1. Introduction
2. The essences of proposed method
1. | 1.If ${c}_{0}=\tfrac{{c}_{4}^{3}\,\left({m}^{2}-1\right)}{32{c}_{6}^{2}{m}^{2}}\,,\,\,\,{c}_{2}=\tfrac{{c}_{4}^{2}\left(5{m}^{2}-1\right)}{16{c}_{6}{m}^{2}},\,\,\,\,{c}_{6}\gt 0,$ then $\begin{eqnarray}g(\zeta )=\mathrm{sn}(\rho \zeta )\,\,\,\,\,\mathrm{or}\,\,\,\,\,g(\zeta )=\displaystyle \frac{1}{m\,\mathrm{sn}(\rho \zeta ))},\end{eqnarray}$ where $\rho =\tfrac{{c}_{4}}{2m}\sqrt{\tfrac{1}{{c}_{6}}}$. |
2. | 2.If ${c}_{0}=\tfrac{{c}_{4}^{3}\,(1-{m}^{2})}{32{c}_{6}^{2}}\,,\,\,\,{c}_{2}=\tfrac{{c}_{4}^{2}\left(5-{m}^{2}\right)}{16{c}_{6}},\,\,\,\,{c}_{6}\gt 0,$ then $\begin{eqnarray}g(\zeta )=m\,\mathrm{sn}(\rho \zeta )\,\,\,\,\,\mathrm{or}\,\,\,\,\,g(\zeta )=\displaystyle \frac{1}{\mathrm{sn}(\rho \zeta ))},\end{eqnarray}$ where $\rho =\tfrac{{c}_{4}}{2}\sqrt{\tfrac{1}{{c}_{6}}}.$ |
3. | 3.If ${c}_{0}=\tfrac{{c}_{4}^{3}}{32{m}^{2}{c}_{6}^{2}}\,,\,\,\,{c}_{2}=\tfrac{{c}_{4}^{2}\left(4{m}^{2}+1\right)}{16{c}_{6}{m}^{2}},\,\,\,\,{c}_{6}\lt 0,$ then $\begin{eqnarray}g(\zeta )=\mathrm{cn}(\rho \zeta )\,\,\,\,\,\mathrm{or}\,\,\,\,\,g(\zeta )=\displaystyle \frac{\sqrt{1-{m}^{2}}\,\mathrm{sn}(\rho \zeta )}{\mathrm{dn}(\rho \zeta ))},\end{eqnarray}$ where $\rho =\tfrac{-{c}_{4}}{2m}\sqrt{\tfrac{-1}{{c}_{6}}}.$ |
4. | 4.If ${c}_{0}=\tfrac{{c}_{4}^{3}{m}^{2}}{32{c}_{6}^{2}({m}^{2}-1)},\,\,\,{c}_{2}=\tfrac{{c}_{4}^{3}\left(5{m}^{2}-4\right)}{16{c}_{6}({m}^{2}-1)},\,\,\,{c}_{6}\lt 0,$ then $\begin{eqnarray}g(\zeta )=\displaystyle \frac{\mathrm{dn}\left(\rho \zeta \right)}{\sqrt{1-{m}^{2}}}\,\,\,\,\,\mathrm{or}\,\,\,\,\,g(\zeta )=\displaystyle \frac{1}{\mathrm{dn}(\rho \zeta ))}\,,\end{eqnarray}$ where $\rho =\tfrac{{c}_{4}}{2}\sqrt{\tfrac{1}{{c}_{6}\,({m}^{2}-1)}}\,.$ |
5. | 5.If ${c}_{0}=\tfrac{{c}_{4}^{3}}{32{c}_{6}^{2}(1-{m}^{2})}\,,\,\,\,{c}_{2}=\tfrac{{c}_{4}^{2}\left(4{m}^{2}-5\right)}{16{c}_{6}({m}^{2}-1)},\,\,\,\,{c}_{6}\gt 0,$ then $\begin{eqnarray}g(\zeta )=\displaystyle \frac{1}{\mathrm{cn}(\zeta \rho )}\,\,\,\,\,\mathrm{or}\,\,\,\,\,g(\zeta )=\displaystyle \frac{\mathrm{dn}(\zeta \rho )}{\sqrt{1-{m}^{2}}\,\mathrm{sn}\left(\rho \zeta \right)}\,,\end{eqnarray}$ where $\rho =\tfrac{{c}_{4}}{2}\sqrt{\tfrac{1}{{c}_{6}\,(1-{m}^{2})}}$. |
6. | 6.If ${c}_{0}=\tfrac{{m}^{2}{c}_{4}^{3}}{32{c}_{6}^{2}},\,\,\,{c}_{2}=\tfrac{{c}_{4}^{2}\left({m}^{2}+4\right)}{16{c}_{6}},\,\,\,\,{c}_{6}\lt 0,$ then $\begin{eqnarray}g(\zeta )=\mathrm{dn}(\rho \zeta )\,\,\,\,\,\mathrm{or}\,\,\,\,\,g(\zeta )=\displaystyle \frac{\sqrt{1-{m}^{2}}}{\mathrm{dn}\left(\rho \zeta \right)}\,,\end{eqnarray}$ where $\rho =\tfrac{-{c}_{4}}{2}\sqrt{\tfrac{-1}{{c}_{6}\,}}$. |
3. Mathematical analysis of the model
4. Chirped soliton solution
Figure 1. (a) 3D profile of solution ( |
Figure 2. (a) 3D profile of solution ( |
Figure 3. (a) 3D profile of the solution ( |
Figure 4. (a) 3D profile of the solution ( |
Figure 5. (a) 3D profile of the solution ( |
5. Linear stability analysis
Figure 6. G(k) for equation ( |