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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo


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Volume 15, Issue 1

Issues

Volume 13 (2015)

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

Mohsen Khaleghi Moghadam
  • Corresponding author
  • Department of Basic Sciences, Sari Agricultural Sciences and Natural Resources University, Sari, Iran
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/ Johnny Henderson
Published Online: 2017-08-19 | DOI: https://doi.org/10.1515/math-2017-0090

Abstract

Triple solutions are obtained for a discrete problem involving a nonlinearly perturbed one-dimensional p(k)-Laplacian operator and satisfying Dirichlet boundary conditions. The methods for existence rely on a Ricceri-local minimum theorem for differentiable functionals. Several examples are included to illustrate the main results.

Keywords: Discrete boundary value problem; p(k)-Laplacian; Three solutions; Variational methods; Critical point theory

MSC 2010: 39A10; 39A12; 39A70; 34B15; 65Q10

1 Introduction

There is an increasing interest in the existence of solutions to boundary value problems for finite difference equations with the p(x)–Laplacian operator in the last decades. These kinds of problems like (1) play a fundamental role in different fields of research, because of their applications in many fields, they can model various phenomena arising from the study of elastic mechanics [33], electrorheological fluids [15] and image restoration [14] and other fields such as biological neural networks, cybernetics, ecology, control systems, economics, computer science, physics, finance, artificial and many others.

Important tools in the study of nonlinear difference equations are fixed point methods in cone; see [2, 19], and upper and lower solution techniques; see, for instance, [24, 25] and references therein. It is well known that critical point theory is an important tool to deal with the problems for differential equations. For background and recent results for nonlinear discrete boundary value problems, we refer the reader to [57, 10, 12, 13, 16, 17, 20, 27] and the seminal papers [3, 4] and references therein.

The aim of this paper is to establish the existence of three solutions for the following discrete boundary-value problem Δ(w(k1)ϕp(k1)(Δu(k1))+q(k)ϕp(k)(u(k))=λf(k,u(k))+μg(k,u(k)),u(0)=u(T+1)=0,(1) for every k ∈ [1, T], where T ≥ 2 is a fixed positive integer, [1, T] is the discrete interval {1, …, T}, f, g : [1, T] × ℝ → ℝ are two continuous functions in the second variable, λ > 0 and μ ≥ 0 are two parameters, w : [0, T] → [1, ∞) is a fix function such that w:=mink[0,T]w(k)andw+:=maxk[0,T]w(k), and Δ u(k) = u(k + 1) − u(k) is the forward difference operator, and ϕp(⋅)(s) = |s|p(⋅)−2s is as one-dimensional discrete p(⋅)-Laplacian operator such that Δ(w(k1)ϕp(k1)(Δu(k1))=w(k1)|Δu(k1)|p(k1)2Δu(k1)w(k)|Δu(k)|p(k)2Δu(k), the function p : [0, T + 1] → [2, ∞) is bounded, we denote for short p+:=maxk[0,T+1]p(k)andp:=mink[0,T+1]p(k), and the function q : [0, T + 1] →[1, ∞) is bounded such that q+:=maxk[0,T+1]q(k).

We note that problem (1) is the discrete variant of a type of the variable exponent anisotropic problem i=1Nxi(wi(x)|uxi|pi(x)2uxi)+q(x)|u|pi(x)2u=λf(k,u)+μg(k,u),xΩ,u=0,xΩ, where Ω ⊂ ℝN, N ≥ 3, is a bounded domain with smooth boundary, f, gC(Ω × ℝ, ℝ) are two given functions that satisfy certain properties, and pi(x), wi(x) ≥ 1 and q(x) ≥ 1 are continuous functions on Ω, with 2 ≤ pi(x), for each x ∈ Ω and every i ∈ {1, 2, …, N}, λ > 0 and μ are real numbers.

In [31] the authors studied (1) depending on a parameter with λ = −1 and μ = 0 by the mountain pass method. This method has been applied for (1) with q(k) = 0 and μ = 0 in [18]. In 2003, by using the upper and lower solution method, Atici and Cabada [1] considered Eq. (1) with w(k) = 1, p(k) = 2, and λ = −1 subject to the boundary value conditions u(0) = u(T), Δ u(0) = Δ u(T). In [11] the authors studied Eq. (1) with p(x) = 2 and w(k) = 1 and μ = 0 and Neumann boundary condition Δ u(0) = Δ u(T) = 0. In [23, 26], the authors considered Eq. (1) in the special case, w(k) = 1 and p(x) = p. In [21], the authors applying variational methods, studied the existence of solutions of the following Kirchhoff-type discrete boundary value problems M(up)Δ(ϕp(Δu(k1))+q(k)ϕp(u(k))=f(k,u(k)),k[1,T]Δu(0)=Δu(T+1)=0, where M : [0, ∞) → ℝ is a continuous function.

In this paper, based on a local minimum theorem (Theorem 2.5) due to Ricceri [30], we ensure an exact interval of the parameter λ, in which the problem (1) admits at least three solutions.

As an example, here, we point out the following special case of our main results.

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|00tf(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 Δ2(u(k1))+u(k)=14T(T+1)f(u(k))+μg(k,u(k)),k[1,T],u(0)=u(T+1)=0,(2) has at least three solutions whose norms are less than r.

The rest of this paper is arranged as follows. In Section 2, we recall some basic definitions and the main tool (Theorem 2.5.), and in Section 3, we provide our main results that contain several theorems, and finally, we illustrate the results by giving examples.

2 Preliminaries

Let us introduce some notations that will be used later. Let T ≥ 2 be a fixed positive integer, [1, T] is the discrete interval {1, …, T}. In order to give the variational formulation of the problem (1), we introduce, T-dimensional Banach space W:={u:[0,T+1]R:u(0)=u(T+1)=0}, equipped with the norm u:=k=1T+1w(k1)|Δu(k1)|p+q(k)|u(k)|p1/p.

If u+=k=1T+1w(k1)|Δu(k1)|p++q(k)|u(k)|p+1/p+, then by Weighted Hölder’s inequality, one can conclude that Lu+u2p+pp+pLu+, where, L={(T+1)max{w+,q+}}p+pp+p.

In the space W we can also consider the Luxemburg norm [8], up()=inf{μ>0:k=1T+1w(k1)|Δu(k1)μ|p(k1)+q(k)|u(k)μ|p(k)1}.

Since W has finite dimension, the two last norms are equivalent. Therefore there exist constants L1 > 0 and L2 > 1 such that L1up()uL2up().(3)

Now, let φ : W → ℝ be given by the formula φ(u):=k=1T+1[w(k1)|Δu(k1)|p(k1)+q(k)|u(k)|p(k)].(4)

It is easy to check that for any uW the following properties hold: up()<1up()p+φ(u)up()p, up()>1up()pφ(u)up()p+.(5)

Let, F(k,t):=0tf(k,ξ)dξ,G(k,t):=0tg(k,ξ)dξ, for every k ∈ [1, T] and t ∈ ℝ.

In the sequel, we will use the following inequality

Lemma 2.1

Let uW, Then uc1u,(6) where u=maxk[1,T]|u(k)|,c1=(2T+2)p1p.

Proof

Let u be in W. Then by the discrete Hölder inequality, for every k ∈ [1, T], we get |u(k)|k=1T+1|u(k)|+k=1T+1|Δu(k1)|(T+1)p1pk=1T+1|u(k)|p1p+(T+1)p1pk=1T+1|Δu(k1)|p1p(T+1)p1pk=1T+1q(k)|u(k)|p1p+k=1T+1w(k1)|Δu(k1)|p1p2p1p(T+1)p1pk=1T+1w(k1)|Δu(k1)|p+q(k)|u(k)|p1p=(2T+2)p1pu, for all k ∈ [1, T]. □

Lemma 2.2

  1. There exists a positive constant C1 such that for all uW withu∥ > 1, k=1T+1[w(k1)p(k1)|Δu(k1)|p(k1)+q(k)p(k)|u(k)|p(k)]upp+C1.

  2. for all uW withu∥ < 1, k=1T+1[w(k1)p(k1)|Δu(k1)|p(k1)+q(k)p(k)|u(k)|p(k)]2pp+pp+Lp+up+.

Proof

Let uW be fixed. By a similar argument as in [28], we set A<:={k[0,T+1],|Δu(k)|<1},B<:={k[0,T+1],|u(k)|<1},A>:={k[0,T+1],|Δu(k)|>1},B>:={k[0,T+1],|u(k)|>1},A=:={k[0,T+1],|Δu(k)|=1},B=:={k[0,T+1],|u(k)|=1}.

We define for each k ∈ [0, T + 1], αk:=p+, if kA<,p, if kA>,1, if kA=,andβk:=p+, if kB<,p, if kB>,1, if kB=.

Thus, if ∥u∥ > 1, we have k=1T+1[w(k1)p(k1)|Δu(k1)|p(k1)+q(k)p(k)|u(k)|p(k)]1p+k=1T+1[w(k1)|Δu(k1)|p(k1)+q(k)|u(k)|p(k)]=1p+[(kA<+kA>+kA=)w(k1)|Δu(k1)|p(k1)+(kB<+kB>+kB=)q(k)|u(k)|p(k)]1p+[k=1T+1w(k1)|Δu(k1)|αk1+k=1T+1q(k)|u(k)|βk]1p+[k=1T+1w(k1)|Δu(k1)|pkA<(w(k1)(|Δu(k1)|p|Δu(k1)|p+))+k=1T+1q(k)|u(k)|pkB<(q(k)[|u(k)|p|u(k)|p+])]1p+[k=1T+1w(k1)|Δu(k1)|pw+kA<(|Δu(k1)|p|Δu(k1)|p+)+k=1T+1q(k)|u(k)|pq+kB<([|u(k)|p|u(k)|p+])]1p+[k=1T+1w(k1)|Δu(k1)|p(T+1)w++k=1T+1q(k)|u(k)|pq+(T+1)]=upp+(T+1)(w++q+)p+.

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 k=1T+1[w(k1)p(k1)|Δu(k1)|p(k1)+q(k)p(k)|u(k)|p(k)]1p+k=1T+1[w(k1)|Δu(k1)|p++q(k)|u(k)|p+]=1p+u+p+2pp+pp+Lp+up+. □

To study the problem (1), we consider the functional Iλ,μ : W → ℝ defined by Iλ,μ(u)=k=1T+1[w(k1)p(k1)|Δu(k1)|p(k1)+q(k)p(k)|u(k)|p(k)]λk=1TF(k,u)μk=1TG(k,u).

An easy computation ensures that Iλ,μ is of class C1 on W with Iλ,μ(u)(v)=k=1T+1[w(k1)ϕp(k1)(Δu(k1))Δv(k1)+q(k)|u(k)|p(k)2u(k)v(k)]k=1T[λf(k,u(k))+μg(k,u(k)]v(k), for all u, vW.

Lemma 2.3

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

Proof

Let u be a critical point of Iλ,μ in W. Then u(0) = u(T + 1) = 0 and for all vW, Iλ,μ(u)(v) = 0. Thus, for every vW, and taking v(0) = v(T + 1) = 0 and summation by parts into account, one has 0=Iλ,μ(u¯)(v)=k=1T+1w(k1)ϕp(k1)(Δu¯(k1))Δv(k1)+k=1T[q(k)|u¯(k)|p(k)2u¯(k)v(k)λf(k,u¯(k))v(k)μg(k,u¯(k))v(k)]=w(T+1)ϕp(k1)(Δu¯(T+1))v(T+1)w(0)ϕp(k1)(Δu¯(0))v(0)k=1T+1Δ(w(k1)ϕp(k1)(Δu¯(k1)))v(k)+k=1T[q(k)|u¯(k)|p(k)2u¯(k)v(k)λf(k,u¯(k))v(k)μg(k,u¯(k))v(k)]=k=1T[Δ(w(k1)ϕp(k1)(Δu¯(k1)))q(k)|u¯(k)|p(k)2u¯(k)+λf(k,u¯(k))+μg(k,u¯(k))]v(k).

Since vW is arbitrary, one has Δ(w(k1)ϕp(k1)(Δu¯(k1))+q(k)ϕp(k)(u¯(k))=λf(k,u¯(k))+μg(k,u¯(k), 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 Iλ,μ in W is exactly a solution of the problem (1).

Also, if uW is a solution of problem (1), one can show that u is a critical point of Iλ,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 k=1T+1w(k1)ϕp(k1)(Δu¯(k1))Δv(k1)=k=1TΔ(w(k1)ϕp(k1)(Δu¯(k1)))v(k), we have Iλ,μ (u)(v) = 0, hence u is a critical point for Iλ,μ. Thus the vice versa holds and the proof is completed. □

Our main tool is a local minimum theorem (Theorem 2.5) due to Ricceri (see [30, Theorem 2]), which is recalled below. We refer to the papers [22, 30, 32] in which Theorem 2.5 has been successfully employed for the existence of at least three solutions for two-point boundary value problems. First, we give the following definition.

Definition 2.4

If X is a real Banach space, we denote by 𝓦X the class of all functionals Φ : X → ℝ possessing the following property: if {un} is a sequence in X converging weakly to uX and lim infn→∞ Φ(un) ≤ Φ(u), then {un} 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 ug(∥u∥) belongs to the class 𝓦X.

Our main tool reads as follows:

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 𝓦X, 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 α=max{0,lim supx+J(x)Φ(x),lim supx0J(x)Φ(x)},β=supxΦ1(]0+[)J(x)Φ(x), 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.

Also we state the following consequence of the Strong comparison principle [3, Lemma 2.3] (see also ([5, Theorem 2.2]) which we will use in the sequel in order to obtain positive solutions to the problem (1), i.e. u(k) > 0 for each k ∈ [1, T].

Lemma 2.6

If uW and Δ(w(k1)ϕp(k1)(Δu(k1)))+q(k)ϕp(k)(u(k))0,k[1,T],u(0)0,u(T+1)0,(7) 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 mink∈ [1, T] u(k) = u(k0) < 0 and Δ u(k0−1) ≤ 0. Hence from (7), one has 0>w(k01)ϕp(k01)(Δu(k01))+q(k0)|u(k0)|p(k0)u(k0)w(k0)ϕp(k0)(Δu(k0)).

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 0w(k)|u(k+1)|p(k)2u(k+1)w(k1)|u(k1)|p(k1)2u(k1)0, from which it follows that, u(k + 1) = u(k − 1) = 0. Thus either u is positive or u ≡ 0. Hence, the result follows. □

By a similar argument used in [5, Theorem 2.3], we obtain the next result which guarantees the same conclusion of the preceding Strong maximum principle, independently of the sign of the operator.

Lemma 2.7

Fix uW such that, if u(k) ≤ 0, it follows that Δ(w(k1)ϕp(k1)(Δu(k1)))+q(k)ϕp(k)(u(k))=0. Then either u is positive or u ≡ 0.

3 Main results

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+ppa¯c1p+}<pk=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 λ1=dp(w(0)+w(T)+k=1Tq(k))pk=T1F(k,d),λ2=1p+Tmax{b¯c1p,Lp+2p+ppa¯c1p+}, 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: Φ(u):=k=1T+1[w(k1)p(k1)|Δu(k1)|p(k1)+q(k)p(k)|u(k)|p(k)],(8) J(u):=k=1TF(k,u(k)),andΨ(u):=k=1TG(k,u(k)),(9) for every uW. So Iλ,μ = Φ − λ Jμ Ψ. An easy computation ensures that Φ, J and Ψ turn out to be of class C1 on W with Φ(u)(v)=k=1T+1[w(k1)ϕp(k1)(Δu(k1))Δv(k1)+q(k)|u(k)|p(k)2u(k)v(k)]=k=1T[Δ(w(k1)ϕp(k1)(Δu(k1))v(k)q(k)|u(k)|p(k)2u(k)v(k)], and J(u)(v)=k=1Tf(k,u(k))v(k),Ψ(u)(v)=k=1Tg(k,u(k))v(k), for all u, vW. By Lemma 2.3, the solutions of the equation Iλ¯,μ¯=Φλ¯Jμ¯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 up()>uL2>1. Also by (5), φ(u)>up()p>upL2p. Therefore by (4), one has Φ(u)>φ(u)p+>up()pp+>upp+L2p+ as u+.

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 vW and 0<Φ(v¯)<φ(v¯)p=1pw(0)dp(0)+w(T)dp(T)+k=1Tdp(k)q(k)<dppw(0)+w(T)+k=1Tq(k),(10) and J(v¯)=k=1TF(k,v¯(k))=k=1TF(k,d).

Therefore, β=supuΦ1(]0,[)J(u)Φ(u)>J(v¯)Φ(v¯)>pk=T1F(k,d)dp(w(0)+w(T)+k=1Tq(k))=1λ1>1λ¯.(11)

On the other hand, from the conditions (B1) and (B2), and bearing (6) in mind, we get k=1TF(k,u(k))<Ta¯c1sus+a¯Tc1p+up+, for u small enough,k=1TF(k,u(k))<Tb¯+b¯Tc1pup, for u large enough.

Hence, by Lemma 2.2(1) lim supuJ(u)Φ(u)<lim supuTb¯+b¯Tc1pupupp+C1=p+Tb¯c1p,(12) and by Lemma 2.2(2) lim supu0J(u)Φ(u)<lim supu0Ta¯c1sus+a¯Tc1p+up+2p+ppp+Lp+up+=p+TLp+2p+ppa¯c1+p.(13)

By (12) and (13), we get, α=max{0,lim supuJ(u)Φ(u),lim supu0J(u)Φ(u)}<p+Tmax{b¯c1p,Lp+2p+ppa¯c1p+}=1λ2<1λ¯.(14)

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(10k)k2+1, hence w(0) = 1, w(10) = 1, w+ = e4.5, L=11e454,k=1Tq(k)=10,p+=4,p=2,c1=(2T+2)p1p=22. Let f(k,t)=2coshtsinh3te1sinh2t,f(k,0)=0,t0,k[1,10],F(k,t)=e1sinh2t, for any k ∈ [1, 10], where t+ = max{0, t}. Therefore, p+Tmax{b¯c1p,Lp+2p+ppa¯c1p+}=40×223e14.50.2148117565,pk=T1F(k,d)dp(w(0)+w(T)+k=1Tq(k))=53e(ln(1+2)20.7892856465, it follows that (B0) holds. Also by the graph of the function h(x):=e1sinh(x)2|x|4.5+x4e19, 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.

The graph of the function h
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 [a,b]3e(ln(1+2))25,e14.540×223, 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 Δ(ek(10k)k2+1ϕp(k1)(Δu(k1))+ϕp(k)(u(k))=λf(k,u+(k))+μg(k,u(k)),u(0)=u(11)=0, k ∈ [1, 10], has at least three solutions whose norms are less than r, hence by Lemma 2.6 whose signs are positive.

The next result reads as follows.

Theorem 3.3

Assume that there exists a constant d > 1 such that,

(C0) p+Tc1p+2p+ppLp+<pk=1TF(k,d)dp+(w(0)+w(T)+k=1Tq(k)).

(C1) max{lim sup|t|+F(k,t)|t|p,limsup|t|0F(k,t)|t|p+}<1; for every k ∈ [1, T].

Then, for each compact interval [a, b] ⊂ Λ := ]λ1, λ2[, where λ1=dp+(w(0)+w(T)+k=1Tq(k))pk=1TF(k,d),λ2=1p+Tc1p+2p+ppLp+, there exists r > 0 with the following property: for every λ ∈ [a, b] and 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 (C0) the interval] λ1, λ2 [is non-empty, so fix λ in]λ1, λ2 [, take X = W, and put Φ, Ψ and J, as given in (8) and (9). So Iλ¯,μ¯=Φλ¯Jμ¯Ψ. By Lemma 2.3, the solutions of the equation Iλ¯,μ¯=Φλ¯Jμ¯Ψ=0 are exactly the solutions for problem (1). We know that Φ, Ψ and J satisfy the regularity assumptions of Theorem 2.5. Hence, to prove our result, it is enough to apply Theorem 2.5.

We show that for λ¯[a,b]Λ:=]λ1,λ2[to be fixed,α<1λ¯,β>1λ¯. Let d > 1 be fixed and put v(k) = d for every k ∈ [1, T] and v(0) = v(T + 1) = 0. Clearly vW and 0<Φ(v¯)<φ(v¯)p<1p(w(0)dp(0)+w(T)dp(T)+k=1Tdp(k)q(k))<dp+p(w(0)+w(T)+k=1Tq(k)).

Therefore β=supuΦ1(]0,[)J(u)Φ(u)>J(v¯)Φ(v¯)>pk=1TF(k,d)dp+(w(0)+w(T)+k=1Tq(k))=1λ1>1λ¯.(15)

In view of (C1), there exist 0 < r1 < 1 < r2 such that F(k,t)<|t|p+, for any (k,t)[1,T]×[r1,r1],F(k,t)<|t|p<|t|p+, for any (k,t)[1,T]×R([r2,r2]).

Since F is continuous, then F(k, t) is bounded for any (k, t) ∈ [1, T] × ([r1, r2] ∪ [−r2, −r1]), so we can choose C2 > 0 and s > p+ such that F(k,t)<|t|p++C2|t|s, for all (k,t)[1,T]×R, and bearing (6) in mind, we get J(u)=k=1TF(k,u(k))<Tc1p+up++TC2c1sus, for all uW, and by Lemma 2.2(2) lim sup||u||0J(u)Φ(u)<lim sup||u||0Tc1p+up++TC2c1sus2pp+pp+Lp+up+=p+2p+ppLp+Tc1p+.(16)

Again since F is continuous, then F(k, t) is bounded for any (k, t) ∈ [1, T] × [−r2, r2], so we can choose C3 > 0 such that F(k,t)<|t|p+C3, for all (k,t)[1,T]×R, and bearing (6) in mind, we get J(u)=k=1TF(k,u(k))<Tc1pup+TC3, for all uW, hence, by Lemma 2.2(1) lim sup||u||J(u)Φ(u)<lim sup||u||Tc1pu||p+TC3||u||pp+C1=p+Tc1p.(17)

By (16) and (17), we get α=max{0,lim sup||u||J(u)Φ(u),lim sup0||u||0J(u)Φ(u)}<p+Tmax{c1p,2p+ppLp+c1p+}=p+Tc1p+2p+ppLp+=1λ2<1λ¯.(18)

By (15) and (18), 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 3.3.

Example 3.4

Let T = 10, and for every k ∈ [1, 10 ], put f(k,t)=0,ift=0,t3[sin(ln|t|)+cos(ln|t|)],if|t|(0,1],t2+(2+e6π)t+e6π,ift[e6π,1],(t)2+(2+e6π)te6π,ift[1,e6π],t[sin(ln|t|)+cos(ln|t|)],if|t|[e6π,), as an odd continuous function on ℝ. Let d=eπ4,p(k)=211k+4,q(k)=1,w(k)=ek(10k)k2+5k+6 for k = 1, 2, 3, … 10. Hence p = 2, p+ = 4, k=1Tq(k)=10,q+=1,w+=e0.8,L=11e0.084andc1=(2T+2)p1p=22. Simple calculations show that F(k,t)=t417[3cos(ln|t|+5sin(ln|t|)],if|t|[0,1]t33+(2+e6π)t22e6πt23+12e6π+317,if|t|[1,e6π],t25[3sin(ln|t|)+cos(ln|t|)]+16e18π15e12π+12e6π23+317,if|t|[e6π,).lim sup|t|+F(k,t)|t|2=105<1,lim sup|t|0F(k,t)|t|4=3417<1.

By using software Maple, one can conclude that pk=1TF(k,d)dp+(w(0)+w(T)+k=1Tq(k))=2012eπ{e3π43+eπ2+e132π2e25π42551+e6π2}7873816.449>p+Tc1p+2p+ppLp+=40223e0.8947902.4.

Hence, by using Theorem 3.3, for each compact interval [a, b] ∈ ] 0.00000012, 0.000001 [there exists r > 0 with the following property: for every λ ∈ [a, b] and every continuous function g : [1, 10] × ℝ → ℝ, there exists δ > 0 such that, for each μ ∈ [0, δ], the problem Δ(ek2+5k+6k(10k)ϕp(k1)(Δu(k1))+ϕp(k)(u(k))=λf(k,u(k))+μg(k,u(k)),u(0)=u(11), for every k ∈ [1, 10], has at least three solutions whose norms are less than r.

Remark 3.5

By Lemma 2.6, the ensured solutions in the conclusions of Theorems 3.1 and 3.3 are either zero or positive. Now, let f(k, 0)+g(k, 0) = 0 for all k ∈ [0, T], by putting f(k,t)+g(k,t)=f(k,t)+g(k,t),t>0,0,t0, and due to Lemma 2.7, the ensured solutions are positive (see [5, Remark 2.1]).

Finally we present the problem (1), in which the function f(k, u) has separable variables and λ = 1.

Theorem 3.6

Let f0 : [1, T] → ℝ be a non-negative, non-zero and essentially bounded function such that k=1Tf0(k)=1p+2p+ppLp+c1p+, and f1 : ℝ → ℝ be a non-negative and continuous function and F1(ξ)=0ξf1(x)dxforeveryξR. Assume that there exists a positive constant d < 1 such that,

(DO) F1(d)dp>p+2p+ppLp+c1p+pw(0)+w(T)+k=1Tq(k)

(D1) max{lim sup|t|+F1(t)|t|p,lim sup|t|0F1(t)|t|p+}<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 Δ(w(k1)ϕp(Δu(k1)))+q(k)ϕp(u(k))=f0(k)f1(u(k))+μg(k,u(k)),u(0)=u(T+1)=0,(19) for every k ∈ [1, T], has at least three solutions whose norms are less than r.

Proof

We can choose λ = 1 ∈ [a, b] such that [a, b] ⊂ (1β,1α). Take X = W, and put Φ, Ψ and J as given in (8) and (9), so I1,μ = Φ − JμΨ and the solutions of the equation I1,μ¯ = Φ′ − J′ − μΨ′ = 0 are exactly the solutions for problem (19). Hence, to prove our result, it is enough to apply Theorem 2.5.

We show that, α < 1, β > 1. Put v(k) = d ∈ (0, 1) for every k ∈ [1, T] and v(0) = v(T + 1) = 0. In view of (D0) and (10), β>J(v¯)Φ(v¯)>pF1(d)k=1Tf0(k)dp(w(0)+w(T)+k=1Tq(k))>p+2p+ppLp+c1p+k=1Tf0(k)=1.(20)

Applying similar argument as in the proof of Theorem 3.3, and taking into account (D1) and Lemma 2.2 lim supu0J(u)Φ(u)p+2p+ppLp+c1p+k=1Tf0(k),lim supuJ(u)Φ(u)p+c1pk=1Tf0(k), therefore, considering c1 > 6 and L ≥ 1, we have α=max{0,lim supuJ(u)Φ(u),lim supu0J(u)Φ(u)}p+2p+ppLp+c1p++k=1Tf0(k)=1.(21)

By (20) and (21), 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. □

Remark 3.7

Theorem 1.1 follows from Theorem 3.6 taking into account that p(k) = 2, f1(t) = f(t) and w(k − 1) = 1 for every k ∈ [1, T + 1], q(k) = 1, f0(k)=14T(T+1) for every k ∈ [1, T].

Finally, we present an example of Theorem 1.1.

Example 3.8

Let T = 4 be a fixed positive integer Consider the problem (2) where f(t)=0,ift=0,t[sin(2ln|t|)+cos(2ln|t|)],if0<te2π,e2πsin(2πln|t|)+t,ife2πt1,t[sin(2ln|t|)+cos(2ln|t|)],if1t, as an odd continuous function on ℝ. Simple calculations show that F(t)=t22[(sin(2ln|t|)],ift<e2π,e2π1+(2π)2[tsin(2πln|t|)2πtcos(2πln|t|)]+t22+K,ife2πt<1,t22[sin(2ln|t|)]+L,if1t, where K=2π1+(2π)2e4π2,L=K2πe2π1+(2π)2+12.

Thus lim sup|t|0,+F(k,t)|t|2=12<1.

Hence, (A1) holds. Let d=e12(e2π,1), so one has, F(d)=e2π12(2π)1+(2π)2+e12+K50.75439760>d2(2T2+6T+4)36.39183958.

Thus, (A2) holds. Therefore, by using Theorem 1.1, there exists r > 0 such that for every continuous function g : [1, T] × ℝ → ℝ, there exists δ > 0 such that, for each μ ∈ [0, δ], the problem (2) has at least three solutions whose norms are less than r.

Acknowledgement

The authors express their gratitude to Professor Biagio Ricceri for his helpful suggestions. The authors are thankful to the anonymous referee for helpful suggestions and comments.

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About the article


Received: 2016-12-10

Accepted: 2017-05-29

Published Online: 2017-08-19


Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 1075–1089, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2017-0090.

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© 2017 Moghadam and Henderson. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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