September 08, 2025
In this paper, we consider the inverse spectral problem of determining the spherically symmetric refractive index in a bounded spherical region of radius \(b\). Instead of the usual case of the refractive index \(\rho\in W^2_2\), by using singular Sturm-Liouville theory, we first discuss the case when the refractive index \(\rho\) is a piecewise \(W^1_2\) function. We prove that if \(\int_0^b \sqrt{\rho(r)} dr<b\), then \(\rho\) is uniquely determined by all special transmission eigenvalues; if \(\int_0^b \sqrt{\rho(r)} dr=b\), then all special transmission eigenvalues with some additional information can uniquely determine \(\rho\). We also consider the mixed spectral problem and obtain that \(\rho\) is uniquely determined from partial information of \(\rho\) and the “almost real subspectrum".
The interior transmission problem appears in scattering theory for inhomogeneous acoustic and electromagnetic media, which was introduced by Kirsch, Colton and Monk [1], [2]. It is a non-selfadjoint problem for two fields \(w\) and \(v\): \[\begin{align} \label{transmission} \begin{cases} \Delta w+\lambda \rho(\mathbf{x})w=0, & \mathbf{x}\in \Omega, \\ \Delta v+\lambda v=0,& \mathbf{x}\in \Omega,\\ w=v, \frac{\partial w}{\partial \mathbf{v}}=\frac{\partial v}{\partial \mathbf{v}},& \mathbf{x}\in \partial \Omega. \end{cases} \end{align}\tag{1}\] Here \(\lambda\) is the spectral parameter, \(\Omega\) is a bounded and simply connected set in \(\mathbb{R}^n\) with smooth boundary, \(\mathbf{v}\) is the outward unit normal to \(\partial \Omega\), \(\rho(\mathbf{x})\) denotes the refractive index of the medium [3], [4]. The complex values of \(\lambda\) for which a nontrivial solution \((w, v)\) exists are called transmission eigenvalues. We refer to [5]–[9] for the results about the existence of transmission eigenvalues and bound of eigenvalues of Laplacian, the relation of the bound of transmission eigenvalues and Laplacian eigenvalues.
An interesting and important issue related to the interior transmission problem is the corresponding inverse spectral problem. Namely, whether we can uniquely determine \(\rho\) in \(\Omega\) if all or the certain subset of the transmission eigenvalue are known. If \(n = 3\), \(\Omega=\Omega_b\) is a ball of radius \(b>0\) centered at the origin and \(\rho(\mathbf{x})\) is spherically symmetric (\(\rho(\mathbf{x}) = \rho(r), r = |\mathbf{x}|\)), then problem 1 can be transformed into the one-dimensional eigenvalue problem [10], [11]. In this paper, we consider inverse problems of recovering \(\rho(r)\) from transmission eigenvalues with spherically symmetric eigenfunctions \((\omega,v)\). Then the inverse problem is equivalent to recovering \(\rho\) from eigenvalues of the special transmission eigenvalue problem \(Q(\rho)\) \[\begin{align} \label{string} \left\{\begin{aligned} &-u^{\prime \prime }=\lambda \rho(r)u,\qquad 0<r<b, \\ &u(0)=0=u'(b)\frac{\sin( \sqrt{\lambda}b)}{\sqrt{\lambda}}-u(b){\cos (\sqrt{\lambda}b)}. \end{aligned}\right. \end{align}\tag{2}\]
Inverse spectral problem for \(Q(\rho)\) was first studied by McLaughlin-Polyakov [11] and then considered by a number of authors [12]–[14], but?. Previous literature on inverse spectral analysis for problem \(Q(\rho)\) always considered that \(\rho\) is a \(W_2^2\) or piecewise \(C^2\) function [14]–[17]. Aktosun-Gintides-Papanicolaou [15] showed that if \(a<b\), then all special transmission eigenvalues uniquely determine \(\rho\); if \(a=b\), then all special transmission eigenvalues together with the additional constant \(\gamma\) uniquely determine \(\rho\). Here \[\begin{align} \label{definitiona} a=\int_0^b \sqrt{\rho(r)} dr. \end{align}\tag{3}\] In particular, when \(a = b\), for the unique determination of the potential, the additional constant \(\gamma\) is necessary to be known [18]. Wei-Xu [19] considered the case when \(a>b\). They showed that \(\rho\) is uniquely determined by all special transmission eigenvalues and normalizing constants corresponding to the partial simple eigenvalues. Later, Yang-Buterin [17] gave the uniqueness theorem from the data involving fractions of the special transmission eigenvalues. See [20]–[28] more results about eigenvalue problems and related inverse problems.
In this paper, we first investigate inverse spectral problems related to 2 for a piecewise \(W_2^1\) refractive index. In this case, problem 2 models a complicated medium for a less smooth refractive index with several layers where the index has jumps between each layer. This uniqueness question shows that less smooth materials with layers can be determined from the scattered far fields [11]. It can also be used for the numerical investigation of inverse problems using a piecewise constant approximation of the refractive index [29]. We consider that \(\rho\) has a jump discontinuous point \(b_1\). Namely, \(\rho\in W_2^1\left((0, b_1)\cup (b_1, b)\right)\) and satisfies \[\begin{align} \rho(b_1+)=b_2\rho(b_1-),\;b_2>0\;\mathrm{and}\;b_2\neq 1,\;\rho(r)>0\;\mathrm{for\;any\;} r\in [0,b_1)\cup (b_1,b]. \label{piecewiseac} \end{align}\tag{4}\] Note that \(b_2\neq 1\) is a natural assumption since \(\rho\) is continuous on the whole interval if \(b_2=1\).
The relaxation of the refractive index makes inverse transmission problems much more complicated. If \(\rho\in W_2^2(0,b)\), by Liouville transformation 7 , we can transform the equation \(-u''=\lambda \rho u\) into a Sturm-Liouville (SL) equation with the potential \(q\in L^2(0,a)\). The mapping \[\begin{align} \nonumber \mathcal{M}: \rho\rightarrow (\rho(b),\rho'(b),q(x) ) \end{align}\] is injective. If \(\rho\) is a piecewise \(W^1_2\) function, the reduction to the potential form is possible, but the potential is a distribution from \(W_2^{-1}(0,a)\) and eigenfunctions have a discontinuous point \(d\). The mapping \(\mathcal{M}\) cannot be extended directly to the case when \(\rho\) is a piecewise \(W^1_2\) function. In this case, we can reformulate the information of \((\rho'(b), q(x))\) as \(\sigma(x)\), where \(\sigma\in L^2(0,a)\) is the anti-derivative function of \(q\). The mapping from \(\rho\) to \((\rho(b), \sigma(x),d, d_1)\) is injective (see Section 2 for definitions of \(d, d_1\)). By Liouville transformation, the Barcilon formula refined by Shen [30] with \(\rho\) to be a piecewise continuously differentiable function is a consequence of certain asymptotic formula. In this paper, by studying the discontinuous SL problem with singular potentials, namely, recovering \((\sigma,d,d_1)\) instead of \(q\) in classical SL theory, we obtain uniqueness theorems even dropping the information of \(\rho'(b)\) (see Theorem 8). See [31]–[34] for some results on the singular SL problem without discontinuities.
The structure of this paper is as follows. In Section 2, we use Liouville transformation to transform 2 into the discontinuous SL problem with singular potentials. Section 3 gives the integral representation of the initial solution. Section 4 introduces the Weyl-Titchmarsh function of the discontinuous SL problem with singular potentials and proves the corresponding uniqueness theorem. We also give the high-energy asymptotic behavior of the Weyl-Titchmarsh function. In Section 5, we use the discontinuous SL problem with singular potentials to study inverse transmission eigenvalue problems by all eigenvalues. Section 6 studies properties of “almost real subspectrum” \(\{\mu_m\}_{m=1}^\infty\), which are real except for finite many eigenvalues. We show that the “almost real subspectrum” \(\{\mu_m\}_{m=1}^\infty\) and some information on the refractive index uniquely determine \(\rho\).
Let \(u(r,\lambda)\) be the solution of \(-u''=\lambda \rho u\) satisfying the initial condition \(u(0,\lambda)=0, u'(0,\lambda)=1.\) It is known [15] that special transmission eigenvalues of 2 coincide with the zeros of its characteristic function \[D (\lambda):=\left\vert \begin{array}{cc} \frac{\sin (\sqrt{\lambda}b )}{\sqrt{\lambda} } & u(b,\lambda) \\ \cos (\sqrt{\lambda}b ) & u^{\prime }(b,{\lambda}) \end{array}\right\vert . \label{eq1464}\tag{5}\] Let \(\{\lambda _{k}\}_{k=1}^{\infty }\) denote the eigenvalues of 2 with account of multiplicity. Then according to Hadamard’s factorization theorem, we have \[D (\lambda)=\gamma \lambda ^{s}\prod_{\lambda _{k}\neq 0}\left(1-\frac{\lambda }{\lambda _{k}}\right), \label{eq1465}\tag{6}\] where \(s\geq 1\) is the multiplicity of the eigenvalue \(\lambda =0\), \(\gamma \in \mathbb{R}\).
By Liouville transformation \[\begin{align} \label{liouville} x=\int_0^r\sqrt{\rho(t) }dt, \end{align}\tag{7}\] we can transform \(-u''=\lambda \rho u\) into the discontinuous SL problem with singular potentials.
Lemma 1. Assume that \(\rho\in W_2^1\left((0, b_1)\cup (b_1, b)\right)\) and satisfies 4 . Then \[\begin{align} \label{zxlambda} z(x,\lambda):=(\rho(r))^{1/4}u(r,\lambda) \end{align}\tag{8}\] satisfies the equation \[\begin{align} -&\frac{d z^{[1]}(x)}{dx}- \sigma(x) z'(x)=\lambda z(x), \quad x\in(0,d)\cup (d,a), \tag{9} \\ &z(d+)=d_1z(d-), \;z^{[1]}(d+)=d_1^{-1}z^{[1]}(d-), \tag{10} \end{align}\] where \(a\) is defined by 3 , \(\sigma\in L^2(0,a)\), \(z^{[1]}(x)=z'(x)-\sigma(x) z(x),\) \[\begin{align} d=\int_0^{b_1} \sqrt{\rho(t)} dt, \;d_1={b_2}^{1/4}, \label{definitione1} \end{align}\tag{11}\] \[\begin{align} \sigma(x)=\frac{1}{4}\frac{\rho'(r)}{(\rho(r))^{3/2}}+g(x). \label{definitionsigma} \end{align}\tag{12}\] Here \(g(x)\) satisfies \[\begin{align} \label{gprime} g'(x)=\frac{1}{16}\frac{(\rho'(r))^2}{(\rho(r))^3}, \quad x\in (0,d)\cup (d,a), \end{align}\tag{13}\] and the jump condition \[\begin{align} \label{gprime1} g(d-)={b_2}^{1/2}g(d+). \end{align}\tag{14}\]
Proof. We first show that \(z(x)\) satisfies 9 . For \(x\in(0,d)\cup (d,a)\), by 8 , we know \[\begin{align} z'(x)=\frac{1}{4}(\rho(r))^{-5/4}\rho'(r)u(r) +(\rho(r))^{-1/4}u'(r). \nonumber \end{align}\] Hence \[\begin{align} z'(x)-\sigma(x) z(x)=(\rho(r))^{-1/4}u'(r)-g(x)(\rho(r))^{1/4}u(r), \label{zprime} \end{align}\tag{15}\] \[\begin{align} \sigma(x) z'(x)=&\frac{1}{16}(\rho'(r))^2(\rho(r))^{-11/4}u(r) +\frac{1}{4}(\rho(r))^{-7/4}\rho'(r)u'(r) \nonumber \\ &+\frac{1}{4}g(x)(\rho(r))^{-5/4}\rho'(r)u(r)+g(x)(\rho(r))^{-1/4}u'(r). \label{ac} \end{align}\tag{16}\] Differentiating 15 with respect to \(x\), we obtain that \[\begin{align} -\frac{d z^{[1]}(x)}{dx}=&\frac{1}{4}(\rho(r))^{-7/4}\rho'(r)u'(r)-(\rho(r))^{-3/4}u''(r)+\frac{1}{16}(\rho'(r))^2(\rho(r))^{-11/4}u(r) \nonumber\\ &+g(x)\left(\frac{1}{4}(\rho(r))^{-5/4}\rho'(r)u(r) +(\rho(r))^{-1/4}u'(r)\right). \label{z11111111} \end{align}\tag{17}\] Subtracting 16 from 17 , by 8 , one has that \[\begin{align} \nonumber -\frac{d z^{[1]}(x)}{dx}- \sigma(x) z'(x)=-(\rho(r))^{-\frac{3}{4}}u''(r)=\lambda z(x). \end{align}\] Therefore, we conclude that 9 holds.
According to 4 , 8 and 15 , \(z(x,\lambda)\) satisfies the following jump condition \[\begin{align} z(d+)=d_1z(d-), \;z^{[1]}(d+)=d_1^{-1}z^{[1]}(d-)+d_2z(d-), \label{jump1} \end{align}\tag{18}\] where \[\begin{align} d_1={b_2}^{1/4}, d_2= g(d-) {b_2}^{-{1}/{4}}-g(d+){b_2}^{{1}/{4}}. \label{definitione2} \end{align}\tag{19}\] Assume that \(d_2=0\). By 19 , \(g(x)\) satisfies the jump condition 14 . The lemma is proved. ◻
Remark 1. Denote \(q=\sigma', q\in W^{-1}_2(0,a).\) Then 9 can be recast in the form of SL equation \(-z''+q(x)z=\lambda z\) in the distribution sense. Hence we call 9 the SL equation with singular potentials. We mention that Albeverio-Hryniv-Mykytyuk [31] showed that some SL operators in impedance form are unitarily equivalent to SL operators with singular potentials.
By Liouville transformation 8 , we can transform 2 into discontinuous SL equation with jump condition 10 . Also the characteristic function \(D(\lambda)\) is transformed into \[\label{d} D (\lambda)=\rho(b)^{1/4}\left\vert \begin{array}{cc} \frac{\sin (\sqrt{\lambda}b )}{\sqrt{\lambda}}&\beta z(a,\lambda) \\ \cos (\sqrt{\lambda}b )& z^{[1]}(a, \lambda)+g(a)z(a,\lambda) \end{array}\right\vert, \beta=\frac{1}{\rho(b)^{1/2}}.\tag{20}\] Since \(g(a)\) is an arbitrary real number, then we transform problem \(Q(\rho)\) into a family of discontinuous SL problems with singular potentials. In order to ensure the uniqueness of the image of Liouville transformation, we choose \(g(a)=0\). Therefore, we can transform problem \(Q(\rho)\) into the problem \[\label{eq215} \left\{ \begin{align} & -\frac{d z^{[1]}(x)}{dx}-\sigma z'(x)=\lambda z(x), \quad x\in(0,d)\cup (d,a), \\ &z(d+)=d_1z(d-), \;z^{[1]}(d+)=d_1^{-1}z^{[1]}(d-), \\ &z(0)=D(\lambda)=0, \end{align}\right.\tag{21}\] where \[D (\lambda)=\rho(b)^{1/4}\left\vert \begin{array}{cc} \frac{\sin (\sqrt{\lambda} b)}{\sqrt{\lambda}}&\beta z(a,\lambda) \\ \cos (\sqrt{\lambda}b )& z^{[1]}(a, \lambda) \end{array}\right\vert. \label{eq146461}\tag{22}\]
The following lemma shows that under the conditions that \(\rho(b)\) is known and \(g(a)=0\), the Liouville transformation is injective.
Lemma 2. Assume that \(\rho\in W_2^1\left((0, b_1)\cup (b_1, b)\right)\) and satisfies 4 , \(\sigma(x)\), \(d\), \(d_1\) are defined by 11 and 12 with \(g(a)=0\). Then \(\rho\) is uniquely determined by \(d\), \(d_1\), \(\rho(b)\) and \(\sigma(x),x\in[0,a]\).
Proof. Denote \(\hat{\rho}(x):=\rho(r)\). By 12 and 13 , \((\hat{\rho}(x))^{1/4}\) and \(g(x)\) satisfy ordinary differential equations \[\begin{align} \tag{23} \frac{d (\hat{\rho}(x))^{1/4}}{dx}&=(\hat{\rho}(x))^{1/4}(\sigma(x)-g(x)),\\ \frac{d g(x)}{dx}&=(\sigma(x)-g(x))^2, \tag{24} \end{align}\] and the following initial conditions \[\begin{align} \nonumber (\hat{\rho}(a))^{1/4}=\rho(b)^{1/4}, g(a)=0. \end{align}\] By the uniqueness theorem for ordinary differential equations, we can uniquely determine \((\hat{\rho}(x))^{1/4}\), \(g(x)\), \(x\in (d,a)\) if \(\rho(b)\) is known. Since \(d\), \(d_1\) are known, by 4 , 11 and 14 , we know that \(b_2\), \((\hat{\rho}(d-))^{1/4}\) and \(g(d-)\) are uniquely determined. Note that \((\hat{\rho}(x))^{1/4}\) and \(g(x)\) also satisfy the ordinary differential equations 23 and 24 on the interval \((0,d)\). Therefore we can uniquely determine \((\hat{\rho}(x))^{1/4}\), \(g(x)\), \(x\in (0,d)\). Since \(x=\int_0^r\sqrt{\rho(s) }ds\), then \(r(x)\) satisfies \[\begin{align} \nonumber \frac{dr}{dx}=\frac{1}{\sqrt{\hat{\rho}(x)}} \end{align}\] and \[r(0)=0.\] By the uniqueness theorem for the differential equation, \(\hat{\rho}(x)\) uniquely determines \(r(x)\) and hence \(\rho(r)\) is uniquely determined. ◻
In this section, we consider the SL equation 9 with discontinuous condition 10 , denote it by \(L(\sigma,d,d_1)\). Here \(\sigma\in L^2(0,a), 0<d<a, d_1\ne 1>0.\)
Let \(s(x,\lambda)\) be the solution of 9 satisfying the initial condition \(s(0,\lambda) =0\), \(s^{[1]}(0,\lambda)=1\) and the discontinuous condition 10 . By 9 , for \(0\le x<d\), \(s\) and \(s^{[1]}\) satisfy the following equations (see also [32]) \[\begin{align} s&(x,\lambda)=\frac{\sin\sqrt{\lambda}x}{\sqrt{\lambda}}- \int_0^x\frac{\sin\sqrt{\lambda} (x-t)}{\sqrt{\lambda}}\sigma(t)s^{[1]}(t,\lambda) dt \nonumber \\ &+\int_0^x{\cos\sqrt{\lambda} (x-t)}\sigma(t)s(t,\lambda) dt -\int_0^x\frac{\sin\sqrt{\lambda} (x-t)}{\sqrt{\lambda}}\sigma(t)^2s(t,\lambda) dt, \label{integrationz} \end{align}\tag{25}\] \[\begin{align} s&^{[1]}(x,\lambda)=\cos \sqrt{\lambda}x- \int_0^x\cos\sqrt{\lambda} (x-t)\sigma(t)s^{[1]}(t,\lambda) dt \nonumber\\ &-\sqrt{\lambda}\int_0^x{\sin\sqrt{\lambda} (x-t)}\sigma(t)s(t,\lambda) dt -\int_0^x{\cos\sqrt{\lambda} (x-t)}\sigma(t)^2s(t,\lambda) dt. \label{integrantionz1aa} \end{align}\tag{26}\] We next show the equations that \(s\) and \(s^{[1]}\) satisfy for \(d<x\le a\). Notice that for \(d<x\le a\), there exist \(A,B\in \mathbb{R}\), so that \[\begin{align} s&(x,\lambda)=A\frac{\sin\sqrt{\lambda}(x-d)}{\sqrt{\lambda}}+B\cos \sqrt{\lambda}(x-d) -\int_d^x\frac{\sin\sqrt{\lambda} (x-t)}{\sqrt{\lambda}}\sigma(t)s^{[1]}(t,\lambda) dt \nonumber \\ &+\int_d^x{\cos\sqrt{\lambda} (x-t)}\sigma(t)s(t,\lambda) dt -\int_d^x\frac{\sin\sqrt{\lambda} (x-t)}{\sqrt{\lambda}}\sigma(t)^2s(t,\lambda) dt, \label{integrationz11} \end{align}\tag{27}\] \[\begin{align} s&^{[1]}(x,\lambda)=A\cos \sqrt{\lambda}(x-d)-B\sqrt{\lambda}{\sin\sqrt{\lambda}(x-d)}- \int_d^x\cos\sqrt{\lambda} (x-t)\sigma(t)s^{[1]}(t,\lambda) dt \nonumber\\ &-\sqrt{\lambda}\int_d^x{\sin\sqrt{\lambda} (x-t)}\sigma(t)s(t,\lambda) dt -\int_d^x{\cos\sqrt{\lambda} (x-t)}\sigma(t)^2s(t,\lambda) dt. \label{integrantionz12} \end{align}\tag{28}\] By 27 28 , one has that \(A=s^{[1]}(d+,\lambda), B=s(d+,\lambda)\). On the other hand, From 25 , 26 and the jump condition 10 , we know that \[\begin{align} s&(d+,\lambda)=d_1\Big(\frac{\sin\sqrt{\lambda}d}{\sqrt{\lambda}}-\int_0^d \frac{\sin\sqrt{\lambda} (d-t)}{\sqrt{\lambda}}\sigma(t)s^{[1]}(t,\lambda) dt \nonumber \\ &+\int_0^d{\cos\sqrt{\lambda} (d-t)}\sigma(t)s(t,\lambda) dt -\int_0^d\frac{\sin\sqrt{\lambda} (d-t)}{\sqrt{\lambda}}\sigma(t)^2s(t,\lambda) dt\Big), \label{integrationz1111} \end{align}\tag{29}\] \[\begin{align} s&^{[1]}(d+,\lambda)=d_1^{-1}\Big(\cos \sqrt{\lambda}d- \int_0^d\cos\sqrt{\lambda} (x-t)\sigma(t)s^{[1]}(t,\lambda) dt \nonumber\\ &-\sqrt{\lambda}\int_0^d{\sin\sqrt{\lambda} (d-t)}\sigma(t)s(t,\lambda) dt -\int_0^d{\cos\sqrt{\lambda} (d-t)}\sigma(t)^2s(t,\lambda)dt \Big). \label{integrantionz111} \end{align}\tag{30}\] Therefore, for \(d<x\le a\), \(s\) and \(s^{[1]}\) satisfy the following equations \[\begin{align} s&(x,\lambda)=\frac{1}{\sqrt{\lambda}}\left(d_1 \sin \sqrt{\lambda} d \cos \sqrt{\lambda} (x-d)+d_1^{-1} \cos \sqrt{\lambda} d \sin \sqrt{\lambda} (x-d)\right)\nonumber\\ &-\frac{1}{\sqrt{\lambda}} \int_{0}^d \left(d_1 \sin \sqrt{\lambda}( d-t) \cos \sqrt{\lambda} (x-d)+d_1^{-1} \cos \sqrt{\lambda} (d-t) \sin \sqrt{\lambda} (x-d)\right)\sigma(t)s^{[1]}(t)dt \nonumber \\ &+\int_{0}^d \left(d_1 \cos \sqrt{\lambda} (d-t) \cos \sqrt{\lambda} (x-d)-d_1^{-1} \sin \sqrt{\lambda}(d-t) \sin \sqrt{\lambda} (x-d) \right)\sigma(t)s(t)dt \nonumber\\ &-\frac{1}{\sqrt{\lambda}}\int_{0}^d \left(d_1 \sin \sqrt{\lambda}( d-t) \cos \sqrt{\lambda} (x-d)+d_1^{-1} \cos \sqrt{\lambda} (d-t) \sin \sqrt{\lambda} (x-d)\right)\sigma^2(t)s(t)dt \nonumber \\ &-\int_d^x\frac{\sin\sqrt{\lambda}(x-t)}{\sqrt{\lambda}}\sigma(t)s^{[1]}(t,\lambda) dt+\int_d^x{\cos\sqrt{\lambda} (x-t)}\sigma(t)z(t,\lambda) dt \nonumber \\ &-\int_d^x\frac{\sin\sqrt{\lambda} (x-t)}{\sqrt{\lambda}}\sigma(t)^2s(t,\lambda) dt, \label{integrationz1} \end{align}\tag{31}\] \[\begin{align} s&^{[1]}(x,\lambda)=\left(-d_1 \sin \sqrt{\lambda} d \sin \sqrt{\lambda} (x-d) +d_1^{-1} \cos \sqrt{\lambda} d \cos \sqrt{\lambda} (x-d) \right)\nonumber\\ &+\int_{0}^d \left(d_1 \sin \sqrt{\lambda}( d-t) \sin \sqrt{\lambda} (x-d)-d_1^{-1} \cos \sqrt{\lambda} (d-t) \cos \sqrt{\lambda} (x-d)\right)\sigma(t)s^{[1]}(t)dt \nonumber \\ &-\sqrt{\lambda}\int_{0}^d \left(d_1 \cos \sqrt{\lambda} (d-t) \sin \sqrt{\lambda} (x-d)+d_1^{-1} \sin \sqrt{\lambda} (d-t) \cos \sqrt{\lambda} (x-d)\right)\sigma(t)s(t)dt \nonumber\\ &+\int_{0}^d \left(d_1 \sin \sqrt{\lambda}( d-t) \sin \sqrt{\lambda} (x-d)-d_1^{-1} \cos \sqrt{\lambda} (d-t) \cos \sqrt{\lambda} (x-d)\right)\sigma^2(t)s(t)dt \nonumber \\ &-\int_d^x{\cos\sqrt{\lambda}(x-t)}\sigma(t)s^{[1]}(t,\lambda) dt -\sqrt{\lambda}\int_d^x{\sin\sqrt{\lambda} (x-t)}\sigma(t)s(t,\lambda) dt \nonumber \\ &-\int_d^x\cos\sqrt{\lambda} (x-t)\sigma(t)^2s(t,\lambda) dt. \label{integrationz2} \end{align}\tag{32}\]
Denote \(Y(x,\lambda)=(z(x,\lambda), z^{[1]}(x,\lambda))^T\). For \(0\le x<d\), 25 and 26 can be written in the matrix form \[\begin{align} \label{11111} Y(x,\lambda)=Y_0(x,\lambda) +\int_0^x A(t,\lambda)Y(t) dt,\;0\le x<d, \end{align}\tag{33}\] where \[\begin{align} Y_0(x,\lambda)=\begin{pmatrix} \frac{\sin\sqrt{\lambda}x}{\sqrt{\lambda}} \\ \cos \sqrt{\lambda}x \end{pmatrix}, \nonumber \end{align}\] \[\begin{align} A(t,\lambda)=\begin{pmatrix} \cos\sqrt{\lambda} (x-t)\sigma(t)-\frac{\sin\sqrt{\lambda} (x-t)}{\sqrt{\lambda}}\sigma(t)^2 & -\frac{\sin\sqrt{\lambda} (x-t)}{\sqrt{\lambda}}\sigma(t) \\ -\sqrt{\lambda}{\sin\sqrt{\lambda} (x-t)}\sigma(t)-{\cos\sqrt{\lambda} (x-t)}\sigma(t)^2&-\cos\sqrt{\lambda} (x-t)\sigma(t) \end{pmatrix}. \nonumber \end{align}\]
Equation 33 can be solved by the method of successive approximations; namely, with \[\begin{align} \label{y0c} Y_n(x,\lambda)\equiv \begin{pmatrix} Y_{1,n}(x,\lambda) \\ Y_{2,n}(x,\lambda) \end{pmatrix} =\int_0^x A(t,\lambda)Y_{n-1}(t,\lambda) dt, \end{align}\tag{34}\] then at least formally, we have \[\begin{align} Y(x,\lambda)=\sum_{n=0}^\infty Y_n(x,\lambda). \label {y} \end{align}\] For \(d<x\le a\), 31 and 32 can be written in the matrix form \[\begin{align} \label{yn11} Y(x,\lambda)=Y_0(x,\lambda) +\int_d^x A(t,\lambda)Y(t) dt,\;d<x\le a \end{align}\tag{35}\] with \[\begin{align} Y_0(x,\lambda)=\begin{pmatrix} Y_{0,1}(x,\lambda) \\ Y_{0,2}(x,\lambda) \end{pmatrix}. \nonumber \end{align}\] Here \(Y_{0,1}(x,\lambda)\) is the sum of the first four terms of 31 , \(Y_{0,2}(x,\lambda)\) is the sum of the first four terms of 32 . By the method of successive approximations, at least formally, \(Y(x,\lambda)\) has the representation [y]. Here \[\begin{align} Y_n(x,\lambda)\equiv \begin{pmatrix} Y_{n,1}(x,\lambda) \\ Y_{n,2}(x,\lambda) \end{pmatrix} \nonumber =\int_d^x A(t,\lambda)Y_{n-1}(t,\lambda) dt. \end{align}\]
Set \[\begin{align} \nonumber \varphi(x,\lambda)= \begin{cases} \frac{\sin\sqrt{\lambda }x}{\sqrt{\lambda }}, &0\le x<d, \\ \frac{1}{\sqrt{\lambda}}\left(d_1 \sin \sqrt{\lambda} d \cos \sqrt{\lambda} (x-d)+d_1^{-1} \cos \sqrt{\lambda} d \sin \sqrt{\lambda} (x-d)\right), &d< x\le a. \end{cases} \end{align}\] and \[\begin{align} \nonumber \phi(x,\lambda)= \begin{cases} \cos \sqrt{\lambda }x, &0\le x<d, \\ -d_1 \sin \sqrt{\lambda} d \sin \sqrt{\lambda} (x-d)+d_1^{-1} \cos \sqrt{\lambda} d \cos \sqrt{\lambda} (x-d),&d< x\le a. \end{cases} \end{align}\] Then we have the following lemma.
Lemma 3. For all \(\lambda\in\mathbb{C}\) and \(x\ne d\), there exist \(K(x, \cdot), N(x, \cdot) \in L^2(0,x)\), so that \[\begin{align} \label{definitionkxt} s(x,\lambda)=\varphi(x,\lambda)+\int_0^{x} K(x,t)\frac{\sin\sqrt{\lambda }t}{\sqrt{\lambda }} dt, \end{align}\tag{36}\] \[\begin{align} \label{definitionnxt} s^{[1]}(x,\lambda)=\phi(x,\lambda)+\int_0^{x} N(x,t)\cos \sqrt{\lambda}t dt. \end{align}\tag{37}\]
Proof. We first show that if \(0\le x<d\), for any \(n\ge1\), \(Y_{1,n}\) and \(Y_{2,n}\) have the following representation (see 34 for definitions of \(Y_{1,n}\) and \(Y_{2,n}\)) \[\begin{align} Y_{n,1}(x,\lambda)&=\int_0^x K_n(x,t)\frac{\sin\sqrt{\lambda}t}{\sqrt{\lambda}} dt, \tag{38}\\ Y_{n,2}(x,\lambda)&=\int_0^x N_n(x,t)\cos\sqrt{\lambda}tdt. \tag{39} \end{align}\]
First we calculate \(Y_{1,1},Y_{1,2}\). By trigonometric addition formulas, using the change of variables and interchanging the order of integration, we know that for \(n=1\), 38 and 39 hold with \[\begin{align} K_1(x,t)=&\frac{1}{2}\sigma\left(\frac{x+t}{2}\right)-\frac{1}{2}\sigma\left(\frac{x-t}{2}\right)-\frac{1}{2}\int_0^t \sigma^2(s)ds \nonumber \\ &+\frac{1}{4}\int_t^x\sigma^2\left(\frac{\tau-t}{2}\right)-\sigma^2\left(\frac{\tau+t}{2}\right)d\tau, \nonumber \end{align}\] \[\begin{align} N_1(x,t)=&-\frac{1}{2}\sigma\left(\frac{x+t}{2}\right)-\frac{1}{2}\sigma\left(\frac{x-t}{2}\right)-\frac{1}{2}\int_0^t \sigma^2(s) ds \nonumber \\ &-\int_t^x \sigma^2(s) ds +\frac{1}{4}\int_t^x\sigma^2\left(\frac{\tau-t}{2}\right)+\sigma^2\left(\frac{\tau+t}{2}\right)d\tau. \nonumber \end{align}\]
Assume that for \(n=j\), 38 and 39 hold. Letting \(n=j+1\) in 34 and substituting the integral representation of \(Y_{1,j},Y_{2,j}\) into 34 , one can see that 38 and 39 hold for \(n=j+1\), where \[\begin{align} K_{j+1}&(x,t)=-\frac{1}{2}\int_{x-t}^x N_j(s,t-x+s)\sigma(s) ds -\frac{1}{2}\int_{\frac{x-t}{2}}^{x-t} N_j(s,x-s-t)\sigma(s) ds \nonumber \\ &+\frac{1}{2}\int_{\frac{x+t}{2}}^x N_j(s,x-s+t)\sigma(s) ds- \frac{1}{2}\int_t^x d\xi\Big(\int_{\xi-t}^\xi K_j(s,t-\xi+s)\sigma^2(s)ds \nonumber \\ &-\int_{\frac{\xi-t}{2}}^{\xi-t} K_j(s,\xi-s-t)\sigma^2(s)ds+\int_{\frac{\xi+t}{2}}^\xi K_j(s,t-s+\xi)\sigma^2(s) ds\Big)\nonumber \\ &+\frac{1}{2}\int_{x-t}^x K_j(s,t-x+s)\sigma(s) ds-\frac{1}{2}\int_{\frac{x-t}{2}}^{x-t} K_j(s,x-s-t)\sigma(s) ds \nonumber \\ &+\frac{1}{2}\int_{\frac{x+t}{2}}^x K_j(s,x-s+t)\sigma(s) ds, \label{kn431} \end{align}\tag{40}\] \[\begin{align} N_{j+1}&(x,t)=-\frac{1}{2}\int_{x-t}^x N_j(s,t-x+s)\sigma(s) ds -\frac{1}{2}\int_{\frac{x-t}{2}}^{x-t} N_j(s,x-s-t)\sigma(s) ds \nonumber \\ &-\frac{1}{2}\int_{\frac{x+t}{2}}^x N_j(s,x-s+t)\sigma(s) ds- \frac{1}{2}\int_t^x d\xi\Big(\int_{\xi-t}^\xi K_j(s,t-\xi+s)\sigma^2(s)ds \nonumber \\ &-\int_{\frac{\xi-t}{2}}^{\xi-t} K_j(s,\xi-s-t)\sigma^2(s)ds-\int_{\frac{\xi+t}{2}}^\xi K_j(s,\xi-s+t)\sigma^2(s) ds\Big)\nonumber \\ &-\int_t^x \sigma(s)^2 K_j(s,t)ds +\frac{1}{2}\int_{x-t}^x K_j(s,t-x+s)\sigma(s) ds \nonumber \\ &-\frac{1}{2}\int_{\frac{x-t}{2}}^{x-t} K_j(s,x-s-t)\sigma(s) ds -\frac{1}{2}\int_{\frac{x+t}{2}}^x K_j(s,x-s+t)\sigma(s) ds. \label{nn431} \end{align}\tag{41}\] By induction, it follows that the series \[\begin{align} K(x,t):=\sum_{n=1}^{\infty} K_n(x,t), N(x,t):=\sum_{n=1}^{\infty} N_n(x,t) \nonumber \end{align}\] converge in \(L^2(0,x)\) and hence \(Y(x,\lambda)\) defined by [y] is indeed a solution of 33 . Moreover, \(K(x,t)\) with respect to \(t\) and the function \(\sigma\) have the same smoothness.
We next show for \(d<x\le a\), 36 and 37 hold. In this case, the computation is much more complicated. We only present the main steps. By the definition of \(Y_0(x,\lambda)\), one obtains that there exist \(K_0(x,\cdot), N_0(x,\cdot)\in L^2(0,x)\), such that \[\begin{align} Y_{0,1}(x,\lambda)&=\frac{1}{\sqrt{\lambda}}\left(d_1 \sin \sqrt{\lambda} d \cos \sqrt{\lambda} (x-d)+d_1^{-1} \cos \sqrt{\lambda} d \sin \sqrt{\lambda} (x-d)\right) \nonumber\\ &+\int_0^x K_0(x,t)\frac{\sin\sqrt{\lambda}t}{\sqrt{\lambda}}dt, \label{y011} \end{align}\tag{42}\] \[\begin{align} Y_{0,2}(x,\lambda)&=\left(-d_1 \sin \sqrt{\lambda} d \sin \sqrt{\lambda} (x-d)+d_1^{-1} \cos \sqrt{\lambda} d \cos \sqrt{\lambda} (x-d)\right) \nonumber \\ &+\int_0^x N_0(x,t)\frac{\sin\sqrt{\lambda}t}{\sqrt{\lambda}}dt. \label{y021} \end{align}\tag{43}\]
We prove that for \(d<x\le a\), 38 and 39 hold. Denote \[\begin{align} \nonumber \sigma_-(x) = \begin{cases} 0, & 0<x<d,\\ \sigma(x), & d<x\le a. \end{cases} \end{align}\] Substituting 42 and 43 into 35 with \(n=1\), we get that 38 and 39 hold for \(n=1\). Assume that for \(n=j\), 38 and 39 hold. Letting \(n=j+1\) in 35 and substituting the integral representation of \(Y_{1,j},Y_{2,j}\) into 34 , we can see that 38 and 39 hold for \(n=j+1\). Here \(K_{j+1}(x,t)\) and \(N_{j+1}(x,t)\) are given by 40 and 41 , respectively, with the function \(\sigma\) replaced by \(\sigma_-\). By induction, we obtain 38 and 39 hold for \(n\ge 1\). Furthermore, the series \[\begin{align} \nonumber K(x,t):=\sum_{n=0}^{\infty} K_n(x,t), N(x,t):=\sum_{n=0}^{\infty} N_n(x,t) \end{align}\] converge in \(L^2(0,x)\) and hence \(Y(x,\lambda)\) defined by [y] is indeed a solution of 35 . The proof is completed. ◻
Define the Weyl-Titchmarsh function of 9 and 10 by \[\begin{align} \nonumber m(x, \lambda)=-\frac{s^{[1]}(x-,\lambda)}{s(x-,\lambda)}. \end{align}\] By 36 and 37 , as \(|\lambda|\rightarrow \infty\) in the sector \(\Lambda_{\delta}:=\{\lambda\in \mathbb{C}| \delta<\arg(\lambda)<\pi-\delta, \delta\in (0,\pi/2)\}\), \(m(x,\lambda)\) has the asymptotic formula \[\begin{align} m(x,\lambda)=i\sqrt{\lambda}(1+o(1)),x\in(0,d)\cup (d,a). \label{aaaaa} \end{align}\tag{44}\] Moreover, \(m(x,\lambda)\) obeys the Riccati equation \[\begin{align} m'(x,\lambda)-m^2(x,\lambda)+2\sigma(x)m(x,\lambda)=\sigma^2(x)+\lambda, x\in (0,d)\cup (d,a), \label{ricatti} \end{align}\tag{45}\] and the jump condition \[\begin{align} \nonumber m(d+,\lambda)=\frac{1}{d_1^{2}}m(d-,\lambda). \end{align}\]
For given \(y\in [0,a]\), let \({s}(x,\lambda;y),{c}(x,\lambda;y)\) be solutions of 9 satisfying the initial conditions \[\begin{align} \label{normalized} s(y,\lambda;y)=c^{[1]}(y,\lambda;y)=0, s^{[1]}(y,\lambda;y)=c(y,\lambda;y)=1 \end{align}\tag{46}\] at \(y\) and the jump condition 10 . Then \[\begin{align} \label{langsiji} W(c(x,\lambda;y),s(x,\lambda;y))\equiv c(x,\lambda;y)s^{[1]}(x,\lambda;y)-c^{[1]}(x,\lambda;y)s(x,\lambda;y)=1. \end{align}\tag{47}\] Obviously, we have \(s(x,\lambda)=s(x,\lambda;0)\) and \[\begin{align} \tag{48} s(x,\lambda)=& s^{[1]}(y,\lambda) s(x,\lambda;y)+s(y,\lambda) c(x,\lambda;y),\\ s^{[1]}(x,\lambda)=& s^{[1]}(y,\lambda) s^{[1]}(x,\lambda;y)+s(y,\lambda) c^{[1]}(x,\lambda;y). \tag{49} \end{align}\] For simplicity, denote \(S(x,\lambda)\equiv s(x,\lambda;a), C(x,\lambda)\equiv c(x,\lambda;a)\).
Define \[\begin{align} \label{definitionpsi} \Psi(x,\lambda)=\frac{s(x,\lambda)}{s(a,\lambda)}. \end{align}\tag{50}\] Then according to 48 ,
\[\begin{align} \label{psi111111} \Psi(x,\lambda)=C(x,\lambda)-{m(a, \lambda)}S(x,\lambda). \end{align}\tag{51}\] By 47 , one has \[\begin{align} \label{cccc111111111} W(\Psi, S)=1. \end{align}\tag{52}\]
By using the method of spectral mappings [22], [32], we obtain the following theorem.
Theorem 2. Assume that \(d_1\ne1\). Then Weyl-Titchmarsh function \(m(a,\lambda)\) uniquely determines \(d, d_1\) and \(\sigma(x), x\in [0,a]\).
Proof. We require that if a certain symbol \(\gamma\) denotes an object related to \(L(\sigma, d,d_1)\), then the corresponding symbol \(\tilde{\gamma}\) denotes the analogous object related to \(L(\tilde{\sigma}, \tilde{d},\tilde{d}_1)\).
Define the matrix \(P(x,\lambda)=[P_{ij}(x,\lambda)]_{j,k=1,2}\) by the formula \[\begin{align} \label{definitionp} P(x,\lambda)\begin{pmatrix} \tilde{S}(x,\lambda)& \tilde{\Psi}(x,\lambda) \\ \tilde{S}^{[1]}(x,\lambda)&\tilde{\Psi}^{[1]}(x,\lambda) \end{pmatrix} =\begin{pmatrix} {S}(x,\lambda)& {\Psi}(x,\lambda) \\ {S}^{[1]}(x,\lambda)&{\Psi}^{[1]}(x,\lambda) \end{pmatrix}. \end{align}\tag{53}\] Then from 52 , \[\begin{align} \label{matrixp} \begin{pmatrix} {P_{11}}(x,\lambda)& {P_{12}}(x,\lambda) \\ {P_{21}}(x,\lambda)&P_{22}(x,\lambda) \end{pmatrix} =\begin{pmatrix} -{S}\tilde{\Psi}^{[1]}+\tilde{S}^{[1]}{\Psi}&-\tilde{S}{\Psi}+{S}\tilde{\Psi} \\ -{S}^{[1]}\tilde{\Psi}^{[1]}+\tilde{S}^{[1]}{\Psi}^{[1]}&-\tilde{S}{\Psi}^{[1]}+{S}^{[1]}\tilde{\Psi} \end{pmatrix}. \end{align}\tag{54}\] By 50 and Lemma 3, as \(|\lambda|\rightarrow \infty\) in the sector \(\Lambda_{\delta}\), one obtains \[\begin{align} \nonumber |P_{11}(x,\lambda)|\le C, |P_{12}(x,\lambda)|=o(1). \end{align}\] If \(m(a, \lambda)=\tilde{m}(a, \lambda)\), by 51 and 54 , \(P_{11}(x,\lambda), P_{12}(x,\lambda)\) are entire functions with respect to \(\lambda\). Using Phragmén-Lindelöf theorem [35] and Liouville theorem, one has that \(P_{11}(x,\lambda)=A(x), P_{12}(x,\lambda)=0\). Therefore, by 53 , \[\begin{align} \nonumber S(x,\lambda)=A(x)\tilde{S}(x,\lambda),\Psi(x,\lambda)=A(x)\tilde{\Psi}(x,\lambda). \end{align}\] Since \(W(\Psi, S)=W(\tilde{\Psi}, \tilde{S})=1\), we know that \(A(x)^2=1\). From the asymptotic behavior of \(S\) and \(\tilde{S}\), we can get that \(A(x)= 1\). Therefore, \(S(x,\lambda)=\tilde{S}(x,\lambda)\), \(\Psi(x,\lambda)=\tilde{\Psi}(x,\lambda)\). From the fact that \(S(0,\lambda)=-s(a,\lambda)\) and 50 , for any \(x\in [0,d)\cup (d,a]\), we have \[\label{sxlambdaequal} s(x,\lambda)=\tilde{s}(x,\lambda).\tag{55}\]
From 55 , one obtains that \(d=\tilde{d},d_1=\tilde{d}_1\). By equations \[\begin{align} -(s'-\sigma s)'-\sigma(s'-\sigma s)-\sigma^2s&=\lambda s, x\in (0,d)\cup (d,a), \nonumber \\ -(s'-\tilde{\sigma} s)'-\tilde{\sigma}(s'-\tilde{\sigma} s)-\tilde{\sigma}^2s&=\lambda s, \nonumber x\in (0,d)\cup (d,a), \end{align}\] one knows that for \(x\in [0,d)\cup (d,a]\), \(((\sigma-\tilde{\sigma})s)'=(\sigma-\tilde{\sigma})s'\). In particular, the function \((\sigma-\tilde{\sigma})s\) is absolutely continuous on the interval \([0, d)\cup(d,a]\). Choosing \(\lambda_0\in \mathbb{C}\) so that for any \(x\in[0,a]\), \[\begin{align} \label{sxlambda} s(x, \lambda_0)\ne 0. \end{align}\tag{56}\] Then on the interval \([0, d)\cup (d,a]\), the function \((\sigma-\tilde{\sigma})\) is absolutely continuous and \((\sigma-\tilde{\sigma})'=0\) almost everywhere. Therefore, there exist \(C_1, C_2\in \mathbb{R}\), such that \[\begin{align} \nonumber \sigma-\tilde{\sigma}= \begin{cases} C_1, & 0\le x<d, \\ C_2, & d<x\le a. \end{cases} \end{align}\] According to \(m(a, \lambda)=\tilde{m}(a, \lambda)\) and 55 , \(s^{[1]}(a,\lambda)=\tilde{s}^{[1]}(a,\lambda)\). By definitions of \(s^{[1]}\) and \(\tilde{s}^{[1]}\), we know that \[(\sigma(a)-\tilde{\sigma}(a))s(a,\lambda)=0.\] Using 56 , one has \(C_2=0\). Then one obtains that \(\sigma(x)= \tilde{\sigma}(x)\) on \((d,a]\). Hence for any \(x\in (d,a]\), one can see that \(s^{[1]}(x, \lambda)=\tilde{s}^{[1]}(x, \lambda)\). Because \(d_1=\tilde{d}_1\), we have \(s^{[1]}(d-, \lambda)=\tilde{s}^{[1]}(d-, \lambda)\). From 56 and the definition of \(s^{[1]}\), one has \[\lim_{ x\rightarrow d-} \sigma(x)-\tilde{\sigma}(x)=0.\] Then \(C_1=0\). Hence, we know that \(\sigma(x)=\tilde{\sigma}(x)\) almost everywhere on \([0,a]\). ◻
Remark 3. Consider the equation 9 with jump condition 18 , denote it by \(L(\sigma, d, d_1,d_2)\). Let \(\sigma_1(x)\equiv 0,\) \[\begin{align} \nonumber \sigma_2(x) = \begin{cases} -2, & 0\le x<d,\\ 0, & d<x\le a. \end{cases} \end{align}\] Then for \(d<x\le a\), \(L(\sigma_1, d,2,1)\) and \(L(\sigma_2, d,2,0)\) have the same Weyl-Titchmarsh function \[\begin{align} \nonumber m(x,\lambda)=-\frac{A(\lambda)\cos\sqrt{\lambda}(x-d)-B(\lambda)\sqrt{\lambda}\sin\sqrt{\lambda}(x-d)}{A(\lambda)\sin\sqrt{\lambda}(x-d)/\sqrt{\lambda}+B(\lambda)\cos\sqrt{\lambda}(x-d)}. \end{align}\] Here \[\begin{align} A(\lambda)=\frac{\cos2\sqrt{\lambda}d}{2}+\frac{\sin\sqrt{\lambda}d}{\sqrt{\lambda}}, B(\lambda)=\frac{2\sin\sqrt{\lambda}d}{\sqrt{\lambda}}. \nonumber \end{align}\] In order to ensure the uniqueness of the inverse spectral problem, we transform \(Q(\rho)\) into the equation 9 with the jump condition 10 .
\(m(a,\lambda)\) has the following high-energy asymptotic behavior.
Lemma 4. Assume that \(\sigma\in L^2(0,a)\) and \(\sigma\) is \(C^n\) near \(a\) for some \(n\in\mathbb{N}\), then as \(|\lambda|\rightarrow \infty\) in the sector \(\Lambda_{\delta}:=\{\lambda\in \mathbb{C}| \delta<\arg(\lambda)<\pi-\delta, \delta\in (0,\pi/2)\}\), \(m(a,\lambda)\) has an asymptotic formula \[\begin{align} m(a,\lambda)=i\sqrt{\lambda}+i\sum_{l=0}^{n} c_l(a)\frac{1}{\lambda^{l/2}}+o\left(\frac{1}{\lambda^{n/2}}\right). \label{asym} \end{align}\tag{57}\] The expansion coefficients \(c_l(a)\) can be recursively computed from \[\begin{align} c_{0}(a)&=-i\sigma(a), c_1(a)=-\frac{1}{2}\sigma'(a), \nonumber\\ c_{l+1}(a)&=-\frac{i}{2}c_{l}'(a)-\frac{1}{2} \sum_{j=1 }^{l-1}c_j(a)c_{l-j}(a), l\ge 1. \label{recu} \end{align}\tag{58}\]
Proof. Assume that \(\sigma\) is \(C^n\) on the interval \([y,a]\). We first compute the high-energy asymptotic form of \(-{s^{[1]}(a,\lambda;y)}/{s(a,\lambda;y)}\), where \(s(a,\lambda;y)\) is normalized according to 46 . By Lemma 3, \(s(x,\lambda;y)\) and \(s^{[1]}(x,\lambda;y)\) have the following representation \[\begin{align} s(x,\lambda;y)=\frac{\sin(\sqrt{\lambda }(x-y))}{\sqrt{\lambda }}+\int_y^{x} K(x,t;y)\frac{\sin(\sqrt{\lambda }(t-y))}{\sqrt{\lambda }} dt, \label{1} \end{align}\tag{59}\] \[\begin{align} s^{[1]}(x,\lambda;y)=\cos(\sqrt{\lambda }(x-y))+\int_y^{x} N(x,t;y)\cos (\sqrt{\lambda}(t-y) )dt. \label{2} \end{align}\tag{60}\] Recall that [36], [37] if \(f\) is continuous on \([y,a]\), then as \(|\lambda|\rightarrow \infty\) in the sector \(\Lambda_{\delta}\), \[\begin{align} \int_y^a f(t)e^{-i\sqrt{\lambda}(t-y)} dt={e^{-i\sqrt{\lambda}(a-y)}}\left(-f(a)\frac{1}{i\sqrt{\lambda}} +o(\frac{1}{\sqrt{\lambda}})\right). \label{reimann1} \end{align}\tag{61}\]
Since \(\sigma\) is \(C^n\) on \([y,a]\), then kernel functions \(K(a,t;y)\) and \(N(a,t;y)\) are also \(C^n\) with respect to \(t\) on \([y,a]\). Letting \(x=a\) in 59 and 60 , integration by parts \(n\) times, using 61 and the estimates \[\begin{align} \nonumber \cos(\sqrt{\lambda}(a-y))= \frac{e^{-i\sqrt{\lambda}(a-y)}}{2}\left(1+O(e^{2i\sqrt{\lambda}(a-y)})\right), \end{align}\] \[\begin{align} \nonumber \sin(\sqrt{\lambda}(a-y))=- \frac{e^{-i\sqrt{\lambda}(a-y)}}{2i}\left(1+O(e^{2i\sqrt{\lambda}(a-y)})\right), \end{align}\] then there exist \(m_l(a), \tau_l(a), l=0, \cdots, n\), so that \[\begin{align} s(a, \lambda;y)=\frac{e^{-i\sqrt{\lambda}(a-y)}}{2\sqrt{\lambda}} \left(i+ \sum_{l=0}^{n}\frac{m_l(a)}{\lambda^{\frac{l+1}{2}}} +o(\lambda^{-\frac{n+1}{2}})\right), \label{zay} \end{align}\tag{62}\] \[\begin{align} s^{[1]}(a, \lambda;y)=\frac{e^{-i\sqrt{\lambda}(a-y)}}{2\sqrt{\lambda}} \left(\sqrt{\lambda}+\sum_{l=0}^{n}\frac{\tau_l(a)}{\lambda^{\frac{l}{2}}} +o(\lambda^{-\frac{n}{2}})\right). \label{z1ay} \end{align}\tag{63}\] From 62 and 63 , one knows that \[\begin{align} -\frac{s^{[1]}(a, \lambda;y)}{s(a, \lambda;y)}&=- \left(\sqrt{\lambda}+\sum_{l=0}^{n}\frac{\tau_l(a)}{\lambda^{\frac{l}{2}}} +o(\lambda^{-\frac{n}{2}})\right)\times \left(i+ \sum_{j=1}^{n+1}\frac{m_{j-1}(a)}{\lambda^{\frac{j}{2}}} +o(\lambda^{-\frac{n+1}{2}})\right)^{-1} \nonumber \\ &=i\sqrt{\lambda}+i\sum_{l=0}^{n} \hat{c}_l(a)\frac{1}{\lambda^{l/2}}+o\left(\frac{1}{\lambda^{n/2}}\right). \label{high1} \end{align}\tag{64}\] Substituting 64 into the Riccati equation 45 , the coefficients \(\hat{c}_l(a), l=0,\cdots,n\), obey the recursion relation 58 .
On the other hand, by 47 , 48 and 49 , one has \[\begin{align} \nonumber \frac{s^{[1]}(a,\lambda;y)}{s(a,\lambda;y)}+m(a,\lambda)=\frac{s(y,\lambda)}{s(a,\lambda ) s(a,\lambda;y ) }. \end{align}\] From the integral representation of \(s\), as \(|\lambda|\rightarrow\infty\) in the sector \(\Lambda_{\delta}\), \[\begin{align} \label{exponential} \frac{s(y,\lambda)}{s(a,\lambda ) s(a,\lambda;y ) }=O(e^{-2(a-y)|\rm{Im} \sqrt{\lambda}|}). \end{align}\tag{65}\] Using 65 , we know that \(m(a,\lambda)\) has a high-energy asymptotic expansion 57 and its coefficients satisfy \({c}_l(a)=\hat{c}_l(a), l=0,\cdots,n\). Since \(\hat{c}_l(a), l=0,\cdots,n,\) satisfy the recursion relation 58 , then \({c}_l(a), l=0,\cdots,n,\) also satisfy the recursion relation 58 . The proof is completed. ◻
In this section, we consider the inverse transmission problem knowing all eigenvalues. Let \(u(r,\lambda)\) be the solution of \(-u''=\lambda\rho u\) satisfying initial conditions \(u(0,\lambda)=0, u'(0,\lambda)=1.\) Then we have the following lemma.
Lemma 5. Assume that \(\rho\in W_2^1\left((0, b_1)\cup (b_1, b)\right)\) and satisfies 4 . Then as \(|\lambda|\rightarrow \infty\) in the sector \(\Lambda_{\delta}:=\{\lambda\in \mathbb{C}| \delta<\arg(\lambda)<\pi-\delta, \delta\in (0,\pi/2)\}\), \[\begin{align} u(r,\lambda)=\frac{1}{\left(\rho(0)\rho(r)\right)^{1/4}\sqrt{\lambda}} \left(\sin(\sqrt{\lambda} x(r))+o(e^{{\rm{Im}}\sqrt{\lambda}x(r) })\right) , \label{urlambda} \end{align}\tag{66}\] \[\begin{align} u'(r,\lambda)=\left(\frac{\rho(r)}{\rho(0)}\right) ^{1/4}\left(\cos(\sqrt{\lambda} x(r))+o(e^{{\rm{Im}}\sqrt{\lambda}x(r) })\right) . \label{asmptoticurlambda1} \end{align}\tag{67}\]
Proof. According to 8 and 15 , we know \[\begin{align} \label{linearproperty} s(x,\lambda)=z(x,\lambda)(\rho(0))^{1/4}. \end{align}\tag{68}\] In light of 8 , 15 and 68 , an application of Lemma 3 and Riemann-Lebesgue lemma yields 66 and 67 . ◻
It is known that \(\rho\) is uniquely determined by the knowledge of two sets of spectra [38]–[40]. If \(\rho\) satisfies 4 , we provide a new proof of two-spectra theorem.
Lemma 6. Assume that \(\rho\in W_2^1\left((0, b_1)\cup (b_1, b)\right)\) and satisfies 4 . Then all zeros of \(u(b,\lambda)\) and \(u'(b,\lambda)\) uniquely determine \(\rho(r)\) on \([0,b]\).
Proof. First note that by (11.7) in [41], the constant \(a\) is uniquely determined by all zeros of \(u(b,\lambda)\). According to 8 and the requirement \(g(a)=0\), we get \[\begin{align} m(a,\lambda)=-\frac{u'(b,\lambda)}{\beta u(b,\lambda)}. \label{aaaxx} \end{align}\tag{69}\] Here \(\beta\) is defined by 20 . By \[\begin{align} \label{ub0} u(b,0)=b,\;u'(b,0)=1, \end{align}\tag{70}\] we know that all zeros of \(u(b,\lambda)\) and \(u'(b,\lambda)\) uniquely determine \(u(b,\lambda)\) and \(u'(b,\lambda)\), respectively. From 44 and 69 , we have that \(\rho(b)\) and hence \(m(a,\lambda)\) are uniquely determined by all zeros of \(u(b,\lambda)\) and \(u'(b,\lambda)\). Using Theorem 2 and Lemma 2, all zeros of \(u(b,\lambda)\) and \(u'(b,\lambda)\) uniquely determine \(\rho(r)\) on \([0,b]\). ◻
Lemma 7. Assume that \(\rho\in W_2^1\left((0, b_1)\cup (b_1, b)\right)\) and satisfies 4 .
(i) If \(\rho(b)\ne1\), then there exists \(A_0>0\), so that in the sector \(\Lambda_{\delta}:=\{\lambda\in \mathbb{C}| \delta<\arg(\lambda)<\pi-\delta,
\delta\in (0,\pi/2)\}\), \(D(\lambda)\) has the following estimate \[\begin{align}
|D(\lambda)|\ge A_0\frac{e^{|{\rm{Im} }\sqrt{\lambda}|(a+b)}}{|\sqrt{\lambda}|}, |\lambda|\rightarrow\infty. \nonumber
\end{align}\] (ii) Assume that \(\rho(b)=1\) and there exist \(m\ge 1, \varepsilon>0\), so that \(\rho\in C^{(m)}(b-\varepsilon,b]\), for \(k=1,\cdots, m-1\), \(\rho^{(k)}(b)=0\) and \(\rho^{(m)}(b)\ne0\). Then there exists \(A_0>0\), so that in the sector \(\Lambda_{\delta}\), \[\begin{align} \label{D}
|D(\lambda)|\ge A_0\frac{e^{|{\rm{Im} }\sqrt{\lambda}|(a+b)}}{|\sqrt{\lambda}|^{m+1}}, |\lambda|\rightarrow\infty.
\end{align}\tag{71}\]
Proof. We first prove that (ii) holds. According to 12 , \(\sigma \in C^{(m-1)}(a-\varepsilon,a]\) and for \(=1,\cdots, m-2\), \(\sigma^{(l)}(a)=0\). From 22 , we know that \[\begin{align} D(\lambda)=-\frac{\sin \sqrt{\lambda}b}{\sqrt{\lambda}}z(a,\lambda)\left(\frac{\sqrt{\lambda}\cos \sqrt{\lambda}b}{\sin \sqrt{\lambda}b}+m(a,\lambda)\right). \label{dlambda} \end{align}\tag{72}\] Notice that in the sector \(\Lambda_{\delta}\), for any \(p\in \mathbb{N}\), one has \[\begin{align} \nonumber -\frac{\sqrt{\lambda}\cos \sqrt{\lambda}b}{\sin \sqrt{\lambda}b}=i\sqrt{\lambda}+o(\frac{1}{\lambda^{p/2}}), |\lambda|\rightarrow\infty. \end{align}\] From the high-energy asymptotics of \(m(a,\lambda)\), we know that there exists \(A_0>0\), so that in the sector \(\Lambda_{\delta}\), \[\begin{align} \nonumber \left|\frac{\sqrt{\lambda}\cos \sqrt{\lambda}b}{\sin \sqrt{\lambda}b}+m(a,\lambda)\right|\ge A_0 \frac{1}{|\sqrt{\lambda}|^{m-1}}, |\lambda|\rightarrow\infty. \end{align}\] Therefore by 72 , one can obtain 71 .
We next show that (i) holds. From 22 , one knows \[\begin{align} \frac{D(\lambda)}{\rho(b)^{1/4}}&=(1-\beta)\frac{\sin \sqrt{\lambda}b}{\sqrt{\lambda}}z^{[1]}(a,\lambda)+ \beta\frac{\sin \sqrt{\lambda}b}{\sqrt{\lambda}}z(a,\lambda)\left(-\frac{\sqrt{\lambda}\cos \sqrt{\lambda}b}{\sin \sqrt{\lambda}b}-m(a,\lambda)\right) \nonumber\\ &\equiv D_1+D_2. \nonumber \end{align}\] By the asymptotic form of \(z^{[1]}(a,\lambda)\), one obtains there exists \(A_0>0\), so that in the sector \(\Lambda_{\delta}\), \[\begin{align} |D_1(\lambda)|\ge A_0\frac{e^{|{\rm{Im}} \sqrt{\lambda}|(a+b)}}{|\sqrt{\lambda}|}, |\lambda|\rightarrow\infty. \label{d1} \end{align}\tag{73}\] By 44 , in the sector \(\Lambda_{\delta}\), \(D_2(\lambda)\) has the following estimate \[\begin{align} D_2(\lambda)= O{(|\lambda|^{-1})e^{|{\rm{Im} }\sqrt{\lambda}|(a+b)}}o(|\lambda|^{1/2})=o{(|\lambda|^{-1/2})e^{|{\rm{Im}} \sqrt{\lambda}|(a+b)}}, |\lambda|\rightarrow\infty. \label{d2} \end{align}\tag{74}\] According to 73 and 74 , we can obtain (i). ◻
When \(a<b\), we prove the following uniqueness theorem.
Theorem 4. Assume that \(\rho\in W_2^1\left((0, b_1)\cup (b_1, b)\right)\) and satisfies 4 and \(a<b\). Then all special transmission eigenvalues uniquely determine \(\rho\).
Proof. We require that if a certain symbol \(\gamma\) denotes an object related to \(Q( \rho)\), then the corresponding symbol \(\tilde{\gamma}\) denotes the analogous object related to \(Q(\tilde{\rho})\).
From 6 , we know \[\begin{align} \nonumber \frac{1}{\gamma}D\left(\frac{k^2\pi^2}{b^2}\right)=\frac{1}{\tilde{\gamma}}\tilde{D}\left(\frac{k^2\pi^2}{b^2}\right), k\in \mathbb{N}. \end{align}\] Then by 5 , \[\begin{align} \frac{1}{\gamma}u\left(b, \frac{k^2\pi^2}{b^2}\right)=\frac{1}{\tilde{\gamma}}\tilde{u}\left(b, \frac{k^2\pi^2}{b^2}\right), k\in \mathbb{N}. \label{u} \end{align}\tag{75}\] Define \[\begin{align} \nonumber f_1(\lambda)=\frac{1}{\gamma}u(b, \lambda)-\frac{1}{\tilde{\gamma}}\tilde{u}(b, \lambda). \end{align}\] From 75 , \(\frac{\sqrt{\lambda}f_1(\lambda)}{\sin \sqrt{\lambda}b}\) is an entire function. Since \(a<b\), using 66 , we obtain that in the sector \(\Lambda_{\delta}\), \[\begin{align} \nonumber \lim_{|\lambda|\rightarrow \infty}\frac{\sqrt{\lambda}f_1(\lambda)}{\sin \sqrt{\lambda}b}=0. \end{align}\] According to Phragmén-Lindelöf theorem and Liouville theorem, one has \(\frac{\sqrt{\lambda}f_1(\lambda)}{\sin \sqrt{\lambda}b}\equiv 0\). Then \(f_1(\lambda)\equiv 0,\) and hence \[\begin{align} \label{ublambda1111} \frac{1}{\gamma}u(b, \lambda)=\frac{1}{\tilde{\gamma}}\tilde{u}(b, \lambda). \end{align}\tag{76}\]
By a similar argument, from \[\begin{align} \nonumber \frac{1}{\gamma}{D}\left(\frac{(2k-1)^2\pi^2}{4b^2}\right)=\frac{1}{\tilde{\gamma}}\tilde{D}\left(\frac{(2k-1)^2\pi^2}{4b^2}\right), k\in \mathbb{N}, \end{align}\] one has \[\begin{align} \frac{1}{\gamma}u'\left(b, \frac{(2k-1)^2\pi^2}{4b^2}\right)=\frac{1}{\tilde{\gamma}}\tilde{u}'\left(b, \frac{(2k-1)^2\pi^2}{4b^2}\right), k\in \mathbb{N}. \label{uprime} \end{align}\tag{77}\] Define \[\begin{align} \nonumber f_2(\lambda)=\frac{1}{\gamma}u'(b, \lambda)-\frac{1}{\tilde{\gamma}}\tilde{u}'(b, \lambda). \end{align}\] From 77 we know that \(\frac{f_2(\lambda)}{\cos \sqrt{\lambda}b}\) is an entire function. Since \(a<b\), according to 67 , in the sector \(\Lambda_{\delta}\), we have \[\begin{align} \nonumber \lim_{|\lambda|\rightarrow \infty}\frac{f_2(\lambda)}{\cos \sqrt{\lambda}b}=0. \end{align}\] By using Phragmén-Lindelöf theorem and Liouville theorem, one obtains \(\frac{f_2(\lambda)}{\cos \sqrt{\lambda}b}\equiv 0\). Then \(f_2(\lambda)\equiv 0\), and hence \[\begin{align} \label{ublambda11111} \frac{1}{\gamma}u'(b, \lambda)=\frac{1}{\tilde{\gamma}}\tilde{u}'(b, \lambda). \end{align}\tag{78}\] By 76 , 78 and Lemma 6, we know that \(\rho\equiv\tilde{\rho}\). The proof is complete. ◻
When \(a=b\), we need more information to uniquely determine \(\rho\).
Theorem 5. Assume that \(\rho\in W_2^1\left((0, b_1)\cup (b_1, b)\right)\) and satisfies 4 and \(a=b\). Assume that one of the following conditions holds:
(i) the constant \(\gamma\) in 6 is known;
(ii) \(\rho(b)\ne1\) is known;
(iii) \(\rho(b)=1\) and \(\rho\in C^{(m)}(b-\varepsilon,b]\) for some \(\varepsilon>0\) and some \(m\in \mathbb{N}\),
for \(k=1,\cdots, m-1\), \(\rho^{(k)}(b)=0\) and \(\rho^{(m)}(b)\ne0\) is known.
Then all special transmission eigenvalues uniquely determine \(\rho\).
Proof. We first prove (i). Since \(a=b\), arguing as in Theorem 4, one can obtain that in the sector \(\Lambda_{\delta}\), \({\sqrt{\lambda}f_1(\lambda)}/({\sin \sqrt{\lambda}b}), {f_2(\lambda)}/({\cos \sqrt{\lambda}b})\) are bounded. By Phragmén-Lindelöf theorem and Liouville theorem, then there exist \(C_1, C_2\in \mathbb{R}\), so that \[\begin{align} \frac{1}{\gamma}u(b, \lambda)-\frac{1}{\tilde{\gamma}}\tilde{u}(b, \lambda)=C_1 \frac{\sin\sqrt{\lambda}b}{\sqrt{\lambda}}, \nonumber \\ \frac{1}{\gamma}u'(b, \lambda)-\frac{1}{\tilde{\gamma}}\tilde{u}'(b, \lambda)=C_2 \cos\sqrt{\lambda} b.\nonumber \end{align}\] Letting \(\lambda=0\), by 70 , we know \(C_1=C_2=\frac{1}{\gamma} -\frac{1}{\tilde{\gamma}}.\) Since \(\gamma=\tilde{\gamma}\), then \(C_1=C_2=0\). According to Lemma 6, \(\rho\equiv\tilde{\rho}\). (i) is proved.
We next show (ii). Define \[\begin{align} H(\lambda)=\beta (z(a,\lambda)\tilde{z}^{[1]}(a,\lambda)-\tilde{z}(a,\lambda)z^{[1]}(a,\lambda)), F(\lambda)=\frac{H(\lambda)}{D(\lambda)}.\label{definitionhz} \end{align}\tag{79}\] Since \(\rho(b)=\tilde{\rho}(b)\), by 22 , we know \[\begin{align} \label{1a} H(\lambda)=\frac{1}{\rho(b)^{1/4}\cos{\sqrt{\lambda}b}}(\tilde{D}(\lambda)z^{[1]}(a,\lambda)-D(\lambda)\tilde{z}^{[1]}(a,\lambda)). \end{align}\tag{80}\]
Assume that \(\mu_m\) is a zero of \(D(\lambda), \tilde{D}(\lambda)\) of multiplicity \(k\) satisfying \({\cos{b\sqrt{\mu_m}}}\ne0\). From 80 , \(\mu_m\) is a zero of \(H(\lambda)\) of multiplicity \(k\). In this case, \(F(\lambda)\) is an entire function.
Assume that \(\mu_m\) is a zero of \(D(\lambda), \tilde{D}(\lambda)\) of multiplicity \(k\) satisfying \({\cos{b\sqrt{\mu_m}}}=0\). By 22 and the fact that \[\begin{align} \nonumber D(\mu_m)=\tilde{D}(\mu_m)=0, \end{align}\] one has \[\begin{align} z^{[1]}(a,\mu_m)=\tilde{z}^{[1]}(a,\mu_m)=0. \nonumber \end{align}\] Then \(\mu_m\) is also a zero of \(H(\lambda)\) of multiplicity \(k\). Therefore in this case, we conclude that \(F(\lambda)\) is also an entire function.
From 79 , one can obtain \[\begin{align} \label{ko} H(\lambda)=\beta z(a,\lambda)\tilde{z}(a,\lambda)(\tilde{m}(a,\lambda)-m(a,\lambda)). \end{align}\tag{81}\] Using 44 , as \(|\lambda|\rightarrow\infty\) in the sector \(\Lambda_{\delta}\), we have \[\begin{align} \label{ko1} \tilde{m}(a,\lambda)-m(a,\lambda)=o(|\lambda|^{1/2}). \end{align}\tag{82}\] By Lemma 3, 81 and 82 , as \(|\lambda|\rightarrow\infty\) in the sector \(\Lambda_{\delta}\), one gets \[\begin{align} \nonumber H(\lambda)=O(|\lambda|^{-1}e^{2a|\rm{Im }\sqrt{\lambda}|})o(|\lambda|^{1/2})=o(|\lambda|^{-1/2}e^{2a|\rm{Im } \sqrt{\lambda}|}). \end{align}\] According to Lemma 7, in the sector \(\Lambda_{\delta}\), \[\begin{align} \nonumber F(\lambda)=o(|\lambda|^{-1/2}e^{2a|\rm{Im} \sqrt{\lambda}|})O(|\lambda|^{1/2}e^{-2a|\rm{Im} \sqrt{\lambda}|})=o(1), |\lambda |\rightarrow \infty. \end{align}\] By Phragmén-Lindelöf theorem and Liouville theorem, we have \(F(\lambda)\equiv0\). Therefore, we have \(m(a,\lambda)\equiv \tilde{m}(a,\lambda)\). According to Theorem 2 and Lemma 2, one knows that \(\rho\equiv \tilde{\rho}.\) (ii) is proved.
Finally, we prove (iii). Let \(\rho(b)=1\) and there exist \(m\ge 1, \varepsilon>0\), so that \(\rho\in C^{(m)}(b-\varepsilon,b]\), for \(k=1,\cdots, m-1\), \(\rho^{(k)}(b)=\tilde{\rho}^{(k)}(b)=0\) and \(\rho^{(m)}(b)=\tilde{\rho}^{(m)}(b)\ne0\) are known. From 12 , one can see that \(\sigma, \tilde{\sigma}\in C^{(m-1)}(a-\varepsilon,a]\) and \(\sigma^{(l)}(a)=\tilde{\sigma}^{(l)}(a), l=1,\cdots, m-1\). From the high-energy asymptotic form 57 of the Weyl-Titchmarsh function, one obtains that as \(|\lambda|\rightarrow\infty\) in the sector \(\Lambda_{\delta}\), \[\begin{align} \label{ko2} \tilde{m}(a,\lambda)-m(a,\lambda)=o(|\lambda|^{-m/2+1/2}). \end{align}\tag{83}\] By Lemma 3, 81 and 83 , as \(|\lambda|\rightarrow\infty\) in the sector \(\Lambda_{\delta}\), \[\begin{align} \nonumber H(\lambda)=O(|\lambda|^{-1}e^{2a|\rm{Im} \sqrt{\lambda}|})o(|\lambda|^{-m/2+1/2})=o(|\lambda|^{-m/2-1/2}e^{2a|\rm{Im } \sqrt{\lambda}|}). \end{align}\] Using Lemma 7, in the sector \(\Lambda_{\delta}\), \[\begin{align} \nonumber F(\lambda)=o(|\lambda|^{-m/2-1/2}e^{2a|\rm{Im }\sqrt{\lambda}|})O(|\lambda|^{m/2+1/2}e^{-2a|\rm{Im} \sqrt{\lambda}|})=o(1), |\lambda |\rightarrow \infty. \end{align}\] According to the Phragmén-Lindelöf theorem and Liouville theorem, we have \(F(\lambda)\equiv0\). Therefore \(m(a,\lambda)\equiv \tilde{m}(a,\lambda)\). By Theorem 2 and Lemma 2, one concludes \(\rho\equiv \tilde{\rho}.\) (iii) is proved. ◻
In this section, we study properties of “almost real subspectrum” \(\{\mu_m\}_{m=1}^\infty\) [11], [12] and recover the refractive index from the “almost real subspectrum” \(\{\mu_m\}_{m=1}^\infty\) and partial information on the refractive index. We always assume that \(\rho(b)=1\) in this section.
We need the following lemma.
Lemma 8. Assume that complex numbers \(a_{mn}(m,n\ge1)\) satisfy \[\begin{align} |a_{mn}|=O\left(\frac{m\beta_m}{m^2-n^2}\right), m\ne n, \label{amn} \end{align}\tag{84}\] where \(\{\beta_m\}_{m=1}^\infty\in \ell^2\). Then there exists \(\{\gamma_n\}_{n=1}^\infty\in \ell^2\), such that \[\begin{align} \prod_{m\ge1, m\ne n}(1+a_{mn}) =1+O(\gamma_n)=1+o(1), n\ge1. \label{l2} \end{align}\tag{85}\] In particular, if \(\beta_m=O(1/m)\), then \(\gamma_n=O(\log n/n)\).
Proof. If \(\beta_m=O(1/m)\), \(\gamma_n=O(\log n/n)\) comes from Lemma E.1 in [42].
We first prove that if \(\{\beta_m\}_{m=1}^\infty\in \ell^2\), then \[\begin{align} \label{fuliye} \begin{Bmatrix} \sum_{m\ge1, m\ne n} \frac{\beta_m}{|m-n|} \end{Bmatrix}_{n=1}^\infty \in \ell^2. \end{align}\tag{86}\] To this end, it suffices to show that the infinite matrix \[\begin{align} A= \begin{pmatrix} 0 & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} &\cdots \\ 1 & 0 & 1 & \frac{1}{2} & \frac{1}{3} &\cdots \\ \frac{1}{2} & 1 & 0 & 1 & \frac{1}{2} & \cdots \\ \frac{1}{3} &\frac{1}{2} &1& 0 & 1& \cdots \\ \frac{1}{4} & \frac{1}{3} & \frac{1}{2} & 1 & 0&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots \end{pmatrix}\nonumber \end{align}\] is a bounded linear operator from \(\ell^2\) to \(\ell^2\). Let \({e}_m\) be the sequence in \(\ell^2\) which has all its terms equal to zero except for a one in the \(m\)-th place. Obviously, the \(n\)-th place for \(A{e}_m\) satisfies \[\begin{align} \nonumber (A{e}_m)_n=\begin{cases} 0, & m=n, \\ \frac{1}{|m-n|}, & m\ne n. \end{cases} \end{align}\] Hence \[\begin{align} \nonumber ||A{e}_m||^2=\sum_{n=1,n\ne m}^\infty \frac{1}{{(m-n)}^2}\le 2\sum_{m=1}^\infty \frac{1}{{m}^2}=\frac{\pi^2}{3}. \end{align}\] Therefore \(A\) is a bounded linear operator from \({\rm{span}}\{{e}_1,{e}_2,\cdots\}\) to \(\ell^2\) with norm \(||A||\le \pi/\sqrt{3}\). Since \({\rm{span}}\{{e}_1,{e}_2,\cdots\}\) is dense in \(\ell^2\), then \(A\) is a bounded linear operator from \(\ell^2\) to \(\ell^2\) with norm \(||A||\le \pi/\sqrt{3}\).
We next prove that 85 holds. By 84 , there exists \(C>0\), so that \[\begin{align} \sum_{m=1,m\ne n}^\infty |a_{mn}|\le C \sum_{m=1,m\ne n}^\infty\left|\frac{m\beta_m}{m^2-n^2}\right|. \label{amn1} \end{align}\tag{87}\] Notice that \[\begin{align} \sum_{m\ge1, m\ne n}\left|\frac{m\beta_m}{m^2-n^2}\right|&\le \sum_{m\ge1, m\ne n}\left|\frac{\beta_m}{m-n}\right|. \label{po} \end{align}\tag{88}\] By 86 , 87 and 88 , one has \[\begin{align} \begin{Bmatrix} \sum_{m=1,m\ne n}^\infty |a_{mn}| \end{Bmatrix}_{n=1}^\infty \in \ell^2. \end{align}\] According to the inequality \[\begin{align} \left|\prod_{m\ge1, m\ne n}(1+a_{mn})-1\right|&\le \prod_{m\ge1, m\ne n} (1+|a_{mn}|)-1 \nonumber \\ &\le e^{\sum_{m\ge1, m\ne n}|a_{mn}|}-1 \nonumber \\ &=O\left(\sum_{m\ge1, m\ne n}|a_{mn}|\right), \nonumber \end{align}\] we conclude that \[\begin{align} \nonumber \begin{Bmatrix} \prod_{m\ge1, m\ne n}(1+a_{mn})-1 \end{Bmatrix}_{n=1}^\infty \in \ell^2. \end{align}\] The lemma is proved. ◻
Consider the function \[\begin{align} \label{d0} D_0(\lambda)=\alpha_1 \frac{\sin \sqrt{\lambda}(b-a)}{\sqrt{\lambda}}-\alpha_2 \frac{\sin \sqrt{\lambda} \xi}{\sqrt{\lambda}}, \end{align}\tag{89}\] where \[\alpha_1=(d_1+d_1^{-1})/2, \alpha_2=(d_1-d_1^{-1})/2, \xi=2d-a+b.\] For \(m\in\mathbb{N}\), denote \[\begin{align} \nonumber x_{1,m}=\frac{(m\pi-\arcsin \frac{\alpha_0}{\alpha_1})^2}{(a-b)^2}, x_{2,m}=\frac{(m\pi+\arcsin \frac{\alpha_0}{\alpha_1})^2}{(a-b)^2}, \end{align}\] where \(\alpha_0=\max\{d_1,d_1^{-1}\}.\) By 89 , \[\begin{align} D_0(x_{1,m})D_0(x_{2,m})=\frac{(-1)^{m}\alpha_0-\alpha_2\sin \sqrt{x_{1,m}}\xi}{\sqrt{x_{1,m}}} \frac{(-1)^{m+1}\alpha_0-\alpha_2\sin \sqrt{x_{2,m}} \xi}{\sqrt{x_{2,m}}}<0. \nonumber \end{align}\] Hence for any \(m\in \mathbb{N}\), \(D_0(\lambda)\) has at least one zero \(\mu_{0,m}\) on the interval \((x_{1,m}, x_{2,m})\), namely \[\begin{align} \mu_{0,m}\in (x_{1,m}, x_{2,m}). \label{large} \end{align}\tag{90}\]
We next show if \(|\xi|\le |a-b|\), then for any \(m\in\mathbb{N}\), \(\mu_{0,m}\) is a simple zero of \(D_0(\lambda)\). Denote \(k=\sqrt{\lambda}, k_m=\sqrt{\mu_{0,m}}\) and \(\eta(k)=kD_0(k)\). Then \[\begin{align} \label{inf} \inf_{m\in \mathbb{N}} \left|\frac{ d {\eta}(k_m)}{d k}\right|>0. \end{align}\tag{91}\] If \(|\xi|=|a-b|\), obviously we can obtain 91 . If \(|\xi|<|a-b|\), we can prove 91 by using the method in [43]. In fact, \[\begin{align} \left|\frac{ d {\eta}(k_m)}{d k}\right|=&|\alpha_1(b-a)\cos k_m (b-a)-\alpha_2 \xi\cos k_m\xi| \nonumber \\ &\ge\alpha_1|b-a|\left(\sqrt{1-\frac{\alpha_2^2}{\alpha_1^2}\sin^2 k_m\xi}-\frac{\alpha_2|\xi|}{\alpha_1|b-a|}\sqrt{1-\sin^2 k_m\xi}\right) \nonumber \\ & \equiv \alpha_1|b-a| A_1(k_m). \nonumber \end{align}\] If \(|\xi|/|b-a|\ge \alpha_2/\alpha_1\), then the minimum of \(A_1(k_m)\) is \(1-({\alpha_2|\xi|}/({\alpha_1|b-a|}))\). If \(|\xi|/|b-a|< \alpha_2/\alpha_1\), then the minimum of \(A_1(k_m)\) is \((1-(\alpha_2/\alpha_1)^2)^{1/2}(1-(|\xi|/|b-a|)^2)^{1/2}\). Therefore, for any \(m\in\mathbb{N}\), \[\begin{align} \left|\frac{ d {\eta}(k_m)}{d k}\right|\ge \alpha_1|b-a|\left(1-\frac{\alpha_2^2}{\alpha_1^2}\right)^{1/2}\left(1-\frac{\xi^2}{|b-a|^2}\right)^{1/2}. \end{align}\]
If \(|\xi|> |a-b|\), \(\mu_{0,m}\) is not necessarily a simple zero of \(D_0(\lambda)\). For example, assume that \[D_0(\lambda)=\frac{2\sin \sqrt{\lambda}-\sin 2\sqrt{\lambda}}{\sqrt{\lambda}}.\] Then \(\mu_{0,m}=m^2\pi^2,m=1,2,\cdots\). For any even \(m\), one has \[\begin{align} \nonumber \frac{ d {\eta}(k_m)}{d k}=0. \end{align}\]
We have the following lemma.
Lemma 9. Assume that \(\sigma\in L^2(0,a)\) and \(a\ne b\). If \(\rho(b)=1\) and \(|\xi|\le|a-b|\), then problem 21 has real eigenvalues \(\{\mu_m\}_{m=m_0+1}^\infty\) satisfying \[\begin{align} \label{pertubation} \sqrt{\mu_m}=\sqrt{\mu_{0,m}}+\kappa_m, \end{align}\tag{92}\] where \(\{\kappa_m\}_{m=m_0+1}^\infty\in \ell^2\).
Proof. By Lemma 3, 22 and 68 , the characteristic function \(D(\lambda)\) has the representation \[\begin{align} \label{dlambdainte} D(\lambda)= \frac{1}{\rho(0)^{1/4}}\left(\alpha_1\frac{\sin( \sqrt{\lambda} (b-a)) }{\sqrt{\lambda}} -\alpha_2\frac{\sin (\sqrt{\lambda}\xi )}{\sqrt{\lambda}}+\int_{b-a}^{b+a} h(t) \frac{\sin(\sqrt{\lambda}t)}{\sqrt{\lambda}}dt\right). \end{align}\tag{93}\] Here \(h\in L^2(b-a,a+b)\). According to 93 and Riemann-Lebesgue lemma, there exists \(m_0\), so that for \(m>m_0\), \(D(x_{1,m})D(x_{2,m})<0\). Therefore for \(m>m_0\), \(D(\lambda)\) has at least one zero \(\mu_{m}\) on \((x_{1,m}, x_{2,m})\). Namely, \[\begin{align} \mu_{m}\in (x_{1,m}, x_{2,m}). \label{large1} \end{align}\tag{94}\]
We next show that \(\left\{\int_{b-a}^{b+a} h(t) \sin \sqrt{\mu}_{m} t dt\right\}_{m=m_0+1}^\infty \in \ell^2\). Without loss of generality, assume that \(a>b\). We first prove that for any \(f\in L^2(0,a-b)\), there holds \[\left\{\int_0^{a-b} f(t) \sin \sqrt{\mu}_{m} t dt\right\}_{m=m_0+1}^\infty \in \ell^2.\] Because \(\left\{\sqrt{\frac{{2}}{a-b}}\sin (n\pi t/(a-b))\right\}_{n=1}^\infty\) is an orthonormal basis in \(L^2(0,a-b)\), then there exists \(\{\beta_n\}_{n=1}^\infty \in \ell^2\), so that \[\begin{align} \nonumber f(t)=\sum_{n=1}^\infty \beta_n \sin \frac{n\pi t}{a-b}. \end{align}\] Therefore \[\begin{align} & \int_0^{a-b} f(t) \sin \sqrt{\mu}_{m}t dt=\sum_{n=1}^\infty \beta_n \int_0^{a-b} \sin\sqrt{\mu}_{m}t \sin \frac{n\pi t}{a-b} tdt \nonumber \\ &=\sum_{n=1}^\infty \frac{\beta_n}{2} \left(-\int_0^{a-b}\cos\left(\sqrt{\mu}_m+\frac{n\pi}{a-b}\right)t+\cos\left(\sqrt{\mu}_m-\frac{n\pi}{a-b}\right)tdt\right) \nonumber \\ &=\sum_{n=1}^\infty \frac{O(\beta_n)}{\sqrt{\mu}_m+\frac{n\pi}{a-b}} +\sum_{n=1, n\ne m}^\infty \frac{O(\beta_n)}{\sqrt{\mu}_m-\frac{n\pi}{a-b}}+O(\beta_m). \label{muinl2} \end{align}\tag{95}\] From 94 and the fact that \(\{\beta_n\}_{n=1}^\infty \in \ell^2\), one has \[\begin{align} \nonumber \sum_{m=m_0+1}^\infty\sum_{n=1}^\infty \frac{|\beta_n|^2}{(\sqrt{\mu}_m +\frac{n\pi}{a-b})^2}\le\sum_{m=m_0+1}^\infty\sum_{n=1}^\infty \frac{|\beta_n|^2}{\mu_m}<\infty. \end{align}\] Then \[\begin{align} \label{plus} \begin{Bmatrix} \sum_{n=1}^\infty \frac{\beta_n}{\sqrt{\mu}_m+\frac{n\pi}{a-b}} \end{Bmatrix}_{m=m_0+1}^\infty \in \ell^2. \end{align}\tag{96}\] According to 86 , \[\begin{align} \label{minus} \begin{Bmatrix} \sum_{n=1,n\ne m}^\infty \frac{\beta_n}{\sqrt{\mu}_m-\frac{n\pi}{a-b}} \end{Bmatrix}_{m=m_0+1}^\infty \in \ell^2. \end{align}\tag{97}\] By 95 , 96 and 97 , we know \[\left\{\int_0^{a-b} f(t) \sin \sqrt{\mu}_{m} t dt\right\}_{m_0+1}^\infty \in \ell^2.\] Let \(\left\lfloor{2a}/(a-b)\right\rfloor\) be the largest integer not exceeding \({2a}/(a-b)\). Define \[\begin{align} \label{st} h_1(t)= \begin{cases} h(t), & b-a<t<b+a, \\ 0, & b+a<t<(a-b)\left\lfloor\frac{2a}{a-b}+1\right\rfloor. \end{cases} \end{align}\tag{98}\] Using the periodic properties of trigonometric functions, for any \(j=0,\cdots, \left\lfloor{2a}/(a-b)\right\rfloor\), we have \[\begin{align} \nonumber \begin{Bmatrix} \int_{b-a+j(a-b)}^{b-a+(j+1)(a-b)} h_1(t)\sin \sqrt{\mu}_{m} t dt \end{Bmatrix}_{m=m_0+1}^\infty \in \ell^2. \end{align}\] Hence \[\begin{align} \nonumber \begin{Bmatrix} \int_{b-a}^{(a-b)\left\lfloor\frac{2a}{a-b}+1\right\rfloor} h_1(t) \sin \sqrt{\mu}_{m} t dt \end{Bmatrix}_{m=m_0+1}^\infty \in \ell^2. \end{align}\] By 98 , one concludes that \[\left\{\int_{b-a}^{b+a} h(t) \sin \sqrt{\mu}_{m} t dt\right\}_{m=m_0+1}^\infty \in \ell^2.\]
We next show that 92 holds. Substituting \(\lambda=\mu_m\) into 93 , using 92 and the Taylor expansion of trigonometric functions, there exists \(\{\beta_m\}_{m=m_0+1}^\infty\in \ell^2\), so that \[\begin{align} \nonumber (\alpha_1(b-a)\cos\sqrt{\mu_{0,m}} (b-a)-\alpha_2\xi\cos\sqrt{\mu_{0,m}}\xi)\kappa_m+O(\kappa_m^2)=\beta_m. \end{align}\] From 91 , we conclude that \(\{\kappa_m\}_{m=m_0+1}^\infty\in \ell^2\). The proof is completed. ◻
Remark 6. Notice that if \(|\xi|\le |a-b|\), then for \(m\) large enough, \(D_0(\lambda)\) has exactly one zero \(\mu_{0,m}\) on the interval \(((m-\frac{1}{2})^2\pi^2/(a-b)^2,(m+\frac{1}{2})^2\pi^2/(a-b)^2).\) By Lemma 9, we conclude that \(D(\lambda)\) also has exactly one zero \(\mu_{m}\) on \(((m-\frac{1}{2})^2\pi^2/(a-b)^2,(m+\frac{1}{2})^2\pi^2/(a-b)^2).\)
Lemma 10. Assume that \(|a-b|\ge|\xi|\) and \(\{\mu_{0,m}\}_{m=1}^\infty\) is the zero of \(D_0(\lambda)\). Assume that the positive sequence \(\{\mu_m\}_{m=m_0+1}^\infty\) satisfies \[\begin{align} \sqrt{\mu_m}=\sqrt{\mu_{0,m}}+\kappa_m, \label{mui} \end{align}\tag{99}\] where \(\{\kappa_m\}_{m=m_0+1}^\infty\in \ell^2\). Then the infinite product \[\begin{align} g(\lambda)=\prod_{m=m_0+1}^\infty \left(1-\frac{\lambda}{\mu_m}\right) \label{g} \end{align}\tag{100}\] is an entire function with respect to \(\lambda\). Moreover, there exist \(C>c>0\), so that in the sector \(\Lambda_{\delta}=\{\lambda\in \mathbb{C}| \delta<\arg(\lambda)<\pi-\delta, \delta\in (0,\pi/2)\}\), \[\begin{align} c|\sin (\sqrt{\lambda}(a-b))||\lambda|^{-m_0-1/2}<|g(\lambda)|<C|\sin( \sqrt{\lambda}(a-b))||\lambda|^{-m_0-1/2} \label{infinite} \end{align}\tag{101}\]
Proof. By 94 , the series \(\sum_{m>m_0} |\lambda/\mu_m|\) converges uniformly on the bounded set of \(\lambda\) plane. Therefore, the infinite product 100 converges uniformly on the bounded set of \(\lambda\) plane and hence \(g(\lambda)\) is an entire function.
Notice that \[\begin{align} \nonumber D_0(\lambda)=(\alpha_1(b-a)-\alpha_2\xi)\prod_{m=1}^\infty \left(1-\frac{\lambda}{\mu_{0,m}}\right). \end{align}\] Therefore \[\begin{align} \frac{g(\lambda)}{D_0(\lambda)}=(\alpha_1(b-a)-\alpha_2\xi)\prod_{m=1}^{m_0}\frac{\mu_{0,m}}{\mu_{0,m}-\lambda} \times\prod_{m=m_0+1}^\infty \frac{\mu_{0,m}(\mu_m-\lambda)}{\mu_{m}(\mu_{0,m}-\lambda)} \label{c0}. \end{align}\tag{102}\] For \(\lambda\in\Lambda_\delta\), there exist \(C_1>c_1>0\), such that \[\begin{align} c_1|\lambda|^{-m_0}<\prod_{m=1}^{m_0}\left|\frac{\mu_{0,m}}{\mu_{0,m}-\lambda}\right|<C_1|\lambda|^{-m_0}. \label{c1} \end{align}\tag{103}\] From 94 and 99 , for \(m>m_0\), \[\begin{align} \nonumber \frac{\mu_{0,m}}{\mu_{m}}=1+\frac{\beta_m}{m}, \end{align}\] where \(\{\beta_m\}_{m=1}^\infty\in \ell^2\). Then by Cauchy-Schwarz inequality, the series \(\sum_{m>m_0} (1-\mu_{0,m}/\mu_{m})\) converges and there exist \(C_2>c_2>0\), such that \[\begin{align} 0<c_2<\prod_{m=m_0+1}^\infty\frac{\mu_{0,m}}{\mu_{m}}<C_2. \label{c2} \end{align}\tag{104}\]
For \(|\lambda|=(n+1/2)^2\pi^2/(a-b)^2\), \(n=1,2,\cdots\), we have \[\begin{align} \nonumber \frac{\mu_m-\lambda}{\mu_{0,m}-\lambda}= \begin{cases} 1+O(\kappa_n),& m=n, \\ 1+O\left(\frac{m\beta_m}{m^2-n^2}\right),& m\ne n. \end{cases} \end{align}\] Thus, applying Lemma 8 to \[\begin{align} \nonumber a_{mn} = \begin{cases} 0, & m\le m_0or ~n\le m_0, \\ \frac{\mu_m-\lambda}{\mu_{0,m}-\lambda}-1, & m, n>m_0 \end{cases} \end{align}\] with \(|\lambda|=(n+1/2)^2\pi^2/(a-b)^2\), \(n=1,2,\cdots\), we have \[\begin{align} \prod_{m=m_0+1}^\infty \frac{\mu_m-\lambda}{\mu_{0,m}-\lambda}=1+o(1), n\rightarrow\infty. \label{i} \end{align}\tag{105}\]
We next show that for any \(\lambda\in \Lambda_\delta\), 105 holds. Note that for any fixed \(\lambda\), there exists \(n_0\), such that \[(n_0-1/2)^2\pi^2/(a-b)^2\le|\lambda|<(n_0+1/2)^2\pi^2/(a-b)^2.\] Therefore, it suffices to prove that there exists \(C>0\), which is independent of \(\lambda, m, n_0\), so that for any \(m>m_0\), \[\begin{align} \label{xianran} \frac{1}{|\lambda-\mu_{0,m}|}\le C\frac{1}{|(n_0+1/2)^2\pi^2/(a-b)^2-\mu_{0,m}|}. \end{align}\tag{106}\] The proof of 106 is obvious and we omit the steps.
By 102 , 103 , 104 and 105 , we can obtain 101 . The lemma is proved. ◻
When \(a\ne b\), problem \(Q(\rho)\) has the “almost real subspectrum” \(\{\mu_m\}_{m=1}^\infty\). Recall that the set of all eigenvalues of \(Q(\rho)\) is denoted by \(\{\lambda_k\}_{k=1}^\infty\).
Theorem 7. Assume that \(\rho\in W_2^1\left((0, b_1)\cup (b_1, b)\right)\) and satisfies 4 and \(\rho(b)= 1\). Assume that \(a\ne b\). Then \(Q(\rho)\) has a subsequence of eigenvalues \(\{\mu_m\}_{m=1}^\infty\) satisfying
(i) there exists \(m_0\in \mathbb{N}\) such that for \(m=1,\cdots, m_0\), we have \(|\mu_m|<(m_0+\frac{1}{2})^2\pi^2/(a-b)^2\);
(ii) for \(m> m_0\), all \(\mu_m\) are real and satisfy \[(m-\frac{1}{2})^2\pi^2/(a-b)^2<|\mu_{m}|<(m+\frac{1}{2})^2\pi^2/(a-b)^2.\]
Moreover, if
(1) \(0< b<a\) and \(\int_{b_1}^b \sqrt{\rho(r)}dr \ge b\),
or
(2) \(b>a\),
then \(\{\mu_m\}_{m=1}^\infty\) is a proper set of \(\{\lambda_k\}_{k=1}^\infty\).
Proof. By 94 , we know that (ii) holds. Let \(N(r)\) be the number of zeros of \(D(\lambda)\) in the circle \(|\lambda|<r\). By Lemma 3 and Lemma 10, there exists \(n_0\), for \(n>n_0\), we can obtain
\[\begin{align} \label{number} N\left(\frac{(n+1/2)^2\pi^2}{(a-b)^2}\right) \ge n. \end{align}\tag{107}\] The proof of 107 is similar to that of [11] and we omit the steps. By 107 , we conclude that (i) holds.
Moreover, if (1) or (2) of Theorem 7 is satisfied, we have \(|\xi|> |a-b|\). Then \(D(\lambda)\) is an entire function of order 1/2 and type greater than \(|a-b|\). From [35], \(D(\lambda)\) has infinite zeros besides \(\{\mu_m\}_{m=1}^\infty\). Therefore \(\{\mu_m\}_{m=1}^\infty\) is a proper set of \(\{\lambda_k\}_{k=1}^\infty\). The proof is completed. ◻
Now in a position to state and prove our main result in this section. The following theorem considers the mixed spectral problem [44]. That is, recover the refractive index from the “almost real subspectrum” \(\{\mu_m\}_{m=1}^\infty\) and partial information on the refractive index. Theorem 8 refines the refractive index in [11] from a \(W_2^2\) function to a piecewise \(W_2^1\) function. Moreover, we drop the condition \(\rho'(b)=0\) in [11].
Theorem 8. Assume that \(\rho\in W_2^1\left((0, b_1)\cup (b_1, b)\right)\) and satisfies 4 and \(\rho(b)= 1\). Suppose that \(a\) is known. Assume that one of the following four conditions holds:
(1) \(0< b<a\) and \(\rho\) is known on the interval \([A,b]\), where \(A\) satisfies \(\int_A^b \sqrt{\rho(r)} dr=(a+b)/2\), \(\int_{b_1}^b \sqrt{\rho(r)}dr \ge b\);
(2) \(0< b<a\) and \(\rho\) is known on the interval \([A,b]\), where \(A\) satisfies \(\int_A^b \sqrt{\rho(r)} dr=(a+b)/2\), \(\int_{b_1}^b \sqrt{\rho(r)}dr < b\), one of the eigenvalues, denoted by \(\mu_0\), in \(\{\lambda_k\}_{k=1}^\infty\setminus \{\mu_m\}_{m=1}^\infty\) is known;
(3) \(a<b\le 3a\) and \(\rho\) is known on the interval \([A,b]\), where \(A\) satisfies \(\int_A^b \sqrt{\rho(r)} dr=(3a-b)/2\), one of the eigenvalues, denoted by \(\mu_0\), in \(\{\lambda_k\}_{k=1}^\infty\setminus \{\mu_m\}_{m=1}^\infty\) is known;
(4) \(3a<b\).
Then \(\rho\) is uniquely determined by \(\{\mu_m\}_{m=1}^\infty\).
Proof. We require that if a certain symbol \(\gamma\) denotes an object related to \(Q(\rho)\), then the corresponding symbol \(\tilde{\gamma}\) denotes the analogous object related to \(Q(\tilde{\rho})\).
Define \[\begin{align} \nonumber G(\lambda)= \begin{cases} \prod_{m=1}^\infty \left(1-\frac{\lambda}{\mu_m}\right), &case ~(1) or ~(4), \\ \prod_{m=0}^\infty \left(1-\frac{\lambda}{\mu_m}\right), &case ~(2) or ~(3). \end{cases} \end{align}\] In case (1), \(|a-b|\ge |\xi|\), by Lemma 10, in the sector \(\Lambda_{\delta}\), \[\begin{align} \label{glambda11} |G(\lambda)|>c|\sin (\sqrt{\lambda}(a-b))||\lambda|^{-1/2}. \end{align}\tag{108}\] In case (2) or (3), letting \(\lambda=iy,y\in\mathbb{R}\), then there exists \(C_0>0\), such that \[\begin{align} \label{glambda1} |G(iy)|\ge C_0\prod_{m=1}^\infty \left|1+\frac{y^2}{(m-1/2)^4\pi^4/(a-b)^4}\right|^{1/2} =C_0\left|{\cos(\sqrt{iy}(a-b))}\right|. \end{align}\tag{109}\] In case (4), letting \(\lambda=iy,y\in\mathbb{R}\), then there exists \(C_0>0\), such that \[\begin{align} \label{glambda111111111} |G(iy)|\ge C_0\left|\frac{{\cos(\sqrt{iy}(a-b))}}{y}\right|. \end{align}\tag{110}\]
Define \[\begin{align} H(\lambda)= z(a,\lambda)\tilde{z}^{[1]}(a,\lambda)-\tilde{z}(a,\lambda)z^{[1]}(a,\lambda), \label{definitionhz1} \end{align}\tag{111}\] Then in \(\Sigma_{\delta}:=\Lambda_{\delta}\cup\{\lambda\in \mathbb{C}| \pi+\delta<\arg(\lambda)<2\pi-\delta, \delta\in (0,\pi/2)\}\), we have \[\begin{align} \label{1234} H(\lambda)= \begin{cases} o\left(\frac{e^{|a-b||\rm{Im}\sqrt{\lambda}|}}{\sqrt{\lambda}}\right), &case ~(1), (2) or ~(3), \\ o\left(\frac{e^{2a|\rm{Im}\sqrt{\lambda}|}}{\sqrt{\lambda}}\right), &case ~(4). \end{cases} \end{align}\tag{112}\] We only prove 112 in case (1), and other cases can be proved similarly. In case (1), \[\begin{align} s\left(a,\lambda; \left(\frac{a-b}{2}\right)-\right)&=\tilde{s}\left(a, \lambda; \left(\frac{a-b}{2}\right)-\right),\nonumber \\ s^{[1]}\left(a, \lambda; \left(\frac{a-b}{2}\right)-\right)&=\tilde{s}^{[1]}\left(a, \lambda; \left(\frac{a-b}{2}\right)-\right),\nonumber \\ c\left(a, \lambda; \left(\frac{a-b}{2}\right)-\right)&=\tilde{c}\left(a, \lambda; \left(\frac{a-b}{2}\right)-\right),\nonumber \\ c^{[1]}\left(a,\lambda; \left(\frac{a-b}{2}\right)-\right)&=\tilde{c}^{[1]}\left(a, \lambda; \left(\frac{a-b}{2}\right)-\right). \nonumber \end{align}\] Here \(s(x,\lambda;y), c(x,\lambda;y)\) are normalized according to 46 . Letting \(x=a, y=((a-b)/2)-\) in 48 and 49 , substituting them into 111 , then using 47 and 68 , one knows \[\begin{align} H(\lambda)=&z\left(\left(\frac{a-b}{2}\right)-,\lambda\right)\tilde{z}^{[1]}\left(\left(\frac{a-b}{2}\right)-,\lambda\right)-\tilde{z}\left(\left(\frac{a-b}{2}\right)-,\lambda\right)z^{[1]}\left(\left(\frac{a-b}{2}\right)-,\lambda\right)\nonumber \\ =&z\left(\left(\frac{a-b}{2}\right)-,\lambda\right)\tilde{z}\left(\left(\frac{a-b}{2}\right)-,\lambda\right)\left(m\left(\left(\frac{a-b}{2}\right)-,\lambda\right)-\tilde{m}\left(\left(\frac{a-b}{2}\right)-,\lambda\right)\right). \nonumber \end{align}\] From Lemma 3 and 44 , in the sector \(\Lambda_{\delta}\), \(H(\lambda)\) has asymptotic form 112 . Since \(H(\lambda)\) is a real entire function, then \(H(\lambda)\) also has asymptotic form 112 in \(\Sigma_{\delta}\).
We next show that \(H(\lambda)\equiv 0\). Define \[\begin{align} \nonumber F(\lambda):=\frac{H(\lambda)}{G(\lambda)}. \end{align}\] Arguing as in Theorem 5, we know \(F(\lambda)\) is an entire function. In case (1), by 108 and 112 , as \(|\lambda|\rightarrow \infty\) in the sector \(\Lambda_\delta\), we have \(F(\lambda)=o(1)\). In cases (2), (3) or (4), by 109 , 110 and 112 , as \(|\lambda|\rightarrow \infty\) on the imaginary axis, one has \(F(\lambda)=o(1)\). Note that in the above four cases, using Phragmén-Lindelöf theorem and Liouville theorem, we can get \(F(\lambda)\equiv0\). From 111 , one can obtain that \(m(a,\lambda)\equiv \tilde{m}(a,\lambda)\). Using Theorem 2 and Lemma 2, we have \(\rho\equiv \tilde{\rho}.\) ◻
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