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Date: 8-10-2016
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To illustrate how actually to solve a control problem, in this last section we introduce some ad hoc calculus and geometry methods for the rocket car problem, Example 5 in(EXAMPLES).
First of all, let us guess that to find an optimal solution we will need only to consider the cases a = 1 or a = −1. In other words, we will focus our attention only upon those controls for which at each moment of time either the left or the right rocket engine is fired at full power.
CASE 1: Suppose first that α ≡ 1 for some time interval, during which
And so
Let t0 belong to the time interval where α ≡ 1 and integrate from t0 to t:
In other words, so long as the control is set for α ≡ 1, the trajectory stays on the curve v2 = 2q + b for some constant b.
CASE 2: Suppose now α ≡ −1 on some time interval. Then as above
And hence
Let t1 belong to an interval where α ≡ −1 and integrate:
Consequently, as long as the control is set for α ≡ −1, the trajectory stays on the curve v2 = −2q + c for some constant c.
GEOMETRIC INTERPRETATION. Formula (1.1) says if α ≡ 1, then (q(t), v(t)) lies on a parabola of the form
v2= 2q + b.
Similarly, (1.2) says if α ≡ −1, then (q(t), v(t)) lies on a parabola
v2= −2q + c.
Now we can design an optimal control α∗(.), which causes the trajectory to jump between the families of right– and left–pointing parabolas, as drawn. Say we start at the black dot, and wish to steer to the origin. This we accomplish by first setting the control to the value α = −1, causing us to move down along the second family of parabolas. We then switch to the control α = 1, and thereupon move to a parabola from the first family, along which we move up and to the left, ending up at the origin. See the picture.
References
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