10.2: Solución por Diagonalización
- Page ID
- 119121
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Otra forma de ver el problema de las odas lineales acopladas de primer orden es desde la perspectiva de la diagonalización matricial. Con
\[\dot{x}=A x \nonumber \]
suponemos que A se puede diagonalizar usando
\[\mathrm{S}^{-1} \mathrm{AS}=\Lambda, \nonumber \]
donde\(\Lambda\) está la matriz de valores propios diagonal, y\(S\) contiene los vectores propios. Podemos cambiar variables en la Ecuación\ ref {10.7} usando
\[x=S y \nonumber \]
y obtener
\[\text { Sy் }=\text { ASy. } \nonumber \]
La multiplicación a la izquierda por\(\mathrm{S}^{-1}\) y usando la ecuación\ ref {10.8} da como resultado
\[\begin{aligned} \dot{\mathrm{y}} &=\mathrm{S}^{-1} \mathrm{ASy} \\ &=\Lambda \mathrm{y} . \end{aligned} \nonumber \]
Las ecuaciones diferenciales de primer orden en las\(y\) variables -ahora están desacopladas y se pueden resolver inmediatamente, y las variables x se pueden recuperar usando la ecuación\ ref {10.9}.
Ejemplo: Resolver el ejemplo anterior
\[\frac{d}{d t}\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{ll} 1 & 1 \\ 4 & 1 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) \nonumber \]
por el método de diagonalización.
Se conocen los valores propios y los vectores propios, y tenemos
\[\Lambda=\left(\begin{array}{rr} 3 & 0 \\ 0 & -1 \end{array}\right), \quad S=\left(\begin{array}{rr} 1 & 1 \\ 2 & -2 \end{array}\right) \nonumber \]
y las\(y\) ecuaciones desacopladas están dadas por
\[\dot{y}_{1}=3 y_{1}, \quad \dot{y}_{2}=-y_{2}, \nonumber \]
con solución
\[y_{1}(t)=c_{1} e^{3 t}, \quad y_{2}=c_{2} e^{-t} \nonumber \]
Transformando de nuevo a las\(x\) variables -usando la ecuación\ ref {10.9}, tenemos
\[\begin{aligned} \left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) &=\left(\begin{array}{rr} 1 & 1 \\ 2 & -2 \end{array}\right)\left(\begin{array}{c} c_{1} e^{3 t} \\ c_{2} e^{-t} \end{array}\right) \\ &=\left(\begin{array}{c} c_{1} e^{3 t}+c_{2} e^{-t} \\ 2 c_{1} e^{3 t}-2 c_{2} e^{-t} \end{array}\right) \end{aligned} \nonumber \]
que concuerda con la solución Ecuación\ ref {10.6}.