Electronic Journal of Differential Equations, Vol. 2009(2009), No. 163, pp. 1–9. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu THREE POSITIVE SOLUTIONS FOR A SYSTEM OF SINGULAR GENERALIZED LIDSTONE PROBLEMS JIAFA XU, ZHILIN YANG Abstract. In this article, we show the existence of at least three positive solutions for the system of singular generalized Lidstone boundary value problems (−1)m x(2m) = a(t)f1 (t, x, −x00 , . . . , (−1)m−1 x(2m−2) , y, −y 00 , . . . , (−1)n−1 y (2n−2) ), n (2n) (−1) y = b(t)f2 (t, x, −x00 , . . . , (−1)m−1 x(2m−2) , y, −y 00 , . . . , (−1)n−1 y (2n−2) ), a1 x(2i) (0) − b1 x(2i+1) (0) = c1 x(2i) (1) + d1 x(2i+1) (1) = 0, a2 y (2j) (0) − b2 y (2j+1) (0) = c2 y (2j) (1) + d2 y (2j+1) (1) = 0. The proofs of our main results are based on the Leggett-Williams fixed point theorem. Also, we give an example to illustrate our results. 1. Introduction In this article, we study the existence positive solutions for the following system of singular generalized Lidstone boundary value problems (−1)m x(2m) = a(t)f1 (t, x, −x00 , . . . , (−1)m−1 x(2m−2) , y, −y 00 , . . . , (−1)n−1 y (2n−2) ), (−1)n y (2n) = b(t)f2 (t, x, −x00 , . . . , (−1)m−1 x(2m−2) , y, −y 00 , . . . , (−1)n−1 y (2n−2) ), a1 x(2i) (0) − b1 x(2i+1) (0) = c1 x(2i) (1) + d1 x(2i+1) (1) = 0 (i = 0, 1, . . . , m − 1), a2 y (2j) (0) − b2 y (2j+1) (0) = c2 y (2j) (1) + d2 y (2j+1) (1) = 0 (j = 0, 1, . . . , n − 1). (1.1) where m, n ≥ 1, a(t), b(t) ∈ C((0, 1), [0, +∞)), a(t) and b(t) are allowed to be singular at t = 0 and/or t = 1; fi ∈ C([0, 1] × Rm+n , R+ )(R+ := [0, +∞)); + ai , bi , ci , di ∈ R+ with ρi := ai ci + ai di + bi ci > 0, i = 1, 2. Singular boundary value problems for ordinary differential equation describe many phenomena in applied mathematics and physical sciences, which can be found 2000 Mathematics Subject Classification. 34A34, 34B18, 45G15, 47H10. Key words and phrases. Singular generalized Lidstone problem; positive solution; cone; concave functional. c 2009 Texas State University - San Marcos. Submitted October 7, 2009. Published December 21, 2009. Supported by grants 10871116 and 10971179 from the NNSF of China. 1 2 J. XU, Z. YANG EJDE-2009/163 in the theory of nonlinear diffusion generated by nonlinear sources and in the thermal ignition of gases, see [1, 4]. Very recently, increasing attention is paid to question of positive solutions for systems of second-order or higher order singular differential equations, see for example [2, 5, 6, 7] and references therein. In [2], by using Leggett-Williams fixed point theorem [3], Kang et al obtained at least three positive solutions to the following singular nonlocal boundary value problems for systems of nonlinear second-order ordinary differential equations. u00 (t) + a1 (t)f1 (t, u(t), v(t)) = 0, 0 < t < 1, 00 v (t) + a2 (t)f2 (t, u(t), v(t)) = 0, 0 < t < 1, Z 1 Z 0 0 α1 u(0) − β1 u (0) = 0, γ1 u(1) + δ1 u (1) = g1 ( u(s)dφ1 (s), α2 v(0) − β2 v 0 (0) = 0, 0 Z 1 γ2 v(1) + δ2 v 0 (1) = g2 ( 1 v(s)dφ1 (s)), 0 Z 1 u(s)dφ2 (s), 0 v(s)dφ2 (s)), 0 (1.2) where ai ∈ C((0, 1), R+ ) is allowed to be singular at t = 0 or t = 1; fi ∈ C([0, 1] × R+ × R+ , R+ ) and gi ∈ C([0, 1] × R+ × R+ , R+ ) are such that a1 (t)f1 (t, 0, 0) or a2 (t)f2 (t, 0, 0) does not vanish identically on any subinterval of (0, 1); αi ≥ 0, R1 βi ≥ 0, γi ≥ 0, δi ≥ 0, and ρi = αi γi + αi δi + βi γi > 0; 0 u(s)dφ1 (s) and R1 v(s)dφ1 (s) denote the Riemann-Stieltjes integrals, i = 1, 2. 0 Motivated by [2], we deal with the system of singular generalized Lidstone problems (1.1). To overcome the difficulties of (1.1) resulting from the derivatives of even orders, as in [8], we use the method of order reduction to transform (1.1) into an equivalent system of integro-integral equations, then prove the existence of positive solutions for the resulting system, thereby establishing that of positive solutions for (1.1)(see the main result in Section 3). The features of this paper mainly include the following aspects. Firstly, our study is on systems of singular generalized Lidstone problems. Secondly, a and b are allowed to be singular at t = 0 and/or t = 1. Finally, the system contains two equations, which can be of different orders. Thus the results presented here are different from those in [2, 5, 6, 7]. The remaining of this paper is organized as follows: Section 2 gives some preliminary results. The main result is stated and proved in Section 3, then followed by an example to illustrate the validity of our main result. 2. Preliminaries Given a cone K in a real Banach space E, a map α is said to be a nonnegative continuous concave functional on K provided that α : K → [0, +∞) is continuous and α(tx + (1 − t)y) ≥ tα(x) + (1 − t)α(y) for all x, y ∈ K and 0 ≤ t ≤ 1. Let 0 < a < b be given and let α be a nonnegative continuous concave functional on K. Define the convex sets Pr and P (α, a, b) by Pr := {x ∈ K : kxk < r}, P (α, a, b) := {x ∈ K : a ≤ α(x), kxk ≤ b}. To prove our main result, we need the following Leggett-Williams fixed point theorem. EJDE-2009/163 THREE POSITIVE SOLUTIONS 3 Lemma 2.1 (see [3]). Let T : Pc → Pc be a completely continuous operator and let α be a nonnegative continuous concave functional on K such that α(x) ≤ kxk for all x ∈ Pc . Suppose there exist 0 < a < b < d ≤ c such that (C1) {x ∈ P (α, b, d) : α(x) > b} = 6 ∅, and α(T x) > b for x ∈ P (α, b, d), (C2) kT xk < a for kxk ≤ a, and (C3) α(T x) > b for x ∈ P (α, b, c) with kT xk > d. Then T has at least three fixed points x1 , x2 , and x3 such that kx1 k < a, b < α(x2 ) and kx3 k > a with α(x3 ) < b. For fixed nonnegative constants a, b, c, d with ρ := ac + ad + bc > 0, we let ( 1 (b + as)(c + d − ct), 0 ≤ s ≤ t ≤ 1, (2.1) k(t, s) := ρ (b + at)(c + d − cs), 0 ≤ t ≤ s ≤ 1, By definition, k ∈ C([0, 1] × [0, 1], R+ ) has the following properties: (i) k(t, s) ≤ k(s, s) for all t, s ∈ [0, 1]. θc+d (ii) For any θ ∈ (0, 1/2), there exists γ ∈ (0, min{ θa+b a+b , c+d }] such that k(t, s) ≥ γk(s, s), ∀t ∈ [θ, 1 − θ], s ∈ [0, 1]. R1 Lemma 2.2. Let f ∈ C[0, 1], h(t) ∈ C(0, 1) and 0 k(s, s)h(s)ds < +∞. The boundary value problem −u00 = h(t)f (t), au(0) − bu0 (0) = 0, cu(1) + du0 (1) = 0, has a unique solution Z 1 u(t) = k(t, s)h(s)f (s)ds, 0 where k(t, s) is given by (2.1). Let ( (b1 + a1 s)(c1 + d1 − c1 t), (b1 + a1 t)(c1 + d1 − c1 s), ( 1 (b2 + a2 s)(c2 + d2 − c2 t), g1 (t, s) := ρ2 (b2 + a2 t)(c2 + d2 − c2 s), 1 k1 (t, s) := ρ1 For i, j = 2, . . . , define Z 1 ki (t, s) := k1 (t, τ )ki−1 (τ, s)dτ, Z gj (t, s) := 0 0 ≤ s ≤ t ≤ 1, 0 ≤ t ≤ s ≤ 1, (2.2) 0 ≤ s ≤ t ≤ 1, 0 ≤ t ≤ s ≤ 1. (2.3) 1 g1 (t, τ )gj−1 (τ, s)dτ 0 and the operators Ai : C[0, 1] → C[0, 1] and Bj : C[0, 1] → C[0, 1] by Z 1 (Ai u)(t) := ki (t, s)u(s)ds, i = 1, 2, . . . , m, 0 Z (Bj v)(t) := 1 gj (t, s)v(s)ds, 0 j = 1, 2, . . . , n. (2.4) 4 J. XU, Z. YANG EJDE-2009/163 For 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1, let Z 1 ξi := min ki (t, s)ds, θ≤t≤1−θ ξ := ηi := max 1 0 1 Z min gj (t, s)ds, θ≤t≤1−θ ki (t, s)ds, 0≤t≤1 0 Z µj := 1 Z νj := max 0≤t≤1 0 min 1≤i≤m−1,1≤j≤n−1 {ξi , µj }, η := gj (t, s)ds, 0 max 1≤i≤m−1,1≤j≤n−1 {ηi , νj }. 3. Main result Let u(t) = (−1)m−1 x(2m−2) and v(t) = (−1)n−1 y (2n−2) . It is easy to see that (1.1) is equivalent to the system of integro-ordinary differential equations Z 1 Z 1 −u00 (t) = a(t)f1 t, km−1 (t, s)u(s)ds, . . . , k1 (t, s)u(s)ds, u(t), 0 Z 0 1 Z 1 gn−1 (t, s)v(s)ds, . . . , 0 0 Z −v 00 (t) = b(t)f2 t, 1 Z 1 km−1 (t, s)u(s)ds, . . . , 0 Z g1 (t, s)v(s)ds, v(t) , k1 (t, s)u(s)ds, u(t), 0 1 Z gn−1 (t, s)v(s)ds, . . . , 0 1 g1 (t, s)v(s)ds, v(t) , 0 subject to the boundary conditions a1 u(0) − b1 u0 (0) = c1 u(1) + d1 u0 (1) = 0, a2 v(0) − b2 v 0 (0) = c2 v(1) + d2 v 0 (1) = 0. Furthermore, the above system is equivalent to the system Z 1 Z 1 Z 1 u(t) = k1 (t, s)a(s)f1 s, km−1 (s, τ )u(τ )dτ, . . . , k1 (s, τ )u(τ )dτ, u(s), Z 0 1 0 0 1 Z gn−1 (s, τ )v(τ )dτ, . . . , 0 Z v(t) = 0 Z 0 1 Z g1 (t, s)b(s)f2 s, 1 Z 0 1 1 km−1 (s, τ )u(τ )dτ, . . . , k1 (s, τ )u(τ )dτ, u(s), 0 Z gn−1 (s, τ )v(τ )dτ, . . . , 0 g1 (s, τ )v(τ )dτ, v(s) ds, 1 g1 (s, τ )v(τ )dτ, v(s) ds. 0 (3.1) Let E := C([0, 1], R) × C([0, 1], R) endowed with the norm k(u, v)k := kuk + kvk, where kuk := max0≤t≤1 |u(t)|, and define the cone K ⊂ E by n o K := (u, v) ∈ E : u(t) ≥ 0, v(t) ≥ 0, t ∈ [0, 1], min (u(t) + v(t)) ≥ γk(u, v)k . θ≤t≤1−θ Clearly, (E, k · k) is a real Banach space and P is a cone on E. Define the operator T : K → K by T (u, v)(t) := (T1 (u, v)(t), T2 (u, v)(t)), EJDE-2009/163 THREE POSITIVE SOLUTIONS 5 where 1 Z k1 (t, s)a(s)f1 s, (Am−1 u)(s), . . . , (A1 u)(s), 0 u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds, Z 1 T2 (u, v)(t) := g1 (t, s)b(s)f2 s, (Am−1 u)(s), . . . , (A1 u)(s), 0 u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds. T1 (u, v)(t) := m+n Now f ∈ C([0, 1] × R+ , R+ ) and g ∈ C([0, 1] × Rm+n , R+ ) imply that T : K → + K is a completely continuous operator. In our setting, the existence of positive solutions for (3.1) is equivalent to that of positive fixed points of T . Lemma 3.1. The operator T maps K into K. Proof. If (u, v) ∈ K, then (T1 (u, v)(t))00 = −a(t)f1 t, (Am−1 u)(t), . . . , (A1 u)(t), u(t), (Bn−1 v)(t), . . . , (B1 v)(t), v(t) ≤ 0, (T2 (u, v)(t))00 = −b(t)f2 t, (Am−1 u)(t), . . . , (A1 u)(t), u(t), (Bn−1 v)(t), . . . , (B1 v)(t), v(t) ≤ 0. So T1 (u, v)(t) and T2 (u, v)(t) are concave on [0,1]. If (u, v) ∈ K, then from the properties of k1 (t, s) and g1 (t, s), we have Z 1 Z 1 k1 (s, s)a(s)f1 s, (Am−1 u)(s), . . . , (A1 u)(s), 0 u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds, kT1 (u, v)(t)k ≤ and g1 (s, s)b(s)f2 s, (Am−1 u)(s), . . . , (A1 u)(s), 0 u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds. kT2 (u, v)(t)k ≤ On the other hand, Z min θ≤t≤1−θ 1 k1 (s, s)a(s)f1 s, (Am−1 u)(s), . . . , (A1 u)(s), u(s), 0 (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds T1 (u, v)(t) ≥ γ ≥ γkT1 (u, v)(t)k, 6 J. XU, Z. YANG EJDE-2009/163 and Z min θ≤t≤1−θ 1 g1 (s, s)b(s)f2 s, (Am−1 u)(s), . . . , (A1 u)(s), u(s), 0 (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds T2 (u, v)(t) ≥ γ ≥ γkT2 (u, v)(t)k. Combining the preceding inequalities, we arrive at min (T1 (u, v)(t) + T2 (u, v)(t)) ≥ θ≤t≤1−θ min (T1 (u, v)(t)) + θ≤t≤1−θ min (T2 (u, v)(t)) θ≤t≤1−θ ≥ γ(kT1 (u, v)(t)k + kT2 (u, v)(t)k). This completes the proof. In this paper, we use the following assumptions: ((H1) a(t) and b(t) do not vanish identically on any subinterval of (0, 1), and there R1 exists t0 ∈ (0, 1) such that a(t0 ) > 0, b(t0 ) > 0 and 0 < 0 k1 (s, s)a(s)ds < R1 +∞, 0 < 0 g1 (s, s)b(s)ds < +∞. Finally, we define the nonnegative continuous concave functional α(u, v) := min (u(t) + v(t)). θ≤t≤1−θ We observe here that, for each (u, v) ∈ K, α(u, v) ≤ k(u, v)k. Let Z 1 Z 1 ξe1 := min k1 (t, s)a(s)ds, ηe1 := max k1 (t, s)a(s)ds, θ≤t≤1−θ µ e1 := 0≤t≤1 0 Z min 1 g1 (t, s)b(s)ds, θ≤t≤1−θ 0 0 Z νe1 := max 0≤t≤1 1 g1 (t, s)b(s)ds . 0 Theorem 3.2. Let 1 ≤ i ≤ m − 1, and 1 ≤ j ≤ n − 1.Assume there exist nonnegative numbers a, b, c such that 0 < a < b ≤ min{γ, ξe1 /e η1 , ξi /ηi , µ e1 /e ν1 , µj /νj }c, and f (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ), g(t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) satisfy the following growth conditions: (H2) f1 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) ≤ c/2e η1 , f2 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) ≤ c/2e ν1 , for all t ∈ [0, 1], xi + yj ∈ [0, ηc], xm + yn ∈ [0, c]. (H3) f1 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) < a/2e η1 , f2 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) < a/2e ν1 , for all t ∈ [0, 1], xi + yj ∈ [0, ηa], xm + yn ∈ [0, a]. (H4) f1 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) ≥ b/2ξe1 , f2 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) ≥ b/2e µ1 , for all t ∈ [θ, 1 − θ], xi + yj ∈ [ξb, ηb/γ], xm + yn ∈ [b, b/γ]. Then (1.1) has at least three positive fixed points (u1 , v1 ), (u2 , v2 ), (u3 , v3 ) such that k(u1 , v1 )k < a, b < minθ≤t≤1−θ (u2 (t) + v2 (t)), and k(u3 , v3 )k > a, with minθ≤t≤1−θ (u3 (t) + v3 (t)) < b. Proof. We note first that T : Pc → Pc is a completely continuous operator. If (u, v) ∈ Pc , then k(u, v)k ≤ c. For 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1, Z 1 Z 1 (Ai u)(t) + (Bj v)(t) = ki (t, s)u(s)ds + gj (t, s)v(s)ds ≤ ηi kuk + νj kvk ≤ ηc. 0 0 EJDE-2009/163 THREE POSITIVE SOLUTIONS 7 From (H2), we have kT (u, v)(t)k = max |T1 (u, v)(t)| + max |T2 (u, v)(t)| 0≤t≤1 0≤t≤1 Z 1 = max k1 (t, s)a(s)f1 s, (Am−1 u)(s), . . . , (A1 u)(s), 0≤t≤1 0 u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds Z 1 + max g1 (t, s)b(s)f2 s, (Am−1 u)(s), . . . , (A1 u)(s), 0≤t≤1 0 u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds Z 1 Z 1 g1 (t, s)b(s)ds ≤ c. k1 (t, s)a(s)ds + c/2e ν1 max ≤ c/2e η1 max 0≤t≤1 0≤t≤1 0 0 Therefore, T : Pc → Pc . In an analogous argument, one can verify that (H3) implies condition (C2) of Lemma 2.1. Clearly, {(u, v) ∈ P (α, b, b/γ) : α(u, v) > b} = 6 ∅. If (u, v) ∈ P (α, b, b/γ), then b ≤ u(t) + v(t) ≤ b/γ, t ∈ [θ, 1 − θ]. For 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1, we have Z 1 Z 1 (Ai u)(t) + (Bj v)(t) = ki (t, s)u(s)ds + gj (t, s)v(s)ds ≥ ξb, 0 1 Z (Ai u)(t) + (Bj v)(t) = Z gj (t, s)v(s)ds ≤ ηb/γ. ki (t, s)u(s)ds + 0 0 1 0 By (H4), α(T (u, v)(t)) = ≥ min (T1 (u, v)(t) + T2 (u, v)(t)) θ≤t≤1−θ min T1 (u, v)(t) + min T2 (u, v)(t) θ≤t≤1−θ Z 1 ≥ min k1 (t, s)a(s)f1 s, (Am−1 u)(s), . . . , (A1 u)(s), θ≤t≤1−θ 0 u(s), (Bn−1 v)(s), . . . , (A1 v)(s), v(s) ds Z 1 + min g1 (t, s)b(s)f2 s, (Am−1 u)(s), . . . , (A1 u)(s), θ≤t≤1−θ 0 u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds Z 1 Z 1 e ≥ b/2ξ1 min k1 (t, s)a(s)ds + b/2e µ1 min g1 (t, s)b(s)ds θ≤t≤1−θ θ≤t≤1−θ 0 θ≤t≤1−θ 0 = b. Therefore, condition (C1) of Lemma 2.1 is satisfied. Finally, we show that (C3) is also satisfied. If (u, v) ∈ P (α, b, c) and kT (u, v)k > b/γ, then α(T (u, v)(t)) = min (T1 (u, v)(t) + T2 (u, v)(t)) ≥ γkT (u, v)k > b. θ≤t≤1−θ Therefore, (C3) of Lemma 2.1 is also satisfied. By Lemma 2.1, the system (1.1) has at least three positive fixed points (u1 , v1 ), (u2 , v2 ), (u3 , v3 ) such that k(u1 , v1 )k < a, 8 J. XU, Z. YANG EJDE-2009/163 b < minθ≤t≤1−θ (u2 (t)+v2 (t)), and k(u3 , v3 )k > a, with minθ≤t≤1−θ (u3 (t)+v3 (t)) < b. An example. Consider a system of nonlinear second-order and fourth-order ordinary differential equations (with m = 1, n = 2): −x00 = a(t)f1 (t, x, y, −y 00 ) = 0, y (4) = b(t)f2 (t, x, y, −y 00 ) = 0, x(0) − x0 (0) = x(1) + x0 (1) = 0, 0 (3.2) 0 y(0) − y (0) = y(1) + y (1) = 0, 00 y (0) − y 000 (0) = y 00 (1) + y 000 (1) = 0. Let u := x and v := −y 00 , then the problem (3.2) is equivalent to the following system of nonlinear integral equations Z 1 Z 1 u(t) = k1 (t, s)a(s)f1 (s, u(s), g1 (s, τ )v(τ )dτ, v(s)), 0 0 (3.3) Z 1 Z 1 v(t) = g1 (t, s)b(s)f2 (s, u(s), 0 g1 (s, τ )v(τ )dτ, v(s)), 0 where 1 k1 (t, s) = g1 (t, s) := 3 ( (1 + s)(2 − t), 0 ≤ s ≤ t ≤ 1, (1 + t)(2 − s), 0 ≤ t ≤ s ≤ 1. We choose a(t) := (1 − t)−1/2 , b(t) := t−1/2 , t 1 + 10 (x1 + x3 )2 , 100 2 t 2 100 + 6[(x1 + x3 ) − 2(x1 + x3 )] + 5 , t + 20 log2 (x1 + x3 ) + 2(x1 + x3 ) + 25 , 100 t 1 542 100 + 4 (x1 + x3 ) + 5 , θ := 1/4, f1 (t, x1 , x2 , x3 ) equals t ∈ [0, 1], t ∈ [0, 1], t ∈ [0, 1], t ∈ [0, 1], 0 ≤ x1 + x3 ≤ 2, x2 ≥ 0, 2 < x1 + x3 < 4, x2 ≥ 0, 4 ≤ x1 + x3 ≤ 16, x2 ≥ 0, x1 + x3 > 16, x2 ≥ 0. and f2 (t, x1 , x2 , x3 ) equals t 1 + 10 (x1 + x3 )2 , t ∈ [0, 1], 100 t 13 2 2 t ∈ [0, 1], 100 + 8 [(x1 + x3 ) − 2(x1 + x3 )] + 5 , t 5 2 + log (x + x ) + 2(x + x ) + , t ∈ [0, 1], 3 1 3 2 1 2 5 100 t 1 192 t ∈ [0, 1], 100 + 4 (x1 + x3 ) + 5 , 0 ≤ x1 + x3 ≤ 2, x2 ≥ 0, 2 < x1 + x3 < 4, x2 ≥ 0, 4 ≤ x1 + x3 ≤ 16, x2 ≥ 0, x1 + x3 > 16, x2 ≥ 0. Then by direct calculation, we obtain ξ ≈ 0.4120, η = 0.5, ξe1 ≈ 0.0417, µ e1 ≈ 0.9273, ηe1 ≈ 0.8889, νe1 ≈ 1.1111. It is easy to check that (H1) holds. Choose γ = 41 , a = 1, b = 4, c = 800. Also, it is easy to verify that f1 and f2 satisfy conditions (H2)-(H4). So the system (3.2) has at least three positive solutions (u1 , v1 ), (u2 , v2 ), (u3 , v3 ) such that k(u1 , v1 )k < 1, 4 < min1/4≤t≤3/4 (u2 (t) + v2 (t)), and k(u3 , v3 )k > 1, with min1/4≤t≤3/4 (u3 (t) + v3 (t)) < 4. EJDE-2009/163 THREE POSITIVE SOLUTIONS 9 References [1] N. Anderson, A. M. Arthurs; Analytic bounding functions for diffusion problems with Michaelis-Menten kinetics, Bull. Math. Biol. 47 (1985) 145-153. [2] P. Kang, Z. Wei; Three positive solutions of singular nonlocal boundary value problems for systems of nonlinear second-order ordinary differential equations, Nonlinear Analysis 70 (2009) 444-451. [3] R. W. Leggett, L. R. Williams; Multiple positive fixed points of nonlinear operators on ordered Banach spaces, Indiana Univ. Math. J. 28 (1979) 673-688. [4] S. H. Lin; Oxygen diffusion in a spherical cell with nonlinear oxygen uptake kinetics, J. Theor. Biol. 60 (1976) 449-457. [5] Y. Li, Y. Guo, G. Li; Existence of positive solutions for systems of nonlinear third-order differential equations, Commun Nonlinear Sci Numer Simulat 14 (2009) 3792-3797. [6] L. Liu, P. Kang, Y. Wu; Benchawan Wiwatanapataphee, Positive solutions of singular boundary value problems for systems of nonlinear fourth order differential equations, Nonlinear Analysis 68 (2008) 485-498. [7] L. Liu, B. Liu, Y. Wu; Positive solutions of singular boundary value problems for nonlinear differential systems, Applied Mathematics and Computation 186 (2007) 1163-1172. [8] Z. Yang; Existence and uniqueness of positive solutions for a higher order boundary value problem, Computers and Mathematics with Applications 54 (2007) 220-228. Jiafa Xu Department of Mathematics, Qingdao Technological University, No 11, Fushun Road, Qingdao, China E-mail address: xujiafa292@sina.com Zhilin Yang Department of Mathematics, Qingdao Technological University, No 11, Fushun Road, Qingdao, China E-mail address: zhilinyang@sina.com