Präsentation feritg.

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