Before showing Theorem 3.1, we need to prove the following six propositions.
Let
and
be the solutions of (1.1) satisfying the following initial conditions:
respectively. Obviously,
and
are two linearly independent solutions of (1.1). By Lemma 2.2 we have
which, together with the assumption of
, implies
For any fixed
,
, and all
, we define
Note that
Let
Hence, we have
and by Lemma 2.2, we get
Proposition 4.1.
For
,
is an eigenvalue of (1.1) with (1.2) if and only if
Moreover,
is a multiple eigenvalue of (1.1) with (1.2) if and only if
Proof.
Since
and
are linearly independent solutions of (1.1), then
is an eigenvalue of the problem (1.1) with (1.2) if and only if there exist two constants
and
, not both zero, such that
satisfies (1.2), which yields
It is evident that (4.15) has a nontrivial solution
if and only if
which together with (4.3), (4.4) and
implies that
It follows from the above relation and the fact that
that
is an eigenvalue of (1.1) with (1.2) if and only if
satisfies
On the other hand, (1.1) has two linearly independent solutions satisfying (1.2) if and only if all the entries of the coefficient matrix of (4.16) are zero. Hence,
is a multiple eigenvalue of (1.1) with (1.2) if and only if (4.15) holds. This completes the proof.
The following result is a direct consequence of the first result of Proposition 4.1.
Corollary 4.2.
For any
,
For
and
, we consider the separated boundary problem (1.1) with (2.14). Let
, be all the eigenvalues of (1.1) with (2.14) and ordered as that in Theorem 2.7. Since
and
are all entire functions in
for each
,
is an entire functions in
. Denote
Proposition 4.3.
Assume that
and
or
and
. For each
,
if
is odd, and
if
is even.
Proof.
It is noted that
is eigenvalue of (1.1) with (2.14) if and only if
. Hence,
is an eigenfunction with respect to
. By Theorem 2.7 and the last two relations in (4.1), we have that
has exactly
generalized zeros in
and
-
(i)
If
, then it follows from
that
By (4.3) and the first relation in (4.23) we have
By the definition of
, the second relation in (4.23), and
, we get
Hence,
Noting
,
, and (4.22), we have that if
is odd, then
and if
is even, then
-
(ii)
If
, then it is noted that
is eigenvalue of (1.1) with (2.14) if and only if
Hence,
is an eigenfunction with respect to
By Theorem 2.7,
has exactly
generalized zeros in
and
Hence
for all 
Next we will show
In the case that
,
and
It follows from (1.1) and (2.4) that
which implies
In the other case that
, then
Further, by the existence and uniqueness theorem of solutions of initial value problems for (1.1) [5, Theorem
], we obtain that 
With a similar argument from above, we get sgn
By referring to
and from (4.3), it follows that
Hence, noting
and
, if
is odd, then
and if
is even, then
This completes the proof.
Proposition 4.4.
Assume that
and
or
and
. There exists a constant
such that
and
.
Proof.
Since
and
are solutions of (1.1), we have
By integration, it follows from (4.1) and (4.36) that
where
is used. In addition, from (4.36), we obtain
which, similarly together with (4.1) and by integration, imply that
On the other hand, it follows from Lemma 2.5 and (4.1) that for all sufficiently large
,
,
, for all
, where
and
in (2.11) are taken as
,
, and
,
, respectively, which satisfy
. So, from (4.37) and (4.39), we obtain that
and by Lemma 2.4, it implies
By Proposition 4.3,
. Therefore, there exists a
such that
. This completes the proof.
Lemma 4.5.
For any
one has
Proof.
Since
and
are solutions of (1.1) with (4.1), then they satisfy (4.36). Differentiating (4.36) with respect to
, we have
By the variation of constants formula [5, Theorem
], we get
Further, it follows from [5, Theorem
] that
From (4.46), (4.47), and (4.11), we have
where
It follows from (4.11) and (4.13) that
where 
Hence,
By (4.10) and (4.12), we have
Differentiating above relation with respect to
, and with (4.12), we have
which together with (4.10) confirm (4.42).
To establish (4.43), from (4.12) and (4.10), we obtain
Thus
That is, (4.43) holds. The identity (4.44) can be verified similarly. This completes the proof.
Corollary 4.6.
If
satisfies
, then
,
, and
.
Proof.
These are direct consequences of (4.43) and (4.44).
Lemma 4.7.
if and only if
for some
and
is an eigenfunction of
.
Proof.
It is directly follows from the definition of
and the initial conditions (4.1).
Lemma 4.8.
Assum that
or
and
is a multiple eigenvalue of (1.1) with (1.2) if and only if
.
Proof.
Assume that
. By (4.15)
is a multiple eigenvalue if and only if
hence, it follows from (4.12) that
Therefore
and 
(i)Suppose that
is a multiple eigenvalue of (1.1) with (1.2). Then
and
By (4.42),
.
(ii)Suppose that
is an eigenvalue of (1.1) with (1.2) and
. Then by (4.14),
. From (4.43) and (4.44) we get
Since
and
are linearly independence solutions of (1.1), we have
It follows from
and
that
. Thus,
is a multiple eigenvalue of (1.1) with (1.2).
The case
can be established by replacing
by
in the above argument. This completes the proof.
Lemma 4.9.
Assume
or
. If
is a multiple eigenvalue of (1.1) with (1.2), then there exists
such that
.
Proof.
Assume that
is a multiple eigenvalue of (1.1) with (1.2). From the proof of Lemma 4.8 we see that
. From (4.9) we have
This means that
for some
This completes the proof.
Proposition 4.10.
Assume that
and
or
and
.
-
(i)
Equations
and
or
hold if and only if
is a multiple eigenvalue of (1.1) with (1.2) with
or
.
-
(ii)
If
or
and
is a multiple eigenvalue of (1.1) with (1.2), then
,
.
-
(iii)
If
or
for some
,
, then
is a simple eigenvalue of (1.1) with (1.2) with
or
.
-
(iv)
Moreover, for every
,
, with
one has
and in the case of
,
Proof.
Parts (i), (ii), and (iii) follow from Lemmas 4.8 and 4.9. It follows from Propositions 4.3 and 4.4 and Corollary 4.6 that
,
with
and
when
. Hence,
with
,
. Similarly, by Proposition 4.3 and Corollary 4.6, we have
which implies
If
, then all the points of
are isolated. In this case, (1.1) can be rewritten as
where
By Theorem 2.7, (1.1) with (2.14) has
eigenvalues:
It follows from (4.65) that
and
are two polynomials of degree
in
and
and
are two polynomials of degree
in
. Then
can be written as
where
and
is a polynomial in
whose order is not larger than
. By Proposition 4.3, if
is odd, then
, and if
is even, then
. It follows that if
is odd, then
as
and if
is even, then
as
. Hence, if
is odd, then there exists a constant
such that
. Similarly, in the other case that
is even, there exists a constant
such that
, and by using Corollary 4.6, we have
Hence,
This completes the proof.
Proposition 4.11.
For any fixed
,
, each eigenvalue of (1.1) with (1.2) is simple.
Proof.
It follows from (4.46) and (4.47) that
where
Then from (4.1),
, and the definition of
, we have
Thus, if
, then
.
is always positive semidefinite or negative semidefinite. Consequently,
is not change sign in
. In this case,
cannot vanish unless
. Because
and
are linearly independent,
if and only if all the entries of the matrix
vanish, namely,
Suppose that
is an eigenvalue of the problem (1.1) with (1.2) and fix
with
. By Proposition 4.1, we have
, then
, and the matrix
is positive definite or negative definite. Hence,
or
for
, since
and
are linearly independent.
If
is a multiple eigenvalue of problem (1.1) with (1.2), then (4.15) holds by Proposition 4.1. By using (4.15), it can be easily verified that (4.74) holds; that is, all the entries of the matrix
are zeros. Then
, which is contrary to
. Hence,
is a simple eigenvalue of (1.1) and (1.2). This completes the proof.
Proposition 4.12.
If
is odd,
, and
then
; if
is even,
, and
then
.
Proof.
Assume
and
with
being odd. It follows from Proposition 4.1 that
As in the proof of Lemma 4.5 and by (4.11) and (4.13),
Hence,
where (4.51) is used and
by the Holder inequality [8, Lemma
(iv)]. Therefore
. Since
and
are linearly independent, which proves the first conclusion, the second conclusion can be shown similarly. This completes the proof.