EXISTENCE AND MULTIPLICITY RESULTS FOR NONLINEAR ...

Opuscula Math. 34, no. 3 (2014), 621–638 http://dx.doi.org/10.7494/OpMath.2014.34.3.621

Opuscula Mathematica

EXISTENCE AND MULTIPLICITY RESULTS FOR NONLINEAR PROBLEMS INVOLVING THE p(x)-LAPLACE OPERATOR Najib Tsouli and Omar Darhouche Communicated by P.A. Cojuhari Abstract. In this paper we study the following nonlinear boundary-value problem −∆p(x) u = λf (x, u) p(x)−2 ∂u

|∇u|

∂ν

+ β(x)|u|

p(x)−2

in Ω,

u = µg(x, u)

on ∂Ω,

where Ω ⊂ RN is a bounded domain with smooth boundary ∂Ω, ∂u is the outer unit normal ∂ν derivative on ∂Ω, λ, µ are two real numbers such that λ2 + µ2 6= 0, p is a continuous function on Ω with inf x∈Ω p(x) > 1 , β ∈ L∞ (∂Ω) with β − := inf x∈∂Ω β(x) > 0 and f : Ω × R → R, g : ∂Ω × R → R are continuous functions. Under appropriate assumptions on f and g, we obtain the existence and multiplicity of solutions using the variational method. The positive solution of the problem is also considered. Keywords: critical points, variational method, p(x)-Laplacian, generalized Lebesgue-Sobolev spaces. Mathematics Subject Classification: 35B38, 35D05, 35J20, 35J60, 35J66.

1. INTRODUCTION This paper is devoted to finding existence and multiplicity results for the following nonlinear problem −∆p(x) u = λf (x, u) in Ω, ∂u |∇u|p(x)−2 + β(x)|u|p(x)−2 u = µg(x, u) on ∂Ω, ∂ν

(1.1)

where Ω ⊂ RN is a bounded smooth domain, ∂u ∂ν is the outer unit normal deriva2 2 tive on ∂Ω, λ, µ ∈ R such that λ + µ 6= 0, p is a continuous function on Ω c AGH University of Science and Technology Press, Krakow 2014

621

622

Najib Tsouli and Omar Darhouche

with p− := inf x∈Ω p(x) > 1 and β ∈ L∞ (∂Ω) with β − := inf x∈∂Ω β(x) > 0. The main interest in studying such problems arises from the presence of the p(x)-Laplace operator div(|∇u|p(x)−2 ∇u), which is a natural extension of the classical p-Laplace operator div(|∇u|p−2 ∇u) obtained in the case when p is a positive constant. However, such generalizations are not trivial since the p(x)-Laplace operator possesses a more complicated structure than p-Laplace operator, for example it is inhomogeneous. In recent years increasing attention has been paid to the study of differential and partial differential equations involving variable exponent conditions. The interest in studying such problems was stimulated by their applications in elastic mechanics, fluid dynamics and the calculus of variations, for information on modelling physical phenomena by equations involving the p(x)-growth condition we refer to [1, 8, 10, 17, 19, 20, 24, 29, 31, 32]. In the past decades a vast amount of literature that deal with the existence for problems of the type −∆p(x) u = f (x, u) with different boundary conditions (Dirichlet, Neumann, Robin, nonlinear, etc.) have appeared. See, for instance [9, 11, 14, 16, 27, 30] and references therein. In [16], the authors have studied the problem (1.1) with g(x, u) ≡ 0. Using the variational approach based on the nonsmooth critical point theory for locally Lipschitz functions, they obtain the existence of at least two nontrivial solutions. This same problem has been studied in [26]. Under appropriate assumptions on f , and using variational methods, we have obtained important results on existence and multiplicity of solutions. In [3], the authors considered the problem (1.1) with λf (x, u) ≡ |u|p(x)−2 u and β(x) ≡ 0. Using Ricceri’s variational principle, they establish the existence of at least three solutions of the problem. If β(x) ≡ 0 and µg(x, u) ≡ 0, the problem (1.1) becomes the nonlinear Neumann boundary value problem. It was studied in [27]. Using the three critical point theorem due to Ricceri, under the appropriate assumptions on f , the authors establish the existence of at least three solutions of this problem. The purpose of this paper is to prove the existence and multiplicity results of solutions to the problem (1.1) under appropriate assumptions on f and g following ideas from [30]. These results extend some of the results in [25] for the p-Laplacian. Next, we make the following assumptions on f and g: (f0 ) f : Ω×R → R satisfies the Carathéodory condition and there exist two constants C1 ≥ 0, C2 > 0 such that |f (x, s)| ≤ C1 + C2 |s|α(x)−1

for all

(x, s) ∈ Ω × R,

where α(x) ∈ C+ (Ω) and α(x) < p∗ (x), for all x ∈ Ω, where ( N p(x) , if p(x) < N, p∗ (x) = N −p(x) +∞, if p(x) ≥ N ; (f1 ) there exist M1 > 0, θ1 > p+ such that 0 < θ1 F (x, s) ≤ sf (x, s) for all |s| ≥ M1 , (f2 ) f (x, s) = o(|s|p

+

−1

), s → 0 for x ∈ Ω uniformly;

x ∈ Ω;

623

Existence and multiplicity results for nonlinear problems. . .

(f3 ) f (x, −s) = −f (x, s) for all x ∈ Ω, s ∈ R; (g0 ) g : ∂Ω × R → R satisfies the Carathéodory condition and there exist two constants C10 ≥ 0, C20 > 0 such that |g(x, s)| ≤ C10 + C20 |s|γ(x)−1

for all (x, s) ∈ ∂Ω × R,

where γ(x) ∈ C+ (∂Ω) and γ(x) < p∂ (x), for all x ∈ ∂Ω, where ( (N −1)p(x) ∂ N −p(x) , if p(x) < N, p (x) = +∞, if p(x) ≥ N ; (g1 ) there exist M2 > 0, θ2 > p+ such that 0 < θ2 G(x, s) ≤ sg(x, s) for all |s| ≥ M2 ,

x ∈ ∂Ω;

+

(g2 ) g(x, s) = o(|s|p −1 ), s → 0 for x ∈ ∂Ω uniformly; (g3 ) g(x, −s) = −g(x, s) for all x ∈ ∂Ω, s ∈ R. Let H be the energy functional corresponding to problem (1.1). The main results of this paper are the following: Theorem 1.1. If (f0 ), (g0 ) hold and α+ , γ + < p− , then problem (1.1) has a weak solution. Theorem 1.2. If (f0 ), (f1 ), (f2 ), (g0 ), (g1 ), (g2 ) hold and α− , γ − > p+ , λ, µ ≥ 0, then problem (1.1) has a nontrivial weak solution. Theorem 1.3. If (f0 ), (f1 ), (f3 ), (g0 ), (g1 ), (g3 ) hold and α− , γ − > p+ , λ, µ ≥ 0, then H has a sequence of critical points (±un ) such that H(±un ) → ∞ as n → ∞. Meanwhile, problem (1.1) has infinite many pairs of weak solutions. Theorem 1.4. Let α(x) ∈ C+ (Ω), γ(x) ∈ C+ (∂Ω) and α(x) < p∗ (x)

for all

x ∈ Ω;

γ(x) < p∂ (x)

for all

x ∈ ∂Ω.

If f (x, u) = |u|α(x)−2 u, g(x, u) = |u|γ(x)−2 u, α− > p+ , and γ + < p− , then we have: (i) for all λ > 0 and µ ∈ R, problem (1.1) has a sequence of weak solutions (±uk ) such that H(±uk ) → ∞ as k → ∞; (ii) for all µ > 0 and λ ∈ R, problem (1.1) has a sequence of weak solutions (±vk ) such that H(±vk ) < 0, and H(±vk ) → 0 as k → ∞. Theorem 1.5. If (f0 ), (g0 ) hold and α+ , γ + < p− , then problem (1.1) has a nonnegative weak solution. Theorem 1.6. If (f0 ), (f1 ), (f2 ), (g0 ), (g1 ), (g2 ) hold and α− , γ − > p+ , λ, µ ≥ 0, then problem (1.1) has a nonnegative nontrivial weak solution. This article is organized as follows. In Section 2, we introduce some necessary preliminary knowledge on variable exponent Lebesgue and Sobolev spaces. In Section 3, we will give the proof of Theorems 1.1–1.4. In Section 4, we will give the proof of Theorems 1.5–1.6.

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2. PRELIMINARIES For completeness, we first recall some facts on the variable exponent spaces Lp(x) (Ω) and W 1,p(x) (Ω). Suppose that Ω is a bounded open domain of RN with smooth boundary ∂Ω and p ∈ C+ (Ω), where o n C+ (Ω) = p ∈ C(Ω) : inf p(x) > 1 . x∈Ω −

+

Denote by p := inf x∈Ω p(x) and p := supx∈Ω p(x). Define the variable exponent Lebesgue space Lp(x) (Ω) by Z n o Lp(x) (Ω) = u : Ω → R is measurable and |u|p(x) dx < +∞ Ω

with the norm |u|Lp(x) (Ω) = |u|p(x)

Z p(x) o n u dx ≤ 1 . = inf τ > 0 : τ Ω

Define the variable exponent Sobolev space W 1,p(x) (Ω) by n o W 1,p(x) (Ω) = u ∈ Lp(x) (Ω) : |∇u| ∈ Lp(x) (Ω) , with the norm Z n o ∇u p(x) u p(x) kuk = inf τ > 0 : + dx ≤ 1 , τ τ Ω

kuk = |∇u|p(x) + |u|p(x) . We refer the reader to [9, 12, 13] for the basic properties of the variable exponent Lebesgue and Sobolev spaces. Lemma 2.1 ([13]). Both (Lp(x) (Ω), | · |p(x) ) and (W 1,p(x) (Ω), k · k) are separable, reflexive and uniformly convex Banach spaces. Lemma 2.2 ([13]). Hölder inequality holds, namely Z 0 |uv|dx ≤ 2|u|p(x) |v|p0 (x) for all u ∈ Lp(x) (Ω), v ∈ Lp (x) (Ω), Ω

where

1 p(x)

+

1 p0 (x)

= 1.

Now, we introduce a norm, which will be used later. For u ∈ W 1,p(x) (Ω), define   Z Z   ∇u(x) p(x) u(x) p(x) kukβ = inf λ > 0 : dx + β(x) dσ ≤ 1 . x λ   λ Ω

∂Ω

Then, by Theorem 2.1 in [9], kukβ is also a norm on W 1,p(x) (Ω) which is equivalent to kuk.


625

Lemma 2.3 (see [13, 14, 30]). (1) If q ∈ C+ (Ω) and q(x) < p∗ (x) for any x ∈ Ω, then the imbedding from W 1,p(x) (Ω) to Lq(x) (Ω) is compact and continuous. (2) If q ∈ C+ (Ω) and q(x) < p∂ (x) for any x ∈ ∂Ω, then the trace imbedding from W 1,p(x) (Ω) to Lq(x) (∂Ω) is compact and continuous. An important role in manipulating the generalized Lebesgue-Sobolev spaces is played by the mapping defined by Z Z p(x) Iβ (u) = |∇u| dx + β(x)|u|p(x) dσx for all u ∈ W 1,p(x) (Ω). Ω

∂Ω

Lemma 2.4 ([9]). −

+

+

−

(1) kukβ ≥ 1 ⇒ kukpβ ≤ Iβ (u) ≤ kukpβ , (2) kukβ ≤ 1 ⇒ kukpβ ≤ Iβ (u) ≤ kukpβ , (3) kukβ → 0 if and only if Iβ (u) → 0 (as k → ∞), (4) |u(x)|p(x) → ∞ if and only if Iβ (u) → ∞ (as k → ∞). Remark 2.5. From (1) and (2) of the previous lemma, one can easily deduce that kukβ < (=; >)1 ⇔ Iβ (u) < (=; >)1.

(2.1)

Theorem 2.6. If f : Ω × R → R is a continuous function satisfying |f (x, s)| ≤ C(1 + |s|α(x)−1 )

for all

(x, s) ∈ Ω × R,

where C ≥ 0 is a constant, α(x)R ∈ C+ (Ω) such that for all Rx ∈ Ω, α(x) < p∗ (x). u Set X = W 1,p(x) (Ω), F (x, u) = 0 f (x, t)dt, and ψ(u) = − Ω F (x, u(x)) dx. Then R ψ(u) ∈ C 1 (X, R) and Dψ(u, ϕ) = hψ 0 (u), ϕi = − Ω f (x, u(x))ϕ dx. Moreover, the operator ψ 0 : X → X ∗ is compact. Proof. It is easily adapted from that of [27, Theorem 2.1]. Theorem 2.7. If g : ∂Ω × R → R is a Carathéodory function and |g(x, s)| ≤ C”(1 + |s|α(x)−1 )

for all

(x, s) ∈ ∂Ω × R,

where C” is a positive constant and α(x) ∈R C+ (∂Ω) such that for all R x ∈ ∂Ω, α(x) < u p∂ (x). Set X = W 1,p(x) (Ω), G(x, u) = 0 g(x, t)dt, Rψ(u) = − ∂Ω G(x, u(x))dσx . Then ψ(u) ∈ C 1 (X, R) and Dψ(u, ϕ) = hψ 0 (u), ϕi = − ∂Ω g(x, u(x))ϕdσx . Moreover, the operator ψ 0 : X → X ∗ is compact. Proof. It is easily adapted from that of [2, Theorem 2.9]. Let X = W 1,p(x) (Ω) and define Z Z 1 β(x) p(x) φ(u) = |∇u|p(x) dx + |u| dσx , p(x) p(x) Ω ∂Ω Z Z ψ(u) = − F (x, u) dx, J(u) = − G(x, u) dσx , Ω

∂Ω

626


where F (x, t) =

Rt

f (x, s)ds, and G(x, t) =

0

Rt

g(x, s)ds. It is easy to see that φ ∈

0

C 1 (X, R) and 0

Z

p(x)−2

|∇u|

(φ (u), v) =

Z ∇u∇v dx +

Ω

β(x)|u|p(x)−2 uv dσx ,

v ∈ X.

∂Ω

Moreover, we have the following proposition. Proposition 2.8 ([16, Proposition 2.2]). (1) φ0 : X → X ∗ is a continuous, bounded and strictly monotone operator. (2) φ0 : X → X ∗ is a mapping of type (S)+ , that is, if un * u in X and lim sup(φ0 (un ) − φ0 (u), un − u) ≤ 0, then un → u in X. n→∞

(3) φ0 : X → X ∗ is a homeomorphism. Under the conditions (f0 ) and (g0 ), and from Theorem 2.6 and Theorem 2.7, ψ and J are continuously Gâteaux differentiable functionals whose Gâteaux derivative is compact, and we have Z Z hψ 0 (u), vi = − f (x, u)v dx, hJ 0 (u), vi = − g(x, u)v dσx . Ω

∂Ω

The energy functional corresponding to problem (1.1) is defined on X as H(u) = φ(u) + λψ(u) + µJ(u). The functional H is of class C 1 (X, R), and the weak solution of problem (1.1) corresponds to the critical point of the functional H. Definition 2.9. We say that u ∈ W 1,p(x) (Ω) is a weak solution of the problem (1.1) if for all v ∈ W 1,p(x) (Ω) Z Z Z Z |∇u|p(x)−2 ∇u∇v dx + β(x)|u|p(x)−2 uv dσx = λ f (x, u)v dx + µ g(x, u)v dσx , Ω

∂Ω

Ω

∂Ω

where dσx is the measure on the boundary ∂Ω. Remark 2.10. In the following sections, the symbols C, D, M denote the generic nonnegative of positive constants, which may not be the same at each occurrence.

3. EXISTENCE AND MULTIPLICITY OF SOLUTIONS In this section, we shall prove Theorems 1.1–1.4. By using the variational principle, we prove the existence and multiplicity of results for problem (1.1).

627


Proof of Theorem 1.1. From (f0 ) and (g0 ), there exist C > 0 such that |F (x, t)| ≤ C(1 + |t|α(x) ), γ(x)

|G(x, t)| ≤ C(1 + |t|

),

(x, t) ∈ Ω × R, (x, t) ∈ ∂Ω × R.

Obviously, H is weakly lower semicontinuous. It suffices to show that H is coercive. Let u ∈ X be such that kukβ > 1. Then Z 1 |∇u|p(x) dx+ H(u) = p(x) Ω Z Z Z 1 + β(x)|u|p(x) dσx − λF (x, u)dx − µG(x, u)dσx ≥ p(x) Ω ∂Ω ∂Ω Z Z − 1 ≥ + kukpβ − |λ| C(1 + |u|α(x) )dx − |µ| C(1 + |u|γ(x) )dσx ≥ p Ω

∂Ω

− + 1 γ+ ≥ + kukpβ − |λ|Ckukα β − |µ|Ckukβ − M. p

So H(u) → ∞ as kukβ → ∞, since α+ , γ + < p− . Then H is coercive and H has a minimum point u in X which is a weak solution of problem (1.1). Corollary 3.1. Under the assumptions in Theorem 1.1, if λ, µ 6= 0, and there exist two positive constants d1 , d2 < p− such that: sgn(λ)F (x, t) >0 t→0 |t|d1 sgn(µ)G(x, t) lim inf >0 t→0 |t|d2 lim inf

for x ∈ Ω uniformly,

(3.1)

for x ∈ ∂Ω uniformly,

(3.2)

then the problem (1.1) has a nontrivial weak solution. Proof. From Theorem 1.1 we know that H has a global minimum point u. It suffices to show that u is nontrivial. From (3.1) and (3.2), for 0 < t < 1 small enough, there exists a positive constant C such that sgn(λ)F (x, t) ≥ C|t|d1 ,

sgn(µ)G(x, t) ≥ C|t|d2 .

Choose u0 ≡ M > 0, then u0 ∈ X. Then we have Z − Z tp H(tu0 ) ≤ − β(x)|M |p(x) dσx − |λ| (sgn(λ))F (x, tM )dx− p Ω ∂Ω Z − |µ| (sgn(µ))G(x, tM )dσx ≤ ∂Ω p−

Z

t ≤ − p

p(x)

β(x)|M |

≤ D1 t

dσx − |λ|

− |λ|D2 t

d1

d1

Z

C|tM | dx − |µ| Ω

∂Ω p−

Z

d2

− |µ|D3 t .

∂Ω

C|tM |d2 dσx ≤

628


Since d1 , d2 < p− , there exists 0 < t0 < 1 small enough such that H(t0 u0 ) < 0. So the global minimum point u of H is nontrivial. Remark 3.2. The conclusion of Corollary 3.1 remains valid if we suppose one of the following conditions: (i) µ = 0, λ 6= 0 and there exist a positive constant d1 < p− such that (3.1) holds, (ii) λ = 0, µ 6= 0 and there exist a positive constant d2 < p− such that (3.2) holds. Remark 3.3. If f (x, u) = sgn(λ)|u|α(x)−2 u and g(x, u) = sgn(µ)|u|γ(x)−2 u with α+ , γ + < p− , then the conditions in Corollary 3.1 can be fulfilled. To prove Theorem 1.2, we need the following lemma. Lemma 3.4. If (f0 ), (f1 ), (g0 ), (g1 ) hold and λ, µ ≥ 0, then H satisfies the (PS) condition. Proof. Suppose that (un ) ⊂ X is a (PS) sequence, i.e. sup |H(un )| ≤ M,

H 0 (un ) → 0 as n → ∞.

Let us show that (un ) is bounded so as to verify it is precompact in X. By Lemma 2.3, and Theorems 2.6, 2.7, we know that ψ and J are booth weakly continuous and their derivative operators are compact. By Proposition 2.8, we deduce that H 0 = φ0 + λψ 0 + µJ 0 is also of type (S + ). For n large enough, we have 1 1 M + 1 ≥ H(un ) − hH 0 (un ), un i + hH 0 (un ), un i = θ Z Z θ 1 β(x) p(x) = |∇un | dx + |un |p(x) dσx − p(x) p(x) Ω ∂Ω Z Z − λ F (x, un )dx − µ G(x, un )dσx − Ω

−

1 θ

∂Ω

Z

|∇un |p(x) dx +

Ω

Z

β(x)|un |p(x) dσx −

∂Ω

Z −λ

Z f (x, un )un dx − µ

Ω

1 + hH 0 (un ), un i ≥ θ − 1 1 ≥ − kun kpβ − p+ θ − 1 1 ≥ − kun kpβ − + p θ

g(x, un )un dσx +

∂Ω

1 kH 0 (un )kX ∗ kun kβ − C ≥ θ 1 kun kβ − C, θ

where θ = min{θ1 , θ2 }. From the inequality above, we know that (un ) is bounded in X since θ > p+ . This completes the proof.

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Proof of Theorem 1.2. We will use the mountain pass theorem (see [4, 28]). By the previous lemma, we know that H satisfies the (PS) condition. So it suffices to verify the geometric conditions in the mountain pass theorem. We have the following compact embedding + + W 1,p(x) (Ω) ,→ Lp (Ω), W 1,p(x) (Ω) ,→ Lp (∂Ω), since p+ < α− ≤ α(x) < p∗ (x) for all x ∈ Ω;

p+ < γ − ≤ γ(x) < p∂ (x) for all x ∈ ∂Ω.

So there exists a constant C > 0 such that |u|Lp+ (Ω) ≤ Ckukβ ,

|u|Lp+ (∂Ω) ≤ Ckukβ

for all u ∈ X.

Conditions (f0 ), (f2 ) and (g0 ), (g2 ) assure that there exists an arbitrary constant 0 < ε < 1 and two positive constants (both denoted by C(ε)) such that +

|F (x, t)| ≤ ε|t|p + C(ε)|t|α(x) p+

|G(x, t)| ≤ ε|t|

γ(x)

+ C(ε)|t|

for all

(x, t) ∈ Ω × R,

for all (x, t) ∈ ∂Ω × R.

So for kukβ small enough (kukβ < 1). We have Z Z 1 p+ H(u) ≥ + kukβ − λ F (x, u)dx − µ G(x, u)dσx ≥ p Ω ∂Ω Z + + 1 ≥ + kukpβ − λ ε|u|p + C(ε)|u|α(x) dx− p Ω Z p+ −µ ε|u| + C(ε)|u|γ(x) dσx ≥ ∂Ω + + − 1 γ− ≥ + kukpβ − (λεC + µεC)kukpβ − λC(ε)Ckukα β − µC(ε)Ckukβ . p

Choose ε > 0 small enough such that 0 < λεC + µεC
0 and δ > 0 such that H(u) ≥ δ for all u ∈ X : kukβ = r.

630


Now, to apply the mountain pass theorem, we must prove that H(tu) → −∞ as t → +∞, for a certain u ∈ X. From conditions (f1 ) and (g1 ) we have for suitable positive constants C, D F (x, s) ≥ C|s|θ1 − D G(x, s) ≥ C|s|

θ2

−D

for all (x, s) ∈ Ω × R, for all (x, s) ∈ ∂Ω × R.

Let u ∈ X and t > 1. We have Z p(x) Z p(x) t t β(x) p(x) p(x) |∇u| |u| dx + dσx − H(tu) = p(x) p(x) Ω ∂Ω Z Z − λ F (x, tu)dx − µ G(x, tu)dσx ≤ Ω p+

≤

t p−

∂Ω

Z

|∇u|p(x) dx +

Ω

β(x)|u|p(x) dσx −

∂Ω

Z −λ

Z

(C|tu|

θ1

Z − D)dx − µ

Ω p+

≤

t p−

∂Ω

Z

|∇u|p(x) dx +

Ω θ1

(C|tu|θ2 − D)dσx ≤

Z

β(x)|u|p(x) dσx −

∂Ω

Z

− t λC

θ1

|u| dx − tθ2 µC

Ω

Z

|u|θ2 dσx + M.

∂Ω

+

The fact that θ1 , θ2 > p implies H(tu) → −∞ as t → +∞. It follows that there exists e ∈ X such that kekβ > r and H(e) < 0. According to the mountain pass theorem, H admits a critical value τ ≥ δ which is characterized by τ = inf sup H(h(t)), h∈Γ t∈[0,1]

where Γ = {h ∈ C([0, 1], X) : h(0) = 0 and h(1) = e} . The proof is complete. Proof of Theorem 1.3. Since X is a separable and reflexive Banach space [7,12], there ∞ ∗ exist {en }∞ n=1 ⊂ X and {fn }n=1 ⊂ X such that ( 1, if n = m, fn (em ) = δn,m = 0, if n 6= m.


X = span{en : n = 1, 2, . . . },

631

∗

X ∗ = spanW {fn : n = 1, 2, . . . }.

For n = 1, 2, . . . denote by Xn = span{en },

Yn = ⊕nj=1 Xj ,

Z n = ⊕∞ j=n Xj .

Then we have the following lemma. Lemma 3.5 ([15, Proposition 3.5]). If α(x) ∈ C+ (Ω), α(x) < p∗ (x) for all x ∈ Ω, and γ(x) ∈ C+ (∂Ω), γ(x) < p∂ (x) for all x ∈ ∂Ω, denote αk = sup |u|Lα(x) (Ω) : kukβ = 1, u ∈ Zk , γk = sup |u|Lγ(x) (∂Ω) : kukβ = 1, u ∈ Zk . Then limk→∞ αk = 0 and limk→∞ γk = 0. Now, we return to the proof of Theorem 1.3. To do that, we will use the Fountain theorem (see [28]). Obviously, H is an even functional and satisfies the (PS) condition. We will prove that if k is large enough, then there exist ρk > rk > 0 such that (A1) bk := inf {H(u) : u ∈ Zk , kukβ = rk } → +∞ as k → +∞, (A2) ak := max {H(u) : u ∈ Yk , kukβ = ρk } ≤ 0 as k → +∞. (A1) For u ∈ Zk such that kukβ = rk > 1, by conditions (f0 ) and (g0 ), we have Z Z Z Z β(x) p(x) 1 p(x) |∇u| dx + |u| dσx − λ F (x, u)dx − µ G(x, u)dσx ≥ H(u) = p(x) p(x) Ω Ω ∂Ω ∂Ω Z Z 1 p− α(x) γ(x) )dx − µ C(1 + |u| ≥ + kukβ − λ C(1 + |u| )dσx ≥ p Ω

∂Ω

n o − + 1 α− ≥ + kukpβ − λC max |u|α Lα(x) (Ω) , |u|Lα(x) (Ω) − p n o + − − µC max |u|γLγ(x) (∂Ω) , |u|γLγ(x) (∂Ω) − M ≥ ≥

− 1 kukpβ − p+

n o + γ+ γ− α− − C(λ, µ) max |u|α Lα(x) (Ω) , |u|Lα(x) (Ω) , |u|Lγ(x) (∂Ω) , |u|Lγ(x) (∂Ω) − M. o n + + γ− γ+ α− , then we have , |u| , |u| = |u|α If max |u|α , |u| α(x) α(x) γ(x) γ(x) Lα(x) (Ω) L (Ω) L (Ω) L (∂Ω) L (∂Ω) H(u) ≥

− + + 1 kukpβ − C(λ, µ)αkα kukα β − M. + p +

1

If we choose rk = (α+ C(λ, µ)αkα ) p− −α+ , we obtain 1 1 p− H(u) ≥ rk − + − M. p+ α

632


Since αk → 0, rk → +∞ and p+ < α− ≤ α+ , we have H(u) → +∞ as k → +∞. In the other three cases, we can deduce in the same way that H(u) → ∞,

since αk → 0, γk → 0 as k → +∞.

So (A1) holds. (A2) Conditions (f1 ) and g1 implies that there exist positive constants C, D such that F (x, s) ≥ C|s|θ1 − D

for all

G(x, s) ≥ C|s|θ2 − D

for all (x, s) ∈ ∂Ω × R.

(x, s) ∈ Ω × R,

Let u ∈ Yk be such thatkukβ = ρk > rk > 1. Then + 1 H(u) ≤ − kukpβ − λ p

Z (C|u|

θ1

Z − D) dx − µ

Ω + 1 ≤ − kukpβ − λC p

(C|u|θ2 − D) dσx ≤

∂Ω

Z

|u|θ1 dx − µC

Ω

Z

|u|θ1 dσx + M.

∂Ω

Since the space Yk has finite dimension, then all norms are equivalents and we obtain H(u) ≤

+ 1 kukpβ − λCkukθβ1 − µCkukθβ2 + M. − p

Finally, H(u) → −∞ as kukβ → +∞, u ∈ Yk since θ1 , θ2 > p+ . So the assertion (A2) is then satisfied and the proof of Theorem 1.3 is complete.

Proof of Theorem 1.4. (i) As in the proof of Theorem 1.3, we will use in a similar way, the Fountain theorem. So, it suffices to verify the (PS) condition. Assume (un ) ⊂ X,

sup H(un ) ≤ M,

H 0 (un ) → 0 as n → +∞.

633


For n large enough, we have 1 1 M + 1 ≥ H(un ) − − hH 0 (un ), un i + − hH 0 (un ), un i = α α Z Z 1 β(x) p(x) = |∇un | dx + |un |p(x) dσx − p(x) p(x) Ω ∂Ω Z Z 1 1 α(x) |un | |un |γ(x) dσx − −λ dx − µ α(x) γ(x) Ω ∂Ω Z Z 1 |∇un |p(x) dx + β(x)|un |p(x) dσx − − − α Ω ∂Ω Z Z α(x) − λ |un | dx − µ |un |γ(x) dσx + Ω

∂Ω

1 + − hH 0 (un ), un i ≥ α − + 1 1 1 ≥ − kun kpβ − Ckun kγβ − − kH 0 (un )kX ∗ kun kβ ≥ + − p α α − + 1 1 1 ≥ − − kun kpβ − Ckun kγβ − − kun kβ . + p α α Since α− > p+ , γ + < p− , we know that (un ) is bounded in X. This completes the proof. (ii) We will use the dual of the Fountain theorem. We need to prove that H satisfies the (P S)∗c condition (see [28]) and there exist ρk > rk > 0 such that for k large enough we have (B1) ak := max {H(u) : u ∈ Yk , kukβ = rk } < 0, (B2) bk := inf {H(u) : u ∈ Zk , kukβ = ρk } ≥ 0, (B3) dk := max {H(u) : u ∈ Yk , kukβ ≤ ρk } → 0 as k → +∞. Let us show that (B1) holds. We assume kukβ < 1 for convenience. For u ∈ Yk , we have Z Z 1 β(x) H(u) = |∇un |p(x) dx + |un |p(x) dσx − p(x) p(x) Ω ∂Ω Z Z 1 1 α(x) −λ |un | dx − µ |un |γ(x) dσx ≤ α(x) γ(x) Ω ∂Ω Z Z − 1 |λ| µ ≤ − kukpβ + − |u|α(x) dx − + |u|γ(x) dσx . p α γ Ω

∂Ω

If we choose rk > 0 small enough, we get ak := max {H(u) : u ∈ Yk , kukβ = rk } < 0, since dim Yk < ∞ and p− > γ + , α− > p+ . So (B1) holds.

634


(B2) Let u ∈ Zk , then H(u) ≥

+ |λ| 1 kukpβ − − + p α

Z

Z µ |u|γ(x) dσx ≥ γ− ∂Ω Z µ − − |u|γ(x) dσx ≥ γ

|u|α(x) dx −

Ω + − C|λ| 1 ≥ + kukpβ − − kukα β p α

∂Ω + − C|λ| µ 1 γ+ γ− ≥ + kukpβ − − kukα β − − max{|u|Lγ(x) (∂Ω) , |u|Lγ(x) (∂Ω) }. p α γ

There exists ρ0 > 0 small enough such that kukβ ≤ ρ0 , since α− > p+ . Then we have H(u) ≥

≤

p+ 1 2p+ kukβ

as 0 < ρ =

+ + − 1 µ kukpβ − − max{|u|γLγ(x) (∂Ω) , |u|γLγ(x) (∂Ω) }. + 2p γ −

+

C|λ| α− α− kukβ

+

If max{|u|γLγ(x) (∂Ω) , |u|γLγ(x) (∂Ω) } = |u|γLγ(x) (∂Ω) , then H(u) ≥ Choose ρk =

2p+ µγk γ−

γ+

1 p+ −γ +

+ + µ + 1 kukpβ − − γkγ kukγβ . + 2p γ

, then H(u) ≥ 0. Since p− > γ + , γk → 0, we get −

+

−

ρk → 0 as k → ∞. The case max{|u|γLγ(x) (∂Ω) , |u|γLγ(x) (∂Ω) } = |u|γLγ(x) (∂Ω) is similar, so (B2) holds. (B3) From the proof above and the fact that Yk ∩Zk 6= ∅, we know that for u ∈ Zk , kuk kβ ≤ ρk small enough H(u) ≥ −

+ µ γ+ γk kukγβ − γ

or

−

− µ γ− γk kukγβ . − γ

Since γk → 0 and ρk → 0 as k → ∞, (B3) holds and obviously we can choose ρk > rk > 0. Now, to verify the (P S)∗c condition, we consider a sequence (unj ) ⊂ X such that nj → ∞,

unj ∈ Ynj ,

H(unj ) → C,

(H|Ynj )0 (unj ) → 0.

Assume kukβ > 1, then for n large enough and λ ≥ 0 we have 1 1 C + 1 ≥ H(unj ) − − hH 0 (unj ), unj i + − hH 0 (unj ), unj i ≥ α α + 1 1 1 p− ≥ − − kunj kβ − Dkunj kγβ − − kunj kβ . + p α α Since α− > p+ and p− > γ + , we deduce that (unj ) is bounded in X. If λ < 0, then for n large enough, we have C + 1 ≥ H(unj ) −

1 1 hH 0 (unj ), unj i + + hH 0 (unj ), unj i. α+ α

635


Going if necessary to a subsequence, we can assume that unj * u weakly in X. As X = ∪nj Ynj , we can choose vnj ∈ Ynj such that vnj → u. Hence lim H 0 (unj )(unj − u) = lim H 0 (unj )(unj − vnj ) + lim H 0 (unj )(vnj − u) =

nj →∞

nj →∞

nj →∞

0

= lim (H|Ynj ) (unj )(unj − vnj ) = 0. nj →∞

Then we can conclude that unj → u since H 0 is of type (S + ). Moreover, we have H 0 (unj ) → H 0 (u). Now, it only remains to prove that H 0 (u) = 0. For an arbitrary wk ∈ Yk , we have for nj ≥ k H 0 (u)wk = (H 0 (u) − H 0 (unj ))wk + H 0 (unj )wk = = (H 0 (u) − H 0 (unj ))wk + (H|Ynj )0 (unj )wk . Going to the limit on the right side of the above equation, one get H 0 (u)wk = 0 for all wk ∈ Yk , so H 0 (u) = 0, this shows that the functional H satisfies the (P S)∗c condition for every c ∈ R. The proof of Theorem 1.4 is complete. 4. EXISTENCE OF NONNEGATIVE SOLUTION AND POSITIVE SOLUTION In this section, we will assume that f and g satisfy the following condition: f (x, 0) = 0 for all

x ∈ Ω,

Define

and g(x, 0) = 0 for all x ∈ ∂Ω. (

f (x, t), 0, ( g(x, t), g+ (x, t) = 0,

f+ (x, t) =

Let F+ (x, t) =

Rt

f+ (x, s)ds and G+ (x, t) =

0

Rt

if t ≥ 0, if t < 0, if t ≥ 0, if t < 0.

g+ (x, s)ds. Consider the following prob-

0

lem:

−∆p(x) u = λf+ (x, u) in Ω, ∂u |∇u|p(x)−2 + β(x)|u|p(x)−2 u = µg+ (x, u) on ∂Ω, ∂ν The energy functional associated with problem (4.1) is Z Z 1 1 H+ (u) = |∇u|p(x) dx + β(x)|u|p(x) dσx − p(x) p(x) Ω Z Z∂Ω − λF+ (x, u)dx − µG+ (x, u)dσx . Ω

∂Ω

(4.1)

636


Proof of Theorem 1.5. By Theorem 1.1, we know that problem (4.1) has a weak solution u. Multiplying the equation in (4.1) by u− := max{−u, 0} and integrating over Ω, in view of the boundary condition, we get Z Z − p(x) |∇u | dx + β(x)|u− |p(x) dσx = 0, Ω

∂Ω

which implies that ku− kβ = 0 and then u− = 0 in X. So we conclude that u is a nonnegative solution of the problem (4.1). By the same arguments, and using Theorem 1.2, we prove Theorem 1.6.

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638 Najib Tsouli [email protected] Department of Mathematics University Mohamed I Oujda, Morocco

Omar Darhouche [email protected] Department of Mathematics University Mohamed I Oujda, Morocco

Received: November 29, 2013. Revised: February 3, 2014. Accepted: February 4, 2014.
