has: $
{\lvert{\nabla\psi}\rvert}^2=g^{11},
$ which immediately implies that we can take $$K_{\Omega,\psi}\leq \dfrac{\sup_{\Omega}g^{11}}{\inf_{\Omega}g^{11}}.$$ Note in particular that if $\Omega$ is rotationally invariant, so that the metric can be put in the form: $$g=\left(
    \begin{array}{cc}
    1 &  0 \\
    0 & \alpha(r)^2
     \end{array}
         \right),$$ for some function $\alpha(r)$, then $K_{\Omega,\psi}=1$. The estimate becomes $$\label{simple}
\lambda_1(\Omega,A)\geq\dfrac{4\pi^2}{L^2}\cdot d(\Phi^A,{\bf Z})^2,$$ where $L$ is the maximum length of a level curve $r={\rm const}$.

Yet more generally, one can fix a smooth closed curve $\gamma$ on a Riemannian surface $M$ and consider the tube of radius $R$ around $\gamma$: $$\Omega=\{x\in M: d(x,\gamma)\leq R\}.$$ It is well-known that if $R$ is sufficiently small (less than the injectivity radius of the normal exponential map) then $\Omega$ is a cylinder with smooth boundary which can be foliated by the level sets of $\psi$, the distance function to $\gamma$. Clearly ${\lvert{\nabla\psi}\rvert}=1$ and holds as well.

A concrete example where we could estimate the width $R$ is the case of a compact surface $M$ of genus $\ge 2$ and curvature $-a^2\le K\le -b^2$, $a \ge b >0$. Let $\gamma$ be a simple closed geodesic. Then, using the Gauss-Bonnet theorem, one can show that $R$ is bounded below by an explicit positive 