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Homework Help: Surface Optimization

  1. Oct 2, 2005 #1
    Ok, so here's the problem:

    Two circular wire hoops of radius [itex]R[/itex] are spaced a distance [itex]2\ell[/itex] apart. Consider a soap film stretching beatween the two hoops. Due to surface tension the film's equlibrium form is a surface of minimal area. Use the calculus of variations to find this minimal surface.

    What happens when [itex]\ell/R[/itex] is small? What happens when [itex]\ell/R[/itex] becomes large? Is there a critical value of [itex]\ell/R[/itex]? If so, explain why; find in numerically; and say what happens when one exceeds it.

    Calculus of Variations is simple: To "optimize" a quantity, [itex]y(x)[/itex], (i.e. find its minima/maxima), you find the value of [itex]x[/itex] which satisfies [itex]y'(x)=0[/itex].

    If the independent is a function which optimizes a quantity (as in the case of this problem),
    [tex]A=\int F(y(x),\,y'(x);\,x)\,dx[/tex]​
    the optimized [itex]y(x)[/itex] is given by the solution to the differential equation below (Euler's Equation):
    [tex]\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)=0[/tex]​


    Here's what I did. It makes most sense to set this problem in cylindrical coordinates. So, the surface area of the soap film is given by

    [tex]A=\int_0^{2\ell}2\pi r(z)\sqrt{1+r'(z)^2}dz.[/tex]​

    Here, we are assuming that there is no angular dependence of [itex]r[/itex]. So in anticipation of applying the calculus of variations (Euler's Equation), I evaluated the following quantities for [itex]F(r(z),\,r'(z);\,z)=r(z)\sqrt{1+r'(z)^2}[/itex]:

    [tex]\frac{\partial F}{\partial r}=\sqrt{1+r'^2}\,,\,\,\,\,\,\,\,\,\frac{\partial F}{\partial r'}=\frac{r(z)r'(z)}{\sqrt{1-r'^2}}\,,\,\,\,\,\text{and}\,\,\,\,\,\,\frac{d}{dz}\left(\frac{\partial F}{\partial r'}\right)=\frac{1-r'^4+rr''}{(1+r'^2)^{3/2}}[/tex]​

    Therefore Euler's equation becomes,
    [tex]\frac{1+r'^2-rr''}{(1+r'^2)^{3/2}}=0.[/tex]​
    If we disallow the denominator from vanishing, we can simplify this:
    [tex]1+\left(\frac{dr}{dz}\right)^2-r(z)\frac{d^2r}{dz^2}=0.[/tex]​

    I have no clue how to approach this. Mathematica gave me some really wierd ugly answer, which when I asked it to FullSimplify, gave an unmanagable expression with hyperbolic sines and cosines. What's the solution to this differential equation, and how do I analytically apply the boundary conditions?
     
  2. jcsd
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