Added Nov 25, 2018.
Problem 12.2.5 (d) from Richard Haberman applied partial differential equations book, 5th edition
Solve for \(u(x,t)\) \[ \omega _t +3 t \omega _x = \omega (x,t) \]
with \(\omega (x,0)=f(x)\).
See my HW 12, Math 322, UW Madison.
Mathematica ✓
ClearAll["Global`*"]; pde = D[w[x, t], t] + 3*t*D[w[x, t], x] == w[x, t]; ic = w[x, 0] == f[x]; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, ic}, w[x, t], {x, t}]], 60*10]];
\begin {align*} & \left \{w(x,t)\to e^{-\sqrt {t^2}} f\left (x-\frac {3 t^2}{2}\right )\right \}\\& \left \{w(x,t)\to e^{\sqrt {t^2}} f\left (x-\frac {3 t^2}{2}\right )\right \}\\ \end {align*}
Maple ✓
restart; pde := diff(w(x,t),t)+3*t*diff(w(x,t),x)=w(x,t); ic:=w(x,0)=f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],w(x,t))),output='realtime'));
\[w \left ( x,t \right ) =f \left ( -{\frac {3\,{t}^{2}}{2}}+x \right ) {{\rm e}^{t}}\]
Hand solution
Solve \begin {equation} \frac {\partial w}{\partial t}+3t\frac {\partial w}{\partial x}=w\left ( x,t\right ) \tag {1} \end {equation}
With initial conditions \(w\left ( x,0\right ) =f\left ( x\right ) \)
Solution
Let \(w\equiv w\left ( x\left ( t\right ) ,t\right ) \) then \begin {equation} \frac {dw}{dt}=\frac {\partial w}{\partial x}\frac {dx}{dt}+\frac {\partial w}{\partial t}\tag {2} \end {equation}
Comparing (1,2) shows that
\begin {align} \frac {dw}{dt} & =w\tag {3}\\ \frac {dx}{dt} & =3t\tag {4} \end {align}
Solving (3) gives
\begin {equation} w=Ce^{t}\nonumber \end {equation}
From initial conditions at \(t=0\), the above becomes \(f\left ( x\left ( 0\right ) \right ) =C\). Hence the above becomes
\begin {equation} w\left ( x,t\right ) =f\left ( x\left ( 0\right ) \right ) e^{t}\tag {5} \end {equation}
From (4)
\begin {align*} x & =\frac {3}{2}t^{2}+x\left ( 0\right ) \\ x\left ( 0\right ) & =x-\frac {3}{2}t^{2} \end {align*}
Substituting the above in (5) gives
\[ w\left ( x\left ( t\right ) ,t\right ) =f\left ( x-\frac {3}{2}t^{2}\right ) e^{t}\]
Alternative solution
Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin {align} \frac {dt}{ds} & =1\tag {1}\\ \frac {dx}{ds} & =3t\tag {2}\\ \frac {dw}{ds} & =w\tag {3} \end {align}
With initial conditions at \(s=0\)\[ t\left ( 0\right ) =t_{1},x\left ( 0\right ) =t_{2},w\left ( 0\right ) =t_{3}\] And \(w\left ( x,0\right ) =f\left ( x\right ) \) becomes \begin {equation} t_{3}=f\left ( t_{2}\right ) ,t_{1}=0\tag {4} \end {equation} Equation (1) gives\begin {align} t & =s+t_{1}\nonumber \\ & =s\tag {5} \end {align}
Equation (2) gives, after replacing \(t\) by \(s\) from (5)\begin {align} \frac {dx}{ds} & =3s\nonumber \\ x & =\frac {3}{2}s^{2}+t_{2}\tag {6} \end {align}
Solving for \(t_{2}\) gives\begin {equation} t_{2}=x-\frac {3}{2}s^{2}\tag {7} \end {equation} Equation (3) gives\begin {align*} \ln w & =s+t_{3}\\ w & =t_{3}e^{s}\\ & =f\left ( t_{2}\right ) e^{s} \end {align*}
Using (7,5) in the above gives the solution\[ w\left ( x,t\right ) =f\left ( x-\frac {3}{2}t^{2}\right ) e^{t}\]
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