# Modified Kudryashov Method to Solve Generalized Kuramoto-Sivashinsky Equation

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## Abstract

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## 1. Introduction

## 2. Analysis of the Modified Kudryashov Method

**Step 1.**- Consider the given NLPDE of the following form $u=u(x,t)$.$$\begin{array}{c}\hfill P\left(u,{u}_{t},{u}_{x},{u}_{tt},{u}_{xx},{u}_{xt},\cdots \right)=0.\end{array}$$
**Step 2.**- Apply the wave transformation $u(x,t)=u\left(\eta \right)$ in Equation (2), where:$$\begin{array}{c}\hfill \eta =\mu (x-\lambda t).\end{array}$$Here, $\mu $ is the wave variable and $\lambda $ is the velocity; both are non-zero constants. Hence, Equation (2) transforms to the following ODE:$$\begin{array}{c}\hfill O\left(u,{u}^{\prime},{u}^{\u2033},u{u}^{\prime},\cdots \right)=0,\end{array}$$
**Step 3.**- Let the initial solution guess of Equation (4) be,$$\begin{array}{c}\hfill u\left(\eta \right)={A}_{0}+\sum _{i=1}^{N}{A}_{i}{\left[Q\left(\eta \right)\right]}^{i},\end{array}$$$$\begin{array}{c}\hfill \frac{dQ\left(\eta \right)}{d\eta}=Q\left(\eta \right)\left[Q\left(\eta \right)-1\right]ln\left(a\right);\phantom{\rule{4pt}{0ex}}a\ne 1,\end{array}$$$$\begin{array}{c}\hfill Q\left(\eta \right)=\frac{1}{1+D{a}^{\eta}},\end{array}$$
**Step 4.**- Substituting Equations (5) and (6) in Equation (4) leads to the polynomial in $Q{\left(\eta \right)}^{i};\phantom{\rule{4pt}{0ex}}i=0,1,2,\cdots $. As $Q{\left(\eta \right)}^{i}\ne 0$, so collecting its coefficients and then equating to zero give the systems of overdetermined algebraic equations, which upon solving give the unknowns of Equations (3) and (5).
**Step 5.**

## 3. MKM Application to Solve the Generalized Kuramoto–Sivashinsky Equation

**Case 1.**- For $\alpha ={\delta}_{2}$ and $\beta =4{\delta}_{1}$ in Equation (1), the unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}={A}_{1}=0,\phantom{\rule{4pt}{0ex}}{A}_{2}=120{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =6{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 1),$$\begin{array}{c}\hfill {u}_{1}(x,t):=\frac{120{\delta}_{3}{a}^{\mu x-6{\delta}_{3}\mu t}}{{\left(1+{a}^{\mu x-6{\delta}_{3}\mu t}\right)}^{3}}.\end{array}$$Further, for the same $\alpha $ and $\beta $ value, the second set of unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=-12{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{1}=0,\phantom{\rule{4pt}{0ex}}{A}_{2}=120{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =-6{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 1),$$\begin{array}{c}\hfill {u}_{2}(x,t):=-\frac{12{\delta}_{3}\left(1+{a}^{3\mu x}{e}^{3\left(6{\delta}_{3}\mu ln\left(a\right)t\right)}+3{a}^{2\mu x}{e}^{2\left(6{\delta}_{3}\mu ln\left(a\right)t\right)}-7{a}^{\mu x}{e}^{6{\delta}_{3}\mu ln\left(a\right)t}\right)}{{\left(1+{a}^{\mu x}{e}^{6{\delta}_{3}\mu ln\left(a\right)t}\right)}^{3}}.\end{array}$$
**Case 2.**- For $\alpha ={\delta}_{2}$ and $\beta =-4{\delta}_{1}$ in Equation (1), the unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=0,\phantom{\rule{4pt}{0ex}}{A}_{1}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{2}=240{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =-6{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 2),$$\begin{array}{c}\hfill {u}_{3}(x,t):=-\frac{120{\delta}_{3}{a}^{2\left(\mu x+6{\delta}_{3}\mu t\right)}}{{\left(1+{a}^{\mu x+6{\delta}_{3}\mu t}\right)}^{3}}.\end{array}$$Further, for the same $\alpha $ and $\beta $ value, the second set of unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=12{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{1}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{2}=240{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =6{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 2),$$\begin{array}{c}\hfill {u}_{4}(x,t):=\frac{12{\delta}_{3}\left({a}^{3\left(\mu x-6{\delta}_{3}\mu t\right)}-7{a}^{2\left(\mu x-6{\delta}_{3}\mu t\right)}+3{a}^{\mu x-6{\delta}_{3}\mu t}+1\right)}{{\left(1+{a}^{\mu x-6{\delta}_{3}\mu t}\right)}^{3}}.\end{array}$$
**Case 3.**- For $\alpha =-19{\delta}_{2}$ and $\beta =0$ in Equation (1), the unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=-60{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{1}=0,\phantom{\rule{4pt}{0ex}}{A}_{2}=180{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =-30{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 3),$$\begin{array}{c}\hfill {u}_{5}(x,t):=-\frac{60{\delta}_{3}{e}^{2\left(30{\delta}_{3}\mu ln\left(a\right)t\right)}\left({a}^{3\mu x}{e}^{30{\delta}_{3}\mu ln\left(a\right)t}+3{a}^{2\mu x}\right)}{{\left(1+{a}^{\mu x}{e}^{30{\delta}_{3}\mu ln\left(a\right)t}\right)}^{3}}.\end{array}$$Further, for the same $\alpha $ and $\beta $ value, the second set of unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}={A}_{1}=0,\phantom{\rule{4pt}{0ex}}{A}_{2}=180{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =30{\delta}_{3}.\end{array}$$
**Case 4.**- For $\alpha =47{\delta}_{2}$ and $\beta =12{\delta}_{1}$ in Equation (1), the unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}={A}_{1}={A}_{2}=0,\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =-60{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 4),$$\begin{array}{c}\hfill {u}_{7}(x,t):=-\frac{120{\delta}_{3}}{{\left(1+{a}^{\mu x+60{\delta}_{3}\mu t}\right)}^{3}}.\end{array}$$Further, for the same $\alpha $ and $\beta $, the second set of unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=120{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{1}={A}_{2}=0,\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =60{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 4),$$\begin{array}{c}\hfill {u}_{8}(x,t):=\frac{120{\delta}_{3}\left(3{a}^{\mu x}{e}^{2\left(60{\delta}_{3}\mu ln\left(a\right)t\right)}+3{a}^{2\mu x}{e}^{60{\delta}_{3}\mu ln\left(a\right)t}+{a}^{3\mu x}\right)}{{\left({a}^{\mu x}+{e}^{60{\delta}_{3}\mu ln\left(a\right)t}\right)}^{3}}.\end{array}$$
**Case 5.**- For $\alpha =47{\delta}_{2}$ and $\beta =-12{\delta}_{1}$ in Equation (1), the unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=0,\phantom{\rule{4pt}{0ex}}{A}_{1}=-360{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{2}=360{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =-60{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 5),$$\begin{array}{c}\hfill {u}_{9}(x,t):=-\frac{120{\delta}_{3}\left(3{a}^{2\mu x}{e}^{2\left(60{\delta}_{3}\mu ln\left(a\right)t\right)}+3{a}^{\mu x}{e}^{60{\delta}_{3}\mu ln\left(a\right)t}+1\right)}{{\left(1+{a}^{\mu x}{e}^{60{\delta}_{3}\mu ln\left(a\right)t}\right)}^{3}}.\end{array}$$Further, for the same $\alpha $ and $\beta $ value, the second set of unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=120{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{1}=-360{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{2}=360{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =60{\delta}_{3}.\end{array}$$
**Case 6.**- For $\alpha =73{\delta}_{2}$ and $\beta =16{\delta}_{1}$ in Equation (1), the unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=180{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{1}=0,\phantom{\rule{4pt}{0ex}}{A}_{2}=-60{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =90{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 6),$$\begin{array}{c}\hfill {u}_{11}(x,t):=\frac{60{\delta}_{3}\left(8{a}^{\mu x}{e}^{2\left(90{\delta}_{3}\mu ln\left(a\right)t\right)}+9{a}^{2\mu x}{e}^{90{\delta}_{3}\mu ln\left(a\right)t}+3{a}^{3\mu x}\right)}{{\left({e}^{90{\delta}_{3}\mu ln\left(a\right)t}+{a}^{\mu x}\right)}^{3}}.\end{array}$$Further, for the same $\alpha $ and $\beta $ value, the second set of unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}={A}_{1}=0,\phantom{\rule{4pt}{0ex}}{A}_{2}=-60{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =-90{\delta}_{3}.\end{array}$$
**Case 7.**- For $\alpha =73{\delta}_{2}$ and $\beta =-16{\delta}_{1}$ in Equation (1), the unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=180{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{1}=-480{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{2}=420{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =90{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 7),$$\begin{array}{c}\hfill {u}_{13}(x,t):=\frac{60{\delta}_{3}\left({a}^{2\mu x}{e}^{90{\delta}_{3}\mu ln\left(a\right)t}+3{a}^{3\mu x}\right)}{{\left({e}^{90{\delta}_{3}\mu ln\left(a\right)t}+{a}^{\mu x}\right)}^{3}}.\end{array}$$Further, for the same $\alpha $ and $\beta $ value, the second set of unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=0,\phantom{\rule{4pt}{0ex}}{A}_{1}=-480{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{2}=420{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =-90{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 7),$$\begin{array}{c}\hfill {u}_{14}(x,t):=-\frac{60{\delta}_{3}\left(8{a}^{2\mu x}{e}^{2\left(90{\delta}_{3}\mu ln\left(a\right)t\right)}+9{a}^{\mu x}{e}^{90{\delta}_{3}\mu ln\left(a\right)t}+3\right)}{{\left(1+{a}^{\mu x}{e}^{90{\delta}_{3}\mu ln\left(a\right)t}\right)}^{3}}.\end{array}$$
**Case 8.**- For $\alpha =\frac{19}{11}{\delta}_{2}$ and $\beta =0$ in Equation (1), the unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=\frac{60}{11}{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{1}=-\frac{720}{11}{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{2}=180{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =\frac{30}{11}{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 8),$$\begin{array}{c}\hfill {u}_{15}(x,t):=\frac{60{\delta}_{3}{a}^{\left(\mu x-\frac{30}{11}{\delta}_{3}\mu t\right)}\left({a}^{2\left(\mu x-\frac{30}{11}{\delta}_{3}\mu t\right)}-9{a}^{\left(\mu x-\frac{30}{11}{\delta}_{3}\mu t\right)}+12\right)}{11{\left(1+{a}^{\left(\mu x-\frac{30}{11}{\delta}_{3}\mu t\right)}\right)}^{3}}.\end{array}$$Further, for the same $\alpha $ and $\beta $ value, the second set of unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=0,\phantom{\rule{4pt}{0ex}}{A}_{1}=-\frac{720}{11}{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{2}=180{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =-\frac{30}{11}{\delta}_{3}.\end{array}$$Therefore, the exact solution of Equation (1) is given by (Figure 8),$$\begin{array}{c}\hfill {u}_{16}(x,t):=-\frac{60{\delta}_{3}\left(1-9{a}^{\left(\mu x+\frac{30}{11}{\delta}_{3}\mu t\right)}+12{a}^{2\left(\mu x+\frac{30}{11}{\delta}_{3}\mu t\right)}\right)}{11{\left(1+{a}^{\left(\mu x+\frac{30}{11}{\delta}_{3}\mu t\right)}\right)}^{3}}.\end{array}$$
**Case 9.**- For $\alpha =-{\delta}_{2}$ and $\beta =4i{\delta}_{1}$ in Equation (1), the unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=0,\phantom{\rule{4pt}{0ex}}{A}_{1}=-60{\mu}^{3}ln{\left(a\right)}^{3}\left(\gamma -i\gamma \right),\phantom{\rule{4pt}{0ex}}{A}_{2}=60(3-i){\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =4i{\delta}_{3}.\end{array}$$Therefore, the exact complex solution of Equation (1) is given by,$$\begin{array}{c}\hfill {u}_{17}(x,t):=\frac{60{\delta}_{3}{a}^{\mu x-4i{\delta}_{3}\mu t}\left(i+1+\left(i-1\right){a}^{\mu x-4i{\delta}_{3}\mu t}\right)}{{\left(1+{a}^{\mu x-4i{\delta}_{3}\mu t}\right)}^{3}}.\end{array}$$The 2D graph of real and imaginary parts of ${u}_{17}(x,t)$ are drawn in Figure 9.Further, for the same $\alpha $ and $\beta $ value, the second set of unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=-8i{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{1}=-60{\mu}^{3}ln{\left(a\right)}^{3}\left(\gamma -i\gamma \right),\phantom{\rule{4pt}{0ex}}{A}_{2}=60(3-i){\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =-4i{\delta}_{3}.\end{array}$$Therefore, the exact complex solution of Equation (1) is given by,$$\begin{array}{c}\hfill {u}_{18}(x,t):=-\frac{8{\delta}_{3}}{{\left(1+{a}^{\mu x+4i{\delta}_{3}\mu t}\right)}^{3}}\left[i(1+{a}^{3\left(\mu x+4i{\delta}_{3}\mu t\right)})+\left(\frac{15-9i}{2}\right){a}^{2\left(\mu x+4i{\delta}_{3}\mu t\right)}-\left(\frac{15+9i}{2}\right){a}^{\mu x+4i{\delta}_{3}\mu t}\right].\end{array}$$
**Case 10.**- For $\alpha =-{\delta}_{2}$ and $\beta =-4i{\delta}_{1}$ in Equation (1), the unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=0,\phantom{\rule{4pt}{0ex}}{A}_{1}=-60{\mu}^{3}ln{\left(a\right)}^{3}(\gamma +i\gamma ),\phantom{\rule{4pt}{0ex}}{A}_{2}=60(3+i){\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =-4i{\delta}_{3}.\end{array}$$Therefore, the exact complex solution of Equation (1) is given by,$$\begin{array}{c}\hfill {u}_{19}(x,t):=-\frac{60{\delta}_{3}{a}^{\mu x+4i{\delta}_{3}\mu t}\left(i-1+(i+1){a}^{\mu x+4i{\delta}_{3}\mu t}\right)}{{\left(1+{a}^{\mu x+4i{\delta}_{3}\mu t}\right)}^{3}}.\end{array}$$The 2D graphs of real and imaginary parts of ${u}_{19}(x,t)$ are drawn in Figure 11.Further, for the same $\alpha $ and $\beta $ value, the second set of unknown coefficients are given by,$$\begin{array}{c}\hfill {A}_{0}=8i{\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{1}=-60{\mu}^{3}ln{\left(a\right)}^{3}(\gamma +i\gamma ),\phantom{\rule{4pt}{0ex}}{A}_{2}=60(3+i){\delta}_{3},\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{\delta}_{3},\phantom{\rule{4pt}{0ex}}\lambda =4i{\delta}_{3}.\end{array}$$Therefore, the exact complex solution of Equation (1) is given by,$$\begin{array}{c}\hfill {u}_{20}(x,t):=\frac{8{\delta}_{3}}{{\left(1+{a}^{\mu x-4i{\delta}_{3}\mu t}\right)}^{3}}\left[i(1+{a}^{3\left(\mu x-4i{\delta}_{3}\mu t\right)})-\left(\frac{15+9i}{2}\right){a}^{2\left(\mu x-4i{\delta}_{3}\mu t\right)}+\left(\frac{15-9i}{2}\right){a}^{\mu x-4i{\delta}_{3}\mu t}\right].\end{array}$$

## 4. Conclusions

## Author Contributions

## Funding

## Conflicts of Interest

## Appendix A. GKSE in the Previous Studies

- $$\begin{array}{c}\hfill {A}_{0}=-\frac{{\beta}^{3}}{576{\gamma}^{2}},\phantom{\rule{4pt}{0ex}}{A}_{1}=\frac{5{\beta}^{2}}{4\gamma},\phantom{\rule{4pt}{0ex}}{A}_{2}=-15\beta ,\phantom{\rule{4pt}{0ex}}{A}_{3}=120\gamma ,\phantom{\rule{4pt}{0ex}}\alpha =\frac{47{\beta}^{2}}{144\gamma},\phantom{\rule{4pt}{0ex}}b=\frac{{\beta}^{2}}{576{\gamma}^{2}},\phantom{\rule{4pt}{0ex}}{C}_{0}=-\frac{5{\beta}^{3}}{144{\gamma}^{2}}.\end{array}$$
- $$\begin{array}{c}\hfill {A}_{0}=\frac{30{\beta}^{3}}{128{\gamma}^{2}},\phantom{\rule{4pt}{0ex}}{A}_{1}=-\frac{30{\beta}^{2}}{16\gamma},\phantom{\rule{4pt}{0ex}}{A}_{2}=-30\beta ,\phantom{\rule{4pt}{0ex}}{A}_{3}=120\gamma ,\phantom{\rule{4pt}{0ex}}\alpha =\frac{{\beta}^{2}}{16\gamma},\phantom{\rule{4pt}{0ex}}b=\frac{{\beta}^{2}}{64{\gamma}^{2}},\phantom{\rule{4pt}{0ex}}{C}_{0}=\frac{3{\beta}^{3}}{32{\gamma}^{2}}.\end{array}$$

- $$\begin{array}{c}\hfill {A}_{0}=\nu -6{a}^{3}\gamma ,\phantom{\rule{4pt}{0ex}}{A}_{1}=-120{a}^{2}b\gamma ,\phantom{\rule{4pt}{0ex}}{A}_{2}=240a{b}^{2}\gamma ,\phantom{\rule{4pt}{0ex}}{A}_{3}=-120{b}^{3}\gamma ,\phantom{\rule{4pt}{0ex}}\alpha ={a}^{2}\gamma ,\phantom{\rule{4pt}{0ex}}\beta =4a\gamma .\end{array}$$
- $$\begin{array}{c}\hfill {A}_{0}=-990\gamma +60\gamma k+\nu ,\phantom{\rule{4pt}{0ex}}{A}_{1}=60\gamma +180\gamma k,\phantom{\rule{4pt}{0ex}}{A}_{2}=60\gamma k,\phantom{\rule{4pt}{0ex}}{A}_{3}=-120\gamma ,\phantom{\rule{4pt}{0ex}}\alpha =365\gamma ,\phantom{\rule{4pt}{0ex}}\beta =-36\gamma -4\gamma k.\end{array}$$

## Appendix B. Studying GKSE by GKM and SGEEM

- For solving Equation (1) by the generalized Kudryashov method [22,23,24], the homogeneous balancing of Equation (8) gives $N=M+3$, which has infinite solutions. For the value $M=1$, this gives $N=4$. Therefore,$$\begin{array}{c}\hfill u\left(\eta \right)=\frac{{A}_{0}+{A}_{1}Q\left(\eta \right)+{A}_{2}{\left(Q\left(\eta \right)\right)}^{2}+{A}_{3}{\left(Q\left(\eta \right)\right)}^{3}+{A}_{4}{\left(Q\left(\eta \right)\right)}^{4}}{{B}_{0}+{B}_{1}Q\left(\eta \right)}.\end{array}$$
- Next, for solving Equation (1) by the sine-Gordon equation expansion method [26], the homogeneous balancing is the same as the MKM given by $N=3$. Thus,$$\begin{array}{c}\hfill u\left(\eta \right)={A}_{0}+{A}_{1}tanh\left(\eta \right)+{B}_{1}\mathrm{sech}\left(\eta \right)+{A}_{2}{tanh}^{2}\left(\eta \right)+{B}_{2}tanh\left(\eta \right)\mathrm{sech}\left(\eta \right)+{A}_{3}{tanh}^{3}\left(\eta \right)+{B}_{3}{tanh}^{2}\left(\eta \right)\mathrm{sech}\left(\eta \right).\end{array}$$Substituting the above equation $u\left(\eta \right)$ in Equation (8) and following the steps in [26] lead to the polynomials in $sin\left(w\right)$, $cos\left(w\right)$, their products and powers. Collecting the coefficients, equating them to zero and solving in Maple result in the continuous execution. Thus, we conclude that Equation (1) cannot be solved by the sine-Gordon equation expansion method either.

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**Figure 1.**Solutions in Case 1, Equations (10) and (11), respectively from left to right for a = 5, μ = 1 and t = 1 in x ∈ [−15, 15] for different values of γ.

**Figure 2.**Solutions in Case 2, Equations (12) and (13), respectively from left to right for a = 5, μ = 1 and t = 1 in x ∈ [−15, 15] for different values of γ.

**Figure 3.**Solutions in Case 3, Equations (14) and (15), respectively from left to right for a = 5, μ = 1 and t = 1 in x ∈ [−40, 40] for u

_{5}(x, t) and in x ∈ [−50, 50] for u

_{6}(x, t) for different values of γ.

**Figure 4.**Solutions in Case 4, Equations (16) and (17), respectively from left to right for a = 5, μ = 1 and t = 1 in x ∈ [−15, 15] for different values of γ.

**Figure 5.**Solutions in Case 5, Equations (18) and (19), respectively from left to right for a = 5, μ = 1 and t = 1 in x ∈ [−15, 15] for different values of γ.

**Figure 6.**Solutions in Case 6, Equations (20) and (21), respectively from left to right for a = 5, μ = 1 and t = 1 in x ∈ [−15, 15] for u

_{11}(x, t) and x ∈ [−20, 20] for u

_{12}(x, t) for different values of γ.

**Figure 7.**Solutions in Case 7, Equations (22) and (23), respectively from left to right for a = 5, μ = 1 and t = 1 in x ∈ [−20, 20] for different values of γ.

**Figure 8.**Solutions in Case 8, Equations (24) and (25), respectively from left to right for a = 5, μ = 1 and t = 1 in x ∈ [−20, 20] for different values of γ.

**Figure 9.**Real and imaginary part of the solution in Case 9, Equation (26), respectively from left to right for a = 5, μ = 1 and t = 1 in x ∈ [−3, 3] for different values of γ.

**Figure 10.**Real and imaginary part of the solution in Case 9, Equation (27), respectively from left to right for a = 5, μ = 1 and t = 1 in x ∈ [−3, 3] for different values of γ.

**Figure 11.**Real and imaginary part of the solution in Case 10, Equation (28), respectively from left to right for a = 5, μ = 1 and t = 1 in x ∈ [−3, 3] for different values of γ.

**Figure 12.**Real and imaginary part of the solution in Case 10, Equation (29), respectively from left to right for a = 7, μ = 1 and t = 1 in x ∈ [−3, 3] for different values of γ.

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Kilicman, A.; Silambarasan, R.
Modified Kudryashov Method to Solve Generalized Kuramoto-Sivashinsky Equation. *Symmetry* **2018**, *10*, 527.
https://doi.org/10.3390/sym10100527

**AMA Style**

Kilicman A, Silambarasan R.
Modified Kudryashov Method to Solve Generalized Kuramoto-Sivashinsky Equation. *Symmetry*. 2018; 10(10):527.
https://doi.org/10.3390/sym10100527

**Chicago/Turabian Style**

Kilicman, Adem, and Rathinavel Silambarasan.
2018. "Modified Kudryashov Method to Solve Generalized Kuramoto-Sivashinsky Equation" *Symmetry* 10, no. 10: 527.
https://doi.org/10.3390/sym10100527