Hindawi Publishing Corporation Advances in Difference Equations Volume 2009, Article ID 360871, 17 pages doi:10.1155/2009/360871
Research Article Nonlinear Discrete Periodic Boundary Value Problems at Resonance Ruyun Ma1 and Huili Ma2 1 2
Department of Mathematics, Northwest Normal University, Lanzhou 730070, China College of Economics and Management, Northwest Normal University, Lanzhou 730070, China
Correspondence should be addressed to Ruyun Ma, ruyun
[email protected] Received 25 June 2009; Revised 4 October 2009; Accepted 6 December 2009 Recommended by Kanishka Perera Let T ∈ N be an integer with T > 2, and let T : {1, . . . , T }. We study the existence of solutions of nonlinear discrete problems Δ2 ut − 1 λk atut gt, ut ht, t ∈ T, u0 uT , u1 uT 1, where a, h : T → R with a > 0, λk is the kth eigenvalue of the corresponding linear eigenvalue problem. Copyright q 2009 R. Ma and H. Ma. 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 Initialed by Lazer and Leach 1 , much work has been devoted to the study of existence result for nonlinear periodic boundary value problem y x m2 yx g x, yx ex, y0 y2π,
x ∈ 0, 2π,
y 0 y 2π,
1.1
where m ≥ 0 is an integer. Results from the paper have been extended to partial differential equations by several authors. The reader is referred, for detail, to Landesman and Lazer 2 , Amann et al. 3 , Br´ezis and Nirenberg 4 , Fuˇc´ık and Hess 5 , and Iannacci and Nkashama 6 for some reference along this line. Concerning 1.1, results have been carried out by many authors also. Let us mention articles by Mawhin and Ward 7 , Conti et al. 8 , Omari and Zanolin 9 , Ding and Zanolin 10 , Capietto and Liu 11 , Iannacci and Nkashama 12 , Chu et al. 13 , and the references therein.
2
Advances in Difference Equations However, relatively little is known about the discrete analog of 1.1 of the form Δ2 ut − 1 λk atut gt, ut ht, u0 uT ,
t ∈ T,
u1 uT 1,
1.2
where T : {1, . . . , T }, a, h : T → R with a > 0, gt, s : T × R → R is continuous in s. The likely reason is that the spectrum theory of the corresponding linear problem Δ2 ut − 1 λk atut 0, u0 uT ,
t ∈ T,
u1 uT 1
1.3
was not established until 14 . In 14 , Wang and Shi showed that the linear eigenvalue problem 1.3 has exactly T real eigenvalues μ0 < μ1 ≤ μ2 < · · · < μT −2 ≤ μT −1 ,
when T is odd,
μ0 < μ1 ≤ μ2 < · · · ≤ μT −2 < μT −1 ,
when T is even.
1.4
Suppose that these above eigenvalues have N 1 different values λk , k 0, 1, . . . , N. Then 1.4 can be rewritten as λ0 < λ1 < · · · < λN .
1.5
For each λk , we denote its eigenspace by Mk . If dim Mk 1, then we assume that Mk : span{ψk } in which ψk is the eigenfunction of λk . If dim Mk 2, then we assume that Mk : span{ψk , ϕk } in which ψk and ϕk are two linearly independent eigenfunctions of λk . It is the purpose of this paper to prove the existence results for problem 1.2 when there occurs resonance at the eigenvalue λk and the nonlinear function g may “touching” the eigenvalue λk1 . To have the wit, we have what follows. Theorem 1.1. Let a, h : T → R with a > 0, gt, s : T × R → R is continuous in s, and for some r ∗ < 0 < R∗ , gt, x ≥ At,
∀x ≥ R∗ ,
gt, x ≤ Bt,
∀x ≤ r ∗ ,
1.6
where A, B : T → R are two given functions. Suppose for some 1 ≤ k ≤ N − 1, dim Mk1 2.
1.7
Assume that for all ε > 0, there exist a constant R Rε > 0 and a function b : T → R such that gt, u ≤ Γt εat|u| bt,
t ∈ T, |u| ≥ R,
1.8
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where Γ : T → R is a given function satisfying 0 ≤ Γt ≤ λk1 − λk ,
t ∈ T,
1.9
and for at least T/2 2 points in 1, T , Γt < λk1 − λk ,
1.10
where r denotes the integer part of the real number r. Then 1.2 has at least one solution provided T
htvt
0
g− tvt,
1.11
vt 0 such that gt, u ≤ αt|u| βt,
t ∈ T, |u| ≥ B0 .
2.10
Then for each real number κ > 0, there is a decomposition gt, x qκ t, x eκ t, x
2.11
0 ≤ xqκ t, x, t ∈ T, x ∈ R, qκ t, u ≤ αt|u| βt κ, t ∈ T, |u| ≥ max{1, B0 },
2.12
of g satisfying
2.13
and there exists a function σκ : T → 0, ∞ depending on r, R, and g such that |eκ t, x| ≤ σκ t,
t ∈ T, x ∈ R.
2.14
3. Existence of Periodic Solutions In this section, we need to give some lemmas first, which have vital importance to prove Theorem 1.1. For convenience, we set as dimMk 1.
ϕk : 0,
3.1
Thus, for any u ∈ D, we have the following Fourier expansion: ut a0
N ai ψi t bi ϕi t ,
t ∈ T.
3.2
i1
Let us write t, ut ut u0 t u
u⊥ t ut − u0 t,
3.3
where ut a0
k−1
ai ϕi t bi ψi t ,
i1
u t ak ϕk t bk ψk t, 0
u t
N ik1
ai ϕi t bi ψi t .
3.4
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Lemma 3.1. Suppose that for 1 ≤ k ≤ N − 1, λk1 is an eigenvalue of 1.3 of multiplicity 2. Let Γ : T → R be a given function satisfying 0 ≤ Γt ≤ λk1 − λk ,
t ∈ T,
3.5
and for at least T/2 2 points in 1, T , Γt < λk1 − λk .
3.6
Then there exists a constant δ δΓ > 0 such that for all u ∈ D, one has T 2 Δ2 ut − 1 λk atut Γtatut ut u0 t − u t ≥ δu⊥ .
3.7
t1
Proof. For u ∈ D, N Δ2 ut − 1 −at ai λi ψi t bi λi ϕi t .
3.8
i1
in D, we have Taking into account the orthogonality of u, u0 , and u T Δ2 ut − 1 λk atut Γtatut ut u0 t − u t t1
T
T 2 Δ2 ut − 1 λk atut ut Γtat ut u0 t
t1
t1
T
Δ2 u t − 1 λk at ut Γtat ut − ut
t1
T
Δ2 u0 t − 1 λk atu0 t u0 t
3.9
t1
T
T 2 −Δut2 λk atu2 t Γtat ut u0 t
t1
T
t1
u2 t − Γtat u2 t ut2 − λk at Δ
t1 T T T ≥ λk − λk−1 atu2 t Δ u2 t. ut 2 − λk at Γtat t1
t1
t1
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Set
Λu λk − λk−1
T
atu2 t.
3.10
t1
Then, Λu ≥ δ1 u 2 ,
3.11
where δ1 is a positive constant less than λk − λk−1 . Let
ΛΓ u
T T u2 t. ut 2 − λk at Γtat Δ t1
3.12
t1
We claim that ΛΓ u ≥ 0 with the equality holding only if u A0 ψk1 B0 ϕk1 , where A0 , B0 ∈ R are constants. In fact, we have from Lemma 2.1 that
ΛΓ u
T T u2 t ut 2 − λk at Γtat Δ t1
−
t1
T T u tΔ2 u t − 1 − λk at Γtat u2 t t1
t1
T N N λi at ai ψi t bi ϕi t ai ψi t bi ϕi t t1 ik1
ik1
T − λk at Γtat t1
≥
N
ai ψi t bi ϕi t
2
ik1
N T N λj at aj ψj t bj ϕj t ai ψi t bi ϕi t t1 ik1
jk1
T − λk1 at
t1
N N
⎞ ⎛ N ai ψi t bi ϕi t ⎝ aj ψj t bj ϕj t ⎠
ik1
ai aj λj
N N
jk1
N N T T atψi tψj t bi bj λj atψi tψj t t1
ik1 jk1
−
N
ai aj λk1
ik1 jk1
ik1 jk1 T t1
atψi tψj t
t1
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−
N N
bi bj λk1
N
N a2j λj − λk1 bj2 λj − λk1
jk1
atψi tψj t
t1
ik1 jk1
T
jk1
N jk1
a2j bj2
λj − λk1 ≥ 0. 3.13
Obviously, ΛΓ u 0 implies that ak2 · · · aN bk2 · · · bN 0, and accordingly u t A0 ψk1 t B0 ϕk1 t for some A0 , B0 ∈ R. Next we prove that ΛΓ u 0 implies u 0. Suppose to the contrary that u / 0. We note that u has at most T/2 1 zeros in T. Otherwise, u must have two consecutive zeros in T, and subsequently, u ≡ 0 in 0, T 1 by 1.3. This is a contradiction. Using 3.6 and the fact that u has at most T/2 1 zeros in T, it follows that
ΛΓ u
T ut 2 λk1 at − λk at − Γtat t1
atλk1 − λk − Γt ut 2
3.14
t∈T, ut / 0
> 0,
u 0. Hence, u 0. which contradicts ΛΓ We claim that there is a constant δ2 δ2 Γ > 0 such that u ≥ δ2 ΛΓ u 2 .
3.15
Assume that the claim is not true. Then we can find a sequence { un } ⊂ D and u ∈ D, such that, by passing to a subsequence if necessary,
0 ≤ ΛΓ un ≤
1 , n
−→ 0, un − u
un 1,
3.16
n −→ ∞.
3.17
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From 3.17, it follows that T T T T 2 2 2 2 un t − Δ ut un t 1 − u n t − ut 1 − u t Δ t1 t1 t1 t1 ≤
T T 2 2 un t 1 − u un t − u 2 t 1 2 t t1
t1
T 2 | un t| | un t 1 − u t 1| | ut 1| | un t − u t| t1
−→ 0. 3.18 By 3.12, 3.16, and 3.17, we obtain, for n → ∞, T T ut 2 , un t 2 −→ Δ λk at Γtat t1
3.19
t1
and hence T T ut 2 , ut 2 ≤ Δ λk at Γtat t1
3.20
t1
that is, ΛΓ u ≤ 0.
3.21
By the first part of the proof, u 0, so that, by 3.19, Tt1 Δ un t 2 → 0, a contradiction with the second equality in 3.16. 2 u 2 u 2 the proof is complete. Set δ min{δ1 , δ2 } > 0 and observing that u⊥ Lemma 3.2. Let Γ be as in Lemma 3.1 and let δ > 0 be associated with Γ by that lemma. Let ε > 0. Let p : T → R be a function satisfying 0 ≤ pt ≤ Γt ε.
3.22
Then for all u ∈ D, one has T 2 Δ2 ut − 1 λk atut ptatut ut u0 t − u t ≥ δ − εu⊥ . t1
3.23
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Proof. Using the computations in the proof of Lemma 3.1 and 3.22, we obtain T Δ2 ut − 1 λk atut ptatut ut u0 t − u t t1
T T 2 Δ2 ut − 1 λk atut ut ptat ut u0 t t1
t1
T Δ2 u t − 1 λk at ut ptat ut − ut t1
T Δ2 u0 t − 1 λk atu0 t u0 t t1
≥
T ut2 ut2 − λk at ptat Δ
3.24
t1
T −Δut2 λk atut2 t1
≥
T T ut2 − εat ut2 ut2 − λk at Γtat Δ t1
t1
T −Δut2 λk atut2 t1
2 ≥ δu⊥ − ε u 2 . So that, using 3.7, 3.8, the relation u t that
N
ik1 ai ψi tbi ϕi t , and Lemma
2.1, it follows
T 2 Δ2 ut − 1 λk atut ptatut ut u0 t − u t ≥ δ − εu⊥ .
3.25
t1
Proof of Theorem 1.1. The proof is motivated by Iannacci and Nkashama 12 . Let δ > 0 be associated to the function Γ by Lemma 3.1. Then, by assumption 1.8, there exist Rδ > 0 and b : T → R, such that gt, u ≤
Γt
δ at|u| bt, 4
3.26
for all t ∈ T and all u ∈ R with |u| ≥ R. Hence, 1.2 is equivalent to Δ2 ut − 1 λk atut q1 t, ut e1 t, ut ht, u0 uT ,
u1 uT 1,
3.27
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where q1 and e1 satisfy 2.12 and 2.14 with κ 1. Moreover, by 2.13 q1 t, u ≤
δ Γt at|u| bt 1, 4
t ∈ T, |u| > max{1, R}.
3.28
Let B > max{1, R}, so that bt 1 δ < at, 4 |u|
t ∈ T, |u| > B.
3.29
It follows from 3.28 and 3.29 that
−1
0 ≤ u q1 t, u ≤
δ at, Γt 2
t ∈ T, |u| ≥ B.
3.30
Define γ : T × R → R by ⎧ ⎪ u−1 q1 t, u, |u| ≥ B, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ −1 u u t, B q B 1 − Γtat, 0 ≤ u < B, 1 γt, u B B ⎪ ⎪ ⎪ ⎪ ⎪ u ⎪ −1 u ⎪ ⎪ 1 Γtat, −B < u ≤ 0. ⎩B q1 t, −B B B
3.31
So we have 0 ≤ γt, u ≤
δ Γt at, 2
t ∈ T, u ∈ R.
3.32
Define f : T × R → R ft, u e1 t, u q1 t, u − γt, uu.
3.33
Then there exists ν : T → 0, ∞ such that ft, u ≤ νt,
t ∈ T, u ∈ R.
3.34
Therefore, 1.2 is equivalent to Δ2 ut − 1 λk atut γt, utut ft, ut ht, u0 uT ,
u1 uT 1.
3.35
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To prove that 1.2 has at least one solution in D, it suffices, according to the LeraySchauder continuation method 15 , to show that all of the possible solutions of the family of equations Δ2 ut − 1 λk atut 1 − η τatut ηγt, utut ηft, ut ηht, u0 uT ,
t ∈ T,
u1 uT 1 3.36
in which η ∈ 0, 1 , τ ∈ 0, λk1 − λk with τ < δ/4, τ fixed are bounded by a constant K0 which is independent of η and u. Notice that, by 3.32, we have δ 0 ≤ 1 − η τat ηγt, u ≤ Γt at, 2
t ∈ T, u ∈ R.
3.37
It is clear that for η 0, 3.36 has only the trivial solution. Now if u ∈ D is a solution of 3.36 for some η ∈ 0, 1, using Lemma 3.2 and Cauchy’s inequality, we obtain 0
T
ut u0 t − u t
Δ2 ut − 1 λk atut
1 − η τat ηγt, ut ut
t1
T
ut u0 t − u t
ηft, ut − ηht
t1
2 δ ⊥ u u0 ν h , ≥ u − ζ u 2
3.38
where
ζ
√
T ! mint∈T at
2 .
3.39
So we conclude that 0≥
2 δ ⊥ u − β u⊥ u0 , 2
3.40
for some constant β > 0, depending only on a, ν and h but not on u or η. Taking α βδ−1 , we get 1/2 ⊥ . u ≤ α α2 2αu0
3.41
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We claim that there exists ρ > 0, independent of u and η, such that for all possible solutions of 3.36
u < ρ.
3.42
Suppose on the contrary that the claim is false. Then there exists {ηn , un } ⊂ 0, 1 × D with un ≥ n and for all n ∈ N, Δ2 un t − 1 λk atun t 1 − ηn τatun t ηn gt, un t ηn ht, un 0 un T ,
un 1 un T 1.
3.43
From 3.41, it can be shown that 0 un −→ ∞,
−1 ⊥ 0 −→ 0, un un
3.44
and accordingly, u⊥n u0n −1 is bounded in D. Setting vn un / un , we have Δ2 vn t − 1 λk atvn t τatvn t gt, un t ht ηn τatvn t − ηn , ηn
un
un vn 0 vn T ,
t ∈ T,
3.45
vn 1 vn T 1.
Define an operator A : D → D by Awt : Δ2 wt − 1 λk atwt τatwt, Aw0 : AwT ,
t ∈ T,
3.46
Aw1 : AwT 1.
Then A−1 : D → D is completely continuous since D is finite dimensional. Now, 3.45 is equivalent to # " g·, un · h· ηn τa·vn · − ηn vn t A ηn t,
un
un −1
t ∈ T.
3.47
By 3.26, it follows that {g·, un ·/ un } is bounded. Using 3.47, we may assume that taking a subsequence and relabeling if necessary vn → v in D, · , ||v|| 1 and v0 vT , v1 vT 1. On the other hand, using 3.41, we deduce immediately that ⊥ vn −→ 0,
n −→ ∞.
3.48
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Therefore, vt ak ϕk t bk ψk t,
t ∈ T.
3.49
Rewrite vn vn0 vn⊥ , and let, taking a subsequence and relabeling if necessary, vn0 −→ v∗ ,
3.50
in D.
Set I {t ∈ T : v∗ t > 0},
I− {t ∈ T : v∗ t < 0}.
3.51
∅ or I− / ∅. Since ut / ≡ 0 in T, I / We claim that lim un t ∞,
n→∞
lim un t −∞,
n→∞
∀t ∈ I ,
3.52
∀t ∈ I− .
3.53
We may assume that I / ∅, and only deal with the case t ∈ I . The other case can be treated by similar method. It follows from 3.50 that 0 vn − v∗ : max vn0 t − v∗ t | t ∈ T −→ 0, ∞
n −→ ∞,
3.54
which implies that for all n sufficiently large, vn0 t ≥
1 ∗ v t > 0, 2
3.55
∀t ∈ I .
On the other hand, we have from 3.44, 3.55, and the fact ||un || ≥ ||u0n || that there exists N > 0 such that for n > N and t ∈ I ,
un t
u0n t
u⊥n t
un
vn0 t
u⊥ t n
un
≥
1
un vn0 t. 2
3.56
This together with 3.55 implies that for n ≥ N, un t ≥ Therefore, 3.52 holds.
1
un v∗ t, 4
t ∈ T .
3.57
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Now let us come back to 3.43. Multiplying both sides of 3.43 by vn0 and summing from 1 to T , we get that T 2 ht − gt, un t vn0 t. 0 ≤ ηn−1 1 − ηn τ vn0 un
3.58
t1
Combining this with 3.52 and 3.53, it follows that T
htv∗ t ≥
t1
g tv∗ t
vt>0
g− tv∗ t.
3.59
vt
