# Variable-Coefficient Exact Solutions for Nonlinear Differential Equations by a New Bernoulli Equation-Based Subequation Method

Authors: Chunxia Qi; Shunliang Huang
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

A new Bernoulli equation-based subequation method is proposed to establish variable-coefficient exact solutions for
nonlinear differential equations. For illustrating the validity of this method, we apply it to the asymmetric (2 + 1)-dimensional NNV
system and the Kaup-Kupershmidt equation. As a result, some new exact solutions with variable functions coefficients for them are successfully obtained.


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## Body

## 1. Introduction

Nonlinear differential equations (NLDEs) can be used to describe many nonlinear phenomena such as fluid mechanics, plasma physics, optical fibers, biology, solid state physics, chemical kinematics, and chemical physics. Recently, research for seeking exact analytical solutions of NLDEs has been a hot topic, and many powerful and efficient methods to find exact solutions have been presented so far. For example, these methods include the known homogeneous balance method [1, 2], the tanh method [3–5], the inverse scattering transform [6], the Backlund transform [7, 8], the Hirotas bilinear method [9, 10], the generalized Riccati subequation method [11, 12], the Jacobi elliptic function expansion [13, 14], the F-expansion method [15], the exp-function expansion method [16, 17], and the (G′/G)-expansion method [18, 19]. However, we notice that most of the existing methods are accompanied with constant coefficients, while very few methods are concerned with variable coefficients.In this paper, by introducing a new ansatz,  we develop a new Bernoulli equation-based sub equation method for obtaining variable-coefficient exact solutions for NLDEs.  First we give the description of the Bernoulli equation-based subequation method. Then we apply the method to solve the asymmetric (2+1)-dimensional NNV system and the Kaup-Kupershmidt equation. Some conclusions are presented at the end of the paper.


## 2. Description of the Bernoulli Equation-Based Subequation Method

We consider the following Bernoulli equation:(1)G′+λG=G2,
where λ≠0 is a complex number and G=G(ξ). The solutions of (1) are denoted by
(2)G(ξ)=λ1+λdeλξ,
where d is an arbitrary constant. In particular, when λ is a real number and d=1/λ, we obtain
(3)G(ξ)=λ2(1-tanh(λξ2)).Whend=1/λ, λ=iλ~, where λ~ is a real number and i is the unit of imaginary number, we obtain
(4)G(ξ)=λ2-λi2tan(λ~ξ2).Suppose that a nonlinear equation, say in two or three independent variablesx, y, t, is given by
(5)P(u,ut,ux,uy,utt,uxt,uxx,uxy,…)=0,
where u=u(x,y,t) is an unknown function and P is a polynomial in u=u(x,y,t) and its various partial derivatives, in which the highest order derivatives and nonlinear terms are involved.Step 1.
Suppose that(6)u(x,y,t)=u(ξ),ξ=ξ(x,y,t),
and then (5) can be turned into the following form:
(7)P~(u,u′,u′′,…)=0.Step 2.
Suppose that the solution of (7) can be expressed by a polynomial in G as follows:
(8)u(ξ)=am(x,y,t)Gm+am-1(x,y,t)Gm-1u(ξ)=+⋯+a0(x,y,t),
where G=G(ξ) satisfies (1) and am(x,y,t),am-1(x,y,t),…,a0(x,y,t) are all unknown functions to be determined later with am(x,y,t)≠0. The positive integer m can be determined by considering the homogeneous balance between the highest order derivatives and nonlinear terms appearing in (7).Step 3.
Substituting (8) into (7) and using (1), collecting all terms with the same order of G together, the left-hand side of (7) is converted to another polynomial in G. Equating each coefficient of this polynomial to zero yields a set of partial differential equations for am(x,y,t),am-1(x,y,t),…,ξ(x,y,t),λ.Step 4.
Solving the equations system in Step3, and using the solutions of (1), we can construct exact coefficient function solutions of (7).Remark 1.
As the partial differential equations in Step3 are usually overdetermined, we may choose some special forms of am,am-1,…,a0 as done in the following.


## 3. Application of the Bernoulli Equation-Based Subequation Method to Some NLDEs

### 3.1. Asymmetric (2+1)-Dimensional NNV System

We consider the (2+1)-dimensional asymmetric NNV system [20–22]:
(9)ut-uxxx+α(uv)x=0,ux+βvy=0,
where α and β are arbitrary nonzero constants.Assume thatu(x,y,t)=U(ξ), where ξ=ξ(x,y,t), and then (9) can be turned into
(10)ξt′U′-(ξx3U′′′+3ξxξxxU′′+ξxxxU′)+αξx(UV)′=0,(11)ξx′U′+βξyV′=0.
Suppose that the solutions of (10)-(11) can be expressed by a polynomial in G as follows:
(12)U(ξ)=∑i=0m‍ai(y,t)Gi,V(ξ)=∑i=0n‍bi(y,t)Gi,
where ai(y,t), bi(y,t) are underdetermined functions and G=G(ξ) satisfies (1). Balancing the order of U′′′ and (UV)′ in (10) and the order of U′ and V′ in (11), we can obtain m+3=m+n+1, m+1=n+1⇒m=2, n=2. So we have
(13)U(ξ)=a2(y,t)G2+a1(y,t)G+a0(y,t),V(ξ)=b2(y,t)G2+b1(y,t)G+b0(y,t).
Substituting (13) into (10)-(11) and collecting all the terms with the same power of G together, equating each coefficient to zero yields a set of underdetermined partial differential equations for ai(x,y,t), bi(x,y,t), i=0,1,2, and ξ(x,y,t). Solving these equations yields the following.Case 1.
Consider(14)ξ(x,y,t)=C1x-∫C1αF1(t)dt+C13λ2t+F2(y),a2(y,t)=-6βC1F2′(y)α,a1(y,t)=6βλC1F2′(y)α,a0(y,t)=0,b2(y,t)=6C12α,b1(y,t)=-6C12λα,b0(y,t)=F1(t),
where C1 is an arbitrary constant and F1(t) and (y) are two arbitrary functions with respect to the variables t and y, respectively.Case 2.
Consider(15)ξ(x,y,t)=C1x+∫-αβC1C2F1(t)+C12(α+C1C2βλ2)βC2dt+C2∫F2(y)dy+C3,a2(y,t)=-6βC1C2F2(y)α,a1(y,t)=6βλC1C2F2(y)α,a0(y,t)=F2(y),b2(y,t)=6C12α,b1(y,t)=-6C12λα,b0(y,t)=F1(t),
where C1, C2, and C3 are arbitrary constants and F1(t) and F2(y) are two arbitrary functions with respect to the variables t and y, respectively.Case 3.
Consider(16)ξ(x,y,t)=-αF1(y)+(α2+4αβ2λ2C1C22)F12(y)2βC2λ2F1(y)xξ(x,y,t)=+C2∫F1(y)dy+C3,a0(y,t)=F1(y),b0(y,t)=C1,a2(y,t)=3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λ2α,a1(y,t)=-3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λα,b2(y,t)=3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ4C22F12(y),b1(y,t)=-3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ3C22F12(y),
where C1, C2, and C3 are arbitrary constants and F1(y) is an arbitrary function.Case 4.
Consider(17)ξ(x,y,t)=-C1αxλ+F1(y),a2(y,t)=6βC1F1′(y)λC1α,a1(y,t)=-6βC1F1′(y)C1α,a0(y,t)=0,b2(y,t)=6C1λ2,b1(y,t)=-6C1λ,b0(y,t)=C1,
where C1 is an arbitrary constant with C1α>0 and F1(y) is an arbitrary function.Substituting the results in the four cases mentioned previously into (13), and combining with the solutions of (1) as denoted in (2), we can obtain a rich variety of exact solutions to the asymmetric (2+1)-dimensional NNV system as follows.Family 1. One has
(18)u1(x,y,t)=-6βC1F2′(y)αu1(x,y,t)=×(λ1+λdeλ[C1x-∫‍C1αF1(t)dt+C13λ2t+F2(y)])2u1(x,y,t)=+6βλC1F2′(y)αu1(x,y,t)=×(λ1+λdeλ[C1x-∫‍C1αF1(t)dt+C13λ2t+F2(y)]),v1(x,y,t)=6C12α(λ1+λdeλ[C1x-∫‍C1αF1(t)dt+C13λ2t+F2(y)])2-6C12λα(λ1+λdeλ[C1x-∫‍C1αF1(t)dt+C13λ2t+F2(y)])+F1(t).Family 2. One has
(19)u2(x,y,t)=-6βC1C2F2(y)α(λ1+λdeλξ)2u2(x,y,t)=+6βλC1C2F2(y)α(λ1+λdeλξ)+F2(y),v2(x,y,t)=6C12α(λ1+λdeλξ)2v2(x,y,t)=-6C12λα(λ1+λdeλξ)+F1(t),
where ξ=C1x+∫((-αβC1C2F1(t)+C12(α+C1C2βλ2))/βC2)dt+C2∫F2(y)dy+C3.Family 3. One has
(20)u3(x,y,t)=3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λ2α×(λ1+λdeλξ)2-3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λα×(λ1+λdeλξ)+F1(y),(21)v3(x,y,t)=3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ4C22F12(y)×(λ1+λdeλξ)2-3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ3C22F12(y)×(λ1+λdeλξ)+C1,
where ξ=-((αF1(y)+(α2+4αβ2λ2C1C22)F12(y))/2βC2λ2F1(y))x+C2∫‍F1(y)dy+C3.Family 4. Consider
(22)u4(x,y,t)=6βC1F1′(y)λC1α(λ1+λdeλξ)2u4(x,y,t)=-6βC1F1′(y)C1α(λ1+λdeλξ),v4(x,y,t)=6C1λ2(λ1+λdeλξ)2-6C1λ(λ1+λdeλξ)+C1,
where ξ=-(C1αx/λ)+F1(y).Particularly, by a combination between Cases1–4 and (3) we can obtain some special hyperbolic function solutions as follows:
(23)u5(x,y,t)=-6βC1F2′(y)α[λ2(1-tanh(λξ2))]2u5(x,y,t)=+6βλC1F2′(y)α[λ2(1-tanh(λξ2))],v5(x,y,t)=6C12α[λ2(1-tanh(λξ2))]2v5(x,y,t)=-6C12λα[λ2(1-tanh(λξ2))]+F1(t),
where ξ=C1x-∫C1αF1(t)dt+C13λ2t+F2(y).Consider(24)u6(x,y,t)=-6βC1C2F2(y)α[λ2(1-tanh(λξ2))]2u6(x,y,t)=+6βλC1C2F2(y)α[λ2(1-tanh(λξ2))]u6(x,y,t)=+F2(y),v6(x,y,t)=6C12α[λ2(1-tanh(λξ2))]2v6(x,y,t)=-6C12λα[λ2(1-tanh(λξ2))]+F1(t),
where ξ=C1x+∫((-αβC1C2F1(t)+C12(α+C1C2βλ2))/βC2)dt+C2∫F2(y)dy+C3.Consider(25)u7(x,y,t)=3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λ2αu7(x,y,t)=×[λ2(1-tanh(λξ2))]2u7(x,y,t)=-3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λαu7(x,y,t)=×[λ2(1-tanh(λξ2))]+F1(y),v7(x,y,t)=3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ4C22F12(y)v7(x,y,t)=×[λ2(1-tanh(λξ2))]2v7(x,y,t)=-3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ3C22F12(y)v7(x,y,t)=×[λ2(1-tanh(λξ2))]+C1,
where ξ=-((αF1(y)+(α2+4αβ2λ2C1C22)F12(y))/2βC2λ2F1(y))x+C2∫F1(y)dy+C3.Consider(26)u8(x,y,t)=6βC1F1′(y)λC1α[λ2(1-tanh(λξ2))]2u8(x,y,t)=-6βC1F1′(y)C1α[λ2(1-tanh(λξ2))],v8(x,y,t)=6C1λ2[λ2(1-tanh(λξ2))]2v8(x,y,t)=-6C1λ[λ2(1-tanh(λξ2))]+C1,
where ξ=(-C1αx/λ)+F1(y).By a combination with (4) we can obtain some special trigonometric function solutions as follows:
(27)u9(x,y,t)=-6βC1F2′(y)α[λ2-λi2tan(λ~ξ2)]2u9(x,y,t)=+6βλC1F2′(y)α[λ2-λi2tan(λ~ξ2)],v9(x,y,t)=6C12α[λ2-λi2tan(λ~ξ2)]2v9(x,y,t)=-6C12λα[λ2-λi2tan(λ~ξ2)]+F1(t),
where ξ=C1x-∫C1αF1(t)dt+C13λ2t+F2(y), λ~ is a real number and λ=iλ~.One has(28)u10(x,y,t)=-6βC1C2F2(y)α[λ2-λi2tan(λ~ξ2)]2u10(x,y,t)=+6βλC1C2F2(y)α[λ2-λi2tan(λ~ξ2)]u10(x,y,t)=+F2(y),v10(x,y,t)=6C12α[λ2-λi2tan(λ~ξ2)]2v10(x,y,t)=-6C12λα[λ2-λi2tan(λ~ξ2)]+F1(t),
where ξ=C1x+∫((-αβC1C2F1(t)+C12(α+C1C2βλ2))/βC2)dt+C2∫‍F2(y)dy+C3.One has(29)u11(x,y,t)=3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λ2αu11(x,y,t)=×[λ2-λi2tan(λ~ξ2)]2u11(x,y,t)=-3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λαu11(x,y,t)=×[λ2-λi2tan(λ~ξ2)]+F1(y),v11(x,y,t)=3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ4C22F12(y)v11(x,y,t)=×[λ2-λi2tan(λ~ξ2)]2v11(x,y,t)=-3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ3C22F12(y)v11(x,y,t)=×[λ2-λi2tan(λ~ξ2)]+C1,
where ξ=-((αF1(y)+(α2+4αβ2λ2C1C22)F12(y))/2βC2λ2F1(y))x+C2∫F1(y)dy+C3.(30)u12(x,y,t)=6βC1F1′(y)λC1α[λ2-λi2tan(λ~ξ2)]2u12(x,y,t)=-6βC1F1′(y)C1α[λ2-λi2tan(λ~ξ2)],v12(x,y,t)=6C1λ2[λ2-λi2tan(λ~ξ2)]2v12(x,y,t)=-6C1λ[λ2-λi2tan(λ~ξ2)]+C1,
where ξ=-(C1αx/λ)+F1(y).Remark 2.
In [20–22], some exact solutions for the asymmetric (2+1)-dimensional NNV system are established using different methods. We note that the established solutions mentioned previously are different from them essentially as they are new exact solutions with variable functions coefficients and have been reported by other authors in the literature.


### 3.2. Kaup-Kupershmidt Equation

We consider the following Kaup-Kupershmidt equation [23–25]:
(31)uxxxxx+ut+45uxu2-752uxuxx-15uuxxx=0.
Suppose that u(x,t)=U(ξ), ξ=ξ(x,t), and, furthermore, by balancing the order in (31), we can suppose that
(32)U(ξ)=a2(x,t)G2+a1(x,t)G+a0(x,t),
where G=G(ξ) satisfies (1). Similar to the process mentioned before, we obtain
(33)ξ(x,t)=2λiarctan(C1(x+C2)2λi),a2(x,t)=128C12(C12(x+C2)2-4λ2)2,a1(x,t)=-64C13(x+C2)(C12(x+C2)2-4λ2)2,a0(x,t)=8C14(x+C2)2(C12(x+C2)2-4λ2)2+λ2.
Then we can obtain a kind of exact solutions of (31) which is unrelated to time variable as follows:
(34)u1(x,t)=128C12(C12(x+C2)2-4λ2)2×(λ1+λde(2/i)arctan(C1(x+C2)/2λi))2-64C13(x+C2)(C12(x+C2)2-4λ2)2×(λ1+λde(2/i)arctan(C1(x+C2)/2λi))+8C14(x+C2)2(C12(x+C2)2-4λ2)2+λ2.
In particular, by the combination with (3)-(4) we can obtain hyperbolic function and trigonometric function solutions as follows:
(35)u2(x,t)=128C12(C12(x+C2)2-4λ2)2[λ2(1-tanh(λξ2))]2u2(x,t)=-64C13(x+C2)(C12(x+C2)2-4λ2)2[λ2(1-tanh(λξ2))]u2(x,t)=+8C14(x+C2)2(C12(x+C2)2-4λ2)2+λ2,(36)u3(x,t)=128C12(C12(x+C2)2-4λ2)2[λ2-λi2tan(λ~ξ2)]2u3(x,t)=-64C13(x+C2)(C12(x+C2)2-4λ2)2λ2[λ2-λi2tan(λ~ξ2)]u3(x,t)=+8C14(x+C2)2(C12(x+C2)2-4λ2)2+λ2,
where ξ=(2/λi)arctan(C1(x+C2)/2λi), λ~ is a real number and λ=iλ~.Remark 3.
The previous established solutions for the Kaup-Kupershmidt equation cannot be obtained by the methods in [23–25] and are new exact solutions to our best knowledge.


## 3.1. Asymmetric (2+1)-Dimensional NNV System

We consider the (2+1)-dimensional asymmetric NNV system [20–22]:
(9)ut-uxxx+α(uv)x=0,ux+βvy=0,
where α and β are arbitrary nonzero constants.Assume thatu(x,y,t)=U(ξ), where ξ=ξ(x,y,t), and then (9) can be turned into
(10)ξt′U′-(ξx3U′′′+3ξxξxxU′′+ξxxxU′)+αξx(UV)′=0,(11)ξx′U′+βξyV′=0.
Suppose that the solutions of (10)-(11) can be expressed by a polynomial in G as follows:
(12)U(ξ)=∑i=0m‍ai(y,t)Gi,V(ξ)=∑i=0n‍bi(y,t)Gi,
where ai(y,t), bi(y,t) are underdetermined functions and G=G(ξ) satisfies (1). Balancing the order of U′′′ and (UV)′ in (10) and the order of U′ and V′ in (11), we can obtain m+3=m+n+1, m+1=n+1⇒m=2, n=2. So we have
(13)U(ξ)=a2(y,t)G2+a1(y,t)G+a0(y,t),V(ξ)=b2(y,t)G2+b1(y,t)G+b0(y,t).
Substituting (13) into (10)-(11) and collecting all the terms with the same power of G together, equating each coefficient to zero yields a set of underdetermined partial differential equations for ai(x,y,t), bi(x,y,t), i=0,1,2, and ξ(x,y,t). Solving these equations yields the following.Case 1.
Consider(14)ξ(x,y,t)=C1x-∫C1αF1(t)dt+C13λ2t+F2(y),a2(y,t)=-6βC1F2′(y)α,a1(y,t)=6βλC1F2′(y)α,a0(y,t)=0,b2(y,t)=6C12α,b1(y,t)=-6C12λα,b0(y,t)=F1(t),
where C1 is an arbitrary constant and F1(t) and (y) are two arbitrary functions with respect to the variables t and y, respectively.Case 2.
Consider(15)ξ(x,y,t)=C1x+∫-αβC1C2F1(t)+C12(α+C1C2βλ2)βC2dt+C2∫F2(y)dy+C3,a2(y,t)=-6βC1C2F2(y)α,a1(y,t)=6βλC1C2F2(y)α,a0(y,t)=F2(y),b2(y,t)=6C12α,b1(y,t)=-6C12λα,b0(y,t)=F1(t),
where C1, C2, and C3 are arbitrary constants and F1(t) and F2(y) are two arbitrary functions with respect to the variables t and y, respectively.Case 3.
Consider(16)ξ(x,y,t)=-αF1(y)+(α2+4αβ2λ2C1C22)F12(y)2βC2λ2F1(y)xξ(x,y,t)=+C2∫F1(y)dy+C3,a0(y,t)=F1(y),b0(y,t)=C1,a2(y,t)=3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λ2α,a1(y,t)=-3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λα,b2(y,t)=3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ4C22F12(y),b1(y,t)=-3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ3C22F12(y),
where C1, C2, and C3 are arbitrary constants and F1(y) is an arbitrary function.Case 4.
Consider(17)ξ(x,y,t)=-C1αxλ+F1(y),a2(y,t)=6βC1F1′(y)λC1α,a1(y,t)=-6βC1F1′(y)C1α,a0(y,t)=0,b2(y,t)=6C1λ2,b1(y,t)=-6C1λ,b0(y,t)=C1,
where C1 is an arbitrary constant with C1α>0 and F1(y) is an arbitrary function.Substituting the results in the four cases mentioned previously into (13), and combining with the solutions of (1) as denoted in (2), we can obtain a rich variety of exact solutions to the asymmetric (2+1)-dimensional NNV system as follows.Family 1. One has
(18)u1(x,y,t)=-6βC1F2′(y)αu1(x,y,t)=×(λ1+λdeλ[C1x-∫‍C1αF1(t)dt+C13λ2t+F2(y)])2u1(x,y,t)=+6βλC1F2′(y)αu1(x,y,t)=×(λ1+λdeλ[C1x-∫‍C1αF1(t)dt+C13λ2t+F2(y)]),v1(x,y,t)=6C12α(λ1+λdeλ[C1x-∫‍C1αF1(t)dt+C13λ2t+F2(y)])2-6C12λα(λ1+λdeλ[C1x-∫‍C1αF1(t)dt+C13λ2t+F2(y)])+F1(t).Family 2. One has
(19)u2(x,y,t)=-6βC1C2F2(y)α(λ1+λdeλξ)2u2(x,y,t)=+6βλC1C2F2(y)α(λ1+λdeλξ)+F2(y),v2(x,y,t)=6C12α(λ1+λdeλξ)2v2(x,y,t)=-6C12λα(λ1+λdeλξ)+F1(t),
where ξ=C1x+∫((-αβC1C2F1(t)+C12(α+C1C2βλ2))/βC2)dt+C2∫F2(y)dy+C3.Family 3. One has
(20)u3(x,y,t)=3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λ2α×(λ1+λdeλξ)2-3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λα×(λ1+λdeλξ)+F1(y),(21)v3(x,y,t)=3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ4C22F12(y)×(λ1+λdeλξ)2-3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ3C22F12(y)×(λ1+λdeλξ)+C1,
where ξ=-((αF1(y)+(α2+4αβ2λ2C1C22)F12(y))/2βC2λ2F1(y))x+C2∫‍F1(y)dy+C3.Family 4. Consider
(22)u4(x,y,t)=6βC1F1′(y)λC1α(λ1+λdeλξ)2u4(x,y,t)=-6βC1F1′(y)C1α(λ1+λdeλξ),v4(x,y,t)=6C1λ2(λ1+λdeλξ)2-6C1λ(λ1+λdeλξ)+C1,
where ξ=-(C1αx/λ)+F1(y).Particularly, by a combination between Cases1–4 and (3) we can obtain some special hyperbolic function solutions as follows:
(23)u5(x,y,t)=-6βC1F2′(y)α[λ2(1-tanh(λξ2))]2u5(x,y,t)=+6βλC1F2′(y)α[λ2(1-tanh(λξ2))],v5(x,y,t)=6C12α[λ2(1-tanh(λξ2))]2v5(x,y,t)=-6C12λα[λ2(1-tanh(λξ2))]+F1(t),
where ξ=C1x-∫C1αF1(t)dt+C13λ2t+F2(y).Consider(24)u6(x,y,t)=-6βC1C2F2(y)α[λ2(1-tanh(λξ2))]2u6(x,y,t)=+6βλC1C2F2(y)α[λ2(1-tanh(λξ2))]u6(x,y,t)=+F2(y),v6(x,y,t)=6C12α[λ2(1-tanh(λξ2))]2v6(x,y,t)=-6C12λα[λ2(1-tanh(λξ2))]+F1(t),
where ξ=C1x+∫((-αβC1C2F1(t)+C12(α+C1C2βλ2))/βC2)dt+C2∫F2(y)dy+C3.Consider(25)u7(x,y,t)=3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λ2αu7(x,y,t)=×[λ2(1-tanh(λξ2))]2u7(x,y,t)=-3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λαu7(x,y,t)=×[λ2(1-tanh(λξ2))]+F1(y),v7(x,y,t)=3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ4C22F12(y)v7(x,y,t)=×[λ2(1-tanh(λξ2))]2v7(x,y,t)=-3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ3C22F12(y)v7(x,y,t)=×[λ2(1-tanh(λξ2))]+C1,
where ξ=-((αF1(y)+(α2+4αβ2λ2C1C22)F12(y))/2βC2λ2F1(y))x+C2∫F1(y)dy+C3.Consider(26)u8(x,y,t)=6βC1F1′(y)λC1α[λ2(1-tanh(λξ2))]2u8(x,y,t)=-6βC1F1′(y)C1α[λ2(1-tanh(λξ2))],v8(x,y,t)=6C1λ2[λ2(1-tanh(λξ2))]2v8(x,y,t)=-6C1λ[λ2(1-tanh(λξ2))]+C1,
where ξ=(-C1αx/λ)+F1(y).By a combination with (4) we can obtain some special trigonometric function solutions as follows:
(27)u9(x,y,t)=-6βC1F2′(y)α[λ2-λi2tan(λ~ξ2)]2u9(x,y,t)=+6βλC1F2′(y)α[λ2-λi2tan(λ~ξ2)],v9(x,y,t)=6C12α[λ2-λi2tan(λ~ξ2)]2v9(x,y,t)=-6C12λα[λ2-λi2tan(λ~ξ2)]+F1(t),
where ξ=C1x-∫C1αF1(t)dt+C13λ2t+F2(y), λ~ is a real number and λ=iλ~.One has(28)u10(x,y,t)=-6βC1C2F2(y)α[λ2-λi2tan(λ~ξ2)]2u10(x,y,t)=+6βλC1C2F2(y)α[λ2-λi2tan(λ~ξ2)]u10(x,y,t)=+F2(y),v10(x,y,t)=6C12α[λ2-λi2tan(λ~ξ2)]2v10(x,y,t)=-6C12λα[λ2-λi2tan(λ~ξ2)]+F1(t),
where ξ=C1x+∫((-αβC1C2F1(t)+C12(α+C1C2βλ2))/βC2)dt+C2∫‍F2(y)dy+C3.One has(29)u11(x,y,t)=3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λ2αu11(x,y,t)=×[λ2-λi2tan(λ~ξ2)]2u11(x,y,t)=-3(αF1(y)+(α2+4αβ2λ2C1C22)F12(y))λαu11(x,y,t)=×[λ2-λi2tan(λ~ξ2)]+F1(y),v11(x,y,t)=3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ4C22F12(y)v11(x,y,t)=×[λ2-λi2tan(λ~ξ2)]2v11(x,y,t)=-3[αF1(y)+(α2+4αβ2λ2C1C22)F12(y)]22αβ2λ3C22F12(y)v11(x,y,t)=×[λ2-λi2tan(λ~ξ2)]+C1,
where ξ=-((αF1(y)+(α2+4αβ2λ2C1C22)F12(y))/2βC2λ2F1(y))x+C2∫F1(y)dy+C3.(30)u12(x,y,t)=6βC1F1′(y)λC1α[λ2-λi2tan(λ~ξ2)]2u12(x,y,t)=-6βC1F1′(y)C1α[λ2-λi2tan(λ~ξ2)],v12(x,y,t)=6C1λ2[λ2-λi2tan(λ~ξ2)]2v12(x,y,t)=-6C1λ[λ2-λi2tan(λ~ξ2)]+C1,
where ξ=-(C1αx/λ)+F1(y).Remark 2.
In [20–22], some exact solutions for the asymmetric (2+1)-dimensional NNV system are established using different methods. We note that the established solutions mentioned previously are different from them essentially as they are new exact solutions with variable functions coefficients and have been reported by other authors in the literature.


## 3.2. Kaup-Kupershmidt Equation

We consider the following Kaup-Kupershmidt equation [23–25]:
(31)uxxxxx+ut+45uxu2-752uxuxx-15uuxxx=0.
Suppose that u(x,t)=U(ξ), ξ=ξ(x,t), and, furthermore, by balancing the order in (31), we can suppose that
(32)U(ξ)=a2(x,t)G2+a1(x,t)G+a0(x,t),
where G=G(ξ) satisfies (1). Similar to the process mentioned before, we obtain
(33)ξ(x,t)=2λiarctan(C1(x+C2)2λi),a2(x,t)=128C12(C12(x+C2)2-4λ2)2,a1(x,t)=-64C13(x+C2)(C12(x+C2)2-4λ2)2,a0(x,t)=8C14(x+C2)2(C12(x+C2)2-4λ2)2+λ2.
Then we can obtain a kind of exact solutions of (31) which is unrelated to time variable as follows:
(34)u1(x,t)=128C12(C12(x+C2)2-4λ2)2×(λ1+λde(2/i)arctan(C1(x+C2)/2λi))2-64C13(x+C2)(C12(x+C2)2-4λ2)2×(λ1+λde(2/i)arctan(C1(x+C2)/2λi))+8C14(x+C2)2(C12(x+C2)2-4λ2)2+λ2.
In particular, by the combination with (3)-(4) we can obtain hyperbolic function and trigonometric function solutions as follows:
(35)u2(x,t)=128C12(C12(x+C2)2-4λ2)2[λ2(1-tanh(λξ2))]2u2(x,t)=-64C13(x+C2)(C12(x+C2)2-4λ2)2[λ2(1-tanh(λξ2))]u2(x,t)=+8C14(x+C2)2(C12(x+C2)2-4λ2)2+λ2,(36)u3(x,t)=128C12(C12(x+C2)2-4λ2)2[λ2-λi2tan(λ~ξ2)]2u3(x,t)=-64C13(x+C2)(C12(x+C2)2-4λ2)2λ2[λ2-λi2tan(λ~ξ2)]u3(x,t)=+8C14(x+C2)2(C12(x+C2)2-4λ2)2+λ2,
where ξ=(2/λi)arctan(C1(x+C2)/2λi), λ~ is a real number and λ=iλ~.Remark 3.
The previous established solutions for the Kaup-Kupershmidt equation cannot be obtained by the methods in [23–25] and are new exact solutions to our best knowledge.


## 4. Conclusions

We have proposed a Bernoulli equation-based subequation method for solving nonlinear differential equations and applied it to find exact solutions with variable functions coefficients of the asymmetric (2+1)-dimensional asymmetric NNV system and the Kaup-Kupershmidt equation. As a result, some new exact solutions for them have been successfully found. Finally, we note that the proposed method can be applied to solve other nonlinear evolution equations.



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*Source: 923408-2013-05-12.xml*