Solving z(r) Equation with Boundary Conditions

[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 realize how might a closed solution be obtained.

Any hints?

Thanx.

 

[Astronuc]

What is the relationship between z(r) and z_r(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] ?

 

[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 realize how might a closed solution be obtained.

Any hints?

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.

 

[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.

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

 

[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..