Let
and
be defined in Section 2, let
be the eigenvalues of the separated boundary value problem (1.1) with (2.7), and let
be the eigenvalues of the coupled boundary value problem (1.1) and (1.2) and arranged in the nondecreasing order
Clearly,
denotes the eigenvalue of the problem (1.1) and (1.2) with
, and
denotes the eigenvalue of the problem (1.1) and (1.2) with
. We now present the main results of this paper.
Theorem 3.1.
Assume that
or
. Then, for every fixed
,
, one has the following inequalities:
Remark 3.2.
If
or
, a similar result can be obtained by applying Theorem 3.1 to
. In fact,
for
and
for
. Hence, the boundary condition (1.2) in the cases of
or
and
, can be written as condition (1.2), where
is replaced by
for
and
for
, and
is replaced by
.
Before proving Theorem 3.1, we prove the following five propositions.
Proposition 3.3.
For
,
is an eigenvalue of (1.1) and (1.2) if and only if
where
Moreover,
is a multiple eigenvalue of (1.1) and (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) and (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 (3.6) has a nontrivial solution
if and only if
which, together with (2.14) and
, implies that
Then (3.3) follows from the above relation and the fact that
. 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 (3.6) are zero. Hence,
is a multiple eigenvalue of (1.1) and (1.2) if and only if (3.5) holds. This completes the proof.
The following result is a direct consequence of the first result of Proposition 3.3.
Corollary 3.4.
For any
,
Proposition 3.5.
Assume that
or
. Then one has the following results.
-
(i)
For each
,
,
if
is odd, and
if
is even.
-
(ii)
There exists a constant
such that
.
-
(iii)
If the boundary value problem (1.1) and (2.7) has exactly
eigenvalues then there exists a constant
such that
and
, where
is odd, and there exists a constant
such that
and
, where
is even.
Proof.
(i) If
is an eigenfunction of the problem (1.1) and (2.7) respect to
then
. By Lemma 2.3 and the initial conditions (2.13), we have that if
then the sequence
,
exhibits
changes of sign and
Case 1.
If
then it follows from
that
By (2.14) and the first relation in (3.11), for each
,
, we have
By the definition of
, (3.11), and 
Hence,
Noting
,
, and (3.10), we have that if
is odd then
and if
is even then
Case 2.
If
then it follows from (2.7) and (2.14) that for each
,
,
From (2.15) and by the definition of
, we get
Hence, noting
,
, and by Lemma 2.5, we have that if
is odd, then
and if
is even, then
(ii) By the discussions in the first paragraph of Section 2,
is a polynomial of degree
in
,
is a polynomial of degree
in
,
is a polynomial of degree
in
, and
is a polynomial of degree
in
. Further,
can be written as
where
and
is a certain constant for
. Then
where
is a polynomial in
whose degree is not larger than
. Clearly, as
,
since
. By the first part of this proposition,
. So there exists a constant
such that
.
-
(iii)
It follows from the first part of this proposition that if
is odd,
and if
is even,
. By (3.22), if
is odd,
as
; if
is even,
as
. Hence, if
is odd, there exists a constant
such that
; if
is even, there exists a constant
such that
. This completes the proof.
Since
and
are both polynomials in
, so is
. Denote
Proposition 3.6.
Assume that
or
. Equations
and
or
hold if and only if
is a multiple eigenvalue of (1.1) and (1.2) with
or
. If
or
for some
, then
is a simple eigenvalue of (1.1) and (1.2) with
or
and for every
, with
one has:
Proof.
Since
and
are solutions of (1.1), we have
Differentiating (3.25) and (3.26) with respect to
, respectively, yields that
It follows from (2.13) that
Thus, by Lemma 2.6 and from (3.27)–(3.28), we have
It follows from (3.29) that
Hence, not indicating
explicitly, we get
where
which is symmetric for any
. Then, we have
Hence, if
or
, we get from (3.33) that
. Then, for any fixed
with
or
, the matrix
is positive semidefinite or negative semidefinite. Therefore, for such a
,
cannot vanish unless
for all
. Because
and
are linearly independent,
is identically zero if and only if all the entries of the matrix
vanish, namely,
which, together with
and
, implies
Then by Proposition 3.3,
is a multiple eigenvalue of (1.1) and (1.2) with
. In addition, (3.34), together with
and
, implies
Then by Proposition 3.3,
is a multiple eigenvalue of (1.1) and (1.2) with
. Conversely, from (3.35) or (3.36), it can be easily verified that (3.34) holds, then
. It follows again from (3.35) or (3.36) that
or
. Thus
and
or
if and only if
is a multiple eigenvalue of (1.1) and (1.2) with
or
.
Further, for every fixed
with
or
, not indicating
explicitly, (3.33) implies that
Therefore, from (3.37) and by the definition of
, we have
and consequently, not indicating
explicitly, we have
for every fixed
with
or
.
Suppose that
or
for some
(
), we have
From the above discussions again,
is a simple eigenvalue of (1.1) and (1.2) with
or
, and
is not identically zero for
.
For this
(
), (3.39) implies that
, and from Proposition 3.5 (i), (ii) that
,
. Hence,
, where
. It follows from Proposition 3.5 (i) that
and
, where
. By Proposition 3.5 (i), (iii),
and there exists
such that
if
is odd, and
and there exists
such that
if
is even. Hence,
where
. This completes the proof.
Proposition 3.7.
For any fixed
,
, each eigenvalue of (1.1) and (1.2) is simple.
Proof.
Fix
,
with
. Suppose that
is an eigenvalue of the problem (1.1) and (1.2). By Proposition 3.3, we have
. It follows from (3.33) that
and the matrix
is positive definite or negative definite. Hence,
for
or
for
since
and
are linearly independent.
If
is a multiple eigenvalue of problem (1.1) and (1.2), then (3.5) holds by Proposition 3.3. By using (3.5), it can be easily verified that (3.34) holds, that is, all the entries of the matrix
are zero. Then
for
, which is contrary to
for
. Hence,
is a simple eigenvalue of (1.1) and (1.2). This completes the proof.
Proposition 3.8.
Assume that
or
. If
is odd,
, and
then
; if
is even,
, and
then
for 
Proof.
We first prove the first result. Suppose that
is odd,
, and
. Then
is a multiple eigenvalue of (1.1) and (1.2) with
by Proposition 3.6. Then by Proposition 3.3, (3.5) holds for
and
, that is,
Differentiating
with respect to
two times, we get
Differentiating (2.14) with respect to
two times and from (3.40), we get
which, together with (3.41), implies that
On the other hand, it follows from (3.29) and (2.14) that, not indicating
explicitly,
Since
and
are linearly independent on
, the above relation implies that
by Hölder's inequality, which proves the first conclusion.
The second conclusion can be shown similarly. Hence, the proof is complete.
Finally, we turn to the proof of Theorem 3.1.
Proof of Theorem 3.1.
By Propositions 3.3–3.8, and the intermediate value theorem, one can obtain the graph of
(see Figure 1), which implies the results of Theorem 3.1. We now give its detailed proof.
By Propositions 3.3–3.6,
,
for all
with
and there exists
such that
Therefore, by the continuity of
and the intermediate value theorem, (1.1) and (1.2) with
has only one eigenvalue
, (1.1) and (1.2) with
has only one eigenvalue
, and (1.1) and (1.2) with
,
has only one eigenvalue
, and they satisfy
Similarly, by Propositions 3.3–3.6, the continuity of
and the intermediate value theorem,
reaches
,
(
,
), and
exactly one time, respectively, between any two consecutive eigenvalues of the separated boundary value problem (1.1) with (2.7). Hence, (1.1) and (1.2) with
;
,
;
has only one eigenvalue between any two consecutive eigenvalues of (1.1) with (2.7), respectively. In addition, by Proposition 3.6, if
or
and
, then
is not only an eigenvalue of (1.1) with (2.7) but also a multiple eigenvalue of (1.1) and (1.2) with
and
.
By Proposition 3.5 (i), if
is odd,
and if
is even,
. It follows (3.22) that if
is odd, then
as
and if
is even, then
as
. Hence, if
is odd, then there exists a constant
such that
, which, together with Proposition 3.6, implies that (1.1) and (1.2) with
;
,
;
has only one eigenvalue
,
, and
, satisfying
(see Figure 2). Similarly, in the other case that
is even, there exists a constant
such that
which, together with Proposition 3.6, implies that (1.1) and (1.2) with
;
,
;
has only one eigenvalue
,
, and
, satisfying
(see Figure 3). Therefore, we get that (1.1) and (1.2) with
,
has
eigenvalues and it is real and satisfies
This completes the proof.
Remark 3.9.
Let
, that is,
,
. Then
. In this case, Propositions 3.5 and 3.8 are the same as those mentioned in [4, Propositions
, 3.3–3.5], respectively, and most of the results of Proposition 3.6 are the same as the results of [4, Proposition
].