Boundary conditions for the Heat Equation

[Leonardo Machado]

Hello guys.

I am studying the heat equation in polar coordinates

$$
u_t=k(u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta})
$$

via separation of variables.
$$u(r,\theta,t)=T(t)R(r)\Theta(\theta)$$

which gives the ODEs

$$T''+k \lambda^2 T=0$$
$$r^2R''+rR+(\lambda^2 r^2-\mu^2)R=0$$
$$\Theta''+\mu^2\Theta=0$$

but i can't properly think about the boundary conditions to this problem. I see every where people resolving it with

$$
|u(0,\theta,t)|<\inf \mapsto |R(0)| < \inf
$$

and

$$u(r*,\theta,t)=0 \mapsto R(r*)=0$$

being r* the border of the disc.

But i understand the radial condition as a termal bath at zero temperature and i really want to change it for a finite value but i don't know how to procede without the zeros...

Any suggestions?

 

[Orodruin]

The boundary conditions depend on what physical situation you wish to describe. Can you be more specific about this?

 

[Chestermiller]

The boundary condition at r = 0 should be zero radial temperature gradient.

 

[Dr Transport]

Time equation should be
$$ T' + k\lambda^2T = 0$$

 

[Orodruin]

The boundary condition at r = 0 should be zero radial temperature gradient.

This is correct only in the case of rotational symmetry of the problem. The general solution will contain contributions from $J_1$, which has non-zero derivative at $r = 0$.

 

[hunt_mat]

You should have the inner boundary condition:
$$\frac{\partial u}{\partial r}\Bigg|_{r=0}=0$$
This is the proper symmetry condition. The outer boundary condition is physics dependent however and can be absolutely anything.

 

[Leonardo Machado]

I have solved it guys!

To operate with inhomogeneous bondary conditions I've used

$$
u(r,\theta,t)=v(r,\theta,t)+u_E(r,\theta)
$$

being u_E the steady state and "v" the solution of the heat equation.