Abstract

In this paper we consider the existence of solutions to following kind of problems where is an open bounded subset of , and , is a function which belongs to a suitable integrable space.

1. Introduction

In this paper, we study the following kind of problems

where is an open bounded subset of , is a function which belongs to a suitable integrable space and is a Carathéodory function, which satisfies the following assumptions: for almost every and every

for every with , where are positive constants and .

We note that there is a difficulty in dealing with (2) under the assumption (3). Indeed the differential operator

is well defined in , but is not coercive on the same space when is unbounded. Therefore, the classical Leray—Lions theorem cannot be apply even in the case the datum belongs to for . In order to overcome this difficulty, we will proceed by approximation by means of truncatures in to get a coercive differential operator on . In particular in [13], the authors analyzed the problems

where with , is a Carathéodory function such that

for all and . Denote

Main results of [13] summarized as:(1) If with , then there exists a solution to problem (6) such that

(2) If , then ;(3) If , then ;(4) If , then .

In the case where , problem (6) may has no solution at all, even if the data is a constant (see [1, 3]). Thus the non-coerciveness of operator has great effect on the existence and regularity of the solution. Some other results of non-coercivity quasilinear elliptic problem, see [424].

Furthermore, in the literature [21], they proved the -regularity for solutions to the following nonlinear elliptic equations with degenerate coercivity

where is an open bounded subset of , satisfies

with , and is a Carathéodory function which satisfies the following assumptions

for almost every , for every , where is a nonnegative function in with , and

where is a real such that and is a decreasing continuous function such that its primitive

is unbounded. Under the hypothesis (11)–(14), the authors proved that (10) has at least a weak solution in the sense that

for all in .

On the other hand, as in [22], it has been shown the existence results of solutions for nonlinear elliptic boundary value problem

where . We recall the following results contained in [22].(1) Let . Then there exists a distributional solution of (18).(2) Let , and if , then there exists a distributional solution of (18).

The aim of this article is to study the existence of solutions to the problem (1). Our main results are following:

Theorem 1. Assume (2)–(4) hold and with

Then problem (1) exists a distributional solution provided

Remark 2. We point out that the existence in is not so usual in the study of elliptic problems. The main difficulty of study is due to the a priori estimate of the sequence of the approximate solutions of (18) in the non-reflexive space . For more results about existence of to elliptic equations, see [2325].

Remark 3. Observe that, if (20) holds.

Remark 4. It can be easily seen that, (20) implies , while . Therefore, in (20), we should require , instead of .

Remark 5. It is worth pointing out that, the existence of solutions to (18) was established provided and with . According to Remark 4, we know that problem (1) has solution if satisfies.

This fact shows that the lack of coercivity of the differential operator has the great influence on the existence of solutions to (1). Especially, for , we obtain the existence of solutions for with , rather than .

For , we have the following theorem.

Theorem 6. Assume (2)–(4) hold. Then problem (1) has a distributional solution if

Remark 7. Note that if , then .

The paper is organized as follows. In Section 2 we collect some definitions and useful tools. The proof of Theorems 1 and 6 is given in Section 3.

2. Preliminaries

In order to prove our main results, we need to introduce some basic definitions and lemmas. We shall used the following notion of distributional solution for problem (1).

Definition 8. We say that is a distributional solutions to problem (1) if

for every .

Definition 9. For all , the truncation function is defined by

We recall the following lemma (see [26]).

Lemma 10. Let be measurable function such that for . Then there exists a unique measurable such that the map is called the weak gradient of . Moreover if , then coincides with the standard distributional gradient of .

Definition 11. Let and , the Zygmund space , is made up of all measurable functions such thatfor on .

Remark 12. Observe that if and , then the (26) becomes

We obtain the existence solutions to (2) by considering the following approximation problems

Lemma 13. Let be solutions to (28) and . Then the sequence is bounded in .

Proof. Let us take as test function in (28), we get

Then we arrive to the conclusion.

In order to prove the existence part of Theorems 1 and 6 we need the following preliminary result. This convergence of the gradients is a consequence of a classical result (see [22]).

Lemma 14. Let be the sequence of solutions to (28), which are bounded in , Assume and (2)–(4) hold. Suppose that(1) is such that for constant ;(2) converges to almost everywhere in ;(3) converges weakly to in .Then, up to a subsequence, converges almost everywhere in to .

3. Proof of Main Results

In this section, we prove the existence results of Theorems 1 and 6 by a classical approach. Consider the following approximate problems

Proof of Theorem 1. Step 1: Letand we know the functionis bounded since . Thus, choose as a test function in the approximate problem (28), we findwhere denotes a strictly positive constant, depending only on and . By the Sobolev embedding theorem, we have

Combining (34), (35), then

Note that since . Moreover the choice of implies

Thus we prove that

This estimate implies that

Now we can estimate . Indeed, apply the Hölder inequality and we obtain

Note that , so the right hand side is bounded. Then we get that the sequence is uniformly bounded in , subsequently there exist such that

As a consequence there exists a subsequence converging to a.e. in .

Step 2: Let us choose as test function in (28)

Using assumption (2) on , we get

and so

This implies that weakly in .

Now we are also going to estimate . Choosing as a test function in (28), we have

where we have use (37) in the last passage. By the Hölder inequality we get

Let be a measurable subset of , and let . Then by (43), (45) and Hölder inequality, we have

The estimates (37) and (46) imply that the sequence is equiintegrable. Thus, by the Dunford—Pettis theorem, and up to subsequences, there exists in such that weakly converges to in . Since is the distributional partial derivative of , then for every in , we have

where . We now pass to the limit in the above identities, using that weakly converges to in , and that strongly converges to in , we find

This implies that , and this result holds for every . Since belongs to for every , belongs to , as desired.

Step 3: We can apply Lemma 14 and conclude that a.e., that is almost everywhere converges to . Since is bounded in , the inequality

and the Vitali theorem imply that converges to in . Hence we are able to pass to the limit in (28). Thus we prove that is a distributional solution of (2) and yields the conclusion of the proof of Theorem 1.

Proof of Theorem 6. Step 1: Let . Choosing as a test function in (28) and by (3), we get.

Then using the Sobolev embedding theorem we have

So combining (50) and (51),

Note that

since , then we prove that the sequence is bounded in , that is

Step 2: We use the following inequality: there exists only depending on such that

for all .

We take as a test function in the approximated problem (28) and by inequality (55), we obtain

Now we can apply Hölder and Sobolev inequalities, since

then

Since , we prove that the sequence is bounded in and so it is compact in .

Step 3: Choosing as a test function in (28), we get

In addition,

Therefore, for every measurable subset , following (46) and we have

Thus we prove that weakly converges to in . Furthermore, the last part of the proof we follow the approach of the Theorem 1 and we can proof that is a distributional solution of (1).

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This research was partially supported by the National Natural Science Foundation of China (No. 11761059), Fundamental Research Funds for the Central Universities (Nos. Yxm2019112, 31920200036), the Program for Young Talent of State Ethnic Affairs Commission of China (No. XBMU-2019-AB-34) and Key Subject of Gansu Province.