I wish to solve Laplace's equation on an annulus such that the radius $r$ has bounds $1 < r < 2$ subjected to the boundary conditions $$u(1, \theta) = 1 \quad u(2,\theta) = \theta(2\pi - \theta).$$
My attempt:
Assume $u(r,\theta) = \Theta(\theta)R(r)$, then we have $$\Delta u = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}\\ \implies r^2 \frac{R''}{R} + r \frac{R'}{R} + \frac{\Theta''}{\Theta} = 0\\ \implies -r^2 \frac{R''}{R} - r \frac{R'}{R} = \frac{\Theta''}{\Theta} = -\lambda.$$ Letting $n= \lambda^2$ and first solving the eigenvalue problem for $\Theta(\theta)$ results in: $$\Theta(\theta) = c_0, \quad n= 0\\ \Theta(\theta) = c_1\cos(n\theta) + c_2\sin(n\theta), \quad n \in \mathbb{N}$$ where $c_0, c_1, c_2$ are constants.
Solving for $R(r)$ then results in: $$R(r) = c_3 + c_4\log(r), \quad n = 0\\ R(r) = c_5r^n + c_6r^{-n}, \quad n \in \mathbb{N}.$$
I am fairly confident about the above, but after this point is where I am doubting myself and some input would be very appreciated.
Using the boundary conditions first for $n = 0$ we have: $$R(1) = c_3 = 1\\ \implies R(r) = 1 + c_4 \log(r)\\ R(2) = 1 + c_4\log(2) = \theta(2\pi - \theta)\\ \implies c_4 = \frac{\theta(2\pi - \theta) - 1}{\log(2)}.$$ For $n \in \mathbb{N}$ we have $$R(1) = c_5 + c_6 = 1\\ R(2) = c_5 2^n + c_6 2^{-n} = \theta(2\pi - \theta).$$ Solving this system I obtain: $$c_5 = \frac{\theta(2\pi - \theta) - 2^{-n}}{2^n - 2^{-n}}, \quad c_6 = \frac{2^n - \theta(2\pi - \theta)}{2^n - 2^{-n}}.$$ Putting it all together: $$u(r, \theta) = c_0\Bigl(1 + \frac{\theta(2\pi - \theta) - 1}{\log(2)} \log(r)\Bigr) \\+ \sum_{n=1}^\infty \Bigl(c_1\cos(n\theta) + c_2\sin(n\theta)\Bigr)\Bigl(\frac{\theta(2\pi - \theta) - 2^{-n}}{2^n - 2^{-n}} 2^n + \frac{2^n - \theta(2\pi - \theta)}{2^n - 2^{-n}}\Bigr).$$
The problem is that the4 coefficients $c_1$ and $c_2$ are still not determined. I know this is usually done through Fourier series, which leads me to believe I used the boundary conditions too early. However, when I tried applying the boundary conditions after multiplying $R(r)\Theta(\theta)$ things were not in a clear Fourier series form. Furthermore I had 6 unknowns and only two boundary conditions.
Any advice/input would be much appreciated. Thank you.
Edit:
I have made a second attempt at the problem with a slightly different approach. This time I assemble the solution first using the forms I found for $R(r)$ and $\Theta(\theta)$. Thus, $$u(r, \theta) = R(r)\Theta(\theta)\\ = c_0\big(c_3 + c_4\log(r)\big) + \sum_{n=1}^\infty \big( c_1\cos(n\theta) + c_2\sin(n\theta) \big)\big(c_5r^n + c_6r^{-n}\big).$$ I then rearrange the above to get it in a full Fourier series type form and combine/redefine constants to get: $$u(r,\theta) = \frac{1}{2}\big(A_0 + B_0\log(r)\big) + \sum_{n=1}^\infty \big(A_nr^n + B_nr^{-n}\big)\cos(n\theta) + \big(C_nr^n + D_nr^{-n}\big)\sin(n\theta).$$
We may now use the formulas corresponding to a full Fourier series along with the boundary conditions. First: $$A_0 + B_0\log(1) = A_0 = \frac{1}{\pi}\int_0^{2\pi} d\theta = 2\\ A_0 + B_0\log(2) = \frac{1}{\pi}\int_0^{2\pi} \theta(2\pi-\theta)d\theta = \frac{4\pi^2}{3}\\ \implies A_0 = 2, \quad B_0 = \frac{4\pi^2-3}{3\log(2)}.$$
So using the above I have: $$u(r,\theta) = 1 + \frac{4\pi^2-3}{6\log(2)}\log(r) + \sum_{n=1}^\infty \big(A_nr^n + B_nr^{-n}\big)\cos(n\theta) + \big(C_nr^n + D_nr^{-n}\big)\sin(n\theta).$$
To determine the coefficients inside the summation I again use the Fourier coefficient formulas to obtain a linear system: $$A_n1^n + B_n1^{-n} = A_n + B_n = \frac{1}{\pi}\int_0^{2\pi} \cos(n\pi)d\theta = 0\\ A_n2^n + B_n2^{-n} = \frac{1}{\pi}\int_0^{2\pi} \theta(2\pi-\theta)\cos(n\pi)d\theta\\ \implies A_n = \frac{1}{(2^n - 2^{-n})\pi}\int_0^{2\pi} \theta(2\pi-\theta)\cos(n\pi)d\theta\\ \implies B_n = -\frac{1}{(2^n - 2^{-n})\pi}\int_0^{2\pi} \theta(2\pi-\theta)\cos(n\pi)d\theta$$.
Likewise, $$C_n1^n + D_n2^{-n} = C_n + D_n = \frac{1}{\pi}\int_0^{2\pi} \sin(n\pi)d\theta = 0\\ C_n2^n + D_n2^{-n} = \frac{1}{\pi}\int_0^{2\pi} \theta(2\pi-\theta)\sin(n\pi)d\theta\\ \implies C_n = \frac{1}{(2^n - 2^{-n})\pi}\int_0^{2\pi} \theta(2\pi-\theta)\sin(n\pi)d\theta\\ \implies D_n = -\frac{1}{(2^n - 2^{-n})\pi}\int_0^{2\pi} \theta(2\pi-\theta)\sin(n\pi)d\theta$$.
Thus we have that $$u(r,\theta) = 1 + \frac{4\pi^2-3}{6\log(2)}\log(r) + \sum_{n=1}^\infty \big(A_nr^n + B_nr^{-n}\big)\cos(n\theta) + \big(C_nr^n + D_nr^{-n}\big)\sin(n\theta)$$ where the coefficients are defined above.
Is there any mistakes in my solution or my approach?