# Brave ODE

1. Jan 25, 2006

### Clausius2

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

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

subject to the boundary conditions

$$z_r(0)=z_{ro}$$ and
$$z(\infty)=0$$

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. Feb 13, 2006

### Staff: Mentor

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

i.e. is the equation -
$$z_r(r)=\frac{1}{r}\frac{d}{dr}\left[\frac{z_r r}{(1+z_r^2)^{1/2}}\right]$$ ?

3. Feb 17, 2006

### Clausius2

Sorry Astro, $$z_r=dz/dr$$ 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 $$\phi=z_r/\sqrt{z_r^2+1}$$ the equation is reduced to $$\phi'+\phi/r=2\sqrt{1-\phi^2}$$, but again I don't find a way of how to solve this.

4. Feb 18, 2006

### Clausius2

My last change of variable is wrong, and the resulting equation too. I've just realised of that.

5. Feb 19, 2006

### arildno

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

Last edited: Feb 19, 2006