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
Note that
and, concerning
, we distinguish two cases. If
is such that
, then
and, if
is such that
, then
In consequence,
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
Moreover,
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
This proves that
Hence,
In consequence,
Finally, we check that
is compact. Let
be a bounded set in
.
-
(i)
First, we check that
is a bounded set in
.
Indeed,
Hence
and then
which implies that
is a bounded set in
.
-
(ii)
Now, we prove that
is an equicontinuous set in
. Following the calculus in (a), we show that
tends to zero as
.
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