In this section, we present the expression and properties of Green's function associated with boundary value problem ().
Lemma 2.1.
Assume that
Then for any
, the unique solution of boundary value problem
is given by
where
Proof.
By Lemma 1.4, we can reduce the equation of problem (2.1) to an equivalent integral equation
By
, there is
. Thus,
Differentiating (2.7), we have
By (2.8) and
we have
Similarly, we can obtain that
Then
By
, we have
Therefore, the unique solution of BVP (2.1) is
where
is defined by (2.4).
From (2.11), we have
It follows that
Substituting (2.13) into (2.11), we obtain
where
and
are defined by (2.3), (2.4), and (2.5), respectively. The proof is complete.
From (2.3), (2.4), and (2.5), we can prove that
and
have the following properties.
Proposition 2.2.
The function
defined by (2.4) satisfies
(i)
is continuous for all
;
(ii)for all
, one has
where
Proof.
-
(i)
It is obvious that
is continuous on
and
when
.
For
, we have
So, by (2.4), we have
Similarly, for
, we have
.
-
(ii)
Since
, it is clear that
is increasing with respect to
for
.
On the other hand, from the definition of
, for given
, we have
Let
Then, we have
and so,
Noticing
, from (2.22), we have
Then, for given
, we have
arrives at maximum at
when
. This together with the fact that
is increasing on
, we obtain that (2.15) holds.
Remark 2.3.
From Figure 1, we can see that
for
. If
, then
Remark 2.4.
From Figure 2, we can see that
is increasing with respect to
.
Remark 2.5.
From Figure 3, we can see that
for
, where
.
Remark 2.6.
Let
. From (2.15), for
, we have
Remark 2.7.
From (2.25), we have
Remark 2.8.
From Figure 4, it is easy to obtain that
is decreasing with respect to
, and
Proposition 2.9.
There exists
such that
Proof.
For
, we divide the proof into the following three cases for
.
Case 1.
If
, then from (i) of Proposition 2.2 and Remark 2.5, we have
It is obvious that
and
are bounded on
. So, there exists a constant
such that
Case 2.
If
, then from (2.4), we have
On the other hand, from the definition of
, we obtain that
takes its maximum
at
. So
Therefore,
. Letting
, we have
Case 3.
If
, from (i) of Proposition 2.2, it is clear that
In view of Remarks 2.6–2.8, we have
From (2.35), there exists a constant
such that
Letting
and using (2.30), (2.33), and (2.36), it follows that (2.28) holds. This completes the proof.
Let
Proposition 2.10.
If
, then one has
(i)
is continuous for all
;
(ii)
.
Proof.
Using the properties of
, definition of
, it can easily be shown that (i) and (ii) hold.
Theorem 2.11.
If
, the function
defined by (2.3) satisfies
(i)
is continuous for all
;
(ii)
for each
, and
where
is defined by (2.16),
is defined in Proposition 2.9.
Proof.
-
(i)
From Propositions 2.2 and 2.10, we obtain that
is continuous for all
, and
.
-
(ii)
From (ii) of Proposition 2.2 and (ii) of Proposition 2.10, we have that
for each
.
Now, we show that (2.38) holds.
In fact, from Proposition 2.9, we have
Then the proof of Theorem 2.11 is completed.
Remark 2.12.
From the definition of
, it is clear that
.