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 ].