# 阿多米安分解法

${\displaystyle L(u)+R(u)+NL(u)=g(x,t)}$

${\displaystyle L^{-1}L(u)=-L^{-1}R(u)-L^{-1}NL(u)+L^{-1}g(x,t)=}$.

${\displaystyle u(x,t)=u(x,0)-L^{-1}NL(u)+L^{-1}g(x,t)}$.

${\displaystyle u=u_{0}+u_{1}+u_{2}+u_{3}+\cdots }$

NL（u）=${\displaystyle A_{0}+A_{1}+A_{2}+\cdots }$

${\displaystyle A_{n}={\frac {1}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} \lambda ^{n}}}f(u(\lambda ))\mid _{\lambda =0},}$

${\displaystyle {\frac {\mathrm {d} ^{n}}{\mathrm {d} \lambda ^{n}}}u(\lambda )\mid _{\lambda =0}=n!u_{n}}$

${\displaystyle u(x,t)=u(x,0)+L_{-1}g(x,t)}$
${\displaystyle u_{1}(x,t)=-L^{-1}Ru_{0}-L^{-1}A_{0}}$
${\displaystyle u_{n}(x,t)=-L^{-1}Ru_{n-1}-L^{-1}A_{n-1}}$

Burgers-Fisher方程：

${\displaystyle {\frac {\partial u}{\partial t}}+u^{2}*{\frac {\partial u}{\partial x}}-{\frac {\partial ^{2}u}{\partial u^{2}}}=u*(1-u^{2})}$
${\displaystyle u[0]=tanh(x)}$
${\displaystyle u[1]=-tanh(x)*(1-tanh(x)^{2})*t}$
${\displaystyle u[2]=-(1/2)*t^{2}*tanh(x)*(-1+tanh(x)^{2})*(2-4*tanh(x)^{2})}$
${\displaystyle u[3]=-(1/3)*t^{3}*tanh(x)*(3-16*tanh(x)^{2}+26*tanh(x)^{4}-13*tanh(x)^{6}-3*tanh(x)^{2}*(1-tanh(x)^{2})+(2*(1-tanh(x)^{2}))*tanh(x)^{4})}$

pa := (-1.*tanh(x)-82360.*tanh(x)^13+73.*tanh(x)^3-1195.*tanh(x)^5+8233.*tanh(x)^7-29990.*tanh(x)^9+63510.*tanh(x)^15-26980.*tanh(x)^17+4862.*tanh(x)^19+63850.*tanh(x)^11)*t^9+(14650.*tanh(x)^13-16170.*tanh(x)^11+tanh(x)+1430.*tanh(x)^17+688.8*tanh(x)^5+10230.*tanh(x)^9-7102.*tanh(x)^15-54.67*tanh(x)^3-3672.*tanh(x)^7)*t^8+(-373.8*tanh(x)^5+1491.*tanh(x)^7-1.*tanh(x)+39.67*tanh(x)^3+3333.*tanh(x)^11+429.*tanh(x)^15-3036.*tanh(x)^9-1881.*tanh(x)^13)*t^7+(132.*tanh(x)^13+187.8*tanh(x)^5-502.*tanh(x)^11+743.5*tanh(x)^9-27.67*tanh(x)^3+tanh(x)-534.6*tanh(x)^7)*t^6+(-135.3*tanh(x)^9+161.1*tanh(x)^7-1.*tanh(x)+42.*tanh(x)^11-85.13*tanh(x)^5+18.33*tanh(x)^3)*t^5+(-37.*tanh(x)^7+33.33*tanh(x)^5+14.*tanh(x)^9-11.33*tanh(x)^3+tanh(x))*t^4+(5.*tanh(x)^7-10.33*tanh(x)^5+6.333*tanh(x)^3-1.*tanh(x))*t^3+(-3.*tanh(x)^3+tanh(x)+2.*tanh(x)^5)*t^2+(-1.*tanh(x)+tanh(x)^3)*t+tanh(x)

${\displaystyle u_{t}=u^{3}u_{xxx}.\,}$
${\displaystyle u[0]=cosh(x)}$
${\displaystyle u[1]=-cosh(x)*sinh(x)*t}$
${\displaystyle u[5]=-t^{5}*cosh(x)*sinh(x)^{5}-(20/3)*t^{5}*cosh(x)^{3}*sinh(x)^{3}-(47/15)*t^{5}*cosh(x)^{5}*sinh(x)}$