constant $R=R(\gamma,a)$, hence the $R$-neighborhood of $\gamma$ is diffeomorphic to the product $S^1 \times (-1,1)$ (see for example [@CF]). If we take $\Omega$ as the Riemannian cylinder of width $R(\gamma,a)$ having one boundary component equal to $\gamma$ then we can foliate $\Omega$ with the level sets of the distance function to $\gamma$ and so $K=1$ and holds, with $L$ given by the length of the other boundary component.

Proof of Theorem \[main2\]: plane annuli {#convex}
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Let $\Omega$ be an annulus in ${{\bf R}^{2}}$, which is starlike with respect to its inner convex boundary component $\Sigma_1$. Assume that $A$ is a closed potential having flux $\Phi^A$ around $\Sigma_1$. Recall that we have to show: $$\label{annuliestimate}
\lambda_1(\Omega,A)\geq \dfrac{4\pi^2}{L^2} \dfrac{\beta m}{B} d(\Phi^A,{\bf Z})^2$$ where $\beta, B$ and $m$ will be recalled below and $L$ is the length of the outer boundary component. If we assume that $\Sigma_2$ is also convex, then we show that $m\geq \beta/B$ and the lower bound takes the form: $$\label{annuliestimatetwo}
\lambda_1(\Omega,A)\geq \dfrac{4\pi^2}{L^2} \dfrac{\beta^2}{B^2} d(\Phi^A,{\bf Z})^2.$$

Before giving the proof let us recall notation. For $x\in\Sigma_1$, the ray $\gamma_x$ is the geodesic segment $\gamma_x(t)=x+tN_x$, where $N_x$ is the exterior normal to $\Sigma_1$ at $x$ and $t\geq 0$. The ray 