# On the Solvability of Discrete Nonlinear Two-Point Boundary Value Problems

Authors: Blaise Kone; Stanislas Ouaro
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We prove the existence and uniqueness of solutions for a family of discrete boundary value problems by using discrete's Wirtinger inequality. The boundary condition is a combination of Dirichlet and Neumann boundary conditions.


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## Body

## 1. Introduction

In this paper, we study the following nonlinear discrete boundary value problem:(1.1)-Δ(a(k-1,Δu(k-1)))=f(k),k∈Z[1,T],u(0)=Δu(T)=0,
where T≥2 is a positive integer and Δu(k)=u(k+1)-u(k) is the forward difference operator. Throughout this paper, we denote by ℤ[a,b] the discrete interval {a,a+1,…,b}, where a and b are integers and a<b.We consider in (1.1) two different boundary conditions: a Dirichlet boundary condition (u(0)=0) and a Neumann boundary condition (Δu(T)=0). In the literature, the boundary condition considered in this paper is called a mixed boundary condition.We also consider the function space(1.2)W={v:Z[0,T+1]⟶R;such thatv(0)=Δv(T)=0},
where W is a T-dimensional Hilbert space [1, 2] with the inner product
(1.3)(u,v)=∑k=1T‍u(k)v(k),∀u,v∈W.
The associated norm is defined by(1.4)‖u‖=(∑k=1T‍|u(k)|2)1/2.
For the data f and a, we assume the following.(H1)
f:ℤ[1,T]→ℝ.(H2)
a(k,·):ℝ→ℝforallk∈ℤ[0,T] and their exists a mapping A:ℤ[0,T]×ℝ→ℝ which satisfies a(k,ξ)=(∂/∂ξ)A(k,ξ), for all k∈ℤ[0,T] and A(k,0)=0, for all k∈ℤ[0,T].(H3)
(a(k,ξ)-a(k,η))·(ξ-η)>0 for all k∈ℤ[0,T] and ξ,η∈ℝ such that ξ≠η.(H4)
|ξ|p(k)≤a(k,ξ)ξ≤p(k)A(k,ξ) for all k∈ℤ[0,T] and ξ∈ℝ.(H5)
p:ℤ[0,T]→(1,+∞).The theory of difference equations occupies now a central position in applicable analysis. We just refer to the recent results of Agarwal et al. [1], Yu and Guo [3], Koné and Ouaro [4], Guiro et al. [5], Cai and Yu [6], Zhang and Liu [7], Mihăilescu et al. [8], Candito and D′Agui [9], Cabada et al. [10], Jiang and Zhou [11], and the references therein. In [7], the authors studied the following problem:(1.5)Δ2y(k-1)+λf(y(k))=0,k∈Z[1,T],y(0)=y(T+1)=0,
where λ>0 is a parameter, Δ2y(k)=Δ(Δy(k)), f:[0,+∞)→ℝ a continuous function satisfying the condition
(1.6)f(0)=-a<0,whereaisapositiveconstant.The problem (1.5) is referred as the “semipositone” problem in the literature, which was introduced by Castro and Shivaji [2]. Semipositone problems arise in bulking of mechanical systems, design of suspension bridges, chemical reactions, astrophysics, combustion, and management of natural resources.The studies regarding problems like (1.1) or (1.5) can be placed at the interface of certain mathematical fields such as nonlinear partial differential equations and numerical analysis. On the other hand, they are strongly motivated by their applicability in mathematical physics as mentioned above.In [11], Jiang and Zhou studied the following problem:
(1.7)Δ2u(k-1)=f(k,u(k)),k∈Z[1,T],u(0)=Δu(T)=0,
where T is a fixed positive integer, f:ℤ[1,T]×ℝ→ℝ is a continuous function. Jiang and Zhou proved an existence of nontrivial solutions for (1.7) by using strongly monotone operator principle and critical point theory.In this paper, we consider the same boundary conditions as in [11] but the main operator is more general than the one in [11] and involves variable exponent.Problem (1.1) is a discrete variant of the variable exponent anisotropic problem
(1.8)-∑i=1N‍∂∂xiai(x,∂u∂xi)=f(x)inΩ,u=0on⁡Γ1∂u∂n=0on⁡Γ2,
where Ω⊂ℝN(N≥3) is a bounded domain with smooth boundary, Γ1∪Γ2=∂Ω, f∈L∞(Ω), pi continuous on Ω¯ such that 1<pi(x)<N and ∑i=1N‍(1/pi-)>1 for all x∈Ω¯ and all i∈ℤ[1,N], where pi-:=essinfx∈Ωpi(x).The first equation in (1.8) was recently analyzed by Koné et al. [12] and Ouaro [13] and generalized to a Radon measure data by Koné et al. [14] for an homogeneous Dirichlet boundary condition (u=0 on ∂Ω). The study of (1.8) will be done in a forthcoming work. Problems like (1.8) have been intensively studied in the last decades since they can model various phenomena arising from the study of elastic mechanics (see [15, 16]), electrorheological fluids (see [15, 17–19]), and image restoration (see [20]). In [20], Chen et al. studied a functional with variable exponent 1≤p(x)≤2 which provides a model for image denoising, enhancement, and restoration. Their paper created another interest for the study of problems with variable exponent.Note that Mihăilescu et al. (see [21, 22]) were the first authors who studied anisotropic elliptic problems with variable exponent. In general, the interested reader can find more information about difference equations in [1–11, 23–25], more information about variable exponent in [12–22, 26].Our goal in this paper is to use a minimization method in order to establish some existence results of solutions of (1.1). The idea of the proof is to transfer the problem of the existence of solutions for (1.1) into the problem of existence of a minimizer for some associated energy functional. This method was successfully used by Bonanno et al. [27] for the study of an eigenvalue nonhomogeneous Neumann problem, where, under an appropriate oscillating behavior of the nonlinear term, they proved the existence of a determined open interval of positive parameters for which the problem considered admits infinitely many weak solutions that strongly converges to zero, in an appropriate Orlicz-Sobolev space. Let us point out that, to our best knowledge, discrete problems like (1.1) involving anisotropic exponents have been discussed for the first time by Mihăilescu et al. (see [8]), in a second time by Koné and Ouaro [4], and in a third time by Guiro et al. [5]. In [8], the authors proved by using critical point theory, existence of a continuous spectrum of eigenvalues for the problem
(1.9)-Δ(|Δu(k-1)|p(k-1)-2Δu(k-1))=λ|u(k)|q(k)-2u(k),k∈Z[1,T],u(0)=u(T+1)=0,
where T≥2 is a positive integer and the functions p:ℤ[0,T]→[2,∞) and q:ℤ[1,T]→[2,∞) are bounded while λ is a positive constant.In [4], Koné and Ouaro proved, by using minimization method, existence and uniqueness of weak solutions for the following problem:
(1.10)-Δ(a(k-1,Δu(k-1)))=f(k),k∈Z[1,T],u(0)=u(T+1)=0,
where T≥2 is a positive integer.The functiona(k-1,Δu(k-1)) which appears in the left-hand side of problem (1.1) is more general than the one which appears in (1.9). Indeed, as examples of functions which satisfy the assumptions (H2)–(H5), we can give the following.  (i)
A(k,ξ)=(1/p(k))|ξ|p(k), where a(k,ξ)=|ξ|p(k)-2ξ, for all k∈ℤ[0,T] and ξ∈ℝ.(ii)
A(k,ξ)=1/p(k)[(1+|ξ|2)p(k)/2-1], where a(k,ξ)=(1+|ξ|2)(p(k)-2)/2ξ, for all k∈ℤ[0,T] and ξ∈ℝ.In [5], Guiro et al. studied the following two-point boundary value problems
(1.11)-Δ(a(k-1,Δu(k-1)))+|u(k)|p(k)u(k)=f(k),k∈Z[1,T],Δu(0)=Δu(T)=0.The functiona(k-1,Δu(k-1)) has the same properties as in [4], but the boundary conditions are different. For this reason, Guiro et al. defined a new norm in the Hilbert space considered in order to get, by using minimization methods, existence of a unique weak solution (which is also a classical solution since the Hilbert space associated is of finite dimension). Indeed, they used the following norm:(1.12)‖u‖=(∑k=1T+1‍|Δu(k-1)|2+∑k=1T‍|u(k)|2)1/2,
which is associated to the Hilbert space
(1.13)W={v:Z[0,T+1]⟶R;suchthatΔv(0)=Δv(T)=0}.
In order to get the coercivity of the energy functional, the authors of [5] assumed the following on the exponent:
(1.14)p:Z[0,T]⟶(2,+∞).
The assumption above allowed them to exploit the convexity property of the map x→xp-/2.Problem (1.11) is a discrete variant of the following problem:
(1.15)-∑i=1N‍∂∂xiai(x,∂u∂xi)=f(x,u)inΩ,∂u∂n=0on⁡Ω,
which was studied by Boureanu and Radulescu in [26] with an additional condition that u≥0. Note that, in [26], the Neumann condition is more general than the one in problem (1.11). In this paper, we use the discrete Wirtinger inequality (see [23]) which allows us to assume that the exponent p:ℤ[0,T]→(1,+∞). The discrete Wirtinger inequality is a discrete variant of the well-known Poincaré-Wirtinger inequality (see [28]). Another difference of the present paper compared to [5] is on the boundary condition.The remaining part of this paper is organized as follows. Section2 is devoted to mathematical preliminaries. The main existence and uniqueness result is stated and proved in Section 3. In Section 4, we discuss some extensions, and, finally, in Section 5, we apply our theoretical results to an example.


## 2. Preliminaries

From now, we will use the following notations:(2.1)p-=min⁡k∈Z[0,T]p(k),p+=max⁡k∈Z[0,T]p(k).
Moreover, it is useful to introduce other norms on W, namely,
(2.2)|u|m=(∑k=1T‍|u(k)|m)1/m∀u∈W,m≥2.
We have the following inequalities (see [6, 8]) which are used in the proof of Lemma 2.1:
(2.3)T(2-m)/(2m)|u|2≤|u|m≤T1/m|u|2∀u∈W,m≥2.
In the sequel, we will use the following auxiliary result.Lemma 2.1 (see [5]).
There exist two positive constantsC1, C2 such that
(2.4)∑k=1T+1‍|Δu(k-1)|p(k-1)≥C1(∑k=1T+1‍|Δu(k-1)|2)p-/2-C2,
for all u∈Wwith∥u∥>1.We have the following result.Lemma 2.2 (discrete Wirtinger's inequality, see Theorem12.6.2, page 860 in [23]).
For any functionu(k), k∈ℤ[0,T] satisfying u(0)=0, the following inequality holds:
(2.5)4sin⁡2(π2(2T+1))∑k=1T‍|u(k)|2≤∑k=0T-1‍|Δu(k)|2.


## 3. Existence and Uniqueness of Weak Solution

In this section, we study the existence and uniqueness of weak solution of (1.1).Definition 3.1.
A weak solution of (1.1) is a function u∈W such that
(3.1)∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)=∑k=1T‍f(k)v(k)foranyv∈W.
Note that, sinceW is a finite dimensional space, the weak solutions coincide with the classical solutions of the problem (1.1).
We have the following result.Theorem 3.2.
Assume that (H1)–(H5) hold. Then, there exists a unique weak solution of (1.1).
The energy functional corresponding to problem (1.1) is defined by J:W→ℝ such that
(3.2)J(u)=∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍f(k)u(k).
We first present some basic properties of J.Proposition 3.3.
The functionalJ is well defined on W and is of class C1(W,ℝ) with the derivative given by
(3.3)〈J′(u),v〉=∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)-∑k=1T‍f(k)v(k),
for all u,v∈W.The proof of Proposition3.3 can be found in [5].We now define the functionalI by
(3.4)I(u)=∑k=1T+1‍A(k-1,Δu(k-1)).We need the following lemma for the proof of Theorem3.2.Lemma 3.4.
The functionalI is weakly lower semicontinuous.Proof.
A is convex with respect to the second variable according to (H2). Thus, it is enough to show that I is lower semicontinuous. For this, we fix u∈W and ϵ>0. Since I is convex, we deduce that, for any v∈W,
(3.5)I(v)≥I(u)+〈I′(u),v-u〉≥I(u)+∑k=1T+1‍a(k-1,Δu(k-1))(Δv(k-1)-Δu(k-1))≥I(u)-∑k=1T+1‍|a(k-1,Δu(k-1))||Δv(k-1)-Δu(k-1)|≥I(u)-∑k=1T+1‍|a(k-1,Δu(k-1))||v(k)-u(k)+u(k-1)-v(k-1)|≥I(u)-∑k=1T+1‍|a(k-1,Δu(k-1))|(|v(k)-u(k)|+|v(k-1)-u(k-1)|).
We define H and B by
(3.6)H=∑k=1T+1‍|a(k-1,Δu(k-1))||v(k)-u(k)|,B=∑k=1T+1‍|a(k-1,Δu(k-1))||v(k-1)-u(k-1)|.
By using Schwartz inequality, we get
(3.7)H≤(∑k=1T+1‍|a(k-1,Δu(k-1))|2)1/2(∑k=1T+1‍|v(k)-u(k)|2)1/2≤(∑k=1T+1‍|a(k-1,Δu(k-1))|2)1/2‖v-u‖.
The same calculus gives
(3.8)B≤(∑k=1T+1‍|a(k-1,Δu(k-1))|2)1/2‖v-u‖.
Finally, we have
(3.9)I(v)≥I(u)-(1+2∑k=1T+1‍|a(k-1,Δu(k-1))|2)1/2‖v-u‖≥I(u)-ϵ
for all v∈W with ∥v-u∥<δ=ϵ/K(T,u), where K(T,u)=(1+2∑k=1T+1‍|a(k-1,Δu(k-1))|2)1/2.
We conclude thatI is weakly lower semicontinuous.We also have the following result.Proposition 3.5.
The functionalJ is bounded from below, coercive, and weakly lower semicontinuous.Proof .
By Lemma3.4, J is weakly lower semicontinuous. We will only prove the coerciveness of the energy functional since the boundedness from below of J is a consequence of coerciveness. The other proofs can be found in [5]. By (H4), we deduce that
(3.10)J(u)=∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍f(k)u(k)≥∑k=1T+1‍1p(k)|Δu(k-1)|p(k-1)-∑k=1T‍|f(k)u(k)|≥1p+∑k=1T+1‍|Δu(k-1)|p(k-1)-(∑k=1T‍|f(k)|2)1/2(∑k=1T‍|u(k)|2)1/2.
To prove the coercivity of J, we may assume that ∥u∥>1, and we get from the above inequality and Lemma 2.1, the following:
(3.11)J(u)≥C1p+(∑k=1T+1‍|Δu(k-1)|2)p-/2-C2-(∑k=1T‍|f(k)|2)1/2(∑k=1T‍|u(k)|2)1/2.
Using Wirtinger's discrete inequality, we obtain
(3.12)J(u)≥C1p+(4sin⁡2(π2(2T+1))∑k=1T‍|u(k)|2)p-/2-C2-(∑k=1T‍|f(k)|2)1/2(∑k=1T‍|u(k)|2)1/2≥C12p-p+sin⁡p-(π2(2T+1))(∑k=1T‍|u(k)|2)p-/2-C2-(∑k=1T‍|f(k)|2)1/2(∑k=1T‍|u(k)|2)1/2≥C12p-p+sin⁡p-(π2(2T+1))‖u‖p--K1‖u‖-C2,
where K1 is positive constant. Hence, since p->1, J is coercive.We can now give the proof of Theorem3.2.Proof of Theorem3.2.
By Proposition3.5, J has a minimizer which is a weak solution of (1.1).
In order to end the proof of Theorem3.2, we will prove the uniqueness of the weak solution.
Letu1 and u2 be two weak solutions of problem (1.1), then we have
(3.13)∑k=1T+1‍a(k-1,Δu1(k-1))Δ(u1-u2)(k-1)=∑k=1T‍f(k)(u1-u2)(k),∑k=1T+1‍a(k-1,Δu2(k-1))Δ(u1-u2)(k-1)=∑k=1T‍f(k)(u1-u2)(k).
Adding the two equalities of (3.13), we obtain
(3.14)∑k=1T+1‍[a(k-1,Δu1(k-1))-a(k-1,Δu2(k-1))]Δ(u1-u2)(k-1)=0.
Using (H3), we deduce from (3.14) that
(3.15)Δu1(k-1)=Δu2(k-1)∀k=1,…,T+1.
Therefore, by using discrete's Wirtinger inequality, we get
(3.16)4sin⁡2(π2(2T+1))∑k=1T‍|(u1-u2)(k)|2≤∑k=1T+1‍|Δ(u1-u2)(k-1)|2=0,
which implies that (∑k=1T‍|(u1-u2)(k)|2)1/2=0. It follows that u1=u2.


## 4. Some Extensions

### 4.1. Extension 1

In this section, we show that the existence result obtained for (1.1) can be extended to more general discrete boundary value problem of the form
(4.1)-Δ(a(k-1,Δu(k-1)))+|u(k)|q(k)-2u(k)=f(k),k∈Z[1,T]u(0)=Δu(T)=0,
where T≥2 is a positive integer, and we assume that (H6)
q:ℤ[1,T]→(1,+∞).By a weak solution of problem (4.1), we understand a function u∈W such that, for any v∈W,
(4.2)∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)+∑k=1T‍|u(k)|q(k)-2u(k)v(k)=∑k=1T‍f(k)v(k).We have the following result.Theorem 4.1.
Under assumptions (H1)–(H6), there exists a unique weak solution of problem (4.1).Proof .
Foru∈W,
(4.3)J(u)=∑k=1T+1‍A(k-1,Δu(k-1))+∑k=1T‍1p(k)|u(k)|q(k)-∑k=1T‍f(k)u(k)
is such that J∈C1(W;ℝ) and is weakly lower semicontinuous, and we have
(4.4)〈J′(u),v〉=∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)+∑k=1T‍|u(k)|q(k)-2u(k)v(k)-∑k=1T‍f(k)v(k),
for all u,v∈W.
This implies that the weak solutions of problem (4.1) coincide with the critical points of J. We then have to prove that J is bounded below and coercive in order to complete the proof.
As(4.5)∑k=1T‍1q(k)|u(k)|q(k)≥0,
then
(4.6)J(u)≥∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍f(k)v(k).
Using Proposition 3.5, we deduce that J is bounded below and coercive.
Letu1 and u2 be two weak solutions of problem (4.1), then we have
(4.7)∑k=1T+1‍a(k-1,Δu1(k-1))Δ(u1-u2)(k-1)+∑k=1T‍|u1(k)|q(k)-2u1(k)(u1(k)-u2(k))=∑k=1T‍f(k)(u1-u2)(k),∑k=1T+1‍a(k-1,Δu2(k-1))Δ(u1-u2)(k-1)+∑k=1T‍|u2(k)|q(k)-2u2(k)(u1(k)-u2(k))=∑k=1T‍f(k)(u1-u2)(k).
Adding these two equalities, we obtain
(4.8)∑k=1T+1‍[a(k-1,Δu1(k-1))-a(k-1,Δu2(k-1))]Δ(u1-u2)(k-1)+∑k=1T‍(|u1(k)|q(k)-2u1(k)-|u2(k)|q(k)-2u2(k))(u1(k)-u2(k))=0.
We deduce that
(4.9)∑k=1T‍(|u1(k)|q(k)-2u1(k)-|u2(k)|q(k)-2u2(k))(u1(k)-u2(k))=0,
which implies that
(4.10)u1(k)-u2(k)=0∀k=1,…,T,
and we get u1=u2.


### 4.2. Extension 2

In this section, we show that the existence result obtained for (1.1) can be extended to more general discrete boundary value problem of the form
(4.11)-Δ(a(k-1,Δu(k-1)))+λ|u(k)|β+-2u(k)=f(k,u(k)),k∈Z[1,T],u(0)=Δu(T)=0,
where T≥2 is a positive integer, λ∈ℝ+, and f:ℤ[1,T]×ℝ→ℝ is a continuous function with respect to the second variable for all (k,z)∈ℤ[1,T]×ℝ.For everyk∈ℤ[1,T] and every t∈ℝ, we put F(k,t)=∫0t‍f(k,τ)dτ.By a weak solution of problem (4.11), we understand a function u∈W such that
(4.12)∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)+λ∑k=1T‍|u(k)|β+-2u(k)v(k)=∑k=1T‍f(k,u(k))v(k),foranyv∈W.
We assume the following.(H7)
f(k,·):ℝ→ℝ is continuous for all k∈ℤ[1,T].(H8)
There exists a positive constantC3 such that |f(k,t)|≤C3(1+|t|β(k)-1), for all k∈ℤ[1,T] and t∈ℝ.(H9)
1<β-<p-.Remark 4.2.
The hypothesis(H8) implies that there exists one constant C′>0 such that |F(k,t)|≤C′(1+|t|β(k)).
We have the following result.Theorem 4.3.
Under assumptions (H2)–(H5) and (H7)–(H9), there exists λ*>0 such that, for λ∈[λ*,+∞), the problem (4.11) has at least one weak solution.Proof.
Letg(u)=∑k=1T‍F(k,u(k)), then g′:W→W is completely continuous, and, thus, g is weakly lower semicontinuous.
Therefore, foru∈W,
(4.13)J(u)=∑k=1T+1‍A(k-1,Δu(k-1))+λβ+∑k=1T‍|u(k)|β+-∑k=1T‍F(k,u(k))
is such that J∈C1(W;ℝ) and is weakly lower semicontinuous, and we have
(4.14)〈J′(u),v〉=∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)+λ∑k=1T‍|u(k)|β+-2u(k)v(k)-∑k=1T‍f(k,u(k))v(k),
for all u,v∈W.
This implies that the weak solutions of problem (4.11) coincide with the critical points of J. We then have to prove that J is bounded below and coercive in order to complete the proof.
Then, foru∈W such that ∥u∥>1,
(4.15)J(u)≥C12p-p+sin⁡p-(π2(2T+1))‖u‖p-+λβ+∑k=1T‍|u(k)|β+-C2-∑k=1T‍F(k,u(k))≥C12p-p+sin⁡p-(π2(2T+1))‖u‖p-+λβ+∑k=1T‍|u(k)|β+-C2-C′∑k=1T‍(1+|u(k)|β(k))≥C12p-p+sin⁡p-(π2(2T+1))‖u‖p-+λβ+∑k=1T‍|u(k)|β+-C2-C′T-C′(∑k=1T‍|u(k)|β(k))≥C12p-p+sin⁡p-(π2(2T+1))‖u‖p-+(λβ+-C′)∑k=1T‍|u(k)|β+-C2-C′T-C′‖u‖β-≥C12p-p+sin⁡p-(π2(2T+1))‖u‖p--C2-C′T-C′‖u‖β-,
where we put λ*=C′β+ with C′ a positive constant.
Furthermore, by the fact that1<β-<p-, it turns out that
(4.16)J(u)≥Cp+‖u‖p--C2-C′T-C′‖u‖β-⟶+∞as‖u‖⟶+∞.
Therefore, J is coercive.


### 4.3. Extension 3

We consider the problem(4.17)-Δ(a(k-1,Δu(k-1)))=f(k,u(k)),k∈N[1,T]u(0)=Δu(T)=0,
where T≥2.We suppose the following.(H10)
There exist two positive constantsC5 and C6 such that f+(k,t)≤C5+C6|t|β-1, for all (k,t)∈ℤ[1,T]×ℝ, where 1<β<p-.Definition 4.4.
A weak solution of problem (4.17) is a function u∈W such that
(4.18)∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)=∑k=1T‍f(k,u(k))v(k),∀v∈W.
We have the following result.Theorem 4.5.
Under the hypothesis (H2)–(H5) and (H10), problem (4.3) admits at least one weak solution.Proof.
We consider(4.19)J(u)=∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍F(k,u(k)),∀u∈H.J is such that J∈C1(W,ℝ) and
(4.20)〈J′(u),v〉=∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)-∑k=1T‍f(k,u(k))v(k),
for all u,v∈W.
Asf=f+-f-, then F-(k,t)=∫0T‍f+(k,τ)dτ.
By(H10), there exists C>0 such that
(4.21)|F+(k,t)|≤C(1+|t|β).
For all u∈W such that ∥u∥>1, we have
(4.22)J(u)=∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍F(k,u(k))=∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍F+(k,u(k))+∑k=1T‍F-(k,u(k))≥∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍F+(k,u(k)).
Therefore, similar to the proof of Theorem 4.3, Theorem 4.5 follows immediately.


## 4.1. Extension 1

In this section, we show that the existence result obtained for (1.1) can be extended to more general discrete boundary value problem of the form
(4.1)-Δ(a(k-1,Δu(k-1)))+|u(k)|q(k)-2u(k)=f(k),k∈Z[1,T]u(0)=Δu(T)=0,
where T≥2 is a positive integer, and we assume that (H6)
q:ℤ[1,T]→(1,+∞).By a weak solution of problem (4.1), we understand a function u∈W such that, for any v∈W,
(4.2)∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)+∑k=1T‍|u(k)|q(k)-2u(k)v(k)=∑k=1T‍f(k)v(k).We have the following result.Theorem 4.1.
Under assumptions (H1)–(H6), there exists a unique weak solution of problem (4.1).Proof .
Foru∈W,
(4.3)J(u)=∑k=1T+1‍A(k-1,Δu(k-1))+∑k=1T‍1p(k)|u(k)|q(k)-∑k=1T‍f(k)u(k)
is such that J∈C1(W;ℝ) and is weakly lower semicontinuous, and we have
(4.4)〈J′(u),v〉=∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)+∑k=1T‍|u(k)|q(k)-2u(k)v(k)-∑k=1T‍f(k)v(k),
for all u,v∈W.
This implies that the weak solutions of problem (4.1) coincide with the critical points of J. We then have to prove that J is bounded below and coercive in order to complete the proof.
As(4.5)∑k=1T‍1q(k)|u(k)|q(k)≥0,
then
(4.6)J(u)≥∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍f(k)v(k).
Using Proposition 3.5, we deduce that J is bounded below and coercive.
Letu1 and u2 be two weak solutions of problem (4.1), then we have
(4.7)∑k=1T+1‍a(k-1,Δu1(k-1))Δ(u1-u2)(k-1)+∑k=1T‍|u1(k)|q(k)-2u1(k)(u1(k)-u2(k))=∑k=1T‍f(k)(u1-u2)(k),∑k=1T+1‍a(k-1,Δu2(k-1))Δ(u1-u2)(k-1)+∑k=1T‍|u2(k)|q(k)-2u2(k)(u1(k)-u2(k))=∑k=1T‍f(k)(u1-u2)(k).
Adding these two equalities, we obtain
(4.8)∑k=1T+1‍[a(k-1,Δu1(k-1))-a(k-1,Δu2(k-1))]Δ(u1-u2)(k-1)+∑k=1T‍(|u1(k)|q(k)-2u1(k)-|u2(k)|q(k)-2u2(k))(u1(k)-u2(k))=0.
We deduce that
(4.9)∑k=1T‍(|u1(k)|q(k)-2u1(k)-|u2(k)|q(k)-2u2(k))(u1(k)-u2(k))=0,
which implies that
(4.10)u1(k)-u2(k)=0∀k=1,…,T,
and we get u1=u2.


## 4.2. Extension 2

In this section, we show that the existence result obtained for (1.1) can be extended to more general discrete boundary value problem of the form
(4.11)-Δ(a(k-1,Δu(k-1)))+λ|u(k)|β+-2u(k)=f(k,u(k)),k∈Z[1,T],u(0)=Δu(T)=0,
where T≥2 is a positive integer, λ∈ℝ+, and f:ℤ[1,T]×ℝ→ℝ is a continuous function with respect to the second variable for all (k,z)∈ℤ[1,T]×ℝ.For everyk∈ℤ[1,T] and every t∈ℝ, we put F(k,t)=∫0t‍f(k,τ)dτ.By a weak solution of problem (4.11), we understand a function u∈W such that
(4.12)∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)+λ∑k=1T‍|u(k)|β+-2u(k)v(k)=∑k=1T‍f(k,u(k))v(k),foranyv∈W.
We assume the following.(H7)
f(k,·):ℝ→ℝ is continuous for all k∈ℤ[1,T].(H8)
There exists a positive constantC3 such that |f(k,t)|≤C3(1+|t|β(k)-1), for all k∈ℤ[1,T] and t∈ℝ.(H9)
1<β-<p-.Remark 4.2.
The hypothesis(H8) implies that there exists one constant C′>0 such that |F(k,t)|≤C′(1+|t|β(k)).
We have the following result.Theorem 4.3.
Under assumptions (H2)–(H5) and (H7)–(H9), there exists λ*>0 such that, for λ∈[λ*,+∞), the problem (4.11) has at least one weak solution.Proof.
Letg(u)=∑k=1T‍F(k,u(k)), then g′:W→W is completely continuous, and, thus, g is weakly lower semicontinuous.
Therefore, foru∈W,
(4.13)J(u)=∑k=1T+1‍A(k-1,Δu(k-1))+λβ+∑k=1T‍|u(k)|β+-∑k=1T‍F(k,u(k))
is such that J∈C1(W;ℝ) and is weakly lower semicontinuous, and we have
(4.14)〈J′(u),v〉=∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)+λ∑k=1T‍|u(k)|β+-2u(k)v(k)-∑k=1T‍f(k,u(k))v(k),
for all u,v∈W.
This implies that the weak solutions of problem (4.11) coincide with the critical points of J. We then have to prove that J is bounded below and coercive in order to complete the proof.
Then, foru∈W such that ∥u∥>1,
(4.15)J(u)≥C12p-p+sin⁡p-(π2(2T+1))‖u‖p-+λβ+∑k=1T‍|u(k)|β+-C2-∑k=1T‍F(k,u(k))≥C12p-p+sin⁡p-(π2(2T+1))‖u‖p-+λβ+∑k=1T‍|u(k)|β+-C2-C′∑k=1T‍(1+|u(k)|β(k))≥C12p-p+sin⁡p-(π2(2T+1))‖u‖p-+λβ+∑k=1T‍|u(k)|β+-C2-C′T-C′(∑k=1T‍|u(k)|β(k))≥C12p-p+sin⁡p-(π2(2T+1))‖u‖p-+(λβ+-C′)∑k=1T‍|u(k)|β+-C2-C′T-C′‖u‖β-≥C12p-p+sin⁡p-(π2(2T+1))‖u‖p--C2-C′T-C′‖u‖β-,
where we put λ*=C′β+ with C′ a positive constant.
Furthermore, by the fact that1<β-<p-, it turns out that
(4.16)J(u)≥Cp+‖u‖p--C2-C′T-C′‖u‖β-⟶+∞as‖u‖⟶+∞.
Therefore, J is coercive.


## 4.3. Extension 3

We consider the problem(4.17)-Δ(a(k-1,Δu(k-1)))=f(k,u(k)),k∈N[1,T]u(0)=Δu(T)=0,
where T≥2.We suppose the following.(H10)
There exist two positive constantsC5 and C6 such that f+(k,t)≤C5+C6|t|β-1, for all (k,t)∈ℤ[1,T]×ℝ, where 1<β<p-.Definition 4.4.
A weak solution of problem (4.17) is a function u∈W such that
(4.18)∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)=∑k=1T‍f(k,u(k))v(k),∀v∈W.
We have the following result.Theorem 4.5.
Under the hypothesis (H2)–(H5) and (H10), problem (4.3) admits at least one weak solution.Proof.
We consider(4.19)J(u)=∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍F(k,u(k)),∀u∈H.J is such that J∈C1(W,ℝ) and
(4.20)〈J′(u),v〉=∑k=1T+1‍a(k-1,Δu(k-1))Δv(k-1)-∑k=1T‍f(k,u(k))v(k),
for all u,v∈W.
Asf=f+-f-, then F-(k,t)=∫0T‍f+(k,τ)dτ.
By(H10), there exists C>0 such that
(4.21)|F+(k,t)|≤C(1+|t|β).
For all u∈W such that ∥u∥>1, we have
(4.22)J(u)=∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍F(k,u(k))=∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍F+(k,u(k))+∑k=1T‍F-(k,u(k))≥∑k=1T+1‍A(k-1,Δu(k-1))-∑k=1T‍F+(k,u(k)).
Therefore, similar to the proof of Theorem 4.3, Theorem 4.5 follows immediately.


## 5. Example

We consider the following problem:(5.1)-Δ(|Δu(0)|2Δu(0))+λ|u(1)|2u(1)=15(u(1))2,-Δ(|Δu(1)|3Δu(1))+λ|u(2)|2u(2)=15(u(2))3,u(0)=0,u(2)=u(3).
Then, T=2, p(0)=4, p(1)=5, β(1)=3, β(2)=4, p-=4, p+=5, β-=3, β+=4,f(1,t)=(1/5)t2, and f(2,t)=(1/5)t3. Thus,
(5.2)F(1,t)=115t3,F(2,t)=120t4.
After computation, we can take C′=1/15 and we deduce that λ*=4/15.Therefore, by Theorem4.3, for any λ≥4/15, Problem (5.1) admits at least one weak solution.



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*Source: 927607-2012-06-25.xml*