I have to solve:
$$ \begin{cases} u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^2}u_{\theta \theta}=0 & [0,1) \times [-\pi,\pi] \\ u(1,\theta)=0 & (-\pi,0) & (1) \\ u(1,\theta)=1 & (0,\pi) & (2) \\ \end{cases} $$
And show that $u$ and $u_{\theta}$ are continuous in $[0, \frac{1}{2}] \times [-\pi,\pi]$. So, as u cannot be continuous in $[-\pi,\pi] \times [0,1]$ due to the discontinuity:
Given the solutions in separated variables $1,r^k(a_k\cos(k\theta)+b_k\sin(k\theta))$, I have tried to define the solution so that it is continuous inside the circle.
Usually I would compute the Fourier coefficients for 1 and 0. But this produces a discontinuity at $\theta=0$ as it gives me:
$$u(r,\theta)=0, \theta \in (-\pi,0]$$ $$ u(r, \theta)=1, \theta \in (0, \pi] $$
But In this case, as I am asked to show that $u$ and $u_{\theta}$ are continuous in $[0, \frac{1}{2}] \times [-\pi,\pi]$, I don't know how to proceed.
Maybe I could define it as $u(r,\theta)=1$ everywhere inside the circle and then as the Boundary conditions define?
Any help would be apreciated.
Proceed as usual. The Fourier coefficients of a piecewise defined functions are still computed as an integral: $$ a_k = \frac{1}{\pi}\int_{-\pi}^{\pi} u(1,\theta) \cos k\theta\,d\theta $$ It's just that to evaluate this, you'll have to split the integral into the sum of two: $$ a_k = \frac{1}{\pi}\int_{-\pi}^{0} u(1,\theta) \cos k\theta\,d\theta + \frac{1}{\pi}\int_{0}^{\pi} u(1,\theta) \cos k\theta\,d\theta =\frac{1}{\pi}\int_{0}^{\pi} 1 \cos k\theta\,d\theta $$ Notice how the cases translate into summation in the formula for coefficients. The coefficients themselves are not case-defined; indeed they are just numbers, independent of $\theta$. So, when you form the Fourier series with these coefficients, you have continuous terms.
Fourier series are never case-defined, even if the function is.
To prove continuity, you need uniform convergence on the specified set. The Weierstrass M-test will do the job: the supremum of the $k$th term is at most some constant times $(1/2)^k$.