 Solution

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.