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In this article, two numerical techniques, namely, the homotopy perturbation and the matrix approach methods have been proposed and implemented to obtain an approximate solution of the linear fractional differential equation. To test the effectiveness of these methods, two numerical examples with known exact solution are illustrated. Numerical experiments show that the accuracy of these methods is in a good agreement with the exact solution. However, a comparison between these methods shows that the matrix approach method provides more accurate results.

Fractional differential equations appear frequently in various fields involving science and engineering, namely, in signal processing, control theory, diffusion, thermodynamics, biophysics, blood flow phenomena, rheology, electrodynamics, electrochemistry, electromagnetism, continuum and statistical mechanics and dynamical systems. For more details on the applications of fractional differential Equations (see [

In general, most of fractional differential equations do not have exact solutions. Instead, analytical and numerical methods become increasingly important for finding solution of fractional differential equations. In recent years, many efficient methods for solving FDEs have been developed. Among these methods are the monotone iterative technique [

The paper is organized as follows: In Section (2) we recall some basic definitions and notions concerning fractional calculus. In Section (3), we introduce the homotopy perturbation method. The matrix approach method is addressed in Section (4). The proposed methods are implemented using numerical examples with known analytical solution by applying MAPLE software in Section (5). Conclusions are given in Section (6).

In this section, we review some necessary definitions and mathematical preliminaries concerning fractional calculus that will be used in this work.

Definition 1. [

j p u ( x ) = 1 Γ ( p ) ∫ 0 x ( x − t ) p − 1 u ( t ) d t , x > 0

Definition 2. [

D 0 , t p u ( t ) = 1 Γ ( n − p ) d n d t n ∫ 0 t u ( τ ) ( t − τ ) p + 1 − n d τ (1)

Definition 3. [

D a t p u ( t ) = lim h → 0 m h = t − a h − p ∑ i = 0 m ( − 1 ) i ( p i ) u ( t − i h ) (2)

where ( p i ) = Γ ( p + 1 ) i ! ( Γ ( p + 1 ) ) .

Definition 4. [

D c t α ( t − c ) p = Γ ( p + 1 ) Γ ( − α + p + 1 ) ( t − c ) p − α (3)

Definition 5. [

D ∗ p u ( t ) = { 1 Γ ( n − α ) ∫ 0 t u n ( τ ) ( t − τ ) α + 1 − n d τ , n − 1 < α < n d n u ( τ ) d t n , α = n ∈ ℕ (4)

Theorem 1. [

D ∗ α t c = { Γ ( c + 1 ) Γ ( c − α + 1 ) t c − α = D α t c , n − 1 < α < n , c > n − 1 , c ∈ ℝ 0 , n − 1 < α < n , c ≤ n − 1 , c ∈ ℕ (5)

Theorem 2. [

D α ( u ( t ) g ( t ) ) = ∑ k = 0 ∞ ( α k ) [ D k g ( t ) ] [ D α − k u ( t ) ] (6)

Proof. See [

Theorem 3. [

D ∗ p ( u ( t ) g ( t ) ) = ∑ i = 0 ∞ ( p i ) [ D ∗ p g ( t ) ] [ D ∗ p − i u ( t ) ] − ∑ i = 0 m − 1 t i − p Γ ( i + 1 − p ) ( u ( t ) g ( t ) ) ( i ) (7)

Proof. See [

The fractional initial value problem in the operator form is:

D α f ( t ) + L f ( t ) = g ( t ) (8)

f ( i ) ( 0 ) = c i , i = 0 , 1 , 2 , ⋯ , n − 1 (9)

where c i is the initial conditions, L is the linear operator which might include other fractional derivative operators D β ( β < α ) , while the function g, the source function is assumed to be in c − 1 if α is an integer, and in c − 1 1 if α is not an integer. The solution f ( t ) is to be determined in c − 1 n .

In virtue of [

( 1 − p ) D α f + p [ D α f + L f ( t ) − g ( t ) ] = 0 (10)

or

D α f + p [ L f ( t ) − g ( t ) ] = 0 (11)

where p ∈ [ 0,1 ] is an embedding parameter. If p = 0 , Equations (10) and (11) become

D α f = 0 (12)

when p = 1 , both Equations (10) and (11) yields the original FDE Equation (8).

The solution of Equation (8) is:

f ( t ) = f 0 ( t ) + p f 1 ( t ) + p 2 f 2 ( t ) + p 3 f 3 ( t ) + ⋯ (13)

Substituting p = 1 in Equation (13) then we get the solution of Equation (8) in the form:

f ( t ) = f 0 ( t ) + f 1 ( t ) + f 2 ( t ) + f 3 ( t ) + ⋯ (14)

Substituting Equation (13) into Equation (11) and collecting all the terms with the same powers of p, we get:

p 0 : D α f 0 = 0 (15)

p 1 : D α f 1 = − L f 0 + g ( t ) (16)

p 2 : D α f 2 = − L f 1 ( t ) (17)

p 3 : D α f 3 = − L f 2 ( t ) (18)

and so on.

Following [

f 0 = ∑ i = 0 n − 1 f ( i ) ( 0 ) t i i ! = ∑ i = 0 n − 1 c i t i i !

f 1 = − Ω α [ L f 0 ( t ) ] + Ω α [ L g ( t ) ]

f 2 = − Ω α [ L f 1 ( t ) ]

f 3 = − Ω α [ L f 2 ( t ) ]

The general form of the HPM solution is:

f n = − Ω α [ L f n − 1 ( t ) ]

The homotopy perturbation solution takes the general form:

f ( t ) = f 0 + f 1 + f 2 + f 3 + ⋯ + f n + ⋯ (19)

Consider a function g ( x ) , defined in [ c , d ] , such that g ( x ) ≡ 0 for x < c and of real order m − 1 ≤ α ≤ m , such as:

D c x α g ( x ) = 1 Γ ( m − α ) ( d d x ) m ∫ c x g ( ϵ ) d ϵ ( x − ϵ ) α − m + 1 , ( c < x < d ) (20)

Let us take equidistant nodes with the step size

l : x i = i l ( i = 0 , 1 , ⋯ , N )

in the interval [ c , d ] , where x 0 = c and x N = d . Using the backward fractional difference approximation for the derivative at the points x i , i = 0 , 1 , ⋯ , N , we have:

c D x i α g ( x ) ≈ ∇ α g ( x i ) l α = l − α ∑ j = 0 i ( − 1 ) j ( α j ) g i − j , i = 0 , 1 , 2 , ⋯ , N (21)

Equation (21) is equivalent to the following matrix form [

( l − α ∇ α g ( x 0 ) l − α ∇ α g ( x 1 ) l − α ∇ α g ( x 2 ) ⋮ l − α ∇ α g ( x N − 1 ) l − α ∇ α g ( x N ) ) = 1 l α ( w 0 α 0 0 0 ⋯ 0 w 1 α w 0 α 0 0 ⋯ 0 w 2 α w 1 α w 0 α 0 ⋯ 0 ⋱ ⋱ ⋱ ⋱ ⋯ ⋱ w N − 1 α ⋱ w 2 α w 1 α w 0 α 0 w N − 2 α w N − 1 α ⋱ w 2 α w 1 α w 0 α ) ( g 0 g 1 g 2 ⋮ g N − 1 g N ) (22)

w i ( α ) = ( − 1 ) i ( α i ) , i = 0 , 1 , ⋯ , N (23)

In Equation (22), the column vector of functions g i ( i = 0 , 1 , ⋯ , N ) is multiplied by the matrix

A N α = 1 l α ( w 0 α 0 0 0 ⋯ 0 w 1 α w 0 α 0 0 ⋯ 0 w 2 α w 1 α w 0 α 0 ⋯ 0 ⋱ ⋱ ⋱ ⋱ ⋯ ⋱ w N − 1 α ⋱ w 2 α w 1 α w 0 α 0 w N − 2 α w N − 1 α ⋱ w 2 α w 1 α w 0 α ) (24)

the result is the column vector of approximated values of the fractional derivatives

c D x i α g ( x ) , i = 0 , 1 , ⋯ , N

The generating function for the matrix is

A α ( z ) = l − α ( 1 − z ) α (25)

Since A N α and A N β are lower triangular matrices then we have

A N α A N β = A N ( α + β )

Theorem 4. [

D c x α ( D c x β g ( x ) ) = D c x β ( D c x α g ( x ) ) = D c x α + β g ( x )

holds if

g ( i ) ( c ) = 0 , i = 0 , 1 , 2 , ⋯ , a − 1 , wher e a = max { n , m } (26)

then we can treat such matrices as discrete analogues of the corresponding left-sided fractional derivatives

c D x α and c D x β

where n − 1 ≤ α < n and m − 1 ≤ β < m .

Consider a function g ( x ) , defined in [ c , d ] , such that g ( x ) ≡ 0 for x > d .

Then its right sided fractional derivative of real order α ( m − 1 ≤ α < m ) is

D c x α g ( x ) = ( − 1 ) m Γ ( m − α ) ( d d x ) m ∫ x d g ( ϵ ) d ϵ ( x − ϵ ) α − m + 1 , ( c < x < d ) (27)

Thus we get the discrete analogue of the right sided fractional differentiation with the step size

l : x i = i l ( i = 0 , 1 , ⋯ , N )

in the interval [ c , d ] , where x 0 = c and x N = d , which is represented by the matrix [

G N α = 1 l α ( w 0 α w 1 α ⋱ ⋱ w N − 1 α w N α 0 w 0 α w 1 α ⋱ ⋱ w N − 1 α 0 0 w 0 α w 1 α ⋱ ⋱ ⋮ ⋮ ⋮ ⋱ ⋱ ⋱ 0 ⋯ 0 0 w 0 α w 1 α 0 ⋯ 0 0 0 w 0 α ) (28)

The generating function for the matrix G N α is the same for A N α

A α ( z ) = l − α ( 1 − z ) α

The transposition of the matrix A N α gives the matrix G N α and the opposite holds:

( A N α ) T = G N α , ( G N α ) T = A N α . (29)

In this section, in order to examine the accuracy of the proposed methods, we solve two numerical examples of fractional differential equations. Moreover, the numerical results will be compared with exact solution.

Example 1. Consider the linear fractional differential equation:

D α x ( t ) + x ( t ) = 2 Γ ( 3 − α ) t 2 − α + t 3 (30)

with initial conditions: x ( 0 ) = 0 , x ′ ( 0 ) = 0 .

The exact solution of Equation (30) with α = 1.9 is:

x ( t ) = t 2

In virtue of Equation (11), we can write Equation (30) in the homotopy form:

D α x ( t ) + p x ( t ) − 2 Γ ( 3 − α ) t 2 − α − t 3 = 0 (31)

the solution of Equation (30) is:

x ( t ) = x 0 ( t ) + p x 1 ( t ) + p 2 x 2 ( t ) + p 3 x 3 ( t ) + ⋯ (32)

Substituting Equation (32) into equation Equation (31) and collecting terms with the same power of p, we get:

{ p 0 : D α x 0 ( t ) = 0 p 1 : D α x 1 ( t ) = − x 0 ( t ) + f ( t ) p 2 : D α x 2 ( t ) = − x 1 ( t ) p 3 : D α x 3 ( t ) = − x 2 ( t ) ⋮ (33)

Applying Ω α and the inverse operator of D α , on both sides of Equation (33) and using the definition of Riemann-Liouville fractional integral operator ( Ω α ) of order α ≥ 0 we obtain:

x 0 ( t ) = ∑ i = 0 1 x ( i ) ( 0 ) t i i ! = x ( 0 ) t 0 0 ! + x ′ ( 0 ) t 1 1 ! = 0

x 1 ( t ) = − Ω α [ x 0 ( t ) + Ω α [ f ( t ) ] ] = − Ω α [ 2 Γ ( 3 − α ) t 2 − α + t 3 ] = Ω α [ 2 Γ ( 3 − α ) t 2 − α ] + Ω α [ t 3 ] = 2 Γ ( 3 − α ) Γ ( 3 − α ) Γ ( 3 − α + α ) t α + 2 − α + Γ ( 4 ) Γ ( 4 + α ) t 3 + α = t 2 + Γ ( 4 ) Γ ( 4 + α ) t 3 + α

x 2 ( t ) = − Ω α [ x 1 ( t ) ] = − Ω α [ t 2 ] − Ω α [ Γ ( 4 ) Γ ( 4 + α ) t 3 + α ] = − Γ ( 3 ) Γ ( 3 + α ) t 2 + α − Γ ( 4 ) Γ ( 4 + α ) Γ ( 4 + α ) Γ ( 4 + α + α ) t 3 + α + α = − 2 Γ ( 3 + α ) t 2 + α − 6 Γ ( 4 + 2 α ) t 3 + 2 α

x 3 ( t ) = − Ω α [ x 2 ( t ) ] = − Ω α [ − 2 Γ ( 3 + α ) t 2 + α − 6 Γ ( 4 + 2 α ) t 3 + 2 α ] = − Ω α [ − 2 Γ ( 3 + α ) t 2 + α ] − Ω α [ − 6 Γ ( 4 + 2 α ) t 3 + 2 α ] = 2 Γ ( 3 + α ) Γ ( 3 + α ) Γ ( 3 + 2 α ) t 2 + 2 α + 6 Γ ( 4 + 2 α ) Γ ( 4 + 2 α ) Γ ( 3 + 3 α ) t 3 + 3 α = 2 Γ ( 3 + 2 α ) t 2 + 2 α + 6 Γ ( 3 + 3 α ) t 3 + 3 α

Hence the solution of Equation (30) is:

x ( t ) = x 0 ( t ) + x 1 ( t ) + x 2 ( t ) + x 3 ( t ) + ⋯ (34)

x ( t ) = t 2 + Γ ( 4 ) Γ ( 4 + α ) t 3 + α − 2 Γ ( 3 + α ) t 2 + α − 6 Γ ( 4 + 2 α ) t 3 + 2 α + ⋯ (35)

when α = 1.9

x ( t ) = t 2 + 6 Γ ( 5.9 ) t 4.9 − 2 Γ ( 4.9 ) t 3.9 − 6 Γ ( 7.8 ) t 6.8 + ⋯ = t 2 + 0.059247439 t 4.9 − 0.096770806 t 3.9 − 0.001776766299 t 6.8 + ⋯ = t 2 − smallterms ≈ t 2 (36)

Now, we implement Algorithm 1 to solve Equation (30) using the matrix approach method.

t k | Exact solution x ( t ) = t 2 | Approximation solution x n ( t ) | Error = | x ( t ) − x n ( t ) | |
---|---|---|---|

0.0 | 0 | 0 | 0 |

0.1 | 0.1000 | 0.008621129726629 | 0.001378870273371 |

0.2 | 0.0400 | 0.037689907370590 | 0.00231009269410 |

0.3 | 0.0900 | 0.086539405801948 | 0.003460594198052 |

0.4 | 0.1600 | 0.154740697067752 | 0.005259302932248 |

0.5 | 0.2500 | 0.241931774005283 | 0.008068225994717 |

0.6 | 0.3600 | 0.347860012395293 | 0.012139987604707 |

0.7 | 0.4900 | 0.472440448226374 | 0.01755955177362 |

0.8 | 0.6400 | 0.615814946797499 | 0.024185053202501 |

0.9 | 0.8100 | 0.77840884576597 | 0.031591154203403 |

1 | 1.0000 | 0.960983804641099 | 0.039016195358901 |

Algorithm 1. Numerical realization using matrix approach method.

For N = 60 and N = 70 ,

Example 2. Consider the linear fractional differential equation:

D α x ( t ) + x ( t ) = 1 , α ∈ ( 1 , 2 ) (37)

with initial conditions: x ( 0 ) = 0 , x ′ ( 0 ) = 0 .

The exact solution of Equation (37) is:

x ( t ) = t 1.1 E 1.1 , 2.1 ( − t 1.1 )

Now, we implement the homotopy perturbation method to solve Equation (37).

In virtue of Equation (11), we can write Equation (37) in the homotopy form:

D α x ( t ) + p x ( t ) − 1 = 0 (38)

The solution of Equation (38) has the form:

x ( t ) = x 0 ( t ) + p x 1 ( t ) + p 2 x 2 ( t ) + p 3 x 3 ( t ) + ⋯ (39)

Substituting Equation (39) into Equation (38) and collecting terms with the same power of p, then we get:

{ p 0 : D α x 0 ( t ) = 0 p 1 : D α x 1 = − x 0 ( t ) + f ( t ) p 2 : D α x 2 ( t ) = − x 1 ( t ) p 3 : D α x 3 ( t ) = − x 2 ( t ) ⋮ (40)

Applying Ω α and the inverse operator of D α , on both sides of Equation (40), then we get: and using the definition of Riemann-Liouville fractional integral operator ( Ω α ) of order α ≥ 0 we obtain:

x 0 ( t ) = ∑ i = 0 1 x ( i ) ( 0 ) t i i ! = x ( 0 ) t 0 0 ! + x ′ ( 0 ) t 1 1 ! = 0

x 1 ( t ) = − Ω α [ x 0 ( t ) + Ω α [ f ( t ) ] ] = Ω α [ 1 ] = t α Γ ( 1 + α )

x 2 ( t ) = − Ω α [ x 1 ( t ) ] = − Ω α [ t α Γ ( 1 + α ) ] = − t 2 α Γ ( 2 α + 1 )

x 3 ( t ) = − Ω α [ x 2 ( t ) ] = − Ω α [ − t 2 α Γ ( 2 α + 1 ) ] = t 3 α Γ ( 3 α + 1 )

⋮

Then the solution of Equation (37) has the general form:

x ( t ) = x 0 ( t ) + x 1 ( t ) + x 2 ( t ) + x 3 ( t ) + ⋯ (41)

x ( t ) = t α Γ ( 1 + α ) − t 2 α Γ ( 2 α + 1 ) + t 3 α Γ ( 3 α + 1 ) + ⋯ (42)

x ( t ) = ∑ i = 1 ∞ ( − 1 ) i + 1 ( 0 ) t α i Γ ( α i + 1 ) (43)

when α = 1.1 , we get

x ( t ) = t 1.1 Γ ( 2.1 ) − t 2.2 Γ ( 3.2 ) − t 3.3 Γ ( 4.3 ) − t 4.4 Γ ( 5.4 ) + ⋯ = t 1.1 0.95135 − t 2.2 0.95135 − t 3.3 0.95135 − t 4.4 0.95135 + ⋯ = 0.95557 t 1.1 − 0.41255 t 2.2 + 0.11293 t 3.3 − 0.02242 t 4.4 + ⋯ (44)

Now, we implement Algorithm 1 to solve Equation (37) using the matrix approach method.

t k | Exact solution = ∑ i = 2 n ( − 1 ) i + 1 t 1.1 i Γ ( 1.1 i + 1 ) | Approximation solution x n ( t ) | Error = | x ( t ) − x n ( t ) | |
---|---|---|---|

0.0 | 0 | 0 | 0 |

0.1 | 0.073357053781371 | 0.058092499960337 | 0.015264553821034 |

0.2 | 0.151282884629052 | 0.135430820233760 | 0.015852064395293 |

0.3 | 0.226984580680193 | 0.211095760530730 | 0.015888820149463 |

0.4 | 0.29890238480688 | 0.283237827667134 | 0.015664557141554 |

0.5 | 0.366411147911488 | 0. 351151361983237 | 0.015259785928251 |

0.6 | 0.429259300754372 | 0.414572350125656 | 0.014686950628716 |

0.7 | 0.487372840288318 | 0.473456997040934 | 0.013915843247384 |

0.8 | 0.540762298480572 | 0.52788452424550 | 0.012877774238022 |

0.9 | 0.589469952960841 | 0.578006370352886 | 0.011463582607954 |

1 | 0.633536032460000 | 0.624016649518593 | 0.009519382941407 |

In this article, two numerical techniques namely, the homotopy perturbation method and the matrix approach method have been proposed and implemented to solve fractional differential equations. The accuracy and the validity of these techniques are tested with some numerical examples. The results show clearly that both techniques are in a good agreement with the analytical solution. According to numerical results mentioned in tables and figures, we conclude that the matrix approach method provides more accurate results than its counterpart and therefore is more advantageous. In addition, we strongly believe that the matrix approach method is regarded to be one of the most effective methods among the other methods mentioned in the literature. It is known for its fast converges and accuracy.

The authors declare that they have no conflict of interest.

Daraghmeh, A., Qatanani, N. and Saadeh, A. (2020) Numerical Solution of Fractional Differential Equations. Applied Mathematics, 11, 1100-1115. https://doi.org/10.4236/am.2020.1111074

1) D α : Fractional Derivative.

2) Γ ( p ) : Gamma Function.

3) Ω : Libera Integral Operator.

4) Ω α : Riemann-Liouville Fractional Integral Operator.

5) { sin α t α , cos α t α , sinh α t α , cosh α t α } : Mittage Leffer Functions.

6) j 0 : Identity Operator.

7) D d t p : Grumwald-Letnikov Fractional Derivative.

8) D 0, t p : Riemann-Liouville Derivative of Order p.

9) D ∗ α : Caputo Fractional Derivative.