1. Introduction
As is well known, nonlinear evolution equations play an important role in the field of mathematical physics. The (2+1)-dimensional Heisenberg ferromagnetic spin chain equation is a nonlinear integrable evolution equation, given by [1]
$\begin{eqnarray}{\rm{i}}{z}_{t}+{\gamma }_{1}{z}_{xx}+{\gamma }_{2}{z}_{yy}+{\gamma }_{3}{z}_{xy}-{\gamma }_{4}{\left|z\right|}^{2}z=0,\end{eqnarray}$
where z(t, x, y) signifies the appropriate continuum approximation of the coherent magnetism amplitude to the bosonic operators at spin–lattice sites, and γi(i = 1, 2, 3, 4) are parameters regarding magnetic coupling coefficients. In recent years, this key nonlinear model has attracted the attention of many scholars and has become their research focus (see [2–11] and references therein). In particular, Osman et al [5], Sahoo and Tripathy [8] and Abdel-Aty [11] obtained new exact soliton solutions of the (2+1)-dimensional Heisenberg ferromagnetic spin chain equation using the new extended Fan sub-equation method, the modified Khater method and the $G^{\prime} /(bG^{\prime} +G+a)$-expansion method, respectively.Fractional differential equations (FDEs), due to the nonlocal and memory effects of the fractional derivative [12–15], have been successfully applied in many aspects of science and technology recently. This paper extends the classical (2+1)-dimensional Heisenberg ferromagnetic spin chain equation to the following time-fractional version:
$\begin{eqnarray}\begin{array}{l}{\rm{i}}{D}_{t}^{\alpha }z+{\gamma }_{1}{z}_{xx}+{\gamma }_{2}{z}_{yy}+{\gamma }_{3}{z}_{xy}-{\gamma }_{4}{\left|z\right|}^{2}z=0,\\ 0\lt \alpha \lt 1.\end{array}\end{eqnarray}$
Many scholars have used different methods to obtain analytical or numerical solutions of this equation in different forms of fractional derivative definitions [16–23]. Among them, Bakcerler et al [18] used the modified extended tanh-function method to attain some analytic traveling wave solutions of the time-fractional (2+1)-dimensional nonlinear Heisenberg ferromagnetic spin chain equation with a conformable fractional derivative. Rani et al [20] applied the generalized Riccati equation mapping method to obtain some new optical soliton solutions for the space time conformable (2+1) dimensional Heisenberg ferromagnetic spin chain equation with Atangana's conformable derivative. Luo [23] investigated the soliton solutions and dynamical analysis of the (2+1)-dimensional Heisenberg ferromagnetic spin chains model with a beta fractional derivative by using the second-order complete discriminant system.Assuming z(t, x, y) = u(t, x, y) + iv(t, x, y), equation (2 ) can be rewritten as the following system:
$\begin{eqnarray}\left\{\begin{array}{l}{D}_{t}^{\alpha }u=-{\gamma }_{1}{v}_{xx}-{\gamma }_{2}{v}_{yy}-{\gamma }_{3}{v}_{xy}+{\gamma }_{4}v({u}^{2}+{v}^{2}),\\ {D}_{t}^{\alpha }v={\gamma }_{1}{u}_{xx}+{\gamma }_{2}{u}_{yy}+{\gamma }_{3}{u}_{xy}-{\gamma }_{4}u({u}^{2}+{v}^{2}).\end{array}\right.\end{eqnarray}$
Compared to numerical methods for FDEs, such as the finite difference method [24] and the finite element method [25], this paper adopts an analytical method: the Lie symmetry analysis method, which was initially advocated by Norwegian mathematician Sophus Lie at the end of the 19th century, and was further developed by Ovsianikov [26] and others [27–31]. The Lie symmetry analysis method can effectively handle various forms of integer-order differential equations [32–34] because it can treat differential equations uniformly, regardless of their forms, transforming some solutions of these equations into other forms of solutions [35]. It was extended to FDEs by Gazizov et al [36] in 2007, and then widely applied to various models of FDEs occurring in different areas of applied science (see [37–48]).It should be noted that this paper adopts the most widely used Riemann–Liouville fractional derivative defined by [12]
$\begin{eqnarray*}\begin{array}{l}{\mathrm{}}_{a}{D}_{t}^{\alpha }f(t,x)={D}_{t}^{n}{\mathrm{}}_{a}{I}_{t}^{n-\alpha }f(t,x)\\ \quad =\,\left\{\begin{array}{ll}\frac{1}{{\rm{\Gamma }}(n-\alpha )}\frac{{\partial }^{n}}{\partial {t}^{n}}{\displaystyle \int }_{a}^{t}\frac{f(s,x)}{{\left(t-s\right)}^{\alpha -n+1}}{\rm{d}}s, & n-1\lt \alpha \lt n,\\ {D}_{t}^{n}f(t,x), & \alpha =n\in {\mathbb{N}},\end{array}\right.\end{array}\end{eqnarray*}$
for t > a. We denote the operator ${}_{0}{D}_{t}^{\alpha }$ as ${D}_{t}^{\alpha }$ for simplicity.This paper is organized as follows. In section 2, we present the Lie symmetry analysis of equation (3 ). In section 3, we obtain the power series solutions of equation (3 ) via similarity reduction and prove their convergence. The conserved vectors for all Lie symmetries admitted by equation (3 ) are constructed in section 4. The conclusion is given in the last section.
2. Lie symmetry analysis of equation (3 )
Let equation (3 ) remain invariant under the following one-parameter (ε) continuous point transformation group:
$\begin{eqnarray}\begin{array}{rcl}{t}^{* } & = & t+\epsilon \tau (t,x,y,u,v)+o(\epsilon ),\\ {x}^{* } & = & x+\epsilon \xi (t,x,y,u,v)+o(\epsilon ),\\ {y}^{* } & = & y+\epsilon \theta (t,x,y,u,v)+o(\epsilon ),\\ {u}^{* } & = & u+\epsilon \eta (t,x,y,u,v)+o(\epsilon ),\\ {v}^{* } & = & v+\epsilon \zeta (t,x,y,u,v)+o(\epsilon ),\\ {D}_{{t}^{* }}^{\alpha }{u}^{* } & = & {D}_{t}^{\alpha }u+\epsilon {\eta }^{\alpha ,t}+o(\epsilon ),\\ {D}_{{t}^{* }}^{\alpha }{v}^{* } & = & {D}_{t}^{\alpha }v+\epsilon {\zeta }^{\alpha ,t}+o(\epsilon ),\\ {D}_{{x}^{* }}{u}^{* } & = & {D}_{x}u+\epsilon {\eta }^{x}+o(\epsilon ),\\ {D}_{{x}^{* }}{v}^{* } & = & {D}_{x}v+\epsilon {\zeta }^{x}+o(\epsilon ),\\ {D}_{{y}^{* }}{u}^{* } & = & {D}_{y}u+\epsilon {\eta }^{y}+o(\epsilon ),\\ {D}_{{y}^{* }}{v}^{* } & = & {D}_{y}v+\epsilon {\zeta }^{y}+o(\epsilon ),\\ {D}_{{x}^{* }}^{2}{u}^{* } & = & {D}_{x}^{2}u+\epsilon {\eta }^{xx}+o(\epsilon ),\\ {D}_{{x}^{* }}^{2}{v}^{* } & = & {D}_{x}^{2}v+\epsilon {\zeta }^{xx}+o(\epsilon ),\\ {D}_{{y}^{* }}^{2}{u}^{* } & = & {D}_{y}^{2}u+\epsilon {\eta }^{yy}+o(\epsilon ),\\ {D}_{{y}^{* }}^{2}{v}^{* } & = & {D}_{y}^{2}v+\epsilon {\zeta }^{yy}+o(\epsilon ),\\ {D}_{{x}^{* }}{D}_{{y}^{* }}{u}^{* } & = & {D}_{x}{D}_{y}u+\epsilon {\eta }^{xy}+o(\epsilon ),\\ {D}_{{x}^{* }}{D}_{{y}^{* }}{v}^{* } & = & {D}_{x}{D}_{y}v+\epsilon {\zeta }^{xy}+o(\epsilon ),\end{array}\end{eqnarray}$
where τ, ξ, θ, η and ζ are infinitesimals and ηα,t, ζα,t, ηx, ζx, ηy, ζy, ηxx, ζxx, ηyy, ζyy, ηxy, ζxy are the corresponding prolongations of orders α, 1 and 2, respectively. The corresponding group generator is defined by $\begin{eqnarray}X=\tau \frac{\partial }{\partial t}+\xi \frac{\partial }{\partial x}+\theta \frac{\partial }{\partial y}+\eta \frac{\partial }{\partial u}+\zeta \frac{\partial }{\partial v}.\end{eqnarray}$
Then, the corresponding prolongation of X has the form $\begin{eqnarray}\begin{array}{rcl}{\rm{P}}{{\rm{r}}}^{(\alpha ,2)}X & = & X+{\eta }^{\alpha ,t}\frac{\partial }{\partial {u}_{t}^{\alpha }}+{\zeta }^{\alpha ,t}\frac{\partial }{\partial {v}_{t}^{\alpha }}+{\eta }^{xx}\frac{\partial }{\partial {u}_{xx}}\\ & & +{\zeta }^{xx}\frac{\partial }{\partial {v}_{xx}}+{\eta }^{yy}\frac{\partial }{\partial {u}_{yy}}+{\zeta }^{yy}\frac{\partial }{\partial {v}_{yy}}\\ & & +{\eta }^{xy}\frac{\partial }{\partial {u}_{xy}}+{\zeta }^{xy}\frac{\partial }{\partial {v}_{xy}},\end{array}\end{eqnarray}$
where $\begin{eqnarray}{\eta }^{x}={D}_{x}(\eta )-{u}_{t}{D}_{x}(\tau )-{u}_{x}{D}_{x}(\xi )-{u}_{y}{D}_{x}(\theta ),\end{eqnarray}$
$\begin{eqnarray}{\zeta }^{x}={D}_{x}(\zeta )-{v}_{t}{D}_{x}(\tau )-{v}_{x}{D}_{x}(\xi )-{v}_{y}{D}_{x}(\theta ),\end{eqnarray}$
$\begin{eqnarray}{\eta }^{y}={D}_{y}(\eta )-{u}_{t}{D}_{y}(\tau )-{u}_{x}{D}_{y}(\xi )-{u}_{y}{D}_{y}(\theta ),\end{eqnarray}$
$\begin{eqnarray}{\zeta }^{y}={D}_{y}(\zeta )-{v}_{t}{D}_{y}(\tau )-{v}_{x}{D}_{y}(\xi )-{v}_{y}{D}_{y}(\theta ),\end{eqnarray}$
$\begin{eqnarray}{\eta }^{xx}={D}_{x}({\eta }^{x})-{u}_{xt}{D}_{x}(\tau )-{u}_{xx}{D}_{x}(\xi )-{u}_{xy}{D}_{x}(\theta ),\end{eqnarray}$
$\begin{eqnarray}{\zeta }^{xx}={D}_{x}({\zeta }^{x})-{v}_{xt}{D}_{x}(\tau )-{v}_{xx}{D}_{x}(\xi )-{v}_{xy}{D}_{x}(\theta ),\end{eqnarray}$
$\begin{eqnarray}{\eta }^{yy}={D}_{y}({\eta }^{y})-{u}_{yt}{D}_{y}(\tau )-{u}_{yx}{D}_{y}(\xi )-{u}_{yy}{D}_{y}(\theta ),\end{eqnarray}$
$\begin{eqnarray}{\zeta }^{yy}={D}_{y}({\zeta }^{y})-{v}_{yt}{D}_{y}(\tau )-{v}_{yx}{D}_{y}(\xi )-{v}_{yy}{D}_{y}(\theta ),\end{eqnarray}$
$\begin{eqnarray}{\eta }^{xy}={D}_{x}({\eta }^{y})-{u}_{xt}{D}_{x}(\tau )-{u}_{xx}{D}_{x}(\xi )-{u}_{xy}{D}_{x}(\theta ),\end{eqnarray}$
$\begin{eqnarray}{\zeta }^{xy}={D}_{x}({\zeta }^{y})-{v}_{xt}{D}_{x}(\tau )-{v}_{xx}{D}_{x}(\xi )-{v}_{xy}{D}_{x}(\theta ),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\eta }^{\alpha ,t} & = & \frac{{\partial }^{\alpha }\eta }{\partial {t}^{\alpha }}+({\eta }_{u}-\alpha {D}_{t}(\tau ))\frac{{\partial }^{\alpha }u}{\partial {t}^{\alpha }}-u\frac{{\partial }^{\alpha }{\eta }_{u}}{\partial {t}^{\alpha }}\\ & & +({\eta }_{v}\frac{{\partial }^{\alpha }v}{\partial {t}^{\alpha }}-v\frac{{\partial }^{\alpha }{\eta }_{v}}{\partial {t}^{\alpha }})\\ & & +\displaystyle \sum _{n=1}^{\infty }[\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right)\frac{{\partial }^{n}{\eta }_{u}}{\partial {t}^{n}}-\left(\genfrac{}{}{0.0pt}{}{\alpha }{n+1}\right){D}_{t}^{n+1}(\tau )]{D}_{t}^{\alpha -n}(u)\\ & & +\displaystyle \sum _{n=1}^{\infty }\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right)\frac{{\partial }^{n}{\eta }_{v}}{\partial {t}^{n}}{D}_{t}^{\alpha -n}(v)\\ & & -\displaystyle \sum _{n=1}^{\infty }\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right){D}_{t}^{n}(\xi ){D}_{t}^{\alpha -n}({u}_{x})\\ & & -\displaystyle \sum _{n=1}^{\infty }\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right){D}_{t}^{n}(\theta ){D}_{t}^{\alpha -n}({u}_{y})+{\mu }_{1}+{\mu }_{2},\end{array}\end{eqnarray}$
with $\begin{eqnarray*}\begin{array}{rcl}{\mu }_{1} & = & \displaystyle \sum _{n=2}^{\infty }\displaystyle \sum _{m=2}^{n}\displaystyle \sum _{k=2}^{m}\displaystyle \sum _{r=0}^{k-1}\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right)\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left(\genfrac{}{}{0.0pt}{}{k}{r}\right)\\ & & \times \frac{{t}^{n-\alpha }{(-u)}^{r}}{k!{\rm{\Gamma }}(n+1-\alpha )}\frac{{\partial }^{m}{u}^{k-r}}{\partial {t}^{m}}\frac{{\partial }^{n-m+k}\eta }{\partial {t}^{n-m}\partial {u}^{k}},\\ {\mu }_{2} & = & \displaystyle \sum _{n=2}^{\infty }\displaystyle \sum _{m=2}^{n}\displaystyle \sum _{k=2}^{m}\displaystyle \sum _{r=0}^{k-1}\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right)\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left(\genfrac{}{}{0.0pt}{}{k}{r}\right)\\ & & \times \frac{{t}^{n-\alpha }{(-v)}^{r}}{k!{\rm{\Gamma }}(n+1-\alpha )}\frac{{\partial }^{m}{v}^{k-r}}{\partial {t}^{m}}\frac{{\partial }^{n-m+k}\eta }{\partial {t}^{n-m}\partial {v}^{k}},\end{array}\end{eqnarray*}$
and $\begin{eqnarray}\begin{array}{rcl}{\zeta }^{\alpha ,t} & = & \frac{{\partial }^{\alpha }\zeta }{\partial {t}^{\alpha }}+({\zeta }_{v}-\alpha {D}_{t}(\tau ))\frac{{\partial }^{\alpha }v}{\partial {t}^{\alpha }}-v\frac{{\partial }^{\alpha }{\zeta }_{v}}{\partial {t}^{\alpha }}\\ & & +({\zeta }_{u}\frac{{\partial }^{\alpha }u}{\partial {t}^{\alpha }}-u\frac{{\partial }^{\alpha }{\zeta }_{u}}{\partial {t}^{\alpha }})\\ & & +\displaystyle \sum _{n=1}^{\infty }[\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right)\frac{{\partial }^{n}{\zeta }_{v}}{\partial {t}^{n}}-\left(\genfrac{}{}{0.0pt}{}{\alpha }{n+1}\right){D}_{t}^{n+1}(\tau )]{D}_{t}^{\alpha -n}(v)\\ & & +\displaystyle \sum _{n=1}^{\infty }\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right)\frac{{\partial }^{n}{\zeta }_{u}}{\partial {t}^{n}}{D}_{t}^{\alpha -n}(u)\\ & & -\displaystyle \sum _{n=1}^{\infty }\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right){D}_{t}^{n}(\xi ){D}_{t}^{\alpha -n}({v}_{x})\\ & & -\displaystyle \sum _{n=1}^{\infty }\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right){D}_{t}^{n}(\theta ){D}_{t}^{\alpha -n}({v}_{y})+{\mu }_{3}+{\mu }_{4},\end{array}\end{eqnarray}$
with $\begin{eqnarray*}\begin{array}{rcl}{\mu }_{3} & = & \displaystyle \sum _{n=2}^{\infty }\displaystyle \sum _{m=2}^{n}\displaystyle \sum _{k=2}^{m}\displaystyle \sum _{r=0}^{k-1}\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right)\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left(\genfrac{}{}{0.0pt}{}{k}{r}\right)\\ & & \times \frac{{t}^{n-\alpha }{(-u)}^{r}}{k!{\rm{\Gamma }}(n+1-\alpha )}\frac{{\partial }^{m}{u}^{k-r}}{\partial {t}^{m}}\frac{{\partial }^{n-m+k}\zeta }{\partial {t}^{n-m}\partial {u}^{k}},\\ {\mu }_{4} & = & \displaystyle \sum _{n=2}^{\infty }\displaystyle \sum _{m=2}^{n}\displaystyle \sum _{k=2}^{m}\displaystyle \sum _{r=0}^{k-1}\left(\genfrac{}{}{0.0pt}{}{\alpha }{n}\right)\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\left(\genfrac{}{}{0.0pt}{}{k}{r}\right)\\ & & \times \frac{{t}^{n-\alpha }{(-v)}^{r}}{k!{\rm{\Gamma }}(n+1-\alpha )}\frac{{\partial }^{m}{v}^{k-r}}{\partial {t}^{m}}\frac{{\partial }^{n-m+k}\zeta }{\partial {t}^{n-m}\partial {v}^{k}},\end{array}\end{eqnarray*}$
where Dt, Dx and Dy are the total derivatives with respect to t, x and y, respectively. The infinitesimal transformations, equation (
$\begin{eqnarray}\tau (t,x,y,u,v){| }_{t=0}=0.\end{eqnarray}$
From the expressions of μi(i = 1, 2, 3, 4), if the infinitesimals η and ζ are linear with respect to the variables u and v, then μi = 0, that is,
$\begin{eqnarray}\frac{{\partial }^{2}\eta }{\partial {u}^{2}}=\frac{{\partial }^{2}\eta }{\partial {v}^{2}}=\frac{{\partial }^{2}\zeta }{\partial {u}^{2}}=\frac{{\partial }^{2}\zeta }{\partial {v}^{2}}=0.\end{eqnarray}$
The invariance criterion of equation (3 ) under equation (4 ) is 22 ) and solving the obtained determining equations, we can get the infinitesimals as follows: 3 ) admitted the four-dimension Lie algebra spanned by
$\begin{eqnarray}\left\{\begin{array}{l}{\rm{P}}{{\rm{r}}}^{(\alpha ,2)}X({D}_{t}^{\alpha }u+{\gamma }_{1}{v}_{xx}+{\gamma }_{2}{v}_{yy}+{\gamma }_{3}{v}_{xy}-{\gamma }_{4}v({u}^{2}+{v}^{2})){| }_{()}=0,\\ {\rm{P}}{{\rm{r}}}^{(\alpha ,2)}X({D}_{t}^{\alpha }v-{\gamma }_{1}{u}_{xx}-{\gamma }_{2}{u}_{yy}-{\gamma }_{3}{u}_{xy}+{\gamma }_{4}u({u}^{2}+{v}^{2})){| }_{()}=0,\end{array}\right.\end{eqnarray}$
which can be rewritten as $\begin{eqnarray}\left\{\begin{array}{l}({\eta }^{\alpha ,t}+{\gamma }_{1}{\zeta }^{xx}+{\gamma }_{2}{\zeta }^{yy}+{\gamma }_{3}{\zeta }^{xy}-2{\gamma }_{4}uv\eta -{\gamma }_{4}({u}^{2}+3{v}^{2})\zeta ){| }_{()}=0,\\ ({\zeta }^{\alpha ,t}-{\gamma }_{1}{\eta }^{xx}-{\gamma }_{2}{\eta }^{yy}-{\gamma }_{3}{\eta }^{xy}+2{\gamma }_{4}uv\zeta +{\gamma }_{4}(3{u}^{2}+{v}^{2})\eta ){| }_{()}=0.\end{array}\right.\end{eqnarray}$
Putting ηα,t, ζα,t, ηxx, ζxx, ηyy, ζyy, ηxy, ζxy into equation ( $\begin{eqnarray}\begin{array}{rcl}\tau & = & {c}_{1}t,\,\,\xi =\frac{\alpha }{2}{c}_{1}x+{c}_{2},\,\,\theta =\frac{\alpha }{2}{c}_{1}y+{c}_{3},\\ \eta & = & -\frac{\alpha }{2}{c}_{1}u+{c}_{4}v,\,\,\zeta =-\frac{\alpha }{2}{c}_{1}v-{c}_{4}u,\end{array}\end{eqnarray}$
where c1, c2, c3 and c4 are arbitrary constants. Therefore, equation ( $\begin{eqnarray}\begin{array}{rcl}{X}_{1} & = & t\frac{\partial }{\partial t}+\frac{\alpha }{2}x\frac{\partial }{\partial x}+\frac{\alpha }{2}y\frac{\partial }{\partial y}-\frac{\alpha }{2}u\frac{\partial }{\partial u}-\frac{\alpha }{2}v\frac{\partial }{\partial v},\\ {X}_{2} & = & v\frac{\partial }{\partial u}-u\frac{\partial }{\partial v},\,\,{X}_{3}=\frac{\partial }{\partial x},\,\,{X}_{4}=\frac{\partial }{\partial y}.\end{array}\end{eqnarray}$
3. Similarity reductions and exact solutions of equation (3 )
In this section, the aimed equations (3 ) can be reduced to solvable (1+1)-dimensional fractional partial differential equations with the left-hand Erdélyi–Kober fractional derivative.
Case 1: ${X}_{1}=t\frac{{\rm{\partial }}}{{\rm{\partial }}t}+\frac{\alpha }{2}x\frac{{\rm{\partial }}}{{\rm{\partial }}x}+\frac{\alpha }{2}y\frac{{\rm{\partial }}}{{\rm{\partial }}y}-\frac{\alpha }{2}u\frac{{\rm{\partial }}}{{\rm{\partial }}u}-\frac{\alpha }{2}v\frac{{\rm{\partial }}}{{\rm{\partial }}v}.$
From the characteristic equation corresponding to the group generator X1, i.e.3 ) as follows:
$\begin{eqnarray}\frac{{\rm{d}}t}{t}=\frac{{\rm{d}}x}{\frac{\alpha }{2}x}=\frac{{\rm{d}}y}{\frac{\alpha }{2}y}=\frac{{\rm{d}}u}{-\frac{\alpha }{2}u}=\frac{{\rm{d}}v}{-\frac{\alpha }{2}v},\end{eqnarray}$
we obtain the similarity variables $x{t}^{-\frac{\alpha }{2}}$, $y{t}^{-\frac{\alpha }{2}}$, $u{t}^{\frac{\alpha }{2}}$ and $v{t}^{\frac{\alpha }{2}}$. Therefore, we get the invariant solutions of equation ( $\begin{eqnarray}u(t,x,y)={t}^{-\frac{\alpha }{2}}f({\omega }_{1},{\omega }_{2}),\,\,v(t,x,y)={t}^{-\frac{\alpha }{2}}g({\omega }_{1},{\omega }_{2}),\end{eqnarray}$
with ${\omega }_{1}=x{t}^{-\frac{\alpha }{2}}$, ${\omega }_{2}=y{t}^{-\frac{\alpha }{2}}$. The similarity transformations $u(t,x,y)\,={t}^{-\frac{\alpha }{2}}f({\omega }_{1},{\omega }_{2})$, $v(t,x,y)={t}^{-\frac{\alpha }{2}}g({\omega }_{1},{\omega }_{2})$ with the similarity variable ${\omega }_{1}=x{t}^{-\frac{\alpha }{2}}$, ${\omega }_{2}=y{t}^{-\frac{\alpha }{2}}$ reduce equation (
$\begin{eqnarray}\left\{\begin{array}{l}({{ \mathcal P }}_{\frac{2}{\alpha },\frac{2}{\alpha }}^{1-\frac{3\alpha }{2},\alpha }f)({\omega }_{1},{\omega }_{2})=-{\gamma }_{1}{g}_{{\omega }_{1}{\omega }_{1}}-{\gamma }_{2}{g}_{{\omega }_{2}{\omega }_{2}}-{\gamma }_{3}{g}_{{\omega }_{1}{\omega }_{2}}+{\gamma }_{4}g({f}^{2}+{g}^{2}),\\ ({{ \mathcal P }}_{\frac{2}{\alpha },\frac{2}{\alpha }}^{1-\frac{3\alpha }{2},\alpha }g)({\omega }_{1},{\omega }_{2})={\gamma }_{1}{f}_{{\omega }_{1}{\omega }_{1}}+{\gamma }_{2}{f}_{{\omega }_{2}{\omega }_{2}}+{\gamma }_{3}{f}_{{\omega }_{1}{\omega }_{2}}-{\gamma }_{4}f({f}^{2}+{g}^{2}),\end{array}\right.\end{eqnarray}$
where $({{ \mathcal P }}_{{\delta }_{1},{\delta }_{2}}^{\iota ,\kappa })$ is the left-hand Erdélyi–Kober fractional differential operator defined by $\begin{eqnarray}\begin{array}{l}(\underset{{\delta }_{1},{\delta }_{2}}{\overset{\iota ,\kappa }{{ \mathcal P }}}\psi )({\omega }_{1},{\omega }_{2}):= {\prod }_{j=0}^{m-1}\left(\iota +j-\frac{1}{{\delta }_{1}}{\omega }_{1}\frac{{\rm{d}}}{{\rm{d}}{\omega }_{1}}-\frac{1}{{\delta }_{2}}{\omega }_{2}\frac{{\rm{d}}}{{\rm{d}}{\omega }_{2}}\right)\\ \quad \times \,\left({{ \mathcal K }}_{{\delta }_{1},{\delta }_{2}}^{\iota +\kappa ,m-\kappa }\psi \right)({\omega }_{1},{\omega }_{2}),\,\kappa \gt 0,\\ m=\left\{\begin{array}{l}[\kappa ]+1,\,\,\kappa \notin {\mathbb{N}},\\ \kappa ,\,\,\,\,\,\kappa \in {\mathbb{N}},\end{array}\right.\end{array}\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{c}({{ \mathcal K }}_{{\delta }_{1},{\delta }_{2}}^{\iota ,\kappa }\psi )({\omega }_{1},{\omega }_{2})\\ \,:=\left\{\begin{array}{cc}\frac{1}{\left.{\rm{\Gamma }}(\kappa \right)}{\displaystyle \int }_{1}^{\infty }{\left(s-1\right)}^{\kappa -1}{s}^{\left.-(\iota +\kappa \right)}\psi ({\omega }_{1}{s}^{\frac{1}{{\delta }_{1}}},{\omega }_{2}{s}^{\frac{1}{{\delta }_{2}}}){\rm{d}}s,\,\, & \kappa \gt 0,\\ \psi ({\omega }_{1},{\omega }_{2}),\,\, & \kappa =0.\end{array}\right.\end{array}\end{eqnarray}$
For 0 < α < 1, the Riemann–Liouville time-fractional derivative of u(t, x, y) can be obtained as follows:
$\begin{eqnarray*}\begin{array}{rcl}\frac{{\partial }^{\alpha }u}{\partial {t}^{\alpha }} & = & \frac{{\partial }^{\alpha }}{\partial {t}^{\alpha }}({t}^{-\frac{\alpha }{2}}f({\omega }_{1},{\omega }_{2}))=\frac{\partial }{\partial t}\left[\frac{1}{{\rm{\Gamma }}(1-\alpha )}\right.\\ & & \left.\times {\displaystyle \int }_{0}^{t}{(t-s)}^{-\alpha }{s}^{\frac{-\alpha }{2}}f(x{s}^{-\frac{\alpha }{2}},y{s}^{-\frac{\alpha }{2}}){\rm{d}}s\right].\end{array}\end{eqnarray*}$
Assuming $r=\frac{t}{s}$, we have $\begin{eqnarray*}\begin{array}{rcl}\frac{{\partial }^{\alpha }u}{\partial {t}^{\alpha }} & = & \frac{\partial }{\partial t}\left[\frac{{t}^{1-\frac{3\alpha }{2}}}{{\rm{\Gamma }}(1-\alpha )}{\displaystyle \int }_{1}^{\infty }{(r-1)}^{-\alpha }{r}^{\frac{3\alpha }{2}-2}f({\omega }_{1}{r}^{\frac{\alpha }{2}},{\omega }_{2}{r}^{\frac{\alpha }{2}}){\rm{d}}r\right]\\ & = & \frac{\partial }{\partial t}\left[{t}^{1-\frac{3\alpha }{2}}({{ \mathcal K }}_{\frac{2}{\alpha },\frac{2}{\alpha }}^{1-\frac{\alpha }{2},1-\alpha }f)({\omega }_{1},{\omega }_{2})\right].\end{array}\end{eqnarray*}$
Because of ${\omega }_{1}=x{t}^{-\frac{\alpha }{2}}$ and ${\omega }_{2}=y{t}^{-\frac{\alpha }{2}}$, the following relation holds: $\begin{eqnarray*}\begin{array}{l}t\frac{\partial }{\partial t}\psi ({\omega }_{1},{\omega }_{2})=t(-\frac{\alpha }{2}x{t}^{-\frac{\alpha }{2}-1}{\psi }_{{\omega }_{1}}-\frac{\alpha }{2}x{t}^{-\frac{\alpha }{2}-1}{\psi }_{{\omega }_{2}})\\ \quad =-\frac{\alpha }{2}{\omega }_{1}\frac{{\rm{d}}}{{\rm{d}}{\omega }_{1}}\psi -\frac{\alpha }{2}{\omega }_{2}\frac{{\rm{d}}}{{\rm{d}}{\omega }_{2}}\psi .\end{array}\end{eqnarray*}$
Hence, we arrive at $\begin{eqnarray*}\begin{array}{rcl}\frac{{\partial }^{\alpha }u}{\partial {t}^{\alpha }} & = & {t}^{-\frac{3\alpha }{2}}\left[\left(1-\frac{3\alpha }{2}-\frac{\alpha }{2}{\omega }_{1}\frac{{\rm{d}}}{{\rm{d}}{\omega }_{1}}-\frac{\alpha }{2}{\omega }_{2}\frac{{\rm{d}}}{{\rm{d}}{\omega }_{2}}\right)\right.\\ & & \left.\times \left({{ \mathcal K }}_{\frac{2}{\alpha },\frac{2}{\alpha }}^{1-\frac{\alpha }{2},1-\alpha }f\right)(\omega )\right]={t}^{-\frac{3\alpha }{2}}({{ \mathcal P }}_{\frac{2}{\alpha },\frac{2}{\alpha }}^{1-\frac{3\alpha }{2},\alpha }f)({\omega }_{1},{\omega }_{2}).\end{array}\end{eqnarray*}$
Similarly, $\begin{eqnarray*}\frac{{\partial }^{\alpha }v}{\partial {t}^{\alpha }}={t}^{-\frac{3\alpha }{2}}({{ \mathcal P }}_{\frac{2}{\alpha },\frac{2}{\alpha }}^{1-\frac{3\alpha }{2},\alpha }g)({\omega }_{1},{\omega }_{2}).\end{eqnarray*}$
Meanwhile, $\begin{eqnarray*}\begin{array}{l}-{\gamma }_{1}{v}_{xx}-{\gamma }_{2}{v}_{yy}-{\gamma }_{3}{v}_{xy}+{\gamma }_{4}v({u}^{2}+{v}^{2})={t}^{-\frac{3\alpha }{2}}\\ \quad \times \,(-{\gamma }_{1}{g}_{{\omega }_{1}{\omega }_{1}}-{\gamma }_{2}{g}_{{\omega }_{2}{\omega }_{2}}-{\gamma }_{3}{g}_{{\omega }_{1}{\omega }_{2}}+{\gamma }_{4}g({f}^{2}+{g}^{2})),\\ \quad {\gamma }_{1}{u}_{xx}+{\gamma }_{2}{u}_{yy}+{\gamma }_{3}{u}_{xy}-{\gamma }_{4}u({u}^{2}+{v}^{2})={t}^{-\frac{3\alpha }{2}}\\ \quad \times \,({\gamma }_{1}{f}_{{\omega }_{1}{\omega }_{1}}+{\gamma }_{2}{f}_{{\omega }_{2}{\omega }_{2}}+{\gamma }_{3}{f}_{{\omega }_{1}{\omega }_{2}}-{\gamma }_{4}f({f}^{2}+{g}^{2})).\end{array}\end{eqnarray*}$
This completes the proof. □Next, we use the power series method to derive the power series solutions of equation (27 ). Assuming 30 )–(33 ) into equation (27 ) arrives at the following system: 3 ) are
$\begin{eqnarray}\begin{array}{rcl}f({\omega }_{1},{\omega }_{2}) & = & \displaystyle \sum _{k=0}^{\infty }{a}_{k}{\omega }^{k},\,\,g({\omega }_{1},{\omega }_{2})=\displaystyle \sum _{k=0}^{\infty }{b}_{k}{\omega }^{k},\\ \omega & = & {C}_{1}{\omega }_{1}+{C}_{2}{\omega }_{2},\end{array}\end{eqnarray}$
with constants ak and bk to be known later, we can get $\begin{eqnarray}\begin{array}{rcl}f^{\prime} (\omega ) & = & \displaystyle \sum _{k=0}^{\infty }(k+1){a}_{k+1}{\omega }^{k},\\ f^{\prime\prime} (\omega ) & = & \displaystyle \sum _{k=0}^{\infty }(k+2)(k+1){a}_{k+2}{\omega }^{k},\\ g^{\prime} (\omega ) & = & \displaystyle \sum _{k=0}^{\infty }(k+1){b}_{k+1}{\omega }^{k},\\ g^{\prime\prime} (\omega ) & = & \displaystyle \sum _{k=0}^{\infty }(k+2)(k+1){b}_{k+2}{\omega }^{k},\end{array}\end{eqnarray}$
and $\begin{eqnarray}\begin{array}{l}({{ \mathcal P }}_{\frac{2}{\alpha }}^{1-\frac{3\alpha }{2},\alpha }f)(\omega )=(1-\frac{3\alpha }{2}-\frac{\alpha }{2}\omega \frac{{\rm{d}}}{{\rm{d}}\omega })(\frac{1}{\left.{\rm{\Gamma }}(1-\alpha \right)}\\ \,\times {\displaystyle \int }_{1}^{\infty }{\left(s-1\right)}^{-\alpha }{s}^{\frac{3\alpha }{2}-2}f(\omega {s}^{\frac{\alpha }{2}}){\rm{d}}s)\\ \,=(1-\frac{3\alpha }{2}-\frac{\alpha }{2}\omega \frac{{\rm{d}}}{{\rm{d}}\omega })(\frac{1}{\left.{\rm{\Gamma }}(1-\alpha \right)}{\displaystyle \int }_{1}^{\infty }{\left(s-1\right)}^{-\alpha }{s}^{\frac{3\alpha }{2}-2}\\ \,\times \displaystyle \sum _{k=0}^{\infty }({a}_{k}{\omega }^{k}{s}^{\frac{k\alpha }{2}}){\rm{d}}s)\\ \,=(1-\frac{3\alpha }{2}-\frac{\alpha }{2}\omega \frac{{\rm{d}}}{{\rm{d}}\omega })(\displaystyle \sum _{k=0}^{\infty }{a}_{k}{\omega }^{k}\frac{1}{\left.{\rm{\Gamma }}(1-\alpha \right)}\\ \,\times {\displaystyle \int }_{1}^{\infty }{\left(s-1\right)}^{-\alpha }{s}^{\frac{k+3}{2}\alpha -2}{\rm{d}}s)\\ \,=(1-\frac{3\alpha }{2}-\frac{\alpha }{2}\omega \frac{{\rm{d}}}{{\rm{d}}\omega })\left(\displaystyle \sum _{k=0}^{\infty }{a}_{k}{\omega }^{k}\frac{B\left(1-\frac{\left(k+1)\alpha \right.}{2},1-\alpha \right)}{\left.{\rm{\Gamma }}(1-\alpha \right)}\right)\\ \,=(1-\frac{3\alpha }{2}-\frac{\alpha }{2}\omega \frac{{\rm{d}}}{{\rm{d}}\omega })(\displaystyle \sum _{k=0}^{\infty }\frac{{\rm{\Gamma }}\left(1-\frac{\left(k+1)\alpha \right.}{2}\right)}{{\rm{\Gamma }}\left(2-\frac{\left(k+3)\alpha \right.}{2}\right)}{a}_{k}{\omega }^{k})\\ \,=\displaystyle \sum _{k=0}^{\infty }\frac{{\rm{\Gamma }}\left(1-\frac{\left(k+1)\alpha \right.}{2}\right)}{{\rm{\Gamma }}\left(1-\frac{\left(k+3)\alpha \right.}{2}\right)}{a}_{k}{\omega }^{k}.\end{array}\end{eqnarray}$
Similarly, $\begin{eqnarray}\begin{array}{l}({{ \mathcal P }}_{\frac{2}{\alpha },\frac{2}{\alpha }}^{1-\frac{3\alpha }{2},\alpha }g)({\omega }_{1},{\omega }_{2})=({{ \mathcal P }}_{\frac{2}{\alpha }}^{1-\frac{3\alpha }{2},\alpha }g)(\omega )\\ \quad =\displaystyle \sum _{k=0}^{\infty }\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{b}_{k}{\omega }^{k}.\end{array}\end{eqnarray}$
Substituting equations ( $\begin{eqnarray*}\begin{array}{l}\displaystyle \sum _{k=0}^{\infty }\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{a}_{k}{\omega }^{k}=-({\gamma }_{1}{C}_{1}^{2}+{\gamma }_{2}{C}_{2}^{2}+{\gamma }_{3}{C}_{1}{C}_{2})\\ \quad \times \displaystyle \sum _{k=0}^{\infty }(k+2)(k+1){b}_{k+2}{\omega }^{k}\\ \quad +{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{b}_{n}+\displaystyle \sum _{l+m+n=k}^{}{b}_{l}{b}_{m}{b}_{n}){\omega }^{k},\\ \displaystyle \sum _{k=0}^{\infty }\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{b}_{k}{\omega }^{k}=({\gamma }_{1}{C}_{1}^{2}+{\gamma }_{2}{C}_{2}^{2}+{\gamma }_{3}{C}_{1}{C}_{2})\\ \quad \times \displaystyle \sum _{k=0}^{\infty }(k+2)(k+1){a}_{k+2}{\omega }^{k}\\ \quad -{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{a}_{n}+\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{b}_{m}{b}_{n}){\omega }^{k}.\end{array}\end{eqnarray*}$
Next, we equate the coefficients of different powers of ω to obtain the explicit expressions of ak and bk. For k≥0, we have $\begin{eqnarray}\left\{\begin{array}{l}{a}_{k+2}=\frac{1}{({\gamma }_{1}{C}_{1}^{2}+{\gamma }_{2}{C}_{2}^{2}+{\gamma }_{3}{C}_{1}{C}_{2})(k+2)(k+1)}\left[\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{b}_{k}\right.\\ \left.+{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{a}_{n}+\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{b}_{m}{b}_{n})\right],\\ {b}_{k+2}=\frac{-1}{({\gamma }_{1}{C}_{1}^{2}+{\gamma }_{2}{C}_{2}^{2}+{\gamma }_{3}{C}_{1}{C}_{2})(k+2)(k+1)}\left[\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{a}_{k}\right.\\ \left.-{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{b}_{n}+\displaystyle \sum _{l+m+n=k}^{}{b}_{l}{b}_{m}{b}_{n})\right],\end{array}\right.\end{eqnarray}$
with the initial conditions a0 = f(0), b0 = g(0), ${a}_{1}=f^{\prime} (0)$ and ${b}_{1}=g^{\prime} (0)$. Therefore, the power series solutions of equation ( $\begin{eqnarray}\begin{array}{l}u(t,x,y)={a}_{0}{t}^{-\frac{\alpha }{2}}+{a}_{1}({C}_{1}x+{C}_{2}y){t}^{-\alpha }\\ \quad +\displaystyle \sum _{k=0}^{\infty }\frac{{({C}_{1}x+{C}_{2}y)}^{k+2}{t}^{-\frac{(k+3)\alpha }{2}}}{({\gamma }_{1}{C}_{1}^{2}+{\gamma }_{2}{C}_{2}^{2}+{\gamma }_{3}{C}_{1}{C}_{2})(k+2)(k+1)}\\ \quad \times \,\left[\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{b}_{k}+{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{a}_{n}+\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{b}_{m}{b}_{n})\right],\\ \quad v(t,x,y)={b}_{0}{t}^{-\frac{\alpha }{2}}+{b}_{1}({C}_{1}x+{C}_{2}y){t}^{-\alpha }\\ \quad -\displaystyle \sum _{k=0}^{\infty }\frac{{({C}_{1}x+{C}_{2}y)}^{k+2}{t}^{-\frac{(k+3)\alpha }{2}}}{({\gamma }_{1}{C}_{1}^{2}+{\gamma }_{2}{C}_{2}^{2}+{\gamma }_{3}{C}_{1}{C}_{2})(k+2)(k+1)}\\ \quad \times \,\left[\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{a}_{k}-{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{b}_{n}+\displaystyle \sum _{l+m+n=k}^{}{b}_{l}{b}_{m}{b}_{n})\right].\end{array}\end{eqnarray}$
The power series solutions, equation (
From equation (
$\begin{eqnarray}\left\{\begin{array}{l}| {a}_{k+2}| \leqslant \frac{1}{| {\gamma }_{1}{C}_{1}^{2}+{\gamma }_{2}{C}_{2}^{2}+{\gamma }_{3}{C}_{1}{C}_{2}| (k+2)(k+1)}\left[\frac{\left|{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})\right|}{\left|{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})\right|}| {b}_{k}| \right.\\ \left.+| {\gamma }_{4}| (\displaystyle \sum _{l+m+n=k}^{}| {a}_{l}\parallel {a}_{m}\parallel {a}_{n}| +\displaystyle \sum _{l+m+n=k}^{}| {a}_{l}\parallel {b}_{m}\parallel {b}_{n}| )\right],\\ | {b}_{k+2}| \leqslant \frac{1}{| {\gamma }_{1}{C}_{1}^{2}+{\gamma }_{2}{C}_{2}^{2}+{\gamma }_{3}{C}_{1}{C}_{2}| (k+2)(k+1)}\left[\frac{\left|{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})\right|}{\left|{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})\right|}| {a}_{k}| \right.\\ \left.+| {\gamma }_{4}| (\displaystyle \sum _{l+m+n=k}^{}| {a}_{l}\parallel {a}_{m}\parallel {b}_{n}| +\displaystyle \sum _{l+m+n=k}^{}| {b}_{l}\parallel {b}_{m}\parallel {b}_{n}| )\right].\end{array}\right.\end{eqnarray}$
From the properties of the gamma function, it can be found that $\frac{\left|{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})\right|}{\left|{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})\right|}\leqslant 1$ for arbitrary k. Thus, equation ( $\begin{eqnarray}\begin{array}{l}| {a}_{k+2}| \leqslant M\left(\left|{b}_{k}| +\displaystyle \sum _{l+m+n=k}^{}| {a}_{l}\parallel {a}_{m}\parallel {a}_{n}\right|\right.\\ \quad \left.+\displaystyle \sum _{l+m+n=k}^{}| {a}_{l}\parallel {b}_{m}\parallel {b}_{n}| \right),\,\,k\geqslant 0,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}| {b}_{k+2}| \leqslant M\left(\left|{a}_{k}| +\displaystyle \sum _{l+m+n=k}^{}| {a}_{l}\parallel {a}_{m}\parallel {b}_{n}\right|\right.\\ \quad \left.+\displaystyle \sum _{l+m+n=k}^{}| {b}_{l}\parallel {b}_{m}\parallel {b}_{n}| \right),\,\,k\geqslant 0,\end{array}\end{eqnarray}$
where $M={\rm{\max }}\{\frac{1}{| {\gamma }_{1}{C}_{1}^{2}+{\gamma }_{2}{C}_{2}^{2}+{\gamma }_{3}{C}_{1}{C}_{2}| (k+2)(k+1)},\frac{| {\gamma }_{4}| }{| {\gamma }_{1}{C}_{1}^{2}+{\gamma }_{2}{C}_{2}^{2}+{\gamma }_{3}{C}_{1}{C}_{2}| (k+2)(k+1)}\}$. Consider another power series
$\begin{eqnarray}P(\omega )=\displaystyle \sum _{k=0}^{\infty }{p}_{k}{\omega }^{k},\,\,Q(\omega )=\displaystyle \sum _{k=0}^{\infty }{q}_{k}{\omega }^{k},\end{eqnarray}$
where p0 = ∣a0∣, p1 = ∣a1∣, q0 = ∣b0∣, q1 = ∣b1∣ and $\begin{eqnarray}\begin{array}{l}{p}_{k+2}=M\left({q}_{k}+\displaystyle \sum _{l+m+n=k}^{}{p}_{l}{p}_{m}{p}_{n}+\displaystyle \sum _{l+m+n=k}^{}{p}_{l}{q}_{m}{q}_{n}\right),\\ k\geqslant 0,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{q}_{k+2}=M\left({p}_{k}+\displaystyle \sum _{l+m+n=k}^{}{p}_{l}{p}_{m}{q}_{n}+\displaystyle \sum _{l+m+n=k}^{}{q}_{l}{q}_{m}{q}_{n}\right),\\ k\geqslant 0.\end{array}\end{eqnarray}$
Therefore, it is easily seen that ∣ak∣≤pk and ∣bk∣≤qk for k = 0, 1, 2, …, that is, the power series, equation ( $\begin{eqnarray}P(\omega )={p}_{0}+{p}_{1}\omega +M\left(\right.Q(\omega )+{P}^{3}(\omega )+P(\omega ){Q}^{2}(\omega )\left)\right.{\omega }^{2},\end{eqnarray}$
$\begin{eqnarray}Q(\omega )={q}_{0}+{q}_{1}\omega +M\left(\right.P(\omega )+{Q}^{3}(\omega )+Q(\omega ){P}^{2}(\omega )\left)\right.{\omega }^{2}.\end{eqnarray}$
Consider the following implicit function with respect to the independent variable ω: $\begin{eqnarray}\begin{array}{l}F(\omega ,P,Q)=P-{P}_{0}-{P}_{1}\omega \\ \quad -M\left(\right.Q(\omega )+{P}^{3}(\omega )+P(\omega ){Q}^{2}(\omega )\left)\right.{\omega }^{2},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}G(\omega ,P,Q)=Q-{q}_{0}-{q}_{1}\omega \\ \quad -M\left(\right.P(\omega )+{Q}^{3}(\omega )+Q(\omega ){P}^{2}(\omega )\left)\right.{\omega }^{2},\end{array}\end{eqnarray}$
which are analytic in a neighborhood of (0, p0, q0), and F(0, p0, q0) = 0, G(0, p0, q0) = 0, $\frac{\partial (F,G)}{\partial (P,Q)}{| }_{(0,{p}_{0},{q}_{0})}=1\ne 0$. Therefore, via the implicit function theorem, the power series, equation (In figure 1, we illustrate the physical features of the power series solutions, equation (35 ), for different chosen parameters. Assuming a0 = b0 = a1 = b1 = 1, ${C}_{1}={C}_{2}=\frac{1}{2}$ and γi = 1(i = 1, 2, 3, 4), these graphs show that as the fractional order changes slightly, the solutions, equation (35 ), exhibit subtle differences for the initial values and constants C1, C2 and γi. Therefore, the fractional form is more suitable for adjusting the evolution of the model in practical situations based on changes in order α.
Figure 1. Dynamical profiles of the truncated power series solution, equation ( |
Case 2: ${X}_{3}=\frac{{\rm{\partial }}}{{\rm{\partial }}x}.$
From the characteristic equation corresponding to the group generator X3, i.e.3 ) as follows: 47 ) into equation (3 ), we have the following reduced equations: 48 ) to fractional ordinary differential equations using the Lie symmetry analysis method again, and obtain the following convergent power series solutions: 35 ) when C1 = 0 and C2 = 1.
$\begin{eqnarray}\frac{{\rm{d}}t}{0}=\frac{{\rm{d}}x}{1}=\frac{{\rm{d}}y}{0}=\frac{{\rm{d}}u}{0}=\frac{{\rm{d}}v}{0},\end{eqnarray}$
we obtain the similarity variables t, y, u and v. Therefore, we get the invariant solutions of equation ( $\begin{eqnarray}u=f(t,y),\,\,v=g(t,y).\end{eqnarray}$
Substituting equation ( $\begin{eqnarray}\left\{\begin{array}{l}{D}_{t}^{\alpha }f=-{\gamma }_{2}{g}_{yy}+{\gamma }_{4}g({f}^{2}+{g}^{2}),\\ {D}_{t}^{\alpha }g={\gamma }_{2}{f}_{yy}-{\gamma }_{4}f({f}^{2}+{g}^{2}),\end{array}\right.\end{eqnarray}$
which is the time-fractional cubic Schrödinger equation [40] with γ2 = 1 and γ4 = − γ. Similar to [40], we can further reduce equation ( $\begin{eqnarray}\begin{array}{l}u(t,y)={a}_{0}{t}^{-\frac{\alpha }{2}}+{a}_{1}y{t}^{-\alpha }+\displaystyle \sum _{k=0}^{\infty }\frac{{y}^{k+2}{t}^{-\frac{(k+3)\alpha }{2}}}{{\gamma }_{2}(k+2)(k+1)}\\ \quad \times \,\left[\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{b}_{k}+{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{a}_{n}+\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{b}_{m}{b}_{n})\right],\\ v(t,y)={b}_{0}{t}^{-\frac{\alpha }{2}}+{b}_{1}y{t}^{-\alpha }-\displaystyle \sum _{k=0}^{\infty }\frac{{y}^{k+2}{t}^{-\frac{(k+3)\alpha }{2}}}{{\gamma }_{2}(k+2)(k+1)}\\ \quad \times \,\left[\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{a}_{k}-{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{b}_{n}+\displaystyle \sum _{l+m+n=k}^{}{b}_{l}{b}_{m}{b}_{n})\right],\end{array}\end{eqnarray}$
which is consistent with equation (Case 3: ${X}_{4}=\frac{{\rm{\partial }}}{{\rm{\partial }}y}.$
From the characteristic equation corresponding to the group generator X4, i.e.3 ) as follows: 51 ) into equation (3 ), we have the following reduced equations: 35 ) when C1 = 1 and C2 = 0.
$\begin{eqnarray}\frac{{\rm{d}}t}{0}=\frac{{\rm{d}}x}{0}=\frac{{\rm{d}}y}{1}=\frac{{\rm{d}}u}{0}=\frac{{\rm{d}}v}{0},\end{eqnarray}$
we obtain the similarity variables t, x, u and v. Therefore, we get the invariant solutions of equation ( $\begin{eqnarray}u=f(t,x),\,\,v=g(t,x).\end{eqnarray}$
Substituting equation ( $\begin{eqnarray}\left\{\begin{array}{l}{D}_{t}^{\alpha }f=-{\gamma }_{1}{g}_{xx}+{\gamma }_{4}g({f}^{2}+{g}^{2}),\\ {D}_{t}^{\alpha }g={\gamma }_{1}{f}_{xx}-{\gamma }_{4}f({f}^{2}+{g}^{2}),\end{array}\right.\end{eqnarray}$
which is the time-fractional cubic Schrödinger equation [40] with γ1 = 1 and γ4 = − γ. Similar to case 2, we can obtain the following convergent power series solutions: $\begin{eqnarray}\begin{array}{l}u(t,x)={a}_{0}{t}^{-\frac{\alpha }{2}}+{a}_{1}x{t}^{-\alpha }+\displaystyle \sum _{k=0}^{\infty }\frac{{x}^{k+2}{t}^{-\frac{(k+3)\alpha }{2}}}{{\gamma }_{1}(k+2)(k+1)}\\ \quad \times \,\left[\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{b}_{k}+{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{a}_{n}+\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{b}_{m}{b}_{n})\right],\\ v(t,x)={b}_{0}{t}^{-\frac{\alpha }{2}}+{b}_{1}x{t}^{-\alpha }-\displaystyle \sum _{k=0}^{\infty }\frac{{x}^{k+2}{t}^{-\frac{(k+3)\alpha }{2}}}{{\gamma }_{1}(k+2)(k+1)}\\ \quad \times \,\left[\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{a}_{k}-{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{b}_{n}+\displaystyle \sum _{l+m+n=k}^{}{b}_{l}{b}_{m}{b}_{n})\right],\end{array}\end{eqnarray}$
which is consistent with equation (Case 4: ${X}_{3}-{X}_{4}=\frac{{\rm{\partial }}}{{\rm{\partial }}x}-\frac{{\rm{\partial }}}{{\rm{\partial }}y}.$
The characteristic equation corresponding to the group generator X3 − X4 is 3 ) 35 ) when C1 = 1 and C2 = 1.
$\begin{eqnarray}\frac{{\rm{d}}t}{0}=\frac{{\rm{d}}x}{1}=\frac{{\rm{d}}y}{-1}=\frac{{\rm{d}}u}{0}=\frac{{\rm{d}}v}{0},\end{eqnarray}$
from which, we obtain the similarity variables t, x + y, u and v. Therefore, we get the invariant solutions of equation ( $\begin{eqnarray}\begin{array}{rcl}u(t,x,y) & = & f({\omega }_{1},{\omega }_{2}),\,\,v(t,x,y)=g({\omega }_{1},{\omega }_{2}),\\ {\omega }_{1} & = & t,\,\,{\omega }_{2}=x+y,\end{array}\end{eqnarray}$
and the reduced equations $\begin{eqnarray}\left\{\begin{array}{l}{D}_{{\omega }_{1}}^{\alpha }f=-({\gamma }_{1}+{\gamma }_{2}+{\gamma }_{3}){g}_{{\omega }_{2}{\omega }_{2}}+{\gamma }_{4}g({f}^{2}+{g}^{2}),\\ {D}_{{\omega }_{1}}^{\alpha }g=({\gamma }_{1}+{\gamma }_{2}+{\gamma }_{3}){f}_{{\omega }_{2}{\omega }_{2}}-{\gamma }_{4}f({f}^{2}+{g}^{2}).\end{array}\right.\end{eqnarray}$
Similar to cases 2 and 3, we can apply the Lie symmetry analysis method again to obtain the following convergent power series solutions: $\begin{eqnarray}\begin{array}{l}u(t,x,y)={a}_{0}{t}^{-\frac{\alpha }{2}}+{a}_{1}(x+y){t}^{-\alpha }+\displaystyle \sum _{k=0}^{\infty }\frac{{(x+y)}^{k+2}{t}^{-\frac{(k+3)\alpha }{2}}}{({\gamma }_{1}+{\gamma }_{2}+{\gamma }_{3})(k+2)(k+1)}\\ \quad \times \,\left[\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{b}_{k}+{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{a}_{n}+\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{b}_{m}{b}_{n})\right],\\ v(t,x,y)={b}_{0}{t}^{-\frac{\alpha }{2}}+{b}_{1}(x+y){t}^{-\alpha }-\displaystyle \sum _{k=0}^{\infty }\frac{{(x+y)}^{k+2}{t}^{-\frac{(k+3)\alpha }{2}}}{({\gamma }_{1}+{\gamma }_{2}+{\gamma }_{3})(k+2)(k+1)}\\ \quad \times \,\left[\frac{{\rm{\Gamma }}(1-\frac{(k+1)\alpha }{2})}{{\rm{\Gamma }}(1-\frac{(k+3)\alpha }{2})}{a}_{k}-{\gamma }_{4}(\displaystyle \sum _{l+m+n=k}^{}{a}_{l}{a}_{m}{b}_{n}+\displaystyle \sum _{l+m+n=k}^{}{b}_{l}{b}_{m}{b}_{n})\right],\end{array}\end{eqnarray}$
which is consistent with equation (4. Conservation laws of equation (3 )
We denote equation (3 ) as 58 ) are given by
$\begin{eqnarray}\left\{\begin{array}{l}{F}_{1}={D}_{t}^{\alpha }u+{\gamma }_{1}{v}_{xx}+{\gamma }_{2}{v}_{yy}+{\gamma }_{3}{v}_{xy}-{\gamma }_{4}v({u}^{2}+{v}^{2})=0,\\ {F}_{2}={D}_{t}^{\alpha }v-{\gamma }_{1}{u}_{xx}-{\gamma }_{2}{u}_{yy}-{\gamma }_{3}{u}_{xy}+{\gamma }_{4}u({u}^{2}+{v}^{2})=0,\end{array}\right.\end{eqnarray}$
of which the formal Lagrangian is given by $\begin{eqnarray}\begin{array}{rcl}{ \mathcal L } & = & p(t,x,y){F}_{1}+q(t,x,y){F}_{2}\\ & = & p(t,x,y)\left({D}_{t}^{\alpha }u+{\gamma }_{1}{v}_{xx}+{\gamma }_{2}{v}_{yy}+{\gamma }_{3}{v}_{xy}\right.\\ & & \left.-{\gamma }_{4}v({u}^{2}+{v}^{2})\right)\\ & & +q(t,x,y)\left({D}_{t}^{\alpha }v-{\gamma }_{1}{u}_{xx}-{\gamma }_{2}{u}_{yy}\right.\\ & & \left.-{\gamma }_{3}{u}_{xy}+{\gamma }_{4}u({u}^{2}+{v}^{2})\right),\end{array}\end{eqnarray}$
where p(t, x, y) and q(t, x, y) are new dependent variables. The Euler–Lagrange operators are $\begin{eqnarray}\begin{array}{rcl}\frac{\delta }{\delta u} & = & \frac{\partial }{\partial u}+{({D}_{t}^{\alpha })}^{* }\frac{\partial }{\partial ({D}_{t}^{\alpha }u)}\\ & & +\displaystyle \sum _{s=1}^{\infty }{(-1)}^{s}{D}_{{i}_{1}}\cdots {D}_{{i}_{s}}\frac{\partial }{\partial {u}_{{i}_{1}\cdots {i}_{s}}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}\frac{\delta }{\delta v} & = & \frac{\partial }{\partial v}+{({D}_{t}^{\alpha })}^{* }\frac{\partial }{\partial ({D}_{t}^{\alpha }v)}\\ & & +\displaystyle \sum _{s=1}^{\infty }{(-1)}^{s}{D}_{{i}_{1}}\cdots {D}_{{i}_{s}}\frac{\partial }{\partial {v}_{{i}_{1}\cdots {i}_{s}}},\end{array}\end{eqnarray}$
where ${({D}_{t}^{\alpha })}^{* }$ is the adjoint operator of ${D}_{t}^{\alpha }$. It is defined by the right-side of the Caputo fractional derivative, i.e. $\begin{eqnarray*}\begin{array}{l}{({D}_{t}^{\alpha })}^{* }f(t,x){\equiv }_{t}^{c}{D}_{T}^{\alpha }f(t,x)\\ \quad =\left\{\begin{array}{ll}\frac{1}{{\rm{\Gamma }}(n-\alpha )}{\displaystyle \int }_{t}^{T}\frac{1}{{(t-s)}^{\alpha -n+1}}\frac{{\partial }^{n}}{\partial {s}^{n}}f(s,x){\rm{d}}s, & n-1\lt \alpha \lt n,\\ {D}_{t}^{n}f(t,x), & \alpha =n\in {\mathbb{N}}.\end{array}\right.\end{array}\end{eqnarray*}$
The adjoint equations to equation ( $\begin{eqnarray}\left\{\begin{array}{l}{F}_{1}^{* }=\frac{\delta { \mathcal L }}{\delta u}={({D}_{t}^{\alpha })}^{* }p+{\gamma }_{1}{q}_{xx}+{\gamma }_{2}{q}_{yy}+{\gamma }_{3}{q}_{xy}-{\gamma }_{4}(2puv-3q{u}^{2}-q{v}^{2})=0,\\ {F}_{2}^{* }=\frac{\delta { \mathcal L }}{\delta v}={({D}_{t}^{\alpha })}^{* }q-{\gamma }_{1}{p}_{xx}-{\gamma }_{2}{p}_{yy}-{\gamma }_{3}{p}_{xy}+{\gamma }_{4}(2quv-3p{v}^{2}-p{u}^{2})=0.\end{array}\right.\end{eqnarray}$
Next, we will use the above adjoint equations and the new conservation theorem [49] to construct conservation laws of equation (3 ). From the classical definition of the conservation laws, a vector C = (Ct, Cx, Cy) is called the conserved vector for the governing equation if it satisfies the conservation equation ${[{D}_{t}{C}^{t}+{D}_{x}{C}^{x}+{D}_{y}{C}^{y}]}_{{F}_{1},{F}_{2}=0}=0$. We can obtain the components of the conserved vector using the generalization of the Noether operators [50]. From the fundamental operator identity, i.e.6 ), ${ \mathcal I }$ is the identity operator, and Wu = η − τut − ξux − θuy, Wv = ζ − τvt − ξvx − θuy are the characteristics for the group generator X. We can get the Noether operators as follows:
$\begin{eqnarray}\begin{array}{l}{\rm{P}}{{\rm{r}}}^{(\alpha ,2)}X+{D}_{t}\tau \cdot { \mathcal I }+{D}_{x}\xi \cdot { \mathcal I }+{D}_{y}\theta \cdot { \mathcal I }\\ \quad ={W}^{u}\cdot \frac{\delta }{\delta u}+{W}^{v}\cdot \frac{\delta }{\delta v}+{D}_{t}{{ \mathcal N }}^{t}+{D}_{x}{{ \mathcal N }}^{x}+{D}_{y}{{ \mathcal N }}^{y},\end{array}\end{eqnarray}$
where Pr(α,2)X is mentioned in equation ( $\begin{eqnarray}\begin{array}{rcl}{{ \mathcal N }}^{t} & = & \tau { \mathcal I }+\displaystyle \sum _{k=0}^{n-1}{(-1)}^{k}{D}_{t}^{\alpha -1-k}({W}^{u}){D}_{t}^{k}\frac{\partial }{\partial ({D}_{t}^{\alpha }u)}\\ & & -{(-1)}^{n}J({W}^{u},{D}_{t}^{n}\frac{\partial }{\partial ({D}_{t}^{\alpha }u)})\\ & & +\displaystyle \sum _{k=0}^{n-1}{(-1)}^{k}{D}_{t}^{\alpha -1-k}({W}^{v}){D}_{t}^{k}\frac{\partial }{\partial ({D}_{t}^{\alpha }v)}\\ & & -{(-1)}^{n}J({W}^{v},{D}_{t}^{n}\frac{\partial }{\partial ({D}_{t}^{\alpha }v)}),\,\,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{{ \mathcal N }}^{x} & = & \xi { \mathcal I }-{W}^{u}({D}_{x}\frac{\partial }{\partial {u}_{xx}}+{D}_{y}\frac{\partial }{\partial {u}_{xy}})\\ & & -{W}^{v}({D}_{x}\frac{\partial }{\partial {v}_{xx}}+{D}_{y}\frac{\partial }{\partial {v}_{xy}})\\ & & +{D}_{x}({W}^{u})\frac{\partial }{\partial {u}_{xx}}+{D}_{y}({W}^{u})\frac{\partial }{\partial {u}_{xy}}\\ & & +{D}_{x}({W}^{v})\frac{\partial }{\partial {v}_{xx}}+{D}_{y}({W}^{v})\frac{\partial }{\partial {v}_{xy}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{{ \mathcal N }}^{y} & = & \theta { \mathcal I }-{W}^{u}({D}_{y}\frac{\partial }{\partial {u}_{yy}}+{D}_{x}\frac{\partial }{\partial {u}_{yx}})\\ & & -{W}^{v}({D}_{y}\frac{\partial }{\partial {v}_{yy}}+{D}_{x}\frac{\partial }{\partial {v}_{yx}})\\ & & +{D}_{y}({W}^{u})\frac{\partial }{\partial {u}_{yy}}+{D}_{x}({W}^{u})\frac{\partial }{\partial {u}_{yx}}\\ & & +{D}_{y}({W}^{v})\frac{\partial }{\partial {v}_{yy}}+{D}_{x}({W}^{v})\frac{\partial }{\partial {v}_{yx}},\end{array}\end{eqnarray}$
where n = [α] + 1, and J is given by $\begin{eqnarray}J(f,g)=\frac{1}{{\rm{\Gamma }}(n-\alpha )}{\int }_{0}^{t}{\int }_{t}^{T}\frac{f(\tau ,x)g(\theta ,x)}{{(\theta -\tau )}^{\alpha +1-n}}{\rm{d}}\theta {\rm{d}}\tau .\end{eqnarray}$
The components of the conserved vector are defined by $\begin{eqnarray}{C}^{t}={{ \mathcal N }}^{t}{ \mathcal L },\,\,{C}^{x}={{ \mathcal N }}^{x}{ \mathcal L },\,\,{C}^{y}={{ \mathcal N }}^{y}{ \mathcal L }.\end{eqnarray}$
Case 1: ${X}_{1}=t\frac{{\rm{\partial }}}{{\rm{\partial }}t}+\frac{\alpha }{2}x\frac{{\rm{\partial }}}{{\rm{\partial }}x}+\frac{\alpha }{2}y\frac{{\rm{\partial }}}{{\rm{\partial }}y}-\frac{\alpha }{2}u\frac{{\rm{\partial }}}{{\rm{\partial }}u}-\frac{\alpha }{2}v\frac{{\rm{\partial }}}{{\rm{\partial }}v}.$
The characteristics of X1 are
$\begin{eqnarray}\begin{array}{rcl}{W}^{u} & = & -\frac{\alpha }{2}u-t{u}_{t}-\frac{\alpha }{2}x{u}_{x}-\frac{\alpha }{2}y{u}_{y},\\ {W}^{v} & = & -\frac{\alpha }{2}v-t{v}_{t}-\frac{\alpha }{2}x{v}_{x}-\frac{\alpha }{2}y{v}_{y}.\end{array}\end{eqnarray}$
Therefore, for 0 < α < 1, $\begin{eqnarray}\begin{array}{rcl}{C}^{t} & = & -p{D}_{t}^{\alpha -1}(\frac{\alpha }{2}u+t{u}_{t}+\frac{\alpha }{2}x{u}_{x}+\frac{\alpha }{2}y{u}_{y})\\ & & -J(\frac{\alpha }{2}u+t{u}_{t}+\frac{\alpha }{2}x{u}_{x}+\frac{\alpha }{2}y{u}_{y},{p}_{t})\\ & & -q{D}_{t}^{\alpha -1}(\frac{\alpha }{2}v+t{v}_{t}+\frac{\alpha }{2}x{v}_{x}+\frac{\alpha }{2}y{v}_{y})\\ & & -J(\frac{\alpha }{2}v+t{v}_{t}+\frac{\alpha }{2}x{v}_{x}+\frac{\alpha }{2}y{v}_{y},{q}_{t}),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{C}^{x} & = & -({\gamma }_{1}{q}_{x}+{\gamma }_{3}{q}_{y})(\frac{\alpha }{2}u+t{u}_{t}+\frac{\alpha }{2}x{u}_{x}+\frac{\alpha }{2}y{u}_{y})\\ & & +({\gamma }_{1}{p}_{x}+{\gamma }_{3}{p}_{y})(\frac{\alpha }{2}v+t{v}_{t}+\frac{\alpha }{2}x{v}_{x}+\frac{\alpha }{2}y{v}_{y})\\ & & +{\gamma }_{1}q(\alpha {u}_{x}+t{u}_{tx}+\frac{\alpha }{2}{u}_{xx}+\frac{\alpha }{2}y{u}_{xy})\\ & & +{\gamma }_{3}q(\alpha {u}_{y}+t{u}_{ty}+\frac{\alpha }{2}x{u}_{xy}+\frac{\alpha }{2}y{u}_{yy})\\ & & -{\gamma }_{1}p(\alpha {v}_{x}+t{v}_{tx}+\frac{\alpha }{2}{v}_{xx}+\frac{\alpha }{2}y{v}_{xy})\\ & & -{\gamma }_{3}p(\alpha {v}_{y}+t{v}_{ty}+\frac{\alpha }{2}x{v}_{xy}+\frac{\alpha }{2}y{v}_{yy}),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{C}^{y} & = & -({\gamma }_{1}{q}_{y}+{\gamma }_{3}{q}_{x})(\frac{\alpha }{2}u+t{u}_{t}+\frac{\alpha }{2}x{u}_{x}+\frac{\alpha }{2}y{u}_{y})\\ & & +({\gamma }_{1}{p}_{y}+{\gamma }_{3}{p}_{x})(\frac{\alpha }{2}v+t{v}_{t}+\frac{\alpha }{2}x{v}_{x}+\frac{\alpha }{2}y{v}_{y})\\ & & +{\gamma }_{1}q(\alpha {u}_{y}+t{u}_{ty}+\frac{\alpha }{2}x{u}_{xy}+\frac{\alpha }{2}y{u}_{yy})\\ & & +{\gamma }_{3}q(\alpha {u}_{x}+t{u}_{tx}+\frac{\alpha }{2}{u}_{xx}+\frac{\alpha }{2}y{u}_{xy})\\ & & -{\gamma }_{1}p(\alpha {v}_{y}+t{v}_{ty}+\frac{\alpha }{2}x{v}_{xy}+\frac{\alpha }{2}y{v}_{yy})\\ & & -{\gamma }_{3}p(\alpha {v}_{x}+t{v}_{tx}+\frac{\alpha }{2}{v}_{xx}+\frac{\alpha }{2}y{v}_{xy}).\end{array}\end{eqnarray}$
Case 2: ${X}_{2}=v\frac{{\rm{\partial }}}{{\rm{\partial }}u}-u\frac{{\rm{\partial }}}{{\rm{\partial }}v}.$
The characteristics of X2 are
$\begin{eqnarray}{W}^{u}=v,\,\,{W}^{v}=-u.\end{eqnarray}$
Therefore, for 0 < α < 1, $\begin{eqnarray}{C}^{t}=p{D}_{t}^{\alpha -1}v+J(v,{p}_{t})-q{D}_{t}^{\alpha -1}u-J(u,{q}_{t}),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{C}^{x} & = & v({\gamma }_{1}{q}_{x}+{\gamma }_{3}{q}_{y})+u({\gamma }_{1}{p}_{x}+{\gamma }_{3}{p}_{y})\\ & & -{\gamma }_{1}q{v}_{x}-{\gamma }_{3}q{v}_{y}-{\gamma }_{1}p{u}_{x}-{\gamma }_{3}p{u}_{y},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{C}^{y} & = & v({\gamma }_{1}{q}_{y}+{\gamma }_{3}{q}_{x})+u({\gamma }_{1}{p}_{x}+{\gamma }_{3}{p}_{y})\\ & & -{\gamma }_{1}q{v}_{y}-{\gamma }_{3}q{v}_{x}-{\gamma }_{1}p{u}_{y}-{\gamma }_{3}p{u}_{x}.\end{array}\end{eqnarray}$
Case 3: ${X}_{3}=\frac{{\rm{\partial }}}{{\rm{\partial }}x}.$
The characteristics of X3 are
$\begin{eqnarray}{W}^{u}=-{u}_{x},\,\,{W}^{v}=-{v}_{x}.\end{eqnarray}$
Therefore, for 0 < α < 1, $\begin{eqnarray}{C}^{t}=-p{D}_{t}^{\alpha -1}{u}_{x}-J({u}_{x},{p}_{t})-q{D}_{t}^{\alpha -1}{v}_{x}-J({v}_{x},{q}_{t}),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{C}^{x} & = & -{u}_{x}({\gamma }_{1}{q}_{x}+{\gamma }_{3}{q}_{y})+{v}_{x}({\gamma }_{1}{p}_{x}+{\gamma }_{3}{p}_{y})\\ & & +{\gamma }_{1}q{u}_{xx}+{\gamma }_{3}q{u}_{xy}-{\gamma }_{1}p{v}_{xx}-{\gamma }_{3}p{v}_{xy},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{C}^{y} & = & -{u}_{x}({\gamma }_{1}{q}_{y}+{\gamma }_{3}{q}_{x})+{v}_{x}({\gamma }_{1}{p}_{x}+{\gamma }_{3}{p}_{y})\\ & & +{\gamma }_{1}q{u}_{xy}+{\gamma }_{3}q{u}_{xx}-{\gamma }_{1}p{v}_{xy}-{\gamma }_{3}p{v}_{xx}.\end{array}\end{eqnarray}$
Case 4: ${X}_{4}=\frac{{\rm{\partial }}}{{\rm{\partial }}y}.$
The characteristics of X4 are
$\begin{eqnarray}{W}^{u}=-{u}_{y},\,\,{W}^{v}=-{v}_{y}.\end{eqnarray}$
Therefore, for 0 < α < 1, $\begin{eqnarray}{C}^{t}=-p{D}_{t}^{\alpha -1}{u}_{y}-J({u}_{y},{p}_{t})-q{D}_{t}^{\alpha -1}{v}_{y}-J({v}_{y},{q}_{t}),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{C}^{x} & = & -{u}_{y}({\gamma }_{1}{q}_{x}+{\gamma }_{3}{q}_{y})+{v}_{y}({\gamma }_{1}{p}_{x}+{\gamma }_{3}{p}_{y})\\ & & +{\gamma }_{1}q{u}_{xy}+{\gamma }_{3}q{u}_{yy}-{\gamma }_{1}p{v}_{xy}-{\gamma }_{3}p{v}_{yy},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{C}^{y} & = & -{u}_{y}({\gamma }_{1}{q}_{y}+{\gamma }_{3}{q}_{x})+{v}_{y}({\gamma }_{1}{p}_{x}+{\gamma }_{3}{p}_{y})\\ & & +{\gamma }_{1}q{u}_{yy}+{\gamma }_{3}q{u}_{xy}-{\gamma }_{1}p{v}_{yy}-{\gamma }_{3}p{v}_{xy}.\end{array}\end{eqnarray}$
5. Conclusion
This paper shows that the Lie symmetry analysis method is effective in solving nonlinear fractional evolution equations. We obtained all the Lie symmetries of the (2+1)-dimensional time-fractional Heisenberg ferromagnetic spin chain equation and used them to reduce the equation, thereby getting some convergent power series solutions. In addition, the new conservation theorem and the generalization of the Noether operators are developed to construct the conservation laws for the equation studied. Inspired by this, our next step is to apply the Lie symmetry analysis method to more nonlinear fractional evolution equations that have significant value in mathematics and physics.
Acknowledgments
This research is partially supported by the State Key Program of the National Natural Science Foundation of China (72031009).
Confict of interest
The authors have no competing interests to declare that are relevant to the content of this article.