#### 2.1.15 (Haberman 12.2.2) $$\omega _t -3 \omega _x = 0$$ and $$\omega (x,0)=\cos x$$

problem number 15

Problem 12.2.2 from Richard Haberman applied partial diﬀerential equations book, 5th edition

Solve for $$u(x,t)$$ $\omega _t -3 \omega _x = 0$ With initial conditions $$\omega (x,0)=\cos x$$.

See my HW 12, Math 322, UW Madison.

Mathematica

$\{\{w(x,t)\to \cos (3 t+x)\}\}$

Maple

$w \left (x , t\right ) = \cos \left (3 t +x \right )$

Hand solution

Solve \begin {equation} w_{t}-3w_{x}=0\tag {1} \end {equation} With I.C. $$w\left ( x,0\right ) =\cos x$$

Solution

Let $$w=w\left ( x\left ( t\right ) ,t\right )$$. 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} & =0\tag {3}\\ \frac {dx}{dt} & =-3\tag {4} \end {align}

Solving (3) gives\begin {align} w & =w\left ( x\left ( 0\right ) \right ) \nonumber \\ & =\cos \left ( x\left ( 0\right ) \right ) \tag {5} \end {align}

We need to ﬁnd $$x\left ( 0\right )$$. From (4)\begin {align*} x & =-3t+x\left ( 0\right ) \\ x\left ( 0\right ) & =x+3t \end {align*}

Hence (5) becomes

$w\left ( x\left ( t\right ) ,t\right ) =\cos \left ( x+3t\right )$

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