# Interpreting the Helmholtz equation for a time-harmonic acoustic pressure field?

An equation such as Gauss' law $$\nabla \cdot E(x) = \frac{\rho(x)}{\varepsilon_0}$$ is easy to interpret in physical terms. If we use the divergence theorem we have $$\int_{\partial \Omega} E\cdot \hat{n} \ dx = \int_{\Omega} \frac{\rho(x)}{\varepsilon_0} dx$$ which says that the electric flux out of the domain $\Omega$ is equal to the total amount of charge $\rho$ inside $\Omega$.

Now what about the Helmholtz equation which governs time-harmonic acoustic pressure waves:

$$\Delta p + k^2 p = 0$$ where $k$ is the wavenumber. Is it possible to give as direct an interpretation for this equation as it was for Gauss' law? Things were really clear in the case of Gauss' law because we could use the divergence theorem to lead to an integral equation that relates the strength of a variable leaving a domain to the amount of another variable inside the domain.

The same approach can be taken in thermodynamics, i.e. relating the amount of heat flux out of a body to the amount of energy inside the body.

Can a similar clear interpretation of the Helmholtz equation by given?

Consider the differential identity

$$\nabla^2 p = \vec{\nabla} \cdot \vec{\nabla}p$$

Now integrate Helmoltz equation in a volume $V$:

$$\int_V d^3x \vec{\nabla} \cdot \vec{\nabla}p = - k^2 \int_V pd^3x$$

Now use Gauss' theorem in the left hand side:

$$-\int_{\partial V} \vec{\nabla}p \cdot \hat{n} d \Sigma = k^2 \int_V p d^3x$$

As you see, the left hand side is the negative flux of pressure gradient through the surface surrounding the integration volume $\Phi(\vec{\nabla}p)$, while the integral $\int_V p d^3x$ represents the total work.