Volume 2021, Issue 1 5589504
Research Article
Open Access

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

Aziz Bouhlal

Corresponding Author

Aziz Bouhlal

Laboratoire de Mathématiques et Applications, Faculty of Sciences, B.P.20, El Jadida, Morocco uss.rnu.tn

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Jaouad Igbida

Jaouad Igbida

Labo DGTIC, Department of Mathematics, CRMEF, El Jadida, Morocco crmef.org

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First published: 27 April 2021
Citations: 1
Academic Editor: Jaume Giné

Abstract

For q, γ > 0, we study existence and regularity of solutions for unbounded elliptic problems whose simplest model is , where fLm(Ω), m ≥ 1.

1. Introduction

Consider the Dirichlet problem for some nonlinear elliptic equations:
()
under the following assumptions. The set Ω is a bounded open subset of N, with N ≥ 3:
()
a : Ω⟶ is a measurable function satisfying the following conditions:
()
for almost every x ∈ Ω, where α and β 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 functional
()
the Euler–Lagrange equation associated to the functional J 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 problem
()
where θ is a positive constant and M : Ω × is a Carathéodory function. More precisely, existence of positive solutions for (7) was shown in [13], for M(x, t) = 1 and 0 < θ ≤ 1, and the uniqueness of positive solution, for M(x, t) = 1 and 0 < θ < 1, in [4]. On the contrary, the existence of positive solutions of (7) is shown in [6] for 0 < θ ≤ 1, provided M is a bounded uniformly elliptic matrix and 0≨fLm(Ω) (m > (2N/N + 2)). Later, in [9], it is proved the existence of solution for (7) with 0 < θ < 1, where M(x, t) = 1 and the data fLm(Ω) with m > (N/2), 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 0 < θ < 1, M(x, t) = 1 + |t|q, and 0≨fLm(Ω). We refer also that, in [5], the author has shown the same result as in [7], in the case 0 < θ < 1 and without any sign restriction over f.
Let us now consider the Dirichlet boundary value problem (7) in the simple case:
()
If we define , then the function v is solution of
()
which is singular on the right-hand side. Let us remark that, in the case of nonnegative f, in [11], the authors considered the elliptic semilinear problems whose model is
()
where γ > 0. More precisely, they have shown that the term (f/|u|γ) has a regularizing effect on the solutions u. In [12], the author has shown the existence of solutions to the following elliptic problem with degenerate coercivity:
()
where p, γ > 0.

The purpose of this paper is to study the same kind of lower order term as in problems (7) and (9) (indeed, (f/|u|γ)) 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 div((a(x) + |u|q)∇u) is not well defined on the whole ; the solutions did not belong, in general, to and the lower order term has a singularity at u = 0. Despite these difficulties, we prove that, in our case too, the lower order term (f/|u|γ) has a regularizing effect.

Our main existence results are as follows.

Theorem 1. Assume that (2) and (3) hold true. If 0≨fLm(Ω) with m > (N/2), then there is a positive solution uL(Ω) of (1), in the sense of distributions, that is,

()
for any test function φ in . Moreover, we have the following summability results for u:
  • (1)

    Let 0 < q < 1:

    • (i)

      If 0 < γ ≤ 1 − q, then .

    • (ii)

      If γ > 1 − q, then .

  • (2)

    Let q = 1:

    • (i)

      If 0 < γ ≤ 1, then .

    • (ii)

      If γ > 1, then .

  • (3)

    Let q > 1, then .

When fLm(Ω), 1 < m < (N/2), we will prove the following regularizing effects.

Theorem 2. We suppose that 0≨fLm(Ω), 1 < m < (N/2) and that (2) and (3) are satisfied. If 0 < q < 1, then, there exists a solution u of (1) in the sense (19), such that

  • (1)

    If γ < 1 − q and (2/2 + q − 1 + γ) ≤ m < (N/2), then , where

    ()

  • (2)

    If γ = 1 − q, then .

  • (3)

    If γ > 1 − q, then .

Notation: throughout this paper, we fix an integer N ≥ 3. For any p > 1, p = (p/p − 1) will be the Hölder conjugate exponent of p, and if 1 ≤ p < N, we will denote by p = (Np/Np) the Sobolev conjugate exponent of p. As usual, let us denote by the Sobolev constant, i.e.,
()
We denote by the Poincaré constant given by
()
For all k > 0, we recall the definition of a truncated function Tk(s) defined by
()
We also consider
()
As usual, we consider the positive and negative part of a measurable function u(x)
()

2. The Approximated Problem

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

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

Lemma 1. Let g be positive function belonging to L(Ω). 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 Sn : L2(Ω)⟶L2(Ω) which associates to every vL2(Ω) the solution to

()

From the results of [13], the operator Sn is well defined and wn is bounded by the results of [14]. We take wn as a test function in (19), and we use Hölder’s inequality and (3) to deduce that

()

Thanks to Poincaré’s inequality, we deduce

()

Hence, there exists an invariant ball for Sn. On the contrary, from the L2(Ω) embedding, it is easily seen that Sn is continuous and compact. The Schauder theorem shows that Sn has a fixed point or equivalently, and there exists a solution to problems

()

Moreover, by the maximum principle, it is clear that the sequence un is nonnegative since g is nonnegative, and we choose Gk(un) as test function in (25) and use (3) to obtain

()
where Ak = {x ∈ Ω : |un| > k}. By the method of Stampacchia (see [14]), the sequence un is bounded in L(Ω). Supposing that un is bounded by dn in L(Ω), 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 un is not 0 in Ω. For this, we are going to prove that it is uniformly away from zero in every compact set in Ω. We will follow a similar technique to that one in [12].

Lemma 2. Assume that (2) and (3) hold true. If 0≨fL1(Ω) and un is the solution of problem (19), then for every n: unun+1 a.e. in Ω. Furthermore, if ω ⊂ ⊂Ω, then, for every n, there exists cω > 0 such that uncω > 0 a.e. in ω.

Proof. Let us consider as a test function in problems (19). Then,

()

Observing that fnfn+1, we have

()

Therefore, by (3), we deduce that

()

Consequently, we obtain , so by Poincaré’s inequality, we have for every k > 0. Thus, unun+1 a.e. x ∈ Ω.

We remark that u1 is bounded; indeed, |u1| ≤ c, for some positive constant c. Then, it follows that

()

Thanks to (3), we have . Thus, we infer that u1 is a supersolution of a linear Dirichlet problem with a strictly positive and bounded, measurable coefficient. The strong maximum principle implies that u1 > 0. In addition, Harnack’s inequality gives the stronger conclusion: for every ω ⊂ ⊂Ω, there exists cω such that u1cω a.e. in ω. Finally, using that the sequence un is increasing, one deduces that uncω a.e. in ω for every n.

2.1. Existence of Bounded Solutions

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

Lemma 3. Let 0≨fLm(Ω) with m > (N/2). Suppose that (2) and (3) hold true. Let {un} be a sequence solutions of (19) with fn = f for every n. Then, the norm of the sequence {un} in L(Ω) is bounded by a constant which depends on q, m, N, α, γ, meas(Ω) and on the norm of f in Lm(Ω).

Proof. The use of Gk(un) as test function in (19) and (3), implies that

()
where Ak = {x ∈ Ω : |un| > k}. Hence, we can use Theorem 4.1 in [14] and obtain a positive constant, say M, that only depends on the parameters: q, N, α, γ, meas(Ω) and such that: for all n.

Lemma 4. We assume that 0≨fLm(Ω) with m > (N/2), and (2) and (3) are satisfied. Let {un} be a sequence solutions of (19) with fn = f for every n. If q < 1 and γ ≤ 1 − q, then the sequence {un} is uniformly bounded in .

Proof. We denote by C 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 obtain

()
and thus (since q ≤ 1),
()
from which the sequence un is bounded in .

Lemma 5. Let 0≨fLm(Ω) with m > (N/2), and we suppose that (2) and (3) are satisfied. If q < 1 and γ > 1 − q and un is a solution to problem (19), then un is uniformly bounded in .

Proof. Let and ω = Suppφ be the support of φ; then, from Lemma 2, there exists cω > 0 such that uncω for a.e. xω.

Choosing as test function in (19) and using (3), we obtain

()
which then implies
()

We can use Young’s inequality with ϵ, and we obtain

()

Using (3), we have

()
for every q > 0 and t ≥ 0 (and for a suitable c0 independent on n).

We then have

()

Applying (38) to (35) and letting ɛ = (α(1 − q)/2), we obtain

()
and this gives that un is bounded in .

Lemma 6. Let q = 1. Suppose that (2) and (3) hold. If 0≨fLm(Ω) with m > (N/2), then the sequence {un} defined by (19) satisfies the following summability:

  • (1)

    If 0 < γ ≤ 1, then un is uniformly bounded in

  • (2)

    If γ > 1, then un is uniformly bounded in

Proof.

  • (1)

    Let us take log(1 + un) as test function in (19) and use (3) to obtain that

    ()

  • (2)

    Let and choose log(1 + un)φ2, as a test function in problem (19). From assumption (19), one has

()
where ω = Suppφ. By Young’s inequalities, it is easy to prove
()

Hence, equality (41) implies that

()

Letting ɛ = (min(1, α)/2), we get that un is bounded in .

Lemma 7. Let q > 1. Assume that (2) and (3) hold true. If 0≨fLm(Ω) with m > (N/2), then the solution un of (19) is uniformly bounded in .

Proof. Let φ be a function in and ω = Suppφ. Take as test function in (19) and use (3) to obtain

()

Using Young’s inequality with ϵ, we have by (3) and Lemma 3 that

()

Taking the above estimate in (44) and letting ɛ = (min(1, α)/2q), we obtain

()
and 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 un and a function such that un weakly converges to u in . Now, we can pass to the limit in the equation satisfied by the approximated solutions un:

()
where fn(x) = (f(x)/1 + (1/n)f(x)).

For the term of the left-hand side, it is sufficient to observe that ∇un converge to ∇u weakly in and a.e. (and weakly − in L(Ω) converges towards [a(x) + uq]. On the contrary, for the limit of the right-hand side of (47), let ω = Suppφ, and one can use Lebesgue’s dominated convergence theorem, since

()

Finally, passing to the limit as n 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 that
()
holds true.

Lemma 8. We suppose that (2), (3), and (50) hold true. Let γ < 1 − q and 0≨fLm(Ω), with

()

Then, the solutions un to problem (19) are uniformly bounded in .

Proof. Let us take as a test function in (19) and use assumption (3) to obtain

()

We can use Hölder’s inequality on the right-hand side with exponent p = (2/2 + q − 1 + γ) = (2N/N(γ + 1 + q) + 2(1 − qγ)) > 1, and Sobolev inequality on the left-hand side to deduce

()

We note that 2 = p(1 − qγ); moreover, (2/2) ≥ (1/p) (thanks to the fact that γ < 1 − q). This last estimate imply that un is uniformly bounded in and in .

We are going to prove now that the sequence un is bounded in . Let λ = (N(1 + q)(m − 1) + γm(N − 2)/N − 2m); using as a test function for problem (19), we can deduce

()

Now, we rewrite

()
and use the Sobolev inequality and the Hölder inequality in (54) to obtain
()

We note that the choice of λ is equivalent to require (2/2)(1 + q + λ) = m(λγ); furthermore, (2/2) ≥ (1/m) and (2/2)(1 + q + λ) = m∗∗(1 + q + γ). Thus, the sequence {un} is uniformly bounded in .

Lemma 9. Under the hypotheses 0≨fL1(Ω), (2), (3), and (50), if γ = 1 − q, then the solutions un are uniformly bounded in .

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

()

Therefore, un is bounded in .

Lemma 10. Let 0≨fL1(Ω). Under hypotheses (2), (3), and (50), if γ > 1 − q, then the solutions un are uniformly bounded in .

Proof. Choosing as test function in (19) and using Hölder and Sobolev inequalities, thanks to (3), we obtain that

()

The above inequality implies that

()

Now, we prove that the sequence un is bounded in . Let and choose , as a test function in problems (19). From assumption (19), one has

()
where ω = Suppφ. We can use Young’s inequality with ϵ and both (37) and (59) to obtain
()

Hence, equality (60) implies that

()

Letting ɛ = (min(1, α)/2), we get that un is bounded in .

Lemma 11. Under the assumptions of Theorem 2, let un be a solution to problem (19). Then, the sequence is uniformly bounded in , for every σ < (N/N − 1).

Proof. We will prove our proof in two steps:

  • Step 1: we want to prove that, for every λ > 1, . Indeed, let λ > 1, and ω = Suppφ is the support of φ. Thanks to (3), we have from (19) with test function

    ()

  • We use Young’s inequality, and since q < 1, we deduce from (37) that

    ()

  • Thus, by the above estimate and since un is uniformly bounded in , this proves Step 1.

  • Step 2: here, we show that is uniformly bounded in for every r < (N/N − 1). For this, let σ < 2, , and ω = Suppφ. We use Hölder inequality with exponent 2/σ and by step 1, and1 we obtain

()

Using the Sobolev inequality, we obtain

()

Noticing that (σ/σ) > (2 − σ/2) and choosing σ such that (q + 1)σ = (σ(λ + q)/2 − σ) yields σ = (N(2 + qλ)/N(q + 1) − (λ + q)). Using Young’s inequality with ϵ, we obtain

()

It is easy to check that the hypotheses λ > 1 imply σ < (N/N − 1) < 2.

Proof. of Theorem 2.

The proof of the theorem is similar to the proof of the previous theorem with just a small change for the convergence of the term on the left side of equation (47). Indeed, using Lemma 11, we have that is weak in for every σ < (N/N − 1). Hence, for every , we can pass to the limit with respect to n in the integral in the left-hand side of (47).

Remark 1. Assume that (2) and (3) are satisfied. We can choose , as test function in (19), using (3), and we obtain that

()

We deduce from (68) that the sequence is bounded in . Therefore, uγ + q + 1/2 belongs to .

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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