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Theory and Modern Applications

Table 2 Comparing the results of Mohyud-Din et al. [17] and Najafi et al. [30] with our results

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

Obtained results

Results of Najafi et al. and Mohyud-Din et al.

If λ = 1, μ = −1, k = 1, \(a_{1}=2\sqrt{3}\), \(b_{0}=2\), \(b_{1}=1\), t = 0, x = 0 then the solution is \(u_{1} ( \eta ) =-2\sqrt{3}\)

If k = 1, \(b_{1}=\frac{1}{2}\), β = −1, η = 0, \(c_{1}=1\) then the solution is \(u_{4} ( \eta ) =-2\sqrt{3}\)

If λ = 1, μ = −1, k = 1, \(a_{1}=2\sqrt{3}\), \(b_{0}=2\), \(b_{1}=1\), t = 0, x = 0 then the solution is \(u_{1} ( \eta ) =-2\sqrt{3}\)

If k = 1, \(b_{1}=\frac{1}{2}\), β = −1, η = 0, \(c_{1}=1\) then the solution is \(u_{10} ( \eta ) =-2\sqrt{3}\)

If \(\lambda =\frac{2}{\sqrt{2}}\), μ = −1, k = 1, \(a_{1}=3\sqrt{6}\), \(b_{0}=\sqrt{2}\), \(b_{1}=1\), \(V=6\sqrt{2}\), t = 1, x = 0 then the solution is \(u_{3} ( \eta ) =\frac{3\sqrt{6}}{5\sqrt{2}}\)

If λ = 1, k = 1, \(b_{0}=1\), β = −1, η = 0, \(c_{1}=1\), μ = 1 then the solution is \(u_{4} ( \eta ) =\frac{3\sqrt{6}}{5\sqrt{2}}\)

If \(\lambda =\frac{2}{\sqrt{2}}\), μ = −1, k = 1, \(a_{1}=3\sqrt{6}\), \(b_{0}=\sqrt{2}\), \(b_{1}=1\), \(V=6\sqrt{2}\), t = 1, x = 0 then the solution is \(u_{3} ( \eta ) =\frac{3\sqrt{6}}{5\sqrt{2}}\)

If λ = 1, k = 1, \(b_{0}=1\), β = −1, η = 0, \(c_{1}=1\), μ = 1 then the solution is \(u_{9} ( \eta ) =\frac{3\sqrt{6}}{5\sqrt{2}} \)

If V = 6, \(a_{1}=1\), t = 0, x = 0, \(b_{1}=\frac{-1}{6}\), \(k=\frac{5}{2}\) then the solution is \(u_{7} ( \eta ) =\sqrt{6}\)

If \(\lambda =\sqrt{-1}\), k = 1, \(b_{0}=0\), β = 1, k = 1 then the solution is \(u_{2} ( \eta ) =\sqrt{6}\)

If V = 6, \(a_{1}=1\), t = 0, x = 0, \(b_{1}=\frac{-1}{6}\), \(k=\frac{5}{2}\) then the solution is \(u_{7} ( \eta ) =\sqrt{6}\)

If \(\lambda =\sqrt{-1}\), k = 1, \(b_{0}=0\), β = 1, k = 1 then the solution is \(u_{8} ( \eta ) =\sqrt{6}\)

If k = −6, \(b_{1}=\frac{12}{\sqrt{6}}\), β = −1, k = −6, \(c_{1}=\frac{\sqrt{6}}{12}\) then the solutions are \(u_{18} ( \eta ) =u_{23} ( \eta ) =-2\sqrt{3}\). It matches with our obtained solution (i) in this table