### Author Topic: HA2 problem 4  (Read 2202 times)

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### HA2 problem 4
« on: January 27, 2015, 09:36:46 PM »
For a solution $u(x, t)$ of the wave equation $u_{tt}=c^2u_{xx}$, the energy density is defined as $e=\frac{1}{2}\bigl(u_t^2+c^2 u_x^2\bigr)$ and the momentum density as $p =c u_t u_x$.

a.  Show that

\frac{\partial e}{\partial t} = c\frac{\partial p}{\partial x} \qquad \text{and} \qquad \frac{\partial p}{\partial t} = c\frac{\partial e}{\partial x}. \label{eq-HA2.11}

b.  Show that both $e(x, t)$ and $p(x,t)$ also satisfy the same wave equation.

#### Yiyun Liu

• Full Member
• Posts: 15
• Karma: 0
##### Re: HA2 problem 4
« Reply #1 on: January 29, 2015, 09:01:43 PM »
$\begin{array}{l} part(a):\\ \\ \frac{{\partial e}}{{\partial t}} = {u_t}{u_{tt}} + {c^2}{u_x}{u_{xt}}\\ \frac{{\partial \rho }}{{\partial x}} = c{u_{tx}}ux + c{u_t}{u_{xx}},since\quad {u_{tt}} = {c^2}{u_{xx}}\\ \Rightarrow \frac{{\partial e}}{{\partial t}} = {c^2}{u_t}{u_{xx}} + {c^2}{u_x}{u_{xt}}\quad (1)\\ c\frac{{\partial \rho }}{{\partial x}} = {c^2}{u_{tx}}ux + {c^2}{u_t}{u_{xx}} = \frac{{\partial e}}{{\partial t}} = (1)\\ \frac{{\partial \rho }}{{\partial t}} = c{u_{tt}}{u_x} + c{u_t}{u_{xt}}\quad (2)\\ \frac{{\partial e}}{{\partial x}} = {u_t}{u_{xt}} + {c^2}{u_x}{u_{xx}}\quad \\ \Rightarrow \frac{{\partial e}}{{\partial x}} = {u_t}{u_{xt}} + {u_x}{u_{tt}}\\ c\frac{{\partial e}}{{\partial x}} = c{u_t}{u_{xt}} + c{u_x}{u_{tt}} = \frac{{\partial \rho }}{{\partial t}} = (2)\\ \\ part(b):\\ \\ from\quad (a)\quad known\quad that\quad c\frac{{\partial \rho }}{{\partial x}} = \frac{{\partial e}}{{\partial t}},\quad c\frac{{\partial e}}{{\partial x}} = \frac{{\partial \rho }}{{\partial t}}\\ thus,\quad c\frac{{{\partial ^2}\rho }}{{\partial xt}} = \frac{{{\partial ^2}e}}{{\partial {t^2}}}\quad (1),\quad \frac{{{\partial ^2}\rho }}{{\partial {t^2}}} = c\frac{{{\partial ^2}e}}{{\partial xt}}\quad (2)\\ likewise,\quad \frac{{{\partial ^2}e}}{{\partial xt}} = c\frac{{{\partial ^2}\rho }}{{\partial {x^2}}}\quad (3),\quad \frac{{{\partial ^2}\rho }}{{\partial xt}} = c\frac{{{\partial ^2}e}}{{\partial {x^2}}}(4)\\ from\quad (1),(4):\quad {e_{tt}} = {c^2}{e_{xx}}\\ from\quad (2)(3):\quad {\rho _{tt}} = {c^2}{\rho _{xx}}\\ As\quad shown,\quad both\quad e(x,t)\quad and\quad \rho (x,t)\quad from\quad the\quad same\quad wave\quad equation. \end{array}$
« Last Edit: January 29, 2015, 09:04:32 PM by Yiyun Liu »