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.
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.
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.
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 .
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
This proves that
Finally, we check that is compact. Let be a bounded set in .
First, we check that is a bounded set in .
which implies that is a bounded set in .
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.
Assume that (4.3) and (4.4) hold. Then the problem (1.1)–(3.4) has at least one solution in
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.
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).
Assume that (4.4) holds. Then the problem (1.1)–(3.4) has a unique solution in provided that
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).
If , condition (4.32) is reduced to