From 4313f2c207d5d60171898ccfd4c3b3d0d2fb4a75 Mon Sep 17 00:00:00 2001
From: Roy Seitz <roy.seitz@ost.ch>
Date: Sun, 18 Apr 2021 17:49:56 +0200
Subject: =?UTF-8?q?Pr=C3=A4sentation=20feritg.?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
vorlesungen/slides/10/intro.tex | 45 ++++++
vorlesungen/slides/10/n-zu-1.tex | 98 +++++++-------
vorlesungen/slides/10/repetition.tex | 40 +-----
vorlesungen/slides/10/so2.tex | 237 ++++++++++++++++----------------
vorlesungen/slides/10/vektorfelder.mp | 241 +++++++++++++++++++++++----------
vorlesungen/slides/10/vektorfelder.tex | 82 +++++++++++
6 files changed, 475 insertions(+), 268 deletions(-)
create mode 100644 vorlesungen/slides/10/intro.tex
create mode 100644 vorlesungen/slides/10/vektorfelder.tex
(limited to 'vorlesungen/slides/10')
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
--
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