Three Solutions for Inequalities Dirichlet Problem Driven by p(x)-Laplacian-Like

Proposition 1 (see [12].)In $$ Poincare′s inequality holds; that is, there exists a positive constant C₀ such that

So |∇u|_p(x) is an equivalent norm in $$.

Proposition 2 (see [10].)If $$ and q(x) < p^*(x) for any $$, then the embedding from W^1,p(x)(Ω) to L^q(x)(Ω) is compact and continuous.

Consider the following function:

We know that (see [1]).

If one denotes $$, then

for all $$.

Proposition 3 (see [19].)Set $$; A is as shown, then

• (1)

    A : X → X^*  is a convex, bounded previously; and strictly monotone operator;

• (2)

    A : X → X^*  is a mapping of type (S) ₊; that is, $$  in X  and limsup _n→∞〈A(u_n), u_n − u〉≤0  implies u_n → u  in X;

• (3)

    A : X → X^*  is a homeomorphism.

Let (X, ∥·∥) be a real Banach space, and let X^* be its topological dual. A function f : X → ℝ is called locally Lipschitz if each point u ∈ X possesses a neighborhood Ω_u such that |f(u₁) − f(u₂)| ≤ L  ∥u₁ − u₂∥ for all u₁, u₂ ∈ Ω_u, for a positive constant L depending on Ω_u. The generalized directional derivative of f at the point u ∈ X in the direction h ∈ X is

The generalized gradient of f at u ∈ X is defined by

which is a nonempty, convex, and w^*-compact subset of X, where 〈·, ·〉 is the duality pairing between X^* and X. One says that u ∈ X is a critical point of f if 0 ∈ ∂f(u).

Theorem 4 (see [22], [23].)Let X be a separable and reflexive real Banach space, and let Φ, Ψ : X → ℝ be two locally Lipschitz functions. Assume that there exists u₀ ∈ X such that Φ(u₀) = Ψ(u₀) = 0 and Φ(u) ≥ 0 for every u ∈ X and that there exist u₁ ∈ X and r > 0 such that

• (1)

  r < Φ(u₁);

• (2)

  sup _Φ(u)<rΨ(u) < r(Ψ(u₁)/Φ(u₁)), and further, one assumes that function Φ − λΨ is sequentially lower semicontinuous and satisfies the (PS)-condition;

• (3)

  lim _{∥u∥→∞}(Φ(u) − λΨ(u)) = +∞ for every $$, where

  $$ ()

Then, there exist an open interval $$ and a positive real number σ such that, for every λ ∈ Λ₁, the function Φ(u) − λΨ(u) admits at least three critical points whose norms are less than σ.

Definition 5. We say that I satisfies (PS)_c-condition if any sequence $$, such that I(u_n) → c and m(u_n) → 0, as n → +∞, has a strongly convergent subsequence, where $$.

By a solution of (P), we mean a function $$ to which there corresponds a mapping Ω∋x → w(x) with w(x) ∈ ∂F(x, u) for almost every x ∈ Ω having the property that, for every $$, the function x → w(x)v(x) ∈ L¹(Ω) and

We know that $$ is compactly embedded into $$ (by N < p⁻ < p^*(x)). So there is a constant c₀ > 0 such that |u|_∞ ≤ c₀∥u∥, for all $$.

Set $$,  $$ and φ(u) = Φ(u) − λΨ(u), for all $$.

We know that the critical points of φ are just the weak solutions of (P).

We consider a non-smooth potential function F : Ω × ℝ → ℝ such that F(x, 0) = 0 a.e. on Ω satisfying the following conditions:

H(j): 

•  

  (h₁) F(·, t) is measurable for all t ∈ ℝ;

•  

  (h₂) F(x, ·) is locally Lipschitz for a.e. x ∈ Ω;

•  

  (h₃) there exist a ∈ L^∞(Ω) ₊, c > 0 such that

  $$ ()
  where w ∈ ∂F(x, t) and 1 < α⁻ ≤ α⁺ < p⁻;

•  

  (h₄) there exists $$ with p⁺ < q⁻ ≤ q(x) < p^*(x), such that lim _|t|→0(F(x, t)/|t|^q(x)) = 0 uniformly a.e. x ∈ Ω;

•  

  (h₅) sup _t∈ℝF(x, t) > 0, for all $$.

Theorem 6. Let (h₁)–(h₅) hold. Then, there are an open interval Λ⊆[0, +∞) and a number σ such that, for every λ belonging to Λ, problem (P) possesses at least three solutions in $$ whose norms are less than σ.

Proof. We observe that Ψ(u) is Lipschitz on L^α(x)(Ω) and, taking into account that α(x) < p^*(x), Ψ is also locally Lipschitz on $$ (see Proposition 2.2 of [15]). Moreover it results in ∂Ψ(u)⊆∫_Ω ∂F(x, u)dx (see [24]). The interpretation of ∂Ψ(u)⊆∫_Ω ∂F(x, u)dx is as follows: to every w ∈ ∂Ψ(u) there corresponds a mapping w(x) ∈ ∂F(x, u) for almost all x ∈ Ω having the property that for every $$ the function w(x)v(x) ∈ L¹(Ω) and 〈w, v〉 = ∫_Ω w(x)v(x)dx (see [24]). The proof is divided into the following five steps.

Step  1. We show that φ is coercive.

By (h₂), for almost all x ∈ Ω, t ↦ F(x, t) is differentiable almost everywhere on ℝ and we have

From (h₃), there exist positive constants a₁, a₂ such that for a.e. x ∈ Ω and t ∈ ℝ.

Note that 1 < α(x) ≤ α⁺ < p⁻ < p^*(x); then by Proposition 2, we have $$ (compact embedding). Furthermore, there exists c₁ such that |u|_α(x) ≤ c₁∥u∥.

So, for |u|_α(x) > 1 and ∥u∥  > 1, we have $$.

Hence,

as ∥u∥ → +∞.

Step  2. We show that φ is weakly lower semicontinuous.

Let u_n⇀u weakly in $$, and by Proposition 2, we obtain the following results:

By Fatou′s lemma, we have Thus,

Step  3. We show that (PS)-condition holds.

Suppose $$ such that |φ(u_n)| ≤ c and m(u_n) → 0 as n → +∞. If $$ is such that $$, then we know that

where the nonlinear operator $$ is defined as for all $$. From the work of Chang [1], we know that if w_n ∈ ∂Ψ(u_n), then $$, where 1/α^′(x) + 1/α(x) = 1.

Since φ is coercive, {u_n} _n≥1 is bounded in $$ and there exists $$ such that a subsequence of {u_n} _n≥1, which is still denoted as {u_n} _n≥1, satisfies u_n⇀u weakly in $$. Next we will prove that u_n → u in $$.

By $$, we have u_n → u in L^α(x)(Ω). Moreover, since $$, we get $$.

Since $$, we obtain

Moreover, since u_n → u in L^α(x)(Ω) and {w_n} _n≥1 are bounded in $$, where 1/α(x) + 1/α^′(x) = 1, one has ∫_Ω w_n(u_n − u)dx → 0. Therefore, But we know that Φ^′ is a mapping of type (S₊) (by Proposition 3). Thus we obtain

Step  4. There exists a $$ such that Ψ(u₁) > 0.

By (h₅), for each $$, there is t_x ∈ ℝ such that F(x, t_x) > 0.

For x ∈ ℝ^N, denote by N_x a neighborhood of x which is the product of N compact intervals. From (h₅) and $$, for any $$, there are $$, $$ and δ₀ > 0, such that $$ for all $$.

Since Ω⊆ℝ^N is bounded, $$ is compact. Then we can find $$ such that $$ and $$ and, also, we can find $$, and n positive numbers δ₁, δ₂, …, δ_n such that

Now, set δ₀ = min {δ₁, δ₂, …, δ_n}, and $$, and

Then, we can find a closed set $$ such that where meas (A) denote the Lebesgue measure of set A. We consider a function $$ such that |u₁(x)|∈[0, t₀] and $$ for all $$. For instance, we can set $$, where $$ and Then, from (27)–(29), we have

Step  5. We show that Φ, Ψ satisfy conditions (1) and (2) of Theorem 4.

Let u₀ = 0; then we can easily find Φ(u₀) = Ψ(u₀) = 0.

From (7) and Proposition 1, we have the following:

if ∥u∥  ≥ 1, then

if ∥u∥  < 1, then From (h₄), there exist η ∈   ]0,1[ and C₃ > 0 such that In view of (h₃), if we put then we have Fix r such that 0 < r < 1. And when $$, by Sobolev Embedding Theorem ($$), we have (for suitable positive constants C₅, C₆) Since q⁻ > p⁺ ≥ p⁻, we have And so, taking into account (32) and (33), From Step 4, there exists $$ such that Ψ(u₁) > 0. Thanks to (32) and (33), we have and so By (32), (33), and (39), there exists $$ such that, for each r ∈   ]0, r₀[, By choosing r ∈  ]0, r₀[, conditions (1) and (2) requested in Theorem 4 are verified and so the proof is complete.