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


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

Maple


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

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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