-y,t-s)n_a(y,s)\, dyds \\&\quad -\gamma _2 \int _{0}^{t}\int _{-\infty }^{\infty }\partial _{x}^{2}g_{22}^{*}(x-y,t-s)n^*(y,s)\, dyds. \end{aligned}$$

We define the decomposition $I(x,t)=I_1(x,t)+I_2(x,t)$ by

$$\begin{aligned} I_1(x,t)=\sum _{j=1}^{2}\int _{0}^{t}\int _{-\infty }^{\infty }\partial _x (g_{1j}-g_{1j}^{*})(x-y,t-s)(n_a-n^*)(y,s)\, dyds \end{aligned}$$

and

$$\begin{aligned} I_2(x,t)&= \sum _{j=1}^{2}\int _{0}^{t}\int _{-\infty }^{\infty }\partial _x (g_{1j}-g_{1j}^{*})(x-y,t-s)n^*(y,s)\, dyds \\&\quad -\gamma _2 \int _{0}^{t}\int _{-\infty }^{\infty }\partial _{x}^{2}g_{22}^{*}(x-y,t-s)n^*(y,s)\, dyds. \end{aligned}$$

We first consider $I_1(x,t)$. By (32), to show that $|I_1(x,t)|$ is bounded by $C(\delta +P(t))^2 \Psi _{1;n}(x,t)$, it suffices to prove the same bound for

$$\begin{aligned} \int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1}(t-s+1)^{-1/2}e^{-\frac{(x-y-\lambda _j(t-s))^2}{C(t-s)}}|(n_a-n^*)(y,s)|\, dyds \quad (j=1,2) \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{t}e^{-\frac{c^2}{\nu }(t-s)}|(n_a-n^*)(x,s)|\, ds. \end{aligned}$$

The term corresponding to $\delta ^{(1)}(x)$ is not needed since $q_{10}=(1/2 \, -1/2)^{T}$. Noting that

$$\begin{aligned} |(n_a-n^*)(x,t)|\le C(\delta +P(t))^2(t+1)^{-1/2}[\psi _n(x,t;c)+\psi _n(x,t;-c)], \end{aligned}$$

Lemmas A.6, A.7, and A.10 imply the desired bounds for the two integrals above.

We next consider $I_2(x,t)$. We have $I_2(x,t)=I_{21}(x,t)+I_{22}(x,t)$ with

$$\begin{aligned} I_{21}(x,t)&=-\sum _{j=1}^{2}\int _{0}^{t}\int _{-\infty }^{\infty }\partial _x (g_{1j}-g_{1j}^{*})(x-y,t-s)(\theta _{1}^{2}/2+\theta _1 \Xi _1)(y,s)\, dyds \\&\quad +\gamma _2 \int _{0}^{t}\int _{-\infty }^{\infty }\partial _{x}^{2}g_{22}^{*}(x-y,t-s)(\theta _{1}^{2}/2+\theta _1 \Xi _1)(y,s)\, dyds \end{aligned}$$

and

$$\begin{aligned} I_{22}(x,t)&=-\sum _{j=1}^{2}\int _{0}^{t}\int _{-\infty }^{\infty }\partial _x (g_{1j}-g_{1j}^{*})(x-y,t-s)(\theta _{2}^{2}/2+\theta _2 \Xi _2)\, dyds \\&\quad +\gamma _2 \int _{0}^{t}\int _{-\infty }^{\infty }\partial _{x}^{2}g_{22}^{*}(x-y,t-s)(\theta _{2}^{2}/2+\theta _2 \Xi _2)\, dyds. \end{aligned}$$

Taking into account (32) and (33), Lemmas A.6, A.7, and A.10 yield $|I_{21}(x,t)|\le C(\delta +P(t))^2 \Psi _{1;n}(x,t)$ (we divide the domain of temporal integration into [0, t/2] and [t/2, t] then use integration by parts before applying the lemmas). For $I_{22}(x,t)$, we proceed as follows: using the technique in the proof of Lemma A.1, we obtain

$$\begin{aligned} I_{22}(x,t)&=-\sum _{j=1}^{2}\int _{0}^{t^{1/2}}\int _{-\infty }^{\infty }\partial _x(g_{1j}-g_{1j}^{*})(x-y,t-s)(\theta _{2}^{2}/2+\theta _2 \Xi _2)(y,s)\, dyds \\&\quad +\gamma _2 \int _{0}^{t^{1/2}}\int _{-\infty }^{\infty }\partial _{x}^{2}g_{22}^{*}(x-y,t-s)(\theta _{2}^{2}/2+\theta _2 \Xi _2)(y,s)\, dyds \\&\quad -\frac{1}{2c}\sum _{j=1}^{2}\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }L_1(g_{1j}-g_{1j}^{*})(x-y,t-s)(\theta _{2}^{2}/2+\theta _2 \Xi _2)(y,s)\, dyds \\&\quad +\frac{\nu }{4c}\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }\partial _{x}^{2}g_{22}^{*}(x-y,t-s)(\theta _{2}^{2}/2+\theta _2 \Xi _2)(y,s)\, dyds \\&\quad +\frac{1}{2c}\sum _{j=1}^{2}\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }(g_{1j}-g_{1j}^{*})(x-y,t-s)L_2(\theta _{2}^{2}/2+\theta _2 \Xi _2)(y,s)\, dyds \\&\quad +\frac{1}{2c}\sum _{j=1}^{2}\int _{-\infty }^{\infty }(g_{1j}-g_{1j}^{*})(x-y,t-t^{1/2})(\theta _{2}^{2}/2+\theta _2 \Xi _2)(y,t^{1/2})\, dyds, \end{aligned}$$

where $L_i=\partial _t+\lambda _i \partial _x-(\nu /2)\partial _{x}^{2}$. Here we used $\lim _{t\rightarrow 0}(g_{1j}-g_{1j}^{*})(x,t)=0$. The sum of the first two terms on the right-hand side can be bounded using Lemmas A.6, A.8, and A.10; the sum of the third and the fourth term can be bounded using Lemmas A.6, A.7, A.10, and the relation

$$\begin{aligned} L_1(g_{1j}-g_{1j}^{*})=(\nu /2)\partial _{x}^{2}g_{2j}. \end{aligned}$$

To bound the sum of the fifth and the sixth term, noting that

$$\begin{aligned} |L_2(\theta _{2}^{2}/2+\theta _2 \Xi _2)(x,t)|\le C\delta ^2 \Theta _4(x,t;-c,\nu ^*), \end{aligned}$$

it suffices to show that the following integrals are bounded by $C(\delta +P(t))^2 \Psi _{1;n}(x,t)$:

$$\begin{aligned} A(x,t)= & {} \int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}(t-s+1)^{-1/2}e^{-\frac{(x-y-c(t-s))^2}{C(t-s)}}\Theta _4(y,s;-c,\nu ^*)\, dyds,\\ B(x,t)= & {} \int _{-\infty }^{\infty }(t-t^{1/2})^{-1}e^{-\frac{(x-y-c(t-\sqrt{t}))^2}{C(t-\sqrt{t})}}\Theta _2(y,t^{1/2};-c,\nu ^*)\, dy, \\ C(x,t)= & {} \int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }(t-s)^{-1}(t-s+1)^{-1/2}e^{-\frac{(x-y+c(t-s))^2}{C(t-s)}}\Theta _4(y,s;-c,\nu ^*)\, dyds, \\ D(x,t)= & {} \int _{-\infty }^{\infty }(t-t^{1/2})^{-1}(t-t^{1/2}+1)^{-1/2}e^{-\frac{(x-y+c(t-\sqrt{t}))^2}{C(t-\sqrt{t})}}\Theta _2(y,t^{1/2};-c,\nu ^*)\, dy, \end{aligned}$$

and

$$\begin{aligned} E(x,t)&=\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }\partial _x g_{22}^{*}(x-y,t-s)L_2(\theta _{2}^{2}/2+\theta _2 \Xi _2)(y,s)\, dyds \\&\quad +\int _{-\infty }^{\infty }\partial _x g_{22}^{*}(x-y,t-t^{1/2})(\theta _{2}^{2}/2+\theta _2 \Xi _2)(y,t^{1/2})\, dy. \end{aligned}$$

We can bound A(x, t) using Lemma A.4, B(x, t) and D(x, t) by Lemma A.3, and C(x, t) by Lemma A.6. Finally, we consider E(x, t). Taking into account $L_2 g_{22}^{*}=0$ and $\lim _{t\rightarrow 0}\partial _x g_{22}^{*}(x-y,t)=\delta ^{(1)}(x-y)$, integration by parts applied to the operator $L_2$ yields

$$\begin{aligned} E(x,t)=\int _{-\infty }^{\infty }\delta ^{(1)}(x-y)(\theta _{2}^{2}/2+\theta _2 \Xi _2)(y,t)\, dy=\partial _x(\theta _{2}^{2}/2+\theta _2 \Xi _2)(x,t). \end{aligned}$$

Hence $|E(x,t)|\le C(\delta +P(t))^2 \Psi _{1;n}(x,t)$. This ends the proof of the lemma. $\square$

The lemma below can be proved similarly. Note that ${\mathcal {N}}_{i,b}(x,t)$ is related to the nonlinear term $(v-1)u_x$ as opposed to $(v-1)^2$ for ${\mathcal {N}}_{i,a}(x,t)$; see (36). As in the bound of ${\mathcal {N}}_{i,a}(x,t)$, the term $(v-1)$ is dealt with the inequality $|v_i(x,s)|\le P(t)\Psi _{i;n}(x,s)$ ($0\le s\le t$); on the other hand, the first derivative $u_x$ is handled using [10, Theorem 2.6 and Remark 2.8] as in the proof of Lemma 3.10. The term ${\mathcal {N}}_{i,c}(x,t)$ can be handled in a similar manner.

Lemma 3.13

Let $n\ge 1$. If $\delta$ defined by (12) is sufficiently small, we have

$$\begin{aligned} |{\mathcal {N}}_{i,b}(x,t)|+|{\mathcal {N}}_{i,c}(x,t)|\le C(\delta +P(t))^2 \Psi _{i;n}(x,t). \end{aligned}$$

Combining Lemmas 3.12 and 3.13, the proof of Lemma 3.6 is complete.

3.2.4 Final Step of the Proof

The remaining step of the proof is standard. By Lemma 3.5 and 3.6, we obtain

$$\begin{aligned} P(t)\le C\delta +C(\delta +P(t))^2 \le C_1 \delta +C_2 P(t)^2 \end{aligned}$$

(37)

for some $C_1,C_2>0$. Here P(t) is defined by (27). When $\delta$ is sufficiently small, the line $y=p$ and the parabola $y=C_1 \delta +C_2 p^2$ intersect at $p=p_1$ and $p_2$, where

$$\begin{aligned} p_1=\frac{1-\sqrt{1-4C_1 C_2 \delta }}{2C_2}, \quad p_2=\frac{1+\sqrt{1-4C_1 C_2 \delta }}{2C_2}. \end{aligned}$$

Note that $C_1 \delta \le p_1<p_2$. By (37), we either have $P(t)\le p_1$ or $P(t)\ge p_2$. Since P(t) is continuous in t, if $P(0)\le p_1$, then $P(t)\le p_1$ for all $t\ge 0$. By (12), taking $C_1$ sufficiently large, we indeed have $P(0)\le C_1 \delta \le p_1$. Therefore, we conclude that

$$\begin{aligned} P(t)\le p_1 \le \frac{1-(1-4C_1 C_2 \delta )}{2C_2}\le 2C_1 \delta . \end{aligned}$$

This ends the proof of Theorem 2.1.

Appendix A. Lemmas on Convolutions Involving a Heat Kernel

Lemma A.1

Suppose that $f=f(x,t)$ is a $C^2$-smooth function on ${\mathbb {R}}\times (0,\infty )$. Let $\lambda \ne \lambda '$ and $\nu >0$, and set $L_{\lambda '}=\partial _t+\lambda ' \partial _x-(\nu /2)\partial _{x}^{2}$. Then for $t\ge 1$, the function I(x, t) defined by

$$\begin{aligned} I(x,t)=\int _{0}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-\lambda (t-s))^2}{2\nu (t-s)}}\partial _x f(y,s)\, dyds \end{aligned}$$

can be written as

$$\begin{aligned} I(x,t)=(\lambda -\lambda ')^{-1}\sqrt{2\pi \nu }f(x,t)+I_1(x,t)+I_2(x,t), \end{aligned}$$

where

$$\begin{aligned} I_1(x,t)=\int _{0}^{t^{1/2}}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-\lambda (t-s))^2}{2\nu (t-s)}}\partial _x f(y,s)\, dyds \end{aligned}$$

and

$$\begin{aligned} I_2(x,t)&=-(\lambda -\lambda ')^{-1}\int _{-\infty }^{\infty }(t-t^{1/2})^{-1/2}e^{-\frac{(x-y-\lambda (t-\sqrt{t}))^2}{2\nu (t-\sqrt{t})}}f(y,t^{1/2})\, dy \\&\quad -(\lambda -\lambda ')^{-1}\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-\lambda (t-s))^2}{2\nu (t-s)}}L_{\lambda '}f(y,s)\, dyds. \end{aligned}$$

Proof

Let

$$\begin{aligned} g_{\lambda }(x,t)=t^{-1/2}e^{-\frac{(x-\lambda t)^2}{2\nu t}}. \end{aligned}$$

Dividing the domain of temporal integration, we get

$$\begin{aligned} I(x,t)&=\int _{0}^{t^{1/2}}\int _{-\infty }^{\infty }g_{\lambda }(x-y,t-s)\partial _x f(y,s)\, dyds \\&\quad +\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }g_{\lambda }(x-y,t-s)\partial _x f(y,s)\, dyds. \end{aligned}$$

The first term on the right-hand side is $I_1(x,t)$. For the second term, integration by parts yields

$$\begin{aligned}&\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }g_{\lambda }(x-y,t-s)\partial _x f(y,s)\, dyds \\&\quad =-(\lambda -\lambda ')^{-1}\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }g_{\lambda }(x-y,t-s) \\&\qquad \cdot [-\partial _s+\partial _s-\lambda \partial _y+\lambda ' \partial _y+(\nu /2)\partial _{y}^{2}-(\nu /2)\partial _{y}^{2}]f(y,s)\, dyds \\&\quad =(\lambda -\lambda ')^{-1}\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }L_{\lambda }g_{\lambda }(x-y,t-s)f(y,s)\, dyds \\&\qquad -(\lambda -\lambda ')^{-1}\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }g_{\lambda }(x-y,t-s)L_{\lambda '}f(y,s)\, dyds \\&\qquad +(\lambda -\lambda ')^{-1}\sqrt{2\pi \nu }f(x,t)-(\lambda -\lambda ')^{-1}\int _{-\infty }^{\infty }g_{\lambda }(x-y,t-t^{1/2})f(x,t^{1/2})\, dyds, \end{aligned}$$

where $L_{\lambda }=\partial _s+\lambda \partial _y-(\nu /2)\partial _{y}^{2}$. Here we used $\lim _{s\rightarrow t}g_{\lambda }(x-y,t-s)=\sqrt{2\pi \nu }\delta (x-y)$. The lemma follows from the equality above by noting that $L_{\lambda }g_{\lambda }=0$. $\quad \square$

In the lemmas below, C and $\nu ^*$ denote generic large constants. We remind the reader that $\Theta _{\alpha }(x,t;\lambda ,\mu )$ is defined by (13).

Lemma A.2

Let $\lambda \ne \lambda '$, $\mu >0$, and $0<\alpha \le 3$. Then we have

$$\begin{aligned}&\int _{0}^{t^{1/2}}\int _{-\infty }^{\infty }(t-s)^{-1}e^{-\frac{(x-y-\lambda (t-s))^2}{\mu (t-s)}}\Theta _{\alpha }(y,s;\lambda ',\mu )\, dyds \\&\quad \le \left\{ \begin{array}{ll} C\Theta _{(\alpha +1)/2}(x,t;\lambda ,\nu ^*) &{} \text{ if } \alpha \ne 3, \\ C\log (t+2)\Theta _{(\alpha +1)/2}(x,t;\lambda ,\nu ^*) &{} \text{ if } \alpha =3. \end{array} \right. \end{aligned}$$

Proof

See the analysis of $I_1(x,t)$ in the proof of [10, Lemma 3.4]. $\quad \square$

Lemma A.3

Let $\lambda ,\lambda ' \in {\mathbb {R}}$, $\mu >0$, and $\alpha >0$ (not necesarily $\lambda \ne \lambda '$). Then for $t\ge 1$, we have

$$\begin{aligned} \int _{-\infty }^{\infty }e^{-\frac{(x-y-\lambda (t-\sqrt{t}))^2}{\mu (t-\sqrt{t})}}\Theta _{\alpha }(y,t^{1/2};\lambda ',\mu )\, dyds\le C\Theta _{(\alpha -1)/2}(x,t;\lambda ,\nu ^*). \end{aligned}$$

Proof

See the analysis of $I_{21}(x,t)$ in the proof of [10, Lemma 3.4]. $\quad \square$

Lemma A.4

Let $\lambda \ne \lambda '$, $\mu >0$, and $\alpha >1$. Then we have

$$\begin{aligned} \begin{aligned}&\int _{t^{1/2}}^{t}\int _{-\infty }^{\infty }(t-s)^{-1/2}e^{-\frac{(x-y-\lambda (t-s))^2}{\mu (t-s)}}\Theta _{\alpha }(y,s;\lambda ',\mu )\, dyds \\&\quad \le C[(x-\lambda (t+1))^2+(t+1)]^{-(\alpha -1)/4}. \end{aligned} \end{aligned}$$

Proof

See the analysis of $I_{22}^{(1)}(x,t)$ in the proof of [10, Lemma 3.4]. $\quad \square$

Lemma A.5

([10, Lemma 3.2]) Let $\lambda \in {\mathbb {R}}$, $\mu >0$, $\alpha \ge 0$, and $\beta >0$. Then we have

$$\begin{aligned} \begin{aligned}&\int _{0}^{t/2}\int _{-\infty }^{\infty }(t-s)^{-1}(t+1-s)^{-\alpha /2}e^{-\frac{(x-y-\lambda (t-s))^2}{\mu (t-s)}}\Theta _{\beta }(y,s;\lambda ,\mu )\, dyds \\&\quad \le \left\{ \begin{array}{ll} C\Theta _{\gamma }(x,t;\lambda ,\nu ^*) &{} \hbox { if}\ \beta \ne 3, \\ C\log (t+2)\Theta _{\gamma }(x,t;\lambda ,\nu ^*) &{} \hbox { if}\ \beta =3, \end{array}\right. \end{aligned} \end{aligned}$$

where $\gamma =\alpha +\min (\beta ,3)-1$. We also have

$$\begin{aligned} \begin{aligned}&\int _{t/2}^{t}\int _{-\infty }^{\infty }(t-s)^{-1}(t+1-s)^{-\alpha /2}e^{-\frac{(x-y-\lambda (t-s))^2}{\mu (t-s)}}\Theta _{\beta }(y,s;\lambda ,\mu )\, dyds \\&\quad \le \left\{ \begin{array}{ll} C\Theta _{\gamma }(x,t;\lambda ,\nu ^*) &{} \hbox { if}\ \alpha \ne 1, \\ C\log (t+2)\Theta _{\gamma }(x,t;\lambda ,\nu ^*) &{} \hbox { if}\ \alpha =1, \end{array}\right. \end{aligned} \end{aligned}$$

where $\gamma =\min (\alpha ,1)+\beta -1$.

We remind the reader that $\psi _n(x,t;\lambda )$ is defined in (8).

Lemma A.6

Let $\lambda \in {\mathbb {R}}$, $\mu >0$, and $\alpha ,\beta \ge 0$. Then we have

$$\begin{aligned} \begin{aligned}&\int _{0}^{t/2}\int _{-\infty }^{\infty }(t-s)^{-1}(t+1-s)^{-\alpha /2}e^{-\frac{(x-y-\lambda (t-s))^2}{\mu (t-s)}}(s+1)^{-\beta /2}\psi _n(y,s;\lambda )\, dyds \\&\quad \le \left\{ \begin{array}{ll} C(t+1)^{-\gamma _1/2}\psi _n(x,t;\lambda ) &{}{} \text{ if }\ \beta \ne 3-\alpha _n, \\ C\log (t+2)(t+1)^{-\gamma _1/2}\psi _n(x,t;\lambda ) &{}{} \text{ if }\ \beta =3-\alpha _n, \end{array}\right. \end{aligned} \end{aligned}$$

where $\gamma _1=\alpha +\min (\beta ,3-\alpha _n)-1$. We also have

$$\begin{aligned} \begin{aligned}&\int _{t/2}^{t}\int _{-\infty }^{\infty }(t-s)^{-1}(t+1-s)^{-\alpha /2}e^{-\frac{(x-y-\lambda (t-s))^2}{\mu (t-s)}}(s+1)^{-\beta /2}\psi _n(y,s;\lambda )\, dyds \\ {}&\quad \le \left\{ \begin{array}{ll} C(t+1)^{-\gamma _2/2}\psi _n(x,t;\lambda ) &{}{} \text{ if }\ \alpha \ne 1, \\ C\log (t+2)(t+1)^{-\gamma _2/2}\psi _n(x,t;\lambda ) &{}{} \text{ if }\ \alpha =1, \end{array}\right. \end{aligned} \end{aligned}$$

where $\gamma _2=\min (\alpha ,1)+\beta -1$.

Proof

A straightforward (but lengthy) adaptation of the proof of [7, Lemma A.7] proves the lemma. $\quad \square$

For $\lambda ,\lambda ' \in {\mathbb {R}}$ and $K>0$, let

$$\begin{aligned}&\chi _K(x,t;\lambda ,\lambda ') \\&\quad {:}{=}\textrm{char}\left\{ \min (\lambda ,\lambda ')(t+1)+K\sqrt{t+1}\le x\le \max (\lambda ,\lambda ')(t+1)-K\sqrt{t+1} \right\} , \end{aligned}$$

where $\textrm{char}\{ S \}$ is the indicator function of a set S.

Lemma A.7

Let $\lambda \ne \lambda '$, $\mu >0$, $\alpha \ge 0$, and $0\le \beta \le 2\alpha _n$ ($\beta \ne 2$).³ Then for $K>0$ large enough, we have

$$\begin{aligned} \begin{aligned}&\int _{0}^{t/2}\int _{-\infty }^{\infty }(t-s)^{-1}(t+1-s)^{-\alpha /2}e^{-\frac{(x-y-\lambda (t-s))^2}{\mu (t-s)}}(s+1)^{-\beta /2}\psi _n(y,s;\lambda ')\, dyds \\&\quad \le \left\{ \begin{array}{ll} C[(t+1)^{-\gamma _1/2}\psi _n(x,t;\lambda )+(t+1)^{-\gamma _{1}'/2}\psi _n(x,t;\lambda ')] &{} \hbox { if}\ \beta \ne 3-\alpha _n \\ C[\log (t+2)(t+1)^{-\gamma _1/2}\psi _n(x,t;\lambda )+(t+1)^{-\gamma _{1}'/2}\psi _n(x,t;\lambda ')] &{} \hbox { if}\ \beta =3-\alpha _n \end{array}\right. \\&\qquad +C|x-\lambda (t+1)|^{-(\alpha _n+\min (\beta ,\alpha _n+1)-1)/2}|x-\lambda '(t+1)|^{-\alpha /2-1/2}\chi _K(x,t;\lambda ,\lambda '), \end{aligned} \end{aligned}$$

where $\gamma _1=\alpha +\min (\beta ,3-\alpha _n)-1$ and $\gamma _{1}'=\alpha +\min (\beta ,2)-1$. We also have

$$\begin{aligned} \begin{aligned}&\int _{t/2}^{t}\int _{-\infty }^{\infty }(t-s)^{-1}(t+1-s)^{-\alpha /2}e^{-\frac{(x-y-\lambda (t-s))^2}{\mu (t-s)}}(s+1)^{-\beta /2}\psi _n(y,s;\lambda ')\, dyds \\&\quad \le \left\{ \begin{array}{ll} C(t+1)^{-\gamma _2/2}[\psi _n(x,t;\lambda )+\psi _n(x,t;\lambda ')] &{} \hbox { if}\ \alpha \ne 1 \\ C\log (t+2)(t+1)^{-\gamma _2/2}[\psi _n(x,t;\lambda )+\psi _n(x,t;\lambda ')] &{} \hbox { if}\ \alpha =1 \end{array}\right. \\&\qquad +C|x-\lambda (t+1)|^{-(\alpha _n+\beta -1)/2}|x-\lambda '(t+1)|^{-\min (\alpha ,1)/2-1/2}\chi _K(x,t;\lambda ,\lambda '), \end{aligned} \end{aligned}$$

where $\gamma _2=\min (\alpha ,1)+\beta -1$.

Proof

This is also proved by an adaptation of the proof of [7, Lemma A.8]. $\quad \square$

Lemma A.8

Let $\lambda \ne \lambda '$, $\mu >0$, $\alpha \ge 0$, and $0\le \beta <3-\alpha _n$. Then we have

$$\begin{aligned}&\int _{0}^{t^{1/2}}\int _{-\infty }^{\infty }(t-s)^{-1-\alpha }e^{-\frac{(x-y-\lambda (t-s))^2}{\mu (t-s)}}(s+1)^{-\beta /2}\psi _n(y,s;\lambda ')\, dyds \\&\quad \le C(t+1)^{-\alpha -(\beta -\alpha _n+1)/4}\psi _n(x,t;\lambda ). \end{aligned}$$

Proof

This is a simple generalization of [7, Lemma A.2]. $\quad \square$

Lemma A.9

Let $\lambda \ne \lambda '$, $\mu >0$, and $\alpha \ge 0$. Then for $t\ge 4$, we have

$$\begin{aligned}&\int _{t^{1/2}}^{t/2}\int _{-\infty }^{\infty }(t-s)^{-1/2-\alpha }e^{-\frac{(x-y-\lambda (t-s))^2}{\mu (t-s)}}(s+1)^{-\beta _{n+1}}\psi _n(y,s;\lambda ')\, dyds \\&\quad \le C(t+1)^{-\alpha }\psi _n(x,t;\lambda ). \end{aligned}$$

Proof

A simple adaptation of the proof of [7, Lemma A.6] proves the lemma. $\square$

Lemma A.10

Let $\lambda \in {\mathbb {R}}$, $\mu >0$, and $\alpha \ge 0$. Then

$$\begin{aligned} \int _{0}^{t}e^{-\frac{t-s}{\mu }}(s+1)^{-\alpha }\psi _n(x,s;\lambda )\, ds\le C(t+1)^{-\alpha }\psi _n(x,t;\lambda ). \end{aligned}$$

Proof

The lemma can be proved by slightly modifying the proof of [10, Lemma 3.9].

$\square$