So we have an half disk as in the figure with boundaries $\Gamma_1$ and $\Gamma_2$.
The variable of interest is $w(\rho,\phi)$ in polar coordinates (the temperature for the heat equation, for example) and must obey to:
$$\Delta_{\rho,\phi} w = 0 $$
with BC: $w(R,\phi) = w_0$ (Dirichlet BC on $\Gamma_1$) and $\boldsymbol{e_y} . \boldsymbol{\nabla} w \vert_{\phi = m \pi} = \frac{\cos \phi}{\rho} \partial_{\phi} w (\rho, {\phi = m \pi}) = f_0$ (Neumann BC on $\Gamma_2$), with $m = 0, 1$.
Because the problem is symmetric with respect to y axis and definite in $\rho = 0$ we have the general solution:
$$ w = A + \sum_{n=1} b_n \sin(n \phi) \rho^n $$
Let's apply the first BC, we obtain:
$$ w_0 = A + \sum_{n=1} b_n \sin(n \phi) R^n $$
Applying scalar product by $\int_{0}^{\pi} \sin(n \phi) d \phi$ we obtain $b_n$ in function of $A$:
$$ b_n = \frac{2 (w_0-A)}{n \pi R^n} \left[ 1 - (-1)^n \right] $$
Fine but we need $A$ now, let's apply the second BC:
$$ \sum_{n=1} n b_n \cos(n m \pi) \cos(m \pi) \rho^{n-1} = \sum_{n=1} n b_n (-1)^{m(n+1)} \rho^{n-1} = f_0 $$
How to reduce this last equation to find $A$ ?
The only idea I had was to scalar product by $\delta(\rho)$ (this certainly leads to an "incomplete" solution), delta function, that gave me:
$$b_1 = f_0$$
Then using the expression of $b_n$ I found $A = w_0 - \frac{\pi R f_0}{4}$, but I tried to plot it on Matlab but the partial derivative condition on flat surface $\Gamma_2$ is not correct.