Definition 3.1.

A function
is said to be a solution of (1.1) if
satisfies (1.1).

In what follows one assumes that
One needs the following auxiliary result.

Lemma 3.2.

. Let

. Then the function defined by

is the unique solution of the boundary value problem

Proof.

Let

be a solution of the problem (3.2). Then integratingly, we obtain

Now, multiply (3.6) by

and integrate over

, to get

Substituting in (3.6) we have

Set

Note that

Our first result reads

Theorem 3.3.

Assume that
is an
-Carathéodory function and the following hypothesis

(A1) There exists

such that

then the BVP (1.1) has a unique solution.

Proof.

Transform problem (1.1) into a fixed-point problem. Consider the operator

defined by

We will show that

is a contraction. Indeed, consider

Then we have for each

showing that,
is a contraction and hence it has a unique fixed point which is a solution to (1.1). The proof is completed.

We now present an existence result for problem (1.1).

Theorem 3.4.

Suppose that hypotheses

(H1) The function
is an
-Carathéodory,

(H2) There exist functions

and

such that

are satisfied. Then the BVP (1.1) has at least one solution. Moreover the solution set

is compact.

Proof.

Transform the BVP (1.1) into a fixed-point problem. Consider the operator
as defined in Theorem 3.3. We will show that
satisfies the assumptions of the nonlinear alternative of Leray-Schauder type. The proof will be given in several steps.

Step 1 (
is continuous).

Let

be a sequence such that

in

Then

Since

is

-Carathéodory and

then

Step 2 (
maps bounded sets into bounded sets in
).

Indeed, it is enough to show that there exists a positive constant
such that for each
one has
.

Let

. Then for each

, we have

By (H2) we have for each

Then for each

we have

Step 3 (
maps bounded set into equicontinuous sets of
).

Let

,

and

be a bounded set of

as in Step 2. Let

and

we have

As
the right-hand side of the above inequality tends to zero. Then
is equicontinuous. As a consequence of Steps 1 to 3 together with the Arzela-Ascoli theorem we can conclude that
is completely continuous.

Step 4 (A priori bounds on solutions).

Let

for some

. This implies by

that for each

we have

If

we have

and consider the operator
From the choice of
, there is no
such that
for some
As a consequence of the nonlinear alternative of Leray-Schauder type [15], we deduce that
has a fixed point
in
which is a solution of the problem (1.1).

Now, prove that

is compact. Let

be a sequence in

, then

As in Steps 3 and 4 we can easily prove that there exists

such that

and the set

is equicontinuous in

hence by Arzela-Ascoli theorem we can conclude that there exists a subsequence of

converging to

in

Using that fast that

is an

-Carathédory we can prove that

Thus
is compact.