Acoustic waves

Linear acoustic waves

Strong formulation

This formulation of acoustic waves is derived under linear approximation assumption of Newtonian fluids. Such fluids follows the Navier–Stokes equations for compressible Newtonian fluids

$$\begin{align} \rho \Bigr(\frac{\partial \boldsymbol{v}}{\partial t} + (\boldsymbol{v} \cdot \nabla) \boldsymbol{v} \Bigr) &= -\nabla p + \nabla \cdot (\mu (\nabla \boldsymbol{v} + (\nabla \boldsymbol{v})^T)) + \nabla (-\tfrac{2}{3} \mu \nabla \cdot \boldsymbol{v}) + \boldsymbol{f} \\[10pt] \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \boldsymbol{v}) &= 0. \end{align}$$

For small density and pressure variations the speed of sound in the fluid $c$ is given by

$$\begin{align} c = \sqrt{\frac{\partial p}{\partial \rho}}. \end{align}$$

We can neglect the viscosity terms in (1) since we are interesteed only in the region within a few acoustic wavelengths. Considering an inviscid fluid and no external forces, (1) rewrites as

$$\begin{align} \rho \Bigr(\frac{\partial \boldsymbol{v}}{\partial t} + (\boldsymbol{v} \cdot \nabla) \boldsymbol{v}\Bigr) = -\nabla p \\[10pt] \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \boldsymbol{v}) = 0, \end{align}$$

which can, for tiny perturbations, be linearised around a mean value:

$$\begin{align} p = \overline{p} + \delta p,\\ \rho = \overline{\rho} + \delta \rho,\\ \boldsymbol{v} = \overline{\boldsymbol{v}} + \delta \boldsymbol{v} = \delta \boldsymbol{v}, \end{align}$$

where the overlined quantities are the mean values (constant in space and time) and the $\delta$ terms are tiny perturbations. Since the fluid is at rest, $\overline{\boldsymbol{v}} = 0$. Injecting equations (6) to (8) into equations (4) and (5) and neglecting nonlinear perturbations gives

$$\begin{align} \overline{\rho} \frac{\partial \delta \boldsymbol{v}}{\partial t} + \delta \rho \displaystyle\frac{\partial \delta \boldsymbol{v}}{\partial t} + \overline{\rho} (\delta \boldsymbol{v} \cdot \nabla) \delta \boldsymbol{v} = - \nabla \delta p\\[10pt] \frac{\partial \delta \rho}{\partial t} + \overline{\rho} \nabla \cdot \delta \boldsymbol{v} = 0. \end{align}$$

By algebraic manipulation (using the product rule and the continuity equation multiplied by $\delta \boldsymbol{v}$) and neglecting second‐order perturbations we obtain useful approximation

$$\begin{align} \overline{\rho}\,(\delta \boldsymbol{v}\!\cdot\nabla)\,\delta \boldsymbol{v} \approx - \delta \rho \frac{\partial \delta \boldsymbol{v}}{\partial t}. \end{align}$$

As a result, the two middle terms in the first relation of (9) cancel and we obtain obtains

$$\begin{align} \overline{\rho} \frac{\partial \delta \boldsymbol{v}}{\partial t} + \nabla\,\delta p &= 0 \\[10pt] \frac{\partial \delta \rho}{\partial t} + \overline{\rho} \nabla \cdot \delta \boldsymbol{v} &= 0. \end{align}$$

Taking the divergence of the first relation and the time derivative of the second we get

$$\begin{align} \overline{\rho} \nabla \cdot \frac{\partial \delta \boldsymbol{v}}{\partial t} + \Delta \delta p &= 0 \\[10pt] \frac{\partial^2 \delta \rho}{\partial t^2} + \overline{\rho} \nabla \cdot \frac{\partial \delta \boldsymbol{v}}{\partial t} &= 0, \end{align}$$

which can be combined into a single equation

$$\begin{align} \frac{\partial^2 \delta \rho}{\partial t^2} - \Delta \delta p = 0. \end{align}$$

With the isentropic approximation (3) the acoustic wave equation can be written as

$$\begin{align} \Delta \delta p - \frac{1}{c^2} \frac{\partial^2 \delta p}{\partial t^2} = 0, \end{align}$$

with $c$ the speed of sound in the fluid and $\delta p$ the pressure variation around the mean pressure.

Weak formulation

From now on we will write the $\delta p$ just as $p$. To form the weak formulation we will multiply both sides by test function $p^{\prime}$ and integrate over the whole domain $\Omega$.

$$\begin{align} \int_{\Omega}\ (\Delta p - \frac{1}{c^2}\frac{\partial^2 p}{\partial t^2})\ p^{\prime}\ d\Omega = 0. \end{align}$$

We can use the Leibniz rule for a nabla operator to rewrite the Laplace term.

$$\begin{align} \int_{\Omega}\ \nabla \cdot (p^{\prime} \nabla p)\ d\Omega - \int_{\Omega}\ \nabla p \cdot \nabla p^{\prime}\ d\Omega - \int_{\Omega}\ \frac{1}{c^2}\frac{\partial^2 p}{\partial t^2} \ p^{\prime}\ d\Omega = 0, \end{align}$$

and then apply the Divergence theorem on the divergence term to obtain the weak formulation

$$\begin{align} \int_{\Gamma}\ (\nabla p \cdot \boldsymbol{n})\ p^{\prime}\ d\Gamma - \int_{\Omega}\ \nabla p \cdot \nabla p^{\prime}\ d\Omega - \int_{\Omega}\ \frac{1}{c^2}\frac{\partial^2 p}{\partial t^2} \ p^{\prime}\ d\Omega = 0. \end{align}$$