A novel analytical technique to obtain the solitary solutions for nonlinear evolution equation of fractional order

Ghaffar, Abdul; Ali, Ayyaz; Ahmed, Sarfaraz; Akram, Saima; Junjua, Moin-ud-Din; Baleanu, Dumitru; Nisar, Kottakkaran Sooppy

doi:10.1186/s13662-020-02751-5

Advances in Continuous and Discrete Models

Theory and Modern Applications

Table 1 Comparing the results of Akber et al. [26] and Liu et al. [22] with our results

From: A novel analytical technique to obtain the solitary solutions for nonlinear evolution equation of fractional order

Obtained results	Results of Liu et al. and Akber et al.
If \(m_{1}=\frac{(V-2k)a_{1}}{Vb_{1}}\sqrt{-\frac{2V-4k}{V}}=1\), then the solution is \(u_{19}=m_{1}\coth ( \frac{x-\frac{Vt}{\varGamma ( \alpha +1 ) } }{2} ) \)	If \(m_{2}=0\), d = 0, \(\sqrt{\varOmega }=1\), \(\varPhi =x-\frac{Vt}{\varGamma ( \alpha +1 ) }\), A = 1, then the solution is \(u_{1}=m_{1}\coth ( \frac{x-\frac{Vt}{\varGamma ( \alpha +1 ) }}{2} ) \)
If \(m_{1}=\frac{(V-2k)a_{1}}{Vb_{1}}\sqrt{-\frac{2V-4k}{V}}=1\), then the solution is \(u_{19}=m_{2}\coth ( \frac{x-\frac{Vt}{\varGamma ( \alpha +1 ) } }{2} ) \)	If \(m_{2}=0\), d = 0, \(\sqrt{\varOmega }=1\), \(\varPhi =x-\frac{Vt}{\varGamma ( \alpha +1 ) }\), A = 2, then the solution is \(u_{6}=m_{2}\coth ( \frac{x-\frac{Vt}{\varGamma ( \alpha +1 ) }}{2} ) \)
If \(\sqrt{-\frac{2V-4K}{V}}=\frac{1}{\sqrt{2}}\) and \(m_{1}=\frac{ a_{0}(V\lambda ^{2}+2V+4K)}{2Vb_{1}}\), then the solution is \(u_{10}=m_{1}\cot ( \frac{\frac{Vt}{\varGamma ( \alpha +1 ) }-x}{ 2} ) \)	If \(m_{2}=0\), d = 0, B = 0, \(\iota \sqrt{ \varOmega }=1\), \(\varPhi =\frac{Vt}{\varGamma ( \alpha +1 ) } -x\), A = 1, then the solution is \(u_{3}=m_{1}\cot ( \frac{\frac{Vt}{\varGamma ( \alpha +1 ) }-x}{2} ) \)
If \(\sqrt{-\frac{2V-4K}{V}}=\frac{1}{\sqrt{2}}\) and \(m_{2}=\frac{ a_{0}(V\lambda ^{2}+2V+4K)}{2Vb_{1}}\), then the solution is \(u_{10}=m_{2}\cot ( \frac{\frac{Vt}{\varGamma ( \alpha +1 ) }-x}{ 2} ) \)	If \(m_{1}=\frac{1}{2}\), d = 0, \(\iota \sqrt{\varOmega }=1\), \(\varPhi =\frac{Vt}{\varGamma ( \alpha +1 ) } -x\), A = 2, then the solution is \(u_{8}=m_{1}\cot ( \frac{\frac{Vt}{\varGamma ( \alpha +1 ) }-x}{2} ) \)
If \(\sqrt{-\frac{2V-4K}{V}}=1\) and \(m_{2}=\frac{a_{1}(2K+V)}{Vb_{1}}\) then the solution is \(u_{19}=m_{2}\coth ( \frac{\frac{Vt}{\varGamma ( \alpha +1 ) }-x }{2} ) \)	If \(m_{1}=\frac{1}{2} \), d = 0, \(\sqrt{\Delta }=1\), \(\varPhi =\frac{Vt}{\varGamma ( \alpha +1 ) } -x\), A = 2, then the solution is \(u_{7}=m_{2}\coth ( \frac{\frac{Vt}{\varGamma ( \alpha +1 ) }-x}{2} ) \)
If \(\sqrt{-\frac{2V-4K}{V}}=1\), \(a_{1}=1\) and \(m_{2}=\frac{2K+V}{Vb_{1}}=\frac{2}{5}\) then the solution is \(u_{20}=\frac{2}{5}\coth ( \frac{\frac{Vt}{\varGamma ( \alpha +1 ) }-x}{2} ) \)	If \(m_{1}=\frac{1}{5}\), \(m_{2}=0\), d = 0, \(\sqrt{\Delta }=1\), \(\varPhi =\frac{Vt}{\varGamma ( \alpha +1 ) }-x\), A = 2, then the solution is \(u_{6}=\frac{2}{5}\coth ( \frac{\frac{Vt}{ \varGamma ( \alpha +1 ) }-x}{2} ) \)
If \(\sqrt{-\frac{2V-4K}{V}}=\frac{1}{\sqrt{2}}\) and \(m_{1}=\frac{a_{1}(-2K+V)}{2Vb_{1}}=1\) then we get \(u_{20}=\cot ( \frac{\frac{Vt}{\varGamma ( \alpha +1 ) }-x}{2} ) \)	If \(m_{1}=\frac{1}{2}\), \(m_{2}=0\), d = 0, \(\sqrt{\Delta }=1\), \(\varPhi =\frac{Vt}{ \varGamma ( \alpha +1 ) }-x\), A = 2, then the solution is \(u_{8}=\cot ( \frac{\frac{Vt}{\varGamma ( \alpha +1 ) }-x}{2} ) \)

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