= Solution
The <Rayleigh-Ritz variational principle> bounds the lowest <eigenvalue> by the <Rayleigh quotient> of any nonzero admissible trial function. For $f=1-(r/a)^p$, take $p>0$, which ensures boundedness, the outer <Dirichlet boundary condition>, and finite radial Dirichlet energy. Direct <integration> gives
$$
\int_0^ar(f')^2dr=\frac p2,\qquad\int_0^arf^2dr=a^2\left[\frac12-\frac2{p+2}+\frac1{2p+2}\right]=\frac{a^2p^2}{2(p+1)(p+2)}.
$$
Therefore
$$
\lambda_1\leq\frac{(p+1)(p+2)}{a^2p}=\frac1{a^2}\left(p+3+\frac2p\right).
$$
The derivative is $1-2/p^2$, so the unique minimizing positive exponent is $p=\sqrt2$. This yields the <power-law variational bound for a circular membrane>
$$
\boxed{\lambda_1\leq\frac{3+2\sqrt2}{a^2},\qquad \nu_1\leq\frac{c(1+\sqrt2)}{2\pi a}\simeq0.384234\,\frac ca.}
$$
The optimizing function has zero derivative at the origin and is admissible for the energy principle despite its unbounded second derivative there. For comparison, the exact radial fundamental has $\sqrt{\lambda_1}a\simeq2.404826$, giving $\nu_1\simeq0.382740\,c/a$. The bound is about $0.39\%$ above this value.
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