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The precise question is, find the velocity field of ideal fluid given that the fluid has a sink of strength $2\pi k$ at the origin (and no other singularities) and that it has velocity $V_\infty$ (a complex number) at infinity.

This is all the given information. I'm not really to sure where to begin. Any pointers would be appreciated! Thanks!

EDIT This is what I've come up with:

We have $$ \phi(z)=-\frac{2\pi k}{|z|}=\phi(x,y)=-\frac{2\pi k}{\sqrt{x^2+y^2}}, $$ which gives the velocity field $$ v(z)=v(x,y)=\nabla\phi(x,y)=\frac{2\pi kx}{(x^2+y^2)^{3/2}}+i\frac{2\pi ky}{(x^2+y^2)^{3/2}} $$ Now, as $z\to\infty$ we have $v\to0$. However, at infinity we must have $$ \lim_{z\to\infty}v(z)\equiv V_\infty $$ So, the velocity field becomes $$ v(x,y)=\frac{2\pi kx}{(x^2+y^2)^{3/2}}+i\frac{2\pi ky}{(x^2+y^2)^{3/2}}+V_\infty=\frac{2\pi k}{(x^2+y^2)^{3/2}}(x+iy)+V_\infty, $$ which gives the result.

Lou
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  • Welcome! What things do you know about the background of the problem? – Vladhagen Dec 12 '13 at 05:44
  • Thanks! I actually took some time to think about it, this is what I've come up with. – Lou Dec 13 '13 at 02:19
  • In terms of background, I'm not sure what you mean. It's a question on the applications of complex analysis. I hope that helps! – Lou Dec 13 '13 at 04:05

2 Answers2

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From what I know about ideal fluids, the potential should be a harmonic function, which $1/|z|$ is not. I think the correct potential is $$\phi(z) = -k\log |z|$$ The constant $2\pi$ appears naturally: $\Delta \phi$ is $ - 2\pi k\delta $ where $\delta$ is the Dirac delta at $0$.

The real gradient can be taken using complex notation: $$ \nabla \phi = \frac{\partial}{\partial \bar z} (2\phi) = \frac{\partial}{\partial \bar z} (-k\log|z|^2 ) = \frac{\partial}{\partial \bar z} (-k\log z - k\log \bar z ) = -\frac{k}{\bar z} = -\frac{kz}{|z|^2}$$ Add $V_\infty$ to get the right behavior at $\infty$, as you did.

Post No Bulls
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This is what I find in my (40 years old) notes: $$ \oint v_r\,ds = \int_0^{2\pi} v_r\,r\,d\theta = v_r\,r\,2\pi = -Q = -2\pi k \qquad \Longrightarrow \qquad v_r = -|u+i\,v| = -\frac{k}{r} $$ So, for that sink, when augmented with the velocity field at infinity: $$ u + i\,v = - \frac{k}{r} e^{i\,\theta} + V_\infty = - \frac{k}{r^2} \left[\,r \cos(\theta) + i\,r\sin(\theta)\,\right] + V_\infty = - \frac{k\,(x+i\,y)}{x^2+y^2} + V_\infty $$ Furthermore: $$ u - i\,v = -\frac{k}{r} e^{-i\,\theta} + \overline{V_\infty} = -\frac{k}{z} + \overline{V_\infty} = \frac{d\phi}{dz} $$ So the complex potential $\phi$ is (still apart from a complex constant): $$ \phi = - k\,\ln(z) + \overline{V_\infty}\, z $$ Which is all a bit different from the other answers.

Han de Bruijn
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