2.1 Eigenvalues on a caterpillar graph

Let us consider the problem $B:=B(q)$ on $G(V,E)$ for equation (2.1) with matching conditions (2.2) and (2.3) in the internal vertices, as well as Neumann conditions at the pendant vertices $v_1,v_2,v_5,v_6$.

Denote by $S_j(x, \lambda )$ and $C_j(x, \lambda ), j = \overline {1, 5}$, the solutions of the equation (2.1) on the edge $e_j$ satisfying the initial conditions

\[ S_j(0, \lambda) = C_j ' (0, \lambda) = 0,\quad S_j ' (0, \lambda) = C_j(0, \lambda) = 1. \]

For each fixed $x \in [0, 1]$, the functions $S_j^{(v)}(x, \lambda )$ and $C_j^{(v)}(x, \lambda ), j = \overline {1, 5}, v= 0, 1$, are entire in $\lambda$ of the order $\frac {1}{2}$. Moreover, one gets (see, e.g., [Reference Freiling and Yurko10, Chap. 1] for details) the following asymptotical formulae as $|\lambda | \to \infty$, uniformly in $x \in [0, 1]$:

(2.4)\begin{align} S_j(x, \lambda)& =\frac{\sin\rho x}{\rho}-\frac{\cos \rho x}{\rho^{2}}[q_j]_x+\frac{\kappa_1(\rho)}{\rho^{2}}, \end{align}

(2.5)\begin{align} S_j'(x, \lambda)& =\cos\rho x+\frac{\sin \rho x}{\rho}[q_j]_x+\frac{\kappa_2(\rho)}{\rho}, \end{align}

(2.6)\begin{align} C_j(x, \lambda)& =\cos\rho x+\frac{\sin \rho x}{\rho}[q_j]_x+\frac{\kappa_3(\rho)}{\rho}, \end{align}

(2.7)\begin{align} C_j'(x, \lambda)& ={-}\rho\sin\rho x+[q_j]_x\cos\rho x+\kappa_4(\rho), \end{align}

where $\lambda = \rho ^{2}$, $[q_j]_x=\frac {1}{2}\int _0^{x}q_j(t)dt$ and $\tau = {\textrm {Im}} \rho$, $\kappa _i(\rho )=o(1)$ ($i=1,2,3,4$) for large real $\rho$, and $\kappa _i\in \mathcal {L}^{x}$ ($\mathcal {L}^{x}$ is the class of entire functions of exponential potential type less than $x$, belonging to $L^{2}(\mathbb {R})$ for real $\rho$).

Put

\[ Y(x,\lambda)=\{y_i(x)\}_{i=\overline{1,5}},\quad y_i(x) = \begin{cases} A_i(\lambda)C_i(x,\lambda),\ \ i=1,2,4,5, \\ A_3(\lambda)C_3(x,\lambda)+B_3(\lambda)S_3(x,\lambda).\end{cases} \]

Then the function $Y(x,\lambda )$ satisfies equations and the boundary conditions. If $\lambda ^{*}$ is an eigenvalue of the problem then the function $Y(x,\lambda ^{*})$ is an eigenfunction. Submitting $Y(x,\lambda )$ into the matching conditions (2.2) and (2.3) we obtain a linear systems about the variables $A_i(\lambda )$ with $i=1,2,4,5$ and $A_3(\lambda )$ and $B_3(\lambda )$ appearing in $Y(x,\lambda )$. $\lambda$ is an eigenvalue of the problem if and only if the determinant $\Delta (\lambda )$ of the coefficients of this linear systems about the variables $A_i(\lambda )$ with $i=1,2,4,5$ and $A_3(\lambda )$ and $B_3(\lambda )$ vanishes.

Moreover, a direct calculation yields the determinant

(2.8)\begin{align} \Delta(\lambda)& =C_1(1, \lambda)C_2(1, \lambda)[C_3'(1, \lambda)C_4(1, \lambda)C_5(1, \lambda)+C_3(1, \lambda)\nonumber\\ & \quad\enspace C_4'(1, \lambda)C_5(1, \lambda) +C_3(1, \lambda)C_4(1, \lambda)C_5'(1, \lambda)]+[C_1'(1, \lambda)\nonumber\\ & \quad\enspace C_2(1, \lambda)+C_1(1, \lambda)C_2'(1, \lambda)][S_3'(1, \lambda)C_4(1, \lambda)C_5(1, \lambda)\nonumber\\ & \qquad+S_3(1, \lambda)C_4'(1, \lambda)C_5(1, \lambda)+S_3(1, \lambda)C_4(1, \lambda)C_5'(1, \lambda)]. \end{align}

Substituting (2.4)–(2.7) into (2.8) we get

(2.9)\begin{equation} \Delta(\lambda)=\Delta^{0}(\lambda)+o\left(\rho^{2}\exp(5|\tau|)\right),\ |\lambda| \to \infty, \end{equation}

where

(2.10)\begin{equation} \Delta^{0}(\lambda)={-}\rho^{2}\sin\rho\cos^{2}\rho(9\cos^{2}\rho-4). \end{equation}

Notice that $\Delta ^{0}(\lambda )$ is the characteristic function for the problem $B^{0} := B(0)$ with the zero potential. It follows from (2.10) that the problem $B^{0}$ has a countable set of eigenvalues $\sigma (B^{0}) = \{\lambda _{ns}^{0}\}_{n\geq 0, s=\overline {1,5}}$ (counting multiplicities), where $\lambda _{ns}^{0}=(\rho _{ns}^{0})^{2}$,

(2.11)\begin{equation} \rho_{n1}^{0}=n\pi; \quad \rho_{n2}^{0}=n\pi-\theta;\quad \rho_{n3}^{0}=n\pi+\theta, \end{equation}

where $\theta =\arccos \frac {2}{3}$ and

(2.12)\begin{equation} \rho_{ns}^{0}=\left(n+\frac{1}{2}\right)\pi\ (\text{multiplicities two}),\quad s=4,5. \end{equation}

Here we used the fact that for the second-order self-adjoint differential operators on graphs the algebraic and geometric multiplicities of an eigenvalue are equal (see definition 3.3 and theorem 3.5 in [Reference Currie7]), so the order of a zero of the function in (2.10) coincides with the multiplicity as an eigenvalue of the problem $B^{0}$.

Note that the function $\Delta (\lambda )$ is entire in $\lambda$ of the order $\frac {1}{2}$, and from the above analysis we know that its zeros coincide with the eigenvalues of the problem $B$. Applying the standard argument, based on Rouche's theorem (see, e.g., theorem 1.1.3 in [Reference Freiling and Yurko10]), we see that the function $\Delta (\lambda )$ has a countable set of eigenvalues $\{\lambda _{ns}\}_{n\geq 0, s=\overline {1,5}}$ (counting multiplicities), where $\lambda _{ns}=\rho _{ns}^{2}$. Combining the arguments in [Reference Carlson and Pivovarchik4, Reference Yang and Yang23] we arrive at the following asymptotic formulas.

Lemma 2.1 The problem $B$ has a countable set of eigenvalues $\sigma (B) = \{\lambda _{ns}\}_{n\geq 0}$, $s=\overline {1,5}:=\{1,\cdots,5\}$. The eigenvalues counting with their multiplicities as $\{\lambda _{ns}\}_{n\geq 0, s=\overline {1,5}}$ in the nondecreasing order are numbered: $\lambda _{n_1,s_1}\leq \lambda _{n_2,s_2}$, if $(n_1,s_1)<(n_2,s_2)$(this means that $n_1< n_2$ or $n_1=n_2, s_1< s_2$). All eigenvalues are real and have the asymptotics

(2.13)\begin{equation} \begin{aligned} \rho_{n1} & =n\pi+\frac{\omega}{n\pi}+o\left(\frac{1}{n}\right); \quad \rho_{n2}=n\pi-\theta+\frac{\omega}{n\pi}+o\left(\frac{1}{n}\right);\\ \rho_{n3} & =n\pi+\theta+\frac{\omega}{n\pi}+o\left(\frac{1}{n}\right), \end{aligned} \end{equation}

where $\omega =\frac {1}{5}\sum _{j=1}^{5}[q_j]_1$, and

(2.14)\begin{equation} \rho_{ns}=\left(n+\frac{1}{2}\right)\pi+\frac{\kappa_s}{\left(n+\frac{1}{2}\right)\pi}+o\left(\frac{1}{n}\right),\quad s=4,5, \end{equation}

where $\kappa _s$ are the roots of the function $f(x):$

\[ f(x)=\displaystyle\sum_{j=4,5}(x-[q_1]_1)(x-[q_j]_1)+\sum_{j=4,5}(x-[q_2]_1)(x-[q_j]_1). \]

2.2 Nodes on a caterpillar graph

At the beginning of this section, we give a lemma.

Lemma 2.2 The components of the eigenfunction $Y(x,\lambda _{n1})$ with

\[ ||Y({\cdot},\lambda_{n1})||_{\bigoplus_{i=1}^{5}L^{2}(0,1)}=1, \]

corresponding to the eigenvalues $\lambda _{n1}$, are not identically zero on different edges.

Proof. In fact, from the matching conditions (2.2) and (2.3), we have

(2.15)\begin{align} ({-}1)^{n}A_1(\lambda_{n1})+o(1)& =({-}1)^{n}A_2(\lambda_{n1})+o(1)=A_3(\lambda_{n1}),\nonumber\\ ({-}1)^{n}A_4(\lambda_{n1})+o(1)& =({-}1)^{n}A_5(\lambda_{n1})+o(1)\nonumber\\ & \quad=({-}1)^{n}A_3(\lambda_{n1})+o(B_3(\lambda_{n1})) \end{align}

and

(2.16)\begin{equation} A_1(\lambda_{n1})([q_1]_1-\omega)+A_2(\lambda_{n1})([q_2]_1-\omega)+o(1)=B_3(\lambda_{n1}). \end{equation}

If $A_1(\lambda _{n1})=0$ then from (2.15) we get $A_i(\lambda _{n1})=o(1)$ for $i=2,3$. Again, from (2.16), we obtain $B_3(\lambda _{n1})=o(1)$, which implies from (2.15), that $A_i(\lambda _{n1})=o(1)$ for $i=4,5$. This leads to a contraction to that $Y(x,\lambda _{n1})$ is an eigenfunction corresponding to the eigenvalue $\lambda _{n1}$, which is notrival. Similarly, if $A_3(\lambda _{n1})=0$ then from (2.15) we get $A_i(\lambda _{n1})=o(1)$ for $i=1,2$. Again, from (2.16), we obtain $B_3(\lambda _{n1})=o(1)$, which implies from (2.15), that $A_i(\lambda _{n1})=o(1)$ for $i=4,5$. This also leads to a contraction. Therefore, for $i=\overline {1,5}$ the quantities $A_i(\lambda _{n1})$ can't be zero. Lemma 2.2 is complete.

Moreover, combining lemma 2.2, (2.15) with (2.16), it yields

(2.17)\begin{equation} \frac{B_3(\lambda_{n1})}{A_3(\lambda_{n1})}=[q_1+q_2]_1-2\omega+o(1). \end{equation}

Using the asymptotic expressions (2.13), (2.4) and (2.6), when $n\rightarrow \infty$, we obtain the asymptotics for the components (modulus $A_i(\lambda _{n1})$) of the eigenfunction $Y(x,\lambda _{n1})$, uniformly in $x\in [0,1]$:

\[ C_i(x,\lambda_{n1})=\cos\rho_{n1}^{0}x+([q_i]_x-\omega x)\frac{\sin\rho_{n1}^{0}x}{\rho_{n1}^{0}}+o\left(\frac{1}{n}\right),\quad i=1,2,4,5 \]

and

\begin{align*} & C_3(x,\lambda_{n1})+\frac{B_3(\lambda_{n1})}{A_3(\lambda_{n1})}S_3(x,\lambda_{n1})\\ & \quad=\cos\rho_{n1}^{0}x+([q_1+q_2]_1-2\omega+[q_i]_x-\omega x)\frac{\sin\rho_{n1}^{0}x}{\rho_{n1}^{0}}+o\left(\frac{1}{n}\right). \end{align*}

Fix $i=1,2,4,5$. There exists $N_0$ such that for $n\geq N_0$ the function $C_i(x,\lambda _{n1})$ (or $C_3(x,\lambda _{n1})+\frac {B_3(\lambda _{n1})}{A_3(\lambda _{n1})}S_3(x,\lambda _{n1})$) has exactly $n$ simple zeros inside the interval $(0,1)$, that is, $0< x_{ni}^{1}<\cdots < x_{ni}^{n}<1$. The sets $X_i:=\{x_{ni}^{j}\}_{n\geq N_0}$ ($i=\overline {1,5}, j=\overline {1,n}$) are called the nodes on the edge $e_i$ with respect to the eigenvalues $\lambda _{n1}$.

Taking asymptotic formulae (2.4) and (2.6) into account, we obtain the asymptotic expressions of nodes as follows.

Lemma 2.3 For large $n$, the following asymptotic formulae for the nodes hold uniformly in $j:$

(2.18)\begin{equation} x_{ni}^{j} = \begin{cases} \dfrac{j-\dfrac{1}{2}}{n}+\dfrac{[q_i]_x-\omega x}{n^{2}\pi^{2}}+o\left(\dfrac{1}{n^{2}}\right),\quad i=1,2,4,5, \\ \dfrac{j-\dfrac{1}{2}}{n}+\dfrac{[q_1+q_2]_1+[q_i]_x-2\omega-\omega x}{n^{2}\pi^{2}}+o\left(\dfrac{1}{n^{2}}\right),\quad i=3.\end{cases} \end{equation}

Proof. Fixing $i=1,2,4,5$, there exists $N_0$ such that for $n\geq N_0$ the function $C_i(x,\lambda _{n1})$ has exactly $n$ simple zeros inside the interval $(0,1)$. Other cases are similar and omitted.

Combining the asymptotics $\rho _{n1}=\sqrt {\lambda _{n1}}=n\pi + o\left (\frac {1}{n}\right )$, $n \to \infty$, with (2.6) and (2.7), we obtain that

(2.19)\begin{equation} C_i(x,\lambda_{n1})=\cos\rho_{n1}x+f_{n,i,1}(x), \ C_i'(x,\lambda_{n1})={-}\rho_{n1}\sin\rho_{n1}x+f_{n,i,2}(x), \end{equation}

where

\[ f_{n,i,1}(x)=O\left(\frac{1}{n}\right),\quad f_{n,i,2}(x)=O\left(1\right)\quad \text{for}\ n\rightarrow\infty, \]

uniformly on $[0,1]$. Therefore we conclude that

\[ f_{n,i,1}'(x)=O\left(1\right),\quad n\rightarrow\infty, \]

uniformly for $x \in \mathbb {R}$.

Consider the equation $C_i(x,\lambda _{n1})=0$ on $(0,1)$, which is equivalent to the equations

(2.20)\begin{equation} x = x_{ni}^{j}(x), \quad x_{ni}^{j}(x):=\frac{\left(j-\frac{1}{2}\right)\pi}{\rho_{n1}}+f_{n,i,1}^{j}(x),\quad j\in\mathbb{N}, \end{equation}

where $f_{n,i,1}^{j}(x)=(-1)^{j}\frac {\arcsin f_{n,i,1}(x)}{\rho _{n1}}$, and

(2.21)\begin{equation} (f_{n,i,1}^{j})^{'}(x)=O\left(\frac{1}{n}\right),\quad n\rightarrow\infty, \end{equation}

uniformly for $j\in \mathbb {N}$ and $x\in [0,1]$. One can continue $f_{n,i,1}^{j}(x)$ on $(-\infty,0)\cup (1,\infty )$ by differentiability in any way to satisfy (2.21) uniformly for $j\in \mathbb {N}$ and $x\in [0,1]$.

Consider the equation (2.20) in $\mathbb {R}$. According to (2.21) and the formula

\[ x_{ni}^{j}(x_1)-x_{ni}^{j}(x_2)=(f_{n,i,1}^{j})'(\theta)(x_1-x_2),\quad \theta\in(x_1,x_2), \]

there exists $N_0$ such that for $n\geq N_0$ the function $x_{ni}^{j}(x)$ is a contracting mapping in $\mathbb {R}$ for all $j\in \mathbb {N}$. Thus, for each $j\in \mathbb {N}$ the equation (2.20) has a unique solution in $\mathbb {R}$, which is denoted by $x_{ni}^{j}$.

Again, taking into account (2.20), we arrive at the formula

(2.22)\begin{equation} x_{ni}^{j}=\frac{\left(j-\frac{1}{2}\right)\pi}{\rho_{n1}}+O\left(\frac{1}{n^{2}}\right), \quad n\to \infty,\quad j \in \mathbb{N}. \end{equation}

Note that, for sufficiently large $n$, we have $x_{ni}^{1} \in \left (0, \frac {\pi }{\rho _{n1}}\right )$. Introduce the nodal length $l_{ni}^{j}:=x_{ni}^{j+1}-x_{ni}^{j}$. Then from (2.22), we have

\[ l_{ni}^{j} = \displaystyle\frac{\pi}{\rho_{n1}} + O\left(\frac{1}{n^{2}}\right), \]

uniformly with respect to $j$. Hence, we obtain for sufficiently large $n$, that $x_{ni}^{j} \in \left (\frac {(j-1)\pi }{\rho _{n1}}, \frac {j\pi }{\rho _{n1}}\right )$ for $j= \overline {1, n}$. Thus, we have proved that the function $C_i(x, \rho _{n1})$ has exactly $n$ nodes for large positive values of $n$.

In order to obtain more precise asymptotic expression, we substitute (2.22) into the equation $C_i(x, \lambda _{n1}) = 0$. In view of (2.22), we have

\[ |\sin\rho_{n1} x | \ge c_0 > 0 \]

for sufficiently large $n$ and $x = x_{ni}^{j}$, where the constant $c_0$ does not depend on $n$ and $j$. Therefore we derive the relation

\[ \cot\rho_{n1} x={-}\frac{[q_i]_x-\omega x}{\rho_{n1}}+o\left(\frac{1}{\rho_{n1}}\right), \]

which is equivalent to

\[ \tan\left(\rho_{n1} x+\frac{\pi}{2}\right)=\frac{[q_i]_x-\omega x}{\rho_{n1}}+o\left(\frac{1}{\rho_{n1}}\right). \]

Using Taylor's expansion for the arctangent, we obtain the following asymptotic formulae for the nodal points as $n\rightarrow \infty$:

(2.23)\begin{equation} x_{ni}^{j}=\frac{\left(j-\frac{1}{2}\right)\pi}{\rho_{n1}}+\frac{[q_i]_x-\omega x}{\rho_{n1}^{2}}+o\left(\frac{1}{\rho_{n1}^{2}}\right). \end{equation}

From the asymptotics of $\rho _{n1}$, we have $\frac {1}{\rho _{n1}}=\frac {1}{n\pi }+o\left (\frac {1}{n^{3}}\right )$ and $\frac {1}{\rho _{n1}^{2}}=\frac {1}{n^{2}\pi ^{2}}+o\left (\frac {1}{n^{4}}\right )$. Combining the latter formulae with (2.23), we arrive at the expected results (2.18).

2.3 Inverse nodal problems on a caterpillar graph

Note that for the fixed $i=\overline {1,5}$ the nodal sets $X_i$ is dense in $(0,1)$. Analyzing the asymptotic expressions $x_{ni}^{j}$ we have the following statements.

Fixed $i=\overline {1,5}$ and $x\in [0,1]$. Suppose that $X_i^{0}\subset X_i$ is dense on $(0,1)$ and choose $\{x_{ni}^{j_{ni}}\}\subset X_i^{0}$ such that $\lim _{n\rightarrow \infty } x_{ni}^{j_{ni}}=x$. Then the following finite limits hold:

(2.24)\begin{equation} \displaystyle\lim_{n\rightarrow\infty} n^{2}\pi^{2}\left(x_{ni}^{j_{ni}}-\frac{j-\frac{1}{2}}{n}\right)\mathop{=}\limits^{\text{exists}}f_i(x),\quad i=\overline{1,5}, \end{equation}

where $f_i(x)=[q_i]_x-\omega x$ for $i=1,2,4,5$, and $f_3(x)=[q_1+q_2]_1+[q_3]_x-2\omega -\omega x$.

Now we can provide a uniqueness theorem and constructive algorithm for the inverse nodal problem on the whole interval $(0,1)$. For stating the theorem, together with $B:=B(q)$ we consider a boundary value problem $\tilde {B}:=B(\tilde {q})$ of the same form but with a different potential $q$. We agree that if a certain symbol $\alpha$ denotes an object related to $B$, then $\tilde {\alpha }$ will denote an analogous object related to $\tilde {B}$.

Note that the zero sets $X_i$ ($i=\overline {1,5}$) are defined as shown before lemma 2.3.

In fact, (2.25) follows from (2.24). If $X_i^{0}=\tilde {X}_i^{0}$ then (2.24) implies that $f_i(x)=\tilde {f}_i(x)$ for $x\in [0,1]$, and consequently $q_i(x)-\omega =\tilde {q}_i(x)-\tilde {\omega }$ a.e. on $(0,1)$. Moreover, theorem 2.1 demonstrates that the nodal data on one edge can determine the potential on the edge up to a constant.

3.1 Eigenvalue and eigenfunction

Section 2 is a special tree graph (called caterpillar graph), however, §3 deals with a general tree graph. All eigenvalues in §2 can be estimated, while in §3 this is not possible, we only obtain a subsequences $\{\lambda _n\}$ of eigenvalues.

A connected graph $G(V,E)$ with edges of the equal length is considered in [Reference Kiss13]. The spectral determinant of Schrödinger operators on $G(V,E)$ with standard matching conditions has a sequence of roots which asymptotically differ by the mean value of the potential from the corresponding sequence of roots of the spectral determinant of the free Schrödinger operator. Precisely, the problem (3.1)–(3.3) has a sequence of eigenvalues

(3.4)\begin{equation} \lambda_n=(2n\pi)^{2}+\langle q\rangle+o(1) \end{equation}

for large integer $n$, and $\langle q\rangle =\frac {1}{|E|}\sum _j\int _0^{1}q_j(t)\,\textrm {d}t$, $|E|$ denotes the number of edges on the tree graphs. Moreover, if $G(V,E)$ is a bipartite graph, the problem (3.1)–(3.3) has a sequence of eigenvalues

(3.5)\begin{equation} \lambda_n=(n\pi)^{2}+\langle q\rangle+o(1). \end{equation}

Denote by $S_j(x, \lambda )$ and $C_j(x, \lambda )$ the solutions of the equation (3.1) on the edge $e_j$ satisfying the initial conditions

\[ S_j(0, \lambda) = C_j ' (0, \lambda) = 0,\quad S_j ' (0, \lambda) = C_j(0, \lambda) = 1. \]

For each fixed $x \in [0, 1]$, the functions $S_j^{(v)}(x, \lambda )$ and $C_j^{(v)}(x, \lambda ), v= 0, 1$, are entire in $\lambda$ of the order $\frac {1}{2}$, and these solutions possess the asymptotic expressions (2.4)–(2.7).

Before proving the main result we recall some preliminaries. From (2.4)–(2.7) we know that for $\lambda >0$ the following estimates hold

(3.6)\begin{equation} C_j(x,\lambda)=\cos(\sqrt\lambda x)+O\left(\frac{1}{\sqrt\lambda}\right),\quad S_j(x,\lambda)=\frac{\sin(\sqrt\lambda x)}{\sqrt{\lambda}}+O\left(\frac{1}{\lambda}\right). \end{equation}

Suppose that $Y(x,\lambda )$ is a vector function whose components $y_j(x,\lambda )$ satisfy (3.1), and which is given the graph $L^{2}$ norm:

\[ \| Y(x,\lambda) \|^{2}=\sum_{j:e_j\in E}\displaystyle\int_0^{1} | y_j(x,\lambda)|^{2}\,\textrm{d}x. \]

Each $y_j(x,\lambda )$ may be written as a linear combination

\[ y_j(x,\lambda)=A_j(\lambda)C_j(x,\lambda)+B_j(\lambda)S_j(x,\lambda). \]

Suppose $\| Y(x,\lambda ) \| =1$. Then there is a $\lambda _0>0$ and a constant $C$ such that (see [Reference Carlson and Pivovarchik5, lemma 2.1])

\[ |A_j(\lambda)| \le C,\quad |B_j(\lambda)/\sqrt\lambda| \le C,\quad \lambda\ge\lambda_0. \]

Suppose that $\lambda$ has the form $\lambda =(2\pi n)^{2}+\langle q\rangle +o(1)$ as $n\to \infty$, then there hold (see [Reference Carlson and Pivovarchik5, equation (2.3)])

(3.7)\begin{equation} \begin{aligned} & C_j(1,\lambda)=1+o(1),\quad C'_j(1,\lambda)=[q_j]_1+o(1),\\ & S_j(1,\lambda)=o(n^{{-}1}),\quad S'_j(1,\lambda)=1+o(1), \end{aligned} \end{equation}

where $[q_j]_1=\frac {1}{2}\int _0^{1}q_j(x)\,\textrm {d}x$.

Suppose that $\{Y(x,\lambda _n)\}$ is a sequence of eigenfunctions for (3.1) with norm $1$, corresponding to the eigenvalue $\lambda _n=(2\pi n)^{2}+\langle q\rangle +o(1)$ as $n\rightarrow \infty$. Write the components $y_j(x,\lambda _n)$ as a linear combination

\[ y_j(x,\lambda_n)=A_j(\lambda_n)C_j(x,\lambda_n)+B_j(\lambda_n)S_j(x,\lambda_n). \]

Recall that the coefficients $A_j(\lambda _n)$ and $B_j(\lambda _n)/\sqrt {\lambda _n}$ are bounded sequences (see [Reference Carlson and Pivovarchik5, lemma 2.1]).

Firstly, we consider the values of $y_j(x,\lambda _n)$ for edges incident on a vertex $v$. The continuity of $Y$ at the vertex $v$ thus implies [Reference Carlson and Pivovarchik5]

(3.8)\begin{equation} A_j(\lambda_n)=A_k(\lambda_n)+o(1), \quad n\rightarrow\infty, \end{equation}

for all edges $j,k$ incident on $v$. Since the graph $G(V,E)$ is connected, equation (3.8) can be extended to all edges $j,k$.

Secondly, the root vertex is regarded as the top of the graph $G(V,E)$, that is to say, an edge $e_j$ is below an edge $e_k\neq e_j$ if a path from $e_j$ to the root passes through $e_k$. We label each vertex $v$ of the graph $G(V,E)$ with the combinatorial distance from the root, and label edges with the larger of the vertex labels on the edge. Let $M$ be the maximum label. If the vertex $v$ has label $M-1$, then all its incoming edges $e_k$ join $v$ to a pendant vertex $v$. If $e_j$ is the outgoing edge for $v$, then the derivative condition at $v$ gives

(3.9)\begin{equation} B_j(\lambda_n)=A_1(\lambda_n)\sum_h[q_h]+o(1), \end{equation}

where the last sum is taken over all edges $e_h$ which are below $e_j$ on $G(V,E)$.

For vertices $v$ with label $M-2$, and outgoing edge $e_j$, the derivative condition at $v$ gives

(3.10)\begin{equation} B_j(\lambda_n)=A_1(\lambda_n)\sum_l[q_l]+o(1), \end{equation}

where the last sum is taken over all edges $e_l$ which are below $e_j$ on $G(V,E)$.

Thirdly, $A_j(\lambda _n)$ are bounded away from zero, otherwise by $||Y(x,\lambda _n)||=1$ we get a contradiction.

3.2 Nodes on a tree

Put

\[ Y(x,\lambda_n)=\{y_j(x,\lambda_n)\} \]

where

\[ y_j(x,\lambda_n) = \begin{cases} A_j(\lambda_n)C_j(x,\lambda_n)\text{ for boundary edges}, \\ A_j(\lambda_n)C_j(x,\lambda_n)+B_j(\lambda_n)S_j(x,\lambda_n)\text{ for other edges}. \end{cases} \]

Using the asymptotic expressions (3.4), (2.4) and (2.6), when $n\rightarrow \infty$, we obtain the asymptotics for the components (modulus nonvanishing $A_j(\lambda _n)$) of the eigenfunction $Y(x,\lambda _n)$, uniformly in $x\in [0,1]$:

\[ C_j(x,\lambda_n)=\cos\sqrt{\lambda_n}x+[q_j]_x\frac{\sin\sqrt{\lambda_n}x}{\sqrt{\lambda_n}}+o\left(\frac{1}{n}\right)\text{ on boundary edges} \]

and on other edges

\begin{align*} & C_j(x,\lambda_n)+\frac{B_j(\lambda_n)}{A_j(\lambda_n)}S_j(x,\lambda_n)\\ & \quad=\cos\sqrt{\lambda_n}x+\left(\sum_l[q_l]_1-l\langle q\rangle+[q_j]_x\right)\frac{\sin\sqrt{\lambda_n}x}{\sqrt{\lambda_n}}+o\left(\frac{1}{n}\right). \end{align*}

Here the sum is taken over all edges $e_l$ which are below $e_j$ on $G(V,E)$.

There exists $N_0$ such that for $n\geq N_0$ the function $C_j(x,\lambda _n)$ (or $C_j(x,\lambda _n)+\frac {B_j(\lambda _n)}{A_j(\lambda _n)}S_j(x,\lambda _n)$) has exactly $n$ simple zeros inside the interval $(0,1)$, that is, $0< x_{nj}^{1}<\cdots < x_{nj}^{n}<1$. The sets $X_j:=\{x_{nj}^{k}\}_{n\geq N_0}$ ($k=\overline {1,n}$) are called the nodes on the edge $e_j$ with respect to the eigenvalues $\lambda _n$.

Taking asymptotic formulae (2.4) and (2.6) into account, we obtain the asymptotic expressions of nodes as follows.

Lemma 3.1 For large $n$, the following asymptotic formulae for the nodes hold uniformly in $j:$

(3.11)\begin{equation} x_{nj}^{k} = \begin{cases} \dfrac{k-\dfrac{1}{2}}{n}+\dfrac{[q_j]_x-\langle q\rangle x}{n^{2}\pi^{2}}+o\left(\dfrac{1}{n^{2}}\right) \text{ on boundary edges},\\ \dfrac{k-\dfrac{1}{2}}{n}+\dfrac{\sum_l[q_l]_1+[q_j]_x-l\langle q\rangle-\langle q\rangle x}{n^{2}\pi^{2}}+o\left(\dfrac{1}{n^{2}}\right) \text{ on other edges}. \end{cases} \end{equation}

3.3 Inverse nodal problems on a tree

Note that for the fixed $e_j$ the nodal sets $X_j$ is dense in $(0,1)$, respectively. Analyzing the asymptotic expression $x_{nj}^{k}$ we have the following statements.

Fixed $e_j$ and $x\in [0,1]$. Suppose that $X_j^{0}\subset X_j$ is dense on $(0,1)$ and choose $\{x_{nj}^{k_{nj}}\}\subset X_j^{0}$ such that $\displaystyle \lim _{n\rightarrow \infty } x_{nj}^{k_{nj}}=x$. Then the following finite limit holds:

(3.12)\begin{equation} \displaystyle\lim_{n\rightarrow\infty} n^{2}\pi^{2}\left(x_{nj}^{k_{nj}}-\frac{k-\frac{1}{2}}{n}\right)\mathop{=}\limits^{\text{exists}}\theta_j(x), \end{equation}

where $\theta _j(x)=[q_j]_x-\langle q\rangle$ for boundary edges, and $\theta _j(x)=\sum _l[q_l]_1+[q_j]_x-l\langle q\rangle -\langle q\rangle x$ for other edges.

Now we can provide a uniqueness theorem and constructive algorithm for the inverse nodal problem on the whole interval $(0,1)$.

Theorem 3.1 Fix the edge $e_j$. Suppose that the following conditions is true:

\[ X_j^{0}\subset X_j\ is\ dense\ on\ (0,1)\ and\ X_j^{0}=\tilde{X}_j^{0}, \]

then $q_j(x)=\tilde {q}_j(x)$ a.e. on $(0,1)$. Therefore, the data $X_j^{0}$ uniquely determines the potential $q_j(x)-\langle q\rangle$ on the edge $e_j$.

The constructive algorithm is as follows:

1. (1) $X_j^{0}$ determining $\theta _j(x)$ from (3.12);

2. (2)

   (3.13)\begin{equation} q_j(x)-\langle q\rangle\mathop{=}\limits^{a.e.}2\theta_j'(x). \end{equation}