An archived instance of discourse for discussion in undergraduate PDE.

Energy in a string

Mark

A string of length $\ell$, density $\rho$ is fixed at the ends to have tension $\tau$. It's motion is governed by the wave equation $\rho u_{t t}=\tau u_{x x}$ with boundary conditions $u(0,t)=u(\ell ,t)=0$. (Its initial configuration is not important for this problem.) We define it's energy to be
$$E=\int _0^{\ell }\left(\frac{1}{2}\rho u_t^2+\frac{1}{2}\tau u_x^2\right)dx.$$

Show that $E$ is constant.

rjensen

Conceptually, conservation of energy implies that energy is constant with respect to time, we can write that condition formally as $E(t) = C$, $\forall t > 0$. A more useful way to express this is, $\frac{d}{dt}E(t) = 0$. Using the definition of energy from above $E$, I will attempt to show $E$ is constant given a wave equation with the boundary conditions above. First we take the derivative with respect to time of both sides:
$$\frac{d}{dt}E=\frac{d}{dt}\int _0^{\ell }\left(\frac{1}{2}\rho u_t^2+\frac{1}{2}\tau u_x^2\right)dx. $$
Since derivatives and integrals are linear operations we can move $\frac{d}{dt}$ into the integrand like so,
$$=\int _0^{\ell }\left(\frac{1}{2}\rho \frac{\partial}{\partial t}u_t^2+\frac{1}{2}\tau \frac{\partial}{\partial t}u_x^2\right)dx. $$
Using the chain rule we compute $\frac{\partial}{\partial t}u_t^2 = 2u_{tt}u_{t}$ and $\frac{\partial}{\partial t}u_x^2 = 2u_{xt}u_{x}$. The $\frac{1}{2}$ and $2$ cancel so our problem now looks like this:
$$\frac{d}{dt}E=\int _0^{\ell }\left(\rho u_{tt}u_t+\tau u_{xt}u_x\right)dx. $$
At this point you may notice (after 2 hours of attempting this integral in any form you can think of) that we have some parts to apply a reverse product rule: $\frac{d}{dx}[f(x,t)g(x,t)]=f_x(x,t)g(x,t)+f(x,t)g_x(x,t)$. In our case the goal is to set it up so that, $f(x,t)=u_x$ and $g(x,t)=u_t$, since $\frac{d}{dx}u_t = u_{tx}$ and $\frac{d}{dx}u_x = u_{xx}$. The only thing missing from the integrand is the $u_{xx}$ term, however the defining wave equation states that $u_{tt} = \frac{\tau}{\rho}u_{xx}$ so we can now substitute for $u_{tt}$. Now we have:
$$=\int _0^{\ell }\left(\frac{\rho\tau}{\rho} u_{xx}u_t+\tau u_{xt}u_x\right)dx=\tau\int _0^{\ell }\left( \frac{d}{dx}[u_{x}u_{t}]\right)dx.$$
Now we can evaluate the integral easily and get:
$$\frac{d}{dt}E(t) = \tau[u_{x}u_{t}]\Big|_0^\ell=\tau[u_x(\ell,t)u_t(\ell,t) - u_x(0,t)u_t(0,t)].$$
The boundary conditions state that $u(0,t)=u(\ell,t)=0$, therefore $u_t(0,t)=u_t(\ell,t)=0$ as well. Using this last piece of information, we can now solve the equation:
$$\frac{d}{dt}E(t) =\tau[u_x(\ell,t)0 - u_x(0,t)0] = 0.$$
Therefore $E(t)$ does not change with respect to time which implies $E$ is constant $\blacksquare$











Mark

Pretty awesome! I do have a couple of small comments.

  • When $u$ is a function of several variables, we typically write $\displaystyle \frac{\partial u}{\partial t}$, rather than $\displaystyle \frac{du}{dt}$.

  • I guess you're using the reverse product rule (also known as integration by parts), rather than the reverse power rule.

I went ahead and edited your post to reflect these issues.