$w_1(r),w_2(r)$ such that $$u(t)=e^{i\phi(t)}\Big(w_1(r)e^{\frac{\pi i}{L}  s(r,t)}+w_2(r)e^{-\frac{\pi i}{L}  s(r,t)}\Big),$$ which, setting $f(t)=e^{-i\phi(t)}u(t)$, we can re-write $$\label{rewrite}
f(t)=w_1(r)e^{\frac{\pi i}{L}  s(r,t)}+w_2(r)e^{-\frac{\pi i}{L} s(r,t)}.$$ Recall that here $\phi(t)=\int_0^tH(\tau)\,d\tau$ and $$s(r,t)=\int_0^t\theta(r,\tau)\,d\tau.$$ We take the real part on both sides of and obtain smooth real-valued functions $F(t), p(r),q(r)$ such that $$F(t)=p(r)\cos({\frac{\pi}{L}}s(r,t))+q(r)\sin(\frac{\pi}{L} s(r,t)).$$ Since $\theta(0,t)=1$ for all $t$, we see $$s(0,t)=t.$$ Clearly $s(r,0)=0$; finally, $s(r,L)=\int_0^L\theta(r,\tau)\,d\tau=L$, being the length of the level circle $\Sigma_r$. Thus, we can apply Lemma \[calculus\] and conclude that $s(r,t)=t$ for all $t$, that is, $$\theta(r,t)=1$$ for all $(r,t)$ and the metric is a Riemannian product.

It might happen that $p(r)=q(r)\equiv 0$. But then the real part of $f(t)$ is zero and we can work in an analogous way with the imaginary part of $f(t)$, which cannot vanish unless $u\equiv 0$.

General estimate of $K_{\Omega,\psi}$ {#estimate K}
-------------------------------------

We can estimate $K_{\Omega,\psi}$ for a Riemannian cylinder $\Omega=[0,a]\times{{\bf S}^{1}}$ if we know the explicit expression of the metric in the normal coordinates $(r,t)$, where $t\in [0,2\pi]$ is arc-length : $$g=
\left(
    \begin{array}{cc}
    g_{11} &  g_{12}  \\
     g_{21} & g_{22}
     \end{array}
         \right).$$ If $g^{ij}$ is the inverse matrix of $g_{ij}$, and if $\psi=\psi(r,t)$ one has: $${\lvert{\nabla\psi}\rvert}^2=g^{11}\Big({\dfrac{{\partial}\psi}{\bdr}}\Big)^2+2g^{12}{\dfrac{{\partial}\psi}{\bdr}}{\dfrac{{\partial}\psi}{\bdt}}+
g^{22}\Big({\dfrac{{\partial}\psi}{\bdt}}\Big)^2.$$ The function $\psi(r,t)=r$ belongs to ${\cal F}_{\Omega}$ and one 