Existence of solutions for a coupled system of. fractional differential equations

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1 Existence of solutions for a coupled system of fractional differential equations Zhigang Hu, Wenbin Liu, Wenjuan Rui Department of Mathematics, China University of Mining and Technology, Xuzhou 228, PR China Abstract. In this paper, by using the coincidence degree theory, we consider the following Neumann boundary value problem for a coupled system of fractional differential equations D α u(t) = f(t, v(t), v (t)), t (, ), + D β v(t) = g(t, u(t), u (t)), t (, ), + u () = u () =, v () = v () =, where D α +, D β + are the standard Caputo fractional derivative, < α 2, < β 2. A new result on the existence of solutions for above fractional boundary value problem is obtained. Keywords: Fractional differential equation; Neumann boundary conditions; Coincidence degree theory; MR Subject Classification: 34B5. Introduction In recent years, the fractional differential equations have received more and more attention. The fractional derivative has been occurring in many physical applications such as a non-markovian dif- This research was supported by the National Natural Science Foundation of China (27364) and the Fundamental Research Funds for the Central Universities (2LKSX9). Corresponding author. Telephone number: (86-56) (W. Rui). Fax number: (86-56) (W. Rui). addresses: wenjuanrui@26.com (W. Rui).

2 fusion process with memory [], charge transport in amorphous semiconductors [2], propagations of mechanical waves in viscoelastic media [3], etc. Phenomena in electromagnetics, acoustics, viscoelasticity, electrochemistry and material science are also described by differential equations of fractional order (see [4]-[9]). Recently, boundary value problems for fractional differential equations have been studied in many papers (see []-[2]). Since Bai et al. considered existence of a positive solution for a singular coupled system of nonlinear fractional differential equations in 24 (see [2]), the coupled system of fractional differential equations have been studied in many papers (see [22]-[3]). In 2, Zhang et al. (see [29]) considered existence results for a coupled system of nonlinear fractional three-point boundary value problems at resonance by using the coincidence degree theory due to Mawhin. In 22, by using the coincidence degree theory, a coupled system of nonlinear fractional 2m-point boundary value problems at resonance have been studied by W. Jiang (see [3]). To the best of our knowledge, the existence of solutions for a coupled system of Neumann boundary value problems has rarely been studied. Motivated by the work above, in this paper, we consider the following Neumann boundary value problem (NBVP for short) for a coupled system of fractional differential equations given by D α u(t) = f(t, v(t), v (t)), t (, ), + D β v(t) = g(t, u(t), u (t)), t (, ), (.) + u () = u () =, v () = v () =, where D α, D β + are the standard Caputo fractional derivative, < α 2, < β 2 and f, g : + [, ] R 2 R is continuous. The rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions and lemmas. In Section 3, we establish a theorem on existence of solutions for NBVP (.) under nonlinear growth restriction of f and g, basing on the coincidence degree theory due to Mawhin (see [3]). Finally, in Section 4, an example is given to illustrate the main result. 2 Preliminaries In this section, we will introduce some notations, definitions and preliminary facts which are used throughout this paper. Let X and Y be real Banach spaces and let L : doml X Y be a Fredholm operator with 2

3 index zero, and P : X X, Q : Y Y be projectors such that ImP = KerL, KerQ = ImL, X = KerL KerP, Y = ImL ImQ. It follows that L doml KerP : doml KerP ImL is invertible. We denote the inverse by K P. If Ω is an open bounded subset of X, and doml Ω, the map N : X Y will be called L compact on Ω if QN(Ω) is bounded and K P (I Q)N : Ω X is compact, where I is identity operator. Lemma 2.. ([3]) Let L : doml X Y be a Fredholm operator of index zero and N : X Y L compact on Ω. Assume that the following conditions are satisfied () Lx λnx for every (x, λ) [(doml \ KerL)] Ω (, ); (2) Nx ImL for every x KerL Ω; (3) deg(qn KerL, KerL Ω, ), where Q : Y Y is a projection such that ImL = KerQ. Then the equation Lx = Nx has at least one solution in doml Ω. Definition 2.2. ([32]) The Riemann-Liouville fractional integral operator of order α > of a function x is given by I α +x(t) = Γ(α) (t s) α x(s)ds, provided that the right side integral is pointwise defined on (, + ). Definition 2.3. ([32]) Assume that x(t) is (n )-times absolutely continuous function, the Caputo fractional derivative of order α > of x is given by D α d n x(t) +x(t) = In α + dt n = Γ(n α) (t s) n α x (n) (s)ds, where n is the smallest integer greater than or equal to α, provided that the right side integral is pointwise defined on (, + ). Lemma 2.4. ([32]) Let α > and n = [ α]. If x (n ) AC[, ], then I α +Dα +x(t) = x(t) + c + c t + c 2 t c n t n where c i = x(i) () i! R, i =,, 2,..., n, here n is the smallest integer greater than or equal to α. 3

4 In this paper, we denote X = C [, ] with the norm x X = max{ x, x } and Y = C[, ] with the norm y Y = y, where x = max t [,] x(t). Then we denote X = X X with the norm (u, v) X = max{ u X, v X } and Y = Y Y with the norm (x, y) Y = max{ x Y, y Y } Obviously, both X and Y are Banach spaces. Define the operator L : doml X Y by L u = D α +u, where doml = {u X D α +u(t) Y, u () = u () = }. Define the operator L 2 : doml 2 X Y by L 2 v = D β v, + where doml 2 = {v X D β v(t) Y, v () = v () = }. + Define the operator L : doml X Y by L(u, v) = (L u, L 2 v), (2.) where doml = {(u, v) X u doml, v doml 2 }. Let N : X Y be defined as N(u, v) = (N v, N 2 u), where N : Y X N v(t) = f(t, v(t), v (t)), and N 2 : Y X N 2 u(t) = g(t, u(t), u (t)). Then NBVP (.) is equivalent to the operator equation L(u, v) = N(u, v), (u, v) doml. 4

5 3 Main result In this section, a theorem on existence of solutions for NBVP (.) will be given. Theorem 3.. Let f, g : [, ] R 2 R be continuous. Assume that (H ) there exist nonnegative functions p i, q i, r i C[, ], (i =, 2) with such that for all (u, v) R 2, t [, ], Γ(α)Γ(β) (Q + R )(Q 2 + R 2 ) Γ(α)Γ(β) > f(t, u, v) p (t) + q (t) u + r (t) v and g(t, u, v) p 2 (t) + q 2 (t) u + r 2 (t) v, where P i = p i, Q i = q i, R i = r i, (i =, 2); (H 2 ) there exists a constant B > such that for all t [, ], u > B, v R either uf(t, u, v) >, ug(t, u, v) > or uf(t, u, v) <, ug(t, u, v) < ; (H 3 ) there exists a constant D > such that for every c, c 2 R satisfying min{c, c 2 } > D either c N (c 2 ) >, c 2 N 2 (c ) > or c N (c 2 ) <, c 2 N 2 (c ) <. Then NBVP (.) has at least one solution. Now, we begin with some lemmas below. Lemma 3.2. Let L be defined by (2.). Then KerL = (KerL, KerL 2 ) = {(u, v) X (u, v) = (u(), v())}, (3.) ImL = (ImL, ImL 2 ) = {(x, y) Y ( s) α 2 x(s)ds =, ( s) β 2 y(s)ds = }. (3.2) 5

6 Proof. By Lemma 2.4, L u = D α + u(t) = has solution u(t) = u() + u ()t. Combining with the boundary value conditions of NBVP (.), one has KerL = {u X u = u()}. For x ImL, there exists u doml such that x = L u Y. By Lemma 2.4, we have Then, we have u(t) = Γ(α) u (t) = Γ(α ) (t s) α u(s)ds + u() + u ()t. By conditions of NBVP (.), we can get that x satisfies (t s) α 2 x(s)ds + u (). ( s) α 2 x(s)ds =. On the other hand, suppose x Y and satisfies ( s)α 2 x(s)ds =. Let u(t) = I α + x(t), then u doml and D α + u(t) = x(t). So that, x ImL. Then we have Similarly, we can get ImL = {x Y ( s) α 2 x(s)ds = }. KerL 2 = {v X v = v()}, ImL 2 = {y Y Then, the proof is complete. ( s) β 2 y(s)ds = }. Lemma 3.3. Let L be defined by (2.). Then L is a Fredholm operator of index zero, and the linear continuous projector operators P : X X and Q : Y Y can be defined as P (u, v) = (P u, P 2 v) = (u(), v()), Q(x, y) = (Q x, Q 2 y) = ((α ) ( s) α 2 x(s)ds, (β ) Furthermore, the operator K P : ImL doml KerP can be written by ( s) β 2 y(s)ds). K P (x, y) = (I α +x(t), Iβ + y(t)). 6

7 Proof. Obviously, ImP = KerL and P 2 (u, v) = P (u, v). It follows from (u, v) = ((u, v) P (u, v)) + P (u, v) that X = KerP + KerL. By simple calculation, we can get that KerL KerP = {(, )}. Then we get X = KerL KerP. For (x, y) Y, we have Q 2 (x, y) = Q(Q x, Q 2 y)) = (Q 2 x, Q 2 2y). By the definition of Q, we can get Q 2 x = Q x(α ) ( s) α 2 x(s)ds = Q x. Similar proof can show that Q 2 2y = Q 2 y. Thus, we have Q 2 (x, y) = Q(x, y). Let (x, y) = ((x, y) Q(x, y)) + Q(x, y), where (x, y) Q(x, y) KerQ = ImL, Q(x, y) ImQ. It follows from KerQ = ImL and Q 2 (x, y) = Q(x, y) that ImQ ImL = {(, )}. Then, we have Y = ImL ImQ. Thus dim KerL = dim ImQ = codim ImL. This means that L is a Fredholm operator of index zero. Now, we will prove that K P is the inverse of L doml KerP. In fact, for (x, y) ImL, we have LK P (x, y) = (D α +(Iα +x), Dβ + (I β + y)) = (x, y). (3.3) Moreover, for (u, v) doml KerP, we have u() =, v() = and K P L(u, v) = (I α +Dα +u(t), Iβ + D β + v(t)) = (u(t) + u() + u ()t, v(t) + v() + v ()t), which together with u () = v () = yields that K P L(u, v) = (u, v). (3.4) Combining (3.3) with (3.4), we know that K P = (L doml KerP ). The proof is complete. 7

8 Lemma 3.4. Assume Ω X is an open bounded subset such that doml Ω, then N is L-compact on Ω. Proof. By the continuity of f and g, we can get that QN(Ω) and K P (I Q)N(Ω) are bounded. So, in view of the Arzelà-Ascoli theorem, we need only prove that K P (I Q)N(Ω) X is equicontinuous. From the continuity of f and g, there exist constants A i >, i =, 2, such that for all (u, v) Ω (I Q )N v A, (I Q 2 )N 2 u A 2. Furthermore, for t < t 2, (u, v) Ω, we have K P (I Q)N(u(t 2 ), v(t 2 )) (K P (I Q)N(u(t ), v(t ))) = (I α +(I Q )N v(t 2 ), I β (I Q + 2 )N 2 u(t 2 )) (I α +(I Q )N v(t ), I β (I Q + 2 )N 2 u(t )) = (I α +(I Q )N v(t 2 ) I α +(I Q )N v(t ), I β (I Q + 2 )N 2 u(t 2 ) I β (I Q + 2 )N 2 u(t )). By and I α +(I Q )N v(t 2 ) I α +(I Q )N v(t ) t2 Γ(α) (t 2 s) α (I Q )N v(s)ds A Γ(α) = [ A Γ(α + ) (tα 2 t α ) 2 (t 2 s) α (t s) α ds + (t 2 s) α ds t (t s) α (I Q )N v(s)ds ] (I α +(I Q )N v) (t 2 ) (I α +(I Q )N v) (t ) = α 2 Γ(α) (t 2 s) α 2 (I Q )N v(s)ds (t s) α 2 (I Q )N v(s)ds [ A t 2 ] (t s) α 2 (t 2 s) α 2 ds + (t 2 s) α 2 ds Γ(α ) A Γ(α) [tα 2 t α + 2(t 2 t ) α ]. Similar proof can show that I β (I Q + 2 )N 2 u(t 2 ) I β (I Q + 2 )N 2 u(t ) 8 t A 2 Γ(β + ) (tβ 2 tβ ),

9 (I β (I Q + 2 )N 2 u) (t 2 ) (I β (I Q + 2 )N 2 u) (t ) A 2 Γ(β) [tβ 2 t β + 2(t 2 t ) β ]. Since t α, t α, t β and t β are uniformly continuous on [, ], we can get that K P (I Q)N(Ω) X is equicontinuous. Thus, we get that K P (I Q)N : Ω X is compact. The proof is complete. Lemma 3.5. Suppose (H ), (H 2 ) hold, then the set Ω = {(u, v) doml \ KerL L(u, v) = λn(u, v), λ (, )} is bounded. Proof. Take (u, v) Ω, then N(u, v) ImL. By (3.2), we have ( s) α 2 f(s, v(s), v (s))ds =, ( s) β 2 g(s, u(s), u (s))ds =. Then, by the integral mean value theorem, there exist constants ξ, η (, ) such that f(ξ, v(ξ), v (ξ)) = and g(η, u(η), u (η)) =. Then we have v(ξ)f(t, v(ξ), v (ξ)) =, u(η)g(t, u(η), u (η)) =. So, from (H 2 ), we get v(ξ) B and u(η) B. Hence That is u(t) = u(η) + u (s)ds B + u. (3.5) η u B + u. (3.6) Similar proof can show that v B + v. (3.7) and By L(u, v) = λn(u, v), we have u(t) = v(t) = λ Γ(α) λ Γ(β) (t s) α f(s, v(s), v (s))ds + u() (t s) β g(s, u(s), u (s))ds + v(). 9

10 Then we get and So, we have u (t) = v (t) = λ Γ(α ) λ Γ(β ) (t s) α 2 f(s, v(s), v (s))ds (t s) β 2 g(s, u(s), u (s))ds. u Γ(α ) Γ(α ) (t s) α 2 f(s, v(s), v (s)) ds (t s) α 2 [p (s) + q (s) v(s) + r (s) v (s) ]ds Γ(α ) [P + Q B + (Q + R ) v ] (t s) α 2 ds Γ(α) [P + Q B + (Q + R ) v ]. (3.8) Similarly, we can get and Together with (3.8) and (3.9), we have v Γ(β) [P 2 + Q 2 B + (Q 2 + R 2 ) u ]. (3.9) u Γ(α) {P + Q B + (Q + R ) Γ(β) [P 2 + Q 2 B + (Q 2 + R 2 ) u ]}. Thus, from Γ(α)Γ(β) (Q+R)(Q2+R2) Γ(α)Γ(β) > and (3.9), we obtain u Γ(β)(P + Q B) + (Q + R )(P 2 + Q 2 B) Γ(α)Γ(β) (Q + R )(Q 2 + R 2 ) Together with (3.6) and (3.7), we get So Ω is bounded. The proof is complete. v Γ(β) [P 2 + Q 2 B + (Q 2 + R 2 )M ] := M 2. (u, v) X max{m + B, M 2 + B} := M. := M Lemma 3.6. Suppose (H 3 ) holds, then the set is bounded. Ω 2 = {(u, v) (u, v) KerL, N(u, v) ImL}

11 Proof. For (u, v) Ω 2, we have (u, v) = (c, c 2 ), c, c 2 R. Then from N(u, v) ImL, we get ( s) α 2 f(s, c 2, )ds =, ( s) β 2 g(s, c, )ds =, which together with (H 3 ) imply c, c 2 D. Thus, we have (u, v) X D. Hence, Ω 2 is bounded. The proof is complete. Lemma 3.7. Suppose the first part of (H 3 ) holds, then the set Ω 3 = {(u, v) KerL λ(u, v) + ( λ)qn(u, v) = (, ), λ [, ]} is bounded. Proof. For (u, v) Ω 3, we have (u, v) = (c, c 2 ), c, c 2 R and λc + ( λ)(α ) λc 2 + ( λ)(β ) ( s) α 2 f(s, c 2, )ds =, (3.) ( s) β 2 g(s, c, )ds =. (3.) If λ =, then c, c 2 D because of the first part of (H 3 ). If λ =, then c = c 2 =. For λ (, ], we can obtain c, c 2 D. Otherwise, if c or c 2 > D, in view of the first part of (H 3 ), one has λc 2 + ( λ)(α ) or λc ( λ)(β ) ( s) α 2 c f(s, c 2, )ds >, ( s) β 2 c 2 g(s, c, )ds >, which contradict to (3.) or (3.). Therefore, Ω 3 is bounded. The proof is complete. Remark 3.8. If the second part of (H 3 ) holds, then the set Ω 3 = {(u, v) KerL λ(u, v) + ( λ)qn(u, v) = (, ), λ [, ]} is bounded. Proof of Theorem 3.. Set Ω = {(u, v) X (u, v) X < max{m, D} + }. It follows from Lemma 3.3 and Lemma 3.4 that L is a Fredholm operator of index zero and N is L-compact on Ω. By Lemma 3.5 and Lemma 3.6, we get that the following two conditions are satisfied

12 () L(u, v) λn(u, v) for every ((u, v), λ) [(doml \ KerL) Ω] (, ); (2) Nx / ImL for every (u, v) KerL Ω. Take H((u, v), λ) = ±λ(u, v) + ( λ)qn(u, v). According to Lemma 3.7 (or Remark 3.8), we know that H((u, v), λ) for (u, v) KerL Ω. Therefore deg(qn KerL, Ω KerL, (, )) = deg(h(, ), Ω KerL, (, )) = deg(h(, ), Ω KerL, (, )) = deg(±i, Ω KerL, (, )). So that, the condition (3) of Lemma 2. is satisfied. By Lemma 2., we can get that L(u, v) = N(u, v) has at least one solution in doml Ω. Therefore NBVP (.) has at least one solution. The proof is complete. 4 Example Example 4.. Consider the following NBVP D 3 2 +u(t) = t2 6 [v(t) ] + (t) 6, t [, ] e v D 5 4 +v(t) = t3 2 [u(t) 8] + 2 sin2 (u (t)), t [, ] u () = u () =, v () = v () =. (4.) Choose p (t) = /6, p 2 (t) = 3/4, q (t) = /6, q 2 (t) = /2, r (t) = r 2 (t) =, B = D =. By simple calculation, we can get that (H ), (H 2 ) and the first part of (H 3 ) hold. By Theorem 3., we obtain that the problem NBVP (4.) has at least one solution. References [] R. Metzler, J. Klafter, Boundary value problems for fractional diffusion equations, Phys. A 278 (2) [2] H. Scher, E. Montroll, Anomalous transit-time dispersion in amorphous solids, Phys. Rev. B 2 (975)

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