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2.13 HW 12

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2.13.1 Problem 12.2.1

Show that the wave equation can be considered as the following system of two coupled first-order PDE

$$\begin{align} \frac{\partial u}{\partial t}-c\frac{\partial u}{\partial x} & =w\tag{1}\\ \frac{\partial w}{\partial t}+c\frac{\partial w}{\partial x} & =0 \tag{2} \end{align}$$

Answer

The wave PDE in 1D is $\frac{\partial ^{2}u}{\partial t^{2}}-c^{2}\frac{\partial ^{2}u}{\partial x^{2}}=0$. Taking time derivative of equation (1) gives (assuming $c$ is constant)

$$\begin{equation} \frac{\partial ^{2}u}{\partial t^{2}}-c\frac{\partial ^{2}u}{\partial x\partial t}=\frac{\partial w}{\partial t} \tag{3} \end{equation}$$ Taking space derivative of equation (1) gives (assuming $c$ is constant) $$\begin{equation} \frac{\partial ^{2}u}{\partial t\partial x}-c\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial w}{\partial x} \tag{4} \end{equation}$$ Multiplying (4) by $c$ $$\begin{equation} c\frac{\partial ^{2}u}{\partial t\partial x}-c^{2}\frac{\partial ^{2}u}{\partial x^{2}}=c\frac{\partial w}{\partial x} \tag{5} \end{equation}$$ Adding (3)+(5) gives $$\begin{align*} \frac{\partial ^{2}u}{\partial t^{2}}-c\frac{\partial ^{2}u}{\partial x\partial t}+c\frac{\partial ^{2}u}{\partial t\partial x}-c^{2}\frac{\partial ^{2}u}{\partial x^{2}} & =\frac{\partial w}{\partial t}+c\frac{\partial w}{\partial x}\\ \frac{\partial ^{2}u}{\partial t^{2}}-c^{2}\frac{\partial ^{2}u}{\partial x^{2}} & =\frac{\partial w}{\partial t}+c\frac{\partial w}{\partial x} \end{align*}$$

But the RHS of the above is zero, since it is equation (2). Therefore the above reduces to

$$\frac{\partial ^{2}u}{\partial t^{2}}-c^{2}\frac{\partial ^{2}u}{\partial x^{2}}=0$$ Which is the wave PDE.

2.13.2 Problem 12.2.2

Solve

$$\begin{equation} \frac{\partial w}{\partial t}-3\frac{\partial w}{\partial x}=0 \tag{1} \end{equation}$$ with $w\left ( x,0\right ) =\cos x$

Answer

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 (2) and (1), we see that if we let $\frac{dx}{dt}=-3$ in the above, then we obtain (1). Hence we conclude that $\frac{dw}{dt}=0$. Therefore, $w\left ( x\left ( t\right ) ,t\right )$ is constant. At time $t=0$, we are given that $$\begin{equation} w\left ( x\left ( 0\right ) ,t\right ) =\cos x\left ( 0\right ) \qquad t=0\tag{3} \end{equation}$$ We just now need to determine $x\left ( 0\right )$. This is found from $\frac{dx}{dt}=-3$, which has the solution $x=x\left ( 0\right ) -3t\,.$ Hence $x\left ( 0\right ) =x+3t$. Therefore (3) becomes $$w\left ( x\left ( t\right ) ,t\right ) =\cos \left ( x+3t\right )$$

2.13.3 Problem 12.2.3

Solve

$$\begin{equation} \frac{\partial w}{\partial t}+4\frac{\partial w}{\partial x}=0 \tag{1} \end{equation}$$ with $w\left ( 0,t\right ) =\sin 3t$

Answer

Let

$$w\equiv w\left ( x,t\left ( x\right ) \right )$$ Hence $$\begin{equation} \frac{dw}{dx}=\frac{\partial w}{\partial x}+\frac{\partial w}{\partial t}\frac{dt}{dx} \tag{2} \end{equation}$$ Comparing (2) and (1), we see that if we let $\frac{dt}{dx}=\frac{1}{4}$ in (2), 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 $$\begin{equation} w\left ( x,t\left ( 0\right ) \right ) =\sin \left ( 3t\left ( 0\right ) \right ) \qquad x=0 \tag{3} \end{equation}$$ We just now need to determine $t\left ( 0\right )$. This is found from $\frac{dt}{dx}=\frac{1}{4}$, which has the solution $t\left ( x\right ) =t\left ( 0\right ) +\frac{1}{4}x\,.$ Hence $t\left ( 0\right ) =t\left ( x\right ) -\frac{1}{4}x$. Therefore (3) becomes $$\begin{align*} w\left ( x,t\left ( x\right ) \right ) & =\sin \left ( 3\left ( t\left ( x\right ) -\frac{1}{4}x\right ) \right ) \\ & =\sin \left ( 3t-\frac{3}{4}x\right ) \end{align*}$$

2.13.4 Problem 12.2.4

Solve

$$\begin{equation} \frac{\partial w}{\partial t}+c\frac{\partial w}{\partial x}=0\tag{1} \end{equation}$$ with $c>0$ and $$\begin{align*} w\left ( x,0\right ) & =f\left ( x\right ) \qquad x>0\\ w\left ( 0,t\right ) & =h\left ( t\right ) \qquad t>0 \end{align*}$$

Answer

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 (2) and (1), 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 $$w\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}f\left ( x-ct\right ) & & x>ct\\ h\left ( t-\frac{1}{c}x\right ) & & x<ct \end{array} \right .$$

2.13.5 Problem 12.2.5

2.13.5.1 Part (a)

Solve

$$\begin{equation} \frac{\partial w}{\partial t}+c\frac{\partial w}{\partial x}=e^{2x} \tag{1} \end{equation}$$ with $w\left ( x,0\right ) =f\left ( x\right )$

Answer 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 (2) and (1), we see that if we let $\frac{dx}{dt}=c$ in the above, then we obtain (1). Hence we conclude that $\frac{dw}{dt}=e^{2x}$. Hence $$w=w\left ( 0\right ) +te^{2x}$$ At $t=0$, $w\left ( 0\right ) =f\left ( x\left ( 0\right ) \right )$, hence $$\begin{equation} w=f\left ( x\left ( 0\right ) \right ) +te^{2x} \tag{3} \end{equation}$$ We just now need to determine $x\left ( 0\right )$. This is found from $\frac{dx}{dt}=c$, which has the solution $x=x\left ( 0\right ) +ct\,.$ Hence $x\left ( 0\right ) =x-ct$. Therefore (3) becomes $$w\left ( x\left ( t\right ) ,t\right ) =f\left ( x-ct\right ) +te^{2x}$$

2.13.5.2 Part (b)

Solve

$$\begin{equation} \frac{\partial w}{\partial t}+x\frac{\partial w}{\partial x}=1 \tag{1} \end{equation}$$ with $w\left ( x,0\right ) =f\left ( x\right )$

Answer 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 (2) and (1), we see that if we let $\frac{dx}{dt}=x$ in the above, then we obtain (1). Hence we conclude that $\frac{dw}{dt}=1$. Hence $$w=w\left ( 0\right ) +t$$ At $t=0$, $w\left ( 0\right ) =f\left ( x\left ( 0\right ) \right )$, hence the above becomes $$w=f\left ( x\left ( 0\right ) \right ) +t$$ We now need to find $x\left ( 0\right )$. From $\frac{dx}{dt}=x$, the solution is  $\ln \left \vert x\right \vert =t+x\left ( 0\right )$ or $x=x\left ( 0\right ) e^{t}$. Hence $x\left ( 0\right ) =xe^{-t}$ and the above becomes $$w=f\left ( xe^{-t}\right ) +t$$

2.13.5.3 Part (c)

Solve

$$\begin{equation} \frac{\partial w}{\partial t}+t\frac{\partial w}{\partial x}=1 \tag{1} \end{equation}$$ with $w\left ( x,0\right ) =f\left ( x\right )$

Answer 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 (2) and (1), we see that if we let $\frac{dx}{dt}=t$ in the above, then we obtain (1). Hence we conclude that $\frac{dw}{dt}=1$. Hence $$w=w\left ( 0\right ) +t$$ At $t=0$, $w\left ( 0\right ) =f\left ( x\left ( 0\right ) \right )$, hence the above becomes $$w=f\left ( x\left ( 0\right ) \right ) +t$$ We now need to find $x\left ( 0\right )$. From $\frac{dx}{dt}=t$, the solution is  $x=x\left ( 0\right ) +\frac{t^{2}}{2}$. Hence $x\left ( 0\right ) =x-\frac{t^{2}}{2}$ and the above becomes $$w=f\left ( x-\frac{t^{2}}{2}\right ) +t$$

2.13.5.4 Part (d)

Solve

$$\begin{equation} \frac{\partial w}{\partial t}+3t\frac{\partial w}{\partial x}=w \tag{1} \end{equation}$$ with $w\left ( x,0\right ) =f\left ( x\right )$

Answer 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 (2) and (1), we see that if we let $\frac{dx}{dt}=3t$ in the above, then we obtain (1). Hence we conclude that $\frac{dw}{dt}=w$. Hence $$\begin{align*} \ln \left \vert w\right \vert & =w\left ( 0\right ) +t\\ w & =w\left ( 0\right ) e^{t} \end{align*}$$

At $t=0$, $w\left ( 0\right ) =f\left ( x\left ( 0\right ) \right )$, hence the above becomes

$$w=f\left ( x\left ( 0\right ) \right ) e^{t}$$ We now need to find $x\left ( 0\right )$. From $\frac{dx}{dt}=3t$, the solution is  $x=x\left ( 0\right ) +\frac{3t^{2}}{2}$. Hence $x\left ( 0\right ) =x-\frac{3t^{2}}{2}$ and the above becomes $$w=f\left ( x-\frac{3t^{2}}{2}\right ) e^{t}$$