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Boundary Value Problems

, 2013:112 | Cite as

Some existence results for a nonlinear fractional differential equation on partially ordered Banach spaces

  • Dumitru Baleanu
  • Ravi P Agarwal
  • Hakimeh Mohammadi
  • Shahram Rezapour
Open Access
Research

Abstract

By using fixed point results on cones, we study the existence and uniqueness of positive solutions for some nonlinear fractional differential equations via given boundary value problems. Examples are presented in order to illustrate the obtained results.

Keywords

Banach Space Fractional Order Fractional Derivative Fractional Calculus Normal Cone 
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

1 Introduction

The field of fractional differential equations has been subjected to an intensive development of the theory and the applications (see, for example, [1, 2, 3, 4, 5, 6] and the references therein). It should be noted that most of papers and books on fractional calculus are devoted to the solvability of linear initial fractional differential equations on terms of special functions. There are some papers dealing with the existence of solutions of nonlinear initial value problems of fractional differential equations by using the techniques of nonlinear analysis such as fixed point results, the Leray-Schauder theorem, stability, etc. (see, for example, [7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19] and the references therein). In fact, fractional differential equations arise in many engineering and scientific disciplines such as physics, chemistry, biology, economics, control theory, signal and image processing, biophysics, blood flow phenomena and aerodynamics (see, for example, [20, 21, 22, 23] and the references therein). The main advantage of using the fractional nonlinear differential equations is related to the fact that we can describe the dynamics of complex non-local systems with memory. In this line of taught, the equations involving various fractional orders are important from both theoretical and applied view points. We need the following notions.

Definition 1.1 ([1, 4])

For a continuous function f : [ 0 , ) R Open image in new window, the Caputo derivative of fractional order α is defined by
D α c f ( t ) = 1 Γ ( n α ) 0 t ( t s ) n α 1 f ( n ) ( s ) d s , Open image in new window

where n 1 < α < n Open image in new window, n = [ α ] + 1 Open image in new window and [ α ] Open image in new window denotes the integer part of α.

Definition 1.2 ([1, 4])

The Riemann-Liouville fractional derivative of order α for a continuous function f is defined by
D α f ( t ) = 1 Γ ( n α ) ( d d t ) n 0 t f ( s ) ( t s ) α n 1 d s ( n = [ α ] + 1 ) , Open image in new window

where the right-hand side is pointwise defined on ( 0 , ) Open image in new window.

Definition 1.3 ([1, 4])

Let [ a , b ] Open image in new window be an interval in ℝ and α > 0 Open image in new window. The Riemann-Liouville fractional order integral of a function f L 1 ( [ a , b ] , R ) Open image in new window is defined by
I a α f ( t ) = 1 γ ( α ) a t f ( s ) ( t s ) 1 α d s Open image in new window

whenever the integral exists.

Suppose that E is a Banach space which is partially ordered by a cone P E Open image in new window, that is, x y Open image in new window if and only if y x P Open image in new window. We denote the zero element of E by θ. A cone P is called normal if there exists a constant N > 0 Open image in new window such that θ x y Open image in new window implies x N y Open image in new window (see [24]). Also, we define the order interval [ x 1 , x 2 ] = { x E | x 1 x x 2 } Open image in new window for all x 1 , x 2 E Open image in new window [24]. We say that an operator A : E E Open image in new window is increasing whenever x y Open image in new window implies A x A y Open image in new window. Also, x y Open image in new window means that there exist λ > 0 Open image in new window and μ > 0 Open image in new window such that λ x y μ x Open image in new window (see [24]). Finally, put P h = { x E | x h } Open image in new window for all h > θ Open image in new window. It is easy to see that P h P Open image in new window is convex and λ P h = P h Open image in new window for all λ > 0 Open image in new window. We recall the following in our results. Let E be a real Banach space and let P be a cone in E. Let ( a , b ) Open image in new window be an interval and let τ and φ be two positive-valued functions such that φ ( t ) τ ( t ) Open image in new window for all t ( a , b ) Open image in new window and τ : ( a , b ) ( 0 , 1 ) Open image in new window is a surjection. We say that an operator A : P P Open image in new window is τ-φ-concave whenever A ( τ ( t ) x ) φ ( t ) A x Open image in new window for all t ( a , b ) Open image in new window and x P Open image in new window [13]. We say that A is φ-concave whenever τ ( t ) = t Open image in new window for all t [13]. We recall the following result.

Theorem 1.1 ([13])

Let E be a Banach space, let P be a normal cone in E, and let A : P P Open image in new window be an increasing and τ-φ-concave operator. Suppose that there exists θ h P Open image in new window such that A h P h Open image in new window. Then there are u 0 , v 0 P h Open image in new window and r ( 0 , 1 ) Open image in new window such that r v 0 u 0 v 0 Open image in new window and u 0 A u 0 A v 0 v 0 Open image in new window, the operator A has a unique fixed point x [ u 0 , v 0 ] Open image in new window, and for x 0 P h Open image in new window and the sequence { x n } Open image in new window with x n = A x n 1 Open image in new window, we have x n x 0 Open image in new window.

2 Main results

We study the existence and uniqueness of a solution for the fractional differential equation
D α u ( t ) + f ( t , u ( t ) ) = 0 Open image in new window

on partially ordered Banach spaces with two types of boundary conditions and two types of fractional derivatives, Riemann-Liouville and Caputo.

2.1 Existence results for the fractional differential equation with the Riemann-Liouville fractional derivative

First, we study the existence and uniqueness of a positive solution for the fractional differential equation

where D α Open image in new window is the Riemann-Liouville fractional derivative of order α. Let E = C [ ε , T ] Open image in new window. Consider the Banach space of continuous functions on [ ε , T ] Open image in new window with the sup norm and set P = { y C [ ε , T ] : min t [ ε , T ] y ( t ) 0 } Open image in new window. Then P is a normal cone.

Lemma 2.1 Let 0 < ε < T Open image in new window, T 1 Open image in new window, t [ ε , T ] Open image in new window, η ( ε , t ) Open image in new window and 0 < α < 1 Open image in new window. Then the problem D α u ( t ) + f ( t , u ( t ) ) = 0 Open image in new window with the boundary value condition u ( η ) = u ( T ) Open image in new window has a solution u 0 Open image in new window if and only if u 0 Open image in new window is a solution of the fractional integral equation
u ( t ) = ε T G ( t , s ) f ( s , u ( s ) ) d s , Open image in new window
where
G ( t , s ) = { t α 1 ( η s ) α 1 t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) ( t s ) α 1 Γ ( α ) , ε s η t T , t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) ( t s ) α 1 Γ ( α ) , ε η s t T , t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) , ε η t s T . Open image in new window
Proof From D α u ( t ) + f ( t , u ( t ) ) = 0 Open image in new window and the boundary condition, it is easy to see that u ( t ) c 1 t α 1 = I ε α f ( t , u ( t ) ) Open image in new window. By the definition of a fractional integral, we get
u ( t ) = c 1 t α 1 ε t ( t s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s . Open image in new window
Thus, u ( η ) = c 1 η α 1 ε η ( η s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s Open image in new window and
u ( T ) = c 1 T α 1 ε T ( T s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s . Open image in new window
Since u ( η ) = u ( T ) Open image in new window, we obtain
c 1 = 1 η α 1 T α 1 ε η ( η s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s 1 η α 1 T α 1 ε T ( T s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s . Open image in new window
Hence,
u ( t ) = t α 1 η α 1 T α 1 ε η ( η s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s t α 1 η α 1 T α 1 ε T ( T s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s ε t ( t s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s = ε T G ( t , s ) f ( s , u ( s ) ) d s . Open image in new window

This completes the proof. □

Now, we are ready to state and prove our first main result.

Theorem 2.2 Let 0 < ε < T Open image in new window be given and let τ and φ be two functions on ( ε , T ) Open image in new window such that φ ( t ) τ ( t ) Open image in new window for all t ( ε , T ) Open image in new window. Suppose that τ : ( ε , T ) ( 0 , 1 ) Open image in new window is a surjection and f ( t , u ( t ) ) C ( [ ε , T ] × [ 0 , ] ) Open image in new window is increasing in u for each fixed t, f ( t , u ( t ) ) 0 Open image in new window and f ( t , τ ( λ ) u ( t ) ) φ ( λ ) f ( t , u ( t ) ) Open image in new window for all t , λ ( ε , T ) Open image in new window and u P Open image in new window. Assume that there exist M 1 > 0 Open image in new window, M 2 > 0 Open image in new window and θ h P Open image in new window such that
M 1 h ( t ) ε T G ( t , s ) f ( s , h ( s ) ) d s M 2 h ( t ) Open image in new window

for all t [ ε , T ] Open image in new window, where G ( t , s ) Open image in new window is the green function defined in Lemma 2.1. Then the problem (2.1) with the boundary value condition (2.2) has a unique positive solution u P h Open image in new window. Moreover, for the sequence u n + 1 = ε T G ( t , s ) f ( s , u n ( s ) ) d s Open image in new window, we have u n u 0 Open image in new window for all u 0 P h Open image in new window.

Proof By using Lemma 2.1, the problem is equivalent to the integral equation
u ( t ) = ε T G ( t , s ) f ( s , u ( s ) ) d s , Open image in new window
where
G ( t , s ) = { t α 1 ( η s ) α 1 t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) ( t s ) α 1 Γ ( α ) , ε s η t T , t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) ( t s ) α 1 Γ ( α ) , ε η s t T , t α 1 ( T s ) α 1 ( η α 1 T α 1 ) Γ ( α ) , ε η t s T . Open image in new window
Define the operator A : P E Open image in new window by A u ( t ) = ε T G ( t , s ) f ( s , u ( s ) ) d s Open image in new window. Then u is a solution for the problem if and only if u = A u Open image in new window. It is easy to check that the operator A is increasing on P. On the other hand,
A ( τ ( λ ) u ) ( t ) = ε T G ( t , s ) f ( s , τ ( λ ) u ( s ) ) d s φ ( λ ) ε T G ( t , s ) f ( s , u ( s ) ) d s = φ ( λ ) A u ( t ) Open image in new window
for all λ [ ε , T ] Open image in new window and u P Open image in new window. Thus, the operator A is τ-φ-concave. Since
M 1 h ( t ) A h ( t ) = ε T G ( t , s ) f ( s , h ( s ) ) d s M 2 h ( t ) Open image in new window

for all t [ ε , T ] Open image in new window, we get A h P h Open image in new window. Now, by using Theorem 1.1, the operator A has a unique positive solution u P h Open image in new window. This completes the proof. □

Here, we give the following example to illustrate Theorem 2.2.

Example 2.1 Let 0 < ε < 1 Open image in new window be given. Consider the periodic boundary value problem
D 1 3 u ( t ) + { g ( t ) + [ u ( t ) ] α } = 0 ( t [ ε , 1 ] ) , u ( η ) = u ( 1 ) , Open image in new window
where η ( ε , t ) Open image in new window, g is continuous on [ ε , 1 ] Open image in new window and min t [ ε , 1 ] g ( t ) > 0 Open image in new window. Put
G ( t , s ) = { t 2 / 3 ( η s ) 2 / 3 t 2 / 3 ( 1 s ) 2 / 3 ( η 2 / 3 1 2 / 3 ) Γ ( 1 / 3 ) ( t s ) 2 / 3 Γ ( 1 / 3 ) , ε s η t 1 , t 2 / 3 ( 1 s ) 2 / 3 ( η 2 / 3 1 2 / 3 ) Γ ( 1 / 3 ) ( t s ) 2 / 3 Γ ( 1 / 3 ) , ε η s t 1 , t 2 / 3 ( 1 s ) 2 / 3 ( η 2 / 3 1 2 / 3 ) Γ ( 1 / 3 ) , ε η t s 1 . Open image in new window
Then ε 1 G ( t , s ) d s = t 2 / 3 ( η ε ) 1 / 3 t 2 / 3 ( 1 ε ) 1 / 3 ( t ε ) 1 / 3 ( η 2 / 3 1 ) Γ ( 4 / 3 ) ( η 2 / 3 1 ) Open image in new window. Now, define τ ( t ) = t Open image in new window, φ ( t ) = t 1 / 3 Open image in new window, γ 1 = min t [ ε , 1 ] g ( t ) Open image in new window, γ 2 = max t [ ε , 1 ] g ( t ) Open image in new window and also f ( t , u ) = g ( t ) + u 1 / 3 Open image in new window for all t. Then τ : ( 0 , 1 ) ( 0 , 1 ) Open image in new window is a surjection and φ ( t ) > τ ( t ) Open image in new window for all t ( ε , 1 ) Open image in new window. For each u 0 Open image in new window, we have
f ( t , τ ( λ ) u ( t ) ) = f ( t , λ u ( t ) ) = g ( t ) + λ 1 / 3 [ u ( t ) ] 1 / 3 λ 1 / 3 ( g ( t ) + [ u ( t ) ] 1 / 3 ) = φ ( λ ) f ( t , u ( t ) ) . Open image in new window
Now, put h 1 Open image in new window, M 1 = ( γ 1 + 1 ) min t [ ε , 1 ] , η [ ε , 1 ] t 2 / 3 ( 1 ε ) 1 / 3 ( t ε ) 1 / 3 ( η 2 / 3 1 ) Γ ( 4 / 3 ) ( η 2 / 3 1 ) Open image in new window and M 2 = ( γ 2 + 1 ) max η [ ε , 1 ] ε 2 / 3 η 1 / 3 Γ ( 4 / 3 ) ( η 2 / 3 1 ) Open image in new window. Then we get
ε 1 G ( t , s ) { g ( s ) + [ h ( s ) ] 1 / 3 } d s ε 1 G ( t , s ) ( γ 2 + 1 ) d s ( γ 2 + 1 ) max t [ ε , 1 ] ε 1 G ( t , s ) d s ( γ 2 + 1 ) ( max η [ ε , 1 ] ε 2 / 3 η 1 / 3 Γ ( 4 / 3 ) ( η 2 / 3 1 ) ) = M 2 h Open image in new window
and
ε 1 G ( t , s ) { g ( s ) + [ h ( s ) ] 1 / 3 } d s ( γ 1 + 1 ) min t [ ε , 1 ] ε 1 G ( t , s ) d s ( γ 1 + 1 ) min t [ ε , 1 ] , η [ ε , 1 ] t 2 / 3 ( 1 ε ) 1 / 3 ( t ε ) 1 / 3 ( η 2 / 3 1 ) Γ ( 4 / 3 ) ( η 2 / 3 1 ) = M 1 h . Open image in new window

Thus, by using Theorem 2.2, the problem has a unique solution in P h = P 1 Open image in new window.

2.2 Existence results for the fractional differential equation with the Caputo fractional derivative

Here, we study the existence and uniqueness of a positive solution for the fractional differential equation
D α c u ( t ) + f ( t , u ( t ) ) = 0 ( t [ 0 , T ] , T 1 , 1 < α < 2 ) , Open image in new window
(2.3)
u ( 0 ) = β 1 u ( η ) , u ( T ) = β 2 u ( η ) ( η ( 0 , t ) , 0 < β 1 < β 2 < 1 ) , Open image in new window
(2.4)
where D α c Open image in new window is the Caputo fractional derivative of order α. Let E = C [ 0 , T ] Open image in new window be the Banach space of continuous functions on [ 0 , T ] Open image in new window with the sup norm and
P = { y C [ 0 , T ] : min t [ 0 , T ] y ( t ) 0 } . Open image in new window

It is known that P is a normal cone. Similar to the proof of Lemma 2.1, we can prove the following result.

Lemma 2.3 Let 1 < α < 2 Open image in new window, T 1 Open image in new window, t [ 0 , T ] Open image in new window, η ( 0 , t ) Open image in new window and 0 < β 1 < β 2 < 1 Open image in new window. Then the problem D α c u ( t ) + f ( t , u ( t ) ) = 0 Open image in new window with the boundary value conditions u ( 0 ) = β 1 u ( η ) Open image in new window and u ( T ) = β 2 u ( η ) Open image in new window has a solution u 0 Open image in new window if and only if u 0 Open image in new window is a solution of the fractional integral equation u ( t ) = 0 T G ( t , s ) f ( s , u ( s ) ) d s Open image in new window, where
G ( t , s ) = { [ β 1 T + t ( β 2 β 1 ) ] ( η s ) α 1 + t ( T s ) α 1 T ( t s ) α 1 T Γ ( α ) , 0 s η t T , t ( T s ) α 1 T ( t s ) α 1 T Γ ( α ) , 0 η s t T , t ( T s ) α 1 T Γ ( α ) , 0 η t s T . Open image in new window
Theorem 2.4 Let T 1 Open image in new window be given and let τ and φ be two positive-valued functions on ( 0 , T ) Open image in new window such that φ ( t ) τ ( t ) Open image in new window for all t ( 0 , T ) Open image in new window. Suppose that τ : ( 0 , T ) ( 0 , 1 ) Open image in new window is a surjection and f ( t , u ( t ) ) C ( [ ε , T ] × [ 0 , ] ) Open image in new window is increasing in u for each fixed t, f ( t , u ( t ) ) = 0 Open image in new window whenever 0 < η < s < t < T Open image in new window and f ( t , u ( t ) ) 0 Open image in new window otherwise, and also f ( t , τ ( λ ) u ( t ) ) φ ( λ ) f ( t , u ( t ) ) Open image in new window for all t , λ ( 0 , T ) Open image in new window and u P Open image in new window. Assume that there exist M 1 > 0 Open image in new window, M 2 > 0 Open image in new window and θ h P Open image in new window such that
M 1 h ( t ) 0 T G ( t , s ) f ( s , h ( s ) ) d s M 2 h ( t ) Open image in new window

for all t [ 0 , T ] Open image in new window, where G ( t , s ) Open image in new window is the green function defined in Lemma 2.3. Then the problem (2.3) with the boundary value conditions (2.4) has a unique positive solution u P h Open image in new window. Moreover, for the sequence u n + 1 = ε T G ( t , s ) f ( s , u n ( s ) ) d s Open image in new window, we have u n u 0 Open image in new window for all u 0 P h Open image in new window.

Proof It is sufficient to define the operator A : P E Open image in new window by
A u ( t ) = 0 T G ( t , s ) f ( s , u ( s ) ) d s . Open image in new window

Now, by using a similar proof of Theorem 2.2, one can show that A u ( t ) 0 Open image in new window for all u P Open image in new window and t [ 0 , T ] Open image in new window, and also the operator A is τ-φ-concave. By using Theorem 1.1, the operator A has a unique positive solution u P h Open image in new window. This completes the proof by using Lemma 2.3. □

Below we present an example to illustrate Theorem 2.4.

Example 2.2 Let α = 3 2 Open image in new window. Consider the periodic boundary value problem
D α c u ( t ) + g ( t ) + [ u ( t ) ] α = 0 ( t [ 0 , 1 ] ) , u ( 0 ) = 1 3 u ( 1 2 ) u ( 1 ) = 1 2 u ( 1 2 ) , Open image in new window
where g is a continuous function on [ 0 , 1 ] Open image in new window with min t [ 0 , 1 ] g ( t ) > 0 Open image in new window. Put β 2 = η = 1 / 2 Open image in new window, β 1 = 1 / 3 Open image in new window and
G ( t , s ) = { [ 1 3 + 1 6 t ] ( 1 2 s ) 1 / 2 + t ( 1 s ) 1 / 2 ( t s ) 1 / 2 Γ ( 3 / 2 ) , 0 s η t 1 , t ( 1 s ) 1 / 2 ( t s ) 1 / 2 Γ ( 3 / 2 ) , 0 η s t 1 , t ( 1 s ) 1 / 2 Γ ( 3 / 2 ) , 0 η t s 1 . Open image in new window
Then 0 1 G ( t , s ) d s = [ 1 3 + 1 6 t ] ( 1 2 ) 3 / 2 + t t 3 / 2 Γ ( 5 / 2 ) Open image in new window. Now, define τ ( t ) = t Open image in new window, φ ( t ) = t α Open image in new window, γ 1 = min t [ 0 , 1 ] g ( t ) Open image in new window, γ 2 = max t [ 0 , 1 ] g ( t ) Open image in new window and f ( t , u ) = g ( t ) + u α Open image in new window. Then it is easy to see that τ : ( 0 , 1 ) ( 0 , 1 ) Open image in new window is a surjection map and φ ( t ) > τ ( t ) Open image in new window for t ( 0 , 1 ) Open image in new window. Also, we have
f ( t , τ ( λ ) u ( t ) ) = f ( t , λ u ( t ) ) = g ( t ) + λ α [ u ( t ) ] α λ α ( g ( t ) + [ u ( t ) ] α ) = φ ( λ ) f ( t , u ( t ) ) Open image in new window
for all u 0 Open image in new window. Now, put h 1 Open image in new window, M 1 = ( γ 1 + 1 ) min t [ 0 , 1 ] 1 3 t ( 1 2 ) 3 / 2 t 3 / 2 Γ ( 5 / 2 ) Open image in new window and also M 2 = ( γ 2 + 1 ) 5 6 ( 1 2 ) 3 / 2 + 1 Γ ( 5 / 2 ) Open image in new window. Then we have
0 1 G ( t , s ) { g ( s ) + [ h ( s ) ] 3 / 2 } d s 0 1 G ( t , s ) ( γ 2 + 1 ) d s ( γ 2 + 1 ) max t [ 0 , 1 ] 0 1 G ( t , s ) d s ( γ 2 + 1 ) 5 6 ( 1 2 ) 3 / 2 + 1 Γ ( 5 / 2 ) = M 2 h Open image in new window
and
0 1 G ( t , s ) { g ( s ) + [ h ( s ) ] 3 / 2 } d s ( γ 1 + 1 ) min t [ 0 , 1 ] 0 1 G ( t , s ) d s ( γ 1 + 1 ) min t [ 0 , 1 ] 1 3 t ( 1 2 ) 3 / 2 t 3 / 2 Γ ( 5 / 2 ) = M 1 h . Open image in new window

Thus, by using Theorem 2.4, the problem has a unique solution in P h = P 1 Open image in new window.

Notes

Acknowledgements

This work is partially supported by the Scientific and Technical Research Council of Turkey. Research of the third and forth authors was supported by Azarbaidjan Shahid Madani University. Also, the authors express their gratitude to the referees for their helpful suggestions which improved final version of this paper.

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© Baleanu et al.; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  • Dumitru Baleanu
    • 1
    • 2
    • 3
  • Ravi P Agarwal
    • 4
  • Hakimeh Mohammadi
    • 5
  • Shahram Rezapour
    • 5
  1. 1.Department of Chemical and Materials Engineering, Faculty of EngineeringKing Abdulaziz UniversityJeddahSaudi Arabia
  2. 2.Department of MathematicsCankaya UniversityBalgat, AnkaraTurkey
  3. 3.Institute of Space SciencesMagurele, BucharestRomania
  4. 4.Department of MathematicsTexas A&M UniversityKingsvilleUSA
  5. 5.Department of MathematicsAzarbaidjan Shahid Madani UniversityAzarshahr, TabrizIran

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