庞特里亚金极小原理及生产贮利润最优建模

2023-08-06 20:15 作者:艾琳娜的糖果屋 0人读过 | 我要投稿

古典变分法虽然强大，但是也有它的局限性，它要求容许函数是可以任意取的，但实际上容许函数往往是有限定的，就好比一元函数求极值一样在给定的区间内求其最大或最小，往往在边界取得而非极值点处。对于此类问题，就需要利用庞特里亚金极小值（极大）原理解决，下面形式上的给出原理的推导，并利用它建立一个生产-贮存-销售模型。

考察如下泛函 $%0AJ%3D%5Cvarphi%20%5Cleft(%20t_2%2Cx_2%20%5Cright)%20%2B%5Cint_%7Bt_1%7D%5E%7Bt_2%7D%7BF%5Cleft(%20t%2Cx%2Cu%20%5Cright)%20%5Cmathrm%7Bd%7Dt%7D%5C%2C%5C%2C%20x%5Cprime%5Cleft(%20t%20%5Cright)%20%3Df%5Cleft(%20t%2Cx%2Cu%20%5Cright)%20%5C%2C%5C%2C%20u%5Cin%20D%5C%2C%5C%2Ct_2%0A%0A$ 自由， $%0Ax_2%0A%0A$ 自由

构造哈密尔顿函数 $%0AH%5Cleft(%20t%2Cx%2Cu%2C%5Clambda%20%5Cright)%20%3DF%5Cleft(%20t%2Cx%2Cu%20%5Cright)%20%2B%5Clambda%20%5Cleft(%20t%20%5Cright)%20f%5Cleft(%20t%2Cx%2Cu%20%5Cright)%20%0A%0A$

假设 $u$ 是使 $J$ 取极小值的最优控制，由约束条件可以得到最优控制条件下的状态方程 $x(t)$

对于微扰 $%0A%5Cdelta%20u%0A%0A$ 泛函变动到 $%0A%5Cvarphi%20%5Cleft(%20t_2%2B%5Cdelta%20t_2%2Cx_2%2B%5Cdelta%20x_2%20%5Cright)%20%2B%5Cint_%7Bt_1%7D%5E%7Bt_2%2B%5Cdelta%20t_2%7D%7B%5Cleft%5C%7B%20H%5Cleft(%20t%2Cx%2B%5Cdelta%20x%2Cu%2B%5Cdelta%20u%2C%5Clambda%20%2B%5Cdelta%20%5Clambda%20%5Cright)%20-%5Clambda%20%5Cleft(%20t%20%5Cright)%20f%5Cleft(%20t%2Cx%2B%5Cdelta%20x%2Cu%2B%5Cdelta%20u%20%5Cright)%20%5Cright%5C%7D%20%5Cmathrm%7Bd%7Dt%7D%0A%0A$

且有 $%0AJ%5Cleft%5B%20x%2B%5Cdelta%20x%2Cu%2B%5Cdelta%20u%20%5Cright%5D%20-J%5Cleft%5B%20x%2Cu%20%5Cright%5D%20%5Cgeqslant%200%0A%0A$

$%0AJ%5Cleft%5B%20x%2B%5Cdelta%20x%2Cu%2B%5Cdelta%20u%20%5Cright%5D%20-J%5Cleft%5B%20x%2Cu%20%5Cright%5D%20%3D%0A%5Cvarphi%20%5Cleft(%20t_2%2B%5Cdelta%20t_2%2Cx_2%2B%5Cdelta%20x_2%20%5Cright)%20-%5Cvarphi%20%5Cleft(%20t_2%2Cx_2%20%5Cright)%20%2B%5C%5C%5Cint_%7Bt_1%7D%5E%7Bt_2%7D%7B%5Cleft%5C%7B%20H%5Cleft(%20t%2Cx%2B%5Cdelta%20x%2Cu%2B%5Cdelta%20u%2C%5Clambda%20%2B%5Cdelta%20%5Clambda%20%5Cright)%20-H%5Cleft(%20t%2Cx%2Cu%2C%5Clambda%20%5Cright)%20%2B%5Clambda%20%5Cleft(%20t%20%5Cright)%20f%5Cleft(%20t%2Cx%2Cu%20%5Cright)%20-%5Clambda%20%5Cleft(%20t%20%5Cright)%20f%5Cleft(%20t%2Cx%2B%5Cdelta%20x%2Cu%2B%5Cdelta%20u%20%5Cright)%20%5Cright%5C%7D%20%5Cmathrm%7Bd%7Dt%7D%5C%5C%2B%5Cleft(%20H-%5Clambda%20x%5Cprime%5Cmid_%7Bt%3Dt_2%7D%5E%7B%7D%20%5Cright)%20%5Cdelta%20t_2%2B%5Cvarepsilon%20%0A%0A%0A$

$%3D%0A%5Cleft(%20%5Cvarphi%20_%7Bt_2%7D%2B%5Cleft%5B%20H-%5Clambda%20x%5Cprime%20%5Cright%5D%20%5Cmid_%7Bt%3Dt_2%7D%5E%7B%7D%20%5Cright)%20%5Cdelta%20t_2%2B%5Cvarphi%20_%7Bx_2%7D%5Cdelta%20x_2%2Bo%5Cleft(%20%5Cdelta%20t_2%20%5Cright)%20%2Bo%5Cleft(%20%5Cdelta%20x_2%20%5Cright)%20-%5Cint_%7Bt_1%7D%5E%7Bt_2%7D%7B%5Clambda%20%5Cleft(%20t%20%5Cright)%20%5Cdelta%20x%5Cprime%5Cmathrm%7Bd%7Dt%7D%2B%5C%5C%5Cint_%7Bt_1%7D%5E%7Bt_2%7D%7B%5Cleft%5C%7B%20H%5Cleft(%20t%2Cx%2B%5Cdelta%20x%2Cu%2B%5Cdelta%20u%2C%5Clambda%20%2B%5Cdelta%20%5Clambda%20%5Cright)%20-H%5Cleft(%20t%2Cx%2Cu%2B%5Cdelta%20u%2C%5Clambda%20%2B%5Cdelta%20%5Clambda%20%5Cright)%20%2BH%5Cleft(%20t%2Cx%2Cu%2B%5Cdelta%20u%2C%5Clambda%20%2B%5Cdelta%20%5Clambda%20%5Cright)%20-H%5Cleft(%20t%2Cx%2Cu%2C%5Clambda%20%5Cright)%20%5Cright%5C%7D%20%5Cmathrm%7Bd%7Dt%7D%2B%5Cvarepsilon%20%0A%0A$

$%0A%3D%5Cleft(%20%5Cvarphi%20_%7Bt_2%7D%2BH%5Cmid_%7Bt%3Dt_2%7D%5E%7B%7D%20%5Cright)%20%5Cdelta%20t_2%2B%5Cleft(%20%5Cvarphi%20_%7Bx_2%7D-%5Clambda%20%5Cmid_%7Bt%3Dt_2%7D%5E%7B%7D%20%5Cright)%20%5Cdelta%20x_2%2B%5Cint_%7Bt_1%7D%5E%7Bt_2%7D%7B%5Cleft(%20H_x%2B%5Clambda%20%5E%7B%5Cprime%7D%5Cleft(%20t%20%5Cright)%20%5Cright)%20%5Cdelta%20x%5Cmathrm%7Bd%7Dt%7D%2B%5C%5C%5Cint_%7Bt_1%7D%5E%7Bt_2%7D%7B%5Cleft%5C%7B%20H%5Cleft(%20t%2Cx%2Cu%2B%5Cdelta%20u%2C%5Clambda%20%2B%5Cdelta%20%5Clambda%20%5Cright)%20-H%5Cleft(%20t%2Cx%2Cu%2C%5Clambda%20%5Cright)%20%5Cright%5C%7D%20%5Cmathrm%7Bd%7Dt%7D%2Bo%5Cleft(%20%5Cdelta%20t_2%20%5Cright)%20%2Bo%5Cleft(%20%5Cdelta%20x_2%20%5Cright)%20%2Bo%5Cleft(%20%5Cdelta%20x%20%5Cright)%20%2B%5Cvarepsilon%20%0A%0A$

现在选定 $%0A%5Clambda%20%5Cleft(%20t%20%5Cright)%20%0A%0A$ 满足方程 $%0AH_x%2B%5Clambda%20%5E%7B%5Cprime%7D%5Cleft(%20t%20%5Cright)%20%3D0%0A%0A$

以及终端的横截条件 $%0A%5Cvarphi%20_%7Bt_2%7D%2BH%5Cmid_%7Bt%3Dt_2%7D%5E%7B%7D%3D0%20%3B%5Cvarphi%20_%7Bx_2%7D-%5Clambda%20%5Cmid_%7Bt%3Dt_2%7D%5E%7B%7D%3D0%0A%0A$

那么可以猜测成立 $%0AH%5Cleft(%20t%2Cx%2Cu%2B%5Cdelta%20u%2C%5Clambda%20%2B%5Cdelta%20%5Clambda%20%5Cright)%20-H%5Cleft(%20t%2Cx%2Cu%2C%5Clambda%20%5Cright)%20%5Cgeqslant%200%0A%0A$ ，这就是极小原理，我们可以对对控制函数分段以及积分分段处理去证明它，这里就不再证明了，极大情况相同。

此不等式的意思就是说哈密尔顿函数关于控制 $u(t)$ 在最优控制的情况下取得极小。

下面通过极小(极大)原理来建立（瞎扯）一个生产—贮存—销售的最优模型。

我们假设贮存函数为 $x(t)$ ,生产速率函数为 $u(t)$ ,销售速率函数为 $v(t)$ ,那么它们之间满足

$%0Ax%5Cprime%5Cleft(%20t%20%5Cright)%20%3Du%5Cleft(%20t%20%5Cright)%20-v%5Cleft(%20t%20%5Cright)%20%0A%0A$ ，其次假定初始情况下贮存量 $%0Ax%5Cleft(%20t_0%20%5Cright)%20%3Dx_0%0A%0A$ ，那么如何选取生产速率函数，使得总利润最大呢？为了分析的方便我们来简化一下销售速率函数，一般情况下销售的大小和生产的大小成正相关，与贮存成负相关由此我们假定。 $%0Av%5Cleft(%20t%20%5Cright)%20%3Dk_1u%5Cleft(%20t%20%5Cright)%20-k_2x%5Cprime%5Cleft(%20t%20%5Cright)%20%5C%2C%5C%2Ck_1%2Ck_2%3E0%0A%0A$

其次在时间段 $%0A%5Bt_0%2Ct_1%5D%0A%0A$ 内产品的销售价格是不变的记为 $p$ ，同时贮存会产生贮存费用，记单位产品产生的贮存费用为 $q$ ，生产原材料的成本也是固定的为 $C$ ，并且实际生产过程中生产速率具有一定的约束，不可能无限增加即，

$%0A0%3Cu_%7B%5Cmin%7D%5Cleqslant%20u%5Cleft(%20t%20%5Cright)%20%5Cleqslant%20u_%7B%5Cmax%7D%0A%0A$ ，那么此时间段内总利润函数可以表示为

$%0AJ%3D%5Cint_%7Bt_0%7D%5E%7Bt_1%7D%7Bpv%5Cleft(%20t%20%5Cright)%20%5Cmathrm%7Bd%7Dt%7D-qx%5Cleft(%20t_1%20%5Cright)%20-C%3D%5Cint_%7Bt_0%7D%5E%7Bt_1%7D%7Bp%5Cleft(%20k_1u%5Cleft(%20t%20%5Cright)%20-k_2x%5Cprime%5Cleft(%20t%20%5Cright)%20%5Cright)%20%5Cmathrm%7Bd%7Dt%7D-qx%5Cleft(%20t_1%20%5Cright)%20-C%5C%2C%5C%2C%20%20%20%0A%0A$

因此问题归结为在约束条件 $%0A%5C%2C%5C%2Cx%5Cprime%5Cleft(%20t%20%5Cright)%20%3D%5Cfrac%7B1-k_1%7D%7B1-k_2%7Du%5Cleft(%20t%20%5Cright)%20%2C0%3Cu_%7B%5Cmin%7D%5Cleqslant%20u%5Cleft(%20t%20%5Cright)%20%5Cleqslant%20u_%7B%5Cmax%7D%0A%0A$ 下求泛函

$%0AJ%3D-qx%5Cleft(%20t_1%20%5Cright)%20-C%2B%5Cint_%7Bt_0%7D%5E%7Bt_1%7D%7Bp%5Cleft(%20k_1u%5Cleft(%20t%20%5Cright)%20-k_2x%5Cprime%5Cleft(%20t%20%5Cright)%20%5Cright)%20%5Cmathrm%7Bd%7Dt%7D%0A%0A$ 的最大值

构造哈密尔顿函数： $%0AH%3Dp%5Cleft(%20k_1u%5Cleft(%20t%20%5Cright)%20-k_2x%5Cprime%5Cleft(%20t%20%5Cright)%20%5Cright)%20%2B%5Cfrac%7B1-k_1%7D%7B1-k_2%7D%5Clambda%20%5Cleft(%20t%20%5Cright)%20u%5Cleft(%20t%20%5Cright)%20%0A$

横截条件： $%0AH_x%3D0%3D-%5Clambda%20%5Cprime%5Cleft(%20t%20%5Cright)%20%0A%0A%3B%5Cvarphi%20_%7Bx_1%7D%3D-q%3D%5Clambda%20%5Cleft(%20t_1%20%5Cright)%20%0A%0A$

由此得到 $%0A%5Clambda%20%5Cleft(%20t%20%5Cright)%20%3D-q%3BH%3D-k_2px%5Cprime%5Cleft(%20t%20%5Cright)%20%2B%5Cleft(%20pk_1-q%5Cfrac%7B1-k_1%7D%7B1-k_2%7D%20%5Cright)%20u%5Cleft(%20t%20%5Cright)%20%0A%0A$

于是最大利润下的生产速率函数、贮存函数的选取应该满足

$%0Au%5Cleft(%20t%20%5Cright)%20%3D%5Cbegin%7Bcases%7D%0A%09u_%7B%5Cmax%7D%26%09%09%5Cfrac%7Bp%7D%7Bq%7D%3E%5Cfrac%7B1%7D%7Bk_1%7D%5Cfrac%7B1-k_1%7D%7B1-k_2%7D%5C%5C%0A%09u_%7B%5Cmin%7D%26%09%09%5Cfrac%7Bp%7D%7Bq%7D%3C%5Cfrac%7B1%7D%7Bk_1%7D%5Cfrac%7B1-k_1%7D%7B1-k_2%7D%5C%5C%0A%5Cend%7Bcases%7D%0A%0A$ $%0Ax%5Cleft(%20t%20%5Cright)%20%3D%5Cbegin%7Bcases%7D%0A%09u_%7B%5Cmax%7D%5Cfrac%7B1-k_1%7D%7B1-k_2%7D%5Cleft(%20t-t_0%20%5Cright)%20%2Bx_0%26%09%09%5Cfrac%7Bp%7D%7Bq%7D%3E%5Cfrac%7B1%7D%7Bk_1%7D%5Cfrac%7B1-k_1%7D%7B1-k_2%7D%5C%5C%0A%09u_%7B%5Cmin%7D%5Cfrac%7B1-k_1%7D%7B1-k_2%7D%5Cleft(%20t-t_0%20%5Cright)%20%2Bx_0%26%09%09%5Cfrac%7Bp%7D%7Bq%7D%3C%5Cfrac%7B1%7D%7Bk_1%7D%5Cfrac%7B1-k_1%7D%7B1-k_2%7D%5C%5C%0A%5Cend%7Bcases%7D%0A%0A$

上述假设中可以发现，极值函数刚好取在某一个边界上，如果假设销售速率函数与贮存量有关，即 $%0A%0A%0A$ $%0Av%5Cleft(%20t%20%5Cright)%20%3Dk_1u%5Cleft(%20t%20%5Cright)%20-k_2x%5Cleft(%20t%20%5Cright)%20%5C%2C%5C%2Ck_1%2Ck_2%3E0%0A%0A$ 此时最大利润可以表示为

$%0AJ%3D-qx%5Cleft(%20t_1%20%5Cright)%20-C%2B%5Cint_%7Bt_0%7D%5E%7Bt_1%7D%7Bp%5Cleft(%20k_1u%5Cleft(%20t%20%5Cright)%20-k_2x%5Cleft(%20t%20%5Cright)%20%5Cright)%20%5Cmathrm%7Bd%7Dt%7D%5C%2C%5C%2C%0A%0A$ 以及约束条件

$%0A%5C%2C%5C%2Cx%5Cprime%5Cleft(%20t%20%5Cright)%20%3D%5Cleft(%201-k_1%20%5Cright)%20u%5Cleft(%20t%20%5Cright)%20%2Bk_2x%5Cleft(%20t%20%5Cright)%20%3B0%3Cu_%7B%5Cmin%7D%5Cleqslant%20u%5Cleft(%20t%20%5Cright)%20%5Cleqslant%20u_%7B%5Cmax%7D%0A%0A$

构造哈密尔顿函数 $%0AH%3D%5Cleft%5C%7B%20pk_1%2B%5Cleft(%201-k_1%20%5Cright)%20%5Clambda%20%5Cleft(%20t%20%5Cright)%20%5Cright%5C%7D%20u%5Cleft(%20t%20%5Cright)%20%2Bk_2x%5Cleft(%20t%20%5Cright)%20%5Cleft%5C%7B%20%5Clambda%20%5Cleft(%20t%20%5Cright)%20-p%20%5Cright%5C%7D%20%0A%0A$

横截条件: $%0AH_x%3Dk_2%5Clambda%20%5Cleft(%20t%20%5Cright)%20-k_2p%3D-%5Clambda%20%5Cprime%5Cleft(%20t%20%5Cright)%20%5C%2C%5C%2C%20%20%5Cvarphi%20_%7Bx_1%7D%3D%5Clambda%20%5Cleft(%20t_2%20%5Cright)%20%3D-q%0A%0A$

解得 $%0A%0A%5Clambda%20%5Cleft(%20t%20%5Cright)%20%3Dp-%5Cleft(%20p%2Bq%20%5Cright)%20e%5E%7Bk_2%5Cleft(%20t_2-t%20%5Cright)%7D%0A%0A%0A$ ,同时假定 $0%3Ck_1%3C1$

得到利润最大时候的生产速率函数和贮存函数

$%0A%0Au%5Cleft(%20t%20%5Cright)%20%3D%5Cbegin%7Bcases%7D%0A%09u_%7B%5Cmin%7D%26%09%09t_0%5Cleqslant%20t%3Ct%5E*%5C%5C%0A%09u_%7B%5Cmax%7D%26%09%09t%5E*%3Ct%5Cleqslant%20t_1%5C%5C%0A%5Cend%7Bcases%7D%3Bt%5E*%3D%5C%2C%5C%2Ct_1-%5Cfrac%7B1%7D%7Bk_2%7D%5Clog%20%5Cfrac%7Bp%7D%7B%5Cleft(%201-k_1%20%5Cright)%20%5Cleft(%20p%2Bq%20%5Cright)%7D%0A%0A$

$%0Ax%5Cleft(%20t%20%5Cright)%20%3D%5Cbegin%7Bcases%7D%0A%09-u_%7B%5Cmin%7D%5Cfrac%7B1-k_1%7D%7Bk_2%7D%2B%5Cleft(%20x_0%2Bu_%7B%5Cmin%7D%5Cfrac%7B1-k_1%7D%7Bk_2%7D%20%5Cright)%20e%5E%7Bk_2%5Cleft(%20t-t_0%20%5Cright)%7D%26%09%09t_0%5Cleqslant%20t%5Cleqslant%20t%5E*%5C%5C%0A%09-u_%7B%5Cmax%7D%5Cfrac%7B1-k_1%7D%7Bk_2%7D%2B%5Cleft(%20x%5Cleft(%20t%5E*%20%5Cright)%20%2Bu_%7B%5Cmax%7D%5Cfrac%7B1-k_1%7D%7Bk_2%7D%20%5Cright)%20e%5E%7Bk_2%5Cleft(%20t-t%5E*%20%5Cright)%7D%26%09%09t%5E*%5Cleqslant%20t%5Cleqslant%20t_1%5C%5C%0A%5Cend%7Bcases%7D%5C%2C%5C%2C%0A%0A$

如果 $k_1%3E1$ ,那么利润最大时的生产速率函数和贮存函数则为

$%0Au%5Cleft(%20t%20%5Cright)%20%3Du_%7B%5Cmax%7D%5C%2C%5C%2C%20%20%20t_0%5Cleqslant%20t%5Cleqslant%20t_1%0A%5C%5C%0Ax%5Cleft(%20t%20%5Cright)%20%3D-u_%7B%5Cmax%7D%5Cfrac%7B1-k_1%7D%7Bk_2%7D%2B%5Cleft(%20x_0%2Bu_%7B%5Cmax%7D%5Cfrac%7B1-k_1%7D%7Bk_2%7D%20%5Cright)%20e%5E%7Bk_2%5Cleft(%20t-t_0%20%5Cright)%7D%5C%2C%5C%2C%20t_0%5Cleqslant%20t%5Cleqslant%20t_1%0A%0A$