# Sound wave equation: Neumann boundary conditions

In this paper it's described the solution of the damped wave equation in cylindrical coordinates

$$\nabla^2\left(c^2\rho_1+\nu\frac{\partial\rho_1}{\partial t}\right)-\frac{\partial^2\rho_1}{\partial t^2}=0$$

where $$\rho_1$$ is the difference of the density relative to the unperturbed state $$\rho_0$$.

The applied boundary condition is

$$\mathbf{v}\big|_{r=r_0}=v_A\cos(\omega t)\mathbf{\hat{r}}$$

where $$v$$ is the velocity of the fluid.

They claim that this boundary condition can we rewritten as

$$\begin{equation} \frac{\partial\rho_1}{\partial r} \bigg|_{r=r_0}=\frac{\rho_0v_A\omega c^2}{\nu^2\omega^2+c^4}\sin(\omega t)-\frac{\rho_0v_A\omega^2 \nu}{\nu^2\omega^2+c^4}\cos(\omega t)\tag{1} \end{equation}$$

just imposing $$\nabla\times \mathbf{v}=\mathbf{0}$$ and using the equations for the conservation of mass and momentum

$$\frac{\partial\rho_1}{\partial t} + \nabla\cdot(\rho_0 \mathbf{v}) =0$$ $$\frac{\partial}{\partial t}(\rho_0 \mathbf{v})+c^2\nabla\rho_1+\nabla \cdot \mathbf{D}_1=\mathbf{0}$$

where $$\mathbf{D}_1$$ is the viscous stress tensor.

It is possible to prove that, if $$\nabla\times \mathbf{v}=0$$, then $$\nabla \cdot \mathbf{D}_1 = -\nu\nabla^2\mathbf{v}$$.

I tried hard but I've not been able to prove equation $$(1)$$. Do you know how to proceed?

Reference:

Euan McLeoda and Craig B. Arnold, Mechanics and refractive power optimization of tunable acoustic gradient lenses, Journal of Applied Physics 2007 102:3

• How could a vectorial expression be equal to a scalar expression? There are several equations which do not make sense. See for example the last equation.
– Cham
Dec 14, 2018 at 18:11
• @Cham I don't see any equality between a scalar and a vector. Which are the expression that do not make sense in your opinion? Please also note that the one you are rading here are exactly the same of the paper. Dec 14, 2018 at 18:13
• Your last equation reads $\nabla \cdot \boldsymbol{\mathrm{D}}_1 = -\nu \nabla^2 \boldsymbol{\mathrm{v}}$. The left member is a scalar (divergence of vector $\boldsymbol{\mathrm{D}}_1$). The right member is a vector (laplacian of vector $\boldsymbol{\mathrm{v}}$).
– Cham
Dec 14, 2018 at 18:19
• @Cham The left member is a vector, sorry. The divergence operator decreases the rank of the tensor by 1. Since the rank of $\mathbf{D}_1$ is 2 then the rank of $\nabla\cdot\mathbf{D}_1$ is 1. Dec 14, 2018 at 18:21
• Ok then. But the notation is misleading.
– Cham
Dec 14, 2018 at 18:21

$$\rho_0\frac{\partial v}{\partial t}+c^2\frac{\partial \rho_1}{\partial r}-\nu\frac{\partial}{\partial t}\frac{\partial \rho_1}{\partial r}\bigg|_{r=r_0} = 0$$
using the Fourier transform method is possible to solve this differential equation for the variable $$\frac{\partial \rho_1}{\partial r}\bigg|_{r=r_0}$$ obtaining then equation $$(1)$$.