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christian
2020-06-28 20:12:55 +02:00
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commit df5ec7ac25
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ana16.tex
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@@ -82,9 +82,51 @@ Notationen: $x' = f(t,x), \dot x = f(t,x), \dv{x}{t} = f(t,x)$ (Dynamischer Proz
\end{align*}
\end{enumerate}
\end{bsp}
%\begin{figure}[h]
% \caption{Veranschaulichung: Richtungsfeld
%\end{figure}
\begin{figure}[h]
\centering
\begin{tikzpicture}[declare function={f(\x) = 2*(\x-0.25);}]
\begin{axis}%
[%minor tick num=4,
%grid style={line width=.1pt, draw=gray!10},
%major grid style={line width=.2pt,draw=gray!50},
%axis lines=middle,
%enlargelimits={abs=0.2},
%ymax=5,
%ymin=0
width=0.6\textwidth, % Overall width of the plot
axis equal image, % Unit vectors for both axes have the same length
view={0}{90}, % We need to use "3D" plots, but we set the view so we look at them from straight up
xmin=0, xmax=1.1, % Axis limits
ymin=0, ymax=1.1,
domain=0:1, y domain=0:1, % Domain over which to evaluate the functions
xtick={0.7}, ytick={0.3525}, % Tick marks
xticklabels={$t_0$},
yticklabels={$y_0$},
xlabel=$t$,
ylabel=$y$,
samples=11, % How many arrows?
cycle list={ % Plot styles
gray,
quiver={
u={1}, v={f(x)}, % End points of the arrows
scale arrows=0.075,
every arrow/.append style={
-latex % Arrow tip
},
}\\
red, samples=31, smooth, thick, no markers, domain=0:1.1\\ % The plot style for the function
}
]
\addplot3 (x,y,0);
\addlegendentry{$f'(t,x)$}
\addplot{(x-0.25)^2+0.15};
\addlegendentry{$y(t)$}
\end{axis}
\end{tikzpicture}
\caption{Veranschaulichung: Richtungsfeld für DGL der Form $y' = f(x)$}
\end{figure}
\begin{definition}[System erster Ordnung]
Sei $D = I\times \Omega \subset \R\times \R^n,\ f\colon D\to \R^n$ stetig. Dann heißt
\begin{equation}
@@ -141,6 +183,30 @@ Eine Lösung von \eqref{DGLOrd1} ist eine differenzierbare Funktion $y:I\to \R^n
\]
Dann existiert eine Lösung $y(t)$ von AWA auf dem Intervall $I \coloneqq [t_0-T,t_0+T]$ mit \[T \coloneqq \min_{y(t)}\left\{\alpha,\frac{\beta}{M}\right\},\; M\coloneqq \max_{(t,x)\in D}\norm{f(t,x)}\]
\end{satz}
%\begin{figure}[h]
% \begin{tikzpicture}
% \begin{axis}%
% [grid=none,
% minor tick num=4,
% grid style={line width=.1pt, draw=gray!10},
% major grid style={line width=.2pt,draw=gray!50},
% axis lines=middle,
% %enlargelimits={abs=0.2},
% ymax=5, ymin=-1.5,
% xmin=2, xmax=7,
% xtick={5}, ytick={2},
% xticklabels={$t_0$},
% yticklabels={$y_0$},
% xlabel=$t$,
% ylabel=$x$,
% ]
% \draw (4,1) rectangle (6,3);
% \node at (5.8,1.3) {$D$};
% \addplot[domain=1:10,samples=50,smooth,red] {2^(x-3)-2};
% \addlegendentry{$y(t)$}
% \end{axis}
% \end{tikzpicture}
%\end{figure}
Reminder:
\begin{enumerate}
\item Gleichmäßige Stetigkeit: \[f\colon D\to \R,\; D\subset \R^n\] ist gleichmäßig stetig in $D$, falls $\forall \epsilon > 0,\;\exists \delta > 0$, sodass $\forall x,x_0\in D$ gilt \[\norm{x-x_0}< \delta \implies \norm{f(x)-f(x_0)}< \epsilon\]
@@ -168,6 +234,48 @@ Reminder:
\end{itemize}
Definiere die stückweise lineare Funktion $y^h(t)$
\[y^h(t)\coloneqq y_{n-1}^h + (t-t_{n-1})f(t_{n-1},y_{n-1}^h),\quad t\in [t_{n-1},t_n],\quad \forall n\ge 1\]
\begin{figure}[h]
\centering
\begin{tikzpicture}[declare function={f1(\x) = 0.5*(2)^(\x-1) + 10/(\x+2);
f2(\x) = 0.5*(2)^(\x-1);
f3(\x) = 0.5*(2)^(\x-1) - 10/(\x+2);
f4(\x) = 0.5*(2)^(\x-1) - 20/(\x+2);}]
\begin{axis}%
[grid=none,
%minor tick num=4,
grid style={line width=.1pt, draw=gray!10},
major grid style={line width=.2pt,draw=gray!50},
axis lines=middle,
%enlargelimits={abs=0.2},
ymax=10, ymin=-1.5,
xmin=-1, xmax=7,
xtick={2,3,4,5},
ytick=\empty,
xticklabels={$t_0$, $t_1$, $t_2$, $t_3$},
%yticklabels={$y_0$, $y_1^{h}$, $y_2^{h}$},
xlabel=$t$,
ylabel=$x$,
]
\addplot[domain=0:10,samples=50,smooth,green] {f1(x)};
\addlegendentry{$y(t,t_0,y_0)$};
\addplot[domain=0:10,samples=50,smooth,blue] {f2(x)};
\addlegendentry{$y(t, t_1, y_1^{h})$};
\addplot[domain=0:10,samples=50,smooth,orange] {f3(x)};
\addlegendentry{$y(t, t_2, y_2^{h})$};
\addplot[domain=0:10,samples=50,smooth,pink] {f4(x)};
\addlegendentry{$y(t, t_3, y_3^{h})$};
\draw (2,{f1(2)}) node[circle,fill,inner sep=0.5pt] {}
-- (3,{f2(3)}) node[circle,fill,inner sep=0.5pt] {}
-- (4, {f3(4)}) node[circle,fill,inner sep=0.5pt] {}
-- (5, {f4(5)}) node[circle,fill,inner sep=0.5pt] {};
\draw[dashed,green] (2, {f1(2)}) -- (0, {f1(2)}) node[label=left:$y_0$](){};
\draw[dashed,blue] (3, {f2(3)}) -- (0, {f2(3)}) node[label=left:$y_1$](){};
\draw[dashed,orange] (4, {f3(4)}) -- (0, {f3(4)}) node[label=left:$y_2$](){};
\draw[dashed,pink] (5, {f4(5)}) -- (0, {f4(5)}) node[label=left:$y_3$](){};
\end{axis}
\end{tikzpicture}
\caption{Eulersches Polygonzugverfahren}
\end{figure}
\begin{enumerate}[1)]
\item \textbf{z.Z.} dass dieses Verfahren durchführbar ist, d.h. $\graph(y^h)\subset D$. Sei $(t,y^h(t))\subset D$ für $t_0 \le t\le t_{k-1}$. Dann gilt
\[
@@ -187,7 +295,52 @@ Reminder:
\end{align*}
Also ist $(t,y^h(t))\in D$ für $t_{k-1} \le t\le t_k$. Mit Annahme folgt $(t,y^h(t))\in D$ für $t_0\le t\le t_k \implies \graph(y^h)\subset D$.
\item \begin{enumerate}[(a)]
\item \textbf{z.Z.} dass die Funktionenfamilie $\{y^h\}_{h>0}$ gleichgradig stetig ist. Seien dafür $t,t'\in I, \ t'\le t$ beliebig mit $t\in [t_{k-1},t_k],\; t'\in [t_{j-1},t_j]$ für ein $t_j\le t_k$.\begin{itemize}
\item \textbf{z.Z.} dass die Funktionenfamilie $\{y^h\}_{h>0}$ gleichgradig stetig ist. Seien dafür $t,t'\in I, \ t'\le t$ beliebig mit $t\in [t_{k-1},t_k],\; t'\in [t_{j-1},t_j]$ für ein $t_j\le t_k$.
\begin{figure}[h]
\centering
\begin{tikzpicture}[declare function={f1(\x) = 0.5*(2)^(\x-1) + 10/(\x+2);
f2(\x) = 0.5*(2)^(\x-1);
f3(\x) = 0.5*(2)^(\x-1) - 10/(\x+2);
f4(\x) = 0.5*(2)^(\x-1) - 20/(\x+2);}]
\begin{axis}%
[grid=none,
%minor tick num=4,
grid style={line width=.1pt, draw=gray!10},
major grid style={line width=.2pt,draw=gray!50},
axis lines=middle,
%enlargelimits={abs=0.2},
ymax=10, ymin=-1.5,
xmin=1, xmax=6,
xtick={2,3,4,5},
ytick={1},
xticklabels={$t_0$, $t_1$, $t_2$, $t_3$},
yticklabels={$y_0$},
xlabel=$t$,
ylabel=$x$,
]
\addplot[domain=0:10,samples=50,smooth] {f2(x)};
\draw (2,{f2(2)}) node[circle,fill,inner sep=0.5pt] {}
(3,{f2(3)}) node[circle,fill,inner sep=0.5pt] {}
(4,{f2(4)}) node[circle,fill,inner sep=0.5pt] {}
(5,{f2(5)}) node[circle,fill,inner sep=0.5pt] {};
\draw[dashed,red] (2.4, {f2(2.4)}) -- (2.4, 0)
node [label={[label distance=-0.8mm]below:$t$}](){};
\draw[dashed,red] (4.7, {f2(4.7)}) -- (4.7, 0)
node [label={[label distance=-0.8mm]below:$t'$}](){};
\draw[dashed,blue] (3.2, {f2(3.2)}) -- (3.2, 0)
node [label={[label distance=-1mm]below:$t$}](){};
\draw[dashed,blue] (3.8, {f2(3.8)}) -- (3.8, 0)
node [label={[label distance=-1mm]below:$t'$}](){};
\draw[dashed,black] (2, {f2(2)}) -- (0, {f2(2)})
node [label={[label distance=-1mm]below:$t$}](){};
%\draw[dashed,blue] (3, {f2(3)}) -- (0, {f2(3)}) node[label=left:$y_1$](){};
%\draw[dashed,orange] (4, {f3(4)}) -- (0, {f3(4)}) node[label=left:$y_2$](){};
%\draw[dashed,pink] (5, {f4(5)}) -- (0, {f4(5)}) node[label=left:$y_3$](){};
\end{axis}
\end{tikzpicture}
\caption{Blau: erster Fall, Rot: zweiter Fall}
\end{figure}
\begin{itemize}
\item $t,t' \in [t_{k-1},t_k]$:
\begin{align*}
y^h(t)-y^h(t')&= y_{k-1}^h + (t-t_{k-1})f(t_{k-1},y_{k-1}^h)\\
@@ -228,4 +381,4 @@ Reminder:
\item \textbf{z.Z.} $y(t)$ erfüllt die Differentialgleichung $y'(t) = f(t,y(t))$ oder äquivalent dazu: $y(t)$ erfüllt die Integralgleichung \[y(t) = y_0 + \int_{t_0}^{t} f(s,y(s))\d s\]
\end{enumerate}
\end{proof}
\end{document}
\end{document}
analysisII.pdf
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