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

Authors: Aziz Bouhlal; Jaouad Igbida
Journal: International Journal of Differential Equations
 (2021)

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## Abstract

Forq,γ>0, we study existence and regularity of solutions for unbounded elliptic problems whose simplest model is −div1+uq∇u=f/uγinΩu=0on∂Ω, where f∈LmΩ, m≥1.


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## Body

## 1. Introduction

Consider the Dirichlet problem for some nonlinear elliptic equations:(1)−divax+uq∇u=fuγ,x∈Ω,u∈H01Ω,under the following assumptions. The set Ω is a bounded open subset of ℝN, with N≥3:(2)q,γ>0.a:Ω⟶ℝ is a measurable function satisfying the following conditions:(3)α≤ax≤β,for almost every x∈Ω, where α and β are positive constant, and(4)0≨f∈LmΩ,withm≥1.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(5)Jv=12∫Ωax+v1−θ∇v2−∫Ωfxv,the Euler–Lagrange equation associated to the functional J is(6)−divax+v1−θ∇v+1−θ2∇v2vθsignv=f.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, [1–9]), namely, with the model problem(7)−divMx,u∇u+∇u2uθsignu=fx,x∈Ω,ux=0,x∈∂Ω,where θ is a positive constant and M:Ω×ℝ⟶ℝ is a Carathéodory function. More precisely, existence of positive solutions for (7) was shown in [1–3], for Mx,t=1 and 0<θ≤1, and the uniqueness of positive solution, for Mx,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≨f∈LmΩ (m>2N/N+2). Later, in [9], it is proved the existence of solution for (7) with 0<θ<1, where Mx,t=1 and the data f∈LmΩ 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, Mx,t=1+tq, and 0≨f∈LmΩ. 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:(8)−2Δu+∇u2u=fx,x∈Ω,ux=0,x∈∂Ω.If we definev=2u/u, then the function v is solution of(9)−Δv=fxv,x∈Ω,vx=0,x∈∂Ω,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(10)−Δu=fuγ,x∈Ω,u=0,x∈∂Ω,where γ>0. More precisely, they have shown that the term f/uγ 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:(11)−div∇u1+up=fuγ,x∈Ω,u=0,x∈∂Ω,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/uγ) 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 divax+uq∇u is not well defined on the whole H01Ω; the solutions did not belong, in general, to H01Ω 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/uγ has a regularizing effect.Our main existence results are as follows.Theorem 1.
Assume that (2) and (3) hold true. If 0≨f∈LmΩ with m>N/2, then there is a positive solution u∈L∞Ω of (1), in the sense of distributions, that is,(12)∫Ωax+uq∇u∇φ=∫Ωfφuγ,for any test function φ in C01Ω. Moreover, we have the following summability results for u:(1)
Let0<q<1:(i)
If0<γ≤1−q, then u∈H01Ω.(ii)
Ifγ>1−q, then u∈Hloc1Ω.(2)
Letq=1:(i)
If0<γ≤1, then u∈H01Ω.(ii)
Ifγ>1, then u∈Hloc1Ω.(3)
Letq>1, then u∈Hloc1Ω.Whenf∈LmΩ, 1<m<N/2, we will prove the following regularizing effects.Theorem 2.
We suppose that0≨f∈LmΩ, 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)
Ifγ<1−q and 2∗/2∗+q−1+γ≤m<N/2, then u∈H01Ω∩Lm∗∗1+q+γΩ, where(13)m∗∗=m∗∗=NmN−2m.(2)
Ifγ=1−q, then u∈H01Ω.(3)
Ifγ>1−q, then u∈L1+q+γ/22∗Ω∩Hloc1Ω.Notation: throughout this paper, we fix an integerN≥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/N−p the Sobolev conjugate exponent of p. As usual, let us denote by S the Sobolev constant, i.e.,(14)S=infu∈H01Ω−0∇u22u2∗2.We denote byP the Poincaré constant given by(15)P=infu∈H01Ω−0∇u22u22.For allk>0, we recall the definition of a truncated function Tks defined by(16)Tks=maxmink,s−k.We also consider(17)Gks=s−Tks.As usual, we consider the positive and negative part of a measurable functionux(18)ux=u+x−u−x,whereu+x=uxχu≥0andu−x=−uxχu<0.


## 2. The Approximated Problem

To prove our existence results, we will use the following approximating problems:(19)−divax+unq∇un=fnun+1/nγ,x∈Ω,where n∈ℕ∗, and(20)fnx=fx1+1/nfx.As in [11], we prove existence of positive solution of the approximated problem.Lemma 1.
Letg be positive function belonging to L∞Ω. Suppose that (2) and (3) are satisfied. Then, there exists a positive solution un∈H01Ω∩L∞Ω of the problem(21)−divax+unq∇un=gun+1/nγ,x∈Ω,un∈H01Ω.Proof.
To prove it, we define the following operatorSn:L2Ω⟶L2Ω which associates to every v∈L2Ω the solution wn∈H01Ω to(22)−divax+Tnwnq∇wn=gv+1/nγ,inΩ,wn=0,on∂Ω.
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(23)α∫Ω∇wn2≤∫Ωax+Tnwnq∇wn2=∫Ωgwnv+1/nγ,≤nγgL∞Ω∫Ωwn≤nγgL∞ΩΩwnL2Ω.
Thanks to Poincaré’s inequality, we deduce(24)αPwnL2Ω2≤nγgL∞ΩΩwnL2Ω.
Hence, there exists an invariant ball forSn. On the contrary, from the H01Ω↪L2Ω 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 un∈H01Ω to problems(25)−divax+Tnunq∇un=gun+1/nγ,inΩ,un=0,on∂Ω.
Moreover, by the maximum principle, it is clear that the sequenceun is nonnegative since g is nonnegative, and we choose Gkun as test function in (25) and use (3) to obtain(26)α∫AkGkun2≤1kγ∫AkgGkun,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 un:=un+dn+1∈L∞Ω∩H01Ω is a solution of (13).
By Lemma1, it follows the existence of a solution un∈L∞Ω∩H01Ω of (19).
Now, we are going to prove that the sequenceun 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∈L1Ω 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 considerTkun−un+1+ as a test function in problems (19). Then,(27)∫Ωax+unq∇un∇Tkun−un+1+=∫Ωfnun+1/nγTkun−un+1+.
Observing thatfn≤fn+1, we have(28)∫Ωfnun+1/nγTkun−un+1+≤∫Ωfn+1un+1+1/n+1γTkun−un+1+=∫Ωax+un+1q∇un+1∇Tkun−un+1+≤∫Ωax+unq∇un+1∇Tkun−un+1+.
Therefore, by (3), we deduce that(29)α∫Ω∇Tkun−un+1+2≤∫Ωax+unq∇Tkun−un+1+2≤0.
Consequently, we obtain∫Ω∇Tkun−un+1+2=0, so by Poincaré’s inequality, we have Tkun−un+1+=0 for every k>0. Thus, un≤un+1 a.e. x∈Ω.
We remark thatu1 is bounded; indeed, u1≤c, for some positive constant c. Then, it follows that(30)−divax+u1q∇u1≥f1c+1γ,x∈Ω.
Thanks to (3), we have α≤ax+u1q≤β+cq. 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.
Let0≨f∈LmΩ 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,measΩ and on the norm of f in LmΩ.Proof.
The use ofGkun as test function in (19) and (3), implies that(31)α∫Ak∇Gkun2≤1kγ∫AkfGkun,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,measΩ and fLmΩ such that: unL∞Ω≤M for all n∈ℕ∗.Lemma 4.
We assume that0≨f∈LmΩ 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 H01Ω.Proof.
We denote byC a positive constant which may only depend on the parameters of our problem, and its value may vary from line to line.
We use1+un1−q−1 as test function in (19) to obtain(32)1−q∫Ωax+unq1+unq∇un2≤C∫Ωfun1−q−γ,and thus (since q≤1),(33)1−qminα,1∫Ω∇un2≤C∫Ωfun1−q−γ≤CunL∞Ω1−q−γ∫Ωf≤C.from which the sequence un is bounded in H01Ω.Lemma 5.
Let0≨f∈LmΩ 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 Hloc1Ω.Proof.
Letφ∈C01Ω 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.
Choosingun+11−q−1φ2 as test function in (19) and using (3), we obtain(34)α1−q∫Ω∇un2φ2+2∫Ωax+unqun+11−q−1∇un∇φφ≤∫Ωfnun+1/nγun+11−q−1φ2≤φL∞Ω2cωγ∫Ωf,which then implies(35)α1−q∫Ω∇un2φ2≤φL∞Ω2cωγ∫Ωf−2∫Ωax+unqun+11−q−1∇un∇φφ.
We can use Young’s inequality withϵ, and we obtain(36)2∫Ωax+unqun+11−q−1∇un∇φφ≤ɛ∫Ω∇un2φ2+Cɛ∫Ωax+unq2un+11−q−12∇φ2.
Using (3), we have(37)ax+tq≤c01+tq,for every q>0 and t≥0 (and for a suitable c0 independent on n).
We then have(38)2∫Ωax+unqun+11−q−1∇un∇φφ≤ɛ∫Ω∇un2φ2+Cɛc02∫Ωun2∇φ2.
Applying (38) to (35) and letting ɛ=α1−q/2, we obtain(39)∫Ω∇un2φ2≤C+C∫Ωun2∇φ2≤C+CunL∞Ω2∫Ω∇φ2≤C,and this gives that un is bounded in Hloc1Ω.Lemma 6.
Letq=1. Suppose that (2) and (3) hold. If 0≨f∈LmΩ with m>N/2, then the sequence un defined by (19) satisfies the following summability:(1)
If0<γ≤1, then un is uniformly bounded in H01Ω(2)
Ifγ>1, then un is uniformly bounded in Hloc1ΩProof.
(1)
Let us takelog1+un as test function in (19) and use (3) to obtain that(40)min1,α∫Ω∇un2≤∫Ωflog1+unun+1/nγ≤∫Ωfun1−γ≤unL∞Ω1−γ∫Ωf≤C.(2)
Letφ∈C01Ω and choose log1+unφ2, as a test function in problem (19). From assumption (19), one has(41)min1,α∫Ω∇un2φ2+2∫Ωax+unqlog1+un∇un∇φφ≤∫Ωflog1+unun+1/nγφ2≤∫Ωfφ2unγ−1≤φL∞Ω2cωγ−1∫Ωf,where ω=Suppφ. By Young’s inequalities, it is easy to prove(42)2∫Ωax+unqlog1+un∇un∇φφ≤ɛ∫Ω∇un2φ2+Cɛ.
Hence, equality (41) implies that(43)min1,α∫Ω∇un2φ2≤φL∞Ω2cωγ−1∫Ωf+ɛ∫Ω∇un2φ2+Cɛ.
Lettingɛ=min1,α/2, we get that un is bounded in Hloc1Ω.Lemma 7.
Letq>1. Assume that (2) and (3) hold true. If 0≨f∈LmΩ with m>N/2, then the solution un of (19) is uniformly bounded in Hloc1Ω.Proof.
Letφ be a function in C01Ω and ω=Suppφ. Take 1−un+11−qφ2 as test function in (19) and use (3) to obtain(44)min1,α2q−1∫Ω∇un2φ2≤q−1min1,α∫Ω1+unq1+unq∇un2φ2≤∫Ωfun+1/nγφ2−2∫Ωax+unq1−un+11−q∇un∇φφ.
Using Young’s inequality withϵ, we have by (3) and Lemma 3 that(45)2∫Ωax+unq1−un+11−q∇un∇φφ≤ɛ∫Ω∇un2φ2+Cɛ∫Ω∇φ2.
Taking the above estimate in (44) and letting ɛ=min1,α/2q, we obtain(46)min1,α2q∫Ω∇un2φ2≤φL∞Ω2cωγ∫Ωf+C,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 Lemmas3 and 4, there exists a subsequence un and a function u∈H01Ω∩L∞Ω such that un weakly converges to u in H01Ω. Now, we can pass to the limit in the equation satisfied by the approximated solutions un:(47)∫Ωax+unq∇un∇φ=∫Ωfnφun+1/nγ,∀φ∈C01Ω,where fnx=fx/1+1/nfx.
For the term of the left-hand side, it is sufficient to observe that∇un converge to ∇u weakly in Lloc2Ω and ax+unq a.e. (and weakly −∗ in L∞Ω converges towards ax+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(48)fnφun+1/nγ≤fφcωγ.
Finally, passing to the limit asn goes to infinity in equation (47), we conclude that(49)∫Ωax+uq∇u∇φ=∫Ωfφuγ,∀φ∈C01Ω.


### 2.2. Further Existence Result

In this section, we suppose (2) and (3) and we assume that(50)0<q<1holds true.Lemma 8.
We suppose that (2), (3), and (50) hold true. Let γ<1−q and 0≨f∈LmΩ, with(51)2∗2∗+q−1+γ≤m<N2.
Then, the solutionsun to problem (19) are uniformly bounded in H01Ω∩Lm∗∗1+q+γΩ.Proof.
Let us take1+un1−q−1 as a test function in (19) and use assumption (3) to obtain(52)1−qmin1,α∫Ω∇un2≤1−q∫Ωax+unq1+unq∇un2≤C∫Ωfun1−q−γ.
We can use Hölder’s inequality on the right-hand side with exponentp=2∗/2∗+q−1+γ=2N/Nγ+1+q+21−q−γ>1, and Sobolev inequality on the left-hand side to deduce(53)Smin1,α1−q∫Ωun2∗2/2∗≤C∫Ωunp′1−q−γ1/p′.
We note that2∗=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 L2∗Ω and in H01Ω.
We are going to prove now that the sequenceun is bounded in Lm∗∗1+q+γΩ. Let λ=N1+qm−1+γmN−2/N−2m; using 1+unλ−1 as a test function for problem (19), we can deduce(54)λmin1,α∫Ω∇un21+un1−λ−q≤∫Ωfn1/n+unγ1+unλ−1,≤C+C∫Ωf1+unγ−λ.
Now, we rewrite(55)4λmin1,α1+q+λ2∫Ω∇1+un1+q+λ/2−12=λmin1,α∫Ω∇un21+un1−λ−qand use the Sobolev inequality and the Hölder inequality in (54) to obtain(56)∫Ω1+un1+q+λ/2−12∗2/2∗≤∫Ωun+1m′λ−γ1/m′.
We note that the choice ofλ 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 Lm∗∗1+q+γΩ.Lemma 9.
Under the hypotheses0≨f∈L1Ω, (2), (3), and (50), if γ=1−q, then the solutions un are uniformly bounded in H01Ω.Proof.
We choose1+un1−q−1 as test function in (19) to obtain, by hypothesis (3), that(57)1−qminα,1∫Ω∇un2≤1−q∫Ωax+unq1+unq∇un2≤C∫Ωf.
Therefore,un is bounded in H01Ω.Lemma 10.
Let0≨f∈L1Ω. Under hypotheses (2), (3), and (50), if γ>1−q, then the solutions un are uniformly bounded in L1+q+γ/22∗Ω∩Hloc1Ω.Proof.
Choosingunγ as test function in (19) and using Hölder and Sobolev inequalities, thanks to (3), we obtain that(58)4γS1+q+γ2∫Ωun1+q+γ/22∗2/2∗≤γ∫Ω∇un2unγ+q−1,≤γ∫Ωax+unq∇un2unγ−1≤∫Ωf.
The above inequality implies that(59)∫Ωun1+q+γ/22∗≤C.
Now, we prove that the sequenceun is bounded in Hloc1Ω. Let φ∈C01Ω and choose 1+un1−q−1φ2, as a test function in problems (19). From assumption (19), one has(60)min1,α∫Ω∇un2φ2++2∫Ωax+unqun+11−q−1∇un∇φφ∫Ωfnun+1/nγun+11−q−1φ2≤φL∞Ω2cωγ−q+1∫Ωf,where ω=Suppφ. We can use Young’s inequality with ϵ and both (37) and (59) to obtain(61)2∫Ωax+unqun+11−q−1∇un∇φφ≤ɛ∫Ω∇un2φ2+Cɛ∫Ωax+unqun+11−q−1∇φ2≤ɛ∫Ω∇un2φ2+Cɛc0∫Ωun2∇φ2≤ɛ∫Ω∇un2φ2+Cɛ.
Hence, equality (60) implies that(62)min1,α∫Ω∇un2φ2≤φL∞Ω2cωγ+q−1∫Ωf+ɛ∫Ω∇un2φ2+Cɛ.
Lettingɛ=min1,α/2, we get that un is bounded in Hloc1Ω.Lemma 11.
Under the assumptions of Theorem2, let un be a solution to problem (19). Then, the sequence unq∇un is uniformly bounded in LlocσΩ, for every σ<N/N−1.Proof.
We will prove our proof in two steps:Step 1: we want to prove that, for everyλ>1, 1+unq−λ∇un2∈Lloc1Ω. Indeed, let λ>1, φ∈C01Ω and ω=Suppφ is the support of φ. Thanks to (3), we have from (19) with test function 1−1/1+unλ−1φ2(63)λ−1∫Ω∇un21+unλ−qφ2+2∫Ωax+unq1−un+11−λ∇un∇φφ≤Cω.We use Young’s inequality, and sinceq<1, we deduce from (37) that(64)λ−1∫Ω∇un21+unλ−qφ2≤Cω+12∫Ω∇un2φ2+12∫Ωax+unq21−un+11−λ2∇φ2≤Cω+12∫Ω∇un2φ2+c02∫Ω1+un2q∇φ2≤Cω+12∫Ω∇un2φ2+C∫Ωun2∇φ2+C.Thus, by the above estimate and sinceun is uniformly bounded in Hloc1Ω, this proves Step 1.Step 2: here, we show thatunq∇un is uniformly bounded in LlocrΩ for every r<N/N−1. For this, let σ<2, 0<φ∈C01Ω, and ω=Suppφ. We use Hölder inequality with exponent 2/σ and by step 1, and1 we obtain(65)∫Ωunqσ∇unσφσ≤∫Ω∇unσ1+unσλ−q/2φσ2N−2/2N−σ1+unσλ+q/2φNσ2−σ/2N−σ≤∫Ω∇un21+unλ−qφσN−2/N−σσ/2∫Ω1+unσλ+q/2−σφσ∗2−σ/2≤Cω∫Ω1+unσλ+q/2−σφσ∗2−σ/2.
Using the Sobolev inequality, we obtain(66)∫Ωunq+1σ∗φσ∗σ/σ∗≤Cω∫Ω1+unσλ+q/2−σφσ∗2−σ/2+Cω.
Noticing thatσ/σ∗>2−σ/2 and choosing σ such that q+1σ∗=σλ+q/2−σ yields σ=N2+q−λ/Nq+1−λ+q. Using Young’s inequality with ϵ, we obtain(67)∫Ωunq+1σ∗φσ∗σ/σ∗≤ɛ∫Ω1+unq+1σ∗φσ∗σ/σ∗+Cω,ɛ.
It is easy to check that the hypothesesλ>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 ax+unq∇un⟶ax+uq∇u is weak in LlocσΩN for every σ<N/N−1. Hence, for every φ∈C01Ω, 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 unγ, as test function in (19), using (3), and we obtain that(68)4γγ+q+12∫Ω∇unγ+q+1/22=γ∫Ω∇un2unγ+q−1,≤γ∫Ωax+unq∇un2unγ−1≤∫Ωf.
We deduce from (68) that the sequence unγ+q+1/2 is bounded in H01Ω. Therefore, uγ+q+1/2 belongs to H01Ω.


## 2.1. Existence of Bounded Solutions

In this section, we will prove existence of bounded weak solutions for (1).Lemma 3.
Let0≨f∈LmΩ 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,measΩ and on the norm of f in LmΩ.Proof.
The use ofGkun as test function in (19) and (3), implies that(31)α∫Ak∇Gkun2≤1kγ∫AkfGkun,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,measΩ and fLmΩ such that: unL∞Ω≤M for all n∈ℕ∗.Lemma 4.
We assume that0≨f∈LmΩ 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 H01Ω.Proof.
We denote byC a positive constant which may only depend on the parameters of our problem, and its value may vary from line to line.
We use1+un1−q−1 as test function in (19) to obtain(32)1−q∫Ωax+unq1+unq∇un2≤C∫Ωfun1−q−γ,and thus (since q≤1),(33)1−qminα,1∫Ω∇un2≤C∫Ωfun1−q−γ≤CunL∞Ω1−q−γ∫Ωf≤C.from which the sequence un is bounded in H01Ω.Lemma 5.
Let0≨f∈LmΩ 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 Hloc1Ω.Proof.
Letφ∈C01Ω 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.
Choosingun+11−q−1φ2 as test function in (19) and using (3), we obtain(34)α1−q∫Ω∇un2φ2+2∫Ωax+unqun+11−q−1∇un∇φφ≤∫Ωfnun+1/nγun+11−q−1φ2≤φL∞Ω2cωγ∫Ωf,which then implies(35)α1−q∫Ω∇un2φ2≤φL∞Ω2cωγ∫Ωf−2∫Ωax+unqun+11−q−1∇un∇φφ.
We can use Young’s inequality withϵ, and we obtain(36)2∫Ωax+unqun+11−q−1∇un∇φφ≤ɛ∫Ω∇un2φ2+Cɛ∫Ωax+unq2un+11−q−12∇φ2.
Using (3), we have(37)ax+tq≤c01+tq,for every q>0 and t≥0 (and for a suitable c0 independent on n).
We then have(38)2∫Ωax+unqun+11−q−1∇un∇φφ≤ɛ∫Ω∇un2φ2+Cɛc02∫Ωun2∇φ2.
Applying (38) to (35) and letting ɛ=α1−q/2, we obtain(39)∫Ω∇un2φ2≤C+C∫Ωun2∇φ2≤C+CunL∞Ω2∫Ω∇φ2≤C,and this gives that un is bounded in Hloc1Ω.Lemma 6.
Letq=1. Suppose that (2) and (3) hold. If 0≨f∈LmΩ with m>N/2, then the sequence un defined by (19) satisfies the following summability:(1)
If0<γ≤1, then un is uniformly bounded in H01Ω(2)
Ifγ>1, then un is uniformly bounded in Hloc1ΩProof.
(1)
Let us takelog1+un as test function in (19) and use (3) to obtain that(40)min1,α∫Ω∇un2≤∫Ωflog1+unun+1/nγ≤∫Ωfun1−γ≤unL∞Ω1−γ∫Ωf≤C.(2)
Letφ∈C01Ω and choose log1+unφ2, as a test function in problem (19). From assumption (19), one has(41)min1,α∫Ω∇un2φ2+2∫Ωax+unqlog1+un∇un∇φφ≤∫Ωflog1+unun+1/nγφ2≤∫Ωfφ2unγ−1≤φL∞Ω2cωγ−1∫Ωf,where ω=Suppφ. By Young’s inequalities, it is easy to prove(42)2∫Ωax+unqlog1+un∇un∇φφ≤ɛ∫Ω∇un2φ2+Cɛ.
Hence, equality (41) implies that(43)min1,α∫Ω∇un2φ2≤φL∞Ω2cωγ−1∫Ωf+ɛ∫Ω∇un2φ2+Cɛ.
Lettingɛ=min1,α/2, we get that un is bounded in Hloc1Ω.Lemma 7.
Letq>1. Assume that (2) and (3) hold true. If 0≨f∈LmΩ with m>N/2, then the solution un of (19) is uniformly bounded in Hloc1Ω.Proof.
Letφ be a function in C01Ω and ω=Suppφ. Take 1−un+11−qφ2 as test function in (19) and use (3) to obtain(44)min1,α2q−1∫Ω∇un2φ2≤q−1min1,α∫Ω1+unq1+unq∇un2φ2≤∫Ωfun+1/nγφ2−2∫Ωax+unq1−un+11−q∇un∇φφ.
Using Young’s inequality withϵ, we have by (3) and Lemma 3 that(45)2∫Ωax+unq1−un+11−q∇un∇φφ≤ɛ∫Ω∇un2φ2+Cɛ∫Ω∇φ2.
Taking the above estimate in (44) and letting ɛ=min1,α/2q, we obtain(46)min1,α2q∫Ω∇un2φ2≤φL∞Ω2cωγ∫Ωf+C,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 Lemmas3 and 4, there exists a subsequence un and a function u∈H01Ω∩L∞Ω such that un weakly converges to u in H01Ω. Now, we can pass to the limit in the equation satisfied by the approximated solutions un:(47)∫Ωax+unq∇un∇φ=∫Ωfnφun+1/nγ,∀φ∈C01Ω,where fnx=fx/1+1/nfx.
For the term of the left-hand side, it is sufficient to observe that∇un converge to ∇u weakly in Lloc2Ω and ax+unq a.e. (and weakly −∗ in L∞Ω converges towards ax+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(48)fnφun+1/nγ≤fφcωγ.
Finally, passing to the limit asn goes to infinity in equation (47), we conclude that(49)∫Ωax+uq∇u∇φ=∫Ωfφuγ,∀φ∈C01Ω.


## 2.2. Further Existence Result

In this section, we suppose (2) and (3) and we assume that(50)0<q<1holds true.Lemma 8.
We suppose that (2), (3), and (50) hold true. Let γ<1−q and 0≨f∈LmΩ, with(51)2∗2∗+q−1+γ≤m<N2.
Then, the solutionsun to problem (19) are uniformly bounded in H01Ω∩Lm∗∗1+q+γΩ.Proof.
Let us take1+un1−q−1 as a test function in (19) and use assumption (3) to obtain(52)1−qmin1,α∫Ω∇un2≤1−q∫Ωax+unq1+unq∇un2≤C∫Ωfun1−q−γ.
We can use Hölder’s inequality on the right-hand side with exponentp=2∗/2∗+q−1+γ=2N/Nγ+1+q+21−q−γ>1, and Sobolev inequality on the left-hand side to deduce(53)Smin1,α1−q∫Ωun2∗2/2∗≤C∫Ωunp′1−q−γ1/p′.
We note that2∗=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 L2∗Ω and in H01Ω.
We are going to prove now that the sequenceun is bounded in Lm∗∗1+q+γΩ. Let λ=N1+qm−1+γmN−2/N−2m; using 1+unλ−1 as a test function for problem (19), we can deduce(54)λmin1,α∫Ω∇un21+un1−λ−q≤∫Ωfn1/n+unγ1+unλ−1,≤C+C∫Ωf1+unγ−λ.
Now, we rewrite(55)4λmin1,α1+q+λ2∫Ω∇1+un1+q+λ/2−12=λmin1,α∫Ω∇un21+un1−λ−qand use the Sobolev inequality and the Hölder inequality in (54) to obtain(56)∫Ω1+un1+q+λ/2−12∗2/2∗≤∫Ωun+1m′λ−γ1/m′.
We note that the choice ofλ 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 Lm∗∗1+q+γΩ.Lemma 9.
Under the hypotheses0≨f∈L1Ω, (2), (3), and (50), if γ=1−q, then the solutions un are uniformly bounded in H01Ω.Proof.
We choose1+un1−q−1 as test function in (19) to obtain, by hypothesis (3), that(57)1−qminα,1∫Ω∇un2≤1−q∫Ωax+unq1+unq∇un2≤C∫Ωf.
Therefore,un is bounded in H01Ω.Lemma 10.
Let0≨f∈L1Ω. Under hypotheses (2), (3), and (50), if γ>1−q, then the solutions un are uniformly bounded in L1+q+γ/22∗Ω∩Hloc1Ω.Proof.
Choosingunγ as test function in (19) and using Hölder and Sobolev inequalities, thanks to (3), we obtain that(58)4γS1+q+γ2∫Ωun1+q+γ/22∗2/2∗≤γ∫Ω∇un2unγ+q−1,≤γ∫Ωax+unq∇un2unγ−1≤∫Ωf.
The above inequality implies that(59)∫Ωun1+q+γ/22∗≤C.
Now, we prove that the sequenceun is bounded in Hloc1Ω. Let φ∈C01Ω and choose 1+un1−q−1φ2, as a test function in problems (19). From assumption (19), one has(60)min1,α∫Ω∇un2φ2++2∫Ωax+unqun+11−q−1∇un∇φφ∫Ωfnun+1/nγun+11−q−1φ2≤φL∞Ω2cωγ−q+1∫Ωf,where ω=Suppφ. We can use Young’s inequality with ϵ and both (37) and (59) to obtain(61)2∫Ωax+unqun+11−q−1∇un∇φφ≤ɛ∫Ω∇un2φ2+Cɛ∫Ωax+unqun+11−q−1∇φ2≤ɛ∫Ω∇un2φ2+Cɛc0∫Ωun2∇φ2≤ɛ∫Ω∇un2φ2+Cɛ.
Hence, equality (60) implies that(62)min1,α∫Ω∇un2φ2≤φL∞Ω2cωγ+q−1∫Ωf+ɛ∫Ω∇un2φ2+Cɛ.
Lettingɛ=min1,α/2, we get that un is bounded in Hloc1Ω.Lemma 11.
Under the assumptions of Theorem2, let un be a solution to problem (19). Then, the sequence unq∇un is uniformly bounded in LlocσΩ, for every σ<N/N−1.Proof.
We will prove our proof in two steps:Step 1: we want to prove that, for everyλ>1, 1+unq−λ∇un2∈Lloc1Ω. Indeed, let λ>1, φ∈C01Ω and ω=Suppφ is the support of φ. Thanks to (3), we have from (19) with test function 1−1/1+unλ−1φ2(63)λ−1∫Ω∇un21+unλ−qφ2+2∫Ωax+unq1−un+11−λ∇un∇φφ≤Cω.We use Young’s inequality, and sinceq<1, we deduce from (37) that(64)λ−1∫Ω∇un21+unλ−qφ2≤Cω+12∫Ω∇un2φ2+12∫Ωax+unq21−un+11−λ2∇φ2≤Cω+12∫Ω∇un2φ2+c02∫Ω1+un2q∇φ2≤Cω+12∫Ω∇un2φ2+C∫Ωun2∇φ2+C.Thus, by the above estimate and sinceun is uniformly bounded in Hloc1Ω, this proves Step 1.Step 2: here, we show thatunq∇un is uniformly bounded in LlocrΩ for every r<N/N−1. For this, let σ<2, 0<φ∈C01Ω, and ω=Suppφ. We use Hölder inequality with exponent 2/σ and by step 1, and1 we obtain(65)∫Ωunqσ∇unσφσ≤∫Ω∇unσ1+unσλ−q/2φσ2N−2/2N−σ1+unσλ+q/2φNσ2−σ/2N−σ≤∫Ω∇un21+unλ−qφσN−2/N−σσ/2∫Ω1+unσλ+q/2−σφσ∗2−σ/2≤Cω∫Ω1+unσλ+q/2−σφσ∗2−σ/2.
Using the Sobolev inequality, we obtain(66)∫Ωunq+1σ∗φσ∗σ/σ∗≤Cω∫Ω1+unσλ+q/2−σφσ∗2−σ/2+Cω.
Noticing thatσ/σ∗>2−σ/2 and choosing σ such that q+1σ∗=σλ+q/2−σ yields σ=N2+q−λ/Nq+1−λ+q. Using Young’s inequality with ϵ, we obtain(67)∫Ωunq+1σ∗φσ∗σ/σ∗≤ɛ∫Ω1+unq+1σ∗φσ∗σ/σ∗+Cω,ɛ.
It is easy to check that the hypothesesλ>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 ax+unq∇un⟶ax+uq∇u is weak in LlocσΩN for every σ<N/N−1. Hence, for every φ∈C01Ω, 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 unγ, as test function in (19), using (3), and we obtain that(68)4γγ+q+12∫Ω∇unγ+q+1/22=γ∫Ω∇un2unγ+q−1,≤γ∫Ωax+unq∇un2unγ−1≤∫Ωf.
We deduce from (68) that the sequence unγ+q+1/2 is bounded in H01Ω. Therefore, uγ+q+1/2 belongs to H01Ω.



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*Source: 5589504-2021-04-27.xml*