# 潘勒韦分析

## Kruskal 简化法原理

   ${\displaystyle F(u_{x},u_{t},u_{xx},\cdot \cdot \cdot )=0;}$


${\displaystyle u(x,t)=\sum _{k=0}^{\infty }(u[j](t)*phi^{j},j=0..N))/phi^{\rho }}$

${\displaystyle u}$ ${\displaystyle \psi ^{-\rho }*u_{0}}$

## 实例

${\displaystyle sys:={\frac {\partial u(x,t)}{\partial t}}+au(x,t){\frac {\partial (u(x,t)}{\partial x}}+b{\frac {\partial ^{2}(u(x,t)}{\partial x^{2}}}=0}$

${\displaystyle u(x,t)=\sum _{k=0}^{\infty }(u[j](t)\phi ^{j},j=0..N))/\phi ^{\rho }}$

${\displaystyle u}$ ${\displaystyle \psi ^{-\rho }u_{0}}$

${\displaystyle \rho \phi u[0]\psi [t]}$ ${\displaystyle -\phi ^{-\rho +1}u[0]^{2}\rho a-\rho (-b-b\rho )u[0]+\phi ^{2}u[0,t]=0}$

ρ=1，u[0] = 2 b/a；

${\displaystyle u(x,t)=2*b/(a*(x-\psi ))+u[j]*(x-\psi )^{(}j-1)}$  代人偏微分方程，

φ的最低次项为

${\displaystyle \phi ^{j-3}*b*(j+1)*(j-2)=0}$

${\displaystyle u(x,t)=\sum _{k=0}^{2}{\frac {u[j](t)\phi ^{j})}{\phi }}}$  再带入原方程得：

${\displaystyle a*phi^{4}*u[2]^{2}-(-u[1]*a+psi[t])*phi^{3}*u[2]+phi^{4}*u[2,t]+phi^{3}*u[1,t]+phi^{2}*u[0,t]+(-u[1]*a+psi[t])*u[0]*phi-u[0]^{2}*a+2*b*u[0]}$

${\displaystyle u[0,t]=0,-(u[1]*a-psi[t])*u[0]=0,-u[0]*(-2*b+a*u[0])}$

${\displaystyle u(x,t)={\frac {2*b}{a*(x-\psi )}}+{\frac {\psi [t]}{a}}+(x-\psi )*u[2]}$

## 参考文献

1. ^ Inna Shingareva, Carlos Lizarrraga-Celyaya Solving Nonlinear Partial Differential Equations with Maple and Mathematica, SpringerWienNewYork