Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 2 / iv

A positive matching root $x_{\ast }$ gives $s=x_{\ast }^2/R^2$. Substituting the root value requested by the question gives the intended expressionAfter the index correction in part (iii), this is indeed the lowest mode in the $\ell =1$ sector, because its condition is $j_0(x)=\sin x/x=0$ and its first positive root is $\pi$. It is also the minimum over $\ell \ge 1$.

However, the literal scalar problem includes $\ell =0$, for which matching gives $\cos x=0$. Its first positive root is $\pi /2$, so the actual minimum for the written scalar equations and stated boundary conditions isAn explicit counterexample to the larger proposed minimum is $B_{\mathrm{in}}=j_0(\pi r/(2R))$, $B_{\mathrm{out}}=2R/(\pi r)$, with no angular dependence. At $R$, both values are $2/\pi$ and both radial derivatives are $-2/(\pi R)$.

To verify minimality, put $u=r{F}_{\ell }$. The radial problem is $-u^{\prime \prime }+\ell (\ell +1)u/r^2=su$, $u(0)=0$, $u^{\prime }(R)+\ell u(R)/R=0$. Its Rayleigh quotient isFor $\ell =0$ this is the Dirichlet-Neumann interval problem, whose lowest eigenvalue is ${\pi }^2/(4R^2)$. The additional nonnegative terms increase with $\ell$, proving the scalar minimum and the minimum-at-$\ell =1$ statement for the restricted family. A monopole-free magnetic multipole interpretation would exclude $\ell =0$ and yield the intended value, but that extra restriction is not supplied by the stated scalar model. The claim about the first zero is not valid for every $f_{\ell }$ as printed.