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In fluid Mechanics,

The superimposed stream function of point source and sink is: $\psi=-\frac{Qcos\theta_1}{4\pi}+\frac{Qcos\theta_2}{4\pi}$

Graphical image of the function

and for a sink - source doublet, we need to show that as: $l\rightarrow 0$

we get:

$\psi=\frac{m}{r}sin^2\theta$
where: $m=\lim_{l\rightarrow0}\frac{Ql}{4\pi}$

I know it's a simple geometric limit problem , but somehow I can't derive the desired result.

Thanks!

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1 Answer 1

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The positions of the sink and source are $(-l/2,0)$ and $(l/2,0),$ respectively.

Hence,

$$\psi = \frac{Q}{4 \pi}(\cos \theta_2 - \cos \theta_1) \\= \frac{Q}{4 \pi}\left( \frac{x-l/2}{\sqrt{(x-l/2)^2 + y^2}}- \frac{x+l/2}{\sqrt{(x+l/2)^2 + y^2}}\right) \\ = \frac{Q}{4 \pi}\left( \frac{x-l/2}{\sqrt{x^2 + y^2 -lx +l^2/4}}- \frac{x+l/2}{\sqrt{x^2 + y^2 +lx + l^2/4}}\right) $$

In terms of polar coordinates, $r = \sqrt{x^2 +y^2}$ and

$$\psi = \frac{Q}{4 \pi}\left( \frac{x-l/2}{r}\left(1 - \frac{lx -l^2/4}{r^2}\right)^{-1/2}- \frac{x+l/2}{r}\left(1 + \frac{lx +l^2/4}{r^2}\right)^{-1/2}\right).$$

Using the Taylor expansion of the square root terms for small $l$ we obtain

$$\psi = \frac{Q}{4 \pi}\left( \frac{x-l/2}{r}\left(1 + \frac{1}{2}\frac{lx}{r^2}+ O(l^2)\right)- \frac{x+l/2}{r}\left(1 - \frac{1}{2}\frac{lx}{r^2}+ O(l^2)\right)\right) \\ =\frac{Q}{4 \pi}\left(-\frac{l}{r}+ \frac{lx^2}{r^3} +O(l^2) \right) \\= \frac{Ql}{4 \pi r}\left(-1+ \frac{x^2}{r^2} \right)+O(l^2),$$

and with $x = r\cos \theta,$

$$\psi = \frac{Ql}{4 \pi r}\left(-1+ \cos^2 \theta \right)+O(l^2) \\ = -\frac{Ql}{4 \pi r}\sin^2 \theta+O(l^2).$$

Taking the limit as $l \to 0$ we get

$$\lim_{l \to 0} \psi = -\frac{m}{r}\sin^2 \theta$$

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