. 2020 Apr 15;6(4):e03727. doi: 10.1016/j.heliyon.2020.e03727

Table 3.

Comparison of the solutions to the space-time fractional (2 + 1)-dimensional breaking soliton equation.

Guner [61] solutions	The obtained solutions
For $A = \frac{3 k^{2}}{2}$ and $c = 4 a w k^{2}$ , then the non-topological solution of Eq. (42) takes into the form $u (x, y, t) = \frac{3 k^{2}}{2} s e c h^{2} (\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{4 a k^{2} w t^{α}}{Γ (1 + α)})$ , and $v (x, y, t) = \frac{3 k w}{2} s e c h^{2} (\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{4 a k^{2} w t^{α}}{Γ (1 + α)})$ ,	Especially, if $A_{2} = 0$ , $μ = 0$ and $A_{1} \neq 0$ , then the soliton solutions (4.3.14) and (4.3.15) become $u (x, y, t) = - \frac{k^{2} λ}{2} (\mp \frac{3}{2} c o t h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}) c s c h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}) - \frac{3}{2} {(c o t h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}))}^{2} + 1)$ , and $v (x, y, t) = - \frac{k w λ}{2} (\mp \frac{3}{2} c o t h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}) c s c h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}) - \frac{3}{2} {(c o t h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}))}^{2} + 1) .$

Guner [61] solutions

The obtained solutions

For

A = \frac{3 k^{2}}{2}

and

c = 4 a w k^{2}

, then the non-topological solution of Eq. (42) takes into the form

u (x, y, t) = \frac{3 k^{2}}{2} s e c h^{2} (\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{4 a k^{2} w t^{α}}{Γ (1 + α)})

, and

v (x, y, t) = \frac{3 k w}{2} s e c h^{2} (\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{4 a k^{2} w t^{α}}{Γ (1 + α)})

Especially, if

A_{2} = 0

μ = 0

and

A_{1} \neq 0

, then the soliton solutions (4.3.14) and (4.3.15) become

u (x, y, t) = - \frac{k^{2} λ}{2} (\mp \frac{3}{2} c o t h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}) c s c h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}) - \frac{3}{2} {(c o t h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}))}^{2} + 1)

, and

v (x, y, t) = - \frac{k w λ}{2} (\mp \frac{3}{2} c o t h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}) c s c h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}) - \frac{3}{2} {(c o t h ((\frac{k x^{α}}{Γ (1 + α)} + \frac{w y^{α}}{Γ (1 + α)} - \frac{a k^{2} w λ t^{α}}{Γ (1 + α)}) \sqrt{- λ}))}^{2} + 1) .

From the above comparison, it is observed that all of the obtained solutions are completely fresh and general than the solutions existing in the literature.