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Electronic Journal of Differential Equations, Vol. 2007(2007), No. 70, pp. 1–6. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp) EXISTENCE OF POSITIVE PSEUDO-SYMMETRIC SOLUTIONS FOR ONE-DIMENSIONAL p-LAPLACIAN BOUNDARY-VALUE PROBLEMS YITAO YANG Abstract. We prove the existence of positive pseudo-symmetric solutions for four-point boundary-value problems with p-Laplacian. Also we present an monotone iterative scheme for approximating the solution. The interesting point here is that the nonlinear term f involves the first-order derivative. 1. Introduction In this paper, we consider the four-point boundary value problem (φp (u0 ))0 (t) + q(t)f (t, u(t), u0 (t)) = 0, 0 u(0) − αu (ξ) = 0, 0 t ∈ (0, 1), 0 u(ξ) − γu (η) = u(1) + γu (1 + ξ − η), (1.1) (1.2) where φp (s) = |s|p−2 s, p > 1, (φp )−1 = φq , p1 + 1q = 1, α, γ ≥ 0, ξ, η ∈ (0, 1) are prescribed and ξ < η. The study of multipoint boundary-value problems for linear second-order ordinary differential equations was initiated by Il’in and Moiseer [7, 8]. Since then, the more general nonlinear multipoint boundary-value problems have been studied by many authors by using the Leray-Schauder continuation theorem, nonlinear alternative of Leray-Schauder and coincidence degree theory, we refer the reader to [1, 2, 3, 6] for some recent results. Recently, Avery and Henderson [4] consider the existence of three positive solutions for the problem (φp (u0 ))0 (t) + q(t)f (t, u(t)) = 0, u(0) = 0, t ∈ (0, 1), u(η) = u(1). (1.3) (1.4) The definition of pseudo-symmetric was introduced in their paper. Based on this definition, Ma [9] studied the existence and iteration of positive pseudo-symmetric solutions for the problem (1.3)-(1.4). However, to the best of our knowledge, no work has been done for BVP (1.1)-(1.2) using the monotone iterative technique. The aim of this paper is to fill the gap in the relevant literatures. We obtain not only the existence of positive solutions for (1.1)-(1.2), but also give an iterative scheme for approximating the solutions. It is worth stating that the first term of our iterative scheme is a constant function or a simple function. Therefore, the 2000 Mathematics Subject Classification. 34B15, 34B18. Key words and phrases. Iterative; pseudo-symmetric positive solution; p-Laplacian. c 2007 Texas State University - San Marcos. Submitted March 12, 2007. Published May 10, 2007. 1 2 Y. YANG EJDE-2007/70 iterative scheme is significant and feasible. At the same time, we give a way to find the solution which will be useful from an application viewpoint. We consider the Banach space E = C 1 [0, 1] equipped with norm kuk := max{kuk0 , ku0 k0 }, where kuk0 = max0≤t≤1 |u(t)|, ku0 k0 = max0≤t≤1 |u0 (t)|. In this paper, a positive solution u(t) of BVP (1.1), (1.2) means a solution u(t) of (1.1), (1.2) satisfying u(t) > 0, for 0 < t < 1. We recall that a function u is said to be concave on [0, 1], if u(λt2 + (1 − λ)t1 ) ≥ λu(t2 ) + (1 − λ)u(t1 ), t1 , t2 , λ ∈ [0, 1]. Definition 1.1. For ξ ∈ (0, 1) a function u ∈ E is said to be pseudo-symmetric if u is symmetric over the interval [ξ, 1]. That is, for t ∈ [ξ, 1] we have u(t) = u(1+ξ −t). Remark 1.2. For ξ ∈ (0, 1), if u ∈ E is pseudo-symmetric, we have u0 (t) = −u0 (1 + ξ − t), t ∈ [ξ, 1]. Define the cone K of E as K = {u ∈ C 1 [0, 1] : u(t) ≥ 0, u is concave on[0, 1] and u is symmetric on [ξ, 1]}. For x, y in K a cone of E, recall that x ≤ y if y − x ∈ K. In the rest of the paper, we make the following assumptions: (H1) q(t) ∈ L1 [0, 1] is nonnegative and q(t) = q(1 + ξ − t), a.e. t ∈ [ξ, 1], and q(t) 6≡ 0 on any subinterval of [0,1]; (H2) f ∈ C([0, 1]×[0, ∞)×R, [0, ∞)) and f (t, x, y) = f (1+ξ−t, x, −y), (t, x, y) ∈ [ξ, 1] × [0, ∞) × R. Moreover, f (t, ·, y) is nondecreasing for (t, y) ∈ [0, ξ+1 2 ]× ξ+1 R, f (t, x, ·) is nondecreasing for (t, x) ∈ [0, 2 ] × [0, ∞). 2. Existence Result Lemma 2.1 ([9]). Each u ∈ K satisfies the following properties: 2 (i) u(t) ≥ 1+ξ kuk0 min{t, 1 + ξ − t}, t ∈ [0, 1]; 2ξ 1+ξ kuk0 , kuk0 = u( 1+ξ 2 ). (ii) u(t) ≥ (iii) t ∈ [ξ, 1+ξ 2 ]; For x ∈ K, we define a mapping T : K → E given by R 1+ξ 2 q(τ )f (τ, x(τ ), x0 (τ ))dτ ) αφ ( q R ξ R 1+ξ t + φ ( 2 q(τ )f (τ, x(τ ), x0 (τ ))dτ )ds, 0 ≤ t ≤ 1+ξ 2 , 0 q s 1+ξ R (T x)(t) = αφ ( 2 q(τ )f (τ, x(τ ), x0 (τ ))dτ ) q ξ Rξ R 1+ξ + 0 φq ( s 2 q(τ )f (τ, x(τ ), x0 (τ ))dτ )ds + R 1 φ (R s q(τ )f (τ, x(τ ), x0 (τ ))dτ )ds, 1+ξ 1+ξ q 2 ≤ t ≤ 1. t (2.1) 2 Obviously, T x ∈ E, and we can prove T x is a solution of the boundary-value problem (φp (u0 ))0 (t) + q(t)f (t, x(t), x0 (t)) = 0, 0 u(0) − αu (ξ) = 0, 0 t ∈ (0, 1), 0 u(ξ) − γu (η) = u(1) + γu (1 + ξ − η). Therefore, each fixed point of T is a solution of problem (1.1)-(1.2). (2.2) (2.3) EJDE-2007/70 EXISTENCE OF POSITIVE PSEUDO-SYMMETRIC SOLUTIONS 3 Lemma 2.2. Suppose (H1), (H2) hold, then T : K → K is completely continuous and nondecreasing. 1+ξ Proof. For t ∈ [ξ, 1+ξ 2 ], we have 1 + ξ − t ∈ [ 2 , 1]. Therefore, (T x)(1 + ξ − t) Z Z 1+ξ 2 q(τ )f (τ, x(τ ), x0 (τ ))dτ ) + = αφq ( 1 Z φq ( + 1+ξ−t Z = αφq ( + t q(τ )f (τ, x(τ ), x0 (τ ))dτ )ds Z 0 ξ Z φq ( q(τ )f (τ, x(τ ), x (τ ))dτ ) + 0 ξ q(τ )f (τ, x(τ ), x0 (τ ))dτ )ds s 1+ξ 2 1+ξ 2 1+ξ 2 Z φq ( 1+ξ 2 s ξ Z Z φq ( 0 ξ Z ξ 1+ξ 2 q(τ )f (τ, x(τ ), x0 (τ ))dτ )ds s q(τ )f (τ, x(τ ), x0 (τ ))dτ )ds s Z = αφq ( 1+ξ 2 Z 0 q(τ )f (τ, x(τ ), x (τ ))dτ ) + ξ t Z φq ( 0 1+ξ 2 q(τ )f (τ, x(τ ), x0 (τ ))dτ )ds s = (T x)(t). (2.4) So that, T x is symmetric on [ξ, 1] with respect to 1+ξ . Obviously, (T x)(t) ≥ 0, Tx 2 is concave on [0, 1]. Therefore, T : K → K. It is easy to see that T : K → K is completely continuous. For x1 (t), x2 (t) ∈ K and x1 (t) ≤ x2 (t), thus x2 (t) − x1 (t) ∈ K. So that x02 (t) − 1+ξ 0 0 0 x1 (t) ≥ 0, t ∈ [0, 1+ξ 2 ] and x2 (t) − x1 (t) ≤ 0, t ∈ [ 2 , 1]. Assumption (H2) implies (T x1 )(t) ≤ (T x2 )(t). Lemma 2.3 ([5]). Let u0 , v0 ∈ E, u0 < v0 and T : [u0 , v0 ] → E be an increasing operator such that u0 ≤ T u0 , T v0 ≤ v0 . Suppose that one of the following conditions is satisfied: (C1) K is normal and T is condensing; (C2) K is regular and T is semicontinuous, i.e., xn → x strongly implies T xn → T x weakly. Then T has a maximal fixed point x∗ and a minimal fixed point x∗ in [u0 , v0 ]; moreover x∗ = lim vn , x∗ = lim un , n→∞ n→∞ where vn = T vn−1 , un = T un−1 (n = 1, 2, 3, . . . ), and u0 ≤ u1 ≤ · · · ≤ un ≤ · · · ≤ vn ≤ · · · ≤ v1 ≤ v0 . Denote the positive quantities Z 1+ξ Z 2 A = 1/ max αφq ( q(τ )dτ ) + ξ 1+ξ 2 Z φq ( 0 1+ξ 2 Z q(τ )dτ )ds, φq ( s 1+ξ 2 q(τ )dτ ) , 0 (2.5) Z B = 1/ ξ 1+ξ 2 Z φq ( s 1+ξ 2 q(τ )dτ )ds. (2.6) 4 Y. YANG EJDE-2007/70 Theorem 2.4. Assume (H1), (H2) hold. If there exist two positive numbers a, b 2b with 1+ξ < a, such that sup f (t, a, a) ≤ φp (aA), inf t∈[ξ, 1+ξ 2 ] t∈[0,1] f (t, 2ξb , 0) ≥ φp (bB). 1+ξ (2.7) Then, (1.1)-(1.2) has at least one positive pseudo-symmetric solution υ ∗ ∈ K with b ≤ kυ ∗ k0 ≤ a, where υ0 (t) = 2b 1+ξ 0 ≤ k(υ ∗ )0 k0 ≤ a, lim T n υ0 = υ ∗ , n→∞ min{t, 1 + ξ − t}, t ∈ [0, 1]. Proof. We denote K[b, a] = {ω ∈ K : b ≤ kωk0 ≤ a, 0 ≤ kω 0 k0 ≤ a}. Next, we first prove T K[b, a] ⊂ K[b, a]. Let ω ∈ K[b, a], then 0 ≤ ω(t) ≤ maxt∈[0,1] ω(t) ≤ kωk0 ≤ a, t ∈ [0, 1], 0 ≤ ω 0 (t) ≤ maxt∈[0,1] |ω 0 (t)| = kω 0 k0 ≤ a, t ∈ [0, 1+ξ 2 ]. By lemma 2.1 2ξ 2ξb (ii), mint∈[ξ, 1+ξ ] ω(t) ≥ 1+ξ kωk0 ≥ 1+ξ , mint∈[ξ, 1+ξ ] ω 0 (t) ≥ ω 0 ( 1+ξ 2 ) = 0. So, by 2 2 assumption (2.7), we have 1+ξ ], (2.8) 0 ≤ f (t, ω(t), ω 0 (t)) ≤ f (t, a, a) ≤ sup f (t, a, a) ≤ φp (aA), t ∈ [0, 2 t∈[0,1] f (t, ω(t), ω 0 (t)) ≥ f (t, 2ξb 2ξb , 0) ≥ inf f (t, , 0) ≥ φp (bB), 1+ξ 1 +ξ t∈[ξ, 1+ξ ] 2 t ∈ [ξ, 1+ξ ]. 2 (2.9) By lemma 2.2, we know T ω ∈ K. So, lemma 2.1 (iii) implies kT ωk0 = (T ω)( 1+ξ 2 ). As a result, kT ωk0 1+ξ ) 2 = (T ω)( Z = αφq ( 1+ξ 2 Z 0 1+ξ 2 q(τ )f (τ, x(τ ), x (τ ))dτ ) + ξ 0 Z ≤ aA(αφq ( 1+ξ 2 1+ξ 2 Z q(τ )dτ ) + ξ 1+ξ 2 q(τ )f (τ, x(τ ), x0 (τ ))dτ )ds s 1+ξ 2 Z φq ( 0 Z φq ( q(τ )dτ )ds) ≤ a; s k(T ω)0 k0 = (T ω)0 (0) Z 1+ξ 2 = φq ( q(τ )f (τ, x(τ ), x0 (τ ))dτ ) 0 1+ξ 2 Z ≤ aAφq ( q(τ )dτ ) ≤ a; 0 kT ωk0 = (T ω)( Z 1+ξ 2 ≥ 1+ξ ) 2 Z 1+ξ 2 φq q(τ )f (τ, x(τ ), x0 (τ ))dτ ds ξ s Z ≥ bB 1+ξ 2 φq ξ Z s 1+ξ 2 q(τ )dτ ds = b. EJDE-2007/70 EXISTENCE OF POSITIVE PSEUDO-SYMMETRIC SOLUTIONS 5 Thus, b ≤ kT ωk0 ≤ a, 0 ≤ k(T ω)0 k0 ≤ a, which implies T K[b, a] ⊂ K[b, a]. 0 2b 2b Let υ0 (t) = 1+ξ min{t, 1 + ξ − t}, t ∈ [0, 1], then kυ0 k0 = b and kυ0 k0 = 1+ξ < a, so υ0 ∈ K[b, a]. Let υ1 = T υ0 , then υ1 ∈ K[b, a], we denote υn+1 = T υn = T n+1 υ0 , (n = 0, 1, 2, . . . ). (2.10) Since T K[b, a] ⊂ K[b, a], we have υn ∈ K[b, a], (n = 0, 1, 2, . . . ). From υ1 ∈ K[b, a], thus υ1 (t) ≥ 2b 2 kυ1 k0 min{t, 1 + ξ − t} ≥ min{t, 1 + ξ − t} = υ0 (t), 1+ξ 1+ξ t ∈ [0, 1], which implies T υ0 ≥ υ0 . K is normal and T is completely continuous. By Lemma 2.3, we have T has a fixed point υ ∗ ∈ K[b, a]. Moreover, υ ∗ = limn→∞ υn . Since kυ ∗ k0 ≥ b > 0 and υ ∗ is a nonnegative concave function on [0, 1], we conclude that υ ∗ (t) > 0, t ∈ (0, 1). Therefore, υ ∗ is a positive pseudo-symmetric solution of (1.1)-(1.2). Corollary 2.5. Assume (H1),(H2) hold. If lim sup l→0 inf t∈[ξ, 1+ξ 2 ] particularly, lim supl→0 inf t∈[ξ, 1+ξ ] 2 f (t, l, 0) 1+ξ ≥ φp ( B), φp (l) 2ξ f (t,l,0) φp (l) lim inf sup l→+∞ t∈[0,1] particularly, lim inf l→+∞ supt∈[0,1] (2.11) = +∞, f (t, l, l) ≤ φp (A), φp (l) f (t,l,l) φp (l) (2.12) = 0. Where A, B are defined as (2.5), 2b (2.6). Then there exist two positive numbers a, b with 1+ξ < a, such that problem (1.1), (1.2) has at least one positive pseudo-symmetric solution υ ∗ ∈ K with b ≤ kυ ∗ k0 ≤ a, 0 ≤ k(υ ∗ )0 k0 ≤ a, lim T n υ0 = υ ∗ , n→∞ where υ0 (t) = 2b 1+ξ min{t, 1 + ξ − t}, t ∈ [0, 1]. Remark 2.6. Problem (1.1)-(1.2) may have two positive pseudo-symmetric solutions ω ∗ , υ ∗ ∈ K, if we make another iteration by choosing ω0 (t) = a and ωn = limn→∞ T n ω0 = ω ∗ . However, ω ∗ and υ ∗ may be the same solution. Example 2.7. We consider the problem (|u0 |3 u0 )0 (t) + 1 0 2 2 t ∈ (0, 1), 1 [(u (t)) + ln((u(t)) + 1)] = 0, − t) 2 1 1 1 5 u(0) − 2u0 ( ) = 0, u( ) − 3u0 ( ) = u(1) + 3u0 ( ). 3 3 2 6 t 1 2 ( 43 (2.13) (2.14) We notice that p = 5, α = 2, ξ = 31 , γ = 3, η = 12 . Obviously, f (t, u, u0 ) = (u0 (t))2 +ln((u(t))2 +1) is nondecreasing for (t, u0 ) ∈ [0, 23 ]×R, f (t, u, u0 ) = (u0 (t))2 + ln((u(t))2 + 1) is nondecreasing for (t, u) ∈ [0, 32 ] × [0, ∞), q(t) = 1 41 1 is t 2 ( 3 −t) 2 nonnegative and pseudo-symmetric about 32 . So, conditions (H1),(H2) are satisfied. 6 Y. YANG EJDE-2007/70 On the other hand, lim sup l→0 f (t, l, 0) ln(l2 + 1) = ∞, = lim sup inf 1 2 φp (l) l4 t∈[ξ, 1+ξ l→0 t∈[ 3 , 3 ] 2 ] inf f (t, l, l) l2 + ln(l2 + 1) = 0. = lim inf sup l→+∞ t∈[0,1] l4 t∈[0,1] φp (l) lim inf sup l→+∞ Therefore, from Corollary 2.5, it follows that (2.13)-(2.14) has at least one positive pseudo-symmetric solution. References [1] R. P. Agarwal and D. O’Regan; Nonlinear boundary-value problems on time scales, Nonlinear Anal. 44 (2001), 527-535. [2] R. P. Agarwal and D. O’Regan; Lidstone continuous and discrete boundary-value problems, Mem. Diff. Equ. Math. Phys. 19 (2000), 107-125. [3] R. P. Agarwal, H. Lü and D. 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[9] Dexiang Ma, Weigao Ge; Existence and iteration of positive pseudo-symmetric solutions for a three-point second-order p-Laplacian BVP, Appl. Math. Lett. (2007), doi: 10.1016 j.aml.2006.05.025. Yitao Yang Institute of Automation, Qufu Normal University, Qufu, Shandong 273165, China E-mail address: [email protected]