Solve
\begin{eqnarray} \left \{\begin {array}{lll} \Delta u(\rho,\phi)=\ln \rho+2\cos^2\phi&,~ 1<r<3,~0\leqslant \phi \leqslant \pi,~0\leqslant \phi < 2\pi\\ \displaystyle u_{\rho}(1,\phi)=0,~~u(e,\phi)=\cos(2\phi)&,~0\leqslant \phi< 2\pi\\ \end{array} \right.. \end{eqnarray}
Attempt. Writing $\ln \rho+2\cos^2\phi=\ln \rho+1+\cos(2\phi)$ we seek one solution of the form $w_1(\rho)$ such that $w_1''+\frac{1}{\rho}\,w_1'=\ln \rho+1$ and one solution the form $w_2(\rho,\phi)$ such that $\Delta w_2=\cos(2\phi)$. Then $v=u-w_1-w_2$ solves the Laplace equation:
\begin{eqnarray} \left \{\begin {array}{lll} \Delta v(\rho,\phi)= 0&,~ 1<r<3,~0\leqslant \phi \leqslant \pi,~0\leqslant \phi < 2\pi\\ \displaystyle v_{\rho}(1,\phi)=-w_1'(1)-(w_2)_{\rho}(1,\phi),\\ v(e,\phi)=\cos(2\phi)-w_1(e)-w_2(e,\phi)&,~0\leqslant \phi< 2\pi\\ \end{array} \right.. \end{eqnarray}
The problem of $w_1$ is relatively easy: one solution is $\rho^2\ln \rho/4$. For the other part of $w_2$, i can't seem to find any obvious solution (candidate $r^2\cos(2\phi)$ solve solves the homogenuous problem, of course).
Thank you in advance!