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.
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\[ w\left ( x\left ( t\right ) ,t\right ) =\cos \left ( x+3t\right ) \]
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*}
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\[ w\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}f\left ( x-ct\right ) & & x>ct\\ h\left ( t-\frac{1}{c}x\right ) & & x<ct \end{array} \right . \]
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\[ w\left ( x\left ( t\right ) ,t\right ) =f\left ( x-ct\right ) +te^{2x}\]
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\[ w=f\left ( xe^{-t}\right ) +t \]
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\[ w=f\left ( x-\frac{t^{2}}{2}\right ) +t \]
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\[ w=f\left ( x-\frac{3t^{2}}{2}\right ) e^{t}\]