doi:10.1155/2009/730484
Research Article
Solutions of 2
n
th-Order Boundary Value Problem
for Difference Equation via Variational Method
Qingrong Zou and Peixuan Weng
School of Mathematics, South China Normal University, Guangzhou 510631, China
Correspondence should be addressed to Peixuan Weng,wengpx@scnu.edu.cn
Received 7 July 2009; Accepted 15 October 2009
Recommended by Kanishka Perera
The variational method and critical point theory are employed to investigate the existence of solutions for 2nth-order difference equationΔnp
k−nΔnyk−n −1n1fk, yk 0 fork∈1, N with boundary value conditiony1−ny2−n · · ·y00, yN1· · ·yNn 0 by constructing a functional, which transforms the existence of solutions of the boundary value problemBVP to the existence of critical points for the functional. Some criteria for the existence of at least one solution and two solutions are established which is the generalization for BVP of the even-order difference equations.
Copyrightq2009 Q. Zou and P. Weng. 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.
1. Introduction
Difference equations have been applied as models in vast areas such as finance insurance, biological populations, disease control, genetic study, physical field, and computer applica-tion technology. Because of their importance, many literature deals with its existence and uniqueness problems. For example, see1–10 .
We notice that the existing results are usually obtained by various analytical techniques, for example, the conical shell fixed point theorem 1, 6 , Banach contraction map method 7 , Leray-Schauder fixed point theorem 2, 10 , and the upper and lower solution method 3 . It seems that the variational technique combining with the critical point theory11 developed in the recent decades is one of the effective ways to study the boundary value problems of difference equations. However because the variational method requires a “symmetrical” functional, it is hard for the odd-order difference equations to create a functional satisfying the “symmetrical” property. Therefore, the even-order difference equations have been investigated in most references.
with some traditional analytical skills:
Δnp
k−nΔnyk−n
−1n1fk, yk
0 k∈1, N , 1.1
y1−n y2−n· · ·y00, yN1· · ·yNn0, 1.2
whereΔny
k Δn−1yk1−Δn−1yk n1is the forward difference operator;pk ∈Rfork ∈
1−n, N andf∈C1, N ×R,R.A variational functional for BVP1.1-1.2is constructed which transforms the existence of solutions of the boundary value problem BVP to the existence of critical points of this functional. In order to prove the existence criteria of critical points of the functional, some lemmas are given inSection 2. Two criteria for the existence of at least one solution and two solutions for BVP1.1-1.2are established inSection 3which is the generalization for BVP of the even-order difference equations. The existence results obtained in this paper are not found in the references, to the best of our knowledge.
For convenience, we will use the following notations in the following sections:
Fk, u
u
0
fk, sds, p max
k∈1−n,N pk, pk∈min1−n,N
pk. 1.3
2. Variational Structure and Preliminaries
We need two lemmas from12 or11 .
Lemma 2.1. LetH be a real reflexive Banach space with a norm · , and letφbe a weakly lower (upper) semicontinuous functional, such that
lim
x → ∞φx ∞
or lim
x → ∞φx −∞
, 2.1
then there existsx0∈Hsuch that
φx0 inf x∈Hφx0
or φx0 sup x∈H
φx0 . 2.2
Furthermore, ifφhas bounded linear Gˆateaux derivative, thenφx0 0.
Lemma 2.2 mountain-pass lemma. Let H be a real Banach space, and let φ : H →
R be continuously differential, satisfying the P-S condition. Assume that x0, x1 ∈ H and Ω
is an open neighborhood of x0, but x1/∈Ω. If max{φx0, φx1} < infx∈∂Ωφx, then c
infh∈Γmaxt∈0,1 φhtis the critical value ofφ,where
Γ {h|h:0,1 −→H, his continuous, h0 x0, h1 x1}. 2.3
This means that there existsx2∈H, s.t.φx2 0,φx2 c.
Lemma 2.3. IfAm×mis a symmetric and positive-defined real matrix,Bm×nis a real matrix,BTis the
transposed matrix ofB.ThenBTABis positive defined if and only ifrankBn.
Proof. SinceAis positive defined, then
BTABis positive-defined⇐⇒ ∀x /0, xTBTABx >0
⇐⇒ ∀x /0, BxTABx>0⇐⇒ ∀x /0, Bx /0⇐⇒rankBn.
2.4
LetHbe a Hilbert space defined by
Hy:1−n, Nn −→R|y1−ny2−n· · ·y00, yN1 · · ·yNn0
Lemma 2.4. There is
Repeating the above process, we obtain
all eigenvalues of BTBare real and positive. Let λ be the minimal eigenvalue of these N
eigenvalues, thenλ >0.ThereforexTBTBxλxTx,that is,N
1−nΔnxk2λx2.
However, how to find the λ in Lemma 2.4 is a skillful and challenging task. The following lemma from13 offers some help for the estimation ofλ.
Lemma 2.5Brualdi13 . IfA aij ∈MNis weak irreducible, then each eigenvalue is contained
InLemma 2.5,Cis the complex number set, and the denotations CA,γ,Pi,Ri can
be found in 13 . SinceBTBis positive defined, all eigenvalues are positive real numbers.
Therefore, byLemma 2.5, let
B calculate the eigenvalues directly.
Define the functionalφonHby
φy
we have
Repeating the above process, we obtain
N
Sincex ∈ H is arbitrary, we know that the solution of BVP1.1-1.2corresponds to the critical point ofφ.
3. Main Results
Now we present our main results of this paper.
Theorem 3.1. If there existM1>0,a1>0,a2∈R, andσ >2s.t.
Fk, ua1|u|σa2, ∀ |u|> M1, 3.1
then BVP1.1-1.2has at least one solution.
Proof. In fact, we can choose a suitablea2<0 such that
Since there existsσ1>0 withσ1yyσ,we have
N
1−n Fk, yk
a1
N
1
ykσa2Nna1σ1σy σa
2Nn. 3.3
Then byLemma 2.4, we obtain
φy p
24
ny2−
a1σσ1yσ−a2Nn. 3.4
Noticing thatσ >2,we have limy →∞φy −∞.FromLemma 2.1, the conclusion of this
lemma follows.
Corollary 3.2. If there existsM2 >0s.t.ufk, u>0for all|u|> M2, and
inf
k∈1−n,N ulim→ ∞
fk, u
|u|r r1, 3.5
where r,r1 satisfy eitherr 1,r1 > 4np orr > 1, r1 > 0, then BVP1.1-1.2has at least one
solution.
Proof. Assume thatr 1,r1 > 4np.Then for1 r1−4np/2 > 0,there exists M3 > M2,
such that|fk, y|r1−1|y|as|y|> M3. We have from the continuity offk, uthat there
is aK > 0 such that−K ≤ fk, u ≤ K for allk ∈ 1, N ,|u| ≤ M3. When y > 0, one has fk, yr1−1y >0 fory∈M3,∞,then
y
0
fk, sds
M3
0
fk, sds
y
M3
fk, sds−KM3 r1−1
2 y
2−r1−1
2 M
2
3; 3.6
wheny <0, one hasfk, yr1−1y <0 fory∈−∞,−M3,then
y
0
fk, sds
−M3
0
fk, sds
y
−M3
fk, sds−KM3r1−1
2 y
2−r1−1
2 M
2
3. 3.7
Letc :−KM3−r1−1/2M23, then we have
*y
0fk, sds r1−1/2y2cfory∈R.
Therefore, we have
φyp
24
ny2−r1−1
2
N
1
yk2−c
4np−r 11
2 y
2− c−1
2y
2−
c, 3.8
which implies limy →∞φy −∞,and byLemma 2.1, the conclusion of this lemma follows.
Assume thatr > 1, r1 > 0.Then for 2 r1/2 > 0, there exists M4 > M2, such
a K > 0 such that−K ≤ fk, u ≤ K for allk ∈ 1, N ,|u| ≤ M4. Wheny > 0, one has fk, yr1/2yr >0,y∈M4,∞,then we have
y
0
fk, sds
M4
0
fk, sds
y
M4
fk, sds
−KM4
r1
2r1y
r1− r1
2r1M r1 4 ,
3.9
wheny <0, one hasfk, y−r1/2|y|r −r1/2−yr <0,y∈−∞,−M4,then we have
y
0
fk, sds
−M4
0
fk, sds
y
−M4
fk, sds
−KM4r1
2
y
−M4
−srd−s
−KM42 r1 r1y
r1− r1
2r1M4 r1.
3.10
Letd : −KM4−r1/2r1M4r1, then we have
*y
0fk, sds r1/2r1|y|r1dfor
|y|> M4.Therefore, byTheorem 3.1, the conclusion of this lemma follows.
Theorem 3.3. Assume thatpk>0,k1−n, . . . , N,and
isupk∈1−n,N limu→0fk, u/ur2< pλ,λ >0is defined inLemma 2.4;
iiFsatisfies3.1inTheorem 3.1orfsatisfies the assumptions inCorollary 3.2. Then BVP1.1-1.2has at least two solutions.
Proof. We first show thatφsatisfies the P-S condition. Let{ym}∞m1⊂Hsatisfy that{φym}
is bounded and limm→ ∞φym 0. If {ym} is unbounded, it possesses a divergent
subseries, say ymk → ∞ as k → ∞. However fromii, we get 3.4 or 3.8, hence
φymk → −∞ask → ∞, which is contradictory to the the fact that{φym}is bounded. Next we use the mountain-pass lemma to finish the proof. Byi, for3 pλ−r2/2>
0, there existsR1 > 0 such thatfk, y/y r23 for|y| R1. Then
*y
0fk, sds r2 3/2y2for|y|R1. Now together withLemma 2.3,we have
φyp
2λy
2−r23
2
N
1−n
|yk|2y2
pλ−r
2−3
2
3
2y
2>0 for yR 1,
3.11
which implies that
φy 3
2R
2
1>0φθ, y∈∂Ω, 3.12
whereθis the zero element inH, andΩ {y ∈ H | y < R1}.Since we have from3.4
φy1 < φθ 0.Using Lemma 2.2, we have shown thatξ infh∈Γmaxt∈0,1 φhtis the
critical value ofφ,withΓdefined as
Γ h|h:0,1 −→H, his continuous, h0 θ, h1 y1
. 3.13
We denoteyas its corresponding critical point.
On the other hand, byTheorem 3.1orCorollary 3.2, we know that there existsy∗∈H,
s.t.φy∗ supy∈Hφy.If y∗/y, the theorem is proved. If on the contrary, y∗ y, that is, supy∈Hφy infh∈Γmaxt∈0,1 φht, that implies for any h ∈ Γ, maxt∈0,1 φht
supy∈Hφy.Takingh1/h2 inΓ with maxt∈0,1 φh1t maxt∈0,1 φh2t supy∈Hφy,
by the continuity of φht, there exist t1, t2 ∈ 0,1 s.t. φh1t1 maxt∈0,1 φh1t, φh2t2 maxt∈0,1 φh2t.Henceh1t1,h2t2are two different critical points ofφ,that
is, BVP1.1-1.2has at least two different solutions.
4. An Example
Consider the 6th-order boundary value problem for difference equation
Δ6y
k−3y3key
2
k−90, k∈1,300 ,
y−2y−1y00, y301y302y3030.
4.1
Let fk, u u3eu2−9
, we have limu→0fk, u/u 0, limu→ ∞fk, u/u ∞. Hence
fk, usatisfies the conditions inTheorem 3.3, the boundary value problem4.1has at least two solutions.
Acknowledgments
This research is partially supported by the NSF of China and NSF of Guangdong Province.
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