% % exponentialfunktion.tex % % (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule % \begin{frame}[t] \setlength{\abovedisplayskip}{5pt} \setlength{\belowdisplayskip}{5pt} \frametitle{Exponentialfunktion} \vspace{-15pt} \begin{columns}[t,onlytextwidth] \only<1-6>{% \begin{column}{0.48\textwidth} \begin{block}{$x(t) \in\mathbb{R}$} \vspace{-10pt} \begin{align*} \frac{d}{dt}x(t) &= ax(t) &a&\in\mathbb{R} \\ x(0) &= c&&\in\mathbb{R} \intertext{\uncover<2->{Lösung:}} \uncover<2->{x(t) &= ce^{at}} \end{align*} \end{block} \end{column}} \begin{column}{0.48\textwidth} \uncover<3->{% \begin{block}{$X(t) \in M_n(\mathbb{R})$} \vspace{-10pt} \begin{align*} \frac{d}{dt}X(t) &= A X(t)&A&\in M_n(\mathbb{R}) \\ X(0)&=C&&\in M_n(\mathbb{R}) \intertext{\uncover<4->{gekoppelte Differentialgleichung für vier Funktionen $x_{ij}(t)$}} \uncover<5->{\dot{x}_{11} &= \rlap{$a_{11} x_{11}(t) + a_{12} x_{21}(t)$}}\\ \uncover<5->{\dot{x}_{12} &= \rlap{$a_{11} x_{12}(t) + a_{12} x_{22}(t)$}}\\ \uncover<5->{\dot{x}_{21} &= \rlap{$a_{21} x_{11}(t) + a_{22} x_{21}(t)$}}\\ \uncover<5->{\dot{x}_{22} &= \rlap{$a_{21} x_{12}(t) + a_{22} x_{22}(t)$}}\\ \intertext{\uncover<6->{Lösung:}} \uncover<6->{X(t) &= \exp(At) C} \end{align*} \end{block}} \end{column} \only<7-9>{% \begin{column}{0.48\textwidth} \begin{block}{Beispiel: Diagonalmatrix} %$D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ \begin{align*} \frac{d}{dt}X&=DX &&\uncover<8->{\Rightarrow &\dot{x}_{ij}(t) &= \lambda_i x_{ij}(t)} \\ X(0)&=C &&\uncover<8->{\Rightarrow&x_{ij}(t)&=c_{ij}} \end{align*} \uncover<9->{% Lösung: \[ x_{ij}(t) =c_{ij}e^{\lambda_i t} \]} \end{block} \end{column}} \uncover<10->{% \begin{column}{0.48\textwidth} \begin{block}{Beispiel: Jordan-Block} \vspace{-10pt} \begin{align*} A&=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix} \rlap{$\displaystyle,\; X(t) = \only<22>{ e^{\lambda t} \begin{pmatrix} 1&t/\lambda\\ 0&1 \end{pmatrix} } \only<23>{ \frac{e^{\lambda t}}{\lambda} \begin{pmatrix} \lambda&t\\ 0&\lambda \end{pmatrix} } C $} \\ \uncover<11->{ \dot{x}_{1i}(t)&=\lambda x_{1i}(t) + \phantom{\lambda}x_{2i}(t),&&x_{1i}(0)&=c_{1i} } \\ \uncover<12->{ \dot{x}_{2i}(t)&=\phantom{\lambda x_{1i}(t)+\mathstrut}\lambda x_{2i}(t),&&x_{2i}(0)&=c_{2i} } \end{align*} \uncover<13->{% Lösung:} \begin{align*} \uncover<14->{ x_{2i}(t)&=c_{2i}e^{\lambda t} } \\ \uncover<15->{ \dot{x}_{1i}(t)&=\lambda x_{1i}(t) + c_{2i}e^{\lambda t} } \\ \only<16-17>{x_{1i\only<16>{,h}}(t)} \only<18->{\dot{x}_{1i}(t)} & \only<16-17>{=c\only<17>{(t)}\lambda e^{\lambda t}} \only<18>{=\dot{c}(t)\lambda e^{\lambda t} + c(t)\lambda^2 e^{\lambda t}} \only<19->{=\lambda x_{1i}(t) + \dot{c}(t)\lambda e^{\lambda t}} \\ \uncover<20->{\Rightarrow \dot{c}(t)&= c_{2i}/\lambda \Rightarrow c(t) = c_{2i}(0) +tc_{2i}/\lambda } \\ \uncover<21->{ x_{1i}(t) & =c_{1i}e^{\lambda t} + t(c_{2i}/\lambda)e^{\lambda t} } \end{align*} \end{block} \end{column}} \end{columns} \end{frame}