Stream function for double sink / source flow

[MichielM]

Homework Statement

Consider a source/sink in the origin of a coordinate system with a strength equal to 2m and two sources/sinks, each with a strength of −m, i.e. with a opposite sign with respect to the source/sink in the origin. The two sources/sinks are positioned along the x-axis at x = δx and x = −δx, respectively. Consider the limit for δx → 0 under the condition that

\lim_{\delta x \rightarrow 0} \delta x^2 m = \mu_2

with \mu_2 finite. Consider this problem in three dimensions and sketch the resulting streamline patterns. Is there a closed streamline, which does not pass through the origin.

Homework Equations

The potential \Phi_e for a single source/sink of strength m in 3D is:
\Phi_e=-\frac{m}{4\pi r}
\bar{u}_e=\nabla\Phi_e=-\frac{m \bar{r}}{4\pi r^3}
with \bar{r}=\bar{x}-\bar{x}_0 where x = (x, y, z) and r = |\bar{x}-\bar{x_0}| = \sqrt{x^2 + y^2 + z^2}

The Attempt at a Solution

Since the poisson equation which leads to the potential solution above is linear, I can add the 2 sinks and 1 source together to get 1 potential:
\Phi_e=-\frac{m}{4 \pi}\left(\frac{-1}{|\bar{x}-\delta\bar{x}|}+\frac{2}{|\bar{x}|}+\frac{-1}{|\bar{x}+\delta\bar{x}|}\right)

I have to take the limit to 0 for δx and I want to introduce \mu_2 so I take:
\Phi_e=\lim_{\delta x \rightarrow 0} -\frac{m \delta x^2}{4 \pi}\frac{1}{\delta x^2}\left(\frac{-1}{|\bar{x}-\delta\bar{x}|}+\frac{2}{|\bar{x}|}+\frac{-1}{|\bar{x}+\delta\bar{x}|}\right)

From this point on I'm highly unsure about my method.

Taking the limit it follows that \lim_{\delta x \rightarrow 0} \delta x^2 m = \mu_2 and \frac{1}{\delta x^2}\left(\frac{-1}{|\bar{x}-\delta\bar{x}|}+\frac{2}{|\bar{x}|}+\frac{-1}{|\bar{x}+\delta\bar{x}|}\right)=\frac{\partial}{\partial x^2}\left(\frac{3}{|\bar{x}|}\right)

So my potential becomes:
\Phi_e=\frac{\mu_2}{4\pi}\frac{\partial}{\partial x^2}\left(\frac{3}{|\bar{x}|}\right)

From here I could proceed to calculate the velocity, but honestly I don't trust my answer enough to take the effort so I first wanted to ask whether anybody could run through my calculation and comment on any mistakes/errors!

Thanks in advance!

 

[nickjer]

I do not know what streamline patterns are, but I can help with the math some. I have no idea where you came up with the solution you got. But I played with it some and got a nice solution for the potential in the end, so I can only guess I did it right.

You will first want to write out |\vec{x}-\vec{\delta x}| out in square roots.

Then you are going to want to expand your potential in a taylor series since you know \delta x is small.

I also recommend expanding it out to 2nd order since you know you want \delta x^2 (because they give you a limit for it). That, and you will see something happen to the first order term.

 

[MichielM]

Ok, I know that the source/sink are positioned on the x-axis so \delta\vec{x}=\delta x. Then I get:

<br /> |\vec{x}-\vec{\delta x}|=\sqrt{x^2+y^2+z^2+\delta x^2-2x\delta x}<br />
and
<br /> |\vec{x}+\vec{\delta x}|=\sqrt{x^2+y^2+z^2+\delta x^2+2x\delta x}<br />

Taylor expanding the \frac{-1}{|\vec{x}+\vec{\delta x}|} and \frac{-1}{|\vec{x}-\vec{\delta x}|} terms yields:

<br /> \frac{-1}{|\vec{x}-\vec{\delta x}|}=\frac{-1}{\sqrt{x^2+y^2+z^2}}-\frac{x}{(x^2+y^2+z^2)^{3/2}}\delta x+0.5\left(\frac{1}{(x^2+y^2+z^2)^{3/2}}-\frac{3x^2}{(x^2+y^2+z^2)^{5/2}}\right)\delta x^2<br />

and

<br /> \frac{-1}{|\vec{x}+\vec{\delta x}|}=\frac{-1}{\sqrt{x^2+y^2+z^2}}+\frac{x}{(x^2+y^2+z^2)^{3/2}}\delta x+0.5\left(\frac{1}{(x^2+y^2+z^2)^{3/2}}-\frac{3x^2}{(x^2+y^2+z^2)^{5/2}}\right)\delta x^2<br />

So the \delta x term drops out of the potential and the 0-order terms cancel out against the \frac{2}{|\vec{x}|} term so I get:

<br /> \Phi_e=\frac{m\delta x^2}{4\pi}\left[\frac{1}{(x^2+y^2+z^2)^{3/2}}-\frac{3x^2}{(x^2+y^2+z^2)^{5/2}}\right]<br />

Taking the limit then turns m\delta x^2 into \mu_2.

The problem I have with this equation is that I expect a derivative to pop up somewhere, because that's how the solution for a 'normal' sink/source doublet(single source, single sink with infinitesimally small distance to each other) is. Take a look at this: http://web.mit.edu/fluids-modules/www/potential_flows/LecturesHTML/lec1011/node26.html"

 

[nickjer]

That last answer looks like the right answer except you lost a negative sign somewhere. Also thanks for the link, I know what you want now. You need to use the trick:

\frac{\partial^2 f(x,y,z)}{\partial x^2}=\lim_{h\to 0}\frac{f(x-h,y,z)-2f(x,y,z)+f(x+h,y,z)}{h^2}

Just treat your:

f(x,y,z)=\frac{1}{|\vec{x}|}=\frac{1}{\sqrt{x^2+y^2+z^2}}

That will give you a very similar expression to what you had originally, except you were originally off by a factor. Hope this helps. It will also give you the same answer you got in the last post (besides a negative sign) if you take the derivatives of your expression, just to check your work.

 

[MichielM]

Then I get:
<br /> \Phi_e=\frac{\mu_2}{4\pi}\frac{\partial^2}{\partial x^2}\left(\frac{1}{|\bar{x}|}\right)=\frac{\mu_2}{4\pi}\frac{\partial^2}{\partial x^2}\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)<br />

Solving the differential part yields for my potential:
<br /> <br /> \Phi_e=-\frac{\mu_2}{4\pi}\left[\frac{1}{(x^2+y^2+z^2)^{3/2}}-\frac{3x^2}{(x^2+y^2+z^2)^{5/2}}\right]<br /> <br />

Just like you said I would. Thanks! That was really helpful!

 

[nickjer]

No problem. Hah, that method was easier than a taylor expansion anyways. Didn't know about that method until you showed me, so we both learned something.