# Boundary conditions for the Heat Equation

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 cant 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 dont know how to procede without the zeros...

Any suggestions?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
The boundary conditions depend on what physical situation you wish to describe. Can you be more specific about this?

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

Dr Transport
Gold Member
Time equation should be
$$T' + k\lambda^2T = 0$$

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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
Homework Helper
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.

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.