La parametrización de Weierstrass-Enneper es una parametrización que hace uso del análisis complejo en la geometría diferencial para encontrar superficies mínimas en coordenadas isotermas respectivamente.
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| La Superficie de Enneper es un ejemplo de superficie mínima isoterma |
Por consiguiente partiendo del análisis complejo ${\large z = u + iv }$
$${\large \begin{align*} u &= \frac{z + \bar{z}}{2} \hspace{0.5cm}&\hspace{0.5cm} v &= \frac{z - \bar{z}}{2i} \\ \frac{\partial}{\partial z} &= \frac{1}{2}\left(\frac{\partial}{\partial u} - i\frac{\partial}{\partial v} \right) \hspace{0.5cm}&\hspace{0.5cm} \frac{\partial}{\partial \bar{z}} &= \frac{1}{2}\left(\frac{\partial}{\partial u} + i\frac{\partial}{\partial v} \right) \end{align*}}$$
Donde u es la parte real y v es la parte imaginaria. Ahora si usamos el análisis complejo para estudiar una superficie en ${\large \mathbb{R}^3 }$
$${\large \begin{align*} S(u,v) &= \left(r_{1}(u,v), r_{2}(u,v), r_{3}(u,v) \right) \\ S(z,\bar{z}) &= \left(r_{1}(z,\bar{z}), r_{2}(z,\bar{z}), r_{3}(z,\bar{z}) \right) \\ S(z,\bar{z}) &= r_{j}(z,\bar{z}) \end{align*} }$$
Las derivadas parciales de la superficie ${\large S(z,\bar{z}) }$ son:
$${\large \begin{align*}\frac{\partial S}{\partial z} &= \frac{\partial r_{j}}{\partial z} = \frac{1}{2}\left(\frac{\partial r_{j}}{\partial u} - i\frac{\partial r_{j}}{\partial v} \right) \\ \frac{\partial S}{\partial \bar{z}} &= \frac{\partial r_{j}}{\partial \bar{z}} = \frac{1}{2}\left(\frac{\partial r_{j}}{\partial u} + i\frac{\partial r_{j}}{\partial v} \right) \end{align*} }$$
El modulo de las derivadas parciales son:
$${\large \begin{align*} \left\| \frac{\partial S}{\partial z} \right\|^2 &= \frac{1}{4}\left( \frac{\partial r_{j}}{\partial u} \right)^2 - \frac{1}{2}\left(\frac{\partial r_{j}}{\partial u} \frac{\partial r_{j}}{\partial v}\right) - \frac{1}{4}\left( \frac{\partial r_{j}}{\partial v} \right)^2 \\ \left\| \frac{\partial S}{\partial \bar{z}} \right\|^2 &= \frac{1}{4}\left( \frac{\partial r_{j}}{\partial u} \right)^2 - \frac{1}{2}\left(\frac{\partial r_{j}}{\partial u} \frac{\partial r_{j}}{\partial v}\right) + \frac{1}{4}\left( \frac{\partial r_{j}}{\partial v} \right)^2 \end{align*} }$$
Si la superficie ${\large S(z,\bar{z}) }$ es una superficie isoterma entonces ${\large E = G}$ y ${\large F = 0}$
$${\large \begin{align*} E &= \frac{\partial r_{j}}{\partial u} \\ F &= \frac{\partial r_{j}}{\partial u} \frac{\partial r_{j}}{\partial v} \\ G &= \frac{\partial r_{j}}{\partial v} \end{align*} }$$
Por consiguiente:
$${\large \begin{align*} \left\| \frac{\partial S}{\partial z} \right\| &= 0 \hspace{0.5cm}&\hspace{0.5cm} \frac{\partial S}{\partial z} &= \phi_{j} \\ \left\| \frac{\partial S}{\partial \bar{z}} \right\| &= 0 \hspace{0.5cm}&\hspace{0.5cm} \frac{\partial S}{\partial \bar{z}} &= \bar{\phi}_{j} \end{align*} }$$
Ahora realizando la derivada parcial de segundo orden para encontrar la relación con la función armónica ${\large S(u,v) }$
$${\large \begin{align*} \frac{\partial }{\partial \bar{z}}\left( \frac{\partial S}{\partial z} \right) &= \frac{\partial}{\partial \bar{z}}\left( \frac{1}{2}\left(\frac{\partial r_{j}}{\partial u} - i\frac{\partial r_{j}}{\partial v} \right) \right) \\ \frac{\partial }{\partial \bar{z}}\left( \frac{\partial S}{\partial z} \right) &= \frac{1}{2}\left( \frac{\partial }{\partial u}\left( \frac{1}{2}\frac{\partial S}{\partial u} - \frac{1}{2}\frac{\partial S}{\partial v}i\right) + i\frac{\partial }{\partial v}\left( \frac{1}{2}\frac{\partial S}{\partial u} - \frac{1}{2}\frac{\partial S}{\partial v}i \right) \right) \\ \frac{\partial }{\partial \bar{z}}\left( \frac{\partial S}{\partial z} \right) &= \frac{1}{4}\left( \frac{\partial^2 S}{\partial u^2} + \frac{\partial^2 S}{\partial v^2} \right) \\ \frac{\partial }{\partial \bar{z}}\left( \frac{\partial S}{\partial z} \right) &= \frac{1}{4} \nabla^2 S(u,v) \\ \\ \frac{\partial }{\partial z}\left( \frac{\partial S}{\partial \bar{z}} \right) &= \frac{\partial}{\partial z}\left( \frac{1}{2}\left(\frac{\partial r_{j}}{\partial u} + i\frac{\partial r_{j}}{\partial v} \right) \right) \\ \frac{\partial }{\partial \bar{z}}\left( \frac{\partial S}{\partial z} \right) &= \frac{1}{2}\left( \frac{\partial }{\partial u}\left( \frac{1}{2}\frac{\partial S}{\partial u} + \frac{1}{2}\frac{\partial S}{\partial v}i\right) - i\frac{\partial }{\partial v}\left( \frac{1}{2}\frac{\partial S}{\partial u} + \frac{1}{2}\frac{\partial S}{\partial v}i \right) \right) \\ \frac{\partial }{\partial \bar{z}}\left( \frac{\partial S}{\partial z} \right) &= \frac{1}{4}\left( \frac{\partial^2 S}{\partial u^2} + \frac{\partial^2 S}{\partial v^2} \right) \\ \frac{\partial }{\partial \bar{z}}\left( \frac{\partial S}{\partial z} \right) &= \frac{1}{4} \nabla^2 S(u,v) \end{align*} }$$
Ahora encontraremos los diferenciales complejos.
$${\large \begin{align*} z &= u + iv \hspace{0.5cm}&\hspace{0.5cm} \bar{z} &= u - iv \\ \partial z &= \partial u + i\partial v \hspace{0.5cm}&\hspace{0.5cm} \partial \bar{z} &= \partial u - i\partial v \end{align*} }$$
Ahora procederemos hallar el diferencial de ${\large r_{j}}$
$${\large \begin{align*} \partial r_{j} &= \phi_{j} \partial z + \bar{\phi}_{j} \partial \bar{z} \\ \partial r_{j} &= \frac{1}{2}\left(\frac{\partial r_{j}}{\partial u} - i\frac{\partial r_{j}}{\partial v} \right) (\partial u + i\partial v) + \frac{1}{2}\left(\frac{\partial r_{j}}{\partial u} + i\frac{\partial r_{j}}{\partial v} \right) (\partial u - i\partial v) \\ \partial r_{j} &= \frac{\partial r_{j}}{\partial u}\partial u + \frac{\partial r_{j}}{\partial v}\partial v \\ \partial r_{j} &= 2 \left( \frac{1}{2} \left( \frac{\partial r_{j}}{\partial u} - i\frac{\partial r_{j}}{\partial v}\right) \right) \cdot (du + idv) \\ \partial r_{j} &= 2\phi_{j} dz \end{align*} }$$
Tomando la parte real de la función compleja
$${\large \begin{align*} \partial r_{j} &= 2\phi_{j} dz \\ \partial r_{j} &= 2Re\left(\phi_{j} dz\right) \\ r_{j} &= 2 Re \left( \int \phi_{j} dz\right) + C_{j} \end{align*} }$$
Partiendo del modulo de la derivada parcial de la superficie S
$${\large \begin{align*} 0 &= \left\| \frac{\partial S}{\partial z} \right\|^2 \\ 0 &= {\phi_{j}}^2 \\ 0 &= {\phi_{1}}^2 + {\phi_{2}}^2 + {\phi_{3}}^2 \\ - {\phi_{3}}^2 &= {\phi_{1}}^2 + {\phi_{2}}^2 \\ - {\phi_{3}}^2 &= (\phi_{1} + i\phi_{2})(\phi_{1} - i\phi_{2}) \end{align*} }$$
Si definimos las siguientes funciones
$${\large \begin{equation*} f = \phi_{1} - i\phi_{2} \hspace{1cm},\hspace{1cm} g = \frac{\phi_{3}}{\phi_{1} - i\phi_{2}} \end{equation*}}$$
Las componentes de ${\large \phi_{j} }$ quedan de la siguiente forma:
$${\large \begin{align*} \phi_{1} &= f(1-g^2)/2 \\ \phi_{2} &= if(1+g^2)/2 \\ \phi_{3} &= fg \end{align*} }$$
Superficies Mínimas en Coordenadas Isotermas
Una superficie mínima es una superficie con curvatura media igual a cero. En el caso de una superficie con coordenadas isotermas, es decir, ${\large E = G}$ y ${\large F = 0}$. Por lo tanto la superficie es armónica con respecto a las variables u y v.
- Superficie de Enneper
La parametrización de Weierstrass - Enneper para la superficie de Enneper es deducida a partir de las funciones complejas ${\large f(z) = 1}$ y ${\large g(z) = z}$
$${\large \begin{align*} r_{1} &= Re\left(z - \frac{z^3}{3} \right) \\ r_{2} &= Re\left(zi - \frac{z^3 i}{3} \right) \\ r_{3} &= Re\left( z^2 \right) \end{align*}}$$
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| ${\large S(u,v) = \left( u - (1/3)u^3 +uv^2, \hspace{0.2cm} (1/3)v^{3} - vu^2 -v, \hspace{0.2cm} u^2 - v^2 \right)} $ |
- Catenoide
La parametrización de Weierstrass - Enneper para el Catenoide es deducida a partir de las funciones complejas ${\large f(z) = e^{z/c}}$ y ${\large g(z) = e^{-z/c}}$
$${\large \begin{align*} r_{1} &= 2cRe\left( \cosh(z/c) \right) \\ r_{2} &= 2cRe\left( i\sinh(z/c) \right) \\ r_{3} &= 2Re\left( z \right) \end{align*}}$$
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| ${\large S(u,v) = (2c\cosh(u/c)\cos(u/c), \hspace{0.2cm} 2c\cosh(u/c)\sin(u/c), \hspace{0.2cm} 2u)}$ |
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