#### 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

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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