6.7 The Helmholtz equation

The Helmholtz equation is the equation acoustics actually solves in the frequency domain. It arises whenever we look at a wave equation and ask, what does the field look like at a single frequency $\omega$ ? — which is almost always the right question for steady-state acoustic problems: room modes, cavity resonances, scattering off an obstacle, radiation from a vibrating object, transmission through an aperture. The time-domain wave equation is the underlying object, but the working equation is Helmholtz.

This lesson is a separate lesson because the Helmholtz equation looks elliptic — no time derivative — yet is born from the hyperbolic wave equation. That dual heritage explains both its mathematical structure (an eigenvalue problem with sinusoidal eigenfunctions) and its role across acoustics.

Derivation: time-harmonic ansatz on the wave equation

Start with the wave equation,

\frac{\partial^2 u}{\partial t^2} \;=\; c^2 \nabla^2 u,

and look for time-harmonic solutions — fields that oscillate at a single frequency $\omega$ :

u(\mathbf{r}, t) \;=\; \phi(\mathbf{r})\, e^{-i \omega t}.

This is the same complex-exponential ansatz we use for driven oscillators (Foundations 3); the sign convention $e^{-i\omega t}$ is conventional in physics and the choice doesn’t matter as long as you stick with it. Substitute and take partial derivatives:

\frac{\partial^2 u}{\partial t^2} \;=\; -\omega^2\, \phi(\mathbf{r})\, e^{-i\omega t}, \qquad \nabla^2 u \;=\; \nabla^2 \phi(\mathbf{r}) \cdot e^{-i\omega t}.

The factor $e^{-i\omega t}$ cancels on both sides, leaving an equation for the spatial amplitude $\phi$ :

\boxed{\;\nabla^2 \phi \;+\; k^2 \phi \;=\; 0, \qquad k \;\equiv\; \omega / c. \;}

This is the Helmholtz equation, and $k$ is the wavenumber at frequency $\omega$ . The time dependence has been factored out; what remains is a purely spatial equation for the field’s envelope at frequency $\omega$ .

A few observations before we go further:

The Helmholtz equation has no time derivative. By the classification table in 6.1, it is elliptic — it is Laplace’s equation with an extra zeroth-order term $k^2 \phi$ added. The boundary-value-problem character of elliptic PDEs is exactly what holds: $\phi$ on a closed domain is determined by data on the boundary.
For each $\omega$ (equivalently each $k$ ) you get a different Helmholtz equation. Sweeping $\omega$ generates a family of spatial problems indexed by frequency. This is why frequency-domain acoustics looks like a stack of independent spatial calculations rather than a single time-marching simulation.
If $\phi$ solves the Helmholtz equation, then so does $\phi^*$ (the complex conjugate). Real-valued physical fields are recovered by taking real parts at the end: $u(\mathbf{r}, t) = \mathrm{Re}[\phi(\mathbf{r}) e^{-i\omega t}]$ .

Why frequency-domain acoustics lives here

Three reasons the Helmholtz equation, rather than the time-domain wave equation, is what gets solved in practice:

Steady-state is the question. Many acoustic problems ask not “how does the field evolve from some initial state?” but “what does the steady-state field look like at frequency $\omega$ once transients have died?” Helmholtz answers exactly that.
Linearity decouples frequencies. Any time-dependent forcing can be Fourier-decomposed into sinusoidal components, each of which generates a Helmholtz problem. Solving the time-domain wave equation with arbitrary forcing is hard; solving Helmholtz at each frequency and summing is straightforward.
Eigenvalue structure. On a bounded domain with homogeneous boundary conditions, Helmholtz reduces to an eigenvalue problem (refresher →): find the values of $k^2$ for which $\nabla^2 \phi + k^2 \phi = 0$ has a non-trivial solution. Those are the cavity modes of the domain — the same modes we have been calling $X_n$ , $X_{mn}$ , and so on. The Helmholtz formulation makes this eigenvalue structure obvious; the wave equation hides it inside the separated time ODE.

Worked example: 2-D rectangular cavity

The canonical Helmholtz problem in acoustics is a 2-D rectangular cavity with rigid or pressure-released boundaries. Walk through the separation-of-variables step by step.

▶ Worked example: every step, no shortcuts

The problem. A rectangular cavity of size $L_x \times L_y$ in the $(x, y)$ -plane, with the acoustic pressure $\phi(x, y)$ obeying

\frac{\partial^2 \phi}{\partial x^2} \;+\; \frac{\partial^2 \phi}{\partial y^2} \;+\; k^2 \phi \;=\; 0,

inside the rectangle, with Dirichlet boundary conditions

\phi(0, y) = \phi(L_x, y) = 0, \qquad \phi(x, 0) = \phi(x, L_y) = 0,

on all four edges. (Physically: pressure-released walls, the model used for an open-ended cavity. The rigid-wall case swaps Dirichlet for Neumann; the algebra is identical with cosines replacing sines.) We want the eigenvalues $k^2$ for which non-trivial solutions exist, and the corresponding mode shapes.

Step 1 — Product ansatz. Try

\phi(x, y) \;=\; X(x)\, Y(y).

The Laplacian splits cleanly:

\frac{\partial^2 \phi}{\partial x^2} \;=\; X''(x)\, Y(y), \qquad \frac{\partial^2 \phi}{\partial y^2} \;=\; X(x)\, Y''(y).

Substitute into the Helmholtz equation:

X''(x)\, Y(y) \;+\; X(x)\, Y''(y) \;+\; k^2\, X(x)\, Y(y) \;=\; 0.

Step 2 — Divide by $X Y$ and separate.

\frac{X''(x)}{X(x)} \;+\; \frac{Y''(y)}{Y(y)} \;+\; k^2 \;=\; 0.

Move the $y$ -piece to the right:

\frac{X''(x)}{X(x)} \;=\; -k^2 \;-\; \frac{Y''(y)}{Y(y)}.

The left depends only on $x$ ; the right depends only on $y$ . Both must equal a common constant — call it $-k_x^2$ :

\frac{X''(x)}{X(x)} \;=\; -k_x^2, \qquad -k^2 - \frac{Y''(y)}{Y(y)} \;=\; -k_x^2.

Rearrange the second equation to read $Y''/Y = -(k^2 - k_x^2)$ . Defining $k_y^2 \equiv k^2 - k_x^2$ , we get the separation identity

k_x^2 \;+\; k_y^2 \;=\; k^2,

and two ODEs:

X''(x) \;+\; k_x^2\, X(x) \;=\; 0, \qquad Y''(y) \;+\; k_y^2\, Y(y) \;=\; 0.

Each is a 1-D Helmholtz equation (equivalently, the spatial part of a 1-D wave equation).

Step 3 — Solve each ODE with its boundary conditions. The $X$ ODE with $X(0) = X(L_x) = 0$ is exactly the clamped-string problem from 6.3. The allowed wavenumbers are quantised:

k_x \;=\; \frac{m \pi}{L_x}, \qquad m = 1, 2, 3, \ldots

with mode shapes $X_m(x) = \sin(m \pi x / L_x)$ . Likewise the $Y$ equation gives

k_y \;=\; \frac{n \pi}{L_y}, \qquad n = 1, 2, 3, \ldots

with $Y_n(y) = \sin(n \pi y / L_y)$ .

Step 4 — Combine. Each pair $(m, n)$ specifies one mode of the cavity:

\phi_{m n}(x, y) \;=\; \sin\!\left(\frac{m \pi x}{L_x}\right) \sin\!\left(\frac{n \pi y}{L_y}\right),

with eigenvalue

k_{m n}^2 \;=\; \left(\frac{m \pi}{L_x}\right)^{\!2} \;+\; \left(\frac{n \pi}{L_y}\right)^{\!2}.

The eigenfrequencies are $\omega_{m n} = c\, k_{m n}$ :

\boxed{\;\omega_{m n} \;=\; c\, \pi\, \sqrt{\left(\frac{m}{L_x}\right)^{\!2} + \left(\frac{n}{L_y}\right)^{\!2}}\;.}

Notice the structure. A 1-D bounded problem produced a single integer index $n$ and evenly-spaced frequencies $\omega_n = n \pi c / L$ . The 2-D cavity produced two integer indices and frequencies that are not evenly spaced — they form a 2-D lattice in the $(k_x, k_y)$ plane, with the actual frequencies given by the distance from the origin. The same idea extends to a 3-D rectangular cavity, where you get three integer indices and the frequencies form a 3-D lattice.

Step 5 — Sanity-check. Three quick checks:

Boundary conditions. Each $\phi_{mn}$ has $\sin(m \pi x / L_x)$ , which vanishes at $x = 0$ and $x = L_x$ . Similarly in $y$ . All four edges satisfied. ✓
Helmholtz equation. $\partial_x^2[\sin(m \pi x / L_x)] = -(m \pi / L_x)^2 \sin(m \pi x / L_x)$ , and similarly for $y$ . Adding gives $-k_{m n}^2 \phi_{m n}$ , so $\nabla^2 \phi_{m n} + k_{m n}^2 \phi_{m n} = 0$ . ✓
Reduces to 1-D. In the limit $L_y \to \infty$ at fixed $L_x$ , the $n$ -spacing of $k_y$ becomes infinitesimal and you recover a continuous distribution along $k_y$ on top of the same 1-D ladder in $k_x$ . ✓

Step 6 — The general time-dependent solution. Re-attaching the time dependence,

u(x, y, t) \;=\; \sum_{m, n} \bigl[\,A_{m n} \cos(\omega_{m n} t) + B_{m n} \sin(\omega_{m n} t)\,\bigr]\, \sin\!\left(\frac{m \pi x}{L_x}\right) \sin\!\left(\frac{n \pi y}{L_y}\right),

with the $A_{m n}, B_{m n}$ determined by Fourier projection of the initial conditions onto the 2-D mode basis, exactly as in 6.5. The mode catalogue has become a doubly-indexed family but every other step is mechanically the same as the 1-D case.

Mode shapes, made visible

m (x mode index) = 1 n (y mode index) = 1

cavity aspect ratio L_y / Lₓ = 1.00

Each mode is a product of one sine in x and one sine in y. The two integer indices (m, n) label which mode you are looking at: m counts the half-wavelengths along x, n counts them along y. The two contributions add in quadrature to give the squared wavenumber, kₘₙ² = (mπ / Lₓ)² + (nπ / L_y)². Changing the aspect ratio changes how the two indices weight the eigenfrequency, but the mode catalogue stays the doubly-indexed lattice you see here.

Pick a mode index pair $(m, n)$ and the aspect ratio $L_y / L_x$ to see the corresponding mode shape and the eigenfrequency. The first integer counts the number of half-wavelengths along $x$ (and therefore the number of interior nodal lines perpendicular to $x$ ); the second does the same for $y$ . The $(1, 1)$ mode is the lowest fundamental — a single positive blob with no interior nodes. The $(2, 1)$ mode has a single nodal line at $x = L_x / 2$ . The $(2, 2)$ mode has a “checkerboard” pattern with two nodal lines.

Aspect ratio matters in a way it didn’t for 1-D. A near-square cavity gives nearly-degenerate frequencies for $(m, n)$ and $(n, m)$ — they coincide exactly when $L_x = L_y$ . A stretched cavity breaks the degeneracy. Real concert halls and listening rooms deliberately avoid simple integer ratios for their dimensions to keep mode frequencies from clustering.

Mode density grows with frequency

In one dimension the modes are evenly spaced and the modal density is constant. In two dimensions the modes are points on a 2-D lattice, and the number of modes with $\omega_{m n} \leq \Omega$ is approximately the area of a quarter-circle of radius $\Omega / c$ in $(k_x, k_y)$ -space, weighted by the lattice density:

N(\Omega) \;\approx\; \frac{1}{4} \cdot \pi (\Omega/c)^2 \cdot \frac{L_x L_y}{\pi^2} \;=\; \frac{L_x L_y\, \Omega^2}{4 \pi c^2}.

Differentiate to get the modal density:

D_{2D}(\omega) \;=\; \frac{d N}{d \omega} \;=\; \frac{L_x L_y\, \omega}{2 \pi c^2}.

The modal density grows linearly with frequency in 2-D. In 3-D it grows quadratically (as we saw in 6.5). The Schroeder frequency, above which a room has so many overlapping modes that the field looks statistically uniform, is set by this growth and is the central organising idea of Sound 7.8.

On unbounded domains: scattering and radiation

The Helmholtz equation on a bounded domain produces eigenvalues and modes. On an unbounded domain it produces wave-like solutions instead, parameterised continuously by direction. The simplest examples in 3-D:

Plane wave: $\phi(\mathbf{r}) = e^{i \mathbf{k} \cdot \mathbf{r}}$ , with $|\mathbf{k}| = k = \omega / c$ . Substituting into Helmholtz: $\nabla^2 e^{i \mathbf{k} \cdot \mathbf{r}} = -k^2 e^{i \mathbf{k} \cdot \mathbf{r}}$ , and the equation is satisfied for any direction of $\mathbf{k}$ .
Spherical wave: $\phi(r) = e^{i k r} / r$ , an outgoing wave from the origin. Substituting into the radial Laplacian $\partial_r^2(r \phi) / r$ verifies the equation everywhere except at the origin (where the source sits).

The “boundary condition at infinity” for an unbounded Helmholtz problem is called the Sommerfeld radiation condition: $\phi$ at large $r$ must look like an outgoing spherical wave $e^{i k r}/r$ , not an incoming one. This is the mathematical statement that energy radiates away from the source — it cannot have arrived from infinity in the steady state. Imposing it is what makes scattering and radiation problems have unique solutions.

The Sound book applies these unbounded solutions in Sound Ch 6 — Sources and radiation: a pulsating sphere radiates the spherical-wave Helmholtz solution, a dipole radiates the gradient of one, and a piston in a baffle radiates an integral of them weighted by the piston surface.

What we use it for

Cavity modes for rectangular rooms and ducts — Sound 7.7 — Rectangular cavity modes is the worked example of this lesson scaled up to 3-D.
Tube modes — Sound 7.6 is the 1-D specialisation, where the Helmholtz problem reduces to a single ODE.
Spherical and cylindrical waves — Sound 6.2 and 6.3 are Helmholtz on those geometries, with Bessel and spherical-Bessel functions playing the role our $\sin(m\pi x/L)$ does here.
Scattering and diffraction — Sound 7.5 is a Helmholtz problem on a half-space with a slit; the incident and scattered fields are both solutions of Helmholtz, matched by boundary conditions.
The Kirchhoff–Helmholtz integral, which expresses the field at any interior point as an integral of $\phi$ and $\partial_n \phi$ over a surrounding surface — the Helmholtz analogue of the d’Alembert formula. We don’t develop it here, but you will see it when you reach the boundary-integral method in numerical acoustics.

The next lesson, 6.8 — The Schrödinger equation, shows that the same separation-of-variables machine applies to a PDE with imaginary time coefficient, with consequences that look very different from the acoustics-flavoured Helmholtz problem but use the identical mathematics.