2.1.18 (Haberman 12.2.5 (d)) \(\omega _t +3 t \omega _x = \omega (x,t)\) and \(\omega (x,0)=f(x)\)

problem number 18

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'));
 

\[w \left (x , t\right ) = f \left (-\frac {3 t^{2}}{2}+x \right ) {\mathrm e}^{t}\]

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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