Rayleigh Quotient for Vibrating Drumhead

Let $\nabla^2u=\nabla\cdot \nabla u$ be the 2D Laplacian of a function $u$. The coordinate system is unimportant for our analysis. Our aim is to derive a Rayleigh quotient for the eigenvalue problem, \[ \begin{align} \nabla^2 u + \lambda u =0\ &\text{in} \ \Omega\\ u|_C=0.& \end{align} \] Here $C$ is the closed curve shown below and $\Omega$ is region enclosed by $C$. $u|_C=0$ just means that the function $u$ is $0$ on the curve $C$.

We begin with Green's Theorem in "flux" or "divergence" form: \[ \oint_C \mathbf F\cdot \mathbf n ds = \int\!\int_\Omega \nabla\cdot \mathbf F dA, \] where $\mathbf F$ is a vector field that is continuously differentiable in $\Omega$. Let $u$ be the solution to the eigenvalue problem and take $\mathbf F = u\nabla u$. With this choice for $\mathbf F$, Green's Theorem implies that \[ \oint_C u \nabla u\cdot \mathbf n ds = \int\!\int_\Omega \nabla\cdot (u\nabla u) dA. \] Two things. First, since $u|_C=0$, we have $\oint_C u_C\nabla u \cdot \mathbf n ds = \oint_C 0 \nabla u\cdot \mathbf n ds = 0 = \int\!\int_\Omega \nabla\cdot (u\nabla u)dA$. Second, $\nabla\cdot (u\nabla u) = \nabla u \cdot \nabla u +u\nabla^2 u$. Putting these two together gives us \[ 0=\int\!\int_\Omega \nabla\cdot (u\nabla u)dA = \int\!\int_\Omega u\nabla^2 u dA + \int\!\int_\Omega |\nabla u|^2dA. \] Consequently, we have \[ (\ast) \quad \int\!\int_\Omega u\nabla^2 u dA = - \int\!\int_\Omega |\nabla u|^2dA. \]

To get the Rayleigh quotient, follow these steps.

  1. Multiply both sides of $\nabla^2 u + \lambda u =0$ by $u$ and then integrate. \[ \int \! \int_\Omega u\nabla^2 u dA+ \lambda \int \! \int_\Omega u^2dA =0 \]
  2. Use $(*)$ to replace the integral on the left. \[ - \int\!\int_\Omega |\nabla u|^2dA + \lambda \int \! \int_\Omega u^2dA =0 \]
  3. Solve for $\lambda$ to get the Rayleigh quotient. \[ \lambda = \frac{\int\!\int_\Omega |\nabla u|^2dA}{\int \! \int_\Omega u^2dA} \]
Updated 4/22/2013. (fjn)