# Study on the Existence and Uniqueness of  Solution of Generalized Capillarity Problem

Authors: Li Wei; Liling Duan; Haiyun Zhou
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

By using the perturbation theories on sums of ranges of nonlinear accretive mappings of Calvert and Gupta (1978), the abstract result on the existence and uniqueness of the solution inLp(Ω) of the generalized Capillarity equation with nonlinear Neumann boundary value conditions, where 2N/(N+1)<p<+∞ and
N≥1 denotes the dimension of RN, is studied. The equation discussed in this paper and the methods here are a continuation of and a complement to the previous corresponding results. To obtain the results, some new techniques are used in this paper.


---

## Body

## 1. Introduction and Preliminary

Since thep-Laplacian operator -Δp with p≠2 arises from a variety of physical phenomena, such as nonNewtonian fluids, reaction-diffusion problems, and petroleum extraction,  it becomes a very popular topic in mathematical fields.We began our study on this topic in 1995. We used a perturbation result of ranges form-accretive mappings in Calvert and Gupta [1] to obtain a sufficient condition in Wei and He [2] so that the following zero boundary value problem, (1.1)-Δpu+g(x,u(x))=f(x),a.e.onΩ,-∂u∂n=0,a.e.onΓ,
has solutions in Lp(Ω), where 2≤p<+∞. Later on, a series work of ours has been done from different angles on this kind of equations, cf. [3–7], and so forth.Especially, in 2008, as a summary of the work done in [2–6], we use some new techniques to work for the following problem with so-called generalized p-Laplacian operator:
(1.2)-div[(C(x)+|∇u|2)(p-2)/2∇u]+ε|u|q-2u+g(x,u(x))=f(x),a.e.inΩ,-〈ϑ,(C(x)+|∇u|2)(p-2)/2∇u〉∈βx(u(x)),a.e.onΓ,
where 0≤C(x)∈Lp(Ω), ε is a nonnegative constant and ϑ denotes the exterior normal derivative of Γ. We showed in Wei and Agarwal [7] that (1.2) has solutions in Ls(Ω) under some conditions, where 2N/(N+1)<p≤s<+∞,  1≤q<+∞ if p≥N, and 1≤q≤Np/(N-p) if p<N, for N≥1.Capillarity equation is another important equation appeared in the capillarity phenomenon and we notice that in Chen and Luo [8], the authors studied the eigenvalue problem for the following generalized Capillarity equations:
(1.3)-div[(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u]=λ(|u|q-2u+|u|r-2u),inΩ,u=0,a.e.on∂Ω.Their work inspired us and one idea came to our mind. Can we borrow the main ideas dealing with the nonlinear elliptic boundary value problems with the generalizedp-Laplacian operator to study the nonlinear generalized Capillarity equation with Neumann boundary conditions?We will answer the question in this paper. By using the perturbation results of ranges form-accretive mappings in Calvert and Gupta [1] again, we will study the following one:
(1.4)-div[(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u]+λ(|u|q-2u+|u|r-2u)+g(x,u(x))=f(x),a.e.inΩ,-〈ϑ,(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u〉∈βx(u(x)),a.e.onΓ.More details on (1.4) will be given in Section 2. Our methods and techniques are different from those in Chen and Luo [8].Now, we list some basic knowledge we need in sequel.LetX be a real Banach space with a strictly convex dual space X*. We use “→” and “w-lim” to denote strong and weak convergence, respectively. For any subset G of X, we denote by intG its interior and G¯ its closure, respectively. Let “X↪↪Y” denote that space X is embedded compactly in space Y and “X↪Y” denote that space X is embedded continuously in space Y. A mapping T:D(T)=X→X* is said to be hemicontinuous on X if w-limt→0T(x+ty)=Tx, for any x,y∈X. Let J denote the duality mapping from X into 2X* defined by
(1.5)J(x)={f∈X*:(x,f)=∥x∥·∥f∥,∥f∥=∥x∥},x∈X,
where (·,·) denotes the generalized duality pairing between X and X*. It is wellknown that J is a single-valued mapping since X* is strictly convex.LetA:X→2X be a given multivalued mapping. We say that A is boundedly-inversely compact if for any pair of bounded subsets G and G′ of X, the subset G⋂‍A-1(G′) is relatively compact in X. The mapping A:X→2X is said to be accretive if (v1-v2,J(u1-u2))≥0, for any ui∈D(A) and vi∈Aui,  i=1,2. The accretive mapping A is said to be m-accretive if R(I+μA)=X, for some μ>0.LetB:X→2X* be a given multi-valued mapping. The graph of B,  G(B), is defined by G(B)={[u,w]∣u∈D(B),w∈Bu}. Then B:X→2X* is said to be monotone if G(B) is a monotone subset of X×X* in the sense that
(1.6)(u1-u2,w1-w2)≥0,
for any [ui,wi]∈G(B),  i=1,2. The monotone operator B is said to be maximal monotone if G(B) is maximal among all monotone subsets of X×X* in the sense of inclusion. The mapping B is said to be strictly monotone if the equality in (1.6) implies that u1=u2. The mapping B is said to be coercive if limn→+∞((xn,xn*)/∥xn∥)=∞ for all [xn,xn*]∈G(B) such that limn→+∞∥xn∥=+∞.Definition 1.1.
Theduality mappingJ:X→2X* is said to be satisfying Condition (I) if there exists a function η:X→[0,+∞) such that
(I)∥Ju-Jv∥≤η(u-v),for∀u,v∈X.Definition 1.2.
LetA:X→2X be an accretive mapping and J:X→X* be a duality mapping. We say that A satisfies Condition (*) if, for any f∈R(A) and a∈D(A), there exists a constant C(a,f) such that
(*)(v-f,J(u-a))≥C(a,f),foranyu∈D(A),v∈Au.Lemma 1.3 (Li and Guo [9]).
LetΩ be a bounded conical domain in RN. Then we have the following results.(a)
Ifmp>N, then Wm,p(Ω)↪CB(Ω); if mp<N and q=Np/(N-mp), then Wm,p(Ω)↪Lq(Ω); if mp=N and p>1, then for 1≤q<+∞,Wm,p(Ω)↪Lq(Ω).(b)
Ifmp>N, then Wm,p(Ω)↪↪CB(Ω); if 0<mp≤N and q0=Np/(N-mp), then Wm,p(Ω)↪↪Lq(Ω),  1≤q<q0.Lemma 1.4 (Pascali and Sburlan [10]).
IfB:X→2X* is an everywhere defined, monotone and hemicontinuous operator, then B is maximal monotone. If B:X→2X* is maximal monotone and coercive, then R(B)=X*.Lemma 1.5 (Pascali and Sburlan [10]).
IfΦ:X→(-∞,+∞] is a proper convex and lower-semicontinuous function, then ∂Φ is maximal monotone from X to X*.Lemma 1.6 (Pascali and Sburlan [10]).
IfA and B are two maximal monotone operators in X such that (intD(A))⋂‍D(B)≠∅, then A+B is maximal monotone.Proposition 1.7 (Calvert and Gupta [1]).
LetX=Lp(Ω) and Ω be a bounded domain in RN. For 2≤p<+∞, the duality mapping Jp:Lp(Ω)→Lp′(Ω) defined by Jpu=|u|p-1sgnu∥u∥p2-p, for u∈Lp(Ω), satisfies Condition (I); for 2N/(N+1)<p≤2 and N≥1, the duality mapping Jp:Lp(Ω)→Lp′(Ω) defined by Jpu=|u|p-1sgnu, for u∈Lp(Ω), satisfies Condition (I), where (1/p)+(1/p′)=1.Lemma 1.8 (see Calvert and Gupta [1]).
LetΩ be a bounded domain in RN and g:Ω×R→R be a function satisfying Caratheodory's conditions such that (i)
g(x,·) is monotonically increasing on R;(ii)
the mappingu∈Lp(Ω)→g(x,u(x))∈Lp(Ω) is well defined, where 2N/(N+1)<p<+∞ and N≥1.
LetJp:Lp(Ω)→Lp'(Ω),  (1/p)+(1/p′)=1 be the duality mapping defined by
(1.7)Jpu={|u|p-1sgnu,if2NN+1<p≤2,|u|p-1sgnu∥u∥p2-p,if2≤p<+∞,
for u∈Lp(Ω). Then the mapping B:Lp(Ω)→Lp(Ω) defined by (Bu)(x)=g(x,u(x)), for any x∈Ω satisfies Condition (*).Theorem 1.9 (Calvert and Gupta [1]).
LetX be a real Banach space with a strictly convex dual X*. Let J:X→X* be a duality mapping on X satisfying Condition (I). Let A,B1:X→2X be accretive mappings such that (i)
either bothA,B1 satisfy Condition (*) or D(A)⊂D(B1) and B1 satisfies Condition (*),(ii)
A+B1 is m-accretive and boundedly inversely compact.
IfB2:X→X be a bounded continuous mapping such that, for any y∈X, there is a constant C(y) satisfying (B2(u+y),Ju)≥-C(y) for any u∈X. Then. (a)
[R(A)+R(B1)]¯⊂R(A+B1+B2)¯.(b)
int[R(A)+R(B1)]⊂intR(A+B1+B2).


## 2. The Main Results

### 2.1. Notations and Assumptions of (1.4)

In the next of this paper, we assume2N/(N+1)<p<+∞,  1≤q,  r<+∞ if p≥N, and 1≤q,  r≤Np/(N-p) if p<N, where N≥1. We use ∥·∥p,  ∥·∥q,  ∥·∥r, and ∥·∥1,p,Ω to denote the norms in Lp(Ω),  Lq(Ω),  Lr(Ω) and W1,p(Ω). Let (1/p)+(1/p')=1,  (1/q)+(1/q')=1, and (1/r)+(1/r′)=1.In (1.4), Ω is a bounded conical domain of a Euclidean space RN with its boundary Γ∈C1, (c.f. [4]). We suppose that the Green's Formula is available. Let |·| denote the Euclidean norm in RN,  〈·,·〉 the Euclidean inner-product and ϑ the exterior normal derivative of Γ.  λ is a nonnegative constant.Letφ:Γ×R→R be a given function such that, for each x∈Γ,(i)
φx=φ(x,·):R→R is a proper, convex, lower-semicontinuous function with φx(0)=0.(ii)
βx=∂φx (: subdifferential of φx) is maximal monotone mapping on R with 0∈βx(0) and for each t∈R, the function x∈Γ→(I+μβx)-1(t)∈R is measurable for μ>0.Letg:Ω×R→R be a given function satisfying Caratheodory's conditions such that for 2N/(N+1)<p<+∞ and N≥1, the mapping u∈Lp(Ω)→g(x,u(x))∈Lp(Ω) is defined. Further, suppose that there is a function T(x)∈Lp(Ω) such that g(x,t)t≥0, for |t|≥T(x), x∈Ω.


### 2.2. Existence and Uniqueness of the Solution of (1.4)

Definition 2.1 (Calvert and Gupta [1]).
Defineg+(x)=liminft→+∞g(x,t) and g-(x)=limsupt→-∞g(x,t).
Further, define a functiong1:Ω×R→R by
(2.1)g1(x,t)={(infs≥tg(x,s))∧(t-T(x)),∀t≥T(x),0,∀t∈[-T(x),T(x)],(sups≤tg(x,s))∨(t+T(x)),∀t≤-T(x).
Then for all x∈Ω,  g1(x,t) is increasing in t and limt→±∞g1(x,t)=g±(x). Moreover, g1:Ω×R→R satisfies Caratheodory's conditions and the functions g±(x) are measurable on Ω. And, if g2(x,t)=g(x,t)-g1(x,t) then g2(x,t)t≥0, for |t|≥T(x), x∈Ω.Proposition 2.2 (see Calvert and Gupta [1]).
For2N/(N+1)<p<+∞ and N≥1, define the mapping B1:Lp(Ω)→Lp(Ω) by (B1u)(x)=g1(x,u(x)), for all u∈Lp(Ω) and x∈Ω, then B1 is a bounded, continuous, and m-accretive mapping.Moreover, Lemma1.8 implies that B1 satisfies Condition (*).DefineB2:Lp(Ω)→Lp(Ω) by (B2u)(x)=g2(x,u(x)), where g2(x,t)=g(x,t)-g1(x,t), then B2 satisfies the inequality:
(2.2)(B2(u+y),Jpu)≥-C(y),
for any u,y∈Lp(Ω), where C(y) is a constant depending on y and Jp:Lp(Ω)→Lp'(Ω) denotes the duality mapping, where (1/p)+(1/p′)=1.Lemma2.3 (Wei and Agarwal [7]).
The mappingΦp:W1,p(Ω)→R defined by
(2.3)Φp(u)=∫Γ‍φx(u|Γ(x))dΓ(x),
for any u∈W1,p(Ω), is a proper, convex, and lower-semicontinuous mapping on W1,p(Ω).Moreover, Lemma1.5 implies that ∂Φp, the subdifferential of Φp, is maximal monotone.Lemma2.4 (Wei and He [2]).
LetX0 denote the closed subspace of all constant functions in W1,p(Ω). Let X be the quotient space W1,p(Ω)/X0. For u∈W1,p(Ω), define the mapping P:W1,p(Ω)→X0 by Pu=(1/meas(Ω))∫Ω‍udx. Then, there is a constant C>0, such that for all u∈W1,p(Ω),
(2.4)∥u-Pu∥p≤C∥∇u∥(Lp(Ω))N.Lemma 2.5.
Define the mappingBp,q,r:W1,p(Ω)→(W1,p(Ω))* by
(2.5)(v,Bp,q,ru)=∫Ω‍〈(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u,∇v〉dx+λ∫Ω‍|u(x)|q-2u(x)v(x)dx+λ∫Ω‍|u(x)|r-2u(x)v(x)dx,
for any u,v∈W1,p(Ω). Then Bp,q,r is everywhere defined, strictly monotone, hemicontinuous, and coercive.Proof.
Step1. Bp,q,r is everywhere defined.
From Lemma1.3, we know that W1,p(Ω)↪CB(Ω), when p>N. And, W1,p(Ω)↪Lq(Ω),  W1,p(Ω)↪Lr(Ω), when p≤N. Thus, for all  v∈W1,p(Ω),  ∥v∥q≤k1∥v∥1,p,Ω,  ∥v∥r≤k2∥v∥1,p,Ω, where k1,k2 are positive constants.
Foru,v∈W1,p(Ω), we have
(2.6)|(v,Bp,q,ru)|≤2∫Ω‍|∇u|p-1|∇v|dx+λ∫Ω‍|u|q-1|v|dx+λ∫Ω‍|u|r-1|v|dx≤2∥|∇u|∥pp/p′∥|∇v|∥p+λ∥v∥q∥u∥qq/q′+λ∥v∥r∥u∥rr/r'≤2∥u∥1,p,Ωp/p′∥v∥1,p,Ω+k1′λ∥v∥1,p,Ω∥u∥1,p,Ωq/q′+k2′λ∥v∥1,p,Ω∥u∥1,p,Ωr/r',
where k1′ and k2′ are positive constants. Thus Bp,q,r is everywhere defined.
Step2. Bp,q,r is strictly monotone.
Foru,v∈W1,p(Ω), we have
(2.7)|(u-v,Bp,q,ru-Bp,q,rv)|=∫Ω‍〈(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u-(1+|∇v|p1+|∇v|2p)|∇v|p-2∇v,∇u-∇v〉dx+λ∫Ω‍(|u|q-2u-|v|q-2v)(u-v)dx+λ∫Ω‍(|u|r-2u-|v|r-2v)(u-v)dx=∫Ω‍{(1+|∇u|p1+|∇u|2p)|∇u|p-(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u∇v-(1+|∇v|p1+|∇v|2p)|∇v|p-2∇u∇v+(1+|∇v|p1+|∇v|2p)|∇v|p}dx+λ∫Ω‍(|u|q-2u-|v|q-2v)(u-v)dx+λ∫Ω‍(|u|r-2u-|v|r-2v)(u-v)dx≥∫Ω‍{(1+|∇u|p1+|∇u|2p)|∇u|p-1-(1+|∇v|p1+|∇v|2p)|∇v|p-1}(|∇u|-|∇v|)dx+λ∫Ω‍(|u|q-1-|v|q-1)(|u|-|v|)dx+λ∫Ω‍(|u|r-1-|v|r-1)(|u|-|v|)dx.
If, we leth(t)=(1+(t/1+t2))t(p-1)/p, for t≥0. Then we know that
(2.8)h′(t)=t(p-1)/p(1+t2)3/2+t-(1/p)(1+t1+t2)p-1p≥0,
since t≥0. And, h′(t)=0 if and only if t=0. Then h(t) is strictly monotone. Thus we can easily know that Bp,q,r is strictly monotone.
Step3.  Bp,q,r is hemicontinuous.
In fact, it suffices to show that, for anyu,v,w∈W1,p(Ω) and t∈[0,1],  (w,Bp,q,r(u+tv)-Bp,q,ru)→0, as t→0.
By Lebesque's dominated convergence theorem, it follows that(2.9)0≤limt→0|(w,Bp,q,r(u+tv)-Bp,q,ru)|≤∫Ω‍limt→0|(1+|∇u+t∇v|p1+|∇u+t∇v|2p)|∇u+t∇v|p-2(∇u+t∇v)-(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u||∇w|dx+λ∫Ω‍limt→0||u+tv|q-2(u+tv)-|u|q-2u||w|dx+λ∫Ω‍limt→0||u+tv|r-2(u+tv)-|u|r-2u||w|dx=0,
and hence Bp,q,r is hemicontinuous.
Step4.  Bp,q,r is coercive.
Now, foru∈W1,p(Ω), Lemma 2.4 implies that ∥u∥1,p,Ω→∞ is equivalent to ∥u-(1/meas(Ω))∫Ω‍udx∥1,p,Ω→∞ and hence we have the following result:
(2.10)(u,Bp,q,ru)∥u∥1,p,Ω=∫Ω‍(1+(|∇u|p/1+|∇u|2p))|∇u|pdx∥u∥1,p,Ω+λ∫Ω‍|u|qdx∥u∥1,p,Ω+λ∫Ω‍|u|rdx∥u∥1,p,Ω=∫Ω‍(|∇u|p+1+|∇u|2p)dx-∫Ω‍(1/1+|∇u|2p)dx∥u∥1,p,Ω+λ∫Ω‍|u|qdx∥u∥1,p,Ω+λ∫Ω‍|u|rdx∥u∥1,p,Ω≥2∫Ω‍|∇u|pdx-∫Ω‍(1/1+|∇u|2p)dx∥u∥1,p,Ω+λ∫Ω‍|u|qdx∥u∥1,p,Ω+λ∫Ω‍|u|rdx∥u∥1,p,Ω→+∞,
as ∥u∥1,p,Ω→+∞, which implies that Bp,q,r is coercive.
This completes the proof.Remark 2.6.
Lemma2.5 is a key result for later use.Definition 2.7.
Define a mappingAp:Lp(Ω)→2Lp(Ω) as follows:
(2.11)D(Ap)={u∈Lp(Ω)∣thereexistsanf∈Lp(Ω),suchthatf∈Bp,q,ru+∂Φp(u)}.
For u∈D(Ap), let Apu={f∈Lp(Ω)∣f∈Bp,q,ru+∂Φp(u)}.Proposition 2.8.
The mappingAp:Lp(Ω)→2Lp(Ω) is m-accretive.Proof.
Step1.    Ap is accretive.
Case1.  If p≥2, the duality mapping Jp:Lp(Ω)→Lp′(Ω) is defined by Jpu=|u|p-1sgnu∥u∥p2-p for u∈Lp(Ω). It then suffices to prove that for any ui∈D(Ap) and vi∈Apui,i=1,2,
(2.12)(v1-v2,Jp(u1-u2))≥0.
To this, we are left to prove that both
(2.13)(|u1-u2|p-1sgn(u1-u2)∥u1-u2∥p2-p,Bp,q,ru1-Bp,q,ru2)≥0,(|u1-u2|p-1sgn(u1-u2)∥u1-u2∥p2-p,∂Φp(u1)-∂Φp(u2))≥0
are available.
Now take for a constantk>0, χk:R→R is defined by χk(t)=|(t∧k)∨(-k)|p-1sgnt. Then χk is monotone, Lipschitz with χk(0)=0 and χk′ is continuous except at finitely many points on R. This gives that
(2.14)(|u1-u2|p-1sgn(u1-u2)∥u1-u2∥p2-p,∂Φp(u1)-∂Φp(u2))=limk→+∞∥u1-u2∥p2-p(χk(u1-u2),∂Φp(u1)-∂Φp(u2))≥0.
Also,
(2.15)(|u1-u2|p-1sgn(u1-u2)∥u1-u2∥p2-p,Bp,q,ru1-Bp,q,ru2)=∥u1-u2∥p2-p×limk→+∞∫Ω‍〈(1+|∇u1|p1+|∇u1|2p)|∇u1|p-2∇u1-(1+|∇u2|p1+|∇u2|2p)|∇u2|p-2∇u2,∇u1-∇u2(1+|∇u1|p1+|∇u1|2p)〉×χk′(u1-u2)dx+λ∥u1-u2∥p2-p∫Ω‍(|u1|q-2u1-|u2|q-2u2)|u1-u2|p-1sgn(u1-u2)dx+λ∥u1-u2∥p2-p∫Ω‍(|u1|r-2u1-|u2|r-2u2)|u1-u2|p-1sgn(u1-u2)dx≥0,
the last inequality is available since χk is monotone and χk(0)=0.
Case2.  If 2N/(N+1)<p<2, the duality mapping Jp:Lp(Ω)→Lp′(Ω) is defined by Jpu=|u|p-1sgnu, for u∈Lp(Ω). It then suffices to prove that for any ui∈D(Ap) and vi∈Apui,i=1,2,
(2.16)(v1-v2,Jp(u1-u2))≥0.
To this, we define the function χn:R→R by
(2.17)χn(t)={|t|p-1sgnt,if|t|≥1n,(1n)p-2t,if|t|≤1n.
Then χn is monotone, Lipschitz with χn(0)=0 and χn′ is continuous except at finitely many points on R. So (χn(u1-u2),∂Φp(u1)-∂Φp(u2))≥0.
Then, forui∈D(Ap),  vi∈Apui,i=1,2, we have(2.18)(v1-v2,Jp(u1-u2))=(|u1-u2|p-1sgn(u1-u2),Bp,q,ru1-Bp,q,ru2)+(|u1-u2|p-1sgn(u1-u2),∂Φp(u1)-∂Φp(u2))=(|u1-u2|p-1sgn(u1-u2),Bp,q,ru1-Bp,q,ru2)+limn→∞(χn(u1-u2),∂Φp(u1)-∂Φp(u2))≥0.
Step2.  R(I+μAp)=Lp(Ω), for every μ>0.
First, define the mappingIp:W1,p(Ω)→(W1,p(Ω))* by Ipu=u and (v,Ipu)(W1,p(Ω))*×W1,p(Ω)=(v,u)L2(Ω) for u,v∈W1,p(Ω), where (·,·)L2(Ω) denotes the inner product of L2(Ω). Then Ip is maximal monotone [7].
Secondly, for anyμ>0, define the mapping Tμ:W1,p(Ω)→2(W1,p(Ω))* by Tμu=Ipu+μBp,q,ru+μ∂Φp(u), for u∈W1,p(Ω). Then similar to that in [7], by using Lemmas 1.4, 1.6, 2.3, and 2.5, we know that Tμ is maximal monotone and coercive, so that R(Tμ)=(W1,p(Ω))*, for any μ>0.
Therefore, for anyf∈Lp(Ω), there exists u∈W1,p(Ω), such that
(2.19)f=Tμu=u+μBp,q,ru+μ∂Φp(u).
From the definition of Ap, it follows that R(I+μAp)=Lp(Ω), for all μ>0.
This completes the proof.Lemma 2.9.
The mappingAp:Lp(Ω)→2Lp(Ω) has a compact resolvent for 2N/(N+1)<p<2 and N≥1.Proof.
SinceAp is m-accretive by Proposition 2.8, it suffices to prove that if u+μApu=f(μ>0) and {f} is bounded in Lp(Ω), then {u} is relatively compact in Lp(Ω). Now define functions χn,ξn:R→R by
(2.20)χn(t)={|t|p-1sgnt,if|t|≥1n(1n)p-2t,if|t|≤1n,ξn(t)={|t|2-(2/p)sgnt,if|t|≥1n,(1n)1-(2/p)t,if|t|≤1n.
Noticing that χn′(t)=(p-1)×(p′/2)p×(ξn′(t))p,  for|t|≥1/n,  where(1/p)+(1/p′)=1 and χn′(t)=(ξn′(t))p,  for|t|≤1/n. We know that (χn(u),∂Φp(u))≥0 for u∈W1,p(Ω) since χn is monotone, Lipschitz with χn(0)=0, and χn′ is continuous except at finitely many points on R. Then
(2.21)(|u|p-1sgnu,Apu)=limn→∞(χn(u),Apu)≥limn→∞(χn(u),Bp,q,ru)=limn→∞∫Ω‍(1+|∇u|p1+|∇u|2p)|∇u|pχn′(u)dx+λlimn→∞∫Ω‍|u|q-2uχn(u)dx+λlimn→∞∫Ω‍|u|r-2uχn(u)dx≥limn→∞∫Ω‍|∇u|pχn′(u)dx≥const·limn→∞∫Ω‍|grad(ξn(u))|pdx≥const∫Ω‍|grad(|u|2-(2/p)sgnu)|pdx.
We now have fromf=u+μApu that
(2.22)∥f∥p∥|u|2-(2/p)sgnu∥p2/2(p-1)p2/2(p-1)p′≥(|u|p-1sgnu,f)=(|u|p-1sgnu,u)+μ(|u|p-1sgnu,Apu)≥∥|u|2-(2/p)sgnu∥p2/2(p-1)p2/2(p-1)+μ·const·∥grad|u|2-(2/p)sgnu∥pp,
which gives that
(2.23)∥|u|2-(2/p)sgnu∥pp/2(p-1)≤∥|u|2-(2/p)sgnu∥p2/2(p-1)p/2(p-1)∥f∥p≤const,
in view of the fact that p<p2/2(p-1) when 2N/(N+1)<p<2 for N≥1. Again from (2.22), we have ∥grad(|u|2-(2/p)sgnu)∥p≤const. Hence {f} bounded in Lp(Ω) implies that {|u|2-(2/p)sgnu} is bounded in W1,p(Ω).
We notice thatW1,p(Ω)↪↪Lp2/2(p-1)(Ω) when N≥2 and W1,p(Ω)↪↪CB(Ω) when N=1 by Lemma 1.3, hence{|u|2-(2/p)sgnu} is relatively compact in Lp2/2(p-1)(Ω). This gives that {u} is relatively compact in Lp(Ω) since the Nemytskii mapping u∈Lp2/2(p-1)(Ω)→|u|p/2(p-1)sgnu∈Lp(Ω) is continuous.
This completes the proof.Remark 2.10.
SinceΦp(u+α)=Φp(u), for any u∈W1,p(Ω) and α∈C0∞(Ω), we have f∈Apu implies that f=Bp,q,ru in the sense of distributions.Proposition 2.11.
Forf∈Lp(Ω), if there exists u∈Lp(Ω) such that f∈Apu, then u is the unique solution of (1.4).Proof.
First, we show that(2.24)-div[(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u]+λ|u|q-2u+λ|u|r-2u=f(x),a.e.x∈Ω
is available.
Nowf∈Apu implies that f=Bp,q,ru+∂Φp(u). For all φ∈C0∞(Ω), by Remark 2.10, we have
(2.25)(φ,f)=(φ,Bp,q,ru+∂Φp(u))=(φ,Bp,q,ru)=∫Ω‍〈(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u,∇φ〉dx+λ∫Ω‍|u|q-2uφdx+λ∫Ω‍|u|r-2uφdx=∫Ω‍-div[(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u]φdx+λ∫Ω‍|u|q-2uφdx+λ∫Ω‍|u|r-2uφdx,
which implies that (2.24) is true.
Secondly, we show that(2.26)-〈ϑ,(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u〉∈βx(u(x)),a.e.x∈Γ.
We will prove (2.26) under the additional condition |βx(u)|≤a|u|p/p′+b(x), where b(x)∈Lp′(Γ) and a∈R. Refer to the result of Brezis [11] for the general case.
Now, from (2.24), f∈Apu implies that f(x)=-div[(1+|∇u|p/1+|∇u|2p)|∇u|p-2∇u]+λ|u(x)|q-2u(x)+λ|u|r-2u∈Lp(Ω). By using Green's Formula, we have that for any v∈W1,p(Ω),
(2.27)∫Γ‍〈ϑ,(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u〉v|ΓdΓ(x)=∫Ω‍div[(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u]vdx+∫Ω‍〈(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u,∇v〉dx.
Then-〈ϑ,(1+|∇u|p/1+|∇u|2p)|∇u|p-2∇u〉∈W-(1/p),p′(Γ)=(W1/p,p(Γ))*, where W1/p,p(Γ) is the space of traces of W1,p(Ω).
Now let the mappingB:Lp(Γ)→Lp′(Γ) be defined by Bu=g(x), for any u∈Lp(Γ), where g(x)=βx(u(x)) a.e. on Γ. Clearly, B=∂Ψ where Ψ(u)=∫Γ‍φx(u(x))dΓ(x) is a proper, convex, and lower-semicontinuous function on Lp(Γ). Now define the mapping K:W1,p(Ω)→Lp(Γ) by K(v)=v|Γ for any v∈W1,p(Ω). Then K*BK:W1,p(Ω)→(W1,p(Ω))* is maximal monotone since both K,B are continuous. Finally, for any u,v∈W1,p(Ω), we have
(2.28)Ψ(Kv)-Ψ(Ku)=∫Γ‍[φx(v|Γ(x))-φx(u|Γ(x))]dΓ(x)≥∫Γ‍βx(u|Γ(x))(v|Γ(x)-u|Γ(x))dΓ(x)=(BKu,Kv-Ku)=(K*BKu,v-u).
Hence we get K*BK⊂∂Φp and so K*BK=∂Φp. Therefore, we have
(2.29)-〈ϑ,(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u〉∈βx(u(x)),a.e.onΓ.
Finally, we will show thatu is unique.
Iff∈Apu and f∈Apv, where u,v∈D(Ap). Then
(2.30)0≤(u-v,Bp,q,ru-Bp,q,rv)=(u-v,∂Φp(v)-∂Φp(u))≤0,
since Bp,q,r is strictly monotone and ∂Φp is maximal monotone, which implies that u(x)=v(x).
This completes the proof.Remark 2.12.
Ifβx≡0 for any x∈Γ, then ∂Φp(u)≡0, for all u∈W1,p(Ω).Proposition 2.13.
Ifβx≡0 for any x∈Γ, then {f∈Lp(Ω)∣∫Ω‍fdx=0}⊂R(Ap).Proof.
We can easily know thatR(Bp,q,r)=(W1,p(Ω))* in view of Lemmas 1.4 and 2.5. Note that for any f∈Lp(Ω) with ∫Ω‍fdx=0, the linear function u∈W1,p(Ω)→∫Ω‍fudx is an element of (W1,p(Ω))*. So there exists a u∈W1,p(Ω) such that
(2.31)∫Ω‍fvdx=∫Ω‍〈(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u,∇v〉dx+λ∫Ω‍|u|q-2uvdx+λ∫Ω‍|u|r-2uvdx,
for any v∈W1,p(Ω). So f=Apu in view of Remark 2.12.
This completes the proof.Definition 2.14 (see [1, 7]).
Fort∈R,  x∈Γ, let βx0(t)∈βx(t) be the element with least absolute value if βx(t)≠∅ and βx0(t)=±∞, where t>0 or <0, respectively, in case βx(t)=∅. Finally, let β±(x)=limt→±∞βx0(t) (in the extended sense) for x∈Γ. β±(x) define measurable functions on Γ, in view of our assumptions on βx.Proposition 2.15.
Letf∈Lp(Ω) such that
(2.32)∫Γ‍β-(x)dΓ(x)<∫Ω‍fdx<∫Γ‍β+(x)dΓ(x).
Then f∈IntR(Ap).Proof.
Letf∈Lp(Ω) and satisfy (2.32), by Proposition 2.8, there exists un∈Lp(Ω) such that, for each n≥1,  f=(1/n)un+Apun. In the same reason as that in [1], we only need to prove that ∥un∥p≤const, for all n≥1.
Indeed, suppose to the contrary that1≤∥un∥p→∞, as n→∞. Let vn=un/∥un∥p. Let ψ:R→R be defined by ψ(t)=|t|p,  ∂ψ:R→R be its subdifferential and for μ>0,  ∂ψμ:R→R denote the Yosida-approximation of ∂ψ.   Let θμ:R→R denote the indefinite integral of [(∂ψμ)′]1/p with θμ(0)=0 so that (θμ′)p=(∂ψμ)′. In view of Calvert and Gupta [1], we have
(2.33)(∂ψμ(vn),∂Φp(un))≥∫Γ‍βx((1+μ∂ψ)-1(un|Γ(x)))×∂ψμ(vn|Γ(x))dΓ(x)≥0.
Now multiplying the equationf=(1/n)un+Apun by ∂ψμ(vn), we get that
(2.34)(∂ψμ(vn),f)=(∂ψμ(vn),1nun)+(∂ψμ(vn),Bp,q,run)+(∂ψμ(vn),∂Φp(un)).
Since ∂ψμ(0)=0, it follows that (∂ψμ(vn),un)≥0. Also, we can know that
(2.35)(∂ψμ(vn),Bp,q,run)=∫Ω‍〈(1+|∇un|p1+|∇un|2p)|∇un|p-2∇un,∇vn〉(∂ψμ)′(vn)dx+λ∫Ω‍|un|q-2un∂ψμ(vn)dx+λ∫Ω‍|un|r-2un∂ψμ(vn)dx≥∫Ω‍|∇u|p∥un∥p(∂ψμ)′(vn)dx=∥un∥pp-1∫Ω‍|grad(θμ(vn))|pdx.
Then we get from (2.33) that
(2.36)∥un∥pp-1∫Ω‍|grad(θμ(vn))|pdx+∫Γ‍βx((1+μ∂ψ)-1(un|Γ(x)))×∂ψμ(vn|Γ(x))dΓ(x)≤(∂ψμ(vn),f).
Since |∂ψμ(t)|≤|∂ψ(t)| for any t∈R and μ>0, we see from ∥vn∥p=1 for n≥1, that ∥∂ψμ(vn)∥p′≤C, for μ>0, where C is a constant which does not depend on n or μ and (1/p)+(1/p′)=1.
From (2.36), we have
(2.37)∫Ω‍|grad(θμ(vn))|pdx≤C∥un∥pp-1,forμ>0,n≥1.
Now we easily know that (θμ′)p=(∂ψμ)′→(∂ψ)′ as μ→0  a.e. on R.
Lettingμ→0, we see from Fatou's lemma and (2.37) that
(2.38)∫Ω|grad(|vn|2-(2/p)sgnvn)|pdx≤C∥un∥pp-1.
From (2.38), we know that |vn|2-(2/p)sgnvn→k (a constant) in Lp(Ω), as n→+∞.
Next we will show thatk≠0 is in Lp(Ω) from two aspects.(i)
Ifp≥2, since ∥|vn|2-(2/p)sgnvn∥p=∥vn∥2p-22-(2/p)≥∥vn∥p2-(2/p)=1, it follows that k≠0 in Lp(Ω),(ii)
if2N/(N+1)<p<2,  ∥|vn|2-(2/p)sgnvn∥p=∥vn∥2p-22-(2/p)≥∥vn∥p2-(2/p)=1, then {|vn|2-(2/p)sgnvn} is bounded in W1,p(Ω). By Lemma 1.3, W1,p(Ω)↪↪CB(Ω) when N=1 and W1,p(Ω)↪↪Lp2/2(p-1)(Ω), when N≥2. So {|vn|2-(2/p)sgnvn} is relatively compact in Lp2/2(p-1)(Ω). Then there exists a subsequence of {|vn|2-(2/p)sgnvn}, for simplicity, we denote it by {|vn|2-(2/p)sgnvn}, satisfying |vn|2-(2/p)sgnvn→g in Lp2/2(p-1)(Ω). Noticing that p≤p2/2(p-1) when 2N/(N+1)<p<2, it follows that k=g  a.e. on Ω. Now,
(2.39)1=∥vn∥pp=∫Ω‍||vn|2-(2/p)sgnvn|p2/2(p-1)dx≤const∫Ω‍||vn|2-(2/p)sgnvn-g|p2/2(p-1)dx+const∥g∥p2/2(p-1)p2/2(p-1),it follows that g≠0 in Lp(Ω) and then k≠0 in Lp(Ω). Assume, now, k>0, we see from (2.36) that
(2.40)∫Γ‍βx((1+μ∂ψ)-1(un|Γ(x)))×∂ψμ(vn|Γ(x))dΓ(x)≤(∂ψμ(vn),f).
Choosing a subsequence so that un|Γ(x)→+∞ a.e. on Γ, we see letting n→+∞ that ∫Γ‍β+(x)dΓ(x)≤∫Ω‍f(x)dx, which is a contradiction with (2.32). Similarly, if k<0, it also leads to a contradiction. Thus f∈intR(Ap).
This completes the proof.Proposition 2.16.
A
p
+
B
1
:
L
p
(
Ω
)
→
L
p
(
Ω
) is m-accretive and has a compact resolvent.Proof.
Using a theorem in Corduneanu [12], we know that Ap+B1:Lp(Ω)→Lp(Ω) is m-accretive.
To show thatAp+B1:Lp(Ω)→Lp(Ω) has a compact resolvent, we only need to prove that if w∈Apu+B1u with {w} and {u} being bounded in Lp(Ω), then {u} is relatively compact in Lp(Ω). Now we discuss it from two aspects.(i)
Ifp≥2, since
(2.41)∫Ω‍|∇u|pdx≤(u,Bp,q,ru)=(u,Apu)-(u,∂Φp(u))≤(u,Apu)+(u,B1u)=(u,w)≤∥u∥p∥u∥p′≤const,
it follows that {u} is bounded in W1,p(Ω), where (1/p)+(1/p')=1. Then {u} is relatively compact in Lp(Ω) since W1,p(Ω)↪↪Lp(Ω);(ii)
if2N/(N+1)<p<2, since w∈Apu+B1u with {w} and {u} being bounded in Lp(Ω), we have w-B1u∈Apu with {w-B1u} and {u} being bounded in Lp(Ω) which gives that {u} is relatively compact in Lp(Ω) since Ap is m-accretive by Proposition 2.8 and has a compact resolvent by Lemma 2.9.
This completes the proof.Theorem 2.17.
Letf∈Lp(Ω) be such that
(2.42)∫Γ‍β-(x)dΓ(x)+∫Ω‍g-(x)dx<∫Ω‍f(x)dx<∫Γ‍β+(x)dΓ(x)+∫Ω‍g+(x)dx,
then (1.4) has a unique solution in Lp(Ω), where 2N/(N+1)<p<+∞ and N≥1.Proof.
We want to use Theorem1.9 to finish our proof. From Propositions 1.7, 2.2, 2.8, and 2.16, we can see that all of the conditions in Theorem 1.9 are satisfied. It then suffices to show that f∈int[R(Ap)+R(B1)] which ensures that f∈R(Ap+B1+B2). Thus Proposition 2.11 tells us (1.4) has a unique solution in Lp(Ω).
Using the similar methods as those in [2, 4, 7], by dividing it into two cases and using Propositions 2.13 and 2.15, respectively, we know that f∈int[R(Ap)+R(B1)].
This completes the proof.Remark 2.18.
Compared to the work done in [1–7], not only the existence of the solution of (1.4) is obtained but also the uniqueness of the solution is obtained.


## 2.1. Notations and Assumptions of (1.4)

In the next of this paper, we assume2N/(N+1)<p<+∞,  1≤q,  r<+∞ if p≥N, and 1≤q,  r≤Np/(N-p) if p<N, where N≥1. We use ∥·∥p,  ∥·∥q,  ∥·∥r, and ∥·∥1,p,Ω to denote the norms in Lp(Ω),  Lq(Ω),  Lr(Ω) and W1,p(Ω). Let (1/p)+(1/p')=1,  (1/q)+(1/q')=1, and (1/r)+(1/r′)=1.In (1.4), Ω is a bounded conical domain of a Euclidean space RN with its boundary Γ∈C1, (c.f. [4]). We suppose that the Green's Formula is available. Let |·| denote the Euclidean norm in RN,  〈·,·〉 the Euclidean inner-product and ϑ the exterior normal derivative of Γ.  λ is a nonnegative constant.Letφ:Γ×R→R be a given function such that, for each x∈Γ,(i)
φx=φ(x,·):R→R is a proper, convex, lower-semicontinuous function with φx(0)=0.(ii)
βx=∂φx (: subdifferential of φx) is maximal monotone mapping on R with 0∈βx(0) and for each t∈R, the function x∈Γ→(I+μβx)-1(t)∈R is measurable for μ>0.Letg:Ω×R→R be a given function satisfying Caratheodory's conditions such that for 2N/(N+1)<p<+∞ and N≥1, the mapping u∈Lp(Ω)→g(x,u(x))∈Lp(Ω) is defined. Further, suppose that there is a function T(x)∈Lp(Ω) such that g(x,t)t≥0, for |t|≥T(x), x∈Ω.


## 2.2. Existence and Uniqueness of the Solution of (1.4)

Definition 2.1 (Calvert and Gupta [1]).
Defineg+(x)=liminft→+∞g(x,t) and g-(x)=limsupt→-∞g(x,t).
Further, define a functiong1:Ω×R→R by
(2.1)g1(x,t)={(infs≥tg(x,s))∧(t-T(x)),∀t≥T(x),0,∀t∈[-T(x),T(x)],(sups≤tg(x,s))∨(t+T(x)),∀t≤-T(x).
Then for all x∈Ω,  g1(x,t) is increasing in t and limt→±∞g1(x,t)=g±(x). Moreover, g1:Ω×R→R satisfies Caratheodory's conditions and the functions g±(x) are measurable on Ω. And, if g2(x,t)=g(x,t)-g1(x,t) then g2(x,t)t≥0, for |t|≥T(x), x∈Ω.Proposition 2.2 (see Calvert and Gupta [1]).
For2N/(N+1)<p<+∞ and N≥1, define the mapping B1:Lp(Ω)→Lp(Ω) by (B1u)(x)=g1(x,u(x)), for all u∈Lp(Ω) and x∈Ω, then B1 is a bounded, continuous, and m-accretive mapping.Moreover, Lemma1.8 implies that B1 satisfies Condition (*).DefineB2:Lp(Ω)→Lp(Ω) by (B2u)(x)=g2(x,u(x)), where g2(x,t)=g(x,t)-g1(x,t), then B2 satisfies the inequality:
(2.2)(B2(u+y),Jpu)≥-C(y),
for any u,y∈Lp(Ω), where C(y) is a constant depending on y and Jp:Lp(Ω)→Lp'(Ω) denotes the duality mapping, where (1/p)+(1/p′)=1.Lemma2.3 (Wei and Agarwal [7]).
The mappingΦp:W1,p(Ω)→R defined by
(2.3)Φp(u)=∫Γ‍φx(u|Γ(x))dΓ(x),
for any u∈W1,p(Ω), is a proper, convex, and lower-semicontinuous mapping on W1,p(Ω).Moreover, Lemma1.5 implies that ∂Φp, the subdifferential of Φp, is maximal monotone.Lemma2.4 (Wei and He [2]).
LetX0 denote the closed subspace of all constant functions in W1,p(Ω). Let X be the quotient space W1,p(Ω)/X0. For u∈W1,p(Ω), define the mapping P:W1,p(Ω)→X0 by Pu=(1/meas(Ω))∫Ω‍udx. Then, there is a constant C>0, such that for all u∈W1,p(Ω),
(2.4)∥u-Pu∥p≤C∥∇u∥(Lp(Ω))N.Lemma 2.5.
Define the mappingBp,q,r:W1,p(Ω)→(W1,p(Ω))* by
(2.5)(v,Bp,q,ru)=∫Ω‍〈(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u,∇v〉dx+λ∫Ω‍|u(x)|q-2u(x)v(x)dx+λ∫Ω‍|u(x)|r-2u(x)v(x)dx,
for any u,v∈W1,p(Ω). Then Bp,q,r is everywhere defined, strictly monotone, hemicontinuous, and coercive.Proof.
Step1. Bp,q,r is everywhere defined.
From Lemma1.3, we know that W1,p(Ω)↪CB(Ω), when p>N. And, W1,p(Ω)↪Lq(Ω),  W1,p(Ω)↪Lr(Ω), when p≤N. Thus, for all  v∈W1,p(Ω),  ∥v∥q≤k1∥v∥1,p,Ω,  ∥v∥r≤k2∥v∥1,p,Ω, where k1,k2 are positive constants.
Foru,v∈W1,p(Ω), we have
(2.6)|(v,Bp,q,ru)|≤2∫Ω‍|∇u|p-1|∇v|dx+λ∫Ω‍|u|q-1|v|dx+λ∫Ω‍|u|r-1|v|dx≤2∥|∇u|∥pp/p′∥|∇v|∥p+λ∥v∥q∥u∥qq/q′+λ∥v∥r∥u∥rr/r'≤2∥u∥1,p,Ωp/p′∥v∥1,p,Ω+k1′λ∥v∥1,p,Ω∥u∥1,p,Ωq/q′+k2′λ∥v∥1,p,Ω∥u∥1,p,Ωr/r',
where k1′ and k2′ are positive constants. Thus Bp,q,r is everywhere defined.
Step2. Bp,q,r is strictly monotone.
Foru,v∈W1,p(Ω), we have
(2.7)|(u-v,Bp,q,ru-Bp,q,rv)|=∫Ω‍〈(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u-(1+|∇v|p1+|∇v|2p)|∇v|p-2∇v,∇u-∇v〉dx+λ∫Ω‍(|u|q-2u-|v|q-2v)(u-v)dx+λ∫Ω‍(|u|r-2u-|v|r-2v)(u-v)dx=∫Ω‍{(1+|∇u|p1+|∇u|2p)|∇u|p-(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u∇v-(1+|∇v|p1+|∇v|2p)|∇v|p-2∇u∇v+(1+|∇v|p1+|∇v|2p)|∇v|p}dx+λ∫Ω‍(|u|q-2u-|v|q-2v)(u-v)dx+λ∫Ω‍(|u|r-2u-|v|r-2v)(u-v)dx≥∫Ω‍{(1+|∇u|p1+|∇u|2p)|∇u|p-1-(1+|∇v|p1+|∇v|2p)|∇v|p-1}(|∇u|-|∇v|)dx+λ∫Ω‍(|u|q-1-|v|q-1)(|u|-|v|)dx+λ∫Ω‍(|u|r-1-|v|r-1)(|u|-|v|)dx.
If, we leth(t)=(1+(t/1+t2))t(p-1)/p, for t≥0. Then we know that
(2.8)h′(t)=t(p-1)/p(1+t2)3/2+t-(1/p)(1+t1+t2)p-1p≥0,
since t≥0. And, h′(t)=0 if and only if t=0. Then h(t) is strictly monotone. Thus we can easily know that Bp,q,r is strictly monotone.
Step3.  Bp,q,r is hemicontinuous.
In fact, it suffices to show that, for anyu,v,w∈W1,p(Ω) and t∈[0,1],  (w,Bp,q,r(u+tv)-Bp,q,ru)→0, as t→0.
By Lebesque's dominated convergence theorem, it follows that(2.9)0≤limt→0|(w,Bp,q,r(u+tv)-Bp,q,ru)|≤∫Ω‍limt→0|(1+|∇u+t∇v|p1+|∇u+t∇v|2p)|∇u+t∇v|p-2(∇u+t∇v)-(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u||∇w|dx+λ∫Ω‍limt→0||u+tv|q-2(u+tv)-|u|q-2u||w|dx+λ∫Ω‍limt→0||u+tv|r-2(u+tv)-|u|r-2u||w|dx=0,
and hence Bp,q,r is hemicontinuous.
Step4.  Bp,q,r is coercive.
Now, foru∈W1,p(Ω), Lemma 2.4 implies that ∥u∥1,p,Ω→∞ is equivalent to ∥u-(1/meas(Ω))∫Ω‍udx∥1,p,Ω→∞ and hence we have the following result:
(2.10)(u,Bp,q,ru)∥u∥1,p,Ω=∫Ω‍(1+(|∇u|p/1+|∇u|2p))|∇u|pdx∥u∥1,p,Ω+λ∫Ω‍|u|qdx∥u∥1,p,Ω+λ∫Ω‍|u|rdx∥u∥1,p,Ω=∫Ω‍(|∇u|p+1+|∇u|2p)dx-∫Ω‍(1/1+|∇u|2p)dx∥u∥1,p,Ω+λ∫Ω‍|u|qdx∥u∥1,p,Ω+λ∫Ω‍|u|rdx∥u∥1,p,Ω≥2∫Ω‍|∇u|pdx-∫Ω‍(1/1+|∇u|2p)dx∥u∥1,p,Ω+λ∫Ω‍|u|qdx∥u∥1,p,Ω+λ∫Ω‍|u|rdx∥u∥1,p,Ω→+∞,
as ∥u∥1,p,Ω→+∞, which implies that Bp,q,r is coercive.
This completes the proof.Remark 2.6.
Lemma2.5 is a key result for later use.Definition 2.7.
Define a mappingAp:Lp(Ω)→2Lp(Ω) as follows:
(2.11)D(Ap)={u∈Lp(Ω)∣thereexistsanf∈Lp(Ω),suchthatf∈Bp,q,ru+∂Φp(u)}.
For u∈D(Ap), let Apu={f∈Lp(Ω)∣f∈Bp,q,ru+∂Φp(u)}.Proposition 2.8.
The mappingAp:Lp(Ω)→2Lp(Ω) is m-accretive.Proof.
Step1.    Ap is accretive.
Case1.  If p≥2, the duality mapping Jp:Lp(Ω)→Lp′(Ω) is defined by Jpu=|u|p-1sgnu∥u∥p2-p for u∈Lp(Ω). It then suffices to prove that for any ui∈D(Ap) and vi∈Apui,i=1,2,
(2.12)(v1-v2,Jp(u1-u2))≥0.
To this, we are left to prove that both
(2.13)(|u1-u2|p-1sgn(u1-u2)∥u1-u2∥p2-p,Bp,q,ru1-Bp,q,ru2)≥0,(|u1-u2|p-1sgn(u1-u2)∥u1-u2∥p2-p,∂Φp(u1)-∂Φp(u2))≥0
are available.
Now take for a constantk>0, χk:R→R is defined by χk(t)=|(t∧k)∨(-k)|p-1sgnt. Then χk is monotone, Lipschitz with χk(0)=0 and χk′ is continuous except at finitely many points on R. This gives that
(2.14)(|u1-u2|p-1sgn(u1-u2)∥u1-u2∥p2-p,∂Φp(u1)-∂Φp(u2))=limk→+∞∥u1-u2∥p2-p(χk(u1-u2),∂Φp(u1)-∂Φp(u2))≥0.
Also,
(2.15)(|u1-u2|p-1sgn(u1-u2)∥u1-u2∥p2-p,Bp,q,ru1-Bp,q,ru2)=∥u1-u2∥p2-p×limk→+∞∫Ω‍〈(1+|∇u1|p1+|∇u1|2p)|∇u1|p-2∇u1-(1+|∇u2|p1+|∇u2|2p)|∇u2|p-2∇u2,∇u1-∇u2(1+|∇u1|p1+|∇u1|2p)〉×χk′(u1-u2)dx+λ∥u1-u2∥p2-p∫Ω‍(|u1|q-2u1-|u2|q-2u2)|u1-u2|p-1sgn(u1-u2)dx+λ∥u1-u2∥p2-p∫Ω‍(|u1|r-2u1-|u2|r-2u2)|u1-u2|p-1sgn(u1-u2)dx≥0,
the last inequality is available since χk is monotone and χk(0)=0.
Case2.  If 2N/(N+1)<p<2, the duality mapping Jp:Lp(Ω)→Lp′(Ω) is defined by Jpu=|u|p-1sgnu, for u∈Lp(Ω). It then suffices to prove that for any ui∈D(Ap) and vi∈Apui,i=1,2,
(2.16)(v1-v2,Jp(u1-u2))≥0.
To this, we define the function χn:R→R by
(2.17)χn(t)={|t|p-1sgnt,if|t|≥1n,(1n)p-2t,if|t|≤1n.
Then χn is monotone, Lipschitz with χn(0)=0 and χn′ is continuous except at finitely many points on R. So (χn(u1-u2),∂Φp(u1)-∂Φp(u2))≥0.
Then, forui∈D(Ap),  vi∈Apui,i=1,2, we have(2.18)(v1-v2,Jp(u1-u2))=(|u1-u2|p-1sgn(u1-u2),Bp,q,ru1-Bp,q,ru2)+(|u1-u2|p-1sgn(u1-u2),∂Φp(u1)-∂Φp(u2))=(|u1-u2|p-1sgn(u1-u2),Bp,q,ru1-Bp,q,ru2)+limn→∞(χn(u1-u2),∂Φp(u1)-∂Φp(u2))≥0.
Step2.  R(I+μAp)=Lp(Ω), for every μ>0.
First, define the mappingIp:W1,p(Ω)→(W1,p(Ω))* by Ipu=u and (v,Ipu)(W1,p(Ω))*×W1,p(Ω)=(v,u)L2(Ω) for u,v∈W1,p(Ω), where (·,·)L2(Ω) denotes the inner product of L2(Ω). Then Ip is maximal monotone [7].
Secondly, for anyμ>0, define the mapping Tμ:W1,p(Ω)→2(W1,p(Ω))* by Tμu=Ipu+μBp,q,ru+μ∂Φp(u), for u∈W1,p(Ω). Then similar to that in [7], by using Lemmas 1.4, 1.6, 2.3, and 2.5, we know that Tμ is maximal monotone and coercive, so that R(Tμ)=(W1,p(Ω))*, for any μ>0.
Therefore, for anyf∈Lp(Ω), there exists u∈W1,p(Ω), such that
(2.19)f=Tμu=u+μBp,q,ru+μ∂Φp(u).
From the definition of Ap, it follows that R(I+μAp)=Lp(Ω), for all μ>0.
This completes the proof.Lemma 2.9.
The mappingAp:Lp(Ω)→2Lp(Ω) has a compact resolvent for 2N/(N+1)<p<2 and N≥1.Proof.
SinceAp is m-accretive by Proposition 2.8, it suffices to prove that if u+μApu=f(μ>0) and {f} is bounded in Lp(Ω), then {u} is relatively compact in Lp(Ω). Now define functions χn,ξn:R→R by
(2.20)χn(t)={|t|p-1sgnt,if|t|≥1n(1n)p-2t,if|t|≤1n,ξn(t)={|t|2-(2/p)sgnt,if|t|≥1n,(1n)1-(2/p)t,if|t|≤1n.
Noticing that χn′(t)=(p-1)×(p′/2)p×(ξn′(t))p,  for|t|≥1/n,  where(1/p)+(1/p′)=1 and χn′(t)=(ξn′(t))p,  for|t|≤1/n. We know that (χn(u),∂Φp(u))≥0 for u∈W1,p(Ω) since χn is monotone, Lipschitz with χn(0)=0, and χn′ is continuous except at finitely many points on R. Then
(2.21)(|u|p-1sgnu,Apu)=limn→∞(χn(u),Apu)≥limn→∞(χn(u),Bp,q,ru)=limn→∞∫Ω‍(1+|∇u|p1+|∇u|2p)|∇u|pχn′(u)dx+λlimn→∞∫Ω‍|u|q-2uχn(u)dx+λlimn→∞∫Ω‍|u|r-2uχn(u)dx≥limn→∞∫Ω‍|∇u|pχn′(u)dx≥const·limn→∞∫Ω‍|grad(ξn(u))|pdx≥const∫Ω‍|grad(|u|2-(2/p)sgnu)|pdx.
We now have fromf=u+μApu that
(2.22)∥f∥p∥|u|2-(2/p)sgnu∥p2/2(p-1)p2/2(p-1)p′≥(|u|p-1sgnu,f)=(|u|p-1sgnu,u)+μ(|u|p-1sgnu,Apu)≥∥|u|2-(2/p)sgnu∥p2/2(p-1)p2/2(p-1)+μ·const·∥grad|u|2-(2/p)sgnu∥pp,
which gives that
(2.23)∥|u|2-(2/p)sgnu∥pp/2(p-1)≤∥|u|2-(2/p)sgnu∥p2/2(p-1)p/2(p-1)∥f∥p≤const,
in view of the fact that p<p2/2(p-1) when 2N/(N+1)<p<2 for N≥1. Again from (2.22), we have ∥grad(|u|2-(2/p)sgnu)∥p≤const. Hence {f} bounded in Lp(Ω) implies that {|u|2-(2/p)sgnu} is bounded in W1,p(Ω).
We notice thatW1,p(Ω)↪↪Lp2/2(p-1)(Ω) when N≥2 and W1,p(Ω)↪↪CB(Ω) when N=1 by Lemma 1.3, hence{|u|2-(2/p)sgnu} is relatively compact in Lp2/2(p-1)(Ω). This gives that {u} is relatively compact in Lp(Ω) since the Nemytskii mapping u∈Lp2/2(p-1)(Ω)→|u|p/2(p-1)sgnu∈Lp(Ω) is continuous.
This completes the proof.Remark 2.10.
SinceΦp(u+α)=Φp(u), for any u∈W1,p(Ω) and α∈C0∞(Ω), we have f∈Apu implies that f=Bp,q,ru in the sense of distributions.Proposition 2.11.
Forf∈Lp(Ω), if there exists u∈Lp(Ω) such that f∈Apu, then u is the unique solution of (1.4).Proof.
First, we show that(2.24)-div[(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u]+λ|u|q-2u+λ|u|r-2u=f(x),a.e.x∈Ω
is available.
Nowf∈Apu implies that f=Bp,q,ru+∂Φp(u). For all φ∈C0∞(Ω), by Remark 2.10, we have
(2.25)(φ,f)=(φ,Bp,q,ru+∂Φp(u))=(φ,Bp,q,ru)=∫Ω‍〈(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u,∇φ〉dx+λ∫Ω‍|u|q-2uφdx+λ∫Ω‍|u|r-2uφdx=∫Ω‍-div[(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u]φdx+λ∫Ω‍|u|q-2uφdx+λ∫Ω‍|u|r-2uφdx,
which implies that (2.24) is true.
Secondly, we show that(2.26)-〈ϑ,(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u〉∈βx(u(x)),a.e.x∈Γ.
We will prove (2.26) under the additional condition |βx(u)|≤a|u|p/p′+b(x), where b(x)∈Lp′(Γ) and a∈R. Refer to the result of Brezis [11] for the general case.
Now, from (2.24), f∈Apu implies that f(x)=-div[(1+|∇u|p/1+|∇u|2p)|∇u|p-2∇u]+λ|u(x)|q-2u(x)+λ|u|r-2u∈Lp(Ω). By using Green's Formula, we have that for any v∈W1,p(Ω),
(2.27)∫Γ‍〈ϑ,(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u〉v|ΓdΓ(x)=∫Ω‍div[(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u]vdx+∫Ω‍〈(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u,∇v〉dx.
Then-〈ϑ,(1+|∇u|p/1+|∇u|2p)|∇u|p-2∇u〉∈W-(1/p),p′(Γ)=(W1/p,p(Γ))*, where W1/p,p(Γ) is the space of traces of W1,p(Ω).
Now let the mappingB:Lp(Γ)→Lp′(Γ) be defined by Bu=g(x), for any u∈Lp(Γ), where g(x)=βx(u(x)) a.e. on Γ. Clearly, B=∂Ψ where Ψ(u)=∫Γ‍φx(u(x))dΓ(x) is a proper, convex, and lower-semicontinuous function on Lp(Γ). Now define the mapping K:W1,p(Ω)→Lp(Γ) by K(v)=v|Γ for any v∈W1,p(Ω). Then K*BK:W1,p(Ω)→(W1,p(Ω))* is maximal monotone since both K,B are continuous. Finally, for any u,v∈W1,p(Ω), we have
(2.28)Ψ(Kv)-Ψ(Ku)=∫Γ‍[φx(v|Γ(x))-φx(u|Γ(x))]dΓ(x)≥∫Γ‍βx(u|Γ(x))(v|Γ(x)-u|Γ(x))dΓ(x)=(BKu,Kv-Ku)=(K*BKu,v-u).
Hence we get K*BK⊂∂Φp and so K*BK=∂Φp. Therefore, we have
(2.29)-〈ϑ,(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u〉∈βx(u(x)),a.e.onΓ.
Finally, we will show thatu is unique.
Iff∈Apu and f∈Apv, where u,v∈D(Ap). Then
(2.30)0≤(u-v,Bp,q,ru-Bp,q,rv)=(u-v,∂Φp(v)-∂Φp(u))≤0,
since Bp,q,r is strictly monotone and ∂Φp is maximal monotone, which implies that u(x)=v(x).
This completes the proof.Remark 2.12.
Ifβx≡0 for any x∈Γ, then ∂Φp(u)≡0, for all u∈W1,p(Ω).Proposition 2.13.
Ifβx≡0 for any x∈Γ, then {f∈Lp(Ω)∣∫Ω‍fdx=0}⊂R(Ap).Proof.
We can easily know thatR(Bp,q,r)=(W1,p(Ω))* in view of Lemmas 1.4 and 2.5. Note that for any f∈Lp(Ω) with ∫Ω‍fdx=0, the linear function u∈W1,p(Ω)→∫Ω‍fudx is an element of (W1,p(Ω))*. So there exists a u∈W1,p(Ω) such that
(2.31)∫Ω‍fvdx=∫Ω‍〈(1+|∇u|p1+|∇u|2p)|∇u|p-2∇u,∇v〉dx+λ∫Ω‍|u|q-2uvdx+λ∫Ω‍|u|r-2uvdx,
for any v∈W1,p(Ω). So f=Apu in view of Remark 2.12.
This completes the proof.Definition 2.14 (see [1, 7]).
Fort∈R,  x∈Γ, let βx0(t)∈βx(t) be the element with least absolute value if βx(t)≠∅ and βx0(t)=±∞, where t>0 or <0, respectively, in case βx(t)=∅. Finally, let β±(x)=limt→±∞βx0(t) (in the extended sense) for x∈Γ. β±(x) define measurable functions on Γ, in view of our assumptions on βx.Proposition 2.15.
Letf∈Lp(Ω) such that
(2.32)∫Γ‍β-(x)dΓ(x)<∫Ω‍fdx<∫Γ‍β+(x)dΓ(x).
Then f∈IntR(Ap).Proof.
Letf∈Lp(Ω) and satisfy (2.32), by Proposition 2.8, there exists un∈Lp(Ω) such that, for each n≥1,  f=(1/n)un+Apun. In the same reason as that in [1], we only need to prove that ∥un∥p≤const, for all n≥1.
Indeed, suppose to the contrary that1≤∥un∥p→∞, as n→∞. Let vn=un/∥un∥p. Let ψ:R→R be defined by ψ(t)=|t|p,  ∂ψ:R→R be its subdifferential and for μ>0,  ∂ψμ:R→R denote the Yosida-approximation of ∂ψ.   Let θμ:R→R denote the indefinite integral of [(∂ψμ)′]1/p with θμ(0)=0 so that (θμ′)p=(∂ψμ)′. In view of Calvert and Gupta [1], we have
(2.33)(∂ψμ(vn),∂Φp(un))≥∫Γ‍βx((1+μ∂ψ)-1(un|Γ(x)))×∂ψμ(vn|Γ(x))dΓ(x)≥0.
Now multiplying the equationf=(1/n)un+Apun by ∂ψμ(vn), we get that
(2.34)(∂ψμ(vn),f)=(∂ψμ(vn),1nun)+(∂ψμ(vn),Bp,q,run)+(∂ψμ(vn),∂Φp(un)).
Since ∂ψμ(0)=0, it follows that (∂ψμ(vn),un)≥0. Also, we can know that
(2.35)(∂ψμ(vn),Bp,q,run)=∫Ω‍〈(1+|∇un|p1+|∇un|2p)|∇un|p-2∇un,∇vn〉(∂ψμ)′(vn)dx+λ∫Ω‍|un|q-2un∂ψμ(vn)dx+λ∫Ω‍|un|r-2un∂ψμ(vn)dx≥∫Ω‍|∇u|p∥un∥p(∂ψμ)′(vn)dx=∥un∥pp-1∫Ω‍|grad(θμ(vn))|pdx.
Then we get from (2.33) that
(2.36)∥un∥pp-1∫Ω‍|grad(θμ(vn))|pdx+∫Γ‍βx((1+μ∂ψ)-1(un|Γ(x)))×∂ψμ(vn|Γ(x))dΓ(x)≤(∂ψμ(vn),f).
Since |∂ψμ(t)|≤|∂ψ(t)| for any t∈R and μ>0, we see from ∥vn∥p=1 for n≥1, that ∥∂ψμ(vn)∥p′≤C, for μ>0, where C is a constant which does not depend on n or μ and (1/p)+(1/p′)=1.
From (2.36), we have
(2.37)∫Ω‍|grad(θμ(vn))|pdx≤C∥un∥pp-1,forμ>0,n≥1.
Now we easily know that (θμ′)p=(∂ψμ)′→(∂ψ)′ as μ→0  a.e. on R.
Lettingμ→0, we see from Fatou's lemma and (2.37) that
(2.38)∫Ω|grad(|vn|2-(2/p)sgnvn)|pdx≤C∥un∥pp-1.
From (2.38), we know that |vn|2-(2/p)sgnvn→k (a constant) in Lp(Ω), as n→+∞.
Next we will show thatk≠0 is in Lp(Ω) from two aspects.(i)
Ifp≥2, since ∥|vn|2-(2/p)sgnvn∥p=∥vn∥2p-22-(2/p)≥∥vn∥p2-(2/p)=1, it follows that k≠0 in Lp(Ω),(ii)
if2N/(N+1)<p<2,  ∥|vn|2-(2/p)sgnvn∥p=∥vn∥2p-22-(2/p)≥∥vn∥p2-(2/p)=1, then {|vn|2-(2/p)sgnvn} is bounded in W1,p(Ω). By Lemma 1.3, W1,p(Ω)↪↪CB(Ω) when N=1 and W1,p(Ω)↪↪Lp2/2(p-1)(Ω), when N≥2. So {|vn|2-(2/p)sgnvn} is relatively compact in Lp2/2(p-1)(Ω). Then there exists a subsequence of {|vn|2-(2/p)sgnvn}, for simplicity, we denote it by {|vn|2-(2/p)sgnvn}, satisfying |vn|2-(2/p)sgnvn→g in Lp2/2(p-1)(Ω). Noticing that p≤p2/2(p-1) when 2N/(N+1)<p<2, it follows that k=g  a.e. on Ω. Now,
(2.39)1=∥vn∥pp=∫Ω‍||vn|2-(2/p)sgnvn|p2/2(p-1)dx≤const∫Ω‍||vn|2-(2/p)sgnvn-g|p2/2(p-1)dx+const∥g∥p2/2(p-1)p2/2(p-1),it follows that g≠0 in Lp(Ω) and then k≠0 in Lp(Ω). Assume, now, k>0, we see from (2.36) that
(2.40)∫Γ‍βx((1+μ∂ψ)-1(un|Γ(x)))×∂ψμ(vn|Γ(x))dΓ(x)≤(∂ψμ(vn),f).
Choosing a subsequence so that un|Γ(x)→+∞ a.e. on Γ, we see letting n→+∞ that ∫Γ‍β+(x)dΓ(x)≤∫Ω‍f(x)dx, which is a contradiction with (2.32). Similarly, if k<0, it also leads to a contradiction. Thus f∈intR(Ap).
This completes the proof.Proposition 2.16.
A
p
+
B
1
:
L
p
(
Ω
)
→
L
p
(
Ω
) is m-accretive and has a compact resolvent.Proof.
Using a theorem in Corduneanu [12], we know that Ap+B1:Lp(Ω)→Lp(Ω) is m-accretive.
To show thatAp+B1:Lp(Ω)→Lp(Ω) has a compact resolvent, we only need to prove that if w∈Apu+B1u with {w} and {u} being bounded in Lp(Ω), then {u} is relatively compact in Lp(Ω). Now we discuss it from two aspects.(i)
Ifp≥2, since
(2.41)∫Ω‍|∇u|pdx≤(u,Bp,q,ru)=(u,Apu)-(u,∂Φp(u))≤(u,Apu)+(u,B1u)=(u,w)≤∥u∥p∥u∥p′≤const,
it follows that {u} is bounded in W1,p(Ω), where (1/p)+(1/p')=1. Then {u} is relatively compact in Lp(Ω) since W1,p(Ω)↪↪Lp(Ω);(ii)
if2N/(N+1)<p<2, since w∈Apu+B1u with {w} and {u} being bounded in Lp(Ω), we have w-B1u∈Apu with {w-B1u} and {u} being bounded in Lp(Ω) which gives that {u} is relatively compact in Lp(Ω) since Ap is m-accretive by Proposition 2.8 and has a compact resolvent by Lemma 2.9.
This completes the proof.Theorem 2.17.
Letf∈Lp(Ω) be such that
(2.42)∫Γ‍β-(x)dΓ(x)+∫Ω‍g-(x)dx<∫Ω‍f(x)dx<∫Γ‍β+(x)dΓ(x)+∫Ω‍g+(x)dx,
then (1.4) has a unique solution in Lp(Ω), where 2N/(N+1)<p<+∞ and N≥1.Proof.
We want to use Theorem1.9 to finish our proof. From Propositions 1.7, 2.2, 2.8, and 2.16, we can see that all of the conditions in Theorem 1.9 are satisfied. It then suffices to show that f∈int[R(Ap)+R(B1)] which ensures that f∈R(Ap+B1+B2). Thus Proposition 2.11 tells us (1.4) has a unique solution in Lp(Ω).
Using the similar methods as those in [2, 4, 7], by dividing it into two cases and using Propositions 2.13 and 2.15, respectively, we know that f∈int[R(Ap)+R(B1)].
This completes the proof.Remark 2.18.
Compared to the work done in [1–7], not only the existence of the solution of (1.4) is obtained but also the uniqueness of the solution is obtained.



---

*Source: 154307-2012-12-31.xml*