We first prove Theorem 2.2 via the following Lemmas.

Lemma 3.1.

If , , and , then is a Fredholm operator of index zero. Furthermore, the linear continuous projector operator can be defined by

and the linear operator can be written by

Furthermore,

Proof.

It is clear that

Obviously, the problem

has a solution satisfying , , if and only if

which implies that

In fact, if (3.5) has solution satisfying , , then from (3.5) we have

According to , , we obtain

then

On the other hand, if (3.6) holds, setting

where is an arbitrary constant, then is a solution of (3.5), and , and from and (3.6), we have

Then . Hence (3.7) is valid.

For , define

Let , and we have

then , thus . Hence, , where , also . So we have , and

Thus, is a Fredholm operator of index zero.

We define a projector by . Then we show that defined in (3.2) is a generalized inverse of .

In fact, for , we have

and, for , we know

In view of , , and , thus

This shows that . Also we have

then . The proof of Lemma 3.1 is finished.

Lemma 3.2.

Under conditions (2.5) and (2.9), there are nonnegative functions satisfying

Proof.

Without loss of generality, we suppose that . Take , then there exists such that

Let

Obviously, , and

Then

and from (2.5) and (3.21), we have

Hence we can take , , 0, and to replace , , , and , respectively, in (2.5), and for the convenience omit the bar above , , and , that is,

Lemma 3.3.

If assumptions (H1), (H2) and , , and hold, then the set for some is a bounded subset of .

Proof.

Suppose that and . Thus and , so that

thus by assumption (H2), there exists , such that . In view of

then, we have

Again for , , then , thus from Lemma 3.1, we know

From (3.29) and (3.30), we have

If (2.5) holds, from (3.31), and (3.26), we obtain

Thus, from and (3.32), we have

From , (3.32), and (3.33), one has

that is,

From (3.35) and (3.33), there exists , such that

Thus

Again from (2.5), (3.35), and (3.36), we have

Then we show that is bounded.

Lemma 3.4.

If assumption (H2) holds, then the set is bounded.

Proof.

Let , then and ; therefore,

From assumption (H2), , so , clearly is bounded.

Lemma 3.5.

If the first part of condition (H3) of Theorem 2.2 holds, then

for all . Let

where is the linear isomorphism given by , for all , . Then is bounded.

Proof.

Suppose that , then we obtain

or equivalently

If , then . Otherwise, if , in view of (3.40), one has

which contradicts . Then = and we obtain ; therefore, is bounded.

The proof of Theorem 2.2 is now an easy consequence of the above lemmas and Theorem 2.1.

Proof of Theorem 2.2.

Let such that . By the Ascoli-Arzela theorem, it can be shown that is compact; thus is -compact on . Then by the above Lemmas, we have the following.

(i) for every .

(ii) for every .

(iii)Let , with as in Lemma 3.5. We know , for . Thus, by the homotopy property of degree, we get

According to definition of degree on a space which is isomorphic to , , and

We have

and then

Then by Theorem 2.1, has at least one solution in , so that the BVP (1.1), (1.2) has at least one solution in . The proof is completed.

Remark 3.6.

If the second part of condition (H3) of Theorem 2.2 holds, that is,

for all , then in Lemma 3.5, we take

and exactly as there, we can prove that is bounded. Then in the proof of Theorem 2.2, we have

since . The remainder of the proof is the same.

By using the same method as in the proof of Theorem 2.2 and Lemmas 3.1–3.5, we can show Lemma 3.7 and Theorem 2.3.

Lemma 3.7.

If , , and , then is a Fredholm operator of index zero. Furthermore, the linear continuous projector operator can be defined by

and the linear operator can be written by

Furthermore,

Notice that

Proof of Theorem 2.3.

Let

Then, for , ; thus , ; hence

thus, from assumption (H4), there exists , such that and in view of , we obtain

From , there exists , such that . Thus, from , one has

We let ; hence from (3.58) and (3.59), we have

thus, by using the same method as in the proof of Lemmas 3.2 and 3.3, we can prove that is bounded too. Similar to the other proof of Lemmas 3.4–3.7 and Theorem 2.2, we can verify Theorem 2.3.

Finally, we give two examples to demonstrate our results.

Example 3.8.

Consider the following boundary value problem:

where ,

and satisfying , , and , then we can choose , , and , for ; thus

Since

and has the same sign as when , we may choose , and then the conditions (H1)–(H3) of Theorem 2.2 are satisfied. Theorem 2.2 implies that BVP (3.61) has at least one solution, .

Example 3.9.

Consider the following boundary value problem:

where ,

and satisfying , and , then we can choose , , and , for ; thus

Similar to Example 3.8, we have

and has the same sign as when , we may choose , and then all conditions of Theorem 2.3 are satisfied. Theorem 2.3 implies that BVP (3.65) has at least one solution .