#### 2.1.16 (Haberman 12.2.4) $$\omega _t +c \omega _x = 0$$ and $$\omega (x,0)=f(x)$$ and $$\omega (0,t)=h(t)$$

problem number 16

Problem 12.2.4 from Richard Haberman applied partial diﬀerential equations book, 5th edition

Solve for $$u(x,t)$$ $\omega _t +c \omega _x = 0$ With $$c>0$$. For $$x>0,t>0$$ if $$\omega (x,0)=f(x)$$ and $$\omega (0,t)=h(t)$$.

See my HW 12, Math 322, UW Madison.

Mathematica

Failed

Maple

$w \left (x , t\right ) = \frac {c h \left (t -\frac {x}{c}\right ) \theta \left (t -\frac {x}{c}\right )+\mathcal {L}^{-1}\left (\left (\int {\mathrm e}^{\frac {s x}{c}} f (x )d x \right ) {\mathrm e}^{-\frac {s x}{c}}, s , t\right )-\mathcal {L}^{-1}\left (\left (\int _{}^{0}{\mathrm e}^{\frac {\mathit {\_a} s}{c}} f \left (\mathit {\_a} \right )d\mathit {\_a} \right ) {\mathrm e}^{-\frac {s x}{c}}, s , t\right )}{c}$ Solution contains unresolved invlaplace calls

Hand solution

\begin {equation} \frac {\partial w}{\partial t}+c\frac {\partial w}{\partial x}=0\tag {1} \end {equation} Let $w\equiv w\left ( x\left ( t\right ) ,t\right )$ Hence \begin {equation} \frac {dw}{dt}=\frac {\partial w}{\partial t}+\frac {\partial w}{\partial x}\frac {dx}{dt}\tag {2} \end {equation} Comparing given (1) and (2), we see that if we let $$\frac {dx}{dt}=c$$ in (2), then we obtain (1). Hence we conclude that $$\frac {dw}{dt}=0$$. Therefore, $$w\left ( x\left ( t\right ) ,t\right )$$ is constant. At $$t=0$$, we are given that \begin {equation} w\left ( x\left ( t\right ) ,t\right ) =f\left ( x\left ( 0\right ) \right ) \qquad t=0\tag {3} \end {equation} We just now need to determine $$x\left ( 0\right )$$. This is found from $$\frac {dx}{dt}=c$$, which has the solution $$x\left ( t\right ) =x\left ( 0\right ) +ct\,.$$ Hence $$x\left ( 0\right ) =x\left ( t\right ) -ct$$. Therefore (3) becomes$w\left ( x,t\right ) =f\left ( x-ct\right )$ This is valid for $$x>ct.$$ We now start all over again, and look at Let $w\equiv w\left ( x,t\left ( x\right ) \right )$ Hence \begin {equation} \frac {dw}{dx}=\frac {\partial w}{\partial x}+\frac {\partial w}{\partial t}\frac {dt}{dx}\tag {4} \end {equation} Comparing (4) and (1), we see that if we let $$\frac {dt}{dx}=\frac {1}{c}$$ in (4), then we obtain (1). Hence we conclude that $$\frac {dw}{dx}=0$$. Therefore, $$w\left ( x,t\left ( x\right ) \right )$$ is constant. At $$x=0$$, we are given that \begin {equation} w\left ( x,t\left ( x\right ) \right ) =h\left ( t\left ( 0\right ) \right ) \qquad x=0\tag {5} \end {equation} We just now need to determine $$t\left ( 0\right )$$. This is found from $$\frac {dt}{dx}=\frac {1}{c}$$, which has the solution $$t\left ( x\right ) =t\left ( 0\right ) +\frac {1}{c}x\,.$$ Hence $$t\left ( 0\right ) =t\left ( x\right ) -\frac {1}{c}x$$. Therefore (5) becomes$w\left ( x,t\right ) =h\left ( t-\frac {1}{c}x\right )$ Valid for $$t>\frac {x}{c}$$ or $$x<ct$$. Therefore, the solution is$w\left ( x,t\right ) =\left \{ \begin {array} [c]{ccc}f\left ( x-ct\right ) & & x>ct\\ h\left ( t-\frac {1}{c}x\right ) & & x<ct \end {array} \right .$

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