Zum Inhalt springen

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

Wir wollen für das Differentialgleichungssystem

${\displaystyle {}{\begin{pmatrix}x'\\y'\end{pmatrix}}={\begin{pmatrix}x^{2}-ty\\txy\end{pmatrix}}=F(t,x,y)\,}$

mit der Anfangsbedingung

${\displaystyle {}{\begin{pmatrix}x(0)\\y(0)\end{pmatrix}}={\begin{pmatrix}1\\1\end{pmatrix}}\,}$

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

${\displaystyle {}P_{0}={\begin{pmatrix}1\\1\end{pmatrix}}\,,}$
${\displaystyle {}P_{1}=P_{0}+{\frac {1}{10}}F{\left(0,P_{0}\right)}={\begin{pmatrix}1\\1\end{pmatrix}}+{\frac {1}{10}}{\begin{pmatrix}1\\0\end{pmatrix}}={\begin{pmatrix}1,1\\1\end{pmatrix}}\,,}$
{\displaystyle {}{\begin{aligned}P_{2}&=P_{1}+{\frac {1}{10}}F{\left({\frac {1}{10}},P_{1}\right)}\\&={\begin{pmatrix}{\frac {11}{10}}\\1\end{pmatrix}}+{\frac {1}{10}}{\begin{pmatrix}{\left({\frac {11}{10}}\right)}^{2}-{\frac {1}{10}}\cdot 1\\{\frac {1}{10}}\cdot {\frac {11}{10}}\cdot 1\end{pmatrix}}\\&={\begin{pmatrix}{\frac {11}{10}}\\1\end{pmatrix}}+{\frac {1}{10}}{\begin{pmatrix}{\frac {111}{100}}\\{\frac {11}{100}}\end{pmatrix}}\\&={\begin{pmatrix}{\frac {1211}{1000}}\\{\frac {1011}{1000}}\end{pmatrix}}\end{aligned}}}

und

{\displaystyle {}{\begin{aligned}P_{3}&=P_{2}+{\frac {1}{10}}F{\left({\frac {2}{10}},P_{2}\right)}\\&={\begin{pmatrix}{\frac {1211}{1000}}\\{\frac {1011}{1000}}\end{pmatrix}}+{\frac {1}{10}}{\begin{pmatrix}{\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{pmatrix}}\\&={\begin{pmatrix}{\frac {1211}{1000}}\\{\frac {1011}{1000}}\end{pmatrix}}+{\frac {1}{10}}{\begin{pmatrix}{\frac {1264321}{1000000}}\\{\frac {2448642}{10000000}}\end{pmatrix}}\\&={\begin{pmatrix}{\frac {133743210}{100000000}}\\{\frac {103548642}{100000000}}\end{pmatrix}}.\end{aligned}}}