1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stream function for double sink / source flow

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data
    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

    [tex]\lim_{\delta x \rightarrow 0} \delta x^2 m = \mu_2[/tex]

    with [tex]\mu_2[/tex] 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.

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

    3. 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:
    [tex]\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)[/tex]

    I have to take the limit to 0 for δx and I want to introduce [tex]\mu_2[/tex] so I take:
    [tex]\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)[/tex]

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

    Taking the limit it follows that [tex]\lim_{\delta x \rightarrow 0} \delta x^2 m = \mu_2[/tex] and [tex]\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)[/tex]

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

    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!
    Last edited: Apr 5, 2010
  2. jcsd
  3. Apr 5, 2010 #2
    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 [tex]|\vec{x}-\vec{\delta x}|[/tex] out in square roots.

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

    I also recommend expanding it out to 2nd order since you know you want [tex]\delta x^2[/tex] (because they give you a limit for it). That, and you will see something happen to the first order term.
  4. Apr 6, 2010 #3
    Ok, I know that the source/sink are positioned on the x-axis so [tex]\delta\vec{x}=\delta x[/tex]. Then I get:

    |\vec{x}-\vec{\delta x}|=\sqrt{x^2+y^2+z^2+\delta x^2-2x\delta x}
    |\vec{x}+\vec{\delta x}|=\sqrt{x^2+y^2+z^2+\delta x^2+2x\delta x}

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

    \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


    \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

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

    \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]

    Taking the limit then turns [tex]m\delta x^2[/tex] into [tex]\mu_2[/tex].

    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"
    Last edited by a moderator: Apr 25, 2017
  5. Apr 6, 2010 #4
    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:

    [tex]\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}[/tex]

    Just treat your:


    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.
  6. Apr 6, 2010 #5
    Then I get:
    \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)

    Solving the differential part yields for my potential:



    Just like you said I would. Thanks! That was really helpful!!
  7. Apr 6, 2010 #6
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook