多德-布洛-米哈伊洛夫方程

多德-布洛-米哈伊洛夫方程不是函数u的多项式形式，因此必须做代换：

${\displaystyle v=e^{u}}$ , 变为：

${\displaystyle v*v_{xt}-v_{t}*v_{x}+\alpha *v^{3}+\gamma =0}$ 

得到函数v(x,t)的行波解： ${\displaystyle v(x,t)=(1/2)*\gamma ^{(}1/3)+(3/2)*\gamma ^{(}1/3)*cot(_{C}1+_{C}2*x-(3/4)*\gamma ^{(}1/3)*t/_{C}2)^{2}}$  ${\displaystyle v(x,t)=(1/2)*\gamma ^{(}1/3)-(3/2)*\gamma ^{(}1/3)*coth(_{C}1+_{C}2*x+(3/4)*\gamma ^{(}1/3)*t/_{C}2)^{2}}$  ${\displaystyle v(x,t)=(1/2)*\gamma ^{(}1/3)+(3/2)*\gamma ^{(}1/3)*tan(_{C}1+_{C}2*x-(3/4)*\gamma ^{(}1/3)*t/_{C}2)^{2}}$  ${\displaystyle v(x,t)=(1/2)*\gamma ^{(}1/3)-(3/2)*\gamma ^{(}1/3)*tanh(_{C}1+_{C}2*x+(3/4)*\gamma ^{(}1/3)*t/_{C}2)^{2}}$  ${\displaystyle v(x,t)=-(1/4)*\gamma ^{(}1/3)-(1/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3)+(-(3/4)*\gamma ^{(}1/3)-(3/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*cot(_{C}1+_{C}2*x+(3/4)*((1/2)*\gamma ^{(}1/3)+(1/2*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*t/_{C}2)^{2}}$  ${\displaystyle v(x,t)=-(1/4)*\gamma ^{(}1/3)-(1/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3)+(-(3/4)*\gamma ^{(}1/3)-(3/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*tan(_{C}1+_{C}2*x+(3/4)*((1/2)*\gamma ^{(}1/3)+(1/2*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*t/_{C}2)^{2}}$  ${\displaystyle v(x,t)=-(1/4)*\gamma ^{(}1/3)-(1/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3)+((3/4)*\gamma ^{(}1/3)+(3/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*coth(_{C}1+_{C}2*x+(3/4)*(-(1/2)*\gamma ^{(}1/3)-(1/2*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*t/_{C}2)^{2}}$  ${\displaystyle v(x,t)=-(1/4)*\gamma ^{(}1/3)-(1/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3)+((3/4)*\gamma ^{(}1/3)+(3/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*tanh(_{C}1+_{C}2*x+(3/4)*(-(1/2)*\gamma ^{(}1/3)-(1/2*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*t/_{C}2)^{2}}$ 

作反变换： ${\displaystyle u(x,t)=ln(v(x,t))}$ 

即得多德-布洛-米哈伊洛夫方程的行波解：

${\displaystyle u(x,t)=ln((1/2)*\gamma ^{(}1/3)+(3/2)*\gamma ^{(}1/3)*cot(_{C}1+_{C}2*x-(3/4)*\gamma ^{(}1/3)*t/_{C}2)^{2})}$  ${\displaystyle u(x,t)=ln((1/2)*\gamma ^{(}1/3)-(3/2)*\gamma ^{(}1/3)*coth(_{C}1+_{C}2*x+(3/4)*\gamma ^{(}1/3)*t/_{C}2)^{)}}$  ${\displaystyle u(x,t)=ln((1/2)*\gamma ^{(}1/3)+(3/2)*\gamma ^{(}1/3)*tan(_{C}1+_{C}2*x-(3/4)*\gamma ^{(}1/3)*t/_{C}2)^{2})}$  ${\displaystyle u(x,t)=ln((1/2)*\gamma ^{(}1/3)-(3/2)*\gamma ^{(}1/3)*tanh(_{C}1+_{C}2*x+(3/4)*\gamma ^{(}1/3)*t/_{C}2)^{2})}$  ${\displaystyle u(x,t)=ln(-(1/4)*\gamma ^{(}1/3)-(1/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3)+(-(3/4)*\gamma ^{(}1/3)-(3/4*I)*{\sqrt {(}}3)*\gamma (1/3))*cot(_{C}1+_{C}2*x+(3/4)*((1/2)*\gamma ^{(}1/3)+(1/2*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*t/_{C}2)^{2})}$  ${\displaystyle u(x,t)=ln(-(1/4)*\gamma ^{(}1/3)-(1/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3)+(-(3/4)*\gamma ^{(}1/3)-(3/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*tan(_{C}1+_{C}2*x+(3/4)*((1/2)*\gamma ^{(}1/3)+(1/2*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*t/_{C}2)^{2})}$  ${\displaystyle u(x,t)=ln(-(1/4)*\gamma ^{(}1/3)-(1/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3)+((3/4)*\gamma ^{(}1/3)+(3/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*coth(_{C}1+_{C}2*x+(3/4)*(-(1/2)*\gamma ^{(}1/3)-(1/2*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*t/_{C}2)^{2})}$  ${\displaystyle u(x,t)=ln(-(1/4)*\gamma ^{(}1/3)-(1/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3)+((3/4)*\gamma ^{(}1/3)+(3/4*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*tanh(_{C}1+_{C}2*x+(3/4)*(-(1/2)*\gamma ^{(}1/3)-(1/2*I)*{\sqrt {(}}3)*\gamma ^{(}1/3))*t/_{C}2)^{2})}$ 

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1. ^ 李志斌编著 《非线性数学物理方程的行波解》 第105-107页，科学出版社 2008

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