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
of the equation
,
, satisfying the boundary conditions (
6) at all non-root boundary vertices. We shall call such a solution
special, and the corresponding Prüfer angle
is said to be the
special Prüfer angle.
We start with introducing the following notions. For an arbitrary edge
we denote by
the closure of the connected component of the graph
containing the edge
e. Thus
is a subtree of
with the root vertex
a and containing along with
e all
that can be reached from
a moving in positive direction. Further, we denote by
the differential operator on the tree
given by the differential expression
, the interface conditions (
4)–(
5) for all interior points of
, the boundary conditions (
6) for all non-root vertices of
, and the Dirichlet boundary condition at the root vertex
a of
. Clearly,
is a self-adjoint operator; we denote also by
the spectrum of
and set
where
is the edge starting from the root
. The set
is bounded below and discrete.
Lemma 6. For every , a special solution of the equation 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
. We shall prove by the reverse induction in the level
k of an edge
that there exists a non-degenerate solution
z of
on the subtree
that satisfies the boundary condition (
6) at all non-root boundary vertices of
. For short, we call such a solution
special for . Also, we shall show that a special solution is unique up to a constant factor and vanishes neither at the interior vertices of
nor at the root vertex
a of
provided it differs from
(i.e., provided
).
The induction starts from
and descends to
. Assume, therefore, that
is any edge of level
l. Then
b is a boundary vertex, whence
. We define a special solution
z for
as a unique solution of the equation
on
subject to the terminal condition
We note that
z does not vanish at
; indeed, otherwise
would be an eigenvalue of
. Clearly, any other solution of
on
satisfying the boundary condition (
6) at
is a multiple of
z so constructed. This gives the base of induction.
Assume that special solutions have already been constructed on the subtrees
for every edge
e of level
k,
, and let
be an edge of level
. Two possibilities occur depending on whether or not
b is a boundary vertex of
. If
, then we define the solution
z of
on
as in the previous paragraph, by fixing the terminal conditions (
11). If
, we denote by
the edges starting from
b and by
special solutions to
on the subtrees
. By the induction assumption,
do not vanish at the vertex
b; we then consider the solution
of the equation
on
subject to the terminal conditions
and
.
Now we construct a function
z on the tree
that is equal to
on the edge
and to
on each subtree
,
. Then
z is non-degenerate, solves the equation
on each edge constituting the tree
and satisfies the interface conditions (
4)–(
5) at every interior vertex of
and the boundary conditions at all non-root boundary points of
. Therefore,
z is the special solution of
on the tree
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
. If
, 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
corresponding to the eigenvalue
, contrary to the assumption that
. This completes the induction step and thus the proof of the lemma.
Corollary 1. Assume that and that y is a non-trivial solution of the equation satisfying the boundary conditions (
6)
at all non-root boundary vertices. Then y is a multiple of the special solution and thus non-degenerate. Proof. It suffices to show that y cannot vanish at interior vertices of : indeed, then y is non-degenerate and thus a multiple of by the above lemma.
Assume, on the contrary, that
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
. In view of (
5), then
y is not identical zero on at least two of the adjacent edges. Denote by
all the edges starting from
v; then the above means that
y does not vanish identically on at least one among the subtrees
. However, then
is an eigenvalue for at least one of the operators
, contrary to the assumption that
. The contradiction derived completes the proof. □
Corollary 2. Every eigenvalue λ of not belonging to Λ is of multiplicity 1 and is the corresponding eigenfunction.
Since the special solution
,
, is unique up to a constant factor, the corresponding special Prüfer angle
is unique modulo
. Clearly,
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
prescribe the boundary values
for
at every non-root boundary vertex
v. We shall drop the requirement that
for edges
ending at interior vertices
b but gain continuity of
in
instead; note that
are not excluded any longer. As usual, for an interior vertex
of valency
d the expression
should be understood as
d limiting values of
along every adjacent edge.
Theorem 3. For every , the special Prüfer angle ϕ can be defined so that
- (A1)
for every fixed , is a continuous strictly decreasing function of ;
- (A2)
there is such that for all and all .
For so defined ϕ the following holds:
- (A3)
for every fixed ;
- (A4)
on and .
Proof. We shall use the backward induction on the level of the edge to prove that the special Prüfer angle can be defined so that it satisfies the stated properties on instead of and with replaced by the root vertex a of .
An edge
of the maximal level (say
l) necessarily ends with a boundary vertex
b. Therefore,
on
e is defined uniquely as a solution of equation (
9) satisfying the initial condition
, and properties (A1)–(A4) for
on
so defined are established in Reference [
4], see also
Appendix A.
Assume statements (A1)–(A4) have already been proved for the subtrees
with edges
e of level
k,
, and let
be an edge of level
. Two possibilities occur depending on whether or not
b is a boundary vertex of
. If
, then
on
is constructed as in the previous paragraph and thus enjoys (A1)–(A4). If
, we denote by
the edges starting from
b; by induction assumption, on the subtrees
the special Prüfer angle
is well defined and satisfies (A1)–(A4). Set
then
g assumes infinite values at the eigenvalues
of
,...,
and by (A1) it is continuous and strictly increasing in between. By virtue of (A2) there is
such that
assumes finite values for
and, moreover,
as
in view of (A3).
Set for . Then is continuous and strictly decreasing for such . Moreover, the properties of show that we can extend this definition by continuity to all , and will strictly decrease on . By construction, if and only if for at least one .
Now we define
on
as a unique solution of Equation (
9) subject to the terminal condition
. Then for
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
as in (
12) but for the tree
instead of
; then, clearly,
Assume first that
for some
. Then
by induction assumption, whence
by the definition of
. Since the Prüfer angle strictly increases through every point
were
, we conclude that
for all
, contrary to the definition of
and continuity of
. Thus
for every
, so that
on the set
It remains to prove that for all and that . Assume, on the contrary, that for some . Since strictly increases through every point where , we conclude that for all . This contradicts continuity of and the definition of and thus shows that on . Finally, the inequality is ruled out by similar reasons.
The proof of (A4) and of the theorem is complete. □
Remark 4. We observe that for any Prüfer angle for the special solution equals the special Prüfer angle 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 for all non-root boundary vertices v, and the same interface conditions at the interior vertices of Γ
for and , cf. (
8)
and the construction of ϕ in the proof of the above theorem. It turns out that (
14) holds even for
; namely, the following holds true.
Lemma 7. Let that be a non-trivial solution of the equation satisfying (
6)
for all non-root boundary vertices. Introduce a Prüfer angle θ for the solution y on every edge where y is non-degenerate; then on all such γ. Proof. On every edge
where
y is non-degenerate the Prüfer angles
and
solve the same Equation (
9) of first order, which is invariant under the shift of
or
by
. Therefore, it suffices to show that the terminal conditions for
and
at the vertex
d are equal modulo
. We shall prove the statement for all the subtrees
taken instead of
and shall use the backward induction on the level of edge
e.
An edge 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 satisfies the terminal condition by construction, and the same is true for , resulting in the identity over e.
Assume the lemma has already been proved for the subtrees
with edges
e of level
k,
, and let
be an edge of level
. The case where
is treated as in the previous paragraph. Assume therefore that
and denote by
the edges starting from
b. Since
only the case where
y is non-degenerate on
is of interest.
If
, then by (
8)
and by the induction assumption the right-hand side of this relation coincides with
giving
by the construction of
. Thus
, which establishes the induction step.
Now assume that ; then . Observe that y cannot be identical zero on all edges starting from b as otherwise y must be zero on as well, and there is nothing to prove. Let therefore y be non-trivial on say the edge . Then is defined over and , so that by the induction assumption. By the construction of we get , thus establishing the induction step and completing the proof. □
Corollary 3. if and only if there is an interior vertex v and an edge e starting from it such that .
Proof. If , then by Lemma 6 the special solution exists and does not vanish at any interior vertex . Clearly, this means that the corresponding special Prüfer angle does not assume values , , at such vertices.
Let now
for some edge
different from the edge
starting from the root
of
. Then there exists an eigenfunction
, that is, a function that is not identically equal to zero over
, solves the equation
on
, and satisfies the boundary conditions (
6) for all non-root boundary vertices of
and the Dirichlet condition
at the root vertex
a of
. We can find an edge
such that
y is non-degenerate on
and
. Then any Prüfer edge
for
y on
satisfies
, and by Lemma 7 we conclude that
as well. The proof is complete. □
Corollary 4. With the number introduced by (
12)
, the following inequalities hold:moreover, is the only eigenvalue of in . Proof. The second inequality follows from Corollary 3 and (A4). Next, properties (A1), (A3), and (A4) show that there exists a unique
such that
. This means that the special solution
satisfies the boundary condition
at the root vertex
. Therefore,
is an eigenvalue of
and
is a corresponding eigenfunction, so that
.
We next show that if
is an eigenvalue of
, then
. Indeed, as
, any corresponding eigenfunction is a multiple of
by Corollary 2 and thus is non-degenerate and verifies the boundary condition
at the root vertex
. Therefore,
; since
strictly decreases from
to 0 as
increases from
to
, we conclude that
. This shows that
is the only eigenvalue of
in
and thus it is the ground eigenvalue
of
. □
Corollary 5. The ground eigenvalue of is simple and the corresponding eigenfunction does not have any zeros in the interior of Γ.
Proof. The fact that is a simple eigenvalue of , with the corresponding eigenfunction , follows from the relation and Corollary 2, while absence of interior zeros of is guaranteed by the inequality 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 is an eigenvalue of if and only if .
Proof. According to Corollary 1, for
any non-trivial solution
y of equation
that satisfies the boundary conditions (
6) at all non-root boundary vertices is a multiple of the special solution
. Therefore, such a
is an eigenvalue of
and if and only if the special solution
satisfies the boundary condition (
6) at the root vertex
if and only if the special Prüfer angle satisfies the relation
. □
6.2. Eigenvalue Multiplicities
The special Prüfer angle can also be used to calculate the multiplicity of non-simple eigenvalues of ; in view of Lemma 4 such eigenvalues necessarily belong to and every corresponding eigenfunction vanishes at some interior vertices.
For every
, we denote by
the subspace of
consisting of all solutions of the equation
on
satisfying the boundary conditions (
6) for all non-root boundary vertices
v in
, and set
Further, we denote by and the dimensions of and respectively. It follows from Lemma 7 that if . We shall prove that otherwise is a proper subspace of and, moreover, establish the formula for and .
To begin with, we set
Also, for
that is of valency
we denote by
the edges starting from
v and set
if none of
vanishes modulo
; otherwise, we let
denote the number of indices
j among
, for which
.
Theorem 4. For every and every the following holds: Moreover, there are that are non-degenerate on e.
Proof. We use the induction over the subtrees , starting from the edges of the largest level and descending to the root edge .
For an edge of the largest level l the subtree is just the edge e. Thus has no interior vertices and is of dimension 1. It follows from the proof of oscillation theorem for an interval that if and only if . This establishes the base of induction.
Assume the lemma has already been proved for the subtrees with edges e of level k, , and let be an edge of level . The case where is treated as in the previous paragraph. Assume therefore that and denote by the edges starting from b. Now we distinguish between two cases:
- (i)
for some it holds that ;
- (ii)
none of the numbers vanishes modulo .
For Case (i), every solution of
on
must vanish at the vertex
b in view of Lemma 7. Therefore, the restrictions of
onto the subtrees
,
, belong to the respective subspaces
. Conversely, if for every
we take a solution
of the equation
on
vanishing at
b (i.e., any element of
), then the function on
constructed this way allows a unique continuation to a solution of
on
. Indeed, one only has to take a solution
of the equation
on the edge
to satisfy the interface conditions (
4)–(
5) at
, and to this end one sets
and
. Therefore,
as claimed. Moreover, if
j is such that
, then
and by the induction assumption there are
that are non-degenerate on
. This means that
can be any real number; therefore,
can be non-zero giving non-degenerate
.
Next, if
, then
and
, which agrees with (
15). Otherwise
must be degenerate on the edge
, which requires that
. Since
F is a linear continuous functional on
that is not identically equal to zero by the arguments in the above paragraph, we conclude that
has codimension 1 in
, thus giving (
15) and finishing the proof for the case (i).
For Case (ii) we first use Lemma 7 to prove that every
vanishes identically on
. Next we denote by
the subspace of solutions
satisfying
and observe that every
vanishes identically on the adjacent edges
and
. And conversely, by taking arbitrary
on
and extending them by zero identically on
, we get an element of
. The dimension of
is therefore equal to
Next we construct a function
satisfying
; such
is clearly non-degenerate on
. As
is not zero modulo
,
by the induction assumption, whence
is a proper subspace of
. Thus there is
such that
. Now we fix one such function for each
and form a solution
of
on
by adjoining to
a unique solution of
on
satisfying the terminal conditions
and
. Notice that
ls denoting the linear span, so that
as in (
15). If
, then the function
constructed above does not belong to
, so that
leading to (
15). The proof is complete. □
Corollary 7. For , we denote by the multiplicity of λ as an eigenvalue of . Then