diff options
Diffstat (limited to '')
| -rw-r--r-- | vorlesungen/slides/10/intro.tex | 45 | ||||
| -rw-r--r-- | vorlesungen/slides/10/n-zu-1.tex | 98 | ||||
| -rw-r--r-- | vorlesungen/slides/10/repetition.tex | 40 | ||||
| -rw-r--r-- | vorlesungen/slides/10/so2.tex | 237 | ||||
| -rw-r--r-- | vorlesungen/slides/10/vektorfelder.mp | 241 | ||||
| -rw-r--r-- | vorlesungen/slides/10/vektorfelder.tex | 82 |
6 files changed, 475 insertions, 268 deletions
diff --git a/vorlesungen/slides/10/intro.tex b/vorlesungen/slides/10/intro.tex new file mode 100644 index 0000000..276bf49 --- /dev/null +++ b/vorlesungen/slides/10/intro.tex @@ -0,0 +1,45 @@ +% +% intro.tex -- Repetition Lie-Gruppen und -Algebren +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% Erstellt durch Roy Seitz +% +% !TeX spellcheck = de_CH +\bgroup + + + +\begin{frame}[t] + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} +% \frametitle{Repetition} +% \vspace{-20pt} + \begin{block}{Offene Fragen} + \begin{itemize}[<+->] + \item Woher kommt die Exponentialfunktion? + \begin{fleqn} + \[ + \exp(At) + = + 1 + + At + + A^2\frac{t^2}{2} + + A^3\frac{t^3}{3!} + + \ldots + \] + \end{fleqn} + \item Wie löst man eine Matrix-DGL? + \begin{fleqn} + \[ + \dot\gamma(t) = A\gamma(t), + \qquad + \gamma(t) \in G \subset M_n + \] + \end{fleqn} + \item Lie-Gruppen und Lie-Algebren + \item Was bedeutet $\exp(At)$? + \end{itemize} + \end{block} +\end{frame} + +\egroup diff --git a/vorlesungen/slides/10/n-zu-1.tex b/vorlesungen/slides/10/n-zu-1.tex index 737df03..09475ad 100644 --- a/vorlesungen/slides/10/n-zu-1.tex +++ b/vorlesungen/slides/10/n-zu-1.tex @@ -7,51 +7,57 @@ % !TeX spellcheck = de_CH \bgroup \begin{frame}[t] -\setlength{\abovedisplayskip}{5pt} -\setlength{\belowdisplayskip}{5pt} -\frametitle{Reicht $1.$ Ordnung?} -\vspace{-20pt} -\begin{columns}[t,onlytextwidth] -\begin{column}{0.48\textwidth} -\begin{block}{Beispiel: DGL 3.~Ordnung} \vspace*{-1ex} - \begin{align*} - x^{(3)} + a_2 \ddot x + a_1 \dot x + a_0 x = 0 \\ - \Rightarrow - x^{(3)} = -a_2 \ddot x - a_1 \dot x - a_0 x - \end{align*} -\end{block} -\begin{block}{Ziel: Nur noch 1.~Ableitungen} - Einführen neuer Variablen: - \begin{align*} - x_0 &\coloneqq x & - x_1 &\coloneqq \dot x & - x_2 &\coloneqq \ddot x - \end{align*} -System von Gleichungen 1.~Ordnung - \begin{align*} - \dot x_0 &= x_1 \\ - \dot x_1 &= x_2 \\ - \dot x_2 &= -a_2 x_2 - a_1 x_1 - a_0 x_0 -\end{align*} -\end{block} -\end{column} -\begin{column}{0.48\textwidth} -\begin{block}{Als Vektor-Gleichung} \vspace*{-1ex} - \begin{align*} - \frac{d}{dt} - \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix} - = \begin{pmatrix} - 0 & 1 & 0 \\ - 0 & 0 & 1 \\ - -a_0 & -a_1 & -a_2 - \end{pmatrix} - \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix} - \end{align*} - - Geht für jede lineare Differentialgleichung! - -\end{block} -\end{column} -\end{columns} + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + %\frametitle{Reicht $1.$ Ordnung?} + %\vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \uncover<1->{ + \begin{block}{Beispiel: DGL 3.~Ordnung} \vspace*{-1ex} + \begin{align*} + x^{(3)} + a_2 \ddot x + a_1 \dot x + a_0 x = 0 \\ + \Rightarrow + x^{(3)} = -a_2 \ddot x - a_1 \dot x - a_0 x + \end{align*} + \end{block} + } + \uncover<2->{ + \begin{block}{Ziel: Nur noch 1.~Ableitungen} + Einführen neuer Variablen: + \begin{align*} + x_0 &\coloneqq x & + x_1 &\coloneqq \dot x & + x_2 &\coloneqq \ddot x + \end{align*} + System von Gleichungen 1.~Ordnung + \begin{align*} + \dot x_0 &= x_1 \\ + \dot x_1 &= x_2 \\ + \dot x_2 &= -a_2 x_2 - a_1 x_1 - a_0 x_0 + \end{align*} + \end{block} + } + \end{column} + \uncover<3->{ + \begin{column}{0.48\textwidth} + \begin{block}{Als Vektor-Gleichung} \vspace*{-1ex} + \begin{align*} + \frac{d}{dt} + \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix} + = \begin{pmatrix} + 0 & 1 & 0 \\ + 0 & 0 & 1 \\ + -a_0 & -a_1 & -a_2 + \end{pmatrix} + \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix} + \end{align*} + + \uncover<4->{Geht für jede lineare Differentialgleichung!} + + \end{block} + \end{column} + } + \end{columns} \end{frame} \egroup diff --git a/vorlesungen/slides/10/repetition.tex b/vorlesungen/slides/10/repetition.tex index c45d47b..7c007ca 100644 --- a/vorlesungen/slides/10/repetition.tex +++ b/vorlesungen/slides/10/repetition.tex @@ -1,5 +1,5 @@ % -% intro.tex -- Repetition Lie-Gruppen und -Algebren +% repetition.tex -- Repetition Lie-Gruppen und -Algebren % % (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule % Erstellt durch Roy Seitz @@ -101,51 +101,19 @@ \begin{align*} \uncover<4->{ X(t) } & - \uncover<4->{= \begin{pmatrix} 0 & t \\ 0 & 0 \end{pmatrix} } + \uncover<4->{= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} } \\ \uncover<5->{ Y(t) } & - \uncover<5->{= \begin{pmatrix} 0 & 0 \\ t & 0 \end{pmatrix} } + \uncover<5->{= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} } \\ \uncover<6->{ H(t) } & - \uncover<6->{= \begin{pmatrix} t & 0 \\ 0 & -t \end{pmatrix} } + \uncover<6->{= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} } \end{align*} \end{column} \end{columns} \end{frame} -\begin{frame}[t] - \setlength{\abovedisplayskip}{5pt} - \setlength{\belowdisplayskip}{5pt} - \frametitle{Repetition} - \vspace{-20pt} - \begin{block}{Offene Fragen} - \begin{itemize}[<+->] - \item Woher kommt die Exponentialfunktion? - \begin{fleqn} - \[ - \exp(At) - = - 1 - + At - + A^2\frac{t^2}{2} - + A^3\frac{t^3}{3!} - + \ldots - \] - \end{fleqn} - \item Wie löst man eine Matrix-DGL? - \begin{fleqn} - \[ - \dot\gamma(t) = A\gamma(t), - \qquad - \gamma(t) \in G \subset M_n - \] - \end{fleqn} - \item Was bedeutet $\exp(At)$? - \end{itemize} - \end{block} -\end{frame} - \egroup diff --git a/vorlesungen/slides/10/so2.tex b/vorlesungen/slides/10/so2.tex index b63a67e..dcbcdc8 100644 --- a/vorlesungen/slides/10/so2.tex +++ b/vorlesungen/slides/10/so2.tex @@ -8,123 +8,134 @@ \bgroup \begin{frame}[t] -\setlength{\abovedisplayskip}{5pt} -\setlength{\belowdisplayskip}{5pt} -\frametitle{Von der Lie-Gruppe zur -Algebra} -\vspace{-20pt} -\begin{columns}[t,onlytextwidth] -\begin{column}{0.48\textwidth} - \begin{block}{Lie-Gruppe} - Darstellung von \gSO2: - \begin{align*} - \mathbb R - &\to - \gSO2 - \\ - t - &\mapsto - \begin{pmatrix} - \cos t & -\sin t \\ - \sin t & \phantom-\cos t - \end{pmatrix} - \end{align*} - \end{block} - \begin{block}{Ableitung am neutralen Element} - \begin{align*} - \frac{d}{d t} - & - \left. - \begin{pmatrix} - \cos t & -\sin t \\ - \sin t & \phantom-\cos t - \end{pmatrix} - \right|_{ t = 0} - \\ - = - & - \begin{pmatrix} -\sin0 & -\cos0 \\ \phantom-\cos0 & -\sin0 \end{pmatrix} - = - \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} - \end{align*} - \end{block} -\end{column} -\begin{column}{0.48\textwidth} - \begin{block}{Lie-Algebra} - Darstellung von \aso2: - \begin{align*} - \mathbb R - &\to - \aso2 - \\ - t - &\mapsto - \begin{pmatrix} - 0 & -t \\ - t & \phantom-0 - \end{pmatrix} - \end{align*} - \end{block} -\end{column} -\end{columns} + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + \frametitle{Von der Lie-Gruppe zur -Algebra} + \vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \uncover<1->{ + \begin{block}{Lie-Gruppe} + Darstellung von \gSO2: + \begin{align*} + \mathbb R + &\to + \gSO2 + \\ + t + &\mapsto + \begin{pmatrix} + \cos t & -\sin t \\ + \sin t & \phantom-\cos t + \end{pmatrix} + \end{align*} + \end{block} + } + \uncover<2->{ + \begin{block}{Ableitung am neutralen Element} + \begin{align*} + \frac{d}{d t} + & + \left. + \begin{pmatrix} + \cos t & -\sin t \\ + \sin t & \phantom-\cos t + \end{pmatrix} + \right|_{ t = 0} + \\ + = + & + \begin{pmatrix} -\sin0 & -\cos0 \\ \phantom-\cos0 & -\sin0 \end{pmatrix} + = + \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} + \end{align*} + \end{block} + } + \end{column} + \begin{column}{0.48\textwidth} + \uncover<3->{ + \begin{block}{Lie-Algebra} + Darstellung von \aso2: + \begin{align*} + \mathbb R + &\to + \aso2 + \\ + t + &\mapsto + \begin{pmatrix} + 0 & -t \\ + t & \phantom-0 + \end{pmatrix} + \end{align*} + \end{block} + } + \end{column} + \end{columns} \end{frame} \begin{frame}[t] -\setlength{\abovedisplayskip}{5pt} -\setlength{\belowdisplayskip}{5pt} -\frametitle{Von der Lie-Algebra zur -Gruppe} -\vspace{-20pt} -\begin{columns}[t,onlytextwidth] -\begin{column}{0.48\textwidth} - \begin{block}{Differentialgleichung} - Gegeben: - \[ - A + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + \frametitle{Von der Lie-Algebra zur -Gruppe} + \vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \uncover<1->{ + \begin{block}{Differentialgleichung} + Gegeben: + \[ + J + = + \dot\gamma(0) = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} + \] + Gesucht: + \[ \dot \gamma (t) = J \gamma(t) \qquad \gamma \in \gSO2 \] + \[ \Rightarrow \gamma(t) = \exp(Jt) \gamma(0) = \exp(Jt) \] + \end{block} + } + \end{column} + \begin{column}{0.48\textwidth} + \uncover<2->{ + \begin{block}{Lie-Algebra} + Potenzen von $J$: + \begin{align*} + J^2 &= -I & + J^3 &= -J & + J^4 &= I & + \ldots + \end{align*} + \end{block} + } + \end{column} + \end{columns} +\uncover<3->{ + Folglich: + \begin{align*} + \exp(Jt) + &= I + Jt + + J^2\frac{t^2}{2!} + + J^3\frac{t^3}{3!} + + J^4\frac{t^4}{4!} + + J^5\frac{t^5}{5!} + + \ldots \\ + &= \begin{pmatrix} + \vspace*{3pt} + 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \ldots + & + -t + \frac{t^3}{3!} - \frac{t^5}{5!} + \ldots + \\ + t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots + & + 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \ldots + \end{pmatrix} = - \dot\gamma(0) = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} - \] - Gesucht: - \[ \dot \gamma (t) = \gamma(t) A \qquad \gamma \in \gSO2 \] - \[ \Rightarrow \gamma(t) = \exp(At) \gamma(0) = \exp(At) \] - \end{block} -\end{column} -\begin{column}{0.48\textwidth} - \begin{block}{Lie-Algebra} - Potenzen von A: - \begin{align*} - A^2 &= -I & - A^3 &= -A & - A^4 &= I & - \ldots - \end{align*} - \end{block} -\end{column} -\end{columns} -Folglich: -\begin{align*} - \exp(At) - &= I + At - + A^2\frac{t^2}{2!} - + A^3\frac{t^3}{3!} - + A^4\frac{t^4}{4!} - + A^5\frac{t^5}{5!} - + \ldots \\ - &= \begin{pmatrix} - \vspace*{3pt} - 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \ldots - & - -t + \frac{t^3}{3!} - \frac{t^5}{5!} + \ldots - \\ - t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots - & - 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \ldots - \end{pmatrix} - = - \begin{pmatrix} - \cos t & -\sin t \\ - \sin t & \phantom-\cos t - \end{pmatrix} -\end{align*} - + \begin{pmatrix} + \cos t & -\sin t \\ + \sin t & \phantom-\cos t + \end{pmatrix} + \end{align*} + } \end{frame} \egroup diff --git a/vorlesungen/slides/10/vektorfelder.mp b/vorlesungen/slides/10/vektorfelder.mp index f488327..e63b2d5 100644 --- a/vorlesungen/slides/10/vektorfelder.mp +++ b/vorlesungen/slides/10/vektorfelder.mp @@ -48,17 +48,17 @@ drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); label.top(btex $x_1$ etex, z2 shifted (10,0)); label.rt(btex $x_2$ etex, z4 shifted (0,10)); -% Draw circles -for x = 0.2 step 0.2 until 1.4: - path p; - p = (x,0); - for a = 5 step 5 until 355: - p := p--(x*cosd(a), x*sind(a)); - endfor; - p := p--cycle; - pickup pencircle scaled 1pt; - draw p scaled unit withcolor red; -endfor; +% % Draw circles +% for x = 0.2 step 0.2 until 1.4: +% path p; +% p = (x,0); +% for a = 5 step 5 until 355: +% p := p--(x*cosd(a), x*sind(a)); +% endfor; +% p := p--cycle; +% pickup pencircle scaled 1pt; +% draw p scaled unit withcolor red; +% endfor; % Define DGL def dglField(expr x, y) = @@ -66,6 +66,10 @@ def dglField(expr x, y) = (-y, x) enddef; +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; + % Draw arrows for each grid point pickup pencircle scaled 0.5pt; for x = -1.5 step 0.1 until 1.55: @@ -78,11 +82,9 @@ endfor; endfig; - - % % Vektorfeld in der Ebene mit Lösungskurve -% X \in sl(2, R) +% Euler(1) % beginfig(2) @@ -101,18 +103,28 @@ drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); label.top(btex $x_1$ etex, z2 shifted (10,0)); label.rt(btex $x_2$ etex, z4 shifted (0,10)); -% Draw flow lines -for y = -1.4 step 0.2 until 1.4: +def dglField(expr x, y) = + (-y, x) +enddef; + +def dglFieldp(expr z) = + dglField(xpart z, ypart z) +enddef; + +def curve(expr z, l, s) = path p; - p = (-1.5,y) -- (1.5, y); - pickup pencircle scaled 1pt; + p := z; + for t = 0 step 1 until l: + p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p))); + endfor; draw p scaled unit withcolor red; -endfor; - -def dglField(expr x, y) = - (y, 0) enddef; +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; +curve(A, 0, 1); + % Draw arrows for each grid point pickup pencircle scaled 0.5pt; for x = -1.5 step 0.1 until 1.55: @@ -125,12 +137,9 @@ endfor; endfig; - - - % % Vektorfeld in der Ebene mit Lösungskurve -% Y \in sl(2, R) +% Euler(2) % beginfig(3) @@ -149,42 +158,82 @@ drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); label.top(btex $x_1$ etex, z2 shifted (10,0)); label.rt(btex $x_2$ etex, z4 shifted (0,10)); -% Draw flow lines -for x = -1.4 step 0.2 until 1.4: +def dglField(expr x, y) = + (-y, x) +enddef; + +def dglFieldp(expr z) = + dglField(xpart z, ypart z) +enddef; + +def curve(expr z, l, s) = path p; - p = (x, -1.5) -- (x, 1.5); - pickup pencircle scaled 1pt; + p := z; + for t = 0 step 1 until l: + p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p))); + endfor; draw p scaled unit withcolor red; +enddef; + +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; +curve(A, 1, 0.5); + +% Draw arrows for each grid point +pickup pencircle scaled 0.5pt; +for x = -1.5 step 0.1 until 1.55: + for y = -1.5 step 0.1 until 1.55: + drawarrow ((x, y) * unit) + --(((x,y) * unit) shifted (8 * dglField(x,y))) + withcolor blue; + endfor; endfor; -def dglField(expr x, y) = - (0, x) -enddef; +endfig; -% def dglFieldp(expr z) = -% dglField(xpart z, ypart z) -% enddef; -% -% def curve(expr z, l) = -% path p; -% p := z; -% for t = 0 step 1 until l: -% p := p--((point (length p) of p) shifted (0.01 * dglFieldp(point (length p) of p))); -% endfor; -% draw p scaled unit withcolor red; -% enddef; % -% numeric outerlength; -% outerlength = 200; -% curve(( 0.1, 0), outerlength); -% curve(( 0.2, 0), outerlength); +% Vektorfeld in der Ebene mit Lösungskurve +% Euler(3) % -% numeric innerlength; -% innerlength = 500; -% -% for a = 0 step 30 until 330: -% curve(0.05 * (cosd(a), sind(a)), innerlength); -% endfor; +beginfig(4) + +numeric unit; +unit := 150; + +z0 = ( 0, 0); +z1 = (-1.5, 0) * unit; +z2 = ( 1.5, 0) * unit; +z3 = ( 0, -1.5) * unit; +z4 = ( 0, 1.5) * unit; + +pickup pencircle scaled 1pt; +drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0)); +drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); +label.top(btex $x_1$ etex, z2 shifted (10,0)); +label.rt(btex $x_2$ etex, z4 shifted (0,10)); + +def dglField(expr x, y) = + (-y, x) +enddef; + +def dglFieldp(expr z) = + dglField(xpart z, ypart z) +enddef; + +def curve(expr z, l, s) = + path p; + p := z; + for t = 0 step 1 until l: + p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p))); + endfor; + draw p scaled unit withcolor red; +enddef; + +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; +curve(A, 3, 0.25); % Draw arrows for each grid point pickup pencircle scaled 0.5pt; @@ -198,12 +247,11 @@ endfor; endfig; - % % Vektorfeld in der Ebene mit Lösungskurve -% H \in sl(2, R) +% Euler(4) % -beginfig(4) +beginfig(5) numeric unit; unit := 150; @@ -215,40 +263,88 @@ z3 = ( 0, -1.5) * unit; z4 = ( 0, 1.5) * unit; pickup pencircle scaled 1pt; -drawarrow (z1 shifted (-25,0))--(z2 shifted (25,0)); +drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0)); drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); -label.top(btex $x_1$ etex, z2 shifted (25,0)); +label.top(btex $x_1$ etex, z2 shifted (10,0)); label.rt(btex $x_2$ etex, z4 shifted (0,10)); def dglField(expr x, y) = - (x, -y) + (-y, x) enddef; def dglFieldp(expr z) = dglField(xpart z, ypart z) enddef; -def curve(expr z, l) = +def curve(expr z, l, s) = path p; p := z; for t = 0 step 1 until l: - p := p--((point (length p) of p) shifted (0.01 * dglFieldp(point (length p) of p))); + p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p))); endfor; draw p scaled unit withcolor red; enddef; -for i = -1 step 2 until 1: - for k = -1 step 2 until 1: - curve((1.3 * i, 1.5 * k), 18); - curve((1.1 * i, 1.5 * k), 35); - curve((0.9 * i, 1.5 * k), 55); - curve((0.7 * i, 1.5 * k), 80); - curve((0.5 * i, 1.5 * k), 114); - curve((0.3 * i, 1.5 * k), 165); - curve((0.1 * i, 1.5 * k), 275); +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; +curve(A, 7, 0.125); + +% Draw arrows for each grid point +pickup pencircle scaled 0.5pt; +for x = -1.5 step 0.1 until 1.55: + for y = -1.5 step 0.1 until 1.55: + drawarrow ((x, y) * unit) + --(((x,y) * unit) shifted (8 * dglField(x,y))) + withcolor blue; endfor; endfor; +endfig; + +% +% Vektorfeld in der Ebene mit Lösungskurve +% Euler(5) +% +beginfig(6) + +numeric unit; +unit := 150; + +z0 = ( 0, 0); +z1 = (-1.5, 0) * unit; +z2 = ( 1.5, 0) * unit; +z3 = ( 0, -1.5) * unit; +z4 = ( 0, 1.5) * unit; + +pickup pencircle scaled 1pt; +drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0)); +drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); +label.top(btex $x_1$ etex, z2 shifted (10,0)); +label.rt(btex $x_2$ etex, z4 shifted (0,10)); + +def dglField(expr x, y) = + (-y, x) +enddef; + +def dglFieldp(expr z) = + dglField(xpart z, ypart z) +enddef; + +def curve(expr z, l, s) = + path p; + p := z; + for t = 0 step 1 until l: + p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p))); + endfor; + draw p scaled unit withcolor red; +enddef; + +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; +curve(A, 99, 0.01); + % Draw arrows for each grid point pickup pencircle scaled 0.5pt; for x = -1.5 step 0.1 until 1.55: @@ -262,5 +358,4 @@ endfor; endfig; - end; diff --git a/vorlesungen/slides/10/vektorfelder.tex b/vorlesungen/slides/10/vektorfelder.tex new file mode 100644 index 0000000..a4612aa --- /dev/null +++ b/vorlesungen/slides/10/vektorfelder.tex @@ -0,0 +1,82 @@ +% +% iterativ.tex -- Iterative Approximation in \dot x = J x +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% Erstellt durch Roy Seitz +% +% !TeX spellcheck = de_CH +\bgroup +\begin{frame}[t] + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + \frametitle{Als Strömungsfeld} + \vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \vfil + \only<1>{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-1.pdf} + } + \only<2>{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-2.pdf} + } + \only<3>{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-3.pdf} + } + \only<4>{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-4.pdf} + } + \only<5>{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-5.pdf} + } + \only<6->{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-6.pdf} + } + \vfil + \end{column} + \begin{column}{0.48\textwidth} + \begin{block}{Differentialgleichung} + \[ + \dot x(t) = J x(t) + \quad + J = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} + \quad + x_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \] + \end{block} + + \only<2>{ + Nach einem Schritt der Länge $t$: + \[ + x(t) = x_0 + \dot x t = x_0 + Jx_0t = (1 + Jt)x_0 + \] + } + + \only<3>{ + Nach zwei Schritten der Länge $t/2$: + \[ + x(t) = \left(1 + \frac{Jt}{2}\right)^2x_0 + \] + } + + \only<4->{ + Nach n Schritten der Länge $t/n$: + \[ + x(t) = \left(1 + \frac{Jt}{n}\right)^nx_0 + \] + } + \only<6->{ + \[ + \lim_{n\to\infty}\left(1 + \frac{At}{n}\right)^n = \exp(At) + \] + } + \end{column} + \end{columns} +\end{frame} +\egroup |