nonuniformly elliptic equations of p-laplacian type

Nonlinear Analysis 61 (2005) 1483–1495www.elsevier.com/locate/na

Nonuniformly elliptic equations ofp-Laplacian type

Duong Minh Duc∗, Nguyen Thanh VuDepartment of Mathematics and Computer Sciences, National University of Hochiminh City, 227 Nguyen Van

Cu, Q5, Hochiminh City, Vietnam

Received 1 October 2004; accepted 9 February 2005

Abstract

This paper deals with the existence of a generalized solution inW1,p0 (�) to a nonuniformly non-

linear elliptic equation of the form−div(a(x, ∇u)) = f (x, u) in a bounded domain� of Rn. Here

a satisfies|a(x, �)|�c0[h0(x) + h1(x)|�|p−1] for any � in Rn, a.e.x ∈ �, h0 ∈ Lp

p−1 (�), andh1 ∈ L1

loc(�).� 2005 Elsevier Ltd. All rights reserved.

MSC:35J20; 35J60; 58E05

Keywords: p-Laplacian; Nontrivial solution; Nonuniform elliptic equations

1. Introduction

Let� be a bounded domain inRn. In the present paper we study the existence of nontrivialsolutions of the following Dirichlet elliptic problem:

(P)

{−div(a(x, ∇u(x))) = f (x, u(x)) in �,

u = 0 on ��,

where|a(x, �)|�c0[h0(x) + h1(x)|�|p−1] for any� in Rn and a.e.x ∈ �, h0(x)�0 andh1(x)�1 for anyx in �.

∗ Corresponding author.E-mail address:[email protected](D.M. Duc).

0362-546X/$ - see front matter� 2005 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2005.02.049

http://www.elsevier.com/locate/na

mailto:[email protected]

1484 D.M. Duc, N. Thanh Vu / Nonlinear Analysis 61 (2005) 1483–1495

If h0 andh1 belong toL∞(�), the problem has been studied in[2–6] and the referencestherein. Here we study the case in whichh0 andh1 belong toLp/(p−1)(�) andL1

loc(�),respectively. The equation now may be nonuniformly elliptic.

A prototype of(P ) is the following nonuniformly elliptic equation{−div(h(x)|∇u|p−2∇u) = f (x, u(x)) in �,

u = 0 on ��,

whereh ∈ L1loc(�).

We reduce the problem to the uniform one by using appropriate weighted Sobolev spaces.Then combining a variation of the mountain-pass theorem in[7] with the results in[5], weprove that problem (P) admits a nontrivial generalized solution in the next section.

In order to state our main theorem, let us introduce our hypotheses on the structureof (P).

Let p be in (1, +∞) and� be a bounded domain inRn havingC2 boundary��. Let

A be a measurable function on� × Rn such thatA(x, 0) = 0 anda(x, �) ≡ �A(x,�)

�� is

a Carathéodory function on� × Rn. Assume that there are positive real numbersc0, k0,k1 and two nonnegative measurable functionsh1, h0 on � such thath1 ∈ L1

loc(�), h0 ∈Lp/(p−1)(�), h1(x)�1 for a.e.x in � and the following conditions hold:

(A1) |a(x, �)|�c0(h0(x) + h1(x)|�|p−1) ∀� ∈ Rn, a.e.x ∈ �.(A2) A is p-uniformly convex, that is,

A(x, t� + (1 − t)�) + k1h1(x)|� − �|p � tA(x, �) + (1 − t)A(x, �),

∀(�, �, t) ∈ Rn × Rn × [0, 1], a.e.x ∈ �.

(A3) A is p-subhomogeneous:

0�a(x, �) · ��pA(x, �) ∀� ∈ Rn, a.e.x ∈ �.

(A4) A(x, �)�k0h1(x)|�|p ∀� ∈ Rn, a.e.x ∈ �.

Regarding the functionf, we assume thatf is a real Carathéodory function on� × R

having the following properties

(F1) |f (x, s)|�c1(1 + |s|q−1) ∀s ∈ R, a.e.x ∈ �, wherec1 is a positive real number,q ∈ (p, +∞) if p�n, andq ∈ (p, p∗) with p∗ = np/(n − p) if p < n.

(F2) There are a constant� > p and a positive real numbers0 such that

0< �F (x, s)�f (x, s)s ∀s ∈ R\(−s0, s0), a.e.x ∈ �,

whereF (x, s) = ∫ s

0 f (x, t) dt .(F3) Let�1 be defined in (2.1). There are� ∈ (0, k0p�1) and a positive real number� such

that

f (x, s)

|s|p−2s�� ∀s ∈ (−�, �)\{0}, a.e.x ∈ �.

D.M. Duc, N. Thanh Vu / Nonlinear Analysis 61 (2005) 1483–1495 1485

Our main theorem is

Theorem 1.1. Under conditions(A1)–(A4) and (F1)–(F3),there exists at least one non-trivial generalized solution inW 1,p

0 (�) to the problem(P).

To see the power of this theorem we compare our assumptions to those considered in[5].Our problem (P) covers the following cases which have been considered in literatures[6,5,p. 1212]:

(a) A(x, �) = 1p|�|p with p�2.

(b) A(x, �) = 1p[(1 + |�|2)p/2 − 1] with p�2.

Moreover, our assumption includes the following situations which could not be handledby [5,6]:

(a) A(x, �) = h(x)p

|�|p with p�2, h ∈ L1loc(�),

(b) A(x, �) = h(x)p

[(1 + |�|2)p/2 − 1] with p�2, h ∈ Lp/(p−1)(�).

2. Proof of Theorem 1.1

We introduce some notations

X ={

u ∈ W1,p0 (�) :

∫�

h1(x)|∇u|p dx < + ∞}

,

‖u‖ =(∫

�|∇u|p dx

)1/p

∀u ∈ W1,p0 (�),

‖u‖X =(∫

�h1(x)|∇u|p dx

)1/p

∀u ∈ X,

�1 = inf

{‖u‖p

X∫� |u|p dx

: u ∈ X\{0}}

. (2.1)

We also need the following notations

E(u) =∫�

A(x, ∇u) dx ∀u ∈ W1,p0 (�),

T (u) =∫�

F (x, u) dx ∀u ∈ W1,p0 (�),

J (u) = E(u) − T (u) ∀u ∈ W1,p0 (�),

‖DJ(u)‖ = sup{|DJ(u)(v)| : v ∈ X, ‖v‖X = 1}.


Remark 2.1. (a)‖u‖X �‖u‖ for anyu ∈ X sinceh1(x)�1 for a.e.x ∈ �.(b) By (A4) and (i) in Lemma 2.3, it is easy to see that

X = {u ∈ W1,p0 (�) : E(u) < + ∞} = {u ∈ W

1,p0 (�) : J (u) < + ∞}.

(c) �1 > 0 by Poincaré inequality, and∫�

|u|p dx � 1

�1‖u‖p

X, ∀u ∈ X.

(d) We have∫� h1(x)|∇v|p dx < ∞ for anyv ∈ C∞

c (�), because|∇v| is in Cc(�) andh1 ∈ L1

loc(�). HenceC∞c (�) is contained inX.

Our main tool is Lemma 2.6 which is a variation of the mountain-pass theorem. In thislemma we use the following concept of weakly continuous differentiability.

Definition 2.2. Let J be a map from a Banach spaceY into R. We say thatJ is weaklycontinuously differentiable onY if and only if following two conditions are satisfied:

(i) For anyu ∈ Y there exists a linear mapDJ(u) fromY into R such that

limt→0

J (u + tv) − J (u)

t= DJ(u)(v) ∀v ∈ Y .

(ii) For anyv ∈ Y , the mapu �→ DJ(u)(v) is continuous onY.

We list here some properties ofA, F andE.

Lemma 2.3. (i) A verifies the growth condition:

|A(x, �)|�c0[h0(x)|�| + h1(x)|�|p] ∀� ∈ Rn, a.e. x ∈ �.

(ii) E(tu + (1 − t)z ) + k1‖u − z‖pX � tE(u) + (1 − t)E(z) ∀u, z ∈ X, t ∈ [0, 1].

(iii) There exists a constantc2 such that

|F (x, s)|�c2(1 + |s|q) ∀s ∈ R, a.e. x ∈ �.

(iv) There exists� ∈ L∞(�) such that�(x) > 0 for a.e. x in� and

F (x, s)��(x)s� ∀s ∈ [s0, ∞), a.e. x ∈ �.

Proof. (i) By (A1) we have

|A(x, �)| =∣∣∣∣∫ 1

0

d

dtA(x, t�) dt

∣∣∣∣ =∣∣∣∣∫ 1

0a(x, t�) · � dt

∣∣∣∣ �∫ 1

0|a(x, t�)| |�| dt

�∫ 1

0c0|h0(x) + h1(x)|�|p−1tp−1| |�| dt �c0[h0(x)|�| + h1(x)|�|p]

(ii) This follows easily from (A2).


(iii) This follows easily from (F1).(iv) For anyx in �, put�(x) = s−�

0 F (x, s0). From (F2),�(x) > 0. By (iii) we have

�(x)�c2(1 + |s0|q)s−�0 a.e.x ∈ �.

By (F2) we have

�t� f (s, t)

F (x, t)= 1

F (x, t)

�F

�t(x, t) ∀t ∈ [s0, +∞), a.e. x ∈ �.

Integrating the above inequality froms0 to swith s �s0, we get

ln

(s

s0

)�

� ln

(F (x, s)

F (x, s0)

)or F (x, s)�s−�

0 F (x, s0)s� = �(x)s�.

The proof is completed. �

The following lemma concerns the smoothness of the functionsT andE.

Lemma 2.4. Let the assumptions of Theorem1.1hold, and X be endowed with the norm‖.‖X.We assert that

(i) T is continuous on X. Moreover, T is weakly continuously differentiable on X and

DT (u)(v) =∫�

f (x, u)v dx ∀u, v ∈ X.

(ii) If {um} is a sequence weakly converging to u inW1,p0 (�), thenT (u)= limm→∞ T (um)

andE(u)� lim inf m→∞ E(um).(iii) E is continuous on X.(iv) E is weakly continuously differentiable on X and

DE(u)(v) =∫�

a(x, ∇u) · ∇v dx ∀u, v ∈ X.

(v) E(u) − E(v)�DE(v)(u − v) ∀u, v ∈ X.

Proof. (i) By Proposition 6 in[6, p. 354]and (F1), the functionT is continuously Fréchetdifferentiable onLq(�) and

DT (u)(v) =∫�

f (x, u)v dx ∀u, v ∈ Lq(�).

By Remark 2.1(a) and the Sobolev embedding theorem (see[1, p. 97]) we obtain (i).(ii) Let {um} be a sequence weakly converging tou in W

1,p0 (�). Arguing as in (i), we get

thatT is continuous onLq(�). By the Rellich–Kondrachov theorem (see[1, p. 144]) wededuce limm→∞ T (um) = T (u).

Applying the Rellich–Kondrachov theorem again, we see that{um} converges stronglyto u in L1(�). By Theorem 4.5 in[8, p. 129]or Theorem 1.6 in[9, p. 9]we have

E(u)� lim infm→∞ E(um).


(iii) Let {wm} be a sequence converging tow in X. We shall prove that there is asubsequence{wmj

}j of {wm} such that{E(wmj)}j converges toE(w). Indeed, since

{h1/p1 |∇wm|}m converges toh1/p

1 |∇w| in Lp(�), there exists a subsequence{h1/p1 |∇wmj

|}j

such that||h1/p1 (|∇wmj+1| − |∇wmj

|)||Lp � 2−j for any positive integerj. Put

g ={

h1/p1 |∇wm1| +

∞∑1

h1/p1 ||∇wmj+1| − |∇wmj

||}p

.

We see thatg is in L1(�) and|∇wmj|p �h1|∇wmj

|p �g a.e. in� for anyj ∈ N. Hence,for anyj ∈ N, a.e.x ∈ � we have

|A(x, ∇wmj(x))|�c0[h0(x)|∇wmj

(x)| + h1(x)|∇wmj(x)|p]

�c0[h0(x)g1/p + g(x)],where the last term on the right-hand side is integrable on�.

On the other hand,{∇wmj(x)}j converges to∇w(x) for almost everywherex in � since

{h1/p1 |∇wmj

|}j converges toh1/p1 |∇w| in Lp(�). Hence, by the Carathéodory property of

A we see that{A(x, ∇wmj(x)} converges toA(x, ∇w(x)) for almost everywherex in �.

By the Lebesgue Dominated convergence theorem, we conclude that

limj→∞

∫�

A(x, ∇wmj) dx =

∫�

A(x, ∇w) dx.

Using the above property ofE, we get (iii).(iv) Firstly we prove

DE(u)(v) =∫�

a(x, ∇u) · ∇v dx ∀u, v ∈ X.

We have for anyu, v in X, anyt in (−1, 1) and anyx in �∣∣∣∣A(x, ∇u(x) + t∇v(x)) − A(x, ∇u(x))

t

∣∣∣∣=

∣∣∣∣∫ 1

0a(x, ∇u(x) + t∇v(x)) · ∇v(x) d

∣∣∣∣�

∫ 1

0c0[h0(x) + h1(x)|∇u(x) + t∇v(x)|p−1]|∇v(x)| d

�c0[h0(x) + h1(x)(|∇u(x)| + |∇v(x)|)p−1]|∇v(x)|�c0h0(x)|∇v(x)|+c0[h1/p

1 (x)|∇v(x)|](h1/p1 (x)|∇u(x)|+h

1/p1 (x)|∇v(x)|)p−1.

Sincehp/(p−1)0 , |∇v|p, h1|∇u|p andh1|∇v|p are integrable on�, the last term on the

right-hand side is integrable on�.Applying the Lebesgue Dominated convergence theorem,we see that

DE(u)(v) = limt→0

∫�

A(x, ∇u + t∇v) − A(x, ∇u)

tdx =

∫�

a(x, ∇u) · ∇v dx.


Fix a vectorv in X. We now prove that the mapu �→ DE(u)(v) is continuous onX.Let w be inX. Suppose by contradiction that the mapu �→ DE(u)(v) is not continuous

at w. Then, there exist a positive and a sequence{wm} in X such that‖wm − w‖X → 0and|DE(wm)(v) − DE(w)(v)| > for anym ∈ N.

Arguing as in (iii) we can find a subsequence{wmj} of {wm} and a functiong in L1(�)

such that|∇wmj|p �h1|∇wmj

|p �g a.e. in� for any j ∈ N. Therefore, for anyj ∈ N,a.e.x ∈ � we get

|a(x, ∇wmj(x))| |∇v(x)|

�c0[h0(x) + h1(x)|∇wmj(x)|p−1]|∇v(x)|

= c0h0(x)|∇v(x)| + c0(h1(x)|∇wmj(x)|p)(p−1)/ph

1/p1 |∇v(x)|

�c0h0(x).|∇v(x)| + c0|g|(p−1)/p(x).h1/p1 (x)|∇v(x)|.

Sinceh0, g(p−1)/p ∈ Lp/(p−1)(�) and |∇v|, h1/p1 |∇v| ∈ Lp(�), the last term on the

right-hand side is integrable on�. On the other hand,∇wmj(x) → ∇w(x) for a.e.x ∈ �,

so thata(x, ∇wmj(x)) ·∇v(x) → a(x, ∇w(x)) ·∇v(x) for a.e.x ∈ � by the Carathéodory

property ofa. By the Lebesgue Dominated convergence theorem, we have∫�

a(x, ∇wmj) · ∇v dx →

∫�

a(x, ∇w) · ∇v dx, i.e.,

DE(wmj)(v) → DE(u)(v).

This contradiction implies the mapu �→ DE(u)(v) is continuous onX. HenceE isweakly differentiable atu.

(v) By (ii) of Lemma 2.3 we obtain the convexity ofE, which implies

E(um + t (u − um)) − E(um)

t= E((1 − t)um + tu) − E(um)

t

� (1 − t)E(um) + tE(u) − E(um)

t

= E(u) − E(um) ∀t ∈ (0, 1).

Lettingt → 0, we haveDE(um)(u−um)�E(u)−E(um). This proves (v) and concludesthe proof of our lemma. �

The following lemma concerns the coercivity ofJ.

Lemma 2.5. (i) There existk3 > 0 andc3 > 0 such that

J (u)�‖u‖pX(k3 − c3‖u‖q−p

X ) ∀u ∈ X.

(ii) There existc4 > 0, k4 ∈ R such that

J (u)�‖u‖X

(c4‖u‖p−1

X − 1

�‖DJ(u)‖

)+ k4 ∀u ∈ X.


Proof. (i) By (F3) we have for a.e.x ∈ �

f (x, s)

{��sp−1 if s ∈ (0, �),

� − �|s|p−1 if s ∈ (−�, 0).

It follows that

F (x, s)� �p

|s|p ∀s ∈ (−�, �), a.e. x ∈ �.

On the other hand, by (iii) of Lemma 2.3 there exists a positive constantc′2 such that

|F (x, s)|�c′2|s|q ∀s ∈ R\(−�, �), a.e. x ∈ �.

Hence

F (x, s)� �p

|s|p + c′2|s|q ∀s ∈ R, a.e.x ∈ �.

Furthermore, by the Sobolev embedding theorem there is a positive real numberc3 suchthat

T (u) =∫�

F (x, u) dx � �p

∫�

|u|p dx + c′2

∫�

|u|q dx

� �p�1

‖u‖pX + c3‖u‖q

X ∀u ∈ X.

Putk3 = k0 − �p�1

> 0, for anyu ∈ X we have

J (u) = E(u) − T (u) =∫�

A(x, ∇u) dx −∫�

F (x, u) dx

�k0

∫�

h1(x)|∇u|p dx − �p�1

‖u‖pX − c3‖u‖q

X = k3‖u‖pX − c3‖u‖q

X.

(ii) Put c4 = k0(1 − p

� ) > 0. From (iv) of Lemma 2.4 and (A3), we infer

E(u) − 1

�DE(u)(u) =

∫�

A(x, ∇u) dx − 1

�

∫�

a(x, ∇u) · ∇u dx

�(1 − p

�

) ∫�

A(x, ∇u) dx �(1 − p

�

)k0

∫�

h1|∇u|p dx

= c4‖u‖pX ∀u ∈ X.

We put�u = {x ∈ � : |u(x)| > s0} for anyu ∈ X. By (F1), (F2) and (iii) of Lemma 2.3,there exists a constantM such that

1

�f (x, u)u − F (x, u)�0 a.e.x ∈ �u

and ∣∣∣∣1

�f (x, u)u − F (x, u)

∣∣∣∣ �M, a.e.x ∈ �\�u.


Putk4 = −M|�|. From (i) of Lemma (2.4), it follows that

1

�DT (u)(u) − T (u) =

∫�u

[1

�f (x, u)u − F (x, u)

]dx

+∫�\�u

[1

�f (x, u)u − F (x, u)

]dx

� − M|�\�u|� − M|�| = k4.

Hence,

J (u) − 1

�DJ(u)(u) =

[E(u) − 1

�DE(u)(u)

]+

[1

�DT (u)(u) − T (u)

]�c4‖u‖p

X + k4,

or J (u)�c4‖u‖pX + 1

�DJ(u)(u) + k4�c4‖u‖p

X − 1

�‖DJ(u)‖‖u‖X + k4.

This proof is completed. �

Our main tool is a variation of the following mountain-pass lemma introduced in[7].This lemma is an application of Theorem 2.1 in[7, p. 433]with f = J andE� = F = X

for any� in D, where we use the notations of Section 2 in[7].

Lemma 2.6(Mountain-pass lemma). Let J be a continuous function from a Banach space(X, ‖u‖X) into R. Let J be weakly continuously differentiable on X and satisfy thePalais–Smale condition. Assume thatJ (0) = 0 and there exist a positive real number randz0 ∈ X such that‖z0‖X > r, J (z0)�J (0) and

� ≡ inf {J (u) : u ∈ X, ‖u‖X = r} > 0.

Put G = { ∈ C([0, 1]), X) : (0) = 0, (1) = z0}. Assume thatG �= ∅. Set� =inf {maxJ ( ([0, 1])) : ∈ G}.Then�� and� is a critical value of J.

However, in order to apply Lemma 2.6 we need to verify the following facts.

Lemma 2.7. Let the assumptions of Theorem1.1hold, and X be endowed with the norm‖.‖X.We have

(i) (X, ‖.‖X) is a Banach space.(ii) J is a continuous function from X toR.

(iii) J is weakly continuously differentiable on X and

DJ(u)(v) =∫�

a(x, ∇u) · ∇v dx −∫�

f (x, u)v dx ∀u, v ∈ X.

(iv) J satisfies the Palais–Smale condition on X.(v) J (0) = 0.


(vi) There exist two positive real number r and� such that

inf {J (u) : u ∈ X, ‖u‖X = r}��.

(vii) There existsz0 ∈ X such that‖z0‖X > r andJ (z0)�0.(viii) The setG ≡ { ∈ C([0, 1]; X) : (0) = 0, (1) = z0} is not empty.

Proof. (i) It is clear thatX is a normed space. Let{um} be a Cauchy sequence inX. Then{‖um‖X}m is bounded and

limm→∞ lim inf

j→∞

∫�

h1|∇uj − ∇um|p dx = 0.

By Remark 2.1(a),{um} is a Cauchy sequence inW 1,p0 (�) and converges to someu in

W1,p0 (�). Therefore{∇um(x)} converges to∇u(x) for a.e.x in �. Applying Fatou’s lemma

we get∫�

h1(x)|∇u|p dx � lim infm→∞

∫�

h1(x)|∇um|p dx = lim infm→∞ ‖um‖p

X < ∞.

Henceu is inX. Applying again Fatou’s lemma we have

limm→∞

∫�

h1(x)|∇u − ∇um|p dx � limm→∞

[lim inf"→∞

∫�

h1(x)|∇uj − ∇um|p dx

]= 0.

Hence{um} converges tou in X. Thus,X is a Banach space.(ii) This comes from (i), (iii) of Lemma 2.4.(iii) This comes from (i), (iv) of Lemma 2.4.(iv) Let {um} be a sequence inX and� be a real number such that limm→∞ J (um) = �

and limm→∞ ‖DJ(um)‖ = 0.Suppose by contradiction that{‖um‖X} is not bounded, then there exists a subsequence

{umj}j of {um} such that‖umj

‖X �j for any j in N. By (ii) of Lemma 2.5, we have

J (umj)�‖umj

‖X

(c4‖umj

‖p−1X − 1

�‖DJ(umj

)‖)

+ k4.

Letting j → ∞, we deduceJ (umj) → ∞, which is a contradiction. Hence{‖um‖X} is

bounded, so that{‖um‖} is also bounded by Remark 2.1(a).Therefore we can (and shall) assume that the sequence{um} converges weakly to some

u in W1,p0 (�). By (ii) of Lemma 2.4 we have

E(u)� limm→∞ E(um) = lim

m→∞(T (um) + J (um)) = T (u) + �.

By Remark 2.1(b) we see thatu ∈ X. Hence{‖um −u‖X} is bounded. Since{‖DJ(um)‖}converges to 0,{DJ(um)(u − um)} converges to 0. Moreover, the sequence{um} convergesstrongly tou in Lq(�) by Rellich–Kondrachov theorem (see[1, p. 144]), andf (x, um) isbounded inLq ′

(�) by the condition(F1) with q ′ = q/(q − 1), so that

limm→∞ DT (um)(u − um) = lim

m→∞

∫�

f (x, um)(u − um) dx = 0.


Hence

limm→∞ DE(um)(u − um) = lim

m→∞[DJ(um)(u − um) + DT (um)(u − um)] = 0,

which together with (v) of Lemma 2.4 imply that

E(u) − limm→∞ E(um) = lim

m→∞[E(u) − E(um)]� limm→∞ DE(um)(u − um) = 0.

Combining this fact with (ii) of Lemma 2.4, we get

limm→∞ E(um) = E(u).

Suppose by contradiction that{um} does not converge strongly tou in X. Then there exista positive real number and a subsequence{umj

} of {um} such that‖umj− u‖X � for any

j ∈ N. By (ii) of Lemma 2.3 we have

12 E(umj

) + 12E(u) − E

(umj

+ u

2

)�k1‖umj

− u‖pX �k1p.

Since limj→∞ E(umj) = E(u), the above inequality implies that

E(u) − lim infj→∞ E

(umj

+ u

2

)

= lim supj→∞

[1

2E(umj

) + 1

2E(u) − E

(umj

+ u

2

)]�k1p.

On the other hand, by (ii) of Lemma 2.4,E(u)� lim inf j→∞ E(umj

+u

2 ) because{umj+u

2 }converges weakly tou. Hence 0�k1p, which is a contradiction. Therefore,{um} convergesstrongly tou in X. Thus,J satisfies the Palais–Smale condition onX.

(v) SinceE(0) = 0 andT (0) = 0, we haveJ (0) = 0.(vi) Sincek3 > 0 andq > p, there exists a positive constantr such that

� = rp(k3 − c3rq−p) > 0.

Let u be inX. By (i) of Lemma 2.5 we have

J (u)�‖u‖pX(k3 − c3‖u‖q−p

X ) = rp(k3 − c3rq−p) = � when‖u‖X = r.

(vii) Let t > 1. Choosev0 in C∞c (�) such thatv0(x)�0 for a.e.x ∈ � andV= {x ∈ � :

v0(x)�s0} has a positive measure. By the condition (F2),F (x, v0(x)) > 0 if x ∈ V . PutVt := {x ∈ � : tv0(x)�s0}. Then it is easy to see thatV ⊂ Vt . By (iv) of Lemma 2.3 wehave ∫

Vt

F (x, tv0) dx �∫

Vt

�(x)(tv0(x))� dx = t�∫

Vt

�v�0 dx � t�

∫V

�v�0 dx = t�I (v0),

whereI (v0) = ∫V

�v�0 dx > 0.


By (iii) of Lemma 2.3 there exists a positive constantM ′ such that|F (x, s)|�M ′ forany s ∈ [0, s0], a.e.x ∈ �. Moreover, by Remark 3.4 in[5, p. 1213 ], the condition (F2)implies that

F (x, ts)�F (x, s)t� ∀s ∈ R\(−s0, s0), a.e.x ∈ �.

On the other hand, by Remark 3.3 in[5, p. 1212 ] the condition (A3) implies thatA(x, t�)�A(x, �)tp for every� ∈ Rn, a.e.x ∈ �. So,E(tv0)� tpE(v0).

Hence

J (tv0) = E(tv0) −∫

Vt

F (x, tv0) dx −∫�\Vt

F (x, tv0) dx

� tpE(v0) − t�I (v0) +∫�\Vt

M ′ dx

� tpE(v0) − t�I (v0) + M ′|�|.Since� > p, we deduceJ (tv0) → −∞ ast → +∞. Hence, there existst1 such that

‖t1v0‖X > r andJ (t1v0)�0. Choosez0 = t1v0, we have‖z0‖X > r andJ (z0)�0.(viii) We consider a function in C[0, 1], X) defined by (t) = tz0 for everyt ∈ [0, 1].

It is clear that ∈ G. Thus,G �= ∅.The proof is completed. �

We are now ready to prove our main result.

Proof of Theorem 1.1. By Lemmas 2.6 and 2.7, there is anu0 in X such that{0< ��J (u0) = inf {maxJ ( ([0, 1]) : ∈ G},DJ(u0)(v) = 0, ∀v ∈ X.

Hence∫�

a(x, ∇u0) · ∇v dx =∫�

f (x, u0)v dx ∀v ∈ X.

We haveu0 �= 0 sinceJ (u0) > 0 = J (0). Moreover,C∞c (�) ⊂ X by Remark 2.1(d).

Henceu0 is a nontrivial generalized solution of (1.1). The proof is completed.�

Remark 2.8. In the proof of Theorem 2.2 we only need the weak continuity and weaklycontinuous differentiability ofT, we therefore can further weaken the condition (F1). Forthe case in whichh1(x) = |x|−ap we can find a convenient condition off in [10].

References

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Methods Appl. 28 (1997) 1623–1632.


[3] Ph. Clement, M. Garcia-Huidobro, R. Manasevich, K. Schmitt, Mountain pass type solutions for quasilinearelliptic equations, Calc. Var. Partial Differential Equations 11 (2000) 33–62.

[4] D.G. Costa, C.A. Magalhaes, Existence results for perturbations of thep-Laplacian, Nonlinear Anal. TheoryMethods Appl. 24 (1995) 409–418.

[5] P. De Nápoli, M.C. Mariani, Mountain pass solutions to equations ofp-Laplacian type, Nonlinear Anal. 54(2003) 1205–1219.

[6] G. Dinca, P. Jebelean, J. Mawhin, Variational and topological methods for Dirichlet problems withp-Laplacian, Protugaliae Math. 58 (2001) 340–377.

[7] D.M. Duc, Nonlinear singular elliptic equations, J. London Math. Soc. (2) 40 (1989) 420–440.[8] E. Giusti, Direct Methods in the Calculus of Variations, World Scientific, New Jersey, 2003.[9] M. Struwe, Variational Methods, Springer, New York, 1996.

[10] B. Xuan, The solvability of Brezis–Nirenberg type problems of singular quasilinear elliptic equation, preprint(arXiv: Math.AP/0403549v1 31 March 2004).

nonuniformly elliptic equations of p-laplacian type

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