Selçuk J. Appl. Math. Selçuk Journal of Vol. 12. No. 2. pp. 81-100, 2011 Applied Mathematics
Existence of Solutions of BVPs for Nonlinear Difference Equations: Non-resonance Case1
Yuji Liu
Department of Mathematics, Hunan Institute of Science and Technology, Yueyang 414000, P.R.China
e-mail: liuyuji888@ sohu.com
Received Date:March 18, 2011 Accepted Date: September 14, 2011
Abstract. Sufficient conditions for the existence of solutions of the following boundary value problem for nonlinear difference equation
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ x(t + 2k) −P2ki=0−1aix(t + i) = f (t, x(t), x(t + 1), · · · , x(t + 2k − 1)), t ∈ [0, T − 1], x(i) = Ai, i ∈ [0, k − 1], x(T + 2k − 1 − i) = Bi, i ∈ [0, k − 1]
at non-resonance case are established. Examples are given to show the efficiency of the result in this paper.
Key words: Solution; nonlinear difference equation; boundary value problem; fixed-point theorem; growth condition.
2000 Mathematics Subject Classification: 34B10, 34B15. 1. Introduction
Recently there has been a large amount of attention paid to the existence of solutions of boundary value problems for the differential equations that arise from various applied problems. Similarly there has been a parallel interest in results for the analogous discrete problems, see the text books [1,2], the papers [3-9,11-14] and the references therein.
Particular significance in these points lies in the fact that when a BVP is dis-cretized, strange and interesting changes can occur in the solutions. For exam-ple, properties such as existence, uniqueness and multiplicity of solutions may
1Supported by the Natural Science Foundation of Guangdong province No:S2011010001900
not be shared between the continuous differential equation and its related dis-crete difference equation [3, p. 520]. Moreover, when investigating difference equations, as opposed to differential equations, basic ideas from calculus are not necessarily available to use, such as the intermediate value theorem, the mean value theorem and Rolle0s theorem. Thus, new challenges are faced and
innovation is required [6].
In [4], the following boundary value problem (BVP) involving second order difference equations and two-point boundary conditions
(1) ( 54yk h2 = f ³ tk, yk,4yhk ´ , k = 1, · · · , n − 1, y0= A, yn= B,
was studied, where A, B ∈ R, n ≥ 2 an integer, f is continuous, scalar-valued function, the step size is h = N/n with N a positive constant, the grid points are tk= kh for k = 0, · · · , n. The differences are given by
4yk = ½ yk+1− yk, k = 0, · · · , n − 1, 0, k = n; 5 4 yk = ½ yk+1− 2yk+ yk−1, k = 1, · · · , n − 1, 0, k = 0 or k = n.
The following two results were proved in [4].
Theorem RT1. Let f be continuous on [0, n] × R2 and α, β and K be
non-negative constants. If there exist c, d ∈ [0, 1) such that
(2) |f(t, u, v)| ≤ α|u|c+ β|v|d+ K, (t, u, v) ∈ [0, n] × R2, then the discrete BVP(1) has at least one solution.
Theorem RT2. Let f be continuous on [0, n] × R2 and α, β and K be non-negative constants. If (3) |f(t, u, v)| ≤ α|u| + β|v| + K, (t, u, v) ∈ [0, n] × R2, and (4) αn 2 8 + βn 2 < 1, then the discrete BVP(1) has at least one solution.
In Theorems RT1 and RT2, the assumptions (2) or (3) allow f to grow either sublinearly or at most linearly. The problem appears naturally:
Problem 1. Under what conditions does BVP(1) has at lest one solution when f grows superlinearly?
To solve problem 1, in [6], the authors investigated the solvability of a class of discrete Dirichlet boundary value problems
( 54yk h2 = f ³ tk, yk,4yhk ´ , k = 1, · · · , n − 1, y0= 0, yn= 0,
by using the lower and upper solution method, where the nonlinear term f (t, u, v) can have a superlinear growth both in u and in v. Moreover, the growth con-ditions on f are one-sided. By computing the priori bounds on solutions to the discrete problem and then obtaining the existence of at least one solution, it is shown that solutions of the discrete problem will converge to solutions of the corresponding ordinary differential equations.
In paper [5,7], the authors studied the following boundary value problems ½ ∆2y k+1= f (tk, yk, ∆yk) , k = 1, · · · , n − 1, y0= A, yn= B, and ( ∆2y k+1 h2 = B(tk)yk+F (tkh)∆yk + g(tk), k = 1, · · · , n − 1, y0= A, yn = B, respectively, where A, B ∈ Rd, f ∈ C({0, 1, · · · , n} × R2d× Rd), 4yk = ½ yk+1− yk, k = 0, · · · , n − 1, 0, k = n; 5 4 yk = ½ yk+1− 2yk+ yk−1, k = 1, · · · , n − 1, 0, k = 0 or k = n.
The following theorem was proved.
Theorem TT [5]. Let α, K, R be nonnegative constants satisfying 2αR < 1 and set β = max{||A||, ||B||}. Suppose that f satisfies
||f(x, y, p)|| ≤ α(< y, f(x, y, p) > +||p||2) + K
for all x ∈ {0, 1, · · · , n}, ||y|| ≤ R, p ∈ Rdwith < y, f (x, y, p) >≥ 0. If β +αβ2
+
Kn
4 ≤ R, then above problem has at least one solution.
In this paper, we study the following boundary value problem for the higher order nonlinear difference equation
(5) ⎧ ⎨ ⎩ x(t + 2k)−P2ki=0−1aix(t + i) = f (t, x(t), x(t + 1), · · · , x(t + 2k − 1)), x(i) = Ai, i ∈ [0, k − 1], x(T + 2k − 1 − i) = Bi, i ∈ [0, k − 1],
where k, m are positive integers, ai(i ∈ [0, 2k − 1] ∈ R, Ai, Bi ∈ R, T ≥ 2k an
integer, [a, b] = {a, a + 1, · · · , b} for nonnegative integers a, b with a ≤ b, f is continuous, scalar-valued function.
It is easy to see that the discretized form of BVP of the continuous type (A) ½ x00(t) = g(t, x(t), x0(t)), t ∈ (0, 1), x(0) = A0, x(1) = B0 is as follows (B)½ x(t + 2) − 2x(t + 1) + x(t) = g(t, x(t), x(t + 1) − x(t)), t ∈ [0, T − 1], x(0) = A0, x(T + 1) = B0.
One can see that BVP(B) is a spacial case of BVP(5) by choosing k = 1, a1= −2, a0= 1 and replacing g(t, x(t), x(t + 1) − x(t)) by f(t, x(t), x(t + 1)) in
(5).
Consider the the corresponding homogenous BVP of BVP(5)
(6) ⎧ ⎨ ⎩ x(t + 2k) −P2ki=0−1aix(t + i) = 0 t ∈ [0, T − 1] x(i) = 0, i ∈ [0, k − 1], x(T + 2k − 1 − i) = 0, i ∈ [0, k − 1].
We call BVP(5) non-resonance if BVP(6) has unique solution x(n) = 0 for all n ∈ [0, 2k + T − 1] and resonance case if BVP(6) has infinitely many solutions. The purpose of this paper is to establish new sufficient conditions guaranteeing the existence of solutions of BVP (5) at non-resonance case. Problem 1 is solved by the way different from those used in known related papers [4-7]. The methods used in this paper to get the priori bound are different from those in [4-7]. This paper is organized as follows. In Section 2, the existence results of BVP(5) at non-resonance case are given. We give examples to illustrate the main result in Section 3.
2. Solvability at Non-resonance Case Define the matrix sequence Cn by
C0 = (ak, ak+1, · · · , a2k−1), C1 = a2k−1C0+ (ak−1, ak, · · · , a2k−2), C2 = a2k−1C1+ a2k−2C0+ (ak−2, ak−1, · · · , a2k−3), C3 = a2k−1C2+ a2k−2C1+ a2k−3C0+ (ak−3, ak−2, · · · , a2k−4), · · · · Ck = a2k−1Ck−1+ a2k−2Ck−2+ · · · + akC0+ (a0, a1, · · · , ak−1), Ck+1 = a2k−1Ck+ a2k−2Ck−1+ · · · + ak−1C0+ (0, a0, · · · , ak−2), Ck+2 = a2k−1Ck+1+ a2k−2Ck+ · · · + ak−2C0+ (0, 0, a0, · · · , ak−3), · · · · C2k−1 = a2k−1C2k−2+ · · · + a1C0+ (0, · · · , 0, a0), C2k = a2k−1C2k−1+ a2k−2C2k−2+ · · · + a0C0,
C2k+1 = a2k−1C2k+ a2k−2C2k−1+ · · · + a0C1,
· · · ·
CT−1 = a2k−1CT−2+ a2k−2CT−3+ · · · + a0CT−2k−1.
Lemma 2.1. Let R(C) denote the rank of the matrix
C = ⎛ ⎜ ⎜ ⎝ CT−k CT−k+1 · · · CT−1 ⎞ ⎟ ⎟ ⎠ k×k .
Then BVP(6) has unique solution x(n) = 0 for all n ∈ [0, T + 2k − 1] if and only if R(C) = k.
Proof. It follows from (6) that
x(2k) = 2kX−1 i=0 aix(i) = 2kX−1 i=k aix(i) = C0 ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ , x(2k + 1) = 2kX−1 i=0 aix(1 + i) = (a2k−1C0+ (ak−1, · · · , a2k−2)) ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ = C1 ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ . Similarly, we get x(2k + 2) = C2 ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ , x(2k + 3) = C3 ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ , · · · · x(3k) = Ck ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ , · · · ·
x(k + T − 1) = CT−k−1 ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ , x(k + T ) = CT−k ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ , · · · · x(2k + T − 1) = CT−1 ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ .
Since x(T + 2k − 1 − i) = 0 for all i ∈ [0, k − 1], we get that
(7) C × ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ = 0.
One can finds easily that x(k) = x(k +1) = · · · = x(2k +T −1) = 0 if and only if R(C) = k. Thus BVP(6) has unique solution x(n) = 0 for all n ∈ [0, T + 2k − 1] if and only if R(C) = k. The proof is complete.
Now, we establish the existence result of BVP(5) at non-resonance case, i.e., R(C) = k. The following abstract existence theorem will be used in the proof of the main result of this section. Its proof can be see in [10].
Lemma 2.2 [10]. Let X and Y be Banach spaces. Suppose L : D(L) ⊂ X → Y is a Fredholm operator of index zero with KerL = {0}, N : X → Y is L−compact on each open bounded subset of X. If 0 ∈ Ω ⊂ X is a open bounded subset and Lx 6= λNx for all x ∈ D(L) ∩ ∂Ω and λ ∈ (0, 1), then there exists at least one x ∈ Ω such that Lx = Nx.
Let X = R2k+T be endowed with the norm ||x|| = max
n∈[0,2k+T −1]|x(n)|
for x = (x0, x1, · · · , x2k+T−1) ∈ R2k+T and Y = RT × Rk× Rk be endowed
with the norm ||(x, u, v)|| = max
½ max
n∈[0,T −1]|x(n)|, maxi∈[0,k−1]|u(i)|,i∈[k+T,2k+T −1]max |v(i)|
¾
for (x, u, v) ∈ Y , where x = (x0, · · · , xT−1, u = (u0, · · · , uk−1) and v =
Let L : X → Y be defined by L¡ x ¢= ⎛ ⎝ x(n + 2k) − P2k−1 i=0 aix(n + i) n ∈ [0, T − 1] x(i) i ∈ [0, k − 1] x(i) i ∈ [T + k, T + 2k − 1] ⎞ ⎠ , x ∈ X, and N : X → Y by N¡ x ¢= ⎛ ⎝ f (n, x(n), x(n + 1), · · · , x(n + 2k − 1)) n ∈ [0, T − 1]Ai i ∈ [0, k − 1] Bi i ∈ [T + k, 2k + T − 1] ⎞ ⎠ , x ∈ X.
Lemma 2.3. Suppose R(C) = k. Then it holds that
(i) x ∈ X is a solution of L(x) = N(x) implies that x is a solution of BVP(5).
(ii) KerL = {0}.
(iii) L is a Fredholm operator of index zero, N is L-compact on each open bounded subset of X.
Proof. It is easy to see that (i) holds by the definitions of L and N . For (ii), suppose x ∈ KerL, then we get (6). It follows that
x(k + T − 1) = CT−k−1 ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ , x(k + T ) = CT−k ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ , · · · · x(2k + T − 1) = CT−1 ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ .
Then (7) holds. Since R(C) = k, we get that ⎛ ⎜ ⎜ ⎝ x(k) x(k + 1) · · · x(2k − 1) ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ 0 0 · · · 0 ⎞ ⎟ ⎟ ⎠ .
Hence x(n) = 0 for all n ∈ [0, 2k + T − 1].
The proof of (iii) is standard and is omitted, one may see [9,11]. The proof is complete.
Theorem 2.1. Suppose that R(C) = k, ak > 0 and
(A) there exist numbers β > 0, θ > 1, nonnegative sequences pi(n), r(n)(i ∈
[0, 2k − 1]), functions g(n, x0, · · · , x2k−1), h(n, x0, · · · , x2k−1) such that
(8) f (n, x0, · · · , x2k−1) = g(n, x0, · · · , x2k−1) + h(n, x0, · · · , x2k−1) and (9) g(n, x0, x1, · · · , x2k−1)xn+k ≥ β|xk|θ+1, and (10) |h(n, x0, · · · , x2k−1)| ≤ 2kX−1 i=0 pi(n)|xi|θ+ r(n), for all n ∈ [0, T − 1], (x0, x1, · · · , x2k−1) ∈ R2k; (B) it holds that (11) kX−1 j=0 2ka2 j ak − ak + 2kX−1 j=k+1 2ka2 j ak +2k ak ≤ 0.
Then BVP(5) has at least one solution if
(12)
2kX−1 i=0
||pi|| < β, where ||pi|| = max
n∈[0,T −1]|p(n)|.
Proof. To apply Lemma 2.2, we consider Ω1= {x ∈ X : Lx = λNx, λ ∈ (0, 1)}.
For x ∈ Ω1, we have (13) ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ x(n + 2k) −P2ki=0−1aix(n + i) = λf (n, x(n), x(n + 1), · · · , x(n + 2k − 1)), n ∈ [0, T − 1], x(i) = λAi, i ∈ [0, k − 1], x(i) = λBi, i ∈ [T + k, T + 2k − 1]. So (14)³ x(n + 2k) −P2ki=0−1aix(n + i) ´ x(n + k) = λx(n + k)f (n, x(n), x(n + 1), · · · , x(n + 2k − 1)) for all n ∈ [0, T − 1]. It is easy to see that
2PTn=0−1³x(n + 2k) −P2ki=0−1aix(n + i) ´ x(n + k) =PTn=0−1¡2x(n + 2k)x(n + k) − 2ak[x(n + k)]2 −2Pi∈[0,2k−1],i6=kaix(n + i)x(n + k) ´
=XT−1 n=0 " − µr 2k akx(n + 2k) − r ak 2kx(n + k) ¶2 −Xk−1 i=0 µ ai √ak 2k x(n + i) + r ak 2kx(n + k) ¶2 −X2k−1 i=k+1 µ ai √ak 2k x(n + i) + r ak 2kx(n + k) ¶2 −2ak[x(n + k)]2+ 2k ak [x(n + 2k)]2+ak 2k[x(n + k)] 2 +Xk−1 i=0 µ 2ka2 i ak [x(n + i)]2+ak 2k[x(n + k)] 2 ¶ +X2k−1 i=k+1 µ 2ka2 i ak [x(n + i)]2+ak 2k[x(n + k)] 2 ¶¸ ≤XTn=0−1 ∙ −ak[x(n + k)]2+ 2k ak [x(n + 2k)]2 +Xk−1 i=0 2ka2 i ak [x(n + i)]2+X2k−1 i=k+1 2ka2 i ak [x(n + i)]2 ¸ =XT−1 n=0 ∙Xk−1 i=0 2ka2 i ak [x(n + i)]2− a k[x(n + k)]2 +X2k−1 i=k+1 2ka2 i ak [x(n + i)]2+2k ak [x(n + 2k)]2 ¸ = 2ka 2 0 ak x(0)2+ µ 2ka2 0 ak +2ka 2 1 ak ¶ x(1)2+ · · · +µXk−1 i=0 2ka2 i ak ¶ x(k − 1)2 +2k akx(2k + T − 1) 2+ µ 2k ak +2ka 2 2k−1 ak ¶ x(2k + T − 2)2+ · · · + µ2k ak +X2k−1 i=T +k+1 2ka2 i ak ¶ x(T + k)2 +X2k−1 i=k à Xk−1 j=0 2ka2j ak − ak +Xi j=k+1 2ka2j ak ! x(i)2 +XT−1 i=2k à Xk−1 j=0 2ka2 j ak − ak +X2k−1 j=k+1 2ka2 j ak +2k ak ! x(i)2 +XT +k−1 i=T à Xk−1 j=i−T +1 2ka2 j ak − a k+ X2k−1 j=k+1 2ka2 j ak +2k ak ! x(i)2
Since (11) and ak > 0 imply that
Xk−1 j=0 2ka2 j ak − ak +Xi j=k+1 2ka2 j ak ≤ 0, i ∈ [k, 2k − 1], Xk−1 j=0 2ka2j ak − ak +X2k−1 j=k+1 2ka2j ak +2k ak ≤ 0,
k−1 X j=i−T +1 2ka2 j ak − a k+ 2kX−1 j=k+1 2ka2 j ak +2k ak ≤ 0, i ∈ [T, T + k − 1]. Then 2 TX−1 n=0 Ã x(n + 2k) − 2kX−1 i=0 aix(n + i) ! x(n + k) ≤ λ2 Ã 2ka2 0 ak A20+ µ 2ka2 0 ak +2ka 2 1 ak ¶ A21+ · · · + Ãk−1 X i=0 2ka2 i ak ! A2k−1 ! +λ2 µ 2k ak B2k+T2 −1+ µ 2k ak +2ka 2 2k−1 ak ¶ B2k+T2 −2+ · · · + Ã 2k ak + 2kX−1 i=T +k+1 2ka2 i ak ! BT +k2 ! + 2kX−1 i=k ⎛ ⎝ k−1 X j=0 2ka2 j ak − ak + i X j=k+1 2ka2 j ak ⎞ ⎠ x(i)2 + TX−1 i=2k ⎛ ⎝ kX−1 j=0 2ka2j ak − ak + 2kX−1 j=k+1 2ka2j ak +2k ak ⎞ ⎠ x(i)2 + T +kX−1 i=T ⎛ ⎝ k−1 X j=i−T +1 2ka2 j ak − ak + 2kX−1 j=k+1 2ka2 j ak +2k ak ⎞ ⎠ x(i)2 ≤ λ2 Ã 2ka20 ak A20+ µ 2ka20 ak +2ka 2 1 ak ¶ A21+ · · · + Ãk−1 X i=0 2ka2i ak ! A2k−1 ! + λ2 µ 2k ak B2k+T2 −1+ µ 2k ak +2ka 2 2k−1 ak ¶ B2k+T2 −2+ · · · + Ã 2k ak + 2kX−1 i=T +k+1 2ka2 i ak ! BT +k2 ! =: λ2M.
So, we get from (14) that λ TX−1 n=0 x(n + k)f (n, x(n), x(n + 1), · · · , x(n + 2k − 1)) ≤ λ2M. It follows that (15) TX−1 n=0 x(n + k)f (n, x(n), x(n + 1), · · · , x(n + 2k − 1)) ≤ λM.
Then (8), (9) and (10) imply that β TX−1 n=0 |x(n + k)|θ+1 ≤ TX−1 n=0 g(n, x(n), x(n + 1), · · · , x(n + 2k − 1))x(n + k) ≤ TX−1 n=0 f (n, x(n), x(n + 1), · · · , x(n + 2k − 1))x(n + k) − TX−1 n=0 h(n, x(n), x(n + 1), · · · , x(n + 2k − 1))x(n + k) ≤ λM − TX−1 n=0 h(n, x(n), x(n + 1), · · · , x(n + 2k − 1))x(n + k) ≤ M + TX−1 n=0 |h(n, x(n), x(n + 1), · · · , x(n + 2k − 1))||x(n + k)| ≤ M + TX−1 n=0 Ã2k−1 X i=0 pi(n)|x(n + i)|θ+ r(n) ! |x(n + k)| ≤ 2kX−1 i=0 ||pi|| TX−1 n=0 |x(n + i)|θ|x(n + k)| +||r|| TX−1 n=0 |x(n + k)| + M.
For xi≥ 0, yi ≥ 0(i = 1, 2, · · · , s), Holder0s inequality implies s X i=1 xiyi≤ à s X i=1 xpi !1/pà s X i=1 yqi !1/q , 1/p + 1/q = 1, q > 0, p > 0. It follows that βXT−1 n=0|x(n + k)| θ+1 ≤ ||pk|| XT−1 n=0|x(n + k)| θ+1+ ||r||Tθ+1θ ∙X T−1 n=0|x(n + 1)| θ+1 ¸ 1 θ+1 +Xk−1 i=0 ||pi|| ∙XT−1 n=0|x(n + i)| θ+1 ¸ θ θ+1∙XT−1 n=0|x(n + k)| θ+1 ¸ 1 θ+1 +X2k−1 i=k+1||pi|| ∙XT−1 n=0|x(n + i)| θ+1 ¸ θ θ+1∙XT−1 n=0|x(n + k)| θ+1 ¸ 1 θ+1 + M
= ||pk|| TX−1 n=0 |x(n + k)|θ+1+ ||r||Tθ+1θ "T−1 X n=0 |x(n + 1)|θ+1 # 1 θ+1 + "T−1 X n=0 |x(n + k)|θ+1 # 1 θ+1kX−1 i=0 ||pi|| "T−1+i X u=i |x(u)|θ+1 # θ θ+1 + "T−1 X n=0 |x(n + k)|θ+1 # 1 θ+1 2kX−1 i=k+1 ||pi|| "T−1+i X u=i |x(u)|θ+1 # θ θ+1 + M ≤ ||pk|| TX−1 n=0 |x(n + k)|θ+1+ ||r||Tθ+1θ "T−1 X n=0 |x(n + 1)|θ+1 # 1 θ+1 + "T−1 X n=0 |x(n + k)|θ+1 # 1 θ+1kX−1 i=0 ||pi|| "T−1+k X u=i |x(u)|θ+1 # θ θ+1 + "T−1 X n=0 |x(n + k)|θ+1 # 1 θ+1 2kX−1 i=k+1 ||pi|| "T−1+i X u=k |x(u)|θ+1 # θ θ+1 + M = ||pk|| TX−1 n=0 |x(n + k)|θ+1+ ||r||Tθ+1θ "T−1 X n=0 |x(n + 1)|θ+1 # 1 θ+1 + "T−1 X n=0 |x(n + k)|θ+1 # 1 θ+1kX−1 i=0 ||pi|| "T−1+k X u=k |x(u)|θ+1+ k−1 X u=i |x(u)|θ+1 # θ θ+1 + "T−1 X n=0 |x(n + k)|θ+1 # 1 θ+1 2k−1 X i=k+1 ||pi|| "T−1+k X u=k |x(u)|θ+1+ TX−1+i u=T +k |x(u)|θ+1 # θ θ+1 +M = ||pk|| TX−1 n=0 |x(n + k)|θ+1+ ||r||Tθ+1θ "T−1 X n=0 |x(n + 1)|θ+1 # 1 θ+1 + "T−1 X n=0 |x(n + k)|θ+1 # 1 θ+1kX−1 i=0 ||pi|| "T−1 X n=0 |x(n + k)|θ+1+ k−1 X u=i |λAi|θ+1 # θ θ+1 + "T−1 X n=0 |x(n + k)|θ+1 # 1 θ+1 2kX−1 i=k+1 ||pi|| "T−1 X n=0 |x(n + k)|θ+1+ TX−1+i u=T +k |λBi|θ+1 # θ θ+1 +M
≤ ||pk|| TX−1 n=0 |x(n + k)|θ+1+ ||r||Tθ+1θ "T−1 X n=0 |x(n + 1)|θ+1 # 1 θ+1 + "T−1 X n=0 |x(n + k)|θ+1 # 1 θ+1kX−1 i=0 ||pi|| "T−1 X n=0 |x(n + k)|θ+1+ kX−1 u=0 |Ai|θ+1 # θ θ+1 + "T−1 X n=0 |x(n + k)|θ+1 # 1 θ+1 2kX−1 i=k+1 ||pi|| "T−1 X n=0 |x(n + k)|θ+1+ T−1+2kX u=T +k |Bi|θ+1 # θ θ+1 +M
We claim that if m > 0, then there exists a constant σ ∈ (0, 1), independent of λ, such that (1 + x)m≤ 1 + (m + 1)x for all x ∈ (0, σ). In fact, let q(x) =
(1 + x)m− (1 + (m + 1)x), we see q(0) = 0, and q0(0+) = −1 < 0. Hence
there exists a constant σ > 0 such that q0(x) < 0 for all x ∈ [0, σ). Then
q(x) ≤ q(0) = 0 for all x ∈ [0, σ). It follows that the claim is valid. We consider two cases.
Case1. PTn=0−1|x(n + k)|θ+1≤ Sk−1
u=0|Ai|θ+1+STu=T +k−1+2k|Bi|θ+1
σ .
At this case, we have
TX−1 n=0 |x(n + k)|θ+1≤ Pk−1 u=0|Ai|θ+1+ PT−1+2k u=T +k |Bi|θ+1 σ =: M1. Case2. PTn=0−1|x(n + k)|θ+1> Sk−1 u=0|Ai|θ+1+STu=T +k−1+2k|Bi|θ+1 σ .
At this case, one sees that σ > Pk−1 u=0|Ai|θ+1+ PT−1+2k u=T +k |Bi|θ+1 PT−1 n=0|x(n + k)|θ+1 . Then σ > Pk−1 u=0|Ai|θ+1 PT−1 n=0|x(n + k)|θ+1 , σ > PT−1+2k u=T +k |Bi|θ+1 PT−1 n=0|x(n + k)|θ+1 . Hence β TX−1 n=0 |x(n + k)|θ+1 ≤ ||pk|| TX−1 n=0 |x(n + k)|θ+1+ ||r||Tθ+1θ "T−1 X n=0 |x(n + 1)|θ+1 # 1 θ+1 + TX−1 n=0 |x(n + k)|θ+1 k−1 X i=0 ||pi|| " 1 + Pk−1 u=0|Ai|θ+1 PT−1 n=0|x(n + k)|θ+1 # θ θ+1 + TX−1 n=0 |x(n + k)|θ+1 2kX−1 i=k+1 ||pi|| " 1 + PT−1+2k u=T +k |Bi|θ+1 PT−1 n=0|x(n + k)|θ+1 # θ θ+1 + M.
Using (1 + x)m≤ 1 + (m + 1)x for all x ∈ (0, σ), one gets that β TX−1 n=0 |x(n + k)|θ+1 ≤ ||pk|| TX−1 n=0 |x(n + k)|θ+1+ ||r||Tθ+1θ "T−1 X n=0 |x(n + 1)|θ+1 # 1 θ+1 + TX−1 n=0 |x(n + k)|θ+1 kX−1 i=0 ||pi|| " 1 + µ 1 + θ 1 + θ ¶ Pk−1 u=0|Ai|θ+1 PT−1 n=0|x(n + k)|θ+1 # + TX−1 n=0 |x(n + k)|θ+1 2kX−1 i=k+1 ||pi|| " 1 + µ 1 + θ 1 + θ ¶ PT−1+2k u=T +k |Bi|θ+1 PT−1 n=0|x(n + k)|θ+1 # +M = TX−1 n=0 |x(n + k)|θ+1 2kX−1 i=0 ||pi|| + ||r||T θ θ+1 "T−1 X n=0 |x(n + 1)|θ+1 # 1 θ+1 + k−1 X i=0 ||pi|| µ 1 + θ 1 + θ ¶kX−1 u=0 |Ai|θ+1+ 2kX−1 i=k+1 ||pi|| µ 1 + θ 1 + θ ¶T−1+2kX u=T +k |Bi|θ+1 +M. Then " β − 2kX−1 i=0 ||pi|| #T−1 X n=0 |x(n + k)|θ+1 ≤ ||r||Tθ+1θ "T−1 X n=0 |x(n + 1)|θ+1 # 1 θ+1 + k−1 X i=0 ||pi|| µ 1 + θ 1 + θ ¶kX−1 u=0 |Ai|θ+1+ 2kX−1 i=k+1 ||pi|| µ 1 + θ 1 + θ ¶T−1+2kX u=T +k |Bi|θ+1 +M.
It follows from (12) that there exists a constant M2> 0 such that TX−1
n=0
|x(n + k)|θ+1≤ M2.
It follows from Case 1 and Case 2 that
TX−1 n=0
Hence |x(n + k)| ≤ (max{M1, M2}) 1
θ+1 for all n ∈ [0, T − 1]. Hence
||x|| = max n∈[0,T +2k−1]|x(n)| ≤ max ½ (max{M1, M2}) 1 θ+1, max i∈[0,k−1]|Ai|, maxi∈[T +k,T +2k−1]|Bi| ¾ . So Ω1 is bounded.
Since f is continuous, Lemma 2.3 implies that KerL = {0}, we know that L is a Fredholm operator of index zero and N is L−compact on each open bounded subset of X. Let Ω0= {x ∈ X : ||x|| < max ½ (max{M1, M2}) 1 θ+1, max i∈[0,k−1]|Ai|, maxi∈[T +k,T +2k−1]|Bi| ¾ + 1}. Then Lx 6= λNx for all λ ∈ (0, 1) and all x ∈ D(L)T∂Ω0. It follows from
Lemma 2.2that BVP (5) has at least one solution. The proof is complete. Theorem 2.2. Suppose that R(C) = k, ak < 0 and
(C) there exist numbers β > 0, θ > 1, nonnegative sequences pi(n), r(n)(i =
0, · · · , 2k − 1), functions g(n, x0, · · · , x2k−1), h(n, x0, · · · , x2k−1) such that
(16) f (n, x0, · · · , x2k−1) = g(n, x0, · · · , x2k−1) + h(n, x0, · · · , x2k−1) and (17) g(n, x0, x1, · · · , x2k−1)xn+k≤ −β|xk|θ+1, and (18) |h(n, x0, · · · , x2k−1)| ≤ 2kX−1 i=0 pi(n)|xi|θ+ r(n), for all n ∈ [0, T − 1], (x0, x1, · · · , x2k−1) ∈ R2k; (D) it holds that k−1 X j=0 2ka2 j ak − ak + i X j=k+1 2ka2 j ak ≥ 0, i ∈ [k, 2k − 1], k−1 X j=0 2ka2j ak − ak + 2kX−1 j=k+1 2ka2j ak +2k ak ≥ 0, k−1 X j=i−T +1 2ka2j ak − ak + 2kX−1 j=k+1 2ka2j ak +2k ak ≥ 0, i ∈ [T, T + k − 1].
Then BVP(5) has at least one solution if
(19)
2kX−1 i=0
||pi|| < β, where ||pi|| = max
Proof. To apply Lemma 2.2, we consider Ω1 = {x ∈ X : Lx = λNx, λ ∈
(0, 1)}. For x ∈ Ω1, we get (13) and (14). It is easy to see that
2 TX−1 n=0 Ã x(n + 2k) − 2kX−1 i=0 aix(n + i) ! x(n + k) = TX−1 n=0 ¡ 2x(n + 2k)x(n + k) − 2ak[x(n + k)]2 −2 X i∈[0,2k−1],i6=k aix(n + i)x(n + k) ⎞ ⎠ = TX−1 n=0 ⎡ ⎣ Ãr 2k −ak x(n + 2k) + r −ak 2k x(n + k) !2 + k−1 X i=0 ⎛ ⎝ ai q −ak 2k x(n + i) − r −ak 2k x(n + k) ⎞ ⎠ 2 + 2kX−1 i=k+1 ⎛ ⎝ ai q −ak 2k x(n + i) − r −ak 2k x(n + k) ⎞ ⎠ 2 − 2ak[x(n + k)]2+ 2k ak [x(n + 2k)]2+ak 2k[x(n + k)] 2 + k−1 X i=0 µ 2ka2 i ak [x(n + i)]2+ak 2k[x(n + k)] 2 ¶ + 2kX−1 i=k+1 µ 2ka2 i ak [x(n + i)]2+ak 2k[x(n + k)] 2 ¶# ≥ TX−1 n=0 ∙ −ak[x(n + k)]2+ 2k ak [x(n + 2k)]2 + k−1 X i=0 2ka2 i ak [x(n + i)]2+ 2kX−1 i=k+1 2ka2 i ak [x(n + i)]2 # = TX−1 n=0 "k−1 X i=0 2ka2 i ak [x(n + i)]2− ak[x(n + k)]2+ 2kX−1 i=k+1 2ka2 i ak [x(n + i)]2 +2k ak [x(n + 2k)]2 ¸
= 2ka 2 0 ak x20+ µ 2ka2 0 ak +2ka 2 1 ak ¶ x21+ · · · + Ãk−1 X i=0 2ka2 i ak ! x2k−1 +2k ak x22k+T−1+ µ2k ak +2ka 2 2k−1 ak ¶ x22k+T−2+ · · · + Ã 2k ak + 2kX−1 i=T +k+1 2ka2 i ak ! x2T +k+ 2kX−1 i=k ⎛ ⎝ k−1 X j=0 2ka2j ak − ak + i X j=k+1 2ka2j ak ⎞ ⎠ x2 i + TX−1 i=2k ⎛ ⎝ k−1 X j=0 2ka2 j ak − ak + 2kX−1 j=k+1 2ka2 j ak +2k ak ⎞ ⎠ x2 i + T +kX−1 i=T ⎛ ⎝ kX−1 j=i−T +1 2ka2 j ak − a k+ 2kX−1 j=k+1 2ka2 j ak +2k ak ⎞ ⎠ x2 i
Since (D) and ak< 0 imply that
2 TX−1 n=0 Ã x(n + 2k) − 2kX−1 i=0 aix(n + i) ! x(n + k) ≥ λ2 Ã 2ka2 0 ak A20+ µ 2ka2 0 ak +2ka 2 1 ak ¶ A21+ · · · + Ãk−1 X i=0 2ka2 i ak ! A2k−1 ! +λ2 µ 2k ak B22k+T−1+ µ 2k ak +2ka 2 2k−1 ak ¶ B2k+T2 −2+ · · · + Ã 2k ak + 2kX−1 i=T +k+1 2ka2 i ak ! BT +k2 ! + 2kX−1 i=k ⎛ ⎝ k−1 X j=0 2ka2 j ak − a k+ i X j=k+1 2ka2 j ak ⎞ ⎠ x2 i + TX−1 i=2k ⎛ ⎝ k−1 X j=0 2ka2 j ak − ak + 2kX−1 j=k+1 2ka2 j ak +2k ak ⎞ ⎠ x2 i + T +kX−1 i=T ⎛ ⎝ k−1 X j=i−T +1 2ka2 j ak − ak + 2kX−1 j=k+1 2ka2 j ak +2k ak ⎞ ⎠ x2 i ≥ λ2³2ka20 ak A 2 0+ ³2ka2 0 ak + 2ka21 ak ´ A2 1+ · · · + ³Pk−1 i=0 2ka2i ak ´ A2 k−1 ´ + λ2³2ka kB 2 2k+T−1+ ³ 2k ak + 2ka22k−1 ak ´ B2k+T2 −2+ · · · +³2ka k + P2k−1 i=T +k+1 2ka2i ak ´ B2 T +k ´ =: λ2M.
So, we get from (14) that λ TX−1 n=0 x(n + k)f (n, x(n), x(n + 1), · · · , x(n + 2k − 1)) ≥ λ2M. Hence TX−1 n=0 x(n + k)f (n, x(n), x(n + 1), · · · , x(n + 2k − 1)) ≥ M.
The remainder of the proof is similar to that of the proof of Theorem L1 and is omitted. The proof is completed.
3. Examples
In this section, we present two examples to illustrate the main result in section 2.
Example 3.1. Consider the following BVP (20)⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ x(n + 2) − 3x(n + 1) − x(n) = β[x(n + 1)]2m+1+ p0(n)[x(n)]2m+1+ p1(n)[x(n + 1)]2m+1+ r(n), n ∈ [0, T − 1], x(0) = A, x(T + 1) = B,
where T ≥ 2 is a positive integer, m ≥ 0, r > 1 integers, β > 0, p0(n), p1(n), r(n)
are sequences. Corresponding to BVP(5), we set k = 1, A0 = A, B0 = B,
a0= 1, a1= 3 > 0, and
f (n, x0, x1) = βx2m+11 + p0(n)x2m+10 + p1(n)x2m+11 + r(n),
g(n, x0, x1) = βx2m+10 ,
and
h(n, x0, x1n) = p0(n)x2m+10 + p1(n)x2m+1i + r(n).
One sees that (A) and (B) in Theorem 2.1 hold. It is easy to see that C0 = (a1) = (3), C1 = a1C0+ (1) = (4), C2 = a1C1+ a0C0= (13), C3 = a1C2+ a0C1, · · · · CT−1 = a1CT−2+ a0CT−3.
One see that R(C) = 1. Then all assumptions in Theorem 2.1 hold. It follows from Theorem 2.1 that BVP(20) has at least one solution if
Example 3.2. Consider the following BVP (21) ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ x(n + 4) − x(n) − 2x(n + 1) + 3x(n + 2) − 4x(n + 3) = −β[x(n + 2)]3+P3 i=0pi(n)[x(n + i)]3+ r(k), n ∈ [0, T − 1], x(0) = A0, x(1) = A1, x(T + 2) = B0, x(T + 3) = B1,
whereT ≥ 4 is a positive integer, β > 0, p0(n), pi(n), r(n) are sequences.
Cor-responding to BVP(5), we set a0 = 1, a1 = 2, a2 = −3 < 0, a3 = 4, k = 2, and f (n, x0, x1, x2, x3) = −βx32+ p0(n)x30+ p1(n)x31+ p2(n)x32+ p3(n)x33+ r(n), g(n, x0, x1, x2, x3) = −βx32, and h(n, x0, x1, x2, x3) = p0(n)x30+ p1(n)x31+ p2(n)x32+ p3(n)x33+ r(n).
It is easy to see that (C) in Theorem 2.2 hold. One can get C by computation and finds that R(C) = 2. Hence (D) in Theorem 2.2 holds. It follows from Theorem 2.2that BVP(21) has at least one solution if
||p0|| + ||p1|| + ||p2|| + ||p3|| < β.
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