International Journal of Differential Equations

International Journal of Differential Equations / 2021 / Article

Research Article | Open Access

Volume 2021 |Article ID 5589504 | https://doi.org/10.1155/2021/5589504

Aziz Bouhlal, Jaouad Igbida, "Existence and Regularity of Solutions for Unbounded Elliptic Equations with Singular Nonlinearities", International Journal of Differential Equations, vol. 2021, Article ID 5589504, 9 pages, 2021. https://doi.org/10.1155/2021/5589504

Existence and Regularity of Solutions for Unbounded Elliptic Equations with Singular Nonlinearities

Academic Editor: Jaume Giné
Received19 Jan 2021
Revised04 Apr 2021
Accepted19 Apr 2021
Published27 Apr 2021

Abstract

For , we study existence and regularity of solutions for unbounded elliptic problems whose simplest model is , where , .

1. Introduction

Consider the Dirichlet problem for some nonlinear elliptic equations:under the following assumptions. The set is a bounded open subset of , with :

is a measurable function satisfying the following conditions:for almost every , where and are positive constant, and

A possible motivation for studying the existence of these types of problems arises from the calculation of variations and stochastic control. For example, if we consider the functionalthe Euler–Lagrange equation associated to the functional is

Several papers deal with existence of solutions to the singular elliptic problems with lower order terms having a quadratic growth with respect to the gradient (for example, [19]), namely, with the model problemwhere is a positive constant and is a Carathéodory function. More precisely, existence of positive solutions for (7) was shown in [13], for and , and the uniqueness of positive solution, for and , in [4]. On the contrary, the existence of positive solutions of (7) is shown in [6] for , provided is a bounded uniformly elliptic matrix and (). Later, in [9], it is proved the existence of solution for (7) with , where and the data with , and does not satisfy any sign assumption. Recently, a problem introduced by L. Boccardo (see [7, 10]) has given a strong impulse to the study of quasilinear problems having the unbounded divergence operator. In particular, in [7], the authors have proved the existence of positive solutions to problem (7) under the assumption that , , and . We refer also that, in [5], the author has shown the same result as in [7], in the case and without any sign restriction over .

Let us now consider the Dirichlet boundary value problem (7) in the simple case:

If we define , then the function is solution ofwhich is singular on the right-hand side. Let us remark that, in the case of nonnegative , in [11], the authors considered the elliptic semilinear problems whose model iswhere . More precisely, they have shown that the term has a regularizing effect on the solutions . In [12], the author has shown the existence of solutions to the following elliptic problem with degenerate coercivity:where .

The purpose of this paper is to study the same kind of lower order term as in problems (7) and (9) (indeed, ) in the case of an elliptic operator with unbounded coefficients. The main difficulties posed by this problem were that the principal part of the differential operator is not well defined on the whole ; the solutions did not belong, in general, to and the lower order term has a singularity at . Despite these difficulties, we prove that, in our case too, the lower order term has a regularizing effect.

Our main existence results are as follows.

Theorem 1. Assume that (2) and (3) hold true. If with , then there is a positive solution of (1), in the sense of distributions, that is,for any test function in . Moreover, we have the following summability results for :(1)Let :(i)If , then .(ii)If , then .(2)Let :(i)If , then .(ii)If , then .(3)Let , then .

When , , we will prove the following regularizing effects.

Theorem 2. We suppose that , and that (2) and (3) are satisfied. If , then, there exists a solution of (1) in the sense (19), such that(1)If and , then , where(2)If , then .(3)If , then .

Notation: throughout this paper, we fix an integer . For any , will be the Hölder conjugate exponent of , and if , we will denote by the Sobolev conjugate exponent of . As usual, let us denote by the Sobolev constant, i.e.,

We denote by the Poincaré constant given by

For all , we recall the definition of a truncated function defined by

We also consider

As usual, we consider the positive and negative part of a measurable function

2. The Approximated Problem

To prove our existence results, we will use the following approximating problems:where , and

As in [11], we prove existence of positive solution of the approximated problem.

Lemma 1. Let be positive function belonging to . Suppose that (2) and (3) are satisfied. Then, there exists a positive solution of the problem

Proof. To prove it, we define the following operator which associates to every the solution toFrom the results of [13], the operator is well defined and is bounded by the results of [14]. We take as a test function in (19), and we use Hölder’s inequality and (3) to deduce thatThanks to Poincaré’s inequality, we deduceHence, there exists an invariant ball for . On the contrary, from the embedding, it is easily seen that is continuous and compact. The Schauder theorem shows that has a fixed point or equivalently, and there exists a solution to problemsMoreover, by the maximum principle, it is clear that the sequence is nonnegative since is nonnegative, and we choose as test function in (25) and use (3) to obtainwhere . By the method of Stampacchia (see [14]), the sequence is bounded in . Supposing that is bounded by in , we have that is a solution of (13).
By Lemma 1, it follows the existence of a solution of (19).
Now, we are going to prove that the sequence is not 0 in . For this, we are going to prove that it is uniformly away from zero in every compact set in . We will follow a similar technique to that one in [12].

Lemma 2. Assume that (2) and (3) hold true. If and is the solution of problem (19), then for every : a.e. in . Furthermore, if , then, for every , there exists such that a.e. in .

Proof. Let us consider as a test function in problems (19). Then,Observing that , we haveTherefore, by (3), we deduce thatConsequently, we obtain , so by Poincaré’s inequality, we have for every . Thus, a.e. .
We remark that is bounded; indeed, , for some positive constant . Then, it follows thatThanks to (3), we have . Thus, we infer that is a supersolution of a linear Dirichlet problem with a strictly positive and bounded, measurable coefficient. The strong maximum principle implies that . In addition, Harnack’s inequality gives the stronger conclusion: for every , there exists such that a.e. in . Finally, using that the sequence is increasing, one deduces that a.e. in for every .

2.1. Existence of Bounded Solutions

In this section, we will prove existence of bounded weak solutions for (1).

Lemma 3. Let with . Suppose that (2) and (3) hold true. Let be a sequence solutions of (19) with for every . Then, the norm of the sequence in is bounded by a constant which depends on and on the norm of in .

Proof. The use of as test function in (19) and (3), implies thatwhere . Hence, we can use Theorem 4.1 in [14] and obtain a positive constant, say , that only depends on the parameters: and such that: for all .

Lemma 4. We assume that with , and (2) and (3) are satisfied. Let be a sequence solutions of (19) with for every . If and , then the sequence is uniformly bounded in .

Proof. We denote by a positive constant which may only depend on the parameters of our problem, and its value may vary from line to line.
We use as test function in (19) to obtainand thus (since ),from which the sequence is bounded in .

Lemma 5. Let with , and we suppose that (2) and (3) are satisfied. If and and is a solution to problem (19), then is uniformly bounded in .

Proof. Let and be the support of ; then, from Lemma 2, there exists such that for a.e. .
Choosing as test function in (19) and using (3), we obtainwhich then impliesWe can use Young’s inequality with , and we obtainUsing (3), we havefor every and (and for a suitable independent on ).
We then haveApplying (38) to (35) and letting , we obtainand this gives that is bounded in .

Lemma 6. Let . Suppose that (2) and (3) hold. If with , then the sequence defined by (19) satisfies the following summability:(1)If , then is uniformly bounded in (2)If , then is uniformly bounded in

Proof. (1)Let us take as test function in (19) and use (3) to obtain that(2)Let and choose , as a test function in problem (19). From assumption (19), one haswhere . By Young’s inequalities, it is easy to proveHence, equality (41) implies thatLetting , we get that is bounded in .

Lemma 7. Let . Assume that (2) and (3) hold true. If with , then the solution of (19) is uniformly bounded in .

Proof. Let be a function in and . Take as test function in (19) and use (3) to obtainUsing Young’s inequality with , we have by (3) and Lemma 3 thatTaking the above estimate in (44) and letting , we obtainand thus, Lemma 7 is proved.

Proof. of Theorem 1.
We start by proving point (1.i), the rest of the proof of the theorem can be proven similarly. According to Lemmas 3 and 4, there exists a subsequence and a function such that weakly converges to in . Now, we can pass to the limit in the equation satisfied by the approximated solutions :where .
For the term of the left-hand side, it is sufficient to observe that converge to weakly in and a.e. (and weakly in converges towards . On the contrary, for the limit of the right-hand side of (47), let , and one can use Lebesgue’s dominated convergence theorem, sinceFinally, passing to the limit as goes to infinity in equation (47), we conclude that

2.2. Further Existence Result

In this section, we suppose (2) and (3) and we assume thatholds true.

Lemma 8. We suppose that (2), (3), and (50) hold true. Let and , withThen, the solutions to problem (19) are uniformly bounded in .

Proof. Let us take as a test function in (19) and use assumption (3) to obtainWe can use Hölder’s inequality on the right-hand side with exponent , and Sobolev inequality on the left-hand side to deduceWe note that ; moreover, (thanks to the fact that ). This last estimate imply that is uniformly bounded in and in .
We are going to prove now that the sequence is bounded in . Let ; using as a test function for problem (19), we can deduceNow, we rewriteand use the Sobolev inequality and the Hölder inequality in (54) to obtainWe note that the choice of is equivalent to require ; furthermore, and . Thus, the sequence is uniformly bounded in .

Lemma 9. Under the hypotheses , (2), (3), and (50), if , then the solutions are uniformly bounded in .

Proof. We choose as test function in (19) to obtain, by hypothesis (3), thatTherefore, is bounded in .

Lemma 10. Let . Under hypotheses (2), (3), and (50), if , then the solutions are uniformly bounded in .

Proof. Choosing as test function in (19) and using Hölder and Sobolev inequalities, thanks to (3), we obtain thatThe above inequality implies thatNow, we prove that the sequence is bounded in . Let and choose , as a test function in problems (19). From assumption (19), one haswhere . We can use Young’s inequality with and both (37) and (59) to obtainHence, equality (60) implies thatLetting , we get that is bounded in .

Lemma 11. Under the assumptions of Theorem 2, let be a solution to problem (19). Then, the sequence is uniformly bounded in , for every .

Proof. We will prove our proof in two steps:Step 1: we want to prove that, for every , . Indeed, let , and is the support of . Thanks to (3), we have from (19) with test function We use Young’s inequality, and since , we deduce from (37) thatThus, by the above estimate and since is uniformly bounded in , this proves Step 1.Step 2: here, we show that is uniformly bounded in for every . For this, let , , and . We use Hölder inequality with exponent and by step 1, and1 we obtain