考慮狀態空間 \[{\bf{\dot x}}\left( t \right) = f\left( {t,{\bf{x}}\left( t \right)} \right),{\bf{x}}\left( 0 \right) = {{\bf{x}}^0}\]則 Picard Iteration 給定 1. Initial guess \[ x(t) = x^0 \]2.. Update Step \[{{\bf{x}}^{k + 1}}\left( t \right) = {{\bf{x}}^0} + \int_0^t {f\left( {\tau ,{{\bf{x}}^k}\left( \tau \right)} \right)d\tau } \] 注意到上述迭代式為 sequence of $\{{\bf x}^k\}$ Example 1: 考慮下列非線性系統 \[{\bf{\dot x}}\left( t \right) = \left[ \begin{array}{l} {{\dot x}_1}\\ {{\dot x}_2} \end{array} \right] = \left[ \begin{array}{l} \cos {x_1}\\ t{x_1} + {e^{ - t}}{x_2} \end{array} \right] \]且 $x_1(0) = 2$ 與 $x_2(0) = -1$。 試透過 Picard iteration 求取 ${\bf x }^1(t)$ Solution: \[\begin{array}{l} {{\bf{x}}^{k + 1}}\left( t \right) = {{\bf{x}}^0} + \int_0^t {f\left( {\tau ,{{\bf{x}}^k}\left( \tau \right)} \right)d\tau } \\ {{\bf{x}}^1}(t) = {{\bf{x}}^0}(t) + \int_0^t {\left[ \begin{array}{l} \cos {x_1}\left( \tau \right)\\ \tau {x_1} + {e^{ - \tau }}{x_2}\left( \tau \right) \end{a
If you can’t solve a problem, then there is an easier problem you can solve: find it. -George Polya