1. 必需条件
该问题以一质点在时间区间t ∈ [ t 0 , t k ] t \in \left[ t_0, t_k \right]t∈[t0,tk]内的运动为背景进行研究。因此,逐次给出以下条件:
- 时间区间
t ∈ [ t 0 , t k ] t \in \left[ t_0, t_k \right]t∈[t0,tk] - 边界条件
S t a r t = { x ( t 0 ) = x 10 x ˙ ( t 0 ) = x 20 ⋮ x ( n − 1 ) ( t 0 ) = x n 0 Start = \begin{cases} x(t_0) &= x_{10} \\ \dot x(t_0) &= x_{20} \\ \vdots \\ x^{(n-1)}(t_0) &= x_{n0} \end{cases}Start=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧x(t0)x˙(t0)⋮x(n−1)(t0)=x10=x20=xn0E n d = { x ( t k ) = x 1 k x ˙ ( t k ) = x 2 k ⋮ x ( n − 1 ) ( t k ) = x n k End = \begin{cases} x(t_k) &= x_{1k} \\ \dot x(t_k) &= x_{2k} \\ \vdots \\ x^{(n-1)}(t_k) &= x_{nk} \end{cases}End=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧x(tk)x˙(tk)⋮x(n−1)(tk)=x1k=x2k=xnk共2 n 2n2n个边界条件。 - 拉格朗日型性能指标
J = ∫ t 0 t k F ( t , x , x ˙ , ⋯ , x ( n ) ) d t → e x t r . J = \int _{t_0} ^{t_k} F \left( t, x, \dot x, \cdots, x^{(n)} \right) dt \rightarrow extr.J=∫t0tkF(t,x,x˙,⋯,x(n))dt→extr.其中F FF中共( n − 1 ) (n-1)(n−1)个导数,算上x xx本身,共n nn个与x xx相关的项。
2. 计算过程
在这个问题中,J JJ极值存在的必要条件不再采用欧拉公式,而是采用欧拉 – 泊松公式:
F x − d d t F x ˙ + d 2 d t 2 F x ¨ − ⋯ + ( − 1 ) n d n d t n F x ( n ) = 0 F_x - \frac{d}{dt} F_{\dot x} + \frac{d^2}{dt^2} F_{\ddot x} - \cdots + (-1)^n \frac{d^n}{dt^n} F_{x^{(n)} } = 0Fx−dtdFx˙+dt2d2Fx¨−⋯+(−1)ndtndnFx(n)=0应当注意:欧拉 – 泊松公式只是J JJ极值存在的必要条件,但不是充分条件。
J JJ极值存在的充分条件称为勒让德条件:
F x ( n ) x ( n ) = ∂ 2 F ∂ x ( n ) x ( n ) F_{x^{(n)} x^{(n)}} = \frac{\partial ^2 F}{\partial x^{(n)} x^{(n)}}Fx(n)x(n)=∂x(n)x(n)∂2F并有:
{ 若 F x ( n ) x ( n ) > 0 , 则 J → m i n 若 F x ( n ) x ( n ) < 0 , 则 J → m a x \begin{cases} 若 \quad F_{x^{(n)} x^{(n)}} > 0, \quad 则J \rightarrow min \\ 若 \quad F_{x^{(n)} x^{(n)}} < 0, \quad 则J \rightarrow max \end{cases}{若Fx(n)x(n)>0,则J→min若Fx(n)x(n)<0,则J→max
3. 例题
□ \square \quad□给出一系列条件:
- t ∈ [ 0 , 1 ] t \in \left[ 0, 1 \right]t∈[0,1];
- 边界条件:
S t a r t = { x ( 0 ) = 0 x ˙ ( 0 ) = 0 Start = \begin{cases} x(0) = 0 \\ \dot x(0) = 0 \end{cases}Start={x(0)=0x˙(0)=0E n d = { x ( 1 ) = 1 x ˙ ( 1 ) = 0 End = \begin{cases} x(1) = 1 \\ \dot x(1) = 0 \end{cases}End={x(1)=1x˙(1)=0 - 性能指标:
J = ∫ 0 1 x ¨ 2 d t → e x t r . J = \int _0 ^ 1 \ddot x ^ 2 dt \rightarrow extr.J=∫01x¨2dt→extr.
下面给出求解过程。
解:
由性能指标得知,
F = x ¨ 2 F = \ddot x ^2F=x¨2则
F x ¨ = 2 x ¨ , F x ¨ x ¨ = 2 > 0 F_{\ddot x} = 2 \ddot x, \quad F_{\ddot x \ddot x} = 2 > 0Fx¨=2x¨,Fx¨x¨=2>0根据勒让德条件知J JJ有最小值。
又:
F x = 0 F x ˙ = 0 F x ¨ = 2 x ¨ \begin{aligned} F_x &= 0 \\ F_{\dot x} &= 0 \\ F_{\ddot x} &= 2 \ddot x \\ \end{aligned}FxFx˙Fx¨=0=0=2x¨则根据欧拉 – 泊松公式
F x − d d t F x ˙ + d 2 d t 2 F x ¨ − ⋯ + ( − 1 ) n d n d t n F x ( n ) = 0 ⟹ 0 − 0 + ( 2 x ¨ ) ′ ′ = 0 ⟹ x ( 3 ) = C 1 ⟹ x ¨ = C 1 t + C 2 ⟹ x ˙ = 1 2 C 1 t 2 + C 2 t + C 3 ⟹ x = 1 6 C 1 t 3 + 1 2 C 2 t 2 + C 3 t + C 4 F_x - \frac{d}{dt} F_{\dot x} + \frac{d^2}{dt^2} F_{\ddot x} - \cdots + (-1)^n \frac{d^n}{dt^n} F_{x^{(n)} } = 0 \\ \Longrightarrow 0 - 0 + (2 \ddot x ) '' = 0 \\ \Longrightarrow x^{(3)} = C_1 \\ \Longrightarrow \ddot x = C_1 t + C_2 \\ \Longrightarrow \dot x = \frac{1}{2} C_1 t^2 + C_2 t + C_3 \\ \Longrightarrow x = \frac{1}{6} C_1 t^3 + \frac{1}{2} C_2 t^2 + C_3 t + C_4Fx−dtdFx˙+dt2d2Fx¨−⋯+(−1)ndtndnFx(n)=0⟹0−0+(2x¨)′′=0⟹x(3)=C1⟹x¨=C1t+C2⟹x˙=21C1t2+C2t+C3⟹x=61C1t3+21C2t2+C3t+C4代入x ( 0 ) = 0 , x ( 1 ) = 1 , x ˙ ( 0 ) = 0 , x ˙ ( 1 ) = 0 x(0) = 0, \quad x(1) = 1, \quad \dot x(0) = 0, \quad \dot x(1) = 0x(0)=0,x(1)=1,x˙(0)=0,x˙(1)=0有:
C 1 = − 12 , C 2 = 6 , C 3 = C 4 = 0 C_1 = -12, \quad C_2 = 6, \quad C_3 = C_4 = 0C1=−12,C2=6,C3=C4=0则得到最优的x xx满足:
x ∘ ( t ) = − 2 t 3 + 3 t 2 x ^{\circ} (t) = -2t^3 + 3t^2x∘(t)=−2t3+3t2此时性能指标也达到最优:
J ∘ = ∫ 0 1 x ¨ ∘ d t = 12. □ J ^{\circ} = \int _0 ^1 \ddot x^{\circ} dt = 12. \quad \squareJ∘=∫01x¨∘dt=12.□