Positive solutions of BVPs for some second-order four-point difference systems

Positive solutions of BVPs for some second-order four-point difference systems Yitao Yang Tianjin University of Technology Department of Applied Mathematics Hongqi Nanlu Extension, Tianjin China yitaoyangqf@63.com Fanwei Meng Qufu Normal University Department of Applied Mathematics 57 West Jingxuan Xilu, Qufu China fwmeng@mail.qfnu.edu.cn Abstract: This paper is concerned with one type of boundary value problems (BVPs). We first construct Green functions for a second-order four-point difference equation, try to find delicate conditions for the existence of positive solutions. Our main tool is a nonlinear alternative of Leray-Schauder type, krasnosel skii s fixed point theorem in a cone Leggett-Williams fixed point theorem. Key Words: Discrete system; Positive solutions; Cone; Nonlinear alternative; Leggett-Williams fixed point theorem; Fixed point. Introduction In this paper, we consider the following discrete system boundary value problem: 2 u (k ) + f (k, u (k), u 2 (k)) = 0, k [, T ], 2 u 2 (k ) + f 2 (k, u (k), u 2 (k)) = 0, k [, T ], with the boundary condition: u (0) = au (l ), u (T ) = bu (l 2 ), u 2 (0) = au 2 (l ), u 2 (T ) = bu 2 (l 2 ), () (2) where T is a fixed positive integer, u(k) = u(k + ) u(k), 2 u(k) = (u(k)), [, T ] = {, 2,..., T Z the set of all integers, l, l 2 [, T ], l < l 2, 0 a < l, 0 b < l 2 0 al + bl 2 < + a, = a( bl 2 ) b( al ) > 0. Many problem in applied mathematics lead to the study of difference system, see [] [2] the references therein. Recently, much attention has been paid to the existence of positive solutions of scalar difference equations [3],[4],[5],[6], [9],[4],[8],[9],[20] discrete difference systems [8],[3],[6]. In [5], Tian considered the multiplicity for fourpoint boundary value problems 2 u(k ) + q(k)f(k, u(k), u(k)) = 0, k N(, T ), u(0) = au(l ), u(t + ) = bu(l 2 ). Sun Li [4] investigated the following discrete system 2 u (k) + f (k, u (k), u 2 (k)) = 0, k [0, T ], 2 u 2 (k) + f 2 (k, u (k), u 2 (k)) = 0, k [0, T ], with the Dirichlet boundary condition u (0) = u (T + 2) = 0, u 2 (0) = u 2 (T + 2) = 0, by using Leggett-Williams fixed point theorem, sufficient conditions are obtained for the existence three positive solutions to the above system. Motivated by the above works, our purpose in this paper is to study problem (),(2). Under suitable conditions on f f 2, we show that the boundary value problems (),(2) have one or two positive solutions. Since the construction of Green functions for difference equations may be more complicated overloaded than that for differential equations, the difficulty of this paper is constructing Green functions for ()(2), which play important roles in the verifying of the existence of positive solutions for the given difference systems. Furthermore, the system (), (2)consists of two second order difference equations. To the authors best knowledge, no paper has constructed Green functions for a second order four-point difference equation (), (2). This paper attempts to fill this gap in the literature. The rest of the paper is organized as follows. First, we shall state two fixed point theorems, the E-ISSN: 2224-2880 926 Issue 0, Volume, October 202

first of which is a nonlinear alternative of Leray- Schauder type, whereas the second is krasnosel skii s fixed point theorem in a cone. We also present the Green s function for problem (), (2). In Section 2, criteria for the existence of one two positive solutions to boundary value problems (), (2) are established. In Section 3, three positive solution for boundary value problems (), (2)are obtained. In section 4, we give the conclusion of my paper.. Several lemmas In order to prove our main results, the following wellknown fixed point theorems are needed. Lemma ([7]) Let X be a Banach space with E X closed convex. Assume U is a relatively open ball of E with 0 U T : Ū E is a continuous compact map. Then, either (a) T has a fixed point in Ū, or (b) there exists u U λ (0, ) such that u = λt u. Lemma 2 ([7]) Suppose X is a Banach space, K X is a cone. Assume Ω, Ω 2 are open subsets of X with θ Ω, Ω Ω 2. Let T : K (Ω 2 \Ω ) K be a completely continuous operator such that either (a) T u u, u K Ω T u u, u K Ω 2, or (b) T u u, u K Ω T u u, u K Ω 2. Then, T has a fixed point in K (Ω 2 \Ω ). Lemma 3 Let := a( bl 2 ) b( al ) 0, then for y : [, T ] R +, the problem 2 u(k ) + y(k) = 0, k [, T ], (3) u(0) = au(l ), u(t ) = bu(l 2 ), (4) has a unique solution u(k) = al + ak T y(j) + a( bl l 2) + abk (l j)y(j) b( al l ) + abk 2 (l 2 j)y(j) k (k j)y(j). Proof: We proceed from (3) obtain 2 u(k ) = y(k), after adding from to i, we have i u(i ) = u(0) y(j), (5) then adding (5) from to k, u(k) = u(0) + ku(0) k i= i y(j) = u(0) + ku(0) k (k j)y(j). (6) From (4) (6), we can see that u(k) = al + ak + a( bl 2) + abk b( al ) + abk k (k j)y(j). T y(j) l (l j)y(j) l 2 (l 2 j)y(j) Lemma 4 Let 0, the Green s function for the boundary value problem is given by G(k, j) = 2 u(k ) = 0, k [, T ], (7) u(0) = au(l ), u(t ) = bu(l 2 ), (8) ( + kb bl 2), j min{k, l T ; [(bj + bl 2) +(j k)(abl 2 abl a)], 0 k j l ; (bk + bl 2)( + aj al ), l j min{k, l 2 T ; (bj + bl 2)( + ak al ), max{k, l j l 2 ; ( + ak al ) + (j k), l 2 j k T ; ( + ak al ), max{k, l 2 j T. Lemma 5 Suppose < l < l 2 < T +, 0 < a < l, 0 < b < l 2, 0 < al + bl 2 + a, > 0. The Green s function G(k, j) satisfies G(k, j) > 0, for 0 < j, k < T +, (9) E-ISSN: 2224-2880 927 Issue 0, Volume, October 202

G(k, j) γ max G(k, j), for 0 k T + l k l 2, < j < T +, (0) where γ is defined as { + bl bl 2 γ = min (b(t + ) + bl 2 )( + al 2 al ), + bl bl 2, a(t + ) + al () Proof: Notice that { min G(k, j) = min k [l,l 2 ] ( + bl bl 2 ), ( + bl bl 2 )( + aj al ), Let γ = ( + bl bl 2 ). max k [0,T +] { G(k, j) = max ( + b(t + ) bl 2), ( + b(t + ) bl 2)( + aj al ), (bj + bl 2)( + aj al ), (a(t + ) + al ) + (j k), (aj + al ) = max{ ( + b(t + ) bl 2)( + al 2 al ), (a(t + ) + al ). { + bl bl 2 = min (b(t + ) + bl 2 )( + al 2 al ), + bl bl, 2 a(t + ) + al it is obvious that 0 < γ <. Therefore, we have G(k, j) γ max G(k, j), 0 k T + for l k l 2, < j < T +. 2 Main result Let the Banach space B = {u : [0, T + ] R be endowed with norm, u 0 = X = B B with norm max u(k), k [0,T +] (u, u 2 ) = max{ u 0, u 2 0, { K = (u, u 2 ) X : u i (k) 0, k [0, T + ], min u i(k) γ u i 0, i =, 2, k [l,l 2 ] where γ is defined as (), then K is a cone in X. The following theorem gives an existence principle for boundary value problems (),(2). This results is used later to establish the existence of one positive solution of (),(2). Theorem 6 Let f i : [, T ] R 2 R, i =, 2 be continuous. Suppose that there exists a constant M, independent of λ, such that u = M, (2) for any solution u = (u, u 2 ) X of the boundary value problem 2 u (k ) + λf (k, u (k), u 2 (k)) = 0, k [, T ], 2 u 2 (k ) + λf 2 (k, u (k), u 2 (k)) = 0, k [, T ], (3) u (0) = au (l ), u (T ) = bu (l 2 ), u 2 (0) = au 2 (l ), u 2 (T ) = bu 2 (l 2 ), (4) where λ (0, ). Then, boundary value problems (),(2) have at least one solution u = (u, u 2 ) X such that u M. Proof: Let the operator T : X X be defined by T (u, u 2 ) = (U (k), U 2 (k)) k [0, T + ], where T U i (k) = G(k, j)f i (j, u (j), u 2 (j)), i =, 2. Then, it is noted that T is continuous completely continuous that solving (3),(4) is equivalent to finding a u X such that u = λt u. In the context of Lemma, let U = {u = (u, u 2 ) X : u < M. In view of (2), we cannot have Conclusion (b) of Lemma, hence, Conclusion (a) of lemma holds, i.e., (),(2) have a solution u Ū with u M. The proof is complete. E-ISSN: 2224-2880 928 Issue 0, Volume, October 202

Our next results offer the existence of one two positive solutions of (), (2). For convenience, the conditions needed are listed as follows. (H ) f i : [, T ] [0, ) [0, ) [0, ) is continuous, i =, 2; (H 2 ) For each i {, 2, assume that f i (k, u, u 2 ) α i (k)ω i (u )ω i2 (u 2 ), for (k, u, u 2 ) [, T ] [0, ) [0, ), where α i : [, T ] (0, ), ω il : [0, ) [0, ), l =, 2 are continuous nondecreasing; (H 3 ) There exists r > 0 such that where r > d i ω i (r)ω i2 (r), i =, 2, T d i = max G(k, j)α i (j); k [0,T +] (H 4 ) For each i = {, 2, there exist τ il : [l, l 2 ] (0, ), l =, 2 such that f i (k, u, u 2 ) τ il (k)ω il (u l ), k [l, l 2 ], u l > 0; (H 5 ) There exists R > r such that for l {, 2, x [γr, R], the following holds for some i (depending on l) in {, 2: x ω il (x)γ l 2 where σ il [0, T + ] is defined as l 2 j=l G(σ il, j)τ il (j) = j=l G(σ il, j)τ il (j) (5) max k [0,T +] l 2 j=l G(k, j)τ il (j); (H 6 ) There exists L (0, r) such that for l {, 2, x [γr, r], inequality (5) holds for some i (depending on l) in, 2. Theorem 7 Suppose that (H ) (H 3 ) hold. Then, boundary value problems (),(2) have a positive solution u = (u, u 2 ) X such that u < r, where r is defined by (H 3 ). Proof: First, we consider the following boundary value problem 2 u (k ) + f (k, u (k), u 2 (k)) = 0, k [, T ], 2 u 2 (k ) + f 2 (k, u (k), u 2 (k)) = 0, k [, T ], (6) u (0) = au (l ), u (T ) = bu (l 2 ), u 2 (0) = au 2 (l ), u 2 (T ) = bu 2 (l 2 ), where f i : [, T ] R 2 R is defined by f i (k, u, u 2 ) = f i (k, u, u 2 ), i =, 2. (7) It is obvious that f i is continuous. We shall show that (6), (7) have a solution. For this, we look at the following problem: 2 u (k ) + λ f (k, u (k), u 2 (k)) = 0, k [, T ], 2 u 2 (k ) + λ f 2 (k, u (k), u 2 (k)) = 0, k [, T ], (8) u (0) = au (l ), u (T ) = bu (l 2 ), (9) u 2 (0) = au 2 (l ), u 2 (T ) = bu 2 (l 2 ), where λ (0, ). Let u = (u, u 2 ) X be any solution of (8), (9). We claim that u r. (20) In fact, it is clear that u (k) = λ T G(k, j) f (j, u (j), u 2 (j)), k [0, T + ], (2) u 2 (k) = λ T G(k, j) f 2 (j, u (j), u 2 (j)), k [0, T + ]. (22) Noting that (2), (H 2 ) (H 3 ), it follows that 0 u (k) = λ T G(k, j) f (j, u (j), u 2 (j)) = λ T G(k, j)f (j, u (j), u 2 (j) ) λ T G(k, j)α (j)ω ( u (j) )ω 2 ( u 2 (j) ) λ T G(k, j)α (j)ω ( u )ω 2 ( u ) T ω ( u )ω 2 ( u ) max G(k, j)α (j) k [0,T +] = d ω ( u )ω 2 ( u ), k [0, T + ]. This immediately leads to u 0 d ω ( u )ω 2 ( u ). (23) Similarly, from (22), (H 2 ) (H 3 ), we have u 2 0 d 2 ω 2 ( u )ω 22 ( u ). (24) E-ISSN: 2224-2880 929 Issue 0, Volume, October 202

If u = u p 0, for some p, 2, then (23), (24) yield u d p ω p ( u )ω p2 ( u ), from which we conclude, by comparing with (H 3 ), that u = r. It now follows from Theorem 6 that boundary value problems (6),(7) have a solution u = (u, u 2 ) X such that u r. Using a similar argument as above, we see that u r. Therefore, u < r. (25) Moreover, for k [0, T + ], we have u (k) = T G(k, j) f (j, u (j), u 2 (j)) = T (26) G(k, j)f (j, u (j), u 2 (j) ), u 2 (k) = T G(k, j) f 2 (j, u (j), u 2 (j)) = T (27) G(k, j)f 2 (j, u (j), u 2 (j) ). it follows immediately that so, u i (k) 0, k [0, T + ], i =, 2, (28) u (k) = T G(k, j)f (j, u (j), u 2 (j) ) = T (29) G(k, j)f (j, u (j), u 2 (j)), u 2 (k) = T G(k, j)f 2 (j, u (j), u 2 (j) ) = T (30) G(k, j)f 2 (j, u (j), u 2 (j)), i.e., u = (u, u 2 ) X is a positive solution of boundary value problems () (2) satisfies u < r. Theorem 8 Suppose that (H ) (H 5 ) hold. Then, boundary value problems (),(2) have two positive solutions u, ū X such that 0 u < r < ū R, where r R are defined by (H 3 ) (H 5 ). Proof: The existence of u is guaranteed by Theorem 7. We shall employ Lemma 2 to prove the existence of ū. Let T : K X be defined by T (u, u 2 ) = (U (k), U 2 (k)), k [0, T + ], where T U i (k) = G(k, j)f i (j, u (j), u 2 (j)), i =, 2. First, we shall show that T maps K into itself. For this, let u = (u, u 2 ) K. Then, it follows immediately that U i (k) 0, k [0, T + ], i =, 2. (3) Then, we obtain for each i {, 2, U i (k) = T G(k, j)f i (j, u (j), u 2 (j)) T max k [0,T +] G(k, j)f i (j, u (j), u 2 (j)), k [0, T + ]. As a result, U i 0 T max k [0,T +] G(k, j)f i (j, u (j), u 2 (j)), i =, 2. (32) Now, in view of (32) Lemma 5, we have for k [l, l 2 ], i =, 2, Therefore, U i (k) = T G(k, j)f i (j, u (j), u 2 (j)) γ T max G(k, j)f i(j, u (j), u 2 (j)) k [0,T +] γ U i 0. min U i(k) γ U i 0, i =, 2. (33) k [l,l 2 ] Combining (3) (33), we obtain T (K) K. Also, the stard arguments yield that T is completely continuous. Let Ω = {u X : u < r Ω 2 = {u X : u < R. We claim that (i) T u u, for u K Ω, (ii) T u u, for u K Ω 2. E-ISSN: 2224-2880 930 Issue 0, Volume, October 202

To justify (i), let u = (u, u 2 ) K Ω, then u = r by (H 2 ) (H 3 ) we have 0 U i (k) = T G(k, j)f i (j, u (j), u 2 (j)) T G(k, j)α i (j)ω i (u (j))ω i2 (u 2 (j)) ω i ( u )ω i2 ( u ) T G(k, j)α i (j) T ω i ( r )ω i2 ( r ) max G(k, j)α i (j) k [0,T +] = d i ω i ( r )ω i2 ( r ) < r = u, k [0, T + ], i =, 2. Therefore, U i 0 u, i =, 2, so Hence, U i 0 u, so T u u. Having obtained (i) (ii), it follows from Lemma 2 that T has a fixed point ū K (Ω 2 \Ω ), i.e., ū is a positive solution of (), (2) r ū R. Using a similar argument as in the proof of Theorem 7, we see that r < ū R. It is noted in Theorem 8 that u may be zero. Our next result guarantees that u 0. So, T u = max{ U 0, U 2 0 u. Next, we prove (ii). Let u = (u, u 2 ) K Ω 2. Theorem 9 Suppose that (H ) (H 6 ) hold. Then, boundary value problems (),(2) have two positive solutions u, ū X such that 0 < L u < r < ū R, u = max{ u 0, u 2 0 = R = u p 0, for some p {, 2. Then, it follows that Thus, we have 0 u p (k) R, k [0, T + ], u p (k) γr, k [l, l 2 ]. γr u p (k) R, k [l, l 2 ]. (34) In view of (34), (H 4 ) (H 5 ), we find that the following holds for some i (depending on p) in {, 2 : U i (σ ip ) = T G(σ ip, j)f i (j, u (j), u 2 (j)) l 2 G(σ ip, j)f i (j, u (j), u 2 (j)) j=l l 2 G(σ ip, j)τ ip (j)ω ip (u p (j)) j=l l 2 G(σ ip, j)τ ip (j)ω ip (γr) j=l l 2 G(σ ip, j)τ ip (j) j=l γr = R γ l 2 G(σ ip, j)τ ip (j) j=l = u. where L, r R are defined by (H 3 ), (H 5 ) (H 6 ), respectively. Proof: The existence of ū is guaranteed by Theorem 8. We shall employ Lemma 2 to show the existence of u. Suppose that the set Ω the map T : K K are the same as in the proof of Theorem 8. Let Ω 3 = {u X : u < L. From the proof of Theorem 8, we see that (i) T u u, for u K Ω ; thus, it remains to prove that (ii) T u u, for u K Ω 3. For this, let Assume that u = (u, u 2 ) K Ω 3. u = u p 0 = L, for some p {, 2. Then, we have γl u p (k) L, k [l, l 2 ]. (35) In view of (35), (H 4 ) (H 6 ), we find that the fol- E-ISSN: 2224-2880 93 Issue 0, Volume, October 202

lowing holds for some i (depending on p) in {, 2 : U i (σ ip ) = T G(σ ip, j)f i (j, u (j), u 2 (j)) l 2 G(σ ip, j)f i (j, u (j), u 2 (j)) j=l l 2 G(σ ip, j)τ ip (j)ω ip (u p (j)) j=l l 2 G(σ ip, j)τ ip (j)ω ip (γl) j=l = l 2 G(σ ip, j)τ ip (j) j=l γl = L γ l 2 G(σ ip, j)τ ip (j) j=l = u. Hence, U i 0 u, so T u u. Having obtained (i) (ii), we conclude from Lemma 2 that T has a fixed point u K (Ω \Ω 3 ), i.e., u is a positive solution of boundary value problems (), (2) L u r. Using a similar argument as in the proof of Theorem 7, we see that L < u r. Example: problem: Consider the following boundary value 2 u (k ) + µ exp(u 2 + u 8 2 ) = 0, k [, 20], 2 u 2 (k ) + µ exp(u 6 + u 3 2 ) = 0, k [, 20], (36) with the boundary condition: u (0) = 6 u (5), u (20) = 00 u (0), u 2 (0) = 6 u 2(5), u 2 (20) = 00 u 2(0), (37) where µ > 0. It is easy to prove that (H ) (H 5 ) are satisfied when µ is small enough. Hence, it follows from Theorem 8 that boundary value problems (38), (39) have two positive solutions when µ is small enough. Remark 0 If conditions (H 2 ) (H 3 ) are replaced by (H 2 ) (H 3 ), respectively, where (H 2 ) for each i {, 2, assume that f i (k, u, u 2 ) α i (k)ω i (u ) + β i (k)ω i2 (u 2 ), for (k, u, u 2 ) [, T ] [0, ) [0, ), where α i, β i : [, T ] (0, ), ω il : [0, ) [0, ), l =, 2 are continuous nondecreasing; (H 3 ) There exists r > 0 such that where r > d i [ω i (r) + ω i2 (r)], i =, 2, { d i = max max k [0,T +] max T G(k, j)α i (j), k [0,T +] T G(k, j)β i (j), i =, 2, then, similar conclusions are true. 3 Three positive solution for boundary value problems (), (2) Let the Banach space B = {u : [0, T + ] R be endowed with norm, u 0 = X = B B with norm max u(k), k [0,T +] (u, u 2 ) = max{ u 0, u 2 0, { P = (u, u 2 ) X : u i (k) 0, k [0, T + ], min u i(k) γ u i 0, i =, 2, k [l,l 2 ] where γ is defined as (), then P is a cone in X. A map α is said to be a nonnegative continuous concave functional on P if is continuous α : P [0, + ) α(tx + ( t)y) tα(x) + ( t)α(y) for all x, y P t [0, ]. For numbers a, b such that 0 < a < b α is a nonnegative continuous concave functional on P we define the following convex sets P a = {x P : x < a, E-ISSN: 2224-2880 932 Issue 0, Volume, October 202

P (α, a, b) = {x P : a α(x), x b. Leggett-Williams fixed point theorem. Let A : P c {x P : x < a be completely continuous α be a nonnegative continuous functional on P such that α(x) x for all x P c. Suppose there exist 0 < d < a < b c such that (i) {x P (α, a, b) : α(x) > a α(ax) > a for x P (α, a, b); (ii) Ax < d for x d; (iii) α(ax) > a for x P (α, a, c) with Ax > b. Then A has at least three fixed pointsx, x 2, x 3 satisfying x < d, a < α(x 2 ), x 3 > d, α(x 3 ) < a. Theorem Suppose that f i : [, T ] [0, + ) [0, + ) [0, + ) is continuous that there exist numbers a d with 0 < d < a such that the following conditions are satisfied: (i) if j [, T ], u, u 2 0 u + u 2 d; then f i (j, u, u 2 ) < d, i =, 2, where D = 2D T max G(k, j); k [0,T +] (ii) there exists i 0 {, 2, such that f i0 (j, u, u 2 ) > a C, j [, T ], u, u 2 0 u + u 2 [a, a γ ], T where C = min G(k, j); k [l,l 2 ] (iii) one of the following conditions holds: (A) (B) lim max f i (j,u,u 2 ) <, i =, 2; u +u 2 j [0,T ] u + u 2 2D there exists a number c such that c > a γ if j [, T ], u, u 2 0, u + u 2 c then f i (j, u, u 2 ) < c, i =, 2. 2D Then the boundary value problem (), (2) has at least three positive solutions. Proof: For u = (u, u 2 ) P, define where α(u) = min u (k) + min u 2(k), k [l,l 2 ] k [l,l 2 ] A(u, u 2 ) = (U (k), U 2 (k)) k [0, T + ], T U i (k) = G(k, j)f i (j, u (j), u 2 (j)), i =, 2, then it is easy to know that α is a nonnegative continuous concave functional on P with α(x) x for x P that A : P P is completely continuous. that For the sake of convenience, set b = a γ. Claim. If there exists a positive number r such f i (j, u, u 2 ) < r 2D, i =, 2 for j [, T ], u, u 2 0, u + u 2 r, then A : P r P r. Suppose that u = (u, u 2 ) P r, then Thus T U i 0 = max G(k, j)f i (j, u (j), u 2 (j)) k [0,T +] < r 2D D = r, i =, 2. 2 Au = u 0 + u 2 0 < r 2 + r 2 = r. Then there exists a number c such that c > b A : P c P c. From Claim with r = d (i) that A : P d P d. Claim 2. We show that {u P (α, a, b) : α(u) > a = α(au) > a for u P (α, a, b) In fact, u = (u (k), u 2 (k)) = ( a + b 4, a + b ) 4 {u P (α, a, b) : α(u) > a. For u = (u (k), u 2 (k)) P (α, a, b), we have b u 0 + u 2 0 u (k) + u 2 (k) min u (k) + min u 2(k) a, k [l,l 2 ] k [l,l 2 ] for all k [l, l 2 ]. Then, in view of (ii), we know that min U T i 0 (k) = min G(k, j)f i0 (j, u (j), u 2 (j)) k [l,l 2 ] k [l,l 2 ] > a T C min G(k, j) = a, k [l,l 2 ] so α(au) = min U (k) + min U 2(k) k [l,l 2 ] k [l,l 2 ] min U i 0 (k) > a. k [l,l 2 ] Claim 3. If u P (α, a, c) Au > b, then α(au) > a. E-ISSN: 2224-2880 933 Issue 0, Volume, October 202

Suppose u = (u, u 2 ) P (α, a, c) Au > b, then, by Lemma 5, we have Thus min k [l,l 2 ] so T U i (k) = min G(k, j)f i (j, u (j), u 2 (j)) k [l,l 2 ] T G(k, j) = min k [l,l 2 ] max G(k, j) 0 k T + max G(k, j)f i(j, u (j), u 2 (j)), 0 k T + γ T max G(k, j)f i(j, u (j), u 2 (j)) 0 k T + = γ U i (k) 0, i =, 2. min U i(k) γ U i (k) 0, i =, 2, k [l,l 2 ] α(au) = min U (k) + min U 2(k) k [l,l 2 ] k [l,l 2 ] γ( U 0 + U 2 0 ) = γ Au > γb = a. Therefore, the hypotheses of Leggett-Williams theorem are satisfied, hence the boundary value problem (), (2) has at least three positive solutions u = (u, u 2 ), v = (v, v 2 ) w = (w, w 2 ) such that w > d with u < d, a < min v (k) + min v 2(k), k [l,l 2 ] k [l,l 2 ] The proof is complete. min w (k) + min w 2(k) < a. k [l,l 2 ] k [l,l 2 ] Example: Consider the following boundary value problem: 2 u (k ) + f (k, u (k), u 2 (k)) = 0, k [, 20], 2 u 2 (k ) + f 2 (k, u (k), u 2 (k)) = 0, k [, 20], (38) with the boundary condition: u (0) = 6 u (5), u (20) = 00 u (0), u 2 (0) = 6 u 2(5), u 2 (20) = 00 u 2(0), (39) k 4485800 + 88 224290 (u +u 2 ), 0 u f i (k, u (k), u 2 (k)) = + u 2. k 4485800 + 88 224290 [50( u + u 2 ) + ], u + u 2. Then, boundary value problem (38), (39) has at least three positive solutions. Proof: Choose d =, a = 600. By computation, we know γ = 57 220, D = 245, C = 90 89 89. Thus f i (k, u (k), u 2 (k)) = k 4485800 + 88 224290 (u + u 2 ) < 89 224290 = d 2D, for k [, 20], 0 u + u 2, k f i (k, u (k), u 2 (k)) = 4485800 + 88 224290 [50( u + u 2 ) + ] > 42400 90 = a C, for k [, 20], 600 u + u 2 352000 57. lim max f i (j,u,u 2 ) = 0 <, i =, 2. u +u 2 j [0,T ] u + u 2 2D To sum up, by an application of Theorem, we know that (38), (39) has at least three positive solutions. 4 Conclusion In this paper, we first construct Green functions for a second-order four-point difference equation, try to find suitable conditions on f f 2, which can guarantee that the boundary value problems (),(2) have one, two or three positive solutions. Our work presented in this paper has the following new features. Firstly, BVP (),(2) is a four-point boundary value problem, which is very difficult when constructing Green functions. Secondly, the main tool used in this paper is Leray-Schauder type [7], krasnosel skii s fixed point theorem in a cone [7] Leggett-Williams fixed point theorem [7] the result obtained is the multiple positive solutions of BVP (),(2). Thirdly, the system (), (2)consists of two second order difference equations. Acknowledgements: The research was supported by Tianjin University of Technology in the case of the first author, it was also supported by the National Natural Science Foundation of China (No. ((778))) the Tianjin City High School Science Technology Fund Planning Project (No. (2009008)) Tianyuan Fund of Mathematics in China (No. (02676)) Yumiao Fund of Tianjin University of Technology (No. (LGYM2002)). The authors thank the referee for his/her careful reading of the paper useful suggestions. E-ISSN: 2224-2880 934 Issue 0, Volume, October 202

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