Table 2 A comparison between the solutions obtained by Uddin et al.34 and the solutions obtained by our study for the space–time fractional modified regularized long-wave equation using IBSEF method.

From: A study of the wave dynamics of the space–time fractional nonlinear evolution equations of beta derivative using the improved Bernoulli sub-equation function approach

Obtained solutions

Uddin et al.34

If \(\alpha \ne \beta , p_{2} = 1, q_{0} = 2, \tau = 1,\) then (4.2.8) solution becomes:

\(V_{{2_{1} }} \left( {x,t} \right) = \frac{{\frac{\alpha }{2\beta } + \frac{1}{{\left( { - \frac{\beta }{\alpha } + \frac{1}{{e^{{2\alpha \left( {\frac{m}{\gamma }\left( {x + \frac{1}{{{{\Gamma \gamma }}}}} \right)^{\gamma } + \frac{k}{\gamma }\left( {t + \frac{1}{{{{\Gamma \gamma }}}}} \right)^{\gamma } } \right)}} }}} \right)}}}}{2}.\)

If \(\alpha \ne \beta , p_{0} = 1, q_{0} = 2, q_{1} = 1,\tau = 1,\) then (4.2.20) solution becomes:

\(V_{{2_{9} }} \left( {x,t} \right) =\) \(\frac{{1 + \frac{1}{{2\left( {\sqrt { - \frac{\beta }{\alpha } + \frac{1}{{e^{{2\alpha \left( {\frac{m}{\gamma }\left( {x + \frac{1}{{{{\Gamma \gamma }}}}} \right)^{\gamma } + \frac{k}{\gamma }\left( {t + \frac{1}{{{{\Gamma \gamma }}}}} \right)^{\gamma } } \right)}} }}} } \right)}}}}{{1 + \frac{2}{{\sqrt { - \frac{\beta }{\alpha } + \frac{1}{{e^{{2\alpha \left( {\frac{m}{\gamma }\left( {x + \frac{1}{{{{\Gamma \gamma }}}}} \right)^{\gamma } + \frac{k}{\gamma }\left( {t + \frac{1}{{{{\Gamma \gamma }}}}} \right)^{\gamma } } \right)}} }}} }}}}.\)

If we put \(q_{0} = q_{1} = 1,\) in solution (4.15) then solution \(u_{{2_{2} }} \left( {x, t} \right)\) becomes:

\(u_{{2_{2} }} \left( {x, t} \right) =\) \(\frac{{\sqrt {\frac{3\upsilon \tau }{{\left( {\tau + 2} \right)}}} \exp \left[ {\left( {x - \frac{2\upsilon }{{\tau + 2}}\frac{{t^{\alpha } }}{\alpha }} \right)} \right] - \sqrt {\frac{3\upsilon \tau }{{\left( {\tau + 2} \right)}}} }}{{\exp \left[ {\left( {x - \frac{2\upsilon }{{\tau + 2}}\frac{{t^{\alpha } }}{\alpha }} \right)} \right] + 1}}.\)

If we set \(q_{0} = q_{1} = q_{ - 1} = 1,\) in solution (4.16) then solution \(u_{{2_{3} }} \left( {x, t} \right)\) becomes:

\(u_{{2_{3} }} \left( {x, t} \right) = \frac{{1 + {\text{exp}}\left[ { - \left( {x - \frac{1}{3}\left( {3\upsilon - 1} \right)\frac{{t^{\alpha } }}{\alpha })} \right)} \right]}}{{1 + {\text{exp}}\left[ { - \left( {x - \frac{1}{3}\left( {3\upsilon - 1} \right)\frac{{t^{\alpha } }}{\alpha })} \right)} \right]}}.\)

If we insert \(q_{0} = q_{1} = q_{ - 1} = 1,\) in solution (4.17) then solution \(u_{{2_{4} }} \left( {x, t} \right)\) becomes:

\(u_{{2_{4} }} \left( {x, t} \right) =\) \(\frac{{\sqrt {\frac{3\upsilon \tau }{{\left( {\tau + 2} \right)}}} - \sqrt {\frac{3\upsilon \tau }{{\left( {\tau + 2} \right)}}} \exp \left[ { - \left( {x - \frac{2\upsilon }{{\tau + 2}}\frac{{t^{\alpha } }}{\alpha }} \right)} \right]}}{{1 + \exp \left[ { - \left( {x - \frac{2\upsilon }{{\tau + 2}}\frac{{t^{\alpha } }}{\alpha }} \right)} \right]}}.\)

Back to article page