#### 6.1. The Special Solution and the Special Prüfer Angle

In this section, we consider a general quantum tree (that might not be generic) and look for a non-degenerate solution

$z(\phantom{\rule{0.166667em}{0ex}}\xb7\phantom{\rule{0.166667em}{0ex}},\lambda )$ of the equation

$\ell y=\lambda y$,

$\lambda \in \mathbb{R}$, satisfying the boundary conditions (

6) at all non-root boundary vertices. We shall call such a solution

special, and the corresponding Prüfer angle

$\varphi (\xb7;\lambda )$ is said to be the

special Prüfer angle.

We start with introducing the following notions. For an arbitrary edge

$e=(a,b)$ we denote by

$\mathrm{\Gamma}\left(e\right)$ the closure of the connected component of the graph

$\mathrm{\Gamma}\backslash \left\{a\right\}$ containing the edge

e. Thus

$\mathrm{\Gamma}\left(e\right)$ is a subtree of

$\mathrm{\Gamma}$ with the root vertex

a and containing along with

e all

$x\in \mathrm{\Gamma}$ that can be reached from

a moving in positive direction. Further, we denote by

$\mathcal{L}\left(e\right)$ the differential operator on the tree

$\mathrm{\Gamma}\left(e\right)$ given by the differential expression

$\tau $, the interface conditions (

4)–(

5) for all interior points of

$\mathrm{\Gamma}\left(e\right)$, the boundary conditions (

6) for all non-root vertices of

$\mathrm{\Gamma}\left(e\right)$, and the Dirichlet boundary condition at the root vertex

a of

$\mathrm{\Gamma}\left(e\right)$. Clearly,

$\mathcal{L}\left(e\right)$ is a self-adjoint operator; we denote also by

$\mathrm{\Lambda}\left(e\right)$ the spectrum of

$\mathcal{L}\left(e\right)$ and set

where

${e}_{0}$ is the edge starting from the root

${v}_{0}$. The set

$\mathrm{\Lambda}$ is bounded below and discrete.

**Lemma** **6.** For every $\lambda \in \mathbb{R}\backslash \mathrm{\Lambda}$, a special solution $z(\xb7;\lambda )$ of the equation $\ell y=\lambda y$ exists, is unique up to a constant factor, and does not vanish at the interior vertices of Γ.

**Proof.** Denote by

l the height of the tree

$\mathrm{\Gamma}$. We shall prove by the reverse induction in the level

k of an edge

$e=(a,b)$ that there exists a non-degenerate solution

z of

$\ell y=\lambda y$ on the subtree

$\mathrm{\Gamma}\left(e\right)$ that satisfies the boundary condition (

6) at all non-root boundary vertices of

$\mathrm{\Gamma}\left(e\right)$. For short, we call such a solution

special for $\mathrm{\Gamma}\left(e\right)$. Also, we shall show that a special solution is unique up to a constant factor and vanishes neither at the interior vertices of

$\mathrm{\Gamma}\left(e\right)$ nor at the root vertex

a of

$\mathrm{\Gamma}\left(e\right)$ provided it differs from

${v}_{0}$ (i.e., provided

$k>1$).

The induction starts from

$k=l$ and descends to

$k=1$. Assume, therefore, that

$e=(a,b)$ is any edge of level

l. Then

b is a boundary vertex, whence

$\mathrm{\Gamma}\left(e\right)=\overline{e}=[a,b]$. We define a special solution

z for

$\mathrm{\Gamma}\left(e\right)$ as a unique solution of the equation

$\tau y=\lambda y$ on

$\overline{e}=[a,b]$ subject to the terminal condition

We note that

z does not vanish at

$x=a$; indeed, otherwise

$\lambda $ would be an eigenvalue of

$\mathcal{L}\left(e\right)$. Clearly, any other solution of

$\tau y=\lambda y$ on

$\mathrm{\Gamma}\left(e\right)$ satisfying the boundary condition (

6) at

$x=b$ is a multiple of

z so constructed. This gives the base of induction.

Assume that special solutions have already been constructed on the subtrees

$\mathrm{\Gamma}\left(e\right)$ for every edge

e of level

k,

$k\le l$, and let

$\gamma =(a,b)$ be an edge of level

$k-1$. Two possibilities occur depending on whether or not

b is a boundary vertex of

$\mathrm{\Gamma}$. If

$b\in \partial \mathrm{\Gamma}$, then we define the solution

z of

$\tau y=\lambda y$ on

$\mathrm{\Gamma}\left(\gamma \right)=\overline{\gamma}$ as in the previous paragraph, by fixing the terminal conditions (

11). If

$b\in I\left(\mathrm{\Gamma}\right)$, we denote by

${\gamma}_{1},\dots ,{\gamma}_{m}$ the edges starting from

b and by

${z}_{1},\dots ,{z}_{m}$ special solutions to

$\ell y=\lambda y$ on the subtrees

$\mathrm{\Gamma}\left({\gamma}_{1}\right),\dots ,\mathrm{\Gamma}\left({\gamma}_{m}\right)$. By the induction assumption,

${z}_{j}$ do not vanish at the vertex

b; we then consider the solution

${y}_{\gamma}$ of the equation

$\tau y=\lambda y$ on

$\overline{\gamma}=[a,b]$ subject to the terminal conditions

$y\left(b\right)=1$ and

${y}^{\left[1\right]}\left(b\right)={\sum}_{j=1}^{m}{z}_{j}^{\left[1\right]}\left(b\right)/{z}_{j}\left(b\right)$.

Now we construct a function

z on the tree

$\mathrm{\Gamma}\left(\gamma \right)$ that is equal to

${y}_{\gamma}$ on the edge

$\gamma $ and to

${z}_{j}/{z}_{j}\left(b\right)$ on each subtree

$\mathrm{\Gamma}\left({\gamma}_{j}\right)$,

$j=1,\dots ,m$. Then

z is non-degenerate, solves the equation

$\tau y=\lambda y$ on each edge constituting the tree

$\mathrm{\Gamma}\left(\gamma \right)$ and satisfies the interface conditions (

4)–(

5) at every interior vertex of

$\mathrm{\Gamma}\left(\gamma \right)$ and the boundary conditions at all non-root boundary points of

$\mathrm{\Gamma}\left(\gamma \right)$. Therefore,

z is the special solution of

$\ell y=\lambda y$ on the tree

$\mathrm{\Gamma}\left(\gamma \right)$ we wanted to construct. Clearly, such solution is defined up to a multiplicative constant and can be parametrized by its value at

b. By construction and the induction assumptions,

z does not vanish at interior vertices of

$\mathrm{\Gamma}\left(\gamma \right)$. If

$a\ne {v}_{0}$, then the special solution

z does not vanish at the root vertex

a as well, as otherwise

z would be an eigenfunction of the operator

$\mathcal{L}\left(\gamma \right)$ corresponding to the eigenvalue

$\lambda $, contrary to the assumption that

$\lambda \notin \mathrm{\Lambda}$. This completes the induction step and thus the proof of the lemma.

**Corollary** **1.** Assume that $\lambda \in \mathbb{R}\backslash \mathrm{\Lambda}$ and that y is a non-trivial solution of the equation $\ell y=\lambda y$ satisfying the boundary conditions (

6)

at all non-root boundary vertices. Then y is a multiple of the special solution $z(\xb7,\lambda )$ and thus non-degenerate. **Proof.** It suffices to show that y cannot vanish at interior vertices of $\mathrm{\Gamma}$: indeed, then y is non-degenerate and thus a multiple of $z(\xb7,\lambda )$ by the above lemma.

Assume, on the contrary, that

$y\left(v\right)=0$ for some interior vertex

v. We can choose such a

v so that

y does not vanish identically on all edges adjacent to

v as otherwise

y would be identical zero on

$\mathrm{\Gamma}$. In view of (

5), then

y is not identical zero on at least two of the adjacent edges. Denote by

${e}_{1},\dots ,{e}_{m}$ all the edges starting from

v; then the above means that

y does not vanish identically on at least one among the subtrees

$\mathrm{\Gamma}\left({e}_{1}\right),\dots ,\mathrm{\Gamma}\left({e}_{m}\right)$. However, then

$\lambda $ is an eigenvalue for at least one of the operators

$\mathcal{L}\left({e}_{1}\right),\dots ,\mathcal{L}\left({e}_{m}\right)$, contrary to the assumption that

$\lambda \notin \mathrm{\Lambda}$. The contradiction derived completes the proof. □

**Corollary** **2.** Every eigenvalue λ of $\mathcal{L}$ not belonging to Λ is of multiplicity 1 and $z(\xb7;\lambda )$ is the corresponding eigenfunction.

Since the special solution

$z(\xb7;\lambda )$,

$\lambda \notin \mathrm{\Lambda}$, is unique up to a constant factor, the corresponding special Prüfer angle

$\varphi (\xb7;\lambda )$ is unique modulo

$\pi $. Clearly,

$\varphi (\xb7;\lambda )$ is continuous along each edge but the limiting values at the interior vertices along different adjacent edges need not be the same. Also, the boundary conditions (

6) for

$z(\xb7;\lambda )$ prescribe the boundary values

$\alpha \left(v\right)\in (0,\pi ]$ for

$\varphi (\xb7;\lambda )$ at every non-root boundary vertex

v. We shall drop the requirement that

${\varphi}_{e}(b;\lambda )\in (0,\pi ]$ for edges

$e=(a,b)$ ending at interior vertices

b but gain continuity of

$\varphi $ in

$\lambda $ instead; note that

$\lambda \in \mathrm{\Lambda}$ are not excluded any longer. As usual, for an interior vertex

$v\in I\left(\mathrm{\Gamma}\right)$ of valency

d the expression

$\varphi (v,\lambda )$ should be understood as

d limiting values of

$\varphi (x,\lambda )$ along every adjacent edge.

**Theorem** **3.** For every $\lambda \in \mathbb{R}$, the special Prüfer angle ϕ can be defined so that

- (A1)
for every fixed $x\in \mathrm{int}\left(\mathrm{\Gamma}\right)\cup \left\{{v}_{0}\right\}$, $\varphi (x;\lambda )$ is a continuous strictly decreasing function of $\lambda \in \mathbb{R}$;

- (A2)
there is $\mu \in \mathbb{R}$ such that $\varphi (x;\lambda )\in (0,\pi )$ for all $x\in \mathrm{int}\left(\mathrm{\Gamma}\right)\cup \left\{{v}_{0}\right\}$ and all $\lambda <\mu $.

For so defined ϕ the following holds:

- (A3)
${lim}_{\lambda \to -\infty}\varphi (x;\lambda )=\pi $ for every fixed $x\in \mathrm{int}\left(\mathrm{\Gamma}\right)\cup \left\{{v}_{0}\right\}$;

- (A4)
$\varphi (x;{\mu}_{*})>0$ on $\mathrm{int}\left(\mathrm{\Gamma}\right)$ and $\varphi ({v}_{0};{\mu}_{*})=0$.

**Proof.** We shall use the backward induction on the level of the edge $e=(a,b)$ to prove that the special Prüfer angle $\varphi $ can be defined so that it satisfies the stated properties on $\mathrm{\Gamma}\left(e\right)$ instead of $\mathrm{\Gamma}$ and with ${v}_{0}$ replaced by the root vertex a of $\mathrm{\Gamma}\left(e\right)$.

An edge

$e=(a,b)$ of the maximal level (say

l) necessarily ends with a boundary vertex

b. Therefore,

$\varphi $ on

e is defined uniquely as a solution of equation (

9) satisfying the initial condition

$\varphi \left(b\right)=\alpha \left(b\right)$, and properties (A1)–(A4) for

$\varphi $ on

$\mathrm{\Gamma}\left(e\right)$ so defined are established in Reference [

4], see also

Appendix A.

Assume statements (A1)–(A4) have already been proved for the subtrees

$\mathrm{\Gamma}\left(e\right)$ with edges

e of level

k,

$k\le l$, and let

$\gamma =(a,b)$ be an edge of level

$k-1$. Two possibilities occur depending on whether or not

b is a boundary vertex of

$\mathrm{\Gamma}$. If

$b\in \partial \mathrm{\Gamma}$, then

$\varphi $ on

$\mathrm{\Gamma}\left(\gamma \right)$ is constructed as in the previous paragraph and thus enjoys (A1)–(A4). If

$b\in I\left(\mathrm{\Gamma}\right)$, we denote by

${\gamma}_{1},\dots ,{\gamma}_{m}$ the edges starting from

b; by induction assumption, on the subtrees

$\mathrm{\Gamma}\left({\gamma}_{1}\right),\dots ,\mathrm{\Gamma}\left({\gamma}_{m}\right)$ the special Prüfer angle

$\varphi $ is well defined and satisfies (A1)–(A4). Set

then

g assumes infinite values at the eigenvalues

${\mu}_{k}$ of

$\mathcal{L}\left({\gamma}_{1}\right)$,...,

$\mathcal{L}\left({\gamma}_{m}\right)$ and by (A1) it is continuous and strictly increasing in between. By virtue of (A2) there is

$\mu =\mu \left(\gamma \right)\in \mathbb{R}$ such that

$g\left(\lambda \right)$ assumes finite values for

$\lambda <\mu $ and, moreover,

$g\left(\lambda \right)\to -\infty $ as

$\lambda \to -\infty $ in view of (A3).

Set $\beta \left(\lambda \right):=arccotg\left(\lambda \right)\in (0,\pi )$ for $\lambda <\mu $. Then $\beta $ is continuous and strictly decreasing for such $\lambda $. Moreover, the properties of ${\varphi}_{{\gamma}_{j}}(b;\lambda )$ show that we can extend this definition by continuity to all $\lambda \in \mathbb{R}$, and $\beta \left(\lambda \right)$ will strictly decrease on $\mathbb{R}$. By construction, $\beta \left(\lambda \right)=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $ if and only if ${\varphi}_{{\gamma}_{j}}(b;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $ for at least one $j\in \{1,\dots ,m\}$.

Now we define

${\varphi}_{\gamma}(\xb7;\lambda )$ on

$[a,b]$ as a unique solution of Equation (

9) subject to the terminal condition

$\varphi (b;\lambda )=\beta \left(\lambda \right)$. Then for

$x\in [a,b]$ property (A1) is ensured by Proposition A2, (A3) follows from Lemma A1, and (A2) is established on Step 1 of its proof (see Reference [

4]).

Define the number

${\mu}_{*}\left(\gamma \right)$ as in (

12) but for the tree

$\mathrm{\Gamma}\left(\gamma \right)$ instead of

$\mathrm{\Gamma}$; then, clearly,

Assume first that

${\mu}_{*}={\mu}_{*}\left({\gamma}_{k}\right)$ for some

$k\in \{1,\dots ,m\}$. Then

${\varphi}_{{\gamma}_{k}}(b,{\mu}_{*})=0$ by induction assumption, whence

${\varphi}_{\gamma}(b,{\mu}_{*})=0$ by the definition of

$\varphi $. Since the Prüfer angle strictly increases through every point

${x}_{*}$ were

${\varphi}_{\gamma}({x}_{*},{\mu}_{*})=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $, we conclude that

${\varphi}_{\gamma}(x,{\mu}_{*})<0$ for all

$x\in [a,b)$, contrary to the definition of

${\mu}_{*}$ and continuity of

$\varphi $. Thus

${\mu}_{*}<{\mu}_{*}\left({\gamma}_{k}\right)$ for every

$k=1,\dots ,m$, so that

$\varphi (x,{\mu}_{*})>0$ on the set

It remains to prove that ${\varphi}_{\gamma}(x,{\mu}_{*})>0$ for all $x\in (a,b)$ and that $\varphi (a,{\mu}_{*})=0$. Assume, on the contrary, that ${\varphi}_{\gamma}({x}_{*},{\mu}_{*})=0$ for some ${x}_{*}\in (a,b)$. Since ${\varphi}_{\gamma}(\xb7,{\mu}_{*})$ strictly increases through every point where ${\varphi}_{\gamma}(x,{\mu}_{*})=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $, we conclude that ${\varphi}_{\gamma}(x,{\mu}_{*})<0$ for all $x\in [a,{x}_{*})$. This contradicts continuity of ${\varphi}_{\gamma}$ and the definition of ${\mu}_{*}$ and thus shows that ${\varphi}_{\gamma}(x,{\mu}_{*})>0$ on $\gamma =(a,b)$. Finally, the inequality ${\varphi}_{\gamma}(a,{\mu}_{*})>0$ is ruled out by similar reasons.

The proof of (A4) and of the theorem is complete. □

**Remark** **4.** We observe that for $\lambda \notin \mathrm{\Lambda}$ any Prüfer angle $\theta (\xb7;\lambda )$ for the special solution $z(\xb7;\lambda )$ equals the special Prüfer angle $\varphi (\xb7;\lambda )$ modulo π, that is,Indeed, both θ and ϕ solve the same differential Equation (

9)

on every edge e of Γ

, satisfy the same boundary conditions $\theta (v;\lambda )=\varphi (v;\lambda )=\alpha \left(v\right)$ for all non-root boundary vertices v, and the same interface conditions at the interior vertices of Γ

for $\mathrm{cot}\theta $ and $\mathrm{cot}\varphi $, cf. (

8)

and the construction of ϕ in the proof of the above theorem. It turns out that (

14) holds even for

$\lambda \in \mathrm{\Lambda}$; namely, the following holds true.

**Lemma** **7.** Let that $y(\xb7;\lambda )$ be a non-trivial solution of the equation $\ell y=\lambda y$ satisfying (

6)

for all non-root boundary vertices. Introduce a Prüfer angle θ for the solution y on every edge $\gamma \in E\left(\mathrm{\Gamma}\right)$ where y is non-degenerate; then $\theta (\xb7;\lambda )\equiv \varphi (\xb7;\lambda )\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $ on all such γ. **Proof.** On every edge

$\gamma =(c,d)$ where

y is non-degenerate the Prüfer angles

$\theta $ and

$\varphi $ solve the same Equation (

9) of first order, which is invariant under the shift of

$\theta $ or

$\varphi $ by

$\pi $. Therefore, it suffices to show that the terminal conditions for

${\theta}_{\gamma}$ and

${\varphi}_{\gamma}$ at the vertex

d are equal modulo

$\pi $. We shall prove the statement for all the subtrees

$\mathrm{\Gamma}\left(e\right)$ taken instead of

$\mathrm{\Gamma}$ and shall use the backward induction on the level of edge

e.

An edge $e=(a,b)$ of the maximal level (say l) necessarily ends with a boundary vertex b. If the solution y is non-degenerate on e, then the Prüfer angle $\theta (\xb7;\lambda )$ satisfies the terminal condition $\varphi (b;\lambda )=\alpha \left(b\right)$ by construction, and the same is true for $\varphi $, resulting in the identity $\theta (\xb7;\lambda )=\varphi (\xb7;\lambda )$ over e.

Assume the lemma has already been proved for the subtrees

$\mathrm{\Gamma}\left(e\right)$ with edges

e of level

k,

$k\le l$, and let

$\gamma =(a,b)$ be an edge of level

$k-1$. The case where

$b\in \partial \mathrm{\Gamma}$ is treated as in the previous paragraph. Assume therefore that

$b\in I\left(\mathrm{\Gamma}\right)$ and denote by

${\gamma}_{1},\dots ,{\gamma}_{m}$ the edges starting from

b. Since

only the case where

y is non-degenerate on

$\gamma $ is of interest.

If

$y\left(b\right)\ne 0$, then by (

8)

and by the induction assumption the right-hand side of this relation coincides with

giving

$\mathrm{cot}{\varphi}_{\gamma}(b;\lambda )$ by the construction of

$\varphi $. Thus

${\theta}_{\gamma}(b;\lambda )={\varphi}_{\gamma}(b;\lambda )\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $, which establishes the induction step.

Now assume that $y\left(b\right)=0$; then ${\varphi}_{\gamma}(b;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $. Observe that y cannot be identical zero on all edges ${\gamma}_{1},\dots ,{\gamma}_{m}$ starting from b as otherwise y must be zero on $\gamma $ as well, and there is nothing to prove. Let therefore y be non-trivial on say the edge ${\gamma}_{1}$. Then $\theta $ is defined over ${\gamma}_{1}$ and ${\theta}_{{\gamma}_{1}}(b;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $, so that ${\varphi}_{{\gamma}_{1}}(b;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $ by the induction assumption. By the construction of $\varphi $ we get ${\varphi}_{\gamma}(b;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $, thus establishing the induction step and completing the proof. □

**Corollary** **3.** $\lambda \in \mathrm{\Lambda}$ if and only if there is an interior vertex v and an edge e starting from it such that ${\varphi}_{e}(v;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $.

**Proof.** If $\lambda \in \mathbb{R}\backslash \mathrm{\Lambda}$, then by Lemma 6 the special solution $z(\xb7;\lambda )$ exists and does not vanish at any interior vertex $v\in I\left(\mathrm{\Gamma}\right)$. Clearly, this means that the corresponding special Prüfer angle $\varphi (\xb7;\lambda )$ does not assume values $\pi n$, $n\in \mathbb{Z}$, at such vertices.

Let now

$\lambda \in \mathrm{\Lambda}\left(e\right)$ for some edge

$e=(a,b)$ different from the edge

${e}_{0}$ starting from the root

${v}_{0}$ of

$\mathrm{\Gamma}$. Then there exists an eigenfunction

$y(\xb7;\lambda )$, that is, a function that is not identically equal to zero over

$\mathrm{\Gamma}\left(e\right)$, solves the equation

$\ell y=\lambda y$ on

$\mathrm{\Gamma}\left(e\right)$, and satisfies the boundary conditions (

6) for all non-root boundary vertices of

$\mathrm{\Gamma}\left(e\right)$ and the Dirichlet condition

$y(a;\lambda )=0$ at the root vertex

a of

$\mathrm{\Gamma}\left(e\right)$. We can find an edge

$\gamma =(c,d)\in E\left(\mathrm{\Gamma}\right(e\left)\right)$ such that

y is non-degenerate on

$\gamma $ and

${y}_{\gamma}(c;\lambda )=0$. Then any Prüfer edge

$\theta $ for

y on

$\gamma $ satisfies

${\theta}_{\gamma}\left(c\right)=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $, and by Lemma 7 we conclude that

${\varphi}_{\gamma}(c;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $ as well. The proof is complete. □

**Corollary** **4.** With the number ${\mu}_{*}$ introduced by (

12)

, the following inequalities hold:moreover, ${\lambda}_{0}$ is the only eigenvalue of $\mathcal{L}$ in $(-\infty ;{\mu}_{*}]$. **Proof.** The second inequality follows from Corollary 3 and (A4). Next, properties (A1), (A3), and (A4) show that there exists a unique

${\lambda}_{*}\le {\mu}_{*}$ such that

$\varphi ({v}_{0};{\lambda}_{*})=\alpha \left({v}_{0}\right)$. This means that the special solution

$z(\xb7;{\lambda}_{*})$ satisfies the boundary condition

at the root vertex

${v}_{0}$. Therefore,

${\lambda}_{*}$ is an eigenvalue of

$\mathcal{L}$ and

$z(\xb7;{\lambda}_{*})$ is a corresponding eigenfunction, so that

${\lambda}_{0}\le {\lambda}_{*}\le {\mu}_{*}$.

We next show that if

$\lambda \in (-\infty ;{\mu}_{*}]$ is an eigenvalue of

$\mathcal{L}$, then

$\lambda ={\lambda}_{*}$. Indeed, as

$\lambda \notin \mathrm{\Lambda}$, any corresponding eigenfunction is a multiple of

$z(\xb7;\lambda )$ by Corollary 2 and thus is non-degenerate and verifies the boundary condition

at the root vertex

${v}_{0}$. Therefore,

$\mathrm{cot}\varphi ({v}_{0};\lambda )=\mathrm{cot}\alpha \left({v}_{0}\right)$; since

$\varphi ({v}_{0};\xb7)$ strictly decreases from

$\pi $ to 0 as

$\lambda $ increases from

$-\infty $ to

${\mu}_{*}$, we conclude that

$\lambda ={\lambda}_{*}$. This shows that

${\lambda}_{*}$ is the only eigenvalue of

$\mathcal{L}$ in

$(-\infty ,{\mu}_{*}]$ and thus it is the ground eigenvalue

${\lambda}_{0}$ of

$\mathcal{L}$. □

**Corollary** **5.** The ground eigenvalue ${\lambda}_{0}$ of $\mathcal{L}$ is simple and the corresponding eigenfunction $z(\xb7;{\lambda}_{0})$ does not have any zeros in the interior of Γ.

**Proof.** The fact that ${\lambda}_{0}$ is a simple eigenvalue of $\mathrm{\Gamma}$, with the corresponding eigenfunction $z(\xb7;\lambda )$, follows from the relation $\lambda \notin \mathrm{\Lambda}$ and Corollary 2, while absence of interior zeros of $z(\xb7;\lambda )$ is guaranteed by the inequality ${\lambda}_{0}\le {\mu}_{*}$ and (A4). □

We stress here the fact that the quantum tree considered here is not assumed generic; thus the simplicity of the ground eigenvalue is not automatic and should have been proved.

**Corollary** **6.** A real number $\lambda \notin \mathrm{\Lambda}$ is an eigenvalue of $\mathcal{L}$ if and only if $\varphi ({v}_{0};\lambda )=\alpha \left({v}_{0}\right)\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $.

**Proof.** According to Corollary 1, for

$\lambda \in \mathbb{R}\backslash \mathrm{\Lambda}$ any non-trivial solution

y of equation

$\ell y=\lambda y$ that satisfies the boundary conditions (

6) at all non-root boundary vertices is a multiple of the special solution

$z(\xb7,\lambda )$. Therefore, such a

$\lambda $ is an eigenvalue of

$\mathcal{L}$ and if and only if the special solution

$z(\xb7,\lambda )$ satisfies the boundary condition (

6) at the root vertex

${v}_{0}$ if and only if the special Prüfer angle satisfies the relation

$\varphi ({v}_{0};\lambda )=\alpha \left({v}_{0}\right)\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $. □

#### 6.2. Eigenvalue Multiplicities

The special Prüfer angle can also be used to calculate the multiplicity of non-simple eigenvalues of $\mathcal{L}$; in view of Lemma 4 such eigenvalues necessarily belong to $\mathrm{\Lambda}$ and every corresponding eigenfunction vanishes at some interior vertices.

For every

$e=(a,b)\in E\left(\mathrm{\Gamma}\right)$, we denote by

$\mathcal{N}(e;\lambda )$ the subspace of

$\mathcal{H}\left(e\right):={L}_{2}\left(\mathrm{\Gamma}\left(e\right)\right)$ consisting of all solutions of the equation

$\ell y=\lambda y$ on

$\mathrm{\Gamma}\left(e\right)$ satisfying the boundary conditions (

6) for all non-root boundary vertices

v in

$\partial \mathrm{\Gamma}\left(e\right)$, and set

Further, we denote by $n(e;\lambda )$ and ${n}_{0}(e;\lambda )$ the dimensions of $\mathcal{N}(e;\lambda )$ and ${\mathcal{N}}_{0}(e;\lambda )$ respectively. It follows from Lemma 7 that ${\mathcal{N}}_{0}(e;\lambda )=\mathcal{N}(e;\lambda )$ if ${\varphi}_{e}(a;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $. We shall prove that otherwise ${\mathcal{N}}_{0}(e;\lambda )$ is a proper subspace of $\mathcal{N}(e;\lambda )$ and, moreover, establish the formula for $n(e;\lambda )$ and ${n}_{0}(e;\lambda )$.

To begin with, we set

Also, for

$v\in I\left(\mathrm{\Gamma}\right)$ that is of valency

$m+1\ge 3$ we denote by

${\gamma}_{1},\dots ,{\gamma}_{m}$ the edges starting from

v and set

$m(v;\lambda )=0$ if none of

${\varphi}_{{\gamma}_{j}}(v;\lambda )$ vanishes modulo

$\pi $; otherwise, we let

$m(v;\lambda )+1$ denote the number of indices

j among

$j=1,\dots ,m$, for which

${\varphi}_{{\gamma}_{j}}(v;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $.

**Theorem** **4.** For every $\lambda \in \mathbb{R}$ and every $e\in \mathrm{\Gamma}$ the following holds: Moreover, there are $y\in \mathcal{N}(e;\lambda )$ that are non-degenerate on e.

**Proof.** We use the induction over the subtrees $\mathrm{\Gamma}\left(e\right)$, starting from the edges of the largest level $l=k$ and descending to the root edge ${e}_{0}$.

For an edge $e=(a,b)$ of the largest level l the subtree $\mathrm{\Gamma}\left(e\right)$ is just the edge e. Thus $\mathrm{\Gamma}\left(e\right)$ has no interior vertices and $\mathcal{N}(e;\lambda )$ is of dimension 1. It follows from the proof of oscillation theorem for an interval that ${n}_{0}(e;\lambda )=1$ if and only if ${\varphi}_{e}(a;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $. This establishes the base of induction.

Assume the lemma has already been proved for the subtrees $\mathrm{\Gamma}\left(e\right)$ with edges e of level k, $k\le l$, and let $\gamma =(a,b)$ be an edge of level $k-1$. The case where $b\in \partial \mathrm{\Gamma}$ is treated as in the previous paragraph. Assume therefore that $b\in I\left(\mathrm{\Gamma}\right)$ and denote by ${\gamma}_{1},\dots ,{\gamma}_{m}$ the edges starting from b. Now we distinguish between two cases:

- (i)
for some $j\in \{1,\dots ,m\}$ it holds that ${\varphi}_{{\gamma}_{j}}(b;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $;

- (ii)
none of the numbers ${\varphi}_{{\gamma}_{j}}(b;\lambda )$ vanishes modulo $\pi $.

For Case (i), every solution of

$\ell y=\lambda y$ on

$\mathrm{\Gamma}\left(\gamma \right)$ must vanish at the vertex

b in view of Lemma 7. Therefore, the restrictions of

$y\in \mathcal{N}(\gamma ;\lambda )$ onto the subtrees

$\mathrm{\Gamma}\left({\gamma}_{j}\right)$,

$j=1,\dots ,m$, belong to the respective subspaces

${\mathcal{N}}_{0}({\gamma}_{j};\lambda )$. Conversely, if for every

$j=1,\dots ,m$ we take a solution

${y}_{j}$ of the equation

$\ell y=\lambda y$ on

$\mathrm{\Gamma}\left({\gamma}_{j}\right)$ vanishing at

b (i.e., any element of

${\mathcal{N}}_{0}({\gamma}_{j};\lambda )$), then the function on

${\cup}_{j}\mathrm{\Gamma}\left({\gamma}_{j}\right)$ constructed this way allows a unique continuation to a solution of

$\ell y=\lambda y$ on

$\mathrm{\Gamma}\left(\gamma \right)$. Indeed, one only has to take a solution

${y}_{\gamma}$ of the equation

$\tau y=\lambda y$ on the edge

$\gamma $ to satisfy the interface conditions (

4)–(

5) at

$v=b$, and to this end one sets

${y}_{\gamma}\left(b\right):=0$ and

${y}_{\gamma}^{\left[1\right]}\left(b\right):={\sum}_{j=1}^{m}{y}_{j}^{\left[1\right]}\left(b\right)$. Therefore,

as claimed. Moreover, if

j is such that

${\varphi}_{{\gamma}_{j}}(b;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $, then

${\mathcal{N}}_{0}({\gamma}_{j};\lambda )=\mathcal{N}({\gamma}_{j};\lambda )$ and by the induction assumption there are

${y}_{j}\in {\mathcal{N}}_{0}({\gamma}_{j};\lambda )$ that are non-degenerate on

${\gamma}_{j}$. This means that

${y}_{j}^{\left[1\right]}\left(b\right)$ can be any real number; therefore,

${y}_{\gamma}^{\left[1\right]}\left(b\right)$ can be non-zero giving non-degenerate

${y}_{\gamma}$.

Next, if

${\varphi}_{\gamma}(a;\lambda )=0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $, then

$\chi (\gamma ;\lambda )=1$ and

${\mathcal{N}}_{0}(\gamma ;\lambda )=\mathcal{N}(\gamma ;\lambda )$, which agrees with (

15). Otherwise

$y\in {\mathcal{N}}_{0}(\gamma ;\lambda )$ must be degenerate on the edge

$\gamma $, which requires that

$F\left(y\right):={\sum}_{j=1}^{m}{y}_{{\gamma}_{j}}^{\left[1\right]}\left(b\right)=0$. Since

F is a linear continuous functional on

$\mathcal{N}(\gamma ;\lambda )$ that is not identically equal to zero by the arguments in the above paragraph, we conclude that

${\mathcal{N}}_{0}(\gamma ;\lambda )$ has codimension 1 in

$\mathcal{N}(\gamma ;\lambda )$, thus giving (

15) and finishing the proof for the case (i).

For Case (ii) we first use Lemma 7 to prove that every

${y}_{j}\in {\mathcal{N}}_{0}({\gamma}_{j};\lambda )$ vanishes identically on

${\gamma}_{j}$. Next we denote by

${\mathcal{N}}^{*}(\gamma ;\lambda )$ the subspace of solutions

$y\in \mathcal{N}(\gamma ;\lambda )$ satisfying

$y\left(b\right)=0$ and observe that every

$y\in {\mathcal{N}}^{*}(\gamma ;\lambda )$ vanishes identically on the adjacent edges

$\gamma $ and

${\gamma}_{1},\dots ,{\gamma}_{m}$. And conversely, by taking arbitrary

${y}_{j}\in {\mathcal{N}}_{0}({\gamma}_{j};\lambda )$ on

$\mathrm{\Gamma}\left({\gamma}_{j}\right)$ and extending them by zero identically on

$\gamma $, we get an element of

${\mathcal{N}}^{*}(\gamma ;\lambda )$. The dimension of

${\mathcal{N}}^{*}(\gamma ;\lambda )$ is therefore equal to

Next we construct a function

${y}^{*}\in \mathcal{N}(\gamma ;\lambda )$ satisfying

${y}^{*}\left(b\right)=1$; such

${y}^{*}$ is clearly non-degenerate on

$\gamma $. As

${\varphi}_{{\gamma}_{j}}(b;\lambda )$ is not zero modulo

$\pi $,

$n({\gamma}_{j};\lambda )>{n}_{0}({\gamma}_{j};\lambda )$ by the induction assumption, whence

${\mathcal{N}}_{0}({\gamma}_{j};\lambda )$ is a proper subspace of

$\mathcal{N}({\gamma}_{j};\lambda )$. Thus there is

${y}_{j}^{*}\in \mathcal{N}({\gamma}_{j};\lambda )$ such that

${y}_{j}^{*}\left(b\right)=1$. Now we fix one such function for each

$j=1,\dots ,m$ and form a solution

${y}^{*}$ of

$\ell y=\lambda y$ on

$\mathrm{\Gamma}\left(\gamma \right)$ by adjoining to

${y}_{1}^{*},\dots ,{y}_{m}^{*}$ a unique solution of

$\tau y=\lambda y$ on

$\gamma $ satisfying the terminal conditions

${y}_{\gamma}\left(b\right):=1$ and

${y}_{\gamma}^{\left[1\right]}\left(b\right):={\sum}_{j=1}^{m}{\left({y}_{j}^{*}\right)}^{\left[1\right]}\left(b\right)$. Notice that

ls denoting the linear span, so that

as in (

15). If

${\varphi}_{\gamma}(a;\lambda )\ne 0\phantom{\rule{3.33333pt}{0ex}}\mathrm{mod}\phantom{\rule{0.277778em}{0ex}}\pi $, then the function

${y}^{*}$ constructed above does not belong to

${\mathcal{N}}_{0}(\gamma ;\lambda )$, so that

${n}_{0}(\gamma ;\lambda )=n(\gamma ;\lambda )-1$ leading to (

15). The proof is complete. □

**Corollary** **7.** For $\lambda \in \mathbb{R}$, we denote by $n\left(\lambda \right)$ the multiplicity of λ as an eigenvalue of $\mathcal{L}$. Then