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Exact solutions for nonlinear evolution equations using novel test function

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Abstract

Based on Bell polynomials approach, in this paper we have used Maple computer algebra package PDEBellII for constructing bilinear equations for some nonlinear evolution equations. Bilinear equations are then used to construct exact solutions using novel test function. Symbolic manipulation program Maple has been used to carry out tedious calculations involved, and a simple Maple code is also given in the form of appendix. The exact solutions obtained using novel test function enrich the solution structure of well-known evolution equations.

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Authors and Affiliations

  1. Yadawindra College of Engineering, Punjabi University, Guru Kashi Campus, Talwandi Sabo, Punjab, 151302, India

    Manjit Singh

  2. Centre for Mathematics and Statistics, School of Basic and Applied Sciences, Central University of Punjab, Bathinda, Punjab, 151001, India

    R. K. Gupta

Authors
  1. Manjit Singh
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  2. R. K. Gupta
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Correspondence to Manjit Singh.

Appendix

Appendix

Maple program for computing coefficients in algebraic expression

Step 1

#Define Hirota derivative;

$$\begin{aligned}&\texttt {Hirota}\_\texttt {Dyt}:=\texttt {(f,g)}\rightarrow \texttt {f}\cdot \texttt {diff(g,y,t)}\\&\quad -\texttt {diff(f,y)}\cdot \texttt {diff(g,t)}\\&\quad -\texttt {diff(f,t)}\cdot \texttt {diff(g,y)}\\&\quad +\texttt {diff(f,y,t)}\cdot \texttt {g};\\&\texttt {Hirota}\_\texttt {Dtz}:=\texttt {(f,g)}\rightarrow \texttt {f}\cdot \texttt {diff(g,t,z)}\\&\quad -\texttt {diff(f,t)}\cdot \texttt {diff(g,z)}\\&\quad -\texttt {diff(f,z)}\cdot \texttt {diff(g,t)}\\&\quad +\texttt {diff(f,t,z)}\cdot \texttt {g};\\&\texttt {Hirota}\_\texttt {Dty}:=\texttt {(f,g)}\rightarrow \texttt {f}\cdot \texttt {diff(g,t,y)}\\&\quad -\texttt {diff(f,t)}\cdot \texttt {diff(g,y)}\\&\quad -\texttt {diff(f,y)}\cdot \texttt {diff(g,t)}\\&\quad +\texttt {diff(f,t,y)}\cdot \texttt {g};\\&\texttt {Hirota}\_\texttt {Dxx}:=\texttt {(f,g)}\rightarrow \texttt {f}\cdot \texttt {diff(g,x,x)}\\&\quad -2\,\texttt {diff(f,x)}\cdot \texttt {diff(g,x)}\\&\quad +\texttt {diff(f,x,x)}\cdot \texttt {g};\\&\texttt {Hirota}\_\texttt {Dyy}:=\texttt {(f,g)}\rightarrow \texttt {f}\cdot \texttt {diff(g,y,y)}\\&\quad -2\,\texttt {diff(f,y)}\cdot \texttt {diff(g,y)}\\&\quad +\texttt {diff(f,y,y)}\cdot \texttt {g};\\&\texttt {Hirota}\_\texttt {Dzz}:=\texttt {(f,g)}\rightarrow \texttt {f}\cdot \texttt {diff(g,z,z)}\\&\quad -2\,\texttt {diff(f,z)}\cdot \texttt {diff(g,z)}+\texttt {diff(f,z,z)}\cdot \texttt {g};\\&\texttt {Hirota}\_\texttt {Dxxxy}:=\texttt {(f,g)}\rightarrow \texttt {f}\cdot \texttt {diff(g,x\$3,y)}\\&\quad -3\cdot \texttt {diff(f,x)}\cdot \texttt {diff(g,x\$2,y)}\\&\quad +3\cdot \texttt {diff(f,x\$2)}\cdot \texttt {diff(g,x,y)}\\&\quad -\texttt {diff(f,x\$3)}\cdot \texttt {diff(g,y)}\\&\quad -\texttt {diff(f,y)}\cdot \texttt {diff(g,x\$3)}\\&\quad +3\cdot \texttt {diff(f,x,y)}\cdot \texttt {diff(g,x\$2)}\\&\quad -3\cdot \texttt {diff(f,x\$2,y)}\cdot \texttt {diff(g,x)}\\&\quad +\texttt {diff(f,x\$3,y)}\cdot \texttt {g};\\&\texttt {Hirota}\_\texttt {Dxxxx}:=\texttt {(f,g)}\rightarrow \texttt {f}\cdot \texttt {diff(g,x\$4)}\\&\quad -4\cdot \texttt {diff(f,x)}\cdot \texttt {diff(g,x\$3)}\\&\quad +6\cdot \texttt {diff(f,x\$2)}\cdot \texttt {diff(g,x\$2)}\\&\quad -4\cdot \texttt {diff(f,x\$3)}\cdot \texttt {diff(g,x)}\\&\quad +\texttt {diff(f,x\$4)}\cdot \texttt {g} \end{aligned}$$

Step 2

#Define new test function;

$$\begin{aligned} \texttt {f}&:= \texttt {e}^{-\texttt {a}_{1}{} \texttt {x}-\texttt {b}_{1}{} \texttt {y}-\texttt {c}_{1}{} \texttt {t}-\texttt {d}_{1}{} \texttt {z}}+\delta _{1}\cdot \texttt {tan}(\texttt {a}_{2}{} \texttt {x}\\&\quad +\texttt {b}_{2}{} \texttt {y}+\texttt {c}_{2}{} \texttt {t}+\texttt {d}_{2}{} \texttt {z})+\delta _{2}\cdot \texttt {tanh}(\texttt {a}_{3}{} \texttt {x}\\&\quad +\texttt {b}_{3}{} \texttt {y}+\texttt {c}_{3}{} \texttt {t}+\texttt {d}_{3}{} \texttt {z})\\&\quad \delta _{3}\cdot \texttt {e}^{\texttt {a}_{1}{} \texttt {x}+\texttt {b}_{1}{} \texttt {y}+\texttt {c}_{1}{} \texttt {t}+\texttt {d}_{1}{} \texttt {z}}; \end{aligned}$$

Step 3

#Substitution of new test function into bilinear equation (23) (for example sake);

$$\begin{aligned}&(\texttt {Hirota}\_\texttt {Dzt}+\texttt {3}\cdot \texttt {Hirota}\_\texttt {Dxx}\\&\quad -\texttt {Hirota}\_\texttt {Dxxxy})\texttt {f}\cdot \texttt {f} \end{aligned}$$

#This command will generate large expression in coefficients of \(e^{\pm \xi _{1}}, e^{\pm \xi _{1}}\tan (\xi _{2})\), \(e^{\pm \xi _{1}}\tanh (\xi _{3}), \tan (\xi _{2})\tanh (\xi _{3})\)

Step 4

#Sorting terms in large expression in previous step;

$$\begin{aligned}&\texttt {sort}(\texttt {expr},\texttt {order}=\texttt {plex}[\texttt {f}_{1},\texttt {f}_{2},\ldots ],\\&\quad \texttt {ascending}); \end{aligned}$$

#This command will sort all terms of expression in appropriate order

Step 5

#Collecting coefficients from expression sorted in previous step

$$\begin{aligned}&\texttt {coeffs}(\texttt {collect}(\texttt {expr},[\texttt {f}_{1},\texttt {f}_{2},\ldots ],\\&\quad \texttt {distributed}),[\texttt {f}_{1},\texttt {f}_{2},\ldots ]); \end{aligned}$$

#This command will produce list of all coefficients in expression.

Step 6

#Solving algebraic equations

$$\begin{aligned} \texttt {solve}(\mathtt {\{list=\sim 0\}},\{a_{i}, b_{i}, c_{i}, d_{i}, \delta _{i}\}),\,\,\texttt {i=1..3} \end{aligned}$$

#This command will generate set of solutions for \(a_{i}, b_{i}, c_{i}, d_{i}, \delta _{i}\)

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Singh, M., Gupta, R.K. Exact solutions for nonlinear evolution equations using novel test function. Nonlinear Dyn 86, 1171–1182 (2016). https://doi.org/10.1007/s11071-016-2955-2

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