Spherical Helmholtz matching to a decaying harmonic exterior ($x{j}_{\ell }^{\prime }(x)+(\ell +1)j_{\ell }(x)=x{j}_{\ell -1}(x)=0$)

A scalar field solving the Helmholtz equation inside a sphere of radius $R$ and Laplace's equation outside has a regular interior mode $b{j}_{\ell }(\sqrt{s}r)P_{\ell }(\cos \theta )$ and a decaying exterior mode $a{r}^{-\ell -1}{P}_{\ell }(\cos \theta )$. Continuity of the field and its radial derivative gives, for $x=\sqrt{s}R>0$,The Spherical Bessel function recurrence turns this into $j_{\ell -1}(x)=0$, not $j_{\ell }(x)=0$. This elimination does not require division by a possibly vanishing boundary value.

For $\ell =0$, $j_0(x)=\sin x/x$ and $j_{-1}(x)=\cos x/x$, so the first positive root is $\pi /2$. For $\ell =1$, the matching condition is $j_0(x)=0$, whose first positive root is $\pi$. The distinction matters when a scalar model permits the monopole sector but a physical multipole model excludes it.

To compare angular sectors, put $u=r{j}_{\ell }(\sqrt{s}r)$. Its equation and boundary conditions are $-u^{\prime \prime }+\ell (\ell +1)u/r^2=su$, $u(0)=0$ and $u^{\prime }(R)+\ell u(R)/R=0$. Its Rayleigh quotient isBoth added terms are nonnegative and increase with $\ell$. Thus the lowest scalar eigenvalue is ${\pi }^2/(4R^2)$ in the $\ell =0$ sector; if only $\ell \ge 1$ is admitted, the lowest is ${\pi }^2/R^2$ in the $\ell =1$ sector. Zeros at $x=0$ caused by the vanishing normalization of $j_{\ell }$ are not additional eigenvalues.