4.2.2: Caso n=2
- Page ID
- 118179
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Considerar el problema de valor inicial
\ begin {eqnarray}
\ label {n2dgl}\ tag {4.2.2.1}
v_ {xx} +v_ {yy} &=&c^ {-2} v_ {tt}\
\ etiqueta {inicial1}\ etiqueta {4.2.2.2}
v (x, y,0) &=&f (x, y)\
\ etiqueta {inicial2}\ etiqueta {4.2.2.2.3}
v_t (x, y,0) &=&g (x, y),
\ end { eqnarray}
donde\(f\in C^3,\ g\in C^2\).
Usando la fórmula para la solución del problema del valor inicial tridimensional derivaremos una fórmula para el caso bidimensional. A la siguiente consideración se le llama método decente de Hadamard.
Dejar\(v(x,y,t)\) ser una solución de (\ ref {n2dgl}) - (\ ref {initial2}), entonces
$$u (x, y, z, t) :=v (x, y, t)\]
es una solución del problema de valor inicial tridimensional con datos iniciales\(f(x,y)\),\(g(x,y)\), independiente de\(z\), ya que\(u\) satisface (\ ref {n2dgl}) - (\ ref {initial2}). De ahí\(u(x,y,z,t)=u(x,y,0,t)+u_z(x,y,\delta z,t)z\), ya que\(0<\delta<1\), y\(u_z=0\), tenemos
$$v (x, y, t) =u (x, y,0, t).\]
La fórmula de Poisson en el caso tridimensional implica
\ begin {eqnarray}
v (x, y, t) &=&\ frac {1} {4\ pi c^2}\ frac {\ parcial} {\ parcial} {\ t parcial}\ izquierda (\ frac {1} {t}\ int_ {\ parcial B_ {ct} (x, y,0)}\ f (\ xi,\ eta)\ dS derecha)\ nonumber
\\ etiqueta {poissonhilf1}\ tag {4.2.2.4}
&&+\ frac {1} {4\ pi c^2 t}\ int_ {\ parcial B_ {ct} (x, y,0)}\ g (\ xi,\ eta)\ dS.
\ end {eqnarray}
Figura 4.2.2.1: Dominios de integración
Los integrandos son independientes de\(\zeta\). La superficie\(S\) está definida por\(\chi(\xi,\eta,\zeta):=(\xi-x)^2+(\eta-y)^2+\zeta^2-c^2 t^2=0\). Entonces la normal exterior\(n\) en\(S\) es\(n=\nabla\chi/|\nabla\chi|\) y el elemento de superficie viene dado por\(dS=(1/|n_3|)d\xi d\eta\), donde la tercera coordenada de\(n\) es
$$n_3=\ pm\ frac {\ sqrt {c^2 t^2- (\ xi-x) ^2- (\ eta-y) ^2}} {ct}.\]
El signo positivo aplica sobre\(S^+\), dónde\(\zeta>0\) y el signo es negativo sobre\(S^-\) dónde\(\zeta<0\), ver Figura 4.2.2.1. Tenemos\(S=S^+\cup\overline{S^-}\).
Set\(\rho=\sqrt{(\xi-x)^2+(\eta-y)^2}\). Luego se deduce de (\ ref {poissonhilf1})
Teorema 4.3. La solución del problema del valor inicial de Cauchy (\ ref {n2dgl}) - (\ ref {initial2}) viene dada por
\ begin {eqnarray*}
v (x, y, t) &=&\ frac {1} {2\ pi c}\ frac {\ parcial} {\ parcial} {\ t parcial}\ int_ {B_ {ct} (x, y)}\\ frac {f (\ xi,\ eta)} {\ sqrt {c^2 t^2-\ rho^2}}\ d\ xi d\ eta\\
&&&+\ frac {1} {2\ pi c}\ int_ {B_ {ct} (x, y)}\\ frac {g (\ xi,\ eta)} {\ sqrt {c^2 t^2-\ rho^2}}\ d\ xi d\ eta. \ end {eqnarray*}
Figura 4.2.2.2: Intervalo de dependencia, caso\(n=2\)
Corolario. En contraste con el caso tridimensional, el dominio de la dependencia es aquí el disco\(B_{ct_o}(x_0,y_0)\) y no solo el límite. Por lo tanto, ver fórmula del Teorema 4.3, si\(f,\ g\) tienen soportes en un dominio compacto\(D\subset\mathbb{R}^2\), entonces estas funciones tienen influencia en el valor\(v(x,y,t)\) para todos los tiempos\(t>T\),\(T\) suficientemente grandes.