128 lines
5.4 KiB
TeX
128 lines
5.4 KiB
TeX
\section{Aufgabe 125}
|
|
|
|
Gegeben sei das eindeutig lösbare lineare Gleichungssystem $\ A\cdot
|
|
\overrightarrow{x}=\overrightarrow{b}$ mit
|
|
|
|
$A=\left(
|
|
\begin{array}
|
|
[c]{cccccc}%
|
|
4 & -1 & 0 & -1 & 0 & 0\\
|
|
-1 & 4 & -1 & 0 & -1 & 0\\
|
|
0 & -1 & 4 & 0 & 0 & -1\\
|
|
-1 & 0 & 0 & 4 & -1 & 0\\
|
|
0 & -1 & 0 & -1 & 4 & -1\\
|
|
0 & 0 & -1 & 0 & -1 & 4
|
|
\end{array}
|
|
\right) $, $\overrightarrow{b}=\left(
|
|
\begin{array}
|
|
[c]{c}%
|
|
2\\
|
|
1\\
|
|
2\\
|
|
2\\
|
|
1\\
|
|
2
|
|
\end{array}
|
|
\right) $
|
|
|
|
\begin{itemize}
|
|
\item[a.] Sei $\overrightarrow{x}^{\left( 0\right) }=\overrightarrow{0}$.
|
|
Berechnen Sie die Näherungsösung $\overrightarrow{x}^{\left(
|
|
3\right) }$\ des Systems, die man nach 3 Schritten des
|
|
Gesamtschrittverfahrens erhält.
|
|
|
|
\item[b.] Zeigen Sie, daß das Gesamtschrittverfahren konvergiert.
|
|
|
|
\item[c.] Führen Sie eine Apeoteriori-Fehlerabschätzung für
|
|
$\overrightarrow{x}^{\left( 3\right) }$\ durch.
|
|
|
|
\item[d.] Führen Sie eine Apriori-Fehlerabschätzung für
|
|
$\overrightarrow{x}^{\left( 10\right) }$\ durch.
|
|
\end{itemize}
|
|
|
|
\subsection{Lösung}
|
|
|
|
\begin{itemize}
|
|
\item[a.] Rechenvorschriften:\newline$x_{1}^{\left( Z\right) }=\frac{1}%
|
|
{4}\left( 2+x_{2}^{\left( Z-1\right) }+x_{4}^{\left( Z-1\right) }\right)
|
|
=\frac{1}{2}+\frac{1}{4}x_{2}^{\left( Z-1\right) }+\frac{1}{4}x_{4}^{\left(
|
|
Z-1\right) }$\newline$x_{2}^{\left( Z\right) }=\frac{1}{4}\left(
|
|
1+x_{1}^{\left( Z-1\right) }+x_{3}^{\left( Z-1\right) }+x_{5}^{\left(
|
|
Z-1\right) }\right) =\frac{1}{4}+\frac{1}{4}x_{1}^{\left( Z-1\right)
|
|
}+\frac{1}{4}x_{3}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
|
|
Z-1\right) }$\newline$x_{3}^{\left( Z\right) }=\frac{1}{4}\left(
|
|
2+x_{2}^{\left( Z-1\right) }+x_{6}^{\left( Z-1\right) }\right) =\frac
|
|
{1}{2}+\frac{1}{4}x_{2}^{\left( Z-1\right) }+\frac{1}{4}x_{6}^{\left(
|
|
Z-1\right) }$\newline$x_{4}^{\left( Z\right) }=\frac{1}{4}\left(
|
|
2+x_{1}^{\left( Z-1\right) }+x_{5}^{\left( Z-1\right) }\right) =\frac
|
|
{1}{2}+\frac{1}{4}x_{1}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
|
|
Z-1\right) }$\newline$x_{5}^{\left( Z\right) }=\frac{1}{4}\left(
|
|
1+x_{2}^{\left( Z-1\right) }+x_{4}^{\left( Z-1\right) }+x_{6}^{\left(
|
|
Z-1\right) }\right) =\frac{1}{4}+\frac{1}{4}x_{2}^{\left( Z-1\right)
|
|
}+\frac{1}{4}x_{4}^{\left( Z-1\right) }+\frac{1}{4}x_{6}^{\left(
|
|
Z-1\right) }$\newline$x_{6}^{\left( Z\right) }=\frac{1}{4}\left(
|
|
2+x_{3}^{\left( Z-1\right) }+x_{5}^{\left( Z-1\right) }\right) =\frac
|
|
{1}{2}+\frac{1}{4}x_{3}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
|
|
Z-1\right) }$\newline\newline%
|
|
\begin{tabular}
|
|
[c]{l||llllll}%
|
|
z & $x_{1}^{\left( Z\right) }$ & $x_{2}^{\left( Z\right) }$ &
|
|
$x_{3}^{\left( Z\right) }$ & $x_{4}^{\left( Z\right) }$ & $x_{5}^{\left(
|
|
Z\right) }$ & $x_{6}^{\left( Z\right) }$\\\hline\hline
|
|
1 & $0.5$ & $0.25$ & $0.5$ & $0.5$ & $0.25$ & $0.5$\\
|
|
2 & $0.6875$ & $0.5625$ & $0.6875$ & $0.6875$ & $0.5625$ & $0.6875$\\
|
|
3 & $0.8125$ & $0.734325$ & $0.8125$ & $0.8125$ & $0.734375$ & $0.8125$%
|
|
\end{tabular}
|
|
\newline
|
|
|
|
$\overrightarrow{x}=\overrightarrow{x}^{\left( 3\right) }=\left(
|
|
0.8125;\text{ }0.734325;\text{ }0.8125;\text{ }0.8125;\text{ }0.734375;\text{
|
|
}0.8125\right) $
|
|
|
|
\item[b.] Berechnung der Kontraktionszahl $\lambda$\newline\newline$%
|
|
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq1}}^{6}}%
|
|
%BeginExpansion
|
|
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq1}}^{6}}
|
|
%EndExpansion
|
|
\left\vert \frac{a_{1j}}{a_{11}}\right\vert =0.5;$ \ \ $%
|
|
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq2}}^{6}}%
|
|
%BeginExpansion
|
|
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq2}}^{6}}
|
|
%EndExpansion
|
|
\left\vert \frac{a_{2j}}{a_{22}}\right\vert =0.75;$ \ \ $%
|
|
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq3}}^{6}}%
|
|
%BeginExpansion
|
|
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq3}}^{6}}
|
|
%EndExpansion
|
|
\left\vert \frac{a_{3j}}{a_{33}}\right\vert =0.5;$\newline\newline$%
|
|
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq4}}^{6}}%
|
|
%BeginExpansion
|
|
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq4}}^{6}}
|
|
%EndExpansion
|
|
\left\vert \frac{a_{4j}}{a_{44}}\right\vert =0.5;$ \ \ $%
|
|
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq5}}^{6}}%
|
|
%BeginExpansion
|
|
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq5}}^{6}}
|
|
%EndExpansion
|
|
\left\vert \frac{a_{5j}}{a_{55}}\right\vert =0.75;$ \ \ $%
|
|
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq6}}^{6}}%
|
|
%BeginExpansion
|
|
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq6}}^{6}}
|
|
%EndExpansion
|
|
\left\vert \frac{a_{6j}}{a_{66}}\right\vert =0.5;$\newline\newline%
|
|
$\lambda=\max\{0.5;0.75\}=\underline{\underline{0.75}}$\newline$\lambda
|
|
<1\Longrightarrow$ \underline{das Gesamtschrittverfahren konvergiert} für
|
|
jeden Startvektor
|
|
|
|
\item[c.] $\underset{i=1,...,6}{\max}\left\vert x_{i}^{\left( 3\right)
|
|
}-x_{i}\right\vert \leq\frac{\lambda}{1-\lambda}$ \ \ \ $\underset
|
|
{i=1,...,6}{\max}\left\vert x_{i}^{\left( 3\right) }-x_{i}^{\left(
|
|
2\right) }\right\vert =\frac{0.75}{0.25}\cdot0.171875=\allowbreak
|
|
\underline{\underline{0.515\,625}}\,$
|
|
|
|
\item[d.] $\underset{i=1,...,6}{\max}\left\vert x_{i}^{\left( 10\right)
|
|
}-x_{i}\right\vert \leq\frac{\lambda^{10}}{1-\lambda}$ \ \ $\underset
|
|
{i=1,...,6}{\max}\left\vert x_{i}^{\left( 1\right) }-x_{i}^{\left(
|
|
0\right) }\right\vert =\frac{0.75^{10}}{0.25}\cdot0.5=\underline{\underline{\allowbreak
|
|
0.112\,627\,029\,\allowbreak4}}$
|
|
\end{itemize} |