[\sqrt{\lambda_-} + \kappa,\sqrt{\lambda_+} -\kappa] \times [0,\kappa^{-1}] $, we have $ {\text{\rm Im}\hspace{0.1cm}}(z_t-z_0) \sim t $.

![Shape of the characteristics near the edge: the solid lines represent the density $ \rho(x) $ and the corresponding characteristics; the dashed lines represent the semicircle distribution and the related characteristics.](characteristics.jpg){height="4cm"}

It is too complicated to work with the ODE satisfied by $ z_t $ in a direct way. The main idea is to compare this characteristics with the corresponding curve for a semicircle distribution (which has a explicit and simple formula). Define the two functions $$g(z):=\dfrac{(1-\xi) + \sqrt{(z^2 - \lambda_-)(z^2 - \lambda_+)}}{2\xi z} , \ \ \ g_{sc}(z):=\dfrac{\sqrt{(z-1)^2 - \xi}}{\xi}.$$ For some $ |z_0-\sqrt{\lambda_+}| < \sqrt{\xi}/10,\ {\text{\rm Im}\hspace{0.1cm}}z_0>0 $, let $ z(t) := z_t $ and $ z_{sc}(t) $ solve the following two initial value problems $$\left\{
\begin{aligned}
& \dfrac{{\mathrm{d}}z}{{\mathrm{d}}t} = g(z)\\
& z(0)=z_0
\end{aligned}
\right.
,\ \ \ \ \ \ \
\left\{
\begin{aligned}
& \dfrac{{\mathrm{d}}z_{sc}}{{\mathrm{d}}t} = g_{sc}(z_{sc})\\
& z_{sc}(0)=z_0
\end{aligned}
\right.
.$$ Note that in $ \Omega=\{z_{sc}(t):0<t<1,|z_0-\sqrt{\lambda_+}| < \sqrt{\xi}/10,{\text{\rm Im}\hspace{0.1cm}}z_0>0 \} $ we have $ g(z) \sim g_{sc}(z) $. This shows that for $ 0<t<1 $, we have $ (z_t-z_0) \sim (z_{sc}(t) -z_0) $. The rest of the proof now follows from 