Further Theorems Inspired by Ambartsumian

Abstract

We continue to derive results in the spirit of classical Ambartsumian Theorem 14.1. In the first part we use heat kernel technique to show that a Schrödinger operator is isospectral to a Laplacian only if the potential is zero. This part is rather technical but does not require any a priori knowledge of heat kernel semigroups. In the second part the theory of almost periodic functions is used to obtain uniqueness results for Laplace and Schrödinger operators.

We continue to derive results in the spirit of classical Ambartsumian Theorem 14.1. In the first part we use heat kernel technique to show that a Schrödinger operator is isospectral to a Laplacian only if the potential is zero. This part is rather technical but does not require any a priori knowledge of heat kernel semigroups. In the second part the theory of almost periodic functions is used to obtain uniqueness results for Laplace and Schrödinger operators.

15.1 Ambartsumian-Type Theorem by Davies

Our goal in this section is to prove that among all standard Schrödinger operators on a metric graph, only the operator with zero potential has the same spectrum as the Laplacian. In other words, the zero potential is unique among all other potentials if one just looks at the spectrum of a quantum graph. In the case where the metric graph is just an interval, this fact is the classical Ambartsumian theorem, but the graphs considered in this section are arbitrary finite compact metric graphs. We are going to assume that such a graph is fixed. In our presentation we are going to follow the proof by E.B. Davies [155], but we adapt it to the case of quantum graphs. The original proof goes as follows: it is first shown that the statement holds under rather general assumptions, then it is shown that the assumptions hold for quantum graphs with standard vertex conditions. Adapting the proof we managed to simplify some arguments and no deep knowledge of the heat kernel approach to spectral theory [154] will be required.

15.1.1 On a Sufficient Condition for the Potential to Be Zero

If you examine the proof of the classical Ambartsumian theorem (Theorem 14.1) you will see that the crucial point is to show that the potential has average value zero. For a single interval this follows from explicit spectral asymptotics. For arbitrary compact graphs this fact follows from the asymptotic analysis of the heat kernel as $t \rightarrow 0.$ Therefore let us first show that proving that the average value of the potential is zero is enough to obtain an Ambartsumian-type theorem.

Theorem 15.1

If

$$\displaystyle \begin{aligned} {} \lambda_1 (L_q^{\mathrm{st}} (\Gamma)) \geq 0 \end{aligned} $$

(15.1)

and

$$\displaystyle \begin{aligned} {} \int_\Gamma q(x) dx \leq 0, \end{aligned} $$

(15.2)

then $q$ is equal to zero almost everywhere.

Proof

Let us use $u (x) \equiv 1$ as a trial function for the quadratic form

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle Q(u,u) \big\vert_{u=1} & = & \displaystyle \int_\Gamma \vert u' (x) \vert^2 dx \big\vert_{u=1} + \int_\Gamma q(x) \vert u(x) \vert^2 dx \big\vert_{u=1} \\[3mm] & = & \displaystyle 0 + \int_\Gamma q(x) dx. \end{array}\end{aligned}$$

Inequality (15.1) implies that $Q(1,1) \geq 0$ and hence

$$\displaystyle \begin{aligned} \int_\Gamma q(x) dx \geq 0 ,\end{aligned}$$

which in combination with (15.2) implies

$$\displaystyle \begin{aligned} \int_\Gamma q(x) = 0 \; \; \mathrm{and} \; \; \lambda_1 (L_q^{\mathrm{st}} (\Gamma)) = 0.\end{aligned}$$

Therefore $u(x) \equiv 1$ is not only the eigenfunction of the standard Laplacian on $\Gamma$ but also an eigenfunction of $L_q^{\mathrm {st}} (\Gamma )$ since the lowest eigenvalue of the Schrödinger operator is simple. Hence, on every edge $u(x) \equiv 1$ satisfies the differential equation

$$\displaystyle \begin{aligned} - \frac{d^2}{dx^2} u (x) + q(x) u(x) = 0 \cdot u(x)\end{aligned}$$

implying

$$\displaystyle \begin{aligned} q(x) = 0\end{aligned}$$

almost everywhere. □

15.1.2 Laplacian Heat Kernel

Our goal in this subsection is to study properties of the heat kernel associated with the standard Laplacian on a finite compact metric graph $\Gamma .$ The heat kernel $H_\Gamma (t,x,y)$ is defined as the kernel of the integral operator solving the heat equation

$$\displaystyle \begin{aligned} {} \hspace{-2pt}\left\{\hspace{-2pt} \begin{array}{ll} \displaystyle \frac{\partial}{\partial t} u(t,x) = - L_q^{\mathrm{st}} u (t,x), & \displaystyle x \,{\in}\, \Gamma, t {\in} (0,\infty), \\ u(0,x) = u_0(x), & x {\in} \Gamma ; \end{array} \right. \Rightarrow u(t,x) = \hspace{-1pt}\int_\Gamma H_\Gamma\hspace{-2pt} (t,x,y) u_0(y) dy. \end{aligned} $$

(15.3)

In the case where all eigenfunctions $\psi _n$ of $L_q^{\mathrm {st}}$ and eigenvalues $\lambda _n$ are known, the solution can be presented as an absolutely converging series

$$\displaystyle \begin{aligned} {} u(t,x) = \sum_{n=1}^\infty e^{- \lambda_n t} \psi_n(x) \langle \psi_n, u_0 \rangle_{L_2(\Gamma)} \end{aligned} $$

(15.4)

leading to an explicit formula for the Heat kernel

$$\displaystyle \begin{aligned} {} H_\Gamma (t,x,y) = \sum_{n=1}^\infty e^{- \lambda_n t} \psi_n (x) \overline{\psi_n(y)}. \end{aligned} $$

(15.5)

Note that the complex conjugation is not needed if the eigenfunctions can be chosen real, for example in the case of standard vertex conditions considered here.

We first study the heat kernel for single interval $[-a,a]$ with Dirichlet boundary conditions imposed at the endpoints and use obtained estimates to analyse Laplacian heat kernel’s behaviour for small times.

15.1.2.1 Heat Kernel for the Dirichlet Laplacian on an Interval

Consider the operator $L^{\mathrm {D}} (-a,a).$ We are interested in the corresponding heat kernel. To obtain an explicit formula we use the eigenfunction expansion for the Dirichlet Laplacian. The normalised eigenfunctions and eigenvalues are

$$\displaystyle \begin{aligned} \psi_n = \frac{1}{\sqrt{a}} \sin \big(\frac{x+a}{2a} \pi n \Big), \; \, \lambda_n = \left(\frac{\pi}{2a} \right)^2 n^2.\end{aligned}$$

The heat kernel is given by (15.5)

$$\displaystyle \begin{aligned} {} H_{[-a,a]} (t, x, y) = \sum_{n=1}^\infty e^{- \left(\frac{\pi}{2a} \right)^2 n^2 t} \frac{1}{a} \sin \Big(\frac{x+a}{2a} \pi n \Big) \sin \Big(\frac{y+a}{2a} \pi n \Big) . \end{aligned} $$

(15.6)

The kernel is a positive continuous function on $(0, \infty ) \times [-a,a] \times [-a,a].$

In what follows we shall need an estimate for $\int _{-a}^a H_{[-a,a]} (t, 0, x) dx .$

Lemma 15.2

The heat kernel $H_{[-a,a]} (t, 0, x)$ for the Dirichlet Laplacian on $[-a,a]$ satisfies the integral estimate ¹

$$\displaystyle \begin{aligned} {} 1 > \int_{-a}^a H_{[-a,a]} (t, 0, x) dx \geq 1 - \frac{4a}{\sqrt{\pi t}} e^{-a^2/(4t)} . \end{aligned} $$

(15.7)

Proof

We start by calculating explicitly

$$\displaystyle \begin{aligned} H_{[-a,a]} (t, 0, x) = \sum_{n=1}^\infty e^{- \left(\frac{\pi}{2a} \right)^2 n^2 t} \frac{1}{a} \sin \frac{\pi n}{2} \sin \frac{x+a}{2a} \pi n\end{aligned}$$

and integrating

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \int_{-a}^a H_{[-a,a]} (t, 0, x) dx & = & \displaystyle \sum_{n=1}^\infty e^{- \left(\frac{\pi}{2a} \right)^2 n^2 t} \frac{1}{a} \sin \frac{\pi n}{2} \int_{-a}^a \sin \frac{x+a}{2a} \pi n dx \\[3mm] & = & \displaystyle \sum_{n=1}^\infty e^{- \left(\frac{\pi}{2a} \right)^2 n^2 t} \frac{1}{a} \sin \frac{\pi n}{2} \; \; \frac{2 a}{\pi n} \left(1- \cos \pi n \right) \\[3mm] & = & \displaystyle \sum_{m=0}^\infty e^{- \left(\frac{\pi}{2a} \right)^2 (2m+1)^2 t} \frac{4 }{\pi (2m+1)} \sin \frac{\pi (2m+1)}{2} \\[3mm] & = & \displaystyle \sum_{m=0}^\infty e^{- \left(\frac{\pi}{2a} \right)^2 (2m+1)^2 t} \frac{4 }{\pi (2m+1)} (-1)^m. \end{array} \end{aligned}$$

All formulas depend on $t/a^2$, hence we can put $a=1$ for a while and substitute $t$ with $t/a^2$ at the final stage. Using this convention the formula for the integral is modified as

$$\displaystyle \begin{aligned} {} \int_{-1}^1 H_{[-1,1]} (t, 0, x) dx = \sum_{m=-\infty}^\infty e^{\displaystyle - \pi^2 (m+1/2)^2 t} \; \; \frac{2 \sin \pi(m+1/2) }{ \pi(m+1/2)} . \end{aligned} $$

(15.8)

To calculate the series we are going to use Poisson summation formula

$$\displaystyle \begin{aligned} {} \sum_{n=-\infty}^\infty f(n) = \sum_{n=-\infty}^\infty \hat{f}(n) , \end{aligned} $$

(15.9)

where $\hat {f}$ is the Fourier transform of $f.$ Consider first an auxiliary function

$$\displaystyle \begin{aligned} g(x) := 2 e^{\displaystyle - \pi^2 x^2 t} \; \; \frac{\sin \pi x }{ \pi x}.\end{aligned}$$

To calculate its Fourier transform $\hat {g} (k)$ we note that $g$ is essentially a product of two functions

$$\displaystyle \begin{aligned} e^{- \pi^2 x^2 t} \; \; \; \mathrm{and} \; \; \; \frac{\sin \pi x}{\pi x},\end{aligned}$$

with explicit formulas for their Fourier transforms

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle \int_{-\infty}^\infty e^{- \pi^2 x^2 t} e^{- 2 \pi i px} dx = \int_{-\infty}^\infty e^{- (\pi x \sqrt{t} + i p/\sqrt{t})^2} dx e^{-p^2/t} = \frac{1}{\sqrt{\pi t}} e^{-p^2/t} ; \\[5mm] \displaystyle \int_{-\infty}^\infty \frac{\sin \pi x}{\pi x} e^{- 2 \pi i p x} dx = \left[ \begin{array}{ll} 1, & \vert p \vert < 1/2,\\[3mm] 0, & \mathrm{otherwise}. \end{array} \right. \end{array}\end{aligned}$$

Taking convolution we get

$$\displaystyle \begin{aligned} \hat{g} (p) = \frac{2}{\sqrt{\pi t}} \int_{p-1/2}^{p+1/2} e^{-s^2/t} ds.\end{aligned}$$

To calculate the series (15.8) consider the function

$$\displaystyle \begin{aligned} f(x) = g(x+1/2)\end{aligned}$$

implying

$$\displaystyle \begin{aligned} \hat{f} (p) = e^{\pi i p} \hat{g} (p) ,\end{aligned}$$

and modify Poisson summation formula (15.9) as

$$\displaystyle \begin{aligned} \sum_{m=-\infty}^{\infty} g(m+1/2) &= \sum_{m=-\infty}^\infty f(m) = \sum_{m=-\infty}^\infty \hat{f} (m) = \sum_{m=-\infty}^\infty \hat{g} (m) e^{\pi i m} \\ &= \sum_{m=-\infty}^\infty \hat{g} (m) (-1)^m .\end{aligned} $$

Then the integral of the heat kernel is given by

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \int_{-1}^1 H_{[-1,1]} (t, 0, x) dx & = & \displaystyle \sum_{m=-\infty}^\infty \hat{g} (m) (-1)^m \\[5mm] & = & \displaystyle \frac{2}{\sqrt{\pi t}} \Bigg( \int_{0}^{1/2} e^{-s^2/t} ds - \int_{1/2}^{3/2} e^{-s^2/t} ds \\[5mm] &&\qquad + \int_{3/2}^{5/2} e^{-s^2/t} ds + \dots \Bigg) \\[5mm] & = & \displaystyle \frac{2}{\sqrt{\pi}} \Bigg( \int_{0}^{1/2 \sqrt{t}} e^{-s^2} ds - \int_{1/2 \sqrt{t}}^{3/2\sqrt{t}} e^{-s^2} ds \\[5mm] &&\qquad + \int_{3/2 \sqrt{t}}^{5/2\sqrt{t}} e^{-s^2} ds + \dots \Bigg). \end{array}\end{aligned}$$

Using Gaußian integral $1 = \frac {2}{\sqrt {\pi }} \int _0^\infty e^{-s^2} ds$ we obtain

$$\displaystyle \begin{aligned} 1 - \int_{-1}^1 H_{[-1,1]} (t, 0, x) dx &= \frac{4}{\sqrt{\pi}} \Bigg( \int_{1/2 \sqrt{t}}^{3/2\sqrt{t}} e^{-s^2} ds + \int_{5/2 \sqrt{t}}^{7/2\sqrt{t}} e^{-s^2} ds \\ &\qquad + \int_{9/2 \sqrt{t}}^{11/2\sqrt{t}} e^{-s^2} ds \dots \Bigg).\end{aligned} $$

We are getting immediately the upper estimate, since every term in the series on the right hand side is positive:

$$\displaystyle \begin{aligned} 1 - \int_{-1}^1 H_{[-1,1]} (t, 0, x) dx > 0 \; \; \Rightarrow \; \; \int_{-1}^1 H_{[-1,1]} (t, 0, x) dx < 1.\end{aligned}$$

This estimate can also be obtained by noting that

$$\displaystyle \begin{aligned} H_{[-a,a]} (t, x, y) \leq H_{(-\infty,\infty)} (t,x,y) = \frac{1}{\sqrt{4\pi t}} e^{-\vert x-y\vert^2/4t},\end{aligned}$$

since the heat kernel is monotone with respect to the domain.

The lower estimate can be obtained with different precisions, for example we may use just the first integral in the series

$$\displaystyle \begin{aligned} 1 - \int_{-1}^1 H_{[-1,1]} (t, 0, x) dx \leq \frac{4}{\sqrt{\pi}} \int_{1/2 \sqrt{t}}^{3/2\sqrt{t}} e^{-s^2} ds < \frac{4}{\sqrt{\pi t}} e^{-1/(4t)} .\end{aligned}$$

□

Lemma 15.2 will allow us to show an explicit two-sided estimate for $H_{[-1,1]} (t, 0, 0).$ A series representation for this function follows directly from (15.6)

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle H_{[-a,a]} (t, 0, x) & = & \displaystyle \sum_{n=1}^\infty e^{- \left(\frac{\pi}{2a} \right)^2 n^2 t} \frac{1}{a} \sin \frac{\pi n}{2} \sin \frac{x+a}{2a} \pi n \\[3mm] & = & \displaystyle \sum_{n=1}^\infty e^{- \left(\frac{\pi}{2a} \right)^2 n^2 t} \frac{1}{2a} \left( \cos \frac{x}{2a} - \cos (\frac{x}{2a} + \pi n) \right) \\[3mm] & = & \displaystyle \sum_{m=0}^\infty e^{- \left(\frac{\pi}{2a} \right)^2 (2m+1)^2 t} \frac{1}{a} \cos \frac{x}{2a} , \end{array}\end{aligned}$$

and in particular

$$\displaystyle \begin{aligned} H_{[-1,1]} (t, 0, 0) = \displaystyle \sum_{m=0}^\infty e^{- \pi^2 (m+1/2)^2 t} = \frac{1}{2} \vartheta [2, 0, e^{-\pi^2 t}],\end{aligned}$$

where $\vartheta$ is the Elliptic Theta function.² It does not help to take the Fourier transform and use Poisson summation formula since the Fourier transform of the Gaussian kernel is a Gaussian kernel.

Lemma 15.3

The heat kernel $H_{[-a,a]} (t, 0, x)$ for the Dirichlet Laplacian on $[-a,a]$ satisfies the estimate ³

$$\displaystyle \begin{aligned} {} \frac{1}{\sqrt{4 \pi t}} \geq H_{[-a,a]} (t,0,0) \geq \frac{1}{\sqrt{4 \pi t}} - \frac{8a}{\pi t } e^{-a^2/(2t)}. \end{aligned} $$

(15.10)

Proof

The estimate is tight for small $t$, which can be illustrated by Fig. 15.1.

Fig. 15.1

Graphs of the functions $\frac {1}{ \sqrt {4 \pi t}}$ (upper), $H_{[-1,1]} (t, 0, 0) = \frac {1}{2} \vartheta [2, 0, e^{-\pi ^2 t}]$ (middle), $\frac {1}{ \sqrt {4 \pi t}} - \frac {8}{\pi t } e^{-1/(2t)}$ (lower)

To prove the estimate we again use the approach developed in [155]. First of all we need the following identity for the heat kernel, which can easily be proven for any compact domain $\Omega$ with the eigenfunctions $\psi _n$ of the Dirichlet Laplacian:

$$\displaystyle \begin{aligned} {} H (2t, 0,0 ) = \int_\Omega H^2(t, 0, x) dx, \end{aligned} $$

(15.11)

where the integral is taken over the domain. We use the standard formula (15.5) for the heat kernel implying

$$\displaystyle \begin{aligned} \displaystyle \int_\Omega H^2 (t, 0, x) dx & = \displaystyle \int_\Omega \sum_{n=1}^\infty e^{-\lambda_n t} \psi_n(0) \psi_n(y) \sum_{m=1}^\infty e^{-\lambda_m t} \psi_m(0) \psi_n(y) dy \\[3mm] & = \displaystyle \sum_{n,m=1}^\infty e^{-\lambda_n t} \psi_n(0) e^{-\lambda_m t} \psi_m(0) \delta_{nm} \\ & = \displaystyle \sum_{m=1}^\infty e^{-2 \lambda_m t} (\psi_m(0))^2 = H(2t, 0,0). \end{aligned} $$

Consider the following two positive functions

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle f(x) & = & \displaystyle \left\{ \begin{array}{cl} \displaystyle H_{[-a,a]} (t,0,x), & \displaystyle \mbox{if } \vert x \vert \leq a, \\[3mm] \displaystyle 0, & \displaystyle \mbox{otherwise;} \end{array} \right. \\[5mm] \displaystyle g(x) & = & \displaystyle H_{(-\infty,\infty)} (t, 0, x) = \frac{1}{\sqrt{4 \pi t}} e^{- x^2/(4t)}, \end{array} \end{aligned}$$

so that

$$\displaystyle \begin{aligned} 0 \leq f(x) \leq g(x) \leq \frac{1}{\sqrt{4 \pi t}}\end{aligned}$$

for all $x \in \mathbb R$ and $\int _{\mathbb R} g(x) dx = 1.$ Using formula (15.11) we get

$$\displaystyle \begin{aligned} \begin{array}{ccl} 0 & \leq & \displaystyle \frac{1}{\sqrt{8 \pi t}}- H_{[-a,a]} (2t, 0,0) \\[3mm] & = & \displaystyle H_{(-\infty,\infty)} (2t, 0,0) - H_{[-a,a]} (2t, 0,0) \\[3mm] & = & \displaystyle \int_{\mathbb R} \big( g^2(x) - f^2(x) \big) dx \\[3mm] & \leq & \displaystyle \int_{\mathbb R} \big( g(x) - f(x) \big) 2 g(x)dx \\[3mm] & \leq & \displaystyle \frac{1}{\sqrt{\pi t }} \int_{\mathbb R} \big( g(x) - f(x) \big) )dx \\[3mm] & = & \displaystyle \frac{1}{\sqrt{\pi t }} \Big( 1 - \int_{-a}^a f(x) dx \Big) \\[3mm] & \leq & \displaystyle \displaystyle \frac{4a}{\pi t } e^{-1/(4t)}, \end{array}\end{aligned}$$

where we used (15.7) on the last step. It remains to make a substitution of $2t$ with $t$ leading to the second estimate in (15.10). □

Lemma 15.3 implies that the heat kernel on the interval has a singularity like $\frac {1}{\sqrt {4 \pi t}}.$ Moreover, the two-sided estimates (15.10) allows one to conclude that the following limit exists and calculate it

$$\displaystyle \begin{aligned} {} \lim_{t \rightarrow 0} \sqrt{4 \pi t} \; H_{[-a.a]} (t, 0, 0) = 1, \end{aligned} $$

(15.12)

since $\lim _{t \rightarrow 0 } \frac {16}{\sqrt {\pi t}} e^{-1/(2t)} = 0.$

15.1.2.2 Heat Kernel for the Standard Laplacian on the Graph

We accomplish this section by proving that the limit (15.12) holds for the standard Laplacian on a metric graph for almost all points. More precisely, the set of exceptional points, where the limit may be violated, coincides with the set of vertices.

Lemma 15.4

Let $[-a.a]$ be an interval on one of the edges on a finite compact metric graph $\Gamma .$ Then the heat kernels $H_{[-a,a]} (t,x,y)$ for the Dirichlet Laplacian on $[-a,a]$ and $H_\Gamma (t,x,y)$ for the standard Laplacian on $\Gamma$ restricted to the interval $[-a,a]$ satisfy the estimate:

$$\displaystyle \begin{aligned} {} H_{[-a,a]} (t,x,y) \leq H_\Gamma (t,x,y); \; \; x,y, \in [-a,a]. \end{aligned} $$

(15.13)

Proof

Consider the graph $\Gamma \cup _{-a,a} \Gamma$ obtained from $\Gamma \cup \Gamma$ by joining pairwise the endpoints of the intervals $[-a,a]$ on both graphs (see Fig. 15.2). We impose standard vertex conditions. The graph is invariant under the symmetry transformation $\tau$ mapping the same points $x$ and $x'$ on the two copies of $\Gamma$ onto each other:

$$\displaystyle \begin{aligned} \tau x = x', \; \; \tau x' = x.\end{aligned}$$

The two copies of $\Gamma$ as subsets of $\Gamma \cup _{-a,a} \Gamma$ will be denoted by $\Gamma$ and $\Gamma '.$

Fig. 15.2

Joining together two copies of a metric graph

Consider the corresponding heat kernel $H_{\Gamma \cup _{-a,a} \Gamma } (t,x,y),$ which of course is positive, since the composite graph is a metric graph with standard vertex conditions at the vertices. Hence the solution to the heat equation on $\Gamma \cup _{-a,a} \Gamma$ is given by

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle u(t,x) & = & \displaystyle \int_{\Gamma \cup_{-a,a} \Gamma} H_{\Gamma \cup_{-a,a} \Gamma} (t,x,y) u_0 (y) dy \\[3mm] & = & \displaystyle \int_{\Gamma} H_{\Gamma \cup_{-a,a} \Gamma} (t,x,y) u_0(y) dy + \int_{\Gamma'} H_{\Gamma \cup_{-a,a} \Gamma} (t,x,y') u_0(y') dy', \end{array}\end{aligned}$$

where $u_0$ is the initial profile (15.3).

The following identities hold for the heat kernels:

$$\displaystyle \begin{aligned} {} \begin{array}{ccll} \displaystyle \hspace{6pt}H_\Gamma (t,x,y) & = & \displaystyle H_{\Gamma \cup_{-a,a} \Gamma} (t,x,y) + H_{\Gamma \cup_{-a,a} \Gamma}(t,x, \tau y), & \displaystyle x,y \in \Gamma; \\[3mm] \displaystyle H_{[-a,a]} (t,x,y) & = & \displaystyle H_{\Gamma \cup_{-a,a} \Gamma} (t,x,y) - H_{\Gamma \cup_{-a,a} \Gamma}(t,x, \tau y), & \displaystyle x,y \in [-a,a]. \end{array} \end{aligned} $$

(15.14)

To see this consider first the heat flow with the even initial profile $u_0 (x) = u_0(\tau x).$ The corresponding solution remains even and therefore restricted to $\Gamma$ coincides with the heat flow on the original graph $\Gamma$ with the initial data $u_0 (x), \; x \in \Gamma .$ To get the heat kernel associated with the Dirichlet operator on $[-a,a]$ consider the heat flow with the odd initial profile $u_0 (x) = - u_0(\tau x).$ The heat flow remains odd and therefore satisfies Dirichlet conditions at the contact vertices between $\Gamma$ and $\Gamma '.$

Taking into account that all kernels appearing in (15.14) are positive we conclude that

$$\displaystyle \begin{aligned} H_\Gamma (t,x,y) - H_{[-a,a]} (t,x,y) = 2 H_{\Gamma \cup_{-a,a} \Gamma}(t,x, \tau y) \geq 0, \; \; x,y \in [-a,a]\end{aligned}$$

and inequality (15.13) is proven. □

Our next step is to prove an upper estimate for the Laplacian heat kernel. We show first that the order of the singularity cannot be higher than those of the heat kernel for Laplacian on the line.

Lemma 15.5

The heat kernel $H_\Gamma (t,x,y)$ for the standard Laplacian on a metric finite compact graph $\Gamma$ satisfies the upper estimate

$$\displaystyle \begin{aligned} {} H_\Gamma (t, x,y) \leq C \frac{1}{\sqrt{t}}, \; \; 0 < t < 1, \end{aligned} $$

(15.15)

where $C$ is a certain constant depending on the graph.

Proof

We use the spectral decomposition for the standard Laplacian on $\Gamma$ and explicit representation for the heat kernel (15.5) leading to

$$\displaystyle \begin{aligned} H_\Gamma (t,x,y) \leq \sum_{n=1}^\infty e^{-\lambda_n t} \max_{x \in \Gamma} \vert \psi_n (x) \vert^2.\end{aligned}$$

We use the fact that the normalised eigenfunctions of the standard Laplacian are uniformly bounded (11.37)

$$\displaystyle \begin{aligned} \vert \psi_n (x) \vert \leq c ,\end{aligned}$$

where $c$ is independent of $x$ and $n$, and the lower estimate for the eigenvalues

$$\displaystyle \begin{aligned} \left(\frac{\pi}{{\mathcal{L}}} \right)^2 (n-M)^2 \leq \lambda_n\end{aligned}$$

to get

where the constant $C$ can be chosen equal to

$$\displaystyle \begin{aligned} C = 2 c^2 \left( 1+ \frac{{\mathcal{L}}}{\sqrt{4\pi}} \right).\end{aligned}$$

□

Note that the constant $C$ appearing in the proof of the lemma is far from being optimal, but it was not our goal to obtain the best constant. On the other hand it is very important to understand that the constant is in general different from $\frac {1}{\sqrt {4 \pi }}$ (appearing in the estimate for the heat kernel for the Dirichlet Laplacian on an interval). For example the heat kernel for the Neumann Laplacian on the interval $[-1,1]$ is given by the infinite series obtained by reflecting the point $y$ with respect to the boundary points $-1$ and 1

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle H^N_{[-1,1]} (t,x,y) & = & \displaystyle \frac{1}{\sqrt{4\pi t}} e^{- \vert x-y\vert^2/(4t)} + \frac{1}{\sqrt{4\pi t}} e^{- \vert x+y+2\vert^2/(4t)} \\[3mm] && \displaystyle + \frac{1}{\sqrt{4\pi t}} e^{- \vert x+y-2\vert^2/(4t)} + \dots , \end{array}\end{aligned}$$

and does not satisfy the upper estimate with the free heat kernel.

The obtained upper estimate is enough to prove that $\sqrt {t} H_\Gamma (t, x,x)$ tends to $1/{\sqrt {4 \pi }}$ for almost any $x.$

Lemma 15.6

Let $\Gamma$ be a finite compact metric graph with the vertex set ${\mathbf {V}}= \cup _{m=1}^M V^m .$ Then for any $z \in \Gamma \setminus {\mathbf {V}}$ the following limit holds

$$\displaystyle \begin{aligned} {} \lim_{t \rightarrow 0} \sqrt{t} \; H_\Gamma (t, x,x) = \frac{1}{\sqrt{4 \pi}}, \; \; x \in \Gamma \setminus {\mathbf{V}}. \end{aligned} $$

(15.16)

Proof (Following [155])

Consider any point $x \in \Gamma \setminus {\mathbf {V}}$ and let $a$ be its half-distance to the nearest vertex. Let us introduce the positive functions

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle f(y) & = & \displaystyle \left\{ \begin{array}{ll} \displaystyle H_{[x-a,x+a]} (t, x,y), & \displaystyle \mathrm{dist} \, (x, y) \leq a , \\[3mm] 0, & \mathrm{otherwise}; \end{array} \right. \\[5mm] \displaystyle g(y) & = & H_\Gamma (t,x,y), \end{array}\end{aligned}$$

where we consider the interval $[x-a, x+a]$ as a subset of $\Gamma .$ Estimate (15.13) implies that

$$\displaystyle \begin{aligned} f(y) \leq g(y).\end{aligned}$$

Moreover, conservation of heat implies

$$\displaystyle \begin{aligned} \int_\Gamma g(y) dy = \int_\Gamma H_\Gamma (t,x,y) dy= 1.\end{aligned}$$

Let $u(t,x)$ be a solution to the heat equation on $\Gamma$, then it holds

due to standard conditions at the vertices. More precisely we use just Kirchhoff conditions. Taking into account (15.7) we obtain

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \int_\Gamma (g(y) - f(y) ) dy & = & \displaystyle 1 - \int_{x-a}^{x+a} H_{[x-a, x+a]} (t,x,y) dy \\[3mm] & \leq & \displaystyle \frac{4 a}{\sqrt{\pi t}} e^{-a^2/(4t)}. \end{array}\end{aligned}$$

Moreover with the help of the identity (15.11) we get

$$\displaystyle \begin{aligned} \begin{array}{ccccl} \displaystyle 0 & \leq & H_\Gamma (2t, x, x) - H_{[x-a, x+a]} (2t, x, x) & = & \displaystyle \int_\Gamma (g(y) - f(y) )^2 dy \\[3mm] &&& \leq & \displaystyle \int_\Gamma (g(y) - f(y) ) 2 g(y) dy \\[3mm] &&& \leq & \displaystyle 2 \frac{C}{\sqrt{t}} \frac{4 a}{\sqrt{\pi t}} e^{-a^2/(4 t)}, \end{array}\end{aligned}$$

where we used (15.15) for $x=y$ and just proven integral inequality. Taking into account that $\frac {1}{\sqrt {t}} e^{-a^2/(4t)} \rightarrow 0$ as $t \rightarrow 0 ,$ we calculate the limit using (15.12). □

The statement of the lemma holds for almost any $x$, since the vertices have measure zero in the metric graph $\Gamma .$ It is very important to notice that the limit does not depend on the point $x \in \Gamma \setminus {\mathbf {V}},$ also the rate of convergence may be different.

15.1.3 On Schrödinger Semigroups

In what follows we shall need formula (15.23) below providing the first order correction for the trace of the heat semigroup in terms of the perturbation $q.$ Also this is a standard fact, we indicate its proof here.

We start by differentiating

$$\displaystyle \begin{aligned} \frac{d}{dt} e^{-L_q t} e^{L t} = e^{-L_q t} (-L_q) e^{Lt} + e^{-L_qt} L e^{Lt} = - e^{-Lt} q e^{Lt}\end{aligned}$$

$$\displaystyle \begin{aligned} \Rightarrow e^{-L_qt} e^{L t} - I = - \int_0^t e^{-L_qs} q e^{Ls} ds,\end{aligned}$$

implying

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle e^{-L_qt} & = & \displaystyle e^{-Lt} - \int_0^t e^{-L_qs} q e^{L(s-t)} ds\\[3mm] & = & \displaystyle \displaystyle e^{-Lt} - \int_0^t e^{-L_q(t-s)} q e^{-Ls} ds , \end{array}\end{aligned}$$

where we changed variables $s \mapsto t-s$.

In a similar way we get

$$\displaystyle \begin{aligned} e^{-L_qt} = \displaystyle e^{-Lt} - \int_0^t e^{-L(t-s)} q e^{-L_qs} ds .\end{aligned}$$

Iterating the equation once we get

(15.17)

The operator $A_1(t)$ defined above can be considered as an integral operator with the following kernel

$$\displaystyle \begin{aligned} L (t,x,y) = \int_{s=0}^t \int_{z \in \Gamma} H_\Gamma (t-s,x,z) q(z) H_\Gamma (s,z,y) dz ds .\end{aligned}$$

For essentially bounded potentials taking into account positivity of the free heat kernel $H_\Gamma$ we get

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \displaystyle \vert L(t,x,y) \vert & \leq & \| q \|{}_\infty \int_{s=0}^t \int_{z \in \Gamma} H_\Gamma (t-s,x,z) H_\Gamma (s,z,y) dz ds\\[3mm] & = & \displaystyle \| q \|{}_\infty \int_{s=0}^t H_\Gamma (t,x,y) ds \\[3mm] & \leq & \| q \|{}_\infty C t^{1/2}, \end{array} \end{aligned} $$

(15.18)

where we used the estimate (15.15) and the fact that $H_\Gamma (t,x,y)$ is a kernel for a semigroup

$$\displaystyle \begin{aligned} \int_{z \in \Gamma} H_\Gamma (t-s,x,z) H_\Gamma (s,z,y) dz = H_\Gamma (t,x,y).\end{aligned}$$

Note that (15.18) is valid for $t < 1$ and almost everywhere with respect to $x$ and $y,$ more precisely, for all $x$ and $y$ not belonging to a vertex.

Continuing iterations one obtains the following formal series

(15.19)

To prove convergence of this formal series one may introduce the integral kernel $L_m (t,x,y)$ associated with $A_m$ and use estimates similar to ones we already carried out in order to get (15.18). Really, using positivity of the heat kernel, the semigroup property and essential boundedness of the potential, we get

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \vert L_m (t,x,y) \vert & \leq & \displaystyle \| q \|{}_\infty^m \int_0^t ds_1 \int_0^{s_1} ds_2 \dots \int_0^{s_{m-1}} ds_m H_\Gamma (t,x,y) \\[3mm] & \leq & \displaystyle \| q \|{}_\infty^m \frac{t^m}{m!} H_\Gamma (t,x,y) \\[3mm] & \leq & \displaystyle C \frac{t^{m-1/2}}{m!} \|q \|{}_\infty^m . \end{array} \end{aligned}$$

It follows that even the perturbed semigroup can be considered as an integral operator with a certain kernel $H_{\Gamma ,q} (t,x,y) , \; x, y \in \Gamma \setminus {\mathbf {V}},$ satisfying the uniform estimate

$$\displaystyle \begin{aligned} H_{\Gamma,q}(t,x,y) \leq C t^{-1/2} \end{aligned} $$

(15.20)

with a maybe different constant $C.$ We return back to formula (15.17) in order to prove that the second integral is of order $t^{3/2}= t^{2-1/2}.$ Really, the kernel $M(t,x,y)$ of the operator $B(t)$ is given by the integral

$$\displaystyle \begin{aligned} {} \begin{array}{cl} & \displaystyle \vert M(t,x,y) \vert \\[3mm] = & \displaystyle \left\vert \hspace{-1pt}\int_0^t\hspace{-2pt} ds \hspace{-2pt}\int_0^s\hspace{-1pt} du \hspace{-1pt}\int_\Gamma\hspace{-1pt} dw \hspace{-1pt}\int_\Gamma dz H_\Gamma (t{-}s, x, z) q(z) H_{\Gamma,q}(s{-}u, z, w) q(w) H_\Gamma (u, w, y)\right\vert \\[3mm] \leq & \displaystyle \| q \|{}_\infty^2 \int_0^t ds \int_0^s du \int_\Gamma dw \int_\Gamma dz H_\Gamma (t-s, x, z) H_{\Gamma,q}(s-u, z, w) H_\Gamma (u, w, y)\\[3mm] \leq & \displaystyle \| q \|{}_\infty^2 e^{\|q \|{}_\infty t} \hspace{-1pt}\int_0^t\hspace{-1pt} ds \hspace{-1pt}\int_0^s\hspace{-1pt} du \hspace{-1pt}\int_\Gamma\hspace{-1pt} dw \hspace{-1pt}\int_\Gamma\hspace{-1pt} dz H_\Gamma (t{-}s, x, z) H_\Gamma(s{-}u, z, w) H_\Gamma (u, w, y) \\[3mm] = & \displaystyle \| q \|{}_\infty^2 e^{\|q \|{}_\infty t} \int_0^t ds \int_0^s du \; H_\Gamma (t,x,y) \\[3mm] = & \displaystyle \| q \|{}_\infty^2 e^{\|q \|{}_\infty t} \; \frac{t^2}{2} \; H_\Gamma (t,x,y) \\[3mm] \leq & \displaystyle C \frac{ \| q \|{}_\infty^2}{2} e^{\|q \|{}_\infty t} t^{3/2} , \end{array} \end{aligned} $$

(15.21)

where we used the estimate

$$\displaystyle \begin{aligned} L - \| q \|{}_\infty \leq L_q \leq L + \| q \|{}_\infty \; \; \Rightarrow \; \; e^{-\| q \|{}_\infty t} e^{-L t} \leq e^{-L_q t} \leq e^{\| q \|{}_\infty t} e^{-L t},\end{aligned}$$

implying in particular that the Schrödinger heat kernel is positive. Now formula (15.17) can be written using integral kernels as follows

$$\displaystyle \begin{aligned} {} \begin{array}{cl} & \displaystyle H_{\Gamma,q}(t,x,y) \\[3mm] = & \displaystyle H_\Gamma (t, x,y) - \int_0^t \int_\Gamma H_\Gamma (t-s, x, z) q(z) H_\Gamma (s,z,y) dz ds \\[3mm] & \displaystyle + \hspace{-1pt}\int_0^t\hspace{-2pt} \int_0^s\hspace{-2pt} \int_\Gamma\hspace{-2pt} \int_\Gamma\hspace{-2pt} H_\Gamma (t{-}s, x, z) q(z) H_{\Gamma,q}(s{-}u, z, w) q(w) H_\Gamma (u, w, y) dz dw du ds . \end{array} \end{aligned} $$

(15.22)

Lemma 15.7

The difference between the traces of the perturbed (Schrödinger) and unperturbed (Laplacian) semigroups satisfies

$$\displaystyle \begin{aligned} {} \mathrm{tr} \left[ e^{-L_qt}\right] - \mathrm{tr}\,\left[e^{-Lt}\right] = - t \int_\Gamma H_\Gamma (t,x,x) q(x) dx + \rho(t), \end{aligned} $$

(15.23)

where $\rho (t) = {\mathcal {O}}(t^{3/2}).$

Proof

We put $x= y$ in formula (15.22) and integrate with respect to $x$

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \mathrm{tr} \left[ e^{-L_qt}\right] & = & \displaystyle \mathrm{tr}\,\left[e^{-Lt}\right] - \int_0^t \int_\Gamma \int_\Gamma H_\Gamma (t-s, x, z) q(z) H_\Gamma (s, z , x) dz dx ds \\[3mm] & & \displaystyle + \rho (t) \\[3mm] & = & \displaystyle \mathrm{tr}\,\left[e^{-Lt}\right] - \int_0^t \int_\Gamma H_\Gamma (t, z, z) q(z) dz ds + \rho (t) \\[3mm] & = & \displaystyle \mathrm{tr}\,\left[e^{-Lt}\right] - t \int_\Gamma H_\Gamma (t, x, x) q(x) dx + \rho (t), \\[3mm]\end{array} \end{aligned} $$

(15.24)

where we used that $H_\Gamma$ is a kernel of a semigroup. Here $\rho (x)$ denotes the integral corresponding to the last term in (15.22). Its estimate as ${\mathcal {O}}(t^{3/2})$ follows directly from (15.21). □

We prepared all tools that are needed to prove the main result of this section—a direct generalisation of Ambartsumian theorem for the case where the fixed metric graph is not necessarily an interval.

15.1.4 A Theorem by Davies

We start by proving a direct analog of the original Ambartsumian theorem.

Theorem 15.8 (Davies [155])

Let $\Gamma$ be a finite compact metric graph. Then the standard Schrödinger operator is isospectral to the standard Laplacian on the same metric graph $\Gamma$

$$\displaystyle \begin{aligned} \lambda_n (L_q^{\mathrm{st}} (\Gamma)) = \lambda_n (L^{\mathrm{st}} (\Gamma)), \end{aligned} $$

(15.25)

for all $n = 1,2, \dots ,$ if and only if $q(x) \equiv 0$ almost everywhere.

Proof

The traces of the Schrödinger and Laplacian semigroups coincide, since the operators are isospectral. Then formula (15.23) reads as follows

$$\displaystyle \begin{aligned} 0 = - t \int_\Gamma H_\Gamma (t,x,x) q(x) dx + \rho(t).\end{aligned}$$

We divide by $\sqrt {t}$ and take the limit as $t \rightarrow 0$. Formula (15.15) implies that $\sqrt {t} H _\Gamma (t,x,x)$ is uniformly bounded, moreover for almost every $x$ it holds (see (15.16))

$$\displaystyle \begin{aligned} \lim_{t \rightarrow 0} \sqrt{t} \; H_\Gamma (t, x,x) = \frac{1}{\sqrt{4 \pi}} ,\end{aligned}$$

implying that

$$\displaystyle \begin{aligned} \int_\Gamma q(x) dx = 0 ,\end{aligned}$$

since $\frac {1}{\sqrt {t}} \rho (t) = O (t).$ It follows from Theorem 15.1 that $q$ is zero almost everywhere, since $\lambda _1 (L^{\mathrm {st}} (\Gamma ) = 0 \Rightarrow \lambda _1 (L_q^{\mathrm {st}} (\Gamma ) = 0$. □

To prove the theorem it was important that the limit (15.16) does not depend on the point $x \in \Gamma .$ This theorem can be strengthened as follows.

Theorem 15.9 (Davies [155])

Let $\Gamma$ be a finite compact metric graph. Assume that the eigenvalues of the standard Schrödinger and Laplace operators satisfy

• 1. $\lambda _1 (L^{\mathrm {st}}_q (\Gamma )) \geq 0;$

• 2. $\displaystyle \limsup \limits _{n \rightarrow \infty } (\lambda _n (L^{\mathrm {st}}_q (\Gamma )) - \lambda _n (L^{\mathrm {st}}_0 (\Gamma ))) \leq 0 ,$

then $q(x)$ is equal to zero almost everywhere.

Proof

To prove the theorem using the same method it is enough to show that

$$\displaystyle \begin{aligned} {} \limsup\limits_{t \rightarrow 0} \frac{1}{\sqrt{t}} \left( \mathrm{tr} \left[ e^{-Lt}\right] - \mathrm{tr}\,\left[e^{-L_qt}\right] \right) \leq 0, \end{aligned} $$

(15.26)

or equivalently

$$\displaystyle \begin{aligned} \limsup\limits_{t \rightarrow 0} \frac{1}{\sqrt{t}} \sum_{n=1}^\infty \left(e^{- \lambda_n (L^{\mathrm{st}}_0)t}- e^{- \lambda_n (L^{\mathrm{st}}_q)t} \right)\leq 0 .\end{aligned}$$

Condition 2 implies that given $\epsilon > 0$ there exists $N = N(\epsilon )$, such that $\lambda _n (L_q) - \lambda _n (L) \leq \epsilon$ for all $n \geq N.$ We use the estimates

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \frac{1}{\sqrt{t}} \sum_{n=1}^{N-1} \left(e^{- \lambda_n (L^{\mathrm{st}}_0)t}- e^{- \lambda_n (L^{\mathrm{st}}_q)t} \right) & \leq & \displaystyle \sqrt{t} \sum_{n=1}^{N-1} \vert \lambda_n (L_q) - \lambda_n (L) \vert; \\[6mm] \displaystyle \frac{1}{\sqrt{t}} \sum_{n=N}^\infty \left(e^{- \lambda_n (L^{\mathrm{st}}_0)t}- e^{- \lambda_n (L^{\mathrm{st}}_q)t} \right) & \leq & \displaystyle \frac{1}{\sqrt{t}} \sum_{n=N}^\infty e^{-\lambda_n (L)} (1- e^{-\epsilon t }) \\[3mm] & \leq & \displaystyle \frac{\epsilon t}{\sqrt{t}} \sum_{n=1}^\infty e^{-\lambda_n (L) t} \\[3mm] & = & \displaystyle \epsilon \sqrt{t} \mathrm{Tr}\, (e^{-H_0t}) \\[3mm] & \leq & \displaystyle C {\mathcal{L}} \epsilon, \end{array}\end{aligned}$$

for a certain $C > 0.$ We conclude that

$$\displaystyle \begin{aligned} \lim_{t \rightarrow 0} \frac{1}{\sqrt{t}} \sum_{n=1}^\infty \left(e^{- \lambda_n (L^{\mathrm{st}}_0)t}- e^{- \lambda_n (L^{\mathrm{st}}_q)t} \right) \leq C {\mathcal{L}} \epsilon ,\end{aligned}$$

where one may need to adjust the constant $C$. Therefore estimate (15.26) holds. □

Proven theorem implies that zero potential, or more generally any constant potential, possesses unique properties allowing one to single out the spectrum of the Laplacian among the spectra of Schrödinger operators on a metric graph. The reason the spectrum of the Laplacian is unique is that it is given by zeroes of a certain generalised trigonometric polynomial. We have already used this fact proving Theorem 14.11.

15.2 On Asymptotically Isospectral Quantum Graphs

Our goal in this section is to prove several geometric versions of Ambartsumian’s theorem without assuming that the underlying graph is just the interval. The main analytic tool will be the theory of almost periodic functions (see for example [92]), but we do not require from the reader any knowledge of this wonderful theory—all results will be proven using well-known facts.

15.2.1 On the Zeroes of Generalised Trigonometric Polynomials

Our analysis is based on the fact that the spectrum of a scaling invariant Laplacian on a finite compact metric graph is given by zeroes of a generalised trigonometric polynomial (see Theorem 6.1). Our first step is to prove that generalised trigonometric polynomials with real $\omega _j$ determine holomorphic almost periodic functions [92] in a strip along the real axis. This is not surprising, since one way to define almost periodic functions is to consider their approximations via generalised trigonometric polynomials.

Lemma 15.10

For any generalised trigonometric polynomial

$$\displaystyle \begin{aligned} p =\sum_{j=1}^{J}p_{j} e^{i \omega_{j} k},\end{aligned}$$

one may find shifts $t (\delta ) \rightarrow \infty$ , as $\delta \rightarrow 0$ , such that

$$\displaystyle \begin{aligned} {} \vert p (k+t (\delta)) - p (k) \vert \leq \delta \end{aligned} $$

(15.27)

for all $k \in \mathbb {C}, \vert \mathit{\mbox{Im}} \, k \vert < 1.$

Proof

Lemma 14.9 implies that for any $\epsilon >0$ one may choose $t (\epsilon )$ such that:

$$\displaystyle \begin{aligned} \vert e^{i(k+t (\epsilon)) \omega_{j}} - e^{i k \omega_{j}} \vert < \epsilon\end{aligned}$$

for all $\omega _{j}$ and $k: \vert \mbox{Im} \, k \vert < 1$. Then

$$\displaystyle \begin{aligned} \vert p (k+t (\epsilon)) - p (k) \vert \leq \left(\sum_{j=1}^{J} \vert p_{j} \vert \right) \epsilon.\end{aligned}$$

Choosing $\displaystyle \epsilon (\delta )=\frac {\delta }{ \sum _{j=1}^{J} \vert p_{j} \vert \, }$, the corresponding sequence $t (\epsilon (\delta ))$ satisfies the claim of the Lemma. □

The above Lemma can be generalised for a finite set of generalised trigonometric polynomials, since the key point in the proof is Lemma 14.9 holds whenever the set of frequencies $\omega ^\ell _j$ is finite.

Lemma 15.11

For any finite set of generalised trigonometric polynomials

$$\displaystyle \begin{aligned} p^\ell=\sum_{j=1}^{J_\ell}p^\ell_j e^{i \omega^\ell_j k},\quad \quad \ell = 1,2, \dots, L, \end{aligned}$$

one may find common shifts $t (\delta ) \rightarrow \infty$ , as $\delta \rightarrow 0$ , such that

$$\displaystyle \begin{aligned} {} \vert p^\ell (k+t (\delta)) - p^\ell(k) \vert \leq \delta \end{aligned} $$

(15.28)

for all $k \in \mathbb {C}, \vert \mathit{\mbox{Im}} \, k \vert < 1.$

We will use the above Lemma for two generalised trigonometric polynomials together with their derivatives, which is permitted since the derivative of a generalised trigonometric polynomial is of course a generalised trigonometric polynomial. The following Theorem is our main analytic tool for this section.

Theorem 15.12

Let p and $q$ be generalised trigonometric polynomials

$$\displaystyle \begin{aligned} p (k) = \sum_{i=1}^{J_1} p_i e^{i \omega_i k}, \; \ q (k) = \sum_{j=1}^{J_2} q_j e^{i \nu_j k} \end{aligned} $$

(15.29)

with real zeros $k_n,\, l_m$ , respectively. If there exists a subsequence $l_{m_n}$ of $l_n$ such that

$$\displaystyle \begin{aligned} {} \lim_{n \rightarrow \infty} (k_n- l_{m_n}) = 0, \end{aligned} $$

(15.30)

then all the zeros of $p$ are zeros of q with at least the same multiplicity.

Proof

Let $k_0$ be any zero of p, and denote its order by $m_0$. We may find an $\epsilon >0$ such that p has no other zeros in the disc $B_{2 \epsilon }(k_0)$ of radius $2\epsilon$ centered at $k_0$, and such that $p,p',q,$ and $q'$ have no zeros on the boundaries of $B_{\epsilon }(k_0)$ and $B_{2 \epsilon }(k_0)$. This choice implies that there are common constants $0<c,C$ such that

$$\displaystyle \begin{aligned} {} c<|p(k)|,|p'(k)|,|q(k)|,|q'(k)|<C, \end{aligned} $$

(15.31)

on the boundaries $\partial B_{\epsilon }(k_0)$, $\partial B_{2 \epsilon }(k_0)$.

Since there are no other zeroes inside the circles we have

$$\displaystyle \begin{aligned} \int_{\partial B_\epsilon (k_0) } \frac{p'(k)}{p(k)} dk = \int_{\partial B_{2 \epsilon} (k_0) } \frac{p'(k)}{p(k)} dk = 2 \pi i m_0, \end{aligned} $$

(15.32)

and

$$\displaystyle \begin{aligned} \int_{\partial B_{2 \epsilon} (k_0) } \frac{q'(k)}{q(k)} dk = 2 \pi i m', \end{aligned} $$

(15.33)

for some $m' \in \mathbb {N}$. We now choose a $\delta$ satisfying

$$\displaystyle \begin{aligned} {} \delta< \min \left\{ \frac{1}{16}\frac{c^2}{C \epsilon}, \frac{c}{2} \right \} \end{aligned} $$

(15.34)

(the specific choice of constants plays a role in the estimates of the integrals below (15.35)) and a common shift $t(\epsilon ,\delta )$ such that

• $|k_n-l_{m_n}|<\epsilon , \quad \text{ for } \quad k_n, \, l_{m_n}>k_0+t(\epsilon ,\delta )-2\epsilon .$

• $|p^{(j)}(k+t(\epsilon ,\delta ))-p^{(j)}(k)|<\delta , \quad j=0,1.$

• $|q^{(j)}(k+t(\epsilon ,\delta ))-q^{(j)}(k)|<\delta , \quad j=0,1.$

We first show that with this choice of $\delta$ and $t(\epsilon ,\delta )$ the integrals over the boundaries of the shifted discs $B_{\epsilon } (k_0+ t(\epsilon , \delta )), B_{2 \epsilon } (k_0+ t(\epsilon , \delta ))$ and the unshifted discs are equal. Consider the difference:

$$\displaystyle \begin{aligned} \begin{array}{rcl} \int_{\partial B_\epsilon (k_0)} \frac{p'(k)}{p(k)} dk & -&\displaystyle \int_{\partial B_\epsilon (k_0+ t(\epsilon, \delta))} \frac{p'(k)}{p(k)} dk \\ & =&\displaystyle \displaystyle \int_{\partial B_\epsilon (k_0)}\left(\frac{p'(k)}{p(k)} - \frac{p'(k+t(\epsilon, \delta))}{p(k+t(\epsilon, \delta))} \right). \end{array} \end{aligned} $$

We estimate:

$$\displaystyle \begin{aligned} {} \begin{array}{cl} & \displaystyle \left\vert \int_{\partial B_\epsilon (k_0)}\left(\frac{p'(k)}{p(k)} - \frac{p'(k+t(\epsilon, \delta))}{p(k+t(\epsilon, \delta))} \right)dk \right\vert \\[5mm] \leq & \displaystyle \int_{\partial B_\epsilon (k_0)} \Bigg\{ \frac{ \vert p'(k) - p'(k+t(\epsilon, \delta)) \vert \; \vert p(k+t(\epsilon, \delta)) \vert }{\vert p(k) \vert \, \vert p(k+t(\epsilon, \delta)) \vert} \\[5mm] & \displaystyle \quad \quad \quad +\frac{ \vert p'(k+t (\epsilon)) \vert \; \vert p (k+t(\epsilon, \delta)) - p(k) \vert}{\vert p(k) \vert \, \vert p(k+t(\epsilon, \delta)) \vert}\Bigg \}dk \\[5mm] \leq & \displaystyle 2 \pi \epsilon \left(\frac{\delta (C+ \delta)}{c(c-\delta)} +\frac{(C+ \delta) \delta}{c (c-\delta)}\right) = \epsilon \frac{4 \pi \delta (C+\delta))}{c(c-\delta)} \\[3mm] \leq & \displaystyle \epsilon \frac{4 \pi \delta 2C}{c^2/2} = \epsilon \frac{16 \pi C}{c^2} \delta < \pi, \end{array} \end{aligned} $$

(15.35)

where we used that (15.34) implies $\delta < c/2$ and $\delta < \frac {1}{16}\frac {c^2}{C \epsilon }.$ The reason to choose $\delta$ satisfying (15.34) is clear now. It follows that the difference between the two integrals is less than $\pi$ and therefore they are equal, so

$$\displaystyle \begin{aligned} \int_{\partial B_\epsilon (k_0+ t(\epsilon, \delta)) } \frac{p'(k)}{p(k)} dk = 2 \pi i m_0, \end{aligned} $$

(15.36)

and $m_0$ zeroes of $p (k)$ lie inside the shifted disc $B_\epsilon (k_0+ t(\epsilon , \delta ))$. In the same way one shows that

$$\displaystyle \begin{aligned} \int_{\partial B_{2 \epsilon} (k_0+ t(\epsilon, \delta))} \frac{p'(k)}{p(k)} dk=2\pi i m_0, \quad \int_{\partial B_{2 \epsilon} (k_0+ t(\epsilon, \delta))} \frac{q'(k)}{q(k)} dk=2 \pi i m',\end{aligned}$$

so p has $m_0$ zeros in $B_{2 \epsilon }(k_0+t(\epsilon ,\delta ))$ all of which are actually contained in $B_{\epsilon }(k_0+t(\epsilon ,\delta ))$, and q has $m'$ zeros in $B_{2 \epsilon }(k_0+t(\epsilon ,\delta ))$.

By the choice of $t(\epsilon ,\delta )$ to every zero of $p$ inside $B_{\epsilon }(k_0 + t(\epsilon ,\delta ))$ corresponds a zero of $q$ lying at a distance less than $\epsilon .$ All such zeroes lie inside the ball of double radius $B_{2\epsilon }(k_0 + t(\epsilon ,\delta ))$, hence $m' \geq m_0.$

Letting $\epsilon \rightarrow 0$ — while choosing suitable $\delta$’s and shifts $t(\epsilon ,\delta )$ — we conclude that $k_0$ is a zero of q of multiplicity at least $m_0$. □

The above theorem holds even for holomorphic almost periodic functions [357]. By applying Theorem 15.12 twice we obtain the following statement as a special case.

Theorem 15.13

Let $p,q$ be generalised trigonometric polynomials

$$\displaystyle \begin{aligned} p (k) = \sum_{i=1}^{\infty} p_i e^{i \omega_i k}, \; \ q (k) = \sum_{j=1}^{\infty} q_j e^{i \nu_j k} \end{aligned} $$

(15.37)

with zeros $k_n,\, l_n$ respectively, if

$$\displaystyle \begin{aligned} {} \lim_{n \rightarrow \infty} (k_n- l_n) = 0, \end{aligned} $$

(15.38)

then the zeros of the functions are identical.

This direction of research has already been continued in [220] where uniqueness theorems for Fourier quasicrystals were established.

15.2.2 Asymptotically Isospectral Quantum Graphs

In this subsection we will apply Theorem 15.13 to the spectral theory of quantum graphs. As was shown in Theorem 6.1 the spectrum of $L^{{\mathbf {S}}}(\Gamma )$ for a finite compact quantum graph and scaling invariant vertex conditions ${{\mathbf {S}}}$ is given by the zeros of a generalised trigonometric polynomial.

We shall also use the notion of asymptotically isospectral quantum graphs given in Definition 11.7. We start by showing that two asymptotically isospectral scaling invariant Laplacians are in fact isospectral. This implies that the spectrum of scaling invariant Laplacians possesses certain rigidity, so that it is determined by the asymptotics.

Theorem 15.14

Let $L^{{\mathbf {S}}_1} (\Gamma _1 )$ and $L^{{\mathbf {S}}_2} (\Gamma _2 )$ be two Laplace operators defined on finite compact metric graphs $\Gamma _1$ and $\Gamma _2$ by certain scaling invariant vertex conditions given by ${\mathbf {S}}_1$ and ${\mathbf {S}}_2$ respectively. If the operators are asymptotically isospectral, then they are isospectral.

Proof

Scaling invariant Laplacians are non-negative operators, which is easily seen from their quadratic forms given by Dirichlet integrals. The positive eigenvalues are given by certain generalised trigonometric polynomials (Theorem 6.1). Then Theorem 15.13 implies that the zeroes of these polynomials coincide, since they are asymptotically close. We have thus proven that all positive eigenvalues coincide.

It remains to show that the multiplicities of the eigenvalue zero coincide.⁴ But this trivially follows from the fact that the lowest non-zero eigenvalues not only coincide but have the same index. □

Taking into account the above theorem one may think that the notion of asymptotic isospectrality is redundant. This is not completely true, for example Schrödinger operator on an interval is asymptotically isospectral but not isospectral to the Laplacian, unless the potential is identically zero. More generally, Theorem 11.9 implies that Schrödinger operator $L_q^{{\mathbf {S}}} (\Gamma )$ is asymptotically isospectral to the reference Laplacian $L^{{\mathbf {S}}(\infty )} (\Gamma ^\infty ) .$

Two asymptotically isospectral Schrödinger operators are not necessarily isospectral—Theorem 15.14 does not hold for Schrödinger operators. On the other hand one may generalise it by proving that the corresponding reference Laplacians are isospectral.

Theorem 15.15

Let $L_{q_1}^{{{\mathbf {S}}}_1}(\Gamma _1)$ and $L_{q_2}^{{{\mathbf {S}}}_2}(\Gamma _2)$ be two Schrödinger operators on finite compact metric graphs $\Gamma _i$ with $q_i \in L_1(\Gamma _i)$ and vertex conditions determined by unitary matrices ${{\mathbf {S}}}_i$ , for $i=1,\, 2$ . Suppose that the operators are asymptotically isospectral, then the corresponding reference Laplacians $L^{{{\mathbf {S}}}_1(\infty )}(\Gamma _1^{\infty })$ and $L^{{{\mathbf {S}}}_1(\infty )}(\Gamma _1^{\infty })$ are isospectral.

Proof

In accordance with Theorem 11.9 the operators

$$\displaystyle \begin{aligned} \begin{array}{ccc} \displaystyle L_{q_1}^{{{\mathbf{S}}}_1}(\Gamma_1) & \quad \mbox{and} \quad & \displaystyle L^{{{\mathbf{S}}}_1(\infty)}(\Gamma_1^{\infty}) , \\[3mm] \displaystyle L_{q_2}^{{{\mathbf{S}}}_2}(\Gamma_2) & \quad \mbox{and} \quad & \displaystyle L^{{{\mathbf{S}}}_2(\infty)}(\Gamma_2^{\infty}) \end{array}\end{aligned}$$

are pairwise asymptotically isospectral. One of the conditions in the current theorem is that the Schrödinger operators $L_{q_1}^{{{\mathbf {S}}}_1}(\Gamma _1)$ and $L_{q_2}^{{{\mathbf {S}}}_2}(\Gamma _2)$ are asymptotically isospectral, hence the reference Laplacians are also asymptotically isospectral. Then Theorem 15.14 implies that they are in fact isospectral. □

The above theorem does not imply that the underlying reference graphs $\Gamma _i^\infty$ coincide, since there exists isospectral scaling invariant or even standard Laplacians. The theorem just implies that the reference Laplacians belong to the same isospectral class.

One may strengthen the above theorem by assuming that two Schrödinger operators are not necessarily asymptotically isospectral, but just have asymptotically close spectra. The only new point is that the multiplicities of zero as the eigenvalue of the two reference Laplacians may be different.

Theorem 15.16

Let $L_{q_1}^{{{\mathbf {S}}}_1}(\Gamma _1)$ and $L_{q_2}^{{{\mathbf {S}}}_2}(\Gamma _2)$ be two Schrödinger operators on finite compact metric graphs $\Gamma _i$ with $q_i \in L_1(\Gamma _i)$ and vertex conditions determined by unitary matrices ${{\mathbf {S}}}_i$ , for $i=1,\, 2$ . Suppose that their spectra

$$\displaystyle \begin{aligned} \{k^2_n\}=\Sigma(L_{q_1}^{{{\mathbf{S}}}_1}(\Gamma_1)), \mathit{\text{ and }} \, \, \, \{l_n^2\}=\Sigma(L_{q_2}^{{{\mathbf{S}}}_2}(\Gamma_2)),\end{aligned}$$

are asymptotically close in the sense that

$$\displaystyle \begin{aligned} {} k_{n+m}-l_n \rightarrow 0, \mathit{\text{ as }} n \rightarrow \infty \end{aligned} $$

(15.39)

for some$m \in \mathbb {N}$. Then all non-zero eigenvalues of the corresponding reference Laplacians$L^{{{\mathbf {S}}}_1(\infty )}(\Gamma _1^{\infty })$and$L^{{{\mathbf {S}}}_2(\infty )}(\Gamma _2^{\infty })$coincide and m is the difference in the multiplicity of the eigenvalue 0 of the operators.

Problem 71

Prove Theorem 15.16 in full details.

The above Theorem cannot be strengthened by showing that the eigenvalue 0 is of the same multiplicity. Consider the circle graph ${{\mathbf {S}}}^1$ of length $2 \pi$ with one vertex with standard conditions and the graph $\Gamma$ consisting of two disjoint intervals of length $\pi$ with standard conditions (i.e. Neumann) at all endpoints. Then $\Sigma (L^{\mathrm {st}}({{\mathbf {S}}}^1))=0,1,1,2^2,2^2,\dots$ while $\Sigma (L^{\mathrm {st}} (\Gamma ))=0,0,1,1,2^ 2,2^2,\dots$. All non-zero eigenvalues coincide, but the multiplicity of the eigenvalue $\lambda = 0$ is determined by the number of connected components in the corresponding graph.

15.2.3 When a Schrödinger Operator Is Isospectral to a Laplacian

Our studies of asymptotically isospectral graphs lead us to the following unexpected generalisation of the Davies theorem (Theorem 15.8).

Theorem 15.17

Let$\Gamma _1$be a finite compact metric graph,$q \in L_{\infty }(\Gamma _1)$and suppose that$\Sigma (L_q^{\mathrm {st}}(\Gamma _1))=\Sigma (L_0^{\mathrm {st}}(\Gamma _2))$for some (may be different) finite compact$\Gamma _2$. Then$q (x) \equiv 0$.

Proof

Theorem 15.15 implies that $\Sigma (L_0^{\mathrm {st}}(\Gamma _1))=\Sigma (L_0^{\mathrm {st}}(\Gamma _2))$, since standard conditions are scaling invariant, and therefore $\Sigma (L_q^{\mathrm {st}}(\Gamma _1))=\Sigma (L_0^{\mathrm {st}}(\Gamma _1))$ and from Theorem 15.8 we obtain $q (x) \equiv 0$. □

Note that we do not claim that the graphs $\Gamma _1$ and $\Gamma _2$ coincide—just the corresponding standard Laplacians are isospectral. On the other hand, there exist metric graphs that are uniquely determined by the spectrum of the corresponding standard Laplacians (see for example Sect. 9.4), for such graphs we may conclude in addition that $\Gamma _1 = \Gamma _2.$

Theorem 15.18

Let$\Gamma$be a finite compact metric graph with rationally independent edge lengths. Then the spectrum of the standard Schrödinger operator$L_q^{\mathrm {st}} (\Gamma )$with$q \in L_1(\Gamma )$determines the unique metric graph$\Gamma$.

The theorem is an easy corollary of Theorems 9.11 and 15.17.

The above theorems show another one time that standard vertex conditions and zero potential are exceptional for the inverse problem.

Problem 72

Is it possible to strengthen the above theorem by assuming that the vertex conditions in the Schrödinger operator are just asymptotically standard (instead of standard conditions)?