作为典型的守恒律方程，Burgers 方程可以完成时间导和空间导的彻底转换，这里记录一下，后面肯定会用到。

\[ u_t + u u_x = 0 \]

一阶导信息

1
u_t -> u_x

一阶导还是显然的：

\[ u_x = -\frac{u_t}{u} \]

二阶导信息

1
u_tt -> u_xt -> u_xx

做一下准备，对原始方程进行如下的求导

\[ \begin{aligned} u_t + u u_x &= 0\\ u_{tt} + u_t u_x + u u_{xt} &=0, (\partial_t)\\ u_{tx} + u_x u_x + u u_{xx} &=0, (\partial_x) \end{aligned} \]

得到

\[ u_{xt} = -\frac{1}{u}\left(u_{tt} + u_t u_x\right) \]

还有

\[ u_{xx} = -\frac{1}{u}\left(u_{tx} + u_x u_x\right) \]

三阶导信息

1
u_ttt -> u_xtt -> u_xxt -> u_xxx

做一下准备，对原始方程进行如下的二阶求导

\[ \begin{aligned} u_t + u u_x &= 0\\ u_{ttt} + u_{tt} u_x + 2 u_t u_{xt} + u u_{xtt} &=0, (\partial_{tt}) \\ u_{xtt} + u_{t} u_{xx} + 2 u_x u_{xt} + u u_{xxt} &=0, (\partial_{tx}) \\ u_{xxt} + 3 u_x u_{xx} + u u_{xxx} &=0, (\partial_{xx}) \\ \end{aligned} \]

因此，可以得到

\[ u_{xtt} = -\frac{1}{u}\left( u_{ttt} + u_{tt} u_x + 2 u_t u_{xt} \right) \]

还有

\[ u_{xxt} = -\frac{1}{u}\left( u_{xtt} + u_t u_{xx} + 2 u_x u_{xt} \right) \]

以及

\[ u_{xxx} = -\frac{1}{u}\left( u_{xxt} + 3 u_x u_{xx} \right) \]

四阶导信息

1
u_tttt -> u_xttt -> u_xxtt -> u_xxxt -> u_xxxx

做一下准备，对原始方程进行如下的三阶求导

\[ \begin{aligned} u_t + u u_x &= 0\\ u^{(0,4)}(x,t)+u^{(0,3)}(x,t) u^{(1,0)}(x,t)+3 u^{(0,2)}(x,t) u^{(1,1)}(x,t)+3 u^{(0,1)}(x,t) u^{(1,2)}(x,t)+u(x,t) u^{(1,3)}(x,t) &=0, (\partial_{ttt}) \\ 2 u^{(1,1)}(x,t)^2+2 u^{(1,0)}(x,t) u^{(1,2)}(x,t)+u^{(1,3)}(x,t)+u^{(0,2)}(x,t) u^{(2,0)}(x,t)+2 u^{(0,1)}(x,t) u^{(2,1)}(x,t)+u(x,t) u^{(2,2)}(x,t) &=0, (\partial_{ttx}) \\ 3 u^{(1,1)}(x,t) u^{(2,0)}(x,t)+3 u^{(1,0)}(x,t) u^{(2,1)}(x,t)+u^{(2,2)}(x,t)+u^{(0,1)}(x,t) u^{(3,0)}(x,t)+u(x,t) u^{(3,1)}(x,t) &=0, (\partial_{txx}) \\ 3 u^{(2,0)}(x,t)^2+4 u^{(1,0)}(x,t) u^{(3,0)}(x,t)+u^{(3,1)}(x,t)+u(x,t) u^{(4,0)}(x,t) &=0, (\partial_{xxx}) \\ \end{aligned} \]

因此，可以得到

\[ u_{xttt} = -\frac{1}{u}\left( u_{tttt} + u_{ttt} u_x + 3 u_{tt} u_{xt} + 3 u_t u_{xtt} \right) \]

还有

\[ u_{xxtt} = -\frac{1}{u}\left( 2 u_{xt}^2 + 2u_{x} u_{xtt} + u_{xttt} + u_{tt} u_{xx} + 2 u_t u_{xxt} \right) \]

还有

\[ u_{xxxt} = -\frac{1}{u}\left( 3u_{xt} u_{xx} + 3 u_{x} u_{xxt} + u_{xxtt} + u_t u_{xxx} \right) \]

还有

\[ u_{xxxx} = -\frac{1}{u}\left( 3 u_{xx}^2 + 4 u_{x} u_{xxx} + u_{xxxt} \right) \]