# 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 '18 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. – Alessandro Zunino Dec 14 '18 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 '18 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. – Alessandro Zunino Dec 14 '18 at 18:21
• Ok then. But the notation is misleading. – Cham Dec 14 '18 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 $$(1)$$.