Journal of Computational and Applied Mathematics. On positive solutions for fourth-order boundary value problem with impulse

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1 Journl of Computtionl nd Applied Mthemtics 225 (2009) Contents lists vilble t ScienceDirect Journl of Computtionl nd Applied Mthemtics journl homepge: On positive solutions for fourth-order boundry vlue problem with impulse Ilky Ysln Krc Deprtment of Mthemtics, Ege University, 3500 Bornov, Izmir, Turkey r t i c l e i n f o b s t r c t Article history: Received 0 Jnury 2007 Received in revised form 6 April 2008 MSC: 3B5 Keywords: Positive solutions Fixed point theorem in cones In this pper, we consider fourth-order boundry vlue problem with impulse. First, we estblish criteri for the existence of one or more thn one positive solution of noneigenvlue problem. Second, we re concerned with determining vlues of λ, for which there exist positive solutions for n eigenvlue problem. In both problems, we shll use the Krsnoselskii fixed point theorem Elsevier B.V. All rights reserved.. Introduction We re interested in proving the existence nd multiplicity results for positive solutions to the fourth-order boundry vlue problem with impulse (BVPI) y () q(x)y = f (x, y), x [, c) (c, b], α y() β y () = 0, γ y(b) + δ y (b) = 0, y (c 0) = d y (c + 0), y (c 0) = d 2 y (.) (c + 0), α 2 y () β 2 y () = 0, γ 2 y (b) + δ 2 y (b) = 0 nd the eigenvlue problem y () q(x)y = λf (x, y) with the sme boundry conditions where λ > 0. Here < 3+b c < 3b < b, y(c 0) is the left-hnd limit of y(x) t c nd y(c + 0) is the right-hnd limit of y(x) t c. We will ssume tht the following conditions re stisfied: (H) q(x) 0 is mesurble function on [, b] nd q(x)dx <. (H2) d > 0, d 2 > 0, α 2, β 2, γ 2, δ 2 0, α 2 + β 2 > 0, γ 2 + δ 2 > 0, if q(x) 0 on [, c) (c, b] then α 2 + γ 2 > 0. (H3) α, β, γ, δ 0, α δ + β γ + α γ (b ) > 0. (H) f (x, ξ) is rel-vlued function continuous with respect to the collection of its rguments x [, c) (c, b] nd ξ R, nd f (x, ξ) 0 for ξ R +, where R + denotes the set of non-negtive rel numbers. Moreover, for ech ξ R there exist finite limits lim (x,ξ) (c,ξ 0 ) x<c f (x, ξ) = f (c 0, ξ 0 ) nd lim f (x, ξ) = f (c + 0, ξ 0 ). (x,ξ) (c,ξ 0 ) x>c Positive solutions of boundry vlue problems for differentil equtions with impulse were erlier studied in [5,6]. For the bsic concepts of impulse differentil equtions we refer to [2,3]. Bereketoglu nd Huseynov [3] studied nonliner < Tel.: E-mil ddress: ilky.krc@ege.edu.tr /$ see front mtter 2008 Elsevier B.V. All rights reserved. doi:0.06/j.cm

2 I.Y. Krc / Journl of Computtionl nd Applied Mthemtics 225 (2009) second-order differentil equtions subject to seprted liner boundry conditions nd to liner impulse conditions by using the Krsnoselskii fixed point theorem. Fourth-order boundry vlue problems hve been studied in recent yers [,,9,0,]. However, our problem nd results in this pper re different from those in [,,9,0,]. Our method is s in [,2]. To the uthor s knowledge, no one hs studied positive solutions for the fourth-order boundry vlue problem with impulse. The key tool in our pproch is the following Krsnoselskii fixed point theorem in cone [7,8]. Theorem.. Let B be Bnch spce, nd let P 0 Ω, Ω Ω 2, nd let A : P ( Ω 2 \ Ω ) P be completely continuous opertor such tht, either (i) Ay y, y P Ω, nd Ay y, y P Ω 2 ; or (ii) Ay y, y P Ω, nd Ay y, u P Ω 2. Then A hs t lest one fixed point in P ( Ω 2 \ Ω ). 2. The preliminry lemms B be cone in B. Assume Ω, Ω 2 re open subsets of B with Denote by θ nd ϕ the solutions of the homogeneous problem y q(x)y = 0, x [, c) (c, b], y(c 0) = d y(c + 0), y (c 0) = d 2 y (c + 0), (2.) stisfying the initil conditions θ() = β 2, θ () = α 2, ϕ(b) = δ 2, ϕ (b) = γ 2. Define the number ρ by β 2 ϕ () + α 2 ϕ(), ρ = [ β 2 ϕ () + α 2 ϕ()], d d 2 if x [, c), if x (c, b]. (2.2) Lemm 2. ([3]). Let conditions (H) nd (H2) hold. The number ρ defined by (2.) is positive for x [, c) (c, b] c ± 0}. Let G(x, s) be the Green function for the boundry vlue problem y (x) q(x)y(x) = 0, x [, c) (c, b], y(c 0) = d y(c + 0), y (c 0) = d 2 y (c + 0) α 2 y() β 2 y () = 0, γ 2 y(b) + δ 2 y (b) = 0 which is given by G(x, s) = ρ ϕ(s)ψ(x), if s x b, ϕ(x)ψ(s), if x s b. Lemm 2.2 ([3]). Let conditions (H) nd (H2) hold. Then G(x, s) 0, for x, s [, c) (c, b]. Lemm 2.3. Assume tht the conditions (H) (H) re stisfied. If h(x) is rel-vlued mesurble function on [, b] such tht h(x) dx <, then the BVPI y () q(x)y = h(x), x [, c) (c, b], α y() β y () = 0, γ y(b) + δ y (b) = 0, y (c 0) = d y (c + 0), y (c 0) = d 2 y (2.3) (c + 0), α 2 y () β 2 y () = 0, γ 2 y (b) + δ 2 y (b) = 0 hs unique solution y(x) = G 2 (ξ, s)h(s)dsdξ

3 358 I.Y. Krc / Journl of Computtionl nd Applied Mthemtics 225 (2009) where nd G (x, s) = G 2 (x, s) = ρ (γ (b s) + δ )(α (x ) + β ), if x s b, α δ + β γ + α γ (b ) (γ (b x) + δ )(α (s ) + β ), if s x b ϕ(s)ψ(x), if s x b, ϕ(x)ψ(s), if x s b. (2.) (2.5) Here ρ is s in (2.5). Proof. Let us consider the following BVP: b y = G 2 (x, s)h(s)ds, x [, c) (c, b], α y() β y () = 0, γ y(b) + δ y (b) = 0. The Green function ssocited with the BVP (2.) is G (x, s). This completes the proof. (2.6) Lemm 2.. Assume tht the conditions (H) (H) re stisfied. If h(x) 0 is mesurble function on [, b] nd h(x)dx <, then the unique solution y of the problem (2.3) stisfies y(x) 0, for x [, c) (c, b]. Proof. It is subsequence of the fcts tht G (x, s) 0, G 2 (x, s) 0 for x, s [, c) (c, b]. Lemm 2.5. Let (H) (H) hold. If h(x) 0 is mesurble function on [, b] nd h(x)dx <, then the unique solution y of the problem (2.3) stisfies [ ) ( 3 + b min y(x) : x, c c, 3b ]} Γ y where y = mxy(x) : x [, c) (c, b]} nd α (b ) + β Γ := min, γ } (b ) + δ. (2.7) α (b ) + β γ (b ) + δ Proof. We hve from (2.) tht nd 0 G (x, s) G (s, s), x, s [, c) (c, b] (2.8) G (x, s) Γ G (s, s), where Γ is s in (2.7). From (2.8), we get y(x) G (ξ, ξ) Thus for x [ 3+b, c) (c, 3b ], y(x) Γ Γ y. G (ξ, ξ) [ ) ( 3 + b x, c c, 3b ], s [, c) (c, b] for ll x [, c) (c, b]. 3. Existence of one or more positive solutions For η > 0, set F(η) = mxf (x, y) : x [, c) (c, b], y [0, η]}, H(η) = minf (x, y) : x [, c) (c, b], y [Γ η, η]}, where Γ is s in (2.7).

4 I.Y. Krc / Journl of Computtionl nd Applied Mthemtics 225 (2009) Denote M = N = G (ξ, ξ) mx x [,c) (c,b], 3b 3+b (3.). (3.2) We work in the Bnch spce B of ll rel-vlued continuous functions y on [, c) (c, b] for which the finite vlues y(c 0) nd y(c + 0) exist with the norm y = mx x [,c) (c,b] y(x). Let [ ]} 3 + b K = y B : y(x) Γ y, x, c ) ( c, 3b (where Γ is s in (2.7)), then K is cone. For ech y K, denote Ty(x) =. By Lemm 2.5, we know tht TK K. Applying the Arzel Ascoli Lemm, we cn esy check tht T is completely continuous. Theorem 3.. Let (H) (H) hold. Suppose there exist two positive numbers η nd η 2 with η η 2 such tht F(η ) η M, H(η 2 ) η 2 N. Then (.) hs t lest one positive solution y stisfying minη, η 2 } y mxη, η 2 }. Proof. We only show the cse tht η < η 2. The other cse cn be treted by the sme method. From Lemm 2.3, we know tht y is solution of (.) if nd only if y solves the fixed point problem y = Ty. We will pply the Krsnoselskii fixed point theorem (see Theorem.) to prove tht T hs fixed point y K with η y η 2. For y K with y = η, we hve tht hence f (x, y(x)) F(η ) η M Ty(x) = η M G (ξ, ξ) η = y. For y K with y = η 2, we hve tht [ ) ( 3 + b Γ η 2 y(x) η 2, x, c c, 3b ] nd so tht [ ) ( 3 + b min f (x, y) : x, c c, 3b ] }, Γ η 2 y(x) η 2 = H(η 2 ) η 2 N Ty mx x [,c) (c,b] η 2 N mx x [,c) (c,b] 3b 3+b 3b 3+b = η 2 = y. Therefore, by the first prt of the Krsnoselskii fixed point theorem, it follows tht T hs fixed point y with η y η 2. (3.3)

5 360 I.Y. Krc / Journl of Computtionl nd Applied Mthemtics 225 (2009) Theorem 3.2. Let (H) (H) hold. If there exist j + positive numbers η, η 2,..., η j+ with η < η 2 < < η j+ such tht either or F(η 2k ) < η 2k M for ll 2k, 2,..., j + }, H(η 2k ) > η 2k N for ll 2k, 2,..., j + }, H(η 2k ) > η 2k M for ll 2k, 2,..., j + }, F(η 2k ) < η 2k N for ll 2k, 2,..., j + }, then (.) hs t lest j positive solutions. Proof. We only prove the result under (3.). In the cse tht (3.5) holds, the results cn be proved by the sme method. Since F nd H re continuous, 0 < N M, we know tht there exist Θ i nd τ i with η i < Θ i < τ i < η i+, i =, 2,..., j such tht F(Θ 2k ) Θ 2k M H(τ 2k ) τ 2k N for ll 2k, 2,..., j + }, H(Θ 2k ) Θ 2k N F(τ 2k ) τ 2k M for ll 2k, 2,..., j + }. From Theorem 3., for ech i, 2,..., j}, (.) hs positive solution y i stisfying η i < Θ i y i τ i < Θ j+. (3.) (3.5) Corollry 3.. Let (H) (H) hold. Assume tht there exist two sequences η i }, Θ i } of (0, + ) such tht (i) lim i + η i = + (ii) lim i + Θ i = + F(η (iii) lim i ) i + η i < M H(Θ (iv) lim i ) i + Θ i > N. Then (.) hs sequence of positive solutions y k } such tht y k s k.. Boundry vlue problem with prmeter In this section we consider the following BVPI with prmeter λ, y () q(x)y = λf (x, y), x [, c) (c, b], αy() βy () = 0, γ y(b) + δy (b) = 0, y (c 0) = d y (c + 0), y (c 0) = d 2 y (c + 0), α 2 y () β 2 y () = 0, γ 2 y (b) + δ 2 y (b) = 0. Define the non-negtive extended rel numbers f 0, f 0, f nd f by f 0 := lim inf min f (x, y), f 0 := lim y 0 + x [,c) (c,b] y sup mx f (x, y), y 0 + x [,c) (c,b] y f (x, y) f := lim inf min, f f (x, y) := lim sup mx, y x [,c) (c,b] y y x [,c) (c,b] y respectively. These numbers cn be regrded s generlized super or subliner conditions on the function f (x, y) t y = 0 nd y =. Thus, if f 0 = f 0 = 0 (+ ), then f (x, y) is superliner (subliner) t y = 0 nd if f = f = 0 (+ ), then f (x, y) is subliner (superliner) t y = +. (.) hs solution y = y(x) if nd only if y solves the opertor eqution y(x) = λ G 2 (ξ, s)f (s, y(s))ds = (T λ y)(x), where G, G 2 re defined by (2.) nd (2.5), respectively. Let us define the cone K s in (3.3). By Lemm 2.5, T λ K K. It is lso esy to check tht T λ : K K is completely continuous. We seek fixed point of T λ in the cone K. Theorem.. Assume tht (H) (H) re stisfied. Then, for ech λ stisfying (.) < λ <, NΓ f Mf 0 there exists t lest one positive solution of (.), where Γ, M, N re s in (2.7), (3.) nd (3.2), respectively. (.2)

6 I.Y. Krc / Journl of Computtionl nd Applied Mthemtics 225 (2009) Proof. Let λ be given s in (.2). Now, let ɛ > 0 be chosen such tht Γ N(f ɛ) λ M(f 0 + ɛ). Now, turning to f 0, there exists n H > 0 such tht f (s, y) (f 0 + ɛ)y for 0 < y H. So, for y K with y = H, we hve from the fct 0 G (x, s) G(s, s) tht T λ y(x) = λ λ(f 0 + ɛ) G (ξ, ξ) λ(f 0 + ɛ)m y y. G 2 (ξ, s)y(s)dsdξ Next, considering f, there exists Hˆ 2 > 0 such tht f (s, y) (f ɛ)y for y Hˆ H 2. Let H 2 = mx2h, ˆ 2 }. Then y K Γ nd y = H 2 imply nd so min y(x) Γ y Hˆ 2, x [ 3+b,c) (c, 3b ] T λ y = λ mx x [,c) (c,b] λ(f ɛ) 3b 3+b λγ (f ɛ) y mx x [,c) (c,b] = λγ (f ɛ)n y y = H 2. G 2 (ξ, s)y(s)dsdξ 3b 3+b Therefore, by the first prt of Theorem (.), it follows tht T λ hs fixed point y stisfying H y H 2. The proof is completed. Theorem.2. Assume tht (H) (H) re stisfied. Then, for ech λ stisfying < λ <, Γ Nf 0 Mf there exists t lest one positive solution of (.) where Γ, M, N re s in (2.5), (3.) nd (3.2) respectively. Proof. Let λ be given s in (.3), nd choose ɛ > 0 such tht Γ N(f 0 ɛ) λ M(f + ɛ). Beginning with f 0, there exists n H > 0 such tht f (s, y) (f 0 ɛ)y for 0 < y H. So, for y K with y = H we hve T λ y = λ mx x [,c) (c,b] λ(f 0 ɛ) 3b 3+b λγ (f 0 ɛ) y mx x [,c) (c,b] = λγ (f 0 ɛ)n y y = H. G 2 (ξ, s)y(s)dsdξ 3b 3+b It remins to consider f. There exists n Hˆ 2 > 0 such tht f (s, y) (f + ɛ)y, for ll y Hˆ 2. There re two cses: () f is bounded, nd (b) f is unbounded. For cse (), suppose R > 0 is such tht f (s, y) R, for ll 0 y. Let H 2 = mx2h, λrm}. (.3)

7 362 I.Y. Krc / Journl of Computtionl nd Applied Mthemtics 225 (2009) Then, for y K with y = H 2, we hve T λ y(x) = λ λr G (ξ, ξ) λrm H 2 = y so tht T λ y y. For cse (b), let g(h) := mxf (x, y) : x [, c) (c, b], y [0, h]}. The function g is non-decresing nd lim h g(h) =. Choose H 2 = mx2h, Hˆ 2 } so tht g(h 2 ) g(h) for 0 h H 2. For y K with y = H 2, T λ y(x) = λ λg(h 2 ) G (ξ, ξ) λ(f + ɛ)mh 2 H 2 = y so tht T λ y y. It follows from Theorem. tht T λ hs fixed point. Hence the problem (.) hs positive solution. The proof is completed. Exmple.. Consider the following boundry vlue problem: y () y = λe x, x [, 2) (2, 3], y() 2y () = 0, 3y(3) + y (3) = 0, y (2 0) = 2 y (2 + 0), y (2 0) = 3 y (2 + 0), y () y () = 0, y (3) + 2y (3) = 0. Tking =, α =, β = 2, b = γ = 3, δ =, we hve Γ = /20. When tking f (x, y) = f (x) = e x, we get f 0 =, f = 0. In the cse of p(x), q(x), d = /2, d 2 = /3, α 2 = β 2 = γ 2 =, c = δ 2 = 2, the solutions of the problems (2.) re e x, if x [, 2), θ(x) = 2 (5ex e 3 x (.5) ), if x (2, 3], nd ϕ(x) = Therefore, we hve = 2 (cosh(x ) + 5 cosh(3 x) + ex + 5e 3 x ), if x [, 2), cosh(3 x) + e 3 x, if x (2, 3]. 6 (3 3ξ)(x + ), if x ξ, (3 3x)(ξ + ), if ξ x, where θ, ϕ re s in (.5) nd (.6) respectively nd ρ = 5e 2 +, if x [, 2), 2 6(5e 2 + ), if x (2, 3]. G 2 (ξ, s) = ρ θ(s)ϕ(ξ), if s ξ, θ(ξ)ϕ(s), if ξ s, By using (.2) nd (.3), we get M = , N = By Theorem.2, for ech λ stisfying 0 < λ <, there exists t lest one positive solution of the problem (.) (see Fig. ). Indeed, if y (x) = z(x), for x [, 2) (2, 3] then the BVP (.) is equivlent to the following system of integrl equtions: 3 y(x) = z(ξ)dξ z(x) = 3 G 2 (x, s)f (s, y(s))ds. (.) (.6)

8 I.Y. Krc / Journl of Computtionl nd Applied Mthemtics 225 (2009) Fig.. The solution of the BVP (.). We obtin nd lso where z(x) = y(x) = r(x), if x [, 2), s(x), if x (2, 3] R(x), if x [, 2), S(x), if x (2, 3], r(x) = (5e 2 + ) (5ex e 2 x 5e x 5e x e x+2 2xe x 30xe x+2 ), s(x) = 2(5e 2 + ) (6ex + 7e 6 x + 55e x+2 xe x 5xe x+2 8e x 6e x 2 ), R(x) = S(x) = e x 6(5e 2 + ) [ 2(5x + 63)ex+ + (369x 33)e x e 2x 2 + 5(3 3x)e x + 6(2x 9)e 2x 5(x + )e x+2 + 7(x + )e x + 32(5x 76)e 2x+2 2(x + )e x+ + 0(x + )e x+5 + 6e 2 (5e 2 + )], e x 6(5e 2 + ) [ 2(5x + 63)ex+ + (369x 33)e x e 2x 2 + 5(3 3x)e x + 32(x 8)e 2x + 2(7x 637)e x+2 + 8(3 3x)e x + 60(3x 7)e 2x (3 3x)e x+ + 0(x + )e x e (8 7e 2 )]. Thus for ech positive λ, there exists one positive solution of problem (.). References [] Dougls R. Anderson, Richrd I. Avery, A fourth-order four-point right focl boundry vlue problem, Rocky Mountin J. Mth. 36 (2) (2006) [2] D.D. Binov, P.S. Simeonov, Impulse Differentil Equtions: Asymptotic Properties of the Solutions, World Scientific, Singpore, 995. [3] Huseyin Bereketoglu, Aydin Huseynov, On positive solutions for nonliner boundry vlue problem with impulse, Czechoslovk Mthemticl Journl 56 (3) (2006) [] Shihu Chen, Wei Ni, Chngping Wng, Positive solution of fourth order ordinry differentil eqution with four-point boundry conditions, Applied Mthemtics Letters 9 (2006) [5] P.W. Eloe, J. Henderson, Positive solutions of boundry vlue problems for ordinry differentil equtions with impulse, Dynmics of Continuous Discrete & Impulsive Systems (998) [6] P.W. Eloe, M. Sokol, Positive solutions nd conjugte points for boundry vlue problem with impulse, Dynmic Systems nd Applictions 7 (998) 9. [7] D. Guo, Some fixed point theorems on cone mps, Kexue Tongbo 29 (98) [8] M.A. Krsnoselskii, Positive Solutions of Opertor Equtions, Noordhoff, Groningen, 96. [9] B. Liu, Positive solutions of fourth-order two point boundry vlue problems, Applied Mthemtics nd Computtion 8 (200) [0] Ruyun M, Zhng Jihui, Fu Shengmo, The method of Lower nd Upper Solutions for Fourth-Order Two-Point Boundry Vlue Problems, Journl of Mthemticl Anlysis nd Applictions 25 (997) [] Ruyun M, Multiple positive solutions for non-liner m-point boundry vlue problems, Applied Mthemtics nd Computtion 8 (200)

9 36 I.Y. Krc / Journl of Computtionl nd Applied Mthemtics 225 (2009) [2] Ruyun M, Bevn Thompson, Positive solutions for nonliner m-point eigenvlue problems, Journl of Mthemticl Anlysis nd Applictions 297 (200) [3] A.M. Smoilenko, N.A. Perestyuk, Impulsive Differentil Equtions, World Scientific, Singpore, 995. [] Qin Zhng, Shihu Chen, Jinhu Lü, Upper nd lower solution method for fourth-order four-point boundry vlue problems, Journl of Computtionl Applied Mthemtics 96 (2006)