1. Introduction
The perturbation of integrable systems (or namely nearly integrable system) is an important topic in physics. An interesting survey on the topic is [1] by Kivshar and Malomed, in which a series of important perturbation problems in physics are discussed in detail through an inverse scattering method and other techniques. It is well-known that the task of the perturbation method is to eliminate the secular terms arising from the expansion of perturbation. For perturbation of solitons, there are several main techniques, such as modified conservation law method [2], Hamiltonian and Lagrangian approach [3, 4], perturbed bilinear method [5], multi-scale method [6, 7], Green's function method [8–12], inverse scattering transformation (IST) [13, 14] and the renormalization group (RG for simplicity) method [15, 16], and so forth. Both in theory and in practice, there are advantages and disadvantages for the above mentioned methods. Among those, for example, the IST method depends on complicated mathematical tricks, while the Green's function method depends strongly on the spectrum theory of the linear operator that is difficult in general, and multiple time scale expansions are needed to assume artificially that the physical problem has multiple time scales. In the RG method, Tu et al’s work [15] depends on the unnatural split of variables, and Kai’s work [16] needs the construct of special solutions of inhomogenous equations.
In [17], a renormalization method is proposed to solve the perturbation of solitons. This renormalization method is based on the Taylor expansion of the solution at a general point (namely TR, for simplicity) [18–20], and combines the renormalization method in [21, 22] with the spectrum theory for linear differential operators. The advantage of the renormalization method is that the secular terms are eliminated automatically, and any assumptions depending on physical experiences and intuitions are not needed, in particular, its topological version can be applied to deal with nonperturbation problems [21–27]. However, the disadvantage of this method is that we need to compute the spectrum and eigenfunctions of linear differential operators with variable coefficients. This is rather difficult in general so the method is not convenient in practice. In the present paper, we overcome the difficulties and provide a new TR method which does not depend on spectrum theory, by which we can solve the perturbation of solitons for some nearly integrable systems with multiple small parameters. The proposed method has all the advantages of the TR method and it is simpler in practice and clearer in mathematical foundation than other routine methods.
The paper is outlined as follows. In section 2 , we give the main steps of the new TR method of solving perturbation of solitons, and we use the new TR method to solve the perturbations of KdV, MKdV and nonlinear Schrodinger equations. In section 3 , we give a short conclusion.
2. The renormalization method for the perturbation of solitons
2.1. The perturbation of a single soliton
We can deal with the following perturbation equation
$\begin{eqnarray}{\partial }_{t}u-N(u)={\epsilon }_{1}{P}_{1}(u)+\cdots +{\epsilon }_{m}{P}_{m}(u),\end{eqnarray}$
where ${\epsilon }_{i}^{\prime} s$ are small parameters for i = 1, ⋯ , m.For simplicity, we only take m = 2 and denote two small parameters as ε1 = ε, ε2 = η to give the formulation. Consider the perturbed equation
$\begin{eqnarray}{\partial }_{t}u-N(u)=\epsilon P(u)+\eta Q(u),\end{eqnarray}$
where N is a nonlinear operator, and P and Q are two perturbation terms. Moreover, $\begin{eqnarray}{\partial }_{t}u-N(u)=0,\end{eqnarray}$
is an integrable equation. In general, this is a vector equation. Here we only consider the case of one scale function for simplicity. It is obvious that our method can be applied to the vector case.Expanding u as a power series of ε2 ) gives
$\begin{eqnarray}u={u}_{0}+\eta {u}_{1}+\cdots \,.\end{eqnarray}$
and substituting it into the perturbed equation ( $\begin{eqnarray}{\partial }_{t}{u}_{0}-N({u}_{0})=\epsilon P({u}_{0}),\end{eqnarray}$
$\begin{eqnarray*}{\partial }_{t}{u}_{1}-\frac{\partial }{\partial {u}_{0}}N({u}_{0}){u}_{1}-\frac{\partial }{\partial {u}_{0x}}N({u}_{0}){u}_{1x}+\cdots \end{eqnarray*}$
$\begin{eqnarray}=\,\epsilon (\frac{\partial }{\partial {u}_{0}}P({u}_{0}){u}_{1}-\frac{\partial }{\partial {u}_{0x}}P({u}_{0}){u}_{1x}+\cdots \,)+Q({u}_{0}),\cdots \,.\end{eqnarray}$
Take u0 = u00 + εu01 + ⋯ and u1 = u10 + εu11 + ⋯ . Then we have
$\begin{eqnarray}{\partial }_{t}{u}_{00}-N({u}_{00})=0,\end{eqnarray}$
$\begin{eqnarray}{\partial }_{t}{u}_{01}-\frac{\partial }{\partial {u}_{00}}N({u}_{00}){u}_{01}-\frac{\partial }{\partial {u}_{00x}}N({u}_{00}){u}_{01x}+\cdots \,=P({u}_{00}),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl} & & {\partial }_{t}{u}_{10}-\frac{\partial }{\partial {u}_{00}}N({u}_{00}){u}_{10}\\ & & -\frac{\partial }{\partial {u}_{00x}}N({u}_{00}){u}_{10x}+\cdots \,=Q({u}_{00}),\cdots \,.\end{array}\end{eqnarray}$
Furthermore, we solve the soliton u00 from the first equation (7 ), and note that u00 includes some parameters vector C. At the same time, u00 and its derivatives all go to zero as x tending to infinity.
Denote the linear operator
$\begin{eqnarray}Lf={\partial }_{t}f-\frac{\partial }{\partial {u}_{00}}N({u}_{00})f-\frac{\partial }{\partial {u}_{00x}}N({u}_{00}){f}_{x}+\cdots \,.\end{eqnarray}$
Then its dual operator L* is given from $\begin{eqnarray*}\begin{array}{rcl}(Lf,g) & = & ({\partial }_{t}f,g)-\left(\frac{\partial }{\partial {u}_{0}}N({u}_{00})f,g\right)\\ & & -\left(\frac{\partial }{\partial {u}_{0x}}N({u}_{00}){f}_{x},g\right)+\cdots \end{array}\end{eqnarray*}$
$\begin{eqnarray}\begin{array}{rcl} & = & {(f,g)}_{t}-(f,{\partial }_{t}g)-\left(f,\frac{\partial N}{\partial {u}_{0}}g\right)\\ & & +(f,{\left(\frac{\partial N}{\partial {u}_{0x}}g\right)}_{x})+\cdots ={(f,g)}_{t}-(f,{L}^{* }g),\end{array}\end{eqnarray}$
that is, $\begin{eqnarray}{L}^{* }g={g}_{t}+\frac{\partial N}{\partial {u}_{0}}g-{\left(\frac{\partial N}{\partial {u}_{0x}}g\right)}_{x}+\cdots \,.\end{eqnarray}$
Then take f = u01, f = u10 respectively and g satisfying L*g = 0. Therefore, we have
$\begin{eqnarray}{({u}_{01},g)}_{t}=(L{u}_{01},g)=(P,g),\end{eqnarray}$
$\begin{eqnarray}{({u}_{10},g)}_{t}=(L{u}_{10},g)=(Q,g).\end{eqnarray}$
When (P, g) and (Q, g) are independent of t, we have $\begin{eqnarray}({u}_{01},g)=(t-{t}_{0})(P,g),\end{eqnarray}$
$\begin{eqnarray}({u}_{10},g)=(t-{t}_{0})(Q,g),\end{eqnarray}$
where t0 is a general point as an integral constant.Taking inner product (u, g) gives
$\begin{eqnarray}\begin{array}{rcl}(u,g) & = & ({u}_{00},g)+\epsilon ({u}_{01},g)+\eta ({u}_{10},g)\\ & & +\cdots =({u}_{00},g)+(t-{t}_{0})\{\epsilon (P,g)\\ & & +\eta (Q,g)\}+\cdots \,.\end{array}\end{eqnarray}$
Furthermore, expand u00, g, P and Q at t0 to give $\begin{eqnarray*}\begin{array}{rcl}(u,g) & = & ({u}_{00}({t}_{0},x,C),g({t}_{0},x,C))\\ & & +\epsilon ({u}_{01},g)+\eta ({u}_{10},g)+\cdots \end{array}\end{eqnarray*}$
$\begin{eqnarray}\begin{array}{rcl} & = & ({u}_{00}({t}_{0},x,C),g({t}_{0},x,C))+(t-{t}_{0})\{\epsilon (P,g)\\ & & +\eta (Q,g)+({u}_{00{t}_{0}},g)+({u}_{00},{g}_{{t}_{0}})\}+\cdots \,.\end{array}\end{eqnarray}$
Because the parameters vector C in u0 is the function of t0, then the renormalization equation is given by
$\begin{eqnarray}\begin{array}{rcl} & & {({u}_{00}({t}_{0},x,C),g({t}_{0},x,C))}_{{t}_{0}}\\ & = & \{\epsilon (P,g)+\eta (Q,g)+({u}_{00{t}_{0}},g)+({u}_{00},{g}_{{t}_{0}})\},\end{array}\end{eqnarray}$
from which we have $\begin{eqnarray}({u}_{00}({t}_{0},x,C),{g}_{C}({t}_{0},x,C){C}_{{t}_{0}})=\epsilon (P,g)+\eta (Q,g),\end{eqnarray}$
where gC means the gradient of g with respect to vector C, and ${g}_{C}{C}_{{t}_{0}}$ is the inner product. Solving it gives C, and hence we get the first order renormalization solution $\begin{eqnarray}u={u}_{00}({t}_{0},x,C).\end{eqnarray}$
Remark 1. In fact a key step in the renormalization method is to solve u01 and u10 for the linear non-homogenous equations (8 ) and (9 ) with variable coefficients. In the TR method with the spectrum theory [17] and Green's function method [12], since we need to construct the Green's function in terms of eigenfunctions, we need to find the eigenvalues and eigenfunctions of the linear operator L, that is Lφ = λφ. If L is not a self conjugate operator, we also need to solve the dual equation L*φ = μφ. In general, for L and L*, respectively, there are continuous eigenvalues λ, μ and discrete eigenvalues λ0, ⋯ , λm and μ0, ⋯ , μm. When λ = 0, there will exist a secular term which needs to be removed by the renormalization method. The details can be found in [17] (see formulas (28)-(41) in [17]).
2.2. Perturbation of multiple solitons
For simplicity, we assume that each single soliton has two parameters a and θ, and denote them as aj and θj for jth soliton in N − solitons solution, j = 1, ⋯ , N. Due to the separability of multiple solitons as time goes to infinity, we can deal independently with the parameters of each soliton. Therefore, we have the equations system
$\begin{eqnarray}\displaystyle \sum _{i=1}^{2N}\frac{{\rm{d}}{q}_{i}}{{\rm{d}}{t}_{0}}\left(\frac{\partial {u}_{00}}{\partial {q}_{i}},{g}_{j}\right)=(\epsilon P+\eta Q,{g}_{j}),j=1,\cdots \,,2N,\end{eqnarray}$
where qi = ai, qN+i = θi for i = 1, ⋯ , N.From this linear equations system, we can solve out $\frac{{\rm{d}}{q}_{i}}{{\rm{d}}{t}_{0}}$, and hence get the solutions of these parameters.
Denote
$\begin{eqnarray}{A}_{ij}=\left(\frac{\partial {u}_{00}}{\partial {q}_{i}},{g}_{i}\right),A={({A}_{ij})}_{2N\times 2N},D=\det A,{D}_{i}=\det {A}_{i},\end{eqnarray}$
where Ai is the matrix of replacing the ith column of A by the vector $V={({v}_{1},\cdots \,,{v}_{2N})}^{{\rm{T}}}$ with vj = (εP + ηQ, gj), j = 1, ⋯ , 2N. Then from the Cramer rule, we get $\begin{eqnarray}\frac{{\rm{d}}{q}_{i}}{{\rm{d}}t}=\frac{{D}_{i}}{D},i=1,\cdots \,,2N.\end{eqnarray}$
Solving the above equations gives the solution of qi for i = 1, ⋯ , 2N.Remark 2. Because that the decomposition of multi-solitons is a complicated thing, and hence it is not easy to provide the analytic results in the computation of integrals in determinants D and Di, so it is hoped that these computations can be given in future work. For example, this issue can be seen from a two-soliton solution for the KdV equation
$\begin{eqnarray*}u(x,t)=-12\frac{\cosh (2x-8t)+\cosh (4x-6t)+3}{{(\cosh (x-28t)+\cosh (3x-36t))}^{2}}.\end{eqnarray*}$
In general, the two-soliton solution for the KdV equation is given by bilinear method $\begin{eqnarray*}u(x,t)=-2\{\mathrm{ln}(1+\exp {\xi }_{1}+\exp {\xi }_{2}+{A}_{12}\exp ({\xi }_{1}+{\xi }_{2}){\}}_{xx},\end{eqnarray*}$
where ${\xi }_{1}={a}_{1}x-{a}_{1}^{3}t+{a}_{1}{\theta }_{1},{\xi }_{2}={a}_{2}x-{a}_{2}^{3}t+{a}_{2}{\theta }_{2}$ and ${A}_{12}\,=\,\frac{{({a}_{1}-{a}_{2})}^{2}}{{({a}_{1}+{a}_{2})}^{2}}$.2.3. Perturbed KdV equation
Consider the perturbation of the KdV equation
$\begin{eqnarray}{u}_{t}-6u{u}_{x}+{u}_{xx}=\epsilon P(u)+\eta Q(u).\end{eqnarray}$
The zeroth and first order equations are respectively $\begin{eqnarray}{u}_{00t}-6{u}_{00}{u}_{00x}+{u}_{00xxx}=0,\end{eqnarray}$
$\begin{eqnarray}{u}_{01t}-6{u}_{00}{u}_{01x}-6{u}_{00x}{u}_{01}+{u}_{01xxx}=P({u}_{00}),\end{eqnarray}$
$\begin{eqnarray}{u}_{10t}-6{u}_{00}{u}_{10x}-6{u}_{00x}{u}_{10}+{u}_{10xxx}=Q({u}_{00}).\end{eqnarray}$
Then we define linear operator L by
$\begin{eqnarray}Lf={f}_{t}-6{u}_{00}{f}_{x}-6{u}_{00x}f+{f}_{xxx},\end{eqnarray}$
and hence the dual of L is $\begin{eqnarray}{L}^{* }g={g}_{t}-6{u}_{00}{g}_{x}+{g}_{xxx},\end{eqnarray}$
where we have used the inner product $(f,g)={\int }_{-\infty }^{+\infty }fg{\rm{d}}x$ and the properties of f, g and their derivatives being zero for x = ∞.From (29 ) and (30 ), we know that if Lf = 0, then we take
$\begin{eqnarray}g={\int }_{-\infty }^{x}f,\end{eqnarray}$
and get f = gx and L*g = 0.Therefore, from
$\begin{eqnarray}(Lf,g)={(f,g)}_{t}-({L}^{* }g,f),\end{eqnarray}$
and L*g = 0 we have $\begin{eqnarray}{({u}_{01},g)}_{t}=(L{u}_{01},g)=(P({u}_{00}),g),\end{eqnarray}$
$\begin{eqnarray}{({u}_{10},g)}_{t}=(L{u}_{10},g)=(Q({u}_{00}),g).\end{eqnarray}$
Since the inner product is a defined integral on x ∈ ( − ∞, + ∞), by the transformation of the variable, we have $\begin{eqnarray}({u}_{01},g)=(t-{t}_{0})(P({u}_{00}),g),\end{eqnarray}$
$\begin{eqnarray}({u}_{10},g)=(t-{t}_{0})(Q({u}_{00}),g),\end{eqnarray}$
where t0 is an arbitrary constant and hence can be considered as a variable.Therefore, we have 26 ) 27 ) with respect to a and θ respectively to give the solutions of Lf = 0, 41 )–(43 ), we get 35 ) and (36 ),
$\begin{eqnarray*}(u,g)=({u}_{00},g)+\epsilon ({u}_{01},g)+\eta ({u}_{10},g)+\cdots \end{eqnarray*}$
$\begin{eqnarray}=\,({u}_{00},g)+(t-{t}_{0})(\epsilon P({u}_{00})+\eta Q({u}_{00}),g)+\cdots \,.\end{eqnarray}$
Expanding (u00, g) at t0 gives $\begin{eqnarray}\begin{array}{rcl} & & ({u}_{00},g)\\ & = & ({u}_{00}({t}_{0},x,a,\theta ),g({t}_{0},x))\\ & & +(t-{t}_{0})(({u}_{00{t}_{0}},g)+({u}_{00{t}_{0}},{g}_{{t}_{0}}))+\cdots \,.\end{array}\end{eqnarray}$
So we have $\begin{eqnarray}\begin{array}{rcl}{({u}_{00}({t}_{0},x,a,\theta ),g({t}_{0},x))}_{{t}_{0}} & = & (\epsilon P({u}_{00})+\eta Q({u}_{00}),g)\\ & & +({u}_{00{t}_{0}},g)+({u}_{00{t}_{0}},{g}_{{t}_{0}}),\end{array}\end{eqnarray}$
from which we have by considering a and θ as the functions of t0, $\begin{eqnarray}\begin{array}{rcl} & & ({u}_{00a}({t}_{0},x,a,\theta ){a}_{{t}_{0}}+{u}_{00\theta }({t}_{0},x,a,\theta ){\theta }_{{t}_{0}},g)\\ & = & (\epsilon P({u}_{00})+\eta Q({u}_{00}),g).\end{array}\end{eqnarray}$
Take the single soliton solution of equation ( $\begin{eqnarray}\begin{array}{rcl}{u}_{00} & = & {u}_{00}(t,x,a,\theta )\\ & = & -2{a}^{2}{{\rm{{\rm{sech}} }}}^{2}(ax-4{a}^{3}t+a\theta )=-2{a}^{2}{{\rm{{\rm{sech}} }}}^{2}\xi ,\end{array}\end{eqnarray}$
where a, θ are two arbitrary parameters, and ξ = ax − 4a3t + aθ. Then take the derivatives of equation ( $\begin{eqnarray}\begin{array}{rcl}{f}_{1} & = & {u}_{00\theta }=4{a}^{3}{{\rm{{\rm{sech}} }}}^{2}\xi ,{f}_{2}={u}_{00a}\\ & = & -4a{{\rm{{\rm{sech}} }}}^{2}\xi +4a\xi {{\rm{{\rm{sech}} }}}^{2}\xi {\rm{\tanh }}\xi -8a(t-{t}_{0}){f}_{1},\end{array}\end{eqnarray}$
where we have used the arbitrariness of θ to introduce t0, and hence give the solutions of L*g = 0 by replacing x by ξ in formula (31) and adding a constant factor and using the property of linearity of L*, $\begin{eqnarray}{g}_{1}=\frac{1}{a}{\int }_{-\infty }^{\xi }{u}_{00\theta },{g}_{2}=\frac{1}{a}{\int }_{-\infty }^{\xi }{u}_{00a}.\end{eqnarray}$
According to three expressions ( $\begin{eqnarray}{g}_{1}\,=\,-2{a}^{2}{{\rm{{\rm{sech}} }}}^{2}\xi {\rm{\tanh }}\xi ,{g}_{2}={g}_{2}^{* }-8a(t-{t}_{0}){g}_{1},\end{eqnarray}$
where we denote $\begin{eqnarray}{g}_{2}^{* }=-2a(\xi {{\rm{{\rm{sech}} }}}^{2}\xi +{\rm{\tanh }}\xi ).\end{eqnarray}$
Therefore, we have from ( $\begin{eqnarray}({u}_{01},{g}_{2}^{* })=(t-{t}_{0})(P,{g}_{2})+8a{(t-{t}_{0})}^{2}(P,{g}_{1}),\end{eqnarray}$
$\begin{eqnarray}({u}_{10},{g}_{2}^{* })=(t-{t}_{0})(Q,{g}_{2})+8a{(t-{t}_{0})}^{2}(Q,{g}_{1}).\end{eqnarray}$
By taking g = g1 and $g={g}_{2}^{* }$, we get
$\begin{eqnarray}\begin{array}{rcl} & & ({u}_{00a}({t}_{0},x,a,\theta ),{g}_{1}){a}_{{t}_{0}}+({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{1}){\theta }_{{t}_{0}}\\ & = & ((\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1}),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl} & & ({u}_{00a}({t}_{0},x,a,\theta ),{g}_{2}^{* }){a}_{{t}_{0}}+({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{2}^{* }){\theta }_{{t}_{0}}\\ & = & ((\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2}^{* }).\end{array}\end{eqnarray}$
Solving the linear equations system yields the renormalization equations $\begin{eqnarray}{a}_{{t}_{0}}=\frac{\left|\begin{array}{cc}(\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1}) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{1})\\ (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2}^{* }) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{2}^{* })\end{array}\right|}{\left|\begin{array}{cc}({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{1}) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{1})\\ ({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{2}^{* }) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{2}^{* })\end{array}\right|},\end{eqnarray}$
$\begin{eqnarray}{\theta }_{{t}_{0}}=\frac{\left|\begin{array}{cc}({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{1}) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1})\\ ({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{2}^{* }) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2}^{* })\end{array}\right|}{\left|\begin{array}{cc}({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{1}) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{1})\\ ({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{2}^{* }) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{2}^{* })\end{array}\right|},\end{eqnarray}$
where we denote $\begin{eqnarray}{u}_{00a}^{* }=-4a{{\rm{{\rm{sech}} }}}^{2}\xi +4a\xi {{\rm{{\rm{sech}} }}}^{2}\xi {\rm{\tanh }}\xi .\end{eqnarray}$
Take (see [1]) P(u) = − u, Q(u) = uxx. Then we get renormalization equations by computing integrals in above formulas,
$\begin{eqnarray}{a}^{{\prime} }=-\frac{2}{3}\epsilon a-\frac{8}{15}\eta {a}^{3},\end{eqnarray}$
$\begin{eqnarray}{\theta }^{{\prime} }=0,\end{eqnarray}$
from which we have for ε ≠ 0 and η = 0, $\begin{eqnarray}\theta ={\theta }_{0},a={a}_{0}{{\rm{e}}}^{-\frac{2}{3}\epsilon t};\end{eqnarray}$
for ε = 0 and η ≠ 0, $\begin{eqnarray}\theta ={\theta }_{0},a=\frac{1}{\sqrt{\frac{1}{{a}_{0}^{2}}+\frac{16}{15}\eta t}};\end{eqnarray}$
for ε ≠ 0 and η ≠ 0, $\begin{eqnarray}\theta ={\theta }_{0},a=\frac{{a}_{0}\sqrt{\frac{2\epsilon }{3}}{{\rm{e}}}^{-\frac{2}{3}\epsilon t}}{\sqrt{1+\frac{8}{15}{a}_{0}^{2}\eta {{\rm{e}}}^{-\frac{4}{3}\epsilon t}}},\end{eqnarray}$
where a0 and θ0 are two arbitrary constants.Correspondingly, the approximate solitary wave solution is given by 55 )–(57 ) at under different conditions.
$\begin{eqnarray}u=-2{a}^{2}{{\rm{{\rm{sech}} }}}^{2}(ax-4{a}^{3}t+a{\theta }_{0}),\end{eqnarray}$
with a satisfying the above formulas in (2.4. Perturbed MKdV equation
Consider the perturbed MKdV equation
$\begin{eqnarray}{u}_{t}+6{u}^{2}{u}_{x}+{u}_{xx}=\epsilon P(u)+\eta Q(u).\end{eqnarray}$
The zeroth and first order equations, respectively, are $\begin{eqnarray}{u}_{00t}+6{u}_{00}^{2}{u}_{00x}+{u}_{00xxx}=0,\end{eqnarray}$
$\begin{eqnarray}{u}_{01t}+6{u}_{00}^{2}{u}_{01x}+12{u}_{00}{u}_{00x}{u}_{01}+{u}_{01xxx}=P({u}_{00}),\end{eqnarray}$
$\begin{eqnarray}{u}_{10t}+6{u}_{00}^{2}{u}_{10x}+12{u}_{00}{u}_{00x}{u}_{10}+{u}_{10xxx}=Q({u}_{00}).\end{eqnarray}$
Then we define linear operator L by
$\begin{eqnarray}Lf={f}_{t}+6{u}_{00}^{2}{f}_{x}+12{u}_{00}{u}_{00x}f+{f}_{xxx},\end{eqnarray}$
and hence the dual of L is $\begin{eqnarray}{L}^{* }g={g}_{t}+6{u}_{00}^{2}{g}_{x}+{g}_{xxx},\end{eqnarray}$
where we have used the inner product $(f,g)={\int }_{-\infty }^{+\infty }fg{\rm{d}}x$ and the properties of f, g and their derivatives being zero for x = ∞.From (63 ) and (64 ), we know that if Lf = 0, then we have L*g = 0 for
$\begin{eqnarray}g={\int }_{-\infty }^{x}f.\end{eqnarray}$
Therefore, from
$\begin{eqnarray}(Lf,g)={(f,g)}_{t}-({L}^{* }g,f),\end{eqnarray}$
and L*g = 0 we have $\begin{eqnarray}{({u}_{01},g)}_{t}=(L{u}_{01},g)=(P({u}_{00}),g),\end{eqnarray}$
$\begin{eqnarray}{({u}_{10},g)}_{t}=(L{u}_{10},g)=(Q({u}_{00}),g).\end{eqnarray}$
Since the inner product is a defined integral on x ∈ ( − ∞, + ∞), by the transformation of variable, we have $\begin{eqnarray}({u}_{01},g)=(t-{t}_{0})(P({u}_{00}),g),\end{eqnarray}$
$\begin{eqnarray}({u}_{10},g)=(t-{t}_{0})(Q({u}_{00}),g),\end{eqnarray}$
where t0 is an arbitrary constant and hence can be considered as a variable.Take the single soliton solution of equation (60 ) 61 ) with respect to a and θ respectively to give the solutions of Lf = 0,
$\begin{eqnarray}{u}_{00}={u}_{00}(t,x,a,\theta )=a{\rm{{\rm{sech}} }}(ax-4{a}^{3}t+a\theta )=a{\rm{{\rm{sech}} }}\xi ,\end{eqnarray}$
where a, θ are two parameters, and ξ = ax − 4a3t + aθ. Then take the derivatives of equation ( $\begin{eqnarray}\begin{array}{rcl}{f}_{1} & = & {u}_{00\theta }=-{a}^{2}{\rm{{\rm{sech}} }}\xi {\rm{\tanh }}\xi ,{f}_{2}={u}_{00a}\\ & = & {\rm{{\rm{sech}} }}\xi -\xi {\rm{{\rm{sech}} }}\xi {\rm{\tanh }}\xi -8a(t-{t}_{0}){f}_{1},\end{array}\end{eqnarray}$
and hence give the solutions of L*g = 0, $\begin{eqnarray}\begin{array}{l}{g}_{1}\,=\,\frac{1}{a}{\displaystyle \int }_{-\infty }^{\xi }{u}_{00\theta }=a{\rm{{\rm{sech}} }}\xi ,\\ {g}_{2}\,=\,\frac{1}{a}{\displaystyle \int }_{-\infty }^{\xi }{u}_{00a}={g}_{2}^{* }-8a(t-{t}_{0}){g}_{1},\end{array}\end{eqnarray}$
where we denote $\begin{eqnarray}{g}_{2}^{* }=\xi {\rm{{\rm{sech}} }}\xi .\end{eqnarray}$
Similarly, we get $\begin{eqnarray}{a}_{{t}_{0}}=\frac{\left|\begin{array}{cc}(\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1}) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{1})\\ (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2}^{* }) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{2}^{* })\end{array}\right|}{\left|\begin{array}{cc}({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{1}) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{1})\\ ({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{2}^{* }) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{2}^{* })\end{array}\right|},\end{eqnarray}$
$\begin{eqnarray}{\theta }_{{t}_{0}}=\frac{\left|\begin{array}{cc}({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{1}) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1})\\ ({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{2}^{* }) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2}^{* })\end{array}\right|}{\left|\begin{array}{cc}({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{1}) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{1})\\ ({u}_{00a}^{* }({t}_{0},x,a,\theta ),{g}_{2}^{* }) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{2}^{* })\end{array}\right|},\end{eqnarray}$
where we denote $\begin{eqnarray}{u}_{00a}^{* }={\rm{{\rm{sech}} }}\xi -\xi {\rm{{\rm{sech}} }}\xi {\rm{\tanh }}\xi .\end{eqnarray}$
Take (see [1]) P(u) = − u, Q(u) = uxx. Then we get renormalization equations by computing integrals in the above formulas,
$\begin{eqnarray}{a}^{{\prime} }=-\frac{2}{7}\epsilon a-\frac{2}{21}\eta {a}^{3},\end{eqnarray}$
$\begin{eqnarray}{\theta }^{{\prime} }=0,\end{eqnarray}$
from which we have for ε ≠ 0 and η = 0, $\begin{eqnarray}\theta ={\theta }_{0},a={a}_{0}{{\rm{e}}}^{-\frac{2}{7}\epsilon t};\end{eqnarray}$
for ε = 0 and η ≠ 0, $\begin{eqnarray}\theta ={\theta }_{0},a=\frac{1}{\sqrt{\frac{1}{{a}_{0}^{2}}+\frac{4}{21}\eta t}};\end{eqnarray}$
for ε ≠ 0 and η ≠ 0, $\begin{eqnarray}\theta ={\theta }_{0},a=\frac{{a}_{0}\sqrt{\frac{2\epsilon }{7}}{{\rm{e}}}^{-\frac{2}{7}\epsilon t}}{\sqrt{1+\frac{2}{21}{a}_{0}^{2}\eta {{\rm{e}}}^{-\frac{4}{7}\epsilon t}}},\end{eqnarray}$
where a0 and θ0 are two arbitrary constants.Correspondingly, the approximate solitary wave solution is given by 80 –82 ) under different conditions.
$\begin{eqnarray}u=a{\rm{{\rm{sech}} }}(ax-4{a}^{3}t+a{\theta }_{0}),\end{eqnarray}$
with a satisfying the above formulas in (2.5. Perturbed NLS equation
Consider the perturbation of the nonlinear Schrodinger equation
$\begin{eqnarray}{\rm{i}}{u}_{t}+{u}_{xx}+2| u{| }^{2}u=\epsilon P(u)+\eta Q(u).\end{eqnarray}$
The zeroth and first order equations are respectively $\begin{eqnarray}{\rm{i}}{u}_{00t}+{u}_{00xx}+2| {u}_{00}{| }^{2}{u}_{00}=0,\end{eqnarray}$
$\begin{eqnarray}{\rm{i}}{u}_{01t}+{u}_{01xx}+2{u}_{00}^{2}{\bar{u}}_{01}+4{u}_{00}{\bar{u}}_{00}{u}_{01}=P({u}_{00}),\end{eqnarray}$
$\begin{eqnarray}{\rm{i}}{u}_{10t}+{u}_{10xx}+2{u}_{00}^{2}{\bar{u}}_{10}+4{u}_{00}{\bar{u}}_{00}{u}_{10}=Q({u}_{00}).\end{eqnarray}$
Then we define linear operator L by
$\begin{eqnarray}Lf={\rm{i}}{f}_{t}+{f}_{xx}+2{u}_{00}^{2}\bar{f}+4{u}_{00}{\bar{u}}_{00}f,\end{eqnarray}$
and hence the dual of L is $\begin{eqnarray}{L}^{* }g={\rm{i}}{g}_{t}+{g}_{xx}+2{u}_{00}^{2}\bar{g}+4{u}_{00}{\bar{u}}_{00}g,\end{eqnarray}$
where we have used the inner product $(f,g)={\int }_{-\infty }^{+\infty }f\bar{g}{\rm{d}}x$ and the properties of f, g and their derivatives being zero for x = ∞. So L = L*.From (88 ) and (89 ), we know that if Lf = 0, then we have L*g = 0 for
$\begin{eqnarray}g={\int }_{-\infty }^{x}f.\end{eqnarray}$
Take the single soliton solution of equation (85 ) 86 ) with respect to a, k,θ and φ to give the solution of L*g = 0,
$\begin{eqnarray}\begin{array}{rcl}{u}_{00} & = & {u}_{00}(t,x,a,k,\theta ,\phi )\\ & = & a{\rm{{\rm{sech}} }}(ax-2akt+a\theta ){{\rm{e}}}^{{\rm{i}}(kx-({k}^{2}-{a}^{2})t+\phi )}\\ & = & a{\rm{{\rm{sech}} }}\xi {{\rm{e}}}^{{\rm{i}}y},\end{array}\end{eqnarray}$
where a, k, θ, φ are four parameters, and z = ax − 2akt + aθ, y = kx − (k2 − a2)t + φ. Then take the derivatives of equation ( $\begin{eqnarray}\begin{array}{rcl}{g}_{1}={u}_{00\phi } & = & a{\rm{i}}{\rm{{\rm{sech}} }}z{{\rm{e}}}^{{\rm{i}}y},{g}_{2}={u}_{00\theta }\\ & = & -{a}^{2}{\rm{i}}{\rm{{\rm{sech}} }}z{\rm{\tanh }}z{{\rm{e}}}^{{\rm{i}}y},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{g}_{3} & = & {u}_{00k}={\rm{i}}z{\rm{{\rm{sech}} }}z{{\rm{e}}}^{{\rm{i}}y}\\ & & +2{a}^{2}(t-{t}_{0}){\rm{{\rm{sech}} }}z{\rm{\tanh }}z{{\rm{e}}}^{{\rm{i}}y},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{g}_{4} & = & {u}_{00a}={\rm{{\rm{sech}} }}z{{\rm{e}}}^{{\rm{i}}y}-z{\rm{{\rm{sech}} }}z{\rm{\tanh }}z{{\rm{e}}}^{{\rm{i}}y}\\ & & +2{a}^{2}{\rm{i}}(t-{t}_{0}){\rm{{\rm{sech}} }}z{{\rm{e}}}^{{\rm{i}}y}.\end{array}\end{eqnarray}$
Furthermore, we take $\begin{eqnarray}{g}_{3}^{* }\,=\,{u}_{00k}={\rm{i}}z{\rm{{\rm{sech}} }}z{{\rm{e}}}^{{\rm{i}}y},\end{eqnarray}$
$\begin{eqnarray}{g}_{4}^{* }\,=\,{u}_{00a}={\rm{{\rm{sech}} }}z{{\rm{e}}}^{{\rm{i}}y}-z{\rm{{\rm{sech}} }}z{\rm{\tanh }}z{{\rm{e}}}^{{\rm{i}}y}.\end{eqnarray}$
Therefore, from $\begin{eqnarray}(Lf,g)={(f,g)}_{t}-({L}^{* }g,f),\end{eqnarray}$
and L*g = 0 we have $\begin{eqnarray}{({u}_{01},g)}_{t}=(L{u}_{01},g)=(P({u}_{00}),g),\end{eqnarray}$
$\begin{eqnarray}{({u}_{10},g)}_{t}=(L{u}_{10},g)=(Q({u}_{00}),g).\end{eqnarray}$
Since the inner product is a defined integral on x ∈ ( − ∞, + ∞), by the transformation of variable, we have $\begin{eqnarray}({u}_{01},g)=(t-{t}_{0})(P({u}_{00}),g),\end{eqnarray}$
$\begin{eqnarray}({u}_{10},g)=(t-{t}_{0})(Q({u}_{00}),g),\end{eqnarray}$
where t0 is an arbitrary constant and hence can be considered as a variable.So we have
$\begin{eqnarray}\begin{array}{rcl} & & {({u}_{00}({t}_{0},x,a,k,\theta ),g({t}_{0},x))}_{{t}_{0}}\\ & = & ((\epsilon P({u}_{00})+\eta Q({u}_{00}),g)+({u}_{00{t}_{0}},g)+({u}_{00{t}_{0}},{g}_{{t}_{0}}),\end{array}\end{eqnarray}$
from which we have considered a, k, θ and φ as the functions of t0, $\begin{eqnarray*}({u}_{00a}({t}_{0},x,k,a,\theta ){a}_{{t}_{0}}+({u}_{00k}({t}_{0},x,k,a,\theta ){k}_{{t}_{0}}+{u}_{00\xi }({t}_{0},x,k,a,\theta ){\theta }_{{t}_{0}},g({t}_{0},x))\end{eqnarray*}$
$\begin{eqnarray}=\,((\epsilon P({u}_{00})+\eta Q({u}_{00}),g).\end{eqnarray}$
By taking g = g1, g = g2, $g={g}_{3}^{* }$ and $g={g}_{4}^{* }$, we get $\begin{eqnarray}\begin{array}{rcl} & & ({u}_{00a},{g}_{1}){a}_{{t}_{0}}+({u}_{00k},{g}_{1}){k}_{{t}_{0}}+({u}_{00\theta },{g}_{1}){\theta }_{{t}_{0}}\\ & & +({u}_{00\phi },{g}_{1}){\phi }_{{t}_{0}}=((\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1}),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl} & & ({u}_{00a},{g}_{2}){a}_{{t}_{0}}+({u}_{00k},{g}_{2}){k}_{{t}_{0}}+({u}_{00\theta },{g}_{2}){\theta }_{{t}_{0}}\\ & & +({u}_{00\phi },{g}_{2}){\phi }_{{t}_{0}}=((\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2}),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl} & & ({u}_{00a},{g}_{3}^{* }){a}_{{t}_{0}}+({u}_{00k},{g}_{3}^{* }){k}_{{t}_{0}}+({u}_{00\theta },{g}_{3}^{* }){\theta }_{{t}_{0}}\\ & & +({u}_{00\phi },{g}_{3}^{* }){\phi }_{{t}_{0}}=((\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{3}^{* }),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl} & & ({u}_{00a},{g}_{4}){a}_{{t}_{0}}+({u}_{00k},{g}_{4}^{* }){k}_{{t}_{0}}+({u}_{00\theta },{g}_{4}^{* }){\theta }_{{t}_{0}}\\ & & +({u}_{00\phi },{g}_{4}^{* }){\phi }_{{t}_{0}}=((\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{4}^{* }).\end{array}\end{eqnarray}$
Solving the linear equations system yields. Similarly, we get
$\begin{eqnarray}{a}_{{t}_{0}}=\frac{\left|\begin{array}{cccc}(\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1}) & ({u}_{00k},{g}_{1}) & ({u}_{00\theta },{g}_{1}) & ({u}_{00\phi },{g}_{1})\\ (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2}) & ({u}_{00k},{g}_{2}) & ({u}_{00\theta },{g}_{2}) & ({u}_{00\phi },{g}_{2})\\ (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{3}^{* }) & ({u}_{00k},{g}_{3}^{* }) & ({u}_{00\theta },{g}_{3}^{* }) & ({u}_{00\phi },{g}_{3}^{* })\\ (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{4}^{* }) & ({u}_{00k},{g}_{4}^{* }) & ({u}_{00\theta },{g}_{4}^{* }) & ({u}_{00\phi },{g}_{4}^{* })\end{array}\right|}{\left|\begin{array}{cccc}({u}_{00a},{g}_{1}) & ({u}_{00k},{g}_{1}) & ({u}_{00\theta },{g}_{1}) & ({u}_{00\phi },{g}_{1})\\ ({u}_{00a},{g}_{2}) & ({u}_{00k},{g}_{2}) & ({u}_{00\theta },{g}_{2}) & ({u}_{00\phi },{g}_{2})\\ ({u}_{00a},{g}_{3}^{* }) & ({u}_{00k},{g}_{3}^{* }) & ({u}_{00\theta },{g}_{3}^{* }) & ({u}_{00\phi },{g}_{3}^{* })\\ ({u}_{00a},{g}_{4}^{* }) & ({u}_{00k},{g}_{4}^{* }) & ({u}_{00\theta },{g}_{4}^{* }) & ({u}_{00\phi },{g}_{4}^{* })\end{array}\right|},\end{eqnarray}$
$\begin{eqnarray}{k}_{{t}_{0}}=\frac{\left|\begin{array}{cccc}({u}_{00a},{g}_{1}) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1}) & ({u}_{00\theta },{g}_{1}) & ({u}_{00\phi },{g}_{1})\\ ({u}_{00a},{g}_{2}) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2}) & ({u}_{00\theta },{g}_{2}) & ({u}_{00\phi },{g}_{2})\\ ({u}_{00a},{g}_{3}^{* }) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{3}^{* }) & ({u}_{00\theta },{g}_{3}^{* }) & ({u}_{00\phi },{g}_{3}^{* })\\ ({u}_{00a},{g}_{4}^{* }) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{4}^{* }) & ({u}_{00\theta },{g}_{4}^{* }) & ({u}_{00\phi },{g}_{4}^{* })\end{array}\right|}{\left|\begin{array}{cccc}({u}_{00a},{g}_{1}) & ({u}_{00k},{g}_{1}) & ({u}_{00\theta },{g}_{1}) & ({u}_{00\phi },{g}_{1})\\ ({u}_{00a},{g}_{2}) & ({u}_{00k},{g}_{2}) & ({u}_{00\theta },{g}_{2}) & ({u}_{00\phi },{g}_{2})\\ ({u}_{00a},{g}_{3}^{* }) & ({u}_{00k},{g}_{3}^{* }) & ({u}_{00\theta },{g}_{3}^{* }) & ({u}_{00\phi },{g}_{3}^{* })\\ ({u}_{00a},{g}_{4}^{* }) & ({u}_{00k},{g}_{4}^{* }) & ({u}_{00\theta },{g}_{4}^{* }) & ({u}_{00\phi },{g}_{4}^{* })\end{array}\right|},\end{eqnarray}$
$\begin{eqnarray}{\theta }_{{t}_{0}}=\frac{\left|\begin{array}{cccc}({u}_{00a},{g}_{1}) & ({u}_{00k},{g}_{1}) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1}) & ({u}_{00\phi },{g}_{1})\\ ({u}_{00a},{g}_{2}) & ({u}_{00k},{g}_{2}) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2}) & ({u}_{00\phi },{g}_{2})\\ ({u}_{00a},{g}_{3}) & ({u}_{00k},{g}_{3}^{* }) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{3}^{* }) & ({u}_{00\phi },{g}_{3}^{* })\\ ({u}_{00a},{g}_{4}) & ({u}_{00k},{g}_{4}^{* }) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{4}^{* }) & ({u}_{00\phi },{g}_{4}^{* })\end{array}\right|}{\left|\begin{array}{cccc}({u}_{00a},{g}_{1}) & ({u}_{00k},{g}_{1}) & ({u}_{00\theta },{g}_{1}) & ({u}_{00\phi },{g}_{1})\\ ({u}_{00a},{g}_{2}) & ({u}_{00k},{g}_{2}) & ({u}_{00\theta },{g}_{2}) & ({u}_{00\phi },{g}_{2})\\ ({u}_{00a},{g}_{3}^{* }) & ({u}_{00k},{g}_{3}^{* }) & ({u}_{00\theta },{g}_{3}^{* }) & ({u}_{00\phi },{g}_{3}^{* })\\ ({u}_{00a},{g}_{4}^{* }) & ({u}_{00k},{g}_{4}^{* }) & ({u}_{00\theta },{g}_{4}^{* }) & ({u}_{00\phi },{g}_{4}^{* })\end{array}\right|},\end{eqnarray}$
$\begin{eqnarray}{\phi }_{{t}_{0}}=\frac{\left|\begin{array}{cccc}({u}_{00a},{g}_{1}) & ({u}_{00k},{g}_{1}) & ({u}_{00\theta },{g}_{1}) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1})\\ ({u}_{00a},{g}_{2}) & ({u}_{00k},{g}_{2}) & ({u}_{00\theta },{g}_{2}) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2})\\ ({u}_{00a},{g}_{3}^{* }) & ({u}_{00k},{g}_{3}^{* }) & ({u}_{00\theta },{g}_{3}^{* }) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{3}^{* })\\ ({u}_{00a},{g}_{4}^{* }) & ({u}_{00k},{g}_{4}^{* }) & ({u}_{00\theta },{g}_{4}^{* }) & (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{4}^{* })\end{array}\right|}{\left|\begin{array}{cccc}({u}_{00a},{g}_{1}) & ({u}_{00k},{g}_{1}) & ({u}_{00\theta },{g}_{1}) & ({u}_{00\phi },{g}_{1})\\ ({u}_{00a},{g}_{2}) & ({u}_{00k},{g}_{2}) & ({u}_{00\theta },{g}_{2}) & ({u}_{00\phi },{g}_{2})\\ ({u}_{00a},{g}_{3}^{* }) & ({u}_{00k},{g}_{3}^{* }) & ({u}_{00\theta },{g}_{3}^{* }) & ({u}_{00\phi },{g}_{3}^{* })\\ ({u}_{00a},{g}_{4}^{* }) & ({u}_{00k},{g}_{4}^{* }) & ({u}_{00\theta },{g}_{4}^{* }) & ({u}_{00\phi },{g}_{4}^{* })\end{array}\right|},\end{eqnarray}$
In order to avoid getting infinity in the determinants in the numerator and denominator, we take a and k as constants, and only take φ and θ as variables to solve the renormalization equations. Then we have
$\begin{eqnarray}{\phi }_{{t}_{0}}=\frac{\left|\begin{array}{cc}(\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1}) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{1})\\ (\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2}) & ({u}_{00\theta }({t}_{0},x,a,\theta ),{g}_{2})\end{array}\right|}{\left|\begin{array}{cc}({g}_{1},{g}_{1}) & ({g}_{1},{g}_{2})\\ ({g}_{2},{g}_{1}) & ({g}_{2},{g}_{2})\end{array}\right|},\end{eqnarray}$
$\begin{eqnarray}{\theta }_{{t}_{0}}=\frac{\left|\begin{array}{cc}({u}_{00\phi }({t}_{0},x,a,\theta ),{g}_{1}) & ((\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{1})\\ ({u}_{00\phi }({t}_{0},x,a,\theta ),{g}_{2}) & ((\epsilon P({u}_{00})+\eta Q({u}_{00}),{g}_{2})\end{array}\right|}{\left|\begin{array}{cc}({g}_{1},{g}_{1}) & ({g}_{1},{g}_{2})\\ ({g}_{2},{g}_{1}) & ({g}_{2},{g}_{2})\end{array}\right|}.\end{eqnarray}$
Take (see [1]) P(u) = − iu and Q(u) = i∣u∣4u. Then we get $\begin{eqnarray}\phi =-\left(\frac{4}{3}{a}^{4}\epsilon +\frac{32}{45}{a}^{10}\eta \right)t+{\phi }_{0},\theta ={\theta }_{0},\end{eqnarray}$
where φ0 and θ0 are the two constants.3. Conclusion
This new direct approach is rather simple in solving the perturbation of solitons of integrable systems. As the typical application, we compute the perturbation of KdV, MKdV and nonlinear Schrodinger equations, and obtain the first order corrections of the parameters. Because the method does not depend on inverse scattering data and spectrum theory, it is very easy in applications. This method can also be applied to solve other nearly integrable systems such as the Benjamin–Ono equation, and so on. Compared with routine methods, this method also has the following advantages: the first is that the mathematical foundation is just based on a simple mathematical theory namely Taylor expansion at a general point, the second is that the secular terms in perturbation series are eliminated automatically, the third is that multiple time scales arise naturally from the final naive perturbation expansion, while any priori physical assumption on the form of the solution is avoided, the fourth is that the Green's function and corresponding spectrum of linear differential operators are not needed. However, this method can not deal with nonlocal solitons such as kink type solution because the boundary terms in inner products can not be eliminated. In addition, we need more considerations and computations in the perturbation of multi-solitons. These problems need further exploration.
Conflict of interest statement
The author declares that there are no competing financial interests.