4.3: El Wronskian
- Page ID
- 116912
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Supongamos que habiendo determinado que dos soluciones de (4.2.1) son\(x = X_1(t)\) y\(x = X_2(t)\), intentamos escribir la solución general a (4.2.1) como (4.2.2). Debemos entonces preguntarnos si esta solución general podrá satisfacer las dos condiciones iniciales dadas por \[\label{eq:1}x(t_0)=x_0,\quad\overset{.}{x}(t_0)=u_0.\]
Aplicando estas condiciones iniciales a (4.2.2), obtenemos \[\begin{align}c_1X_1(t_0)+c_2X_2(t_0)&=x_0,\nonumber \\ c_1\overset{.}{X}_1(t_0)+c_2\overset{.}{X}_2(t_0)&=u_0,\label{eq:2}\end{align}\]) que se observa que es un sistema de dos ecuaciones lineales para las dos incógnitas\(c_1\) y\(c_2\). La solución de\(\eqref{eq:2}\) por métodos estándar da como resultado\[c_1=\frac{x_0\overset{.}{X}_2(t_0)-u_0X_2(t_0)}{W},\quad c_2=\frac{u_0X_1(t_0)-x_0\overset{.}{X}_1(t_0)}{W},\nonumber\] donde\(W\) se llama Wronskian y es dada por \[\label{eq:3}W=X_1(t_0)\overset{.}{X}_2(t_0)-\overset{.}{X}_1(t_0)X_2(t_0).\]
Evidentemente, el wronskiano no debe ser igual a cero\((W\neq 0)\) para que exista una solución.
Por ejemplo, las dos soluciones\[X_1(t)=A\sin\omega t,\quad X_2(t)=B\sin\omega t,\nonumber\] tienen un Wronskian cero en\(t=t_0\), como puede demostrarse mediante la computación\[\begin{aligned} W&=(A\sin\omega t_0)(B\omega\cos\omega t_0)-(A\omega\cos\omega t_0)(B\sin\omega t_0) \\ &=0;\end{aligned}\] mientras que las dos soluciones\[X_1(t)=\sin\omega t,\quad X_2(t)=\cos\omega t,\nonumber\] con\(\omega\neq 0\), tienen un Wronskian distinto de cero en\(t=t_0\),\[\begin{aligned}W&=(\sin\omega t_0)(-\omega\sin\omega t_0)-(\omega\cos\omega t_0)(\cos\omega t_0) \\ &=-\omega.\end{aligned}\]
Cuando el Wronskian no es igual a cero, decimos que las dos soluciones\(X_1(t)\) y\(X_2(t)\) son linealmente independientes. El concepto de independencia lineal se toma prestado del álgebra lineal, y de hecho, el conjunto de todas las funciones que satisfacen (4.2.1) se puede mostrar para formar un espacio vectorial bidimensional.