Integral boundary value problem for nonlinear differential equations of fractional order on an unbounded domain∗† Guotao Wanga , Alberto Cabadab , Lihong Zhanga,‡ a School of Mathematics and Computer Science, Shanxi Normal University, Linfen, Shanxi 041004, People’s Republic of China. b Departamento de An´alise Matem´atica, Facultade de Matem´aticas, Universidade de Santiago de Compostela, Santiago de Compostela, Spain.
Abstract This paper investigates the existence of solutions for nonlinear fractional differential equations with integral boundary conditions on an unbounded domain. An example illustrating how the theory can be applied in practice is also included. Keywords: Fractional differential equations; unbounded solution; integral boundary conditions; semi-infinite interval; fixed point.
1
Introduction
Fractional derivatives provide an excellent tool for the description of memory and hereditary properties of various materials and processes. These characteristics of the fractional derivatives make the fractional-order models more realistic and practical than the classical integer-order models. For more details and applications, we refer the reader to the books [1]–[5]. For some recent development on the topic, see [6]–[15] and the references therein. Boundary value problems (BVPs) on infinite intervals arise naturally in the study of radially symmetric solutions of nonlinear elliptic equations and various physical phenomena [16]–[17]. For BVP of integer order on infinite intervals, there are a lot of works, see [18]–[22] and references ∗
First and third authors supported by the Natural Science Foundation for Young Scientists of Shanxi Province,China (No.2012021002-3). † Second author partially supported by FEDER and Ministerio de Educaci´ on y Ciencia, Spain, project MTM2010-15314 ‡ E-mail address:
[email protected] (G. Wang),
[email protected] (A. Cabada),
[email protected] (L. Zhang).
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therein. For BVPs of fractional order on infinite intervals, there appear some excellent results dealing with nonlinear fractional differential equations [23]–[31]. In [23], by employing Schauder’s fixed point theorem combined with the diagonalization method, Arara, Benchohra, Hamidi and Nieto successfully discussed the existence of the bounded solution of the following problem on infinite intervals: C
α D0+ y(t) = f t, y(t)
for t ∈ J = [0, +∞),
with y(0) = y0 ,
α where 1 < α ≤ 2 and y is bounded on J, for a known f ∈ C(J × R, R) and y0 ∈ R. Here, C D0+ is Caputo fractional derivative of order α. Zhao and Ge [24] used the Leray Schauder nonlinear alternative theorem to show the existence of positive solutions to the following fractional order differential equation:
α D0+ u(t) + f t, u(t) = 0
for t ∈ J,
α−1 = βu(ξ), with u(0) = 0 and lim D0+ t→+∞
α is the standard Riemann– where 1 < α ≤ 2, f ∈ C(J × R, [0, +∞)), 0 ≤ ξ, η < ∞ and D0+ Liouville fractional derivative. Inspired by the papers mentioned above, we will discuss the existence of the solutions for nonlinear fractional differential equations with integral boundary conditions on an unbounded domain. Precisely, we consider the following problem Z τ Dα u(t) + f t, u(t) = 0 for t ∈ J, with u(0) = 0 and Dα−1 u(+∞) = λ u(t)dt, (1.1) 0
where 1 < α ≤ 2, f ∈ C(J × R, R), 0 ≤ λ, τ < ∞ and Dα denotes Riemann–Liouville fractional derivative of order α. Our results assume the following conditions. (H1 ) Γ(α + 1) > λτ α . (H2 ) There exists a nonnegative measurable function b defined on J = [0, +∞) such that |f (t, x) − f (t, y)| ≤ b(t) |x − y| for t ∈ J x, y ∈ R, and Z α
+∞
(1 + tα−1 ) b(t) dt ≤ Γ(α + 1) − λτ α .
0
(H3 ) The function a(t) := |f (t, 0)| (t ∈ J) satisfies Z +∞ a(t) dt < +∞. 0
Integral boundary value problem are an important and significant branch of the theory of BVPs. Recently, integral boundary value problems of fractional differential equations on finite intervals are gaining much importance and attention; see, for instance, [32]–[42]. To the authors’ knowledge, no one has studied integral boundary value problem of fractional differential equations 1.1 on infinite intervals, and we will fill this gap in the literature. The purpose of this paper is to improve and generalize the results mentioned to some degree. 2
Remark 1.1 Problem (1.1) can be considered as some fractional models. For example, in a fractional oscillation model describing damped oscillations with viscoelastic intrinsic damping of the oscillator, Dα u(t) + b(t)u(t) = g(t), with boundary conditions u(0) = 0
and
Dα−1 u(+∞) = λ
Z
τ
u(t)dt, 0
where the integral boundary condition can be regraded as a continuous distribution of the values of the unknown function on an arbitrary finite segment of the interval. For the forthcoming analysis, we define the space X = {u ∈ C[0, +∞) : sup |u(t)|/(1 + tα−1 ) < +∞}, t∈J
equipped with the norm kukX = sup t∈J
|u(t)| . 1 + tα−1
Obviously X is a Banach space.
2
Preliminaries
For the convenience of the reader, in this section we first present some useful definitions and lemmas. Definition 2.1 ([2]) The Riemann–Liouville fractional derivative of order δ for a continuous function f is defined by Z d n t 1 δ D f (t) = ( ) (t − s)n−δ−1 f (s)ds, n = [δ] + 1, Γ(n − δ) dt 0 provided the right hand side is pointwise defined on (0, ∞). Definition 2.2 ([2]) The Riemann–Liouville fractional integral of order δ for a function f is defined as Z t 1 I δ f (t) = (t − s)δ−1 f (s)ds, delta > 0, Γ(δ) 0 provided that this integral exists. Lemma 2.1 ([43]) Let U ⊂ X be a bounded set. Then U is relatively compact in X if the following conditions hold: (i) for any u(t) ∈ U the function u(t)/(1 + tα−1 ) is equicontinuous on any compact subinterval of J. 3
(ii) for any ε > 0, there exists a constant T = T (ε) > 0 such that u(t1 ) u(t2 ) 1 + tα−1 − 1 + tα−1 < ε for any t1 , t2 ≥ T and u ∈ U . 1 2 Lemma 2.2 ([2]) Assume that u ∈ C(0, 1) ∩ L(0, 1) with a fractional derivative of order α > 0 that belongs to C(0, 1) ∩ L(0, 1). Then I0α+ D0α+ u(t) = u(t) − C1 tα−1 − C2 tα−2 − · · · − CN tα−N , for some C1 , C2 , . . . , CN ∈ R with N = [α] + 1. Remark 2.1 For α ≥ 0 and β > −1, we have tβ t(β−α) α D = and Γ(1 + β) Γ(1 + β − α)
I
α
t(β+α) tβ = . Γ(1 + β) Γ(1 + β + α)
First, we calculate an explicit expression for the Green’s function related to the associated linear problem. R∞ Lemma 2.3 Let h ∈ C([0, +∞)) with 0 h(s)ds < ∞. If 1 < α ≤ 2 and Γ(α + 1) 6= λτ α , then the fractional boundary value problem Z τ α α−1 D u(t) + h(t) = 0 for t ∈ J, with u(0) = 0 and D u(+∞) = λ u(t)dt (2.1) 0
has a unique solution Z u(t) =
+∞
G(t, s)h(s) ds, 0
where α α−1 + [Γ(α + 1) − λ(τ − s)α ]tα−1 −[Γ(α + 1) − λτ ](t − s) , Γ(α)[Γ(α + 1) − λτ α ] [Γ(α + 1) − λ(τ − s)α ]tα−1 , Γ(α)[Γ(α + 1) − λτ α ] G(t, s) = −[Γ(α + 1) − λτ α ](t − s)α−1 + Γ(α + 1)tα−1 , α] Γ(α)[Γ(α + 1) − λτ αtα−1 , Γ(α + 1) − λτ α
s ≤ t, s ≤ τ, 0 ≤ t ≤ s ≤ τ, (2.2) 0 ≤ τ ≤ s ≤ t, s ≥ t, s ≥ τ,
Proof. By Lemma 2.2, we can reduce (2.1) to the integral equation Z t 1 (t − s)α−1 h(s) ds + c1 tα−1 + c2 tα−2 , u(t) = − Γ(α) 0 with constants c1 , c2 ∈ R. Firstly, the condition u(0) = 0 implies that c2 = 0. Rτ Secondly, from (2.3) and the condition Dα−1 u(+∞) = λ 0 u(t) dt, we have Z ∞ Z τ α−1 D u(∞) = − h(s) ds + c1 Γ(α) = λ u(t) dt 0 0 Z τ λ λc1 τ α =− (τ − s)α h(s) ds + . Γ(α + 1) 0 α 4
(2.3)
(2.4)
In consequence, we deduce that Z ∞ Z τ λ α α h(s) ds − (τ − s) h(s) ds . c1 = Γ(α + 1) − λτ α 0 Γ(α + 1) 0 Substituting c1 and c2 into (2.3), we have Z ∞ (t − s)α−1 αtα−1 u(t) = − h(s) ds h(s) ds + Γ(α) Γ(α + 1) − λτ α 0 0 Z τ Z ∞ αλtα−1 (τ − s)α G(t, s)h(s) ds, − h(s) ds = Γ(α + 1) − λτ α 0 Γ(α + 1) 0 Z
t
where G(t, s) is defined by (3.6).
(2.5)
Lemma 2.4 If (H1 ) holds, then the Green’s function G(t, s) satisfies G(t, s) ≥ 0 and G(t, s) α ≤ . α−1 1+t Γ(α + 1) − λτ α
(2.6)
Proof. We see using (H1 ) that G(t, s) ≥ 0. If s ≤ t, s ≤ τ and 0 ≤ τ ≤ s ≤ t, then G(t, s) αΓ(α)tα−1 α ≤ ≤ . 1 + tα−1 Γ(α)(Γ(α + 1) − λτ α )(1 + tα−1 ) Γ(α + 1) − λτ α For s ≥ t, s ≥ τ and 0 ≤ t ≤ s ≤ τ , the conclusion is obvious.
3
Main Results This section is devoted to proving our existence result for the problem (1.1).
Theorem 3.1 If (H1 ), (H2 ) and (H3 ) hold, then the problem (1.1) has at least one solution. Proof. First, notice that from condition (H2 ) we have that |f (t, x)| ≤ a(t) + b(t) |x|,
t ∈ J, x ∈ R.
Define now the operator A by Z (Au)(t) =
∞
G(t, s)f (s, u(s))ds,
(3.1)
0
so that the problem (1.1) has a solution u if and only if u solves the operator equation u = Au. Therefore, in what follows, we will set out to verify that the operator A has a fixed point. To this end, we divide the proof in different steps: Step 1: Choose a constant R such that R∞ α 0 a(s) ds R∞ R≥ Γ(α + 1) − λτ α − α 0 b(s)(1 + sα−1 ) ds 5
and let U = { u(t) ∈ X : kukX ≤ R }. It is obvious that if u is a continuous function on [0, ∞) then Au ∈ C([0, ∞)) too. To show that A(U ) ⊂ U , let u ∈ U and t ∈ [0, ∞). We have Z ∞ |Au(t)| 1 = G(t, s)f (s, u(s)) ds α−1 α−1 1+t 1+t 0 Z ∞ α (3.2) ≤ a(s) + b(s)|u(s)| ds α Γ(α + 1) − λτ 0 Z ∞ α α−1 a(s) + b(s)(1 + s )kuk ds ≤ R. ≤ X Γ(α + 1) − λτ α 0 Hence, kAukX ≤ R and thus A(U ) ⊂ U . Step 2: A : U → U is continuous. If un , u ∈ U (n = 1, 2, . . . ) such that kun − ukX → 0 as n → +∞, then, for all t ∈ [0, ∞), Z ∞ Aun (t) Au(t) 1 G(t, s) f (s, un (s)) − f (s, u(s)) ds 1 + tα−1 − 1 + tα−1 = 1 + tα−1 0 Z ∞ α ≤ b(s) |un (s) − u(s)| ds Γ(α + 1) − λτ α 0 (3.3) Z ∞ α b(s) (1 + sα−1 ) kun − ukX ds ≤ Γ(α + 1) − λτ α 0 ≤ kun − ukX , so we conclude that kAun − AukX → 0 as n → +∞. Hence, A is continuous. Step 3: Let V be a subset of U . Following Lemma 2.1, we verify that A(V ) is a relatively compact set in two steps. First, let I ⊂ J be a compact interval, t1 , t2 ∈ I and t1 < t2 . Then for any u ∈ V , we have Z Au(t2 ) Au(t1 ) ∞ G(t2 , s) G(t1 , s) 1 + tα−1 − 1 + tα−1 = α−1 − α−1 f (s, u(s)) ds 1 + t 1 + t 0 2 1 2 1 (3.4) Z ∞ G(t2 , s) G(t , s) 1 α−1 ≤ )kukX ds. 1 + tα−1 − 1 + tα−1 a(s) + b(s)(1 + s 0 2 1 G(t,s) Since it is continuous on J × J, we have that 1+t α−1 is a uniformly continuous function on the compact set I × I. Moreover, for s ≥ t, we have that this function depends only on t, so consequently it is uniformly continuous on I × (J\I). Therefore, we have that for all s ∈ J and t1 , t2 ∈ I the following property holds: G(t2 , s) G(t , s) 1 For all ε > 0 there is δ(ε) > 0 such that if |t1 − t2 | < δ then α−1 − α−1 < ε. 1 + t2 1 + t1
This property, together with (3.4) and the fact that Z ∞ a(s) + b(s)(1 + sα−1 ) R ds < ∞, 0
imply that Au(t)/(1 + tα−1 ) is equicontinuous on I.
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(3.5)
Second, to verify condition (ii) in Lemma 2.1, we use the following property: λ(τ α − (τ − s)α ) , 0 ≤ s ≤ τ, G(t, s) 1 Γ(α) lim = × λτ α t→∞ 1 + tα−1 Γ(α + 1) − λτ α , τ ≤ s. Γ(α)
(3.6)
From this property, it is not difficult to verify that for any ε > 0 given, there exists a constant T = T (ε) > 0 such that G(t1 , s) G(t2 , s) 1 + tα−1 − 1 + tα−1 < ε for any t1 , t2 ≥ T and s ∈ J. 1 2 Now, from (3.4) and (3.5), we have that the same property holds for Au(t)/(1 + tα−1 ), uniformly for u ∈ V . Hence, A(V ) is equiconvergent at ∞. Consequently, Lemma 2.1 yields that A(V ) is relatively compact. Therefore, by Schauder’s fixed point theorem, the operator A has a fixed point in U , that is, the problem (1.1) has at least one solution in X.
4
Example
Consider the following boundary value problem on an unbounded domain: √ 4 π t ln(1 + |u(t)|) 3/2 √ + D u(t) + − arctan t e−t = 0 for t ∈ [0, +∞), 2 (1 + t)(3 + t)2 Z τ 1/2 u(0) = 0, D u(+∞) = λ u(t) dt,
(4.1)
0
where α = 3/2, λ > 0 and τ > 0. By a simple computation, we have √ √ 4 4 t [ln(1 + |x|) − ln(1 + |y|)] t |x − y| √ |f (t, x) − f (t, y)| = , ≤ 1 2 (1 + t)(3 + t) (1 + t 2 )(3 + t)2
(4.2)
for t ∈ J, x, y ∈ R. Noting that a(t) := |f (t, 0)| =
π − arctan t e−t , 2
√ 4 t √ b(t) = , (1 + t)(3 + t)2
we have √ 3 π Γ(α + 1) = , 4
Z
+∞
Z a(t) dt ≤
0
+∞
−t
πe 0
Therefore, if
√ λ
Z dt = π,
+∞
α
α−1
(1 + t 0
√ √ 4 3 π 3π 3 τ < − √ ≈ 0.598445 4 4 2
√ 4 3π )b(t) dt = √ . 4 2
(4.3)
(for instance, if λ = 1/2 and τ = 1, see figure 4) then all conditions of Theorem 3.1 are satisfied. By Theorem 3.1, the BVP (4.1) has at least one solution. 7
Figure 1: Set of (λ, τ ) ∈ R+ × R+ satisfying (4.3).
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