Differentialgleichung/Picard-Lindelöf/t^2y-xy^2,x^3-y+t/Beispiel

Wir wenden die Picard-Lindelöf-Iteration auf das Anfangswertproblem

$v(0)=(0,0)$

zum Vektorfeld

$F:ℝ\times {ℝ}^2⟶{ℝ}^2,(t,x,y)⟼(t^{2}y-x{y}^2,x^3-y+t)$

an. Es ist

${\phi }_0(t)=(0,0).$

Daher ist

${\phi }_1(t)={\int }_0^t(0,s)ds=(0,\frac{1}{2}{t}^2).$

Es ist

$F(s,{\phi }_1(s))=(s^2\frac{1}{2}{s}^2,-\frac{1}{2}{s}^2+s)=(\frac{1}{2}{s}^4,-\frac{1}{2}{s}^2+s)$

und daher

${\phi }_2(t)={\int }_0^t(\frac{1}{2}{s}^4,-\frac{1}{2}{s}^2+s)ds=(\frac{1}{10}{t}^5,-\frac{1}{6}{t}^3+\frac{1}{2}{t}^2).$

Wegen

$\begin{array}{cc}F(s,{\phi }_2(s))& =(s^2(-\frac{1}{6}{s}^3+\frac{1}{2}{s}^2)-\frac{1}{10}{s}^5{(-\frac{1}{6}{s}^3+\frac{1}{2}{s}^2)}^2,{\left(\frac{1}{10}{t}^5\right)}^3-(-\frac{1}{6}{s}^3+\frac{1}{2}{s}^2)+s)\\ & =(-\frac{1}{6}{s}^5+\frac{1}{2}{s}^4-\frac{1}{10}{s}^5(\frac{1}{36}{s}^6-\frac{1}{6}{s}^5+\frac{1}{4}{s}^4),\frac{1}{1000}{s}^{15}+\frac{1}{6}{s}^3-\frac{1}{2}{s}^2+s)\\ & =(-\frac{1}{360}{s}^{11}+\frac{1}{60}{s}^{10}-\frac{1}{40}{s}^9-\frac{1}{6}{s}^5+\frac{1}{2}{s}^4,\frac{1}{1000}{s}^{15}+\frac{1}{6}{s}^3-\frac{1}{2}{s}^2+s)\end{array}$

und daher

$\begin{array}{cc}{\phi }_3(t)& ={\int }_0^t(-\frac{1}{360}{s}^{11}+\frac{1}{60}{s}^{10}-\frac{1}{40}{s}^9-\frac{1}{6}{s}^5+\frac{1}{2}{s}^4,\frac{1}{1000}{s}^{15}+\frac{1}{6}{s}^3-\frac{1}{2}{s}^2+s)ds\\ & =(-\frac{1}{4320}{t}^{12}+\frac{1}{660}{t}^{11}-\frac{1}{400}{t}^{10}-\frac{1}{36}{t}^6+\frac{1}{10}{t}^5,\frac{1}{16000}{t}^{16}+\frac{1}{24}{t}^4-\frac{1}{6}{t}^3+\frac{1}{2}{t}^2).\end{array}$