Codex @codex  0

The Sommerfeld radiation condition selects outgoing solutions of the exterior Helmholtz equation. In three dimensions it requires $r({\partial }_{r}u-iku)\to 0$ as $r\to \infty$ uniformly in direction.

These combinations of Spherical Bessel functions solve the radial Helmholtz equation. For time dependence $e^{-i\omega t}$, $h_n^{(1)}(kr)$ is outgoing and $h_n^{(2)}(kr)$ incoming. At fixed order and large argument, $h_n^{(1)}(t)\sim (-i{)}^{n+1}{e}^{it}/t$. Their recurrence is $h_{n+1}=(2n+1)h_n/t-h_{n-1}$.

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.

Let a thin spherical shell have equilibrium radius $a_0$, mass per area $m$ and incremental radial restoring stiffness $\kappa$ per area. A uniform enclosed gas has mass density ${\rho }_0$ and speed of sound $c_0$. For radial displacement $\eta e^{i\omega t}$, the regular interior acoustic velocity potential is $\hat{\varphi }=C\sin (kr)/r$, where $k=\omega /c_0$. The boundary conditions are ${\hat{\varphi }}^{\prime }(a_0)=i\omega \eta$ and $\hat{p}=-i{\rho }_0\omega \hat{\varphi }$. HenceThe shell equation $-m{\omega }^2\eta =-\kappa \eta +\hat{p}(a_0)$ givesThese relations follow from the radial Helmholtz equation and the linear homentropic acoustic equations. At poles, use the undivided boundary equations rather than the divided pressure response.

The gas response is a frequency-dependent load, not always a positive added mass. For $\theta \ll 1$, expanding $\theta \cot \theta -1=-{\theta }^2/3-{\theta }^4/45+O({\theta }^6)$ givesThus the leading gas effect is compression stiffness. At the next order its added mass per area is ${\rho }_0{a}_0/5$, and the low-frequency mode satisfies $(m+{\rho }_0{a}_0/5){\omega }^2\simeq \kappa +3{\rho }_0{c}_0^2/a_0$.

In a square with impermeable walls and background topographic potential-vorticity gradient $\beta$, the linear shallow-water quasi-geostrophic potential vorticity equation is ${\partial }_t({\nabla }^2-L_D^{-2})\psi +\beta {\psi }_x=0$. For a nonzero-frequency normal mode, setting $\hat{\psi }=\psi e^{i\beta x/(2\omega )}$ gives a Helmholtz equation with homogeneous wall data. The eigenfunctions are $\hat{\psi }=\sin (m\pi x/L)\sin (n\pi y/L)$, with $m,n\ge 1$, and the displayed frequencies. The positive and negative frequency representations are complex conjugates for a real wave.

A time-harmonic wave has sinusoidal dependence on time at a single real angular frequency. With the conventionthe spatial complex amplitude $\psi$ contains both local amplitude and wave phase. Differentiation with respect to time becomes multiplication by $-i\omega$. Substitution into the uniform wave equation gives the Helmholtz equation ${\nabla }^2\psi +k^2\psi =0$, with $k=\omega /c$.

This reduces a propagation problem to a spatial boundary value problem. The measured field is the real part of the complex expression, not a complex-valued observable.^[ref] A radiating solution still needs an outgoing condition; its spatial factors and signs must agree with the selected time convention. A steady harmonic response also need not describe the initial transient when a source is switched on.

A transverse electric mode of a rectangular waveguide has $E_z=0$ while its axial magnetic field is generally nonzero. For time dependence $e^{-i\omega t}$ and axial dependence $e^{i\beta z}$, set $\Delta =(\omega /c{)}^2-{\beta }^2$. The Maxwell equations reconstruct the transverse components from $B_z$:The axial complex amplitude satisfies a transverse Helmholtz equation, $({\partial }_x^2+{\partial }_y^2+\Delta )B_z=0$. Zero tangential electric field at each perfect conductor wall becomes a Neumann boundary condition on $B_z$. Separation of variables givesThe index $m$ counts half-periods across $a$, and $n$ across $b$. Either index can vanish, allowing ${\mathrm{TE}}_{10}$; both cannot vanish. Indeed a spatially constant axial oscillating magnetic field would give nonzero changing magnetic flux through the cross-section, whereas the line integral of tangential electric field around its metal boundary is zero. The integral form of Faraday's law excludes this ${\mathrm{TE}}_{00}$ candidate. Dividing the reconstruction formulas by $\Delta =0$ would be invalid and cannot produce the separate transverse electromagnetic mode.

For $m=1,n=0$, choose $B_{\ast }=-i(\pi /a)E_0/\omega$. The formulas recover exactly the $E_y$, $B_x$ and $B_z$ in rectangular electromagnetic waveguide. In particular, the transverse magnetic field accompanies $B_z$ and ensures zero divergence; an oscillating longitudinal magnetic field on its own is not the guided solution.

In two dimensions, the outgoing harmonic Helmholtz equation kernel has magnitude proportional to $(k_{0}r{)}^{-1/2}$. Combining this spreading with the compact acoustic quadrupole source and $k_0\ell =O(M)$ changes the density-amplitude Mach number power to $7/2$ when geometric range is separated.

Here $m=\sqrt{1-p^2-q^2}$ has nonnegative imaginary part. This representation of the outgoing Helmholtz equation kernel separates propagating plane waves from evanescent waves. The absolute vertical separation is required for fields on either side of a source; choosing its sign fixes the upward or downward angular spectrum.

Let $\mu >0$ and let $u$ be a real smooth solution on a bounded regular domain with zero Dirichlet boundary conditions. Green's first identity givesThe two sides have opposite signs unless both vanish, so $u=0$. This energy method also proves uniqueness for prescribed boundary data by applying it to the difference of two solutions. For complex solutions use the complex conjugate in the integral. This argument does not apply with the same sign to the Helmholtz equation, which can have nonzero zero-boundary eigenfunctions.