entirely contained in $\Omega$.

For $x\in\Sigma_1$, we denote by $\theta_x$ the angle between $\gamma'_x$ and the outer normal to $\Sigma_2$ at the point $Q(x)$, and we let $$m\doteq\min_{x\in\Sigma_1}{\cos\theta_x}.$$ Note that as $\Omega$ is starlike w.r.t. $\Sigma_1$, one has $\theta_x\in [0,\frac{\pi}2]$ and then $m\geq 0$.

[$\bullet\quad$]{}To have a positive lower bound, we will assume that $m>0$ (that is, $\Omega$ is [*strictly*]{} starlike w.r.t. $\Sigma_1$).

We also define $$\label{annulus}
\twosystem
{\beta=\min\{r(x): x\in\Sigma_1\}}
{B=\max\{r(x): x\in\Sigma_1\}}$$

We then have the following result.

\[main2\] Let $\Omega$ be an annulus in ${{\bf R}^{2}}$, which is strictly-starlike with respect to its inner (convex) boundary component $\Sigma_1$. Assume that $A$ is a closed potential having flux $\Phi^A$ around $\Sigma_1$. Then: $$\lambda_1(\Omega,A)\geq \dfrac{4\pi^2}{L^2} \dfrac{\beta m}{B} d(\Phi^A,{\bf Z})^2$$ where $\beta$ and $B$ are as in , and $L$ is the length of the outer boundary component. If $\Sigma_2$ is also convex, then $m\geq \beta/B$ and the lower bound takes the form: $$\lambda_1(\Omega,A)\geq \dfrac{4\pi^2}{L^2} \dfrac{\beta^2}{B^2} d(\Phi^A,{\bf Z})^2.$$

In section \[sharpness\], we will explain why we need to control $\dfrac{\beta}{B}$, $L$, and why we need to impose the starlike condition. If $\beta=B$ and $\Sigma_2$ is the circle of length $L$ we get the estimate $$\lambda_1(\Omega,A)\geq \dfrac{4\pi^2}{L^2} d(\Phi^A,{\bf Z})^2$$ which is the first 