Added May 23, 2019.
From Math 5587 midterm I, Fall 2016, practice exam, problem 9.
Solve for \(u(x,t)\) with \(u(x,0)=0\) and \(u_t(x,0)=x\) by factoring the PDE into two transport PDE \begin {align*} u_{xx} - u_{xt} - 12 u_{tt} = 0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {x,2}] - D[D[u[x,t],x],t] - 12*D[u[x, t], {t,2}] == 0; ic = {u[x,0]==0, Derivative[0,1][u][x,0]==x}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde,ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to t x-\frac {t^2}{24}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),x$2)-diff(diff(u(x,t),x),t) - 12 * diff(u(x,t),t$2)=0; ic := u(x,0)=0, eval( diff(u(x,t),t),t=0)=x; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =-{\frac {{t}^{2}}{24}}+xt\]
Hand solution
Solve \(u_{xx}-u_{xt}-12u_{tt}=0\) with \(u\left ( x,0\right ) =x\) and \(u_{t}\left ( x,0\right ) =x\). Writing the PDE as \[ \left ( \frac {\partial }{\partial x}+3\frac {\partial }{\partial t}\right ) \left ( \frac {\partial }{\partial x}-4\frac {\partial }{\partial t}\right ) u=0 \] Let \begin {equation} \frac {\partial u}{\partial x}-4\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), \[ w\left ( x,0\right ) =u_{x}\left ( x,0\right ) -4u_{t}\left ( x,0\right ) \] But \(u\left ( x,0\right ) =0\) hence \(u_{x}\left ( x,0\right ) =0\), and \(u_{t}\left ( x,0\right ) =x\). Therefore the above gives \[ w\left ( x,0\right ) =-4x \] 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 \(x\left ( 0\right ) =\xi ,t\left ( 0\right ) =0,w\left ( 0\right ) =-4\xi \). The above equations give\begin {align*} x & =s+x\left ( 0\right ) =s+\xi \\ t & =3s\\ w & =u\left ( 0\right ) =-4\xi \end {align*}
Solving for \(\xi ,s\) from the first 2 equations. \(\xi =x-s=x-\frac {t}{3}\). From the last equation above\[ w\left ( x,y\right ) =-4\left ( x-\frac {t}{3}\right ) \] Using the above into (1) gives\begin {equation} \frac {\partial u}{\partial x}-4\frac {\partial u}{\partial t}=-4\left ( x-\frac {t}{3}\right ) \tag {3} \end {equation} with \(u\left ( x,0\right ) =0\). The characteristics for (3) are\begin {align*} \frac {dx}{ds} & =1\\ \frac {dt}{ds} & =-4\\ \frac {du}{ds} & =-4\left ( x-\frac {t}{3}\right ) \end {align*}
With \(x\left ( 0\right ) =\xi ,t\left ( 0\right ) =0,u\left ( 0\right ) =0\). The above equations give\begin {align*} x & =s+x\left ( 0\right ) =s+\xi \\ t & =-4s \end {align*}
Solving the above for \(s,\xi \) gives \(\xi =x-s=x+\frac {t}{4}\). Therefore \(\frac {du}{ds}=-4\left ( x-\frac {t}{3}\right ) \) becomes \begin {align*} \frac {du}{ds} & =-4\left ( s+\xi -\frac {\left ( -4s\right ) }{3}\right ) \\ & =-4\left ( s+\xi +\frac {4s}{3}\right ) \\ & =-\frac {28}{3}s-4\xi \end {align*}
Solving the above gives\begin {align*} u & =-\frac {28}{3}\frac {s^{2}}{2}-4\xi s+u\left ( 0\right ) \\ & =-\frac {28}{3}\frac {s^{2}}{2}-4\xi s \end {align*}
Converting to \(x,t\)\begin {align*} u\left ( x,y\right ) & =-\frac {28}{3}\frac {1}{2}\left ( -\frac {t}{4}\right ) ^{2}-4\left ( x+\frac {t}{4}\right ) \left ( -\frac {t}{4}\right ) \\ & =-\frac {1}{24}\left ( t^{2}-24xt\right ) \\ & =-\frac {1}{24}t^{2}+xt \end {align*}
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