Volume 2012, Article ID 927607,16pages doi:10.1155/2012/927607
Research Article
On the Solvability of Discrete Nonlinear Two-Point
Boundary Value Problems
Blaise Kone
1and Stanislas Ouaro
21Laboratoire d’Analyse Math´ematique des Equations (LAME), Institut Burkinab´e des Arts et M´etiers,
Universit´e de Ouagadougou, 03 BP 7021, Ouagadougou 03, Burkina Faso
2Laboratoire d’Analyse Math´ematique des Equations (LAME), UFR Sciences Exactes et Appliqu´ees,
Universit´e de Ouagadougou, 03 BP 7021, Ouagadougou 03, Burkina Faso
Correspondence should be addressed to Stanislas Ouaro,[email protected]
Received 22 March 2012; Accepted 27 April 2012
Academic Editor: Paolo Ricci
Copyrightq2012 B. Kone and S. Ouaro. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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.
1. Introduction
In this paper, we study the following nonlinear discrete boundary value problem:
−Δak−1,Δuk−1 fk, k∈Z1, T,
u0 ΔuT 0, 1.1
whereT ≥2 is a positive integer andΔuk uk1−ukis the forward difference operator. Throughout this paper, we denote byZa, bthe discrete interval{a, a1, . . . , b}, whereaand
bare integers anda < b.
We consider in 1.1 two different boundary conditions: a Dirichlet boundary condition u0 0 and a Neumann boundary condition ΔuT 0. In the literature, the boundary condition considered in this paper is called a mixed boundary condition.
We also consider the function space
whereWis a T-dimensional Hilbert space1,2with the inner product
u, v T
k1
ukvk, ∀u, v∈W. 1.3
The associated norm is defined by
u
T
k1
|uk|2
1/2
. 1.4
For the datafanda, we assume the following.
H1f :Z1, T → R.
H2ak,·:R → Rfor allk∈Z0, Tand their exists a mappingA:Z0, T×R → R
which satisfiesak, ξ ∂/∂ξAk, ξ, for allk ∈ Z0, TandAk,0 0, for all
k∈Z0, T.
H3 ak, ξ−ak, η·ξ−η>0 for allk∈Z0, Tandξ, η∈Rsuch thatξ /η.
H4|ξ|pk ≤ak, ξξ≤pkAk, ξfor allk∈Z0, Tandξ∈R.
H5 p:Z0, 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 Guo3, Kon´e and Ouaro4, Guiro et al.5, Cai and Yu6, Zhang and Liu7, Mih˘ailescu et al.8, Candito and DAgui9, Cabada et al.10, Jiang and Zhou11, and the references therein. In7, the authors studied the following problem:
Δ2yk−1 λfyk0, k∈Z1, T,
y0 yT1 0,
1.5
whereλ > 0 is a parameter,Δ2yk ΔΔyk,f : 0,∞ → Ra continuous function
satisfying the condition
f0 −a <0, whereais a positive constant. 1.6
The problem 1.5 is referred as the “semipositone” problem in the literature, which was introduced by Castro and Shivaji2. Semipositone problems arise in bulking of mechanical systems, design of suspension bridges, chemical reactions, astrophysics, combustion, and management of natural resources.
In11, Jiang and Zhou studied the following problem:
Δ2uk−1 fk, uk, k∈Z1, T,
u0 ΔuT 0, 1.7
where T is a fixed positive integer, f : Z1, T×R → R is a continuous function. Jiang and Zhou proved an existence of nontrivial solutions for1.7by 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 in11and involves variable exponent.
Problem1.1is a discrete variant of the variable exponent anisotropic problem
−N
i1
∂ ∂xiai
x, ∂u ∂xi
fx inΩ,
u0 onΓ1
∂u
∂n 0 onΓ2,
1.8
whereΩ⊂RNN≥3is a bounded domain with smooth boundary,Γ
1∪Γ2∂Ω,f∈L∞Ω,
pi continuous on Ω such that 1 < pix < N and iN11/p−i > 1 for allx ∈ Ω and all i∈Z1, N, wherepi−: ess infx∈Ωpix.
The first equation in1.8was recently analyzed by Kon´e et al.12and Ouaro13
and generalized to a Radon measure data by Kon´e et al.14for 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 fluidssee15,17–19, and image restorationsee20. In20, Chen et al. studied a functional with variable exponent 1 ≤ px ≤ 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 Mih˘ailescu et al.see21,22were the first authors who studied anisotropic elliptic problems with variable exponent. In general, the interested reader can find more information about difference equations in1–11, 23–25, more information about variable exponent in12–22,26.
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 Mih˘ailescu et al.
8, the authors proved by using critical point theory, existence of a continuous spectrum of eigenvalues for the problem
−Δ|Δuk−1|pk−1−2Δuk−1λ|uk|qk−2uk, k∈Z1, T,
u0 uT1 0,
1.9
whereT ≥ 2 is a positive integer and the functionsp : Z0, T → 2,∞andq :Z1, T →
2,∞are bounded whileλis a positive constant.
In 4, Kon´e and Ouaro proved, by using minimization method, existence and uniqueness of weak solutions for the following problem:
−Δak−1,Δuk−1 fk, k∈Z1, T,
u0 uT1 0,
1.10
whereT ≥2 is a positive integer.
The functionak−1,Δuk−1which appears in the left-hand side of problem1.1
is more general than the one which appears in1.9. Indeed, as examples of functions which satisfy the assumptionsH2–H5, we can give the following.
iAk, ξ 1/pk|ξ|pk, whereak, ξ |ξ|pk−2ξ, for allk∈Z0, Tandξ∈R.
iiAk, ξ 1/pk1|ξ|2pk/2−1, whereak, ξ 1|ξ|2pk−2/2ξ, for allk ∈
Z0, Tandξ∈R.
In5, Guiro et al. studied the following two-point boundary value problems
−Δak−1,Δuk−1 |uk|pkuk fk, k∈Z1, T,
Δu0 ΔuT 0.
1.11
The functionak−1,Δuk−1has the same properties as in4, 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 solutionwhich is also a classical solution since the Hilbert space associated is of finite dimension. Indeed, they used the following norm:
u
T1
k1
|Δuk−1|2
T
k1
|uk|2
1/2
, 1.12
which is associated to the Hilbert space
In order to get the coercivity of the energy functional, the authors of 5 assumed the following on the exponent:
p:Z0, T−→2,∞. 1.14
The assumption above allowed them to exploit the convexity property of the mapx → xp−/2.
Problem1.11is a discrete variant of the following problem:
−N
i1
∂ ∂xi
ai
x, ∂u ∂xi
fx, u inΩ,
∂u
∂n 0 on Ω,
1.15
which was studied by Boureanu and Radulescu in26with an additional condition thatu≥
0. Note that, in26, the Neumann condition is more general than the one in problem1.11. In this paper, we use the discrete Wirtinger inequalitysee23which allows us to assume that the exponentp:Z0, T → 1,∞. The discrete Wirtinger inequality is a discrete variant of the well-known Poincar´e-Wirtinger inequalitysee28. Another difference of the present paper compared to5is on the boundary condition.
The remaining part of this paper is organized as follows. Section 2 is devoted to mathematical preliminaries. The main existence and uniqueness result is stated and proved inSection 3. InSection 4, we discuss some extensions, and, finally, inSection 5, we apply our theoretical results to an example.
2. Preliminaries
From now, we will use the following notations:
p− min
k∈Z0,Tpk, p
max
k∈Z0,Tpk. 2.1
Moreover, it is useful to introduce other norms onW, namely,
|u|m
T
k1
|uk|m
1/m
∀u∈W, m≥2. 2.2
We have the following inequalitiessee6,8which are used in the proof ofLemma 2.1:
T2−m/2m|u|2≤ |u|m≤T1/m|u|2 ∀u∈W, m≥2. 2.3
Lemma 2.1see5. There exist two positive constantsC1,C2such that
T1
k1
|Δuk−1|pk−1≥C1
T1
k1
|Δuk−1|2
p−/2
−C2, 2.4
for allu∈W withu>1.
We have the following result.
Lemma 2.2discrete Wirtinger’s inequality, see Theorem 12.6.2, page 860 in23. For any
functionuk,k∈Z0, Tsatisfyingu0 0, the following inequality holds:
4 sin2
π
22T1 T
k1
|uk|2≤
T−1
k0
|Δuk|2. 2.5
3. Existence and Uniqueness of Weak Solution
In this section, we study the existence and uniqueness of weak solution of1.1.
Definition 3.1. A weak solution of1.1is a functionu∈Wsuch that
T1
k1
ak−1,Δuk−1Δvk−1
T
k1
fkvk for anyv∈W. 3.1
Note that, sinceW is a finite dimensional space, the weak solutions coincide with the classical solutions of the problem1.1.
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 problem1.1is defined byJ:W → Rsuch that
Ju T1
k1
Ak−1,Δuk−1−
T
k1
fkuk. 3.2
We first present some basic properties ofJ.
Proposition 3.3. The functionalJis well defined onWand is of classC1W,Rwith the derivative
given by
Ju, v T1
k1
ak−1,Δuk−1Δvk−1−
T
k1
fkvk, 3.3
The proof ofProposition 3.3can be found in5. We now define the functionalIby
Iu T1
k1
Ak−1,Δuk−1. 3.4
We need the following lemma for the proof ofTheorem 3.2.
Lemma 3.4. The functionalIis weakly lower semicontinuous.
Proof. Ais convex with respect to the second variable according toH2. Thus, it is enough
to show thatIis lower semicontinuous. For this, we fixu∈W and >0. SinceIis convex, we deduce that, for anyv∈W,
Iv≥Iu Iu, v−u
≥Iu
T1
k1
ak−1,Δuk−1Δvk−1−Δuk−1
≥Iu−
T1
k1
|ak−1,Δuk−1||Δvk−1−Δuk−1|
≥Iu−
T1
k1
|ak−1,Δuk−1||vk−uk uk−1−vk−1|
≥Iu−
T1
k1
|ak−1,Δuk−1||vk−uk||vk−1−uk−1|.
3.5
We defineHandBby
H
T1
k1
|ak−1,Δuk−1||vk−uk|,
B
T1
k1
|ak−1,Δuk−1||vk−1−uk−1|.
3.6
By using Schwartz inequality, we get
H≤
T1
k1
|ak−1,Δuk−1|2
1/2T1
k1
|vk−uk|2
1/2
≤
T1
k1
|ak−1,Δuk−1|2
1/2
v−u.
The same calculus gives
B≤
T1
k1
|ak−1,Δuk−1|2
1/2
v−u. 3.8
Finally, we have
Iv≥Iu−
12
T1
k1
|ak−1,Δuk−1|2
1/2
v−u ≥Iu− 3.9
for all v ∈ W with v − u < δ /KT, u, where KT, u
12 Tk11|ak−1,Δuk−1|21/2.
We conclude thatIis weakly lower semicontinuous.
We also have the following result.
Proposition 3.5. The functionalJis bounded from below, coercive, and weakly lower semicontinuous.
Proof . ByLemma 3.4,Jis 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 in5. ByH4, we deduce that
Ju T1
k1
Ak−1,Δuk−1−
T
k1
fkuk
≥T1
k1
1
pk|Δuk−1|
pk−1−T
k1
fkuk
≥ 1
p T1
k1
|Δuk−1|pk−1−
T
k1
fk2
1/2T
k1
|uk|2
1/2
.
3.10
To prove the coercivity of J, we may assume that u > 1, and we get from the above inequality andLemma 2.1, the following:
Ju≥ C1 p
T1
k1
|Δuk−1|2
p−/2
−C2−
T
k1
fk2
1/2T
k1
|uk|2
1/2
Using Wirtinger’s discrete inequality, we obtain
Ju≥ C1 p
4sin2
π
22T1 T
k1
|uk|2
p−/2
−C2−
T
k1
fk2
1/2T
k1
|uk|2
1/2
≥ C12p−
p sin p−
π
22T1 T
k1
|uk|2
p−/2
−C2−
T
k1
fk2
1/2T
k1
|uk|2
1/2
≥ C12p
−
p sin p−
π
22T1
up−−K1u −C2,
3.12
whereK1is positive constant. Hence, sincep−>1,Jis coercive.
We can now give the proof ofTheorem 3.2.
Proof ofTheorem 3.2. ByProposition 3.5,Jhas a minimizer which is a weak solution of1.1.
In order to end the proof ofTheorem 3.2, we will prove the uniqueness of the weak solution.
Letu1andu2be two weak solutions of problem1.1, then we have
T1
k1
ak−1,Δu1k−1Δu1−u2k−1
T
k1
fku1−u2k,
T1
k1
ak−1,Δu2k−1Δu1−u2k−1
T
k1
fku1−u2k.
3.13
Adding the two equalities of3.13, we obtain
T1
k1
ak−1,Δu1k−1−ak−1,Δu2k−1Δu1−u2k−1 0. 3.14
UsingH3, we deduce from3.14that
Δu1k−1 Δu2k−1 ∀k1, . . . , T 1. 3.15
Therefore, by using discrete’s Wirtinger inequality, we get
4 sin2
π
22T1 T
k1
|u1−u2k|2≤
T1
k1
|Δu1−u2k−1|20, 3.16
which implies that Tk1|u1−u2k|2
1/2
4. Some Extensions
4.1. Extension 1
In this section, we show that the existence result obtained for1.1can be extended to more general discrete boundary value problem of the form
−Δak−1,Δuk−1 |uk|qk−2uk fk, k∈Z1, T
u0 ΔuT 0,
4.1
whereT ≥2 is a positive integer, and we assume that
H6q:Z1, T−→1,∞.
By a weak solution of problem 4.1, we understand a function u ∈ W such that, for any
v∈W,
T1
k1
ak−1,Δuk−1Δvk−1
T
k1
|uk|qk−2ukvk
T
k1
fkvk. 4.2
We have the following result.
Theorem 4.1. Under assumptions (H1)–(H6), there exists a unique weak solution of problem4.1.
Proof . Foru∈W,
Ju T1
k1
Ak−1,Δuk−1
T
k1
1
pk|uk|
qk−T
k1
fkuk 4.3
is such thatJ ∈C1W;Rand is weakly lower semicontinuous, and we have
Ju, v T1
k1
ak−1,Δuk−1Δvk−1
T
k1
|uk|qk−2ukvk−
T
k1
fkvk,
4.4
for allu, v∈W.
This implies that the weak solutions of problem4.1coincide with the critical points ofJ. We then have to prove thatJis bounded below and coercive in order to complete the proof.
As
T
k1
1
qk|uk|
then
Ju≥ T1
k1
Ak−1,Δuk−1−
T
k1
fkvk. 4.6
UsingProposition 3.5, we deduce thatJis bounded below and coercive. Letu1andu2be two weak solutions of problem4.1, then we have
T1
k1
ak−1,Δu1k−1Δu1−u2k−1
T
k1
|u1k|qk−2u1ku1k−u2k
T
k1
fku1−u2k,
T1
k1
ak−1,Δu2k−1Δu1−u2k−1
T
k1
|u2k|qk−2u2ku1k−u2k
T
k1
fku1−u2k.
4.7
Adding these two equalities, we obtain
T1
k1
ak−1,Δu1k−1−ak−1,Δu2k−1Δu1−u2k−1
T
k1
|u1k|qk−2u1k− |u2k|qk−2u2k
u1k−u2k 0.
4.8
We deduce that
T
k1
|u1k|qk−2u1k− |u2k|qk−2u2k
u1k−u2k 0, 4.9
which implies that
u1k−u2k 0 ∀k1, . . . , T, 4.10
4.2. Extension 2
In this section, we show that the existence result obtained for1.1can be extended to more general discrete boundary value problem of the form
−Δak−1,Δuk−1 λ|uk|β−2uk fk, uk, k∈Z1, T,
u0 ΔuT 0,
4.11
whereT ≥2 is a positive integer,λ ∈R, andf :Z1, T×R → Ris a continuous function with respect to the second variable for allk, z∈Z1, T×R.
For everyk∈Z1, Tand everyt∈R, we putFk, t 0tfk, τdτ.
By a weak solution of problem4.11, we understand a functionu∈Wsuch that
T1
k1
ak−1,Δuk−1Δvk−1 λ T
k1
|uk|β−2ukvk
T
k1
fk, ukvk, for anyv∈W.
4.12
We assume the following.
H7fk,·:R → Ris continuous for allk∈Z1, T.
H8There exists a positive constantC3 such that|fk, t| ≤ C31|t|βk−1, for allk ∈
Z1, Tandt∈R.
H9 1< β−< p−.
Remark 4.2. The hypothesis H8 implies that there exists one constant C > 0 such that
|Fk, t| ≤C1|t|βk.
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 problem4.11has at least one weak solution.
Proof. Letgu Tk1Fk, uk, theng :W → Wis completely continuous, and, thus,g
is weakly lower semicontinuous. Therefore, foru∈W,
Ju T1
k1
Ak−1,Δuk−1 λ
β T
k1
|uk|β−
T
k1
is such thatJ ∈C1W;Rand is weakly lower semicontinuous, and we have
Ju, v T1
k1
ak−1,Δuk−1Δvk−1 λ T
k1
|uk|β−2ukvk
−T
k1
fk, ukvk,
4.14
for allu, v∈W.
This implies that the weak solutions of problem4.11coincide with the critical points ofJ. We then have to prove thatJis bounded below and coercive in order to complete the proof.
Then, foru∈Wsuch thatu>1,
Ju≥ C12 p−
p sin p−
π
22T1
up− λ
β T
k1
|uk|β−C2−
T
k1
Fk, uk
≥ C12p
−
p sin p−
π
22T1
up− λ
β T
k1
|uk|β−C2−C
T
k1
1|uk|βk
≥ C12p
−
p sin p−
π
22T1
up− λ
β T
k1
|uk|β−C2−CT−C
T
k1
|uk|βk
≥ C12p
−
p sin p−
π
22T1
up−
λ β −C
T
k1
|uk|β−C2−CT−Cuβ
−
≥ C12p
−
p sin p−
π
22T1
up−−C
2−CT−Cuβ
−
,
4.15
where we putλ∗CβwithC a positive constant.
Furthermore, by the fact that 1< β− < p−, it turns out that
Ju≥ C pu
p−−C
2−CT−Cuβ
−
−→∞ asu −→∞. 4.16
Therefore,Jis coercive.
4.3. Extension 3
We consider the problem
−Δak−1,Δuk−1 fk, uk, k∈N1, T
u0 ΔuT 0, 4.17
We suppose the following.
H10There exist two positive constantsC5 andC6 such thatfk, t ≤ C5C6|t|β−1, for
allk, t∈Z1, T×R, where 1< β < p−.
Definition 4.4. A weak solution of problem4.17is a functionu∈Wsuch that
T1
k1
ak−1,Δuk−1Δvk−1
T
k1
fk, ukvk, ∀v∈W. 4.18
We have the following result.
Theorem 4.5. Under the hypothesis (H2)–(H5) andH10, problem4.3admits at least one weak
solution.
Proof. We consider
Ju T1
k1
Ak−1,Δuk−1−
T
k1
Fk, uk, ∀u∈H. 4.19
Jis such thatJ ∈C1W,Rand
Ju, v T1
k1
ak−1,Δuk−1Δvk−1−
T
k1
fk, ukvk, 4.20
for allu, v∈W.
Asff−f−, thenF−k, t 0Tfk, τdτ. ByH10, there existsC >0 such that
|Fk, t| ≤C1|t|β. 4.21
For allu∈Wsuch thatu>1, we have
Ju T1
k1
Ak−1,Δuk−1−
T
k1
Fk, uk
T1
k1
Ak−1,Δuk−1−
T
k1
Fk, uk T
k1
F−k, uk
≥T1
k1
Ak−1,Δuk−1−
T
k1
Fk, uk.
4.22
5. Example
We consider the following problem:
−Δ|Δu0|2Δu0λ|u1|2u1 1
5u1
2,
−Δ|Δu1|3Δu1λ|u2|2u2 1
5u2
3,
u0 0, u2 u3.
5.1
Then, T 2,p0 4,p1 5,β1 3,β2 4,p− 4,p 5,β− 3,β 4,f1, t
1/5t2, andf2, t 1/5t3. Thus,
F1, t 1
15t
3, F2, t 1
20t
4. 5.2
After computation, we can takeC 1/15 and we deduce thatλ∗4/15.
Therefore, byTheorem 4.3, for anyλ ≥ 4/15, Problem5.1admits at least one weak solution.
Acknowledgment
The authors want to express their deepest thanks to the editor and anonymous referees for comments and suggestions on the paper.
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