Nonlocal problems for the generalized Bagley-Torvik fractional differential equation Svatoslav Staněk Workshop on differential equations Malá Morávka, 28. 5. 212 () s 1 / 32
Overview 1) Introduction 2) Preliminaries 3) Formulation of our problem 4) Operators 5) Existence results 6) Uniqueness results 7) Literature () s 2 / 32
1. INTRODUCTION In modelling the motion of a rigid plate immersing in a Newtonian fluid, Torvik and Bagley (1984) considered the fractional differential equation u (t) + AD 3 2 u(t) = au(t) + ϕ(t), A, a R, A, (1) subject to the initial homogeneous conditions where u() =, u () =, (2) D 3 2 u(t) = 1 Γ( 1 2 ) d 2 dt 2 is the Riemann-Liouville fractional derivative of order 3 2. (t s) 1 2 u(s) ds In the literature equation (1) is called the Bagley-Torvik equation. Numerical solution of problem (1), (2) was given by Podlubny (1999), analytical solutions by Kilbas, Srivastava, Trujillo (26), Ray, Bera (25). () s 3 / 32
Numerical solutions of the problem u (t) + A c D α u(t) = au(t) + ϕ(t), u() = y, u () = y 1, d 2 c D α 1 u(t) = Γ(2 α) dt 2 (t s) 1 α (u(s) u() u ()s) ds, α (1, 2), (Caputo fractional derivative of order α) were discussed for α = 3 by Cenesiz, Keskin, Kurnaz (21) and by Diethelm, Ford 2 (22), and by Edwards, Ford, Simpson (22) for α (1, 2). Applying the Adomian decomposition method, analytical solutions of the above problem were obtained by Deftardar-Gejji, Jafari (25) for α (1, 2). Analytical and numerical solutions of the boundary value problem u (t) + A c D 3 2 u(t) = au(t) + ϕ(t), u() = y, y(t ) = y 1. were discussed by Al-Mdallal, Syam, Anwar (21). Wang, Wang (21) investigated general solutions of the equations u (t) + A c D 3 2 u(t) + u(t) =, u (t) + AD 2 3 u(t) + u(t) =. () s 4 / 32
Existence and uniqueness results for the generalized Bagley-Torvik fractional differential equation u (t) + A c D α u(t) = f (t, u(t), c D µ u(t), u (t)), A R \ {}, subject to the boundary conditions u () =, u(t ) + au (T ) =, a R, where α (1, 2), µ (, 1), f Car([, T ] R 3 ) were given by S.S. (212). Note that c D α 1 d 2 u(t) = Γ(2 α) dt 2 c D µ 1 d u(t) = Γ(1 µ) dt (t s) 1 α (u(s) u() u ()s) ds, α (1, 2), (t s) µ (u(s) u()) ds, µ (, 1). () s 5 / 32
2. PRELIMINARIES The Riemann-Liouville fractional integral I γ v of v : [, T ] R of order γ > is defined as Properties of fractional integral: I γ v(t) = 1 Γ(γ) (t s) γ 1 v(s) ds I γ : C[, T ] C[, T ], I γ : L 1 [, T ] L 1 [, T ], γ (, 1), I γ : AC[, 1] AC[, 1], γ (, 1), I γ : L 1 [, T ] AC[, T ], γ [1, 2), I β I γ v(t) = I β+γ v(t) for t [, T ], where v L 1 [, T ], β, γ >, β + γ 1 (semigroup property) d dt I γ+1 v(t) = I γ v(t) for a.e. t [, T ], where v L 1 [, T ] and γ >. () s 6 / 32
LEMMA 1. Let w C[, 1], b C 1 [, T ], α (1, 2) and let ϕ 1 AC[, 1] be such that ϕ 1 () =. Suppose that w(t) = b(t)i 2 α w(t) + ϕ 1 (t) for t [, 1]. Then for each n N there exists ϕ n AC[, 1] such that ϕ n () = and the equality w(t) = b n (t)i n(2 α) w(t) + ϕ n (t) for t [, 1] holds. COROLLARY. Let the assumptions of Lemma 1 hold. Then w AC[, 1]. Proof. Choose n N such that n(2 α) > 2. Then I n(2 α) w = I 1 I n(2 α) 1 w C 1 [, 1]. Since w(t) = b n (t)i n(2 α) w(t) {z } C 1 [,T ] + ϕ n(t) for t [, 1], we have w AC[, 1]. {z } AC[,T ] () s 7 / 32
The following result is a generalization of the Gronwall lemma for integrals with singular kernels (Henry (1989)). LEMMA 2. Let < γ < 1, b L 1 [, T ] be nonnegative and let K be a positive constant. Suppose w L 1 [, T ] is nonnegative and w(t) b(t) + K Then w(t) b(t) + LK t t where L = L(γ) is a positive constant. (t s) γ 1 w(s) ds for a.e. t [, T ]. (t s) γ 1 b(s) ds for a.e. t [, T ], L = KΓ(γ)E γγ(kγ(γ) max{1, T }), X x n E βγ (x) = Mittag-Leffler function Γ(nβ + γ) n= () s 8 / 32
Fractional derivatives The Caputo fractional derivative c D β v of order β >, β N, of x : [, T ] R is defined by c D β 1 x(t) = Γ(n β) dt n d n Xn 1 (t s) n β 1 x(s) where n = [β] + 1 and where [β] means the integral part of β. k=! x (k) () s k ds, k! The Riemann-Liouville fractional derivative D β v of v : [, T ] R of order β > is given by where n = [β] + 1. D β 1 x(t) = Γ(n β) dt n d n (t s) n β 1 x(s) ds «= dn dt I n β x(t), n () s 9 / 32
3. FORMULATION OF OUR PROBLEM Let A be the set of functionals φ : C[, 1] R which are (i) continuous, (ii) lim c R, c ± φ(c) = ±, (We identify the set of constant functions on [, 1] with R) (iii) there exists a positive constant L = L(φ) such that x C[, 1], x(t) > L for t [, 1] φ(x). (φ A, φ(x) = for some x C[, 1] ξ [, 1] : x(ξ) L) EXAMPLE. Let p, g j C(R), p be bounded, lim v ± g j (v) = ±, j =, 1,..., n, and let t 1 < t 2 < < t n 1. Then the functionals «belong to the set A. φ 1(x) = g max x(t) t [,1] φ 3(x) = p( x ) + Z 1 «, φ 2(x) = g min x(t), t [,1] nx g j (x(t j )) g (x(t)) dt, φ 4(x) = () s 1 / 32 j=1
LEMMA 3. Let φ A. Then there exists a positive constant L = L(φ) such that the estimate c < L holds for each λ > and each solution c R of the equation λφ(c) φ( c) =. () s 11 / 32
We investigate the Bagley-Torvik fractional functional differential equation u (t) + a(t) c D α u(t) = (Fu)(t) (3) together with the nonlocal boundary conditions u () =, φ(u) =, (φ A). (4) Here α (1, 2), c D is the Caputo fractional derivative, a C 1 [, 1] and F : C 1 [, 1] L 1 [, 1] is continuous. Note that c D α 1 d 2 u(t) = Γ(2 α) dt 2 (t s) 1 α (u(s) u() u ()s) ds, α (1, 2) We say that a function u AC 1 [, 1] is a solution of problem (3), (4) if u satisfies (4) and (3) holds for a.e. t [, 1]. () s 12 / 32
We work with the following conditions on the function a and the operator F in (3). (H 1) a C 1 [, 1] and a(t) for t [, 1], (H 2) F : C 1 [, 1] L 1 [, 1] is continuous and for a.e. t [, 1] and all x C 1 [, 1], the estimate (Fx)(t) ϕ(t)ω( x + x ) holds, where ϕ L 1 [, 1] and ω C[, ) are nonnegative, ω is nondecreasing and ω(v) lim =. v v () s 13 / 32
4. OPERATORS In order to prove the solvability of problem (3), (4), we define an operator F acting on [, 1] C 1 [, 1] R by the formula F(λ, x, c) = (F 1(λ, x, c), F 2(x, c)), where and F 1(λ, x, c)(t) = c + F 2(x, c) = c φ(x), (Qx)(s) ds + λ (Qx)(t) = a(t)i 2 α x (t) + (t s)(fx)(s) ds, a (s)i 2 α x (s) ds. (5) Here the function a and the operator F are from equation (3) and φ A is from the boundary conditions (4) () s 14 / 32
Properties of Q, F 1 and F 2 Let (H 1) hold. Then Q : C 1 [, 1] C[, 1] and Q is completely continuous. Let (H 1) and (H 2) hold. Then F 1 : [, 1] C 1 [, 1] R C 1 [, 1] and F 1 is completely continuous. Let φ A. Then F 2 : C 1 [, 1] R R and F 2 is completely continuous. F 1(λ, x, c)(t) = c + F 2(x, c) = c φ(x), (Qx)(s) ds + λ (Qx)(t) = a(t)i 2 α x (t) + (t s)(fx)(s) ds, a (s)i 2 α x (s) ds. () s 15 / 32
LEMMA 4. Let (H 1 ) and (H 2 ) hold. Then (a) F : [, 1] C 1 [, 1] R C 1 [, 1] R and F is completely continuous, (b) if (x, c) is a fixed point of F(1,, ), then x is a solution of problem (3), (4) and c = x(). Proof. (b) Let (x, c) be a fixed point of F(1,, ). Then x C 1 [, 1], x(t) = c + (Qx)(s) ds + and φ(x) =. Differentiating (6) gives x (t) = a(t)i 2 α x (t) + (t s)(fx)(s) ds, t [, 1], (6) a (s)i 2 α x (s) ds + (Fx)(s) ds, t [, 1]. (7) Therefore, x () =, and so x satisfies the boundary conditions (4). Since R t a (s)i 2 α x (s) ds C 1 [, 1] and R t (Fx)(s) ds AC[, 1], (7) shows that x (t) = a(t)i 2 α x (t) + ψ(t), t [, 1], where ψ AC[, 1] and ψ() =. Hence, by Corollary, x AC[, 1]. It follows from the R.-L. factional integrals that I 2 α x AC[, 1]. () s 16 / 32
Next we have» d 1 x (t) + dt a(t) = (Fx)(t) x (t) a(t) a (s)i 2 α x (s) ds + L 1 [, 1] for a.e. t [, 1]. «(Fx)(s) ds Since, by (7), the equality I 2 α x (t) = 1 a(t) holds for t [, 1], we have x (t) + a (s)i 2 α x (s) ds + «(Fx)(s) ds d dt I 2 α x (t) = (Fx)(t) x (t) a(t) for a.e. t [, 1]. Consequently, x (t) + a(t) d dt I 2 α x (t) = (Fx)(t) for a.e. t [, 1]. {z } c D α x(t) () s 17 / 32
Since I 2 α x (t) = I 3 α x (t) = I 1 I 2 α x (t), we have d dt I 2 α x (t) = I 2 α x (t) a.e. on [, 1]. Since x L 1 [, 1], it follows that c D α x(t) = I 2 α x (t) for a.e. t [, 1]. Hence d dt I 2 α x (t) = c D α x(t) a.e. on [, 1], and therefore x is a solution of (3). As a result x is a solution of problem (3), (4), and (6) gives c = x(). () s 18 / 32
LEMMA 5. Let (H 1) and (H 2) hold. Then there exists a positive constant S such that for each λ [, 1] and each fixed point (x, c) of the operator F(λ,, ) the estimate x < S, x < S, c < S holds. () s 19 / 32
5. EXISTENCE RESULTS We need the following result (Deimling (1985)). LEMMA 6. Let X be a Banach space and let Ω X be open bounded and symmetric with respect to Ω. Let F : Ω X be a compact operator and G = I F, where I is the identical operator on X. If x Fx for x Ω and G( x) λg(x) on Ω for all λ 1, then deg(i F, Ω, ). () s 2 / 32
THEOREM 1. Let (H 1) and (H 2) hold. Then problem (3), (4) has at least one solution. Proof. We have to show that F(1,, ) has a fixed point (x, c). Then x is a solution of problem (3), (4) and c = x(). Let S be a positive constant from Lemma 5 and let L = L(φ) be from Lemma 3 (note that c < L holds for each λ > and each solution c R of λφ(c) φ( c) = ). Let W = max{s, L} and Ω = {(x, c) C 1 [, 1] R : x < W, x < W, c < W }. We prove by Lemma 6 that deg{i F(,, ), Ω, } =, where I is the identical operator on C 1 [, 1] R. Note that G(x, c) = (x, c) F(, x, c) = x(t) c «(Qx)(s) ds, φ(x). Let (x, c) be a fixed point of F(λ,, ) for some λ [, 1]. Then, by Lemma 5, (x, c) Ω, and therefore, by the homotopy property, deg(i F(1,, ), Ω, ) = deg(i F(,, ), Ω, ). Hence deg(i F(1,, ), Ω, ). The last relation implies that F(1,, ), Ω, ) has a fixed point. () s 21 / 32
EXAMPLE. Let ϕ 1, ϕ 2 L 1 h(v) [, 1], h C[, ), p C(R), lim v = and v =. Define an operator F : C 1 [, 1] L 1 [, 1] by lim v p(v) v (Fx)(t) = ϕ 1(t) h( x ) + p(x(s)) ds «+ ϕ 2(t). Then F satisfies condition (H 2). To check it we take ϕ(t) = ϕ 1(t) + ϕ 2(t) and ω(v) = h(v) + p(v), where h(v) = max{h(ν) : ν v}, p(v) = max{p(ν) : ν v}, v [, ). () s 22 / 32
The special case of (3) is the fractional differential equation u (t) + a(t) c D α u(t) = f (t, u(t), c D γ u(t), u (t)), (8) where α (1, 2), γ (, 1) and f satisfies the condition (H 3) f Car([, 1] R 3 ) and for a.e. t [, 1] and all (x, y, z) R 3 the estimate f (t, x, y, z) ϕ(t)ρ( x + y + z ) holds, where ϕ L 1 [, 1] and ρ C[, ) are nonegative, ρ is nondecreasing and lim v ρ(v) v =. () s 23 / 32
The following theorem gives an existence result for problem (8), (4). THEOREM 2. Let (H 1) and (H 3) hold. Then problem (8), (4) has at least one solution. Proof. Let F be an operator acting on C 1 [, 1] and given by (Fx)(t) = f (t, x(t), c D γ x(t), x (t)). F satisfies condition (H 2) for ω(v) = ρ follows from Theorem 1. 2v Γ(2 γ). The solvability of problem (8), (4) () s 24 / 32
6. UNIQUENESS RESULTS Let B be the set all functionals φ : C[, 1] R which are (i) continuous, (ii) increasing, that is, x, y C[, 1] x(t) < y(t) for t [, 1] φ(x) < φ(y). EXAMPLE. Let g j C(R) be increasing (j =, 1,..., n), and let t t 1 < < t n 1. Then the functionals «φ 1(x) = g max t [,1] x(t) φ 3(x) = Z 1, φ 2(x) = g min t [,1] x(t) g (x(t)) dt, φ 4(x) = «, nx g j (x(t j )) j=1 belong to B. () s 25 / 32
We discuss equation (8), where f (t, x, y, z) = ϕ(t)p(t, x, y, z), that is, the equation u (t) + a(t) c D α u(t) = ϕ(t)p(t, u(t), c D γ u(t), u (t)), (9) where α (1, 2), γ (, 1). Together with (9) the boundary conditions u () =, φ(u) =, (φ B) (1) and u () =, φ(u) =, (φ A B) (11) equation are investigated. () s 26 / 32
u (t) + a(t) c D α u(t) = ϕ(t)p(t, u(t), c D γ u(t), u (t)) THEOREM 3. Let (S 1) a C 1 [, 1], ϕ L 1 [, 1] are such that a <, a on [, 1] and ϕ > a.e. on [, 1], (S 2) p C([, 1] R 3 ) and p(t, x, y, z) is increasing in the variable x and nondecreasing in the variables y and z, (S 3) the exists κ > such that for each ρ R the estimate p(t, ρ + x 1, y 1, z 1) p(t, ρ + x 2, y 2, z 2) k ρ( x 1 x 2 + y 1 y 2 + z 1 z 2 ) holds for x j, y j, z j [ κ, κ], where k ρ C[, 6κ], k ρ is nondecreasing and lim sup v + k ρ(v) v <, hold. Then problem (9), (1) has at most one solution. () s 27 / 32
EXAMPLE. Let q 1 C 1 (R), q 2, q 3 C(R) C 1 [ 1, 1], q 1 be increasing and q 2, q 3 be nondecreasing. Let p j C([, 1] R 2 ) (j = 1, 2, 3) be positive and bounded. Then the function p(t, x, y, z) = p 1(t, y, z)q 1(x) + p 2(t, x, z)q 2(y) + p 3(t, x, y)q 3(z) satisfies conditions (S 2) and (S 3) with κ = 1. () s 28 / 32
THEOREM 4. Let (S 1) (S 3) and (S 4) for t [, 1] and (x, y, z) R 3 the estimate p(t, x, y, z) h( x + y + z ) is fulfilled, where h C[, ), h is nondecreasing and h(v) lim =. v v hold. Then problem (9), (11) has a unique solution. EXAMPLE. Let q 1 C 1 (R), q 2, q 3 C(R) C 1 [ 1, 1], q 1 be increasing and q 2, q 3 be nondecreasing. Let p j C([, 1] R 2 ) (j = 1, 2, 3) be positive and bounded. Besides, lim v 1 v max{ q j( v), q j (v) } = for j = 1, 2, 3. Then the function p(t, x, y, z) = p 1(t, y, z)q 1(x) + p 2(t, x, z)q 2(y) + p 3(t, x, y)q 3(z) satisfies conditions (S 2) (S 4). () s 29 / 32
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