We study the existence of solutions for a class of fractional differential equations. Due to the singularity of the possible solutions, we introduce a new and proper concept of periodic boundary value conditions. We present Green's function and give some existence results for the linear case and then we study the nonlinear problem.
Fractional calculus is a generalization of ordinary differentiation and integration to arbitrary noninteger order. The subject is as old as the differential calculus, and goes back to time when Leibnitz and Newton invented differential calculus. The idea of fractional calculus has been a subject of interest not only among mathematicians but also among physicists and engineers. See, for instance, [1–6].
Fractional-order models are more accurate than integer-order models, that is, there are more degrees of freedom in the fractional-order models. Furthermore, fractional derivatives provide an excellent instrument for the description of memory and hereditary properties of various materials and processes due to the existence of a "memory" term in a model. This memory term insures the history and its impact to the present and future. For more details, see .
Fractional calculus appears in rheology, viscoelasticity, electrochemistry, electromagnetism, and so forth. For details, see the monographs of Kilbas et al. , Kiryakova , Miller and Ross , Podlubny , Oldham and Spanier , and Samko et al. , and the papers of Diethelm et al. [14–16], Mainardi , Metzler et al. , Podlubny et al. , and the references therein. For some recent advances on fractional calculus and differential equations, see [1, 3, 20–24].
In this paper we consider the following nonlinear fractional differential equation of the form
where is the standard Riemann-Liouville fractional derivative, is continuous, and .
This paper is organized as follows. in Section 2 we recall some definitions of fractional integral and derivative and related basic properties which will be used in the sequel. In Section 3, we deal with the linear case where is a continuous function. Section 4 is devoted to the nonlinear case.
2. Preliminary Results
In this section, we introduce notations, definitions, and preliminary facts which are used throughout this paper.
Let the Banach space of all continuous real functions defined on with the norm Define for , . Let , be the space of all functions such that which turn out to be a Banach space when endowed with the norm
By we denote the space of all real functions defined on which are Lebesgue integrable.
The Riemann-Liouville fractional primitive of order of a function is given by
provided the right side is pointwise defined on , and where is the gamma function.
For instance, exists for all , when ; note also that when , then and moreover
Let , if with , then , with If , then is bounded at the origin, whereas if with , then we may expect to be unbounded at the origin.
Recall that the law of composition holds for all
The Riemann-Liouville fractional derivative of order of a continuous function is given by
We have for all .
Let . If one assumes , then the fractional differential equation
has , as unique solutions.
From this lemma we deduce the following law of composition.
Assume that is in with a fractional derivative of order that belongs to . Then
for some .
If with and , then .
3. Linear Problem
In this section, we will be concerned with the following linear fractional differential equation:
where , and is a continuous function.
Before stating our main results for this section, we study the equation
for some .
Note that and . However, since has a singularity at for
It is easy to show that . Hence we should look for solutions, not in but in . We cannot consider the usual initial condition , but Hence, to study the periodic boundary value problem, one has to consider the following boundary condition of periodic type
From (3.3), we have
that leads to the following.
The periodic boundary value problem (3.2)–(3.4) has a unique solution if and only if
The previous result remains true even if . In this case, (3.2) is reduced to the ordinary differential equation
with the periodic boundary condition
and the condition (3.6) is reduced to the classical one:
Now, for different from , consider the homogenous linear equation
The solution is given by
Indeed, we have
since the series representing is absolutely convergent.
Using the identities
Note that the solution can be expressed by means of the classical Mittag-Leffler special functions . Indeed
The previous formula remains valid for . In this case,
which is the classical solution to the homogeneous linear differential equation
Now, consider the nonhomogeneous problem (3.1). We seek the particular solution in the following form:
It suffices to show that
Using the change of variable
Hence, the general solution of the nonhomogeneous equation (3.1) takes the form
Now, consider the periodic boundary value problem (3.1)–(3.4). Its unique solution is given by (3.26) for some . Also is in and
From (3.26), we have
which leads to
since for any , we have
Then the solution of the problem (3.1)–(3.4) is given by
Thus we have the following result.
The periodic boundary value problem (3.1)–(3.4) has a unique solution given by
For , given, is bounded on .
For , (3.1) is
and the boundary condition (3.4) is
In this situation Green's function is
4. Nonlinear Problem
In this section we will be concerned with the existence and uniqueness of solution to the nonlinear problem (1.1)–(3.4). To this end, we need the following fixed point theorem of Schaeffer.
Assume to be a normed linear space, and let operator be compact. Then either
(i)the operator has a fixed point in , or
(ii)the set is unbounded.
If is a solution of problem (1.1)–(3.4), then it is given by
where is Green's function defined in Theorem 3.2.
Define the operator by
Then the problem (1.1)–(3.4) has solutions if and only if the operator equation has fixed points.
Suppose that the following hold:
(i)there exists a constant such that
(ii)there exists a constant such that
Then the operator is well defined, continuous, and compact.
(a) We check, using hypothesis (4.3), that , for every . Indeed, for any , , we have
From the previous expression, we deduce that, if , then
Indeed, note that the integral is bounded by
A similar argument is useful to study the behavior of the last three terms of the long inequality above. On the other hand, if we denote by the second term in the right-hand side of that inequality, then it is satisfied that
and, concerning , we distinguish two cases. If is such that , then
and, if is such that , then
The first term in the right-hand side of the previous inequality clearly tends to zero as . On the other hand, denoting by the integer part function, we have
The finite sum obviously has limit zero as . The infinite sum is equal to
and its limit as is zero. Note that is bounded above by .
The previous calculus shows that , for , hence we can define .
(b) Next, we prove that is continuous.
Note that, for and for every , we have, using hypothesis (4.4),
Using the definition of , we get
Using that for , for , and for we obtain
Note that the Beta function, also called the Euler integral of the first kind,
where and , satisfies that . In particular, . On the other hand, using the change of variable , we deduce that
This proves that
Finally, we check that is compact. Let be a bounded set in .
(i) First, we check that is a bounded set in .
which implies that is a bounded set in .
(ii) Now, we prove that is an equicontinuous set in . Following the calculus in (a), we show that tends to zero as .
Then is equicontinuous in the space , where , for .
As a consequence of (i) and (ii), is a bounded and equicontinuous set in the space .
Hence, for a sequence in , has a subsequence converging to , that is,
Taking , we get
which means that , which proves that is compact.
Assume that (4.3) and (4.4) hold. Then the problem (1.1)–(3.4) has at least one solution in
Consider the set .
Let be any element of , then for some . Thus for each , we have
As in Lemma 4.2, (i), we have
which implies that the set is bounded independently of . Using Lemma 4.2 and Theorem 4.1, we obtain that the operator has at least a fixed point.
In Lemma 4.2, condition (4.3) is used to prove that the operator is continuous. Hence, in Lemma 4.2 and, in consequence, in Theorem 4.3, we can assume the weaker condition.
(i)For each fixed, there exists such that
instead of (4.3).
However, to prove the existence and uniqueness of solution given in the following theorem, we need to assume the Lipschitzian character of (condition (4.3).
Assume that (4.4) holds. Then the problem (1.1)–(3.4) has a unique solution in provided that
We use the Banach contraction principle to prove that the operator has a unique fixed point.
Using the calculus in (b) Lemma 4.2, is a contraction by condition (4.32). As a consequence of Banach fixed point theorem, we deduce that has a unique fixed point which gives rise to a unique solution of problem (1.1)–(3.4).
If , condition (4.32) is reduced to
The research of J. J. Nieto and R. Rodríguez-López has been partially supported by Ministerio de Educacion y Ciencia and FEDER, project MTM2007-61724, and by Xunta de Galicia and FEDER, project PGIDIT06PXIB207023PR.
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