Rayleigh quotient on circular region of radius 2

Question

I' m struggling with the following problem. We have the eigenvalue problem: $$u'' + \lambda u = 0$$ with associated boundary condition: $$u' + 3u = 0$$ Now by using the Rayleigh quotient for $0 \leq r \leq 2$and the trial function: $$\psi_0(r)= \alpha - r $$ where $\alpha$ is chosen so that $\psi_0(r)$ satisfies the boundary condition at $r=2$. Find the lowest eigenvalue $\lambda_1$

Own work: I started finding the correct $\alpha$ that suites the BC, hence: $$ -1 + 3 (\alpha -2) = 0$$ Hence $\alpha = \frac{7}{3}$

An upper bound to $\lambda_1$ is given by the Rayleigh quotient, $$J[\psi_0] = \frac {D[\psi_0]}{H[\psi_0]}$$ where

$$D[\psi_0] = \int_{\Omega}|\bigtriangledown \psi_0|^2 dx + \int_{\partial \Omega} \psi_{0}(2)^2 dx \quad H[\psi_0] = \int_\Omega \psi_0^2 dx$$ (in polar form) $$\bigtriangledown = \frac{\partial}{\partial r} + \frac {1}{r} \frac{\partial}{\partial \theta}$$ Thus $|\bigtriangledown \psi_0| = |\psi'_0(r)|=-1 $ and $$\int_\Omega f(r)dx = \int_0^2 2\pi r f(r)dr$$ so that:$$J[\psi_0] = \frac {\int_0^2 (r + \frac{1}{9}r)dr}{\int_0^2 r(\alpha -r)^2dr}$$

(Already cancelled the $2\pi$ out and left in $\alpha$)

These integrals can be evaluated as: $$\int_0^2 \frac{10}{9}r dr = \big[\frac {10}{18} r^2 \big]_{0}^{2} = \frac {20}{9}$$ and $$\int_0^2 r(\alpha -r)^2dr = \big[\frac{1}{2} r^2 \alpha^2 - \frac{2}{3} \alpha r^3+ \frac{1}{4} r^4 \big]_{0}^{2} = 2\alpha^2 - \frac{16 \alpha}{3}+4 = \frac {22}{9}$$

Which would give $$J[\psi_0]= \frac{10}{11}$$

Which is not correct $\big(\lambda_1 \leq \frac {12}{11} \big)$

Perhaps someone can help me, I think I go wrong by defining $D[\psi_0]$

Answer 1

\begin{equation} J(\psi_0) \geq \lambda_1 \end{equation} Where in this case $\psi_0$ is the trialfunction.

In order to understand the Rayleigh-quotient, now assume the same eigenvalue problem as the example above, but now by using the Rayleigh quotient method.

$$ \Delta u + \lambda u = 0$$ with associated boundary condition: $$\frac {\partial u}{\partial r} + 3u = 0$$ Now by using the Rayleigh quotient for $0 \leq r \leq 2$ and the trial function:

\begin{equation} \label{eq:trial} \psi_0(r)= \alpha - r \end{equation}

where $\alpha$ is chosen so that $\psi_0(r)$ satisfies the boundary condition at $r=2$. to find the lowest eigenvalue $\lambda_1$

The correct $\alpha$ that suits the boundary conditions, $$ -1 + 3 (\alpha -2) = 0$$ Hence $\alpha = \frac{7}{3}$

An upper bound to $\lambda_1$ is given by the Rayleigh quotient \cite{courantvol22008methods},

\begin{equation} J[\psi_0] = \frac {\mathfrak{D} [\psi_0]}{H[\psi_0]} \end{equation}

where

\begin{equation} \label{eq:quadratic} \mathfrak {D} [\psi_0] = \int_{\Omega}|\bigtriangledown \psi_0|^2 dx + \int_{\partial \Omega} \psi_0(2)^2 ds \quad H[\psi_0] = \int_\Omega \psi_0^2 dx \end{equation}

In the next chapter, a more detailled discussion will be given concerning these quadratic functionals.

(in polar form) $$\bigtriangledown \psi =\underline e_r \frac{\partial \psi}{\partial r} + \frac {1}{r} \underline e_{\theta} \frac{\partial \psi}{\partial \theta}$$ Where:

$$\underline e_r = \left( \begin{array}{c} \cos \theta \\ \sin \theta \end{array} \right), \quad \underline e_\theta = \left( \begin{array}{c}- \sin \theta \\ \cos \theta \end{array} \right) $$

Thus $|\bigtriangledown \psi_0| = |\psi'_0(r)|=-1 $ and $$\int_\Omega f(r)dx = \int_0^2 2\pi r f(r)dr$$ so that:$$J[\psi_0] = \frac {2 \pi \int_0^2 (r dr) +3 (\alpha - 2)^2 \int_{\partial \Omega} ds }{2 \pi \int_0^2 r(\alpha -r)^2dr}$$

(For readability, the $\alpha$ is left in.)

which leads to:

$$J[\psi_0] = \frac {2 \pi \int_0^2 (r dr) +3 (\alpha - 2)^2 \ 4\pi }{2 \pi \int_0^2 r(\alpha -r)^2dr}$$

$$J[\psi_0] = \frac {2 \pi \int_0^2 (r dr) + \frac{4\pi}{3} }{2 \pi \int_0^2 r(\alpha -r)^2dr}$$

These integrals can be evaluated as: $$2 \pi \int_0^2 (r dr)= \frac {36 \pi}{9}$$

and $$2 \pi\int_0^2 r(\alpha -r)^2dr = 2 \pi \big[\frac{1}{2} r^2 \alpha^2 - \frac{2}{3} \alpha r^3+ \frac{1}{4} r^4 \big]_{0}^{2} = 2\alpha^2 - \frac{16 \alpha}{3}+4 = \frac {44 \pi}{9}$$

Which would give $$J[\psi_0]= \frac{\frac {36}{9} + \frac {12}{9}} {\frac{44}{9}} = \frac{12}{11}$$ Hence an upper bound for $\lambda_1 = \frac {12}{11}$

Answer 2

First, the gradient in polar coordinates is $\frac{\partial}{\partial r} + \frac{1}{r}\frac{\partial}{\partial \theta}$, see for example here, but that doesn't change anything in your answer. Secondly you have a computation error after plugging in the value of $\alpha$, you should get $\frac{22}{9}$ for the integral. Integrating a positive function and getting a negative answer should raise some warning flags.

This gives a bound of $\frac{9}{11}$. Where does the $\frac{12}{11}$ you mention at the end come from?