Added May 25, 2019.
From HW 3, UMN Math 5587, Fall 2016, problem 2.
Solve for \(u(x,t)\) with \(u(x,0)=e^x\) and \(u_t(x,0)=0\) by factoring the PDE into two transport PDE \begin {align*} u_{xx} - 3 u_{xt} - 4 u_{tt} = 0 \end {align*}
Mathematica ✗
ClearAll["Global`*"]; pde = D[u[x, t], {x,2}] - 3*D[D[u[x,t],x],t] - 4*D[u[x, t], {t,2}] == 0; ic = {u[x,0]==Exp[x], Derivative[0,1][u][x,0]==0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde,ic}, u[x, t], {x, t}], 60*10]];
Failed
Maple ✓
restart; pde := diff(u(x,t),x$2)-3*diff(diff(u(x,t),x),t) - 4 * diff(u(x,t),t$2)=0; ic := u(x,0)=exp(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'));
\[u \left ( x,t \right ) =1/5\,{{\rm e}^{-t+x}}+4/5\,{{\rm e}^{x+t/4}}\]
Hand solution
Solve \(u_{xx}-3u_{xt}-4u_{tt}=0\) with \(u\left ( x,0\right ) =e^{x}\) and \(u_{t}\left ( x,0\right ) =x\). Writing the PDE as \[ \left ( \frac {\partial }{\partial x}+\frac {\partial }{\partial t}\right ) \left ( \frac {\partial }{\partial x}-4\frac {\partial }{\partial t}\right ) u=0 \] 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}+\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 \(u\left ( x,0\right ) =e^{x}\) hence \(u_{x}\left ( x,0\right ) =e^{x}\), and \(u_{t}\left ( x,0\right ) =0\). Therefore the above gives \[ w\left ( x,0\right ) =e^{x}\] 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} & =1\\ \frac {dw}{ds} & =0 \end {align*}
With \(x\left ( 0\right ) =\xi ,t\left ( 0\right ) =0,w\left ( 0\right ) =e^{\xi }\). The above equations give\begin {align*} x & =s+x\left ( 0\right ) =s+\xi \\ t & =s\\ w & =u\left ( 0\right ) =e^{\xi } \end {align*}
Since \(\xi =x-s=x-t\), then the last equation above gives\[ w\left ( x,y\right ) =e^{x-t}\] Using the above into (1) gives\begin {equation} \frac {\partial u}{\partial x}-4\frac {\partial u}{\partial t}=e^{x-t}\tag {3} \end {equation} with \(u\left ( x,0\right ) =e^{x}\). The characteristics for (3) are\begin {align*} \frac {dx}{ds} & =1\\ \frac {dt}{ds} & =-4\\ \frac {du}{ds} & =e^{x-t} \end {align*}
With \(x\left ( 0\right ) =\xi ,t\left ( 0\right ) =0,u\left ( 0\right ) =e^{\xi }\). The above two 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 \(\frac {du}{ds}=e^{x-t}\) becomes \begin {align*} \frac {du}{ds} & =e^{s+\xi -\left ( -4s\right ) }\\ & =e^{5s+\xi } \end {align*}
Solving the above gives\begin {align*} u & =\frac {1}{5}e^{5s+\xi }+u\left ( 0\right ) \\ & =\frac {1}{5}e^{5s+\xi }+e^{\xi } \end {align*}
Converting to \(x,t\), using \(s=\frac {-t}{4}\) and \(\xi =x+\frac {t}{4}\) gives\begin {align*} u\left ( x,y\right ) & =\frac {1}{5}e^{5\left ( \frac {-t}{4}\right ) +\left ( x+\frac {t}{4}\right ) }+e^{\left ( x+\frac {t}{4}\right ) }\\ & =\frac {1}{5}e^{x-t}+e^{\left ( x+\frac {t}{4}\right ) } \end {align*}
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