# Proving energy conservation for wave equation

Hi guys I have a midterm tommorow and I was doing this practice problem that I need help on.

So any hints or solutions would be appreciated. Thank you for your time

Problem

The head of timpani is constituted by some kind of elastic membrane which is stretched over a circular bowl; in a mathematical language, we are studying the displacement $u$ of the membrane as a function $u(x, y, t)$ deﬁned in $D = \{(x,y,t) \ |\ x^2 + y^2 \leq R, \ -\infty < t < \infty\}$, where $R$ is the radius of the bowl. The function u satisﬁes the two-dimensional Wave Equation

$u_{tt} = c^2(u_{xx} + u_{yy})$

with Dirichlet boundary conditions $u = 0$ on the boundary. Prove conservation of energy E(t), which is deﬁned as

$E(t) = \frac{1}{2} \iint\limits_D u_{t}^2 + c^2 (u_{x}^2 + u_{y}^2) dxdy$

[Hint: Proceed as we did for the one-dimensional case and use the divergence theorem to “integrate by parts” ] which I don't understand

Assume now that we take into account friction between air and the membrane; the displacement of the membrane now satisﬁes the damped Wave Equation, which reads:

$u_{tt} - c^2 (u_{xx} + u_{yy}) = -vu_{t}$

with the same boundary conditions. Show that, in this case, energy is always a strictly decreasing function except if $u(x, y, t) = 0$, for which it is constant equal to $0$. [Hint: What condition must $u_t$ satisfy in order for E to be non-decreasing? What PDE will u satisfy? Conclude using the maximum principle.

My attempt:

So first I found the derivative of E(t) and if the derivative of E(t) = 0 then I know the energy is conserved and I used integral by parts in 3 dimension to solve that

to get $\iint 2u_{t}u_{tt} = 0$ because it is decreasing. Since the derivative is 0 E is constant over time

therefore the energy is concerved.

But I do not understand the last part please any help or hints would be greatly appreciated.

Thank you

-
any help? this is really confusing – david ricardo Oct 3 '13 at 18:01

First part: multiply the equation by $u_t$ and integrate over the area of the membrane. The result is then integrated by parts (explained below)

$$0=\int [u_t u_{tt}-c^2 u_t (u_{xx}+u_{yy})]dx dy=\int [u_t u_{tt}+c^2 (u_{tx}u_{x}+u_{ty}u_{y})]dx dy\\ =\frac{1}{2}\frac{d}{dt}\int [u_{t}^2+c^2 (u_{x}^2+u_{y}^2)]dx dy$$

and the energy is conserved. The integration by parts is an application of the divergence theorem in 2D (or Green's theorem). I write it in vector notation:

$$\int \nabla\cdot(u_t\nabla u)dx dy = \int u_t \nabla u\cdot d\mathbf{n}$$

where $d\mathbf{n}$ is a line element directed outwards. The RHS vanishes on account of the boundary conditions.

Second part: repeat with the damped equation and instead find that now

$$\frac{dE}{dt}=-\nu\int u_t^2\, dx dy$$

which is negative.

-