Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 9

Variational formulation. Take real $p,q,w$ with $p,w>0$ in the interval, and impose homogeneous Dirichlet boundary conditions, or other suitable conditions that cancel the boundary terms. The functionalhas Euler-Lagrange equationwhich is equivalent to $-(p{y}^{\prime }{)}^{\prime }+qy-\lambda wy=0$. The same construction is constrained stationarity of the energy $\int [p(y^{\prime }{)}^2+q{y}^2]$ subject to $\int w{y}^2=1$, with $\lambda$ a Lagrange multiplier. For fixed zero endpoint values, the variation vanishes at both ends; homogeneous zero-flux conditions also cancel the boundary term $[2p{y}^{\prime }\delta y{]}_a^b$.

Multiplying the differential equation by $y$ and using integration by parts gives the Rayleigh quotientFor a nonzero admissible trial function, $R[y]\ge {\lambda }_0$ by the Rayleigh-Ritz variational principle. The Rayleigh-Ritz method chooses trial functions satisfying the endpoint conditions, often with adjustable coefficients, and minimizes this quotient over that trial family. Expanding $y=\sum c_n{y}_n$ in a normalized eigenbasis shows why the lower bound holds: $R[y]=\sum {\lambda }_n{c}_n^2/\sum c_n^2$, a weighted average of eigenvalues no smaller than the least one.

Radial quadratic trial. The radial problem becomeswhere the printed endpoint notation $y$ refers to $\psi$. Thus $p=w=r^2$ and $q=0$. At the origin the vanishing $r^2$ and regularity of $\psi$ make the boundary term vanish; the weight must be retained in both numerator and denominator.

For a quadratic $\psi =A+Br+C{r}^2$, the derivative condition sets $B=0$ and the outer condition gives $C=-A$. Overall amplitude cancels in the Rayleigh quotient, so take $\tilde{\psi }=1-r^2$. ThenwhileConsequently

Why the estimate is higher. For a regular radial function put $u=r\psi$. Then $u(0)=u(1)=0$, and direct differentiation transforms the equation to $u^{\prime \prime }+\lambda u=0$. Its first mode is $u=\sin (\pi r)$, corresponding to $\psi =\sin (\pi r)/r$ with its continuous value at the origin. Thus ${\lambda }_0={\pi }^2$. The quadratic trial is not proportional to that Spherical Bessel function, so it has nonzero components in higher modes. Its Rayleigh quotient is therefore strictly larger than ${\pi }^2$, in agreement with $21/2>{\pi }^2$.