# A New Approach for the Approximate Analytical Solution of Space-Time Fractional Differential Equations by the Homotopy Analysis Method

Authors: Ali Demir; Mine Aylin Bayrak; Ebru Ozbilge
Journal: Advances in Mathematical Physics
 (2019)

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## Abstract

The motivation of this study is to construct the truncated solution of space-time fractional differential equations by the homotopy analysis method (HAM). The first space-time fractional differential equation is transformed into a space fractional differential equation or a time fractional differential equation before the HAM. Then the power series solution is constructed by the HAM. Finally, taking the illustrative examples into consideration we reach the conclusion that the HAM is applicable and powerful technique to construct the solution of space-time fractional differential equations.


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## Body

## 1. Introduction

The fractional calculus is a very attractive concept for scientists in various fields since it has a wide application in mathematics, physics, biology, and so on [1–4]. Various fractional derivatives are defined to get the solution of fractional differential equations [5, 6]. In order to get the solution, different methods are introduced and applied [7–11]. These methods include the finite difference method [12], the homotopy perturbation method (HPM) [13, 14], and the generalized differential transform method [15].Since the mathematical models include space and time variables in the modelling of mathematical and physical problems, fractional derivatives are required. The HAM provides as simple way to construct truncated solution of various differential equations. The advantage of the HAM is that the accuracy of the HAM does not depend on small parameter in the considered equation [16–18]. Therefore, compare the other technique HAM has many advantage in the application [19–21].The motivation of this paper is to determine the truncated solution of the following space-time FPDE by implementing the HAM with a new algorithm by which space-time fractional differential equation is reduced to either a space fractional differential equation or a time fractional differential equation making the problem easier to deal with by the HAM:(1)Dtαu=Dxβu+F,(2)ux,0=φx,0<x≤L(3)u0,t=μ1t,0<t≤T(4)ux0,t=μ2t,0<t≤Twhere 0<α⩽1,1<β⩽2 and the source function F depends on either x or t.


## 2. Preliminaries

This section is about basic definitions and fundamental properties of the fractional calculus.Riemann-Liouville fractional integral operator of orderα(α≥0) is introduced as(5)Jαfx=1Γα∫0xx-tα-1ftdt,α>0,x>0(6)J0fx=fx.Some of the properties of the operator Jα, which we will need in our work, are as follows [1, 2]:(7)JαJβfx=Jα+βfx(8)JαJβfx=JβJαfx(9)Jαxγ=Γγ+1Γα+1+γxγ+αIf m-1<α≤m,m∈N, then(10)DαJαfx=fx(11)JαDαfx=fx-∑i=0m-1fi0+xii!,x>0The Caputo fractional derivative of order α is defined as(12)Dαfx=Jm-αDmfx=1Γm-α∫0xx-tm-α-1fmtdt,m-1<α≤m,x>0where Dm is the classical differential operator of order m.The definitions of the Mittag-Leffler function and the generalized Mittag-Leffler function are given, respectively, as follows:(13)Eαz=∑k=0∞zkΓαk+1,α>0(14)Eα,βz=∑k=0∞zkΓαk+β,α,β>0


## 3. Basic Ideas of the Homotopy Analysis Method (HAM)

Let us consider the differential equation(15)Nux,t=0,where N is a nonlinear operator, x and t are independent variables, and u(x,t) is an unknown function. Liao [17] introduced the so-called zero-order deformation equation by the homotopy method as follows:(16)1-pLϕx,t;p-u0x,t=pħHx,tNϕx,t;p,with the initial conditions(17)ϕkx,0;p=gkx,k=0,1,2,…,m-1,where p∈[0,1], ħ≠0 are parameters, H(x,t) is nonzero auxiliary function, and L=Dtα,(n-1<α≤n) is an auxiliary linear operator with the following feature:(18)Lϕx,t=0,ϕx,t=0Obviously, for p=0 and p=1, we have(19)ϕx,t;0=u0x,t,ϕx,t;1=ux,t,where u0(x,t) is an initial guess of u(x,t) and ϕ(x,t;p) is an unknown function. As p goes to 1, the solution ϕ(x,t;p) changes from u0(x,t) to u(x,t). By Taylor series of ϕ(x,t;p) based on p is written as(20)ϕx,t;p=u0x,t+∑m=1∞umx,tpm,where(21)umx,t=1m!∂mϕx,t;p∂pmp=0If the auxiliary linear operator, the initial guess, the auxiliary operator h, and the auxiliary function are properly chosen, the series (20) converges at p=1, then we have(22)ux,t=u0x,t+∑m=1∞umx,t.Define the vector(23)un→=u0x,t,u1x,t,…,unx,t.Differentiating (16) m-times with respect to the embedding parameter p and then setting p=0 and finally dividing them by m!, we obtain the mth-order deformation equation, with assumption H(x,t)=1(24)Lumx,t-χmum-1x,t=ħRmum-1→subject to the following initial conditions:(25)umkx,0=0,k=0,1,2,…,m-1,where(26)Rmum-1→=1m-1!∂m-1Nϕx,t;p∂pm-1p=0and(27)χm=0,m≤11,m>1We can apply the operator Jα to both sides of (24) to obtain(28)JαDαumx,t-χmum-1x,t=ħJαRmum-1→Using the Caputo derivative and the initial conditions, we obtain(29)umx,t=χmum-1x,t+ħJαRmum-1→Finally, we will approximate the HAM solution (22) by the truncated series(30)ϕmx,t=∑k=0m-1ukx,tThe mth-order deformation (24) is linear and thus can be easily solved, especially by means of symbolic computation software such as Matlab and Maple.


## 4. Space-Time FPDE by the HAM

To construct the solution of problem (1)-(4) by the HAM, it is transformed into a space fractional differential problem or time fractional differential problem. The transformation we apply in this paper is motivated by the properties (7)-(9). Therefore let us take the problems with and without source functions into consideration:(A) Transforming into Time-Fractional Differential Equation(A1) No Source Function. In order to make the boundary condition (4) equal to zero, the transformation v(x,t)=u(x,t)-u(x,0) is applied to (1)-(2):(31)Dtαv=Dxβv+Dxβφx,(32)vx,0=0.In order to make the order of derivative on the right-hand side, integer, the transformation v=Dx2-βw is applied to (31)-(32):(33)DtαDx2-βw=Dxxw+Dxβφx,(34)wx,0=Dxβ-2vx,0=0.Substituting (33) into (26), Rm(um-1→) can be given by(35)Rmwm-1→=DtαDx2-βwm-1-Dxxwm-1-DxβφxThe solution of mth-order deformation (26) for m≥1 becomes(36)wmx,t=χmwm-1x,t+ħJαRmwm-1→By using (36) with the initial condition given by (34), we now successively obtain(37)w0x,t=wx,0=0(38)w1x,t=-ħDxβφxtαΓ1+α(39)w2x,t=-ħDxβφxtαΓ1+α-ħ2Dx2φxtαΓ1+α+ħ2Dxβ+2φxt2αΓ1+2α(40)w3x,t=-ħDxβφxtαΓ1+α-ħ2Dx2φxtαΓ1+α+ħ2Dxβ+2φxt2αΓ1+2α-ħ2Dx2φxtαΓ1+α-ħ3Dx4-βφxtαΓ1+α+ħ3Dx4φxt2αΓ1+2α+ħ2Dxβ+2φxt2αΓ1+2α+ħ3Dx4φxt2αΓ1+2α-ħ3Dxβ+4φxt3αΓ1+3αand so on. Hence, by using first four terms of the approximate solution w(x,t) can be obtained in the form of the series below:(41)wx,t=∑k=03wkx,tBy the transformations u(x,t), the solution of problem (1)-(2) is obtained.(A2) Including the Source Function. In order to make the boundary condition (4) equal to zero, the transformation v(x,t)=u(x,t)-u(x,0) is applied to (1)-(2):(42)Dtαv=Dxβv+Dxβφx+fx,(43)vx,0=0.In order to make the order of derivative on the right-hand side, integer, the transformation v=Dx2-βw is applied to (42)-(43):(44)DtαDx2-βw=Dxxw+Dxβφx+fx,(45)wx,0=Dxβ-2vx,0=0.Substituting (26) into (44), Rm(um-1→) can be given by(46)Rmwm-1→=DtαDx2-βwm-1-Dxxwm-1-1-χmFxThe solution of mth-order deformation (46) for m≥1 becomes(47)wmx,t=χmwm-1x,t+ħJαRmwm-1→By using (47) with the initial condition given by (45), we now successively obtain(48)w0x,t=0(49)w1x,t=-ħDxβφx+fxtαΓ1+α(50)w2x,t=-ħDxβφx+fxtαΓ1+α-ħ2Dx2φxtαΓ1+α-ħ2Dx2-βfxtαΓ1+α+ħ2Dxβ+2φxt2αΓ1+2α+ħ2Dx2fxt2αΓ1+2α(51)w3x,t=-ħDxβφx+fxtαΓ1+α-ħ2Dx2φxtαΓ1+α-ħ2Dx2-βfxtαΓ1+α+ħ2Dxβ+2φxt2αΓ1+2α+ħ2Dx2fxt2αΓ1+2α-ħ2Dx2φxtαΓ1+α-ħ2Dx2-βfxtαΓ1+α-ħ3Dx4-βφxtαΓ1+α-ħ3Dx4-2βfxtαΓ1+α+ħ3Dx4φxt2αΓ1+2α-ħ2Dx4-βfxt2αΓ1+2α+ħ2Dxβ+2φxt2αΓ1+2α+ħ2Dx2fxt2αΓ1+2α+ħ3Dx4φxt2αΓ1+2α+ħ3Dx4-βfxt2αΓ1+2α-ħ3Dxβ+4φxt3αΓ1+3α-ħ3Dx4fxt3αΓ1+3αand so on. Hence, the truncated solutions of w(x,t) can be constructed in the series form by using the first four terms as follows:(52)wx,t=∑k=03wkx,tBy the transformations u(x,t), the solution of problem (1)-(2) is determined.(B) Transforming into Space-Fractional Differential Equation(B1) No Source Function. By the transformation u=Dt1-αv to (1), (3), and (4) are transformed into the problem below:(53)Dtv=DxβDt1-αv,(54)v0,t=Dtα-1μ1t,(55)vx0,t=Dtα-1μ2t.Substituting (26) into (53), Rm(um-1→) can be given by(56)Rmvm-1→=DxβDt1-αvm-1-Dtvm-1The solution of mth-order deformation (56) for m≥1 becomes(57)vmx,t=χmvm-1x,t+ħJβRmvm-1→By using (54)-(55), we now successively obtain(58)v0x,t=vx,0=Dtα-1μ1t+xDtα-1μ2t(59)v1x,t=-ħDtαμ1txβΓ1+β-ħDtαμ2tx1+βΓ2+β(60)v2x,t=-ħDtαμ1txβΓ1+β-ħDtαμ2tx1+βΓ2+β-ħ2Dtμ1txβΓ1+β-ħ2Dtμ2tx1+βΓ2+β+ħ2Dt1+αμ1tx2βΓ1+2β+ħ2Dt1+αμ2tx1+2βΓ2+2β(61)v3x,t=-ħDtαμ1txβΓ1+β-ħDtαμ2tx1+βΓ2+β-ħ2Dtμ1txβΓ1+β-ħ2Dtμ2tx1+βΓ2+β+ħ2Dt1+αμ1tx2βΓ1+2β+ħ2Dt1+αμ2tx1+2βΓ2+2β-ħ2Dtμ1txβΓ1+β-ħ2Dtμ2tx1+βΓ2+β-ħ3Dtμ1txβΓ1+β-ħ3Dtμ2tx1+βΓ2+β+ħ3Dt2μ1tx2βΓ1+2β+ħ3Dt2μ2tx1+2βΓ2+2β+ħ2Dtα+1μ1tx2βΓ1+2β+ħ2Dtα+1μ2tx1+2βΓ2+2β+ħ3Dt2μ1tx2βΓ1+2β+ħ3Dt2μ2tx1+2βΓ2+2β-ħ3Dtα+2μ1tx3βΓ1+3β-ħ3Dtα+2μ2tx1+3βΓ2+3βand so on. Hence, the truncated solutions of v(x,t) can be constructed in the series form by using the first four terms below:(62)vx,t=∑k=03vkx,tBy the transformation u(x,t)=Dt1-αv(x,t), the solution of problem (1), (3), and (4) is determined.(B2) Including Source Function. Let us consider the problem below:(63)Dtv=DxβDt1-αv+gt,(64)v0,t=Dtα-1μ1t,(65)vx0,t=Dtα-1μ2t.Substituting (26) into (63), Rm(um-1→) can be given by(66)Rmvm-1→=DxβDt1-αvm-1-Dtvm-1+1-χmgtThe solution of mth-order deformation (66) for m≥1 becomes(67)vmx,t=χmvm-1x,t+ħJβRmvm-1→By using (64)-(65), we now successively obtain(68)
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## 5. Examples

Example 1.
Let us take the homogeneous problem.
(A1) Including No Source Function (73) D t α u = D x β u , (74) u x , 0 = cos ⁡ x , (75) u 0 , t = exp ⁡ - t , (76) u x 0 , t = 0 .Transforming into time-fractional differential equation, (73)-(74) are reduced as follows:(77)DtαDx2-βw=Dxxw-12x2-βE1,3-βix+E1,3-β-ix,(78)wx,0=0.Using the HAM and by the transformations u(x,t), we obtain the following solution of problem (73)-(76):(79)ux,t=cos⁡x+3ħcos⁡x+3ħ2i2xβ-1E1,βix-E1,β-ix-ħ32x2β-2E1,2β-1ix+E1,2β-1-ixtαΓ(1+α3ħ2cos⁡x+2ħ3i2xβ-1E1,βix-E1,β-ixt2αΓ1+2α+ħ3cos⁡xt3αΓ1+3αIn a particular case for α=1 and β=2, the HAM solution of (73)-(76) coincides with the exact solution u(x,t)=exp⁡-tcos⁡(x).
(B1) Including No Source Function (80) D t α u = D x β u , (81) u x , 0 = exp ⁡ x , (82) u 0 , t = exp ⁡ t , (83) u x 0 , t = exp ⁡ t .Transforming into space-fractional differential equation, (80), (82), and (83) are reduced to the problem below:(84)Dtv=DxβDt1-αv,(85)v0,t=Dtα-1exp⁡t=t-αE1,1-αt,(86)vx0,t=Dtα-1exp⁡t=t-αE1,1-αt,.Using the HAM and applying the transformation u(x,t)=Dt1-αv(x,t), we obtain the following solution of problem (80), (82), and (83):(87)
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3
βwhich coincides with the exact solution.In Tables1 and 2, the absolute errors have been worked out for various values of α and β. As can be seen from the tables, for the fixed values of α and β as x increases, the absolute error decreases, while as t increases, the absolute error also increases. Moreover as α increases to 1, β increases to 2 and the absolute error decreases. In addition, the absolute error takes the minimum values at the auxiliary values ħ=-0.94, ħ=-1.02 for case A and case B, respectively.Table 1
The absolute errors for case A in Example1.
ħ = - 0.94 ħ = - 1  t x α = 0.5 α = 0.75 α = 1 α = 0.5 α = 0.75 α = 1 β = 1.5 β = 1.75 β = 2 β = 1.5 β = 1.75 β = 2  0.3 0.2 0.68841 0.03769 0.00004 0.81871 0.04127 0.00031 0.4 0.27514 0.01312 0.00004 0.37117 0.00200 0.00029 0.6 0.08483 0.04706 0.00003 0.16617 0.03354 0.00026 0.8 0.03374 0.07006 0.00003 0.03721 0.05572 0.00022 0.6 0.2 1.02204 0.10261 0.00044 1.16631 0.08697 0.00472 0.4 0.55654 0.06402 0.00041 0.67960 0.07240 0.00443 0.6 0.33977 0.02800 0.00037 0.45762 0.04762 0.00397 0.8 0.19646 0.00200 0.00031 0.31112 0.02465 0.00335 0.9 0.2 1.21882 0.11304 0.00636 1.34963 0.05854 0.02261 0.4 0.76521 0.12078 0.00598 0.89617 0.11277 0.02125 0.6 0.55660 0.10408 0.00536 0.69660 0.12060 0.01904 0.8 0.41428 0.08328 0.00452 0.56229 0.11704 0.01607Table 2
The absolute errors for case B in Example1.
ħ=-1.02 ħ=-1  t x α = 0.5 α = 0.75 α = 1 α = 0.5 α = 0.75 α = 1 β = 1.5 β = 1.75 β = 2 β = 1.5 β = 1.75 β = 2  0.3 0.2 0.07111 0.02540 0.00000 0.07111 0.02540 0.00000 0.4 0.19034 0.07464 0.00000 0.19032 0.07464 0.00000 0.6 0.34423 0.14093 0.00000 0.34412 0.14092 0.00000 0.8 0.53422 0.22399 0.00000 0.53372 0.22390 0.00001 0.6 0.2 0.09598 0.03428 0.00000 0.09598 0.03428 0.00000 0.4 0.25693 0.10075 0.00000 0.25691 0.10076 0.00000 0.6 0.46467 0.19024 0.00000 0.46452 0.19023 0.00000 0.8 0.72113 0.30235 0.00000 0.72045 0.30223 0.00001 0.9 0.2 0.12957 0.04627 0.00000 0.12957 0.04627 0.00000 0.4 0.34681 0.13600 0.00000 0.34679 0.13601 0.00000 0.6 0.62724 0.25680 0.00000 0.62703 0.25678 0.00000 0.8 0.97342 0.40813 0.00000 0.97250 0.40797 0.00001Example 2.
Let us take the nonhomogeneous problem.
(A2) Including Source Function f ( x ) (88) D t α u = D x β u + 3 cos ⁡ x , (89) u x , 0 = 4 cos ⁡ x , (90) u 0 , t = 3 + exp ⁡ - t , (91) u x 0 , t = 0 .Reduction to time-fractional differential equation, (88)-(89) transform to the following:(92)DtαDx2-βw=Dxxw-12x2-βE1,3-βix+E1,3-β-ix+3cos⁡x,(93)wx,0=0.Using the HAM and by the transformations u(x,t), we obtain the following solution of problem (88)-(91):(94)ux,t=4cos⁡x⁡+12ħcos⁡x-9ħi2xβ-1E1,βix-E1,β-ix+6ħ2ixβ-1E1,βix-E1,β-ix+9ħ22x2β-2E1,2β-1ix+E1,2β-1-ix-2ħ2x2β-2E1,2β-1ix+E1,2β-1-ix+3ħ3i2x3β-2E1,3β-1ix-E1,3β-1-ixtαΓ1+α12ħ2cos⁡x-9ħ2i2xβ-1E1,βix-E1,β-ix+8ħ3i2xβ-1E1,βix-E1,β-ix+3ħ3x2β-2E1,2β-1ix+E1,2β-1-ixt2αΓ1+2α4ħ3cos⁡x-3ħ3i2xβ-1E1,βix-E1,β-ixt3αΓ1+3αFor α=1 and β=2, the HAM solution of (88)-(91) coincides with the exact solution u(x,t)=exp⁡-t+3cos⁡(x).
(B2) Including Source Function g t (95) D t α u = D x β u + 3 t 2 , (96) u x , 0 = exp ⁡ x , (97) u 0 , t = exp ⁡ t + t 3 , (98) u x 0 , t = exp ⁡ t .Transforming (95), (97), and (98) into the following problem, we have(99)Dtv=DxβDt1-αv+3t2,(100)v0,t=Dtα-1exp⁡t+t3=t-αE1,1-αt+6t4-αΓ5-α,(101)vx0,t=Dtα-1exp⁡t=t-αE1,1-αt.Using the HAM and the transformation u(x,t)=Dt1-αv(x,t), we obtain the following solution of problem (95), (97), and (98):(102)
u
0
x
,
t
=
exp
⁡
t
+
t
3
+
x
exp
⁡
t
u
1
x
,
t
=
-
ħ
exp
⁡
t
+
3
t
2
x
β
Γ
1
+
β
-
ħ
exp
⁡
t
x
1
+
β
Γ
2
+
β
+
3
ħ
t
2
x
β
Γ
1
+
β
u
2
x
,
t
=
-
ħ
exp
⁡
t
+
3
t
2
x
β
Γ
1
+
β
-
ħ
exp
⁡
t
x
1
+
β
Γ
2
+
β
+
3
ħ
t
2
x
β
Γ
1
+
β
-
ħ
2
t
2
-
α
E
1,3
-
α
t
+
6
t
1
+
α
Γ
2
+
α
x
β
Γ
1
+
β
-
ħ
2
t
2
-
α
E
1,3
-
α
t
+
6
ħ
2
t
1
+
α
Γ
2
+
α
x
β
Γ
1
+
β
+
ħ
2
exp
⁡
t
+
6
t
x
2
β
Γ
1
+
2
β
+
ħ
2
exp
⁡
t
x
1
+
2
β
Γ
2
+
2
β
-
6
ħ
2
t
x
2
β
Γ
1
+
2
β
u
3
x
,
t
=
-
ħ
exp
⁡
t
+
3
t
2
x
β
Γ
1
+
β
-
ħ
exp
⁡
t
x
1
+
β
Γ
2
+
β
+
3
ħ
t
2
x
β
Γ
1
+
β
-
ħ
2
t
2
-
α
E
1,3
-
α
t
+
6
t
1
+
α
Γ
2
+
α
x
β
Γ
1
+
β
-
ħ
2
t
2
-
α
E
1,3
-
α
t
x
1
+
β
Γ
2
+
β
+
6
ħ
2
t
1
+
α
Γ
2
+
α
x
β
Γ
1
+
β
+
ħ
2
exp
⁡
t
+
6
t
x
2
β
Γ
1
+
2
β
+
ħ
2
exp
⁡
t
x
1
+
2
β
Γ
2
+
2
β
-
6
ħ
2
t
x
2
β
Γ
1
+
2
β
-
ħ
2
t
2
-
α
E
1,3
-
α
t
+
6
t
1
+
α
Γ
2
+
α
x
β
Γ
1
+
β
-
ħ
2
t
2
-
α
E
1,3
-
α
t
x
1
+
β
Γ
2
+
β
+
6
ħ
2
t
1
+
α
Γ
2
+
α
x
β
Γ
1
+
β
-
ħ
3
t
3
-
α
E
1,4
-
α
t
+
6
t
2
α
Γ
1
+
2
α
x
β
Γ
1
+
β
-
ħ
3
t
3
-
α
E
1,4
-
α
t
x
1
+
β
Γ
2
+
β
+
6
ħ
3
t
2
α
Γ
1
+
2
α
x
β
Γ
1
+
β
+
ħ
3
t
3
-
α
E
1,4
-
α
t
+
6
t
α
Γ
1
+
α
x
2
β
Γ
1
+
2
β
+
ħ
3
t
3
-
α
E
1,4
-
α
t
x
1
+
2
β
Γ
2
+
2
β
-
6
ħ
3
t
α
Γ
1
+
α
x
2
β
Γ
1
+
2
β
+
ħ
2
exp
⁡
t
+
6
t
x
2
β
Γ
1
+
2
β
+
ħ
2
exp
⁡
t
x
1
+
2
β
Γ
2
+
2
β
-
6
ħ
2
t
x
2
β
Γ
1
+
2
β
+
ħ
3
t
3
-
α
E
1,4
-
α
t
+
6
t
α
Γ
1
+
α
x
2
β
Γ
1
+
2
β
+
ħ
3
t
3
-
α
E
1,4
-
α
t
x
1
+
2
β
Γ
2
+
2
β
+
6
ħ
3
t
α
Γ
1
+
α
x
2
β
Γ
1
+
2
β
-
ħ
3
exp
⁡
t
+
6
x
3
β
Γ
1
+
3
β
-
ħ
3
exp
⁡
t
x
1
+
3
β
Γ
2
+
3
β
+
6
ħ
3
x
3
β
Γ
1
+
3
βwhich is the exact solution.In Tables3 and 4, the absolute errors have been worked out for various values of α and β. As is obvious from the tables, as α increases to 1, β increases to 2 and the absolute error decreases. In addition, the absolute error takes the minimum values at the auxiliary values ħ=-0.94, ħ=-0.57 for case A and case B, respectively.Table 3
The absolute errors for case A in Example2.
ħ=-0.94 ħ=-1  t x α = 0.5 α = 0.75 α = 1 α = 0.5 α = 0.75 α = 1 β = 1.5 β = 1.75 β = 2 β = 1.5 β = 1.75 β = 2  0.3 0.2 0.80914 0.14812 0.00004 0.34918 0.08528 0.00031 0.4 1.16838 0.03354 0.00004 1.17892 0.01225 0.00029 0.6 0.71218 0.09107 0.00003 0.82383 0.09475 0.00026 0.8 0.28227 0.19786 0.00003 0.43360 0.19139 0.00022 0.6 0.2 0.23377 0.10803 0.00044 1.19358 0.03953 0.00472 0.4 1.06874 0.01958 0.00041 0.89593 0.05309 0.00443 0.6 0.80431 0.11677 0.00037 0.83515 0.15221 0.00397 0.8 0.44477 0.24245 0.00031 0.57549 0.25288 0.00335 0.9 0.2 1.63325 0.07079 0.00636 3.10231 0.32344 0.02261 0.4 0.65994 0.08755 0.00598 0.25862 0.23737 0.02125 0.6 0.67465 0.27502 0.00536 0.57511 0.28500 0.01904 0.8 0.46155 0.47543 0.00452 0.52316 0.34909 0.01607Table 4
The absolute errors for case B in Example2.
ħ=-0.57 ħ=-1  t x α = 0.5 α = 0.75 α = 1 α = 0.5 α = 0.75 α = 1 β = 1.5 β = 1.75 β = 2 β = 1.5 β = 1.75 β = 2  0.3 0.2 0.10133 0.03198 0.00243 0.13890 0.04687 0.00825 0.4 0.29402 0.10097 0.01034 0.39635 0.15200 0.03468 0.6 0.57660 0.20716 0.02621 0.75465 0.31273 0.08418 0.8 0.96440 0.35981 0.05427 1.22052 0.53736 0.16431 0.6 0.2 0.17372 0.06593 0.01563 0.30196 0.10909 0.03275 0.4 0.50798 0.21636 0.06467 0.87541 0.36506 0.13518 0.6 0.99839 0.45298 0.15447 1.68492 0.76466 0.31946 0.8 1.67026 0.79509 0.29737 2.75087 1.32983 0.60520 0.9 0.2 0.27067 0.11810 0.03777 0.48542 0.20136 0.07345 0.4 0.78697 0.39088 0.15460 1.40930 0.67743 0.30029 0.6 1.53126 0.81511 0.36227 2.71061 1.41561 0.69954 0.8 2.53281 1.41600 0.68048 4.41814 2.44668 1.30201


## 6. Conclusion

In this research, the truncated solution of space-time FPDEs is constructed by the HAM and special transformations which allows us to implement the HAM to deal with harder problems. It is obvious that applying the HAM to the equation including two different fractional derivatives is harder than applying the HAM to the equation including one fractional derivative. Therefore, the transformation in this paper allows us to find the solution of the problem by the HAM much more easier. Examples manifest that when the order of space and time derivatives is fractional, the result of the HAM method gives better results for smaller values ofħ in some interval unlike the integer order of space and time fractional derivatives.



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*Source: 5602565-2019-05-06.xml*