# On a Class of Anisotropic Nonlinear Elliptic Equations with Variable Exponent

Authors: Guoqing Zhang; Hongtao Zhang
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

Based on truncation technique and priori estimates, we prove the existence and uniqueness of weak solution for a class of anisotropic nonlinear elliptic equations with variable exponentp(x)→ growth. Furthermore, we also obtain that the weak solution is locally bounded and regular; that is, the weak solution is Lloc∞(Ω) and C1,α(Ω).


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## Body

## 1. Introduction

Since the variable exponent spaces have reflected into a various range of applications such as non-Newtonian fluids, plasticity, image processing, and nonlinear elasticity [1–4], some authors began to study the various properties of variable exponent space and some nonlinear problems with variable exponent growth. Edmunds et al. [5], Fan and Zhao [6] obtained that the variable exponent space Lp(x)(Ω) and Wm,p(x)(Ω) are reflexive Banach spaces under suitable conditions on p(x). Later, Edmunds and Rákosník [7], Fan et al. [8] proved some continuous and compact Sobolev embedding theorems for the variable exponent spaces Wk,p(x)(Ω). For the anisotropic variable exponents spaces, in 2008, Mihăilescu et al. [9] studied the eigenvalue problems for a class of anisotropic quasilinear elliptic equations with variable exponents. In 2011, Boureanu et al. [10] proved the existence of multiple solutions for a class of anisotropic elliptic equations with variable exponents. Recently, Stancu-Dumitru [11, 12] has studied the existence of nontrivial solutions for a class of nonhomogeneous anisotropic problem. In particular, Fan [13] established some embedding theorems for anisotropic variable exponent Sobolev spaces.In this paper, we investigate the following anisotropic nonlinear elliptic equation:(1)-∂xiai(x,u,∇u)+∑i=1N‍Ti(x,∇u)=f-∂xihi,x∈Ω,u=0,x∈∂Ω,
where Ω⊂ℝN (N>2) with Lipschitz conditions boundary, ai and Ti (i=1,2,…,N) are Carathéodory functions, hi and f satisfy suitable conditions, and ∂xiai(x,u,∇u) and ∂xihi(x) are Einstein Sum; that is, ∂xiai(x,u,∇u)=∑i=1N(∂ai(x,u,∇u)/∂xi‍), ∂xihi(x)=∑i=1N‍(∂hi(x)/∂xi). We usually use critical theory to obtain the existence of weak solutions. However, since the problem (1) has no variational structure, we cannot define the energy functional for the problem (1). Therefore, based on truncation technique and priori estimates, we prove the existence and uniqueness of weak solutions for the problem (1) in W01,p(x)→(Ω). Furthermore, we obtain that the weak solution for the problem (1) is locally bounded.In particular, for the special case(2)-∂xiai(x,∇u)=f-∂xihi,x∈Ω,u=0,x∈∂Ω,
we obtain that the weak solution is C1,α(Ω). To our knowledge, the above two problems have not been deeply studied in the anisotropic variable exponent Sobolev spaces.The paper is organized as follows. In Section2, we recall some results on anisotropic variable exponent Sobolev spaces and state our main results. In Section 3, we prove the existence, uniqueness and locally bounded of weak solution for the problem (1). In Section 4, the regularity of weak solutions for the problem (2) is proved.


## 2. Preliminary and Main Results

This section is dedicated to a general overview on theW1,p(x)→(Ω) and Lp(x)(Ω); for a deeper treatment on these spaces, see [5, 7, 8, 13, 14].Let(3)C+(Ω-)∶={h:h∈C(Ω-),h(x)>1∀x∈Ω¯}.
Set
(4)h+=supx∈Ω-h(x)andh-=infx∈Ω-h(x),
for any p(x)∈C+(Ω-), and we define the variable exponent Lebesgue space
(5)Lp(·)(Ω)={∫Ω‍u:ameasurablerealmmm,-valuedfunctionsuchthat∫Ω‍|u(x)|p(x)dx<∞}.
We define the norm of Lp(x)(Ω):
(6)∥u∥Lp(x)(Ω)=inf{μ>0;∫Ω‍|u(x)μ|p(x)dx≤1}.
From [6], we have the following:(1)
(Lp(x)(Ω),∥·∥p(x)) is a Banach space;(2)
Lp(x)(Ω) is reflexive, if and only if 1<p-≤p+<∞;(3)
Hölder inequality.For allu, v∈Lp(x)(Ω),
(7)|∫Ω‍uvdx|≤(1p-+1(p′)-)∥u∥Lp(x)(Ω)∥v∥Lp′(x)(Ω),
where 1/p(x)+1/p′(x)=1, Lp′(x)(Ω) is the conjugate space of Lp(x)(Ω).Now, we recall some results on anisotropic variable exponent Sobolev spaceW1,p(x)→(Ω) [13]; set
(8)L+∞(Ω)∶={p∈L∞(Ω):p(x)≥1fora.e.x∈Ω}.
Denote p(x)→=(p1(x),p2(x),…,pN(x))∈(L+∞(Ω))N and define
(9)∀x∈Ω,p-(x)=min{p1(x),p2(x),…,pN(x)},p+(x)=max{p1(x),p2(x),…,pN(x)}.
The anisotropic variable exponent Sobolev space
(10)W1,p(x)→(Ω)={u∈Lp+(x)(Ω):Diu∈Lpi(x)(Ω)fori=1,2,…,N}
is a Banach space with respect to the norm
(11)∥u∥W1,p(x)→(Ω)=∥u∥p+(x)+∑i=1N‍∥Diu∥pi(x).
If pi(x)∈L+∞(Ω), for i=1,2,…,N, pi->1, then W1,p(x)→(Ω) is reflexive. We define W01,p(x)→(Ω) as the closure of C0∞(Ω) with respect to the norm
(12)∥u∥W01,p(·)→(Ω)=∑i=1N‍∥∂u∂xi∥Lpi(x)(Ω),
and W01,p(x)→(Ω) is a reflexive Banach space (see [9]).Let(13)p-(x)=N∑i=1N(1/pi(x)),p-*(x)={Np-(x)N-p-(x)ifp-(x)<N,+∞ifp-(x)≥N,p∞(x)=max{p-*(x),p+(x)}.
Hence, we have the following embedding theorem for W01,p(x)→(Ω).LetΩ⊂ℝN be a bounded domain and p(·)→=(p1(·),p2(·),…,pN(·))∈(C+0(Ω-))N.Theorem 1 (see [13, Theorem 2.5]).
(i)
Ifq∈C+0(Ω-) and
(14)q(x)<p∞(x)∀x∈Ω-,
thenW01,p(·)→(Ω)↪↪Lq(·)(Ω). The embedding is compact.
(ii)
Ifp-(x)>N for all x∈Ω-, then there exists β∈(0,1) such that W01,p(·)→(Ω)↪↪C0,β(Ω-).
The embedding is also compact.Theorem 2 (see [13, Theorem 2.6]).
LetΩ⊂ℝN be a bounded domain and  p(·)→=(p1(·),p2(·),…,pN(·))∈(C+0(Ω-))N. Suppose that
(15)p+(x)<p-*(x)∀x∈Ω-.
Then one has(16)∥u∥Lp+(·)(Ω)≤C∑i=1N‍∥∂u∂xi∥Lpi(·)(Ω)∀u∈W01,p(·)→(Ω),
where C is a positive constant independent of u∈W01,p(·)→(Ω).Remark 3.
From [13], we know that if p+(x)<p-*(x), forallu∈W01,p(·)→(Ω), for every i=1,2,…,N, then ||u||Lpi(x)(Ω)≤C||∂xiu||Lpi(x)(Ω), where C is a positive constant independent of u∈W01,p(·)→(Ω).
Assume thatai:Ω×ℝ×ℝN→ℝ and Ti:Ω×ℝN→ℝ are Carathéodory functions and satisfy the following:(A1)
for a.e.x∈Ω-, for all ξ∈ℝN, s∈ℝ, we have
(17)λ∑i=1N‍|ξi|pi(x)≤∑i=1N‍ai(x,s,ξ)ξi
and |ai(x,s,ξ)|≤β[|s|p∞(x)/pi′(x)+|ξi|pi(x)-1], where λ,β>0;(A2)
for a.e.x∈Ω-, forallξi≠ξi′, γ>0 and ε>0, ai satisfies
(18)[ai(x,s,ξ)-ai(x,s,ξ′)](ξi-ξi′)≥γ(ε+|ξi|+|ξi′|)pi(x)-2|ξi-ξi′|2;(A3)
for a.e.x∈Ω-, for all ξ∈ℝN, ζ∈ℝN, i,j=1,2,…,N,
(19)c1|ζ|2≤∑i=1N‍∑j=1N‍aξji(x,s,ξ)ζiζj≤c2(1+∑i=1N‍|ξi|pi(x)-2)|ζ|2,
where aξji(x,s,ξ)=∂ai(x,s,ξ)/∂ξj;(A4)
for a.e.x∈Ω-, i=1,2,…,N, for some k∈{1,2,…,N}, for all ζ∈ℝN, s∈ℝ,
(20)c3|ζ|2≤∑i=1N‍axki(x,s,ξ)ζi≤c4(|ζ|2+|s|q(x)),
where axk(x,s,ξ)i=∂ai(x,s,ξ)/∂xk;(T1)
for a.e.x∈Ω-,for all ξ∈ℝN, s∈ℝ,
(21)∑i=1N‍|Ti(x,ξ)|≤c5|s|q(x)/2≤∑i=1N‍mi|ξi|pi(x)-1,
where mi∈Lbi(x)(Ω) with 1/bi(x)=1/p∞′(x)-1/pi′(x), c1,…,c5 are positive constants.
We enumerate the hypotheses concerningf and hi,(F1)
f∈Lp∞′(x)(Ω);(F2)
for a.e.x∈Ω-, s∈ℝ, for some k∈{1,2,…,N}, |fxk|≤c6|s|q(x)/2, where fxk=∂f/∂xk;(H1)
fori=1,2,…,N,  hi∈Lpi′(x)(Ω);(H2)
for a.e.x∈Ω-, i=1,2,…,N, for some k∈{1,2,…,N}, ζ∈ℝN, s∈ℝ,
(22)c7|ζ|2≤∑i=1N‍hxkiζi≤c8(|ζ|2+|s|q(x)),
where hxki=∂hi/∂xk, c6, c7, c8 are positive constants.Remark 4.
Now, we give a simple example. Letk=1, x0=(3,0), x=(x1,x2)∈Ω=B(x0,1)⊂ℝ2, u(x)=x12+x22, ζ=x13/2+x23/2,  h1(x)=x12+2x1x2,  h2(x)=(1/2)x12+x1x2+x22, T1(x,ξ)=x1+x2+∑i=12‍ξi, T2(x,ξ)=x1x2+|ξ|2, f(x)=(1/2)x12+x1x2. By a simple calculation, we obtain that Ti satisfies (T1), f satisfies (F2), and hi satisfies (H2), where i=1,2.Now, we define the weak solution of the problem (1). A function u∈W01,p(x)→(Ω) is a weak solution of the problem (1), if forallφ∈W01,p(x)→(Ω),
(23)∑i=1N‍∫Ω‍[ai(x,u,∇u)φxi+Ti(x,∇u)φ]=∫Ω‍[fφ+∑i=1N‍hiφxi],
where φxi=∂φ/∂xi.Theorem 5.
LetΩ⊂ℝN (N>2) be a bounded open subset, for every i=1,2,…,N, x∈Ω, assume p(x)¯<N, q∈C+0(Ω) and q(x)<p∞=p-*(x), pi(x) satisfies
(24)2≤pi(x)≤2Nri(x)Nri(x)-2ri(x)-2N
and ∑i=1N‍(1/pi(x))<2, where ri(x)≥max{N,p∞(x)pi(x)/(p∞(x)-pi(x))}. ai satisfies the hypotheses (A1), (A2), the hypotheses (F1) and (H1) hold, and Ti satisfies (T1) and the following Lipschitz condition:
(25)|Ti(x,ξ)-Ti(x,ξ′)|≤ki(x)(|ξi|+|ξi′|)δi(x)|ξi-ξi′|,
where ki(x)∈Lri(x)(Ω), 0≤δi(x)≤pi(x)/N-pi(x)/ri(x)+(pi(x)-2)/2. Then there exists a unique weak solution u for the problem (1). Furthermore, u∈Lloc∞(Ω).Theorem 6.
LetΩ⊂ℝN (N>2) be a bounded open subset. Suppose (A3), (A4) and (F2), (H2) hold, for i=1,2,…,N, 2≤pi(x)≤2N/(N-2), q∈C+0(Ω) and q(x)<p-*(x). If u is a weak solution of the problem (2), then u∈C1,α(Ω), for all 0<α<1.


## 3. The Proof of Theorem5

We consider the following problem:(26)-∂xiai(x,vn,∇vn)+∑i=1N‍Tni(x,∇vn)=f-∂xihi,x∈Ω,u=0,x∈∂Ω,
where Tni(x,∇v) is the truncation at levels ±n of Ti. Due to [15], we obtain that there exists a weak solution vn∈W01,p(x)→(Ω) for the problem (26).Lemma 7.
Ifp(x)¯<N, (A1), (A2), (T1), (F1), and (H1) hold. Assume vn∈W01,p(x)→(Ω) is a weak solution of the problem (26); then one has
(27)∑i=1N‍∫Ω‍|∂xivn|pi(x)≤C.Proof.
LetK∈ℝ+; K will be chosen later. There exist m measurable subsets Ω1,…,Ωm of Ω and m functions v1,…,vm such that if i≠j, Ωi⋂‍Ωj=Ø. For t∈{1,2,…,m-1}, |Ωt|=K, if t=m, |Ωt|≤K. Let {x∈Ω:|∇vt|≠0}⊂Ωt, ∇v=∇vt a.e. in Ωt, ∇(v1+v2+⋯+vt)vt=(∇v)vt, v1+v2+⋯+vt=v in Ω. For t∈{1,2,…,m}, if vt≠0, sign(v)=sign(vt).
Choosevt as test function of the problem (26) and fix t∈{1,2,…,m}. Using Young inequality, Hölder inequality, the embedding theorem of the W01,p(x)→(Ω), and the hypotheses (A1), (F1), and (H1), we have
(28)∑i=1N‍∫Ω‍|∂xivt|pi(x)≤c1(∥f∥Lp∞′(x)At1/N+∑i=1N‍∫Ω‍|Ti(x,∇v)||vt|mmmmm+∑i=1N‍∫Ω‍|hi|pi′(x)),
where c1>0 is a constant, At=∏i=1N‍||∂xivt||Lpi(x)(Ω). By (T1), Young inequality and Ho¨lder inequality, we obtain, for c2>0,
(29)|∑i=1N‍∫Ω‍Ti(x,∇v)vt|≤∑i=1N‍mi∫Ω‍|∂xiv|pi(x)-1|vt|≤∑i=1N‍mi∑s=1t‍∫Ωs‍|∂xivs|pi(x)-1|vt|≤c2∑i=1N‍∑s=1t‍mmmmmm,×[∫Ωs‍|∂xivs|pi(x)|Ωs|1/pi--1/p∞+mmvvvvvvvmm+Atp++/N|Ωs|1/pi--1/p∞+∫Ωs‍|∂xivs|pi(x)]≤c2∑i=1N‍K1/pi--1/p∞+mmmm,,×[∑s=1t-1‍∫Ωs‍|∂xivs|pi(x)+Atp++/N∫Ωt‍|∂xivt|pi(x)mmmmmm+∑s=1t-1‍∫Ωs‍|∂xivs|pi(x)+Atp++/N],
where pi-=infx∈Ω-pi(x), p++=supx∈Ω-p+(x), p∞+=supx∈Ω-p∞(x).
Combining (28) with (29), we have
(30)∑i=1N‍∫Ω‍|∂xivt|pi(x)≤c1{∥f∥Lp∞′(x)At1/N+∑i=1N‍∫Ω‍|hi|pi′(x)mmmm+c2∑i=1N‍K1/pi--1/p∞+mmmmmmm×[Atp++/N∫Ωt‍|∂xivt|pi(x)mmmmmmmm+∑s=1t-1‍∫Ωs‍|∂xivs|pi(x)+Atp++/N]}.
LetK satisfy c1c2∑i=1N‍K1/pi--1/p∞+<1; then we have
(31)∑i=1N‍∫Ω‍|∂xivt|pi(x)≤c3{∥f∥Lp∞′(x)At1/N+∑i=1N‍∫Ω‍|hi|pi′(x)mmmmm+∑s=1t-1‍∑i=1N‍K1/pi--1/p∞+(∑j=1N‍∫Ωs‍|∂xivs|pj(x))mmmmm+∑i=1N‍K1/pi--1/p∞+Atp++/N},
where c3>0 is a constant. Let t=1; we obtain
(32)∫Ω‍|∂xiv1|pi(x)≤∑i=1N‍∫Ω|∂xiv1|pi(x)≤c3[∥f∥Lp∞′(x)A11/N+∑i=1N‍∫Ω‍|hi|pi′(x)mm.mm+∑i=1N‍K1/pi--1/p∞+A1p++/N].
Now, we recall the following classical inequality, for a1,a2,…,aN are positive numbers,
(33)∏i=1N‍ai1/N≤1N∑i=1N‍ai.
From (33), we have
(34)A1=∏i=1N‍∥∂xiv1∥Lpi(x)(Ω)≤(1N∑i=1N‍∥∂xiv1∥Lpi(x)(Ω))N.
Combining (32) with (34), there exists a constant c4 such that
(35)∑i=1N‍∫Ω‍|∂xiv1|pi(x)≤c4.
Furthermore, put (35) in (31) and iterate on t; we have
(36)∑i=1N‍∫Ω‍|∂xivt|pi(x)≤c5[∥f∥Lp∞′(x)At1/N+∑i=1N‍∫Ω‍|hi|pi′(x)mm,nnn+1+∑i=1N‍K1/pi--1/p∞+A1p++/N].
Therefore, we obtain
(37)∑i=1N‍∫Ω‍|∂xivt|pi(x)≤c6,
for some constant c6>0.Lemma 8.
Ifp(x)¯<N, and ai satisfies the hypotheses (A1), (A2), the hypotheses (T1), (F1), and (H1) hold. Then there exists at least a weak solution for the problem (1).Proof.
By (27), we obtain that there exists a sequence ∂xiun which is bounded in Lpi(x)(Ω); we have for i=1,2,…,N,  ∂xiun⇀∂xiu weakly in Lpi(x)(Ω) and un→u strongly in Lpi(x)(Ω). We argue as in [16] to prove
(38)∂xiun⟶∂xiua.e.inΩfori=1,2,…,N.
From (38), we obtain
(39)ai(x,un,∇un)⟶ai(x,u,∇u)a.e.inΩ,Tni(x,∇un)⟶Tni(x,∇u)a.e.inΩ.
By the hypotheses (A1) and (T1), for κi(x)∈[1,pi′(x)] and c>0, we have
(40)∫Ω‍|ai(x,un,∇un)|κi(x)≤c[(∫Ω‍|u|p∞(x))κi(x)/pi′(x)mmmm+(∫Ω‍|∂xiu|pi(x))κi(x)/pi′(x)]|Ω|(pi′+-κi-)/pi′-,∫Ω‍|Tni(x,∇un)|κi(x)≤c[∫Ω‍|∂xiu|pi(x)]κi(x)/pi′(x)|Ω|(pi′+-κi-)/pi′-.
By Vitali Theorem, we have
(41)ai(x,un,∇un)⟶ai(x,u,∇u)stronglyinLκi(x)(Ω),Tni(x,∇un)⟶Ti(x,∇u)stronglyinLκi(x)(Ω),
for any κi(x)∈[1,pi′(x)]. Hence, we complete the proof of Lemma 8.Lemma 9.
Ifp(x)¯<N, p∞(x)=p-*(x), ai satisfies the hypotheses (A1), (A2), the hypotheses (F1) and (H1) hold, for i=1,2,…,N, x∈Ω, pi(x) satisfies
(42)2≤pi(x)≤2Nri(x)Nri(x)-2ri(x)-2N,
where ri(x)≥max{N,p∞(x)pi(x)/(p∞(x)-pi(x))}, Ti satisfies (T1) and (25). Then there exists a unique weak solution for the problem (1).Proof.
Suppose that there exist two distinct weak solutionsu and v for the problem (1). Denote φ=max{u-v,0}, and for t∈[1,supφ], let Ωt={x∈Ω:t<φ<supφ}. Otherwise, if t∈[0,1), we choose m=1/t and let Ωm={x∈Ω:m<φ<supφ}. We choose
(43)φt={φ-t,φ>t,0,otherwise,
as test function. From (A2) and (25), we have
(44)∑i=1N‍∫Ωt‍|∂xiφt|2(ε+|∂xiu|+|∂xiv|)pi(x)-2≤1γ∑i=1N‍∫Ωt‍ki(x)(|∂xiu|+|∂xiv|)δi(x)|∂xiφt|φt
when δ+(x)≥(pi(x)-2)/2. By Young inequality, for c independent on t, we obtain
(45)∑i=1N‍∫Ωt‍|∂xiφt|2(ε+|∂xiu|+|∂xiv|)pi(x)-2≤c∑i=1N‍∫Ωt‍ki(x)2(|∂xiu|+|∂xiv|)2δ+(x)-pi(x)+2φt2.
Then we have(46)1C(∫Ωt‍φt2*)2/2*≤∏i=1N‍(∫Ωt‍|∂xiφt|2)1/N≤c1∏i=1N‍(∫Ωt‍|∂xiφt|2(ε+|∂xiu|+|∂xiv|)pi(x)-2)1/N≤c1∑i=1N‍∫Ωt‍ki(x)2(|∂xiu|+|∂xiv|)2δ+(x)-pi(x)+2φt2≤c1(∫Ωt‍φt2*)2/2*×∑i=1N‍(∫Ωt‍ki(x)N(|∂xiu|+|∂xiv|)(2δ+(x)-pi(x)+2)(N/2))2/N,
where c1 is a constant independent on t. Hence, we have
(47)1C≤∑i=1N‍(∫Ωt‍ki(x)N(|∂xiu|+|∂xiv|)(2δ+(x)-pi(x)+2)(N/2))2/N.
SinceN/ri(x)+((2δi(x)-pi(x)+2)/pi(x))(N/2)≤1, we have
(48)limt→supφ∫Ωt‍ki(x)N(|∂xiu|+|∂xiv|)(2δ+(x)-pi(x)+2)(N/2)=0.
Therefore, (48) leads to a contradiction.
On the other hand, ifδ+(x)<(pi(x)-2)/2, from (44), and Young inequality, for some constant c2 (independent on t), we have
(49)∫Ωt‍|∂xiφt|2≤c2∑i=1N‍∫Ωt‍ki(x)2(|∂xiu|+|∂xiv|)2δ+(x)φt2.
Argue as in (46), for some constant c3 (independent on t), we have
(50)1C≤c3∑i=1N‍(∫Ωt‍ki(x)N(|∂xiu|+|∂xiv|)Nδ+(x))2/N.
SinceN/ri(x)+Nδi(x)/pi(x)≤1, we obtain a contradiction.Lemma 10.
If (A1), (T1), (F1), and (H1) hold, fori=1,2,…,N, 2≤pi(x) and ∑i=1N‍(1/pi(x))<2, q∈C+0(Ω) and q(x)<p∞=p-*(x), and u is a weak solution of the problem (1), then u∈Lloc∞(Ω).Proof.
We denoteE=W01,p(x)→(Ω). Let t>0, φ=max{u-t,0}, and choose φ as test function to the problem (1). We have
(51)∑i=1N‍∫Ω‍[ai(x,u,∇u)φxi+Ti(x,∇u)φ]=∫Ω‍[fφ+∑i=1N‍hiφxi].
Let x0∈Ω, fix k>1, B(x0,R) be a ball, and let δ∈(0,1), δR<s<t<R, and Ak,t={x∈B(x0,t),u(x)>k}. Using (51), (A1), (F1), (H1), and Hölder inequality, we obtain
(52)∑i=1N‍∫Ak,s‍|∂xiu|pi(x)≤∑i=1N‍∫Ak,t‍|∂xiu|pi(x)≤c[∑i=1N‍∥f∥Lp∞′(Ak,t)∥u-k∥Lp∞(Ak,t)mm,+∑i=1N‍∫Ak,t‍|Ti(x,∇u)||u-k|m,m+∑i=1N‍∫Ak,t‍hi∂xiu].
For2≤pi(x) and (T1), by Young inequality, we have
(53)∑i=1N‍∫Ak,t‍|Ti(x,∇u)||u-k|≤∑i=1N‍(12∫Ak,t‍|u-k|2+12∫Ak,t‍|Ti(x,∇u)|2)≤12∑i=1N‍∫Ak,t‍|u-k|pi(x)+12∑i=1N‍∫Ak,t‍|u|q(x)≤c∫Ak,t‍|u-kt-s|p(x)¯*+12∑i=1N‍∫Ak,t‍|u|q(x),∑i=1N‍∫Ak,t‍hi∂xiu≤12∑i=1N‍∫Ak,t‍hi2+12∑i=1N‍∫Ak,t‍|∂xiu|2≤12∑i=1N‍∫Ak,t‍hi2+12∑i=1N‍∫Ak,t‍|∂xiu|pi(x).
As 2≤pi(x) and ∑i=1N‍(1/pi(x))<2, if p(x)¯<N, there exists ε=1/2-p(x)¯/p(x)¯*>0 such that
(54)c(∫Ak,t‍hi2)1/2|Ak,t|1/2≤c|Ak,t|p(x)¯/p(x)¯*+ε.
If p(x)¯≥N, a similar estimate is true; just only choose a suitable p(x)¯*.
Due toq∈C+0(Ω) and q(x)<p∞(x)=p(x)¯*, we have
(55)∥f∥Lp∞′∥u-k∥Lp∞≤c∫Ak,t‍|u|q(x).
From
(56)∫Ak,t‍|u|q(x)=∫Ak,t‍|(u-k)+k|q(x)≤c∫Ak,t‍|u-k|q(x)+ckq+|Ak,t|≤c∫Ak,t‍|u-kt-s|p(x)¯*+ckq+|Ak,t|,
and (52)–(55), we obtain
(57)∑i=1N‍∫Ak,s‍|∂xiu|pi(x)≤12∑i=1N‍∫Ak,t‍|∂xiu|pi(x)+c∫Ak,t‍|u-kt-s|p(x)¯*+ckq+|Ak,t|+c|Ak,t|p(x)¯/p(x)¯*+ε.
Using Lemma 3.1 in [17], and (57), we have
(58)∑i=1N‍∫Ak,δR‍|∂xiu|pi(x)≤c∫Ak,R‍|u-kR-δR|p(x)¯*+ckq+|Ak,R|mmmm+c|Ak,R|p(x)¯/p(x)¯*+ε.
By Lemma 2.4  in [18], (58), we obtain that u is bounded from above on B(x0,R/2). Note that -u is also a weak solution of the problem (1), where a~i(x,u,∇u)=ai(x,-u,-∇u) and T~i(x,∇u)=Ti(x,-∇u). Hence, -u is also bounded from above on B(x0,R/2), and u∈L∞(B(x0,R/2)). This implies that u∈Lloc∞(Ω).Proof of Theorem5.
From Lemmas8, 9, and 10, we obtain that there exists a uniqueness weak solution u∈W01,p(x)→(Ω) for the problem (1), and what is more, u∈Lloc∞(Ω). Hence we complete the proof of Theorem 5.


## 4. The Proof of Theorem6

In this section, we prove the regularity of weak solutions for the problem (2).Proof of Theorem6.
We denoteE=W01,p(x)→(Ω). For understanding, we write ai(u,Du)=ai(u,∇u), and hi=hi. If g is any function with compact support in Ω, for 1≤k≤N, for small enough m, we define
(59)Δmkg(x)=g(x+mek)-g(x)m,
where ek=(0,0,…,1k,…,0). Hence, if for any ϕ∈E, ϕ has compact support in Ω and |m|<(1/2)dist(suppϕ,∂Ω); from the definition of weak solution for the problem (1), we have
(60)∑i=1N‍∫Ω‍Δmkai(x,Du)ϕxi-∫Ω‍Δmkfϕ-∑i=1N‍∫Ω‍Δmkhiϕxi=0.
For the definition ofΔmk and [19], we have
(61)Δmkai(x,Du)=1m∫01‍ddt[ai(x+tmek,Du+tmΔmkDu)]dt=∫01‍[axki+∑j=1N‍aξjiDj(Δmku)]dt≐ei+gijDj(Δmku),
where ei=∫01‍axkidt, gij=∫01‍∑j=1N‍aξjidt, and
(62)Δmkf=1m∫01‍ddtf(x+tmek)dt=∫01‍fxkdt≐F,Δmkhi=1m∫01‍ddthi(x+tmek)dt=∫01‍hxkidt≐Hi.
Then, combining (60) with (61)-(62), we obtain
(63)∑i=1N‍∫Ω‍eiDiϕ+∑i,j=1N‍∫Ω‍gijDj(Δmku)Diϕ-∫Ω‍Fϕ-∑i=1N‍∫Ω‍HiDiϕ=0.
Due to (A4), (F2), (H2), (63), Young inequality, we have
(64)∑i,j=1N‍∫Ω‍gijDj(Δmku)Diϕ=∫Ω‍Fϕ+∑i=1N‍∫Ω‍HiDiϕ-∑i=1N‍∫Ω‍eiDiϕ≤c∫Ω‍|Dϕ|2+c∫Ω‍|u|q(x)+c∫Ω‍|u|q(x)/2ϕ≤c∫Ω‍|Dϕ|2+c∫Ω‍|u|q(x)+c∫Ω‍ϕ2;
that is,
(65)∑i,j=1N‍∫Ω‍gijDj(Δmku)Diϕ≤c∫Ω‍|Dϕ|2+c∫Ω‍|u|q(x)+c∫Ω‍ϕ2.
Choose ϕ=ψ2Δmku, where ψ∈C01(Ω-), |ψ|≤1, and (65), we obtain
(66)∑i,j=1N‍∫Ω‍gijψ2Di(Δmku)Dj(Δmku)≤-2∑i,j=1N‍∫Ω‍ψugijDiψDj(Δmku)+c∫Ω‍|Dψ2Δmku|2mmmmm,m+c∫Ω‍|u|q(x)+c∫Ω‍(ψ2Δmku)2.
From (66), (A3), Cauchy inequality and Young inequality, we have(67)∑i,j=1N‍∫Ω‍|gijψ2Di(Δmku)Dj(Δmku)|≤2∑i,j=1N‍∫Ω‍|ψugijDiψDj(Δmku)|mmmmm+c∫Ω‍|Dψ2Δmku|2+c∫Ω‍|u|q(x)+c∫Ω‍|ψ|4|Δmku|2≤2∑i,j=1N‍[∫Ω‍gijψ2Di(Δmku)Di(Δmku)]1/2mmmmm×[∫Ω‍gij(Δmku)2DiψDjψ]1/2+c∫Ω‍|Dψ2Δmku|2mmmmmm+c∫Ω‍|u|q(x)+c∫Ω‍|ψ|4|Δmku|2≤2∑i,j=1N‍[ε2(∫Ω‍gijψ2Di(Δmku)Di(Δmku))+12εmmmmmm×(∫Ω‍gij(Δmku)2DiψDjψ)]+c∫Ω‍|u|q(x)+c∫Ω‍ψ2|D(Δmku)|2+c∫Ω‍|Δmku|2|Dψ|2+c∫Ω‍|Δmku|2.
By (A3), we obtain
(68)∫Ω‍ψ2|D(Δmku)|2≤c∫Ω‍[1+∑i=1N‍(|Diu(x)|pi(x)-2mmmmmmm+|Diu(x+mek)|pi(x)-2)∑i=1N‍]×|Δmku|2|Dψ|2+c∫Ω‍|u|q(x).
Assume pN=p++=supx∈Ω-p+(x); we have for i=1,2,…,N, pi(x)≤pN. For q(x)<p-*(x), let k=N and take m→0; then we have
(69)∫Ω‍ψ2|D(DNu)|2≤c∫Ω‍(1+∑i=1N‍|Diu|pi(x)-2)|Dψ|2|DNu|2.
Hence, DNu∈Wloc1,2(Ω). If we take ϕ=ψ2(ΔmNu-r)+, where r>0, |ψ|<1, (ΔmNu-r)+=max{ΔmNu-r,0}. Since DNu∈Wloc1,2(Ω), DNu∈Lloc2N/(N-2)(Ω), we have
(70)∫Ω‍ψ2|D(DNu-r)+|2≤c∫Ω‍(1+∑i=1N‍|Diu|pi(x)-2)|Dψ|2|(DNu-r)+|2≤c∫Ω‍(1+∑i=1N-1‍|Diu|pi(x)-2)|Dψ|2|(DNu-r)+|2+c∫Ω‍|DNu|pN-2|Dψ|2|(DNu-r)+|2≤c∫Ω‍|(DNu-r)+|pN|Dψ|2+crpN-2∫Ω‍|(DNu-r)+|2+c∑i=1N‍∫Ω‍(|(DNu-r)+|2N/(N-2)|Dψ|2N/(N-2))(N-2)/Nmmmmmm×(∫{DNu>k}⋂‍suppψ‍(1+|Diu|pi(x)-2)N/2)2/N.
For 2≤pi(x)≤2N/(N-2), we have
(71)(∫{DNu>k}⋂‍suppψ‍(1+|Diu|pi(x)-2)N/2)2/N≤c|{DNu>k}⋂‍suppψ|2/N-2/p--+1.
If we fix B(x0,R), suppψ⊂B(x0,R), where R<1; let r≥1, 0<δ<1, and take ψ∈C01(B(x0,R)), ψ≡1 on B(x0,δR); 0≤ψ≤1, |Dψ|≤c/(R-δR), then we have
(72)∫Ar,δR‍|D(DNu-r)+|2≤c∫Ar,R‍(|DNu-r|R-δR)2N/(N-2)+c(∫Ar,R‍(|DNu-r|R-δR)2N/(N-2))(N-2)/N×|Ar,R|2/N-2/p--+1+cr2N/(N-2)|Ar,R|.
Hence, (72) implies that DNu is in Lloc∞(Ω). So we obtain u∈Wloc1,∞(Ω)⋂‍Wloc2,2(Ω). By De Giorgi-Moser regularity theorem, for any ω⊂⊂Ω and 0<α<1, we have DNu∈C0,α(ω); then u∈C1,α(Ω), for 0<α<1.



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*Source: 247628-2013-08-18.xml*