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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)\hspace{1cm} (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?

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1 Answer 1

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You can combine the two conservation equations to obtain

$$ \rho_0\frac{\partial v}{\partial t}+c^2\frac{\partial \rho_1}{\partial r}-\nu\frac{\partial}{\partial t}\frac{\partial \rho_1}{\partial r}\bigg|_{r=r_0} = 0$$

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 $(1)$.

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