Throughout this paper, we assume the following:
(H1)
;
(H2)
, and
;
(H3)
is continuous and does not vanish identically on any subinterval of
, and
;
(H4)
is continuous.
Lemma 2.1.
Suppose that (H1) and (H2) hold. Then
(i) the initial value problem
has a unique solution
and
;
(ii) the initial value problem
has a unique solution
and
.
Proof.
We only prove (i). (ii) can be treated in the same way.
Suppose that
and
is a solution of (2.1), that is,
Let
Multiplying both sides of (2.3) by
, then
Since
and
, integrating (2.5) on
, we have
Moreover, integrating (2.6) on
,
, we have
Let
Clearly,
, and (2.7) reduces to
By using Fubini's theorem, we have
Therefore,
which implies that
is a solution of integral equation (2.11).
Conversely, if
is a solution of (2.11) with
, by reversing the above argument we could deduce that the function
is a solution of (2.1) and satisfy
and
. Therefore, to prove that (2.1) has a unique solution,
, and
is equivalent to prove that (2.11) has a unique solution
.
To do this, we endow the following norm in
:
Let
be operator defined by
Since
then,
is well defined. Set
Then, for any
,
and subsequently,
Thus,
Since
,
has a unique fixed point
by Banach contraction principle. That is, (2.11) has a unique solution
.
Remark 2.2.
Lemma 2.1 generalizes Theorem
of [1], where
.
Lemma 2.3.
Suppose that (H1) and (H2) hold. Then
(i)
is nondecreasing in
;
(ii)
is nonincreasing in
.
Proof.
We only prove (i). (ii) can be treated in the same way.
Suppose on the contrary that
is not nondecreasing in
. Then there exists
such that
This together with the equation
implies that
which is a contradiction!
Remark 2.4.
From Lemmas 2.1 and 2.3, there exist positive constants
,
,
, and
such that
In fact, since
we have that
and
. Then, there exist constants
and
, such that
that is
In the following, we will show that
. Suppose on the contrary, if there exist
, such that
then,
, which is a contradiction!
The other inequality can be treated in the same manner.
Lemma 2.5.
Suppose that (H1), (H2), and (H3) hold. Then
Proof.
We only prove the first equality; the other can be treated in the same way. From Remark 2.4 and (H3), we have
Lemma
of [2] together with the facts that
and (H3) implies that
Combining (2.27) and (2.28), we have
Lemma 2.6.
Suppose that (H1), (H2), and (H3) hold. Then the problem
has a unique solution
where
Moreover,
on
.
Proof.
By Lemma 2.3 and (2.32), we have
This together with Remark 2.4 implies that the right side of (2.31) is well defined.
Now we check that the function
satisfies (2.30). In fact,
Therefore,
Equation (2.34) and Lemma 2.5 imply that
Since
for
, then
Let
with the norm
and let
be a cone in
defined by
Lemma 2.7.
Suppose that (H1)–(H3) hold and
is a positive solution of (2.30). Then
where
Furthermore, for any
, there exists corresponding
such that
Proof.
In fact, if
, then
and if
, then
Combining this and
, we have
Take
Then Lemma 2.3 guarantees that
, and Lemma 2.7 guarantees that (2.43) holds.
Remark 2.8.
From Lemma 2.7 and Remark 2.4, we have
Now, for any
, we can define the operator
by
Lemma 2.9.
Let (H1)–(H4) hold. Then
is a completely continuous operator.
Proof.
From (H3) and (H4) and Lemma 2.6, it is easy to see that
, and
is continuous by the Lebesgue
s dominated convergence theorem.
Let
be any bounded set. Then (H3) and (H4) imply that
is a bounded set in
.
Since
then this together with the similar proof of Lemma
of [2] yields
From this fact, it is easy to verify that
is equicontinuous. Therefore, by the Arzela-Ascoli theorem,
is a completely continuous operator.