The edge r = a of a circular plate is kept at temperature f(theta). The plate is insulated so that there is no loss of heat from either surface. Find the temperature distribution in steady state.
I'm stuck at the beginning itself unable to figure of out what should be the boundary conditions. The only one that I am able to see is u(a,theta)=f(theta) what are the other conditions and how many do we need?
The 2-D heat equation is:
$$\frac{\partial T}{\partial t} - \alpha \Delta T = 0 $$
And for steady state we know that $\partial T / \partial t = 0$ which simplifies our equation to
$$\Delta T = 0$$
which in polar coordinates is
$$\frac{1}{r}\frac{\partial}{\partial r}\left(\frac{\partial T}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2T}{\partial\theta^2} = 0 $$
We need boundary conditions: the obvious is our temperature constraint on the edge:
$$T(a,\theta) = f(\theta)$$
However, we need more boundary conditions. This is the tricky part. We know that $T$ must be a cyclic function because $\theta$ is bounded between $-\pi$ and $\pi$. Thus, We need the heat flow and temperature to be the same at the start of the cycle as it is at the end of the cycle. So we make the restrictions that $T_\theta(r,-\pi)=T_\theta(r,\pi)$ and $T(r,-\pi)=T(r,\pi)$. Or more particularly, that $\Theta'(-\pi)=\Theta'(\pi)$ and $\Theta(-\pi)=\Theta(\pi)$ for $T(r,\theta)=\Theta(\theta)R(r)$. The rest should work the way any other separation of variable PDE is solved. Let me know if you are still stuck after this.