Carátula del artículo
Existence and stability analysis of solutions for a new kind of boundary value problems of nonlinear fractional differential equations*
Liu Weiwei mathlls@163.com
Qufu Normal University, China
Lishan Liu mathlls@163.com
Qufu Normal University, China
Nonlinear Analysis: Modelling and Control, vol. 27, no. 6, pp. 1068-1090, 2022
Vilniaus Universitetas
Received: 05 November 2021
Revised: 09 September 2022
Published: 01 November 2022
1 Introduction
Fractional calculus (FC) has a history of more than 300 years, there are some applications of FC with in various fields of mathematics it self. During the last few decades,FC has obtained vigorous development in the applied sciences and gained considerable popularity. Compared with classical integer-order models, fractional derivatives and integrals are more suitable to describe the memory and hereditary properties of various materials, fractional derivatives are more advantageous in simulating mechanical and electrical properties of real materials and describing rheological properties of rocks and many other fields. Based on the description of their properties in terms of fractional derivatives, fractional differential equations (FDEs) are generated naturally, and how to solve these equations is also very necessary. For example, some new models that involve FDEs have been applied successfully, e.g., in mechanics (theory of viscoelasticity and viscoplasticity [6, 24]), (bio-)chemistry (modelling of polymers and proteins [11]),electrical engineering (transmission of ultrasound waves [3, 28]), medicine (modelling of human tissue under mechanical load [31]). . . Accordingly, more and more researchers and scholars devote themselves to the study of various problem of FDEs. In particular,study of boundary value problems for nonlinear FDEs is particularly concerned among these problems.
The aim of this paper is to investigate the following fractional differential equation:
(1)
subjected to boundary condition
(2)
where
is nonnegative, is measurable, is an integer. is singular at is the standard Riemann–Liouville fractional derivative of order defined by
where denotes the Gamma function, and denotes the integer part of number,provided that the right side is pointwise defined on; see [14, 25].
When , there has been a great deal of literature on the fractional differential equation of such boundary conditions; see [2, 4, 9, 15, 16, 30]. As for,for example, in [2],
where satisfies the Carathéodory conditions on is positive and singular at . Based on regularization and sequential techniques, the existence of a positive solution was obtained.
By Schauder fixed point theorem and the Banach contraction principle, Rehman et al.[26] investigated existence and uniqueness of solutions for a class of nonlinear multipointboundary value problems for fractional differential equation
where
.
In [15], the author studied the nonlinear fractional differential equation
subjected with the boundary conditions. Under Carathéodory conditions,using the Leray–Schauder continuation principle, the existence of at least onesolution was obtained.
KyuNam et al. [16] discussed the existence and uniqueness of solutions for a class ofintegral boundary value problems of nonlinear multiterm fractional differential equation
where, and are continuous. The existence results are established by the Banachfixed point theorem, and approximate solutions are determined by the Daftardar-Gejji and Jafari iterative method (DJIM) and the Adomian decomposition method (ADM).
In this article, we will deal with singular nonlinear fractional differential equation with a new boundary condition, which is a generalization of many previous researches.
To the best of our knowledge, when , neither nor was studied like this type of boundary condition. Furthermore, the nonlinear term contained not only lower derivative of order , but also another lower derivative of order . In comparison with the above literature, our results about the difference include both and, this has never been seen before.
The stability of differential equations has grown to be one of the considerable areas inthe field of mathematical analysis, and we find many different types of stability, such asexponential [7,17,23], Mittag-Leffler [20,27], Hyers–Ulam (HU) stability and other typesof stability [13,18,19,21,22]. Among these kinds of stability, Hyers–Ulam stability and itsvarious types provide a bridge between the exact and numerical solutions, so researchers devoted their work to the study of different kinds of HU stability for nonlinear fractionaldifferential equation; see [5, 10, 12, 29].
The paper is organized as follows. In Section 2, we will present some useful lemma sand give some valuable preliminary results. In Section 3, we prove the existence and uniqueness of positive solution to problem (1), (2) by using Leray–Schauder nonlinear alternative theorem and Boyd–Wong’s contraction principles. In Section 4, we investigate various kinds of HU stability of solutions.
2 Auxiliary results
To simplify our statements, we introduce the following spaces:
-
be the Banach space of all continuous functions from to equipped with the norm defined by .
-
end owed the norm
where .
First, we introduce some fundamental facts of the fractional calculus theory, which are used in this paper; see [14, 25].
Property 1.Let
.
-
(i) If, then;
-
(ii) If has a summable derivative , then
where ;
-
(iii) , where
;
-
(iv), where
, provided that
exists;
-
(v) Let
. Assume that
is a uniformly convergent sequence of continuous functions on
and that
exists for every
. Moreover, assume that
converges uniformly on
for every
. Then for every
, we have
Next, for the sake of readers’ convenience, we present some necessary lemmas, which will be used in the main results that follow.
Lemma 1. Suppose
and
,
Then the linear fractional boundary value problem
has a unique solution given by
, where
(4)
and
in which
and
Proof. It is very well known that the equation is equivalent to the following integral equation:
The boundary condition implies that. Using the property of Riemann–Liouville fractional derivative, we know
Combining the boundary condition in (3), it follows that
Therefore the unique solution of problem (3) is given by
Lemma 2. Let
be the Green function related to problem (3), which is given by expression (4). Then for
have the following properties:
-
(i) are nonnegative and continuous on;
-
(ii) for all for all, and
-
(iii) For,
-
(iv) Let , then for
Proof. From the condition we easily get
In addition, we also have a relation.
-
(i) For is obvious. For ,
In the same way, we get is nonnegative, then we have . It is obvious that are continuous on .
-
(ii) The conclusion is obvious, we omit it.
-
(iii) First, we introduce an inequality. For and, we have
(5)
When , using inequality (5), we obtain
When ,
In addition,
Again, we can get the property of , that is,
Therefore,
-
(iv) By substituting the two inequalities (iii) into formula (4) we naturally come to the conclusion.
Consider the problem , where operator is defined by
(6)
where is defined in (4). In order to prove that problem (1), (2) has a solution, we just have to show that operator has a fixed point.
Take the fractional derivative of order for , we have
(7)
where
(8)
(9)
Similarly, we have
where
(10)
Then
(11)
By simple deduction we can get the following properties.
Lemma 3. Let
. The functions
defined in (8), (9) and (10) have the following properties:
-
(i) are nonnegative and continuous on ;
-
(ii) For any ,
For any ,
where
and
is defined in Lemma 2;
-
(iv) , and
Lemma 4. Integral operators
have properties:
-
(i) are continuous;
-
(ii) is continuous.
Proof. (i) Since , we only show the continuity of . First, we prove that is well defined, that is, for any , we have. Notice that because of the condition .
For any ,
then we naturally get the continuity of as
Let be convergent sequence in the space, i.e., there exists a function such that. In order to show that operator is continuous, we have to prove. In fact,
where
According to the inequality above, we have as .
-
(ii) For any, denote .
(12)
where
From this inequality we know is continuous on if one supplies the definition of on.
For any ,
(13)
(14)
(15)
Now, we will evaluate these formulae (13)–(15), respectively.
Choose a constant satisfying, then
where
In the end, as for (15), let , then
Taking all the conclusions above into (13)–(15), we can get as, which implies that is continuous on and.
Combining (12), we easily infer the continuity of integral operator .
We define a normed vector space
equipped with the norm.
Lemma 5.
is Banach space.
Proof. Let be Cauchy sequence in. Clearly, are also Cauchy sequence in the space . Therefore, converge to some . By similar work of Su and Liu [30], we have .
Let . Evidently, is Cauchy sequence in ,then there exists such that in . That is to say, for any , there is a positive integer ,
Choose arbitrary positive number , we will prove uniformly converges to on. In fact, for all, we have
By property of fractional calculus, for every, we have, Hence, for all. So, and, i.e.,.
The proof is completed.
Let. Apparently, is a cone of.
3 Existence and uniqueness results
This section deals with existence and uniqueness of solutions for problem (1), (2). Thenonlinear term satisfies the following assumptions:
-
C1) is continuous, does not vanish on any compact interval of. Furthermore, there exist nonnegative functions and continuous and nondecreasing functions such that for any,
-
(C2) There exists a positive number such that
where .
-
(C3) There exist and are upper semicontinuous from the right and nondecreasing such that for any, we have
-
(C4) Denote
It satisfies for all , where.
-
(C4´) Denote
it satisfies for all .
Lemma 6. Suppose that (C1) holds, and
. Then
is completely continuous.
Proof. Let us first show that the operator is well defined. For any , from (C1),since, we know. Where upon, according to Hölder inequality and Lemma 2, we have and, and we deduce that from Lemma 4. Then by (6) we get . Similarly, using Lemma 4 and formulae (7), (11), we get the conclusions:. Furthermore, we naturally get from thefact that and Green’s functions are nonnegative. So,.
Suppose that in cone , then there exists a constant such that. In order to get the conclusion that operator is continuous, let us start with the fact that in , where. By condition (C1) it follows that
Moreover, on the basis of the continuity of, we deduce that ,for all . Taking advantage of Lebesgue dominated convergence theorem, we know. Thereupon, by Lemma 4, we have convergence to in . In addition, by Hölder inequality, we canalso get andas . Synthesizingthe above conclusions, from the expression of
(6), we have in . Analogously, we can deduce in and in. To wit, in.
In the end, Ascoli–Arzela theorem guarantees operator is compact. That is to say, we can deduce that is bounded and equicontinuous for any bounded subset. The proof can be obtained by the conventional procedure, so we omit this step.
From the above we conclude that the operator is completely continuous.
Theorem 1. (See [1].) Let E be a Banach space with
closed and convex. Assume that
is relatively open subset of
with
and
is a continuous compactmap. Then either
-
(i) A has a fixed point in or
-
ii) There exists and with.
Theorem 2. Assume that (C1), (C2) hold, and
. Then BVP (1), (2) has at least one positive solution.
Proof. By applying nonlinear alternative of Leray–Schauder-type fixed point theorem (Theorem 1), we will prove that has a fixed point. Let is given in condition (C2). Consider the following integral equation:
(16)
where . We claim that any solution of (16) for any must satisfies . Otherwise, assume that is a solution of (16) for some such that. Hence, from condition (C1) and Lemma 2(iv), for any , we have
so,
Similarly, in view of Lemma 3 and (7), we have
then
According to (11), we have
then
Therefore,
This is a contradiction and the claim is proved. Leray–Schauder nonlinear alternative theorem guarantees that operator has a fixed point. Since does not vanish on any compact interval of, we know must be positive.
Next, our uniqueness result for problem (1), (2) relies on Boyd–Wong’s contraction principle [8].
Theorem 3. Let
be a complete metric space and suppose
satisfies
where
is upper semicontinuous function from the right (i.e.,
, and for
for
.Then
has a unique fixed point
.
Theorem 4. Assume that (C3), (C4) hold, and
. Then BVP (1), (2) has a unique positive solution.
Proof. For any and , by using Lemma 2(iv) and condition (C3), we get
so
Similarly,
Synthesizing the above three inequalities and combining with condition (C4), we get
Then Boyd–Wong’s contraction principle can be applied and has a unique fixed pointwhich is the unique solution of problem (1), (2).
Example.
(17)
Let. A simple calculation yields
and, then .
Let , then is continuous, and does not vanish on any compact interval of . Let, then for .
Furthermore,,
Take ,
Hence, conditions (C1), (C2) in Theorem 2 hold, then problem (17) has a positive solution.
4 Stability analysis
In this section, we consider the Banach space equipped with the norm. Let us introduce some definitions related to Ulam stability.
Suppose that function is nonnegative and. Consider thein equalities given below:
(18)
(19)
(20)
Definition 1. We say that is a solution of inequality (18): if there is which depends on , such that and, meanwhile satisfies boundary condition (2), where
Remark 1. The solution of inequalities (19), (20) can be defined as well.
Definition 2. BVP (1), (2) is Hyers–Ulam stable: if there is a constant such thatfor any and for each solution of inequality (18), there exists a unique solution of BVP (1), (2) satisfying
(21)
Definition 3. BVP (1), (2) is generalized Hyers–Ulam stable: if there is a function with such that for any and for each solution of inequality (18), there exists a unique solution of BVP (1), (2) with
Definition 4. BVP (1), (2) is Hyers–Ulam–Rassias (HUR) stable w.r.t. nonnegative function if there is a constant such that for any and foreach solution of inequality (19), there is a unique solution of BVP (1), (2) satisfying
Definition 5. BVP (1), (2) is generalized Hyers–Ulam–Rassias stable w.r.t. if there is a constant such that for each solution of inequality (20),there is a unique solution of BVP (1), (2) satisfying
Theorem 5. Assume that (C3), (C40) hold, and
Let
. If
, then BVP (1), (2) is HU stable.
Proof. For any , suppose be the solution of inequality (18), then, where. Just like the proof method in Theorem 4, we also know that BVP (1), (2) has a unique solution under the new normand can be expressed by by Lemma 1.
On the basis of Lemma 2, we have
Similarly, we can get
Hence,, then
(22)
Let , clearly,
In other words, conclusion (21) in Definition 2 is satisfied, i.e., BVP (1), (2) is Hyers–Ulam stable.
Theorem 6. Suppose that (C3), (C40) hold, and
Let
. If
, then BVP (1), (2) is generalized HU stable.
Proof. For any , for convenience, let us assume . Suppose be the solution of inequality (18), be a unique solution of BVP (1), (2). Following the method of Theorem 5, we know that (22) holds. Thereupon, for any ,
It is obvious that . Then by Definition 3, BVP (1), (2) is generalized HUstable.
Theorem 7. Let the conditions of Theorem 6 be satisfied, moreover, there exists a nonnegative function
satisfying
for all
. Then BVP (1), (2) is HUR stable.
Proof. For any , suppose be the solution of inequality (19), then, where is a unique solution of BVP (1), (2). On the similar way of Theorem 5, one can prove
So, combining with these three inequalities, we have
(23)
For any , from (23), we get
Let
then
By Definition 4, BVP (1), (2) is HUR stable.
Remark 2. Under the condition of Theorem 7, imitating the process, we can prove that BVP (1), (2) is generalized HUR stable.
Acknowledgments
The authors would like to thank the referee for his/her valuable comments and suggestions.