#LyX 2.3 created this file. For more info see http://www.lyx.org/ \lyxformat 544 \begin_document \begin_header \save_transient_properties true \origin unavailable \textclass book \begin_preamble \input{../defs} \end_preamble \use_default_options true \maintain_unincluded_children false \language spanish \language_package default \inputencoding auto \fontencoding global \font_roman "default" "default" \font_sans "default" "default" \font_typewriter "default" "default" \font_math "auto" "auto" \font_default_family default \use_non_tex_fonts false \font_sc false \font_osf false \font_sf_scale 100 100 \font_tt_scale 100 100 \use_microtype false \use_dash_ligatures true \graphics default \default_output_format default \output_sync 0 \bibtex_command default \index_command default \paperfontsize default \spacing single \use_hyperref false \papersize default \use_geometry false \use_package amsmath 1 \use_package amssymb 1 \use_package cancel 1 \use_package esint 1 \use_package mathdots 1 \use_package mathtools 1 \use_package mhchem 1 \use_package stackrel 1 \use_package stmaryrd 1 \use_package undertilde 1 \cite_engine basic \cite_engine_type default \biblio_style plain \use_bibtopic false \use_indices false \paperorientation portrait \suppress_date false \justification true \use_refstyle 1 \use_minted 0 \index Index \shortcut idx \color #008000 \end_index \secnumdepth 3 \tocdepth 3 \paragraph_separation indent \paragraph_indentation default \is_math_indent 0 \math_numbering_side default \quotes_style french \dynamic_quotes 0 \papercolumns 1 \papersides 1 \paperpagestyle default \tracking_changes false \output_changes false \html_math_output 0 \html_css_as_file 0 \html_be_strict false \end_header \begin_body \begin_layout Section Teorema de Liouville \end_layout \begin_layout Standard Sean \begin_inset Formula $f,g:I\to\mathbb{R}$ \end_inset diferenciables con \begin_inset Formula $f^{2}+g^{2}\equiv1$ \end_inset , \begin_inset Formula $t_{0}\in I$ \end_inset y \begin_inset Formula $\theta_{0}\in\mathbb{R}$ \end_inset con \begin_inset Formula $f(t_{0})=\cos\theta_{0}$ \end_inset y \begin_inset Formula $g(t_{0})=\sin\theta_{0}$ \end_inset , entonces existe una única función diferenciable \begin_inset Formula $\theta:I\to\mathbb{R}$ \end_inset con \begin_inset Formula $\theta(t_{0})=\theta_{0}$ \end_inset y, para todo \begin_inset Formula $t\in I$ \end_inset , \begin_inset Formula $f(t)=\cos\theta(t)$ \end_inset y \begin_inset Formula $g(t)=\sin\theta(t)$ \end_inset . \series bold Demostración: \series default Sea \begin_inset Formula \[ \theta(t):=\theta_{0}+\int_{t_{0}}^{t}(f(u)g'(u)-f'(u)g(u))du, \] \end_inset \begin_inset Formula $\theta$ \end_inset es derivable una vez por el teorema fundamental del cálculo y su derivada es \begin_inset Formula ${\cal C}^{\infty}$ \end_inset , por lo que \begin_inset Formula $\theta$ \end_inset es diferenciable. Sea \begin_inset Formula $h:I\to\mathbb{R}$ \end_inset dada por \begin_inset Formula $h(t)\coloneqq (f(t)-\cos\theta(t))^{2}+(g(t)-\sin\theta(t))^{2}$ \end_inset , entonces \begin_inset Formula $h(t_{0})=0$ \end_inset y \begin_inset Formula \begin{align*} \frac{1}{2}h'= & (f-\cos\theta)(f'+\theta'\sin\theta)+(g-\sin\theta)(g'-\theta'\cos\theta)\\ = & (f-\cos\theta)(f'+(fg'-f'g)\sin\theta)+(g-\sin\theta)(g'-(fg'-f'g)\cos\theta)\\ = & ff'+f(fg'-f'g)\sin\theta-f'\cos\theta-(fg'-f'g)\sin\theta\cos\theta+\\ & +gg'-g'\sin\theta-g(fg'-f'g)\cos\theta+(fg'-f'g)\sin\theta\cos\theta, \end{align*} \end_inset pero derivando \begin_inset Formula $f^{2}+g^{2}=1$ \end_inset queda \begin_inset Formula $2ff'+2gg'=0$ \end_inset , \begin_inset Formula $ff'+gg'=0$ \end_inset , luego \begin_inset Formula \begin{align*} \frac{1}{2}h(t) & =(f(fg'-f'g)-g')\sin\theta+(-f'-g(fg'-f'g))\cos\theta\\ & =(g'(f^{2}-1)-ff'g)\sin\theta+(f'(-1+g^{2})-fgg')\cos\theta\\ & =(g^{2}g'-ff'g)\sin\theta+(f^{2}f'-fgg')\cos\theta\\ & =g(gg'-ff')\sin\theta+f(ff'-gg')\cos\theta=0. \end{align*} \end_inset Para la unicidad, sea \begin_inset Formula $\hat{\theta}$ \end_inset otra función que cumple las condiciones, \begin_inset Formula $\hat{\theta}$ \end_inset se diferencia de \begin_inset Formula $\theta$ \end_inset en cada punto en un múltiplo de \begin_inset Formula $2\pi$ \end_inset , pero como \begin_inset Formula $\hat{\theta}-\theta$ \end_inset es continua con dominio conexo, su rango debe ser conexo y estar en la componente conexa de \begin_inset Formula $\{2k\pi\}_{k\in\mathbb{Z}}$ \end_inset en la que está \begin_inset Formula $(\hat{\theta}-\theta)(t_{0})=0$ \end_inset , que es \begin_inset Formula $\{0\}$ \end_inset , luego \begin_inset Formula $\hat{\theta}=\theta$ \end_inset . \end_layout \begin_layout Standard Sean \begin_inset Formula $S$ \end_inset una superficie regular orientada por \begin_inset Formula $N$ \end_inset y \begin_inset Formula $\alpha:I\to S$ \end_inset una curva regular, \begin_inset Formula $e_{1},e_{2},V\in\mathfrak{X}(\alpha)$ \end_inset unitarios con \begin_inset Formula $e_{2}(t)=Je_{1}(t)=N(\alpha(t))\wedge e_{1}(t)$ \end_inset para todo \begin_inset Formula $t\in I$ \end_inset , entonces \begin_inset Formula $(e_{1}(t),e_{2}(t))$ \end_inset es una base ortonormal de \begin_inset Formula $T_{\alpha(t)}S$ \end_inset y existe \begin_inset Formula $\theta(t)$ \end_inset diferenciable tal que \begin_inset Formula $V=\cos\theta e_{1}+\sin\theta e_{2}$ \end_inset , y decimos que \begin_inset Formula $\theta$ \end_inset es el \series bold ángulo de rotación \series default de \begin_inset Formula $V$ \end_inset respecto a \begin_inset Formula $e_{1}$ \end_inset . \end_layout \begin_layout Standard La curvatura geodésica de \begin_inset Formula $\alpha$ \end_inset (no necesariamente p.p.a.) es \begin_inset Formula \[ \kappa_{g}^{\alpha}(s)=\frac{\langle\alpha''(u),J\alpha'(u)\rangle}{\Vert\alpha'(u)\Vert^{3}}. \] \end_inset \end_layout \begin_layout Standard \series bold Teorema de Liouville: \series default Sean \begin_inset Formula $(U,X)$ \end_inset una parametrización ortogonal de \begin_inset Formula $S$ \end_inset con primera forma fundamental \begin_inset Formula $E,F,G$ \end_inset , \begin_inset Formula $\alpha:I\to X(U)$ \end_inset una curva regular p.p.a. con curvatura geodésica \begin_inset Formula $\kappa_{g}$ \end_inset , \begin_inset Formula $\tilde{\alpha}\coloneqq (u,v)\coloneqq X^{-1}\circ\alpha:I\to U$ \end_inset , \begin_inset Formula $e_{1}:I\to\mathbb{R}^{3}$ \end_inset dado por \begin_inset Formula \[ e_{1}(s):=\frac{1}{\sqrt{E(\tilde{\alpha}(s))}}X_{u}(\tilde{\alpha}(s)), \] \end_inset \begin_inset Formula $\theta:I\to\mathbb{R}$ \end_inset el ángulo de rotación de \begin_inset Formula $\alpha'$ \end_inset respecto a \begin_inset Formula $e_{1}$ \end_inset , \begin_inset Formula $\alpha_{v}(u)\coloneqq \beta_{u}(v)\coloneqq X(u,v)$ \end_inset , \begin_inset Formula $(\kappa_{g})_{1}(u,v)$ \end_inset la curvatura geodésica de \begin_inset Formula $\alpha_{v}$ \end_inset en \begin_inset Formula $u$ \end_inset y \begin_inset Formula $(\kappa_{g})_{2}(u,v)$ \end_inset la de \begin_inset Formula $\beta_{u}$ \end_inset en \begin_inset Formula $v$ \end_inset , entonces \begin_inset Formula \[ \kappa_{g}=\theta'+\frac{1}{2\sqrt{EG}}\left(-u'E_{v}(\tilde{\alpha})+v'G_{u}(\tilde{\alpha})\right)=\theta'+(\kappa_{g})_{1}(\tilde{\alpha})\cos\theta+(\kappa_{g})_{2}(\tilde{\alpha})\sin\theta. \] \end_inset \series bold Demostración: \series default En efecto, \begin_inset Formula $e_{1}$ \end_inset es tangente y unitario, ya que \begin_inset Formula \begin{align*} e_{1}(s) & =\frac{X_{u}}{\Vert X_{u}\Vert}(\tilde{\alpha}(s)). \end{align*} \end_inset Entonces \begin_inset Formula $e_{2}(s)\coloneqq Je_{1}(s)$ \end_inset es también tangente y unitario y ortogonal a \begin_inset Formula $\frac{\partial X}{\partial u}$ \end_inset , luego \begin_inset Formula \[ e_{2}(s)=\frac{X_{v}}{\Vert X_{v}\Vert}(\tilde{\alpha}(s))=\frac{1}{\sqrt{G(\tilde{\alpha}(s))}}X_{v}(\tilde{\alpha}(s)). \] \end_inset Con esto, \begin_inset Formula \begin{align*} \left\langle \frac{De_{1}}{ds},e_{1}\right\rangle & =\langle e_{1}',e_{1}\rangle=\frac{1}{2}\frac{d}{ds}\langle e_{1},e_{1}\rangle=0,\\ \left\langle \frac{De_{1}}{ds},e_{2}\right\rangle & =\langle e_{1}',e_{2}\rangle=\frac{d}{ds}\langle e_{1},e_{2}\rangle-\langle e_{1},e_{2}'\rangle=-\langle e_{1},e_{2}'\rangle=-\left\langle \frac{De_{2}}{ds},e_{1}\right\rangle ,\\ \left\langle \frac{De_{2}}{ds},e_{2}\right\rangle & =\frac{1}{2}\frac{d}{ds}\langle e_{2},e_{2}\rangle=0, \end{align*} \end_inset luego si \begin_inset Formula $\omega\coloneqq \langle e_{1}',e_{2}\rangle=-\langle e_{1},e_{2}'\rangle$ \end_inset \begin_inset Formula \begin{align*} \frac{De_{1}}{ds}(s) & =\left\langle \frac{De_{1}}{ds},e_{1}\right\rangle e_{1}+\left\langle \frac{De_{2}}{ds},e_{2}\right\rangle e_{2}=\omega(s)e_{2}(s), & \frac{De_{2}}{ds}(s) & =-\omega(s)e_{1}(s). \end{align*} \end_inset Por tanto, como \begin_inset Formula $\alpha'=\cos\theta e_{1}+\sin\theta e_{2}$ \end_inset , \begin_inset Formula \begin{align*} \frac{D\alpha'}{ds} & =-\theta'\sin\theta e_{1}+\cos\theta\omega e_{2}+\theta'\cos\theta e_{2}-\sin\theta\omega e_{1}=(\theta'+\omega)(\cos\theta e_{2}-\sin\theta e_{1})\\ & =(\theta'+\omega)J\alpha'(s). \end{align*} \end_inset Por otro lado, \begin_inset Formula $\frac{D\alpha'}{ds}(s)=\kappa_{g}(s)J\alpha'(s)$ \end_inset , luego \begin_inset Formula $\kappa_{g}(s)=\theta'(s)+\omega(s)$ \end_inset . Derivando la fórmula de \begin_inset Formula $e_{1}$ \end_inset , \begin_inset Formula \[ e_{1}'=\frac{d}{ds}\left(\frac{1}{\sqrt{E}}\right)X_{u}(\tilde{\alpha})+\frac{1}{\sqrt{E}}\left(u'X_{uu}(\tilde{\alpha})+v'X_{uv}(\tilde{\alpha})\right). \] \end_inset Entonces, como \begin_inset Formula $X_{uu}(\tilde{\alpha})=\Gamma_{11}^{1}X_{u}+\Gamma_{11}^{2}X_{v}+eN$ \end_inset y \begin_inset Formula $X_{uv}=\Gamma_{12}^{1}X_{u}+\Gamma_{12}^{2}X_{v}+fN$ \end_inset , \begin_inset Formula \begin{align*} \omega & =\langle e_{1}',e_{2}\rangle=\frac{1}{\sqrt{EG}}\left\langle u'({\textstyle \Gamma_{11}^{1}X_{u}+\Gamma_{11}^{2}X_{v}+eN)}+v'({\textstyle \Gamma_{12}^{2}X_{u}+\Gamma_{12}^{2}X_{v}+fN}),X_{v}\right\rangle \\ & =\frac{1}{\sqrt{EG}}(u'\Gamma_{11}^{2}G+v'\Gamma_{12}^{2}G), \end{align*} \end_inset pero como \begin_inset Formula \begin{align*} \Gamma_{11}^{2} & =\frac{-F\frac{E_{u}}{2}+EF_{u}-E\frac{E_{v}}{2}}{EG-F^{2}}=-\frac{E\frac{E_{v}}{2}}{EG}=-\frac{E_{v}}{2G}, & \Gamma_{12}^{2} & =\frac{-F\frac{E_{v}}{2}+E\frac{G_{u}}{2}}{EG-F^{2}}=\frac{E\frac{G_{u}}{2}}{EG}=\frac{G_{u}}{2G}, \end{align*} \end_inset queda \begin_inset Formula \[ \omega=\frac{1}{2\sqrt{EG}}(-u'E_{v}+v'G_{u}), \] \end_inset la primera expresión. Por otro lado, \begin_inset Formula \begin{align*} \alpha'_{v}(u) & =X_{u}, & J\alpha'_{v}(u) & =N\wedge X_{u}=\Vert X_{u}\Vert\frac{X_{v}}{\Vert X_{v}\Vert}=\sqrt{\frac{E}{G}}X_{v}, & \alpha''_{v}(u) & =X_{uu},\\ \beta'_{u}(v) & =X_{v}, & J\beta'_{u}(v) & =N\wedge X_{v}=-\Vert X_{v}\Vert\frac{X_{u}}{\Vert X_{u}\Vert}=-\sqrt{\frac{G}{E}}X_{u}, & \beta''_{u}(v) & =X_{vv}, \end{align*} \end_inset y como \begin_inset Formula \[ \Gamma_{22}^{1}=\frac{G(F_{v}-\frac{G_{u}}{2})-F\frac{G_{v}}{2}}{EG-F^{2}}=\frac{-G\frac{G_{u}}{2}}{EG}=-\frac{G_{u}}{2E}, \] \end_inset queda \begin_inset Formula \begin{align*} (\kappa_{g})_{1}(u,v) & =\frac{\langle\alpha''_{v}(u),J\alpha'_{v}(u)\rangle}{\Vert\alpha'_{v}(u)\Vert^{3}}=\frac{\langle\Gamma_{11}^{1}X_{u}+\Gamma_{11}^{2}X_{v}+eN,\sqrt{\frac{E}{G}}X_{v}\rangle}{\Vert X_{u}\Vert^{3}}=\sqrt{\frac{E}{G}}\frac{\Gamma_{11}^{2}G}{E\sqrt{E}}=-\frac{E_{v}}{2E\sqrt{G}},\\ (\kappa_{g})_{2}(u,v) & =\frac{\langle\beta''_{u}(v),J\beta'_{u}(v)\rangle}{\Vert\beta'_{u}(v)\Vert^{3}}=\frac{\langle\Gamma_{22}^{1}X_{u}+\Gamma_{22}^{2}X_{v}+gN,-\sqrt{\frac{G}{E}}X_{u}\rangle}{\Vert X_{v}\Vert^{3}}=-\sqrt{\frac{G}{E}}\frac{\Gamma_{22}^{1}E}{G\sqrt{G}}=\frac{G_{u}}{2G\sqrt{E}}. \end{align*} \end_inset Con esto, \begin_inset Formula $E_{v}=-2E\sqrt{G}(\kappa_{g})_{1}$ \end_inset y \begin_inset Formula $G_{u}=2G\sqrt{E}(\kappa_{g})_{2}$ \end_inset , luego \begin_inset Formula \[ \kappa_{g}=\theta'+\frac{1}{2\sqrt{EG}}\left(u'2E\sqrt{G}(\kappa_{g})_{1}+v'2G\sqrt{E}(\kappa_{g})_{2}\right)=\theta'+u'\sqrt{E}(\kappa_{g})_{1}+\sqrt{G}v'(\kappa_{g})_{2}, \] \end_inset y queda ver que \begin_inset Formula $u'\sqrt{E}=\cos\theta$ \end_inset y \begin_inset Formula $v'\sqrt{G}=\sin\theta$ \end_inset , pero \begin_inset Formula \begin{align*} \alpha' & =(X\circ\tilde{\alpha})'=u'X_{u}+v'X_{v}\\ & =\cos\theta e_{1}+\sin\theta e_{2}=\cos\theta\frac{1}{\sqrt{E}}X_{u}+\sin\theta\frac{1}{\sqrt{G}}X_{v}, \end{align*} \end_inset y usando que \begin_inset Formula $(X_{u},X_{v})$ \end_inset es base despejamos y se obtiene el resultado. \end_layout \begin_layout Section Teorema de rotación de las tangentes \end_layout \begin_layout Standard Sea \begin_inset Formula $S$ \end_inset una superficie regular, un \series bold polígono curvado \series default es la imagen \begin_inset Formula $\Gamma$ \end_inset de un segmento de curva \begin_inset Formula $\alpha:[0,\ell]\to S$ \end_inset regular a trozos p.p.a. (en cada trozo) \series bold cerrado \series default ( \begin_inset Formula $\alpha(0)=\alpha(\ell)$ \end_inset ) y \series bold simple \series default \begin_inset Formula $(\forall s,s'\in[0,\ell],(\alpha(s)=\alpha(s')\implies s=s'\lor\{s,s'\}=\{0,\ell\})$ \end_inset ). Si \begin_inset Formula $\Gamma$ \end_inset es la frontera de una región \begin_inset Formula $R$ \end_inset de \begin_inset Formula $S$ \end_inset simplemente conexa, \begin_inset Formula $\alpha$ \end_inset está \series bold positivamente orientada \series default si, para \begin_inset Formula $s\in[0,\ell]$ \end_inset que no sea un vértice, \begin_inset Formula $J\alpha'(s)$ \end_inset apunta al interior de \begin_inset Formula $R$ \end_inset ( \begin_inset Formula $\exists\delta>0:\forall t\in(0,\delta),\alpha(s)+tJ\alpha'(s)\in R$ \end_inset ). \end_layout \begin_layout Standard Sea \begin_inset Formula $0=s_{0}<\dots