We study the following Laplacian problem with singular term: , , , , where is a bounded domain, . We obtain the existence of solutions in .
1. Introduction
After Kováčik and Rákosník first discussed the spaces and spaces in [1], a lot of research has been done concerning these kinds of variable exponent spaces, for example, see [2–5] for the properties of such spaces and [6–9] for the applications of variable exponent spaces on partial differential equations. Especially in spaces, there are a lot of studies on Laplacian problems; see [8, 9]. In the recent years, the theory of problems with growth conditions has important applications in nonlinear elastic mechanics and electrorheological fluids (see [10–14]).
In this paper, we study the existence of the weak solutions for the following Laplacian problem:
where is a bounded domain, , is Lipschitz continuous on , and
Let be the set of all Lebesgue measurable functions . For all , we denote , , and denote by the fact that .
We impose the following condition on :
(F), , and for , there exist such that , whenever .
A typical example of (1.1) is the following problem involving subcritical SobolevHardy exponents of the form
where , , , . In fact, take , , and , then it is easy to verify that (F) is satisfied.
Our object is to obtain the existence of solutions in the following four cases:
(1);
(2);
(3);
(4).
When , the solution of the Laplacian equations without singularity has been studied by many researchers. The study of problem (1.1) with variable exponents is a new topic.
The paper is organized as follows. In Section 2, we present some necessary preliminary knowledge of variable exponent Lebesgue and Sobolev spaces. In Section 3, we prove our main results.
2. Preliminaries
In this section we first recall some facts on variable exponent Lebesgue space and variable exponent Sobolev space , where is an open set; see [1–4, 8, 15] for the details.
Let and
The variable exponent Lebesgue space is the class of functions such that . is a Banach space endowed with the norm (2.1).
For a given we define the conjugate function as
Theorem 2.1.
Let . Then the inequality
holds for every and with the constant depending on and only.
Theorem 2.2.
The dual space of is if and only if . The space is reflexive if and only if
Theorem 2.3.
Suppose that satisfies (2.4). Let , , then necessary and sufficient condition for is that for almost all one has , and in this case, the imbedding is continuous.
Theorem 2.4.
Suppose that satisfies (2.4). Let . If , then
(1) if and only if ,
(2)if , then ,
(3)if , then ,
(4) if and only if ,
(5) if and only if .
We assume that is a given positive integer.
Given a multiindex , we set and , where is the generalized derivative operator.
The generalized Sobolev space is the class of functions on such that for every multiindex with , endowed with the norm
By we denote the subspace of which is the closure of with respect to the norm (2.5).
In this paper we use the following equivalent norm on :
Then we have the inequality .
Theorem 2.5.
The spaces and are separable reflexive Banach spaces, if satisfies (2.4).
Theorem 2.6.
Suppose that satisfies (2.4). Let . If , then
(1) if and only if ,
(2)if , then ,
(3)if , then ,
(4) if and only if ,
(5) if and only if .
We denote the dual space of by , then we have the following.
Theorem 2.7.
Let . Then for every there exists a unique system of functions such that
The norm of is defined as
Theorem 2.8.
Let be a domain in with cone property. If is Lipschitz continuous and , is measurable and satisfies a.e. , then there is a continuous embedding .
Theorem 2.9.
Let be a bounded domain. If and , then
where is a constant depending on .
Next let us consider the weighted variable exponent Lebesgue space. Let , and for . Define
with the norm
then is a Banach space.
Theorem 2.10.
Suppose that satisfies (2.4). Let . If ,, then
(1)for , if and only if ,
(2) if and only if ,
(3)if , then ,
(4)if , then ,
(5) if and only if ,
(6) if and only if .
Theorem 2.11.
Assume that the boundary of possesses the cone property and . Suppose that , and , for . If and then there is a compact embedding .
Theorem 2.12.
Let be a measurable subset. Let be a Caracheodory function and satisfies
where , , , is a constant, then the Nemytsky operator from to defined by is a continuous and bounded operator.
3. Existence and Multiplicity of Solutions
Let
The critical points of , that is,
for all , are weak solutions of problem (1.1). So we need only to consider the existence of nontrivial critical points of .
Denote by , and the generic positive constants. Denote by the Lebesgue measure of .
To study the existence of solutions for problem (1.1) in the first case, we additionally impose the following conditions.
(A1) and , .
(B1) There exists a function , such that and for , .
(C1)there exist such that for , , where and .
(D1), for all .
Theorem 3.1.
Under assumptions (F) and (A1)–(C1), problem (1.1) admits a nontrivial solution.
Proof.
First we show that any sequence is bounded. Let and , such that and in . By (A1) and (B1), , and . Let and , then , , , and . Let and . Then
By (B1), we get
By (F), we get , so there exist such that on . Note , so we have
By Young's inequality, for , we get
Take sufficiently small so that .
Note that , by Young's inequality again and for , we get
Take sufficiently small so that .
From the above remark, we have
As , and , we have . Since and , by Theorem 2.6 we have
when and is sufficiently large. Then it is easy to see that is bounded in . Next we show that possesses a convergent subsequence (still denoted by ).
Note that
Because is bounded in , there exists a subsequence (still denoted by ), such that weakly in . By Theorem 2.11, there are compact embeddings and , then in and . So we get
Hence as .
By (F), we have
and similarly for every ,
Since
and in and , we obtain and . Similarly,
Because is bounded, we get , as . From the above remark, we conclude , as .
Thus , as . Then we get . As in the proof of Theorem in [6, 7], we divide into the following two parts:
On , we have
Then as .
On , we have
so , as .
Thus we get . Then in .
From (F) and (B1) we have , for all , . So we get , for all , . for all , take , then . Since and , there exists such that , and , for . Let , such that for , for , and in . Then we have
where . So if is sufficiently large, we obtain .
From (F) and (C1), we have , then . So we get
Let , then . By Theorem 2.11, , and . When is sufficiently small, , and . For any , as , for any , we can find such that and whenever . Take , then . is an open covering of . As is compact, we can pick a finite subcovering for from the covering . If we define on . We can use all the hyperplanes, for each of which there exists at least one hypersurface of some lying on it, to divide into finite open hypercubes which mutually have no common points. It is obvious that and for each there exists at least one such that . Let , then and we have
If is sufficiently small such that
we have .
The mountain pass theorem guarantees that has a nontrivial critical point .
Since is a separable and reflexive Banach space, there exist and such that
For , denote , , .
Theorem 3.2.
Under assumptions (F), (A1)–(D1), problem (1.1) admits a sequence of solutions such that .
Proof.
Let . We first show that is weakly strongly continuous. Let weakly in . By the compact embedding , we have and a.e. on . By the inequality and the Vitali Theorem, we get .
Note that
When ,
When , is bounded. So we get
By the compact embedding , we get in . So by Theorem 2.12 we obtain , that is, . Hence we obtain that is weakly strongly continuous. By Proposition in [8], as for . For all , there exists a positive integer such that for all . Assume for each . Define in the following way:
Note that as . Hence for with , we get
So
Note that and . Since the dimension of is finite, any two norms on are equivalent, then . If , it is immediate that . If , then . As in the proof of Theorem 3.1 we can find hypercubes which mutually have no common points such that and , where . Then we need only to consider the case: for every . We have
Let , . Let and . Denote . Let be sufficiently large such that . There at least exists one such that . We have
and as . Hence we obtain that as . Thus for each , there exists such that for . From Theorem 3.1 satisfies condition. In view of (D1), by Fountain Theorem (see [16]), we conclude the result.
In the second case, we additionally impose the following condition:
(A2) and .
Theorem 3.3.
Under assumptions (F), (A2), (B1), and (C1) there exist such that when , problem (1.1) admits a nontrivial solution.
Proof.
It is obvious that . Let be such that . Since , there exists and such that , for all . Thus for all . Let be as defined in Theorem 3.1. By (C1), , and , when . Then for any and , we have
If is sufficiently small, .
From (F) and (C1), we have and . By Theorems 2.8 and 2.11, there exist positive constants such that , , . When is sufficiently small, we have , , and . As in the proof of Theorem 3.1 we can find hypercubes which mutually have no common points such that , and , where . Then
Since , for all , when , . Fix such that . Then we have
Let . When , . As in the proof of Theorem in [17], denote , we have and . Let . Applying Ekeland's variational principle to the functional , we find such that , , , and . Thus we get a sequence such that and . It is clear that is bounded in . As in the proof of Theorem 3.1, we get a subsequence of , still denoted by , such that in . So and .
Theorem 3.4.
Under assumptions (F), (A2), and (B1)–(D1), problem (1.1) has a sequence of solutions such that .
Proof.
First we show that any sequence is bounded. Let and , such that and in . By (B1), there exist such that . From (F), (A2), and (B1)–(D1), we have
Thus we have
By Young's inequality, for , we get
Take , and sufficiently small so that , and , then
Therefore by Theorems 2.6 and 2.9, we get that is bounded in . Then as in the proof of Theorem 3.1 possesses a convergent subsequence (still denoted by ). By Theorem 3.2, we can also have
As in the proof of Theorem 3.1 we can find hypercubes which mutually have no common points such that and , where . Since the dimension of is finite, any two norms on are equivalent. Then we need only to consider the cases and for every . We have
Hence as . As in the proof of Theorem 3.2, we complete the proof.
In the third case, we additionally impose the following condition:
(A3), and ,
(B3) there exist such that for .
Theorem 3.5.
Under assumptions (F), (A3), and (B3), problem (1.1) admits a nontrivial solution.
Proof.
By Young's inequality, for , we get . By (F), we have . Thus
Take sufficiently small so that . From Theorem 2.11, . If , is bounded. Then we need only to consider the case . As in the proof of Theorem 3.1 we can find hypercubes which mutually have no common points such that and , where . Then we have
where , and , . As in the proof of Theorem 3.2, we obtain that is coercive, that is, as . Thus has a critical point such that and further is a weak solution of (1.1).
Next we show that is nontrivial. Let be the same as that in Theorem 3.3. By (B3), . Then
If is sufficiently small, .
In the fourth case, we additionally impose the following condition:
(A4) and .
Theorem 3.6.
Under assumptions (F), (A4), and (D1), problem (1.1) admits a sequence of solutions such that .
Proof.
First we show that any sequence is bounded. Let and , such that and in . Denote and . We have and .
We can get
By Young's inequality, for , we get
Take and sufficiently small so that and . Then
As in the proof of Theorem 3.5, , when . Thus, we conclude that is bounded in . Then as in the proof of Theorem 3.1 possesses a convergent subsequence (still denoted by ). By Theorem 3.2, we can also get
As in the proof of Theorem 3.1 we can find hypercubes which mutually have no common points such that and , where . Since the dimension of is finite, any two norms on are equivalent. Then we need only to consider the cases and for every . We have
Hence we obtain as . As in the proof of Theorem 3.2, we complete the proof.
Acknowledgments
This work is supported by the Science Research Foundation in Harbin Institute of Technology (HITC200702), The Natural Science Foundation of Heilongjiang Province (A200704), and the Program of Excellent Team in Harbin Institute of Technology.
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