7.3.1: Problemas de Valor Límite: Problema de Dirichlet
- Page ID
- 118102
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)El problema de Dirichlet (problema del primer valor límite) es encontrar una solución\(u\in C^2(\Omega)\cap C(\overline{\Omega})\) de
\ begin {eqnarray}
\ label {D1}\ tag {7.3.1.1}
\ triángulo u&=&0\\\ mbox {in}\\ Omega\
\ etiqueta {D2}\ tag {7.3.1.2}
u&=&\ Phi\\ mbox {on}\\ parcial\ Omega,
\ end {eqnarray}
donde\(\Phi\) se da y continua encendido\(\partial\Omega\).
Proposición 7.4. Supongamos que\(\Omega\) está limitado, entonces una solución al problema de Dirichlet se determina de manera única.
Prueba. Principio máximo.
OBSERVACIÓN. El resultado anterior falla si quitamos en la condición de límite (\ ref {D2}) un punto del límite como muestra el siguiente ejemplo. Dejar\(\Omega\subset\mathbb{R}^2\) ser el dominio
$$
\ Omega=\ {x\ in B_1 (0):\ x_2>0\},
\]
Figura 7.3.1.1: Contraejemplo
Supongamos que\(u\in C^2(\Omega)\cap C(\overline{\Omega}\setminus\{0\})\) es una solución de
\ begin {eqnarray*}
\ triángulo u&=&0\\\ mbox {in}\\ Omega\\
u&=&0\\\ mbox {on}\\\ parcial\ Omega\ setmenos\ {0\}.
\ end {eqnarray*}
Este problema tiene soluciones\(u\equiv 0\) y\(u=\mbox{Im}(z+z^{-1})\), donde\(z=x_1+ix_2\). En cuanto a otro ejemplo ver un ejercicio.
En contraste con este comportamiento de la ecuación de Laplace, uno tiene singularidad si $\ triángulo u=0$ es reemplazado por la ecuación de superficie mínima
$$
\ frac {\ partial} {\ partial x_1}\ left (\ frac {u_ {x_1}} {\ sqrt {1+|\ nabla u|^2}}\ derecha) +
\ frac {\ parcial} {\ parcial x_2}\ izquierda (\ frac {u_ {x_2}} {\ sqrt {1+|\ nabla u|^2}}\ derecha) =0.
\]