1. Introduction
As we know, fractional calculus (FC) is a generalized version of classical calculus that relates to the derivative and integral of arbitrary order. FC is as old as classical calculus and has received much attention in the last few decades. Many studies with the theoretical discussion have been done within the framework of FC. So far, different fractional derivatives with singular and non-singular kernels have been proposed, some of these fractional derivatives are Riemann–Liouville derivative [1], Caputo derivative [2], Atangana–Baleanu derivative [3], and Caputo–Fabrizio (CF) derivative [4]. Veeresha and Prakasha [5] studied the time-fractional Zakharov–Kuznetsov equations in the Caputo sense and derived its approximate solutions with the help of the $q$-homotopy analysis transform method (HAMT). Yavuz et al [6] considered the time-fractional Schrödinger–KdV equation in the Atangana–Baleanu context and found its approximate solution through the modified Laplace decomposition method. Aydogan et al [7] explored the Rabies mathematical model in the CF sense and derived its approximate solutions through the Adomian decomposition method. For more studies, see [8–20].
The Sharma–Tasso–Olver–Burgers equation is a new nonlinear model that has been introduced by Yan and Lou in [21] and is expressed as
$\begin{eqnarray*}\begin{array}{l}{u}_{t}\left(x,t\right)+{c}_{1}\left(3{u}_{x}^{2}\left(x,t\right)+3{u}^{2}\left(x,t\right){u}_{x}\left(x,t\right)\right.\\ \left.\,+\,3u\left(x,t\right){u}_{xx}\left(x,t\right)+{u}_{xxx}\left(x,t\right)\right)\\ \,+\,{c}_{2}\left(2u\left(x,t\right){u}_{x}\left(x,t\right)+{u}_{xx}\left(x,t\right)\right)=0,\end{array}\end{eqnarray*}$
where ${c}_{1}$ and ${c}_{2}$ are free constants. It should be noted that the STOB equation is a combination of the STO and Burgers equations. Yan and Lou [21] obtained soliton molecules of the STOB equation using the velocity resonant mechanism. Miao et al [22] derived lump and interaction solutions of the STOB equation by adopting different test functions. Very recently, Hosseini et al [23] found kink solitons of the STOB equation through the exponential method. The main aim of the present study is to consider the time-fractional STOB equation in the CF context and obtain its valid approximations through adopting a mixed approach [24–30] composed of the homotopy analysis method and the Laplace transform. The time-fractional STOB equation in the CF context can be written as follows $\begin{eqnarray}\begin{array}{l}{}_{0}{}^{CF}D_{t}^{\alpha }u\left(x,t\right)+{c}_{1}\left(3{u}_{x}^{2}\left(x,t\right)+3{u}^{2}\left(x,t\right){u}_{x}\left(x,t\right)\right.\\ \left.\,+\,3u\left(x,t\right){u}_{xx}\left(x,t\right)+{u}_{xxx}\left(x,t\right)\right)\\ \,+\,{c}_{2}\left(2u\left(x,t\right){u}_{x}\left(x,t\right)+{u}_{xx}\left(x,t\right)\right)=0,0\lt \alpha \leqslant 1.\end{array}\end{eqnarray}$
The mixed approach namely the HAMT is a powerful analytical method that has been widely used to find valid approximations for nonlinear fractional equations. Aljhani et al [27] applied the HAMT to obtain a valid approximation of a fractional HIV model in the Caputo context. Baleanu et al [28] in their paper dealt with finding a valid approximation to a fractional model of COVID-19 in the CF sense using the HAMT. The authors of [29] derived a valid approximation to a fractional model of anthrax in the CF context through the HAMT. Very newly, Hosseini et al [30] utilized the HAMT to acquire valid approximations of a time-fractional nonlinear water wave equation in the Caputo sense.With a close look at the research done in the above references, it can be concluded that the study of nonlinear fractional equations in the CF context is of particular importance in many branches of science. Therefore, here a series of required introductory definitions in the CF context that will be used in the next sections are provided.
Let $u\left(t\right)\in {H}^{1}\left(a,b\right),\,b\gt a$ and $\alpha \in \left(0,1\right).$ The fractional derivative of $u\left(t\right)$ of order $\alpha $ in the CF context is expressed as [4]
$\begin{eqnarray*}{}_{a}{}^{CF}D_{t}^{\alpha }u\left(t\right)=\displaystyle \frac{M\left(\alpha \right)}{1-\alpha }\displaystyle {\int }_{a}^{t}u^{\prime} \left(\varepsilon \right){{\rm{e}}}^{-\displaystyle \frac{\alpha }{1-\alpha }\left(t-\varepsilon \right)}{\rm{d}}\varepsilon ,\end{eqnarray*}$
with the properties $M\left(0\right)=M\left(1\right)=1.$The fractional integral of $u\left(t\right)$ of order $\alpha $ in the CF sense is defined as [31]
$\begin{eqnarray*}\begin{array}{l}{}_{0}{}^{CF}I_{t}^{\alpha }u\left(t\right)=\displaystyle \frac{2\left(1-\alpha \right)}{\left(2-\alpha \right)M(\alpha )}u\left(t\right)+\displaystyle \frac{2\alpha }{\left(2-\alpha \right)M(\alpha )}\\ \,\,\,\,\,\times \,\displaystyle {\int }_{0}^{t}u\left(\varepsilon \right){\rm{d}}\varepsilon ,t\geqslant 0.\end{array}\end{eqnarray*}$
For a given expression like ${}_{0}{}^{CF}D_{t}^{\alpha }u\left(t\right),$ its Laplace transform can be written as [4]
$\begin{eqnarray*} {\mathcal L} \left[{}_{0}{}^{CF}D_{t}^{\alpha }u\left(t\right)\right]\left(s\right)=\displaystyle \frac{s {\mathcal L} \left[u\left(t\right)\right]\left(s\right)-u(0)}{s+\alpha \left(1-s\right)},\end{eqnarray*}$
and generally, $\begin{eqnarray*}\begin{array}{l} {\mathcal L} \left[{{}_{0}{}^{CF}D}_{t}^{\left(\alpha +n\right)}u\left(t\right)\right]\left(s\right)\\ \,=\displaystyle \frac{{s}^{n+1} {\mathcal L} \left[u\left(t\right)\right]\left(s\right)-{s}^{n}u\left(0\right)-{s}^{n-1}u^{\prime} \left(0\right)-\cdots -{u}^{\left(n\right)}\left(0\right)}{s+\alpha \left(1-s\right)}.\end{array}\end{eqnarray*}$
The outline of this paper is as follows: In section 2 , the existence and uniqueness of the solution of the time-fractional STOB equation in the CF context are investigated by demonstrating the Lipschitz condition for $\varphi (x,\,t;u)$ as the kernel and giving some theorems. In section 3 , valid approximations of the time-fractional STOB equation in the CF context are derived through adopting the HAMT. In section 4 , the effectiveness of the HATM in handling the time-fractional STOB equation in the CF context is shown. Additionally, the CF operator effect on the dynamics of the obtained solitons is investigated in detail. The outcomes of the present investigation are given in the last section.
2. The theoretical discussion on the existence and uniqueness of the solution
Supposing1 ) can be rewritten as
$\begin{eqnarray*}\begin{array}{l}\varphi \left(x,t;u\right)=-\left({c}_{1}\left(3{u}_{x}^{2}\left(x,t\right)+3{u}^{2}\left(x,t\right){u}_{x}\left(x,t\right)\right.\right.\\ \left.\,+\,3u\left(x,t\right){u}_{xx}\left(x,t\right)+{u}_{xxx}\left(x,t\right)\right)\\ \left.\,+\,{c}_{2}\left(2u\left(x,t\right){u}_{x}\left(x,t\right)+{u}_{xx}\left(x,t\right)\right)\right),\end{array}\end{eqnarray*}$
recommends that equation ( $\begin{eqnarray*}{}_{0}{}^{CF}D_{t}^{\alpha }u\left(x,t\right)=\varphi \left(x,t;u\right).\end{eqnarray*}$
Assume that $u(x,\,t)$ and $v(x,\,t)$ are the bounded functions, namely $\parallel u\parallel \leqslant \mu $ and $\parallel v\parallel \leqslant \upsilon .$ Then, $\varphi \left(x,t;u\right)$ is a kernel satisfying the Lipschitz condition.
Proof.
$\begin{eqnarray*}\parallel \varphi \left(x,t;u\right)-\varphi \left(x,t;v\right)\parallel \end{eqnarray*}$
$\begin{eqnarray*}\begin{array}{l}=\,\parallel 3{c}_{1}\left({u}_{x}^{2}-{v}_{x}^{2}\right)+3{c}_{1}\left({u}^{2}{u}_{x}-{v}^{2}{v}_{x}\right)\\ +3{c}_{1}\left(u{u}_{xx}-v{v}_{xx}\right)+{c}_{1}\left({u}_{xxx}-{v}_{xxx}\right)\\ +{c}_{2}\left(2u{u}_{x}-2v{v}_{x}\right)+{c}_{2}\left({u}_{xx}-{v}_{xx}\right)\parallel \end{array}\end{eqnarray*}$
$\begin{eqnarray*}\begin{array}{l}=\parallel 3{c}_{1}\displaystyle \frac{\partial }{\partial x}\left(u-v\right)\displaystyle \frac{\partial }{\partial x}\left(u+v\right)+{c}_{1}\displaystyle \frac{\partial }{\partial x}\left({u}^{3}-{v}^{3}\right)\\ +3{c}_{1}\left(u\displaystyle \frac{{\partial }^{2}u}{\partial {x}^{2}}-u\displaystyle \frac{{\partial }^{2}v}{\partial {x}^{2}}+u\displaystyle \frac{{\partial }^{2}v}{\partial {x}^{2}}-v\displaystyle \frac{{\partial }^{2}v}{\partial {x}^{2}}\right)\\ +{c}_{1}\displaystyle \frac{{\partial }^{3}}{\partial {x}^{3}}\left(u-v\right)+{c}_{2}\displaystyle \frac{\partial }{\partial x}\left({u}^{2}-{v}^{2}\right)+{c}_{2}\displaystyle \frac{{\partial }^{2}}{\partial {x}^{2}}\left(u-v\right)\parallel \end{array}\end{eqnarray*}$
$\begin{eqnarray*}\begin{array}{l}=\,\parallel 3{c}_{1}\displaystyle \frac{\partial }{\partial x}\left(u-v\right)\displaystyle \frac{\partial }{\partial x}\left(u+v\right)+{c}_{1}\left({u}^{2}+uv+{v}^{2}\right)\\ \times \,\displaystyle \frac{\partial }{\partial x}\left(u-v\right)+{c}_{1}\left(u-v\right)\displaystyle \frac{\partial }{\partial x}\left({u}^{2}+uv+{v}^{2}\right)\\ +3{c}_{1}u\displaystyle \frac{{\partial }^{2}}{\partial {x}^{2}}\left(u-v\right)+3{c}_{1}\left(u-v\right)\displaystyle \frac{{\partial }^{2}v}{\partial {x}^{2}}\\ +{c}_{1}\displaystyle \frac{{\partial }^{3}}{\partial {x}^{3}}\left(u-v\right)+{c}_{2}\left(u+v\right)\displaystyle \frac{\partial }{\partial x}\left(u-v\right)\\ +{c}_{2}\left(u-v\right)\displaystyle \frac{\partial }{\partial x}\left(u+v\right)+{c}_{2}\displaystyle \frac{{\partial }^{2}}{\partial {x}^{2}}\left(u-v\right)\parallel \\ \leqslant \,3| {c}_{1}| \parallel \displaystyle \frac{\partial }{\partial x}\left(u-v\right)\parallel \parallel \displaystyle \frac{\partial }{\partial x}\left(u+v\right)\parallel +\left|{c}_{1}\right|\parallel {u}^{2}+uv+{v}^{2}\parallel \parallel \displaystyle \frac{\partial }{\partial x}\left(u-v\right)\parallel \\ +\left|{c}_{1}\right|\parallel u-v\parallel \parallel \displaystyle \frac{\partial }{\partial x}\left({u}^{2}+uv+{v}^{2}\right)\parallel +3\left|{c}_{1}\right|\parallel u\parallel \parallel \displaystyle \frac{{\partial }^{2}}{\partial {x}^{2}}\left(u-v\right)\parallel \\ +3\left|{c}_{1}\right|\parallel u-v\parallel \parallel \displaystyle \frac{{\partial }^{2}v}{\partial {x}^{2}}\parallel +\left|{c}_{1}\right|\parallel \displaystyle \frac{{\partial }^{3}}{\partial {x}^{3}}\left(u-v\right)\parallel +\left|{c}_{2}\right|\parallel u\\ +v\parallel \parallel \displaystyle \frac{\partial }{\partial x}\left(u-v\right)\parallel +\left|{c}_{2}\right|\parallel u-v\parallel \parallel \displaystyle \frac{\partial }{\partial x}\left(u+v\right)\parallel +\left|{c}_{2}\right|\parallel \displaystyle \frac{{\partial }^{2}}{\partial {x}^{2}}\left(u-v\right)\parallel \end{array}\end{eqnarray*}$
$\begin{eqnarray*}\begin{array}{l}\leqslant 3\left|{c}_{1}\right|A\parallel u-v\parallel \parallel u+v\parallel +\left|{c}_{1}\right|B\parallel {u}^{2}+uv+{v}^{2}\parallel \parallel u-v\parallel \\ +\left|{c}_{1}\right|C\parallel u-v\parallel \parallel {u}^{2}+uv+{v}^{2}\parallel +3\left|{c}_{1}\right|D\parallel u\parallel \parallel u-v\parallel +3\left|{c}_{1}\right|E\parallel u-v\parallel \parallel v\parallel \\ +\left|{c}_{1}\right|F\parallel u-v\parallel +\left|{c}_{2}\right|G\parallel u+v\parallel \parallel u-v\parallel +\left|{c}_{2}\right|H\parallel u-v\parallel \parallel u+v\parallel \\ +\left|{c}_{2}\right|K\parallel u-v\parallel \leqslant 3\left|{c}_{1}\right|A\left(\mu +\upsilon \right)\\ \parallel \parallel u-v\parallel +\left|{c}_{1}\right|B\left({\mu }^{2}+\mu \upsilon +{\upsilon }^{2}\right)\parallel u-v\parallel +\left|{c}_{1}\right|\\ \times C\left({\mu }^{2}+\mu \upsilon +{\upsilon }^{2}\right)\parallel u-v\parallel +3\left|{c}_{1}\right|D\mu \parallel u-v\parallel \\ +3\left|{c}_{1}\right|E\upsilon \parallel u-v\parallel +\left|{c}_{1}\right|F\parallel u-v\parallel +\left|{c}_{2}\right|G\left(\mu +\upsilon \right)\\ \times \parallel u-v\parallel +\left|{c}_{2}\right|H\left(\mu +\upsilon \right)\parallel u-v\parallel +\left|{c}_{2}\right|K\parallel u-v\parallel \end{array}\end{eqnarray*}$
$\begin{eqnarray*}\begin{array}{l}=\left(\left(3\left|{c}_{1}\right|A+\left|{c}_{2}\right|\left(G+H\right)\right)\left(\mu +\upsilon \right)+\left|{c}_{1}\right|\left(B+C\right)\right.\\ \times \left({\mu }^{2}+\mu \upsilon +{\upsilon }^{2}\right)+3\left|{c}_{1}\right|(D\mu +E\upsilon )+\left|{c}_{1}| F\right|\\ \left.+\left|{c}_{2}\right|K\right)\parallel u-v\parallel .\end{array}\end{eqnarray*}$
Now, assuming $\begin{eqnarray*}\begin{array}{l}\lambda =\left.3\left|{c}_{1}\right|A+\left|{c}_{2}\right|\left(G+H\right)\right)\left(\mu +\upsilon \right)+\left|{c}_{1}\right|\left(B+C\right)\\ \times \left({\mu }^{2}+\mu \upsilon +{\upsilon }^{2}\right)+3\left|{c}_{1}\right|\left(D\mu +E\upsilon \right)+\left|{c}_{1}\right|F+\left|{c}_{2}\right|K,\end{array}\end{eqnarray*}$
yields $\begin{eqnarray*}\parallel \varphi \left(x,t;u\right)-\varphi (x,t;v)\parallel \leqslant \lambda \parallel u-v\parallel ,\end{eqnarray*}$
which is the desired result.If at $t={t}_{0},$ we have
$\begin{eqnarray*}0\leqslant \displaystyle \frac{2\left(1-\alpha \right)}{\left(2-\alpha \right)M(\alpha )}\lambda +\displaystyle \frac{2\alpha }{\left(2-\alpha \right)M(\alpha )}\lambda {t}_{0}\lt 1,\end{eqnarray*}$
then, the solution of the time-fractional STOB equation in the CF context exists.See [25] and its references.
If at $t={t}_{0},$ we have.
$\begin{eqnarray*}0\leqslant \displaystyle \frac{2\left(1-\alpha \right)}{\left(2-\alpha \right)M(\alpha )}\lambda +\displaystyle \frac{2\alpha }{\left(2-\alpha \right)M(\alpha )}\lambda {t}_{0}\lt 1,\end{eqnarray*}$
then, the solution of the time-fractional STOB equation in the CF context is unique.See [25] and its references.
3. Analytical solutions to the time-fractional STOB equation in the CF context
The authors' purpose in the current section is to establish analytical solutions of the time-fractional STOB equation in the CF context. In this respect, first solitons of the STOB equation are derived with the help of ansatz method. The HATM is then employed to extract approximate solutions of the time-fractional STOB equation in the CF context. The computations are handled by Maple software.
3.1. The STOB equation and its solitons
To start, we consider a test function as [32–34]2 ) into the SOTB equation results in
$\begin{eqnarray}u\left(x,t\right)={A}_{0}+{A}_{1}\,\tanh \left(kx+wt\right),\end{eqnarray}$
where ${A}_{0}$ and ${A}_{1}$ are unknowns. Inserting equation ( $\begin{eqnarray*}\begin{array}{l}\left(-4{k}^{3}{c}_{1}+6{k}^{2}{A}_{1}{c}_{1}-3k{A}_{0}^{2}{c}_{1}-3k{A}_{1}^{2}{c}_{1}\right.\\ \left.-2k{A}_{0}{c}_{2}-w\right)\cos \,\,{{\rm{h}}}^{2}\left(kx+wt\right)+\left(6{k}^{2}{A}_{0}{c}_{1}\right.\\ \left.-6k{A}_{0}{A}_{1}{c}_{1}+2{k}^{2}{c}_{2}-2k{A}_{1}{c}_{2}\right)\cos \,\,{\rm{h}}\left(kx+wt\right)\\ \sin \,\,{\rm{h}}\left(kx+wt\right)+6{k}^{3}{c}_{1}-9{k}^{2}{A}_{1}{c}_{1}+3k{A}_{1}^{2}{c}_{1}=0.\end{array}\end{eqnarray*}$
From the above expression, the following system of nonlinear algebraic equations is obtained $\begin{eqnarray*}-4{k}^{3}{c}_{1}+6{k}^{2}{A}_{1}{c}_{1}-3k{A}_{0}^{2}{c}_{1}-3k{A}_{1}^{2}{c}_{1}-2k{A}_{0}{c}_{2}-w=0,\end{eqnarray*}$
$\begin{eqnarray*}6{k}^{2}{A}_{0}{c}_{1}-6k{A}_{0}{A}_{1}{c}_{1}+2{k}^{2}{c}_{2}-2k{A}_{1}{c}_{2}=0,\end{eqnarray*}$
$\begin{eqnarray*}6{k}^{3}{c}_{1}-9{k}^{2}{A}_{1}{c}_{1}+3k{A}_{1}^{2}{c}_{1}=0.\end{eqnarray*}$
The above system is solved to get the following cases:Case 1
$\begin{eqnarray*}{A}_{1}=k,\,w=-{k}^{3}{c}_{1}-3k{A}_{0}^{2}{c}_{1}-2k{A}_{0}{c}_{2}.\end{eqnarray*}$
Therefore, the following soliton to the SOTB equation is derived $\begin{eqnarray*}\begin{array}{l}{u}_{1}\left(x,t\right)\\ =\,{A}_{0}+k\,\tanh \left(kx-\left({k}^{3}{c}_{1}+3k{A}_{0}^{2}{c}_{1}+2k{A}_{0}{c}_{2}\right)t\right),\end{array}\end{eqnarray*}$
where for ${A}_{0}=1,$ ${c}_{1}=1,$ ${c}_{2}=1,$ and $k=1$ is changed into $\begin{eqnarray}{u}_{1}\left(x,t\right)=1+\,\tanh \left(x-6t\right),\end{eqnarray}$
with a series representation as $\begin{eqnarray*}{u}_{1}\left(x,t\right)=1-\displaystyle \frac{{{\rm{e}}}^{-2x}-1}{{{\rm{e}}}^{-2x}+1}-24\displaystyle \frac{{{\rm{e}}}^{-2x}}{{\left({{\rm{e}}}^{-2x}+1\right)}^{2}}t+\cdots .\end{eqnarray*}$
Case 2
$\begin{eqnarray*}{A}_{0}=-\displaystyle \frac{{c}_{2}}{3{c}_{1}},{A}_{1}=2k,w=-\displaystyle \frac{k\left(12{k}^{2}{c}_{1}^{2}-{c}_{2}^{2}\right)}{3{c}_{1}}.\end{eqnarray*}$
Thus, the following soliton to the SOTB equation is obtained $\begin{eqnarray}\begin{array}{l}{u}_{2}\left(x,t\right)=-\displaystyle \frac{{c}_{2}}{3{c}_{1}}+2k\,\tanh \,\left(kx-\displaystyle \frac{k\left(12{k}^{2}{c}_{1}^{2}-{c}_{2}^{2}\right)}{3{c}_{1}}t\right),\end{array}\end{eqnarray}$
where for ${c}_{1}=1,$ ${c}_{2}=1,$ and $k=1$ can be written $\begin{eqnarray*}{u}_{2}\left(x,t\right)=-\,\displaystyle \frac{1}{3}+2\,\tanh \left(x-\displaystyle \frac{11}{3}t\right),\end{eqnarray*}$
with the following series representation $\begin{eqnarray*}{u}_{2}\left(x,t\right)=-\displaystyle \frac{1}{3}-2\displaystyle \frac{{{\rm{e}}}^{-2x}-1}{{{\rm{e}}}^{-2x}+1}-\displaystyle \frac{88{{\rm{e}}}^{-2x}}{3{\left({{\rm{e}}}^{-2x}+1\right)}^{2}}t+\cdots .\end{eqnarray*}$
3.2. The time-fractional STOB equation in the CF context and its approximate solutions
By considering ${c}_{1}=1$ and ${c}_{2}=1$ (selected by the authors) in equation (1 ) and adopting the Laplace transform, we find5 ) recommends considering the following nonlinear operator
$\begin{eqnarray}\begin{array}{l} {\mathcal L} \left[u\left(x,t\right)\right]-\displaystyle \frac{u\left(x,0\right)}{s}+\displaystyle \frac{s+\alpha \left(1-s\right)}{s} {\mathcal L} \left[3{u}_{x}^{2}\left(x,t\right)\right.\\ +\,3{u}^{2}\left(x,t\right){u}_{x}\left(x,t\right)+3u\left(x,t\right){u}_{xx}\left(x,t\right)+{u}_{xxx}\left(x,t\right)\\ \left.+\,2u\left(x,t\right){u}_{x}\left(x,t\right)+{u}_{xx}\left(x,t\right)\right]\,=0.\end{array}\end{eqnarray}$
Equation ( $\begin{eqnarray*}\begin{array}{l}{\rm{\Phi }}\left[\psi \left(x,t;p\right)\right]= {\mathcal L} \left[\psi \left(x,t;p\right)\right]-\frac{\psi \left(x,0;p\right)}{s}\\ +\frac{s+\alpha \left(1-s\right)}{s} {\mathcal L} \left[3{\psi }_{x}^{2}\left(x,t;p\right)+3{\psi }^{2}\left(x,t;p\right)\right.{\psi }_{x} \left(x,t;p\right)\\ +3\psi \left(x,t;p\right){\psi }_{xx}\left(x,t;p\right)+{\psi }_{xxx}\left(x,t;p\right)\\ \left.+2\psi \left(x,t;p\right){\psi }_{x}\left(x,t;p\right)+{\psi }_{xx}\left(x,t;p\right)\right].\end{array}\end{eqnarray*}$
Based on the HAM [35], the following mth order deformation equation is formally constructed $\begin{eqnarray*} {\mathcal L} \left[{u}_{m}\left(x,t\right)-{\chi }_{m}{u}_{m-1}(x,t)\right]=h{{\mathscr{Q}}}_{m}\left({\vec{u}}_{m-1}\right),\end{eqnarray*}$
where $\begin{eqnarray*}{{\mathscr{Q}}}_{m}\left({\vec{u}}_{m-1}\right)=\displaystyle \frac{1}{\left(m-1\right)!}{\left.\displaystyle \frac{{\partial }^{m-1}{\rm{\Phi }}\left[\psi \left(x,t;p\right)\right]}{\partial {p}^{m-1}}\right|}_{p=0},\end{eqnarray*}$
and $\begin{eqnarray*}{\chi }_{m}=\left\{\begin{array}{c}0\, & m\leqslant 1,\\ 1\, & m\gt 1.\end{array}\right.\end{eqnarray*}$
In the first step, suppose that the initial condition of the time-fractional STOB equation in the CF context when ${c}_{1}=1$ and ${c}_{2}=1$ is as follows $\begin{eqnarray*}u\left(x,0\right)=1-\displaystyle \frac{{{\rm{e}}}^{-2x}-1}{{{\rm{e}}}^{-2x}+1}.\end{eqnarray*}$
Considering the above initial condition and solving the resultant equations successively, yields $\begin{eqnarray*}{u}_{0}\left(x,t\right)=1-\displaystyle \frac{{{\rm{e}}}^{-2x}-1}{{{\rm{e}}}^{-2x}+1},\end{eqnarray*}$
$\begin{eqnarray*}{u}_{1}\left(x,t\right)=\displaystyle \frac{24h\left(\alpha t-\alpha +1\right){{\rm{e}}}^{-2x}}{{\left({{\rm{e}}}^{-2x}+1\right)}^{2}},\cdots .\end{eqnarray*}$
Consequently, the following series solution to the time-fractional STOB equation in the CF context is gained $\begin{eqnarray*}{u}_{1}\left(x,t\right)=1-\displaystyle \frac{{{\rm{e}}}^{-2x}-1}{{{\rm{e}}}^{-2x}+1}+\displaystyle \frac{24h\left(\alpha t-\alpha +1\right){{\rm{e}}}^{-2x}}{{\left({{\rm{e}}}^{-2x}+1\right)}^{2}}+\cdots .\end{eqnarray*}$
In the special case when $\alpha =1$ and $h=-1,$ one can obtain $\begin{eqnarray*}{u}_{1}\left(x,t\right)=1-\displaystyle \frac{{{\rm{e}}}^{-2x}-1}{{{\rm{e}}}^{-2x}+1}-24\displaystyle \frac{{{\rm{e}}}^{-2x}}{{\left({{\rm{e}}}^{-2x}+1\right)}^{2}}t+\cdots ,\end{eqnarray*}$
with a closed-form representation as $\begin{eqnarray*}{u}_{1}\left(x,t\right)=1+\,\tanh \left(x-6t\right).\end{eqnarray*}$
Now, assume that the initial condition of the time-fractional STOB equation in the CF context when ${c}_{1}=1$ and ${c}_{2}=1$ is as follows $\begin{eqnarray*}u\left(x,0\right)=-\displaystyle \frac{1}{3}-2\displaystyle \frac{{{\rm{e}}}^{-2x}-1}{{{\rm{e}}}^{-2x}+1}.\end{eqnarray*}$
By considering the above initial condition and solving the resultant equations successively, one can find $\begin{eqnarray*}{u}_{0}\left(x,t\right)=-\displaystyle \frac{1}{3}-2\displaystyle \frac{{{\rm{e}}}^{-2x}-1}{{{\rm{e}}}^{-2x}+1},\end{eqnarray*}$
$\begin{eqnarray*}{u}_{1}\left(x,t\right)=\displaystyle \frac{88h\left(\alpha t-\alpha +1\right){{\rm{e}}}^{-2x}}{3{\left({{\rm{e}}}^{-2x}+1\right)}^{2}},\cdots .\end{eqnarray*}$
Consequently, a series solution to the time-fractional STOB equation in the CF context is acquired as $\begin{eqnarray*}\begin{array}{l}{u}_{2}\left(x,t\right)=-\displaystyle \frac{1}{3}-2\displaystyle \frac{{{\rm{e}}}^{-2x}-1}{{{\rm{e}}}^{-2x}+1}+\displaystyle \frac{88h\left(\alpha t-\alpha +1\right){{\rm{e}}}^{-2x}}{3{\left({{\rm{e}}}^{-2x}+1\right)}^{2}}\\ +\cdots .\end{array}\end{eqnarray*}$
In the special case when $\alpha =1$ and $h=-1,$ one can derive $\begin{eqnarray*}{u}_{2}\left(x,t\right)=-\displaystyle \frac{1}{3}-2\displaystyle \frac{{{\rm{e}}}^{-2x}-1}{{{\rm{e}}}^{-2x}+1}-\displaystyle \frac{88{{\rm{e}}}^{-2x}}{3{\left({{\rm{e}}}^{-2x}+1\right)}^{2}}t+\cdots ,\end{eqnarray*}$
with the following closed-form representation $\begin{eqnarray*}{u}_{2}\left(x,t\right)=-\displaystyle \frac{1}{3}+2\,\tanh \left(x-\displaystyle \frac{11}{3}t\right).\end{eqnarray*}$
(Convergence analysis) [36]: The series $u\left(x,t\right)\,={u}_{0}\left(x,t\right)+\displaystyle {\sum }_{k=1}^{+\infty }{u}_{k}\left(x,t\right)$ is convergence if $\exists \,0\lt r\lt 1$ such that $\parallel {u}_{k+1}\parallel \leqslant r\parallel {u}_{k}\parallel $ for all $k\geqslant {k}_{0}$ and for some ${k}_{0}\in {\mathbb{N}}.$
Moreover, the estimated error can be given as
$\begin{eqnarray*}\parallel u-\displaystyle \sum _{k=0}^{m}{u}_{k}\parallel \leqslant \displaystyle \frac{{r}^{m+1}}{1-r}\parallel {u}_{0}\parallel .\end{eqnarray*}$
4. Simulations and discussion
In this section, several two- and three-dimensional plots are considered to illustrate the CF operator effect on the dynamics of the obtained solitons. Figure 1 shows the two-dimensional plot of the 4th order approximation for $t=0.001,$ $h=-1,$ and $\alpha =1,$ $0.98,$ and $0.96$ (Selected by the authors). The CF operator effect on the dynamics of the kink soliton is formally observed from this figure. Additionally, by referring to figure 1 again, it can be observed that the results of the 4th order approximation and the exact solution (equation (3 )) are in compromise. The three-dimensional plots of the 4th order approximation have been given in figure 2 for $h=-1$ and $\alpha =1,$ $0.98,$ and $0.96.$ Table 1 presents the absolute error of the 4th order approximation ($h=-1$) and the exact solution (equation (3 )) when ${x}_{i}=0.5i,\,i=0,1,\mathrm{...},10$ and $t=0.001.$ The results confirm the potential of the HATM in handling the time-fractional STOB equation in the CF context. The two-dimensional plot of the 4th order approximation for $t=0.001,$ $h=-1,$ and $\alpha =1,$ $0.98,$ and $0.96$ has been illustrated in figure 3. From this figure, the CF operator effect on the dynamics of the kink soliton is clearly seen. Furthermore, by looking at figure 3 again, it can be seen that the results of the 4th order approximation and the exact solution (equation (4 )) are in compromise. The three-dimensional plots corresponding to the 4th order approximation have been represented in figure 4 for $h=-1$ and $\alpha =1,$ $0.98,$ and $0.96.$ Table 2 demonstrates the absolute error of the 4th order approximation ($h=-1$) and the exact solution (equation (4 )) when ${x}_{i}=0.5i,\,i=0,1,\mathrm{...},10$ and $t=0.001.$ Based on the results presented in table 2, the utility of the HATM in dealing with the time-fractional STOB equation in the CF context is confirmed.
Figure 1. The first exact solution ($t=0.001$) against the 4th order approximation for $t=0.001,$ h = −1, and $\alpha =1,$ $0.98,$ and $0.96.$ |
Figure 2. (a) The first exact solution against (b) the 4th order approximation for $\alpha =1$ and $h=-1$ (c) the 4th order approximation for $\alpha =0.98$ and $h=-1$ (d) the 4th order approximation for $\alpha =0.96$ and $h=-1$. |
Table 1. The absolute error of the 8th order approximation and the first exact solution. |
| ${x}$ | The 8th order approximation when ${t}=0.001,$ ${h}=-1,$ and ${\alpha }=1$ | The first exact solution when ${t}=0.001$ and ${\alpha }=1$ | The absolute error |
|---|---|---|---|
| $0$ | $0.994000072$ | $0.994000072$ | $0$ |
| $0.5$ | $1.457385408$ | $1.457385408$ | $0$ |
| $1$ | $1.759062773$ | $1.759062773$ | $0$ |
| $1.5$ | $1.904058107$ | $1.904058106$ | $1\times {10}^{-9}$ |
| $2$ | $1.963601214$ | $1.963601214$ | $0$ |
| $2.5$ | $1.986453796$ | 1. $986453797$ | $1\times {10}^{-9}$ |
| $3$ | $1.994995203$ | $1.994995203$ | $0$ |
| $3.5$ | $1.998155921$ | $1.998155921$ | $0$ |
| $4$ | $1.999321205$ | $1.999321206$ | $1\times {10}^{-9}$ |
| $4.5$ | $1.999750232$ | $1.999750232$ | $0$ |
| $5$ | $1.999908108$ | $1.999908108$ | $0$ |
Figure 3. The second exact solution ($t=0.001$) against the 4th order approximation for $t=0.001,$ $h=-1,$ and $\alpha =1,$ $0.96,$ and $0.92.$ |
Figure 4. (a) The second exact solution against (b) the 4th order approximation for $\alpha =1$ and $h=-1$ (c) the 4th order approximation for $\alpha =0.96$ and $h=-1$ (d) the 4th order approximation for $\alpha =0.92$ and $h=-1$. |
Table 2. The absolute error of the 8th order approximation and the second exact solution. |
| ${x}$ | The 8th order approximation when ${t}=0.001,$ ${h}=-1,$ and ${\alpha }=1$ | The second exact solution when ${t}=0.001$ and ${\alpha }=1$ | The absolute error |
|---|---|---|---|
| $0$ | $-0.340666633$ | $-0.340666633$ | $0$ |
| $0.5$ | $0.5851239350$ | $0.5851239349$ | $1\times {10}^{-10}$ |
| $1$ | $1.186766556$ | $1.186766556$ | $0$ |
| $1.5$ | $1.475633586$ | $1.475633585$ | $1\times {10}^{-9}$ |
| $2$ | $1.594201885$ | $1.594201886$ | $1\times {10}^{-9}$ |
| $2.5$ | $1.639699546$ | $1.639699546$ | $0$ |
| $3$ | $1.656703559$ | $1.656703559$ | $0$ |
| $3.5$ | $1.662995664$ | $1.662995664$ | $0$ |
| $4$ | $1.665315396$ | $1.665315397$ | $1\times {10}^{-9}$ |
| $4.5$ | $1.666169455$ | $1.666169456$ | $1\times {10}^{-9}$ |
| $5$ | $1.666483738$ | $1.666483739$ | $1\times {10}^{-9}$ |
5. Conclusion
As it was stated, exploring the dynamics of solitons in nonlinear time-fractional partial differential equations has gained considerable interest, in the last decades. The key objective of the current study was to consider the time-fractional STOB equation in the CF context and acquire its valid approximations through utilizing a mixed approach composed of the HAM and the Laplace transform. The existence and uniqueness of the solution of the time-fractional STOB equation in the CF context were analyzed by proving the Lipschitz condition for $\varphi (x,\,t;u)$ as the kernel and providing some theorems. To demonstrate the CF operator effect on the dynamics of the obtained solitons, several two- and three-dimensional plots were portrayed. It was observed that the mixed approach is capable of generating valid approximations to the time-fractional STOB equation in the CF context.