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Introduce a Lagrange multiplier for the normalization and vary $J[y]={\int }_0^1[p(y^{\prime }{)}^2+q{y}^2-\lambda w{y}^2]dx$. With variation $\eta$ satisfying $\eta (1)=0$, integration by parts givesThe upper endpoint term vanishes, and the lower term vanishes because $p(0)=0$ with the stipulated regular differentiable behavior. More generally the natural endpoint condition is $p{y}^{\prime }\to 0$ there. The fundamental lemma of the calculus of variations gives $(p{y}^{\prime }{)}^{\prime }-qy+\lambda wy=0$, the stated Sturm-Liouville problem.

The Rayleigh-Ritz variational principle estimates the lowest eigenvalue by minimizing $R[y]=\int [p(y^{\prime }{)}^2+q{y}^2]dx/\int w{y}^{2}dx$ over an admissible finite-dimensional trial family. Restricting the family can only raise the minimum.

For the specific radial problem, $p=w=x$ and $q=0$. The trial $y=A(1-x^2)$ has ${\int }_0^{1}x{y}^{2}dx=A^2/6$. Normalization gives $\boxed{A=\sqrt{6}}$, with the negative sign an equivalent choice. The numerator is ${\int }_0^{1}x(2Ax{)}^{2}dx=A^2$, soThis is the quadratic case of the power-law variational bound for a circular membrane. The exact first eigenvalue $5.7815$ is smaller because the polynomial trial is not the exact Bessel function of the first kind eigenfunction. Expanding a trial in the exact orthonormal eigenbasis expresses its Rayleigh quotient as a weighted average of eigenvalues, so no admissible trial can lie below the lowest one.

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 givesThereforeThe derivative is $1-2/p^2$, so the unique minimizing positive exponent is $p=\sqrt{2}$. This yields the power-law variational bound for a circular membraneThe 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 $\sqrt{{\lambda }_1}a\simeq 2.404826$, giving ${\nu }_1\simeq 0.382740c/a$. The bound is about $0.39\%$ above this value.