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Consider the acoustic wave equation:

$$\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\right)p=0$$

where $p$ is the pressure perturbation in the medium.

A particular solution of this equation, for the case of spherical waves, is:

$$p(r,t)=\frac{f(r-c\,t)}{r}$$

where $f(r-c\,t)$ is an arbitrary function of argument $r-c\,t$.

By substituting $p(r,t)$ in the wave equation we get:

$$\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\right)\frac{f(r-c\,t)}{r}=0$$

I intend now to show that for $r\rightarrow0$, the above equation reduces to the Laplace equation, i.e. :

$$\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\right)\frac{f(r-c\,t)}{r}=0\xrightarrow[r\rightarrow0]{}\nabla^2\left[\frac{f(r-c\,t)}{r}\right]=0 $$

I really don't know how to show this. Can you help me?


Note:

I acknowledge this behavior of the wave equation from the pag. 147 of the book "Sound and Sources of Sound" (Downling and Williams). You can see what I mean from the uploaded image. The authors say that it was observed in Chapter 2, but I didn't find any derivation or explanation there.

enter image description here

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migrated from physics.stackexchange.com Jan 12 '18 at 13:08

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Maybe the authors of the book explained wrongly this behaviour of the wave equation.

After some analysis, I now really think that they instead wanted to mean:

$$\iiint_{V_\varepsilon}\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\right)\frac{f(r-c\,t)}{r}\,dV\xrightarrow[\varepsilon\rightarrow0]{}\iiint_{V_\varepsilon}\nabla^2\left[\frac{f(r-c\,t)}{r}\right]\,dV $$

(where $V_\varepsilon$ is the volume of a sphere centred in $\vec{x}=0$, with radius $\varepsilon$)

instead of

$$\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\right)\frac{f(r-c\,t)}{r}\xrightarrow[r\rightarrow0]{}\nabla^2\left[\frac{f(r-c\,t)}{r}\right] $$

I achieved these conclusions after studying the following steps that are presented in the book. It's considered a sphere centred $\vec{x}=0$, with radius $\varepsilon$, and the integration of the wave equation in its volume.

Applying the divergence theorem and by the definiton of the centroid of a function, we get:

$$\iiint_{V_\varepsilon}\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\right)\frac{f(r-c\,t)}{r}\,dV=\left\langle\frac{1}{c^2}\frac{\partial^2p}{\partial t^2}\right\rangle V-\iint_{\partial V}\left(\nabla p \cdot\vec{n}\right)\,dS$$

where $\left\langle\frac{1}{c^2}\frac{\partial^2p}{\partial t^2}\right\rangle$ is the centroid value of $\frac{1}{c^2}\frac{\partial^2p}{\partial t^2}$ in the sphere volume.

We then get:

$$ \begin{split} \iiint_{V_\varepsilon}\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\right)\frac{f(r-c\,t)}{r}\,dV &=\left\langle\frac{1}{c^2}\frac{\partial^2p}{\partial t^2}\right\rangle \frac{4}{3}\pi\varepsilon^3-\frac{\partial p}{\partial r}4\pi\varepsilon^2\\ \xrightarrow[\varepsilon\rightarrow0]{}-\frac{\partial p}{\partial r}4\pi\varepsilon^2&=\iiint_{V_\varepsilon}\nabla^2\left[\frac{f(r-c\,t)}{r}\right]\,dV \end{split} $$

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