Appendix
For the self-adjoint classical Sturm–Liouville operators, an interesting uniqueness result is to assume that the potential q satisfies a local smoothness condition so that some eigenvalues and norming constants can be missing. While in [19,20,21] the key technique relies on the high-energy asymptotic expansion of the Weyl m-function [22], in our non-self-adjoint setting, the key to prove the uniqueness problems (Theorems 1, 5, Remark 11, Corollaries 1–4) will be Proposition 1, to be established below.
Definition 1
For \(i=1,2,\) let \(y_{i,r}(x,\lambda )\) and \({\widetilde{y}} _{i,r}(x,\lambda )\) be solutions of (1.1) corresponding to the potential q and \({\widetilde{q}},\) respectively, where \( y_{i,r}(x,\lambda )\) and \({\widetilde{y}}_{i,r}(x,\lambda )\) satisfy the initial conditions
$$\begin{aligned} y_{1,r}(r,\lambda )= & {} y_{2,r}^{\prime }(r,\lambda )=1, y_{2,r}(r,\lambda )=y_{1,r}^{\prime }(r,\lambda )=0,\ \\ {\widetilde{y}}_{1,r}(r,\lambda )= & {} {\widetilde{y}}_{2,r}^{\prime }(r,\lambda )=1,{\widetilde{y}}_{2,r}(r,\lambda )={\widetilde{y}}_{1,r}^{\prime }(r,\lambda )=0,\ r\in \left[ 0,\pi \right) . \end{aligned}$$
For simplicity, denote \(y_{1}(x,\lambda ):=y_{1,0}(x,\lambda ),\, y_{2}(x,\lambda ):=y_{2,0}(x,\lambda ),\,{\widetilde{y}}_{1}(x,\lambda ):= {\widetilde{y}}_{1,0}(x,\lambda ),\,{\widetilde{y}}_{2}(x,\lambda ):=\widetilde{ y}_{2,0}(x,\lambda ).\)
Proposition 1
Let \(x_{0}\in \left( r,\pi \right] \) for \(r\in \left[ 0,\pi \right) \) and assume that \(q,\,{\widetilde{q}}\in \)\(C^{m}\left[ x_{0}-\delta ,x_{0}\right] \) for some sufficiently small \(\delta >0\) (such that \(x_{0}-\delta \ge r)\) and some \(m\in {\mathbb {N}} _{0}.\) If \(q_{-}^{(j)}(x_{0})={\widetilde{q}}_{-}^{(j)}(x_{0})\) for \( j=0,1,\ldots ,m,\) then
$$\begin{aligned}&y_{1,r}(x_{0},\lambda ){\widetilde{y}}_{1,r}^{\prime }(x_{0},\lambda )-y_{1,r}^{\prime }(x_{0},\lambda ){\widetilde{y}}_{1,r}(x_{0},\lambda ) \nonumber \\&\quad =o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \left( x_{0}-r\right) \right) }{\left| \sqrt{\lambda }\right| ^{m+1}} \right) , \end{aligned}$$
(3.40)
$$\begin{aligned}&y_{1,r}(x_{0},\lambda ){\widetilde{y}}_{2,r}^{\prime }(x_{0},\lambda )-y_{1,r}^{\prime }(x_{0},\lambda ){\widetilde{y}}_{2,r}(x_{0},\lambda ) \nonumber \\&\quad =o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \left( x_{0}-r\right) \right) }{\left| \sqrt{\lambda }\right| ^{m+2}} \right) , \end{aligned}$$
(3.41)
$$\begin{aligned}&{\widetilde{y}}_{1,r}^{\prime }(x_{0},\lambda )y_{2,r}(x_{0},\lambda )- {\widetilde{y}}_{1,r}(x_{0},\lambda )y_{2,r}^{\prime }(x_{0},\lambda ) \nonumber \\&\quad =o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \left( x_{0}-r\right) \right) }{\left| \sqrt{\lambda }\right| ^{m+2}} \right) , \end{aligned}$$
(3.42)
$$\begin{aligned}&y_{2,r}(x_{0},\lambda ){\widetilde{y}}_{2,r}^{\prime }(x_{0},\lambda )-y_{2,r}^{\prime }(x_{0},\lambda ){\widetilde{y}}_{2,r}(x_{0},\lambda ) \nonumber \\&\quad =o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \left( x_{0}-r\right) \right) }{\left| \sqrt{\lambda }\right| ^{m+3}} \right) \end{aligned}$$
(3.43)
as \(\left| \lambda \right| \rightarrow \infty \) in \(\Lambda _{\zeta }:=\left\{ \lambda \in {\mathbb {C}}{:}\,\zeta<\text {Arg}(\lambda )<\pi -\zeta \text { for }\zeta >0\right\} .\)
Remark 13
For \(f\in \)\(C^{m}\left[ x_{0}-\delta ,x_{0}\right] ,\) we adopt following notations in this section:
$$\begin{aligned} f_{-}^{(0)}(x_{0}):= & {} f(x_{0}),\text { }f_{-}^{(1)}(x_{0}):=\lim \limits _{x \rightarrow x_{0}^{-}}\frac{f(x)-f(x_{0})}{x-x_{0}}, \\ f_{-}^{(j)}(x_{0}):= & {} \lim \limits _{x\rightarrow x_{0}^{-}}\frac{ f^{(j-1)}(x)-f_{-}^{(j-1)}(x_{0})}{x-x_{0}}\text { for }j=2,3,\ldots ,m. \end{aligned}$$
In addition, \(f\in \)\(C^{m}\left[ x_{0}-\delta ,x_{0}\right] \) implies \(\lim \nolimits _{x\rightarrow x_{0}^{-}}f^{(j)}(x)=f_{-}^{(j)}(x_{0})\) for \( j=0,1,\ldots ,m.\)
The proof of Proposition 1 will be given at the end of this “Appendix” after the proof of the following lemma.
Lemma 7
Let \(x_{0}\in \left( 0,\pi \right] \) and \(q,\,{\widetilde{q}} \in C^{m}\left[ 0,x_{0}\right] \) for some \(m\in {\mathbb {N}} _{0}.\) If
$$\begin{aligned} q_{-}^{(j)}(x_{0})={\widetilde{q}}_{-}^{(j)}(x_{0}) \end{aligned}$$
(3.44)
for \(j=0,1,\ldots ,m,\) then
$$\begin{aligned} y_{2}(x_{0},\lambda ){\widetilde{y}}_{2}^{\prime }(x_{0},\lambda )-y_{2}^{\prime }(x_{0},\lambda ){\widetilde{y}}_{2}(x_{0},\lambda )= & {} o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x_{0}\right) }{\left| \sqrt{\lambda }\right| ^{m+3}}\right) , \end{aligned}$$
(3.45)
$$\begin{aligned} y_{1}(x_{0},\lambda ){\widetilde{y}}_{1}^{\prime }(x_{0},\lambda )-y_{1}^{\prime }(x_{0},\lambda ){\widetilde{y}}_{1}(x_{0},\lambda )= & {} o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x_{0}\right) }{\left( \sqrt{\lambda }\right) ^{m+1}}\right) , \end{aligned}$$
(3.46)
$$\begin{aligned} y_{1}(x_{0},\lambda ){\widetilde{y}}_{2}^{\prime }(x_{0},\lambda )-y_{1}^{\prime }(x_{0},\lambda ){\widetilde{y}}_{2}(x_{0},\lambda )= & {} o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x_{0}\right) }{\left| \sqrt{\lambda }\right| ^{m+2}}\right) , \end{aligned}$$
(3.47)
$$\begin{aligned} {\widetilde{y}}_{1}^{\prime }(x_{0},\lambda )y_{2}(x_{0},\lambda )-\widetilde{y }_{1}(x_{0},\lambda )y_{2}^{\prime }(x_{0},\lambda )= & {} o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x_{0}\right) }{ \left| \sqrt{\lambda }\right| ^{m+2}}\right) \end{aligned}$$
(3.48)
as \(\left| \lambda \right| \rightarrow \infty \) in the sector \( \Lambda _{\zeta }.\)
We shall prove Lemma 7 by analyzing the asymptotic expansion of the fundamental solutions (see Lemmas 8, 9). Now we first give some preliminary facts and notations.
Recall the solution \(y_{2}\) defined by Definition 1, then it follows from [26] that
$$\begin{aligned} y_{2}(x,\lambda )=\sum \limits _{p=0}^{\infty }S_{p}(x,\lambda ), y_{2}^{\prime }(x,\lambda )=\sum \limits _{p=0}^{\infty }C_{p}(x,\lambda ), \end{aligned}$$
(3.49)
where \(S_{0}(x,\lambda )=\frac{\sin (\sqrt{\lambda }x)}{\sqrt{\lambda }},\, C_{0}(x,\lambda )=\cos \left( \sqrt{\lambda }x\right) ,\) and for \(p\ge 1,\)
$$\begin{aligned} S_{p}(x,\lambda )= & {} \int _{0}^{x}\frac{\sin (\sqrt{\lambda }\left( x-t\right) )}{\sqrt{\lambda }}q(t)S_{p-1}(t,\lambda )\mathrm {d}t, \end{aligned}$$
(3.50)
$$\begin{aligned} C_{p}(x,\lambda )= & {} \int _{0}^{x}\cos \left( \sqrt{\lambda }\left( x-t\right) \right) q(t)S_{p-1}(t,\lambda )\mathrm {d}t. \end{aligned}$$
(3.51)
In what follows, we adopt the following notations: \(\left( \pm \right) _{j}=\left\{ \begin{array}{ll} -1&{}\quad \text {if}\,j=4s,4s+1, \\ 1&{}\quad \text {if}\,j=4s+2,4s+3, \end{array} \right. \) and
$$\begin{aligned} \nu _{2s}(x,\lambda ):=\frac{\sin (\sqrt{\lambda }x)}{\left( 2\sqrt{\lambda } \right) ^{2s}},\nu _{2s+1}(x,\lambda ):=\frac{\cos (\sqrt{\lambda }x) }{\left( 2\sqrt{\lambda }\right) ^{2s+1}},s\in {\mathbb {N}} _{0}. \end{aligned}$$
Then we have the following statement relating to \(S_{p}\) defined by (3.50).
Lemma 8
Assume that q\(\in \)\(C^{m}\left[ 0,\delta \right] \) for some \( \delta >0\) and some \(m\in {\mathbb {N}}.\) Denote \(\sigma \left( x\right) :=\int _{0}^{x}q(t)\mathrm {d}t.\) Then for \( x\in \left[ 0,\delta \right] ,\) we have
$$\begin{aligned}&S_{1}(x,\lambda )=\sum \limits _{j=1}^{m+1}\frac{\nu _{j}(x,\lambda )}{\sqrt{ \lambda }}f_{1,j}\left( x\right) +\frac{\left( \pm \right) _{m+2}}{\sqrt{ \lambda }}\int _{0}^{x}\nu _{m+1}(x-2t,\lambda )q^{\left( m\right) }(t)dt, \qquad \qquad \end{aligned}$$
(3.52)
$$\begin{aligned}&S_{2}(x,\lambda )=\sum \limits _{j=1}^{m+2}\frac{\nu _{j}(x,\lambda )}{\sqrt{ \lambda }}f_{2,j}\left( x\right) +B_{2}\left( x,\lambda \right) , \end{aligned}$$
(3.53)
$$\begin{aligned}&S_{p}(x,\lambda )=\sum \limits _{j=1}^{m+2}\frac{\nu _{j}(x,\lambda )}{\sqrt{ \lambda }}f_{p,j}\left( x\right) +B_{p}\left( x,\lambda \right) \text { for } p=3,\ldots ,m+2 \end{aligned}$$
(3.54)
where
$$\begin{aligned} B_{2}\left( x,\lambda \right)= & {} -\frac{\left( \pm \right) _{m+3}}{\sqrt{ \lambda }}\int _{0}^{x}\nu _{m+2}(x-2t,\lambda )\left( \sum \limits _{j=1}^{m+1}\left( \pm \right) _{j}\left( q(t)f_{1,j}(t)\right) ^{\left( m+1-j\right) }\right) dt \\&+\,\frac{\left( \pm \right) _{m+2}}{\sqrt{\lambda }}\int _{0}^{x}\frac{\sin \sqrt{\lambda }\left( x-t\right) }{\sqrt{\lambda }}q(t)\int _{0}^{t}\nu _{m+1}(t-2s,\lambda )q^{\left( m\right) }(s)dsdt, \\ B_{p}\left( x,\lambda \right)= & {} -\frac{\left( \pm \right) _{m+3}}{\sqrt{ \lambda }}\int _{0}^{x}\nu _{m+2}(x-2t,\lambda )\sum \limits _{j=1}^{m+1}\left( \pm \right) _{j}\left( q(t)f_{p-1,j}(t)\right) ^{\left( m+1-j\right) }(t)dt \\&+\,\int _{0}^{x}\frac{\sin (\sqrt{\lambda }\left( x-t\right) )}{\sqrt{\lambda }}q(t)\left[ \frac{\nu _{m+2}(t,\lambda )}{\sqrt{\lambda }} f_{p-1,m+2}(t)+B_{p-1}\left( t,\lambda \right) \right] \mathrm {d}t \end{aligned}$$
for \(p=3,\ldots m+2,\) and the functions \(f_{p,j}\left( x\right) \) are defined by the recurrence relations
$$\begin{aligned} f_{1,j}\left( x\right)= & {} \left( \pm \right) _{j}\left( \sigma ^{\left( j-1\right) }\left( x\right) -(-1)^{j-1}\sigma ^{\left( j-1\right) }\left( 0\right) \right) , \\ f_{p,p}\left( x\right)= & {} (-1)^{p}\int _{0}^{x}q(t)f_{p-1,p-1}(t)dt\ \text {for }p=2,\ldots ,m+2, \\ f_{p,j}\left( x\right)= & {} -\sum \limits _{s=1}^{j-2}\left( \pm \right) _{s}\left( \pm \right) _{j}\left( \left( qf_{p-1,s}\right) ^{\left( j-s-2\right) }\left( x\right) -(-1)^{j-1}\left( qf_{p-1,s}\right) ^{\left( j-s-2\right) }\left( 0\right) \right) \\&+\,(-1)^{j}\int _{0}^{x}q(t)f_{p-1,j-1}(t)dt\ \text {for }j>p\ \text {and } p=2,\ldots ,m+2, \\ f_{p,j}\left( x\right)= & {} 0\text { for }j<p. \end{aligned}$$
Moreover\(,\,f_{p,j}\in C^{m+p-j+1}\left[ 0,\delta \right] .\)
Proof
In order to prove this lemma, we will follow the technique in [27, Lemma 4.2 ]. We first note that
$$\begin{aligned} \frac{\sin (\sqrt{\lambda }\left( x-t\right) )}{\sqrt{\lambda }}\nu _{j}(t,\lambda )=\left( -1\right) ^{j+1}\nu _{j+1}(x,\lambda )+\nu _{j+1}(x-2t,\lambda ) \end{aligned}$$
(3.55)
and for \(f\in C^{1}\left[ 0,x\right] ,\)
$$\begin{aligned}&\int _{0}^{x}\nu _{j}(x-2t,\lambda )f(t)dt \nonumber \\&\quad =\nu _{j+1}(x,\lambda )\left( f\left( x\right) -\left( -1\right) ^{j}f\left( 0\right) \right) +\left( -1\right) ^{j+1}\int _{0}^{x}\nu _{j+1}(x-2t,\lambda )f^{\prime }(t)dt.\nonumber \\ \end{aligned}$$
(3.56)
In view of (3.55) and (3.56), one can easily deduce the expression (3.52). Now we turn to deduce the expressions for the other functions \(S_{j}.\) Suppose that \( f_{j}\in C^{m+1-j}\left[ 0,x\right] ,\) then from (3.55), we know that for \(j=1,\ldots ,m+1,\)
$$\begin{aligned}&\int _{0}^{x}\frac{\sin (\sqrt{\lambda }\left( x-t\right) )}{\sqrt{\lambda } }\nu _{j}(t,\lambda )f_{j}(t)dt \nonumber \\&\quad =\left( -1\right) ^{j+1}\nu _{j+1}(x,\lambda )\int _{0}^{x}f_{j}(t)dt+\int _{0}^{x}\nu _{j+1}(x-2t,\lambda )f_{j}(t)dt,\qquad \end{aligned}$$
(3.57)
Moreover, integrating by parts the second summand on the right-hand side of the above equality \(m+1-j\) times and using (3.55), it follows that for \(j=1,\ldots ,m,\)
$$\begin{aligned}&\int _{0}^{x}\frac{\sin (\sqrt{\lambda }\left( x-t\right) )}{\sqrt{\lambda } }\nu _{j}(t,\lambda )f_{j}(t)dt \nonumber \\= & {} \left( -1\right) ^{j+1}\nu _{j+1}(x,\lambda )\int _{0}^{x}f_{j}(t)dt\nonumber \\&-\sum _{s=j+2}^{m+2}\nu _{s}(x,\lambda )\left( \pm \right) _{j}\left( \pm \right) _{s}(f_{j}^{\left( s-j-2\right) }\left( x\right) -(-1)^{s-1}f_{j}^{\left( s-j-2\right) }\left( 0\right) ) \nonumber \\&\qquad -\left( \pm \right) _{j}\left( \pm \right) _{m+3}\int _{0}^{x}\nu _{m+2}(x-2t,\lambda )f_{j}^{\left( m+1-j\right) }(t)dt. \end{aligned}$$
(3.58)
Therefore, by virtue of (3.57) (for \(j=m+1)\) and (3.58), for \(x\in \left( 0,\delta \right] \) we have
$$\begin{aligned}&\sum \limits _{j=1}^{m+1}\int _{0}^{x}\frac{\sin (\sqrt{\lambda }\left( x-t\right) )}{\sqrt{\lambda }}\nu _{j}(t,\lambda )f_{j}(t)dt \nonumber \\= & {} \nu _{2}(x,\lambda )\int _{0}^{x}f_{1}(t)dt +\sum _{j=3}^{m+2}\nu _{j}(x,\lambda )\left( \left( -1\right) ^{j}\int _{0}^{x}f_{j-1}(t)dt\right. \nonumber \\&\qquad \left. -\sum \limits _{s=1}^{j-2}\left( \pm \right) _{s}\left( \pm \right) _{j}(f_{s}^{\left( j-s-2\right) }\left( x\right) -(-1)^{j-1}f_{s}^{\left( j-s-2\right) }\left( 0\right) )\right) \nonumber \\&\qquad -\left( \pm \right) _{m+3}\int _{0}^{x}\nu _{m+2}(x-2t,\lambda )\sum \limits _{j=1}^{m+1}\left( \pm \right) _{j}f_{j}^{\left( m+1-j\right) }(t)dt. \end{aligned}$$
(3.59)
Now in view of (3.50) and (3.52), we obtain that for \(x\in \left( 0,\delta \right] ,\)
$$\begin{aligned} S_{2}(x,\lambda )= & {} \frac{1}{\sqrt{\lambda }}\sum \limits _{j=1}^{m+1} \int _{0}^{x}\frac{\sin (\sqrt{\lambda }\left( x-t\right) )}{\sqrt{\lambda }} \nu _{j}(t,\lambda )q(t)f_{1,j}(t)\mathrm {d}t \\&+\,\frac{\left( \pm \right) _{m+2}}{\sqrt{\lambda }}\int _{0}^{x}\frac{\sin \sqrt{\lambda }\left( x-t\right) }{\sqrt{\lambda }}q(t)\int _{0}^{t}\nu _{m+1}(t-2s,\lambda )q^{\left( m\right) }(s)dsdt. \end{aligned}$$
Making use of (3.59) with \(f_{j}(t)\) replaced by \( q(t)f_{1,j}(t)\) and in virtue of the fact \(qf_{1,j}\in C^{m+1-j}\left[ 0,\delta \right] \) for j\(=1,\ldots ,m+1,\) we obtain (3.53). Next, from (3.50) and (3.53), it follows that for \(x\in \left( 0,\delta \right] ,\)
$$\begin{aligned} S_{3}(x,\lambda )= & {} \frac{1}{\sqrt{\lambda }}\sum \limits _{j=1}^{m+1} \int _{0}^{x}\frac{\sin (\sqrt{\lambda }\left( x-t\right) )}{\sqrt{\lambda }} \nu _{j}(t,\lambda )q(t)f_{2,j}(t)\mathrm {d}t \\&+\,\int _{0}^{x}\frac{\sin (\sqrt{\lambda }\left( x-t\right) )}{\sqrt{\lambda }}q(t)\left[ \frac{\nu _{m+2}(t,\lambda )}{\sqrt{\lambda }} f_{2,m+2}(t)+B_{2}\left( t,\lambda \right) \right] \mathrm {d}t. \end{aligned}$$
Then the expression (3.54) for \(S_{3}\) can be proved by using (3.59) and letting \(f_{j}(t):=q(t)f_{2,j}(t).\) The proof of the relation (3.54) for \(p=4,\ldots ,m+2\) can be carried out in the same way. \(\square \)
As a consequence of Lemma 8, we have the following assertion relating to \(C_{p}\) defined by (3.51).
Lemma 9
Assume that q\(\in \)\(C^{m}\left[ 0,\delta \right] \) for some \( \delta >0\) and some \(m\in {\mathbb {N}}.\) Denote \(\sigma \left( x\right) :=\int _{0}^{x}q(t)\mathrm {d}t.\) Then for \( x\in \left[ 0,\delta \right] ,\) we have
$$\begin{aligned} C_{1}(x,\lambda )= & {} \frac{-f_{1,1}\left( x\right) }{2}\frac{\nu _{0}(x,\lambda )}{\sqrt{\lambda }}+\sum \limits _{j=1}^{m}\frac{\nu _{j}(x,\lambda )}{\sqrt{\lambda }}\left[ f_{1,j}^{\prime }\left( x\right) + \frac{(-1)^{j+1}f_{1,j+1}\left( x\right) }{2}\right] \\&+\,\frac{\left( \pm \right) _{m+2}}{\sqrt{\lambda }}\int _{0}^{x}\frac{d\nu _{m+1}(x-2t,\lambda )}{dx}q^{\left( m\right) }(t)dt, \\ C_{2}(x,\lambda )= & {} \sum \limits _{j=1}^{m+1}\frac{\nu _{j}(x,\lambda )}{ \sqrt{\lambda }}\left[ f_{2,j}^{\prime }\left( x\right) +\frac{ (-1)^{j+1}f_{2,j+1}\left( x\right) }{2}\right] +D_{2}\left( x,\lambda \right) , \\ C_{p}(x,\lambda )= & {} \sum \limits _{j=1}^{m+1}\frac{\nu _{j}(x,\lambda )}{ \sqrt{\lambda }}\left[ f_{p,j}^{\prime }\left( x\right) +\frac{ (-1)^{j+1}f_{p,j+1}\left( x\right) }{2}\right] \\&\quad +\,D_{p}\left( x,\lambda \right) \ \text {for }p=3,\ldots ,m+2 \end{aligned}$$
where \(f_{p,j}\left( x\right) \) are the functions defined in Lemma 8, and
$$\begin{aligned} D_{2}\left( x,\lambda \right)= & {} -\frac{\left( \pm \right) _{m+3}}{\sqrt{ \lambda }}\int _{0}^{x}\frac{d\nu _{m+2}(x-2t,\lambda )}{dx}\left( \sum \limits _{j=1}^{m+1}\left( \pm \right) _{j}\left( q(t)f_{1,j}(t)\right) ^{\left( m+1-j\right) }\right) dt \\&+\,\frac{\left( \pm \right) _{m+2}}{\sqrt{\lambda }}\int _{0}^{x}\cos \sqrt{ \lambda }\left( x-t\right) q(t)\int _{0}^{t}\nu _{m+1}(t-2s,\lambda )q^{\left( m\right) }(s)dsdt, \\ D_{p}\left( x,\lambda \right)= & {} -\frac{\left( \pm \right) _{m+3}}{\sqrt{ \lambda }}\int _{0}^{x}\frac{d\nu _{m+2}(x-2t,\lambda )}{dx} \sum \limits _{j=1}^{m+1}\left( \pm \right) _{j}\left( qf_{p-1,j}\right) ^{\left( m+1-j\right) }(t)dt \\&+\,\int _{0}^{x}\cos \sqrt{\lambda }\left( x-t\right) q(t)\left[ \frac{\nu _{m+2}(t,\lambda )}{\sqrt{\lambda }}f_{p-1,m+2}(t)+B_{p-1}\left( t,\lambda \right) \right] \mathrm {d}t \end{aligned}$$
for \(p=3,\ldots ,m+2.\)
Lemma 10
Assume that q\(\in \)\(C^{m}\left[ 0,\delta \right] \) for some \(\delta >0\) and some \(m\in {\mathbb {N}} _{0}.\) Then for \(x\in \left[ 0,\delta \right] ,\,y_{2}(x,\lambda )\) and \( y_{2}^{\prime }(x,\lambda )\) can be rewritten as the following form:
$$\begin{aligned} y_{2}(x,\lambda )= & {} \frac{\sin \left( \sqrt{\lambda }x\right) }{\sqrt{ \lambda }}+\sum \limits _{j=1}^{m+2}a_{j}\left( x\right) \frac{\nu _{j}(x,\lambda )}{\sqrt{\lambda }} \nonumber \\&+\,\frac{\left( \pm \right) _{m+2}}{\sqrt{\lambda }}\int _{0}^{x}\nu _{m+1}(x-2t,\lambda )q^{\left( m\right) }(t)dt \nonumber \\&+\,\sum \limits _{p=2}^{m+2}B_{p}\left( x,\lambda \right) +\sum \limits _{p=m+3}^{\infty }S_{p}\left( x,\lambda \right) , \end{aligned}$$
(3.60)
and
$$\begin{aligned} y_{2}^{\prime }(x,\lambda )= & {} \cos \left( \sqrt{\lambda }x\right) +\sum \limits _{j=0}^{m+1}b_{j}\left( x\right) \frac{\nu _{j}(x,\lambda )}{ \sqrt{\lambda }} \nonumber \\&+\,\frac{\left( \pm \right) _{m+2}}{\sqrt{\lambda }}\int _{0}^{x}\frac{d\nu _{m+1}(x-2t,\lambda )}{dx}q^{\left( m\right) }(t)dt \nonumber \\&+\,\sum \limits _{p=2}^{m+2}D_{p}\left( x,\lambda \right) +\sum \limits _{p=m+3}^{\infty }C_{p}\left( x,\lambda \right) , \end{aligned}$$
(3.61)
where
$$\begin{aligned} a_{j}\left( x\right) =\sum \limits _{p=1}^{m+2}f_{p,j}\left( x\right) \ \text { for }j=1,\ldots m+1,a_{m+2}\left( x\right) =\sum \limits _{p=2}^{m+2}f_{p,m+2}\left( x\right) , \end{aligned}$$
and
$$\begin{aligned} b_{0}\left( x\right)= & {} \frac{-f_{1,1}\left( x\right) }{2}, b_{m+1}\left( x\right) =\sum \limits _{p=2}^{m+2}\left( f_{p,m+1}^{\prime }\left( x\right) +\frac{(-1)^{m+2}f_{p,m+2}\left( x\right) }{2}\right) , \\ b_{j}\left( x\right)= & {} \sum \limits _{p=1}^{m+2}\left( f_{p,j}^{\prime }\left( x\right) +\frac{(-1)^{j+1}f_{p,j+1}\left( x\right) }{2}\right) , j=1,2,\ldots ,m. \end{aligned}$$
Proof
For \(m\in {\mathbb {N}},\) the expressions (3.60) and (3.61) can be directly obtained from (3.49), Lemmas 8 and 9. For \(m=0,\) the proof can be carried out in the same way even simpler. \(\square \)
Remark 14
For \(g\in L^{1}\left[ 0,x\right] ,\) one notes that the following identities
$$\begin{aligned} \int _{0}^{x}\sin \left( \sqrt{\lambda }t\right) g(t)dt= & {} o\left( \exp \left( \left| \text {Im}\sqrt{\lambda }\right| x\right) \right) , \end{aligned}$$
(3.62)
$$\begin{aligned} \int _{0}^{x}\cos \left( \sqrt{\lambda }t\right) g(t)dt= & {} o\left( \exp \left( \left| \text {Im}\sqrt{\lambda }\right| x\right) \right) \end{aligned}$$
(3.63)
hold as \(\left| \lambda \right| \rightarrow \infty \) [26, P15]. By virtue of (3.62) and (3.63), it is easy to deduce that as \(\left| \lambda \right| \rightarrow \infty ,\)
$$\begin{aligned} B_{p}\left( x,\lambda \right) =o\left( \frac{\exp \left( \left| \text {Im} \sqrt{\lambda }\right| x\right) }{\left| \sqrt{\lambda }\right| ^{m+3}}\right) ,B_{p}^{\prime }\left( x,\lambda \right) =o\left( \frac{\exp \left( \left| \text {Im}\sqrt{\lambda }\right| x\right) }{ \left| \sqrt{\lambda }\right| ^{m+2}}\right) \end{aligned}$$
and
$$\begin{aligned} D_{p}\left( x,\lambda \right) =o\left( \frac{\exp \left( \left| \text {Im} \sqrt{\lambda }\right| x\right) }{\left| \sqrt{\lambda }\right| ^{m+2}}\right) ,D_{p}^{\prime }\left( x,\lambda \right) =o\left( \frac{\exp \left( \left| \text {Im}\sqrt{\lambda }\right| x\right) }{ \left| \sqrt{\lambda }\right| ^{m+1}}\right) \end{aligned}$$
hold for \(p=2,3,\ldots ,m+2.\)
Remark 15
Note that
$$\begin{aligned} S_{p}(x,\lambda )= & {} \int _{0\le t_{1}\le \cdots \le t_{p+1}:=x}\prod \limits _{i=1}^{p}s_{\lambda }(t_{i+1}-t_{i}\,)s_{\lambda }(t_{1})q(t_{i})dt_{1}\cdots dt_{p}, \\ C_{p}(x,\lambda )= & {} \int _{0\le t_{1}\le \cdots \le t_{p+1}:=x}c_{\lambda }(t_{p+1}-t_{p})\prod \limits _{i=1}^{p-1}s_{\lambda }(t_{i+1}-t_{i}\,)s_{\lambda }(t_{1})q(t_{i})dt_{1}\cdots dt_{p}, \end{aligned}$$
where \(s_{\lambda }(x):=\frac{\sin (\sqrt{\lambda }x)}{\sqrt{\lambda }},c_{\lambda }(x):=\cos (\sqrt{\lambda }x),\) and
$$\begin{aligned} \left| \frac{\sin (\sqrt{\lambda }x)}{\sqrt{\lambda }}\right| \le \frac{\exp \left( \left| \text {Im}\sqrt{\lambda }\right| x\right) }{ \left| \sqrt{\lambda }\right| },\left| \cos (\sqrt{ \lambda }x)\right| \le \exp \left( \left| \text {Im}\sqrt{\lambda } \right| x\right) . \end{aligned}$$
Thus for \(\lambda \in {\mathbb {C}} \) and \(\left| \lambda \right| \) being large enough, one has
$$\begin{aligned} \left| S_{p}(x,\lambda )\right|\le & {} \frac{\exp \left( \left| \text {Im}\sqrt{\lambda }\right| x\right) }{\left| \sqrt{\lambda } \right| ^{m+4}}\frac{\left( \int _{0}^{x}\left| q(t)\right| dt\right) ^{p}}{p!},p\ge m+3, \\ \left| C_{p}(x,\lambda )\right|\le & {} \frac{\exp \left( \left| \text {Im}\sqrt{\lambda }\right| x\right) }{\left| \sqrt{\lambda } \right| ^{m+3}}\frac{\left( \int _{0}^{x}\left| q(t)\right| dt\right) ^{p}}{p!},\text { }p\ge m+3. \end{aligned}$$
This directly yields that as \(\left| \lambda \right| \rightarrow \infty ,\)
$$\begin{aligned} \sum \limits _{p=m+3}^{\infty }S_{p}(x,\lambda )=O\left( \frac{\exp \left( \left| \text {Im}\sqrt{\lambda }\right| x\right) }{\left| \sqrt{ \lambda }\right| ^{m+4}}\right) ,\text { }\sum \limits _{p=m+3}^{\infty }C_{p}(x,\lambda )=O\left( \frac{\exp \left( \left| \text {Im}\sqrt{ \lambda }\right| x\right) }{\left| \sqrt{\lambda }\right| ^{m+3}} \right) . \end{aligned}$$
Similarly, one can also obtain that \(\sum \limits _{p=m+3}^{\infty }C_{p}^{\prime }(x,\lambda )=O\left( \frac{\exp \left( \left| \text {Im} \sqrt{\lambda }\right| x\right) }{\left| \sqrt{\lambda }\right| ^{m+2}}\right) \) as \(\left| \lambda \right| \rightarrow \infty .\)
Now we turn to prove Lemma 7.
Proof of Lemma 7
We only aim to prove the relation (3.45), since the other statements can be treated similarly. We first denote
$$\begin{aligned} g\left( x\right) :=\left\{ \begin{array}{l} q\left( x\right) ,x\in \left[ 0,x_{0}\right] , \\ s\left( x\right) ,x\in \left( x_{0},x_{0}+\delta \right] , \end{array} \right. {\widetilde{g}}\left( x\right) :=\left\{ \begin{array}{l} {\widetilde{q}}\left( x\right) ,x\in \left[ 0,x_{0}\right] , \\ s\left( x\right) ,x\in \left( x_{0},x_{0}+\delta \right] , \end{array} \right. \end{aligned}$$
where \(s\left( x\right) =\sum \limits _{j=0}^{m}q_{-}^{(j)}(x_{0})\left( x-x_{0}\right) ^{j}\) and \(\delta \) is some positive constant. Then by (3.44) it is easy to see that \(g,\,{\widetilde{g}}\)\(\in \)\( C^{m}\left[ 0,x_{0}+\delta \right] \) and
$$\begin{aligned} g^{\left( j\right) }(x_{0})=q_{-}^{(j)}(x_{0})={\widetilde{g}}^{\left( j\right) }(x_{0})\text { for }j=0,1,\ldots ,m. \end{aligned}$$
(3.64)
For \(i=1,2,\) let \(w_{2}(x,\lambda )\) and \({\widetilde{w}}_{2}(x,\lambda )\) be the fundamental solutions of the equations
$$\begin{aligned} -\,y^{\prime \prime }+g\left( x\right) y=\lambda y\text { and}-y^{\prime \prime }+{\widetilde{g}}\left( x\right) y=\lambda y,x\in \left( 0,x_{0}+\delta \right) \end{aligned}$$
respectively, where \(w_{2}(x,\lambda )\) and \({\widetilde{w}}_{2}(x,\lambda )\) are determined by the initial conditions
$$\begin{aligned} w_{2}(0,\lambda )={\widetilde{w}}_{2}(0,\lambda )=0, w_{2}^{\prime }(0,\lambda )={\widetilde{w}}_{2}^{\prime }(0,\lambda )=1. \end{aligned}$$
By (3.60), (3.61), Lemmas 8, 9, Remarks 14 and 15, it is easy to see that there exist functions \(r_{k},u_{k},z_{k}\in C^{1}\left[ 0,x_{0}+\delta \right] \) such that for \(x\in \left[ 0,x_{0}+\delta \right] ,\)
$$\begin{aligned}&w_{2}(x,\lambda ){\widetilde{w}}_{2}^{\prime }(x,\lambda )-w_{2}^{\prime }(x,\lambda ){\widetilde{w}}_{2}(x,\lambda ) \nonumber \\&\quad =\sum \limits _{k=0}^{m+3}r_{k}(x)\frac{\sin \left( \sqrt{\lambda }x\right) \cos \left( \sqrt{\lambda }x\right) }{\left( \sqrt{\lambda }\right) ^{k}} +\sum \limits _{k=0}^{m+3}u_{k}(x)\frac{\cos ^{2}\left( \sqrt{\lambda } x\right) }{\left( \sqrt{\lambda }\right) ^{k}} \nonumber \\&\qquad +\,\sum \limits _{k=0}^{m+3}z_{k}(x)\frac{\sin ^{2}\left( \sqrt{\lambda } x\right) }{\left( \sqrt{\lambda }\right) ^{k}}+\frac{\left( \pm \right) _{m+2}}{\sqrt{\lambda }}I_{1}\left( x,\lambda \right) +I_{2}\left( x,\lambda \right) \end{aligned}$$
(3.65)
where
$$\begin{aligned} I_{1}\left( x,\lambda \right)= & {} \frac{\sin \left( \sqrt{\lambda }x\right) }{ \sqrt{\lambda }}\int _{0}^{x}\frac{d\nu _{m+1}(x-2t,\lambda )}{dx}\left( {\widetilde{g}}^{\left( m\right) }(t)-g^{\left( m\right) }(t)\right) dt \nonumber \\&+\,\cos \left( \sqrt{\lambda }x\right) \int _{0}^{x}\nu _{m+1}(x-2t,\lambda )\left( g^{\left( m\right) }(t)-{\widetilde{g}}^{\left( m\right) }(t)\right) dt \nonumber \\= & {} \left\{ \begin{array}{c} \int _{0}^{x}\frac{\cos \left( 2\sqrt{\lambda }t\right) }{\left( 2\sqrt{ \lambda }\right) ^{m+1}}\left( g^{\left( m\right) }(t)-{\widetilde{g}}^{\left( m\right) }(t)\right) dt\text { if }m\ \text {is even}, \\ \int _{0}^{x}\frac{\sin \left( 2\sqrt{\lambda }t\right) }{\left( 2\sqrt{ \lambda }\right) ^{m+1}}\left( {\widetilde{g}}^{\left( m\right) }(t)-g^{\left( m\right) }(t)\right) dt\text { if }m\ \text {is odd}, \end{array} \right. \end{aligned}$$
(3.66)
and as \(\left| \lambda \right| \rightarrow \infty ,\,I_{2}\left( x,\lambda \right) =o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{ \lambda }\right| x\right) }{\left| \sqrt{\lambda }\right| ^{m+3}} \right) ,\,I_{2}^{\prime }\left( x,\lambda \right) =o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x\right) }{\left| \sqrt{\lambda }\right| ^{m+2}}\right) \).
In view of (3.65) and the fact \(g={\widetilde{g}}\) on \( \left[ x_{0},x_{0}+\delta \right] ,\) for \(x\in \left[ x_{0},x_{0}+\delta \right] ,\) one has
$$\begin{aligned}&\left( w_{2}(x,\lambda ){\widetilde{w}}_{2}^{\prime }(x,\lambda )-w_{2}^{\prime }(x,\lambda ){\widetilde{w}}_{2}(x,\lambda )\right) ^{\prime } \\&\quad =r_{0}(x)\sqrt{\lambda }\cos \left( 2\sqrt{\lambda }x\right) -\left( u_{0}(x)-z_{0}(x)\right) \sqrt{\lambda }\sin \left( 2\sqrt{\lambda }x\right) \\&\qquad +\,\sum \limits _{k=0}^{m+2}\frac{u_{k}^{\prime }(x)+z_{k}^{\prime }(x)}{ 2\left( \sqrt{\lambda }\right) ^{k}}+\sum \limits _{k=0}^{m+2}\left( r_{k+1}(x)+\frac{u_{k}^{\prime }(x)-z_{k}^{\prime }(x)}{2}\right) \frac{\cos \left( 2\sqrt{\lambda }x\right) }{\left( \sqrt{\lambda }\right) ^{k}} \\&\qquad -\,\sum \limits _{k=0}^{m+2}\left( u_{k+1}(x)-z_{k+1}(x)-\frac{r_{k}^{\prime }(x)}{2}\right) \frac{\sin \left( 2\sqrt{\lambda }x\right) }{\left( \sqrt{ \lambda }\right) ^{k}}+o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{ \lambda }\right| x\right) }{\left| \sqrt{\lambda }\right| ^{m+2}} \right) \\&\quad =0. \end{aligned}$$
This forces that for \(x\in \left[ x_{0},x_{0}+\delta \right] ,\, u_{0}(x)-z_{0}(x)=r_{0}(x)\equiv 0,\)
$$\begin{aligned} r_{k+1}(x)+\frac{u_{k}^{\prime }(x)-z_{k}^{\prime }(x)}{2}=0,\text { } u_{k+1}(x)-z_{k+1}(x)-\frac{r_{k}^{\prime }(x)}{2}=0\ \text {for }k=0,\ldots ,m+2, \end{aligned}$$
and thus for \(k=0,1,\ldots ,m+3\) and \(x\in \left[ x_{0},x_{0}+\delta \right] ,\) one has
$$\begin{aligned} u_{k}(x)-z_{k}(x)=r_{k}(x)\equiv 0. \end{aligned}$$
(3.67)
Therefore, by (3.65) and (3.67) we infer that
$$\begin{aligned}&w_{2}(x_{0},\lambda ){\widetilde{w}}_{2}^{\prime }(x_{0},\lambda )-w_{2}^{\prime }(x_{0},\lambda ){\widetilde{w}}_{2}(x_{0},\lambda ) \nonumber \\&\quad =\sum \limits _{k=0}^{m+3}\frac{u_{k}(x_{0})}{\left( \sqrt{\lambda }\right) ^{k}}+\frac{\left( \pm \right) _{m+2}}{\sqrt{\lambda }}I_{1}\left( x_{0},\lambda \right) +o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{ \lambda }\right| x_{0}\right) }{\left| \sqrt{\lambda }\right| ^{m+3}}\right) . \end{aligned}$$
(3.68)
Next, we aim to show that
$$\begin{aligned} I_{1}\left( x_{0},\lambda \right) =o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x_{0}\right) }{\left| \sqrt{\lambda } \right| ^{m+2}}\right) \end{aligned}$$
(3.69)
as \(\left| \lambda \right| \rightarrow \infty \) in the sector \( \Lambda _{\zeta }.\) Due to the definition (3.66) of \(I_{1}\left( x,\lambda \right) \), it is sufficient to prove
$$\begin{aligned} \int _{0}^{x_{0}}\cos \left( 2\sqrt{\lambda }t\right) \left( g^{\left( m\right) }(t)-{\widetilde{g}}^{\left( m\right) }(t)\right) dt= & {} o\left( \frac{ \exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x_{0}\right) }{ \left| \sqrt{\lambda }\right| }\right) , \end{aligned}$$
(3.70)
$$\begin{aligned} \int _{0}^{x_{0}}\sin \left( 2\sqrt{\lambda }t\right) \left( g^{\left( m\right) }(t)-{\widetilde{g}}^{\left( m\right) }(t)\right) dt= & {} o\left( \frac{ \exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x_{0}\right) }{ \left| \sqrt{\lambda }\right| }\right) . \end{aligned}$$
(3.71)
In fact, by (3.64) \(\left( \text {for }j=m\right) \) and the fact \(g,\,{\widetilde{g}}\)\(\in \)\(C^{m}\left[ 0,x_{0}+\delta \right] \) we infer that given any \(\epsilon >0,\) there exists a sufficiently small constant \(\delta _{0}>0\) such that
$$\begin{aligned} \left| g^{\left( m\right) }(t)-{\widetilde{g}}^{\left( m\right) }(t)\right| <\epsilon \text { on }\left[ x_{0}-\delta _{0},x_{0}\right] , \end{aligned}$$
and thus for \(\lambda \in \Lambda _{\zeta }\) and \(\left| \lambda \right| \) being sufficiently large, we obtain
$$\begin{aligned}&\left| \int _{0}^{x_{0}}\cos \left( 2\sqrt{\lambda }t\right) \left( g^{\left( m\right) }(t)-{\widetilde{g}}^{\left( m\right) }(t)\right) dt\right| \\&\quad \le \max _{t\in \left[ 0,x_{0}\right] }\left| g^{\left( m\right) }(t)- {\widetilde{g}}^{\left( m\right) }(t)\right| \int _{0}^{x_{0}-\delta _{0}}\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| t\right) dt\\&\qquad +\epsilon \int _{x_{0}-\delta _{0}}^{x_{0}}\exp \left( 2\left| \text {Im} \sqrt{\lambda }\right| t\right) dt \\&\quad \le \max _{t\in \left[ 0,x_{0}\right] }\left| g^{\left( m\right) }(t)- {\widetilde{g}}^{\left( m\right) }(t)\right| \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \left( x_{0}-\delta _{0}\right) \right) }{2\left| \text {Im}\sqrt{\lambda }\right| }+\epsilon \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x_{0}\right) }{ 2\left| \text {Im}\sqrt{\lambda }\right| } \\&\quad \le \frac{\max \limits _{t\in \left[ 0,x_{0}\right] }\left| g^{\left( m\right) }(t)-{\widetilde{g}}^{\left( m\right) }(t)\right| }{2\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \delta _{0}\right) }\frac{ \exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x_{0}\right) }{ \left| \text {Im}\sqrt{\lambda }\right| }+\frac{\epsilon }{2\sin \frac{\zeta }{2}}\frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda } \right| x_{0}\right) }{\left| \sqrt{\lambda }\right| } \\&\quad \le \frac{\epsilon }{\sin \frac{\zeta }{2}}\frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x_{0}\right) }{\left| \sqrt{\lambda } \right| }, \end{aligned}$$
where we have used \(\left| \cos \left( 2\sqrt{\lambda }t\right) \right| \le \exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \left| t\right| \right) \) for \(\lambda \in {\mathbb {C}},t\in {\mathbb {R}} \) and
$$\begin{aligned} \left| \text {Im}\sqrt{\lambda }\right| \ge \left| \sqrt{\lambda }\right| \sin \frac{\zeta }{2}\text { for }\lambda \in \Lambda _{\zeta }. \end{aligned}$$
(3.72)
This proves the equality (3.70). Note that (3.71) can be treated similarly, and thus (3.69) is proved. Now by (3.68) and (3.69) we have that
$$\begin{aligned} w_{2}(x_{0},\lambda ){\widetilde{w}}_{2}^{\prime }(x_{0},\lambda )-w_{2}^{\prime }(x_{0},\lambda ){\widetilde{w}}_{2}(x_{0},\lambda )=\sum \limits _{k=0}^{m+3}\frac{u_{k}(x_{0})}{2\left( \sqrt{\lambda }\right) ^{k}}+o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda } \right| x_{0}\right) }{\left| \sqrt{\lambda }\right| ^{m+3}} \right) \end{aligned}$$
as \(\left| \lambda \right| \rightarrow \infty \) in the sector \( \Lambda _{\zeta }.\) This together with (3.72) directly yields that
$$\begin{aligned} w_{2}(x_{0},\lambda ){\widetilde{w}}_{2}^{\prime }(x_{0},\lambda )-w_{2}^{\prime }(x_{0},\lambda ){\widetilde{w}}_{2}(x_{0},\lambda )=o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| x_{0}\right) }{\left| \sqrt{\lambda }\right| ^{m+3}}\right) \end{aligned}$$
as \(\left| \lambda \right| \rightarrow \infty \) in the sector \( \Lambda _{\zeta }.\) Now (3.45) is proved by noting that \( w_{2}(x_{0},\lambda )=y_{2}(x_{0},\lambda ),\,{\widetilde{w}} _{2}(x_{0},\lambda )={\widetilde{y}}_{2}(x_{0},\lambda ),\,w_{2}^{\prime }(x_{0},\lambda )=y_{2}^{\prime }(x_{0},\lambda ),\,{\widetilde{w}} _{2}^{\prime }(x_{0},\lambda )={\widetilde{y}}_{2}^{\prime }(x_{0},\lambda ).\)\(\square \)
Proof of Proposition 1
Note that
$$\begin{aligned}&y_{2,r}(x_{0},\lambda ){\widetilde{y}}_{2,r}^{\prime }(x_{0},\lambda )-y_{2,r}^{\prime }(x_{0},\lambda ){\widetilde{y}}_{2,r}(x_{0},\lambda ) \nonumber \\&\quad =B_{1}(\lambda )\left[ y_{1,x_{0}-\delta }(x_{0},\lambda ){\widetilde{y}} _{1,x_{0}-\delta }^{\prime }(x_{0},\lambda )-y_{1,x_{0}-\delta }^{\prime }(x_{0},\lambda ){\widetilde{y}}_{1,x_{0}-\delta }(x_{0},\lambda )\right] \nonumber \\&\qquad +\,B_{2}(\lambda )\left[ y_{1,x_{0}-\delta }(x_{0},\lambda ){\widetilde{y}} _{2,x_{0}-\delta }^{\prime }(x_{0},\lambda )-y_{1,x_{0}-\delta }^{\prime }(x_{0},\lambda ){\widetilde{y}}_{2,x_{0}-\delta }(x_{0},\lambda )\right] \nonumber \\&\qquad +\,B_{3}(\lambda )\left[ {\widetilde{y}}_{1,x_{0}-\delta }^{\prime }(x_{0},\lambda )y_{2,x_{0}-\delta }(x_{0},\lambda )-{\widetilde{y}} _{1,x_{0}-\delta }(x_{0},\lambda )y_{2,x_{0}-\delta }^{\prime }(x_{0},\lambda )\right] \nonumber \\&\qquad +B_{4}(\lambda )\left[ y_{2,x_{0}-\delta }(x_{0},\lambda ){\widetilde{y}} _{2,x_{0}-\delta }^{\prime }(x_{0},\lambda )-y_{2,x_{0}-\delta }^{\prime }(x_{0},\lambda ){\widetilde{y}}_{2,x_{0}-\delta }(x_{0},\lambda )\right] \nonumber \\ \end{aligned}$$
(3.73)
where
$$\begin{aligned} B_{1}(\lambda )= & {} y_{2,r}\left( x_{0}-\delta ,\lambda \right) {\widetilde{y}} _{2,r}\left( x_{0}-\delta ,\lambda \right) =O\left( \left| \lambda ^{-1}\right| \exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \left( x_{0}-\delta -r\right) \right) \right) , \\ B_{2}(\lambda )= & {} y_{2,r}\left( x_{0}-\delta ,\lambda \right) {\widetilde{y}} _{2,r}^{\prime }\left( x_{0}-\delta ,\lambda \right) =O\left( \left| \sqrt{\lambda }\right| ^{-1}\exp \left( 2\left| \text {Im}\sqrt{ \lambda }\right| \left( x_{0}-\delta -r\right) \right) \right) , \\ B_{3}(\lambda )= & {} {\widetilde{y}}_{2,r}\left( x_{0}-\delta ,\lambda \right) y_{2,r}^{\prime }\left( x_{0}-\delta ,\lambda \right) =O\left( \left| \sqrt{\lambda }\right| ^{-1}\exp \left( 2\left| \text {Im}\sqrt{ \lambda }\right| \left( x_{0}-\delta -r\right) \right) \right) , \\ B_{4}(\lambda )= & {} {\widetilde{y}}_{2,r}^{\prime }\left( x_{0}-\delta ,\lambda \right) y_{2,r}^{\prime }\left( x_{0}-\delta ,\lambda \right) =O\left( \exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \left( x_{0}-\delta -r\right) \right) \right) , \end{aligned}$$
which can be obtained from (3.74). Moreover, from Lemma 7, it is easy to see
$$\begin{aligned} y_{1,x_{0}-\delta }(x_{0},\lambda ){\widetilde{y}}_{1,x_{0}-\delta }^{\prime }(x_{0},\lambda )-y_{1,x_{0}-\delta }^{\prime }(x_{0},\lambda ){\widetilde{y}} _{1,x_{0}-\delta }(x_{0},\lambda )= & {} o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \delta \right) }{\left| \sqrt{ \lambda }\right| ^{m+1}}\right) , \\ y_{1,x_{0}-\delta }(x_{0},\lambda ){\widetilde{y}}_{2,x_{0}-\delta }^{\prime }(x_{0},\lambda )-y_{1,x_{0}-\delta }^{\prime }(x_{0},\lambda ){\widetilde{y}} _{2,x_{0}-\delta }(x_{0},\lambda )= & {} o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \delta \right) }{\left| \sqrt{ \lambda }\right| ^{m+2}}\right) , \\ {\widetilde{y}}_{1,x_{0}-\delta }^{\prime }(x_{0},\lambda )y_{2,x_{0}-\delta }(x_{0},\lambda )-{\widetilde{y}}_{1,x_{0}-\delta }(x_{0},\lambda )y_{2,x_{0}-\delta }^{\prime }(x_{0},\lambda )= & {} o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \delta \right) }{\left| \sqrt{\lambda }\right| ^{m+2}}\right) , \\ y_{2,x_{0}-\delta }(x_{0},\lambda ){\widetilde{y}}_{2,x_{0}-\delta }^{\prime }(x_{0},\lambda )-y_{2,x_{0}-\delta }^{\prime }(x_{0},\lambda ){\widetilde{y}} _{2,x_{0}-\delta }(x_{0},\lambda )= & {} o\left( \frac{\exp \left( 2\left| \text {Im}\sqrt{\lambda }\right| \delta \right) }{\left| \sqrt{ \lambda }\right| ^{m+3}}\right) \end{aligned}$$
as \(\left| \lambda \right| \rightarrow \infty \) in \(\Lambda _{\zeta }.\) Thus the equality (3.43) can be directly obtained from (3.73). The statements (3.40)–(3.42) can be proved similarly. \(\square \)
Remark 16
If q and \({\widetilde{q}}\) are both assumed to be in \( L_{ {\mathbb {C}} }^{1}\left[ 0,\pi \right] ,\) then one can easily find that relations (3.40)–(3.43) still hold by taking \(m=-1\). In fact, the equality (3.43) with \(m=-1\) can be obtained by the following asymptotic form [28]:
$$\begin{aligned}&y_{2,r}(x,\lambda ) \nonumber \\&\quad =\frac{\sin (\sqrt{\lambda }\left( x-r\right) )}{\sqrt{\lambda }}-Q\left( x\right) \frac{\cos (\sqrt{\lambda }\left( x-r\right) )}{2\lambda }+o\left( \frac{\exp \left( \left| \text {Im}\sqrt{\lambda }\right| \left( x-r\right) \right) }{\left| \lambda \right| }\right) , \nonumber \\&y_{2,r}^{\prime }(x,\lambda ) \nonumber \\&\quad =\cos (\sqrt{\lambda }\left( x-r\right) )+Q\left( x\right) \frac{\sin ( \sqrt{\lambda }\left( x-r\right) )}{2\sqrt{\lambda }}\nonumber \\&\qquad +\,o\left( \frac{\exp \left( \left| \text {Im}\sqrt{\lambda }\right| \left( x-r\right) \right) }{\left| \sqrt{\lambda }\right| }\right) , \end{aligned}$$
(3.74)
where \(Q\left( x\right) =\int _{r}^{x}q(t)dt.\) (3.40)–(3.42) can be treated similarly.
