前面一章最优控制理论 二、哈密尔顿函数法 ,我们已经利用Hamilton函数法对连续控制系统的Euler方程和横截条件进行了推导。这篇文章讲述如何求解
Hamilton函数法回顾
对于一个包含终端等式约束、Lagrange型性能指标的最优控制问题:
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\begin{aligned} \dot{x}&=f[x(t), u(t), t]\\ \min_{u(t)}J&=\int_{t_{0}}^{t_{f}} L[x(t), u(t), t] d t \\ \quad x\left(t_{0}\right)&=x_0\quad t_{0} \leq t \leq t_{f}\\ &\psi(x(t_f),t_f)=0 \tag 1\end{aligned}
x ˙ u ( t ) min J x ( t 0 ) = f [ x ( t ) , u ( t ) , t ] = ∫ t 0 t f L [ x ( t ) , u ( t ) , t ] d t = x 0 t 0 ≤ t ≤ t f ψ ( x ( t f ) , t f ) = 0 ( 1 ) Hamilton函数法的Euler方程和横截条件为:
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\begin{aligned} \dot{x}&=f(x,u,t)\\ x(t_0)&=x_0,\psi(x(t_f),t_f)=0\\ \dot{\lambda}&=-\frac{\partial H}{\partial x}=-\frac{\partial L}{\partial x}-\lambda^T\frac{\partial f}{\partial x}\\ \lambda(t_f)&=\mu^T\frac{\partial\psi}{\partial x}\\ 0&=\frac{\partial H}{\partial u}=\frac{\partial L}{\partial u}+\lambda^T\frac{\partial f}{\partial u} \end{aligned}\tag{2}
x ˙ x ( t 0 ) λ ˙ λ ( t f ) 0 = f ( x , u , t ) = x 0 , ψ ( x ( t f ) , t f ) = 0 = − ∂ x ∂ H = − ∂ x ∂ L − λ T ∂ x ∂ f = μ T ∂ x ∂ ψ = ∂ u ∂ H = ∂ u ∂ L + λ T ∂ u ∂ f ( 2 ) 公式
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( 2 ) 的前四个构成一个两点边值问题,控制
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二维火箭上升问题
二维火箭上升,推力加速度
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\begin{aligned} \min_{\beta(t)}J&=-V_x(t_f)=V_x(0)-\int_{0}^{t_{f}} a\cos(\beta) d t \\ \quad x\left(t_{0}\right)&=[0,0,0,0]^T\quad (0 \leq t \leq t_{f})\\ \psi(x(t_f),t_f)&=\begin{bmatrix}y(t_f)-h\\V_y(t_f)\end{bmatrix}=0 \end{aligned}
β ( t ) min J x ( t 0 ) ψ ( x ( t f ) , t f ) = − V x ( t f ) = V x ( 0 ) − ∫ 0 t f a cos ( β ) d t = [ 0 , 0 , 0 , 0 ] T ( 0 ≤ t ≤ t f ) = [ y ( t f ) − h V y ( t f ) ] = 0 其中参数设定
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t_f=100s,h=30000m,g=10m/s^2,a=20m/s^2
t f = 1 0 0 s , h = 3 0 0 0 0 m , g = 1 0 m / s 2 , a = 2 0 m / s 2 ,
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H=-a \cos \beta+\lambda_{r}^{T} v+\lambda_{3} a \cos \beta +\lambda_{4}(a \sin \beta-g)
H = − a cos β + λ r T v + λ 3 a cos β + λ 4 ( a sin β − g )
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{\dot{\lambda}=-\frac{\partial H}{\partial x}}\implies {\lambda_{1}=0 \quad \lambda_{2}=0\quad\lambda_{3}=-\lambda_{1} \quad \lambda_{4}=-\lambda_{2}}\\ \lambda\left(t_{f}\right)=\mu^{\top} \frac{\partial \psi}{\partial x}=\mu_{1}\left[\begin{array}{l} 0\\1\\0\\0 \end{array}\right]+\mu_{2}\left[\begin{array}{l} 0 \\0 \\0\\1 \end{array}\right]\\ \lambda_{1}\left(t_{f}\right)=0, \lambda_{2}\left(t_{f}\right)=\mu_{1} \quad\lambda_{3}\left(t_{1}\right)=0 \quad \lambda_{4}\left(t_{1}\right)=\mu_{2}^{*} \\ \lambda_{1}=0\quad \lambda_{2}=\mu_{1} \quad\lambda_{3}=0, \quad \lambda_{4}=-\lambda_2 t+\mu_{2}
λ ˙ = − ∂ x ∂ H ⟹ λ 1 = 0 λ 2 = 0 λ 3 = − λ 1 λ 4 = − λ 2 λ ( t f ) = μ ⊤ ∂ x ∂ ψ = μ 1 ⎣ ⎢ ⎢ ⎡ 0 1 0 0 ⎦ ⎥ ⎥ ⎤ + μ 2 ⎣ ⎢ ⎢ ⎡ 0 0 0 1 ⎦ ⎥ ⎥ ⎤ λ 1 ( t f ) = 0 , λ 2 ( t f ) = μ 1 λ 3 ( t 1 ) = 0 λ 4 ( t 1 ) = μ 2 ∗ λ 1 = 0 λ 2 = μ 1 λ 3 = 0 , λ 4 = − λ 2 t + μ 2
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\frac{\partial H}{\partial\beta}=a\{\sin\beta-\lambda_3\sin\beta+\lambda_4\cos\beta\}=0\\ \implies\tan\beta=\frac{\lambda_4}{\lambda_3-1}=-\lambda_4
∂ β ∂ H = a { sin β − λ 3 sin β + λ 4 cos β } = 0 ⟹ tan β = λ 3 − 1 λ 4 = − λ 4
共有6个未知变量,构成两点边值问题。
两点边值问题求解
两点边值问题(Two Point Boundary Value Problem)是对常微分方程组(
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function sol = RocketAscentIndirectBVP
%Edited by NICAI001
a=20;tf=100;h=30000;g=10;
ti=linspace(0,tf,101);
solinit=bvpinit(ti,@init);
opts=bvpset('Stats','on','RelTol',1e-6);% ,'BCJacobian',@pFpXf
sol=bvp4c(@ode,@bc,solinit,opts);
t=sol.x; y=sol.y;
figure(1)
plot(y(1,:),y(2,:),'k-'),title('trj'),xlabel('x/m'),ylabel('y/m');
figure(2)
plot(t,y(3:4,:));xlabel('t/s'),ylabel('velocity');
legend('V_x','V_y')
figure(3)
plot(t,-57.3*atan(y(6,:)));xlabel('t/s'),ylabel('\beta°');
function dydx=ode(t,y,p)
u=-atan(y(6));
dydx=[y(3);
y(4);
a*cos(u);
a*sin(u)-g;
0;
-y(5)];
end
function [dbcdya,dbcdyb,dbcdp]=pFpXf(t,y,p)
end
function res=bc(ya,yb,p)
res=[ya(1);
ya(2);
ya(3);
ya(4);
yb(2)-h;
yb(4)];
% yb(4)];
end
function y=init(t,p)
y=[400*(t); % y(-1)=-1; y(1)=1;
300*(t);
8*t;
6*t;
1.2*ones(size(t));
-0.1*(120*ones(size(t))-t)];
end
end