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Brave ODE

  1. Jan 25, 2006 #1

    Clausius2

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    I am looking for the general solutions of this equation in [tex]z(r)[/tex]
    If someone remembers well, this equation arises in surface tension physics.

    [tex]z(r)=\frac{1}{r}\frac{d}{dr}\left[\frac{z_r r}{(1+z_r^2)^{1/2}}\right][/tex]

    subject to the boundary conditions

    [tex]z_r(0)=z_{ro}[/tex] and
    [tex]z(\infty)=0[/tex]

    I only come up with rough approximations expanding the RHS around r=0, but I don't realise how might a closed solution be obtained.

    Any hints?

    Thanx.
     
  2. jcsd
  3. Feb 13, 2006 #2

    Astronuc

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    What is the relationship between z(r) and zr(r)?

    i.e. is the equation -
    [tex]z_r(r)=\frac{1}{r}\frac{d}{dr}\left[\frac{z_r r}{(1+z_r^2)^{1/2}}\right][/tex] ?
     
  4. Feb 17, 2006 #3

    Clausius2

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    Sorry Astro, [tex]z_r=dz/dr[/tex] as in usual notation. The original equation is the right one. I am trying to solve the equation of the surface of a thin film of water over an sphere. In fact if one tries the change of variable [tex]\phi=z_r/\sqrt{z_r^2+1}[/tex] the equation is reduced to [tex]\phi'+\phi/r=2\sqrt{1-\phi^2}[/tex], but again I don't find a way of how to solve this.
     
  5. Feb 18, 2006 #4

    Clausius2

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    My last change of variable is wrong, and the resulting equation too. I've just realised of that.
     
  6. Feb 19, 2006 #5

    arildno

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    Well, I thought struck me, I'm sure it's dumb:
    If you set [itex]z_{r}=Sinh(u(r))[/itex] and differentiate your equation, you get:
    [tex]Sinh(u)=\frac{d}{dr}\frac{1}{r}\frac{d}{dr}(rTanh(u))[/tex]
    Perhaps you can solve for u(r) now, but I have to admit I doubt it..
     
    Last edited: Feb 19, 2006
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