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A SIXTH-ORDER HERMITE COLLOCATION TECHNIQUE FOR NUMERICAL STUDY OF NONLINEAR FISHER AND BURGERS–FISHER EQUATIONS

Published online by Cambridge University Press:  14 March 2025

ARCHNA KUMARI
Affiliation:
Department of Mathematics, Sant Longowal Institute of Engineering and Technology, Longowal 148106, India; e-mail: archankaundal@gmail.com
VIJAY KUMAR KUKREJA*
Affiliation:
Department of Mathematics, Sant Longowal Institute of Engineering and Technology, Longowal 148106, India; e-mail: archankaundal@gmail.com
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Abstract

This study aims to formulate a highly accurate numerical method, specifically a seventh-order Hermite technique with an error term of sixth order, to solve the Fisher and Burgers–Fisher equations. This technique employs a combination of orthogonal collocation on the finite element method and hepta Hermite basis functions. By ensuring continuity of the dependent variable and its first three derivatives across the entire solution domain, it achieves a remarkable level of accuracy and smoothness. The space discretization is handled through the application of hepta Hermite polynomials, while the time discretization is managed by the Crank–Nicholson scheme. The stability and convergence analysis of the scheme are discussed in detail. To validate the accuracy of the proposed technique, three examples are taken. The results obtained from these examples are thoroughly analysed and compared against the exact solutions and reliable data from the existing literature. It is established that the proposed technique is easy to implement and gives better results as compared with existing ones.

Information

Type
Research Article
Copyright
© The Author(s), 2025. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc.
Figure 0

Table 1 Comparison of absolute error of Example 6.1 for $\alpha =0.001,~\beta =0.001,~\varepsilon =1$, $h=0.1$ and $\Delta t=0.0001$.

Figure 1

Table 2 Comparison of absolute error of Example 6.1 for $\varepsilon =1$, $\alpha =0.01,~ \beta =0.01$, $h=0.1$ and $\Delta t=0.0002$.

Figure 2

Table 3 Comparison of absolute error of Example 6.1 for $\varepsilon =1$, $\alpha =1,~ \beta =0$, $h=0.1$ and $\Delta t=0.0001$.

Figure 3

Table 4 Absolute error of Example 6.1 for $\alpha =0.01,~\beta =0.01,~\varepsilon =1$, $h=0.5$ and $\Delta t=0.0001$.

Figure 4

Table 5 Comparison of absolute error of Example 6.1 for $\varepsilon =1$, $h=0.1$ and $\Delta t=0.000\,01$.

Figure 5

Table 6 Comparison of maximum absolute error of Example 6.1 for $\varepsilon =1$, $h=0.1$ and $\Delta t=0.000\,01$.

Figure 6

Table 7 Time rate convergence of Example 6.1 for $\varepsilon =1$, $h=0.1$.

Figure 7

Table 8 Space rate convergence of Example 6.1 for $\varepsilon =2^{-17}$, $\Delta t=0.001$, $\alpha =0.01,~ \beta =0.01,~ t=1$.

Figure 8

Figure 1 $\varepsilon $-effect, time-effect and comparison of exact and numerical solutions of Example 6.1 with $h=0.1$, $\Delta t=0.001$, $\alpha =0.01$ and $\beta =0.01$.

Figure 9

Table 9 Comparison of $L_2$ and $L_{\infty }$ errors of Example 6.2 for $\Delta t=0.1$.

Figure 10

Table 10 Comparison of $L_{2}$ and $L_{\infty } $ error of Example 6.2 with $N=64$, $\Delta t=0.000\,005$, $\varepsilon =1$, $\beta =10\,000$.

Figure 11

Table 11 $L_{2}$ and $L_{\infty } $ error of Example 6.2 with $\varepsilon =1$, $\beta =6$ and $N=20$.

Figure 12

Table 12 Time rate convergence of Example 6.2 for $ \varepsilon =1$, $N=21$.

Figure 13

Figure 2 Comparison of exact and numerical solutions of Example 6.2 with $h=0.025$, $\Delta t=0.000\,01$ and $\varepsilon =1$.

Figure 14

Figure 3 Surface plot of solution of Example 6.2 with $h=0.025$, $\Delta t=0.000\,01$, $\varepsilon =1$ and $\beta =10\,000$.

Figure 15

Table 13 Comparison of absolute error of Example 6.3 at $t=4$ with $h=0.25$, $\Delta t=0.01$, $s=0.5$, $z=1$, $v=1$.

Figure 16

Table 14 Time rate convergence of Example 6.3 for $\varepsilon =1$, $h=0.25$.

Figure 17

Figure 4 Comparison of exact and numerical solutions of Example 6.3 with $h=0.5$, $\Delta t=0.01$, $\varepsilon =1$, $s=0.5$, $z=1$ and $v=1$.

Figure 18

Figure 5 Absolute error of Example 6.3 with $h=0.5$, $\Delta t=0.01$, $\varepsilon =1$, $s=0.5$, $z=1$ and $v=1$.