I'm studying a problem connected to finding a minimal surface with a given curve as a boundary in the usual hyperbolic space $\mathbb{H}^5$ with metric


The boundary curve is $y=0$, $x_1^2+x_2^2=R^2$, $x_3=$ constant, $x_4=$ constant, so it is a circle in the $x_1-x_2$ plane of the conformal boundary of $\mathbb{H}^5$. The area functional

$$S= \int\limits_{surface} \sqrt{\det(g_{ind})}$$

can be simplified in this case. First, I use polar coordinates in the $x_1-x_2$ plane, so $dx_1^2+dx_2^2=dr^2+r^2d\varphi^2$, and take $\varphi$ and $r$ as parameters on the surface. From the symmetry of the problem I can argue that $y(r,\varphi)=y(r)$, and the other two coordinates are constants. Therefore, the induced metric is $$g_{ind} : \frac{1}{y^2}((1+y'^2)dr^2+r^2d\varphi^2)$$ and $$S= \int\int \sqrt{\frac{r^2(1+y'^2)}{y^4}}drd\varphi=2\pi\int \frac{r}{y^2}\sqrt{1+y'^2}dr \tag{1}$$

Now, I have considered another coordinate system $$ y=\frac{R}{\gamma} \ , \text{where} \ \gamma=\cosh u\cosh\zeta-\cos\theta\sinh u $$ $$r=\frac{R\cosh u \sinh \zeta}{\gamma}$$ $$\tilde{r}=\frac{R\sinh u \sin\theta}{\gamma}$$ where $\tilde{r}$ and $\tilde{\varphi}$ are polar coordinates in the $x_3-x_4$ plane. The hyperbolic space $\mathbb{H}^5$ now has the metric $$ds^2=du^2+\cosh^2 u(d\zeta^2+\sinh^2\zeta d\varphi^2)+\sinh^2u ds^2_{S^2} \ , \ ds^2_{S^2}=d\theta^2+\sin^2\theta d\tilde{\varphi}^2$$ I take now $\zeta$ and $\varphi$ as parameters on the surface. Again, I can argue that $u(\zeta,\varphi)=u(\zeta)$, and $\theta $ and $\tilde{\varphi}$ are constants. Therefore, the induced metric is $$g_{ind} : (u'^2+\cosh^2 u)d\zeta^2+\cosh^2 u\sinh^2\zeta d\varphi^2$$ and $$S=2\pi\int\sqrt{u'^2+\cosh^2 u}\cosh u\sinh\zeta d\zeta\tag{2}$$

Now, I want to relate these two functionals. Explicitly I want to get (2) starting from (1).

My general idea is to change $dr$ by $\frac{dr}{d\zeta} d\zeta$ so that $$\int \frac{r}{y^2}\sqrt{1+\left(\frac{dy}{dr}\right)^2}dr=\int \sqrt{\frac{r^2}{y^4}\left(1+\left(\frac{dy}{dr}\right)^2\right)}\frac{dr}{d\zeta} d\zeta$$ $$=\int \sqrt{\frac{r^2}{y^4}\left(\left(\frac{dr}{d\zeta}\right)^2+\left(\frac{dy}{dr}\right)^2\left(\frac{dr}{d\zeta}\right)^2\right)} d\zeta$$ Then I have to get the following $$\frac{r^2}{y^4}\left(\left(\frac{dr}{d\zeta}\right)^2+\left(\frac{dy}{dr}\right)^2\left(\frac{dr}{d\zeta}\right)^2\right)=\left(\left(\frac{du}{d\zeta}\right)^2+\cosh^2 u\right)\cosh^2 u\sinh^2\zeta $$ But I didn't succeed. I am having trouble in many steps, for example: Can I say that $$\frac{dy}{dr}\frac{dr}{d\zeta}=\frac{dy}{d\zeta} ?$$

when I calculated $\frac{dy}{d\zeta}=\frac{d}{d\zeta}(R/\cosh u(\zeta)\cosh\zeta-\cos\theta\sinh u(\zeta) )$ I got something proportional to $u'+$something. So then $(\frac{dy}{d\zeta})^2$ will have not only $u'^2$ but also $u'$.

Thanks in advance.


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