diff --git a/ana16.tex b/ana16.tex index 3607fad..89f2a41 100644 --- a/ana16.tex +++ b/ana16.tex @@ -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} \ No newline at end of file +\end{document} diff --git a/analysisII.pdf b/analysisII.pdf index b7163b8..c2d6e93 100644 Binary files a/analysisII.pdf and b/analysisII.pdf differ