(x,y)' ist (x^2-ty,txy)/Polygonzugverfahren/Beispiel

Wir wollen für das Differentialgleichungssystem

$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{c}{x}^2-ty\\ txy\end{array}\right)=F(t,x,y)$

mit der Anfangsbedingung

$\left(\begin{array}{c}x(0)\\ y(0)\end{array}\right)=\left(\begin{array}{c}1\\ 1\end{array}\right)$

gemäß Verfahren einen approximierenden Streckenzug berechnen. Wir wählen die Schrittweite  $s=\frac{1}{10}$.  Somit ist

$P_0=\left(\begin{array}{c}1\\ 1\end{array}\right),$

$P_1=P_0+\frac{1}{10}F(0,P_0)=\left(\begin{array}{c}1\\ 1\end{array}\right)+\frac{1}{10}\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}1,1\\ 1\end{array}\right),$

$\begin{array}{cc}{P}_2& =P_1+\frac{1}{10}F(\frac{1}{10},P_1)\\ & =\left(\begin{array}{c}\frac{11}{10}\\ 1\end{array}\right)+\frac{1}{10}\left(\begin{array}{c}{\left(\frac{11}{10}\right)}^2-\frac{1}{10}\cdot 1\\ \frac{1}{10}\cdot \frac{11}{10}\cdot 1\end{array}\right)\\ & =\left(\begin{array}{c}\frac{11}{10}\\ 1\end{array}\right)+\frac{1}{10}\left(\begin{array}{c}\frac{111}{100}\\ \frac{11}{100}\end{array}\right)\\ & =\left(\begin{array}{c}\frac{1211}{1000}\\ \frac{1011}{1000}\end{array}\right)\end{array}$

und

$\begin{array}{cc}{P}_3& =P_2+\frac{1}{10}F(\frac{2}{10},P_2)\\ & =\left(\begin{array}{c}\frac{1211}{1000}\\ \frac{1011}{1000}\end{array}\right)+\frac{1}{10}\left(\begin{array}{c}{\left(\frac{1211}{1000}\right)}^2-\frac{2}{10}\cdot \frac{1011}{1000}\\ \frac{2}{10}\cdot \frac{1211}{1000}\cdot \frac{1011}{1000}\end{array}\right)\\ & =\left(\begin{array}{c}\frac{1211}{1000}\\ \frac{1011}{1000}\end{array}\right)+\frac{1}{10}\left(\begin{array}{c}\frac{1264321}{1000000}\\ \frac{2448642}{10000000}\end{array}\right)\\ & =\left(\begin{array}{c}\frac{133743210}{100000000}\\ \frac{103548642}{100000000}\end{array}\right).\end{array}$