Existence and Regularity of Solutions for Unbounded Elliptic Equations with Singular Nonlinearities
Abstract
For q, γ > 0, we study existence and regularity of solutions for unbounded elliptic problems whose simplest model is , where f ∈ Lm(Ω), m ≥ 1.
1. Introduction
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≨f ∈ Lm(Ω) with m > (N/2), then there is a positive solution u ∈ L∞(Ω) of (1), in the sense of distributions, that is,
- (1)
Let 0 < q < 1:
- (i)
If 0 < γ ≤ 1 − q, then .
- (ii)
If γ > 1 − q, then .
- (i)
- (2)
Let q = 1:
- (i)
If 0 < γ ≤ 1, then .
- (ii)
If γ > 1, then .
- (i)
- (3)
Let q > 1, then .
When f ∈ Lm(Ω), 1 < m < (N/2), we will prove the following regularizing effects.
Theorem 2. We suppose that 0≨f ∈ Lm(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 .
2. The Approximated Problem
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 v ∈ L2(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
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≨f ∈ L1(Ω) and un is the solution of problem (19), then for every n ∈ ℕ∗: un ≤ un+1 a.e. in Ω. Furthermore, if ω ⊂ ⊂Ω, then, for every n ∈ ℕ∗, there exists cω > 0 such that un ≥ cω > 0 a.e. in ω.
Proof. Let us consider as a test function in problems (19). Then,
Observing that fn ≤ fn+1, we have
Therefore, by (3), we deduce that
Consequently, we obtain , so by Poincaré’s inequality, we have for every k > 0. Thus, un ≤ un+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 u1 ≥ cω a.e. in ω. Finally, using that the sequence un is increasing, one deduces that un ≥ cω a.e. in ω for every n ∈ ℕ∗.
2.1. Existence of Bounded Solutions
In this section, we will prove existence of bounded weak solutions for (1).
Lemma 3. Let 0≨f ∈ Lm(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 ∈ 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
Lemma 4. We assume that 0≨f ∈ Lm(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 ∈ 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
Lemma 5. Let 0≨f ∈ Lm(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 un ≥ cω for a.e. x ∈ ω.
Choosing as test function in (19) and using (3), we obtain
We can use Young’s inequality with ϵ, and we obtain
Using (3), we have
We then have
Applying (38) to (35) and letting ɛ = (α(1 − q)/2), we obtain
Lemma 6. Let q = 1. Suppose that (2) and (3) hold. If 0≨f ∈ Lm(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
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≨f ∈ Lm(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
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:
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
Lemma 8. We suppose that (2), (3), and (50) hold true. Let γ < 1 − q and 0≨f ∈ Lm(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 − q − γ)) > 1, and Sobolev inequality on the left-hand side to deduce
We note that 2∗ = p′(1 − q − γ); 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
We note that the choice of λ is equivalent to require (2/2∗)(1 + q + λ) = m′(λ − γ); 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≨f ∈ L1(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≨f ∈ L1(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
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 + q − λ)/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).
Conflicts of Interest
The authors declare that they have no conflicts of interest.
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Data Availability
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