Article Contents
Article Contents

Numerical investigation of space fractional order diffusion equation by the Chebyshev collocation method of the fourth kind and compact finite difference scheme

• * Corresponding author: Hamid Safdari
• This paper develops a numerical scheme for finding the approximate solution of space fractional order of the diffusion equation (SFODE). Firstly, the compact finite difference (CFD) with convergence order $\mathcal{O}(\delta \tau ^{2})$ is used for discretizing time derivative. Afterwards, the spatial fractional derivative is approximated by the Chebyshev collocation method of the fourth kind. Furthermore, time-discrete stability and convergence analysis are presented. Finally, two examples are numerically investigated by the proposed method. The examples illustrate the performance and accuracy of our method compared to existing methods presented in the literature.

Mathematics Subject Classification: 34K37, 91G80, 97N50.

 Citation:

• Figure 1.  Plots of the approximate solution (left side) and absolute error (right side) of Example 5.1 at $T = 1$, $M = 400$ and $N = 5$

Figure 2.  The maximum absolute error and error norm $L_{2}$ of Example 5.1 at $T = 1$, $N = 5$ and $M = 200,400,600, \ldots, 3000$

Figure 3.  Error histories of Example 5.1 at $T = 1$, $N = 5$ and $M = 100,200,400,800, 1600$

Figure 4.  Error histories of Example 5.1 at $T = 1$, $M = 400$ and $N = 3, 5, 7, 9$

Figure 5.  Error histories of Example 5.2 at $T = 1$, with $M = 100,200,400,800, 1600,$ $N = 5$ (left side) and $N = 7$ (right side)

Table 1.  The absolute error of Example 5.1 at $T = 1$

 $x$ with $N=7$ in with $N=7$ in with $N=3$ in our method with $N=3$ [15] [27] [30] $0$ $2.81\times 10^{-5}$ $0$ $0$ $4.77\times 10^{-17}$ $0.1$ $4.26\times 10^{-5}$ $4.66\times 10^{-5}$ $5.46\times 10^{-6}$ $3.17\times 10^{-9}$ $0.2$ $5.39\times 10^{-5}$ $7.74\times 10^{-5}$ $8.51\times 10^{-6}$ $5.85\times 10^{-9}$ $0.3$ $6.12\times 10^{-5}$ $5.00\times 10^{-5}$ $9.60\times 10^{-6}$ $7.97\times 10^{-9}$ $0.4$ $6.48\times 10^{-5}$ $2.30\times 10^{-5}$ $9.18\times 10^{-6}$ $9.44\times 10^{-9}$ $0.5$ $6.45\times 10^{-5}$ $2.74\times 10^{-5}$ $7.69\times 10^{-6}$ $1.02\times 10^{-8}$ $0.6$ $5.98\times 10^{-5}$ $4.38\times 10^{-5}$ $5.60\times 10^{-6}$ $1.01\times 10^{-8}$ $0.7$ $5.23\times 10^{-5}$ $3.87\times 10^{-5}$ $3.33\times 10^{-6}$ $9.12\times 10^{-9}$ $0.8$ $4.48\times 10^{-5}$ $1.01\times 10^{-5}$ $1.34\times 10^{-6}$ $7.17\times 10^{-9}$ $0.9$ $3.91\times 10^{-5}$ $3.35\times 10^{-5}$ $8.39\times 10^{-8}$ $4.16\times 10^{-9}$ $1.0$ $2.81\times 10^{-5}$ $0$ $0$ $7.55\times 10^{-17}$

Table 2.  The absolute error of Example 5.1 at $T = 2$

 $x$ with $N=5$ in with $N=5$ in with $N=3$ in our method with $N=3$ $[15] [27] [30]$ 0  2.74\times 10^{-5}  0  0  1.86\times 10^{-17}  0.1  4.20\times 10^{-5}  4.47\times 10^{-6}  3.33\times 10^{-6}  1.28\times 10^{-8}  0.2  3.76\times 10^{-5}  2.78\times 10^{-7}  5.65\times 10^{-6}  2.05\times 10^{-8}  0.3  8.44\times 10^{-5}  5.81\times 10^{-6}  7.05\times 10^{-6}  2.40\times 10^{-8}  0.4  3.27\times 10^{-5}  1.02\times 10^{-5}  7.64\times 10^{-6}  2.40\times 10^{-8}  0.5  3.61\times 10^{-5}  1.17\times 10^{-5}  7.52\times 10^{-6}  2.15\times 10^{-8}  0.6  1.94\times 10^{-5}  1.08\times 10^{-5}  6.80\times 10^{-6}  1.72\times 10^{-8}  0.7  2.95\times 10^{-5}  8.54\times 10^{-6}  5.59\times 10^{-6}  1.21\times 10^{-8}  0.8  4.92\times 10^{-5}  6.06\times 10^{-6}  3.98\times 10^{-6}  6.93\times 10^{-9}  0.9  2.83\times 10^{-5}  3.67\times 10^{-6}  2.08\times 10^{-6}  2.62\times 10^{-9}  1.0  7.73\times 10^{-5}  0  0  8.24\times 10^{-18} $Table 3. The absolute error of Example 5.1 at$ T = 10 $$ x  N=3  N=5  N=7  0  5.82\times 10^{-21}  5.93\times 10^{-22}  4.43\times 10^{-21}  0.2  1.01\times 10^{-9}  4.74\times 10^{-9}  2.28\times 10^{-9}  0.4  8.21\times 10^{-9}  8.11\times 10^{-9}  4.21\times 10^{-9}  0.6  1.28\times 10^{-9}  1.17\times 10^{-9}  1.15\times 10^{-9}  0.8  3.76\times 10^{-9}  7.93\times 10^{-10}  2.71\times 10^{-10}  1.0  4.34\times 10^{-21}  3.78\times 10^{-21}  1.14\times 10^{-22} $Table 4. The convergence order, the errors$ L_{2} $and$ L_{\infty} $for Example 5.1 with$ T = 1 $and$ N = 3 $$ \delta \tau  L_{\infty}  C_{\delta \tau}  L_{2}  C_{\delta \tau}  \frac{1}{100}  1.62773\times 10^{-7}  3.76647\times 10^{-7}  \frac{1}{200}  4.06928\times 10^{-8}  2.00002  9.41607\times 10^{-8}  2.00002  \frac{1}{400}  1.01732\times 10^{-8}  2.00000  2.35401\times 10^{-8}  2.00000  \frac{1}{800}  2.54329\times 10^{-9}  2.00000  5.88503\times 10^{-9}  2.00000  \frac{1}{1600}  6.35828\times 10^{-10}  1.99999  1.47127\times 10^{-9}  1.99999  \mathrm{TCO}  2  2 $Table 5. The convergence order, the errors$ L_{2} $and$ L_{\infty} $for Example 5.1 with$ T = 10 $and$ N = 3 $$ \delta \tau  L_{\infty}  C_{\delta\tau}  L_{2}  C_{\delta \tau}  \frac{1}{100}  1.63402\times 10^{-7}  3.10926\times 10^{-7}  \frac{1}{200}  4.08673\times 10^{-8}  1.99941  7.77632\times 10^{-8}  1.99941  \frac{1}{400}  1.02179\times 10^{-8}  1.99985  1.94428\times 10^{-8}  1.99985  \frac{1}{800}  2.55453\times 10^{-9}  1.99996  4.86082\times 10^{-9}  1.99996  \frac{1}{1600}  6.38636\times 10^{-10}  1.99999  1.21521\times 10^{-9}  1.99999  \mathrm{TCO}  2  2 $Table 6. The convergence order, the errors$ L_{2} $and$ L_{\infty} $for Example 5.2 with$ N = 7 $at$ T = 1 $$ \delta\tau  L_{\infty}  C_{\delta\tau}  L_{2}  C_{\delta \tau}  \frac{1}{100}  1.71816\times 10^{-6}  3.73349\times 10^{-6}  \frac{1}{200}  4.29538\times 10^{-7}  2.00000  9.33372\times 10^{-7}  2.00000  \frac{1}{400}  1.07384\times 10^{-7}  2.00000  2.33343\times 10^{-7}  2.00000  \frac{1}{800}  2.68460\times 10^{-8}  2.00000  5.83360\times 10^{-8}  2.00000  \frac{1}{1600}  6.71143\times 10^{-9}  2.00002  1.45842\times 10^{-8}  1.99998  \mathrm{TCO}  2  2 $Table 7. The comparison of maximum error of our proposed method and [32] for Example 5.2, at$ T = 1 $ Max error-CN [32] Max error-ext CN [32] the present method with N=3$ 6.84895\times 10^{-4}  2.82750 \times 10^{-5}  9.95930\times 10^{-8} \$

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Tables(7)