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Analytical and approximate solutions of Fractional Partial Differential-Algebraic Equations

Hatıra Günerhan
Hatıra Günerhan
 und 
Ercan Çelik
Ercan Çelik
   | 30. März 2020
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Online veröffentlicht: 30. März 2020
Seitenbereich: 109 - 120
Eingereicht: 12. Mai 2019
Akzeptiert: 29. Juli 2019
DOI: https://doi.org/10.2478/amns.2020.1.00011
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© 2020 Hatıra Günerhan et al., published by Sciendo
This work is licensed under the Creative Commons Attribution 4.0 International License.

Fig. 1

Exact solution of v1(x, t).
Exact solution of v1(x, t).

Fig. 2

Values of v1(x, t) for α = 1.
Values of v1(x, t) for α = 1.

Fig. 3

Values of v1(x, t) for α = 0.5.
Values of v1(x, t) for α = 0.5.

Fig. 4

Values of v1(x, t) for α = 0.75.
Values of v1(x, t) for α = 0.75.

Fig. 5

Exact solution of v2(x, t).
Exact solution of v2(x, t).

Fig. 6

Values of v2(x, t) for α = 1.
Values of v2(x, t) for α = 1.

Fig. 7

Values of v2(x, t) for α = 0.5
Values of v2(x, t) for α = 0.5

Fig. 8

Values of v2(x, t) for α = 0.75.
Values of v2(x, t) for α = 0.75.

Fig. 9

Exact solution of v3(x, t).
Exact solution of v3(x, t).

Fig. 10

Values of v3(x, t) for α = 1.
Values of v3(x, t) for α = 1.

Fig. 11

Values of v3(x, t) for α = 0.5.
Values of v3(x, t) for α = 0.5.

Fig. 12

Values of v3(x, t) for α = 0.75.
Values of v3(x, t) for α = 0.75.

Numerical solution of v1(x, t)

xtv1FDTMfor α = 0.5v1FDTMfor α = 0.75v1FDTMfor α = 1v1Exact
0.010.010.000089597199410.000096629114150.00009900498330.00009900498337
0.020.020.00034310683240.00037763801660.00039207946940.0003920794693
0.030.030.00074723324260.00083263262730.00087340098020.0008734009802
0.040.040.0012931798640.0014530537080.0015372631030.001537263103
0.050.050.0019742251810.0022314084420.0023780735610.002378073561
0.060.060.0027849189050.0031609629630.0033903523210.003390352321
0.070.070.0037206867200.0042355723710.0045687297180.004568729718
0.080.080.0047776006410.0054495723030.0059079446180.005907944617
0.090.090.0059522316510.0067977035730.0074028426010.007402842601
0.10.10.0072415485600.0082750566510.0090483741810.009048374180

Numerical solution of v3(x, t)

xtv3FDTMfor α = 0.5v3FDTMfor α = 0.75v3FDTMfor α = 1v3Exact
0.010.010.00000998334160.000003161750649.99983333 · 10−79.999833334 · 10−7
0.020.020.00005638016910.000021263156730.000007999466680.00000799946667
0.030.030.00015510631820.000064819738590.000026995950180.00002699595018
0.040.040.00031787092940.00014291761570.000063982934690.00006398293470
0.050.050.00055437015180.00026385051720.00012494792320.0001249479232
0.060.060.00087302456140.00043536310020.00021587042330.0002158704233
0.070.070.0012813461130.00066478045200.00034271995200.0003427199520
0.080.080.0017861538080.00095908787390.00051145404150.0005114540414
0.090.090.0023937136740.0013249845490.00072801624850.0007280162485
0.10.10.0031098359290.0017689218630.00099833416640.0009983341665

The operations for the two-dimensional differential transform method

Transformed functionOriginal function
Uα,β = Vα,β +Wα,βu(x,y) = v(x,y)w(x,y)
Uα,β = λVα,βu(x,y) = λv(x,y)
Uα,β(k,h)=∑r=0k∑s=0hVα,β(r,h−s)Wα,β(k−r,s)[U_{\alpha ,\beta } (k,h) = \sum\nolimits_{r = 0}^k \sum\nolimits_{s = 0}^h V_{\alpha ,\beta } (r,h - s)W_{\alpha ,\beta } (k - r,s)u(x,y) = v(x,y)w(x,y)
Uα,β(k,h)=δ(k−n)δ(h−m)={1,k=n,h=m0,k≠n,h≠m[U_{\alpha ,\beta } (k,h) = \delta (k - n)\delta (h - m) = \{ \begin{array}{*{20}c} {1,k = n,h = m} \\ {0,k \ne n,h \ne m} \\ {} \\\end{array}u(x,y) = (x − x0)mα (y − y0)nβ
Uα,β(k,h)=Γ(α(k+1)+1)Γ(αk+1)Vα,β(k+1,h)[U_{\alpha ,\beta } (k,h) = \frac{{\Gamma (\alpha (k + 1) + 1)}}{{\Gamma (\alpha k + 1)}}V_{\alpha ,\beta } (k + 1,h)Vα,β(k+ 1, h)u(x,y)=Dx0αv(x,y)[u(x,y) = D_{x_0 }^\alpha v(x,y)
Uα,β(k,h)=Γ(β(h+1)+1)Γ(βh+1)Vα,β(k,h+1)[U_{\alpha ,\beta } (k,h) = \frac{{\Gamma (\beta (h + 1) + 1)}}{{\Gamma (\beta h + 1)}}V_{\alpha ,\beta } (k,h + 1)Vα,β(k,h+ 1)u(x,y)=Dy0βv(x,y)[u(x,y) = D_{y_0 }^\beta v(x,y)
Uα,β(k,h)=Γ(α(k+1)+1)Γ(αk+1)Γ(β(h+1)+1)Γ(βh+1)[U_{\alpha ,\beta } (k,h) = \frac{{\Gamma (\alpha (k + 1) + 1)}}{{\Gamma (\alpha k + 1)}}\frac{{\Gamma (\beta (h + 1) + 1)}}{{\Gamma (\beta h + 1)}} . Vα,β(k+ 1, h+ 1), 0< α,β ≤ 1u(x,y)=Dx0αDy0βv(x,y)[u(x,y) = D_{x_0 }^\alpha D_{y_0 }^\beta v(x,y)

Numerical solution of v2(x, t)

xtv2FDTMfor α = 0.5v2FDTMfor α = 0.75v2FDTMfor α = 1v2Exact
0.010.010.00009583789020.00009857498860.000099501247920.0000995012479
0.020.020.00037683059370.00039048806490.00039601993350.0003960199335
0.030.030.00083685434040.00087118442600.00088660074560.0008866007456
0.040.040.0014714633030.0015368151440.0015683178770.001568317877
0.050.050.0019742251810.0022314084420.0023780735610.002378073561
0.060.060.0032501220110.0034093380650.0034936039210.003493603921
0.070.070.0043879939610.0046100740860.0047314665400.004731466540
0.080.080.0056878625250.0059833223780.0061490524100.006149052411
0.090.090.0071471722120.0075264042320.0044435796030.004443579603
0.10.10.0087634945800.0092367530300.00954122942450.0095412294245
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