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

For separation of variables, write $B=F(r)\Theta (\theta )$. The exterior Laplace equation separates intoWith $\mu =\cos \theta$, angular regularity at the poles selects $\lambda =\ell (\ell +1)$ and the Legendre polynomial $P_{\ell }(\mu )$. The radial equation is $(r^2{F}^{\prime }{)}^{\prime }-\ell (\ell +1)F=0$, an Euler-Cauchy equation with solutions $r^{\ell },r^{-\ell -1}$. Decay at infinity excludes the first branch for every $\ell$, including its constant $\ell =0$ case. ThereforeThe expansion uses axisymmetric spherical harmonics. The written scalar boundary-value problem permits the $\ell =0$ term; any additional physical exclusion of that sector must be stated separately.

Separate the interior Helmholtz equation in the same angular basis. Each radial coefficient $F_{\ell }(r)$ satisfiesFor $s>0$, set $x=\sqrt{s}r$. The equation becomes $x^2{f}_{\ell }^{\prime \prime }+2x{f}_{\ell }^{\prime }+[x^2-\ell (\ell +1)]f_{\ell }=0$. Its regular solution is a Spherical Bessel function $j_{\ell }(x)$, up to a multiplicative constant. The second independent solution is singular as $r^{-\ell -1}$ at the origin and is excluded. HenceOne may denote $j_{\ell }$ by $f_{\ell }$ as in the source. Its leading behaviour $j_{\ell }(x)\sim x^{\ell }/(2\ell +1)!!$ ensures regularity. Primes on $j_{\ell }$ below refer to its dimensionless argument; differentiating with respect to $r$ introduces the factor $\sqrt{s}$.

Put $x=\sqrt{s}R>0$. Orthogonality of the Legendre polynomials lets us match each degree separately. Continuity of the field and its radial derivative givesEliminate $A_{\ell }$ by multiplying the first equation by $(\ell +1)/R$ and adding it to the second. A nonzero mode therefore requiresThis argument avoids dividing by a potentially zero boundary value. Apply the supplied recurrence with $\nu =\ell -1$:Consequently spherical Helmholtz matching to a decaying harmonic exterior giveswhere $j_{-1}(x)=\cos x/x$ for the $\ell =0$ sector.

The original PDF has an index error here: it prints $f_{\ell }$ rather than $f_{\ell -1}$ in the resulting eigenvalue condition. This cannot be repaired by silently choosing another normalization. If $j_{\ell }(x)$ itself were zero at a positive matching root, the boundary equation would also give $j_{\ell }^{\prime }(x)=0$, contradicting uniqueness for this nonzero solution of a regular differential equation at that point.

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.