In this section we will be concerned with the existence and uniqueness of solution to the nonlinear problem (1.1)–(3.4). To this end, we need the following fixed point theorem of Schaeffer.

Theorem 4.1.

Assume
to be a normed linear space, and let operator
be compact. Then either

(i)the operator
has a fixed point in
, or

(ii)the set
is unbounded.

If

is a solution of problem (1.1)–(3.4), then it is given by

where
is Green's function defined in Theorem 3.2.

Define the operator

by

Then the problem (1.1)–(3.4) has solutions if and only if the operator equation
has fixed points.

Lemma 4.2.

Suppose that the following hold:

(i)there exists a constant

such that

(ii)there exists a constant

such that

Then the operator
is well defined, continuous, and compact.

Proof.

- (a)
We check, using hypothesis (4.3), that

, for every

. Indeed, for any

,

, we have

From the previous expression, we deduce that, if

, then

Indeed, note that the integral

is bounded by

A similar argument is useful to study the behavior of the last three terms of the long inequality above. On the other hand, if we denote by

the second term in the right-hand side of that inequality, then it is satisfied that

and, concerning

, we distinguish two cases. If

is such that

, then

and, if

is such that

, then

The first term in the right-hand side of the previous inequality clearly tends to zero as

. On the other hand, denoting by

the integer part function, we have

The finite sum obviously has limit zero as

. The infinite sum is equal to

and its limit as
is zero. Note that
is bounded above by
.

The previous calculus shows that

, for

, hence we can define

.

- (b)
Next, we prove that
is continuous.

Note that, for

and for every

, we have, using hypothesis (4.4),

Using the definition of

, we get

Using that

for

,

for

, and

for

we obtain

Note that the Beta function, also called the Euler integral of the first kind,

where

and

, satisfies that

. In particular,

. On the other hand, using the change of variable

, we deduce that

Finally, we check that

is compact. Let

be a bounded set in

.

- (i)
First, we check that
is a bounded set in
.

which implies that

is a bounded set in

.

- (ii)

Then
is equicontinuous in the space
, where
, for
.

As a consequence of (i) and (ii),
is a bounded and equicontinuous set in the space
.

Hence, for a sequence

in

,

has a subsequence converging to

, that is,

Taking

, we get

which means that
, which proves that
is compact.

Theorem 4.3.

Assume that (4.3) and (4.4) hold. Then the problem (1.1)–(3.4) has at least one solution in

Proof.

Consider the set
.

Let

be any element of

, then

for some

. Thus for each

, we have

As in Lemma 4.2, (i), we have

which implies that the set
is bounded independently of
. Using Lemma 4.2 and Theorem 4.1, we obtain that the operator
has at least a fixed point.

Remark 4.4.

In Lemma 4.2, condition (4.3) is used to prove that the operator
is continuous. Hence, in Lemma 4.2 and, in consequence, in Theorem 4.3, we can assume the weaker condition.

(i)For each

fixed, there exists

such that

instead of (4.3).

However, to prove the existence and uniqueness of solution given in the following theorem, we need to assume the Lipschitzian character of
(condition (4.3).

Theorem 4.5.

Assume that (4.4) holds. Then the problem (1.1)–(3.4) has a unique solution in

provided that

Proof.

We use the Banach contraction principle to prove that the operator
has a unique fixed point.

Using the calculus in (b) Lemma 4.2,
is a contraction by condition (4.32). As a consequence of Banach fixed point theorem, we deduce that
has a unique fixed point which gives rise to a unique solution of problem (1.1)–(3.4).

Remark 4.6.

If

, condition (4.32) is reduced to