Problem:
x1’=cos(x3);
x2’=sin(x3);
x3’=u;
|u| ≤0.5;
u-
control;
x=( x1,
x2)- state, a point on a plane.
Find the
minimum time T* of moving the system from point on a plane x(0,0) into the
point x(T)= (4,3);
x3(t)=
0∫t u(t)dt;
(and x(T)= 0∫T u(t)dt);
In fact
the system is 2 dimensional since the point is moving on a plane (x1,x2).
T*~
5.1277. x3[T*]=0.7605;
But you
can solve the system in 3d space (x1, x2, x3).
Then point is moving from (0,0,0) to( 4,5,0.7605).
The
statement of problem you can see in [1] and in [2].
To make
the problem clearer we can consider a geometric interpretation:
The
velocity vector (x1’, x2’) makes angle x3 with
Ox1 axis; the velocity modulo equals 1. In the start moment the x3(0)=
0, short-cut between points (0,0) and (4,3)equals 5, then optimal time will be
more than 5. The best moving is along a straight line (the velocity vector
coincides with the direction of the straight line) so we will try to achieve
this.
Namely:
set u= 0.5 and maintain this value until velocity vector coincides with the
direction of the straight line. This moment is control switch point.
Solving
equation: x3’=0.5, x3(0)= 0 we obtain x3(t)=
0.5t;
The
moment of achieving of straight line (0,0)->(4,3) occurs when tg(x2’/x1’)=sin(0.5t)/cos(0.5t)=0.75,
i.e. control switch point t1=2arctg(0.75) thus T*=t1+ρ, ρ
– distance from (x1(t),x2(t)) to(4,3).
After
control switch point u=0 (it does not change tg(x3)); then t1~1.5,
x3(T*)=x3(t1)= arctg(0.75). Thus, classic
variant of optinmal time control can be formulated as: find the minimal time
for moving controlled system from point (0,0,0) to point (4, 3, arctg(0.75)) in
3 dimension space.
References.
1. Power W., Shich C. "Convergence of gradient-type methods for free final time problems"// AIAA Journal of Numerical Analysis 1976. 14, N11, 1598-1603.
2. Евтушенко Ю.Г. «Методы решения экстремальных задач и их применение в системах оптимизации».