# New Exact Solutions of Kolmogorov Petrovskii Piskunov Equation, Fitzhugh Nagumo Equation, and Newell-Whitehead Equation

Authors: Yu-Ming Chu; Shumaila Javeed; Dumitru Baleanu; Sidra Riaz; Hadi Rezazadeh
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

This work presents the new exact solutions of nonlinear partial differential equations (PDEs). The solutions are acquired by using an effectual approach, the first integral method (FIM). The suggested technique is implemented to obtain the solutions of space-time Kolmogorov Petrovskii Piskunov (KPP) equation and its derived equations, namely, Fitzhugh Nagumo (FHN) equation and Newell-Whitehead (NW) equation. The considered models are significant in biology. The KPP equation describes genetic model for spread of dominant gene through population. The FHN equation is imperative in the study of intercellular trigger waves. Similarly, the NW equation is applied for chemical reactions, Faraday instability, and Rayleigh-Benard convection. The proposed technique FIM can be applied to find the exact solutions of PDEs.


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## Body

## 1. Introduction

The nonlinearity in the world prevails thoroughly; thus, it is significant to develop nonlinear models including partial differential equations [1–4]. Nonlinear conformable PDEs attracted the interest of many researchers because of their vast applications in different fields, for example, in chemistry, acoustic, fluid dynamics, image processing, biology, physics, vibration, and control [5–7]. Nonlinear conformable PDEs have a great potential to apply in several fields; thus, researchers put noteworthy attention for their analytical and numerical solutions [8–12]. Different effective and reliable techniques are proposed like the homotopy analysis method [13], homotopy perturbation technique [14], extended hyperbolic tangent method [15, 16], hyperbolic function method [17], subequation method [18], and exponential rational function method [19] to get solutions.Feng presented an effective technique to obtain travelling wave solutions of NPDEs, known as the FIM method [20–22]. FIM is based on the ring theory and commutative algebra. FIM provides first integral of explicit form having polynomials as coefficients by applying the division theorem. Contrary to other methods, the benefits of FIM are to produce exact and explicit solutions without complicated and lengthy calculations [23–25]. Despite several advantages, FIM can only be applied to integrable PDEs.The focus of this paper is to find the exact solutions of conformable biological models. It includes KPP and its derived models, namely, FHN and NW. KPP is a general form of equation and we can obtain different equations from KPP, for example, FHN, NW, and Cahn Allen. The considered models are significant in biology. KPP describes the genetic model for spread of dominant gene through population. The graphical solutions of the KPP equation can be used for diallel analysis as diallel analysis requires graphical solutions of genes. In diallel analysis, graphical representation of genes is required and some further calculations enable researchers to have point estimation of recessive genes and dominant genes instead of providing an interval of estimation [26]. The FHN equation is used in the study of intercellular trigger waves. Trigger waves are pulses and oscillatory waves; these waves switch from one stable steady state to another [27]. Similarly, the NW equation is applied for Faraday instability, chemical reactions, Rayleigh-Benard convection, and biological systems [28]. Various techniques have been established for solving the KPP equation, for instance, the discrimination algorithm [29], the homotopy perturbation technique [30], the differential transform method [31], the (G′/G)-expansion method [32], and the homotopy analysis method [33]. Generally, the solutions of KPP equations are based on series solutions or numerical solutions. In this work, an effective technique named as FIM was adopted to acquire the exact solutions of KPP, FHN, and NW equations. The work is novel as the exact solutions of considered models using FIM are not presented before in the literature.This paper consists of the following sections. Conformable derivative is described in Section2; the proposed technique FIM is discussed in Section 3; the solutions of conformable KPP, FHN, and NW equations are presented in Section 4, and Section 5 contains summary and further recommendations.


## 2. Preliminaries

### 2.1. Derivative: Conformable

Conformable derivative is defined by Khalil et al. [34, 35].Definition 1.
The conformable derivative for functionh:0,∞→R of order β is given as(1)Tβht=limε→0ht+εt1−β−htε,whereas β∈0,1 having t>0. If function h is β-differentiable in 0,p, here p>0 and limt→0+hβt exists, then at 0, conformable derivative is represented as hβ0=limt→0+hβt.The conformable integral for functionh is given as(2)Iβpht=∫pthxx1−βdx,where p≥0 and β∈0,1.Khalil et al. further proposed the succeeding theorem [34–36].Theorem 2.
Let the functionsv and u at the point t>0 be β-differentiable; for β∈0,1 we have the following properties [34].(1)
Tβcu+dv=cTβu+dTβv∀c,d∈ℝ(2)
Tβtr=rtr−β∀r∈ℝ(3)
TβD=0∀vt=D (constant functions)(4)
Tβvu=vTβu+uTβv(5)
Tβu/v=vTβu−uTβv/v2(6)
Furthermore, if we have a differentiable functionu, then Tβut=t1−βdu/dtThe conformable derivative of any differential function at origin is zero; despite this flaw, several studies have been made on conformable derivative, as it explains higher order integration, sequential differentiation and integration, connection of differentiation and integration, property of linearity, derivative of constant function, quotient and product rule, chain rule, and power rule [34, 36–39]. Consequently, many researchers are working on the applicability of conformable derivative for real-world problems, such as Jacobi elliptical function expansion method used to solve conformable Boussinesq and combined Kdv-mKdv equation [40], conformable space-time fractional (2+1) dimensional dispersive long wave equation [41], conformable heat equation [42], and conformable perturbed nonlinear Schrodinger equation [43].


## 2.1. Derivative: Conformable

Conformable derivative is defined by Khalil et al. [34, 35].Definition 1.
The conformable derivative for functionh:0,∞→R of order β is given as(1)Tβht=limε→0ht+εt1−β−htε,whereas β∈0,1 having t>0. If function h is β-differentiable in 0,p, here p>0 and limt→0+hβt exists, then at 0, conformable derivative is represented as hβ0=limt→0+hβt.The conformable integral for functionh is given as(2)Iβpht=∫pthxx1−βdx,where p≥0 and β∈0,1.Khalil et al. further proposed the succeeding theorem [34–36].Theorem 2.
Let the functionsv and u at the point t>0 be β-differentiable; for β∈0,1 we have the following properties [34].(1)
Tβcu+dv=cTβu+dTβv∀c,d∈ℝ(2)
Tβtr=rtr−β∀r∈ℝ(3)
TβD=0∀vt=D (constant functions)(4)
Tβvu=vTβu+uTβv(5)
Tβu/v=vTβu−uTβv/v2(6)
Furthermore, if we have a differentiable functionu, then Tβut=t1−βdu/dtThe conformable derivative of any differential function at origin is zero; despite this flaw, several studies have been made on conformable derivative, as it explains higher order integration, sequential differentiation and integration, connection of differentiation and integration, property of linearity, derivative of constant function, quotient and product rule, chain rule, and power rule [34, 36–39]. Consequently, many researchers are working on the applicability of conformable derivative for real-world problems, such as Jacobi elliptical function expansion method used to solve conformable Boussinesq and combined Kdv-mKdv equation [40], conformable space-time fractional (2+1) dimensional dispersive long wave equation [41], conformable heat equation [42], and conformable perturbed nonlinear Schrodinger equation [43].


## 3. Methodology

Here, the methodology of FIM is presented.Step 1.
Conformable PDE is given as follows:(3)H∂βu∂tβ,∂βu∂z1β,∂βu∂z2β,⋯,∂βu∂zrβ,∂2βu∂t2β,∂2βu∂z1β∂z1β,∂2βu∂z2β∂z2β⋯=0.Step 2.
Now using the following transformation(4)uz1,z2,⋯,zr,t=UY.
Specifically in case of conformable derivative, the next transformation is applied as(5)Y=m1z1β+m2z2β+⋯+mrzrβ±ptββ.
The transformation defined in equation (5) will convert conformable PDE in nonlinear ODE.(6)FUY,U′Y,U′′Y,⋯=0,where U′Y=dUY/dY and transformed variable is denoted by Y.Step 3.
We will take other independent variables as(7)UY=ZY,UYY=YY.
As a result, FIM will provide a system of ODEs (nonlinear) as(8)∂Z∂Y=YY,∂Y∂Y=GZY,YY.Step 4.
We attain general solutions after integrating equation (8). There is no precise or sound technique to obtain first integrals in case of plane independent (autonomous) system, so it is difficult to get even one first integral. To determine the first integral, the division theorem is utilized. Hence, a first integral is derived (cf. equation (8)) with the help of the division theorem. In this way, nonlinear ordinary differential equations (ODEs) can be reduced into a first-order ODE system (integrable) through the division theorem. Afterwards, solving the obtained system (cf. equation (8)), the exact solutions can be acquired.The theorem for complex domainℂ and two variables is given asDivision Theorem.
Consider polynomialsGz,y and Rz,y, in complex domain ℂ, where Gzy is irreducible. If at all zero points of Rz,y, Gz,y vanishes, then another polynomial Hz,y exists in ℂz,y and the following equality holds:(9)Rz,y=Gz,yHz,y.


## 4. Implementation of FIM: Conformable KPP Equation and Its Derived Equations

The exact solutions of KPP, FHN, and NW equations are presented in this section.


### 4.1. Conformable Space-Time Fractional KPP Equation

Andrey Kolmogorov, Ivan Petrovsky, and Nikolai Piskunov proposed a nonlinear PDE called the Kolmogorov Petrovsky Piskunov (KPP) equation to describe the genetic model for spread of dominant gene through population. Later, the KPP equation is applied in different natural sciences like in physics as combustion, in biology as propagation of nerve impulses, in chemical kinetics as propagation of concentration waves, and in plasma as evolution of set of duffing oscillators.Consider conformable space-time fractional KPP equation defined as [32, 44](10)∂βu∂tβ−∂2βu∂x2β+μu+ηu2+δu3=0,x>0,t>0,where μ, η, and δ are constants and β∈0,1.First, we use conformable derivative with the following transformation:(11)Y=mxββ+ptββ,uY=ux,t,where the transformation variable is Y. The transformation represented in equation (11) will provide the following conversions:(12)∂β.∂tβ=pd.dY,∂2β.∂x2β=m2d2.dY2.Here,m and p are constants. Then, we get ODE by using equation (12) in equation (10):(13)pdudY−m2d2udY2+μu+ηu2+δu3=0.Now, we acquire a 2D system from equation (7) as(14)dZdY=Y,(15)dYdY=μm2Z+ηm2Z2+δm2Z3+pm2Y.Afterwards, the division theorem will provide first integral. According to FIM,Z and Y are supposed to be nontrivial solutions of the system given (cf. equations (14) and (15)). Now, the division theorem provides us an irreducible polynomial RZ,Y=∑r=0narZYr in ℂZ,Y given as(16)RZY,YY=∑r=0narZYYYr=0,wherearZ≠0 and r=0,1,⋯,n. Now, we have a polynomial of form wZ+qZY in ℂZ,Y such that(17)∂R∂Y=∂R∂Z∂Z∂Y+∂R∂Y∂Y∂Y=wZ+qZY∑r=0narZYr.Usingn=1 in equation (17) and equating coefficients for Yrr=0,1, then we have following equations:(18)a1′Z=a1ZqZ,(19)a0′Z=wZa1Z+qZa0Z−a1Zpm2,(20)wZa0Z=a1Zμm2Z+a1Zηm2Z2+a1Zδm2Z3.Here,arZ are polynomials in Z. As equation (18) shows a1Z has constant nature, hence qZ=0 and we can take a1Z=1. We conclude that degwZ can only be 0 or 1 by using a1Z and qZ in equations (19) and (20) and after balancing the functions wZ and a0Z degrees. Now, we can take wZ=A1Z+A0; therefore, equation (19) takes the following form:(21)a0Z=12A1Z2+A0Z−pm2Z+A2,where A2 is an integrating constant.Afterwards, the substitutions of values ofa0Z, wZ in equation (20) provide a system of nonlinear algebraic equations by equating the power of Z. Thus, as a result, we obtain various values of constants given as follows.(22)A2=0,A1=2δm,A0=ηm2δ+η2−4μδm2δ,p=ηm22δ+3m22δη2−4μδ.Case 1.
The following constants are acquired as follows:Substituting equations (21) and (22) into equation (16), we get(23)YY=−12A1Z2−A0Z+pm2Z.Substitution of equation (23) into equation (14) provides the first solution of conformable fractional KPP equation.(24)u1x,t=−2μδδη2−4μδ+δη−2γμδe−2/4mδη2−4μδ−ηmxβ/β+ptβ/β.(25)A2=0,A1=2δm,A0=ηm2δ−η2−4μδm2δ,p=ηm22δ−3m22δη2−4μδ.Case 2.
We getSubstituting equations (21) and (25) into equation (16), we obtain(26)YY=−12A1Z2−A0Z+pm2Z.Substitution of equation (26) into equation (14) provides the second solution of conformable fractional KPP equation.(27)u2x,t=2μδδη2−4μδ−δη+2γμδe2/4mδη2−4μδ+ηmxβ/β+ptβ/β.(28)A2=0,A1=−2δm,A0=−ηm2δ+η2−4μδm2δ,p=−ηm22δ+3mη2−4μδ22δ.Case 3.
We haveSubstituting equations (21) and (28) into equation (16), we get(29)YY=−12A1Z2−A0Z+pm2Z.Substitution of equation (29) into equation (14) provides the third solution of conformable fractional KPP equation.(30)u3x,t=2μδδη2−4μδ−δη+2γμδe−2/4mδη2−4μδ+ηmxβ/β+ptβ/β.(31)A2=0,A1=−2δm,A0=−ηm2δ−η2−4μδm2δ,p=−ηm22δ−3mη2−4μδ22δ.Case 4.
We obtainSubstituting equations (21) and (31) into equation (16), we obtain(32)YY=−12A1Z2−A0Z+cm2Z.Substitution of equation (32) into equation (14) provides the fourth solution of conformable fractional KPP equation.(33)u4x,t=−2μδδη2−4μδ+δη−2γμδe2/4mδη2−4μδ−ηmxβ/β+ptβ/β.(34)A0=0,A1=2δm,A2=μm2δ,p=−ηm2δ.Case 5.
We getSubstituting equations (21) and (34) into equation (16), we obtain(35)YY=−12A1Z2+pm2Z−A2.Substitution of equation (35) into equation (14) provides the fifth solution of conformable fractional KPP equation.(36)u5x,t=−η−tanmxβ/β+ptβ/β+γ8μδ−2η2/4mδ4μδ−η22δ.(37)A0=0,A1=−2δm,A2=−μm2δ,p=ηm2δ.Case 6.
We haveSubstituting equations (21) and (37) into equation (16), we obtain(38)YY=−12A1Z2+cm2Z−A2.Substitution of equation (38) into equation (14) provides the second solution of conformable fractional KPP equation.(39)u6x,t=−η+tanmxβ/β+ptβ/β+γ8μδ−2η2/4mδ4μδ−η22δ.The solutionsu1,u2,u3,u4,u5,u6 are presented in Figure 1. For larger values of β, the solutions attain more height which is depicted in Figure 2. Figure 3 shows the graphical presentation in 2D plot of genes for the KPP equation. The graphical solutions of the KPP equation can be used in diallel analysis as observed in [26].Figure 1
Solutions of conformable fractional KPP equation Cases1–6 using η=1.93−2.25, m=1, δ=1, γ=1, and β=0.8.Figure 2
Solutions of conformable fractional KPP equation (Case1) considering η=2.1, m=1, δ=1, μ=1, γ=1, β=1, β=0.8, β=0.6, and β=0.4.Figure 3
2D plot of conformable fractional KPP equation (Case1) using η=2.1, m=1, δ=1, μ=1, γ=1, and β=0.8.


### 4.2. Conformable Time-Fractional FHN Equation

Richard Fitzhugh proposed a model for transmission of impulses in nerve axon in 1961. Nagumo et al. made the identical circuit in succeeding years and presented the model of an excitable system. FHN is derived from KPP on substitutingμ=ξ, η=−ξ+1 [45, 46].Consider conformable time-space fractional FHN equation as(40)∂βu∂tβ−∂2u∂x2−u1−uu−ξ=0,where β∈0,1 and ξ∈0,0.5.First, we use the conformable derivative with the following transformation:(41)Y=x+ptββ,uY=ux,t,where the transformation variable isY. The transformation represented in equation (41) will provide the following conversions:(42)∂β.∂tβ=pd.dY,∂2.∂x2=d2.dY2.Here,p is a constant. Then, we get an ODE by using equation (42) in equation (40):(43)pdudY−d2udY2−u1−uu−ξ=0.Now, we obtain a 2D system from equation (7) as(44)dZdY=Y,(45)dYdY=Z3−Z2+ξZ−ξZ2+pY.Afterwards, the division theorem will give first integrals. According to FIM,Z and Y are supposed to be nontrivial solutions of the system given (cf. equations (44) and (45)). Now, the division theorem provides us an irreducible polynomial RZ,Y=∑r=0narZYr in ℂZ,Y as(46)RZY,YY=∑r=0narZYYYr=0,where arZ≠0 and r=0,1,⋯,n. Now, we have a polynomial of form wZ+qZY in ℂZ,Y such that(47)∂R∂Y=∂R∂Z∂Z∂Y+∂R∂Y∂Y∂Y=wZ+qZY∑r=0narZYr.Usingn=1 in equation (47) and equating coefficients of Yrr=0,1, then we have the following equations:(48)a1′Z=a1ZqZ,(49)a0′Z=wXa1Z+qZa0Z−pa1Z,(50)wZa0Z=−a1ZZ2+ξa1ZZ+a1ZZ3−ξa1ZZ2.Here,arZ are polynomials in Z. As equation (48) shows a1Z has constant nature, thus qZ=0 and we can take a1Z=1. We conclude that degwZ can only be 0 or 1 by using a1Z and qZ in equations (48) and (49) and after balancing the functions wZ and a0Z degrees. Now, we can take wZ=A1Z+A0; therefore, equation (49) takes the following form:(51)a0Z=12A1Z2−A0Z−pZ+A2,where A2 is an integrating constant.Afterwards, the substitutions of the values ofa0Z, wZ in equation (50) provide a system of nonlinear algebraic equations by equating the power of Z. Now, as a result, we have various constants given as follows.(52)A1=2,A0=0,A2=μ2,p=1+ξ2.Case 7.
We getSubstituting equations (51) and (52) into equation (46), we get the following equation.(53)YY=−22Z2+1+ξ2Z−ξ2.Substitution of equation (53) into equation (44) provides the first solution of conformable fractional FHN equation.(54)u7x,t=−ξ+e2ξ/2x+ptβ/β−2/2x+ptβ/β+γξ2/2−γ2/2−1+eξ2/2x+ptβ/β−2/2x+ptβ/β+γξ2/2−γ2/2.(55)A1=−2,A0=0,A2=−ξ2,p=−1+ξ2.Case 8.
We getSubstituting equations (51) and (55) into equation (46), we get the subsequent equation.(56)YY=22Z2−1+ξ2Z+ξ2.Substitution of equation (56) into equation (44) provides the second solution of conformable fractional FHN equation.(57)u8x,t=−ξ+e2ξ/2x+ptβ/β−2/2x+ptβ/β+γξ2/2−γ2/2−1+eξ2/2x+ptβ/β−2/2x+ptβ/β+γξ2/2−γ2/2.(58)A1=2,A2=0,A0=−2ξ,p=12−ξ2.Case 9.
We haveSubstituting equations (51) and (58) into equation (46), we obtain the following equation.(59)YY=−22Z2+12Z.Substitution of equation (59) into equation (44) provides the third solution of conformable fractional FHN equation.(60)u9x,t=11+γe−2/2x+ptβ/β.(61)A1=2,A2=0,A0=−2,p=−2+ξ12.Case 10.
We obtainSubstituting equations (51) and (61) into equation (46), we get(62)YY=−22Z2+ξ2Z.Substitution of equation (62) into equation (44) provides the fourth solution of conformable fractional FHN equation.(63)u10x,t=ξ1+γξe−ξ2/2x+ptβ/β.(64)A1=−2,A2=0,A0=2ξ,p=−12+ξ2.Case 11.
We getSubstituting equations (51) and (64) into equation (46), we obtain(65)YY=22Z2−12Z.Substitution of equation (65) into equation (44) provides the fifth solution of conformable fractional FHN equation.(66)u11x,t=11+γe2/2x+ptβ/β.(67)A1=−2,A2=0,A0=2,p=−ξ2+2.Case 12.
We haveSubstituting equations (51) and (67) into equation (46), we get(68)YY=22Z2−ξ2Z.Substitution of equation (68) into equation (44) provides the sixth solution of conformable fractional FHN equation.(69)u12x,t=ξ1+γξeξ2/2x+ptβ/β.The solutionsu7,u8,u9,u10,u11,u12 are presented in Figure 4. For smaller value of β, solutions attain more height which is depicted in Figure 5.Figure 4
Conformable fractional FHN equation Cases7–12 using β=0.8, γ=1, and ξ=0.5.Figure 5
Conformable fractional FHN equation (Case10) considering γ=1, ξ=0.45, β=1, β=0.6, and β=0.3.


### 4.3. Conformable Space-Time Fractional NW Equation

NW has wide applications in mechanical and chemical engineering, ecology, and biology [34]. NW can be derived from KPP by substituting μ=−1, η=0, and δ=1.Consider conformable space-time fractional NW equation as(70)∂βu∂tβ=K∂αu∂xα+au−buq,where β∈0,1 and α∈0,2, β>0, α>1, a,b,K>0 and q is a positive integer. For a=1, b=1, K=1, and α=2, equation (70) becomes(71)∂βu∂tβ=∂2u∂x2+u−u3.First, we use the conformable derivative with the following transformation:(72)Y=x+ptββuY=ux,t,where the transformation variable isY. The transformation represented in equation (72) will provide the following conversions.(73)∂β.∂tβ=pd.dY,∂2.∂x2=d2.dY2,wherep is a constant. Then, we get ODE by using equation (73) in equation (71).(74)pdudY−d2udY2−u+u3=0.Thus, we obtain a 2D system from equation (7) as(75)dZdY=Y,(76)dYdY=Z3−Z+pY.Afterwards, the division theorem will give first integrals. According to FIM,Z and Y are supposed to be nontrivial solutions of the system given (cf. equations (75) and (76)). Hence, the division theorem provides us irreducible polynomial RZ,Y=∑r=0narZYr in ℂZ,Y given as(77)RZY,YY=∑r=0narZYYYr=0,where arZ≠0 and r=0,1,⋯,n. Now, we have a polynomial of form wZ+qZY in ℂZ,Y such that(78)∂R∂Y=∂R∂Z∂Z∂Y+∂R∂Y∂Y∂Y=wZ+qZY∑r=0narZYr.Usingn=1 in equation (78) and equating coefficients of Yrr=0,1, then we have following equations:(79)a1′Z=a1ZqZ,(80)a0′Z=wZ−p,(81)wZa0Z=a1ZZ3−a1ZZ.Here,arZ are polynomials in Z. As equation (79) shows a1Z has constant nature, thus qZ=0 and we can take a1Z=1. We conclude that degwZ can only be 0 or 1 by using a1Z and qZ in equations (80) and (81) and after balancing the functions wZ and a0Z degrees. Now, we can take wZ=A1Z+A0; therefore, equation (80) takes the following form:(82)a0Z=12A1Z2+A0Z−pZ+A2,where A2 is an integrating constant.Afterwards, the substitutions of the values ofa0Z, wZ in equation (81) provide a system of nonlinear algebraic equations by equating coefficients of power of Z. Thus, we have various constants as given below.(83)A1=2,A0=0,A2=−12,p=0.Case 13.
We acquireSubstituting equations (82) and (83) into equation (77), we get(84)YY=−22Z2+12.Substitution of equation (84) into equation (75) provides the first solution of conformable fractional NW equation.(85)u13x,t=tanh22x+ptββ+γ22.(86)A1=−2,A0=0,A2=12,p=0.Case 14.
We getSubstituting equations (82) and (86) into equation (77), we obtain(87)YY=22Z2−12.Substitution of equation (87) into equation (75) provides the second solution of conformable fractional NW equation.(88)u14x,t=−tanh22x+ptββ+γ22.(89)A1=2,A2=0,A0=2,p=32.Case 15.
We haveSubstituting equations (82) and (89) into equation (77), we get(90)YY=−22Z2+12Z.Substitution of equation (90) into equation (75) provides the third solution of conformable fractional NW equation.(91)u15x,t=11+γe−2/2x+ptβ/β.(92)A1=−2,A2=0,A0=2,p=32.Case 16.
We obtainSubstituting equations (82) and (92) into equation (77), we get(93)YY=22Z2+12Z.Substitution of equation (93) into equation (75) provides the fourth solution of conformable fractional NW equation.(94)u16x,t=1−1+γe−2/2x+ptβ/β.(95)A1=2,A2=0,A0=−2,p=−32.Case 17.
We getSubstituting equations (82) and (95) into equation (77), we obtain(96)YY=−22Z2−12Z.Substitution of equation (96) into equation (75) provides the fifth solution of conformable fractional NW equation.(97)u17x,t=1−1+γe2/2x+ptβ/β.(98)A1=−2,A2=0,A0=−2,p=−32.Case 18.
We haveSubstituting equations (82) and (98) into equation (77), we obtain(99)YY=22Z2−12Z.Substitution of equation (99) into equation (75) provides the sixth solution of conformable fractional NW equation.(100)u18x,t=11+γe2/2x+ptβ/β.The solutionsu13,u14,u15,u16,u17,u18 are presented in Figure 6. In Figure 7, the results reveal that the peaks are getting sharper and sharper by reducing the value of β.Figure 6
Conformable fractional NW equation Cases13–18 using β=0.8 and γ=1.Figure 7
Conformable fractional NW equation (Case15) considering γ=1, β=1, β=0.8, β=0.4, and β=0.1.


## 4.1. Conformable Space-Time Fractional KPP Equation

Andrey Kolmogorov, Ivan Petrovsky, and Nikolai Piskunov proposed a nonlinear PDE called the Kolmogorov Petrovsky Piskunov (KPP) equation to describe the genetic model for spread of dominant gene through population. Later, the KPP equation is applied in different natural sciences like in physics as combustion, in biology as propagation of nerve impulses, in chemical kinetics as propagation of concentration waves, and in plasma as evolution of set of duffing oscillators.Consider conformable space-time fractional KPP equation defined as [32, 44](10)∂βu∂tβ−∂2βu∂x2β+μu+ηu2+δu3=0,x>0,t>0,where μ, η, and δ are constants and β∈0,1.First, we use conformable derivative with the following transformation:(11)Y=mxββ+ptββ,uY=ux,t,where the transformation variable is Y. The transformation represented in equation (11) will provide the following conversions:(12)∂β.∂tβ=pd.dY,∂2β.∂x2β=m2d2.dY2.Here,m and p are constants. Then, we get ODE by using equation (12) in equation (10):(13)pdudY−m2d2udY2+μu+ηu2+δu3=0.Now, we acquire a 2D system from equation (7) as(14)dZdY=Y,(15)dYdY=μm2Z+ηm2Z2+δm2Z3+pm2Y.Afterwards, the division theorem will provide first integral. According to FIM,Z and Y are supposed to be nontrivial solutions of the system given (cf. equations (14) and (15)). Now, the division theorem provides us an irreducible polynomial RZ,Y=∑r=0narZYr in ℂZ,Y given as(16)RZY,YY=∑r=0narZYYYr=0,wherearZ≠0 and r=0,1,⋯,n. Now, we have a polynomial of form wZ+qZY in ℂZ,Y such that(17)∂R∂Y=∂R∂Z∂Z∂Y+∂R∂Y∂Y∂Y=wZ+qZY∑r=0narZYr.Usingn=1 in equation (17) and equating coefficients for Yrr=0,1, then we have following equations:(18)a1′Z=a1ZqZ,(19)a0′Z=wZa1Z+qZa0Z−a1Zpm2,(20)wZa0Z=a1Zμm2Z+a1Zηm2Z2+a1Zδm2Z3.Here,arZ are polynomials in Z. As equation (18) shows a1Z has constant nature, hence qZ=0 and we can take a1Z=1. We conclude that degwZ can only be 0 or 1 by using a1Z and qZ in equations (19) and (20) and after balancing the functions wZ and a0Z degrees. Now, we can take wZ=A1Z+A0; therefore, equation (19) takes the following form:(21)a0Z=12A1Z2+A0Z−pm2Z+A2,where A2 is an integrating constant.Afterwards, the substitutions of values ofa0Z, wZ in equation (20) provide a system of nonlinear algebraic equations by equating the power of Z. Thus, as a result, we obtain various values of constants given as follows.(22)A2=0,A1=2δm,A0=ηm2δ+η2−4μδm2δ,p=ηm22δ+3m22δη2−4μδ.Case 1.
The following constants are acquired as follows:Substituting equations (21) and (22) into equation (16), we get(23)YY=−12A1Z2−A0Z+pm2Z.Substitution of equation (23) into equation (14) provides the first solution of conformable fractional KPP equation.(24)u1x,t=−2μδδη2−4μδ+δη−2γμδe−2/4mδη2−4μδ−ηmxβ/β+ptβ/β.(25)A2=0,A1=2δm,A0=ηm2δ−η2−4μδm2δ,p=ηm22δ−3m22δη2−4μδ.Case 2.
We getSubstituting equations (21) and (25) into equation (16), we obtain(26)YY=−12A1Z2−A0Z+pm2Z.Substitution of equation (26) into equation (14) provides the second solution of conformable fractional KPP equation.(27)u2x,t=2μδδη2−4μδ−δη+2γμδe2/4mδη2−4μδ+ηmxβ/β+ptβ/β.(28)A2=0,A1=−2δm,A0=−ηm2δ+η2−4μδm2δ,p=−ηm22δ+3mη2−4μδ22δ.Case 3.
We haveSubstituting equations (21) and (28) into equation (16), we get(29)YY=−12A1Z2−A0Z+pm2Z.Substitution of equation (29) into equation (14) provides the third solution of conformable fractional KPP equation.(30)u3x,t=2μδδη2−4μδ−δη+2γμδe−2/4mδη2−4μδ+ηmxβ/β+ptβ/β.(31)A2=0,A1=−2δm,A0=−ηm2δ−η2−4μδm2δ,p=−ηm22δ−3mη2−4μδ22δ.Case 4.
We obtainSubstituting equations (21) and (31) into equation (16), we obtain(32)YY=−12A1Z2−A0Z+cm2Z.Substitution of equation (32) into equation (14) provides the fourth solution of conformable fractional KPP equation.(33)u4x,t=−2μδδη2−4μδ+δη−2γμδe2/4mδη2−4μδ−ηmxβ/β+ptβ/β.(34)A0=0,A1=2δm,A2=μm2δ,p=−ηm2δ.Case 5.
We getSubstituting equations (21) and (34) into equation (16), we obtain(35)YY=−12A1Z2+pm2Z−A2.Substitution of equation (35) into equation (14) provides the fifth solution of conformable fractional KPP equation.(36)u5x,t=−η−tanmxβ/β+ptβ/β+γ8μδ−2η2/4mδ4μδ−η22δ.(37)A0=0,A1=−2δm,A2=−μm2δ,p=ηm2δ.Case 6.
We haveSubstituting equations (21) and (37) into equation (16), we obtain(38)YY=−12A1Z2+cm2Z−A2.Substitution of equation (38) into equation (14) provides the second solution of conformable fractional KPP equation.(39)u6x,t=−η+tanmxβ/β+ptβ/β+γ8μδ−2η2/4mδ4μδ−η22δ.The solutionsu1,u2,u3,u4,u5,u6 are presented in Figure 1. For larger values of β, the solutions attain more height which is depicted in Figure 2. Figure 3 shows the graphical presentation in 2D plot of genes for the KPP equation. The graphical solutions of the KPP equation can be used in diallel analysis as observed in [26].Figure 1
Solutions of conformable fractional KPP equation Cases1–6 using η=1.93−2.25, m=1, δ=1, γ=1, and β=0.8.Figure 2
Solutions of conformable fractional KPP equation (Case1) considering η=2.1, m=1, δ=1, μ=1, γ=1, β=1, β=0.8, β=0.6, and β=0.4.Figure 3
2D plot of conformable fractional KPP equation (Case1) using η=2.1, m=1, δ=1, μ=1, γ=1, and β=0.8.


## 4.2. Conformable Time-Fractional FHN Equation

Richard Fitzhugh proposed a model for transmission of impulses in nerve axon in 1961. Nagumo et al. made the identical circuit in succeeding years and presented the model of an excitable system. FHN is derived from KPP on substitutingμ=ξ, η=−ξ+1 [45, 46].Consider conformable time-space fractional FHN equation as(40)∂βu∂tβ−∂2u∂x2−u1−uu−ξ=0,where β∈0,1 and ξ∈0,0.5.First, we use the conformable derivative with the following transformation:(41)Y=x+ptββ,uY=ux,t,where the transformation variable isY. The transformation represented in equation (41) will provide the following conversions:(42)∂β.∂tβ=pd.dY,∂2.∂x2=d2.dY2.Here,p is a constant. Then, we get an ODE by using equation (42) in equation (40):(43)pdudY−d2udY2−u1−uu−ξ=0.Now, we obtain a 2D system from equation (7) as(44)dZdY=Y,(45)dYdY=Z3−Z2+ξZ−ξZ2+pY.Afterwards, the division theorem will give first integrals. According to FIM,Z and Y are supposed to be nontrivial solutions of the system given (cf. equations (44) and (45)). Now, the division theorem provides us an irreducible polynomial RZ,Y=∑r=0narZYr in ℂZ,Y as(46)RZY,YY=∑r=0narZYYYr=0,where arZ≠0 and r=0,1,⋯,n. Now, we have a polynomial of form wZ+qZY in ℂZ,Y such that(47)∂R∂Y=∂R∂Z∂Z∂Y+∂R∂Y∂Y∂Y=wZ+qZY∑r=0narZYr.Usingn=1 in equation (47) and equating coefficients of Yrr=0,1, then we have the following equations:(48)a1′Z=a1ZqZ,(49)a0′Z=wXa1Z+qZa0Z−pa1Z,(50)wZa0Z=−a1ZZ2+ξa1ZZ+a1ZZ3−ξa1ZZ2.Here,arZ are polynomials in Z. As equation (48) shows a1Z has constant nature, thus qZ=0 and we can take a1Z=1. We conclude that degwZ can only be 0 or 1 by using a1Z and qZ in equations (48) and (49) and after balancing the functions wZ and a0Z degrees. Now, we can take wZ=A1Z+A0; therefore, equation (49) takes the following form:(51)a0Z=12A1Z2−A0Z−pZ+A2,where A2 is an integrating constant.Afterwards, the substitutions of the values ofa0Z, wZ in equation (50) provide a system of nonlinear algebraic equations by equating the power of Z. Now, as a result, we have various constants given as follows.(52)A1=2,A0=0,A2=μ2,p=1+ξ2.Case 7.
We getSubstituting equations (51) and (52) into equation (46), we get the following equation.(53)YY=−22Z2+1+ξ2Z−ξ2.Substitution of equation (53) into equation (44) provides the first solution of conformable fractional FHN equation.(54)u7x,t=−ξ+e2ξ/2x+ptβ/β−2/2x+ptβ/β+γξ2/2−γ2/2−1+eξ2/2x+ptβ/β−2/2x+ptβ/β+γξ2/2−γ2/2.(55)A1=−2,A0=0,A2=−ξ2,p=−1+ξ2.Case 8.
We getSubstituting equations (51) and (55) into equation (46), we get the subsequent equation.(56)YY=22Z2−1+ξ2Z+ξ2.Substitution of equation (56) into equation (44) provides the second solution of conformable fractional FHN equation.(57)u8x,t=−ξ+e2ξ/2x+ptβ/β−2/2x+ptβ/β+γξ2/2−γ2/2−1+eξ2/2x+ptβ/β−2/2x+ptβ/β+γξ2/2−γ2/2.(58)A1=2,A2=0,A0=−2ξ,p=12−ξ2.Case 9.
We haveSubstituting equations (51) and (58) into equation (46), we obtain the following equation.(59)YY=−22Z2+12Z.Substitution of equation (59) into equation (44) provides the third solution of conformable fractional FHN equation.(60)u9x,t=11+γe−2/2x+ptβ/β.(61)A1=2,A2=0,A0=−2,p=−2+ξ12.Case 10.
We obtainSubstituting equations (51) and (61) into equation (46), we get(62)YY=−22Z2+ξ2Z.Substitution of equation (62) into equation (44) provides the fourth solution of conformable fractional FHN equation.(63)u10x,t=ξ1+γξe−ξ2/2x+ptβ/β.(64)A1=−2,A2=0,A0=2ξ,p=−12+ξ2.Case 11.
We getSubstituting equations (51) and (64) into equation (46), we obtain(65)YY=22Z2−12Z.Substitution of equation (65) into equation (44) provides the fifth solution of conformable fractional FHN equation.(66)u11x,t=11+γe2/2x+ptβ/β.(67)A1=−2,A2=0,A0=2,p=−ξ2+2.Case 12.
We haveSubstituting equations (51) and (67) into equation (46), we get(68)YY=22Z2−ξ2Z.Substitution of equation (68) into equation (44) provides the sixth solution of conformable fractional FHN equation.(69)u12x,t=ξ1+γξeξ2/2x+ptβ/β.The solutionsu7,u8,u9,u10,u11,u12 are presented in Figure 4. For smaller value of β, solutions attain more height which is depicted in Figure 5.Figure 4
Conformable fractional FHN equation Cases7–12 using β=0.8, γ=1, and ξ=0.5.Figure 5
Conformable fractional FHN equation (Case10) considering γ=1, ξ=0.45, β=1, β=0.6, and β=0.3.


## 4.3. Conformable Space-Time Fractional NW Equation

NW has wide applications in mechanical and chemical engineering, ecology, and biology [34]. NW can be derived from KPP by substituting μ=−1, η=0, and δ=1.Consider conformable space-time fractional NW equation as(70)∂βu∂tβ=K∂αu∂xα+au−buq,where β∈0,1 and α∈0,2, β>0, α>1, a,b,K>0 and q is a positive integer. For a=1, b=1, K=1, and α=2, equation (70) becomes(71)∂βu∂tβ=∂2u∂x2+u−u3.First, we use the conformable derivative with the following transformation:(72)Y=x+ptββuY=ux,t,where the transformation variable isY. The transformation represented in equation (72) will provide the following conversions.(73)∂β.∂tβ=pd.dY,∂2.∂x2=d2.dY2,wherep is a constant. Then, we get ODE by using equation (73) in equation (71).(74)pdudY−d2udY2−u+u3=0.Thus, we obtain a 2D system from equation (7) as(75)dZdY=Y,(76)dYdY=Z3−Z+pY.Afterwards, the division theorem will give first integrals. According to FIM,Z and Y are supposed to be nontrivial solutions of the system given (cf. equations (75) and (76)). Hence, the division theorem provides us irreducible polynomial RZ,Y=∑r=0narZYr in ℂZ,Y given as(77)RZY,YY=∑r=0narZYYYr=0,where arZ≠0 and r=0,1,⋯,n. Now, we have a polynomial of form wZ+qZY in ℂZ,Y such that(78)∂R∂Y=∂R∂Z∂Z∂Y+∂R∂Y∂Y∂Y=wZ+qZY∑r=0narZYr.Usingn=1 in equation (78) and equating coefficients of Yrr=0,1, then we have following equations:(79)a1′Z=a1ZqZ,(80)a0′Z=wZ−p,(81)wZa0Z=a1ZZ3−a1ZZ.Here,arZ are polynomials in Z. As equation (79) shows a1Z has constant nature, thus qZ=0 and we can take a1Z=1. We conclude that degwZ can only be 0 or 1 by using a1Z and qZ in equations (80) and (81) and after balancing the functions wZ and a0Z degrees. Now, we can take wZ=A1Z+A0; therefore, equation (80) takes the following form:(82)a0Z=12A1Z2+A0Z−pZ+A2,where A2 is an integrating constant.Afterwards, the substitutions of the values ofa0Z, wZ in equation (81) provide a system of nonlinear algebraic equations by equating coefficients of power of Z. Thus, we have various constants as given below.(83)A1=2,A0=0,A2=−12,p=0.Case 13.
We acquireSubstituting equations (82) and (83) into equation (77), we get(84)YY=−22Z2+12.Substitution of equation (84) into equation (75) provides the first solution of conformable fractional NW equation.(85)u13x,t=tanh22x+ptββ+γ22.(86)A1=−2,A0=0,A2=12,p=0.Case 14.
We getSubstituting equations (82) and (86) into equation (77), we obtain(87)YY=22Z2−12.Substitution of equation (87) into equation (75) provides the second solution of conformable fractional NW equation.(88)u14x,t=−tanh22x+ptββ+γ22.(89)A1=2,A2=0,A0=2,p=32.Case 15.
We haveSubstituting equations (82) and (89) into equation (77), we get(90)YY=−22Z2+12Z.Substitution of equation (90) into equation (75) provides the third solution of conformable fractional NW equation.(91)u15x,t=11+γe−2/2x+ptβ/β.(92)A1=−2,A2=0,A0=2,p=32.Case 16.
We obtainSubstituting equations (82) and (92) into equation (77), we get(93)YY=22Z2+12Z.Substitution of equation (93) into equation (75) provides the fourth solution of conformable fractional NW equation.(94)u16x,t=1−1+γe−2/2x+ptβ/β.(95)A1=2,A2=0,A0=−2,p=−32.Case 17.
We getSubstituting equations (82) and (95) into equation (77), we obtain(96)YY=−22Z2−12Z.Substitution of equation (96) into equation (75) provides the fifth solution of conformable fractional NW equation.(97)u17x,t=1−1+γe2/2x+ptβ/β.(98)A1=−2,A2=0,A0=−2,p=−32.Case 18.
We haveSubstituting equations (82) and (98) into equation (77), we obtain(99)YY=22Z2−12Z.Substitution of equation (99) into equation (75) provides the sixth solution of conformable fractional NW equation.(100)u18x,t=11+γe2/2x+ptβ/β.The solutionsu13,u14,u15,u16,u17,u18 are presented in Figure 6. In Figure 7, the results reveal that the peaks are getting sharper and sharper by reducing the value of β.Figure 6
Conformable fractional NW equation Cases13–18 using β=0.8 and γ=1.Figure 7
Conformable fractional NW equation (Case15) considering γ=1, β=1, β=0.8, β=0.4, and β=0.1.


## 5. Conclusion

The purpose of this paper was to find new exact solutions of some conformable biological models. The considered models were KPP, FHN, and NW. FIM was employed to obtain the solutions of fractional KPP, FHN, and NW equations. The proposed method was found brief and direct. The results indicate that FIM is one of the best techniques to calculate exact solutions of nonlinear fractional order problems appearing in biology, physics, and engineering.



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