The problem is the following:
Let $B_1 \subset \mathbb{R}^2$ be the open unit disk. Solve $$\begin{cases} \Delta u = f \, &x \in B_1\\ u = 1 \, &x \in \partial B_1 \end{cases}$$
I'm kind of lost with this problem. First I tried to find the solution for the homogenous problem ($f \equiv 0$) using polar coordinates. The Laplacian in the case is written as $$\Delta (r, \theta) = u_{rr}+\frac{1}{r}u_r + \frac{1}{r^2}u_{\theta \theta}$$ So I tried to look for solutions of the form $u(r, \theta) = R(r) \Theta (\theta)$. Substituting in the equation $\Delta u = 0$ gives $$\frac{r^2 R''(r)+rR'(r)}{R(r)} = -\frac{\Theta(\theta)}{\Theta(\theta)} = \lambda$$ where $\lambda$ is a constant. The problem arises when trying the different values of $\lambda$ ($=, <$ or $>0$) since the boundary conditions are not "usual". Anyway here is my try for some cases.
If $\lambda = 0$, then $\Theta(\theta) = 0$ so $\Theta(\theta) = A_1\theta + A_2$. Now since in the boundary $\partial B_1$ the function is equal to $1$, then $\Theta(0) = A_2 = 1$ so $\Theta(2 \pi) =2 A_1 \pi + 1 = 1$, so $A_1 = 0$ and we get the solution $\Theta(\theta) = constant$. Also the radial part when $\lambda = 0$ reads $r^2R''(r)+r'R(r) = 0$ so the solution is $$R(r) = B_1 \log (r) + B_2$$ and since in the boundary this must be equal to $1$ I get $B_2 = 1$, so $R(r) = B_1 \log(r) + 1$. Plugging then $R(r)\Theta(\theta)$ in the equation $\Delta u(r, \theta) = 0$ the only possible value of $B_1$ is $0$, so the solution when $\lambda = 0$ is $u \equiv 1$.
When I do the computations for the cases $\lambda > 0$ and $\lambda <0$ the solutions for $\Theta$ are respectively $$\Theta(\theta) = C_1\cos(\mu \theta) + C_2 \sin (\mu \theta)$$ and $$\Theta(\theta) = D_1e^{\mu\theta} + D_2e^{-\mu \theta}$$ where $\lambda = \mu^2$ in the first case and $\mu = -\lambda^2$ in the second.
I'd appreciate son help with this cases or if there is another shorter path that doesn't involve this series of calculations. Thank you.