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.