This section contains exact solutions of the considered models of fractional order.

### Exact solutions of the space-time fractional Cahn–Allen equation

Cahn–Allen equation appears in numerous applications of science including quantum mechanics, mathematical biology, and plasma physics [33].

Consider the Cahn–Allen equation fractional in space and time

$$ \frac{\partial^{\alpha}u}{\partial t^{\alpha}}-\frac{\partial^{2\alpha }u}{\partial x^{2\alpha}}+u^{3}-u=0, $$

(13)

where \(\alpha\in(0,1)\). Firstly, we apply the R–L definition of fractional derivative. The following transformation is introduced:

$$ \begin{aligned} &\xi=\frac{kx^{\alpha}}{\Gamma(1+\alpha)}+ \frac{ct^{\alpha}}{\Gamma(1+\alpha)}, \\ &u(\xi)=u(x,t), \end{aligned} $$

(14)

where *ξ* is the transformation variable and *k*, *c* are the constants. Using Eq. (14) in Eq. (13), we convert our problem into an ODE:

$$ c\frac{du}{d\xi}-k^{2}\frac{d^{2}u}{d\xi^{2}}+u^{3}-u=0. $$

(15)

Then using Eq. (10), the 2-D autonomous system is attained

$$\begin{aligned} &\frac{dX}{d\xi}=Y, \end{aligned}$$

(16a)

$$\begin{aligned} &\frac{dY}{d\xi}=\frac{1}{k^{2}} \bigl[cY+X^{3}-X \bigr]. \end{aligned}$$

(16b)

Now, to find the first integral of Eqs. (16a)–(16b), we implement the division theorem. In accordance with the FIM, it is assumed that nontrivial solutions of the above system (cf. Eqs. (16a)–(16b)) are *X* and *Y* respectively. Thus, irreducible polynomial \(Q(X,Y)= \sum_{j=0}^{n}a_{j}(X)Y^{j}\) exists in \(\mathbb {C}[X,Y]\) and the following holds:

$$ Q\bigl(X(\xi),Y(\xi)\bigr)= \sum_{j=0}^{n}a_{j} \bigl(X(\xi)\bigr)Y(\xi)^{j}=0 $$

(17)

for \(j=0,1,\ldots,n\), and \(a_{n}(X)\neq0\). Now a polynomial \(r(X)+s(X)Y\) exists in \(\mathbb{C}[X,Y]\), so

$$ \frac{\partial Q}{\partial\xi}=\frac{\partial Q}{\partial X}\frac {\partial X}{\partial\xi}+ \frac{\partial Q}{\partial Y}\frac{\partial Y}{\partial\xi}= \bigl(r(X)+s(X)Y \bigr) \Biggl( \sum _{j=0}^{n}a_{j}(X)Y^{j} \Biggr). $$

(18)

Suppose \(n=1\). On equating coefficients of \(Y^{j}\) (\(j=0,1\)) in Eq. (18) on both sides, we get

$$\begin{aligned}& a_{1}'(X)=a_{1}(X)s(X), \end{aligned}$$

(19)

$$\begin{aligned}& a_{0}'(X)=-\frac{c}{k^{2}}a_{1}(X)+r(X)a_{1}(X)+s(X)a_{0}(X), \end{aligned}$$

(20)

$$\begin{aligned}& r(X)a_{0}(X)=\frac{1}{k^{2}}a_{1}(X) \bigl(X^{3}-X\bigr). \end{aligned}$$

(21)

As \(a_{j}(X)\) are polynomials of *X*, then from Eq. (19) we come to know that the polynomial \(a_{1}(X)\) is constant in nature, therefore \(s(X)=0\). Let us consider \(a_{1}(X)=1\), for convenience. After substituting afore values, we balance the degrees of the functions \(r(X)\) and \(a_{0}(X)\) and deduce the \(\operatorname{deg}(r(X))\) equal to 0 or 1. Assume that \(r(X)=A_{1}X+A_{0}\), therefore Eq. (20) gives

$$ a_{0}(X)=\frac{1}{2}A_{1}X^{2}+ \biggl(A_{0}-\frac{c}{k^{2}} \biggr)X+B. $$

(22)

Here, *B* is the integration constant.

Replacing the values of \(a_{0}\), \(a_{1}\), *r*, and *s* in Eq. (21), we obtain a nonlinear system of algebraic equations by putting all coefficients equal to zero for the same powers of *X*. After calculations, we get:

*Case* 1:

$$ A_{1}=\frac{\sqrt{2}}{k},\qquad A_{0}= \frac{\sqrt{2}}{k},\qquad B=0,\qquad c=\frac{3k\sqrt {2}}{2},\quad k=k. $$

(23)

Applying the conditions given in Eq. (23) and Eq. (22) in Eq. (17), we have

$$ Y_{1}(\xi)=-\frac{\sqrt{2}}{2k} \bigl[X^{2}-X \bigr]. $$

(24)

Combining Eq. (24) with Eq. (16a), the solution of fractional Cahn–Allen equation with R–L derivative is obtained as follows:

$$ u_{1}(x,t)=\frac{1}{1+\gamma e^{\frac{-\sqrt{2}}{2k} (\frac{kx^{\alpha}+ct^{\alpha}}{\Gamma(1+\alpha)} )}}. $$

(25)

*Case* 2:

$$ A_{1}=\frac{\sqrt{2}}{k},\qquad A_{0}=- \frac{\sqrt{2}}{k},\qquad B=0,\qquad c=-\frac{3k\sqrt {2}}{2},\quad k=k. $$

(26)

Applying the conditions given in Eq. (26) and Eq. (22) in Eq. (17), we have

$$ Y_{2}(\xi)=-\frac{\sqrt{2}}{2k} \bigl[X^{2}+X \bigr]. $$

(27)

Combining Eq. (27) with Eq. (16a), the solution of fractional Cahn–Allen equation with R–L derivative is obtained as follows:

$$ u_{2}(x,t)=\frac{1}{-1+\gamma e^{\frac{\sqrt{2}}{2k} (\frac{kx^{\alpha}+ct^{\alpha}}{\Gamma(1+\alpha)} )}}. $$

(28)

*Case* 3:

$$ A_{1}=-\frac{\sqrt{2}}{k},\qquad A_{0}= \frac{\sqrt{2}}{k},\qquad B=0,\qquad c=\frac{3k\sqrt {2}}{2},\quad k=k. $$

(29)

Using the conditions given in Eq. (29) and Eq. (22) in Eq. (17), we have

$$ Y_{3}(\xi)=\frac{\sqrt{2}}{2k} \bigl[X^{2}+X \bigr]. $$

(30)

Combining Eq. (30) with Eq. (16a), the solution of fractional Cahn–Allen equation with R–L derivative is obtained as follows:

$$ u_{3}(x,t)=\frac{1}{-1+\gamma e^{-\frac{\sqrt{2}}{2k} (\frac {kx^{\alpha}+ct^{\alpha}}{\Gamma(1+\alpha)} )}}. $$

(31)

*Case* 4:

$$ A_{1}=-\frac{\sqrt{2}}{k},\qquad A_{0}=- \frac{\sqrt{2}}{k},\qquad B=0,\qquad c=-\frac{3k\sqrt {2}}{2},\quad k=k. $$

(32)

Using the conditions given in Eq. (32) and Eq. (22) in Eq. (17), we have

$$ Y_{4}(\xi)=\frac{\sqrt{2}}{2k} \bigl[X^{2}-X \bigr]. $$

(33)

Combining Eq. (33) with Eq. (16a), the solution of fractional Cahn–Allen equation with R–L derivative is obtained as follows:

$$ u_{4}(x,t)=\frac{1}{1+\gamma e^{\frac{\sqrt{2}}{2k} (\frac{kx^{\alpha}+ct^{\alpha}}{\Gamma(1+\alpha)} )}}. $$

(34)

Now we apply the conformable definition of fractional derivative. The following transformation is introduced:

$$ \begin{aligned} &\xi=\frac{kx^{\alpha}}{\alpha}+ \frac{ct^{\alpha}}{\alpha}, \\ &u(\xi)=u(x,t), \end{aligned} $$

(35)

where *ξ* is the transformation variable and *k*, *c* are the constants. Afterwards, applying the same procedure given from Eq. (15) to Eq. (22), we get four different solutions:

*Case* 1: For \(A_{1}=\frac{\sqrt{2}}{k}\), \(A_{0}=\frac{\sqrt {2}}{k}\), \(B=0\), \(c=\frac{3k\sqrt{2}}{2}\), \(k=k\),

$$ u_{5}(x,t)=\frac{1}{1+\gamma e^{\frac{-\sqrt{2}}{2k} (\frac{kx^{\alpha}}{\alpha}+\frac{ct^{\alpha}}{\alpha} )}}. $$

(36)

*Case* 2: For \(A_{1}=\frac{\sqrt{2}}{k}\), \(A_{0}=-\frac{\sqrt {2}}{k}\), \(B=0\), \(c=-\frac{3k\sqrt{2}}{2}\), \(k=k\),

$$ u_{6}(x,t)=\frac{1}{-1+\gamma e^{\frac{\sqrt{2}}{2k} (\frac{kx^{\alpha}}{\alpha}+\frac{ct^{\alpha}}{\alpha} )}}. $$

(37)

*Case* 3: For \(A_{1}=-\frac{\sqrt{2}}{k}\), \(A_{0}=\frac{\sqrt {2}}{k}\), \(B=0\), \(c=\frac{3k\sqrt{2}}{2}\), \(k=k\),

$$ u_{7}(x,t)=\frac{1}{-1+\gamma e^{-\frac{\sqrt{2}}{2k} (\frac {kx^{\alpha}}{\alpha}+\frac{ct^{\alpha}}{\alpha} )}}. $$

(38)

*Case* 4: For \(A_{1}=-\frac{\sqrt{2}}{k}\), \(A_{0}=-\frac{\sqrt {2}}{k}\), \(B=0\), \(c=-\frac{3k\sqrt{2}}{2}\), \(k=k\),

$$ u_{8}(x,t)=\frac{1}{1+\gamma e^{\frac{\sqrt{2}}{2k} (\frac{kx^{\alpha}}{\alpha}+\frac{ct^{\alpha}}{\alpha} )}}. $$

(39)

It is important to note that the solutions \(u_{1}\), \(u_{2}\), \(u_{3}\), \(u_{4}\) are acquired by using the R–L derivative and \(u_{5}\), \(u_{6}\), \(u_{7}\), \(u_{8}\) are obtained by using the conformable derivative.

In Figs. 1–4, graphs of exact solutions of fractional Cahn–Allen equation are presented by using the R–L and conformable derivatives.

Figure 5 shows the effects of *α* on solutions \(u_{5}(x,t)\) using the conformable definition. Figure 5 reveals that the steeper peaks are originating from origin for smaller values of *α*, whereas wider peaks are found for values of *α* closer to 1.

### Exact solutions of the space-time fractional Drinfeld’s Sokolov–Wilson system

One of the widely used models is a DSW system introduced by Drinfeld, Sokolov, and Wilson. Zha and Zhi [34] solved this system using an improved F-expansion method. Inc utilized the Adomian decomposition method to find solutions of the DSW system [35]. The solitary solution of this model was obtained by Zhang using variational approach [36].

Consider the space-time fractional DSW system [37]

$$\begin{aligned} &\frac{\partial^{\alpha}u}{\partial t^{\alpha}}+pv\frac{\partial^{\alpha }v}{\partial x^{\alpha}}=0, \end{aligned}$$

(40a)

$$\begin{aligned} &\frac{\partial^{\alpha}v}{\partial t^{\alpha}}+q\frac{\partial^{3 \alpha }v}{\partial x^{3 \alpha}}+ru\frac{\partial^{\alpha}v}{\partial x^{\alpha}}+sv \frac{\partial^{\alpha}u}{\partial x^{\alpha}}=0,\quad x>0, t>0. \end{aligned}$$

(40b)

In Eqs. (40a)–(40b), *α* is the order of the fractional derivative and \(\alpha\in(0,1)\).

Firstly, we apply the R–L definition of fractional derivative. The following transformation is introduced:

$$ \begin{aligned} &\xi=\frac{kx^{\alpha}+ct^{\alpha}}{\Gamma(1+\alpha)}, \\ &u(x,t)=u(\xi), \\ &v(x,t)=v(\xi), \end{aligned} $$

(41)

where *ξ* is the transformation variable and *k*, *c* are the constants. Using Eq. (41) into Eqs. (40a)–(40b), we convert the problem into ODE:

$$\begin{aligned} &c\frac{du}{d\xi}+kpv \frac{dv}{d\xi}=0, \end{aligned}$$

(42a)

$$\begin{aligned} &c\frac{dv}{d\xi}+k^{3}q \frac{d^{3}v}{d\xi^{3}}+kru \frac{dv}{d\xi }+ksv\frac{du}{d\xi}=0. \end{aligned}$$

(42b)

Considering Eq. (42a) and rearranging,

$$ \frac{du}{d\xi}=-\frac{kp}{c} v \frac{dv}{d\xi}. $$

(43)

Integrating Eq. (43) w.r.t *ξ* and taking integration constant to be equal to zero, we get

$$ u=-\frac{kp}{2c}v^{2}. $$

(44)

Embedding Eq. (43) and Eq. (44) into Eq. (42b), we get a nonlinear ODE:

$$ 2ck^{3}q\frac{d^{3}v}{d\xi^{3}}-pk^{2}(r+2s)v^{2} \frac{dv}{d\xi }+2c^{2}\frac{dv}{d\xi}=0. $$

(45)

Integrating Eq. (45) with respect to *ξ* and taking integration constant equal to zero, we arrive at the following ODE:

$$ 2ck^{3}q\frac{d^{2}v}{d\xi^{2}}-\frac{1}{3} pk^{2}(r+2s)v^{3}+2c^{2}v=0. $$

(46)

Then using Eq. (10), the 2-D autonomous system is attained

$$\begin{aligned} &\frac{dX}{d\xi}=Y, \end{aligned}$$

(47a)

$$\begin{aligned} &\frac{dY}{d\xi}=\frac{p(r+2s)}{6cqk}X^{3}-\frac{c}{qk^{3}}X. \end{aligned}$$

(47b)

Now, to find the first integral of Eqs. (47a)–(47b), we implement the division theorem. In accordance with the FIM, it is assumed that nontrivial solutions of the above system (cf. Eqs. (47a)–(47b)) are *X* and *Y* respectively. Thus, the irreducible polynomial \(Q(X,Y)= \sum_{j=0}^{n}a_{j}(X)Y^{j}\) exists in \(\mathbb {C}[X,Y]\) and the following holds:

$$ Q\bigl(X(\xi),Y(\xi)\bigr)= \sum_{j=0}^{n}a_{j} \bigl(X(\xi)\bigr)Y(\xi)^{j}=0 $$

(48)

for \(j=0,1,\ldots,n\), and \(a_{n}(X)\neq0\). Now, a polynomial \(r(X)+s(X)Y\) exists in \(\mathbb{C}[X,Y]\), so

$$ \frac{\partial Q}{\partial\xi}=\frac{\partial Q}{\partial X}\frac {\partial X}{\partial\xi}+ \frac{\partial Q}{\partial Y}\frac{\partial Y}{\partial\xi}= \bigl(r(X)+s(X)Y \bigr) \Biggl( \sum _{j=0}^{n}a_{j}(X)Y^{j} \Biggr). $$

(49)

Suppose \(n=1\). On equating coefficients of \(Y^{j}\) (\(j=0,1\)) in Eq. (49) on both sides, we get

$$\begin{aligned} &a_{1}'(X)=a_{1}(X)s(X), \end{aligned}$$

(50)

$$\begin{aligned} &a_{0}'(X)=r(X)a_{1}(X)+s(X)a_{0}(X), \end{aligned}$$

(51)

$$\begin{aligned} &r(X)a_{0}(X)=a_{1}(X) \biggl( \frac{p(r+2s)}{6cqk} \biggr)X^{3}-\frac{c}{qk^{3}}X. \end{aligned}$$

(52)

As \(a_{j}(X)\) are polynomials of *X*, then from Eq. (50) we come to know that the polynomial \(a_{1}(X)\) is constant in nature, therefore \(s(X)=0\). Let us consider \(a_{1}(X)=1\), for convenience. After substituting these values, we balance the degrees of the functions \(r(X)\) and \(a_{0}(X)\) and deduce the \(\operatorname{deg}(r(X))\) equal to 0 or 1. Assume that \(r(X)=A_{1}X+A_{0}\), therefore Eq. (51) gives

$$ a_{0}(X)=\frac{1}{2}A_{1}X^{2}+A_{0}X+B. $$

(53)

Here, *B* is the integration constant.

Replacing the values of \(a_{0}\), \(a_{1}\), *r*, and *s* in Eq. (52), we get a nonlinear system of algebraic equations by putting all coefficients equal to zero for the same powers of *X*. After some calculations, we get

$$ \begin{aligned} &A_{1}=\pm\sqrt{ \frac{p(r+2s)}{3cqk}}, \\ &A_{0}=0, \\ &B=-\frac{c}{qk^{3}A_{1}}. \end{aligned} $$

(54)

Applying the conditions given in Eq. (54) and Eq. (53) in Eq. (48), we have

$$ Y(\xi)=-\frac{1}{2}A_{1}X^{2}(\xi)-B. $$

(55)

Combining Eq. (55) with Eq. (47a), the solutions of fractional DSW system with R–L derivative are obtained as follows:

$$\begin{aligned}& v_{1}(x,t)=\frac{-\sqrt{2A_{1}B}}{A_{1}} \tan \biggl[ \frac{\sqrt {2A_{1}B}}{2} \biggl(\frac{kx^{\alpha}+ct^{\alpha}}{\Gamma(1+\alpha)} \biggr)+\frac{\sqrt{2A_{1}B}}{2}\gamma \biggr], \end{aligned}$$

(56)

$$\begin{aligned}& u_{1}(x,t)=\frac{-pkB}{cA_{1}} \tan^{2} \biggl[\frac{\sqrt {2A_{1}B}}{2} \biggl(\frac{kx^{\alpha}+ct^{\alpha}}{\Gamma(1+\alpha)} \biggr)+\frac{\sqrt{2A_{1}B}}{2} \gamma \biggr]. \end{aligned}$$

(57)

Now we apply the conformable definition of fractional derivative. The following transformation is introduced:

$$ \begin{aligned} &\xi=\frac{kx^{\alpha}}{\alpha}+ \frac{ct^{\alpha}}{\alpha}, \\ &u(x,t)=u(\xi), \\ &v(x,t)=v(\xi), \end{aligned} $$

(58)

where *ξ* is the transformation variable and *k*, *c* are the constants. Afterwards, adopting the same procedure given in Eqs. (42a)–(42b) to Eq. (53), we get two different solutions:

$$\begin{aligned}& v_{2}(x,t)=\frac{-\sqrt{2A_{1}B}}{A_{1}} \tan \biggl[ \frac{\sqrt {2A_{1}B}}{2} \biggl(\frac{kx^{\alpha}}{\alpha}+\frac{ct^{\alpha}}{\alpha } \biggr)+ \frac{\sqrt{2A_{1}B}}{2}\gamma \biggr], \end{aligned}$$

(59)

$$\begin{aligned}& u_{2}(x,t)=\frac{-pkB}{cA_{1}} \tan^{2} \biggl[\frac{\sqrt {2A_{1}B}}{2} \biggl(\frac{kx^{\alpha}}{\alpha}+\frac{ct^{\alpha}}{\alpha } \biggr)+ \frac{\sqrt{2A_{1}B}}{2}\gamma \biggr]. \end{aligned}$$

(60)

It is important to note that the solutions \(u_{1}\), \(v_{1}\) are acquired by using the R–L derivative and \(u_{2}\), \(v_{2}\) are obtained by using the conformable derivative.

In Figs. 6–7, graphs of exact solutions of fractional DSW system are shown by using the R–L and conformable derivatives.

Figure 8 shows the effects of *α* on the solution \(v_{1}(x,t)\) using the R–L definition. Figure 8 depicts that the curved nonlinearity is observed for smaller values of *α*, whereas less nonlinear trends are found for *α* closer to 1.