SYSTEMS OF LINEAR FRACTIONAL DIFFERENTIAL EQUATIONS ∗
M. ABU HAMMAD1 AND R. KHALIL1 1
University of Jordan, Jordan.
AUTHORS’ CONTRIBUTIONS This work was carried out in collaboration between both authors. Both authors read and approved the final manuscript.
Original Research Article ABSTRACT In this paper we discuss systems of conformable linear differential equations with constant coefficients. We give full solution for homogeneous and non-homogeneous systems.
Keywords: Conformable fractional derivative; systems of fractional differential equations. AMS Classification Number: 26A33.
1
Introduction
There are many definitions available in the literature for fractional derivatives. The main ones are the Riemann Liouville definition and the Caputo definition, see [1] . (i) Riemann - Liouville Definition. For α ∈ [n − 1, n), the α derivative of f is Daα (f )(t) =
1 dn Γ(n − α) dtn
Zt
f (x) dx. (t − x)α−n+1
a
(ii) Caputo Definition. For α ∈ [n − 1, n), the α derivative of f is Daα (f )(t) =
1 Γ(n − α)
Zt
f (n) (x) dx. (t − x)α−n+1
a
Such definitions have many setbacks that one can verify easily such as (i) The Riemann-Liouville derivative does not satisfy Daα (1) = 0 (Daα (1) = 0 for the Caputo derivative), if α is not a natural number. *Corresponding author: E-mail:
[email protected];
(ii) All fractional derivatives do not satisfy the known formula of the derivative of the product of two functions: α Daα (f g) = f Dα a (g) + gD a (f ). (iii) All fractional derivatives do not satisfy the known formula of the derivative of the quotient of two functions: α gDα a (f ) − f D a (g) Daα (f /g) = . g2 (iv) All fractional derivatives do not satisfy the chain rule: Daα (f ◦ g)(t) =f (α) (g(t)) g (α) (t). (v) All fractional derivatives do not satisfy: Dα Dβ f = Dα+β f , in general. (vi) All fractional derivatives, specially Caputo definition, assumes that the function f is differentiable. Recently, the authors in [2], gave a new definition of fractional derivative which is a natural extension to the usual first derivative. So many papers since then were written, and many equations were solved using such definition. We refer to [3], [4], [5], [6],[7], [8] and [9] and references there in for recent results on conformable fractional derivative. The definition goes as follows: Given a function f : [0, ∞) −→ R. Then for all t > 0,
α ∈ (0, 1), let 1−α
) − f (t) , ε is called the conformable fractional derivative of f of order α. Tα (f )(t) = lim
f (t + εt
ε→0
Tα
Let f (α) (t) stands for Tα (f )(t). If f is α−differentiable in some (0, b), b > 0, and lim f (α) (t) exists, then define t→0+
f (α) (0) = lim f (α) (t). t→0+
According to this definition, we have the following properties, [2], 1. Tα (1) = 0, 2. Tα (tp ) = ptp−α for all p ∈ R, 3. Tα (sin at) = at1−α cos at,
a ∈ R,
4. Tα (cos at) = −at1−α sin at, 5. Tα (eat ) = at1−α eat ,
a∈R
a ∈ R.
Further, many functions behave as in the usual derivative. Here are some formulas 1 Tα ( tα ) = 1 α 1 α
1 α
Tα (e α t ) = e α t , 1 1 Tα (sin tα ) = cos( tα ), α α 1 1 Tα (cos tα ) = − sin( tα ). α α
In this paper we use the conformable fractional derivative to study systems of nonhomogeneous linear fractional differential equations.
2
Homogenous Solution
For simplicity, we will consider 2x2-systems. So let y1 , y2 be any two functions of the variable x such that
(α)
= a11 y1 + a12 y2 + f1
(α)
= a21 y1 + a22 y2 + f2
y1 y2
with α ∈ (0, 1]. This system can be written in the form
Y (α) = AY + F............(1) where A =
a11 a21
a12 a22
, Y =
y1 y2
, and F =
f1 f2
Theorem 2.1. Let A has two real distinct real eigenvalues λ1 and λ2 say, with corresponding eigenvectors E1 and E2 respectively. Then the system Y (α) = AY.................(2) has two independent solutions Y1 = E1 e and Y2 = E2 e
λ 1 tα α
λ 2 tα α
and the general solution of the system (2) is Yh = c1 Y1 + c2 Y2 where Yh is the solution of the homogenous system (2). Proof. (α)
Y1
=
λ 1 tα λ 1 tα λ1 tα dα dα λ1αtα α α α E e = E e = E λ e = AE e = AY1 1 1 1 1 1 dtα dtα
Similarly for Y2 . Since system (2) is linear and homogenous then the result follows. Example 1. Consider the system Y (α) =
1 1
3 −1
Y
Solution. The eigenvalues ofthe matrix are λ1 = 2, and λ2 = −2. The corresponding eigenvectors −1 3 . and E2 = are E1 = 1 1 tα tα tα tα −1 3 −1 3 Hence Y1 = e2 α , and Y2 = e−2 α , and Yh = c1 e2 α + c2 e−2 α . 1 1 1 1 Theorem 2.2. Let A has a complex eigenvalue λ = c + id with corresponding eigenvectors E. −
−−
Then λ = c − id is an eigenvector with eigenvector E . Then the system Y (α) = AY.................(2) has two independent solutions: Y1 = e and Y2 = e
ctα α
ctα α
(G cos d
tα tα + H sin d ) α α
(H cos d
tα tα − G sin d ) α α
where
−−
−−
E− E E+ E , H=i G= 2 2 Further Yh = c1 Y1 + c2 Y2 . Proof. Follows the same lines as in Theorem 2.1, and the facts: Tα (sin
1 α 1 t ) = cos( tα ) α α
and
1 1 α t ) = − sin( tα ) α α 0 −1 Example 2. Consider the system Y (α) = Y. The eigenvalues are i and −i. The 1 0 −− i −i 0 −1 eigenvectors are E = , and E = . So G = , and H = . 1 1 1 0 Tα (cos
So Y1 =
0 1
cos
tα + α
−1 0
−1 0
sin
tα = Y1 = α
and Y2 =
cos
tα − α
0 1
0 1
cos
sin
tα + α
tα = α
−1 0
sin
α
− cos tα α − sin tα
tα = α
α
− sin tα α cos tα
Theorem 2.3. Let A has a repeated real eigenvalue λ say, with an eigenvector E. Then the system Y (α) = AY has two linearly independent solutions Y1 = Ee
λtα α
and Y2 =
b1 b2
tα λtαα e + α
c1 c2
e
λtα α
and c1 b1 are to be determined through substituting Y2 in the system (2). Further and c2 b2 Yh = c1 Y1 + c2 Y2 .
where
Proof. The proof follows the same lines as in the classical case. Example 3. Consider the system Y (α) =
−2 3
−3 4
Y
The of A are 1 and 1. An eigenvector for A corresponding to the eigenvalue 1 is eigenvalues 1 . Hence −1 α t 1 Y1 = eα −1 For Y2 , we put Y2 =
b1 b2
tα tαα e + α
c1 c2
e
tα α
. Using the properties fractional of conformable derivative when substituting in the system we get b1 3 c1 0 = , and = . Hence b2 −3 c2 −1 α α α t t tα 0 3 e + eα. Y2 = −1 −3 α
3
Particular Solution
Consider the system Y (α) = AY + F. In section 2, we have found the solution of the homogenous system Y (α) = AY. In this section we try to find a particular solution of the nonhomogeneous system Y (α) = AY + F, say Yp . Then the general solution will be Yg = Yh + Yp Let Y1 and Y2 be the two independent solutions of the system Y (α) = AY, where A is a 2x2 matrix. Form the matrix φ(t) = Y1 Y2 , whose columns are Y1 and Y2 . Since Y1 and Y2 are independent, then φ(t) is invertible. Let us denote the conformable integral of any function g by I(α) (g), [2], where Rt (x) dx. I(α) (g) = xf1−α a
Theorem 3.1. A particular solution of the system Y (α) = AY + F is Yp = φ(t)I(α) (φ−1 (t)F (t)) Proof. The proof is just the method of variation of parameters. Now, c1 Yh = c1 Y1 + c2 Y2 = φ(t) . c2
So we replace the parameters c1 and c2 by functions say u1 and u2 . Following the classical procedure to get Yp = φ(t)I(α) (φ−1 (t)F (t). (α)
Now let us check that such Yp satisfies Yp
= AYp + F :
Since the conformable derivative satisfies the multiplication rule for differentiation, we get: (φ(t)I(α) (φ−1 (t)F (t))(α)
=
φ(α) (t)I(α) (φ−1 (t)F (t) + φ(t)φ−1 (t)F (t)
=
φ(α) (t)I(α) (φ−1 (t)F (t) + F (t).............(∗) (α)
= AYi , then it follows that φ(α) (t) = Aφ(t).
But since both Y1 and Y2 satisfy the equation Yi Consequently, φ(α) (t)I(α) (φ−1 (t)F (t)
(α)
Thus we get from equation (∗) that Yp
=
Aφ(t)I(α) (φ−1 (t)F (t)
=
AYp
= AYp + F.
Example 4. Consider the system Y (α) =
4 3
2 −1
15 4
Y −
te−2
tα α
1 . −1 The eigenvalues of the matrix are −2 and 5, and the corresponding homogeneous solution is ! ! tα tα e−2 α 2e5 α + c2 Yh = c 1 tα tα −3e−2 α −3e5 α ! tα tα e−2 α 2e5 α Now, the fundamental matrix φ(t) = . The inverse of φ(t) is tα tα −3e−2 α −3e5 α ! tα tα 1 −3 tαα e5 α −2e5 α −1 . φ (t) = e tα tα 7 e−2 α 3e−2 α with Y (0) =
Now, Zt Yp = φ(t)
(φ−1 (s)F (s)) ds, s1−α
0
where
F (t) = −
15 4
te−2
tα α
One can see that φ−1 (s)F (s) =
−s α −7 sα
−7se
. ! .
Hence (φ−1 (s)F (s)) = s1−α
!
−sα −7sα e−7
sα α
.
Consequently, Zt φ(t)
Zt (φ−1 (s)F (s)) ds = φ(t) s1−α
0
!
−sα −7sα e−7
sα α
ds.
0
Competing Interests Authors have declared that no competing interests exist.
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