We consider the equation $$\Delta u-q(x)u=0$$ in the circle 2D $B=\{|x| \leq 1\}$ with the boundary condition $u=1: \ |x|=1$, and the function $q$ is $$ q(x)= \begin{cases} \omega: \ |x| < \epsilon,\\ 0: \ |x| \geq \epsilon \end{cases} $$ where $\omega$ and $\epsilon$ are positive constants. In the case of radially symetric solution, the solution of this problem is: $$ u(r)= \begin{cases} v(r)= C \left(1+ \sum_{n=1}^{+\infty} \dfrac{(\sqrt{\omega} r)^{2n}}{\Pi_{i=1}^n (2i)^2}\right) &: r < \epsilon\\ w(r)= \dfrac{a-1}{\ln \epsilon} \ln r+1 &:r > \epsilon \end{cases} $$ where $a= \dfrac{\gamma}{\gamma -1}$ whith $ \gamma = \dfrac{1}{\varepsilon \ln \varepsilon} \dfrac{1+ \sum_{n=1}^{+\infty} \dfrac{ (\sqrt{\omega}\varepsilon)^{2n}}{\Pi_{i=1}^n (2i)^2}} { \sum_{n=1}^{\infty} 2n \dfrac{ (\sqrt{\omega} \varepsilon)^{2n-1}}{\Pi_{i=1}^n (2i)^2} } $
My question is: what's the singularities of the solution $u$ when $\omega \epsilon^2=1$? Thank you in advance for the help.