# L∞-Solutions for Some Nonlinear Degenerate Elliptic Equations

Authors: Albo Carlos Cavalheiro
Journal: International Journal of Differential Equations
 (2011)

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## Abstract

We are interested in the existence of solutions for Dirichlet problem associated to the degenerate quasilinear elliptic equations-∑j=1nDj[ω2(x)𝒜j(x,u,∇u)]+ω1(x)g(x,u(x))+H(x,u,∇u)ω2(x)=f(x),onΩ in the setting of the weighted Sobolev spaces W01,p(Ω,ω1,ω2).


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## Body

## 1. Introduction

In this paper we prove the existence of (weak) solutions in the weighted Sobolev spacesW01,p(Ω,ω1,ω2) for the homogeneous Dirichlet problem: (P)Lu(x)=f(x),on⁡Ω,u(x)=0,on⁡∂Ω,
where L is the partial differential operator: (1.1)Lu(x)=-div⁡[ω2(x)A(x,u,∇u)]+g(x,u)ω1(x)+H(x,u,∇u)ω2(x),
where Ω is a bounded open set in ℝN (N≥2), ω1 and ω2 are two weight functions, and the functions 𝒜:Ω×ℝ×ℝN→ℝN, g:Ω×ℝ→ℝ, and H:Ω×ℝ×ℝN→ℝ are Carathéodory functions.By aweight, we will mean a locally integrable function ω on ℝN such that ω(x)>0 for a.e. x∈ℝN. Every weight ωi (i=1,2) gives rise to a measure on the measurable subsets on ℝN through integration. This measure will be denoted by μi. Thus, μi(E)=∫Eωi(x)dx (i=1,2) for measurable sets E⊂ℝN.In general, the Sobolev spacesWk,p(Ω) without weights occur as spaces of solutions for elliptic and parabolic partial differential equations. For degenerate partial differential equations, that is, equations with various types of singularities in the coefficients, it is natural to look for solutions in weighted Sobolev spaces (see [1–4]).A class of weights, which is particularly well understood, is the class ofAp-weights (or Muckenhoupt class) that was introduced by Muckenhoupt (see [5]). These classes have found many useful applications in harmonic analysis (see [6, 7]). Another reason for studying Ap-weights is the fact that powers of distance to submanifolds of ℝN often belong to Ap (see [8]). There are, in fact, many interesting examples of weights (see [4] for p-admissible weights).Equations like (1.1) have been studied by many authors in the nondegenerate case (i.e., with ω(x)≡1) (see, e.g., [9] and the references therein). The degenerate case with different conditions has been studied by many authors. In [2] Drábek et al. proved that under certain condition, the Dirichlet problem associated with the equation -div(a(x,u,∇u))=h, h∈[W01,p(Ω,ω)]* has at least one solution u∈W01,p(Ω,ω), and in [1] the author proved the existence of solution when the nonlinear term H(x,η,ξ) is equal to zero.Firstly, we prove anL∞ estimate for the bounded solutions of (P): we assume that f/ω1∈Lq(Ω,ω1), with r2/(r2-1)<q<∞ (where r2>1 as in Theorem 2.5), and we prove that any u∈W01,p(Ω,ω1,ω2)∩L∞(Ω) that solves (P) satisfies ∥u∥L∞(Ω)≤C, where C depends only on the data, that is, Ω, N,p,q,α1,α2,C0,C1 and ∥f/ω1∥Lq(Ω,ω1). After that, we prove the existence of solution for problem (P) if f/ω1∈Lq(Ω,ω1), with p′r2/(r2-1)<q<∞.Note that, in the proof of our main result, many ideas have been adapted from [9–11].The following theorem will be proved in Section3.Theorem 1.1.
Letω1 and  ω2 be  Ap-weights,  1<p<∞, with  ω1≤ω2. Suppose the following.   (H1)
x↦𝒜(x,η,ξ) is measurable in Ω for all (η,ξ)∈ℝ×ℝN: (η,ξ)↦𝒜(x,η,ξ) is continuous in ℝ×ℝN for almost all x∈Ω.(H2)
[𝒜(x,η,ξ)-𝒜(x,η′,ξ′)]·(ξ-ξ′)>0, whenever ξ,ξ′∈ℝN, ξ≠ξ′.(H3)
𝒜(x,η,ξ)·ξ≥α1|ξ|p, with 1<p<∞, where α1>0.(H4)
|𝒜(x,η,ξ)|≤K2(x)+h1(x)|η|p/p′+h2(x)|ξ|p/p′, where K2,h1, and h2 are positive functions, with h1 and h2∈L∞(Ω), and K2∈Lp′(Ω,ω2) (1/p+1/p′=1).(H5)
x↦g(x,η) is measurable in Ω for all η∈ℝ:η↦g(x,η) is continuous in ℝ for almost all x∈Ω.(H6)
|g(x,η)|≤K1(x)+h3(x)|η|p/p′, where K1 and h3 are positive functions, with h3∈L∞(Ω) and K1∈Lp′(Ω,ω1).(H7)
g(x,η)η≥α0|η|p, for all η∈ℝ, where α0>0.(H8)
x↦H(x,η,ξ) is measurable in Ω for all (η,ξ)∈ℝ×ℝN: (η,ξ)↦H(x,η,ξ) is continuous in ℝ×ℝN for almost all x∈Ω.(H9)
|H(x,η,ξ)|≤C0+C1|ξ|p, where C0 and C1 are positive constants.(H10)
f/ω1∈Lq(Ω,ω1), with r2/(r2-1)<q<∞ (where r2>1 as in Theorem 2.5) and ω2/ω1∈Lq(Ω,ω1).
Letu∈W01,p(Ω,ω1,ω2)∩L∞(Ω) be a solution of problem (P). Then there exists a constant C>0, which depends only on Ω, n,p,α1,α0,C0,C1 and ∥f/ω1∥Lq(Ω,ω1), such that ∥u∥L∞(Ω)≤C.The main result of this article is given in the next theorem, which is proved in Section4.Theorem 1.2.
Assume that  (H1)–(H9) hold true and suppose that
(H11)
f/ω1∈Lq(Ω,ω1), with p′r2/(r2-1)<q<∞;(H12)
H(x,η,ξ)η≥0, for all η∈ℝ.
Then there exists at least one solutionu∈W01,p(Ω,ω1,ω2)∩L∞(Ω) of the problem P.Theorem1.2 will be proved by approximating problem (P) with the following problems:(Pm)-div⁡[ω2A(x,u,∇u)]+g(x,u)ω1+Hm(x,u,∇u)ω2=f(x),on⁡Ω,u(x)=0,on⁡∂Ω,
where  Hm(x,η,ξ)=H(x,η,ξ)/1+(1/m)|H(x,η,ε)|,  for m∈ℕ. Note that |Hm|≤|H| and that |Hm|≤m.


## 2. Definitions and Basic Results

Letω be a locally integrable nonnegative function in ℝN and assume that 0<ω(x)<∞ almost everywhere. We say that ω belongs to the Muckenhoupt class Ap, 1<p<∞, or that ω is an Ap-weight, if there is a constant C=Cp,ω such that(2.1)(1|B|∫Bω(x)dx)(1|B|∫Bω1/1-p(x)dx)p-1≤Cp,ω
for all balls B⊂ℝN, where |·| denotes the N-dimensional Lebesgue measure in ℝN. If 1<q≤p, then Aq⊂Ap (see [4, 7, 12] or [13] for more information about Ap-weights). The weight ω satisfies the doubling condition if μ(2B)≤Cμ(B), for all balls B⊂ℝn, where μ(B)=∫Bω(x)dx and 2B denotes the ball with the same center as B which is twice as large. If ω∈Ap, then ω is doubling (see Corollary  15.7 in [4]).As an example ofAp-weight, the function ω(x)=|x|α, x∈ℝN, is in Ap if and only if -N<α<N(p-1) (see Corollary  4.4, Chapter IX in [7]). If φ∈BMO(ℝN), then ω(x)=eαφ(x)∈A2 for some α>0 (see [6]).Definition 2.1.
Letω be a weight, and let Ω⊂ℝN be open. For 0<p<∞, we define Lp(Ω,ω) as the set of measurable functions f on Ω such that
(2.2)‖f‖Lp(Ω,ω)=(∫Ω|f(x)|pω(x)dx)1/p<∞.Remark 2.2.
Ifω∈Ap, 1<p<∞, then since ω-1/(p-1) is locally integrable, we have Lp(Ω,ω)⊂Lloc1(Ω) for every open set Ω (see Remark  1.2.4 in [13]). It thus makes sense to talk about weak derivatives of functions in Lp(Ω,ω).Definition 2.3.
LetΩ⊂ℝN be open, 1<p<∞, and let ω1 and ω2 be Ap-weights, 1<p<∞. We define the weighted Sobolev space W1,p(Ω,ω1,ω2) as the set of functions u∈Lp(Ω,ω1) with weak derivatives Dju∈Lp(Ω,ω2), for j=1,…,N. The norm of u in W1,p(Ω,ω1,ω2) is given by
(2.3)‖u‖W1,p(Ω,ω1,ω2)=(∫Ω|u(x)|pω1(x)dx+∑j=1N∫Ω|Dju(x)|pω2(x)dx)1/p.
The space W01,p(Ω,ω1,ω2) is the closure of C0∞(Ω) with respect to the norm ∥u∥W1,p(Ω,ω1,ω2). The dual space of W01,p(Ω,ω1,ω2) is the space
(2.4)[W01,p(Ω,ω1,ω2)]*=W-1,p′(Ω,ω1,ω2)={T=f-div⁡g:g=(g1,…,gN),fω1∈Lp′(Ω,ω1),gjω2∈Lp′(Ω,ω2)}.Remark 2.4.
(a) Ifω∈Ap, 1<p<∞, then C∞(Ω) is dense in W1,p(Ω,ω)=W1,p(Ω,ω,ω) (see Corollary  2.16 in [13]).
(b) Ifω1≤ω2, then
(2.5)W01,p(Ω,ω2)⊂W01,p(Ω,ω1,ω2)⊂W01,p(Ω,ω1).In this paper we use the following four results.Theorem 2.5 (The Weighted Sobolev Inequality).
LetΩ be an open bounded set in  ℝN(N≥2) and  ω2∈Ap(1<p<∞). There exist constants  CΩ and  δ positive such that for all  u∈C0∞(Ω) and all  r2 satisfying  1≤r2≤N/(N-1)+δ,
(2.6)‖u‖Lp*(Ω,ω2)≤CΩ‖∇u‖Lp(Ω,ω2),
where p*=pr2.Proof.
See Theorem  1.3 in [3].The following lemma is due to Stampacchia (see [14], Lemma  4.1).Lemma 2.6.
Letα,  β,  C, and  k0 be real positive numbers, where  β>1.
Letφ:ℝ+→ℝ+ be a decreasing function such that
(2.7)φ(l)≤C(l-k)α[φ(k)]β
for all l>k≥k0. Then φ(k0+d)=0, where dα=C[φ(k0)]β-12αβ/(β-1).Lemma 2.7.
Ifω∈Ap, then  (|E|/|B|)p≤Cp,ω(μ(E)/μ(B)), whenever  B is a ball in  ℝN and  E is a measurable subset of  B.Proof.
See Theorem  15.5Strong doubling ofAp-weights in [4].By Lemma2.7, if μ(E)=0, then |E|=0.Lemma 2.8.
Letω1 and  ω2 be  Ap-weights,  1<p<∞,  ω1≤ω2, and a sequence  {un},  un∈W01,p(Ω,ω1,ω2) satisfies the following:        (i)
un⇀u in W01,p(Ω,ω1,ω2) and μ2-a.e. in Ω;(ii)
∫Ω〈𝒜(x,un,∇un)-𝒜(x,un,∇u),∇(un-u)〉ω2dx→0 with n→∞.
Thenun→u in W01,p(Ω,ω1,ω2).Proof.
The proof of this lemma follows the line of Lemma  5 in [10].Definition 2.9.
We say thatu∈W01,p(Ω,ω1,ω2)∩L∞(Ω) is a (weak) solution of problem (P) if
(2.8)∫Ωω2A(x,u,∇u)⋅∇φdx+∫Ωg(x,u)φω1dx+∫ΩH(x,u,∇u)φω2dx=∫Ωfφdx,
for all φ∈W01,p(Ω,ω1,ω2)∩L∞(Ω).


## 3. Proof of Theorem1.1

Setλ=(C1/α1)+1 and define for k>0 the functions ϕ∈C1(ℝ) and Gk∈W1,∞(ℝ) by(3.1)ϕ(s)={eλs-1,ifs≥0,-e-λs+1,ifs≤0,Gk(s)={s-k,ifs≥k,0,if-k≤s≤k,s+k,ifs≤-k.Ifu∈W01,p(Ω,ω1,ω2)∩L∞(Ω) is a solution of problem (P), define the set A(k)={x∈Ω:|u(x)|>k}. We will use the test functions v(x)=ϕ(Gk(u(x))). Since u∈W01,2(Ω,ω1,ω2)∩L∞(Ω) and Ω is a bounded set in ℝN, then v(x)=ϕ(Gk(u(x)))∈W01,p(Ω,ω1,ω2)∩L∞(Ω) and
(3.2)v(x)=ϕ((|u|-k)+)χA(k)sign⁡(u),∇v=ϕ′((|u|-k)6)χA(k)∇u,
where χA(k) is the characteristic function of the set A(k) (by sign(u(x)) we mean the function equal to +1 for u(x)>0 and to −1 for u(x)<0).Sinceu∈W01,p(Ω,ω1,ω2)∩L∞(Ω), we have that v∈W01,p(Ω,ω1,ω2)∩L∞(Ω).Using the functionv in (2.8) we obtain(3.3)∫Ωω2A(x,u,∇u)⋅∇vdx+∫Ωg(x,u)vω1dx+∫ΩH(x,u,∇u)vω2dx=∫Ωfvdx.
We have the following estimates.(i)
By (H3) we obtain(3.4)∫Ωω2A(x,u,∇u)⋅∇vdx=∫A(k)ϕ′((|u|-k)+)A(x,u,∇u)⋅∇uω2dx≥α1∫A(k)|∇u|pϕ′((|u|-k)+)ω2dx.(ii)
By (H7) we obtain(3.5)∫Ωg(x,u)vω1dx=∫A(k)g(x,u)ϕ((|u|-k)+)ω1dx≥α0∫A(k)|u|p-1ϕ((|u|-k)+)ω1dx.(iii)
Using (H9) we obtain(3.6)|∫ΩH(x,u,∇u)vω2dx|≤∫Ω|H(x,u,∇u)||v|ω2dx≤∫A(k)(C0+C1|∇u|p)ϕ((|u|-k)+)ω2dx.
And we also have(3.7)|∫Ωfvdx|≤∫A(k)|f|ϕ((|u|-k)+)dx.
Hence in (3.3) we obtain(3.8)α1∫A(k)|∇u|pϕ′((|u|-k)+)ω2dx+α0∫A(k)|u|p-1ϕ((|u|-k)+)ω1dx≤∫A(k)(C0+C1|∇u|p)ϕ((|u|-k)+)ω2dx+∫A(k)|f|ϕ((|u|-k)+)dx.Sinceλ=(C1/α1)+1, we have for s≥0(3.9)α1ϕ′(s)-C1ϕ(s)=α1λeλs-C1(eλs-1)=(α1λ-C1)eλs+C1=α1eλs+C1≥α1eλs=α1λp[λeλs/p]p=α1λp[ϕ′(sp)]p.
Hence in (3.8) we obtain(3.10)∫A(k)[α1|∇u|pϕ′((|u|-k)+)-C1|∇u|pϕ((|u|-k)+)]ω2dx+α0∫A(k)|u|p-1ϕ((|u|-k)+)ω1dx≤∫A(k)(|f|+C0ω2)ϕ((|u|-k)+)dx.
Using (3.9) and k<|u(x)| if x∈A(k), we obtain(3.11)α1λp∫A(k)|ϕ′((|u|-k)+p)∇u|pω2dx+α0kp-1∫A(k)ϕ((|u|-k)+)ω1dx≤∫A(k)(|f|+C0ω2)ϕ((|u|-k)+)dx.
Let us define the function ψk by ψk(x)=ϕ((|u(x)|-k)+/p). We have that ψk∈W01,p(Ω,ω1,ω2) and(3.12)∇ψk=1pϕ′((|u|-k)+p)χA(k)sign⁡(u)∇u.
We have the following.(a)
For alls≥0, eλs-1≥(eλs/p-1)p.(b)
There exists, a constantC2>0 (C2=C2(λ,p)) such that for all s≥1(3.13)eλs-1≤C2(eλs/p-1)p,λeλs≤C2λ(eλs/p-1)p.This implies the following.(I1)
ϕ((|u|-k)+)=eλ(|u|-k)+-1≥(eλ(|u|-k)+/p-1)p=|ψk|p a.e. on Ω.(I2)
Ifx∈A(k+1), then(3.14)ϕ((|u|-k)+)=eλ(|u|-k)+-1≤C2(eλ(|u|-k)+/p-1)=C2|ψk|p,ϕ′((|u|-k)+)=(|u|-k)+eλ(|u|-k)+≤C2λ(eλ(|u|-k)+/p-1)p=C2λ|ψk|p.
Combining (I1) and (I2) with (3.11) and (3.12) we obtain(3.15)α1ppλp∫Ω|∇ψk|pω2dx+α0kp-1∫Ω|ψk|pω1dx≤∫A(k)(|f|+C0ω2)ϕ((|u|-k)+)dx≤∫A(k+1)(|f|+C0ω2)C2|ψk|pdx+∫A(k)-A(k+1)(|f|+C0ω2)ϕ((|u|-k)+)dx.
Define the function h=|f|+C0ω2. Since f/ω1∈Lq(Ω,ω1) and ω2/ω1∈Lq(Ω,ω1), we have that h/ω1∈Lq(Ω,ω1). Hence(3.16)∫Ωh|ψk|pdx=∫Ωhω1|ψk|pω11/qω11/q′dx≤‖hω1‖Lq(Ω,ω1)‖ψk‖Lpq′(Ω,ω1)p.
If x∈A(k)-A(k+1), we have k<|u|<k+1. Hence(3.17)ϕ((|u|-k)+)=eλ(|u|-k)+-1≤eλ-1,
and we obtain(3.18)∫A(k)-A(k+1)(|f|+C0ω2)ϕ((|u|-k)+)dx≤∫A(k)-A(k+1)(eλ-1)hdx≤eλ∫A(k)hdx.By Theorem2.5, (3.15), and (3.18) we have(3.19)α1ppλp1CΩp(∫Ω|ψk|p*ω2dx)p/p*+α0k0p-1∫Ω|ψk|pω1dx≤C2∫Ωh|ψk|pdx+eλ∫A(k)hdx.
Therefore, there exist positive constants C3 and C4 (depending only on Ω, α1, p, λ, C2, and CΩ) such that(3.20)C3(∫Ω|ψk|p*ω2dx)p/p*+C4α0k0p-1∫Ω|ψk|pω1dx≤∫Ωh|ψk|pdx+∫A(k)hdx.
Since r2/(r2-1)<q, then q′<r2 and p<pq′<p*. For 0<θ<1 such that 1/pq′=(θ/p)+(1-θ)/p*, using an interpolation inequality, Young’s inequality (with 0<γ<∞), and Hölder’s inequality with exponents q and q′ we thus obtain(3.21)∫Ωh|ψk|pdx≤‖hω1‖Lq(Ω,ω1)‖ψk‖Lpq′(Ω,ω1)p≤‖hω1‖Lq(Ω,ω1)‖ψk‖Lp(Ω,ω1)θp‖ψk‖Lp*(Ω,ω1)(1-θ)p≤(1-θ)γ1/(1-θ)‖ψk‖Lp*(Ω,ω1)p+θγ-1/θ‖hω1‖Lq(Ω,ω1)1/θ‖ψk‖Lp(Ω,ω1)p.
Hence in (3.20) we obtain(3.22)C3(∫Ω|ψk|p*ω2dx)p/p*+C4α0k0p-1∫Ω|ψk|pω1dx≤(1-θ)γ1/(1-θ)‖ψk‖Lp*(Ω,ω1)p+θγ-1/θ‖hω1‖Lq(Ω,ω1)1/θ‖ψk‖Lp(Ω,ω1)p+∫A(k)hdx≤(1-θ)γ1/(1-θ)‖ψk‖Lp*(Ω,ω2)p+θγ-1/θ‖hω1‖Lq(Ω,ω1)1/θ‖ψk‖Lp(Ω,ω1)p+∫A(k)hdx.
Now, we can choose γ in order to have (1-θ)γ1/(1-θ)=C3/2 and k0 such that(3.23)C4α0k0p-1=θγ-1/θ‖hω1‖Lq(Ω,ω1)1/θ.
We obtain, from (3.22), that for every k≥k0 it results(3.24)C32(∫Ω|ψk|p*ω2dx)p/p*≤∫A(k)hdx≤‖hω1‖Lq(Ω,ω1)[μ1(Ak)]1/q′,
where μ1(Ak)=∫Akω1dx. Hence for all k≥k0 we have(3.25)∫Ω|ψk|p*ω2dx≤(2C3‖hω1‖Lq(Ω,ω1)[μ1(A(k))]1/q′)p*/p=(2C3)p*/p‖hω1‖Lq(Ω,ω1)p*/p[μ1(A(k))]p*/pq′=C5[μ1(A(k))]p*/pq′.
Let us now take l>k≥k0. Then we have(3.26)μ1(A(l))[λ(l-kp)]p*≤μ1(A(k))|ϕ(l-kp)|p*≤∫A(k)|ψk|p*ω1dx≤∫Ω|ψk|p*ω2dx.
Therefore for all l>k≥k0 we obtain (by (3.25) and (3.26))(3.27)(l-k)p*μ1(A(l))≤pp*λp*C5[μ1(A(k))]p*/pq′=C6[μ1(A(k))]p*/pq′,
that is, μ1(A(l))≤(C6/(l-k)p*)[μ1(A(k))]p*/pq′.Letφ(k)=μ1(A(k)). Since β=p*/pq′>1, by Lemma 2.6 there exists a constant C7>0 such that(3.28)μ1(A(k))=0,∀k≥C7.Using Lemma2.7 we have |A(k)|=0 for all k≥C7. Therefore any solution u of problem (P) satisfies the estimate ∥u∥L∞(Ω)≤C7.


## 4. Proof of Theorem1.2

Step 1.
Let us define form∈ℕ the approximation
(4.1)Hm(x,η,ξ)=H(x,η,ξ)1+(1/m)|H(x,η,ξ)|.
We have that |Hm(x,η,ξ)|≤|H(x,η,ξ)|, |Hm(x,η,ξ)|<m, and Hm(x,η,ξ) satisfies the conditions (H9) and (H12). We consider the approximate problem
(Pm)-div⁡[ω2A(x,u,∇u)]+g(x,u)ω1+Hm(x,u,∇u)ω2,on⁡Ω,u(x)=0,on⁡∂Ω.
We say that u∈W01,p(Ω,ω1,ω2) is (weak) solution of problem (Pm) if
(4.2)∫ΩA(x,u,∇u)⋅∇φω2dx+∫Ωg(x,u)φω1dx+∫ΩHm(x,u,∇u)φω2dx=∫Ωfφdx,
for all φ∈W01,p(Ω,ω1,ω2). We will prove that there exists at least one solution um of the problem (Pm). For u,v,φ∈W01,p(Ω,ω1,ω2) we define
(4.3)B(u,v,φ)=∫Ωω2A(x,u,∇v)⋅∇φdx,Bm(u,φ)=∫Ωg(x,u)φω1dx+∫ΩHm(x,u,∇u)φω2dx,T(φ)=∫Ωfφdx.
Then u∈W01,p(Ω,ω1,ω2) is a (weak) solution of problem (Pm) if
(4.4)B(u,u,φ)+Bm(u,φ)=T(φ),forallφ∈W01,p(Ω,ω1,ω2).
Let a(u,v,φ)=B(u,v,φ)+Bm(u,φ).(i)
Using (H4) we obtain(4.5)|B(u,v,φ)|≤(‖K2‖Lp′(Ω,ω2)+‖h1‖L∞(Ω)‖u‖W01,p(Ω,ω1,ω2)p/p′+‖h2‖L∞(Ω)‖v‖W01,p(Ω,ω1,ω2)p/p′)×‖φ‖W01,p(Ω,ω1,ω2).(ii)
Using (H6) and|Hm(x,η,ξ)|≤m, we obtain(4.6)|Bm(u,φ)|≤(‖K1‖Lp′(Ω,ω1)+‖h3‖L∞(Ω)‖u‖W01,p(Ω,ω1,ω2)p′/p+mCΩ[μ2(Ω)]1′/p)×‖φ‖W01,p(Ω,ω1,ω2). Hence,
(4.7)|a(u,v,φ)|≤(‖K2‖Lp′(Ω,ω2)+(‖h1‖L∞(Ω)+‖h3‖L∞(Ω))‖u‖W01,p(Ω,ω1,ω2)p/p′+‖K1‖Lp′(Ω,ω1)+‖h2‖L∞(Ω)‖v‖W01,p(Ω,ω1,ω2)p/p′+mCΩ[μ2(Ω)]1/p′)‖φ‖W01,p(Ω,ω1,ω2).
For each (u,v)∈W01,p(Ω,ω1,ω2)×W01,p(Ω,ω1,ω2) we have that a(u,v,.) is linear and continuous. Hence, there exists a linear and continuous operator
(4.8)A(u,v):W01,p(Ω,ω1,ω2)⟶[W01,p(Ω,ω1,ω2)]*
such that 〈A(u,v),φ〉=a(u,v,φ). We set
(4.9)Ã(u)=A(u,u),∀u∈W01,p(Ω,ω1,ω2).
The operator Ã:W01,p(Ω,ω1,ω2)→[W01,p(Ω,ω1,ω2)]* is semimonotone; that is, by similar arguments as in the proof of Theorem  2 in [11] we have the following.(i)
〈A(u,u)-A(u,v),u-v〉≥0 for all u,v∈W01,p(Ω,ω1,ω2).(ii)
For eachu∈W01,p(Ω,ω1,ω2), the operator v↦A(u,v) is hemicontinuous and bounded from W01,p(Ω,ω1,ω2) to [W01,p(Ω,ω1,ω2)]* and,for each v∈W01,p(Ω,ω1,ω2) the operator u↦A(u,v) is hemicontinuous and bounded from W01,p(Ω,ω1,ω2) to [W01,p(Ω,ω1,ω2)]*.(iii)
Ifun⇀u in W01,p(Ω,ω1,ω2) and 〈A(un,un)-A(un,u),un-u〉→0, then A(un,u)⇀A(u,v) in [W01,p(Ω,ω1,ω2)]* as n→∞ for all v∈W01,p(Ω,ω1,ω2).(iv)
Ifv∈W01,p(Ω,ω1,ω2), un⇀u in W01,p(Ω,ω1,ω2), and A(un,v)⇀ṽ in [W01,p(Ω,ω1,ω2)]*, then 〈A(un,v),un〉→〈ṽ,u〉 as n→∞.(v)
The operatorÃ:W01,p(Ω,ω1,ω2)→[W01,p(Ω,ω1,ω2)]* is bounded.Hence the operator Ã:W01,p(Ω,ω1,ω2)→[W01,p(Ω,ω1,ω2)]* is pseudomonotone (see [15]).(vi)
By (H3), (H7), and (H12) we have(4.10)〈Ã(u),u〉≥α1∫Ω|∇u|pω2dx+α0∫Ω|u|pω1dx≥α‖u‖W01,p(Ω,ω1,ω2)p,
where α=min{α0,α1}. Since p>1, we have
(4.11)〈Ã(u),u〉‖u‖W01,p(Ω,ω1,ω2)⟶∞on⁡‖u‖W01,p(Ω,ω1,ω2)⟶∞;
that is, the operator Ã is coercive. Then, by Theorem  27.B in [15], for each T∈[W01,p(Ω,ω1,ω2)]*, the equation
(4.12)Ãu=T,u∈W01,p(Ω,ω1,ω2)
has a solution. Therefore, the problem (Pm), has a solution um∈W01,p(Ω,ω1,ω2).Step 2.
We will show thatum∈L∞(Ω) and ∥um∥L∞(Ω)≤C, where C is independent of m. If u∈W01,p(Ω,ω1,ω2) is a solution of problem (Pm), we define
(4.13)un(x)=Tn(u(x))={u(x),if|u(x)|≤n,n,ifu(x)>n,-n,ifu(x)<-n.
We have Diun=Diu if |u(x)|≤n. For k>0, let us define the function ψn(x)=sign(un(x))max{|un(x)|-k,0}. We have ψn∈W01,p(Ω,ω1,ω2)∩L∞(Ω).
Now consider the function(4.14)Φ(t)={t+k,ift≤-k,0,if|t|≤k,t-k,ift≥k.
Since Φ is a Lipschitz function and Φ(0)=0, then Φ(ψn)∈W01,p(Ω,ω1,ω2). Moreover, DiΦ(ψn)=Φ′(ψn)Diψ and
(4.15)Φ′(un)∇un⟶Φ′(u)∇u,μ2-a.e.inΩ.
We also have, for all measurable subset E⊂Ω,
(4.16)∫E|Φ′(un)∇un|pω2dx≤∫E|∇un|pω2dx.
By applying Vitali’s Convergence Theorem, with ψ=Φ(u), we obtain
(4.17)∇ψn⟶∇ψinLp(Ω,ω2).
Since ω1≤ω2, we obtain
(4.18)‖ψn-ψ‖Lp(Ω,ω1)≤‖ψn-ψ‖Lp(Ω,ω2)≤CΩ‖∇ψn-∇ψ‖Lp(Ω,ω2).
Hence ψn→ψin Lp(Ω,ω1). Since u∈W01,p(Ω,ω1,ω2) is a solution of problem (Pm) and ψn∈W01,p(Ω,ω1,ω2)∩L∞(Ω), we have
(4.19)∫ΩA(x,u,∇u)⋅∇ψnω2dx+∫Ωg(x,u)ψnω1dx+∫ΩHm(x,u,∇u)ψnω2dx=∫Ωfψndx.
Using (H4), (H6), |Hm(x,η,ξ)|≤m, (4.17), and (4.18), we obtain in (4.19) as n→∞(4.20)∫ΩA(x,u,∇u)⋅∇ψω2dx+∫Ωg(x,u)ψω1dx+∫ΩHm(x,u,∇u)ψω2dx=∫Ωfψdx.
Using φ=ψχA(k) in (4.2) (where A(k)={x∈Ω:|u(x)|>k}) we obtain
(4.21)∫A(k)A(x,u,∇u)⋅∇ψω2dx+∫A(k)g(x,u)ψω1dx+∫A(k)Hm(x,u,∇u)ψω2dx=∫A(k)fψdx.
Since
(4.22)ψ=Φ(u)={u+k,ifu≤-k,0,if|u|≤k,u-k,ifu≥k,
we obtain the following.(i)
By (H7) we haveg(x,η)η≥0 for all η∈ℝ, and(4.23)∫A(k)g(x,u)ψω1dx=∫{u≤-k}g(x,u)(u+k)ω1dx+∫{u≥k}g(x,u)(u-k)ω1dx≥0.(ii)
Using (H12) we haveHm(x,η,ξ)η≥0 for all η∈ℝ, and(4.24)∫A(k)Hm(x,u,∇u)ψω2dx=∫{u≤-k}Hm(x,u,∇u)(u+k)ω2dx+∫{u≥k}Hm(x,u,∇u)(u-k)ω2dx≥0.
We have ∇ψ=∇u in A(k). Using (H3), (i), and (ii) we obtain in (4.21)
(4.25)α1∫A(k)|∇u|pω2dx≤∫A(k)fψdx.
By Theorem  2.1.14 in [13] there is a positive constant C such that
(4.26)∫ω|ψ|pω1dx≤∫Ω|ψ|pω2dx≤C∫Ω|∇ψ|pω2dx.
Then we obtain
(4.27)∫A(k)fψdx≤(∫A(k)|fω1|p′ω1)1/p′(∫A(k)|ψ|pω1dx)1/p≤C(∫A(k)|fω1|p′ω1)1/p′(∫A(k)|∇ψ|pω2dx)1/p=C(∫A(k)|fω1|p′ω1)1/p′(∫A(k)|∇u|pω2dx)1/p.
Using (4.27) and Young’s inequality we obtain in (4.25) (for all ε>0)
(4.28)α1∫A(k)|∇u|pω2dx≤C(∫A(k)|fω1|p′ω1)1/p′(∫A(k)|∇u|pω2dx)1/p≤C[ε∫A(k)|∇u|pω2dx+C(ε)∫A(k)|fω1|p′ω1dx],
where C(ε)=(εp)-p′/p/p′. We can choose ε>0 so that Cε=α1/2, and there exists a constant C8 such that
(4.29)∫A(k)|∇u|pω2dx≤C8∫A(k)|fω1|p′ω1dx.
Using Sobolev’s inequality (Theorem 2.5) and Hölder’s inequality with exponents q and q′ we obtain (since q>p′(r/(r-1))>p′)
(4.30)(∫A(k)(|u|-k)p*ω1dx)p/p*=(∫A(k)|ψ|p*ω1dx)p/p*≤(∫A(k)|ψ|p*ω2dx)p/p*≤C∫A(k)|∇ψ|pω2dx=C∫A(k)|∇u|pω2dx≤CC8∫A(k)|fω1|p′ω2dx≤C9(∫Ω|fω1|qω1dx)p′/q[μ1(Ω)]1-(p′/q).
Let us now take l>k>0 and observe that A(l)⊂A(k). Then, from the previous inequality, it follows that
(4.31)μ1(A(l))(l-k)p*=∫A(l)(l-k)p*ω1dx≤∫A(l)(|u|-k)p*ω1dx≤∫A(k)(|u|-k)p*ω1dx≤C9(∫Ω|fω1|qω1dx)p′p*/qp[μ1(A(k))](1-(p′/q))(p*/p).
Hence we obtain
(4.32)μ1(A(l))≤C9(∫Ω|f/ω1|qω1dx)(p′p*)/qp(l-k)p*[μ1(A(k))](1-(p′/q))(p*/p).
Since (1-(p′/q))(p*/p)>1, by Lemma 2.6 there exists a constant C10>0 such that μ1(A(k))=0 for all k≥C10, and using Lemma 2.7 we obtain |A(k)|=0. Therefore if um is a solution of problem (Pm), we have ∥um∥L∞(Ω)≤C10 and C10 is independent of m.Step 3.
Sinceum∈W01,p(Ω,ω1,ω2)∩L∞(Ω) and ∥um∥L∞(Ω)≤C10, then the sequence {um} is relative compact in the strong topology of W01,p(Ω,ω1,ω2) (by apply the analogous results of [10] and Lemma 2.8). Then, by extracting a subsequence {umk} which strongly converges in W01,p(Ω,ω1,ω2) (i.e., there exists u∈W01,p(Ω,ω1,ω2) such that umk→u in W01,p(Ω,ω1,ω2)), we have for all φ∈W01,p(Ω,ω1,ω2)∩L∞(Ω)(4.33)∫ΩA(x,umk,∇umk)⋅∇φω2+∫Ωg(x,umk)φω1dx+∫ΩHmk(x,umk,∇umk)φω2dx⟶∫ΩA(x,u,∇u)⋅∇φω2+∫Ωg(x,u)φω1dx+∫ΩH(x,u,∇u)φω2dx.
Therefore u∈W01,p(Ω,ω1,ω2)∩L∞(Ω) is the solution of problem (P).Example 4.1.
LetΩ={(x,y)∈ℝ2:x2+y2<1}, and consider the weights ω1(x,y)=(x2+y2)1/2 and ω2=(x2+y2)1/4 (ω1 and ω2∈A2), and the functions 𝒜:Ω×ℝ×ℝ2→ℝ2, g:Ω×ℝ→ℝ, and H:Ω×ℝ×ℝ2→ℝ defined by
(4.34)A((x,y),η,ξ)=h2(x,y)ξ,g((x,y),η)=η(cos⁡2(xy)+1),H((x,y),η,ξ)=|ξ|2sin⁡(xy)+arctan(η),
where h2(x,y)=2ex2+y2. Let us consider the partial differential operator
(4.35)Lu(x,y)=-div⁡[ω2(x,y)A((x,y),u,∇u)]+ω1(x,y)g((x,y),u)+ω2(x,y)H((x,y),η,ξ),
and f(x,y)=(x2+y2)3/2qcos(1/(x2+y2)), with 2r2/(r2-1)<q<6. Therefore, by Theorem 1.2, the problem
(P)Lu(x,y)=f(x,y),on⁡Ω,u(x,y)=0,on⁡∂Ω
has a solution u∈W01,2(Ω,ω1,ω2)∩L∞(Ω).



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*Source: 356356-2011-12-15.xml*