Past exam of the short options course of the University of Oxford / 2017 / 6 / b

Part (a)(ii) now givesIf $\lambda =0$, continuity makes $y^{\prime }$ identically zero. Thus $y$ is constant, and the boundary condition forces $y=0$, contradicting ${\int }_0^{\pi }{y}^2=1$. Hence

The extremal equation is $y^{\prime \prime }+\lambda y=0$. For $\lambda >0$, its general solution is $A\cos (\sqrt{\lambda }x)+B\sin (\sqrt{\lambda }x)$. The left endpoint forces $A=0$, while the right endpoint and $B\ne 0$ require $\sin (\sqrt{\lambda }\pi )=0$. ThereforeIn particular, each extremal has the requested Fourier sine series form $y={\sum }_{k\ge 1}{a}_k\sin (kx)$. More strongly, for an individual extremal only the coefficient at $k=n$ can be nonzero. An arbitrary combination of distinct modes is an admissible function, but does not solve this one-eigenvalue equation.

The sine functions obey ${\int }_0^{\pi }\sin (nx)\sin (mx)dx=(\pi /2){\delta }_{nm}$. The Parseval identity for the Fourier sine series givesConsequently

For $y=\pm \sqrt{2/\pi }\sin x$, both the constraint and endpoint values hold, the extremal equation has $\lambda =1$, and direct integration gives $J[y]=1$.

For any admissible function, its derivative has Fourier cosine series coefficients $n{a}_n$ and zero constant coefficient. This follows by integration by parts: ${\int }_0^{\pi }{y}^{\prime }\cos (nx)dx=n{\int }_0^{\pi }y\sin (nx)dx$, and ${\int }_0^{\pi }{y}^{\prime }=0$. Thus the Parseval identity givesFor finite sums the same identity follows directly from the cosine product formula in the hint; the $L^2$ identity justifies the general case. Equality holds only if $a_n=0$ for all $n\ge 2$, and normalization then fixes $a_1=\pm \sqrt{2/\pi }$. ThereforeThis is the minimum Dirichlet energy at fixed L2 norm on an interval, equivalently the sharp Wirtinger inequality for this interval.