POSITIVE SOLUTIONS FOR BOUNDARY VALUE PROBLEM OF SINGULAR FRACTIONAL FUNCTIONAL DIFFERENTIAL EQUATION

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1 International Journal of Pure and Applied Mathematics Volume 92 No. 2 24, ISSN: 3-88 (printed version); ISSN: (on-line version) url: doi: PAijpam.eu POSITIVE SOLUTIONS FOR BOUNDARY VALUE PROBLEM OF SINGULAR FRATIONAL FUNTIONAL DIFFERENTIAL EQUATION Limei Song School of Mathematics Jiaying University Meizhou, Guangdong, 545, P.R. HINA Abstract: In this paper, we consider a class of singular fractional order functional differential equations with delay. By means of a fixed point theorem on cones, some sufficient conditions for the existence of at least one or two positive solutions for the boundary value problem are established. We also give examples to illustrate the applicability of our main results. AMS Subject lassification: 26A33, 34B8 Key Words: singular problem, fractional functional differential equation, boundary value problem, fixed point theorem, Positive solution. Introduction Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications (see[], [2], [3], [4], [5]). In [6], [7], [8], [9], [], they studied the existence and multiplicity of positive solutions to fractional differential equations, and obtained some results. However, to the best of the author knowledge, there are few articles studying Received: October, 23 c 24 Academic Publications, Ltd. url:

2 7 L. Song the functional differential equations of fractional order. In [], the authors studied the existence of positive solution for boundary value problem of fourthorder FDE. Motivated by the work above, this paper investigates the existence of positive solutions for singular fractional order functional differential equation with boundary conditions D α u(t)+r(t)f(u + t ) =, < t <, 2 < α 3 u(t) = φ(t), τ t, u() = u () = u () =, where D α is the standard Riemann-Liouville fractional derivative of order + α; φ(t) ([ τ,],[,+ )), φ() = ; for t [,],u t (θ) = u(t + θ),θ [ τ,], τ < 2 is a constant. () 2. Preliminaries Definition. ([6]) The Riemann-Liouville fractional derivative of order α > of a continuous function y : (, ) R is given by D α + y(t) = Γ(n α) ( d dt )n t y(s) (t s) α n+ds, provided that the right side is pointwise defined on (, ), where n = [α] +, [α] denotes the integer part of the number α. Lemma 2. ([]) Let h(t) [,] be a given function. Then the boundary value problem { D α + u(t)+h(t) =, < t <, 2 < α 3, u() = u () = u () =, has a unique solution where G(t,s) = u(t) = G(t, s)h(s)ds, { t α ( s) α 2 (t s) α Γ(α), s t, t α ( s) α 2 Γ(α), t s. Lemma 3. ([]) The function G(t,s) defined by (2) satisfies (2) (i) G(t,s) >, t,s (,);

3 POSITIVE SOLUTIONS FOR BOUNDARY VALUE PROBLEM... 7 (ii) G(t,s) max G(t,s) = G(,s), t,s [,]; t (iii) min G(t,s) ( 4 )α G(,s), s (,). 4 t 3 4 Note that ( 4 )α 6, so we have min G(t,s) 6G(,s) for s (,). 4 t 3 4 Let = ([ τ,],r) beabanach spacewithanorm ϕ = sup ϕ(θ) τ θ and + = {ϕ ;ϕ(θ),θ [ τ,]}. Define E = {t [,] : 4 t+θ 3 4, τ θ } = [ 4 +τ, 3 4 ]. We assume the following: (A ) f is a nonnegative continuous functional defined on +, and f maps bounded set in + into bounded set in R. (A 2 ) r(t) is a nonnegative measurable function defined on [,], and satisfies < G(, s)r(s)ds < G(, s)r(s)ds < +. E We would mention that r(t) is allowed to be zero on some subset of E and has singularity at points t = and t =. Suppose that u(t) is a solution of BVP(), then u(t) = { G(t,s)r(s)f(u s)ds, t, φ(t), τ t. Suppose that x(t) is the solution of BVP() with f, then {, t, x(t) = φ(t), τ t. (3) (4) Let x(t) = u(t) x(t), then we have from (3)and(4) that { x(t) = G(t,s)r(s)f(x s + x s )ds, t,, τ t. Let K be a cone in Banach space [ τ,] defined by K = {x [ τ,] x(t) =,t [ τ,]; min x(t) 6 x ; 4 t 3 4

4 72 L. Song x(t) >,t (,)}, where x := max{ x(t) : τ t }. Define T : K [ τ,] as (Tx)(t) = { G(t,s)r(s)f(x s + x s )ds, t,, τ t. (5) We define x [,] = sup{ x(t) ; t }, then we have x = x [,] and Tx = Tx [,] for any x K. For < t <, x K, we obtain from (5) and Lemma 3 that (Tx)(t) >, and Tx max G(t,s)r(s)f(x s + x s )ds, t min 4 t 3 4 (T x)(t) = min 4 t Tx. G(t,s)r(s)f(x s + x s )ds max G(t,s)r(s)f(x s + x s )ds t For τ t, x K, one has (Tx)(t) =. That is T(K) K. Then we have the following lemma. Lemma 4. The operator T : K K is completely continuous. Proof. We can obtain the continuity of T from the continuity of f. In face, suppose x (n),x K and x (n) x as n, then we get x (n) s x s = sup x (n) (s+θ) x(s+θ), s [,]. τ θ Thus, for t [ τ,] we have from (5) and Lemma 3 that (Tx (n) )(t) (Tx)(t) max s f(x(n) s + x s ) f(x s + x s ) G(, s)r(s)ds. This implies that Tx (n) Tx as n. Now let Ω K be a bounded subset of K and M > is the constant such that x M for x Ω. Suppose that x = M 2, then x + x

5 POSITIVE SOLUTIONS FOR BOUNDARY VALUE PROBLEM M + M 2 =: M for x Ω. Define a set S = {ϕ + ; ϕ M}. Let L = max f(ϕ)+, then we have t,ϕ S Tx L G(, s)r(s)ds < +, which implies the boundedness of T(Ω). Furthermore, we have for t, (Tx) (t) = t α 2 Γ(α ) Γ(α ) t ( s) α 2 r(s)f(x s + x s )ds (t s) α 2 r(s)f(x s + x s )ds. So, (Tx) (t) 2L Γ(α ) ( s)α 2 r(s)ds =: L For τ < t <, we have (Tx) (t) =. Thus with x Ω, ǫ >, let δ = ǫ L, for t,t 2 [ τ,], t t 2 < δ, we get (Tx)(t ) (Tx)(t 2 ) L t t 2 < ǫ. By means of the Arzela-Ascoli theorem, T : K K is completely continuous. Lemma 5. ([2]) Let X be a Banach space, and let K X be a cone in X. Assume that Ω,Ω 2 are open subsets of X with Ω Ω Ω 2, and let T : K K be a completely continuous operator such that, either (I) Tx x, x K Ω and Tx x, x K Ω 2, or (II) Tx x, x K Ω and Tx x, x K Ω 2 holds. Then T has a fixed point in K (Ω 2 \Ω ). 3. Main Results For convenience, we introduce the following notations. f = f(ϕ) f(ϕ) f = limsup, f = limsup, ϕ ϕ ϕ + ϕ f(ϕ) f(ϕ) liminf, f = liminf, ϕ, ϕ ϕ ϕ, ϕ + ϕ

6 74 L. Song A = 3 [ G(,s)r(s)ds], B = [ G(,s)r(s)ds], 6 E p = max φ(t), Ω α = {x [ τ,]; x < α}. τ t In the next, we let = {ϕ + ; < γ ϕ ϕ(θ),θ [ τ,]}, where < γ 6 is a constant. Theorem 6. Assume that one of the following conditions is satisfied: (H ) f < A, f > Bγ (particularly, f =, f = + ), φ(t) ; (H 2 ) f > Bγ, f < A (particularly, f = +, f = ). Then BVP () has at least one positive solution. Proof. Suppose that (H ) is satisfied. By φ(t), we know x t =,t [,]. From f < A, there is a ρ > such that f(ϕ) A ϕ, ϕ +, ϕ ρ. For any x K, x = ρ, we deduce that x s x = ρ,s [,] and thus (Tx)(t) A x s G(, s)r(s)ds A x G(,s)r(s)ds < x, ( t ) which implies Tx = Tx [,] < x, x K Ω ρ. Since f > Bγ, there exists a ρ 2 > ρ such that f(ϕ) Bγ ϕ, ϕ, ϕ > γ ρ 2. For any x K, x = ρ 2, we have γ x s γ x x min x(t) x(t), t [ 6 4, 3 ], s E. 4 4 t 3 4 which implies that x s for s E and x s γ x = γ ρ 2, s E. (6) Note that when s E we have x s =. Thus, we obtain (Tx)( 2 ) = G( 2,s)r(s)f(x s)ds 6 E G(,s)r(s)f(x s)ds 6 Bγ x s E G(,s)r(s)ds γ ρ 2 E G(,s)r(s)ds = ρ 2 = x, 6 Bγ which leads to Tx x, x K Ω ρ2. According to the first part of Lemma 5, it follows that T has a fixed point x K (Ω ρ2 \ Ω ρ ) such that < ρ x ρ 2. (7)

7 POSITIVE SOLUTIONS FOR BOUNDARY VALUE PROBLEM Now, suppose that (H 2 ) is satisfied. Since f > Bγ, there is a ρ > such that f(ϕ) Bγ ϕ,ϕ, ϕ ρ. For x K, x = ρ, we have x s x = ρ,s [,]. Furthermore, by a similar argument as (6), we have x s, x s γ x = γ ρ, s E. Thus, we have an analogous result to (7): (Tx)(t ) ρ = x, which implies that Tx = Tx [,] x, x K Ω ρ. On the other hand, since f < A, there is a N > ρ such that f(ϕ) A ϕ,ϕ +, ϕ > N. hoose a positive constant ρ 2 such that ρ 2 > max{p,3max{f(ϕ); ϕ N +p } G(, s)r(s)ds}. For x K, x = ρ 2, we have from the facts: x(t),x(t),t [ τ,], that for s [,], Thus, we have x s + x s x s > N, x s > N; x s + x s x s + x s N + x, x s N. (Tx)(t) + x s >N x s N A( x + x ) G(,s)r(s)f(x s + x s )ds G(,s)r(s)f(x s + x s )ds G(, s)r(s)ds + max{f(ϕ); ϕ N +p } G(, s)r(s)ds < 3 x + 3 p + 3 ρ 2 < ρ 2 = x, t, which implies that Tx < x, x K Ω ρ2. By the second part of Lemma 5, it follows that T has a fixed point x K (Ω ρ2 \Ω ρ ) such that < ρ x ρ 2. Suppose that x(t) is the fixed point of T in K (Ω ρ2 \Ω ρ ), then { x(t) = G(t,s)r(s)f(x s + x s )ds, t,, τ t. Let u(t) = x(t) + x(t). Since < ρ x = x [,] ρ 2, x(t) K and x(t), we conclude that u(t) is a positive solution of BVP(). This completes the proof.

8 76 L. Song Theorem 7. If the following conditions are satisfied: (H 3 ) f > Bγ, f > Bγ (particularly, f = +, f = + ); (H 4 ) p > such that for ϕ, ϕ p+p, one has f(ϕ) < ηp, where η = ( G(,s)r(s)ds). Then BVP ()has at least two positive solution u, u 2 K such that < u [,] < p < u 2 [,]. Proof. Since f > Bγ, there exists a < R < p such that f(ϕ) Bγ ϕ, ϕ, < ϕ R. For x K, x = R, similar to (6) one has x s, R x s γ x = γ R, s E. Hence, we obtain an analogous inequality: Tx x, x K Ω R. Since f > Bγ, there is a R 2 > p such that f(ϕ) Bγ ϕ, ϕ, ϕ γ R 2. For x K, x = R 2, similar to (6) one has x s, R 2 x s γ x = γ R 2, s E. Hence, we obtain an analogous inequality: Tx x, x K Ω R2. Now, by (H 4 ), for x K, x = p, since x s + x s x s + x s p+p, we have (Tx)(t) ηp G(,s)r(s)f(x s + x s )ds G(,s)r(s)ds = p = x, t, which implies that Tx x, x K Ω p. According to Lemma 5, it follows that T has two fixed points x, x 2 such that x K (Ω p \Ω R ),x 2 K (Ω R2 \ Ω p ), that is < x < p < x 2. Since x i (t) K, we have x i (t) >, t (,),i =,2. Let u (t) = x (t)+ x(t),u 2 (t) = x 2 (t)+ x(t), then u, u 2 are positive solutions of BVP() satisfying < u [,] < p < u 2 [,]. This completes the proof. Similarly, we have the following result. Theorem 8. If the following conditions are satisfied: (H 5 ) f < A, f < A (particularly, f =, f = ), φ(t) ; (H 6 ) p > such that for ϕ, γ p ϕ p, one has f(ϕ) Bp. Then BVP ()has at least two positive solution u u 2 K such that < u [,] < p < u 2 [,].

9 POSITIVE SOLUTIONS FOR BOUNDARY VALUE PROBLEM An Example Example 9. onsider the BVP D 5 2 +u(t)+(+t 2 )u 2(t 3 ) =, t, u(t) = sinπt, 3 t, u() = u () = u () =. (8) Then τ = 3,f(ϕ) = ϕ 2( 3 ), E = [ 7 2, 3 4 ]. As ϕ + we have 2 f(ϕ) = ϕ 2( 3 ) ϕ = ϕ 2 ϕ ϕ ϕ, that is to say that f = holds. On the other hand, suppose ϕ, then ϕ(θ) γ ϕ, thus, as ϕ, we get f(ϕ) = ϕ 2( 3 ) γ 2 ϕ 2 = γ 2 ϕ ϕ ϕ ϕ 2 +, which means that f = + holds. According to Theorem 6, it follows that BVP (8) has at least a positive solution u(t). Example. onsider the BVP D 5 2 +u(t)+c[u 2(t 3 2(t )+u3 3 )] =, t, u(t) = φ(t), 3 t, u() = u () = u () =. where c > is a constant, φ(t) is continuous on [ 3,], φ(t),φ() = and f(ϕ) = ϕ 2( 3 ) + 2( ϕ3 3 ), τ = 3, E = [ 7 2, 3 4 ]. Suppose ϕ, then ϕ(θ) γ ϕ for θ [ 3,] thus, as ϕ or ϕ +, we get (9) f(ϕ) ϕ = ϕ 2( 3 2( )+ϕ3 3 ) ϕ γ 2 ϕ 2 +γ 3 2 ϕ 3 2 ϕ = γ 2 ϕ 2 +γ 3 2 ϕ 2 +, Deducingη = ( G(,s)r(s)ds) = 45 π 6c, then p > and ϕ p+p, we have f(ϕ) (p+p ) 2 +(p+p ) 3 2 = (p+p ) 2(+ p + p p )p.

10 78 L. Song Define H(p) = (p+p ) 2(+ p + p p ), then lim H(p) = +, lim H(p) = +. () p p + Suppose that c and p satisfy p (2+ p ) < 45 π 6, then H(p 2c ) = 2 p (2+ p ) < 45 π 6c = η holds. By the continuity of H(p) and (), we can find a p > such that f(ϕ) < ηp for ϕ p + p. By Theorem 7 we know that BVP (9) has at least two positive solutions. References [] K.S. Miller, B. Ross, An introduction to the fractional calculus and fractional differential equation,wiley-interscience, USA (993). [2] K.B. Oldham, J. Spanier, The fractional alculus, Academic Press, USA (974). [3] I. Podlubny, Fractional differential equations, Mathematics in Science and Engineering, Academic Press, USA (999). [4] V. Lakshmikantham, S. Leela, J. Vasundhara Devi, Theory of fractional Dynamic systems, ambridge Academic Publishers, ambridge (29). [5] V. Lakshmikantham, Theory of fractional functional differential equations, Nonlinear Analysis, 69 (28), [6] Z.B. Bai, H.S. Lü, Positive solutions for boundary value problem of nonlinear fractional differential equation, J. Math. Anal. Appl., 3 (25), [7] Z.B. Bai, T.T. Qiu, Existence of positive solution for singular fractional differential equation, Appl. Math. omput., 25 (29), [8] X. Liu, M. Jia, Multiple solutions for fractional differential equations with nonlinear boundary conditions, omput. Math. Appl., 59 (2), [9] S. Zhang, Positive solutions to singular boundary value problem for nonlinear fractional differential equation, omput. Math. Appl., 59 (2), 3-39.

11 POSITIVE SOLUTIONS FOR BOUNDARY VALUE PROBLEM [] L.M. Song, Existence of positive solutions to boundary value problem for a nonlinear fractional differential equations, Journal of South hina Normal University: Natural Science Edition, 44 (22), [] L.M. Song, P.X. Weng, Existence of positive solutions for boundary value problem of fourth-order nonlinear functional differential equation, Applied Mathematics a Journal of hinese Universities (Ser.A), 26 (2), [2] M.A. Krasnosel skii, Positive solutions of operator equations, P. Noordhoff, Netherlands (964).

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