[系統理論] Picard Iteration

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{array} \right]} d\tau \\
 \Rightarrow {{\bf{x}}^1}(t) = \left[ \begin{array}{l}
2\\
 - 1
\end{array} \right] + \int_0^t {\left[ \begin{array}{l}
\cos 2\\
\tau 2 + {e^{ - \tau }}\left( { - 1} \right)
\end{array} \right]} d\tau \\
 \Rightarrow {{\bf{x}}^1}(t) = \left[ \begin{array}{l}
2\\
 - 1
\end{array} \right] + \left[ \begin{array}{l}
t\cos 2\\
{t^2} + \left( {{e^{ - t}} - 1} \right)
\end{array} \right] = \left[ \begin{array}{l}
2 + t\cos 2\\
{t^2} + {e^{ - t}} - 2
\end{array} \right]

\end{array}\]

Exercise: 讀者可自行嘗試透過 Picard Iteration 求解 $x^2(t)$ 與 $x^3(t)$


Example 2: Saturation 
考慮非線性系統 $\dot{x} = f(x)$ 且 $x(0) =1$ 其中
\[f\left( x \right) = \left\{ \begin{array}{l}
1,\begin{array}{*{20}{c}}
{}&{}
\end{array}x > 2\\
\frac{1}{2}x,\begin{array}{*{20}{c}}
{}&{}
\end{array}\left| x \right| \le 2\\
 - 1,\begin{array}{*{20}{c}}
{}&{}
\end{array}x <  - 2
\end{array} \right.\]試透過 Picard iteration 求取 $ x^1(t)$ 與 $x^2(t)$
Solution:
首先計算 $x^1(t)$:
\[\begin{array}{l}
{x^1}\left( t \right) = {x^0}\left( t \right) + \int_0^t {f\left( {{x^0}\left( \tau  \right)} \right)d\tau } \\
 \Rightarrow {x^1}\left( t \right) = 1 + \int_0^t {f\left( 1 \right)d\tau } \\
 \Rightarrow {x^1}\left( t \right) = 1 + \frac{1}{2}t
\end{array}\]
接著我們計算 $x^2(t)$:
\[\begin{array}{l}
{x^2}\left( t \right) = {x^0}\left( t \right) + \int_0^t {f\left( {{x^1}\left( \tau  \right)} \right)d\tau } \\
 \Rightarrow {x^2}\left( t \right) = 1 + \left\{ \begin{array}{l}
\int_0^t {\frac{1}{2}\left( {1 + \frac{1}{2}\tau } \right)d\tau ,\begin{array}{*{20}{c}}
{}&{}
\end{array}0 \le t \le 2} \\
\frac{1}{2}\int_0^t {1d\tau } ,\begin{array}{*{20}{c}}
{}&{}
\end{array}t > 2
\end{array} \right.\\
 \Rightarrow {x^2}\left( t \right) = 1 + \left\{ \begin{array}{l}
\frac{1}{2}\left( {t + \frac{1}{4}{t^2}} \right),\begin{array}{*{20}{c}}
{}&{}
\end{array}0 \le t \le 2\\
\frac{1}{2}t,\begin{array}{*{20}{c}}
{}&{}
\end{array}t > 2
\end{array} \right.\\
 \Rightarrow {x^2}\left( t \right) = \left\{ \begin{array}{l}
1 + \frac{t}{2} + \frac{1}{8}{t^2},\begin{array}{*{20}{c}}
{}&{}
\end{array}0 \le t \le 2\\
1 + \frac{t}{2},\begin{array}{*{20}{c}}
{}&{}
\end{array}t > 2
\end{array} \right.
\end{array}\]