Special Issue Article
On solving fractional mobile/immobile equation
Advances in Mechanical Engineering 2017, Vol. 9(1) 1–12 Ó The Author(s) 2017 DOI: 10.1177/1687814016688616 journals.sagepub.com/home/ade
Hossein Pourbashash1, Dumitru Baleanu2,3 and Maysaa Mohamed Al Qurashi4
Abstract In this article, a numerical efficient method for fractional mobile/immobile equation is developed. The presented numerical technique is based on the compact finite difference method. The spatial and temporal derivatives are approximated based on two difference schemes of orders O(t2a ) and O(h4 ), respectively. The proposed method is unconditionally stable and the convergence is analyzed within Fourier analysis. Furthermore, the solvability of the compact finite difference approach is proved. The obtained results show the ability of the compact finite difference. Keywords Mobile/immobile equation, time fractional, compact finite difference, Fourier analysis, stability, convergence, solvability
Date received: 4 September 2016; accepted: 29 November 2016 Academic Editor: Praveen Agarwal
u(x, 0) = g(x),
Introduction The governing equation of transport was derived based on Fick’s law and is commonly called the advectiondispersion equation (ADE).1 The ADE will predict a breakthrough curve (BTC) that can be described by a Gaussian distribution function from an instantaneously releasing solute source.1 The interested readers can find more details in previous studies.2–7 Also, the mobile/ immobile model is considered in previous studies.8–12 Here, the time fractional mobile/immobile equation is studied to the following form12,13 ∂u(x, t) ∂a u(x, t) ∂2 u(x, t) ∂u(x, t) + = ð1Þ ∂t ∂ta ∂x2 ∂x + f (x, t), (x, T ) 2 ½0 h 3 ½0 T , 0\a\1
and the initial condition is
where ∂a ( )=∂ta is the Caputo fractional derivative of order 0\a\1. For getting more information on fractional PDEs the interested readers can refer to.21–25 Some numerical methods have been developed for the solution of equation (1) such as finite difference (FD) method12,13 and meshless method.14 Also, the fractional equation is studied by several methods, for example, high-order FD scheme for modified anomalous fractional sub-diffusion equation15,16 and FD method for a class of fractional sub-diffusion equations.17 The main aim of this article is to see the performances of the compact FD for the fractional mobile/ 1
Department of Mathematics, University of Garmsar, Garmsar, Iran Department of Mathematics, Cankaya University, Ankara, Turkey 3 Institute of Space Sciences, Magurele-Bucharest, Romania 4 Department of Mathematics, King Saud University, Riyadh, Saudi Arabia
u(0, t) = f1 (t) 0\t\T
ð3Þ
2
when the boundary conditions are
u(h, t) = f2 (t)
0\x\h
ð2Þ
Corresponding author: Hossein Pourbashash, Department of Mathematics, University of Garmsar, Garmsar 3581755796, Iran. Email:
[email protected]
Creative Commons CC-BY: This article is distributed under the terms of the Creative Commons Attribution 3.0 License (http://www.creativecommons.org/licenses/by/3.0/) which permits any use, reproduction and distribution of the work without further permission provided the original work is attributed as specified on the SAGE and Open Access pages (https://us.sagepub.com/en-us/nam/ open-access-at-sage).
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immobile equation. The article is organized as follows. In section ‘‘Compact FD scheme,’’ we develop a highorder FD scheme. Section ‘‘Stability analysis’’ presents the stability analysis for the proposed difference scheme. In section ‘‘Convergence analysis,’’ the convergence analysis is studied. Some numerical results are presented in section ‘‘Numerical results.’’ Finally, a brief conclusion is written in section ‘‘Conclusion.’’
∂2 u(x, t) ∂u(x, t) ∂u(x, t) + = f (x, t) w(x, t) ð7Þ + ∂x2 ∂x ∂t
we can write equation (7) as follows ! k ∂u h2 2 k j 1+ + (r1 )kj d u + dx ukj = fjk wkj 12 x j ∂t ! !! ∂ukj ∂ukj h2 2 k k k k d f wj dx fj wj + 12 x j ∂t ∂t
Compact FD scheme
ð8Þ
Let h = L=M and t = T =N be the step sizes of spatial and temporal variables, respectively. So, we can define spatial and temporal nodes as xj = jh, for j = 0, 1, 2, . . . , M, and tk = kt, for k = 0, 1, 2, . . . , N : The exact and approximate solutions at the point (xj , tk ) are denoted by ukj and Ujk , respectively. We introduce the following notations dx ukj =
where there is a constant c1 such that k (r1 )j c1 (h4 ) Based on Lemma 2 in Sun and Wu18 we have Wjk
ukj+ 1 ukj1
ð9Þ
ðtk ∂u(xj , t) 1 = (tk t)a dt G(1 a) ∂t 0
t a = G(2 a) " # k1 X k m 0 (bkm1 bkm )uj bk1 uj + (r2 )kj b0 uj
h k k k u j + 1 2uj + uj1 d2x ukj = h2 k ∂uj ukj ujk1 = dt ukj = + O(t) ∂t t
m=1
ð10Þ
where ukj = u(xj , tk ).
where there exists a constant c2 such that k (r2 )j c2 (t2a )
Lemma 126. Assume the below equation
ð11Þ
2
g
d y(x) dy(x) = f (x), b 2 dx dx
x 2 (0, L)
ð4Þ
h2 2 k 1+ dx uj + dx ukj = fjk Wjk dt ukj 12 h2 2 k dx fj Wjk dt ukj dx fjk Wjk dt ukj + + Rkj 12
with y(0) = y0 ,
Substituting equation (10) into equation (8) gives
y(L) = yL
A CFD scheme for it is as follows b2 h2 2 d yj bdx yj = fj g+ 12g x h2 2 b dx fj dx fj + O(h4 ) + 12 g
ð12Þ
where
ð6Þ
AUk =
k1 X 1 BUk1 + m (bkm1 bkm )BUm t m=1 0
0
so equation (1) can be written as follows
ð13Þ
Denoting m = ta =(G(2 a)), the proposed difference scheme is
Let ðt 1 ∂u(x, z) w(x, t) = (t z)a dz G(1 a) ∂z
k Rj C(t + h4 )
ð5Þ
k
+ mbk1 BU + BF where
ð14Þ
Pourbashash et al.
3
1 h2 1 m 1 hm h 2 h2 5m 5 + + + + , + A = tri 2 1 + 1+ + 12 12 h 2h 12 12t 24 24t h2 6 6t 1 h2 1 m 1 hm h + + 2 1+ + 12 h 2h 12 12t 24 24t 2
5 6 6 6 6 1 h 6 + 6 24 6 12 B=6 6 6 6 6 4
5m 5 m 1 + 2 m1 + 6 6t 12 12t 2m 2 + .0 = 4m1 + 3 3t
3
1 h 12 24 .. .
..
.
..
..
.
.
1 h + 12 24
7 7 7 7 7 7 7 1 h 7 7 12 24 7 7 5 5 6
lj 2m1 +
and if (m=12) m1 + (1=12t)\0 lj 2m1 +
5m 5 m 1 1 + +2 m1 + = m + .0 6 6t 12 12t t
Since m1 , m.0 thus A is non-singular matrix. Lemma 2. The matrix A is non-singular. Proof. For j = 1, 2, . . . , M 1, the eigenvalues of A are 5m 5 lj = 2m1 + + 6 6t s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m 1 hm h 1 m1 + + + + 12 12t 24 24t 2h sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi m 1 hm h 1 m1 + + 3 12 12t 24 24t 2h jp cos M
Theorem 1. For the appeared scheme in equation (14), there is a unique solution. Proof. We must solve linear system of equations AU k = b at each tk = kt to obtain the numerical solution. Since for any m and m1 , the coefficient matrix A is invertible so the solution of scheme in equation (14) exists and is unique.
Stability analysis ~ n be the approximated solution of equation (14). Let U j Consider ~ k, rkj = Ujk U j
that is 5m 5 + lj = 2m1 + 6 6t ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v" u # u m 1 2 hm h 1 2 t +2 m1 + + 12 12t 24 24t 2h jp cos M
m 1 m1 + 12 12t
k1 X 1 Ark = Brk1 + m (bkm1 bkm )Brm + mbk1 Br0 t m=1
ð15Þ where rk0 = rk0 = 0
2
hm h 1 2 + 0 24 24t 2h
the eigenvalues of A can be written as l = a + bi with non-zero real part (a 6¼ 0). For the case
k = 0, 1, . . . , N
With this definition and regarding equation (14), we can obtain the round-off error equation as
In case of
j = 0, 1, . . . , M,
m 1 m1 + 12 12t
2
hm h 1 2 + .0 24 24t 2h
if (m=12) m1 + (1=12t) 0 we have
where
rk = rk1 , rk2 , . . . , rkM1 Now we define the grid function19 8 > < rkj
h h xj \x xj + 2 2 rk (x) = > : 0 0 x h or L h \x L 2 2 Then rk (x), k = 1, 2, . . . , N , can be expanded in a Fourier series
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Advances in Mechanical Engineering
rk (x) =
‘ X
8 u 8mm1 4 u 8 u u sin = mm1 sin2 cos2 0 mm1 sin2 3 3 2 2 3 2 2 8m1 2 u 8m1 4 u 8m1 2 u 2u sin sin = sin cos 0 3t 3t 3t 2 2 2 2 2 2 2 2u 2u 2u m sin sin = sin (m1 1) 0 3t 1 2 3t 2 3 2 2 2 2 u 2u 2u mm1 sin m sin = m sin2 (m1 1) 0 ð19Þ 3 2 3 2 3 2
dk (l)ei2plx=L
l = ‘
where ðL 1 k r (x)ei2plx=L dx dk (l) = L 0
The relation (18) holds if and only if u 2 u 0 4m1 sin2 + 2 4m1 sin2 2 2 m 2u 1 1 u m sin + sin2 3 2 t 3t 2 1 1 hm h sin u + sin u + 2 sin2 u 2 sin u h h 12 12t
Now introduce the following norm19 k r = 2
M 1 X j=1
2 hrkj
!12
2L 312 ð k 2 = 4 r (x) dx5 0
Note that we can obtain ‘ X k 2 r = jdk (l)j2 2
ð16Þ
If we put
l = ‘
m 2 2 u 2 u sin u = msin2 msin4 6 3 2 3 2 1 2 u 2 u sin2 u = sin2 sin4 6t 3t 2 3t 2
by using Parseval equality ðL
‘ X k 2 r (x) dx = jdk (l)j2 l = ‘
0
We can assume the solution of equation (15) as rkj = dk eisjh where s = 2pl=L. Now, substituting this expression into equation (15) yields
dk =
then relation (18) holds if and only if u 2 8 u 8 u 14 u m1 msin2 m1 msin4 + m1 m sin2 4m1 sin2 + 2 3 2 3 2 3 2 8 8 1 14 2u 4u 2u 2u + m sin m1 sin + 2 sin + m1 sin 3t 1 2 3t 2 h 2 3t 2 2 u 2 u 2 u 2 u + m msin2 msin2 + m sin4 + sin4 3 1 2 3 2 3 2 3t 2
!" # k 1 X sin2 u2 ih sin u 1 1 m(bkm1 bkm )dm + mbk1 d0 + dk1 3 12 t m=1 4m1 sin2
where
m1 =
ð17Þ
u m u 1 1 u i ihm ih + m sin2 + sin2 + sin u sin u sin u 2 3 2 t 3t 2 h 12 12t
2
which completes the proof.
1 h , 1+ 12 h2
m=
1 ta G(2 a)
Lemma 3. The below inequality holds
Theorem 2. Let dk be the solutions of equation (17), so jdk j jd0 j,
m 2 u ihm 1 1 ih 2u sin u + sin sin u m sin 3 2 12 t 3t 2 12t 1 4m sin2 u + m m sin2 u + 1 1 sin2 u + i sin u ihm sin u ih sin u 1 2 3 2 t 3t 2 h 12 12t
Proof. Because of h 1, t 1, and G(2 a).0 we can see that m.0. Furthermore m1 .1, too. So, it is easy to see that
k = 1, 2, . . . , N
ð20Þ
ð18Þ
Proof. In case of k = 1 according to equation (17), we have
Pourbashash et al.
5 ! sin2 u2 ih sin u m + 1t d0 1 3 12
d1 = 4m1 sin2
u m u 1 1 u i ihm ih + m sin2 + sin2 + sin u sin u sin u 2 3 2 t 3t 2 h 12 12t
or 0 B @1 d1 = 0
0
u B B @4m1 sin2 + m@1 2
1 u 1 2 ih sin uC m + d0 A 12 t 3
sin2
1 0 11 u 2u sin 1B sin u hm sin u h sin u CC 2C 2 A + @1 AA + i t h 12 12t 3 3
sin2
Also, we can write sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2u 2 h sin u 2 1 1 sin m+ + jd0 j 3 2 12 t ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi jd1 j = v ! !!2 u 2u 2u u sin u hm sin u h sin u 2 t 4m sin2 u + m 1 sin 2 + 1 1 sin 2 + 1 3 3 2 t h 12 12t Now, Lemma 3 yields
so we have k r r0 2 2
jd1 j jd0 j Suppose jdn j jd0 j,
n = 1, 2, . . . , k 1
Convergence analysis
Here, we analyzed the compact difference scheme (14) Now, equation (17) yields in terms of convergence. We will show the relation (14) 2u sin 2 ih sin u 1 1 mðb0 bk1 Þjd0 j + mbk1 jd0 j + jd0 j 3 12 t jdk1 j 4m1 sin2 u + m m sin2 u + 1 1 sin2 u + i sin u ihm sin u ih sin u 2 3 2 t 3t 2 h 12 12t so jdk j jd0 j which completes the proof.
has the spatial accuracy of fourth order. For this end, we need some lemmas and theorems that will be expressed. First, we consider definitions of ek (x) and Rk (x) in Chen et al.20 Thus, for k = 0, 1, . . . , N, ek (x) and Rk (x) are
Theorem 3. The CFD scheme (14) is unconditionally stable.
ek (x) =
Proof. Using inequality (20) in equation (16) yields ‘ ‘ X X k 2 2 2 r = d (l) j j jd0 (l)j2 = r0 2 k 2 l=‘
l=‘
Rk (x) =
‘ X
hk (l)e
l = ‘ ‘ X l = ‘
where
i2plx L
jk (l)e
i2plx L
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Advances in Mechanical Engineering ‘ X k 2 R = jjk (l)j2 2
ðL 1 k i2plx hk (l) = e (x)e L dx L
ð27Þ
l = ‘
0
We can obtain
ðL
1 i2plx Rk (x)e L dx jk (l) = L
k Rj C(t + h4 )
0
Now, we define the following notations20 ekj = u(xj , tk ) Ujk
Subtracting equation (14) from equation (12), we obtain ð21Þ Aek =
where
ek = ek1 , ek2 , . . . , ekM1 ,
Rk = Rk1 , Rk2 , . . . , RkM1
M1 X j=1
M1 X
k R = 2
j=1
2 hekj
!12
2L 312 ð k 2 = 4 e (x) dx5
2 hRkj
2L 312 ð 2 = 4 Rk (x) dx5
ð22Þ
ð23Þ
Using the Parseval equality
ðL
j = 1, 2, . . . , M 1
ekj = hk ei(sjh) Rkj = jk ei(sjh) where s = 2lp=L. By substituting the obtained relations into equation (29), we yield
!" # k1 X sin2 u2 ih sin u 1 1 m(bkm1 bkm )hm + hk1 + jk 3 12 t m=1 u m u 1 1 u i ihm ih sin u sin u 4m1 sin + m sin2 + sin2 + sin u + 2 3 2 t 3t 2 h 12 12t
ð30Þ
2
‘ X k 2 e (x) dx = jhk (l)j2
where u = sh. Because of e0 = 0, we can write ð24Þ
l = ‘
0
e0j = 0,
k = 1, 2, . . . , N 1
Now, assume that ekj and Rkj are as follows
0
hk =
ð29Þ
ek0 = ekM = 0,
0
!12
k 1 X 1 k1 Be + m (bkm1 bkm )Bem + Rk t m=1
with
and introduce the following norms k e = 2
ð28Þ
h0 [h0 (l) = 0 Furthermore, the first equality in equation (23) and inequality of equation (28) yield
and ðL
‘ X k 2 R (x) dx = jjk (l)j2
ð25Þ
l = ‘
0
pffiffiffiffiffiffiffi pffiffiffi k R MhC1 t + h4 = C1 L t + h4 2 Also there is a positive constant C2 as20
Also, we have
jjk j[jjk ðnÞj C2 tjj1 j[C2 t jj1 ðnÞj, k 2 e = 2
‘ X
jhk (l)j2
ð31Þ
k = 1, 2, . . . , N ð32Þ
ð26Þ
l = ‘
Lemma 4. The below inequality holds 1 1 u m u 1 1 u i ihm ih 4m1 sin2 + m sin2 + sin2 + sin u + sin u sin u 2 3 2 t 3t 2 h 12 12t
Pourbashash et al.
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Proof. Since 0\a, t 1, we can obtain m = 1=(ta G(2 a)).1. Now the above relation holds if and only if
Theorem 5. Suppose u(x, t) is the solution of equation (1), then the compact FD scheme described in equation
1
4m1 sin2
2 1 u m u 1 1 u 2 1 hm h + m sin2 + sin2 sin u sin u sin u + 2 3 2 t 3t 2 h 12 12t
Now, we have 1 2 2m 2 2 1 2m + 2 + =9 t 3 3t 2 m 1 1 =9 m + 3 t 3t 2 m 2u 1 1 2u 9 m sin + sin 3 2 t 3t 2 2 m 2u 1 1 2u 9 m sin + sin 3 2 t 3t 2 2 1 hm h sin u sin u + 9 sin u h 12 12t This completes the proof. Theorem 4. If hk (k = 1, 2, . . . , N ) be the solutions of equation (30), then a positive constant C2 exists such that
(14) is L2 -convergent and its convergence order is O(t + h4 ). Proof. According to Theorem 4, equations (26), (27), and (31), we have pffiffiffi k
e ð1 + 3t Þk C2 R1 C1 LC2 e3kt t + h4 2 2 Since kt T, we have k e C(t + h4 ) 2 where pffiffiffi C = C1 C2 Le3T
Numerical results
Here, we present some test problem to verify the developed method. To demonstrate the accuracy of this jhk j C2 ð1 + 3tÞ jj1 j, k = 1, 2, . . . , N method, we use L‘ norm for computing errors. Also, Proof. Applying equations (30), (32), and Lemma 6, we the computational orders of the presented method (denoted by C-Order) is calculated using the following can write formula jj 1 j jh1 j = 4m1 sin2 u + m m sin2 u + 1 1 sin2 u + i sin u + ihm sin u ih sin u 2 3 2 t 3t 2 h 12 12t k
3tC2 jj1 j ð1 + 3tÞC2 jj1 j Now, suppose that jhn j ð1 + 3tÞn C2 jj1 j,
n = 1, 2, . . . , k 1
From Lemmas 3 and 4 " # k 1 sin2 u2 ih sin u X 1 m(bkm1 bkm )jhm j + jhk1 j + jjk j 1 3 12 m = 1 t jhk j = 4m sin2 u + m m sin2 u + 1 1 sin2 u + i sin u + ihm sin u ih sin u 1 2 3 2 t 3t 2 h 12 12t sin2 u2 ih sin u 1 + 3t jj1 jC2 1 (1 + 3t)k1 jj1 jC2 m + 3 12 t 4m sin2 u + m m sin2 u + 1 1 sin2 u + i sin u + ihm sin u ih sin u 1 2 3 2 t 3t 2 h 12 12t
k1 (1 + 3t) + 3t jj1 jC2 (1 + 3t)k jj1 jC2
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Advances in Mechanical Engineering
Table 1. Evaluated computational orders and errors with t = 0:02 and g = 1=30, for Test problem 1. h
1=4 1=8 1=16 1=32 1=64 1=128 1=256 1=512
a = 0:25
a = 0:75
CPU time (s)
L‘
C-order
L‘
C-order
7:15113101 1:33443102 7:00413104 4:21693105 2:61813106 1:63943107 1:02413108 6:390531010
5:7439 4:2518 4:0539 4:0096 3:9972 4:0480 4:0023
7:07213101 1:33443102 7:00523104 4:21773105 2:61233106 1:63983107 1:02443108 6:412031010
5:7279 4:2516 4:0539 4:0131 3:9937 4:0007 3:9978
00:3280 00:5000 00:8280 01:5310 02:7040 05:2030 10:2500 10:2500
CPU: central processing unit.
Figure 1. Approximated solutions and obtained errors for different values of g using presented method from (a) to (h) for Test problem 1.
Pourbashash et al.
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Figure 2. (Left) Surface plot of approximated solution; (right) exact and approximated solution with a = 0:65, g = 0:001, t = 1=50, and h = 1=128 for Test problem 1.
Table 2. Test problem 1 error values obtained for different parameters. h=t
g = 1=10
1=0 1=15 1=25 1=50 1=100
g = 1=100
a = 0:4
a = 0:9
a = 0:4
a = 0:9
5:23393104 1:02973104 1:33233105 8:29543107 5:17973108
5:22623104 1:02803104 1:33003105 8:28083107 5:17053108
1:07423101 7:35523103 1:14903103 8:10483105 5:00493106
1:07273101 7:35583103 1:14913103 8:10563105 8:00543106
Table 3. Comparison of numerical solutions and errors obtained with h = t = 1=100 for Test problem 1. x
0:1 0:2 0:3 0:4 0:5 0:6 0:7 0:8 0:9
Exact solution
0:162000 0:512000 0:882000 1:152000 1:250000 1:152000 0:882000 0:512000 0:162000
Method of Zhang et al.12
Present method
Numerical solution
L‘
Numerical solution
L‘
0:161843 0:510599 0:879024 1:147702 1:245027 1:147196 0:878184 0:509725 0:161279
1:56293104 1:40063103 2:97523103 4:29773103 4:97223103 4:80343103 3:81523103 2:27463103 7:20753104
0:161999 0:511999 0:881999 1:151999 1:249999 1:151999 0:881999 0:511999 0:161999
1:15603109 2:10293109 2:83253109 3:33123109 3:58133109 3:55793109 3:23033109 2:56103109 1:50313109
Table 4. Evaluated computational orders and error values with t = 1=50 for Test problem 2. h
1=4 1=8 1=16 1=32 1=64 1=128 1=256
a = 0:1
a = 0:8
CPU time (s)
L‘
C-order
L‘
C-order
1:37923103 8:62063105 5:41513106 3:38773107 2:11763108 1:32363109 8:237631011
3:9999 3:9937 3:9976 3:9998 3:9999 4:0061
1:40643103 8:79033105 5:51843106 3:45453107 2:15943108 1:34953109 8:181931011
3:9999 3:9936 3:9977 3:9998 4:0001 4:0438
CPU: central processing unit.
00:0982 00:3910 00:7500 01:5620 02:9840 05:3750 10:7960
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Advances in Mechanical Engineering
Table 5. Error values obtained with different values of a and t for Test problem 4. h=t
a = 0:25
a = 0:45
a = 0:65
a = :85
1=10 1=15 1=25 1=50
3:53033105 7:01953106 9:08793107 5:81613108
3:54143105 7:04353106 9:12073107 5:71113108
3:56463105 7:09173106 9:18493107 5:75203108
3:60123105 7:16743106 9:28573107 5:68983108
log
log
E1 E2
h1 h2
where Ei is the error value that corresponds to grid with mesh size hi .
Test problem 1 We consider the time fractional mobile/immobile equation with the form described in equation (1), where f (x, t) = 1a t 2(x 0:5) 2 4(x 0:5)2 +1 + G(2 a) g2 g g (x 0:5)2 exp g where exact solution is a Gaussian pulse with t height centered at x = 0:5, that is (x 0:5)2 u(x, t) = t exp g boundary and initial conditions can be obtained from the exact solution. Table 1 demonstrates the L‘ error, computational order and total central processing unit (CPU) time (in second). The computational order of Table 1 is closed to the theoretical results. Figure 1 is based on a = 0:35, h = 1=256, and t = 1=10. Also, the used parameter in Figure 2 is a = 0:65, g = 0:001, h = 1=128, and t = 1=50. Also, in Table 2, we can see error obtained with h = t and different values of a for this problem.
Test problem 2 Again, we consider the time fractional mobile/immobile equation with the form described in equation (1) with source term
t1a f (x, t) = 10x2 (1 x)2 1 + G(2 a)
+ 10(t + 1) 2 + 14x 18x2 + 4x3
Figure 3. Graphs of exact and approximate solution, absolute error, and surface plot of approximated solution with parameters a = 0:45, t = 0:025, and h = 1=256, for Test problem 2.
when boundary and initial conditions are of the following form u(0, t) = u(1, t) = 0 u(x, 0) = 10x2 (1 x)2
Pourbashash et al. In this case the exact solution is as follows (Figure 3) u(x, t) = 10(1 + t)x2 (1 x)2 In Table 3, we compare the errors obtained with the method of this article and method of Zhang et al.12 where a = 1 0:5 exp( x). We see that the present method has good results in comparison with the method of Zhang et al.12 Computational order, L‘ error and CPU time (in seconds) are shown in Table 4. Also, Table 5 presents the error of numerical results for this problem with different values of parameters a and t. Figure 3 demonstrates graphs of exact and approximate solution, absolute error, and surface plot of approximated solution with parameters a = 0.45, t = 0.025, and h = 1/256, for Test problem 2.
Conclusion In this article, we built a compact difference scheme for the solution of time fractional mobile/immobile equation. We proved the unconditional stability property and convergence by Fourier analysis. It was shown that the numerical simulations obey the theoretical results. Examples are given and when the results obtained using this method with exact solutions are compared, this method shows applicability and efficiency. Acknowledgements The authors are very grateful to reviewers for carefully reading this article and for their comments and suggestions which have improved the article.
Declaration of conflicting interests The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.
Funding The author(s) received no financial support for the research, authorship, and/or publication of this article.
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