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The Local Borg–Marchenko Uniqueness Theorem for Potentials Locally Smooth at the Right Endpoint

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Abstract

In this note we are concerned with the local Borg–Marchenko uniqueness theorem for potentials locally smooth at the right endpoint. We establish the so-called high-energy asymptotics of the difference \((m_{1}-m_{2})(z)\) of two Weyl–Titchmarsh functions \(m_{j}(z)\) corresponding to two Schrödinger operators \(H_j=-d^2/dx^2+q_j\), for \(j=1,2\) and \(q_1=q_2\) a.e. on [0, a], in \(L_{2}(0,b)\) with \(0<a<b\le \infty \), where the potentials \( q_j\) are sufficiently smooth in a right neighbourhood of the point a and their right derivatives at a coincide up to a certain order. Moreover, we also provide a new proof of the local Borg–Marchenko theorem.

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Acknowledgements

The authors would like to express their great appreciation to Professor Christiane Tretter and the anonymous referees for careful reading of the manuscript and giving us very helpful suggestions for the improvement of the original manuscript. The research was supported in part by the NNSF (11571212) of China.

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Correspondence to Guangsheng Wei.

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Appendix A: Proof of Lemma 2.3

Appendix A: Proof of Lemma 2.3

In this appendix we give a new proof of Lemma 2.3. In the following we always let \(z=\lambda ^{2}\) with \(\mathrm {Im}\lambda \ge 0\) for symbolic simplification. Note that \(|z|\rightarrow \infty \) in the sector \(\Lambda (\delta )\) is equivalent to \(|\lambda |\rightarrow \infty \) in the sector \( \Lambda (\delta /2)\).

Before proving Lemma 2.3, for \(q\in C^{2k}[a,c]\) \((0\le a<c<\infty ),\) we use [11, Lemma 1.4.1] to give a solution, denoted by \(\hat{y} (x,\lambda ),\) of (1.4) defined on [ac],  which has the following form:

$$\begin{aligned} \hat{y}(x,\lambda )=e^{i\lambda (x-a)}\left( v_{0}(x)+\frac{v_{1}(x)}{ 2i\lambda }+\ldots +\frac{v_{2k}(x)}{(2i\lambda )^{2k}}+\frac{ v_{2k+1}(x,\lambda )}{(2i\lambda )^{2k+1}}\right) \end{aligned}$$
(A.1)

where \(v_{0}(x)=1,\) \(v_{i}(x)=\int _{a}^{x}H(v_{i-1})(t)dt\) for \( i=1,2,\ldots ,2k+1\) and

$$\begin{aligned} v_{2k+1}(x,\lambda )= & {} v_{2k+1}(x)+\frac{1}{2i\lambda } \int _{a}^{x}q(t)v_{2k+1}(t)dt \nonumber \\&-\int _{a}^{x}\left\{ v_{2k+1}^{\prime }(x+a-\xi )+\frac{1}{2i\lambda } K(x,\xi )\right\} e^{-2i\lambda (\xi -a)}d\xi \nonumber \\ \end{aligned}$$
(A.2)

in which both \(K(x,\xi )\) and \(v_{2k+1}^{\prime }(x+a-\xi )\) are continuous on [ac] in the variable \(\xi \) for every \(x\in [a,c]\). The proof of (A.1) is similar to that of [11, Lemma 1.4.1].

Lemma A.1

For \(q\in C^{2k}[a,c]\), let \(\hat{y}(x,\lambda )\) be the solution of (1.4) defined by (A.1). Then

$$\begin{aligned} \hat{y}(c,\lambda )=\frac{e^{-i\lambda (c-a)}}{(2i\lambda )^{2k+2}} v_{2k+1}^{\prime }(a)+o\left( \frac{e^{(c-a)\mathrm {Im}\lambda }}{\lambda ^{2k+2}}\right) \end{aligned}$$
(A.3)

as \(\left| \lambda \right| \rightarrow \infty \) in the sector \( \Lambda (\delta )\).

Proof

Note that, for any given \(c_{0},c_{1},c_{2}>0,\) the inequality \(e^{c_{0} \mathrm {Im}\lambda }\ge c_{1}|\lambda |^{c_{2}}\) remains true for sufficiently large \(|\lambda |\) with \(\lambda \in \Lambda (\delta )\). From (A.1) and (A.2) we have

$$\begin{aligned} \hat{y}(c,\lambda )&= -\frac{e^{i\lambda (c-a)}}{(2i\lambda )^{2k+1}} \int _{a}^{c}v_{2k+1}^{\prime }(c+a-\xi )e^{-2i\lambda (\xi -a)}d\xi +o\left( \frac{e^{(c-a)\mathrm {Im}\lambda }}{\lambda ^{2k+2}}\right) \\&= -\frac{e^{-i\lambda (c-a)}}{(2i\lambda )^{2k+2}}\int _{a}^{c}(2i\lambda )v_{2k+1}^{\prime }(c+a-\xi )e^{2i\lambda (c-\xi )}d\xi +o\left( \frac{ e^{(c-a)\mathrm {Im}\lambda }}{\lambda ^{2k+2}}\right) \end{aligned}$$

as \(\left| \lambda \right| \rightarrow \infty \) in \(\Lambda (\delta )\). In order to prove (A.3) we only need to prove

$$\begin{aligned} g(\lambda ):=\int _{a}^{c}2i\lambda v_{2k+1}^{\prime }(c+a-\xi )e^{2i\lambda (c-\xi )}d\xi +v_{2k+1}^{\prime }(a)=o(1) \end{aligned}$$
(A.4)

as \(\left| \lambda \right| \rightarrow \infty \) in \(\Lambda (\delta )\).

Let \(\epsilon =1/\sqrt{\mathrm {Im}\lambda }\) and \(\lambda \in \Lambda (\delta )\). Consider the function

$$\begin{aligned} g_{1}(\lambda ):=\int _{c-\epsilon }^{c}2i\lambda v_{2k+1}^{\prime }(c+a-\xi )e^{2i\lambda (c-\xi )}d\xi +v_{2k+1}^{\prime }(a). \end{aligned}$$
(A.5)

Taking \(\int _{c-\epsilon }^{c}2i\lambda e^{2i\lambda (c-\xi )}d\xi =-1+e^{2i\lambda \epsilon }\) into account, we get

$$\begin{aligned} \left| g_{1}(\lambda )\right|&= \left| \int _{c-\epsilon }^{c}2i\lambda \left( v_{2k+1}^{\prime }(c+a-\xi )-v_{2k+1}^{\prime }(a)\right) e^{2i\lambda (c-\xi )}d\xi +v_{2k+1}^{\prime }(a)e^{2i\lambda \epsilon }\right| \\&\le \int _{c-\epsilon }^{c}2\left| \lambda \right| \left| v_{2k+1}^{\prime }(c+a-\xi )-v_{2k+1}^{\prime }(a)\right| e^{-2(c-\xi ) \mathrm {Im}\lambda }d\xi \\&\quad +\left| v_{2k+1}^{\prime }(a)\right| e^{-2\sqrt{\mathrm {Im}\lambda }}. \end{aligned}$$

By the fact that \(\mathrm {Im}\lambda \ge \left| \lambda \right| \sin \delta \) for all \(\lambda \in \Lambda (\delta )\), it follows that

$$\begin{aligned}&\int _{c-\epsilon }^{c}2\left| \lambda \right| \left| v_{2k+1}^{\prime }(c+a-\xi )-v_{2k+1}^{\prime }(a)\right| e^{-2(c-\xi ) \mathrm {Im}\lambda }d\xi \\&\quad \le \max _{t\in [c-\epsilon ,c]}\left| v_{2k+1}^{\prime }(c+a-t)-v_{2k+1}^{\prime }(a)\right| \frac{\left| \lambda \right| }{\mathrm {Im}\lambda }\left( 1-e^{-2\epsilon \mathrm {Im}\lambda }\right) \\&\quad \le \frac{1}{\sin \delta }\max _{t\in [c-1/\sqrt{\mathrm {Im}\lambda },c]}\left| v_{2k+1}^{\prime }(c+a-t)-v_{2k+1}^{\prime }(a)\right| \left( 1-e^{-2\sqrt{\mathrm {Im}\lambda }}\right) , \end{aligned}$$

which implies, for \(\lambda \rightarrow \infty \) in \(\Lambda (\delta ),\) \( g_{1}(\lambda )=o(1)\).

On the other hand, we have

$$\begin{aligned} g_{2}(\lambda ):= & {} \left| \int _{a}^{c-\epsilon }2i\lambda v_{2k+1}^{\prime }(c+a-\xi )e^{2i\lambda (c-\xi )}d\xi \right| \nonumber \\\le & {} C_{0}\int _{a}^{c-\epsilon }2\left| \lambda \right| e^{-2(c-\xi )\mathrm {Im}\lambda }d\xi \nonumber \\\le & {} \frac{C_{0}}{\sin \delta }\left( e^{-2\sqrt{\mathrm {Im}\lambda } }-e^{-2(c-a)\mathrm {Im}\lambda }\right) , \end{aligned}$$
(A.6)

where \(C_{0}=\max _{\xi \in [a,c-\epsilon ]}\left| v_{2k+1}^{\prime }(c+a-\xi )\right| \), which implies \(g_{2}(\lambda )=o(1)\) as \(\lambda \rightarrow \infty \) in \(\Lambda (\delta )\).

The above discussion shows that \(g(\lambda )=g_{1}(\lambda )+g_{2}(\lambda )=o(1)\) as \(\lambda \rightarrow \infty \) in \(\Lambda (\delta )\).

It follows form (A.1) and (A.2) that

$$\begin{aligned} \hat{y}(c,-\lambda )=e^{-i\lambda (c-a)}(1+o(1)), \end{aligned}$$
(A.7)

as \(\left| \lambda \right| \rightarrow \infty \) in the sector \( \Lambda (\delta )\).

We are now in a position to prove Lemma 2.3.

Proof of Lemma 2.3

Let \(c=a+\varepsilon \) in Lemma 2.3. For \(m_{+}(a,z)\) defined by (2.2), by the same proof of Lemma 2.1 and (2.7), we have

$$\begin{aligned} \frac{1}{m_{+}(a,z)}= & {} \frac{1}{y_{N}^{2}(c,\lambda )}\left( \frac{1}{ m_{+}(c,z)+m_{-,a}(c,z)}\right) -\frac{y_{D}(c,\lambda )}{y_{N}(c,\lambda )} \nonumber \\= & {} -\frac{y_{D}(c,\lambda )}{y_{N}(c,\lambda )}\left( 1-\frac{1}{ y_{N}(c,\lambda )y_{D}(c,\lambda )}\left( \frac{1}{m_{+}(c,z)+m_{-,a}(c,z)} \right) \right) , \nonumber \\ \end{aligned}$$
(A.8)

where \(y_{N}(x,\lambda ),\) \(y_{D}(x,\lambda )\) are the solutions of (1.4) under the initial conditions \(y_{N}(a)=1,y_{N}^{\prime }(a)=0;\) \( y_{D}(a)=0,y_{D}^{\prime }(a)=1\) and \(m_{-,a}(c,z)=-y_{N}^{\prime }(c,\lambda )/y_{N}(c,\lambda )\).

By [11, p.58] we arrive at

$$\begin{aligned} \frac{y_{N}(c,\lambda )}{y_{D}(c,\lambda )}=\frac{\hat{y}(c,\lambda )\left( i\lambda -\sigma (a,-\lambda )\right) +\hat{y}(c,-\lambda )\left( i\lambda +\sigma (a,\lambda )\right) }{\hat{y}(c,\lambda )-\hat{y}(c,-\lambda )}, \end{aligned}$$
(A.9)

where

$$\begin{aligned} \sigma (a,\lambda )=\sum _{j=1}^{2k}\frac{\sigma _{j}(a)}{(2i\lambda )^{j}}, \end{aligned}$$
(A.10)

(see [11, p.55-56]) with \(\sigma _{1}(a)=q(a)\), \(\sigma _{2}(a)=-q^{\prime }(a)\) and

$$\begin{aligned} \sigma _{j+1}(a)=-\sigma _{j}^{\prime }(a)-\sum _{p=1}^{j-1}\sigma _{j-p}(a)\sigma _{p}(a)\text { \ }(j=2,3,\ldots ,2k). \end{aligned}$$
(A.11)

From [11, p.56], we have \(v_{2k+1}^{\prime }(a)=\sigma _{2k+1}(a)\). From (A.3) and (A.7), we easily see that

$$\begin{aligned} \frac{\hat{y}(c,\lambda )}{\hat{y}(c,-\lambda )}&= \frac{\sigma _{2k+1}(a)/(2i\lambda )^{2k+2}+o\left( 1/\lambda ^{2k+2}\right) }{1+o(1)} \\&= \frac{\sigma _{2k+1}(a)}{(2i\lambda )^{2k+2}}+o\left( \frac{1}{\lambda ^{2k+2}}\right) , \end{aligned}$$

which implies

$$\begin{aligned} \hat{y}(c,\lambda )-\hat{y}(c,-\lambda )= & {} -\hat{y}(c,-\lambda )\left( 1- \frac{\hat{y}(c,\lambda )}{\hat{y}(c,-\lambda )}\right) \nonumber \\= & {} -\hat{y}(c,-\lambda )\left( 1-\frac{\sigma _{2k+1}(a)}{(2i\lambda )^{2k+2} }+o\left( \frac{1}{\lambda ^{2k+2}}\right) \right) \end{aligned}$$
(A.12)

and

$$\begin{aligned}&\hat{y}(c,\lambda )(i\lambda -\sigma (a,-\lambda ))+\hat{y}(c,-\lambda )(i\lambda +\sigma (a,\lambda )) \nonumber \\&\quad = \hat{y}(c,-\lambda )\left( i\lambda +\sigma (a,\lambda )+\frac{\hat{y} (c,\lambda )}{\hat{y}(c,-\lambda )}(i\lambda -\sigma (a,-\lambda ))\right) \nonumber \\&\quad = \hat{y}(c,-\lambda )\left( i\lambda +\sigma (a,\lambda )+\frac{\sigma _{2k+1}(a)}{2(2i\lambda )^{2k+1}}+o\left( \frac{1}{\lambda ^{2k+1}}\right) \right) \end{aligned}$$
(A.13)

as \(|\lambda |\rightarrow \infty \) in the sector \(\Lambda (\delta /2)\). From (A.9)–(A.10) and (A.12)–(A.13) we have

$$\begin{aligned} \frac{y_{N}(c,\lambda )}{y_{D}(c,\lambda )}= & {} -\frac{i\lambda +\sigma (a,\lambda )+\sigma _{2k+1}(a)/\left( 2(2i\lambda )^{2k+1}\right) +o\left( 1/\lambda ^{2k+1}\right) }{1-\sigma _{2k+1}(a)/(2i\lambda )^{2k+2}+o\left( 1/\lambda ^{2k+2}\right) } \nonumber \\= & {} -i\lambda -\sigma (a,\lambda )-\frac{\sigma _{2k+1}(a)}{(2i\lambda )^{2k+1}}+o\left( \frac{1}{\lambda ^{2k+1}}\right) \nonumber \\= & {} -i\lambda -\sum _{j=1}^{2k+1}\frac{\sigma _{j}(a)}{(2i\lambda )^{j}} +o\left( \frac{1}{\lambda ^{2k+1}}\right) \end{aligned}$$
(A.14)

as \(\left| \lambda \right| \rightarrow \infty \) in \(\Lambda (\delta /2)\). Let \(C_{j}(a)=\sigma _{j}(a)/(2i)^{j}\). From (A.11) and (A.14) it is easy to see that (2.9) holds and

$$\begin{aligned} \frac{y_{N}(c,\lambda )}{y_{D}(c,\lambda )}=-i\lambda -\sum _{j=1}^{2k+1} \frac{C_{j}(a)}{\lambda ^{j}}+o\left( \frac{1}{\lambda ^{2k+1}}\right) \end{aligned}$$
(A.15)

as \(\left| \lambda \right| \rightarrow \infty \) in \(\Lambda (\delta /2)\).

On the other hand, it is known [9] that the solutions \( y_{N}(x,\lambda )\) and \(y_{D}(x,\lambda )\) on [ac] have the following asymptotic expansions:

$$\begin{aligned} y_{N}(x,\lambda )= & {} e^{-i(x-a)\lambda }\left( \frac{1}{2}+O\left( \frac{1}{ \lambda }\right) \right) , \end{aligned}$$
(A.16)
$$\begin{aligned} y_{D}(x,\lambda )= & {} \frac{ie^{-i(x-a)\lambda }}{\lambda }\left( \frac{1}{2} +O\left( \frac{1}{\lambda }\right) \right) , \end{aligned}$$
(A.17)

and it is also known [1] that both \(m_{+}(c,z)\) and \(m_{-,a}(c,z)\) have the same asymptotics: \(i\lambda +o(1)\) as \(\left| \lambda \right| \rightarrow \infty \) in the sector \(\Lambda (\delta /2)\). Consequently, together with (A.16) and (A.17) we have

$$\begin{aligned} \frac{1}{y_{N}(c,\lambda )y_{D}(c,\lambda )}\left( \frac{1}{ m_{+}(c,z)+m_{-,a}(c,z)}\right) =o\left( \frac{1}{\lambda ^{2k+2}}\right) \end{aligned}$$
(A.18)

as \(\left| \lambda \right| \rightarrow \infty \) in the same sector. From (A.8), (A.15) and (A.18) we infer

$$\begin{aligned} m_{+}(a,z)&= -\frac{y_{N}(c,\lambda )}{y_{D}(c,\lambda )}\left( \frac{1}{ 1+o\left( 1/\lambda ^{2k+2}\right) }\right) \\&= -\frac{y_{N}(c,\lambda )}{y_{D}(c,\lambda )}\left( 1+o\left( \frac{1}{ \lambda ^{2k+2}}\right) \right) \\&= i\lambda +\sum _{j=1}^{2k+1}\frac{C_{j}(a)}{\lambda ^{j}}+o\left( \frac{1}{ \lambda ^{2k+1}}\right) \end{aligned}$$

as \(\left| \lambda \right| \rightarrow \infty \) in \(\Lambda (\delta /2),\) which yields (2.8) since \(z=\lambda ^{2}\).

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Bai, Y., Wei, G. The Local Borg–Marchenko Uniqueness Theorem for Potentials Locally Smooth at the Right Endpoint. Integr. Equ. Oper. Theory 91, 31 (2019). https://doi.org/10.1007/s00020-019-2529-z

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