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CONTROLLABILITY, BANG-BANG PRINCIPLE-BANG-BANG PRINCIPLE.
المؤلف: Lawrence C. Evans
المصدر: An Introduction to Mathematical Optimal Control Theory
الجزء والصفحة: 26-30
6-10-2016
331
For this section, we will again take A to be the cube [−1, 1] m in Rm.
DEFINITION. A control α(.) ∈ A is called bang-bang if for each time t ≥ 0 and each index i = 1, . . . ,m, we have |αi(t)| = 1, where
THEOREM 1.1 (BANG-BANG PRINCIPLE). Let t > 0 and suppose x0 ∈C(t), for the system
Then there exists a bang-bang control α(.) which steers x0 to 0 at time t.
To prove the theorem we need some tools from functional analysis, among hem the Krein–Milman Theorem, expressing the geometric fact that every bounded onvex set has an extreme point.
1.1 SOME FUNCTIONAL ANALYSIS. We will study the “geometry” of certain infinite dimensional spaces of functions.
NOTATION:
DEFINITION. Let αn ∈ L∞ for n = 1, . . . and α ∈ L∞. We say αn converges to α in the weak* sense, written
We will need the following useful weak* compactness theorem for L∞:
ALAOGLU’S THEOREM. Let αn ∈ A, n = 1, . . . . Then there exists a subsequence αnk and α ∈ A, such that
DEFINITIONS. (i) The set K is convex if for all x, xˆ ∈ K and all real numbers 0 ≤ λ ≤ 1,
λx + (1 − λ) xˆ ∈ K.
(ii) A point z ∈ K is called extreme provided there do not exist points x, xˆ ∈ K and 0 < λ < 1 such that
z = λx + (1 − λ) xˆ.
KREIN-MILMAN THEOREM. Let K be a convex, nonempty subset of L∞, which is compact in the weak ∗ topology.
Then K has at least one extreme point.
1.2 APPLICATION TO BANG-BANG CONTROLS.
The foregoing abstract theory will be useful for us in the following setting. We will take K to be the set of controls which steer x0 to 0 at time t, prove it satisfies the hypotheses of Krein–Milman Theorem and finally show that an extreme point is a bang-bang control.
So consider again the linear dynamics
Take x0 ∈ C(t) and write
K = {α(.) ∈ A |α(.) steers x0 to 0 at time t}.
LEMMA 1.3 (GEOMETRY OF SET OF CONTROLS). The collection K of admissible controls satisfies the hypotheses of the Krein–Milman Theorem.
Proof. Since x0 ∈ C(t), we see that K = ∅.
Next we show that K is convex. For this, recall that α(.) ∈ K if and only if
Now take also αˆ ∈ K and 0 ≤ λ ≤ 1. Then
Lastly, we confirm the compactness. Let αn ∈ K for n = 1, . . . . According to Alaoglu’s Theorem there exist nk → ∞ and α ∈ A such that αnk∗⇀ α. We need
to show that α ∈ K.
Now αnk ∈ K implies
by definition of weak-* convergence. Hence α ∈ K.
We can now apply the Krein–Milman Theorem to deduce that there exists an extreme point α∗ ∈ K. What is interesting is that such an extreme point corresponds to a bang-bang control.
THEOREM 1.4 (EXTREMALITY AND BANG-BANG PRINCIPLE). The control α∗(.) is bang-bang.
Proof. 1. We must show that for almost all times 0 ≤ s ≤ t and for eachi = 1, . . . ,m, we have
|αi∗ (s)| = 1.
Suppose not. Then there exists an index i ∈ {1, . . . ,m} and a subset E ⊂ [0, t] of positive measure such that |αi∗(s)| < 1 for s ∈ E. In fact, there exist a number ε > 0 and a subset F ⊆ E such that
the function β in the ith slot. Choose any real-valued function β(.) ≡ 0, such that IF (β(.)) = 0
and |β(.)| ≤ 1. Define
α1(.) := α∗(.) + εβ(.)
α2(.) := α∗(.) − εβ(.),
where we redefine β to be zero off the set F
2. We claim that
α1(.),α2(.) ∈ K.
To see this, observe that
Note also α1(.) ∈ A. Indeed,
But on the set F, we have |α∗i (s)| ≤ 1 − ε, and therefore
|α1(s)| ≤ |α∗ (s)| + ε|β(s)| ≤ 1 − ε + ε = 1.
Similar considerations apply for α2. Hence α1,α2 ∈ K, as claimed above.
3. Finally, observe that
and this is a contradiction, since α∗ is an extreme point of K.
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