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Given the surface of the vibrating object $\partial \Omega$, I am trying to simulate its outer sound pressure field $p(x)$ by the equivalent source method.[1]

For the Helmholtz equation $$ (\nabla^2 + k^2) p = 0 $$

I can solve it in harmonic BCs, but how to solve it in damped harmonic BCs?

1. Harmnoic oscillator case

It is easy to solve

Consider position $\mathrm x \in \mathcal \partial \Omega \subset R^3$, If it is a harmonic oscialltor, whose acceleration $\mathrm a(t; \mathrm x)$ are: $$ \mathrm a(t) = \vec d e^{i w t} $$ then the Neumann BCs of Helmholtz can be easily got: $$ \frac{\partial p}{\partial \mathrm n} = - \rho \mathrm a(t) \cdot \mathrm n $$ where $\mathrm n$ is the outer normal of the vibrating surface.

I can then use the Equivalent source method[1] to solve it.

The Equivalent source method[1] is nothing more than finding a linear combination of basis functions (monopoles) to fit the BCs.

In this case, If I solve the question at the start, I can use it forever, because the amplitude of BCs is not changed. The solution will be a constant, undamped sound pressure field.

2. Damped oscillator case (question)

But, if the boundary vibration is damping (which is the common sense, nothing is vibrating forever in daily life, but just tuning for a while), the boundary acceleration $\mathrm a(t)$ become a damped harmonic oscillator. $$ a(t) = \vec d e^{- \xi t} e^{i w t} $$ where $\xi $ is a damped factor (determined by the internal friction or viscosity of this object).

How can I use the equivalent source method in this case? If I solve it at one moment $t$, then at next time $t + \Delta t$, the boundary conditions are damped, It seems I need to resolve it, really costly.

reference

[1] Kondapalli, P.S., Shippy, D.J. and Fairweather, G., 1992. Analysis of acoustic scattering in fluids and solids by the method of fundamental solutions. The Journal of the Acoustical Society of America, 91(4), pp.1844-1854.

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1 Answer 1

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Short answer: Replace $\omega$ with $\omega-i\xi$.

Longer Answer:

The Helmholtz equation is derived from the wave equation, which may be written as $$\nabla^2P -\frac{1}{c^2}\frac{\partial^2P}{\partial t^2} = 0,$$ where $P$ is the time-domain acoustic pressure and $c$ is the wave speed. There are two approaches to obtaining the Helmholtz equation from this equation. The first is to take the Fourier transform, but the other is to assume a particular form of the solution. To demonstrate the second approach, we assume that the acoustic pressure takes the form $P=pe^{-i\omega t},$ and substitute this into the wave equation, yielding the Helmholtz equation $$\nabla^2p + \frac{\omega^2}{c^2}p = 0.$$ Usually this approach is taken when there is a driving force of some sort that has the time dependence assumed.

Following this logic, if the source condition has a time dependence that may be written as $$e^{-i\omega t}e^{-\xi t}=e^{-i(\omega-i\xi)t}=e^{-i\tilde\omega t},$$ where $\tilde\omega\equiv\omega-i\xi$, then we would assume that we may write $$P=pe^{-i\tilde\omega t}.$$ Substitution into the wave equation would then yield the Helmholtz equation $$\nabla^2p + \frac{\tilde\omega^2}{c^2}p = 0.$$ Thus, the only difference between dealing with a system with a harmonic source condition and a decaying harmonic source condition is the use of $\tilde\omega$ versus the use of $\omega$.

Your post is sufficiently vague that I cannot provide too many details on a specific case. I will assume that you would be able to solve your system with a time-harmonic source, and denote the solution as $p=p(\vec x,k)$, where $k=\omega/c$. Then, the solution to the case of a decaying time-harmonic source would simply be $p=p(\vec x,\tilde k)$, where $\tilde k=\tilde\omega/c=(\omega -i\xi)/c$.

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  • $\begingroup$ Sorry for my unclear expression which may cause misleading. But your answer is exactly what I want. Would you please point out which place I can clarify, and then I can accept your answer? $\endgroup$
    – Xudong
    Jun 21, 2022 at 15:37
  • $\begingroup$ Mostly I was just saying that I can't give you a solution, because there isn't a complete problem. For example, you could say that you are looking at a long tube with a source at one end; this would allow you to give an explicit solution. Your question is fine as is. $\endgroup$
    – Michael M
    Jun 21, 2022 at 16:14

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