# A Simplification for Exp-Function Method When the Balanced Nonlinear Term Is a Certain Product

Authors: Hong-Zhun Liu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The Exp-function method plays an important role in searching for analytic solutions of many nonlinear differential equations. In this paper, we prove that the balancing procedure in the method is unnecessary when the balanced nonlinear term is a product of the dependent variable under consideration and its derivatives. And in this case, the ansatz of the method can be simplified to be with less parameters so as to be easy to calculate.


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## Body

## 1. Introduction

In 2006, He and Wu firstly proposed the so-called Exp-function method to search for solitary solutions and periodic solutions of nonlinear partial differential equations (PDE) [1]. This method soon drew the attention of many researchers and was successfully applied to many nonlinear problems [2–31]. Among them, it is worth mentioning that Zhu firstly applied this method to difference-differential equations, which shows that the method is also effective in this case [8, 9]. After Zhu, Dai et al. generalized the Exp-function method to find new exact traveling wave solutions of nonlinear PDE and nonlinear differential-difference equations [19]. Recently, Ma et al. extended the Exp-function method to multiple Exp-function method for constructing multiple wave solutions [32, 33]. He elucidated how to solve fractional differential equations with local fractional derivatives via the fractional complex transformation and the Exp-function method [34].For convenience, we first introduce the Exp-function method in brief.


### 1.1. Outline of the Exp-Function Method

Suppose that we consider a (1+1)-dimensional nonlinear PDE in the form(1)E(u,ut,ux,uxx,uxt,utt,…)=0.
Using traveling wave transformation
(2)u=u(η),η=kx+ωt,
we get a nonlinear ordinary differential equation (ODE)
(3)P(u,u′,u′′,…)=0,
where the prime, as it is in the following, denotes the derivation with respect to η.The Exp-function method is based on the assumption that the solutions of (3) can be expressed in the following form:
(4)u(η)=∑n=-cdanexp(nη)∑n=-pqbnexp(nη)=a-cexp(-cη)+⋯+adexp(dη)b-pexp(-pη)+⋯+bqexp(qη),
where c, d, p, and q are positive integers to be determined and an and bn are constants to be specified.Then we can express the highest order nonlinear and linear terms in (3) in terms of (4). In the resulting terms, determine d and q through balancing the highest order Exp-function and c and p by balancing the lowest order one.Substituting (4) along with the determined c, d, p, and q into (3), we can obtain an equation for exp(η). Setting all the coefficients of the different powers of exp(η) to zero leads to a set of algebraic equations for an, bn, k, and ω. Determine values of an, bn, k, and ω by solving this algebraic equations and put these values into (4). Thus we may obtain nontrivial exact traveling wave solutions of (1).


### 1.2. An Open Problem

Among the Exp-function method, the balancing computation is laborious but prior. However, we observe that the balancing procedure of the Exp-function method in studied examples always leads to the same casec=p and d=q [21]. This fact has partly been proved in [16, 22]. In [16], Ali has proved it by assuming the highest order linear and nonlinear terms as u(n) and uru(s) (s<n), respectively. In [22], making use of the same approach as was done by Ebaid proved the fact for nonlinear terms in the form uγ (γ⩾2), u(s)uk (s,k⩾1), [u(s)]Ω (s⩾1,Ω⩾2), and [u(s)]Ωuλ  (s,Ω,λ⩾1), respectively, along with linear term u(r)(r⩾1). Ebaid claimed in the abstract and section “Conclusions” of his article that the case c=p and d=q is the only relation that “can be obtained through applying the Exp-function ansatz for all possible cases of nonlinear ODEs.”“However, one cannot construct a general form for the highest order nonlinear term because there are many possibilities other than the ones considered” [21]; Aslan and Marinakis concluded that these authors just took some special cases of the nonlinear term into account and hence the problem still remained open. In this paper, we will construct a special case to show that Ebaid's claim is not true; namely, the case c=p and d=q is not the only relation for some special differential equations and hence the problem is still open.In what follows, we will discuss the relations ofc, d, p, and q in a more effective and concise way.


## 1.1. Outline of the Exp-Function Method

Suppose that we consider a (1+1)-dimensional nonlinear PDE in the form(1)E(u,ut,ux,uxx,uxt,utt,…)=0.
Using traveling wave transformation
(2)u=u(η),η=kx+ωt,
we get a nonlinear ordinary differential equation (ODE)
(3)P(u,u′,u′′,…)=0,
where the prime, as it is in the following, denotes the derivation with respect to η.The Exp-function method is based on the assumption that the solutions of (3) can be expressed in the following form:
(4)u(η)=∑n=-cdanexp(nη)∑n=-pqbnexp(nη)=a-cexp(-cη)+⋯+adexp(dη)b-pexp(-pη)+⋯+bqexp(qη),
where c, d, p, and q are positive integers to be determined and an and bn are constants to be specified.Then we can express the highest order nonlinear and linear terms in (3) in terms of (4). In the resulting terms, determine d and q through balancing the highest order Exp-function and c and p by balancing the lowest order one.Substituting (4) along with the determined c, d, p, and q into (3), we can obtain an equation for exp(η). Setting all the coefficients of the different powers of exp(η) to zero leads to a set of algebraic equations for an, bn, k, and ω. Determine values of an, bn, k, and ω by solving this algebraic equations and put these values into (4). Thus we may obtain nontrivial exact traveling wave solutions of (1).


## 1.2. An Open Problem

Among the Exp-function method, the balancing computation is laborious but prior. However, we observe that the balancing procedure of the Exp-function method in studied examples always leads to the same casec=p and d=q [21]. This fact has partly been proved in [16, 22]. In [16], Ali has proved it by assuming the highest order linear and nonlinear terms as u(n) and uru(s) (s<n), respectively. In [22], making use of the same approach as was done by Ebaid proved the fact for nonlinear terms in the form uγ (γ⩾2), u(s)uk (s,k⩾1), [u(s)]Ω (s⩾1,Ω⩾2), and [u(s)]Ωuλ  (s,Ω,λ⩾1), respectively, along with linear term u(r)(r⩾1). Ebaid claimed in the abstract and section “Conclusions” of his article that the case c=p and d=q is the only relation that “can be obtained through applying the Exp-function ansatz for all possible cases of nonlinear ODEs.”“However, one cannot construct a general form for the highest order nonlinear term because there are many possibilities other than the ones considered” [21]; Aslan and Marinakis concluded that these authors just took some special cases of the nonlinear term into account and hence the problem still remained open. In this paper, we will construct a special case to show that Ebaid's claim is not true; namely, the case c=p and d=q is not the only relation for some special differential equations and hence the problem is still open.In what follows, we will discuss the relations ofc, d, p, and q in a more effective and concise way.


## 2. Main Result

### 2.1. Some Terminology

To begin with, we recall some terminology in [35].A monomial in a collection of variablesx1,…,xn is a product
(5)x1α1x2α2…xnαn,
where the αi are nonnegative integers.The total degree of (5) is the sum of the exponents: α1+⋯+αn.A polynomial is said to be homogeneous if all the monomials appearing in it with nonzero coefficients have the same total degree.For instance,u2u′u′′3 is a product of u and its derivatives u′ and u′′, and its total degree is 6. u2+u′2+uu′′ is homogeneous.


### 2.2. Two Introducing Ansatz Function

For convenience, we assumev(η) is expressed in the form
(6)v(η)=∑n=-c1d1knexp(nη)∑n=-p1q1lnexp(nη)=k-c1exp(-c1η)+⋯+kd1exp(d1η)l-p1exp(-p1η)+⋯+lq1exp(q1η),
where c1, d1, p1, and q1 are positive integers to be determined and kn and ln are constants to be specified.The following three formulas will be used in this section:(7)u′(η)=(p-c)a-cb-pexp(-(c+p)η)+⋯+(d-q)adbqexp((d+q)η)b-p2exp(-2pη)+⋯+bq2exp(2qη),(8)u(η)·v(η)=a-ck-c1exp(-(c+c1)η)+⋯+adkd1exp((d+d1)η)b-pl-p1exp(-(p+p1)η)+⋯+bqlq1exp((q+q1)η),(9)u(η)v(η)=a-cl-p1exp(-(c+p1)η)+⋯+adlq1exp((d+q1)η)b-pk-c1exp(-(p+c1)η)+⋯+bqkd1exp((q+d1)η).Before presenting our definitions, we recall an important fact that the constantsa-c, ad, b-p, and bq in the ansatz of (4) can be assumed to be nonzero during the process of balancing the linear term of highest order with the highest order nonlinear term of certain ODE, and so do the constants k-c1, kd, l-p1, and lq in ansatz (6). Hence, in this paper, we assume that all above eight constants are nonzero.Therefore, for ansatz (4), we can define the ansatz functions L(·) and R(·) as follows:
(10)L(u)=L(a-cexp(-cη)+⋯+adexp(dη)b-pexp(-pη)+⋯+bqexp(qη))=-c-(-p)=p-c,R(u)=R(a-cexp(-cη)+⋯+adexp(dη)b-pexp(-pη)+⋯+bqexp(qη))=d-q.
In particular, we define L(C)=R(C)=0 for arbitrary constant C.For example, we have(11)L(exp(-η)+exp(2η)exp(-2η)+exp(2η))=1,R(exp(-η)+exp(2η)exp(-2η)+exp(2η))=0.
According to the definitions, we can find that in ansatz (4) c=p equals L(u)=0 and d=q equals R(u)=0. Therefore, the open problem for ansatz (4) is equal to whether the relations L(u)=0 and L(u)=0 hold.


### 2.3. Properties of the Ansatz Functions

Assumingc≠p, d≠q, c1≠p1, and d1≠q1 in ansatz (4) and (6), from (7), we obtain
(12)L(u′)=2p-(c+p)=p-c=L(u),R(u′)=(d+q)-2q=d-q=R(u).
So we have
(13)L(u)=L(u′)=L(u′′)=⋯=L(u(n)),R(u)=R(u′)=⋯=R(u(n)),
for arbitrary nonnegative integer n.And from (8) and (9), we have
(14)L(u·υ)=(p+p1)-(c+c1)L(u·v)=(p-c)+(p1-c1)L(u·v)=L(u)+L(υ),R(u·υ)=(d+d1)-(q+q1)L(u·v)=(d-q)+(d1-q1)L(u·v)=R(u)+R(υ),L(uυ)=(p+c1)-(c+p1)L(uv)=(p-c)-(p1-c1)L(uv)=L(u)-L(υ),R(uυ)=(d+q1)-(q+d1)L(uv)=(d-q)-(d1-q1)L(uv)=R(u)-R(υ).
Hence, we have
(15)L(uκ)=κ·L(u),R(uκ)=κ·R(u),
for any integer κ.


### 2.4. Theorem and Proof

In this section, we assume the balanced nonlinear term is a product of dependent variableu and its derivatives; namely,
(16)ui1(u′)i2⋯(u(m))im,
where ij (j=1,…,m) are nonnegative integers. The fact that the product (16) is a nonlinear term implies i1+i2+⋯+im⩾2. In other words, the total degree of (16) is at least 2.Theorem 1.
Suppose that the balanced nonlinear term in (3) is a product of u and its derivatives in the form of (16) and the balancing linear term is u(s), where s is a nonnegative integer; then the Exp-function ansatz (4) admits L(u)=0 and R(u)=0.Proof.
By contradiction, suppose thatL(u)≠0. Then we have
(17)L(ui1(u′)i2⋯(u(m))im)=L(ui1)+L((u′)i2)+⋯+L((u(m))im)=i1·L(u)+i2·L(u′)+⋯+im·L(u(m))=i1·L(u)+i2·L(u)+⋯+im·L(u)=(i1+i2+⋯+im)·L(u),L(u(s))=L(u).
Balancing linear and nonlinear terms requires
(18)L(ui1(u′)i2…(u(m))im)=L(u(s)).
So we obtain
(19)(i1+i2+⋯+im-1)·L(u)=0.
Since i1+i2+⋯+im-1≠0, we arrive at the result L(u)=0. This is a contradiction.
The resultR(u)=0 can be obtained in a similar way; here we omit the details.Remark. Since our forms of the linear and nonlinear terms are in a more general setting, Theorem 1 covers the results presented by Ali and Ebaid, respectively.


### 2.5. A Simplification for the Exp-Function Method

According to Theorem1, if the balanced nonlinear term is a product, ansatz (4) can be reduced to the following equivalent form:
(20)u=α0+α1exp(η)+⋯+ατexp(τη)1+β1exp(η)+⋯+βτexp(τη),
where τ is a free positive integer with τ⩾2 and αi and βi are constants to be specified. Ansatz (20) is concise and easy to calculate and makes the Exp-function method more straightforward.For example, Naher et al. applied the Exp-function method to constructing the traveling wave solutions of the (3+1)-dimensional modified KdV-Zakharov-Kuznetsev equation in the form [25]
(21)ut+αu2ux+uxxx+uxyy+uxzz=0.
The traveling wave transformation
(22)u(x,y,z,t)=u(η),η=x+y+z-Vt
carries (21) into an ODE
(23)-Vu′+αu2u′+3u′′′=0.
Since the nonlinear term u2u′ is a product of u and u′, we can immediately assume ansatz (4) in the form
(24)u=α0+α1exp(η)+α2exp(2η)1+β1exp(η)+β2exp(2η),
which is equivalent to the case c=p=1 and d=q=1 and in accord with Naher's (namely, (3.8) in [25]).


## 2.1. Some Terminology

To begin with, we recall some terminology in [35].A monomial in a collection of variablesx1,…,xn is a product
(5)x1α1x2α2…xnαn,
where the αi are nonnegative integers.The total degree of (5) is the sum of the exponents: α1+⋯+αn.A polynomial is said to be homogeneous if all the monomials appearing in it with nonzero coefficients have the same total degree.For instance,u2u′u′′3 is a product of u and its derivatives u′ and u′′, and its total degree is 6. u2+u′2+uu′′ is homogeneous.


## 2.2. Two Introducing Ansatz Function

For convenience, we assumev(η) is expressed in the form
(6)v(η)=∑n=-c1d1knexp(nη)∑n=-p1q1lnexp(nη)=k-c1exp(-c1η)+⋯+kd1exp(d1η)l-p1exp(-p1η)+⋯+lq1exp(q1η),
where c1, d1, p1, and q1 are positive integers to be determined and kn and ln are constants to be specified.The following three formulas will be used in this section:(7)u′(η)=(p-c)a-cb-pexp(-(c+p)η)+⋯+(d-q)adbqexp((d+q)η)b-p2exp(-2pη)+⋯+bq2exp(2qη),(8)u(η)·v(η)=a-ck-c1exp(-(c+c1)η)+⋯+adkd1exp((d+d1)η)b-pl-p1exp(-(p+p1)η)+⋯+bqlq1exp((q+q1)η),(9)u(η)v(η)=a-cl-p1exp(-(c+p1)η)+⋯+adlq1exp((d+q1)η)b-pk-c1exp(-(p+c1)η)+⋯+bqkd1exp((q+d1)η).Before presenting our definitions, we recall an important fact that the constantsa-c, ad, b-p, and bq in the ansatz of (4) can be assumed to be nonzero during the process of balancing the linear term of highest order with the highest order nonlinear term of certain ODE, and so do the constants k-c1, kd, l-p1, and lq in ansatz (6). Hence, in this paper, we assume that all above eight constants are nonzero.Therefore, for ansatz (4), we can define the ansatz functions L(·) and R(·) as follows:
(10)L(u)=L(a-cexp(-cη)+⋯+adexp(dη)b-pexp(-pη)+⋯+bqexp(qη))=-c-(-p)=p-c,R(u)=R(a-cexp(-cη)+⋯+adexp(dη)b-pexp(-pη)+⋯+bqexp(qη))=d-q.
In particular, we define L(C)=R(C)=0 for arbitrary constant C.For example, we have(11)L(exp(-η)+exp(2η)exp(-2η)+exp(2η))=1,R(exp(-η)+exp(2η)exp(-2η)+exp(2η))=0.
According to the definitions, we can find that in ansatz (4) c=p equals L(u)=0 and d=q equals R(u)=0. Therefore, the open problem for ansatz (4) is equal to whether the relations L(u)=0 and L(u)=0 hold.


## 2.3. Properties of the Ansatz Functions

Assumingc≠p, d≠q, c1≠p1, and d1≠q1 in ansatz (4) and (6), from (7), we obtain
(12)L(u′)=2p-(c+p)=p-c=L(u),R(u′)=(d+q)-2q=d-q=R(u).
So we have
(13)L(u)=L(u′)=L(u′′)=⋯=L(u(n)),R(u)=R(u′)=⋯=R(u(n)),
for arbitrary nonnegative integer n.And from (8) and (9), we have
(14)L(u·υ)=(p+p1)-(c+c1)L(u·v)=(p-c)+(p1-c1)L(u·v)=L(u)+L(υ),R(u·υ)=(d+d1)-(q+q1)L(u·v)=(d-q)+(d1-q1)L(u·v)=R(u)+R(υ),L(uυ)=(p+c1)-(c+p1)L(uv)=(p-c)-(p1-c1)L(uv)=L(u)-L(υ),R(uυ)=(d+q1)-(q+d1)L(uv)=(d-q)-(d1-q1)L(uv)=R(u)-R(υ).
Hence, we have
(15)L(uκ)=κ·L(u),R(uκ)=κ·R(u),
for any integer κ.


## 2.4. Theorem and Proof

In this section, we assume the balanced nonlinear term is a product of dependent variableu and its derivatives; namely,
(16)ui1(u′)i2⋯(u(m))im,
where ij (j=1,…,m) are nonnegative integers. The fact that the product (16) is a nonlinear term implies i1+i2+⋯+im⩾2. In other words, the total degree of (16) is at least 2.Theorem 1.
Suppose that the balanced nonlinear term in (3) is a product of u and its derivatives in the form of (16) and the balancing linear term is u(s), where s is a nonnegative integer; then the Exp-function ansatz (4) admits L(u)=0 and R(u)=0.Proof.
By contradiction, suppose thatL(u)≠0. Then we have
(17)L(ui1(u′)i2⋯(u(m))im)=L(ui1)+L((u′)i2)+⋯+L((u(m))im)=i1·L(u)+i2·L(u′)+⋯+im·L(u(m))=i1·L(u)+i2·L(u)+⋯+im·L(u)=(i1+i2+⋯+im)·L(u),L(u(s))=L(u).
Balancing linear and nonlinear terms requires
(18)L(ui1(u′)i2…(u(m))im)=L(u(s)).
So we obtain
(19)(i1+i2+⋯+im-1)·L(u)=0.
Since i1+i2+⋯+im-1≠0, we arrive at the result L(u)=0. This is a contradiction.
The resultR(u)=0 can be obtained in a similar way; here we omit the details.Remark. Since our forms of the linear and nonlinear terms are in a more general setting, Theorem 1 covers the results presented by Ali and Ebaid, respectively.


## 2.5. A Simplification for the Exp-Function Method

According to Theorem1, if the balanced nonlinear term is a product, ansatz (4) can be reduced to the following equivalent form:
(20)u=α0+α1exp(η)+⋯+ατexp(τη)1+β1exp(η)+⋯+βτexp(τη),
where τ is a free positive integer with τ⩾2 and αi and βi are constants to be specified. Ansatz (20) is concise and easy to calculate and makes the Exp-function method more straightforward.For example, Naher et al. applied the Exp-function method to constructing the traveling wave solutions of the (3+1)-dimensional modified KdV-Zakharov-Kuznetsev equation in the form [25]
(21)ut+αu2ux+uxxx+uxyy+uxzz=0.
The traveling wave transformation
(22)u(x,y,z,t)=u(η),η=x+y+z-Vt
carries (21) into an ODE
(23)-Vu′+αu2u′+3u′′′=0.
Since the nonlinear term u2u′ is a product of u and u′, we can immediately assume ansatz (4) in the form
(24)u=α0+α1exp(η)+α2exp(2η)1+β1exp(η)+β2exp(2η),
which is equivalent to the case c=p=1 and d=q=1 and in accord with Naher's (namely, (3.8) in [25]).


## 3. A Counter-Example

In [22], Ebaid claimed in his abstract that the case c=p and d=q was the only relation that could be obtained by applying this method to any nonlinear ODE. In this section, we construct a counter-example to show that the case c=p and d=q is not the only relation. In fact, the claim does not applied to each nonlinear homogeneous ODE.For example, we can create the following ODE:(25)u2+u′2+uu′′=0,
which can be rewritten as
(26)u+u′2u+u′′=0.
We have
(27)L(u′2u)=L(u),L(u′′)=L(u),R(u′2u)=R(u),R(u′′)=R(u).
From (27), the relations L(u′2/u)=L(u′′) and R(u′2/u)=R(u′′) hold automatically. Hence by balancing nonlinear term u′2/u and linear term u′′, we cannot determine the relations of c, d, p, and q. That is to say, all of them are free constants. Hence it is possible that the relation c≠p and d≠q holds. In other words, the relation c=p and d=q is not all-inclusive.


## 4. Conclusion

In summary, we present an entire novel approach to prove that the balancing procedure in Exp-function method is unnecessary when the balanced nonlinear term is a product of the dependent variable under consideration and its derivatives. Our results cover the results presented by Ali and Ebaid. We believe that our work can serve as an answer to the open problem proposed by Aslan et al.



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