# Existence of Solutions for Unbounded Elliptic Equations with Critical Natural Growth

Authors: Aziz Bouhlal; Abderrahmane El Hachimi; Jaouad Igbida; El Mostafa Sadek; Hamad Talibi Alaoui
Journal: International Journal of Differential Equations
 (2018)

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## Abstract

We investigate existence and regularity of solutions to unbounded elliptic problem whose simplest model is{-div[(1+uq)∇u]+u=γ∇u2/1+u1-q+finΩ,u=0on∂Ω,}, where 0<q<1, γ>0 and f belongs to some appropriate Lebesgue space. We give assumptions on f with respect to q and γ to show the existence and regularity results for the solutions of previous equation.


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## Body

## 1. Introduction

In this paper, we consider the Dirichlet problem for some nonlinear elliptic problems such as(1)-divax+uq∇u+u=Hx,u,∇u+f,x∈Ω,u∈H01Ω,under the following assumptions: Ω is a bounded open subset of RN, where N≥3,  0<q<1,  and f∈Lm with m≥2 and a:Ω→R is a measurable function satisfying the following conditions:(2)α≤ax≤β,for almost every x∈Ω, where α and β are positive real constants. H(x,s,ξ) is a Carathéodory-type function satisfying to:(3)Hx,s,ξ≤γξ21+s1-qfor some γ>0.In [1], Arcoya, Boccardo, and Leonor obtained the existence and regularity results for the following elliptic problem with degenerate coercivity:(4)-divα∇u1+u2+u=γ∇u21+u3+f,x∈Ω,u∈H01Ω,where α,γ>0, f∈Lm(Ω) with m≥2, and Ω is a bounded subset of RN, N≥3.The purpose of the present paper is to study the same kind of lower order terms as in problem (4) in the case of an elliptic operator with unbounded coefficients such as (1).There are several papers concerned with existence and regularity of the solution for the following problem:(5)-divMx,u∇u+gx,u,∇u=fxx∈Ω,ux=0x∈∂Ω.We refer the intersting articles: Boccardo, Murat and Puel [2], Bensoussan, Boccardo and Murat [3], and Boccardo, Gallout [4]. In all these works g is a nonlinear lower term having natural growth with respect to ∇u, data f in suitable Lebesgue spaces, and M(x,u) is a Carathéodory-type bounded function subject to certain structural inequalities.Another motivation for studying these problem arises from the calculus of variations in the case where0≤f∈Lm(Ω) with m≥N/2 and(6)gx,u,∇u=∇u2uθ,where θ∈(0,1), which is considered by Puel in [5].We point out that in [6] the authors considered M(x,u) as a bounded function and(7)gx,u,∇u=Qx,u∇u2uθ,where θ∈(0,1]. The function Q(x,s):Ω×R→RN2 is symmetric, measurable with respect to x and continuous with respect to s with the following uniform ellipticity condition: for x∈Ω, and s∈R, (8)μξ2≤Qx,sξξ≤νξ2,0<μ≤ν.We shall prove the following main results on existence and regularity of solutions for problem (1).Theorem 1.
Letα~=min{1,α}. Assuming that the functions a and H satisfy (2) and (3) then, if f belong to Lm(Ω), with(9)m>2γα~+1+q,there exists a distributional solution u∈W01,1(Ω) of problem (1) such that(10)Hx,u,∇u∈L1Ω,ax+uq∇u∈L1Ω,∫Ωax+uq∇u∇ψ+∫Ωuψ=∫ΩHx,u,∇uψ+∫Ωfψ,∀ψ∈C0∞Ω.Furthermore, any solution of the problem (1) belongs to H01(Ω).In the next result, we consider the case wheref has a high summability.Theorem 2.
Letα~=min{1,α}, and assume that (2) and (3) hold true. If u the solution given by Theorem 1 and f belongs to Lm(Ω), with(11)m>max2γα~+1+q,N2γα~+1,then u belongs to H01(Ω)∩L∞(Ω).The rest of the paper is organized as follows: Section2 is devoted to give some a priori estimates for the approximated problem associated with problem (1); while in Section 3, we give the detailed proofs of Theorems 1 and 2.


## 2. The Approximated Problem

In this section, we use the hypotheses (2) and (3) and we suppose that(12)α~m-1-γ>0,where α~=min{1,α} holds true. To prove Theorem 1 and Theorem 2, we will use the following approximating problems associated with problem (1):(13)-divax+unq∇un+un=Hnx,un,∇un+fn,x∈Ω,where(14)fnx=fx1+1/nfx,and(15)Hnx,s,ξ=Hx,s,ξ1+1/nξ2.By the results of [2, 4] there exists a weak solution un in H01(Ω)∩L∞(Ω) of problem (13) in the sense that(16)∫Ωax+unq∇un∇φ+∫Ωunφ=∫ΩHnx,un,∇unφ+∫Ωfnφfor every φ∈H01(Ω)∩L∞(Ω).The following lemma will be very useful, as it gives us an a priori estimate on the summability of the solutions to problems (13).Lemma 3.
Ifun is a solution to problem (13), then for every k≥0,(17)∫ΩGkunm≤∫un≥kfm.Moreover, there exist R>0 depending on fLm(Ω), α, q, and γ, such that(18)unH01Ω≤R.Remark 4.
(i) Let{un} be a sequence of solutions un of (13). As a consequence of Lemma 3, there exists u∈H01(Ω) such that, up to a subsequence, un converges weakly to u in H01(Ω) and a.e. in Ω.
(ii) By the previous lemma we deduce from (3) that(19)Hnx,un,∇unL1Ω≤γ∫Ω∇un21+un1-q≤γR2.Proof of Lemma3.
In order to prove (17), we claim that by assumption (2) and q<1, there exist positive constant c0 such that(20)α~1+tq≤ax+tq≤c01+tq,∀t∈R.Choosing φ=|Gk(un)|m-1sgn(un) in (16) and using (20), we obtain(21)α~m-1∫Ω1+unq∇un2Gkunm-2+∫ΩunGkunm-1≤γ∫Ω∇un21+un1-qGkunm-1+∫ΩfnGkunm-1.Thus, joining the terms involving the gradient, we get(22)∫Ωα~m-1-γGkun1+un1+unq∇un2Gkunm-2+∫ΩGkunm≤∫ΩfnGkunm-1.Using (12) we deduce that(23)∫ΩGkunm≤∫ΩfGkunm-1,and the Hölder inequality on the right hand side yields(24)∫ΩGkunm≤∫un≥kfm1/m∫ΩGkunm1-1/m,which implies (17).
Let us choose nowφ=[(1+|un|)m-1-1]sgn(un) as a test function in (16), and we obtain(25)α~m-1-γ∫Ω1+unm-2+q∇un2≤fLmΩ∫Ω1+unm1-1/m.Since m≥2, the previous calculations imply(26)∫Ω∇un2≤c∫Ω1+unm1-1/m.Using (17) with k=0, (18) follows.Lemma 5.
Letun be the sequence of solutions to problems (13) and let the function u given by Remark 4. Then un strongly converges to u in Lm(Ω). Moreover ∇un strongly converges to ∇u in L1(Ω)N.Remark 6.
Note that (25) implies that there exists δ>0 independent of n such that(27)∫Ω1+unm-2+q∇un2≤δ.By using the previous lemma, we deduce that(28)1+unm-2+q/2∇un→1+um-2+q/2∇uweaklyinL2ΩN.Proof of Lemma5.
We use (17) written for k=0:(29)∫Ωunm≤∫Ωfm≤c.Since un almost everywhere converges to u, we have from Fatou’s lemma that(30)∫Ωum≤c.Hence u belongs to Lm(Ω). Using assumption (17), for any k>0 we have(31)∫Eunm≤∫E∩un≤kunm+∫E∩un≥kunm≤kmmeasE+∫un≥kfm.As before, we first choose k such that the second integral is small, uniformly with respect to n, and then the measure of E small enough such that the first term is small. The almost everywhere convergence of un to u and Vitali’s theorem imply that un strongly converges to u in Lm(Ω).
For the second convergence, we will follow the same technique as in [1] (see also [7]). Let h,k>0. In the sequel C will denote a constant independent of n,h,k. Let us consider Th[un-Tk(u)] as a test function in problems (16). Then,(32)∫Ωax+un2∇un∇Thun-Tku+∫ΩunThun-Tku≤fL1Ω+Hnx,un,∇unL1Ωh.Moreover, thanks to the Lm(Ω) convergence of un, the second integral in (32) converges (as n diverges) to a positive number. Thus, it yields to(33)α∫Ω∇Thun-Tku2≤fL1Ω+γR2h-∫Ωax+unq∇Tku∇Thun-Tku.Let K=h+k, observing that ∇Th[un-Tk(u)]=0 if un>K, then(34)∫Ωax+unq∇Tku∇Thun-Tku=∫Ωax+TKunq∇Tku∇Thun-Tku.Since Th[un-Tk(u)] converges to Th[u-Tk(u)] weakly in (L2(Ω))N and [a(x)+|TK(un)|q]∇Tk(u) strongly converges to [a(x)+|TK(u)|q]∇Tk(u) in (L2(Ω))N, we have(35)limn→+∞∫Ωax+unq∇Tku∇Thun-Tku=0,thus, yielding(36)∫Ω∇Thun-Tku2≤Ch+εn,where ε(n) denote any quantity that vanishes as n diverges. Hence, by Hölder’s inequality, we deduce that(37)∫un-u≤h,u≤k∇un-u=∫Ω∇Thun-Tku≤Ω1/2Ch+εn.Fix, now, ϵ>0 there exist h0 such that, for h<h0, we have(38)Ω1/2Ch<ϵ.Thanks to the weak convergence of un in H01(Ω) and the absolute continuity of the integral, there exists k0 independent from n such that, for k>k0, we have(39)∫u>k∇un+∫u>k∇u≤ϵ.In addition, by Dunford Pettis Theorem, we deduce that there exists n(h,ϵ) such that, for n>n(h,ϵ), we have(40)∫un-u>h∇un-u≤ϵ.We can write(41)∫Ω∇un-u=∫un-u≤h,u≤k∇un-u+∫un-u≤h,u>k∇un-u+∫un-u>h∇un-u.Using (37), (39), and (40), for h<h0 and n>n(h,ϵ), we have(42)∫Ω∇un-u≤3ϵ+εn.This proves the strong convergence of ∇un to ∇u in L1(Ω)N.The following lemma yields some a priori estimate on{un}.Lemma 7.
Letu be the function given by Remark 4. Then uq∇u belongs to Lr(Ω), for every r<N/N-1.Proof.
For everyλ>1, we take 1-1/(1+un)λ-1sign(un) as a test function in (16). Droping positive terms yields (43)α~λ-1∫Ω1+unq∇un21+unλ≤fL1Ω+Hnx,un,∇unL1Ω. Hence, using q<1, it follows that(44)∫Ω∇un21+unλ-q≤fL1Ω+γR2α~λ-1.On the other hand, for every λ>1; we have(45)∫Ωunqr∇unr≤∫Ω∇unr1+unrλ-q/21+unrλ+q/2,≤fL1Ω+γR2α~λ-1r/2∫Ω1+unrλ+q/2-r2-r/2.Then, we obtain(46)∫Ωunq+1r∗r/r∗≤c∫Ωunrλ+q/2-r2-r/2.Let us choose r such that (q+1)r∗=r(λ+q)/2-r, that is(47)r=N2+q-λNq+1-λ+q.Since λ>1, we then have an estimate on unq∇un in Lr(Ω), for every r<N/N-1.The next result will be used in the proof of Theorem2.Lemma 8.
Suppose that (2), (3), and (11) hold true. Let f∈Lm(Ω) and {un} be a solution of (13) with fn=f for every n∈N. Then the norms of {un} in L∞(Ω) and in H01(Ω) are bounded by a constant which depends on q,m,N,α,γ,meas(Ω) and on the norm of f in Lm(Ω).Proof.
Sincem>N/2(γ/α~+1), we have 1/2(γ/α~+1)<m/N. Let us choose σ>0 such that(48)12γα~+q+1<σ<mN+q2.The use of(49)1+un2σ-q-1-1+k2σ-q-1+signun,as test function in (16), (3), and (20), implies that(50)2σ-q-1α~∫Ak∇un21+un2σ-2+∫Akun1+un2σ-q-1-1+k2σ-q-1≤γ∫Ak∇un21+un1-q1+un2σ-q-1+∫Akfn1+un2σ-q-1,where(51)Ak=x∈Ω:un>k.By Young and Hölder’s inequalities, we find(52)2σ-q-1α~-γ∫Ak∇un21+un2σ-2≤C1∫Ak1+un2σ-q+C2∫Akf2σ-q≤CmmeasAk1-2σ-q/m.Then, using Sobolev’s inequality gives (53)2σ-q-1α~-γS2σ2∫Ak1+unσ-1+kσ2∗2/2∗≤CmmeasAk1-2σ-q/m,where S denotes the best constant in Sobolev inequality. Now, we set(54)1+unσ=vnand(55)1+kσ=h. and the fact that Ak={x∈Ω:vn>h}., the last inequality gives(56)2σ-q-1α~-γS2σ2∫Akvn-h2∗2/2∗≤CmmeasAk1-2σ-q/m.Note that σ<m/N+q/2 implies that [1-2σ-q/m]2∗/2>1. Then Stampacchia’s technique implies the following relation for some positive constant C3,(57)vnL∞Ω=1+unσL∞Ω≤C3,that is, unL∞(Ω) is bounded.


## 3. Proof of the Main Results

We are now ready to prove the main result of this paper. We first observe that condition (9) implies (12). Hence the results of the previous section hold true. In order to prove the result, we have to pass to the limit in (16). To this aim, let g be a function in C1(R) such that(58)gs=1+sα~ρ-γifs≥011-sα~ρ-γifs<0,where(59)ρ=m-q-22.Observe that, by (9), g is positive, increasing, and it verifies(60)α~ρg′s-γgs1+s>0,∀s∈R.We will use, for k>0 and s∈R,(61)Rks=1-T1Gks,to define a test function. Remark that Rk≥0, -k-1≤Rk(s)≤k+1 and(62)Rk′s=1if-k-1≤s≤-k-1ifk≤s≤k+10otherwise.First of all, note that the a.e. convergence of ∇un (see Lemma 5), Remark 6, and (20) imply both that(63)ax+unqgρun∇un→ax+uqgρu∇uweaklyinL2ΩNand(64)ax+unq1gρun∇un→ax+uq1gρu∇uweaklyinL2ΩN,where ρ is defined in (59).The proof of the result will be achieved in two steps.Step 1 (The first inequality).
We fixψ∈H01(Ω)∩L∞(Ω), with ψ≥0, and take(65)ϕ=gρungρuRkuψAs test function in (16), we have that(66)∫Ωax+unq∇un∇ψgρungρuRku-ρ∫Ωax+unq∇un∇ugρungρ+1ug′uRkuψ+∫Ωax+unq∇un∇ugρungρuRk′uψ+ρ∫Ωax+unq∇un∇ungρ-1ungρug′unRkuψ-∫ΩHnx,un,∇ungρungρuRkuψ+∫ΩungρungρuRkuψ=∫ΩfngρungρuRkuψ.Remark now that, by the assumptions on a, H, relation (60) and the fact that ψ≥0, then we have(67)ρax+unq∇un∇ungρ-1ungρug′unRkuψ-Hnx,un,∇ungρungρuRkuψ≥1+unq∇un2gρ-1ungρuRkuψα~ρg′un-γgun1+un≥0.Therefore, using the almost everywhere convergence of both ∇un and un, and applying Fatou’s lemma, we get (68)liminfn→∞ρ∫Ωax+unq∇un∇ungρ-1ungρug′unRkuψ-∫ΩHnx,un,∇ungρungρuRkuψ≥ρ∫Ωax+uq∇u∇ug′uguRkuψ-∫ΩHx,u,∇uRkuψ.Furthermore, by using Lebesgue’s theorem and (63), we obtain(69)limn→∞∫Ωax+unq∇un∇ψgρungρuRku=∫Ωax+uq∇u∇ψRku,and(70)limn→∞ρ∫Ωax+unq∇un∇ugρungρ+1ug′uRkuψ=ρ∫Ωax+uq∇u∇ug′uguRkuψ.Similarly, using the convergence (un-fn)→(u-f) in Lm(Ω), we have(71)limn→∞∫Ωun-fngρungρuRkuψ=∫Ωu-fRkuψ.Now, from (62), we get(72)limn→∞∫Ωax+unq∇un∇ugρungρuRk′uψ=∫Ωax+uq∇u∇uRk′u=∫k≤u≤k+1ax+uq∇u∇u.Passing to the limit in (66) when n tends to infinity and gathering together (68)-(72), weobtain (73)∫Ωax+uq∇u∇ψRku+∫ΩuRkuψ+∫k≤u≤k+1ax+uq∇u∇u-∫ΩHx,u,∇uRkuψ≤∫ΩfRkuψ.Choosing T1(Gk(un)) in (16), we get(74)∫k≤un≤k+1ax+unq∇un∇un≤∫k≤unf+∫k≤unHx,un,∇un≤∫k≤unf+γR21+k1-q.By Fatou’s lemma, we have(75)limk→∞∫k≤u≤k+1ax+uq∇u∇u=0.In order to pass to the limit as k tends to infinity in the inequality (73), we recall that H(x,u,∇u)∈L1(Ω) and ax+uq∇u∈L1(Ω),. We obtain(76)∫Ωax+uq∇u∇ψ+∫Ωuψ≤∫ΩHx,u,∇uψ+∫Ωfψ,for every ψ∈H01(Ω)∩L∞(Ω), with ψ≥0; that is, u is a subsolution of problem (1).Step 2 (The second inequality).
Letψ be in H01(Ω)∩L∞(Ω), with ψ≤0, and g be given by (58), and choose(77)ϕ=gρugρunRkuψasa test function in (16). We obtain(78)∫Ωax+unq∇un∇ψgρugρunRku+ρ∫Ωax+unq∇un∇ugρ-1ugρung′uRkuψ+∫Ωax+unq∇un∇ugρugρunRk′uψ-ρ∫Ωax+unq∇un∇ungρugρ+1ung′unRkuψ-∫ΩHnx,un,∇ungρugρunRkuψ+∫ΩungρugρunRkuψ=∫ΩfngρugρunRkuψ.We observe that, by (60) and the fact that ψ≤0, we have(79)-ρax+unq∇un∇ungρugρ+1ung′unRkuψ-Hnx,un,∇ungρugρunRkuψ≥-1+unq∇un2gρugρ+1unRkuψα~ρg′un-γgun1+un≥0.Applying the same argument of Step 1 and using (64) instead of (63), we deduce that(80)∫Ωax+uq∇u∇ψ+∫Ωuψ≤∫ΩHx,u,∇uψ+∫Ωfψ,for every ψ∈H01(Ω)∩L∞(Ω), with ψ≤0.Consequently, summarizing Steps1 and 2, we have(81)∫Ωax+uq∇u∇ψ+∫Ωuψ≤∫ΩHx,u,∇uψ+∫Ωfψ,for every ψ∈H01(Ω)∩L∞(Ω).Finally, interchangingψ and -ψ we conclude that(82)∫Ωax+uq∇u∇ψ+∫Ωuψ=∫ΩHx,u,∇uψ+∫Ωfψ,for every ψ∈H01(Ω)∩L∞(Ω).



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