Rayleigh quotient on circular region of radius 2 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 $D[\psi_0]$
 A: \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 $\lambda_1 = \frac {12}{11}$
A: 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?
