多德-布洛-米哈伊洛夫方程

多德-布洛-米哈伊洛夫方程(Dodd-Bullough-Mikhailov equation)是一個非線性偏微分方程。

$$u_{xt}+\alpha*e^u+\gamma*e^{-2*u} = 0$$

## 行波解

多德-布洛-米哈伊洛夫方程不是函數u的多項式形式，因此必須做代換：

$v=e^u$, 變為：

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

得到函數v(x,t)的行波解：

$$v(x, t) = (1/2)*\gamma^(1/3)+(3/2)*\gamma^(1/3)*cot(_C1+_C2*x-(3/4)*\gamma^(1/3)*t/_C2)^2$$ $$v(x, t) = (1/2)*\gamma^(1/3)-(3/2)*\gamma^(1/3)*coth(_C1+_C2*x+(3/4)*\gamma^(1/3)*t/_C2)^2$$ $$v(x, t) = (1/2)*\gamma^(1/3)+(3/2)*\gamma^(1/3)*tan(_C1+_C2*x-(3/4)*\gamma^(1/3)*t/_C2)^2$$ $$v(x, t) = (1/2)*\gamma^(1/3)-(3/2)*\gamma^(1/3)*tanh(_C1+_C2*x+(3/4)*\gamma^(1/3)*t/_C2)^2$$ $$v(x, t) = -(1/4)*\gamma^(1/3)-(1/4*I)*\sqrt(3)*\gamma^(1/3)+(-(3/4)*\gamma^(1/3)-(3/4*I)*\sqrt(3)*\gamma^(1/3))*cot(_C1+_C2*x+(3/4)*((1/2)*\gamma^(1/3)+(1/2*I)*\sqrt(3)*\gamma^(1/3))*t/_C2)^2$$ $$v(x, t) = -(1/4)*\gamma^(1/3)-(1/4*I)*\sqrt(3)*\gamma^(1/3)+(-(3/4)*\gamma^(1/3)-(3/4*I)*\sqrt(3)*\gamma^(1/3))*tan(_C1+_C2*x+(3/4)*((1/2)*\gamma^(1/3)+(1/2*I)*\sqrt(3)*\gamma^(1/3))*t/_C2)^2$$ $$v(x, t) = -(1/4)*\gamma^(1/3)-(1/4*I)*\sqrt(3)*\gamma^(1/3)+((3/4)*\gamma^(1/3)+(3/4*I)*\sqrt(3)*\gamma^(1/3))*coth(_C1+_C2*x+(3/4)*(-(1/2)*\gamma^(1/3)-(1/2*I)*\sqrt(3)*\gamma^(1/3))*t/_C2)^2$$ $$v(x, t) = -(1/4)*\gamma^(1/3)-(1/4*I)*\sqrt(3)*\gamma^(1/3)+((3/4)*\gamma^(1/3)+(3/4*I)*\sqrt(3)*\gamma^(1/3))*tanh(_C1+_C2*x+(3/4)*(-(1/2)*\gamma^(1/3)-(1/2*I)*\sqrt(3)*\gamma^(1/3))*t/_C2)^2$$

作反變換：

$$u(x,t)=ln(v(x,t))$$

即得多德-布洛-米哈伊洛夫方程的行波解：

$$u(x, t) =ln( (1/2)*\gamma^(1/3)+(3/2)*\gamma^(1/3)*cot(_C1+_C2*x-(3/4)*\gamma^(1/3)*t/_C2)^2)$$ $$u(x, t) =ln( (1/2)*\gamma^(1/3)-(3/2)*\gamma^(1/3)*coth(_C1+_C2*x+(3/4)*\gamma^(1/3)*t/_C2)^)$$ $$u(x, t) =ln( (1/2)*\gamma^(1/3)+(3/2)*\gamma^(1/3)*tan(_C1+_C2*x-(3/4)*\gamma^(1/3)*t/_C2)^2)$$ $$u(x, t) =ln( (1/2)*\gamma^(1/3)-(3/2)*\gamma^(1/3)*tanh(_C1+_C2*x+(3/4)*\gamma^(1/3)*t/_C2)^2)$$ $$u(x, t) =ln( -(1/4)*\gamma^(1/3)-(1/4*I)*\sqrt(3)*\gamma^(1/3)+(-(3/4)*\gamma^(1/3)-(3/4*I)*\sqrt(3)*\gamma(1/3))*cot(_C1+_C2*x+(3/4)*((1/2)*\gamma^(1/3)+(1/2*I)*\sqrt(3)*\gamma^(1/3))*t/_C2)^2)$$ $$u(x, t) =ln( -(1/4)*\gamma^(1/3)-(1/4*I)*\sqrt(3)*\gamma^(1/3)+(-(3/4)*\gamma^(1/3)-(3/4*I)*\sqrt(3)*\gamma^(1/3))*tan(_C1+_C2*x+(3/4)*((1/2)*\gamma^(1/3)+(1/2*I)*\sqrt(3)*\gamma^(1/3))*t/_C2)^2)$$ $$u(x, t) =ln( -(1/4)*\gamma^(1/3)-(1/4*I)*\sqrt(3)*\gamma^(1/3)+((3/4)*\gamma^(1/3)+(3/4*I)*\sqrt(3)*\gamma^(1/3))*coth(_C1+_C2*x+(3/4)*(-(1/2)*\gamma^(1/3)-(1/2*I)*\sqrt(3)*\gamma^(1/3))*t/_C2)^2)$$ $$u(x, t) =ln( -(1/4)*\gamma^(1/3)-(1/4*I)*\sqrt(3)*\gamma^(1/3)+((3/4)*\gamma^(1/3)+(3/4*I)*\sqrt(3)*\gamma^(1/3))*tanh(_C1+_C2*x+(3/4)*(-(1/2)*\gamma^(1/3)-(1/2*I)*\sqrt(3)*\gamma^(1/3))*t/_C2)^2)$$

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