Added May 23, 2019.
From Math 5587 midterm I, Fall 2016, practice exam, problem 8.
Solve for \(u(x,t)\) with \(u(x,0)=x\) and \(u_t(x,0)=0\) by factoring the PDE into two transport PDE \begin {align*} u_{xx} + u_{xt} - 6 u_{tt} = 0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {x,2}] + D[D[u[x,t],x],t] - 6*D[u[x, t], {t,2}] == 0; ic = {u[x,0]==x, Derivative[0,1][u][x,0]==0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde,ic}, u[x, t], {x, t}], 60*10]];
\[\{\{u(x,t)\to x\}\}\]
Maple ✓
restart; pde := diff(u(x,t),x$2)+diff(diff(u(x,t),x),t) - 6 * diff(u(x,t),t$2)=0; ic := u(x,0)=x, eval( diff(u(x,t),t),t=0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =x\]
Hand solution
Solve \(u_{xx}+u_{xt}-6u_{tt}=0\) with \(u\left ( x,0\right ) =x\) and \(u_{t}\left ( x,0\right ) =0\). Writing the PDE as \[ \left ( \frac {\partial }{\partial x}+3\frac {\partial }{\partial t}\right ) \left ( \frac {\partial }{\partial x}-2\frac {\partial }{\partial t}\right ) u=0 \] Let \begin {equation} \frac {\partial u}{\partial x}-2\frac {\partial u}{\partial t}=w\left ( x,t\right ) \tag {1} \end {equation} Then the PDE becomes\begin {equation} \frac {\partial w}{\partial x}+3\frac {\partial w}{\partial t}=0\tag {2} \end {equation} From (1), \[ w\left ( x,0\right ) =u_{x}\left ( x,0\right ) -2u_{t}\left ( x,0\right ) \] But \(u\left ( x,0\right ) =x\) hence \(u_{x}\left ( x,0\right ) =1\), and \(u_{t}\left ( x,0\right ) =0\). Therefore the above gives \[ w\left ( x,0\right ) =1 \] Hence we need to solve (2) for \(w\left ( x,t\right ) \) with the above initial condition. The characteristics for (2) are\begin {align*} \frac {dx}{ds} & =1\\ \frac {dt}{ds} & =3\\ \frac {dw}{ds} & =0 \end {align*}
With \(x\left ( 0\right ) =\xi ,t\left ( 0\right ) =0,w\left ( 0\right ) =1\). The above equations give\begin {align*} x & =s+x\left ( 0\right ) =s+\xi \\ t & =3s\\ w & =u\left ( 0\right ) =1 \end {align*}
Hence \[ w\left ( x,y\right ) =1 \] From (1)\begin {equation} \frac {\partial u}{\partial x}-2\frac {\partial u}{\partial t}=1\tag {3} \end {equation} with \(u\left ( x,0\right ) =x\). The characteristics for (3) are\begin {align*} \frac {dx}{ds} & =1\\ \frac {dt}{ds} & =-2\\ \frac {du}{ds} & =1 \end {align*}
With \(x\left ( 0\right ) =\xi ,t\left ( 0\right ) =0,u\left ( 0\right ) =\xi \). The above equations give\begin {align*} x & =s+x\left ( 0\right ) =s+\xi \\ t & =-2s\\ u & =s+u\left ( 0\right ) =s+\xi \end {align*}
But \(s=-\frac {t}{2}\) and \(\xi =x-s=x-\left ( -\frac {t}{2}\right ) =x+\frac {t}{2}\), hence \begin {align*} u\left ( x,t\right ) & =-\frac {t}{2}+x+\frac {t}{2}\\ & =x \end {align*}
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