International Journal of Applied Mathematical Research, 7 (2) (2018) 56-61
International Journal of Applied Mathematical Research
Website: www.sciencepubco.com/index.php/IJAMR
doi: 10.14419/ijamr.v7i2. 10116
Research paper
General solution of second order fractional
differential equations
Mousa Ilie1,2, Jafar Biazar2,3*, Zainab Ayati4
1Department
of Mathematics, Guilan Science and Research Branch, Islamic Azad University, Rasht, Iran
of Mathematics, Rasht Branch, Islamic Azad University, Rasht, Iran
3
Department of Applied Mathematics, Faculty of Mathematical Sciences, University of Guilan,
P.O. Box. 41335-1914, Guilan, Rasht, Iran
4Department of Engineering sciences, Faculty of Technology and Engineering East of Guilan,
University of Guilan, P.C. 44891-63157, Rudsar-Vajargah, Iran
*Corresponding author E-mail: biazar@iaurasht.ac.ir
2Department
Abstract
Fractional differential equations are often seeming perplexing to solve. Therefore, finding comprehensive methods for solving them sounds
of high importance. In this paper, a general method for solving second order fractional differential equations has been presented based on
conformable fractional derivative. This method realizes on determining a general solution of homogeneous and a particular solution of a
second order linear fractional differential equations. Furthermore, a general solution has been developed for fractional Euler’s equation.
For more explanation of each part, some examples have been solved.
Keywords: Linear fractional differential equations; Conformable fractional derivative; Constant coefficients approach; Euler’s equation; Variation of
parameters; Lagrange method; Undetermined coefficients;
1. Introduction
2. Basic definitions
Fractional differential equations are studied in various fields of
physics and engineering, specifically in signal processing, control
engineering, electromagnetism, biosciences, fluid mechanics, electrochemistry, diffusion processes, dynamic of viscoelastic material,
continuum and statistical mechanics and propagation of spherical
flames. There are many fractional differential equations which can’t
be solved analytically. Due to this fact, finding an approximate solution of fractional differential equations is clearly an important
task. In recent years, many effective methods have been proposed
for the approximate solution fractional differential equations, such
as Adomian decomposition method [3,4], homotopy perturbation
method [5-8], homotopy analysis method [9,10], variational iteration method [11], generalized differential transform method [12]
and other methods [13-29].
The organization of the paper is as follows: In Section 2, Basic definitions, such as conformable fractional derivative, and conformable fractional integral, will be presented. In Section 3, Basic theoretical of the method, will be described. In Section 4, the methods
such as, use of a known solution to find another one or D’Alambert
approach, homogeneous equation with constant coefficients, Euler’s dimensional equation, will be expanded. In Section 5, the
methods such as, variation of parameters or Lagrange technique,
undetermined coefficients, will be explained. Finally, conclusion is
presented in section 6.
The purpose of this section is to recall some preliminaries of the
proposed method.
2.1. Conformable fractional derivative
Given a function 𝑓: [0, ∞) → ℝ. Then the conformable fractional
derivative of 𝑓 of order 𝛼 is defined by
Τα (f)(x) = lim
ε→0
f(x+εx1-α )-f(x)
(1)
ε
For all 𝑥 > 0, 𝛼 ∈ (0,1). If 𝑓 is 𝛼 - differentiable in some
(0, 𝑎), 𝑎 > 0, and provided that lim+ Τ𝛼 (𝑓)(𝑥) exists, then define
Τ𝛼 (𝑓)(0) = lim+ Τ𝛼 (𝑓)(𝑥).
𝑥→0
𝑥→0
If the conformable derivative of 𝑓 of order 𝛼 exists, then we simply
say that 𝑓 is 𝛼- differentiable (see [1,2]).
One can easily show that Τ𝛼 satisfies all the properties in the following properties (see [1]):
Let α ∈ (0,1] and 𝑓, 𝑔 be 𝛼-differentiable at a point 𝑥 > 0, Then
A. For 𝑎, 𝑏 ∈ ℝ Τ𝛼 (𝑎𝑓 + 𝑏𝑔) = 𝑎 Τ𝛼 (𝑓) + 𝑏 Τ𝛼 (𝑔),
B. For all 𝑝 ∈ ℝ Τ𝛼 (𝑥 𝑝 ) = 𝑝𝑥 𝑝−𝛼 ,
C. For all constant functions 𝑓(𝑥) = 𝜆, Τ𝛼 (𝜆) = 0,
D. Τ𝛼 (𝑓. 𝑔) = 𝑔. Τ𝛼 (𝑓) + 𝑓 . Τ𝛼 (𝑔),
𝑓
E. Τ𝛼 ( ) =
𝑔
F. Τ𝛼 (𝑓) =
𝑔.Τ𝛼 (𝑓)−𝑓 .Τ𝛼 (𝑔)
𝑔2
1−𝛼 𝑑𝑓
𝑥
.
𝑑𝑥
,
2.2. Conformable fractional integral
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International Journal of Applied Mathematical Research
Given a function 𝑓: [𝑎, ∞) → ℝ, 𝑎 ≥ 0. Then the conformable
fractional integral of 𝑓 is defined by
𝑥 𝑓(𝑡)
I𝛼𝑎 (𝑓)(𝑥) = ∫𝑎
𝑡
1−𝛼 𝑑𝑡
(2)
Where the integral is the usual Riemann improper integral, and 𝛼 ∈
(0,1) ( see [1,2]).
For the sake of simplicity, let’s considerI𝛼0 (𝑓)(𝑥) = Ι𝛼 (𝑓)(𝑥).
One of the most useful results is the following (see [1]):
For all 𝑥 ≥ 𝑎, and any continuous function in the domain of Ι𝛼𝑎 , we
have Τ𝛼 (Ι𝛼𝑎 𝑓(𝑥)) = 𝑓(𝑥).
3. Basic theoretical of the method
Let’s consider the general second order linear fractional differential
equation based on conformable fractional derivative as follows
Τ𝛼 Τ𝛼 (𝑢(𝑥)) + 𝑃(𝑥)Τ𝛼 (𝑢(𝑥)) + 𝑄(𝑥)𝑢(𝑥) = 𝑅(𝑥),
(3)
Where 𝑃(𝑥), 𝑄(𝑥), and 𝑅(𝑥) are 𝛼 − differentiable functions
and 𝑢(𝑥) is an unknown function. If 𝑅(𝑥) is identically zero, then
fractional equation (3) reduces to the homogeneous fractional equation
Τ𝛼 Τ𝛼 (𝑢(𝑥)) + 𝑃(𝑥)Τ𝛼 (𝑢(𝑥)) + 𝑄(𝑥)𝑢(𝑥) = 0.
(4)
Theorem 3.1. If 𝑢ℎ (𝑥, 𝐶1 , 𝐶2 ) is the general solution of fractional
equation (4) and 𝑢𝑝 (𝑥) is any particular solution of fractional equation (3), then 𝑢ℎ (𝑥, 𝐶1 , 𝐶2 ) + 𝑢𝑝 (𝑥) is a general solution of fractional equation (3).
Proof. Suppose that 𝑢(𝑥) is a solution of (3), since 𝑢𝑝 (𝑥) is any
particular solution of fractional equation (3), then an easy calculation shows that 𝑢(𝑥) − 𝑢𝑝 (𝑥) is a solution of (4):
Τ𝛼 Τ𝛼 (𝑢(𝑥) − 𝑢𝑝 (𝑥)) + 𝑃(𝑥)Τ𝛼 (𝑢(𝑥) − 𝑢𝑝 (𝑥)) +
𝑄(𝑥) (𝑢(𝑥) − 𝑢𝑝 (𝑥)) = (Τ𝛼 Τ𝛼 (𝑢(𝑥)) + 𝑃(𝑥)Τ𝛼 (𝑢(𝑥))
−(Τ𝛼 Τ𝛼 (𝑢𝑝 (𝑥)) + 𝑃(𝑥)Τ𝛼 (𝑢𝑝 (𝑥)) + 𝑄(𝑥)𝑢𝑝 (𝑥)) =
𝑅(𝑥) − 𝑅(𝑥) = 0.
Since 𝑢ℎ (𝑥, 𝐶1 , 𝐶2 ) is a general solution to (4), it results that 𝑢(𝑥) −
𝑢𝑝 (𝑥) = 𝑢ℎ (𝑥, 𝐶1 , 𝐶2 ) or 𝑢(𝑥) = 𝑢ℎ (𝑥, 𝐶1 , 𝐶2 ) + 𝑢𝑝 (𝑥) , for a
suitable choice of the constants 𝐶1 and 𝐶2 (see [23]).
Theorem 3.2. If 𝑢1 (𝑥) and 𝑢2 (𝑥) are any two solutions of fractional equation (4), then 𝐶1 𝑢1 (𝑥) + 𝐶2 𝑢2 (𝑥) is also a solution for
any constants 𝐶1 and 𝐶2 .
Proof. The statement follows immediately from the fact that
Τ𝛼 Τ𝛼 (𝐶1 𝑢1 (𝑥) + 𝐶2 𝑢2 (𝑥) ) + 𝑃(𝑥)Τ𝛼 (𝐶1 𝑢1 (𝑥) +
𝐶2 𝑢2 (𝑥) ) + 𝑄(𝑥)(𝐶1 𝑢1 (𝑥) + 𝐶2 𝑢2 (𝑥) ) =
𝐶1 (Τ𝛼 Τ𝛼 (𝑢1 (𝑥) ) + 𝑃(𝑥)Τ𝛼 (𝑢1 (𝑥) ) + 𝑄(𝑥)𝑢1 (𝑥)) +
𝐶2 (Τ𝛼 Τ𝛼 (𝑢2 (𝑥) ) + 𝑃(𝑥)Τ𝛼 (𝑢2 (𝑥) ) + 𝑄(𝑥)𝑢2 (𝑥)) =
𝐶1 . 0 + 𝐶2 . 0 = 0.
Since by assumption, 𝑢1 (𝑥) and 𝑢2 (𝑥) are solutions of (4) (see
[23]).
Definition. The fractional Wronskian of two functions 𝑓(𝑥) and
𝑔(𝑥), is defined by (see [22,23]),
𝑓(𝑥)
𝑔(𝑥)
|
𝑊𝛼 (𝑓(𝑥), 𝑔(𝑥)) = |
Τ𝛼 (𝑓(𝑥)) Τ𝛼 (𝑔(𝑥))
= 𝑓(𝑥). Τ𝛼 (𝑔(𝑥)) − 𝑔(𝑥). Τ𝛼 (𝑓(𝑥)).
Theorem 3.3. If 𝑢1 (𝑥) and 𝑢2 (𝑥) are any two solutions of fractional equation (4) on an interval [𝑎, 𝑏], then their fractional Wronskian 𝑊 = 𝑊𝛼 (𝑢1 (𝑥), 𝑢2 (𝑥)) is either identically zero or never
zero on [𝑎, 𝑏].
Proof. We begin by observing that
Τ𝛼 (𝑊) = 𝑢1 Τ𝛼 Τ𝛼 (𝑢2 ) − 𝑢2 Τ𝛼 Τ𝛼 (𝑢2 )
Next, since 𝑢1 (𝑥) and 𝑢2 (𝑥) are both solutions of fractional equation (4), we have
Τ𝛼 Τ𝛼 (𝑢1 (𝑥) ) + 𝑃(𝑥)Τ𝛼 (𝑢1 (𝑥) ) + 𝑄(𝑥)𝑢1 (𝑥) = 0
Τ𝛼 Τ𝛼 (𝑢2 (𝑥) ) + 𝑃(𝑥)Τ𝛼 (𝑢2 (𝑥) ) + 𝑄(𝑥)𝑢2 (𝑥) = 0.
First equation multiplying by 𝑢2 subtract to the second equation by
𝑢1 result in
(𝑢1 Τ𝛼 Τ𝛼 (𝑢2 ) − 𝑢2 Τ𝛼 Τ𝛼 (𝑢2 )) + 𝑃(𝑥)(𝑢1 Τ𝛼 (𝑢2 ) −
𝑢2 Τ𝛼 (𝑢1 )) = 0
or
Τ𝛼 (𝑊) + 𝑃(𝑥)𝑊 = 0.
The general solution of this first order fractional differential equation based on conformable fractional derivative is (see [17])
𝑊 = 𝑊𝛼 (𝑥0 )𝑒 −Ι𝛼 (𝑃(𝑥)) ,
and since the exponential factor is never zero, the proof is completed (see [23]).
Theorem 3.4. If 𝑢1 (𝑥) and 𝑢2 (𝑥) are any two solutions of fractional equation (4) on an interval [𝑎, 𝑏], then they are linearly dependent on this interval if and only if their fractional Wronskian
𝑊 = 𝑊𝛼 (𝑢1 (𝑥), 𝑢2 (𝑥)), is identically zero.
Proof. We begin by assuming that 𝑢1 (𝑥) and 𝑢2 (𝑥) are linearly dependent, and we show
𝑊𝛼 (𝑢1 (𝑥), 𝑢2 (𝑥)) = 0.
First, if either of the functions is identically zero on [𝑎, 𝑏], then the
conclusion is clear. Therefore, we may therefor assume without loss
of generality, that neither of them is identically zero, and their linear
dependence result that each of those is a constant multiple of the
other one. Accordingly, we have 𝑢2 = 𝐶𝑢1 , for some constant, so
Τ𝛼 (𝑢2 ) = 𝐶Τ𝛼 (𝑢1 ).
By elimination 𝐶, from this equation, we obtain
𝑢1 Τ𝛼 (𝑢2 ) − 𝑢2 Τ𝛼 (𝑢1 ) = 0,
which proves this half of the theorem. We now assume that the
𝑊𝛼 (𝑢1 (𝑥), 𝑢2 (𝑥)) = 0, and prove linearly dependent. If 𝑢1 (𝑥) is
identically zero on [𝑎, 𝑏], then the functions are linearly dependent.
We may therefore assume that 𝑢1 (𝑥), does not vanish identically
on [𝑎, 𝑏], from which it follows by continuity that 𝑢1 (𝑥) does not
vanish at all on some subinterval [𝑐, 𝑑] of [𝑎, 𝑏]. Since the Wronskian is identically zero on [𝑎, 𝑏], we can divide it by 𝑢12 to get
𝑢1 Τ𝛼 (𝑢2 )−𝑢2 Τ𝛼 (𝑢1 )
𝑢12
= 0, on [𝑐, 𝑑].
𝑢
This can be written in the form Τ𝛼 ( 2) = 0, and by conformable
fractional integrating we obtain
𝑢2
𝑢1
𝑢1
= 𝐶, or 𝑢2 = 𝐶𝑢1 , for some
constant 𝐶 , and all 𝑥, in [𝑐, 𝑑]. Finally, since 𝑢2 , and 𝐶𝑢1 , have
equal value in [𝑐, 𝑑], they have equal conformable fractional derivative, so 𝑢2 = 𝐶𝑢1 , all 𝑥, in [𝑎, 𝑏], which concludes the argument
(see [23]).
Theorem 3.5. Let 𝑢1 (𝑥) and 𝑢2 (𝑥), be linearly dependent of the
homogeneous fractional equation (4), on the interval [𝑎, 𝑏]. Then
𝐶1 𝑢1 (𝑥) + 𝐶2 𝑢2 (𝑥), is the general solution of the fractional equation (4) on this interval.
Proof. Let 𝑢(𝑥), be any solution of (4) on [𝑎, 𝑏]. We must show
that constant 𝐶1 , 𝐶2 , can be found so that 𝑢(𝑥) = 𝐶1 𝑢1 (𝑥) +
𝐶2 𝑢2 (𝑥), for all 𝑥 in [𝑎, 𝑏]. Since 𝐶1 𝑢1 (𝑥) + 𝐶2 𝑢2 (𝑥), and 𝑢(𝑥)
are both solutions of (4) on [𝑎, 𝑏], it suffices to show that for some
point 𝑥0 , in [𝑎, 𝑏], we can find 𝐶1 , 𝐶2 so that
𝐶1 𝑢1 (𝑥0 ) + 𝐶2 𝑢2 (𝑥0 ) = 𝑢(𝑥0 ), and
𝐶1 Τ𝛼 (𝑢1 (𝑥0 )) + 𝐶2 Τ𝛼 (𝑢2 (𝑥0 )) = Τ𝛼 (𝑢(𝑥0 )).
For this system to be solvable for 𝐶1 , 𝐶2 , it suffices that the following determinant be non-zero.
𝑢 (𝑥 )
𝑢2 (𝑥0 )
| 1 0
|
Τ𝛼 (𝑢1 (𝑥0 )) Τ𝛼 (𝑢2 (𝑥0 ))
= 𝑢1 (𝑥0 ). Τ𝛼 (𝑢2 (𝑥0 )) − 𝑢2 (𝑥0 ). Τ𝛼 (𝑢1 (𝑥0 ))
This leads us to investigate the function Wronskian of
𝑢1 (𝑥), 𝑢2 (𝑥) at 𝑥0 . According to theorems 3.3 and 3.4 it is clear
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International Journal of Applied Mathematical Research
that 𝑊𝛼 (𝑢1 (𝑥0 ), 𝑢2 (𝑥0 )), have a value different from zero (see
[23]).
4. Determining a general solution of a homogeneous fractional equation
This section is motivated to obtain general solution of homogeneous fractional differential equations and then to solve Euler’s equation.
4.1. The use of a known solution to find another one or
D’Alambert approach
We assume that 𝑢1 (𝑥), is a known nonzero solution of Eq. (4),
𝑢2 (𝑥) = 𝑣(𝑥)𝑢1 (𝑥), is a solution of (4), where as 𝑣(𝑥) an unknown function (see [20]). So
We are now in a position to give a complete discussion of the homogeneous equation of Eq. (4) for the special case in which 𝑝 and
𝑞 are constants (see [23]).
Τ𝛼 Τ𝛼 (𝑢(𝑥)) + 𝑝 Τ𝛼 (𝑢(𝑥)) + 𝑞 𝑢(𝑥) = 0.
(6)
Our starting point is the fact that the exponential function e
,
has the property that its conformable fractional derivative are all
constant multiples of the function itself. It leads us to consider (see
[19])
1 𝛼
)
𝑢(𝑥) = 𝑒 𝑚(𝛼𝑥
(7)
as a possible solution for Eq. (6), we have
1 𝛼
Τ𝛼 (𝑢(𝑥)) = 𝑚𝑒 𝑚(𝛼𝑥 ) ,
(8)
and
1 𝛼
Τ𝛼 (𝑢2 (𝑥)) = 𝑣(𝑥). Τ𝛼 (𝑢1 (𝑥)) + 𝑢1 (𝑥). Τ𝛼 (𝑣(𝑥))
Τ𝛼 Τ𝛼 (𝑢(𝑥)) = 𝑚2 𝑒 𝑚(𝛼𝑥 ) .
Τ𝛼 Τ𝛼 (𝑢2 (𝑥)) = 𝑣(𝑥). Τ𝛼 Τ𝛼 (𝑢1 (𝑥)) +
and (9) into (6) yields to
2Τ𝛼 (𝑢1 (𝑥)). Τ𝛼 (𝑣(𝑥)) + 𝑢1 (𝑥). Τ𝛼 Τ𝛼 (𝑢1 (𝑥)) , Substituting Eqs. (7), (8),
1
𝑚( 𝑥 𝛼 )
2
(𝑚 + 𝑝 𝑚 + 𝑞)𝑒 𝛼
By substituting the above results into Eq. (4), we get
=0
𝑣(𝑥) (Τ𝛼 Τ𝛼 (𝑢1 (𝑥)) + 𝑃(𝑥)Τ𝛼 (𝑢1 (𝑥)) + 𝑄(𝑥)𝑢1 (𝑥))
+𝑢1 (𝑥)Τ𝛼 Τ𝛼 (𝑣(𝑥)) + Τ𝛼 (𝑣(𝑥))(𝑃(𝑥)𝑢1 (𝑥) + 2Τ𝛼 (𝑢1 (𝑥))
= 0.
Since u1 (x) is a solution of Eq. (4), It reduces
𝑢1 (𝑥)Τ𝛼 Τ𝛼 (𝑣(𝑥)) + Τ𝛼 (𝑣(𝑥))(𝑃(𝑥)𝑢1 (𝑥) + 2Τ𝛼 (𝑢1 (𝑥)) = 0
or
Τ𝛼 Τ𝛼 (𝑣(𝑥))
Τ𝛼 (𝑣(𝑥))
= −2
Τ𝛼 (𝑢1 (𝑥))
− 𝑃(𝑥).
𝑢1 (𝑥)
A fractional integration now gives
ln( Τ𝛼 (𝑣(𝑥))) = −2ln(𝑢1 (𝑥)) − Ι𝛼 (P(x)),
so
1
Τ𝛼 (𝑣(𝑥)) = 2
𝑒 −Ι𝛼 (𝑃(𝑥))
𝑢1 (𝑥)
and
𝑣(𝑥) = Ι𝛼 (
1
𝑢12 (𝑥)
𝑒 −Ι𝛼(𝑃(𝑥)) ).
Consequently, the general solution of homogeneous fractional
equation of (4) is as follows (see [23]),
1
𝑒 −Ι𝛼(𝑃(𝑥)) ) 𝑢1 (𝑥).
𝑢ℎ (𝑥) = 𝐶1 𝑢1 (𝑥) + 𝐶2 (Ι𝛼 ( 2
𝑢1 (𝑥)
3
Example 4.1.1. We know that u1 (x) = 3 √x is a solution of the
following homogeneous equation
3
3
9 √𝑥 2 Τ2 Τ2 (𝑢(𝑥)) − 6 √𝑥 Τ2 (𝑢(𝑥)) + 2 𝑢(𝑥) = 0.
3 3
According to (5), we have
3
𝑣(𝑥) = 3 √𝑥 .
Therefore
𝑢2 (𝑥) =
3
9 √𝑥 2 .
So the general solution is as follows,
3
𝑢ℎ (𝑥) = 𝐶1 √𝑥 + 𝐶2 √𝑥 2 .
Example 4.1.2. We know that u1 (x) = x is a solution of
2𝑥𝛵1 Τ1 (𝑢(𝑥)) + √𝑥Τ1 (𝑢(𝑥)) − 2𝑢(𝑥) = 0.
2 2
2
According to D’Alambert approach 𝑣(𝑥) and second solution
𝑢2 (𝑥) are obtained as follows,
𝑣(𝑥) = −
𝑥 −2
2
(9)
(10)
1
α
m( xα )
is never zero, (7) holds if and only if 𝑚 satisfies
and since e
the following auxiliary equation (see [23]),
𝑚2 + 𝑝 𝑚 + 𝑞 = 0.
(11)
It is clear that the roots 𝑚1 and 𝑚2 of Eq. (11) are distinct real
numbers if and only if 𝑝2 − 4𝑞 > 0.
In this case, we get the two solutions
1 𝛼
)
𝑢1 (𝑥) = 𝑒 𝑚1(𝛼𝑥
Since the ratio
1 𝛼
and 𝑢2 (𝑥) = 𝑒 𝑚2 (𝛼𝑥 ) .
1
m ( xα )
e 1 α
1
m ( xα )
e 2 α
1 α
)
= e(m1-m2 )(αx
is not constant,
these solutions are linearly independent and
1 𝛼
)
𝑢ℎ (𝑥) = 𝐶1 𝑒 𝑚1(𝛼𝑥
1 𝛼
)
+ 𝐶2 𝑒 𝑚2 (𝛼𝑥
is the general solution of Eq. (6).
,
1 𝛼
If 𝑚1 = 𝑚2 , then we obtain only one solution 𝑢1 (𝑥) = 𝑒 𝑚1(𝛼𝑥 ) .
Therefore, we can easily find a second linearly independent solution
by the D’Alambert method as the following form
1 𝛼
1
𝑢2 (𝑥) = ( 𝑥 𝛼 )𝑒 𝑚1 (𝛼𝑥 )
𝛼
and the general solution of Eq. (6) is
1 𝛼
1
𝑢ℎ (𝑥) = (𝐶1 + 𝐶2 ( 𝑥 𝛼 )) 𝑒 𝑚1 (𝛼𝑥 ) .
𝛼
If the roots 𝑚1 and 𝑚2 are distinct complex numbers, then they
can be written in the form 𝑎 ± 𝑖𝑏 and our two real solutions of Eq.
(6) are as follows
1 𝛼
1
𝑢1 (𝑥) = 𝑒 𝑎(𝛼𝑥 ) (cos 𝑏( 𝑥 𝛼 ))
𝛼
1
1 𝛼
𝑎( 𝑥 𝛼 )
𝑢2 (𝑥) = 𝑒 𝛼 (sin 𝑏( 𝑥 )).
𝛼
So the solution of Eq. (6) will be obtained as the following
3
3
1
α
m( xα )
, 𝑢2 (𝑥) = −
1
2𝑥
.
Therefor a general solution of above equation has the following
form,
𝑢ℎ (𝑥) = 𝐶1 𝑥 + 𝐶2 𝑥 −1 .
4.2. The homogeneous fractional equation with constant
coefficients
1 𝛼
)
𝑢ℎ (𝑥) = 𝑒 𝑎(𝛼𝑥
1
1
(𝐶1 cos 𝑏( 𝑥 𝛼 ) + 𝐶2 sin 𝑏( 𝑥 𝛼 )).
𝛼
𝛼
Example.4.2.1. Consider the following homogenous fractional differential equation
Τ𝛼 Τ𝛼 (𝑢(𝑥)) − 3Τ𝛼 (𝑢(𝑥)) + 𝑢(𝑥) = 0.
The general solution of the above equation is as follow
1 𝛼
)
𝑢ℎ (𝑥) = 𝐶1 𝑒 (𝛼𝑥
1 𝛼
+ 𝐶2 𝑒 2(𝛼𝑥 ) .
Example.4.2.2. The general solution of homogeneous equation of
Τ𝛼 Τ𝛼 (𝑢(𝑥)) + 4Τ𝛼 (𝑢(𝑥)) + 4𝑢(𝑥) = 0.
Will be obtained as follows
1 𝛼
1
𝑢ℎ (𝑥) = (𝐶1 + 𝐶2 ( 𝑥 𝛼 )) 𝑒 −2(𝛼𝑥 ) .
𝛼
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International Journal of Applied Mathematical Research
Example.4.2.3. Let’s consider following homogeneous fractional
differential equation
Τ𝛼 Τ𝛼 (𝑢(𝑥)) − 2Τ𝛼 (𝑢(𝑥)) + 3𝑢(𝑥) = 0.
Using signature approach result in
𝑢ℎ (𝑥) = 𝑒
1
𝛼
( 𝑥𝛼)
1
1
(𝐶1 cos √2( 𝑥 𝛼 ) + 𝐶2 sin √2( 𝑥 𝛼 )).
𝛼
𝛼
4.3. Euler’s equidimensional fractional equation
The homogeneous fractional differential equation,
1
𝑥>0
𝛼
(12)
where p, q are constant, is called Euler’s fractional equation (see
[23]). By using the change independent variable (see [19])
1
𝑧 = ln( 𝑥 𝛼 ),
we have
(13)
𝛼
1
Τ𝛼 (𝑢(𝑧)) = ( 𝑥 𝛼 )−1
𝛼
1
𝑑𝑢
𝑑𝑧
,
Τ𝛼 Τ𝛼 (𝑢(𝑧)) = − ( 𝑥 𝛼 )
𝛼
−2 𝑑𝑢
𝑑𝑧
1
+ ( 𝑥𝛼)
𝛼
−2 𝑑 2 𝑢
𝑑𝑧 2
.
(14)
Substituting Eqs. (13) and (14) into Eq. (12), leads to
𝑑2𝑢
𝑑𝑧
(𝑝 − 1)
2 +
𝑑𝑢
𝑑𝑧
+ 𝑞𝑢 = 0.
(15)
That equation (15) is an ordinary differential equation with constant
coefficient, and according to this approach the auxiliary equation
has the following form
𝑚2 + (𝑝 − 1)𝑚 + 𝑞 = 0
(16)
Suppose 𝑚1 and 𝑚2 are roots of Eq.’s (16) (see [23]). If they are
distinct real numbers, then the following solution 0f (12) can be obtained,
1
1
𝑢ℎ (𝑥) = 𝐶1 ( 𝑥 𝛼 )𝑚1 + 𝐶2 ( 𝑥 𝛼 )𝑚2 .
𝛼
𝛼
If 𝑚1 = 𝑚2 , we derive
1
1
𝑢ℎ (𝑥) = (𝐶1 + 𝐶2 ln ( 𝑥 𝛼 )) ( 𝑥 𝛼 )𝑚1 .
𝛼
𝛼
And if 𝑚1 and 𝑚2 are distinct complex numbers then the general
solution of Eq. (12) will be derive as follows
1
1
1
𝑢ℎ (𝑥) = ( 𝑥 𝛼 )𝑎 [𝐶1 cos 𝑏 ln ( 𝑥 𝛼 ) + 𝐶2 sin 𝑏 ln ( 𝑥 𝛼 )].
𝛼
𝛼
𝛼
Example.4.3.1. Let’s consider the following homogeneous fractional differential equation
1
1
( 𝑥 𝛼 )2 Τ𝛼 Τ𝛼 (𝑢(𝑥)) − 2( 𝑥 𝛼 )Τ𝛼 (𝑢(𝑥)) + 2𝑢(𝑥) = 0.
𝛼
𝛼
The auxiliary equation is as follows
𝑚 2 − 3 𝑚 + 2 = 0,
with the roots are 𝑚1 = 1, and 𝑚2 = 2 .
So the general solution is
1
1
𝑢ℎ (𝑥) = 𝐶1 ( 𝑥 𝛼 ) + 𝐶2 ( 𝑥 𝛼 )2 .
𝛼
𝛼
Example.4.3.2. consider following homogeneous equation
1
1
( 𝑥 𝛼 )2 Τ𝛼 Τ𝛼 (𝑢(𝑥)) − 3( 𝑥 𝛼 )Τ𝛼 (𝑢(𝑥)) + 4𝑢(𝑥) = 0.
𝛼
The root of auxiliary are
𝑚1 = 𝑚2 = 2.
𝛼
Thus the general solution is as follows,
1
1
𝑢ℎ (𝑥) = (𝐶1 + 𝐶2 ln ( 𝑥 𝛼 )) ( 𝑥 𝛼 )2 .
𝛼
𝛼
Example.4.3.3. Let’s consider Euler’s fractional equation as follows
1
𝛼 2
1
𝛼
( 𝑥 ) Τ𝛼 Τ𝛼 (𝑢(𝑥)) + 3( 𝑥 )Τ𝛼 (𝑢(𝑥)) + 2𝑢(𝑥) = 0.
𝛼
𝛼
The root of auxiliary equation are
𝑚1 = −1 + 𝑖, 𝑚2 = −1 − 𝑖.
Consequently, we have,
In this section, we have introduced variation of parameters and undetermined coefficients methods for determining a particular solution of nonhomogeneous fractional equations.
5.1. Variation of parameters or Lagrange approach
1
( 𝑥 𝛼 )2 Τ𝛼 Τ𝛼 (𝑢(𝑥)) + 𝑝 ( 𝑥 𝛼 ) Τ𝛼 (𝑢(𝑥)) + 𝑞𝑢(𝑥) = 0,
𝛼
5. Determining a particular solution of nonhomogeneous fractional equation
1
1
1
𝑢ℎ (𝑥) = ( 𝑥 𝛼 )−1 [𝐶1 cos(ln ( 𝑥 𝛼 )) +𝐶2 sin(ln ( 𝑥 𝛼 ))].
𝛼
𝛼
𝛼
Assume that 𝑢1 (𝑥) , 𝑢2 (𝑥) are two linearly independent solution
homogeneous fractional differential equation of the second order
fractional differential equation (3), we suppose that the particular
solution 𝑢𝑝 (𝑥), is
(19)
𝑢𝑝 (𝑥) = 𝜗1 (𝑥)𝑢1 (𝑥) + 𝜗2 (𝑥)𝑢2 (𝑥),
where 𝜗1 (𝑥) , 𝜗2 (𝑥) are two unknown functions (see [23]). By
computing the conformable fractional derivative of (19), we derive
Τ𝛼 (𝑢𝑝 (𝑥)) = (𝑢1 Τ𝛼 (𝜗1 (𝑥)) + 𝑢2 Τ𝛼 (𝜗2 (𝑥)))
(20)
+(𝜗1 Τ𝛼 (𝑢1 (𝑥)) + 𝜗2 Τ𝛼 (𝑢2 (𝑥)).
To avoid using second conformable fractional derivative, we suppose that
(21)
𝑢1 Τ𝛼 (𝜗1 (𝑥)) + 𝑢2 Τ𝛼 (𝜗2 (𝑥)) = 0.
So
Τ𝛼 (𝑢𝑝 (𝑥)) = 𝜗1 Τ𝛼 (𝑢1 (𝑥)) + 𝜗2 Τ𝛼 (𝑢2 (𝑥),
(22)
Therefore,
Τ𝛼 Τ𝛼 (𝑢𝑝 (𝑥)) = 𝜗1 Τ𝛼 Τ𝛼 (𝑢1 (𝑥)) + Τ𝛼 (𝜗1 (x))Τ𝛼 (𝑢1 (𝑥))
+Τ𝛼 (𝜗2 (𝑥))Τ𝛼 (𝑢2 (𝑥)) + 𝜗2 Τ𝛼 Τ𝛼 (𝑢2 (𝑥)).
(23)
By substituting (19) , (22) and (23) into equation (3), and some manipulation, we get
𝜗1 (Τ𝛼 Τ𝛼 (𝑢1 (𝑥)) + 𝑃(𝑥)Τ𝛼 (𝑢1 (𝑥)) + 𝑄(𝑥)𝑢1 (𝑥))
+𝜗2 (Τ𝛼 Τ𝛼 (𝑢2 (𝑥)) + 𝑃(𝑥)Τ𝛼 (𝑢2 (𝑥)) + 𝑄(𝑥)𝑢2 (𝑥))
+Τ𝛼 (𝜗1 (x))Τ𝛼 (𝑢1 (𝑥)) + Τ𝛼 (𝜗2 (𝑥))Τ𝛼 (𝑢2 (𝑥)) = 𝑅(𝑥). (24)
Since 𝑢1 (𝑥) , 𝑎𝑛𝑑 𝑢2 (𝑥) are solutions of (4), the two expressions
in parentheses are equal to zero, and (24) reduces to
Τ𝛼 (𝜗1 (x))Τ𝛼 (𝑢1 (𝑥)) + Τ𝛼 (𝜗2 (𝑥))Τ𝛼 (𝑢2 (𝑥)) = 𝑅(𝑥). (25)
By considering (21) and (25) together, we obtain the following results
−𝑢2 (𝑥)𝑅(𝑥)
𝜗1 (𝑥) = Ι𝛼 (
),
𝑊𝛼 (𝑢1 (𝑥), 𝑢2 (𝑥))
𝑢1 (𝑥)𝑅(𝑥)
)
𝜗2 (𝑥) = Ι𝛼 (
𝑊𝛼 (𝑢1 (𝑥), 𝑢2 (𝑥))
so
−𝑢2 (𝑥)𝑅(𝑥)
𝑢1 (𝑥)𝑅(𝑥)
𝑢𝑝 (𝑥) = 𝑢1 (𝑥). Ι𝛼 (
) + 𝑢2 (𝑥). Ι𝛼 (
).
𝑊𝛼 (𝑢1 (𝑥), 𝑢2 (𝑥))
𝑊𝛼 (𝑢1 (𝑥), 𝑢2 (𝑥))
Example.5.1.1. Let’s consider the following fractional equation
3
3
9 √𝑥 2 Τ2 Τ2 (𝑢(𝑥)) − 6 √𝑥 Τ2 (𝑢(𝑥)) + 2 𝑢(𝑥) = 9𝑥 2 √𝑥 2 .
3
3 3
3
(26)
By using example.4.1.1. the homogeneous solutions of (26) are as
follows,
3
3
𝑢1 (𝑥) = 3 √𝑥 , 𝑢2 (𝑥) = 9√𝑥 2 ,
and 𝑢𝑝 (𝑥) = −
39
56
𝑥 3 a particular solution of (26).
So the equation (26) has a general solution such as
3
3
𝑢(𝑥) = 𝐶1 √𝑥 + 𝐶2 √𝑥 2 −
39
56
𝑥3.
Example.5.1.2. we want to find the general solution of
2𝑥 Τ1 Τ1 (𝑢(𝑥)) + √𝑥Τ1 (𝑢(𝑥)) − 2𝑢(𝑥) = 4𝑥 3 .
2 2
2
By example.4.1.2. the homogeneous solutions are
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International Journal of Applied Mathematical Research
1
1
𝑢1 (𝑥) = 𝑥, 𝑎𝑛𝑑 𝑢2 (𝑥) = − , and 𝑢𝑝 (𝑥) = 𝑥 3 , Is a particular
4
2𝑥
solution of it. So the general solution of equation can be presented
as the following form
𝑢(𝑥) = 𝐶1 𝑥 + 𝐶2 𝑥 −1 + 0.25𝑥 3 .
5.2. Undetermined coefficients
Undetermined coefficients is a procedure for finding 𝑢𝑝 (𝑥) when
(3) has the form
Τ𝛼 Τ𝛼 (𝑢(𝑥)) + 𝑝 Τ𝛼 (𝑢(𝑥)) + 𝑞 𝑢(𝑥) = 𝑅(𝑥),
(27)
where 𝑝 , 𝑞 are constant and
2
𝑛
1
1
1
𝑅(𝑥) = (𝑎0 + 𝑎1 ( 𝑥 𝛼 ) + 𝑎2 ( 𝑥 𝛼 ) + ⋯ + 𝑎𝑛 ( 𝑥 𝛼 ) )
𝛼
𝛼
𝛼
1 𝛼
)
or
sin 𝛾 ( 𝑥 𝛼 )
𝛼
2
𝑛
1
1
1
R(x) = (𝑎0 + 𝑎1 ( 𝑥 𝛼 ) + 𝑎2 ( 𝑥 𝛼 ) + ⋯ + 𝑎𝑛 ( 𝑥 𝛼 ) )
𝛼
𝛼
𝛼
1 𝛼
)
1
𝑒 𝛽(𝛼𝑥
cos 𝛾 ( 𝑥 𝛼 ) .
𝛼
We choose a particular solution in the following form
2
𝑛
1
1
1
𝑢𝑝 (𝑥) = [ (𝐴0 + 𝐴1 ( 𝑥 𝛼 ) + 𝐴2 ( 𝑥 𝛼 ) + ⋯ + 𝐴𝑛 ( 𝑥 𝛼 ) )
𝛼
𝛼
𝛼
1 𝛼
)
𝑒 𝛽(𝛼𝑥
1
sin 𝛾 ( 𝑥 𝛼 ) +
𝛼
2
𝑛
1
1
1
(𝐵0 + 𝐵1 ( 𝑥 𝛼 ) + 𝐵2 ( 𝑥 𝛼 ) + ⋯ + 𝐵𝑛 ( 𝑥 𝛼 ) )
𝛼
𝛼
𝛼
1 𝛼
)
𝑒 𝛽(𝛼𝑥
1
1
sin 𝛾 ( 𝑥 𝛼 )] ( 𝑥 𝛼 )
𝛼
𝛼
𝑚
(28)
that 𝐴0 , 𝐴1 , … , 𝐴𝑛 , 𝐵0 , 𝐵1 , … , 𝐵𝑛 , unknown coefficients and 𝑚
is the lowest non-negative integer number, that removes homogeneous solutions, in choosing 𝑢𝑝 (𝑥). By substituting (28) into (27)
unknown coefficients will be obtained.
Example.5.2.1. let’s consider the following nonhomogeneous fractional equation
3
Τ2 Τ2 (𝑢(𝑥)) − 2Τ2 (𝑢(𝑥)) = 18√𝑥 2 − 10 .
3 3
3
(29)
Homogenous solutions of (29) are as follows
43
2
𝑢1 (𝑥) = 1 , 𝑢2 (𝑥) = 𝑒 3 √𝑥 ,
And a particular solution of it has the following form,
3
𝑢𝑝 (𝑥) = 3√𝑥 2 −
27
4
2
3
(√𝑥 2 ) .
So the general solution of (29) is
43
2
𝑢(𝑥) = 𝐶1 + 𝐶2 𝑒 3 √𝑥 + 3√𝑥 2 −
3
27
4
3
2
(√𝑥 2 ) .
Example.5.2.2. The general solution of
Τ1 Τ1 (𝑢(𝑥)) − 2Τ1 (𝑢(𝑥)) + 𝑢(𝑥) = 2√𝑥 𝑒 2√𝑥 ,
2 2
2
can be presented as the following form
𝑢(𝑥) = 𝐶1 𝑒 2√𝑥 + 𝐶2 √𝑥 𝑒 2√𝑥 +
4
3
𝑥 √𝑥 𝑒 2√𝑥 .
Example.5.2.3. Consider nonhomogeneous fractional equation as
follows
Τ1 Τ1 (𝑢(𝑥)) + 4 𝑢(𝑥) = 4 cos 4√𝑥 + 32𝑥 − 8√𝑥 ,
2 2
This equation has a general solution such as,
𝑢(𝑥) = 𝐶1 cos 4√𝑥 + 𝐶2 sin 4√𝑥 + 2√𝑥 sin 4√𝑥 + 8𝑥 − 2√𝑥 − 2 .
In this article, some methods such as, the use of a known solution
to find another one, homogenous equation with constant coefficients and Euler’s equidimensional equation have been introduced
for determining a general solution of homogenous fractional equations. The variation of parameters or Lagrange method, undetermined coefficients approach, for specifying a particular solution of
second order linear fractional differential equations, are presented.
This approach leads to an exact solution, thus there is no need of
using any numerical approach.
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𝑒 𝛽(𝛼𝑥
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