In this section, we state and prove the first existence result for (1.1). The proof is based on the following nonlinear alternative of Leray-Schauder, which can be found in [39]. This part can be regarded as the scalar version of the results in [4].

Lemma 3.1.

Assume
is a relatively compact subset of a convex set
in a normed space
. Let
be a compact map with
. Then one of the following two conclusions holds:

(a)
has at least one fixed point in

(b)thereexist
and
such that

Theorem 3.2.

Suppose that
satisfies (A) and
satisfies the following.

(H_{1})There exist constants
and
such that

(H_{2})There exist continuous, nonnegative functions
and
such that

is nonincreasing and
is nondecreasing in
.

(H_{3})There exists a positive number
such that
and

Then for each
, (1.1) has at least one positive periodic solution
with
for all
and
.

Proof.

The existence is proved using the Leray-Schauder alternative principle, together with a truncation technique. The idea is that we show that

has a positive periodic solution

satisfying

for

and

If this is true, it is easy to see that

will be a positive periodic solution of (1.1) with

since

Since (
) holds, we can choose
such that
and

Let

. Consider the family of equations

where

and

Problem (3.7) is equivalent to the following fixed point problem:

where

is defined by

We claim that any fixed point
of (3.9) for any
must satisfy
. Otherwise, assume that
is a fixed point of (3.9) for some
such that
. Note that

By the choice of

,

. Hence, for all

, we have

Thus we have from condition (

), for all

,

This is a contradiction to the choice of
and the claim is proved.

From this claim, the Leray-Schauder alternative principle guarantees that

has a fixed point, denoted by

, in

, that is, equation

has a periodic solution
with
. Since
for all
and
is actually a positive periodic solution of (3.17).

In the next lemma, we will show that there exists a constant
such that

for
large enough.

In order to pass the solutions
of the truncation equations (3.17) to that of the original equation (3.4), we need the following fact:

for some constant

and for all

. To this end, by the periodic boundary conditions,

for some

. Integrating (3.17) from 0 to

, we obtain

The fact
and (3.19) show that
is a bounded and equicontinuous family on
. Now the Arzela-Ascoli Theorem guarantees that
has a subsequence,
, converging uniformly on
to a function
. Moreover,
satisfies the integral equation

Letting

, we arrive at

where the uniform continuity of
on
is used. Therefore,
is a positive periodic solution of (3.4).

Lemma 3.3.

There exist a constant
and an integer
such that any solution
of (3.17) satisfies (3.18) for all
.

Proof.

The lower bound in (3.18) is established using the strong force condition (

) of

. By condition (

), there exists

small enough such that

Take
such that
and let
. For
, let

We claim first that
. Otherwise, suppose that
for some
. Then from (3.24), it is easy to verify

Integrating (3.17) from 0 to

, we deduce that

This is a contradiction. Thus
.

Now we consider the minimum values
. Let
. Without loss of generality, we assume that
, otherwise we have (3.18). In this case,

for some

. As

, there exists

(without loss of generality, we assume

) such that

and

for

By (3.24), it can be checked that

Thus for
, we have
As
,
for all
and the function
is strictly increasing on
. We use
to denote the inverse function of
restricted to
.

In order to prove (3.18) in this case, we first show that, for
,

Otherwise, suppose that
for some
. Then there would exist
such that
and

Multiplying (3.17) by

and integrating from

to

, we obtain

By the facts

and

one can easily obtain that the right side of the above equality is bounded. As a consequence, there exists

such that

On the other hand, by the strong force condition (
), we can choose
large enough such that

for all
. So (3.30) holds for

Finally, multiplying (3.17) by
and integrating from
to
, we obtain

(We notice that the estimate (3.30) is used in the second equality above). In the same way, one may readily prove that the right-hand side of the above equality is bounded. On the other hand, if

by (

),

if
Thus we know that
for some constant
.

From the proof of Theorem 3.2 and Lemma 3.3, we see that the strong force condition (
) is only used when we prove (3.18). From the next theorem, we will show that, for the case
, we can remove the strong force condition (
), and replace it by one weak force condition.

Theorem 3.4.

Assume that (
) and (
)–(
) are satisfied. Suppose further that

(H_{4})for each constant
, there exists a continuous function
such that
for all
.

Then for each
with
(1.1) has at least one positive periodic solution
with
for all
and
.

Proof.

We only need to show that (3.18) is also satisfied under condition (

) and

The rest parts of the proof are in the same line of Theorem 3.2. Since (

) holds, there exists a continuous function

such that

for all

. Let

be the unique periodic solution to the problems (2.1)–(2.2) with

. That is

Corollary 3.5.

Assume that
satisfies (
) and
. Then

(i)if
then for each
(1.5) has at least one positive periodic solution for all
;

(ii)if
, then for each
(1.5) has at least one positive periodic solution for each
here
is some positive constant.

(iii)if
, then for each
with
(1.5) has at least one positive periodic solution for all
;

(iv)if
, then for each
with
(1.5) has at least one positive periodic solution for each
.

Proof.

We apply Theorems 3.2 and 3.4. Take

then (

) is satisfied, and the existence condition (

) becomes

for some

. Note that condition (

) is satisfied when

, while (

) is satisfied when

. So (1.5) has at least one positive periodic solution for

Note that
if
and
if
. Thus we have (i)–(iv).