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Existence of Solutions for a Weighted Laplacian Impulsive Integrodifferential System with Multipoint and Integral Boundary Value Conditions
Journal of Inequalities and Applications volume 2010, Article number: 392545 (2010)
Abstract
By the LeraySchauder's degree, the existence of solutions for a weighted Laplacian impulsive integrodifferential system with multipoint and integral boundary value conditions is considered. The sufficient results for the existence are given under the resonance and nonresonance cases, respectively. Moreover, we get the existence of nonnegative solutions at nonresonance.
1. Introduction
In this paper, we consider the existence of solutions for the following weighted Laplacian integrodifferential system:
where , , , , with the following impulsive boundary value conditions
where and , is called the weighted Laplacian; , ; , and ; is nonnegative, with ; ; and are linear operators defined by , , , where .
If and , we say the problem is nonresonant, but if and , we say the problem is resonant.
Throughout the paper, means function which uniformly convergent to 0 (as ); for any , will denote the th component of ; the inner product in will be denoted by ; will denote the absolute value and the Euclidean norm on . Denote , , , , , where , . Denote the interior of , . Let
satisfies , for all , and ;
For any , denote . Obviously, is a Banach space with the norm , and is a Banach space with the norm . Denote with the norm , for all , where .
For simplicity, we denote and by and , respectively, and denote
In recent years, there has been an increasing interest in the study of differential equations with nonstandard growth conditions. These problems have many interesting applications (see [1–4]). Many results have been obtained on these kinds of problems, for example [5–17]. If (a constant), (1.1)–(1.4) becomes the well known Laplacian problem. If is a general function, one can see easily that in general, while , so represents a nonhomogeneity and possesses more nonlinearity, thus is more complicated than . For example, we have the following.
(a)In general, the infimum of eigenvalues for the Laplacian Dirichlet problems is zero, and only under some special conditions (see [10]). When () is an interval, the results in [10] show that if and only if is monotone. But the property of is very important in the study of Laplacian problems, for example, in [18], the authors use this property to deal with the existence of solutions.
(b)If and (a constant) and , then is concave, this property is used extensively in the study of onedimensional Laplacian problems (see [19]), but it is invalid for . It is another difference between and .
Recently, there are many works devoted to the existence of solutions to the Laplacian impulsive differential equation boundary value problems, for example [20–28]. Many methods had been applied to deal with these problems, for example subsupersolution method, fixed point theorem, monotone iterative method, coincidence degree, variational principles (see [29]), and so forth. Because of the nonlinearity of , results about the existence of solutions for Laplacian impulsive differential equation boundary value problems are rare (see [30]). In [31], using coincidence degree method, the present author investigate the existence of solutions for Laplacian impulsive differential equation with multipoint boundary value conditions. Integral boundary conditions for evolution problems have various applications in chemical engineering, thermoelasticity, underground water flow and population dynamics, there are many papers on the differential equations with integral boundary value problems, for example, [32–35].
In this paper, when is a general function, we investigate the existence of solutions and nonnegative solutions for the weighted Laplacian impulsive integrodifferential system with multipoint and integral boundary value conditions. Our results contain both the cases of resonance and nonresonance, and the method is based upon LeraySchauder's degree. Moreover, this paper will consider the existence of (1.1) with (1.2), (1.4) and the following impulsive condition:
where , the impulsive condition (1.8) is called linear impulsive condition (LI for short), and (1.3) is called nonlinear impulsive condition (NLI for short). In generaly, Laplacian impulsive problems have two kinds of impulsive conditions, that is, LI and NLI.
Let , the function is assumed to be Caratheodory, by this we mean the following:
(i)for almost every the function is continuous;
(ii)for each the function is measurable on ;
(iii)for each there is a such that, for almost every and every with , , , , one has
We say a function is a solution of (1.1) if with absolutely continuous on , , which satisfies (1.1) a.e. on .
In this paper, we always use to denote positive constants, if it cannot lead to confusion. Denote
We say satisfies sub growth condition, if satisfies
where , and .
This paper is organized as four sections. In Section 2, we present some preliminary and give the operator equation which has the same solutions of (1.1)–(1.4). In Section 3, we give the existence of solutions and nonnegative solutions for system (1.1)–(1.4) at nonresonance. Finally, in Section 4, we give the existence of solutions for system (1.1)–(1.4) at resonance.
2. Preliminary
For any , denote . Obviously, has the following properties.
Lemma 2.1 (see [31]).
is a continuous function and satisfies the following.
(i)For any , is strictly monotone, satisfying
(ii)There exists a function , as , such that
It is well known that is an homeomorphism from to for any fixed . Denote
It is clear that is continuous and sends bounded sets to bounded sets.
In this section, we will do some preparation and give the operator equation which has the same solutions of (1.1)–(1.4). At first, let us now consider the following simple impulsive problem with boundary value condition (1.4)
where ; .
We will discuss (2.4) with (1.4) in the cases of resonance and nonresonance, respectively.
2.1. The Case of Nonresonance
Suppose and . If is a solution of (2.4) with (1.4), we have
Denote , , . It is easy to see that is dependent on and . Define operator as
By solving for in (2.5) and integrating, we find
which together with the boundary value condition (1.4) implies
Denote with the norm , for all , then is a Banach space.
We define as
then is continuous. Throughout the paper, we denote . It is easy to see the following.
Lemma 2.2.
The function is continuous and sends bounded sets to bounded sets. Moreover, for any , we have
We denote the Nemytskii operator associated to defined by
We define as
where , .
It is clear that is continuous and sends bounded sets of to bounded sets of , and hence it is compact continuous.
If is a solution of (2.4) with (1.4), we have
For fixed , we define as
Define as
Lemma 2.3.

(i)
The operator is continuous and sends equiintegrable sets in to relatively compact sets in .

(ii)
The operator is continuous and sends bounded sets in to relatively compact sets in .
Proof.
(i) It is easy to check that , for all , for all . Since and
it is easy to check that is a continuous operator from to .
Let be an equiintegrable set in , then there exists , such that
We want to show that is a compact set.
Let be a sequence in , then there exists a sequence such that . For any , we have
Hence the sequence is uniformly bounded and equicontinuous. By AscoliArzela theorem, there exists a subsequence of (which we rename the same) which is convergent in . According to the bounded continuous of the operator , we can choose a subsequence of (which we still denote by ) which is convergent in , then is convergent in .
Since
it follows from the continuity of and the integrability of in that is convergent in . Thus is convergent in .

(ii)
It is easy to see from (i) and Lemma 2.2.
This completes the proof.
Let us define as .
It is easy to see that is compact continuous.
Lemma 2.4.
Suppose and , then is a solution of (1.1)–(1.4) if and only if is a solution of the following abstract operator equation
Proof.
Suppose is a solution of (1.1)–(1.4). From the definition of and , similar to the discussion before Lemma 2.2, we know that is a solution of (2.20).
Conversely, if is a solution of (2.20), then (1.2) is satisfied.
From (2.20), we have
It follows from (2.21) that (1.3) is satisfied.
From (2.21) and the definition of , we have
From (2.20) and the definition of , it is easy to check that
It follows from (2.22) and (2.23) that (1.4) is satisfied.
Hence is a solutions of (1.1)–(1.4). This completes the proof.
2.2. The Case of Resonance
Suppose and . If is a solution of (2.4) with (1.4), we have
Denote , , . It is easy to see that is dependent on and .
The boundary value condition (1.4) implies that
For any , we denote
Lemma 2.5.
The function has the following properties.
(i)For any fixed , the equation
has unique solution .
(ii)The function , defined in (i), is continuous and sends bounded sets to bounded sets. Moreover, for any , we have
where
Proof.

(i)
From Lemma 2.1, it is immediate that
(2.30)
and hence, if (2.27) has a solution, then it is unique.
Set
Suppose , it is easy to see that there exists some such that, the absolute value of the th component of satisfies
Thus the th component of keeps sign on , then it is not hard to check that the th component of keeps the same sign of .
Thus . Let us consider the equation
According to the preceding discussion, all the solutions of (2.33) belong to . Therefore
it means the existence of solutions of .
In this way, we define a function , which satisfies .

(ii)
By the proof of (i), we also obtain sends bounded sets to bounded sets, and
(2.35)
It only remains to prove the continuity of . Let is a convergent sequence in and , as . Since is a bounded sequence, it contains a convergent subsequence . Suppose as . Since , letting , we have , which together with (i) implies , it means is continuous. This completes the proof.
We define as
where , .
It is clear that is continuous and sends bounded sets of to bounded sets of , and hence it is a compact continuous mapping.
Let us define
and as
Similar to the proof of Lemma 2.3, we have the following lemma.
Lemma 2.6.
The operator is continuous and sends equiintegrable sets in to relatively compact sets in .
Denote
Lemma 2.7.
Suppose and , then is a solution of (1.1)–(1.4) if and only if is a solution of the following abstract operator equation
Proof.
Suppose is a solution of (1.1)–(1.4), it is clear that is a solution of (2.40).
Conversely, if is a solution of (2.40), then (1.2) is satisfied and
Thus .
From (2.40) and (2.41), we have
According to (2.42), we get that (1.3) is satisfied. Since , we have
It follows from the definition of that
then .
Hence is a solutions of (1.1)–(1.4). This completes the proof.
3. Existence of Solutions in the Case of Nonresonance
In this section, we will apply LeraySchauder's degree to deal with the existence of solutions and nonnegative solutions for system (1.1)–(1.4) at nonresonance.
When satisfies sub growth condition, we have the following.
Theorem 3.1.
Suppose and , satisfies sub growth condition, and operators and satisfy the following condition
then problem (1.1)−(1.4) has at least one solution.
Proof.
First we consider the following problem:
Denote
where is defined in (2.11).
We know that (S_{1}) has the same solution of the following operator equation when ,
It is easy to see that operator is compact continuous for any . It follows from Lemmas 2.2 and 2.3 that is compact continuous from to for any .
We claim that all the solutions of (3.3) are uniformly bounded for . In fact, if it is false, we can find a sequence of solutions for (3.3) such that as , and for any .
From Lemma 2.2, we have
Thus
From (S_{1}), we have
It follows from (2.12) and Lemma 2.2 that
Denote . The above inequality holds
It follows from (3.1) and (3.5) that
For any , we have
which implies that , ; . Thus
It follows from (3.8) and (3.11) that is uniformly bounded.
Thus, we can choose a large enough such that all the solutions of (3.3) belong to . Therefore the LeraySchauder degree is well defined for , and
It is easy to see that is a solution of if and only if is a solution of the following usual differential equation
Obviously, system (S_{2}) possesses a unique solution . Since , we have
which implies that (1.1)–(1.4) has at least one solution. This completes the proof.
Theorem 3.2.
Suppose and , satisfies sub growth condition, and operators and satisfy the following
where , and , , then problem (1.1) with (1.2), (1.4), and (1.8) has at least one solution.
Proof.
Obviously,
From Theorem 3.1, it suffices to show that
(a)Suppose , where is a large enough positive constant. From the definition of , we have
Since , we have . Thus (3.16) is valid.
(b)Suppose , we have
There are two cases.
Case 1 ().
Since , we have , and then
Thus (3.16) is valid.
Case 2 ().
Since , we have , and
Thus (3.16) is valid. Thus problem (1.1) with (1.2), (1.4), and (1.8) has at least one solution. This completes the proof.
Let us consider
where is a parameter, and
where , are Caratheodory.
We have the following.
Theorem 3.3.
Suppose and , satisfies sub growth condition, and we assumethat
then problem (3.21) with (1.2)–(1.4) has at least one solution when the parameter is small enough.
Proof.
Denote
We consider the existence of solutions of the following equation with (1.2)–(1.4)
Denote
where is defined in (2.11).
We know that (3.25) with (1.2)–(1.4) has the same solution of .
Obviously, . So . As in the proof of Theorem 3.1, we know that all the solutions of are uniformly bounded, then there exists a large enough such that all the solutions of belong to . Since is compact continuous from to , we have
Since are Caratheodory, we have
Thus
Obviously, . We obtain
Thus, when is small enough, we can conclude that
Thus has no solution on for any , when is small enough. It means that the LeraySchauder degree is well defined for any , and
Since , from the proof of Theorem 3.1, we can see that the right hand side is nonzero. Thus (3.21) with (1.2)–(1.4) has at least one solution. This completes the proof.
Theorem 3.4.
Suppose and , satisfies sub growth condition, and we assumethat
where , and , , then problem (3.21) with (1.2), (1.4), and (1.8) has at least one solution when the parameter is small enough.
Proof.
As it is similar to the proof of Theorems 3.2 and 3.3, we omit it here.
In the following, we will consider the existence of nonnegative solutions. For any , the notation means for any .
Theorem 3.5.
Suppose , , we also assume
, for all;
for any , , for all .
Then every solution of (1.1)–(1.4) is nonnegative.
Proof.
Let be a solution of (1.1)–(1.4), integrating (1.1) from 0 to , we have
where . The boundary value condition holds
Conditions (1^{0})(2^{0}) mean . Obviously, for any for all , we have
It follows from conditions (1^{0})(2^{0}) and (3.36) that is increasing on , namely , for all with . Thus the boundary value condition holds , then .
Since is increasing and , we have , for all .
Thus every solution of (1.1)–(1.4) is nonnegative. The proof is completed.
Corollary 3.6.
Under the conditions of Theorem 3.1, we also assume
, for all with ;
for any , , for all with ;
for any and , , .
Then (1.1)–(1.4) has a nonnegative solution.
Proof.
Define , where
Denote
then satisfies Caratheodory condition, and for any .
For any , we denote
then and are continuous, and satisfy
It is not hard to check that
, for uniformly, where , and ;
, for all ;
, for all .
Let us consider
It follows from Theorems 3.1 and 3.5 that (3.41) have a nonnegative solution . Since , we have . Thus is a nonnegative solution of (1.1)−(1.4). This completes the proof.
4. Existence of Solutions in the Case of Resonance
In the following, we will consider the existence of solutions for system (1.1)–(1.4) at resonance.
Theorem 4.1.
Suppose and , is an open bounded set in such that the following conditions hold.
For each the problem
has no solution on .
The equation
has no solution on .
The Brouwer degree .
Then problem (1.1)–(1.4) have a solution on .
Proof.
Let us consider the following impulsive equation
For any , if is a solution to (4.1) or is a solution to (4.3), we have necessarily
It means that (4.1) and (4.3) have the same solutions for .
We denote defined by
where is defined by (2.11). Denote
Set
then the fixed point of is a solution for (1.1)–(1.4). Also problem (4.3) can be rewritten in the equivalent form
Since is Caratheodory, it is easy to see that is continuous and sends bounded sets into equiintegrable sets. It is easy to see that is compact continuous. From Lemma 2.6, we can conclude that is continuous and compact for any . We assume that (4.8) does not have a solution on for , otherwise we complete the proof. Now from hypothesis (1^{0}) it follows that (4.8) has no solutions for . For , (4.3) is equivalent to the following usual problem
If is a solution to this problem, we must have
As this problem is a usual differential equation, we have
where is a constant. Therefore keeps the same sign of . From , we have . From the continuity of , there exist , such that , . Hence , it holds , a constant. Thus (4.10) holds
which together with hypothesis (2^{0}) implies that . Thus we have proved that (4.8) has no solution on . Therefore the LeraySchauder degree is well defined for , and from the homotopy invariant property of that degree we have
Now it is clear that the problem
is equivalent to problem (1.1)–(1.4), and (4.13) tells us that problem (4.14) will have a solution if we can show that
It is not hard to check that . Thus
By the properties of the LeraySchauder degree, we have
where the function is defined in (4.2) and denotes the Brouwer degree. By hypothesis (3^{0}), this last degree is different from zero. This completes the proof.
Our next theorem is a consequence of Theorem 4.1. As an application of Theorem 4.1, let us consider the following system
with (1.2), (1.3), and (1.4), where is Caratheodory, is continuous, and for any fixed , holds , for all , .
Theorem 4.2.
Suppose that the following conditions hold
for all and all , where satisfies ;
, for uniformly;
, for all , where ;
, for all , where ;
for large enough , the equation
has no solution on , where ;
the Brouwer degree for large enough , where .
Then problem (4.18) with (1.2), (1.3), and (1.4) has at least one solution.
Proof.
For any and , we denote
At first, we consider the following problem
As in the proof of Theorem 4.1, we know that (4.21) has the same solutions of
where is defined in (2.39).
We claim that all the solutions of (4.21) are uniformly bounded for . In fact, if it is false, we can find a sequence of solutions for (4.21) such that as , and for any .
Since are solutions of (4.21), we have
.Since , we have
It follows from Lemma 2.5 that
From (3^{0}), (4^{0}), (4.23) and (4.25), we can see that
From (4.26), we have
Denote , then and . Thus possesses a convergent subsequence (which still denoted by ), then there exists a vector such that and . Without loss of generality, we assume that . Since , there exist such that
Obviously
Note that (as ) and , it follows from (4.27), (4.28), and (3^{0}) that
By (4.27), (4.29), and (4.30) we have for uniformly, which implies
where , satisfies , .
From (1.4), we have
Note that , it follows from (4.31), (4^{0}) and the continuity of that
which contradicts to (4.32). This implies that there exists a big enough such that all the solutions of (4.21) belong to , then we have
In order to obtaining the existence of solutions (4.18) with (1.2), (1.3), and (1.4), we only need to prove that .
Now we consider the following equation
where .
Similar to the preceding discussion, for any , all the solutions of (4.35) are uniformly bounded.
If is a solution of the following usual equation with (1.4)
we have
As , we have , it means that is a solution of
By hypothesis (5^{0}), (4.35) has no solutions on , from Theorem 4.1, we obtain that (4.18) with (1.2), (1.3), and (1.4) has at least one solution. This completes the proof.
Corollary 4.3.
If is Caratheodory, conditions (2^{0}), (3^{0}) and (4^{0}) of Theorem 4.2 are satisfied, condition (3^{0}) of Corollary 3.6 is also satisfied, , where are positive functions satisfying ; then (4.18) with (1.2), (1.3), and (1.4) has at least one solution.
Proof.
Denote
From condition (4^{0}), we have
Note that and are nonnegative. From the above inequality, we can see that all the solutions of are uniformly bounded for . Thus is well defined for and
and it is easy to see that has a unique solution in and
According to Theorem 4.2, we get that (4.18) with (1.2), (1.3), and (1.4) has at least a solution. This completes the proof.
Let us consider
where is a parameter, and
where are Caratheodory.
From Theorem 4.2, similar to the proof of Theorem 3.3, we have the following.
Theorem 4.4.
If conditions of (1^{0}) and (3^{0})–(6^{0}) of Theorem 4.2 are satisfied, then problem (4.43) with (1.2), (1.3), and (1.4) has at least one solution when the parameter is small enough.
Theorem 4.5.
If conditions of (1^{0})–(3^{0}) and (5^{0})(6^{0}) of Theorem 4.2 are satisfied, and satisfy
where
then problem (4.18) with (1.2), (1.3), and (1.8) has at least one solution.
Proof.
Similar to the proof of Theorem 3.2, the condition (4^{0}) of Theorem 4.2 is satisfied. Thus problem (4.18) with (1.2), (1.3) and (1.8) has at least a solution.
Similar to the proof of Theorem 3.2 and Corollary 4.3, we have the following.
Corollary 4.6.
If is Caratheodory, (4.45), (4.46) and conditions (2^{0}) and (3^{0}) of Theorem 4.2 are satisfied, condition (3^{0}) of Corollary 3.6 is also satisfied, , where are positive functions satisfying ; then (4.43) with (1.2), (1.3), and (1.8) has at least one solution when the parameter is small enough.
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Acknowledgments
This paper is partly supported by the National Science Foundation of China (10701066, 10926075, and 10971087), China Postdoctoral Science Foundation funded project (20090460969), and the Natural Science Foundation of Henan Education Committee (200875565).
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Dong, R., Guo, Y., Zhao, Y. et al. Existence of Solutions for a Weighted Laplacian Impulsive Integrodifferential System with Multipoint and Integral Boundary Value Conditions. J Inequal Appl 2010, 392545 (2010). https://doi.org/10.1155/2010/392545
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DOI: https://doi.org/10.1155/2010/392545