Past exam of the options II course of the University of Oxford / 2012 / G / G2 / Solution

Take a smooth variation $u+\epsilon v$ with $v=0$ on the boundary. Differentiating under the integral gives the first variationIntegration by parts removes the derivatives of $v$, and the boundary contribution vanishes. The fundamental lemma of the calculus of variations therefore givesFor a regular constraint $J[u]=c$, the Lagrange multiplier rule says that an extremal satisfies $\delta I=\lambda \delta J$ for a constant $\lambda$. Equivalently, apply the preceding Euler–Lagrange equation to $F-\lambda G$. Regularity means that $\delta J$ is not identically zero as a linear functional on the admissible variations. For the quadratic constraint below this holds at every nonzero admissible $u$; a zero constraint leaves only $u=0$.

In polar coordinates the constrained integrand isIts Euler–Lagrange equation is $-2\lambda ru-{\partial }_r(2r{u}_r)-{\partial }_{\theta }(2u_{\theta }/r)=0$, orThe field is single-valued and periodic in $\theta$. Substitution $u=f(r)\cos (n\theta )$ with $n\in \{0,1,2,\dots \}$ givesThe centre is part of the disk, so one must also select the regular radial solution: for $n\ge 1$, smoothness gives $f(r)=O(r^n)$ and the regular derivative behaviour; for $n=0$, $f$ is bounded and $f^{\prime }(0)=0$. A singular solution at zero does not define an admissible smooth field on the disk.

For fixed $n$, let $c_n={\int }_0^{2\pi }{\cos }^2(n\theta )d\theta$, so $c_0=2\pi$ and $c_n=\pi$ for $n\ge 1$. The angular derivative term has integral $n^2{c}_n$, including zero when $n=0$. ConsequentlyThe one-dimensional Euler–Lagrange equation for $\hat{I}-\lambda \hat{J}$ isexactly the same radial equation. The common factor $c_n$ cancels.

For two regular solutions with the same $n$ and different multipliers, multiply the equation for $f_1$ by $f_2$, multiply that for $f_2$ by $f_1$, subtract and integrate. The result isAt $a$, both functions vanish; at zero, regularity makes the boundary term tend to zero. Since the multipliers are distinct,This is weighted Sturm–Liouville orthogonality with weight $r$.

Finally, multiply one radial equation by $f$ and integrate by parts. The same endpoint conditions yieldMultiplication by $c_n$ provesFor any nonzero admissible solution the energy on the left before subtraction is positive, so its multiplier is the positive Rayleigh quotient $\lambda =\hat{I}/\hat{J}$.