DSpace at VNU: A MULTIPLICITY RESULT FOR A CLASS OF EQUATIONS OF p-LAPLACIAN TYPE WITH SIGN-CHANGING NONLINEARITIES

The authors obtained the existence of a weak solution by using a variant of the... Mountain pass theorem introduced in [3].. Using the Mountain pass theorem in [3] combined with Ekeland’

doi:10.1017/S001708950900514X Printed in the United Kingdom

A MULTIPLICITY RESULT FOR A CLASS OF EQUATIONS

OF p-LAPLACIAN TYPE WITH SIGN-CHANGING

NONLINEARITIES

NGUYEN THANH CHUNG

Department of Mathematics and Informatics, Quang Binh University, 312 Ly Thuong Kiet,

Dong Hoi, Quang Binh, Vietnam e-mail: ntchung82@yahoo.com

and QU ˆO´ C ANH NG ˆO

Department of Mathematics, College of Science, Vietnam National University, Hanoi, Vietnam Department of Mathematics, National University of Singapore, Science Drive 2, Singapore 117543

e-mail: bookworm vn@yahoo.com

(Received 11 July 2008; accepted 22 December 2008)

Abstract. Using variational arguments we study the non-existence and multi-plicity of non-negative solutions for a class equations of the form

−div(a(x, ∇u)) = λf (x, u) in ,

where  is a bounded domain in ⺢ N , N  3, f is a sign-changing Carath´eodory

function on × [0, +∞) and λ is a positive parameter.

2002 Mathematics Subject Classification 35J20, 35J60, 35J65, 58E05.

1 Introduction. This paper deals with the non-existence and multiplicity of non-negative, non-trivial solutions to the following problem,

where is a bounded domain in ⺢ N , function a satisfies

|a(x, ξ)|  c0(h0(x) + h1(x) |ξ| p−1) for all ξ ∈ ⺢ N , a.e x ∈ , h0(x)  0, h1(x)  1 for a.e x ∈ , p  2 and λ > 0 is a parameter When h0 and h1belong to L∞(), the problem has been studied by many

authors (see [2, 8, 10] for details) Here we study the situation that h0∈ L−1p () and

h1∈ L1

loc() Then problem (1.1)–(1.2) now may be non-uniform in the sense that the

functional associated to the problem may be infinity for some u.

We point out the fact that in [4, 13], D M Duc and N T Vu have studied the

following Dirichlet elliptic problem,

where the nonlinear term f verifies the so-called Ambrosetti–Rabinowitz condition.

The authors obtained the existence of a weak solution by using a variant of the

Mountain pass theorem introduced in [3] Then, H Q Toan and Q.-A Ng ˆo [12] gave

some multiplicity results in the case when f (x , u) = h(x)|u| r−1u + g(x)|u| s−1u Using

the Mountain pass theorem in [3] combined with Ekeland’s variational principle in [5]

they proved that problem (1.3)–(1.4) has at least two weak solutions

Motivated by K Perera [10] and M Mih˘ailescu and V R˘adulescu [7], the goal of

this work is to investigate the problem (1.1)–(1.2) with positive parameterλ and the

sign-changing nonlinearity f We also do not require that the nonlinear term f verifies

the Ambrosetti–Rabinowitz condition as in [4, 12].

In order to state our main result, let us introduce the following hypotheses on problem (1.1)–(1.2)

Assume that N  3 and 2  p < N Let  be a bounded domain with a

smooth boundary∂ Consider that a :  × ⺢ N → ⺢N , a = a(x, ξ), is the continuous

derivative with respect toξ of the continuous function A :  × ⺢ N → ⺢, A = A(x, ξ), that is, a(x , ξ) = ∂A(x, ξ)/∂ξ and A(x, 0) = 0 for a.e x ∈  Assume that there are

positive constant c0 and two non-negative measurable functions h0, h1 such that

h0∈ L p/p−1(), h1 ∈ L1

loc(), h1(x)  1, a.e x ∈  Suppose that a and A satisfy the

following hypotheses

(A 1) |a(x, ξ)|  c0(h0(x) + h1(x) |ξ| p−1) for allξ ∈ ⺢ N , a.e x ∈ .

(A 2) The following inequality holds

0 (a(x, ξ) − a(x, ψ)) · (ξ − ψ)

for allξ, ψ ∈ ⺢ N , a.e x ∈ , with equality if and only if ξ = ψ.

(A 3) There exists a positive constant k0such that

A



x , ξ + ψ

2



1

2A(x , ξ) + 1

2A(x , ψ) − k0h1(x) |ξ − ψ| p

for allξ, ψ ∈ ⺢ N , a.e x ∈ , that is, A is p-uniformly convex.

(A 4) There exists a constant k1> 0 such that the following inequalities hold true

k1h1(x) |ξ| p  a(x, ξ) · ξ  pA(x, ξ)

for allξ ∈ ⺢ N , a.e x ∈ .

EXAMPLE1

1 Let

A(x , ξ) = h(x)

p |ξ| p , a(x, ξ) = h(x)|ξ| p−2ξ,

with p  2 and h ∈ L1

loc() Then we get the operator div(h(x)|∇u| p−2∇u), and

if h(x) ≡ 1 in  we conclude the well-known p-Laplacian operator

pu : = div(|∇u| p−2∇u)

as in [8, 10].

2 Let

A(x, ξ) = h(x)

p

 (1+ |ξ|2)p− 1

with p  2, h ∈ L −1p () Then

a(x , ξ) = h(x)(1 + |ξ|2)p−22 ξ.

We obtain the generalised mean curvature operator

div

h(x)(1 + |∇u|2)p−22 ∇u.

It should be observed that the above examples have not been considered in [2, 8, 10]

yet For more information and connection on these operators, the reader may consult

either [2] or [8] and the references therein.

As in [10], we assume that function f :  × [0, +∞) → ⺢ is a sign-changing

Carath´eodory function and satisfies the following hypotheses:

(F 1) f (x , 0) = 0, |f (x, t)|  Ct p−1 for all t ∈ [0 + ∞), a.e x ∈ , and for some constant C > 0.

(F 2) There exist two positive constants t0, t1> 0 such that F(x, t)  0 for 0  t  t0

and F(x , t1)> 0.

(F 3) lim sup

t→∞

F (x ,t)

t p  0 uniformly in x, where F(x, t) =t

0f (x , s) ds.

Let W1,p() be the usual Sobolev space and W1,p

0 () be the closure of C∞

0 ()

under the norm

u =



 |∇u| p dx

1

.

We now consider the following subspace of W01,p():

H :=



u ∈ W1,p

0 () :





h1(x) |∇u| p dx < +∞

.

Then H is an infinite dimensional Banach space with respect to the norm (see [4])

u H=





h1(x) |∇u| p dx

1

.

We define the functionalλ : H→ ⺢ by

where

(u) =



 A(x , ∇u) dx, I(u) = λ



 F (x , u) dx, u ∈ H. (1.6)

Since h0∈ L p /p−1(), then the value λ (u) may be infinity for some u ∈ W1,p

0 (), that

is, the functional may not be defined throughout W01,p() In order to overcome this

difficulty, we choose the subspace H of W1,p().

DEFINITION1 We say that u ∈ H is a weak solution of problem (1.1)–(1.2) if and

only if



 a(x , ∇u)∇ϕ dx −λ





for allϕ ∈ C∞

0 ().

Then we have the following remark which plays an important role in our arguments REMARK1

(i) By (A4) and (i) in Proposition 2, it is easy to see that

H= u ∈ W1,p

0 () : (u) < ∞= u ∈ W1,p

0 () : λ (u) < ∞.

(ii) Since h1(x)  1, a.e x ∈ , we have u  u H for all u ∈ H Thus, the

continuous embeddings

H → W1,p

0 () → L i(), p  i  p∗ hold true

(iii) Since

 h1(x)|∇u| p dx < +∞ for any u ∈ C∞

0 () and h1∈ L1

loc(), we have

C0∞() ⊂ H.

The main result for the existence of solutions of (1.1) can be formulated as follows THEOREM1 Under hypotheses (A1)–(A4) and (F1), there exists a positive constant

λ such that for all λ ∈ (0, λ), problem (1.1)–(1.2) has no weak solution.

THEOREM 2 Under hypotheses (A1)–(A4) and (F1)–(F3), there exists a positive

constant λ such that for all λ  λ, problem (1.1)–(1.2) has at least two distinct non-negative, non-trivial weak solutions.

To prove Theorem 2, we first prove that the functional associated to the problem (1.1)–(1.2) is bounded from below and coercive, and thus the first weak solution is obtained due to a variant of the minimum principle which we will prove in the next section (see Theorem 4) To obtain the second solution to the problem (1.1)–(1.2), we shall use a variant of the mountain pass theorem due to Duc (see Proposition 1)

2 Auxiliary results. Due to the presence of h1, functional may not be

continuously Fr´echet differentiable functionals on H This means that we cannot apply

the classical Mountain pass theorem by Ambrosetti–Rabinowitz (see [1] for details).

To overcome this difficulty, we shall use a weak version of the Mountain pass theorem

introduced by Duc [3] Now we introduce the following concept of weakly continuously

differentiability due to Duc

DEFINITION 2 LetF be a map from a Banach space X to ⺢ We say that F is

weakly continuously differentiable on X if and only if the following two conditions are

satisfied:

(i) For any u ∈ X there exists a linear map DF(u) from X to ⺢ such that

lim

t→0

F (u + tv) − F(u)

for everyv ∈ X.

(ii) For anyv ∈ X, the map u → DF(u) (v) is continuous on X.

REMARK2 If we suppose further thatv → DF(u) (v) is continuous linear mapping

on X , then F is Gˆateaux differentiable.

DEFINITION3 We call u a generalised critical point (critical point, for short) of F

if D F(u) = 0 c is called a generalised critical value (critical value, for short) of F if F(u) = c for some critical point u of F.

Denote by C1

w (X ) the set of weakly continuously differentiable functionals on X

It is clear that C1(X ) ⊂ C1

w (X ) where we denote by C1(X ) the set of all continuously

Fr´echet differentiable functionals on X Now let F ∈ C1

w (X ); we put

DF(u) = sup{|DF(u)(h)|h ∈ Y, h = 1}

for any u ∈ X, where DF(u) may be +∞.

DEFINITION 4 We say thatF satisfies the Palais-Smale condition at level c ∈ ⺢

(denoted by (PS)c) if any sequence{u n } ⊂ X for which

F(un)→ c and DF(u n)→ 0 in X 

possesses a convergent subsequence If this is true at every level c then we simply say

thatF satisfies the Palais-Smale condition (denoted by (PS)).

DEFINITION5 We say thatF satisfies the Cerami condition at level c ∈ ⺢ (denoted

by (C)c) if any sequence{u n } ⊂ X for which

F(un)→ c and (1 + x n )DF (u n)→ 0 in X 

possesses a convergent subsequence If this is true at every level c, then we simply say

thatF satisfies the Cerami condition (denoted by (C)).

In the proof of our main theorems, we shall use the following results which is

proved in [9] We will recall its proof for completeness.

THEOREM3 (see [9]) Let F ∈ C1

w (X ) where X is a Banach space Assume that

(i) F is bounded from below, c = inf F,

(ii) F satisfies the (PS) condition.

Then c is a critical value of F (i.e there exists a critical point u0∈ X such that F(u0)= c)

Proof of Theorem 3 Let c be an arbitrary real number Before proving the theorem,

we need the following notation:

F c = {u ∈ X|F(u)  c}.

Let us assume, by negation, that c is not a critical value of F Then, Theorem 2.2 in [13]

implies the existence ofε > 0 and η ∈ C([0, +∞) × X, X) satisfying η(1, F c +ε)⊂ F c −ε.

This is a contradiction sinceF c −ε = ∅ due to the fact that c = inf F. 

REMARK3 By Corollary 2.1.1 in [6], ifF : X → ⺢ is a locally Lipschitz, bounded

from below function and it satisfies the (C) condition, thenF is coercive This leads us

to state the following lemma

LEMMA1 If F : X → ⺢ is a locally Lipschitz, bounded from below function and it satisfies the (C) condition then it satisfies the (PS) condition.

Proof Let {u n}n ⊂ X be a sequence such that F(u n ) is bounded and D F(un)→ 0

in X  By Remark 3,F is coercive, and this helps us to deduce that {un}nis bounded in

X Hence also (1 + u n )DF(u n)→ 0 in X , and becauseF satisfies the (C) condition,

it follows that {u n}n has a strongly convergent subsequence This completes the

Similar to Theorem 3, we have the following new result

THEOREM4 Let F be continuous on X and be of class C1

w (X ) where X is a Banach space Assume that

(i) F is bounded from below, c = inf F,

(ii) F satisfies the (C) condition.

Then c is a critical value of F (i.e there exists a critical point u0∈ X such that F(u0)= c).

The proof of Theorem 4 follows from Lemma 1, so we omit it Next we provide a

variant Mountain pass theorem due to Duc [3].

PROPOSITION1 (see [3]) Let F ∈ C1

w (X ) where X is a Banach space and satisfies (PS) condition Assume that F(0) = 0 and there exist a positive constant ρ and z0∈ X

such that

(i) z0X > ρ and F(z0) 0.

(ii) α = inf {F(u) : u ∈ X, uX = ρ} > 0.

Assume that the set

G = {ϕ ∈ C([0, 1], X) : ϕ(0) = 0, ϕ(1) = z0}

is not empty Put

β := inf {max F(ϕ([0, 1])) : ϕ ∈ G} Then β  α and β is a critical value of F.

For the use of Proposition 1, we refer the reader to [3, 12, 13] We end

this section by studying some certain properties of the functional λ given

by (1.5) but we first recall some results which will be used throughout this work

PROPOSITION2 (see [4]).

(i) A verifies the growth condition

|A(x, ξ)|  c0(h0(x) |ξ| + h1(x) |ξ| p)

for all ξ ∈ ⺢ N , a.e x ∈ .

(ii) A(x , ξ) is convex with respect to ξ Moreover, by (A3) for all u , v ∈ H we have



u + v

2



 1

2 (u) +1

2 (v) − k0u − v p

Using the method as in [4] with some simple computations we obtain the following

proposition which concerns the smoothness of the functional λ

(i) If {u m } is a sequence weakly converging to u in W1,p

0 (), then (u)  lim inf

m→∞ (um)

and

lim

m→∞I(u m)= I(u).

(ii) The functionals and I are continuous on H.

(iii) Functional λ is weakly continuously differentiable on H and we have

D λ (u)( ϕ) =



 a(x, ∇u)∇ϕ dx −λ



 f (x, u)ϕ dx for all u , ϕ ∈ H.

3 Proofs of the theorems.

Proof of Theorem 1 Let us denote by S the best constant in the Sobolev embedding

W01,p() → L p(), i.e.

S= inf

W01,p()\{0}



 |∇u| p dx

1



 |u| p dx

Then, if u is a weak solution of problem (1.1)–(1.2), multiplying (1.1) by u and

integrating by parts combined with conditions (A4) and (F1) gives

k1



 |∇u| p dx  k1





h1(x) |∇u| p dx





 a(x , ∇u)∇u dx = λ



 f (x , u)u dx  Cλ



 |u| p dx

(3.2)

Hence, choosingλ = k1S/C, where S is given by (3.1), we conclude the proof. 

We will prove Theorem 2 by using critical point theory Set f (x , t) = 0 for all t < 0

and consider the energy functional λ : H→ ⺢ which is given by (1.5)

LEMMA2 If u is a critical point of  λ then u is non-negative in .

Proof Observe that if u is a critical point of  λ , denoting by u−the negative part

of u, i.e u−(x) = min {u(x), 0} we have

0= D λ (u)(u−)=



 a(x , ∇u)∇u−dx −λ



 f (x , u)u−dx

 k1



 h1(x)|∇u−|p dx = k1u−p

H ,

(3.3)

which yields that u  0 for a.e x in  Thus, non-trivial critical points of the functional

 λare non-negative, non-trivial solutions of problem (1.1)–(1.2) 

The following lemma shows that the functional λsatisfies all of the assumptions

of Theorem 3 Then problem (1.1)–(1.2) admits a weak solution u1∈ H as a global minimiser and u1 0

LEMMA3 The functional  λ is bounded from below and satisfies the (PS) condition

on H.

Proof By conditions (F1) and (F3), there exists a constant C λ = C(λ) > 0 such

that

λF(x, t)  k1S

for all t ∈ ⺢ and a.e x ∈  Hence,

 λ (u)=



 A(x, ∇u) dx −λ



 F (x, u) dx

k1

p



 h1(x)|∇u| p dx−







k1S

2p |u| p + C λ



dx

 k1

2p u p

H − C λ ||,

(3.5)

where || denotes the Lebesgue measure of  in ⺢ N Thus, the functional  λ is

coercive and hence bounded from below on H.

Let{u m } be a Palais-Smale sequence in H, i.e.

| λ (u m)|  c for all m,  λ (u m)→ 0 in H  (3.6) Since λ is coercive on H, {u m } is bounded in H By Remark 1 (ii), {u m} is bounded in

W01,p() It follows that there exists u ∈ W1,p

0 () such that, passing to a subsequence,

still denoted by{u m }, it converges weakly to u in W1,p

0 () We shall prove that {um}

converges strongly to u in H.

Indeed, we observe by Remark 1(i), Proposition 3(i) and (3.6) that u ∈ H Hence, {u m − u H } is bounded This and (3.6) imply that D λ (u m )(u m − u) converges to 0 as

m→ ∞

Using condition (F1) combined with H ¨older’s inequality we deduce that

0



 |f (x, u m)||um − u| dx  C



 |u m|p−1|u m − u| dx

 C u mp−1

L p() u m − u L p()

(3.7)

Since the embedding W01,p() → L p() is compact, {um } converges strongly to u in

L p() Therefore, relation (3.7) implies that

lim

Combining relations (3.6) and (3.8) with the fact that

D (um )(u m − u) = D λ (u m )(u m − u) + DI(u m )(u m − u),

we conclude that

lim

On the other hand, the convex property of functional (see Proposition 2(ii)) implies

that

(u) − lim

m→∞ (um)= lim

m→∞( (u) − (um)) lim

m→∞D (um )(u − u m)= 0. (3.10) Combining this with Proposition 3(i), we have

lim

We now assume by contradiction that{u m } does not converge strongly to u in H,

and then there exist a constant  > 0 and a subsequence {um k } of {u m} such that

u m k − u H   Using Proposition 2(ii) we get

1

2 (u) +1

2 (um k)−



um k + u

2



 k0u m k − u p

H  k0 p (3.12)

Letting k→ ∞, relation (3.12) gives

lim sup

k→∞



um k + u

2



We remark that sequence{u mk +u

2 } also converges weakly to u in W1,p

0 () So, using

Proposition 3(i) again we get

(u)  lim inf

k→∞



u m k + u

2



which contradicts (3.13) Therefore,{u m } converges strongly to u in H. 

LEMMA4 There exists a positive constant λ such that for all λ  λ, infH λ < 0, and hence u1≡ 0, i.e the solution u1is not trivial.

Proof Let 0⊂  be a compact subset large enough and a function ϕ0∈ C∞

0 ()

such that ϕ0(x) = t1 in 0 and 0 ϕ0(x)  t1 in \0, where t1 is chosen as in assumption (F2): then we have





F (x , ϕ0) dx



0

F (x , ϕ0) dx −Ct p

1|\0| > 0. (3.15)

Thus, λ(ϕ0)< 0 for λ  λ with λ large enough This implies that infH  λ < 0 and

In the next part of this paper, we shall show the existence of the second solution

u = u by using the Mountain pass theorem introduced in [3] To this purpose, we

first fixλ  λ and set

f(x, t) =

⎧

⎪

⎪

f (x , t) for 0 t  u1(x) ,

f (x , u1(x)) for t > u1(x) ,

(3.16)

and F(x , t) =t

0f(x, s) ds Define the functional  λ : H→ ⺢ by

 λ (u)=



 A(x , ∇u) dx −λ





With the same arguments as those used for the functional λ, we can show that  λis

weakly continuously differentiable on H and

D  λ (u)( ϕ) =



 a(x, ∇u)∇ϕ dx −λ



 f(x, u)ϕ dx for all u , ϕ ∈ H.

LEMMA5 If u ∈ H is a critical point of  λ then u  u1 So, u is a solution of problem

(1.1)–(1.2) in the order interval [0 , u1].

Proof By the definitions of λand λ , if u is a critical point of λ then u 0 as before and by condition (A2) we have

0= (D  λ (u) − D λ (u1))((u − u1)+)

=





(a(x , ∇u) − a(x, ∇u1))· ∇(u − u1)+dx

− λ



(f (x , u) − f (x, u1))(u − u1)+dx

=



{u>u1 }(a(x , ∇u) − a(x, ∇u1))· (∇u − ∇u1) dx  0.

(3.18)

According to (3.18) and condition (A2), the equality holds if and only if∇u = ∇u1 It follows that∇u(x) = ∇u1(x) for all x ∈ 1:= {y ∈  : u(y) > u1(y)} Hence,



1

|∇(u − u1)|p dx = 0 and thus



 |∇(u − u1)+|p dx = 0.

Combining this with Remark 1(ii), we conclude that (u − u1)+L p()= 0 and then

LEMMA 6 There exist a constant ρ ∈ (0, u1H ) and a constant α > 0 such that

λ (u)  α for all u ∈ H with u H = ρ.

Proof We set u = {x ∈  : u(x) > min {u1(x) , t0}}, where t0 is given as in (F2) Then, by (3.16) and condition (F1), we have F(x , u(x))  0 on \u Hence,

 λ (u)  k1u p

H − λ





The following lemma shows that the functional λsatisfies all of the assumptions

of. .. the functional λ