Main result. Variable Final Time

104 CHAPTER 8. CONINUOUS-TIME OPTIMAL CONTROL

8.3 Variable Final Time

8.3.1 Main result.

In the problem considered up to now the final time t f is assumed to be fixed. In many important cases the final time is itself a decision variable. One such case is the minimum-time problem where we want to transfer the state of the system from a given initial state to a specified final state in minimum time. More generally, consider the optimal control problem 8.27. Maximize Z t f t f t, xt, utdt subject to dynamics: ˙xt = f t, x, t, ut, , t ≤ t ≤ t f , initial condition: g xt = b , final condition: g f xtf = b f , control constraint: u· ∈ U , final-time constraint: t f ∈ t , ∞ . 8.27 We analyze 8.27 by converting the variable time interval [t , t f ] into a fixed-time interval [0, 1]. This change of time-scale is achieved by regarding t as a new state variable and selecting a new time variable s which ranges over [0, 1]. The equation for t is dts ds = αs , 0 ≤ s ≤ 1 , with initial condition t0 = t . Here αs is a new control variable constrained by αs ∈ 0, ∞. Now if x· is the solution of ˙xt = f t, xt, ut , t ≤ t ≤ t f , xt = x 8.28 and if we define zs = xts, vs = uts , 0 ≤ s ≤ 1 , then it is easy to see that z· is the solution of dz ds s = αsf s, zs, vs , 0 ≤ s ≤ 1 z0 = x . 8.29 Conversely from the solution z· of 8.29 we can obtain the solution x· of 8.28 by xt = zst , t ≤ t ≤ t f , where s· : [t , t f ] → [0, 1] is the functional inverse of st; in fact, s· is the solution of the differential equation ˙st = 1αst, st = 0. 8.3. VARIABLE FINAL TIME 105 With these ideas in mind it is natural to consider the fixed-final-time optimal control problem 8.30, where the state vector t, z ∈ R 1+m , and the control α, v ∈ R 1+p : Maximize Z 1 f ts, zs, vsαsds subject to dynamics: ˙zs, ˙ts = f ts, zs, vsαs, αs, initial constraint: g z0 = b , t0 = t , final constraint: g f z1 = b f , t1 ∈ R , control constraint: vs, αs ∈ Ω × 0, ∞ for 0 ≤ s ≤ 1 and v·, α· piecewise continuous. 8.30 The relation between problems 8.27 and 8.30 is established in the following result. Lemma 1: i Let x ∗ ∈ T , u ∗ · ∈ U, t ∗ f ∈ t , ∞ and let x ∗ t = φt, t , x ∗ , u ∗ · be the corresponding trajectory. Suppose that x ∗ t ∗ f ∈ T f , and suppose that u ∗ ·, x ∗ , t ∗ f is optimal for 8.27. Define z ∗ , v ∗ ·, and α ∗ · by z ∗ = x ∗ v ∗ s = u ∗ t + st ∗ f − t α ∗ s = t ∗ f − t , 0 ≤ s ≤ 1 , , 0 ≤ s ≤ 1 . Then v ∗ ·, α ∗ ·, z ∗ is optimal for 8.30. ii Let z ∗ ∈ T , and let v ∗ ·, α ∗ · be an admissible control for 8.30 such that the correspond- ing trajectory t ∗ ·, z ∗ · satisfies the final conditions of 8.30. Suppose that v ∗ ·, α ∗ ·, z ∗ is optimal for 8.30. Define x ∗ , u ∗ · ∈ U, and t ∗ f by x ∗ = z ∗ , u ∗ t = v ∗ s ∗ t , t ≤ t ≤ t ∗ f , t ∗ f = t ∗ 1 , where s ∗ · is functional inverse of t ∗ ·. Then u ∗ ·, z ∗ , t ∗ f is optimal for 8.27. Exercise 1: Prove Lemma 1. Theorem 1: Let u ∗ · ∈ U, let x ∗ ∈ T , let t ∗ f ∈ 0, ∞, and let x ∗ t = φt, t , x ∗ , u ∗ ·, t ≤ t ≤ t f , and suppose that x ∗ t ∗ f ∈ T f . If u ∗ ·, x ∗ , t ∗ f is optimal for 8.27, then there exists a function ˜ p ∗ = p ∗ , p ∗ : [t , t ∗ f ] → R 1+m , not identically zero, and with p ∗ t ≡ constant and p ∗ t ≥ 0, satisfying augmented adjoint equation: · ˜ p ∗ t = −[ ∂ ˜ f ∂ ˜ x t, x ∗ t, u ∗ t] ′ ˜ p ∗ t , 8.31 initial condition: p ∗ t ⊥T x ∗ , 8.32 106 CHAPTER 8. CONINUOUS-TIME OPTIMAL CONTROL final condition: p ∗ t ∗ f ⊥T f x ∗ t ∗ f . 8.33 Also the maximum principle ˜ Ht, x ∗ t, ˜ p ∗ t, u ∗ t = ˜ M t, x ∗ t, ˜ p ∗ t , 8.34 holds for all t ∈ [t , t f ] except possibly for a finite set. Furthermore, t ∗ f must be such that ˆ Ht ∗ f , x ∗ t ∗ f , ˜ p ∗ t ∗ f , u ∗ t ∗ f = 0 . 8.35 Finally, if f and f do not explicitly depend on t, then ˆ M t, x ∗ t, ˜ p ∗ t ≡ 0. Proof: By Lemma 1, z ∗ = x ∗ , v ∗ s = u ∗ t + st ∗ f − t and α ∗ s = t ∗ f − t for 0 ≤ s ≤ 1 constitute an optimal solution for 8.30. The resulting trajectory is z ∗ s = x ∗ t + st ∗ f − t , t ∗ s = t + st ∗ f − t , 0 ≤ s ≤ 1 , so that in particular z ∗ 1 = x ∗ t ∗ f . By Theorem 1 of Section 2, there exists a function ˜ λ ∗ = λ ∗ , λ ∗ , λ ∗ n+1 : [0, 1] → R 1+n+1 , not identically zero, and with λ ∗ s ≡ constant and λ ∗ s ≥ 0, satisfying adjoint equation:       ˙λ ∗ t ˙λ ∗ t ˙λ ∗ n+1 t       = −        {[ ∂f ∂z t ∗ s, z ∗ s, v ∗ s] ′ λ ∗ s +[ ∂f ∂z t ∗ s, z ∗ s, v ∗ s] ′ λ ∗ s}α ∗ s {[ ∂f ∂t t ∗ s, z ∗ s, v ∗ s] ′ λ ∗ s +[ ∂f ∂t t ∗ s, z ∗ s, v ∗ s] ′ λ ∗ s}α ∗ s        8.36 initial condition: λ ∗ 0⊥T z ∗ 8.37 final condition: λ ∗ 1⊥T f z ∗ 1 , λ ∗ n+1 1 = 0 . 8.38 Furthermore, the maximum principle λ ∗ sf t ∗ s, z ∗ s, v ∗ sα ∗ s +λ ∗ s ′ f t ∗ s, z ∗ s, v ∗ sα ∗ s + λ ∗ n+1 sα ∗ s = sup{[λ ∗ sf t ∗ s, z ∗ s, wβ +λ ∗ s ′ f t ∗ s, z ∗ s, wβ + λ ∗ n+1 sβ]|w ∈ Ω, β ∈ 0, ∞} 8.39 holds for all s ∈ [0, 1] except possibly for a finite set. Let s ∗ t = t − t t ∗ f − t , t ≤ t ≤ t ∗ f , and define ˜ p ∗ = p ∗ , p ∗ : [t , t ∗ f ] → R 1+n by p ∗ t = λ ∗ s ∗ t, p ∗ t = λ ∗ s ∗ t, t ≤ t ≤ t ∗ f . 8.40 First of all, ˜ p ∗ is not identically zero. Because if ˜ p ∗ ≡ 0, then from 8.40 we have λ ∗ , λ ∗ ≡ 0 and then from 8.36, λ ∗ n+1 ≡ constant, but from 8.38, λ ∗ n+1 1 = 0 so that we would have ˜ λ ∗ ≡ 0 8.3. VARIABLE FINAL TIME 107 which is a contradiction. It is trivial to verify that ˜ p ∗ · satisfies 8.31, and, on the other hand 8.37 and 8.38 respectively imply 8.32 and 8.33. Next, 8.39 is equivalent to λ ∗ sf t ∗ s, z ∗ s, v ∗ s +λ ∗ s ′ f t ∗ s, z ∗ s, v ∗ s + λ ∗ n+1 s = 0 8.41 and λ ∗ sf t ∗ s, z ∗ s, v ∗ s + λ ∗ s ′ f t ∗ s, z ∗ s, v ∗ s = Sup {[λ ∗ sf t ∗ s, z ∗ s, w + λ ∗ s ′ f t ∗ s, z ∗ s, w]|w ∈ Ω}. 8.42 Evidently 8.42 is equivalent to 8.34 and 8.35 follows from 8.41 and the fact that λ ∗ n+1 1 = 0. Finally, the last assertion of the Theorem follows from 8.35 and the fact that ˜ M t, x ∗ t, ˜ p ∗ t ≡ constant if f , f are not explicitly dependent on t. ♦

8.3.2 Minimum-time problems