Solitary wave solutions of Fitzhugh–Nagumo-type equations with conformable derivatives

The Fitzhugh–Nagumo equation is an important non-linear reaction–diffusion equation used to model the transmission of nerve impulses. This equation is used in biology as population genetics; the Fitzhugh–Nagumo equation is also frequently used in circuit theory. In this study, we give solutions to the fractional Fitzhugh–Nagumo (FN) equation, the fractional Newell–Whitehead–Segel (NWS) equation, and the fractional Zeldovich equation. We found the exact solutions of these equations by conformable derivatives. We have obtained the exact solutions within the time-fractional conformable derivative for these equations.

1 Introduction

Fractional differential equations (FDEs) are generalizations of known differential equations (ODEs). FPDEs are used effectively in many fields of science [1–4]. These equations are significant models to interpret plasma physics, relativistic physics, quantum mechanics, non-linear optics, etc. So, FPDE studies are getting more and more important. Recent developments and applications in fractional calculus have been discussed by many authors [5–9]. Many fractional models can be converted to a FODE, allowing us to use the power-series technique to find all open-series analytical solutions.

The fractional Fitzhugh–Nagumo equation is an important non-linear reaction–diffusion equation used to model the transmission of nerve impulses. This equation is used in biology as population genetics; the fractional Fitzhugh–Nagumo equation is also frequently used in circuit theory.

There are numerous effective ways to find solutions of PDEs. These methods are the $\left(\frac{G^{\prime }}{G}\right)$-expansion [10], the sub-equation (11) and (12), the exp-function [13, 14], the first integral [15], the functional variable [16, 17], the modified simplest [18, 19], the Kudryashov [20, 21], the extended simple [22], and the extended tanh–coth methods [23, 24]. These techniques allow for the calculation of PDE solutions in a variety of different formats.

These equations we deal with in this study are effective equations that play a fundamental role in many phenomena such as plasma physics and optics. The non-linear phenomena of wave happen in different fields such as optical fiber, physics, and biology. It is necessary to gain the exact solutions of such models for the better understanding of non-linear wave phenomena.

In the second part, the conformable derivative is introduced. In the third part, the extended tanh–coth method is given. In the other sections, we found the exact solutions of the (1 + 1) dimensional time-fractional Fitzhugh–Nagumo (FN) equation, the Newell–Whitehead (NW) equation, and the Zeldovich equation via this method.

2 Conformable derivative

Definition 1. The basic limit definition of this derivative is [25]:

$D_t^{\alpha }f\left(t\right)=\left\{\begin{array}{c}\hfill \frac{1}{\mathrm{\Gamma }\left(1-\alpha \right)}\frac{d}{dt}{\int }_0^t{\left(t-\xi \right)}^{-\alpha }\left(f\left(\xi \right)-f\left(0\right)\right)d\xi ,0<\alpha <1,\hfill \\ \hfill {\left(f^{\left(n\right)}\left(t\right)\right)}^{\left(\alpha -n\right)},n\le \alpha <n+1,n\ge 1.\hfill \end{array}\right.(1)$

Some properties of this derivative are given in [26, 27].

Definition 2. Let g: (0, ∞) → R be a function. The conformable derivative of g for order α is defined by

$T_{\alpha }\left(g\left(k\right)\right)=\underset{\epsilon \to 0}{\lim }\frac{g\left(k+\epsilon k^{1-\alpha }\right)-g\left(k\right)}{\epsilon },(2)$

for all k > 0, α ∈ (0, 1).

Theorem 3. If a function g: [0, ∞) → R is α − differentiable at t₀ > 0, α ∈ (0, 1] and g is continuous at t₀.

Theorem 4. Let f and g be α − differentiable at a point t > 0, α ∈ (0, 1]:

$\begin{array}{cc}\hfill T_{\alpha }\left(\alpha f+bg\right)& =a{T}_{\alpha }\left(f\right)+b{T}_{\alpha }\left(g\right),\mathit{for\hspace{0.17em}all\hspace{0.17em}}a,b\in R.\hfill \\ \hfill T_{\alpha }\left(t^p\right)& =p{t}^{p-\alpha },\mathit{for\hspace{0.17em}all\hspace{0.17em}}p\in R.\hfill \\ \hfill T_{\alpha }\left(\lambda \right)& =0,\mathit{\hspace{0.17em}for\hspace{0.17em}constant\hspace{0.17em}functions\hspace{0.17em}}f\left(t\right)=\lambda .\hfill \\ \hfill T_{\alpha }\left(fg\right)& =f{T}_{\alpha }\left(g\right)+g{T}_{\alpha }\left(f\right).\hfill \\ \hfill T_{\alpha }\left(\frac{f}{g}\right)& =\frac{g{T}_{\alpha }\left(f\right)-f{T}_{\alpha }\left(g\right)}{g^2}.\hfill \end{array}(3)$

If g is differentiable,

$T_{\alpha }\left(g\right)\left(t\right)=t^{1-\alpha }\frac{dg}{dt}\left(t\right).(4)$

Non-linear conformable partial differential equations (NCPDEs) with one independent variable are as follows:

$P\left(\frac{{\partial }^{\alpha }u}{\partial t^{\alpha }},\frac{\partial u}{\partial x},\frac{{\partial }^{2\alpha }u}{\partial t^{2\alpha }},\frac{{\partial }^{2}u}{\partial x^2},\dots \right)=0,0<\alpha \le 1,(5)$

$u\left(x,t\right)=U\left(\xi \right),\xi =cx-v\frac{t^{\alpha }}{\alpha },(6)$

$\frac{{\partial }^{\alpha }}{\partial t^{\alpha }}=-v\frac{\partial }{\partial \xi },\frac{{\partial }^{2\alpha }}{\partial t^{2\alpha }}=v^2\frac{{\partial }^2}{\partial {\xi }^2},\frac{\partial }{\partial x}=\frac{\partial }{\partial \xi },\frac{{\partial }^2}{\partial x^2}=\frac{{\partial }^2}{\partial {\xi }^2},\dots (7)$

The non-linear fPDE (5) can be converted to a non-linear ODE using Eq. 6:

$Q\left(U,U^{\prime },U^{\prime \prime },U^{\prime \prime \prime },\dots \right)=0.(8)$

Let us now discuss how we will approach solving the Fitzhugh–Nagumo equation, the Newell–Whitehead equation, and the Zeldovich equation.

3 The extended tanh–coth method

The tanh–coth approach can be summed up as follows in [28]:

The wave variable $\xi =cx-v\frac{t^{\alpha }}{\alpha }$ into a PDE

$P\left(u,u_t,u_x,u_{xx},\dots \right)=0.(9)$

u (x, t) = U(ξ) is a traveling wave solution. A non-linear ODE can be converted from the non-linear PDE (9):

$Q\left(U,U^{\prime },U^{\prime \prime },U^{\prime \prime \prime },\dots \right)=0.(10)$

From the derivatives of the independent variable Y,

$Y=\mathit{tanh}\left(\xi \right),Y^{\prime }=1-Y^2,(11)$

$\begin{array}{ll}\hfill \frac{d}{d\xi }& =\left(1-Y^2\right)\frac{d}{dY},\hfill \\ \hfill \frac{d^2}{d{\xi }^2}& =\left(1-Y^2\right)\left(-2Y\frac{d}{dY}+\left(1-Y^2\right)\frac{d^2}{d{Y}^2}\right).\hfill \end{array}(12)$

The tanh method can be expressed as follows:

$U\left(\xi \right)=S\left(Y\right)=\sum _{k=0}^m{a}_k{Y}^k.(13)$

The following is an extension of Eq. 13 [29]:

$U\left(\xi \right)=S\left(Y\right)=\sum _{k=0}^m{a}_k{Y}^k+\sum _{k=1}^m{b}_k{Y}^{-k}.(14)$

4 Solutions of the Fitzhugh–Nagumo equation

$\frac{{\partial }^{\alpha }u}{\partial t^{\alpha }}=\frac{{\partial }^{2}u}{\partial x^2}+u\left(1-u\right)\left(u-\mu \right),t>0,0<\alpha \le 1,x\in R.(15)$

The FN equation is an important non-linear reaction–diffusion equation used to model the transmission of nerve impulses. Also used in biology as population genetics, this equation is also frequently used in circuit theory, where x is the space variable and t is a time variable.

Since the FN equation is (1 + 1)-dimensional and time-fractional, using the wave variable $\xi =cx-v\frac{t^{\alpha }}{\alpha }$, Eq. 15 can be converted into a non-linear ODE (Eq. 16).

$v{U}^{\prime }+c^2{U}^{\prime \prime }+U\left(1-U\right)\left(U-\mu \right)=0,(16)$

where c and v are constants. By balancing (U)³ with U″ in Eq. 16, we obtain the following:

$\begin{array}{c}3m=m+2,\hfill \end{array}(17)$

$\begin{array}{c}\hfill m=1.\hfill \end{array}(18)$

The solution form is as follows:

$U\left(\xi \right)=a_0+a_{1}Y+b_1{Y}^{-1},Y=\tanh \left(\xi \right),Y^{\prime }=1-Y^2.(19)$

Here a₀ and a₁, b₁ are arbitrary constants. Eq. 19 is substituted into Eq. 16, and if the coefficients of Y are used to set the system of equations to zero, then

$\begin{array}{cc}\hfill -a_1^3+2a_1{c}^2& =0.\hfill \\ \hfill -3a_0{a}_1^2+a_1^2\mu +a_1^2-a_{1}v& =0.\hfill \\ \hfill -3a_0^2{a}_1+2a_0{a}_1\mu -3a_1^2{b}_1-2a_1{c}^2+2a_0{a}_1-a_1\mu & =0.\hfill \\ \hfill -a_0^3+a_0^2\mu -6a_0{a}_1{b}_1+2a_1{b}_1\mu +a_0^2-a_0\mu +2a_1{b}_1+a_{1}v+b_{1}v& =0.\hfill \\ \hfill -3a_0^2{b}_1+2a_0{b}_1\mu -3a_1{b}_1^2-2b_1{c}^2+2a_0{b}_1-b_1\mu & =0.\hfill \\ \hfill -3a_0{b}_1^2+b_1^2\mu +b_1^2-b_{1}v& =0.\hfill \\ \hfill -3a_0{b}_1^2+b_1^2\mu +b_1^2-b_{1}v& =0.\hfill \end{array}(20)$

By solving this system of equations, we get the following cases.

Case 1:

$\begin{array}{cccccc}\hfill a_0& =\frac{1}{2},\hfill & \hfill a_1& =0,\hfill & \hfill b_1& =\pm \frac{1}{2},\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{4},\hfill & \hfill \mu & =\mu ,\hfill & \hfill v& =\pm \left(\frac{\mu }{2}-\frac{1}{4}\right).\hfill \end{array}(21)$

Therefore, the exact solution is as follows:

$u_1\left(x,t\right)=\frac{1}{2}\pm \frac{1}{2}{\left(\tanh \left(\pm \frac{\sqrt{2}}{4}x-\left(\pm \frac{\mu }{2}-\frac{1}{4}\right)\frac{t^{\alpha }}{\alpha }\right)\right)}^{-1}.(22)$

Case 2:

$\begin{array}{cccccc}\hfill a_0& =a_0,\hfill & \hfill a_1& =0,\hfill & \hfill b_1& =\pm a_0,\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{2}{a}_0,\hfill & \hfill \mu & =2a_0,\hfill & \hfill v& =\pm \left(a_0-a_0^2\right).\hfill \end{array}(23)$

The exact solution is as follows:

$u_2\left(x,t\right)=a_0\pm a_0{\left(\tanh \left(\pm \frac{\sqrt{2}}{2}{a}_{0}x\pm \left(a_0-a_0^2\right)\frac{t^{\alpha }}{\alpha }\right)\right)}^{-1}.(24)$

Case 3:

$\begin{array}{cccccc}\hfill a_0& =a_0,\hfill & \hfill a_1& =0,\hfill & \hfill b_1& =\pm \left(a_0-1\right),\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{2}\left(a_0-1\right),\hfill & \hfill \mu & =2a_0-1,\hfill & \hfill v& =\pm a_0\left(a_0-1\right).\hfill \end{array}(25)$

The exact solution is as follows:

$u_3\left(x,t\right)=a_0\pm a_0{\left(\tanh \left(\pm \frac{\sqrt{2}}{2}\left(a_0-1\right)x\pm a_0\left(a_0-1\right)\frac{t^{\alpha }}{\alpha }\right)\right)}^{-1}.(26)$

Case 4:

$\begin{array}{cccccc}\hfill a_0& =\frac{1}{2},\hfill & \hfill a_1& =\pm \frac{1}{2},\hfill & \hfill b_1& =0,\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{4},\hfill & \hfill \mu & =\mu ,\hfill & \hfill v& =\pm \left(\frac{\mu }{2}-\frac{1}{4}\right).\hfill \end{array}(27)$

The exact solution is as follows:

$u_4\left(x,t\right)=\frac{1}{2}\pm \frac{1}{2}\left(\tanh \left(\pm \frac{\sqrt{2}}{4}x\pm \left(\frac{\mu }{2}-\frac{1}{4}\right)\frac{t^{\alpha }}{\alpha }\right)\right).(28)$

Case 5:

$\begin{array}{cccccc}\hfill a_0& =a_0,\hfill & \hfill a_1& =\pm a_0,\hfill & \hfill b_1& =0,\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{2},\hfill & \hfill \mu & =2a_0,\hfill & \hfill v& =\pm \left(a_0^2-a_0\right).\hfill \end{array}(29)$

The exact solution is as follows:

$u_5\left(x,t\right)=a_0\pm a_0\left(\tanh \left(\pm \frac{\sqrt{2}}{2}x\pm \left(a_0^2-a_0\right)\frac{t^{\alpha }}{\alpha }\right)\right).(30)$

Case 6:

$\begin{array}{cccccc}\hfill a_0& =a_0,\hfill & \hfill a_1& =\pm \left(a_0-1\right),\hfill & \hfill b_1& =0,\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{2}\left(a_0-1\right),\hfill & \hfill \mu & =2a_0-1,\hfill & \hfill v& =\pm \left(a_0^2-a_0\right).\hfill \end{array}(31)$

The exact solution is as follows:

$u_6\left(x,t\right)=a_0\pm \left(a_0-1\right)\left(\tanh \left(\pm \frac{\sqrt{2}}{2}\left(a_0-1\right)x\pm \left(a_0^2-a_0\right)\frac{t^{\alpha }}{\alpha }\right)\right).(32)$

Case 7:

$\begin{array}{cccccc}\hfill a_0& =a_0,\hfill & \hfill a_1& =\pm \frac{a_0}{2},\hfill & \hfill b_1& =\pm \frac{a_0}{2},\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{4}{a}_0,\hfill & \hfill \mu & =2a_0,\hfill & \hfill v& =\pm \frac{1}{2}\left(a_0^2-a_0\right).\hfill \end{array}(33)$

The exact solution is as follows:

$\begin{array}{ll}\hfill u_7\left(x,t\right)& =a_0\pm \frac{a_0}{2}\left(\tanh \left(\pm \frac{\sqrt{2}}{4}{a}_{0}x\pm \frac{1}{2}\left(a_0^2-a_0\right)\frac{t^{\alpha }}{\alpha }\right)\right)\hfill \\ \hfill & \pm \frac{a_0}{2}{\left(\tanh \left(\pm \frac{\sqrt{2}}{4}{a}_{0}x\pm \frac{1}{2}\left(a_0^2-a_0\right)\frac{t^{\alpha }}{\alpha }\right)\right)}^{-1}.\hfill \end{array}(34)$

Case 8:

$\begin{array}{cccccc}\hfill a_0& =a_0,\hfill & \hfill a_1& =\pm \frac{1}{2}\left(a_0-1\right),\hfill & \hfill b_1& =\pm \frac{1}{2}\left(a_0-1\right),\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{4}\left(a_0-1\right),\hfill & \hfill \mu & =2a_0-1,\hfill & \hfill v& =\pm \frac{1}{2}\left(a_0^2-a_0\right).\hfill \end{array}(35)$

The exact solution is as follows:

$\begin{array}{ll}\hfill u_8\left(x,t\right)& =a_0\pm \frac{\left(a_0-1\right)}{2}\left[\left(\tanh \left(\pm \frac{\sqrt{2}}{4}\left(a_0-1\right)x\pm \frac{1}{2}\left(a_0^2-a_0\right)\frac{t^{\alpha }}{\alpha }\right)\right)\right.\hfill \\ \hfill & +\left.{\left(\tanh \left(\pm \frac{\sqrt{2}}{4}\left(a_0-1\right)x\pm \frac{1}{2}\left(a_0^2-a_0\right)\frac{t^{\alpha }}{\alpha }\right)\right)}^{-1}\right].\hfill \end{array}(36)$

Case 9:

$\begin{array}{cccccc}\hfill a_0& =\frac{1}{2},\hfill & \hfill a_1& =\pm \frac{1}{4},\hfill & \hfill b_1& =\pm \frac{1}{4},\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{8},\hfill & \hfill \mu & =\mu ,\hfill & \hfill v& =\pm \left(\frac{\mu }{4}-\frac{1}{8}\right).\hfill \end{array}(37)$

The exact solution is as follows:

$u_9\left(x,t\right)=\frac{1}{2}\pm \frac{1}{4}\left(\tanh \left(\pm \frac{\sqrt{2}}{8}x\pm \left(\frac{\mu }{4}-\frac{1}{8}\right)\frac{t^{\alpha }}{\alpha }\right)\right)\pm \frac{1}{4}{\left(\tanh \left(\pm \frac{\sqrt{2}}{8}x\pm \left(\frac{\mu }{4}-\frac{1}{8}\right)\frac{t^{\alpha }}{\alpha }\right)\right)}^{-1}.(38)$

5 Solutions of the Newell–Whitehead equation

If μ = −1 in Eq. 15, the NW equation is obtained by

$\frac{{\partial }^{\alpha }u}{\partial t^{\alpha }}=\frac{{\partial }^{2}u}{\partial x^2}+u-u^3,t>0,0<\alpha \le 1,x\in R.(39)$

Using the wave variable $\xi =cx-v\frac{t^{\alpha }}{\alpha }$, Eq. 39 can be converted to Eq. 40.

$v{U}^{\prime }+c^2{U}^{\prime \prime }+U-U^3=0,(40)$

where c and v are constants. By balancing, we obtain

$\begin{array}{c}3m=m+2,\hfill \end{array}(41)$

$\begin{array}{c}\hfill m=1.\hfill \end{array}(42)$

The solution form is as follows:

$U\left(\xi \right)=a_0+a_{1}Y+b_1{Y}^{-1},Y=\tanh \left(\xi \right),Y^{\prime }=1-Y^2.(43)$

Here, a₀ and a₁, b₁ are arbitrary constants. Eq. 43 is substituted into Eq. 41, and if the coefficients of Y are used to set the system of algebraic equations to zero, then

$\begin{array}{cc}\hfill -a_1^3+2a_1{c}^2& =0.\hfill \\ \hfill -3a_0{a}_1^2-a_{1}v& =0.\hfill \\ \hfill -3a_0^2{a}_1-3a_1^2{b}_1-2a_1{c}^2+a_1& =0.\hfill \\ \hfill -a_0^3-6a_0{a}_1{b}_1+a_{1}v+b_{1}v+a_0& =0.\hfill \\ \hfill -3a_0^2{b}_1-3a_1{b}_1^2-2b_1{c}^2+b_1& =0.\hfill \\ \hfill -3a_0{b}_1^2-b_{1}v& =0.\hfill \\ \hfill -b_1^3+2b_1{c}^2& =0.\hfill \end{array}(44)$

By solving this system of equations, we get the following cases.

Case 1:

$\begin{array}{cccccc}\hfill a_0& =\pm \frac{1}{2},\hfill & \hfill a_1& =0,\hfill & \hfill b_1& =\frac{1}{2},\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{4},\hfill & \hfill & \hfill & \hfill v& =\pm \frac{3}{4}.\hfill \end{array}(45)$

Therefore, the exact solution is as follows:

$u_1\left(x,t\right)=\pm \frac{1}{2}+\frac{1}{2}{\left(\tanh \left(\pm \frac{\sqrt{2}}{4}x\pm \frac{3}{4}\frac{t^{\alpha }}{\alpha }\right)\right)}^{-1}.(46)$

Case 2:

$\begin{array}{cccccc}\hfill a_0& =\pm \frac{1}{2},\hfill & \hfill a_1& =\frac{1}{2},\hfill & \hfill b_1& =0,\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{4},\hfill & \hfill & \hfill & \hfill v& =\pm \frac{3}{4}.\hfill \end{array}(47)$

The exact solution is as follows:

$u_2\left(x,t\right)=\pm \frac{1}{2}+\frac{1}{2}\left(\tanh \left(\pm \frac{\sqrt{2}}{4}x\pm \frac{3}{4}\frac{t^{\alpha }}{\alpha }\right)\right).(48)$

Case 3:

$\begin{array}{cccccc}\hfill a_0& =\pm \frac{1}{2},\hfill & \hfill a_1& =\frac{1}{4},\hfill & \hfill b_1& =\frac{1}{4},\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{8},\hfill & \hfill & \hfill & \hfill v& =\pm \frac{3}{8}.\hfill \end{array}(49)$

The exact solution is as follows:

$u_3\left(x,t\right)=\pm \frac{1}{2}+\frac{1}{4}\left(\tanh \left(\pm \frac{\sqrt{2}}{8}x\pm \frac{3}{4}\frac{t^{\alpha }}{\alpha }\right)\right)+\frac{1}{4}{\left(\tanh \left(\pm \frac{\sqrt{2}}{8}x\pm \frac{3}{4}\frac{t^{\alpha }}{\alpha }\right)\right)}^{-1}.(50)$

6 Solutions of the Zeldovich equation

If μ = 0 in Eq. 15, the Zeldovich equation is obtained:

$\frac{{\partial }^{\alpha }u}{\partial t^{\alpha }}=\frac{{\partial }^{2}u}{\partial x^2}+u-u^3,t>0,0<\alpha \le 1,x\in R.(51)$

Using the wave variable $\xi =cx-v\frac{t^{\alpha }}{\alpha }$, Eq. 51 can be converted to Eq. 52.

$v{U}^{\prime }+c^2{U}^{\prime \prime }+U^2-U^3=0,(52)$

where c and v are constants. From Eq. 52, we obtain

$\begin{array}{c}3m=m+2,\hfill \end{array}(53)$

$\begin{array}{c}\hfill m=1.\hfill \end{array}(54)$

The solution form is as follows:

$U\left(\xi \right)=a_0+a_{1}Y+b_1{Y}^{-1},Y=\tanh \left(\xi \right),Y^{\prime }=1-Y^2.(55)$

Here, a₀ and a₁, b₁ are the arbitrary constants. Eq. 55 is substituted into Eq. 52, and if the coefficients of Y are used to set the system of equations to zero, then

$\begin{array}{cc}\hfill a_1^3-2a_1{c}^2& =0.\hfill \\ \hfill 3a_0{a}_1^2-a_1^2-a_{1}v& =0.\hfill \\ \hfill 3a_0^2{a}_1+3a_1^2{b}_1+2a_1{c}^2-2a_0{a}_1& =0.\hfill \\ \hfill a_0^3+6a_0{a}_1{b}_1-a_0^2-2a_1{b}_1+a_{1}v+b_{1}v& =0.\hfill \\ \hfill 3a_0^2{b}_1+3a_1{b}_1^2+2b_1{c}^2-2a_0{b}_1& =0.\hfill \\ \hfill 3a_0{b}_1^2-b_1^2-b_{1}v& =0.\hfill \\ \hfill b_1^3-2b_1{c}^2& =0.\hfill \end{array}(56)$

By solving this system of equations, we get the following cases.

Case 1:

$\begin{array}{cccccc}\hfill a_0& =a_0,\hfill & \hfill a_1& =0,\hfill & \hfill b_1& =\pm a_0\sqrt{\frac{1-a_0}{3a_0-1}},\hfill \\ \hfill c& =c,\hfill & \hfill & \hfill & \hfill v& =\pm 3a_0^2\sqrt{\frac{1-a_0}{3a_0-1}}-a_0\sqrt{\frac{1-a_0}{3a_0-1}}.\hfill \end{array}(57)$

Therefore, the exact solution is as follows:

$u_1\left(x,t\right)=a_0\pm a_0\sqrt{\frac{1-a_0}{3a_0-1}}{\left(\tanh \left(cx\pm 3a_0^2\sqrt{\frac{1-a_0}{3a_0-1}}-a_0\sqrt{\frac{1-a_0}{3a_0-1}}\frac{t^{\alpha }}{\alpha }\right)\right)}^{-1}.(58)$

Case 2:

$\begin{array}{cccccc}\hfill a_0& =\frac{1}{2},\hfill & \hfill a_1& =\pm \frac{1}{2},\hfill & \hfill b_1& =0,\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{4},\hfill & \hfill & \hfill & \hfill v& =\pm \frac{1}{4}.\hfill \end{array}(59)$

The exact solution is as follows:

$u_2\left(x,t\right)=\frac{1}{2}\pm \frac{1}{2}\left(\tanh \left(\pm \frac{\sqrt{2}}{4}x\pm \frac{1}{4}\frac{t^{\alpha }}{\alpha }\right)\right).(60)$

Case 3:

$\begin{array}{cccccc}\hfill a_0& =\frac{1}{2},\hfill & \hfill a_1& =\pm \frac{1}{4},\hfill & \hfill b_1& =\pm \frac{1}{4},\hfill \\ \hfill c& =\pm \frac{\sqrt{2}}{8},\hfill & \hfill & \hfill & \hfill v& =\pm \frac{1}{8}.\hfill \end{array}(61)$

The exact solution is as follows:

$u_3\left(x,t\right)=\frac{1}{2}\pm \frac{1}{4}\left(\tanh \left(\pm \frac{\sqrt{2}}{8}x\pm \frac{1}{8}\frac{t^{\alpha }}{\alpha }\right)\right)\pm \frac{1}{4}{\left(\tanh \left(\pm \frac{\sqrt{2}}{8}x\pm \frac{1}{8}\frac{t^{\alpha }}{\alpha }\right)\right)}^{-1}.(62)$

7 Conclusion

In this article, we present the extended tanh–coth method for solving non-linear space–time conformable PDEs. We found the exact and traveling wave solutions of some important space–time fPDEs via the extended tanh–coth method. We know many of the results obtained are new solutions that do not exist in the literature. The hyperbolic and trigonometric function solutions are significant to explain a variety of physical phenomena. This suggests that the extended tanh–coth method is more effective in finding the solutions of non-linear fPDEs. The 3D plots of the acquired solutions are presented by choosing appropriate values to the parameters in Figures 1–4. These are the advantages of the extended tanh–coth method. The offered method can be utilized to assist complicated models applicable to a wide variety of physical situations. We hope that the telecommunications industry and other such forms of waveguides will find this study to be beneficial.

FIGURE 1

FIGURE 1. Figure of u4 in (Eq. 4.14) with μ = 2, 0 < t < 1, and 0 < x < 10.

FIGURE 2

FIGURE 2. Figure of u5 in (Eq. 4.16) with a0 = 1/2, 0 < t < 10, and 0 < x < 10.

FIGURE 3

FIGURE 3. Figure of u7 in (Eq. 4.20) with a0 = 1/2, 0 < t < 10, and 0 < x < 10.

FIGURE 4

FIGURE 4. Figure of u9 in (Eq. 4.24) with μ = 2, 0 < t < 1, and 0 < x < 10.

Moreover, we come to the understanding that the newly obtained hyperbolic function and trigonometric function solutions in this study may help explain some complex physical aspects in non-linear physical sciences and are related to such physical properties. In future studies, these approaches can be easily used for other NFDEs, NFDE systems, fractional complex equations, fractional difference equations, etc.

Data availability statement

The original contributions presented in the study are included in the article/Supplementary Material; further inquiries can be directed to the corresponding author.

Author contributions

ACC: Original Draft, Methodology, Validation; AB: Investigation, Supervision, Writing—Review and Editing; OAA: Conceptualization, Methodology, Writing; MA: Original Draft, Validation, Founding acquisition.

Funding

The work in this study was supported, in part, by the Open Access Program from the American University of Sharjah. This study represents the opinions of the authors and does not mean to represent the position or opinions of the American University of Sharjah

Conflict of interest

The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.

Publisher’s note

All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors, and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.

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