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Two circles of unit radius, each normal to the line through their centers are a distance d apart. A soap film is formed between themas shown below; energetic considerations require the filem to assume a shape of minimum surface area.

a) show that for any value of d less than a critical value $d_{c}$, there are two surfaces satisfying the appropriate Euler-Lagrange equation, while for d > $d_{c}$ there is no surface. evaluate $d_{c}$ for a typical d < $d_{c}$, sketch the 2 surfaces. which has the lesser area?

b) show that in a certain range $d_{o}$ < d < $d_{c}$, the minimum surface as given by the Euler-Lagrange equation has an area larger than 2$\pi$, so hat the surface is "metastable"; that is, the surface is stable against small perturbations but not against arbitrary perturbation. evaluate $d_{o}$.

c) what happens when d is in creased beyond $d_{c}$?

Here is an drawing of the surface.

Here is what i have so far.

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  • $\begingroup$ ok i edited the question with what i have been able to do so far. $\endgroup$
    – Bigb
    Mar 29, 2013 at 23:29
  • $\begingroup$ Please make the effort to write your development as MathJax. Desciphering handwriting is enough of my job to not wanting to do it in my hobby here. $\endgroup$
    – vonbrand
    Mar 30, 2013 at 0:22
  • $\begingroup$ haha i understand i will type it up when i get a chance im about to have to get offline for the night. $\endgroup$
    – Bigb
    Mar 30, 2013 at 0:38

2 Answers 2

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The free energy of the film is to be minimized by surface area minimization. If we assume that d is along x-axis due to axial-symmetry minimal surface is a surface of revolution given by $$A[y]=2\pi \int_{x_1}^{x_2} y \sqrt{1+y'^2}\ dx$$ Correspending Euler Lagrange equation $$\frac{\partial g}{\partial y}-\frac{d}{dx}\bigg(\frac{\partial g}{\partial y'}\bigg)=0$$ where ommiting constants $$\frac{\partial g}{\partial y}=\sqrt{1+y'^2}$$ $$\frac{d}{dx}\bigg(\frac{\partial g}{\partial y'}\bigg)=\frac{d}{dx}\bigg(\frac{y\ y'}{\sqrt{1+y'^2}}\bigg)=\frac{y'^2}{\sqrt{1+y'^2}}+\frac{y''}{\sqrt{1+y'^2}}-\frac{y\ y'^2\ y''}{\big(\sqrt{1+y'^2}\big)^3}$$ If you replace the partials and collect the terms you have $$\frac{1}{\sqrt{1+y'^2}}-\frac{y\ y''}{\big(\sqrt{1+y'^2}\big)^3}=0$$ Further simplification by multiplying $y'$ and collecting terms $$\frac{y'}{\sqrt{1+y'^2}}-\frac{y\ y'\ y''}{\big(\sqrt{1+y'^2}\big)^3}=\frac{d}{dx}\bigg(\frac{y}{\sqrt{1+y'^2}}\bigg)=0$$ The minimization problem reduces to find a solution to a first-order differential equation solution. $$\frac{y}{\sqrt{1+y'^2}}=\alpha$$ which can be recast as $$\frac{dy}{dx}=\sqrt{\frac{y^2}{\alpha^2}-1}$$ and separation of variables $$\int dx=\int \frac{dy}{\sqrt{\frac{y^2}{\alpha^2}-1}}$$ Necessary substitution as $y=\alpha\ \cosh t$ and $$\int dx=\alpha \int dt\Rightarrow x+\beta=\alpha\ t$$ and by back-substitution $$y=\alpha \cosh \frac{x+\beta}{\alpha}$$

The boundary conditions must be integrated to the solution. To be able to use symmetry I assume origin lies in the middle of length d. The curve must pass through points $(\frac{-d}{2},1)$ and $(\frac{d}{2},1)$. Due to symmetry we can say that $\beta=0$. The following condition must be satisfied $$1=\alpha \cosh \frac{d}{2\alpha}$$ I guess from this point you can analyze required points.

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  • $\begingroup$ @vonbrand Thanks for correction $\endgroup$
    – AnilB
    Mar 30, 2013 at 23:23
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Here's a very helpful link which I believe answers this question:

http://ocw.mit.edu/courses/mathematics/18-086-mathematical-methods-for-engineers-ii-spring-2006/readings/am72.pdf

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  • $\begingroup$ yes, looks helpful $\endgroup$
    – Gerg
    Mar 31, 2013 at 1:47

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