equipped the norm
then Ω is a Banach space.
We first give some lemmas as follows:
Lemma 3.1 Problem (1.1)-(1.2) is equivalent to the following integral equation
Proof The sufficiency is obvious, we only need to prove the necessity.
is a solution of the problem (1.1)-(1.2). Integrating both sides of (1.1) of α
order from 0 to t
with respect to t
, it follows that
According to (1.2) and (3.4), we have
Combining (3.4) and (3.5), we obtain
According to (3.3), it is easy to show that (3.2) holds. The proof is completed. □
Lemma 3.2 For any, is continuous, andfor any.
Proof The continuity of for is obvious.
we only need to show that
, the rest of the proof is similar or obvious. From the definition of
, we have
for . The proof is completed. □
The new Green’s function has the following properties:
Lemma 3.3is continuous for
Lemma 3.4 For any, is nonincreasing with respect to. Especially, for any, for, andfor. That is, where
, we have from Lemma 3.1, (3.6) and (3.9) that the integral Equation (3
.2) can be rewritten as follows:
and (3.10) is equivalent to the following
We can divide our proof into the following two steps:
First, we replace
by any real number μ
, then (3.12) can be rewritten as
It suffices to show that for any given real number μ, (3.13) has a solution , which implies that Equation (1.1) has a solution which satisfies the first boundary value condition .
Second, we show that there exists a μ such that the solution of (3.13) satisfies , which implies that the solution of (1.1) also satisfies the boundary value condition .
In this section, we will prove the first step. For convenience sake, we define an operator T
on the set Ω as follows:
Lemma 3.5 Suppose that, and (2.4) hold, then the operator T is completely continuous in Ω.
Proof It is easy to show that the operator T maps Ω into itself. We divide the proof into the following three steps.
Step 1. is continuous with respect to .
In fact, suppose that
is a sequence in Ω, and
is continuous with respect to
, and it is obvious that
is uniformly continuous with respect to
from Lemma 3.3, then for any
, there exists an integer N
which follows from (3.14)-(3.15) that
Thus, the operator T is continuous in Ω.
Step 2. T maps bounded set in Ω into bounded set.
is a bounded set with
. Then, we have from (2.4) and (3.14) that
This gives that the operator T maps bounded set into bounded set in Ω.
Step 3. T is equicontinuous in Ω.
It suffices to show that for any
. We consider the following three cases:
We only prove the case (i), the rest two cases are similar. Since B
is bounded, then there exists a
. According to (3.14), we have
According to Step 1-Step 3, the operator T is completely continuous in Ω. The proof is completed. □
Further, we have
Lemma 3.6 Suppose that, and (2.4) and (2.6) holds, then, for any real number μ, the integral Equation (3.13) has at least one solution.
We only need to show that the operator T
is priori bounded. Let
Define a set
To show the existence of a fixed point of T by Lemma 2.3, we need to verify that the second possibility in Lemma 2.3 cannot happen.
In fact, assume that there exists
. It follows that
Here we have the use of the inequality
It is obvious that (3.18) contradicts our assumption that . Therefore, by Lemma 2.3, it follows that T has a fixed point . Hence, the integral Equation (3.14) has at least a solution . The proof is completed. □