91Ó°ÊÓ

Show that the wave equation can be considered as the following system of two coupled first-order partial differential equations: $$ \begin{gathered} \frac{\partial u}{\partial t}-c \frac{\partial u}{\partial x}=w \\ \frac{\partial w}{\partial t}+c \frac{\partial w}{\partial x}=0 \end{gathered} $$

Solve using the method of characteristics: $$ \begin{gathered} \frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}} \\ u(x, 0)=0 \quad u(0, t)=h(t) \\ \frac{\partial u}{\partial t}(x, 0)=0 \quad u(L, t)=0 . \end{gathered} $$

Consider the wave equation on a semi-infinite interval \(\frac{\partial^{2} u}{\partial t^{2}}-c^{2} \frac{\partial^{2} u}{\partial x^{2}} \quad\) for \(03 \end{array} \quad \frac{\partial u}{\partial t}(x, 0)=0\right. $$ Determine the solution. Sketch the solution for various times. (Assume that \(u\) is continuous at \(x=0, t=0\).)

Consider the three-dimensional wave equation $$ \frac{\partial^{2} u}{\partial t^{2}}=c^{2} \nabla^{2} u \text {. } $$ Assume that the solution is spherically symmetric, so that $$ \nabla^{2} u=\left(1 / \rho^{2}\right)(\partial / \partial \rho)\left(\rho^{2} \partial u / \partial \rho\right) $$ (a) Make the transformation \(u=(1 / \rho) w(\rho, t)\) and verify that $$ \frac{\partial^{2} w}{\partial t^{2}}=c^{2} \frac{\partial^{2} w}{\partial \rho^{2}} . $$ (b) Show that the most general spherically symmetric solution of the wave equation consists of the sum of two spherically symmetric waves, one moving outward at speed \(c\) and the other inward at speed \(c\). Note the decay of the amplitude.

Consider the traffic flow problem $$ \frac{\partial \rho}{\partial t}+c(\rho) \frac{\partial \rho}{\partial x}=0 . $$ Assume \(u(\rho)=u_{\max }\left(1-\rho / \rho_{\max }\right)\). Solve for \(\rho(x, t)\) if the initial conditions are (a) \(\rho(x, 0)=\rho_{\max }\) for \(x<0\) and \(\rho(x, 0)=0\) for \(x>0\). This corresponds to the traffic density that results after an infinite line of stopped traffic is started by a red light turning green. $$ \text { (b) } \rho(x, 0)=\left\\{\begin{array}{ll} \rho_{\max } & x<0 \\ \frac{\rho_{\max }}{2} & 0a \end{array} \quad \text { (c) } \quad \rho(x, 0)= \begin{cases}\frac{3 \rho_{\max }}{5} & x<0 \\ \frac{\rho_{\max }}{5} & x>0\end{cases}\right. $$

Solve subject to the initial condition \(\rho(x, 0)=f(x)\) : *(a) \(\frac{\partial \rho}{\partial t}+c \frac{\partial \rho}{\partial x}=e^{-3 x}\) (b) \(\frac{\partial \rho}{\partial t}+3 x \frac{\partial \rho}{\partial x}=4\) *(c) \(\frac{\partial \rho}{\partial t}+t \frac{\partial \rho}{\partial x}=5\) (d) \(\frac{\partial \rho}{\partial t}+5 t \frac{\partial \rho}{\partial x}=3 \rho\) *(e) \(\frac{\partial \rho}{\partial t}-t^{2} \frac{\partial \rho}{\partial x}=-\rho\) (f) \(\frac{\partial \rho}{\partial t}+t^{2} \frac{\partial \rho}{\partial x}=0\) *(g) \(\frac{\partial \rho}{\partial t}+x \frac{\partial \rho}{\partial x}=t\).

For the first-order "quasi-linear" partial differential equation $$ a \frac{\partial u}{\partial x}+b \frac{\partial u}{\partial y}=c \text {, } $$ wherc \(a, b\), and \(c\) are functions of \(x, y\) and \(u\), show that the method of characteristics (if necessary, see Section 11.6) yields $$ \frac{d x}{a}=\frac{d y}{b}=\frac{d u}{c} . $$