How to simplify a PDE sistem with Gauss Egregium Theorem?

Question

I have two parametrizations of two surfaces with the same Gaussian Curvature. $$X(u, v)=(u \cos (v), u \sin (v), v) \\ Y(r, s)=(r \cos(s), r \sin(s), \log(r))$$

Nevertheless, they are not isometric. I got this PDEs system: $$\begin{matrix} \left(\frac{1+u^2}{u^2}\right){r_{u}}^2+u^2{s_ u}^2 & = & 1 \\ \left(\frac{1+u^2}{u^2}\right)r_{u}r_v+u^2s_us_ v & = & 0 \\ \left(\frac{1+u^2}{u^2}\right){r_{v}}^2+u^2{s_v}^2 & = & 1+u^2 \end{matrix} $$

I'm pretty sure that Egregium theorem allows me to deduce that $r_v=0$. But I don´t know how to get this equation.

Answer 1

We can see from the first fundamental form that both parametrizations are conformal coordinates. So it is easy to write the metric for their tangent planes as Pfaff forms. We need now the sistem: $$\begin{matrix} Q & = & \omega_1^2+\omega_2^2 \\ d\omega_1 & = & \omega_1 \wedge \omega_2 \\ d\omega_2 & = & \omega_3 \wedge \omega_1 \\ d\omega_3 & = & K(\omega_1 \wedge \omega_2) \end{matrix}$$ Where $K$ is the gaussian curvature. Doing both systems we get that $K_X=-(u^2+1)^{-2}=K_Y=-(r^2+1)^{-2}$. Therefore $r(u,v)=u$. With this information is very easy to get an easier system: $$\begin{matrix} 1+u^2{s_u}^{2}+\frac{1}{u^2} & = & 1 \\ u^2 s_u s_v & = & 0 \\ u^2{s_v}^2 & = & u^2+1 \end{matrix}$$ The first and the second equations give us the contradiction we need to prove that there is no isometry.