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Subject: Re: Problem solving bvp of Meniscus
Date: Fri, 28 Oct 2011 18:06:13 +0000 (UTC)
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Ataías Pereira Reis <ataiasreis@gmail.com> wrote in message <1207681.241.1319819354868.JavaMail.geo-discussion-forums@yqgn17>...
> I've been trying to solve the following equation:
> y''/(1+y'^2)^(3/2) = y + a
> 
> and a is a constant value that should be determined after solving the equation.
> 
> That's a boundary value problem, with the conditions:
> 
> z(infinity) = 0, z'(infinity) = 0 and z'(0) = -cotg(theta)
> .........
 - - - - - - - - - -
  If by z(infinity) = 0, z'(infinity) = 0 you mean that these conditions apply to y = z, then you would also need y"(infinity) = 0.  Any other value would make y'(infinity) = 0 impossible.  This in turn implies that the quantity 'a' must be zero to satisfy your original differential equation.   Hence you have just the equation

  y" = y*(1+y'^2)^(3/2)

  This is not really a boundary value problem if you do the following.  Define

 v = y' = dy/dt

Then

 y" = dv/dt = (dv/dy)*(dy/dt) = dv/dy * v = 1/2*dv^2/dy

which gives you the equation

 1/2*dv^2/dy = y*(1+v^2)^(3/2)

or written as differentials

 1/2*(1+v^2)^(-3/2)*dv^2 = y*dy

Integrating this gives

 -(1+v^2)^(-1/2) = y^2/2 + C

where C is the constant of integration.  The condition y(infinity) = 0 and v(infinity) = 0 then implies that this constant must be C = -1, giving

 y'^2 = v^2 = 1/(1-y^2/2)^2 - 1 = (y^2-y^4/4)/(1-y^2/2)^2

or

 y' = (+ or -) sqrt((y^2-y^4/4)/(1-y^2/2)^2)

This is then the equation you hand to one of matlab's ode functions.

  The only additional condition you need now is y(0) or y'(0) and the sign will immediately determine the sign needed in the equation for y'.

Roger Stafford