Multi-solitary wave solutions to the general time fractional Sharma–Tasso–Olver equation and the time fractional Cahn–Allen equation

Figures & data

Figure 1. Sketch of kink wave of $u_{1_2}(x,t)$ if $\lambda =-1,$ $d=1,$ $\sigma =1,$ $\alpha =\frac{1}{2},$$-10\le x\le 10,$ and 0$\le t\le 10.$

Figure 2. Sketch of multiple periodic wave of $u_{1_6}(x,t)$ if $\lambda =1,$ $d=1,$ $\sigma =0,$ $\alpha =\frac{1}{2},-10\le x\le 10,$ and 0$\le t\le 10.$

Figure 3. Shape of periodic wave of $u_{1_5}(x,t)$ if $\lambda =1,$ $d=-1,$ $\sigma =1,$ $\alpha =\frac{1}{2},$$-10\le x\le 10,$ and 0$\le t\le 10.$

Figure 4. Shape of singular kink type wave of $u_{1_7}(x,t)$ if $\mu =1,$ $A_1=2,$ $A_2=1,$ $\alpha =\frac{1}{2},$$-10\le x\le 10,$ and 0$\le t\le 10.$

Table 1. Comparison between Roy et al. (2018) solutions and our solutions to the STO equation

Roy et al. (2018)	Obtained solutions
If $c_0=0$ and $c_{00}\ne 0$ the solution becomes: $u_{1_1}\left(x,t\right)=\frac{\sqrt{\sigma }}{2p}\mathrm{ }\text{cot}\mathrm{h}\left\{\frac{\sqrt{\sigma }}{2p}\left(x+\frac{\omega t^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)}\right)\right\}.$	If $\sigma =0$ in Equation (4.8)(4.8) $$u_{1_3}\left(x,t\right)=\frac{\sqrt{-\lambda }}{2}\times \coth \left(\sqrt{-\lambda }(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)})\right)\pm \frac{\sqrt{-\lambda \sigma }}{2}\times \mathit{cosech}\left(\sqrt{-\lambda }\right(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)}\left)\right).$$(4.8) then our solution $u_{1_3}(x,t)$ becomes: $u_{1_3}\left(x,t\right)=\frac{\sqrt{-\lambda }}{2}\times \coth \left(\sqrt{-\lambda }(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)})\right).$
If $c_0\ne 0$ and $c_{00}=0$ the solution becomes: $u_{2_1}\left(x,t\right)=\frac{\sqrt{\sigma }}{2p}\mathrm{ }\hspace{0.17em}\text{tan}\hspace{0.17em}\mathrm{h}\left\{\frac{\sqrt{\sigma }}{2p}\left(x+\frac{\omega t^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)}\right)\right\}.$	If $\sigma =0$ in Equation (4.7)(4.7) $$u_{1_2}\left(x,t\right)=\frac{\sqrt{-\lambda }}{2}\times \tanh \left(\sqrt{-\lambda }(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)})\right)\pm \frac{\sqrt{-\lambda \sigma }}{2}\times \mathit{sech}\left(\sqrt{-\lambda }\right(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)}\left)\right).$$(4.7) then our solution $u_{1_3}\left(x,t\right)$ becomes: $u_{1_2}\left(x,t\right)=\frac{\sqrt{-\lambda }}{2}\times \tanh \left(\sqrt{-\lambda }(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)})\right).$
If $c_0=0$ and $c_{00}\ne 0$ then the solution becomes: $u_{3_1}\left(x,t\right)=\mathrm{i}\frac{\sqrt{\sigma }}{2p}\mathrm{ }\text{cot}\left\{\frac{\sqrt{-\sigma }}{2p}\left(x+\frac{\omega t^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)}\right)\right\}.$	If $\sigma =0$ in Equation (4.11)(4.11) $$u_{1_6}\left(x,t\right)=\frac{\sqrt{\lambda }}{2}\times \text{cot}\left(\sqrt{\lambda }(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)})\right)\pm \frac{\sqrt{\lambda \sigma }}{2}\times \text{cosec}\left(\sqrt{\lambda }\right(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)}\left)\right).$$(4.11) then our solution $u_{1_6}(x,t)$ becomes: $u_{1_6}\left(x,t\right)=\frac{\sqrt{\lambda }}{2}\times \text{cot}\left(\sqrt{\lambda }(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)})\right).$
If $c_0\ne 0$ and $c_{00}=0$ then the solution becomes: $u_{3_1}\left(x,t\right)=-\mathrm{i}\frac{\sqrt{\sigma }}{2p}\mathrm{ }\text{cot}\left\{\frac{\sqrt{-\sigma }}{2p}\left(x+\frac{\omega t^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)}\right)\right\}.$	If $\sigma =0$ in Equation (4.10)(4.10) $$u_{1_5}\left(x,t\right)=\frac{\sqrt{\lambda }}{2}\times \text{tan}\left(\sqrt{\lambda }(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)})\right)\pm \frac{\sqrt{\lambda \sigma }}{2}\times \text{sec}\left(\sqrt{\lambda }\right(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)}\left)\right).$$(4.10) then our solution $u_{1_5}(x,t)$ becomes: $u_{1_5}\left(x,t\right)=\frac{\sqrt{\lambda }}{2}\times \text{tan}\left(\sqrt{\lambda }(x+\frac{\frac{d\lambda }{4}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)})\right).$

Table 2. Comparison between Batool and Ghazala (2018) solutions and our solutionsto the CA equation

Batool and Ghazala (2018)	Obtained solutions
If $A_1\ne 0$ and $A_2=0$ then the solution in Eq. (18) becomes: $u_8\left(\xi \right)$ = $\frac{1}{2}+\frac{1}{2}\tanh (\frac{x}{2\sqrt{2}}+\frac{3t^{\beta }}{4\mathrm{\Gamma }\left(1+\alpha \right)}).$	If $\sigma =0$ in Equation (4.18)(4.18) $$u_{2_2}\left(x,t\right)=\frac{1}{2}\pm \frac{1}{2}\times \tanh \left(\sqrt{-\lambda }\xi \right)\pm \mathrm{ }\times \frac{\sqrt{\lambda \sigma }}{2}s\mathit{ech}(\sqrt{-\lambda }\xi ),$$(4.18) then our solution $u_{2_2}(x,t)$ becomes: $u_{2_2}(x,t)$ = $\frac{1}{2}+\frac{1}{2}\times \tanh \left(\sqrt{-\lambda }\mathrm{ }(\sqrt{-\frac{1}{2\lambda }}x-\frac{3\mathrm{ }\sqrt{-\frac{1}{4\lambda }}{t}^{\alpha }}{\mathrm{\Gamma }(1+\alpha )})\right).$
If $A_1\ne 0$ and $A_2=0$ then the solution in Eq. (18) becomes: $u_{12}(\xi )$ = $\frac{1}{2}-\frac{1}{2}\tanh (\frac{x}{2\sqrt{2}}-\frac{3t^{\beta }}{4\mathrm{\Gamma }\left(1+\alpha \right)}).$	If $\sigma =0$ in Equation (4.18)(4.18) $$u_{2_2}\left(x,t\right)=\frac{1}{2}\pm \frac{1}{2}\times \tanh \left(\sqrt{-\lambda }\xi \right)\pm \mathrm{ }\times \frac{\sqrt{\lambda \sigma }}{2}s\mathit{ech}(\sqrt{-\lambda }\xi ),$$(4.18) then our solution $u_{2_2}(x,t)$ becomes: $u_{2_2}(x,t)$ = $\frac{1}{2}-\frac{1}{2}\times \tanh \left(\sqrt{-\lambda }\left(\sqrt{-\frac{1}{2\lambda }}x-\frac{3\mathrm{ }\sqrt{-\frac{1}{4\lambda }}{t}^{\alpha }}{\mathrm{\Gamma }\left(1+\alpha \right)}\right)\right).$

Roy, R., Akbar, M. A., & Wazwaz, A. M. (2018). Exact wave solutions for the non-linear time fractional Sharma-Tasso-Olver equation and the fractional Klein-Gordon equation by mathematical physics. Optical and Quantum Electronics, 50(1), 219–251.

 Web of Science ®Google Scholar

Batool, F., & Ghazala, A. (2018). New solitary wave solutions of the time fractional Cahn-Allen equation via the improved (G′/G)-expansion method. The European Physical Journal Plus, 133(171), 1–11.

 Google Scholar