Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 10

Self-adjoint form. Grouping the two derivative terms gives $[(1+ϵx)u^{\prime }{]}^{\prime }-ϵu=-\lambda u$. Thus a regular Sturm-Liouville operator isThe boundary conditions are mixed Neumann boundary conditions at zero and Dirichlet boundary conditions at $\pi$. For $ϵ\ge 0$, $p>0$, $q\ge 0$ and $w>0$, as required. The boundary term $[p(f{g}^{\prime }-f^{\prime }g){]}_0^{\pi }$ vanishes for two functions satisfying these conditions, confirming self-adjointness on that domain.

Constrained variation. Introduce a Lagrange multiplier $\Lambda$ for the normalization and varyIts first variation isThe endpoint term vanishes because $v^{\prime }(0)=0$ and $\delta v(\pi )=0$. The Euler-Lagrange equation is thereforewhich is exactly the original eigenvalue problem for the boxed choices of $p,q,w$. Multiply by $v$ and integrate by parts. The same endpoint conditions giveThus the stationary functional values are the eigenvalues, with $\Lambda =\lambda$. More generally, before normalization the relevant Rayleigh quotient is

Unperturbed ground mode. At $ϵ=0$, the equation is $-u^{\prime \prime }=\lambda u$. The quadratic form is nonnegative, and zero eigenvalue would force a constant function incompatible with the endpoint at $\pi$. For positive $\lambda$, $u=A\cos (\sqrt{\lambda }x)+B\sin (\sqrt{\lambda }x)$. The derivative condition forces $B=0$, while the other endpoint gives $\cos (\pi \sqrt{\lambda })=0$. HenceThe lowest eigenvalue and normalized eigenfunction are

Trial-function bound. The Rayleigh-Ritz method says that the minimum eigenvalue is no larger than the quotient of any admissible trial function. The same $U_0$ still satisfies both boundary conditions and retains unit norm. Its derivative is $U_0^{\prime }=-\sqrt{2/\pi }\sin (x/2)/2$, givingUsing ${\sin }^2(x/2)=(1-\cos x)/2$,Therefore the affine-parameter Sturm-Liouville eigenvalue bound isThis is an upper bound for every $ϵ\ge 0$, not merely a small-parameter approximation. As a lower-bound check, $p\ge 1$ and the unperturbed variational minimum imply ${\mathcal{R}}_{ϵ}[v]\ge 1/4+ϵ$ for every admissible normalized $v$. Thus