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Added Sept 12, 2019.
Taken from Peter Olver textbook, Introduction to Partial differential equations.
Solve for \(u(t,x)\) in \(u_t-3 u_x = 0\) with IC \(u(0,x)=e^{-x^2}\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[t, x], {t}] -3* D[u[t, x], {x}] == 0; ic = u[0,x]==Exp[-x^2]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde,ic}, u[t, x], {t, x}], 60*10]];
\[\left \{\left \{u(t,x)\to e^{-(3 t+x)^2}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(t, x), t) - 3*diff(u(t, x),x) =0; ic:=u(0,x)=exp(-x^2); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(t,x))),output='realtime'));
\[u \left (t , x\right ) = {\mathrm e}^{-\left (3 t +x \right )^{2}}\]
Hand solution
Solve \begin {equation} u_{t}-3u_{x}=0 \tag {1} \end {equation} With initial conditions \(u\left ( x,0\right ) =e^{-x^{2}}\)
Solution
Let \(u=u\left ( x\left ( t\right ) ,t\right ) \). Then \begin {equation} \frac {du}{dt}=\frac {\partial u}{\partial x}\frac {dx}{dt}+\frac {\partial u}{\partial t} \tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dt} & =0\tag {3}\\ \frac {dx}{dt} & =-3 \tag {4} \end {align}
Eq (3) gives \(u=u\left ( x\left ( 0\right ) \right ) \). Using the given initial conditions, this becomes \begin {equation} u=e^{-x\left ( 0\right ) ^{2}} \tag {5} \end {equation} Eq (4) is now used to find \(x\left ( 0\right ) \). Soving (4) gives \(x=x\left ( 0\right ) -3t\). Hence \(x\left ( 0\right ) =x-3t\). Therefore (5) becomes\[ u\left ( x,t\right ) =e^{-\left ( x-3t\right ) ^{2}}\] The following is an animation of the solution
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Source code used for the above
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