This paper uses a fixed point theorem in cones to investigate the multiple positive solutions of a boundary value problem for secondorder impulsive singular differential equations on the halfline. The conditions for the existence of multiple positive solutions are established.
1. Introduction
Consider the following nonlinear singular SturmLiouville boundary value problems for secondorder impulsive differential equation on thehalfline:
where , , , , , with on and ; with , in which . , where and are, respectively, the left and right limits of at , , .
The theory of singular impulsive differential equations has been emerging as an important area of investigation in recent years. For the theory and classical results, we refer the monographs to [1, 2] and the papers [3–19] to readers. We point out that in a secondorder differential equation , one usually considers impulses in the position and the velocity . However, in the motion of spacecraft one has to consider instantaneous impulses depending on the position that result in jump discontinuities in velocity, but with no change in position [20]. The impulses only on the velocity occur also in impulsive mechanics [21].
In recent paper [3], by using the Krasnoselskii's fixed point theorem, Kaufmann has discussed the existence of solutions for some secondorder boundary value problem with impulsive effects on an unbounded domain. In [22] Sun et al. and [23] Liu et al., respectively, discussed the existence and multiple positive solutions for singular SturmLiouville boundary value problems for secondorder differential equation on the halfline. But the Multiple positive solutions of this case with both singularity and impulses are not to be studied. The aim of this paper is to fill up this gap.
The rest of the paper is organized as follows. In Section 2, we give several important lemmas. The main theorems are formulated and proved in Section 3. And in Section 4, we give an example to demonstrate the application of our results.
2. Several Lemmas
Lemma 2.1 (see [23]).
If conditions and are satisfied, then the boundary value problem
has a unique solution for any . Moreover, this unique solution can be expressed in the form
where is defined by
Remark 2.2.
It is easy to prove that has the following properties:
(1) is continuous on
(2) is continuous differentiable on , except ,
(3),
(4),
(5),
(6)for all , , , where
Obviously, .
For the interval , and the corresponding in Remark 2.2, we define = : , and exist, . = exists. , and . It is easy to see that is a Banach space with the norm , and is a positive cone in . For details of the cone theory, see [1]. is called a positive solution of BVP (1.1) if for all and satisfies (1.1).
As we know that the AscoliArzela Theorem does not hold in infinite interval , we need the following compactness criterion:
Lemma 2.3 (see [22]).
Let . Then is relatively compact in if the following conditions hold.
(i) is uniformly bounded in .
(ii)The functions from are equicontinuous on any compact interval of .
(iii)The functions from are equiconvergent, that is, for any given , there exists a such that , for any , .
The main tool of this work is a fixed point theorem in cones.
Lemma 2.4 (see [4]).
Let X be a Banach space and is a positive cone in . Assume that are open subsets of with , . Let be a completely continuous operator such that
(i) for all .
(ii)there exists a such that , for all and .
Then has a fixed point in .
Remark 2.5.
If (i) is satisfied for and (ii) is satisfied for , then Lemma 2.4 is still true.
Lemma 2.6 (see [3]).
The function is a solution of the BVP (1.1) if and only if satisfies the equation
The proof of this result is based on the properties of the Green function, so we omit it as elementary.
Define
Obviously, the BVP (1.1) has a solution if and only if is a fixed point of the operator defined by (2.6).
Let us list some conditions as follows.
There exist two nonnegative functions: , such that . , may be singular at . , , are continuous.
,
Lemma 2.7.
If are satisfied, then for any bounded open set , is a completely continuous operator.
Proof.
For any bounded open set , there exists a constant such that for any .
First, we show that is well defined. Let . From , we have , where , , and
Hence, is well defined. For any , we have
Thus, by the Lebesgue dominated convergence theorem and the fact that is continuous on , we have, for any , ,
Therefore, . By the property (3) of , it is easy to get .
On the other hand, by (2.6) we have, for any and ,
Then by , the property (5) of Remark 2.2 and the Lebesgue dominated convergence theorem, we have
Thus .
For any , we get
So
On the other hand, for we obtain
Thus .
Next, we prove that is continuous. Let in , then We prove that . For any , by , there exists a constant such that
On the other hand, by the continuities of on and the continuities of on , for the above , there exists a such that, for any , ,
From , for the above , there exists a sufficiently large number such that, when , we have
Therefore, by (2.15)–(2.17), we have, for ,
This implies that the operator is continuous.
Finally we show that is a compact operator. In fact for any bounded set , there exists a constant such that for any . Hence, we obtain
Therefore, is uniformly bounded in .
Given , for any , as the proof of (2.9), we can get that are equicontinuous on . Since is arbitrary, are locally equicontinuous on . By (2.6), , , and the Lebesgue dominated convergence theorem, we have
Hence, the functions from are equiconvergent. By Lemma 2.3, we have that is relatively compact in . Therefore, is completely continuous. This completed the proof of Lemma 2.7.
3. Main Results
For convenience and simplicity in the following discussion, we use the following notations:
Theorem 3.1.
Let hold. Then the BVP (1.1) has at least two positive solutions satisfying if the following conditions hold:
there exists a such that for all , a.e. .
Proof.
By the definition of and , for any , there exist such that
Define the open sets
Let , then . Now we prove that
If not, then there exist and such that . Let then for any we have
This implies , a contradiction. Therefore, (3.4) holds.
That by the definition of and , for any there exist such that
Define the open sets:
As the proof of (3.4), we can get that
On the other hand, for any , choose in such that
By the definition of , , for the above , there exists , when ; thus, we have
Define
Then, for any and , we can obtain
Therefore,
Thus, we can obtain the existence of two positive solutions and satisfying by using Lemma 2.4 and Remark 2.5, respectively.
Using a similar proof of Theorem 3.1, we can get the following conclusions.
Theorem 3.2.
Let hold. Then the BVP (1.1) has at least two positive solutions satisfying if the following conditions hold:
there exists such that , for all , a.e. .
Corollary 3.3.
In Theorems 3.1 and 3.2, if conditions and are replaced by and , respectively, then the conclusions also hold.
or or
, , , .
Remark 3.4.
Notice that, in the above conclusions, we suppose that the singularity only exist in , that is, as . If we permit as or and as , then the discussion will be much more complex. Now we state the corresponding results.
Let us define the following.
There exist four nonnegative functions , such that , and is nondecreasing on . , , are continuous.
, , where ,
Theorem 3.5.
Suppose hold, then the BVP (1.1) has at least two positive solutions satisfying if hold.
Proof.
Define , for all . We only need to proove is a completely continuous operator. Then the rest of the proof is the same as that Theorem 3.1. Notice that
and change to , , , then the same as the proof of Lemma 2.7, it is easy to compute that is a completely continuous operator.
Corresponding to Theorem 3.2 and Corollary 3.3, there are Theorem 3.6 and Corollary 3.7. We just list here without proof.
Theorem 3.6.
Suppose hold, then the BVP (1.1) has at least two positive solutions satisfying , if hold.
Corollary 3.7.
In Theorems 3.5 and 3.6, if conditions and are replaced by and , respectively, then the conclusions also hold.
4. Example
To illustrate how our main results can be used in practice we present the following example.
Example 4.1.
Consider the following boundary value problem:
Conclusion 1.
BVP (4.1) has at least two positive solutions , satisfying .
Proof.
Let , , , . Then by simple computation we have
where . Furthermore, and
Let . Then . Thus are satisfied. It is easy to get that . Let . Then
Hence, are satisfied. Therefore, by Corollary 3.3, problem (4.1) has at least two positive solutions , satisfying . The proof is completed.
Acknowledgment
This work is supported by the National Nature Science Foundation of P. R.China (10871063) and Scientific Research Fund of Hunan Provincial Education Department (07A038), partially supported by Ministerio de Educacion y Ciencia and FEDER, Project MTM200761724, and by Xunta de Galicia and FEDER, project no.PGIDIT06PXIB207023PR.
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