The Extended Tanh Method and the Exp-Function Method to Solve a Kind of Nonlinear Heat Equation

Abstract

We study a kind of nonlinear heat equation with temperature-dependent thermal properties by the aid of the extended Tanh method and the Exp-function method. We obtain abundant new exact solutions of the equation. By comparing both of the methods, we find that the Exp-function method gives more solutions in this problem.

1. Introduction

The classical heat equation 𝜕𝑢=𝜕𝜕𝑡²𝑢𝜕𝑥², (1.1) also known as the diffusion equation, describes in typical applications of the evolution in time of the density 𝑢=𝑢(𝑥,𝑡) of some quantities such as heat and chemical concentration [1,page 44]. In this case, the thermal diffusivity and thermal conductivity of the medium are assumed to be constant. However, in some media such as gases, the parameters are proportional to the temperature of the medium giving rise to a nonlinear heat equation of the following form [2]:𝐶(𝑥)𝜕𝑢𝜕𝜕𝑡=𝜆𝜕𝑥(𝑘𝑢𝜕𝑢𝜕𝑥), (1.2) where 𝐶=𝐶(𝑥) is the conductivity, 𝑘 is diffusivity, and 𝜆 is a constant. When the diffusivity is proportional to 𝑢^𝛼, a more general nonlinear heat equation reads as 𝐶(𝑥)𝜕𝑢𝜕𝜕𝑡=𝜆𝜕𝑥(𝑢^𝛼𝜕𝑢𝜕𝑥). (1.3) In a recent paper [3], using the Adomian decomposition method, the author discussed the following nonlinear heat equation with temperature dependent diffusivity:𝜕𝑢=𝜕𝜕𝑡𝜕𝑥(𝑓(𝑢)𝜕𝑢𝜕𝑥), (1.4) where 𝑓(𝑢)=𝑢^𝑚and 𝑚=2,−2,1/2.

In this paper we are interested in the following nonlinear heat equation:𝜕𝑢=𝜕𝜕𝑡𝜕𝑥(𝑢⁻¹𝜕𝑢𝜕𝑥) (1.5) and discuss its traveling wave solutions. As we know, a solution 𝑢 of the form𝑢(𝑥,𝑡)=𝑈(𝜉),𝜉=𝑘(𝑥−𝑐𝑡) (1.6) is called a traveling wave (with wavefront normal to 𝑘, velocity 𝑐/|𝑘|, and profile 𝑈) [1, page 172].

Here we employ, for the first time, the extended Tanh method and Exp-function method for solving (1.5), and abundant new exact solutions of (1.5) are presented. We compare both of the methods and find that the Exp-function method is more efficient than the extended Tanh method in this problem.

2. The Extended Tanh Method

We now describe the extended Tanh method for the given partial differential equations. The Tanh method was defined by Malfliet [4] and Fan and Hon [5]. The Tanh method was successfully applied to nonlinear evolution equations [6, 7], and so on. The extended Tanh method was presented in [8] to solve breaking solitary equation. Wazwaz summarized the main steps introduced for using this method as follows [9].

We consider first a general form of nonlinear partial differential equation involving the two variables 𝑡,𝑥𝑃(𝑢,𝑢_𝑡,𝑢_𝑥,𝑢_𝑥𝑥,…)=0. (2.1) In this paper we only discuss the traveling wave solutions.

(1) To find the traveling wave solution of (2.1), make the transformation 𝑢(𝑥,𝑡)=𝑈(𝜉),𝜉=𝑘(𝑥−𝑐𝑡), (2.2) where 𝑘,𝑐 are constants to be determined later. From this reason, we use the following changes: 𝜕𝑑𝜕𝑡=−𝑘𝑐,𝜕𝑑𝜉𝑑𝜕𝑥=−𝑘,𝜕𝑑𝜉₂𝜕𝑥²=𝑘²𝑑²𝑑𝜉²,𝜕³𝜕𝑥³=𝑘³𝑑³𝑑𝜉³,…, (2.3) and so on for the other derivates. Using (2.3) changes the NLPDE (2.1) to an ODE 𝑃(𝑈,𝑈^′,𝑈^′′,𝑈^′′′,…)=0. (2.4)  

(2) If all terms of the resulting ODE contain derivatives in 𝜉, then by integrating this equation, by considering the constant of integration to be zero, we obtain a simplified ODE.

(3) We then introduce a new independent variable 𝑌=tanh(𝜉)or𝑌=coth(𝜉) (2.5) that leads to the change of derivates 𝑑𝑑𝜉=(1−𝑌²)𝑑,𝑑𝑑𝑌₂𝑑𝜉²=(1−𝑌²𝑑)(−2𝑌𝑑𝑌+(1−𝑌²)𝑑²𝑑𝑌²𝑑),₃𝑑𝜉³=(1−𝑌²)((6𝑌²𝑑−2)𝑑𝑌−6𝑌(1−𝑌²)𝑑²𝑑𝑌²+(1−𝑌²)²𝑑³𝑑𝑌³), (2.6) where other derivatives can be derived in a similar manner. We use a new independent variable [9] 𝑌=tan(𝜉)or𝑌=−cot(𝜉) (2.7) that leads to the change of derivates 𝑑𝑑𝜉=(1+𝑌²)𝑑,𝑑𝑑𝑌₂𝑑𝜉²=(1+𝑌²𝑑)(−2𝑌𝑑𝑌+(1+𝑌²)𝑑²𝑑𝑌²𝑑),₃𝑑𝜉³=(1+𝑌²)((6𝑌²𝑑+2)𝑑𝑌+6𝑌(1+𝑌²)𝑑²𝑑𝑌²+(1+𝑌²)²𝑑³𝑑𝑌³), (2.8) where other derivatives can be derived.

(4) Introduce the ansatz 𝑈(𝜉)=^𝑚_𝑠=−𝑛𝑎_𝑠𝑌^𝑠, (2.9) where 𝑚,𝑛 are nonnegative integers, in most cases, that will be determined. Substituting (2.6) and (2.7) into the ODE (2.4) yields an equation in powers of 𝑌.

(5) To determine the parameter𝑚,𝑛, we usually balance linear derivative term of the highest order in the resulting equation with the highest order nonlinear terms [8, 9]. With 𝑚,𝑛 determined, equate the coefficients of powers of 𝑌 to zero in the resulting equation. This will give a system of algebraic equations involving the𝑎_𝑠, (𝑠=−𝑛,…,0,1,…,𝑚). Having determined these parameters, knowing that it is a positive integer in most cases, using (2.9) we obtain an analytic solution in a closed form.

It is worthy notice if 𝑛=0 in (2.9), then the extended Tanh method reduces to the Tanh method, so the Tanh method is a special case of the extended Tanh method.

3. The Exp-Function Method

Recently, He and Wu [10] proposed a straightforward and concise method called Exp-function method to obtain exact solutions of NLEEs. The Exp-function method leads to both generalized solitary solutions and periodic solutions [11–14] and was successfully applied to KdV equation with variable coefficients [15], to the combine KdV-mKdV equations with variable coefficients [16], to difference-differential equations [17, 18], and so forth. This paper applies the Exp-function method with the help of Mathematica computation to a kind of nonlinear heat equation with temperature-dependent thermal properties; abundant new exact solutions are hereby constructed. We consider a general nonlinear PDE in the form𝑃(𝑢,𝑢_𝑡,𝑢_𝑥,𝑢_𝑥𝑥,…)=0. (3.1) Using a transformation𝑢(𝑥,𝑡)=𝑈(𝜉),𝜉=𝛼(𝑥−𝛽𝑡), (3.2) where 𝛼,𝛽 are constants, we can rewrite (3.1) in the following nonlinear ODE:𝑄(𝑈,𝑈^′,𝑈^′′,𝑈^′′′,…)=0, (3.3) where the prime denotes the derivation with respect to 𝜉.

According to Exp-function method, we assume that the solution can be expressed in the form [10]∑𝑈(𝜉)=^{𝑑𝑛=−𝑐}𝑎_𝑛exp(𝑛𝜉)∑^𝑞_𝑚=−𝑝𝑏_𝑚exp(𝑚𝜉), (3.4) where 𝑐,𝑑,𝑝, and q are positive integers which could be freely chosen and 𝑎_𝑛 and 𝑏_𝑚 are unknown constants to be determined. In order to determine the values of 𝑐 and 𝑝, we balance the linear derivative term of highest order in (3.3) with the highest order nonlinear term [10]. Similarly, to determine the values of 𝑑 and 𝑞, we balance the linear derivative term of lowest order in (3.3) with the lowest order nonlinear term [10].

4. The Extended Tanh Method to the Nonlinear Heat Equation (1.5)

As described in Section 2, we make the transformation𝑢(𝑥,𝑡)=𝑈(𝜉),𝜉=𝑟(𝑥−𝑐𝑡), (4.1) and (1.5) becomes−𝑐𝑈²𝑈^′+𝑟(𝑈^′)²−𝑟𝑈𝑈^′′=0. (4.2) By balancing the nonlinear terms 𝑈²𝑈^′,(𝑈^′)²,𝑈𝑈^′′,we have 2𝑚+(𝑚+1)=2(𝑚+1)=𝑚+(𝑚+2),−2𝑛−(𝑛+1)=−2(𝑛+1)=−𝑛−(𝑛+1), (4.3) which yields 𝑚=𝑛=1. Therefore by the use of the Tanh method, we may choose a solution of (4.2) in the form𝑈=𝑈(𝜉)=𝑎₋₁𝑌⁻¹+𝑎₀+𝑎₁𝑌, (4.4) where 𝑌=tanh(𝜉)or 𝑌=coth(𝜉) and 𝑎_𝑗(𝑗=−1,0,1) are constants to be determined later. Substituting (4.4) into (4.2) we have 𝑌⁻⁴(𝐴₀+𝐴₁𝑌+𝐴₂𝑌²+𝐴₃𝑌³+𝐴₄𝑌⁴+𝐴₅𝑌⁵+𝐴₆𝑌⁶+𝐴₇𝑌⁷+𝐴₈𝑌⁸)=0, (4.5) where𝐴₀≡−𝑟𝑎²₋₁+𝑐𝑎³₋₁,𝐴₁≡−2𝑟𝑎₋₁𝑎₀+2𝑐𝑎²₋₁𝑎₀,𝐴₂≡−𝑐𝑎³₋₁+𝑐𝑎₋₁𝑎²₀−4𝑟𝑎₋₁𝑎₁+𝑐𝑎²₋₁𝑎₁,𝐴₃≡2𝑟𝑎₋₁𝑎₀−2𝑐𝑎²₋₁𝑎₀,𝐴₄≡𝑟𝑎²₋₁−𝑐𝑎₋₁𝑎²₀+8𝑟𝑎₋₁𝑎₁−𝑐𝑎²₋₁𝑎₁−𝑐𝑎²₀𝑎₁+𝑟𝑎²₁−𝑐𝑎₋₁𝑎²₁,𝐴₅≡2𝑟𝑎₀𝑎₁−2𝑐𝑎₀𝑎²₁,𝐴₆≡−4𝑟𝑎₋₁𝑎₁+𝑐𝑎²₀𝑎₁+𝑐𝑎₋₁𝑎²₁−𝑐𝑎³₁,𝐴₇≡−2𝑘𝑎₀𝑎₁+2𝑐𝑎₀𝑎²₁,𝐴₈=−𝑘𝑎²₁+𝑐𝑎³₁. (4.6)

Solving the following algebraic equation system with the aid of the Mathematica Package{𝐴₀=0,𝐴₁=0,𝐴₂=0,𝐴₃=0,𝐴₄=0,𝐴₅=0,𝐴₆=0,𝐴₇=0,𝐴₈=0}, (4.7) we then get the following results:(1){𝑎₋₁=(𝑟/𝑐),𝑎₀=−(2𝑟/𝑐),𝑎₁=(𝑟/𝑐)},(2){𝑎₋₁=(𝑟/𝑐),𝑎₀=(2𝑟/𝑐),𝑎₁=(𝑟/𝑐)},(3){𝑎₋₁=0,𝑎₀=−(𝑟/𝑐),𝑎₁=(𝑟/𝑐)},(4){𝑎₋₁=(𝑟/𝑐),𝑎₀=−(𝑟/𝑐),𝑎₁=0},(5){𝑎₋₁=0,𝑎₀=(𝑟/𝑐),𝑎₁=(𝑟/𝑐)},(6){𝑎₋₁=(𝑟/𝑐),𝑎₀=(𝑟/𝑐),𝑎₁=0},

where 𝑟,𝑐 are nonzero free parameters. Substituting these results into (4.4) and then changing to exponential form, we obtain the following exact solutions:𝑢₁₁𝑟(𝑥,𝑡)=𝑐(coth[𝑟(𝑥−𝑐𝑡)]−2+tanh[𝑟(𝑥−𝑐𝑡)])=4𝑟,𝑢𝑐(−1+exp[4𝑟(𝑥−𝑐𝑡)])₁₂𝑟(𝑥,𝑡)=𝑐(coth[𝑟(𝑥−𝑐𝑡)]+2+tanh[𝑟(𝑥−𝑐𝑡)])=4𝑟exp[4𝑟(𝑥−𝑐𝑡)],𝑢𝑐(−1+exp[4𝑟(𝑥−𝑐𝑡)])₁₃𝑟(𝑥,𝑡)=𝑐(−1+tanh[𝑟(𝑥−𝑐𝑡)])=−2𝑟,𝑢𝑐(1+exp[2𝑟(𝑥−𝑐𝑡)])₁₄𝑟(𝑥,𝑡)=𝑐(−1+coth[𝑟(𝑥−𝑐𝑡)])=2𝑟,𝑢𝑐(−1+exp[2r(𝑥−𝑐𝑡)])₁₅𝑟(𝑥,𝑡)=𝑐(1+tanh[𝑟(𝑥−𝑐𝑡)])=2𝑟exp[2𝑟(𝑥−𝑐𝑡)],𝑢𝑐(1+exp[2r(𝑥−𝑐𝑡)])₁₆𝑟(𝑥,𝑡)=𝑐(1+coth[𝑟(𝑥−𝑐𝑡)])=2𝑟exp[2𝑟(𝑥−𝑐𝑡)].𝑐(−1+exp[2𝑟(𝑥−𝑐𝑡)]) (4.8) If we choose the solution forms of (2.7) and insert them into (4.4) and (4.2), we have 𝑌⁻⁴(𝐵₀+𝐵₁𝑌+𝐵₂𝑌²+𝐵₃𝑌³+𝐵₄𝑌⁴+𝐵₅𝑌⁵+𝐵₆𝑌⁶+𝐵₇𝑌⁷+𝐵₈𝑌⁸)=0, (4.9) where 𝐵₀≡−𝑟𝑎²₋₁+𝑐𝑎³₋₁,𝐵₁≡−2𝑟𝑎₋₁𝑎₀+2𝑐𝑎²₋₁𝑎₀,𝐵₂≡𝑐𝑎³₋₁+𝑐𝑎₋₁𝑎²₀−4𝑟𝑎₋₁𝑎₁+𝑐a²₋₁𝑎₁,𝐵₃≡−2𝑟𝑎₋₁𝑎₀+2𝑐𝑎²₋₁𝑎₀,𝐵₄≡𝑟𝑎²₋₁+𝑐𝑎₋₁𝑎²₀−8𝑟𝑎₋₁𝑎₁+𝑐𝑎²₋₁𝑎₁−𝑐𝑎²₀𝑎₁+𝑟𝑎²₁−𝑐𝑎₋₁𝑎²₁,𝐵₅≡−2𝑟𝑎₀𝑎₁−2𝑐𝑎₀𝑎²₁,𝐵₆≡−4𝑟𝑎₋₁𝑎₁−𝑐𝑎²₀𝑎₁−𝑐𝑎₋₁𝑎²₁−𝑐𝑎³₁,𝐵₇≡−2𝑟𝑎₀𝑎₁−2𝑐𝑎₀𝑎²₁,𝐵₈≡−𝑟𝑎²₁−𝑐𝑎³₁. (4.10)

Solving the following algebraic equation system with the aid of the Mathematica Package{𝐵₀=0,𝐵₁=0,𝐵₂=0,𝐵₃=0,𝐵₄=0,𝐵₅=0,𝐵₆=0,𝐵₇=0,𝐵₈=0}, (4.11) we then get the following results:(1){𝑎₋₁=(𝑟/𝑐),𝑎₀=−(2𝑖𝑟/𝑐),𝑎₁=−(𝑟/𝑐)},(2){𝑎₋₁=(𝑟/𝑐),𝑎₀=(2𝑖𝑟/𝑐),𝑎₁=−(𝑟/𝑐)},(3){𝑎₋₁=0,𝑎₀=−(𝑖𝑟/𝑐),𝑎₁=−(𝑟/𝑐)},(4){𝑎₋₁=0,𝑎₀=(𝑖𝑟/𝑐),𝑎₁=−(𝑟/𝑐)},(5){𝑎₋₁=(𝑟/𝑐),𝑎₀=−(𝑖𝑟/𝑐),𝑎₁=0},(6){𝑎₋₁=(𝑟/𝑐),𝑎₀=(𝑖𝑟/𝑐),𝑎₁=0},

where 𝑟,c are nonzero free parameters and 𝑖²=−1. Substituting these results into (4.4) and changing to exponential form, we obtain the following exact solutions:𝑢₂₁𝑟(𝑥,𝑡)=𝑐(cot[𝑟(𝑥−𝑐𝑡)]−2𝑖−tan[𝑟(𝑥−𝑐𝑡)])=4𝑟𝑖exp[4𝑟𝑖(𝑥−𝑐𝑡)],𝑢𝑐(−1+exp[4𝑟𝑖(𝑥−𝑐𝑡)])₂₂𝑟(𝑥,𝑡)=𝑐(cot[𝑟(𝑥−𝑐𝑡)]+2𝑖−tan[𝑟(𝑥−𝑐𝑡)])=4𝑟𝑖,𝑢𝑐(−1+exp[4𝑟𝑖(𝑥−𝑐𝑡)])₂₃𝑟(𝑥,𝑡)=𝑐(−𝑖−tan[𝑟(𝑥−𝑐𝑡)])=−2𝑟𝑖,𝑢𝑐(1+exp[2𝑟𝑖(𝑥−𝑐𝑡)])₂₄𝑟(𝑥,𝑡)=𝑐(𝑖−tan[𝑟(𝑥−𝑐𝑡)])=2𝑟𝑖exp[2𝑟𝑖(𝑥−𝑐𝑡)],𝑢𝑐(1+exp[2𝑟𝑖(𝑥−𝑐𝑡)])₂₅𝑟(𝑥,𝑡)=𝑐(−𝑖+cot[𝑟(𝑥−𝑐𝑡)])=2𝑟𝑖,𝑢𝑐(−1+exp[2𝑟𝑖(𝑥−𝑐𝑡)])₂₆𝑟(𝑥,𝑡)=𝑐(𝑖+cot[𝑟(𝑥−𝑐𝑡)])=2𝑟𝑖exp[2𝑟𝑖(𝑥−𝑐𝑡)].𝑐(−1+exp[2𝑟𝑖(𝑥−𝑐𝑡)]) (4.12)

5. The Exp-Function Method to the Nonlinear Heat Equation (1.5)

In this section, the Exp-function method is applied to the nonlinear heat equation (1.5).

Using the transformation𝑢(𝑥,𝑡)=𝑈(𝜉),𝜉=𝑘(𝑥−𝑐𝑡), (5.1) (1.5) becomes−𝑐𝑈²𝑈^′+𝑘(𝑈^′)²−𝑘𝑈𝑈^′′=0. (5.2) Here we assume that the solution of (5.2) can be expressed in the following form [10]:∑𝑈(𝜉)=^{𝑛𝑖=−𝑚}𝑎_𝑖exp(𝑖𝜉)∑^𝑡_𝑗=−𝑠𝑏_𝑚=𝑎exp(𝑗𝜉)^−𝑚exp(−𝑚𝜉)+⋯+𝑎_𝑛exp(𝑛𝜉)𝑏_−sexp(−𝑠𝜉)+⋯+𝑏_𝑡exp(𝑡𝜉), (5.3) where 𝑎_𝑖,𝑏_𝑗(𝑖,𝑗∈ℤ) are unknown constants and 𝑚,𝑛,𝑠,𝑡 are nonnegative integers to be further determined. Here take notice of nonlinear term in (5.2), and we can balance 𝑈²𝑈^′,(𝑈^′)² and 𝑈𝑈^′′ by the idea of the Exp-function method [10] to determine the values of 𝑚,𝑛𝑠,𝑡. By simple calculation, we have 𝑈𝑈^′=𝑐₁exp[−(2𝑚+3𝑠)𝜉]+⋯+𝑐₂exp[(2𝑛+3𝑡)𝜉]𝑑₁exp[−5𝑠𝜉]+⋯+𝑑₂,exp[5𝑡𝜉](𝑈^′)²=𝑐₃exp[−(2𝑚+3𝑠)𝜉]+⋯+𝑐₄exp[(2𝑛+3𝑡)𝜉]𝑑₃exp[−5𝑠𝜉]+⋯+𝑑₄,exp[5𝑡𝜉]𝑈𝑈^′′=𝑐₅exp[−(2𝑚+3𝑠)𝜉]+⋯+𝑐₆exp[(2𝑛+3𝑡)𝜉]𝑑₅exp[−5𝑠𝜉]+⋯+𝑑₆.exp[5𝑡𝜉] (5.4) According to (5.4), we find that 𝑚,𝑛,𝑠,𝑡 are arbitrary nonnegative integers. This provides great freedom to choose 𝑚,𝑛, 𝑠,𝑡and may be get more abundant solutions of (1.5). For simplicity, we only discuss the following one case, that is 𝑚=𝑠=1 and 𝑛=𝑡=1. In this case, (5.3) reduces to 𝑎𝑈=𝑈(𝜉)=⁻¹exp(−𝜉)+𝑎₀+𝑎₁exp(𝜉)𝑏₋₁exp(−𝜉)+𝑏₀+𝑏₁exp(𝜉). (5.5)

Substituting (5.5) into (5.2) by help of Mathematica package computation yields𝐴(𝐴₁𝑒^𝜉+𝐴₂𝑒^2𝜉+𝐴₃𝑒^3𝜉+𝐴₄𝑒^4𝜉+𝐴₅𝑒^5𝜉+𝐴₆𝑒^6𝜉+𝐴₇𝑒^7𝜉)=0, (5.6) where 𝐴=(𝑏₋₁+𝑏₀𝑒^𝜉+𝑏₁𝑒^2𝜉)⁴,𝐴₁=−𝑐𝑎²₋₁𝑎₀𝑏₋₁−𝑘𝑎₋₁𝑎₀𝑏²₋₁+𝑐𝑎³₋₁𝑏₀+𝑘𝑎²₋₁𝑏₋₁𝑏₀,𝐴₂=−2𝑐𝑎₋₁𝑎²₀𝑏₋₁−2𝑐𝑎²₋₁𝑎₁𝑏₋₁−4𝑘𝑎₋₁𝑎₁𝑏²₋₁+2𝑐𝑎²₋₁𝑎₀𝑏₀+2𝑐𝑎³₋₁𝑏₁+4𝑘𝑎²₋₁𝑏₋₁𝑏₁,𝐴₃=−𝑐𝑎²₀𝑏₋₁−6𝑐𝑎₋₁𝑎₀𝑎₁𝑏₋₁−𝑘𝑎₀𝑎₁𝑏²₋₁+𝑐𝑎₋₁𝑎²₀𝑏₀+𝑐𝑎²₋₁𝑎₁𝑏₀+𝑘𝑎²₀𝑏₋₁𝑏₀−6𝑘𝑎₋₁𝑎₁𝑏₋₁𝑏₀−𝑘𝑎₋₁𝑎₀𝑏²₀+5𝑐𝑎²₋₁𝑎₀𝑏₁+6𝑘𝑎₋₁𝑎₀𝑏₋₁𝑏₁+𝑘𝑎²₋₁𝑏₀𝑏₁,𝐴₄=−4𝑐𝑎²₀𝑎₁𝑏₋₁−4𝑐𝑎₋₁𝑎²₁𝑏₋₁−4𝑘𝑎₋₁𝑎₁𝑏²₀+4𝑐𝑎₋₁𝑎²₀𝑏₁+4𝑐𝑎²₋₁𝑎₁𝑏₁+4𝑘𝑎²₀𝑏₋₁𝑏₁,𝐴₅=−5𝑐𝑎₀𝑎²₁𝑏₋₁−𝑐𝑎²₀𝑎₁𝑏₀−𝑐𝑎₋₁𝑎²₁𝑏₀+𝑘𝑎²₁𝑏₋₁𝑏₀−𝑘𝑎₀𝑎₁𝑏²₀+𝑐𝑎³₀𝑏₁+6𝑐𝑎₋₁𝑎₀𝑎₁𝑏₁+6𝑘𝑎₀𝑎₁𝑏₋₁𝑏₁+𝑘𝑎²₀𝑏₀𝑏₁−6𝑘𝑎₋₁𝑎₁𝑏₀𝑏₁−𝑘𝑎₋₁𝑎₀𝑏²₁,𝐴₆=−2𝑐𝑎³₁𝑏₋₁−2𝑐𝑎₀𝑎²₁𝑏₀+2𝑐𝑎²₀𝑎₁𝑏₁+2𝑐𝑎₋₁𝑎²₁𝑏₁+4𝑘𝑎²₁𝑏₋₁𝑏₁−4𝑘𝑎₋₁𝑎₁𝑏²₁,𝐴₇=−𝑐𝑎³₁𝑏₀+𝑐𝑎₀𝑎²₁𝑏₁+𝑘𝑎²₁𝑏₀𝑏₁−𝑘𝑎₀𝑎₁𝑏²₁. (5.7)

Equating the coefficients of 𝑒^𝑛𝜉(𝑛=1,2,…,7) to zero, we get a set of algebraic equations{𝐴₁=0,𝐴₂=0,𝐴₃=0,𝐴₄=0,𝐴₅=0,𝐴₆=0,𝐴₇=0}. (5.8) Solving the above system by using Mathematica Package, we can get the solution as follows:(1){𝑎₋₁=0,𝑎₀=𝑎₀,𝑎₁=0,𝑏₋₁=0,𝑏₀=−(𝑐𝑎₀/𝑘),𝑏₁=𝑏₁}.(2){𝑎₋₁=0,𝑎₀=0,𝑎₁=𝑎₁,𝑏₋₁=0,𝑏₀=𝑏₀,𝑏₁=(𝑐𝑎₁/𝑘)}.(3){𝑎₋₁=𝑎₋₁,𝑎₀=0,𝑎₁=0,𝑏₋₁=−(𝑐𝑎₋₁/𝑘),𝑏₀=𝑏₀,𝑏₁=0}.(4){𝑎₋₁=𝑎₋₁,𝑎₀=0,𝑎₁=0,𝑏₋₁=−(𝑐𝑎₋₁/2𝑘),𝑏₀=0,𝑏₁=𝑏₁}.(5){𝑎₋₁=0,𝑎₀=0,𝑎₁=𝑎₁,𝑏₋₁=0,𝑏₀=0,𝑏₁=(𝑐𝑎₁/2𝑘)}.(6){𝑎₋₁=0,𝑎₀=𝑎₀,𝑎₁=0,𝑏₋₁=𝑏₋₁,𝑏₀=(𝑐𝑎₀/𝑘),𝑏₁=0}.(7){𝑎₋₁=𝑎₋₁,𝑎₀=𝑎₀,𝑎₁=0,𝑏₋₁=−(𝑐𝑎₋₁/𝑘),𝑏₀=𝑏₀,𝑏₁=𝑎₀(𝑐𝑎₀+𝑘𝑏₀)/𝑘𝑎₋₁}.(8){𝑎₋₁=0,𝑎₀=𝑎₀,𝑎₁=𝑎₁,𝑏₋₁=−(𝑎₀(𝑐𝑎₀−𝑘𝑏₀)/𝑘𝑎₁),𝑏₀=𝑏₀,𝑏₁=(𝑐𝑎₁/𝑘)}.(9){𝑎₋₁=𝑎₋₁,𝑎₀=𝑎₀,𝑎₁=0,𝑏₋₁=−(𝑐𝑎₋₁/𝑘),𝑏₀=−(𝑐𝑎₀/2𝑘),𝑏₁=(𝑐𝑎²₀/2𝑘𝑎₋₁)}.(10){𝑎₋₁=0,𝑎₀=𝑎₀,𝑎₁=𝑎₁,𝑏₋₁=−(𝑐𝑎²₀/2𝑘𝑎₁),𝑏₀=(𝑐𝑎₀/2𝑘),𝑏₁=(𝑐𝑎₁/𝑘)}.

Substituting cases (1)–(10) into (5.5) yields𝑢₃₁=𝑎₀−(𝑐𝑎₀/𝑘)+𝑏₁=exp(𝜉)𝑘𝑎₀−𝑐𝑎₀+𝑘𝑏₁.𝑢exp(𝑘(𝑥−𝑐𝑡))₃₂=𝑎₁exp(𝜉)(𝑐𝑎₁/𝑘)exp(𝜉)+𝑏₀=𝑘𝑎₁exp(𝑘(𝑥−𝑐𝑡))𝑘𝑏₀+𝑐𝑎₁.𝑢exp(𝑘(𝑥−𝑐𝑡))₃₃=𝑎₋₁exp(−𝜉)−(𝑐𝑎₋₁/𝑘)exp(−𝜉)+𝑏₀=𝑘𝑎₋₁exp[−𝑘(𝑥−𝑐𝑡)]−𝑐𝑎₋₁exp[−𝑘(𝑥−𝑐𝑡)]+𝑘𝑏₀.𝑢₃₄=𝑎₋₁exp(−𝜉)−(𝑐𝑎₋₁/2𝑘)exp(−𝜉)+𝑏₁=exp(𝜉)2𝑘𝑎₋₁exp[−𝑘(𝑥−𝑐𝑡)]−𝑐𝑎₋₁exp[−𝑘(𝑥−𝑐𝑡)]+2𝑘𝑏₁.𝑢exp[𝑘(𝑥−𝑐𝑡)]₃₅=𝑎₁exp(𝜉)𝑏₋₁exp(−𝜉)+(𝑐𝑎₋₁=/2𝑘)exp(𝜉)2𝑘𝑎₁exp[𝑘(𝑥−𝑐𝑡)]2𝑘𝑏₋₁exp[−𝑘(𝑥−𝑐𝑡)]+𝑐𝑎₁.𝑢exp[𝑘(𝑥−𝑐𝑡)]₃₆=𝑎₀𝑏₋₁exp(−𝜉)+(𝑐𝑎₀=/𝑘)𝑘𝑎₀𝑘𝑏₋₁exp[−𝑘(𝑥−𝑐𝑡)]+𝑐𝑎₀.𝑢₃₇=𝑎₋₁exp(−𝜉)+𝑎₀−(𝑐𝑎₋₁/𝑘)exp(−𝜉)+𝑏₀+(𝑎₀(𝑐𝑎₀+𝑘𝑏₀)/𝑘𝑎₋₁=)exp(𝜉)𝑘𝑎²₋₁exp[−𝑘(𝑥−𝑐𝑡)]+𝑘𝑎₋₁𝑎₀−𝑐𝑎²₋₁exp[−𝑘(𝑥−𝑐𝑡)]+𝑘𝑎₋₁𝑏₀+𝑎₀(𝑐𝑎₀+𝑘𝑏₀.𝑢)exp[𝑘(𝑥−𝑐𝑡)]₃₈=𝑎₀+𝑎₁exp(𝜉)−(𝑎₀(𝑐𝑎₀−𝑘𝑏₀)/𝑘𝑎₁)exp(−𝜉)+𝑏₀+(𝑐𝑎₁=/𝑘)exp(𝜉)𝑘𝑎₁𝑎₀+𝑘𝑎²₁exp[𝑘(𝑥−𝑐𝑡)]𝑎₀(𝑘𝑏₀−𝑐𝑎₀)exp[−𝑘(𝑥−𝑐𝑡)]+𝑘𝑎₁𝑏₀+𝑐𝑎²₁.𝑢exp[𝑘(𝑥−𝑐𝑡)]₃₉=𝑎₀+𝑎₋₁exp(−𝜉)−(𝑐𝑎₋₁/𝑘)exp(−𝜉)−(𝑐𝑎₀/2𝑘)+(𝑐𝑎²₀/2𝑘𝑎₋₁=)exp(𝜉)2𝑘𝑎₋₁𝑎₀+2𝑘𝑎²₋₁exp[−𝑘(𝑥−𝑐𝑡)]−2𝑐𝑎²₋₁exp[−𝑘(𝑥−𝑐𝑡)]−𝑐𝑎₋₁𝑎₀+𝑐𝑎²₀.𝑢exp[𝑘(𝑥−𝑐𝑡)]_3,10=𝑎₀+𝑎₁exp(𝜉)−(𝑐𝑎²₀/2𝑘𝑎₁)exp(−𝜉)+(𝑐𝑎₀/2𝑘)+(𝑐𝑎₁=/𝑘)exp(𝜉)2𝑘𝑎₁𝑎₀+2𝑘𝑎²₁exp[𝑘(𝑥−𝑐𝑡)]−𝑐𝑎²₀exp[−𝑘(𝑥−𝑐𝑡)]+𝑐𝑎₁𝑎₀+2𝑐𝑎²₁.exp[𝑘(𝑥−𝑐𝑡)] (5.9)

6. Comparison and Discussion

In this section we make comparison between the Tanh method’s solutions and the Exp-function method solutions of (1.5). We can obtain the following results.(1)If we set 𝑘=4𝑟,𝑎₀=1,𝑏₁=(𝑐/4𝑟) in the equation 𝑢₃₁, then 𝑢₃₁=𝑢₁₁,(2)If we set 𝑘=4𝑟,𝑎₁=1,𝑏₀=−(𝑐/4𝑟) in the equation 𝑢₃₂, then 𝑢₃₂=𝑢₁₂,(3)If we set 𝑘=2𝑟,𝑎₀=−1,𝑏₁=−(𝑐/2𝑟) in the equation 𝑢₃₁, then 𝑢₃₁=𝑢₁₃,(4)If we set 𝑘=2𝑟,𝑎₀=1,𝑏₁=(𝑐/2𝑟) in the equation 𝑢₃₁, then 𝑢₃₁=𝑢₁₄,(5)If we set 𝑘=2𝑟,𝑎₁=1,𝑏₀=(𝑐/2𝑟) in the equation 𝑢₃₂, then 𝑢₃₂=𝑢₁₅,(6)If we set 𝑘=2𝑟,𝑎₁=1,𝑏₀=−(𝑐/2𝑟) in the equation 𝑢₃₂, then 𝑢₃₂=𝑢₁₆,(7)If we set 𝑘=4𝑟𝑖,𝑎₁=1,𝑏₀=−(𝑐/4𝑟) in the equation 𝑢₃₂, then 𝑢₃₂=𝑢₂₁,(8)If we set 𝑘=4𝑟𝑖,𝑎₀=1,𝑏₁=(𝑐/4𝑟𝑖) in the equation 𝑢₃₁, then 𝑢₃₁=𝑢₂₂,(9)If we set 𝑘=2𝑟𝑖,𝑎₀=1,𝑏₁=−(𝑐/2𝑟𝑖) in the equation 𝑢₃₁, then 𝑢₃₁=𝑢₂₃,(10) If we set 𝑘=2𝑟𝑖,𝑎₁=1,𝑏₀=(𝑐/2𝑟𝑖) in the equation 𝑢₃₂, then 𝑢₃₂=𝑢₂₄,(11) If we set 𝑘=2𝑟𝑖,𝑎₀=1,𝑏₁=(𝑐/2𝑟𝑖) in the equation 𝑢₃₁, then 𝑢₃₁=𝑢₂₅,(12) If we set 𝑘=2𝑟𝑖,𝑎₁=1,𝑏₀=−(𝑐/2𝑟𝑖) in the equation 𝑢₃₂, then 𝑢₃₂=𝑢₂₆.

where 𝑖²=−1. The above obtained results show that the Exp-function method can obtain more abundant explicit solutions than Tanh method for (1.5). If we use the method of separation of variables [1, page 167], the rational solution of (1.5) can be constructed as follows: 𝑢(𝑥,𝑡)=2𝑎²(𝑘𝑡+𝑏)𝑘(𝑎𝑥+𝑐)², (6.1) where 𝑘,𝑎,𝑏,𝑐 are constants.

7. Conclusion

Nonlinear phenomena appear in a wide variety of scientific fields, such as applied mathematics, physics and engineering problems. However, solving nonlinear differential equations corresponding to the nonlinear problems are often complicate. Especially, obtaining their explicit solutions is even more difficult. Up to now, a lot of new methods for solving nonlinear differential equations are developed, for example, Bäcklund transformation method, inverse scattering method, Darboux transformation method, Hirota’s bilinear method, and so forth. But, generally speaking, all of the above methods have their own advantages and shortcomings, respectively. In this paper, by applying the Exp-function method and the extended Tanh method with the help of Mathematica computation to the nonlinear heat equation with temperature-dependent thermal properties, we obtain abundant exact solutions. The obtained results show that the Exp-function method and the extended Tanh method are simple and effective methods to solve nonlinear differential equations. By comparison, we find that the Exp-function method is more effective in finding exact solutions than the extended Tanh method for (1.5).

Acknowledgment

The author would like to thank the referee for the helpful suggestions which improved the exposition of this paper.