# Some Traveling Wave Solutions of a Hyperbolic Model for Chemotaxis in 1-D

Authors: Qin Zhang; Wen-yan Chen
Journal: ISRN Applied Mathematics
 (2011)

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## Abstract

The traveling wave solution of a hyperbolic model for chemotaxis in one space dimension is studied in this paper. By using some transformations of dependent variables and independent variables, we apply the tanh method and improved tanh method to the model, from which some traveling wave solutions in explicit form are presented.


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## Body

## 1. Introduction

Chemotaxis is a  widespread phenomenon in biological systems describing the response of individuals to an external  chemical together with its gradient. It has attracted significant interest due to its critical role in a wide range of biological phenomena. Chemotaxis was studied mathematically beginning with the early papers of Patlak [1] and Keller and Segel [2, 3]. Since then the mathematical literature on the modeling of chemotaxis has grown rapidly. The parabolic Keller-Segel model of chemotaxis [2] has proved a cornerstone for much of this work. For a review of the recent literature, see Hillen and Painter [4]. However, the parabolic Keller-Segel model allows arbitrarily large speed of the population, it is physically unrealistic for small time. Motivated by the fact, Hillen and Stevens [5] proposed a hyperbolic model for chemotaxis in 1-D, which deal with finite speed. The model is(1.1)ut++(γ(S,St,Sx)u+)x=-μ+(S,St,Sx)u++μ-(S,St,Sx)u-,ut--(γ(S,St,Sx)u-)x=μ+(S,St,Sx)u+-μ-(S,St,Sx)u-,τSt=D0Sxx+f(S,u++u-),τ≥0,
where u±(x,t) denotes the particle densities of right (+)/left (−) moving particles, γ(S,St,Sx) denotes the particle speed, and μ±(S,St,Sx) are turning rates (rates of change of direction from + to − and vice versa).S(x,t) denotes the concentration of the chemical external signal. The production and degradation of the external chemical signal is modeled by the reaction term f.For the special case withD=0, constant speed γ and μ± depending only on S and Sx,(1.2)ut++γux+=-μ+(S,Sx)u++μ-(S,Sx)u-,ut--γux-=μ+(S,Sx)u+-μ-(S,Sx)u-,St=f(S,u++u-).
Segel in [9] first used it to analyze a very specific scenario. Later Rivero et al. [6] and Ford et al. [7, 8] used it to describe experimental data. In the works [5–9], the issues of local and global in time existence of solutions were considered theoretically and numerically. Hillen and Levine [10] studied finite time blowup of the system (1.2) with(1.3)f=S(u++u-),
and three specific forms of turning rates, one of which is(1.4)μ±(S,Sx)=γ2D1(γ∓αSSx).Keller and Segel in 1971 [11] considered an experimental setup that results in visible bands of bacteria traveling up a capillary tube filled with a mixture of oxygen and nutrient rich substrate. It raised the question of existence of a traveling wave solution to the Keller-Segel model of chemotaxis. Liu in 2008 [12] constructed some explicit solutions of the system (1.2) with the choice of f and μ± as in (1.3) and (1.4).In this paper, we try to find some exact traveling wave solutions of the system(1.5)ut++γux+=-μ+(S,Sx)u++μ-(S,Sx)u-,ut--γux-=μ+(S,Sx)u+-μ-(S,Sx)u-,St=D0Sxx+f(S,u++u-),
where D0≥0,γ>0 are constants, f(S,u++u-) and μ±(S,Sx) are defined above (1.3) and (1.4). It is clear that system (1.5) is a special case of the system (1.1) and a more generalized case of (1.2).The system (1.5) can be transformed into an equivalent system for the total particle density, u=u++u-, and particle flux, v=u+-u-. The resulting system for (u,v) is (3.1) below. Using tanh method and improved tanh method, we search for traveling wave solutions of the system (3.1). When D0≠0, three different kinds of traveling wave solutions are found. To the best of our knowledge, those traveling wave solutions are new and have not appeared in literature. When D0=0, the solutions obtained here are different from those known ones in [12].


## 2. The Two Methods

The tanh method and improved tanh method are powerful techniques to search for traveling wave solutions arising from one dimensional nonlinear wave and evolution equations. Let us first review the main features of two methods. They will be used in this paper. For more details, see [13–17] and reference therein.For both methods, we first use the wave variableξ=k(x-pt)+ξ0 to carry a PDE in two independent variables(2.1)H(u,ut,ux,uxx,…)=0
into an ODE(2.2)Q(u,u′,u′′,…)=0,
where k and p are the wave number and the wave speed, respectively.


### 2.1. The tanh Method

In the standard tanh method, the tanh is used as a new variable and the solution of (2.2) is expressed as a finite series of tanh(2.3)u(ξ)=∑i=0MaiTi,T=tanh⁡ξ,
where M is a positive integer that will be determined. Substituting (2.3) into the reduced ODE (2.2) results in an algebraic equation in powers of T. To determine the parameter M, we usually balance the linear terms of highest order in the resulting equation with the highest order nonlinear terms. With M determined, we collect all coefficients of powers of T in the resulting equation where these coefficients have to vanish. This will give a system of algebraic equations involving the parameters k,p, and ai. Having determined these parameters, we obtain an analytic solution u(x,t) in a closed form.The tanh method is a computerizable method, in which generating an algebraic system and solving it are two key procedures and laborious to do by hand. But they can be implemented on a computer with the help of computer algebra software such as Mathematica.


### 2.2. The Improved tanh Method

The crucial point of the improved tanh method is to replace thetanhξ in (2.3) by the solution of Riccati equation(2.4)F′=A+BF+CF2,C≠0,
where ′∶=d/dξ, and A, B, C are some constants.Supposing thatF is a solution of (2.4), it is easy to verify(1)
ifB2-4AC>0, (2.4) possesses a solution
(2.5)F(ξ)=-B2C-B2-4AC2Ctanh(B2-4AC2ξ);(2)
ifB2-4AC=0, (2.4) possesses a solution
(2.6)F(ξ)=-B2C-1Cξ;(3)
ifB2-4AC<0, (2.4) possesses a solution
(2.7)F(ξ)=-B2C+4AC-B22Ctan(4AC-B22ξ).Based on (2.5), the solution of (2.2) can be expressed as in (2.3), and based on (2.6) and (2.7), the solution of (2.2) can be expressed as(2.8)u(ξ)=∑i=0Mai1ξi,(2.9)u(ξ)=∑i=0Maitaniξ,
where parameters M,k,p, and ai are to be determined.


## 2.1. The tanh Method

In the standard tanh method, the tanh is used as a new variable and the solution of (2.2) is expressed as a finite series of tanh(2.3)u(ξ)=∑i=0MaiTi,T=tanh⁡ξ,
where M is a positive integer that will be determined. Substituting (2.3) into the reduced ODE (2.2) results in an algebraic equation in powers of T. To determine the parameter M, we usually balance the linear terms of highest order in the resulting equation with the highest order nonlinear terms. With M determined, we collect all coefficients of powers of T in the resulting equation where these coefficients have to vanish. This will give a system of algebraic equations involving the parameters k,p, and ai. Having determined these parameters, we obtain an analytic solution u(x,t) in a closed form.The tanh method is a computerizable method, in which generating an algebraic system and solving it are two key procedures and laborious to do by hand. But they can be implemented on a computer with the help of computer algebra software such as Mathematica.


## 2.2. The Improved tanh Method

The crucial point of the improved tanh method is to replace thetanhξ in (2.3) by the solution of Riccati equation(2.4)F′=A+BF+CF2,C≠0,
where ′∶=d/dξ, and A, B, C are some constants.Supposing thatF is a solution of (2.4), it is easy to verify(1)
ifB2-4AC>0, (2.4) possesses a solution
(2.5)F(ξ)=-B2C-B2-4AC2Ctanh(B2-4AC2ξ);(2)
ifB2-4AC=0, (2.4) possesses a solution
(2.6)F(ξ)=-B2C-1Cξ;(3)
ifB2-4AC<0, (2.4) possesses a solution
(2.7)F(ξ)=-B2C+4AC-B22Ctan(4AC-B22ξ).Based on (2.5), the solution of (2.2) can be expressed as in (2.3), and based on (2.6) and (2.7), the solution of (2.2) can be expressed as(2.8)u(ξ)=∑i=0Mai1ξi,(2.9)u(ξ)=∑i=0Maitaniξ,
where parameters M,k,p, and ai are to be determined.


## 3. Traveling Wave Solutions

A transformation of the dependent variables is necessary, before the just-described methods can be applied to the hyperbolic model (1.5). As in [10], we rewrite system (1.5) as a system for u=u++u- and v=u+-u-:(3.1)ut+γvx=0,vt+γux=γD1⋅αSxSu-γ2D1v,St=D0Sxx+Su.
Eliminating v in system (3.1), it is clear that (3.1) is of the form(3.2)-1γutt+γuxx=αγD1(uSxS)x+γD1ut,St=D0Sxx+Su.
Taking into account the transformation(3.3)u=u(ξ),S=eφ(ξ),ξ=k(x-pt)+ξ0,
where k>0,p>0, and ξ0 is an arbitrary constant, the (3.2) can be put into the format(3.4)k(γ-p2γ)u′′=αγkD1(u′φ′+uφ′′)-γpD1u′,(3.5)-kpφ′=D0k2(φ′2+φ′′)+u,
where p2≠γ2.


### 3.1. Using tanh Method

Let(3.6)u(ξ)=∑i=0Maitanhiξ.
It is noted that (3.7)∫tanhxdx=ln⁡⁡|cosh⁡⁡x|+C,∫tanh2xdx=x-tanh⁡x+C,∫tanhkxdx=-tanhk-1xk-1+∫tanhk-2xdx,
as a result, we guess the φ has the form(3.8)φ(ξ)=b-2ξ+b-1ln⁡⁡|cosh⁡⁡ξ|+∑i=0M-2bitanhiξ.Substitutingu(ξ) and φ(ξ) into (3.4)-(3.5), we obtain M=2, and  the solutions of system (3.4)-(3.5) is of the form(3.9)u=a0+a1tanh⁡ξ+a2tanh2ξ,φ=b-2ξ+b-1ln⁡⁡|cosh⁡⁡ξ|.
With the help of the symbolic software Mathematica, substituting (3.9) into (3.4) and (3.5)and setting the coefficients of tanhiξ to be zero, we obtain(3.10)2kD1(γ2-p2)a2-αγ2k(a1b-2+a0b-1)+γ2pa1=0,-kD1(γ2-p2)a1-αγ2k(a2b-2+a1b-1)+γ2pa2=0,-3k(2D1(γ2-p2)+αγ2b-1)a2=0,kpb-2+D0k2(b-22+b-1)+a0=0,kpb-1+2D0k2b-2b-1+a1=0,D0k2(b-12-b-1)+a2=0.
If D0≠0 and p satisfies(3.11)Δ=(γ2-p2)2D12(4D02+4αD0+5α2)+2(γ2-p2)D1α3γ2-α2γ4D0(α+D0)>0,
solving the algebraic equations (3.10) with the aid of Mathematica, we arrive at(3.12)k=γ2pΔ2D0D1|(4D1(γ2-p2)+αγ2)(γ2-p2)|,b-1=-2D1(γ2-p2)αγ2,b-2=1αkD0⋅2pD0D1(γ2-p2)+pD0αγ2-pD1α(γ2-p2)4D1(γ2-p2)+αγ2,a0=2k2D0D1(γ2-p2)αγ2-p(pαγ2D0-pαD1(γ2-p2)+2pD0D1(γ2-p2))αD0(αγ2+4D1(γ2-p2))-D0(pαγ2D0-pαD1(γ2-p2)+2pD0D1(γ2-p2))2α2D02(αγ2+4D1(γ2-p2))2,a1=2kD1(γ2-p2)×pα2γ2+2pαγ2D0+2pαD1(γ2-p2)+4pD0D1(γ2-p2)α2γ2(αγ2+4D1(γ2-p2)),a2=-2k2D0D1(γ2-p2)(αγ2+2D1(γ2-p2))(αγ2)2,
which leads to the solutions of system (3.1)(3.13)u(x,t)=a0+a1tanh⁡(k(x-pt)+ξ0)+a2tanh2(k(x-pt)+ξ0),S(x,t)=|cosh⁡⁡(k(x-pt)+ξ0)|b-1⋅exp⁡⁡(b-2(k(x-pt)+ξ0)),v(x,t)=pγu(x,t).
If D0=0, we found a2=0, and (3.14)b-1=-D1(γ2-p2)αγ2,b-2=p2αk,a0=-p22α,a1=kpD1(γ2-p2)αγ2.
Consequently, we obtain the following solutions of system (3.1)(3.15)u(x,t)=-p22a+D1kp(γ2-p2)αγ2tanh⁡(k(x-pt)+ξ0),S(x,t)=(cosh⁡⁡(k(x-pt)+ξ0))-D1(γ2-p2)/αγ2⋅exp⁡⁡(p2ak(k(x-pt)+ξ0)),v(x,t)=-p32aγ+D1kp2(γ2-p2)αγ3tanh⁡(k(x-pt)+ξ0),
which is different from the solutions obtained in [12].


### 3.2. Using Improved tanh Method

Case 1.
We seek solutions in the form of (2.8) u(ξ)=∑i=0Mai/ξi. Motivated by the expression of (3.5), we turn to the assumption φ(ξ)=b-2ξ+b-1ln|ξ|+∑j=0M-2bj/ξj and also imply that M=2. Therefore, we have
(3.16)u(ξ)=a0+a11ξ+a21ξ2,φ(ξ)=b-2ξ+b-1ln⁡⁡|ξ|.
By substituting (3.16) into (3.4) and (3.5), and setting the coefficients of 1/ξj(j=0,1,2) to be zero, we set
(3.17)αγk(a1b-2+a0b-1)-γpa1=0,2kD1(γ2-p2)a1+2αγ2k(a1b-1+a2b-2)-2γ2pa2=0,6kD1(γ2-p2)a2+3αγ2ka2b-1=0,kpb-2+D0k2b-22+a0=0,kpb-1+2D0k2b-2b-1+a1=0,D0k2(b-12-b-1)+a2=0.
If D0≠0 and
(3.18)Δ=1+α3±α(2D0+α)D02+αD0+α25α2D1+4D02D1+4αD0D1>0,
solving algebraic equations (3.17) gives
(3.19)b-1=-2D1(γ2-p2)αγ2,b-2=1αkD0⋅2pD0D1(γ2-p2)+pD0αγ2-pD1α(γ2-p2)4D1(γ2-p2)+αγ2,a0=-p2((2D0D1-D1α)(γ2-p2)+αγ2D0)α2D0(4D1(γ2-p2)+αγ2)2×((3αD1+2D0D1)(γ2-p2)+α2γ2+αγ2D0),a1=2kpD1(γ2-p2)α2γ2(αγ2+4D1(γ2-p2))×(α2γ2+2αγ2D0+2αD1(γ2-p2)+4D0D1(γ2-p2)),a2=-2k2D0D1(γ2-p2)(αγ2)2(αγ2+2D1(γ2-p2)),p=γΔ,
where Δ is specified by (3.18).
This in turn gives the solutions of system (3.1)
(3.20)u(x,t)=a0+a1k(x-pt)+ξ0+a2(k(x-pt)+ξ0)2,S(x,t)=|k(x-pt)+ξ0|b-1⋅exp⁡⁡(b-2(k(x-pt)+ξ0)),v(x,t)=pγ(a0+a1k(x-pt)+ξ0+a2(k(x-pt)+ξ0)2),
where k is an arbitrary constant and ai(i=0,1,2), b-j(j=1,2),p are given in (3.19).Case 2.
We seek solutions in the form of (2.9). Let u(ξ)=∑i=0Maitaniξ and φ(ξ)=b-2ξ+b-1ln|cosξ|+∑j=0M-2bjtanjξ. Substituting u(ξ),φ(ξ) into (3.4) and (3.5), balancing the linear terms of highest order with the highest order nonlinear terms, directly results in M=2 and
(3.21)u=a0+a1tan⁡ξ+a2tan2⁡ξ,φ=b-2ξ+b-1ln⁡⁡|cos⁡⁡ξ|.
To progress further substituting (3.21) into (3.4) and (3.5) and setting the coefficients of taniξ(i=0,1,2) in the resulting equations to be zero, we obtain a set of algebraic equations
(3.22)2kD1(γ2-p2)a2-αγ2k(a1b-2-a0b-1)+γ2pa1=0,kD1(γ2-p2)a1-αγ2k(-a1b-1+a2b-2)+γ2pa2=0,6a2kD1(γ2-p2)+3αγ2ka2b-1=0,-kpb-2-D0k2(b-22-b-1)=a0,kpb-1+2D0k2b-1b-2=a1,-D0k2(b-12-b-1)=a2.
If D0≠0 and p satisfies
(3.23)Δ=-p4(5D12α2+4D0D12α+4D02D12)+p2γ2(2D1α3+10α2D12+8αD0D12+8D02D12)+γ4(α3D0+α2D02-2α3D1-5α2D12-4αD0D12-4D02D12)>0,
we know
(3.24)b-1=-2D1(γ2-p2)αγ2,(3.25)b-2=pαkD0⋅2D0D1(γ2-p2)+αγ2D0-D1α(γ2-p2)4D1(γ2-p2)+αγ2,(3.26)a0=-p2αD0⋅2D0D1(γ2-p2)+αγ2D0-αD1(γ2-p2)4D1(γ2-p2)+αγ2-p2α2D0⋅(2D0D1(γ2-p2)+αγ2D0-αD1(γ2-p2)4D1(γ2-p2)+αγ2)2-2k2D0D1(γ2-p2)αγ2,(3.27)a1=-2kpD1(γ2-p2)(α+2D0)(αγ2+2D1(γ2-p2))α2γ2(αγ2+4D1(γ2-p2)),(3.28)a2=-2k2D0D1(γ2-p2)(αγ2+2D1(γ2-p2))(αγ2)2,(3.29)k=pγ22ΔD02D12(γ2-p2)2(4D1(γ2-p2)+αγ2)2,
where Δ is given by (3.23). To this end, we therefore derive the solutions of system (3.1) as follows:
(3.30)u(x,t)=a0+a1tan⁡(k(x-pt)+ξ0)+a2tan2⁡(k(x-pt)+ξ0),S(x,t)=|cos⁡⁡(k(x-pt)+ξ0)|b-1eb-2(k(x-pt)+ξ0),v(x,t)=(a0+a1tan⁡(k(x-pt)+ξ0)+a2tan2⁡(k(x-pt)+ξ0))pγ,
where b-j(j=1,2) and ai(i=0,1,2) and k are given by (3.24)–(3.29), respectively.


## 3.1. Using tanh Method

Let(3.6)u(ξ)=∑i=0Maitanhiξ.
It is noted that (3.7)∫tanhxdx=ln⁡⁡|cosh⁡⁡x|+C,∫tanh2xdx=x-tanh⁡x+C,∫tanhkxdx=-tanhk-1xk-1+∫tanhk-2xdx,
as a result, we guess the φ has the form(3.8)φ(ξ)=b-2ξ+b-1ln⁡⁡|cosh⁡⁡ξ|+∑i=0M-2bitanhiξ.Substitutingu(ξ) and φ(ξ) into (3.4)-(3.5), we obtain M=2, and  the solutions of system (3.4)-(3.5) is of the form(3.9)u=a0+a1tanh⁡ξ+a2tanh2ξ,φ=b-2ξ+b-1ln⁡⁡|cosh⁡⁡ξ|.
With the help of the symbolic software Mathematica, substituting (3.9) into (3.4) and (3.5)and setting the coefficients of tanhiξ to be zero, we obtain(3.10)2kD1(γ2-p2)a2-αγ2k(a1b-2+a0b-1)+γ2pa1=0,-kD1(γ2-p2)a1-αγ2k(a2b-2+a1b-1)+γ2pa2=0,-3k(2D1(γ2-p2)+αγ2b-1)a2=0,kpb-2+D0k2(b-22+b-1)+a0=0,kpb-1+2D0k2b-2b-1+a1=0,D0k2(b-12-b-1)+a2=0.
If D0≠0 and p satisfies(3.11)Δ=(γ2-p2)2D12(4D02+4αD0+5α2)+2(γ2-p2)D1α3γ2-α2γ4D0(α+D0)>0,
solving the algebraic equations (3.10) with the aid of Mathematica, we arrive at(3.12)k=γ2pΔ2D0D1|(4D1(γ2-p2)+αγ2)(γ2-p2)|,b-1=-2D1(γ2-p2)αγ2,b-2=1αkD0⋅2pD0D1(γ2-p2)+pD0αγ2-pD1α(γ2-p2)4D1(γ2-p2)+αγ2,a0=2k2D0D1(γ2-p2)αγ2-p(pαγ2D0-pαD1(γ2-p2)+2pD0D1(γ2-p2))αD0(αγ2+4D1(γ2-p2))-D0(pαγ2D0-pαD1(γ2-p2)+2pD0D1(γ2-p2))2α2D02(αγ2+4D1(γ2-p2))2,a1=2kD1(γ2-p2)×pα2γ2+2pαγ2D0+2pαD1(γ2-p2)+4pD0D1(γ2-p2)α2γ2(αγ2+4D1(γ2-p2)),a2=-2k2D0D1(γ2-p2)(αγ2+2D1(γ2-p2))(αγ2)2,
which leads to the solutions of system (3.1)(3.13)u(x,t)=a0+a1tanh⁡(k(x-pt)+ξ0)+a2tanh2(k(x-pt)+ξ0),S(x,t)=|cosh⁡⁡(k(x-pt)+ξ0)|b-1⋅exp⁡⁡(b-2(k(x-pt)+ξ0)),v(x,t)=pγu(x,t).
If D0=0, we found a2=0, and (3.14)b-1=-D1(γ2-p2)αγ2,b-2=p2αk,a0=-p22α,a1=kpD1(γ2-p2)αγ2.
Consequently, we obtain the following solutions of system (3.1)(3.15)u(x,t)=-p22a+D1kp(γ2-p2)αγ2tanh⁡(k(x-pt)+ξ0),S(x,t)=(cosh⁡⁡(k(x-pt)+ξ0))-D1(γ2-p2)/αγ2⋅exp⁡⁡(p2ak(k(x-pt)+ξ0)),v(x,t)=-p32aγ+D1kp2(γ2-p2)αγ3tanh⁡(k(x-pt)+ξ0),
which is different from the solutions obtained in [12].


## 3.2. Using Improved tanh Method

Case 1.
We seek solutions in the form of (2.8) u(ξ)=∑i=0Mai/ξi. Motivated by the expression of (3.5), we turn to the assumption φ(ξ)=b-2ξ+b-1ln|ξ|+∑j=0M-2bj/ξj and also imply that M=2. Therefore, we have
(3.16)u(ξ)=a0+a11ξ+a21ξ2,φ(ξ)=b-2ξ+b-1ln⁡⁡|ξ|.
By substituting (3.16) into (3.4) and (3.5), and setting the coefficients of 1/ξj(j=0,1,2) to be zero, we set
(3.17)αγk(a1b-2+a0b-1)-γpa1=0,2kD1(γ2-p2)a1+2αγ2k(a1b-1+a2b-2)-2γ2pa2=0,6kD1(γ2-p2)a2+3αγ2ka2b-1=0,kpb-2+D0k2b-22+a0=0,kpb-1+2D0k2b-2b-1+a1=0,D0k2(b-12-b-1)+a2=0.
If D0≠0 and
(3.18)Δ=1+α3±α(2D0+α)D02+αD0+α25α2D1+4D02D1+4αD0D1>0,
solving algebraic equations (3.17) gives
(3.19)b-1=-2D1(γ2-p2)αγ2,b-2=1αkD0⋅2pD0D1(γ2-p2)+pD0αγ2-pD1α(γ2-p2)4D1(γ2-p2)+αγ2,a0=-p2((2D0D1-D1α)(γ2-p2)+αγ2D0)α2D0(4D1(γ2-p2)+αγ2)2×((3αD1+2D0D1)(γ2-p2)+α2γ2+αγ2D0),a1=2kpD1(γ2-p2)α2γ2(αγ2+4D1(γ2-p2))×(α2γ2+2αγ2D0+2αD1(γ2-p2)+4D0D1(γ2-p2)),a2=-2k2D0D1(γ2-p2)(αγ2)2(αγ2+2D1(γ2-p2)),p=γΔ,
where Δ is specified by (3.18).
This in turn gives the solutions of system (3.1)
(3.20)u(x,t)=a0+a1k(x-pt)+ξ0+a2(k(x-pt)+ξ0)2,S(x,t)=|k(x-pt)+ξ0|b-1⋅exp⁡⁡(b-2(k(x-pt)+ξ0)),v(x,t)=pγ(a0+a1k(x-pt)+ξ0+a2(k(x-pt)+ξ0)2),
where k is an arbitrary constant and ai(i=0,1,2), b-j(j=1,2),p are given in (3.19).Case 2.
We seek solutions in the form of (2.9). Let u(ξ)=∑i=0Maitaniξ and φ(ξ)=b-2ξ+b-1ln|cosξ|+∑j=0M-2bjtanjξ. Substituting u(ξ),φ(ξ) into (3.4) and (3.5), balancing the linear terms of highest order with the highest order nonlinear terms, directly results in M=2 and
(3.21)u=a0+a1tan⁡ξ+a2tan2⁡ξ,φ=b-2ξ+b-1ln⁡⁡|cos⁡⁡ξ|.
To progress further substituting (3.21) into (3.4) and (3.5) and setting the coefficients of taniξ(i=0,1,2) in the resulting equations to be zero, we obtain a set of algebraic equations
(3.22)2kD1(γ2-p2)a2-αγ2k(a1b-2-a0b-1)+γ2pa1=0,kD1(γ2-p2)a1-αγ2k(-a1b-1+a2b-2)+γ2pa2=0,6a2kD1(γ2-p2)+3αγ2ka2b-1=0,-kpb-2-D0k2(b-22-b-1)=a0,kpb-1+2D0k2b-1b-2=a1,-D0k2(b-12-b-1)=a2.
If D0≠0 and p satisfies
(3.23)Δ=-p4(5D12α2+4D0D12α+4D02D12)+p2γ2(2D1α3+10α2D12+8αD0D12+8D02D12)+γ4(α3D0+α2D02-2α3D1-5α2D12-4αD0D12-4D02D12)>0,
we know
(3.24)b-1=-2D1(γ2-p2)αγ2,(3.25)b-2=pαkD0⋅2D0D1(γ2-p2)+αγ2D0-D1α(γ2-p2)4D1(γ2-p2)+αγ2,(3.26)a0=-p2αD0⋅2D0D1(γ2-p2)+αγ2D0-αD1(γ2-p2)4D1(γ2-p2)+αγ2-p2α2D0⋅(2D0D1(γ2-p2)+αγ2D0-αD1(γ2-p2)4D1(γ2-p2)+αγ2)2-2k2D0D1(γ2-p2)αγ2,(3.27)a1=-2kpD1(γ2-p2)(α+2D0)(αγ2+2D1(γ2-p2))α2γ2(αγ2+4D1(γ2-p2)),(3.28)a2=-2k2D0D1(γ2-p2)(αγ2+2D1(γ2-p2))(αγ2)2,(3.29)k=pγ22ΔD02D12(γ2-p2)2(4D1(γ2-p2)+αγ2)2,
where Δ is given by (3.23). To this end, we therefore derive the solutions of system (3.1) as follows:
(3.30)u(x,t)=a0+a1tan⁡(k(x-pt)+ξ0)+a2tan2⁡(k(x-pt)+ξ0),S(x,t)=|cos⁡⁡(k(x-pt)+ξ0)|b-1eb-2(k(x-pt)+ξ0),v(x,t)=(a0+a1tan⁡(k(x-pt)+ξ0)+a2tan2⁡(k(x-pt)+ξ0))pγ,
where b-j(j=1,2) and ai(i=0,1,2) and k are given by (3.24)–(3.29), respectively.


## 4. Conclusion

In the present paper, we have considered the exact traveling wave solutions of a one-space dimensional hyperbolic model for chemotaxis (1.5). Using a transformation of dependent variables and eliminating the dependent variable v in the resultant system (3.1), the system (1.5) is reduced to a couple system of two equations (3.2). Applying the transformation (3.3) to unite the independent variables x and t, the (3.2) is convert to a coupled ordinary differential system. Then, the tanh method and improved tanh method are used, with the help of symbolic software MATHMATICA, to search for exact traveling wave solutions of system (3.1). Finally, three kinds of traveling wave solutions including soliton, rational, and triangular solutions are found, all of which are different from those known ones in the literature.



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*Source: 137406-2011-10-23.xml*