Q:

Laplace’s equation in the Polar Coordinate System

Laplace’s equation in the Polar Coordinate System:
$$\frac{\partial^2}{\partial{x^2}}+\frac{\partial}{\partial{y^2}}=\frac{\partial^2}{\partial{r^2}}+\frac{1}{r}\frac{\partial}{\partial{r}}+\frac{\partial^2}{\partial{\theta^2}}=\nabla^2$$
$$\nabla^2u=0 \Rightarrow u_{rr}+\frac{1}{r}u_r+u_{\theta \theta}=0$$
We suppose that we have a circular disc, that means 
$$0 \leq r <a$$
$$0 \leq \theta \leq 2 \pi$$
Boundary conditions: $u(r=a, \theta)=h(\theta), 0 \leq \theta \leq 2 \pi$
The boundary condition has to be continuous everywhere, so it should be $h(0)=h(2 \pi)$.
$u(r, \theta)=R(r)\Theta(\theta)$
Replacing this at the equation, we get the following problems:
$$\left\{\begin{matrix}
\Theta''+\lambda \Theta=0\\ 
\Theta(0)=\Theta(2\pi)
\end{matrix}\right.$$
$$r^2R''+rR'-\lambda R=0, 0 \leq r <a, (*)$$
Since the function $\Theta$ is periodic with period $2 \pi$ we have $k=n$
$$\Theta=A \cos{(k \theta)}+B \sin{(k \theta)}, k=n, n=0,1,2,\dots$$
Therefore $$\Theta_n=A_n \cos{(n \theta)} + B_n \sin{(n \theta)}, 0 \leq \theta \leq 2 \pi$$
$$\lambda_n =n^2$$
$$(*) \Rightarrow r^2R''+rR'-n^2R=0$$
$$$$
Could you explain if the following two sentences?
1. The boundary condition has to be continuous everywhere, so it should be $h(0)=h(2 \pi)$.
2. Since the function $\Theta$ is periodic with period $2 \pi$ we have $k=n$

A:

Both come from the same fact, that angle $2\pi$ radians and angle $0$ are the same.
EDIT:

The black dot is $r=a, \theta = 0$ and also $r = a, \theta = 2 \pi$.
In order for the boundary condition to be continuous, it has to have the
same limiting value as you approach a point from any direction.  As
you approach the black dot along the circle from above, $\theta \to 0$.
As you approach the black dot along the circle from below, $\theta \to 
2 \pi$.  

