<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">AM</journal-id><journal-title-group><journal-title>Applied Mathematics</journal-title></journal-title-group><issn pub-type="epub">2152-7385</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/am.2020.1111074</article-id><article-id pub-id-type="publisher-id">AM-104089</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Physics&amp;Mathematics</subject></subj-group></article-categories><title-group><article-title>
 
 
  Numerical Solution of Fractional Differential Equations
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Adnan</surname><given-names>Daraghmeh</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Naji</surname><given-names>Qatanani</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Aya</surname><given-names>Saadeh</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib></contrib-group><aff id="aff1"><addr-line>Department of Mathematics, Faculty of Science, An-Najah National University, Nablus, Palestine</addr-line></aff><pub-date pub-type="epub"><day>02</day><month>11</month><year>2020</year></pub-date><volume>11</volume><issue>11</issue><fpage>1100</fpage><lpage>1115</lpage><history><date date-type="received"><day>26,</day>	<month>September</month>	<year>2020</year></date><date date-type="rev-recd"><day>10,</day>	<month>November</month>	<year>2020</year>	</date><date date-type="accepted"><day>13,</day>	<month>November</month>	<year>2020</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  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.
 
</p></abstract><kwd-group><kwd>Fractional Calculus</kwd><kwd> Fractional Differential Equations</kwd><kwd> Homotopy Perturbation Method</kwd><kwd> Matrix Approach Method</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>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 [<xref ref-type="bibr" rid="scirp.104089-ref1">1</xref>] [<xref ref-type="bibr" rid="scirp.104089-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.104089-ref3">3</xref>] [<xref ref-type="bibr" rid="scirp.104089-ref4">4</xref>] [<xref ref-type="bibr" rid="scirp.104089-ref5">5</xref>] ). The concept of fractional calculus is now considered as a partial technique in many branches of science including physics (Oldham and Spanier [<xref ref-type="bibr" rid="scirp.104089-ref6">6</xref>] ). Srivastava et al. [<xref ref-type="bibr" rid="scirp.104089-ref7">7</xref>] gave the model of under actuated mechanical system with fractional order derivative and Sharma [<xref ref-type="bibr" rid="scirp.104089-ref8">8</xref>] studied advanced generalized fractional kinetic equation in Astrophysics. Caputo [<xref ref-type="bibr" rid="scirp.104089-ref9">9</xref>] reformulated the more “classic” definition of the Riemann-Liouville fractional derivatives in order to use integer order initial conditions to solve his fractional order differential equation. Kowankar and Gangel [<xref ref-type="bibr" rid="scirp.104089-ref10">10</xref>] reformulated the Riemann-Liouville fractional derivative in order to differentiate nowhere differentiable fractal functions.</p><p>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 [<xref ref-type="bibr" rid="scirp.104089-ref11">11</xref>] [<xref ref-type="bibr" rid="scirp.104089-ref12">12</xref>], topological degree theory [<xref ref-type="bibr" rid="scirp.104089-ref13">13</xref>], and fixed point theorems [<xref ref-type="bibr" rid="scirp.104089-ref14">14</xref>] [<xref ref-type="bibr" rid="scirp.104089-ref15">15</xref>] [<xref ref-type="bibr" rid="scirp.104089-ref16">16</xref>]. Moreover, numerical solutions are obtained by the following methods: the Adomian decomposition method and the variational iteration method [<xref ref-type="bibr" rid="scirp.104089-ref17">17</xref>], homotopy perturbation method [<xref ref-type="bibr" rid="scirp.104089-ref18">18</xref>] [<xref ref-type="bibr" rid="scirp.104089-ref19">19</xref>], Haar wavelet operational method [<xref ref-type="bibr" rid="scirp.104089-ref20">20</xref>], Neural networks [<xref ref-type="bibr" rid="scirp.104089-ref21">21</xref>], and so forth. Very recently, Hamdan et al. [<xref ref-type="bibr" rid="scirp.104089-ref22">22</xref>] applied Haar Wavelet and the product integration methods to solve the fractional Volterra integral equation of the second kind. In addition, Saadeh [<xref ref-type="bibr" rid="scirp.104089-ref23">23</xref>] in her master thesis has employed several numerical methods for solving fractional differential equations. These methods are the Adomian decomposition method, Homotopy perturbation method, Variational iteration method and Matrix approach method. A comparison between these methods has been carried out. In spirit, our numerical methods, namely the homotopy perturbation and the matrix approach methods are an improvement to those methods presented in the master thesis by one of the authors of this article. A comparison between these methods is carried out by solving some test examples using MAPLE software.</p><p>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).</p></sec><sec id="s2"><title>2. Mathematical Preliminaries and Notions</title><p>In this section, we review some necessary definitions and mathematical preliminaries concerning fractional calculus that will be used in this work.</p><p>Definition 1. [<xref ref-type="bibr" rid="scirp.104089-ref24">24</xref>] The Riemann-Liouville fractional integral operator of order p &gt; 0 , m − 1 &lt; p ≤ m , m ∈ ℕ of a function u ( x ) is defined as</p><p>j p u ( x ) = 1 Γ ( p ) ∫ 0 x ( x − t ) p − 1 u ( t ) d t ,     x &gt; 0</p><p>Definition 2. [<xref ref-type="bibr" rid="scirp.104089-ref25">25</xref>]: (Riemann-Liouville Derivative): Let n − 1 &lt; p &lt; n ∈ ℤ + . The Riemann-Liouville derivative of fractional order p is defined as:</p><p>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)</p><p>Definition 3. [<xref ref-type="bibr" rid="scirp.104089-ref26">26</xref>]: The Grumwald-Letnikov fractional derivative with fractional order p if u ( t ) ∈ ℂ n [ 0, t ] , is defined as:</p><p>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)</p><p>where ( p i ) = Γ ( p + 1 ) i ! ( Γ ( p + 1 ) ) .</p><p>Definition 4. [<xref ref-type="bibr" rid="scirp.104089-ref6">6</xref>]: Grunwald-Letnikov fractional derivative of the power function g ( t ) = ( t − c ) p is given as:</p><p>D c t α ( t − c ) p = Γ ( p + 1 ) Γ ( − α + p + 1 ) ( t − c ) p − α (3)</p><p>Definition 5. [<xref ref-type="bibr" rid="scirp.104089-ref27">27</xref>]: The Caputo derivative of fractional order p of a function u ( t ) is defined as</p><p>D ∗ p u ( t ) = { 1 Γ ( n − α ) ∫ 0 t u n ( τ ) ( t − τ ) α + 1 − n d τ , n − 1 &lt; α &lt; n d n u ( τ ) d t n , α = n ∈ ℕ (4)</p><p>Theorem 1. [<xref ref-type="bibr" rid="scirp.104089-ref28">28</xref>]: The Caputo fractional derivative of the power function satisfies:</p><p>D ∗ α t c = { Γ ( c + 1 ) Γ ( c − α + 1 ) t c − α = D α t c , n − 1 &lt; α &lt; n , c &gt; n − 1 , c ∈ ℝ 0 , n − 1 &lt; α &lt; n , c ≤ n − 1 , c ∈ ℕ (5)</p><p>Theorem 2. [<xref ref-type="bibr" rid="scirp.104089-ref26">26</xref>]: Leibniz rule for Riemann-Liouville fractional derivative: Let t &gt; 0 , α ∈ ℝ , m &gt; α &gt; m − 1 , and m ∈ ℕ . If u ( t ) , g ( t ) , and their derivatives are continuous on [ 0, t ] , then the following holds:</p><p>D α ( u ( t ) g ( t ) ) = ∑ k = 0 ∞ ( α k ) [ D k g ( t ) ] [ D α − k u ( t ) ] (6)</p><p>Proof. See [<xref ref-type="bibr" rid="scirp.104089-ref26">26</xref>] for more details.</p><p>Theorem 3. [<xref ref-type="bibr" rid="scirp.104089-ref29">29</xref>]: Leibniz rule for Caputo fractional derivative: Let t &gt; 0 , p ∈ ℝ , m &gt; p &gt; m − 1 , and m ∈ ℕ . If u ( t ) , g ( t ) , and their derivatives are continuous on [ 0, t ] , then the following holds</p><p>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)</p><p>Proof. See [<xref ref-type="bibr" rid="scirp.104089-ref29">29</xref>] for more details.</p></sec><sec id="s3"><title>3. Homotopy Perturbation Method (HPM)</title><p>The fractional initial value problem in the operator form is:</p><p>D α f ( t ) + L f ( t ) = g ( t ) (8)</p><p>f ( i ) ( 0 ) = c i ,     i = 0 , 1 , 2 , ⋯ , n − 1 (9)</p><p>where c i is the initial conditions, L is the linear operator which might include other fractional derivative operators D β ( β &lt; 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 .</p><p>In virtue of [<xref ref-type="bibr" rid="scirp.104089-ref16">16</xref>], we can write Equation (8) in the homotopy form</p><p>( 1 − p ) D α f + p [ D α f + L f ( t ) − g ( t ) ] = 0 (10)</p><p>or</p><p>D α f + p [ L f ( t ) − g ( t ) ] = 0 (11)</p><p>where p ∈ [ 0,1 ] is an embedding parameter. If p = 0 , Equations (10) and (11) become</p><p>D α f = 0 (12)</p><p>when p = 1 , both Equations (10) and (11) yields the original FDE Equation (8).</p><p>The solution of Equation (8) is:</p><p>f ( t ) = f 0 ( t ) + p f 1 ( t ) + p 2 f 2 ( t ) + p 3 f 3 ( t ) + ⋯ (13)</p><p>Substituting p = 1 in Equation (13) then we get the solution of Equation (8) in the form:</p><p>f ( t ) = f 0 ( t ) + f 1 ( t ) + f 2 ( t ) + f 3 ( t ) + ⋯ (14)</p><p>Substituting Equation (13) into Equation (11) and collecting all the terms with the same powers of p, we get:</p><p>p 0 : D α f 0 = 0 (15)</p><p>p 1 : D α f 1 = − L f 0 + g ( t ) (16)</p><p>p 2 : D α f 2 = − L f 1 ( t ) (17)</p><p>p 3 : D α f 3 = − L f 2 ( t ) (18)</p><p>and so on.</p><p>Following [<xref ref-type="bibr" rid="scirp.104089-ref6">6</xref>], we can write the first three terms of the homotopy perturbation method solution as:</p><p>f 0 = ∑ i = 0 n − 1     f ( i ) ( 0 ) t i i ! = ∑ i = 0 n − 1     c i t i i !</p><p>f 1 = − Ω α [ L f 0 ( t ) ] + Ω α [ L g ( t ) ]</p><p>f 2 = − Ω α [ L f 1 ( t ) ]</p><p>f 3 = − Ω α [ L f 2 ( t ) ]</p><p>The general form of the HPM solution is:</p><p>f n = − Ω α [ L f n − 1 ( t ) ]</p><p>The homotopy perturbation solution takes the general form:</p><p>f ( t ) = f 0 + f 1 + f 2 + f 3 + ⋯ + f n + ⋯ (19)</p></sec><sec id="s4"><title>4. Matrix Approach Method</title><sec id="s4_1"><title>4.1. Left-Sided Fractional Derivative</title><p>Consider a function g ( x ) , defined in [ c , d ] , such that g ( x ) ≡ 0 for x &lt; c and of real order m − 1 ≤ α ≤ m , such as:</p><p>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 &lt; x &lt; d ) (20)</p><p>Let us take equidistant nodes with the step size</p><p>l : x i = i l   ( i = 0 , 1 , ⋯ , N )</p><p>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:</p><p>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)</p><p>Equation (21) is equivalent to the following matrix form [<xref ref-type="bibr" rid="scirp.104089-ref11">11</xref>]:</p><p>( 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)</p><p>w i ( α ) = ( − 1 ) i ( α i ) ,     i = 0 , 1 , ⋯ , N (23)</p><p>In Equation (22), the column vector of functions g i ( i = 0 , 1 , ⋯ , N ) is multiplied by the matrix</p><p>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)</p><p>the result is the column vector of approximated values of the fractional derivatives</p><p>c D x i α g ( x ) ,     i = 0 , 1 , ⋯ , N</p><p>The generating function for the matrix is</p><p>A α ( z ) = l − α ( 1 − z ) α (25)</p><p>Since A N α and A N β are lower triangular matrices then we have</p><p>A N α A N β = A N ( α + β )</p><p>Theorem 4. [<xref ref-type="bibr" rid="scirp.104089-ref26">26</xref>]: Since the following</p><p>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 )</p><p>holds if</p><p>g ( i ) ( c ) = 0 ,     i = 0 , 1 , 2 , ⋯ , a − 1 ,     wher e     a = max { n , m } (26)</p><p>then we can treat such matrices as discrete analogues of the corresponding left-sided fractional derivatives</p><p>c D x α     and     c D x β</p><p>where n − 1 ≤ α &lt; n and m − 1 ≤ β &lt; m .</p></sec><sec id="s4_2"><title>4.2. Right-Sided Fractional Derivative</title><p>Consider a function g ( x ) , defined in [ c , d ] , such that g ( x ) ≡ 0 for x &gt; d .</p><p>Then its right sided fractional derivative of real order α ( m − 1 ≤ α &lt; m ) is</p><p>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 &lt; x &lt; d ) (27)</p><p>Thus we get the discrete analogue of the right sided fractional differentiation with the step size</p><p>l : x i = i l   ( i = 0 , 1 , ⋯ , N )</p><p>in the interval [ c , d ] , where x 0 = c and x N = d , which is represented by the matrix [<xref ref-type="bibr" rid="scirp.104089-ref6">6</xref>]:</p><p>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)</p><p>The generating function for the matrix G N α is the same for A N α</p><p>A α ( z ) = l − α ( 1 − z ) α</p><p>The transposition of the matrix A N α gives the matrix G N α and the opposite holds:</p><p>( A N α ) T = G N α ,     ( G N α ) T = A N α . (29)</p></sec></sec><sec id="s5"><title>5. Numerical Examples and Results</title><p>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.</p><p>Example 1. Consider the linear fractional differential equation:</p><p>D α x ( t ) + x ( t ) = 2 Γ ( 3 − α ) t 2 − α + t 3 (30)</p><p>with initial conditions: x ( 0 ) = 0 , x ′ ( 0 ) = 0 .</p><p>The exact solution of Equation (30) with α = 1.9 is:</p><p>x ( t ) = t 2</p><p>In virtue of Equation (11), we can write Equation (30) in the homotopy form:</p><p>D α x ( t ) + p x ( t ) − 2 Γ ( 3 − α ) t 2 − α − t 3 = 0 (31)</p><p>the solution of Equation (30) is:</p><p>x ( t ) = x 0 ( t ) + p x 1 ( t ) + p 2 x 2 ( t ) + p 3 x 3 ( t ) + ⋯ (32)</p><p>Substituting Equation (32) into equation Equation (31) and collecting terms with the same power of p, we get:</p><p>{ 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)</p><p>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:</p><p>x 0 ( t ) = ∑ i = 0 1     x ( i ) ( 0 ) t i i ! = x ( 0 ) t 0 0 ! + x ′ ( 0 ) t 1 1 ! = 0</p><p>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</p><p>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</p><p>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</p><p>Hence the solution of Equation (30) is:</p><p>x ( t ) = x 0 ( t ) + x 1 ( t ) + x 2 ( t ) + x 3 ( t ) + ⋯ (34)</p><p>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)</p><p>when α = 1.9</p><p>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)</p><p>Now, we implement Algorithm 1 to solve Equation (30) using the matrix approach method.</p><p><xref ref-type="table" rid="table1">Table 1</xref> displays the exact and numerical results using the matrix approach method with α = 1.9 and N = 51 . The maximum error with N = 51 is 0.039016195358901. <xref ref-type="fig" rid="fig1">Figure 1</xref>(a) compares both the exact and numerical solutions for the fractional differential Equation (30). Moreover, <xref ref-type="fig" rid="fig1">Figure 1</xref>(b) shows the absolute error between exact and numerical solutions.</p><table-wrap id="table1" ><label><xref ref-type="table" rid="table1">Table 1</xref></label><caption><title> The exact and numerical solutions using the matrix approach method where N = 51 </title></caption><table><tbody><thead><tr><th align="center" valign="middle" >t k</th><th align="center" valign="middle" >Exact solution x ( t ) = t 2</th><th align="center" valign="middle" >Approximation solution x n ( t )</th><th align="center" valign="middle" >Error = | x ( t ) − x n ( t ) |</th></tr></thead><tr><td align="center" valign="middle" >0.0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td></tr><tr><td align="center" valign="middle" >0.1</td><td align="center" valign="middle" >0.1000</td><td align="center" valign="middle" >0.008621129726629</td><td align="center" valign="middle" >0.001378870273371</td></tr><tr><td align="center" valign="middle" >0.2</td><td align="center" valign="middle" >0.0400</td><td align="center" valign="middle" >0.037689907370590</td><td align="center" valign="middle" >0.00231009269410</td></tr><tr><td align="center" valign="middle" >0.3</td><td align="center" valign="middle" >0.0900</td><td align="center" valign="middle" >0.086539405801948</td><td align="center" valign="middle" >0.003460594198052</td></tr><tr><td align="center" valign="middle" >0.4</td><td align="center" valign="middle" >0.1600</td><td align="center" valign="middle" >0.154740697067752</td><td align="center" valign="middle" >0.005259302932248</td></tr><tr><td align="center" valign="middle" >0.5</td><td align="center" valign="middle" >0.2500</td><td align="center" valign="middle" >0.241931774005283</td><td align="center" valign="middle" >0.008068225994717</td></tr><tr><td align="center" valign="middle" >0.6</td><td align="center" valign="middle" >0.3600</td><td align="center" valign="middle" >0.347860012395293</td><td align="center" valign="middle" >0.012139987604707</td></tr><tr><td align="center" valign="middle" >0.7</td><td align="center" valign="middle" >0.4900</td><td align="center" valign="middle" >0.472440448226374</td><td align="center" valign="middle" >0.01755955177362</td></tr><tr><td align="center" valign="middle" >0.8</td><td align="center" valign="middle" >0.6400</td><td align="center" valign="middle" >0.615814946797499</td><td align="center" valign="middle" >0.024185053202501</td></tr><tr><td align="center" valign="middle" >0.9</td><td align="center" valign="middle" >0.8100</td><td align="center" valign="middle" >0.77840884576597</td><td align="center" valign="middle" >0.031591154203403</td></tr><tr><td align="center" valign="middle" >1</td><td align="center" valign="middle" >1.0000</td><td align="center" valign="middle" >0.960983804641099</td><td align="center" valign="middle" >0.039016195358901</td></tr></tbody></table></table-wrap><disp-formula id="scirp.104089-formula5"><graphic  xlink:href="//html.scirp.org/file/5-7404563x134.png"  xlink:type="simple"/></disp-formula><p>Algorithm 1. Numerical realization using matrix approach method.</p><p>For N = 60 and N = 70 , <xref ref-type="fig" rid="fig2">Figure 2</xref>(a) and <xref ref-type="fig" rid="fig3">Figure 3</xref>(a) compare both the exact and numerical solutions for the fractional differential Equation (30). Moreover, <xref ref-type="fig" rid="fig2">Figure 2</xref>(b) and <xref ref-type="fig" rid="fig3">Figure 3</xref>(b) show the absolute error between exact and numerical solutions.</p><p>Example 2. Consider the linear fractional differential equation:</p><p>D α x ( t ) + x ( t ) = 1 ,     α ∈ ( 1 , 2 ) (37)</p><p>with initial conditions: x ( 0 ) = 0 , x ′ ( 0 ) = 0 .</p><p>The exact solution of Equation (37) is:</p><p>x ( t ) = t 1.1 E 1.1 , 2.1 ( − t 1.1 )</p><p>Now, we implement the homotopy perturbation method to solve Equation (37).</p><p>In virtue of Equation (11), we can write Equation (37) in the homotopy form:</p><p>D α x ( t ) + p x ( t ) − 1 = 0 (38)</p><p>The solution of Equation (38) has the form:</p><p>x ( t ) = x 0 ( t ) + p x 1 ( t ) + p 2 x 2 ( t ) + p 3 x 3 ( t ) + ⋯ (39)</p><p>Substituting Equation (39) into Equation (38) and collecting terms with the same power of p, then we get:</p><p>{ 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)</p><p>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:</p><p>x 0 ( t ) = ∑ i = 0 1     x ( i ) ( 0 ) t i i ! = x ( 0 ) t 0 0 ! + x ′ ( 0 ) t 1 1 ! = 0</p><p>x 1 ( t ) = − Ω α [ x 0 ( t ) + Ω α [ f ( t ) ] ] = Ω α [ 1 ] = t α Γ ( 1 + α )</p><p>x 2 ( t ) = − Ω α [ x 1 ( t ) ] = − Ω α [ t α Γ ( 1 + α ) ] = − t 2 α Γ ( 2 α + 1 )</p><p>x 3 ( t ) = − Ω α [ x 2 ( t ) ] = − Ω α [ − t 2 α Γ ( 2 α + 1 ) ] = t 3 α Γ ( 3 α + 1 )</p><p>⋮</p><p>Then the solution of Equation (37) has the general form:</p><p>x ( t ) = x 0 ( t ) + x 1 ( t ) + x 2 ( t ) + x 3 ( t ) + ⋯ (41)</p><p>x ( t ) = t α Γ ( 1 + α ) − t 2 α Γ ( 2 α + 1 ) + t 3 α Γ ( 3 α + 1 ) + ⋯ (42)</p><p>x ( t ) = ∑ i = 1 ∞ ( − 1 ) i + 1 ( 0 ) t α i Γ ( α i + 1 ) (43)</p><p>when α = 1.1 , we get</p><p>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)</p><p>Now, we implement Algorithm 1 to solve Equation (37) using the matrix approach method. <xref ref-type="table" rid="table2">Table 2</xref> contains the exact and numerical results using the matrix approach method with α = 1.1 and N = 51 . The maximum error with N = 51 is 0.015888820145463. <xref ref-type="fig" rid="fig4">Figure 4</xref>(a) compares both the exact and numerical solutions for the (37). Moreover, <xref ref-type="fig" rid="fig4">Figure 4</xref>(b) shows the absolute error between exact and numerical solutions. For N = 60 and N = 70 , <xref ref-type="fig" rid="fig5">Figure 5</xref>(a) and <xref ref-type="fig" rid="fig6">Figure 6</xref>(a) compare both the exact and numerical solutions for the fractional differential Equation (30). Moreover, <xref ref-type="fig" rid="fig5">Figure 5</xref>(b) and <xref ref-type="fig" rid="fig6">Figure 6</xref>(b) show the absolute error between exact and numerical solutions.</p><table-wrap id="table2" ><label><xref ref-type="table" rid="table2">Table 2</xref></label><caption><title> The exact and numerical solutions using the matrix approach method where N = 51 </title></caption><table><tbody><thead><tr><th align="center" valign="middle" >t k</th><th align="center" valign="middle" >Exact solution = ∑ i = 2 n ( − 1 ) i + 1 t 1.1 i Γ ( 1.1 i + 1 )</th><th align="center" valign="middle" >Approximation solution x n ( t )</th><th align="center" valign="middle" >Error = | x ( t ) − x n ( t ) |</th></tr></thead><tr><td align="center" valign="middle" >0.0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td></tr><tr><td align="center" valign="middle" >0.1</td><td align="center" valign="middle" >0.073357053781371</td><td align="center" valign="middle" >0.058092499960337</td><td align="center" valign="middle" >0.015264553821034</td></tr><tr><td align="center" valign="middle" >0.2</td><td align="center" valign="middle" >0.151282884629052</td><td align="center" valign="middle" >0.135430820233760</td><td align="center" valign="middle" >0.015852064395293</td></tr><tr><td align="center" valign="middle" >0.3</td><td align="center" valign="middle" >0.226984580680193</td><td align="center" valign="middle" >0.211095760530730</td><td align="center" valign="middle" >0.015888820149463</td></tr><tr><td align="center" valign="middle" >0.4</td><td align="center" valign="middle" >0.29890238480688</td><td align="center" valign="middle" >0.283237827667134</td><td align="center" valign="middle" >0.015664557141554</td></tr><tr><td align="center" valign="middle" >0.5</td><td align="center" valign="middle" >0.366411147911488</td><td align="center" valign="middle" >0. 351151361983237</td><td align="center" valign="middle" >0.015259785928251</td></tr><tr><td align="center" valign="middle" >0.6</td><td align="center" valign="middle" >0.429259300754372</td><td align="center" valign="middle" >0.414572350125656</td><td align="center" valign="middle" >0.014686950628716</td></tr><tr><td align="center" valign="middle" >0.7</td><td align="center" valign="middle" >0.487372840288318</td><td align="center" valign="middle" >0.473456997040934</td><td align="center" valign="middle" >0.013915843247384</td></tr><tr><td align="center" valign="middle" >0.8</td><td align="center" valign="middle" >0.540762298480572</td><td align="center" valign="middle" >0.52788452424550</td><td align="center" valign="middle" >0.012877774238022</td></tr><tr><td align="center" valign="middle" >0.9</td><td align="center" valign="middle" >0.589469952960841</td><td align="center" valign="middle" >0.578006370352886</td><td align="center" valign="middle" >0.011463582607954</td></tr><tr><td align="center" valign="middle" >1</td><td align="center" valign="middle" >0.633536032460000</td><td align="center" valign="middle" >0.624016649518593</td><td align="center" valign="middle" >0.009519382941407</td></tr></tbody></table></table-wrap></sec><sec id="s6"><title>6. Conclusion</title><p>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.</p></sec><sec id="s7"><title>Conflicts of Interest</title><p>The authors declare that they have no conflict of interest.</p></sec><sec id="s8"><title>Cite this paper</title><p>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</p></sec><sec id="s9"><title>List of Nomenclatures</title><p>1) D α : Fractional Derivative.</p><p>2) Γ ( p ) : Gamma Function.</p><p>3) Ω : Libera Integral Operator.</p><p>4) Ω α : Riemann-Liouville Fractional Integral Operator.</p><p>5) { sin α t α , cos α t α , sinh α t α , cosh α t α } : Mittage Leffer Functions.</p><p>6) j 0 : Identity Operator.</p><p>7) D d t p : Grumwald-Letnikov Fractional Derivative.</p><p>8) D 0, t p : Riemann-Liouville Derivative of Order p.</p><p>9) D ∗ α : Caputo Fractional Derivative.</p></sec></body><back><ref-list><title>References</title><ref id="scirp.104089-ref1"><label>1</label><mixed-citation publication-type="other" xlink:type="simple">Carpinteri, A. and Mainardi, F. (2014) Fractals and Fractional Calculus in Continuum Mechanics. 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