Triple solutions for a Dirichlet boundary value problem involving a perturbed discrete p(k)-Laplacian operator

Theorem 1.1

Assume T ≥ 2 is a fixed integer number and there exists a positive constant d < 1 such that,

(A0) ∫0df(t)dt>d2(2T2+6T+4).

(A1) max{limsup|t|→+∞∫0tf(ξ)dξt2,limsup|t|→0∫0tf(ξ)dξt2}<1.

Then, there exists r > 0 with the following property: for every continuous function g : [1, T] × ℝ → ℝ, there exists δ > 0 such that, for each μ ∈ [0, δ], the problem

has at least three solutions whose norms are less than r.

Lemma 2.1

Let u ∈ W, Then

where

Proof

Let u be in W. Then by the discrete Hölder inequality, for every k ∈ [1, T], we get

for all k ∈ [1, T]. □

Lemma 2.2

1. There exists a positive constant C₁ such that for all u ∈ W with ∥u∥ > 1,

   ∑k=1T+1[w(k−1)p(k−1)|Δu(k−1)|p(k−1)+q(k)p(k)|u(k)|p(k)]≥∥u∥p−p+−C1.

2. for all u ∈ W with ∥u∥ < 1,

   ∑k=1T+1[w(k−1)p(k−1)|Δu(k−1)|p(k−1)+q(k)p(k)|u(k)|p(k)]≥2p−−p+p−p+Lp+∥u∥p+.

Proof

Let u ∈ W be fixed. By a similar argument as in [28], we set

We define for each k ∈ [0, T + 1],

Thus, if ∥u∥ > 1, we have

If ∥u∥ < 1, then A^> = A⁼ = B^> = B⁼ = ∅. It follows that |Δ u(k − 1 |, |u(k)| < 1 for each k ∈ [1, T + 1]. Hence, we have

 □

Lemma 2.3

The critical points of I_λ,μ are exactly the solutions of the problem (1).

Proof

Let u be a critical point of I_λ,μ in W. Then u(0) = u(T + 1) = 0 and for all v ∈ W, Iλ,μ′ (u)(v) = 0. Thus, for every v ∈ W, and taking v(0) = v(T + 1) = 0 and summation by parts into account, one has

Since v ∈ W is arbitrary, one has

for every k ∈ [1, T]. Therefore, u is a solution of (1). So by bearing in mind that u is arbitrary, we conclude that every critical point of the functional I_λ,μ in W is exactly a solution of the problem (1).

Also, if u ∈ W is a solution of problem (1), one can show that u is a critical point of I_λ,μ. Indeed, by multiplying the difference equation in problem (1) by v(k) as an arbitrary element of W and summing and using the fact that

we have Iλ,μ′ (u)(v) = 0, hence u is a critical point for I_λ,μ. Thus the vice versa holds and the proof is completed. □

Definition 2.4

If X is a real Banach space, we denote by 𝓦_X the class of all functionals Φ : X → ℝ possessing the following property: if {u_n} is a sequence in X converging weakly to u ∈ X and lim inf_n→∞ Φ(u_n) ≤ Φ(u), then {u_n} has a subsequence converging strongly to u. For instance, if X is uniformly convex and g : [0, +∞[→ ℝ is a continuous, strictly increasing function, then, by a classical result, the functional u → g(∥u∥) belongs to the class 𝓦_X.

Theorem 2.5

([30, Theorem 2]). Let X be a separable and reflexive real Banach space and Φ : X → ℝ be a coercive, sequentially weakly lower semicontinuous C¹ functional, belonging to 𝓦_X, bounded on each bounded subset of X and whose derivative admits a continuous inverse on X^*; J : X → ℝ a C¹ functional with compact derivative. Assume that Φ has a strict local minimum x₀ with Φ(x₀) = J(x₀) = 0. Finally, setting

assume that α < β. Then, for each compact interval [a, b]⊂] 1β,1α [(with the conventions 10=+∞,1+∞ = 0 ) there exists r > 0 with the following property: for every λ ∈ [a, b] and every C¹ functional Ψ : X → ℝ with compact derivative, there exists δ > 0 such that, for each μ ∈ [0, δ], the equation Φ′(x) = λ J′(x) + μ Ψ′(x) has at least three solutions whose norms are less than r.

Lemma 2.6

If u ∈ W and

then either u is positive or u ≡ 0.

Proof

First, we can prove that u ≥ 0. Assume that u(k)≱ 0 for any k ∈ [0, T + 1]. Then, there would exist k₀ ∈ [0, T + 1] such that min_{k∈ [1, T]} u(k) = u(k₀) < 0 and Δ u(k₀−1) ≤ 0. Hence from (7), one has

Thus u(k₀+1) < u(k₀). This is a contradiction. Therefore one can conclude u(k) ≥ 0 for any k ∈ [0, T + 1]. Finally, arguing as in the proof of [3, Lemma 2.2], we observe that if u(k) = 0 satisfying (7), then by positivity of w(k) and w(k − 1) and non-negativity of u(k + 1), u(k − 1), one has

from which it follows that, u(k + 1) = u(k − 1) = 0. Thus either u is positive or u ≡ 0. Hence, the result follows. □

Lemma 2.7

Fix u ∈ W such that, if u(k) ≤ 0, it follows that

Then either u is positive or u ≡ 0.

Theorem 3.1

Assume that there exist constants M > 0, m > 0, d ∈ (0, 1), and a, b > 0 and s > p⁺ such that

(B0) p+Tmax{b¯c1p−,Lp+2p+−p−p−a¯c1p+}<p−∑k=1TF(k,d)dp−(w(0)+w(T)+∑k=1Tq(k)).

(B1) F(k, t) < a(|t|^s + |t|^p+); for every (k, |t|) ∈ [1, T] × [0, m].

(B2) F(k, t) < b(1 + |t|^p−); for every (k, |t|) ∈ [1, T] × [M, ∞).

Then, for each compact interval [a, b] ⊂ Λ :=]λ₁, λ₂[, where

there exists r > 0 with the following property: for every λ ∈ [a, b] and for every continuous function g : [1, T] × ℝ → ℝ, there exists δ > 0 such that, for each μ ∈ [0, δ], the problem (1) has at least three solutions whose norms are less than r.

Proof

Our aim it to apply Theorem 2.5 to our problem. To this end, we observe that due to (B0) the interval ]λ₁, λ₂[ is non-empty, so fix λ in [a, b] ⊂]λ₁, λ₂[, take X = W, and put Φ, Ψ as follows:

for every u ∈ W. So I_λ,μ = Φ − λ J − μ Ψ. An easy computation ensures that Φ, J and Ψ turn out to be of class C¹ on W with

and

for all u, v ∈ W. By Lemma 2.3, the solutions of the equation Iλ¯,μ¯′=Φ′−λ¯J′−μ¯Ψ′=0 are exactly the solutions for problem (1). Hence, to prove our result, it is enough to apply Theorem 2.5. We know Φ is a nonnegative continuously Gâteaux differentiable and sequentially weakly lower semicontinuous functional whose Gâteaux derivative admits a continuous inverse on X^*, and Ψ is a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact. Also the functional Φ is coercive. Indeed, by (3) for ∥u∥ > L₂, we have ∥u∥p(⋅)>∥u∥L2>1. Also by (5), φ(u)>∥u∥p(⋅)p−>∥u∥p−L2p−. Therefore by (4), one has

In addition, Φ has a strict local minimum 0 with Φ(0) = J(0) = 0.

We show that for λ ∈ [a, b] ⊂ Λ :=]λ₁, λ₂[ to be fixed, α<1λ¯,β>1λ¯. Let d ∈ (0, 1) be fixed and put v(k) = d for every k ∈ [1, T] and v(0) = v(T + 1) = 0. Clearly v ∈ W and

and

Therefore,

On the other hand, from the conditions (B1) and (B2), and bearing (6) in mind, we get

Hence, by Lemma 2.2(1)

and by Lemma 2.2(2)

By (12) and (13), we get,

By (11) and (14), we deduce that α < β. All the assumptions of Theorem 2.5 are satisfied, so by applying that theorem, the conclusion follows. Hence, the proof is complete. □

Now, we present an example to illustrate the results of Theorem 3.1.

Example 3.2

Choose m = 0.198, M = 250, d=ln⁡(1+2)∈(0,1),s=4.5,p(k)=2k11+2, a = e⁻¹⁹, b = e⁻¹¹, T = 10, q(k) = 1, w(k)=ek(10−k)k2+1, hence w(0) = 1, w(10) = 1, w⁺ = e^4.5, L=11e454,∑k=1Tq(k)=10,p+=4,p−=2,c1=(2T+2)p−−1p−=22. Let

for any k ∈ [1, 10], where t⁺ = max{0, t}. Therefore,

it follows that (B0) holds. Also by the graph of the function

on [−0.198, 0.198], where it is under x axis (see Figure 1), one can conclude (B1) holds. It is clear that (B2) holds.

Figure 1

The graph of the function h

We see that all assumptions of Theorem 3.1 are satisfied, hence Theorem 3.1 implies that for each compact interval

there exists r > 0 with the following property: for every λ ∈ [a, b] and every positive continuous function g : [1, 10] × ℝ → ℝ, there exists δ > 0 such that, for each μ ∈ [0, δ], the problem