考慮 LQR 問題如下 成本函數 \[{V_N}({x_0},u): = \sum\limits_{i = 1}^{N - 1} {\underbrace {l(x\left( i \right),u\left( i \right))}_{{\rm{stage}}\begin{array}{*{20}{c}} {} \end{array}{\rm{cost}}}} + \underbrace {{l_N}({x_N})}_{{\rm{terminal}}\begin{array}{*{20}{c}} {} \end{array}{\rm{cost}}}\]其中 \[\left\{ \begin{array}{l} u: = \{ u(0),u(1),...,u(N - 1)\} \\ l(x,u): = \frac{1}{2}({x^T}Qx + {u^T}Ru)\\ {l_N}(x): = \frac{1}{2}{x^T}{P_f}x \end{array} \right.\]另外假設 $Q \ge 0$ positive semi-definite 且 $R >0$ positive definite。 我們的目標:找到 $u^*=\{u^*(0),u^*(1),...,u^*(N-1)\}$ 使得 \[\begin{array}{l} \mathop {\min }\limits_u {V_N}\left( {x\left( 0 \right),u} \right)\\ s.t.\;\;{x^ + } = Ax + Bu \end{array}\] 現在利用 Backward Dynamic Programming,我們從最後的狀態 $x(N)$ 逐步回推最佳解,亦即觀察 \[{V_N}({x_0},u): = l({x_0},{u_0}) + l({x_1},{u_1}) + ... + \underbrace {l({x_{N - 1}},{u_{N - 1}}) + {l_N}({x_N})}_{{u_{N - 1}}\begin{array}{*{20}{c}} {} \end{array}{\rm{affects}}\begin{array}{*{20}{c}} {} \end{array}{
If you can’t solve a problem, then there is an easier problem you can solve: find it. -George Polya