Computational Methods in Optimal Control Lecture 8. hp-collocation

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1 Computational Methods in Optimal Control Lecture 8. hp-collocation William W. Hager July 26, 2018

2 10,000 Yen Prize Problem (google this) Let p be a polynomial of degree at most N and let 1 < τ 1 < τ 2 <... < τ N < 1 be the Gauss quadrature points. Suppose that p( 1) = 0 and p (τ i ) 1 for all 1 i N. Show that p(τ i ) 2 for all 1 i N. This gives D 1 2: Let p P N with p( 1) = 0. Let p be the vector with components p i = p(τ i ), and let ṗ be the vector with components ṗ i = ṗ(τ i ). Since Dp = ṗ, we have p = D 1 ṗ. Since D 1 = max{ D 1 y : 1 y 1}, it follows that D 1 is the maximum value p(τ i ) over all polynomials for which p( 1) = 0 and ṗ(τ i ) 1. NOTE: the polynomial that achieves the maximum value for p(τ i ) is p(τ) = 1+τ.

3 Model Problem for hp Pseudospectral Method minimize C(x(1)) subject to ẋ(t) = f(x(t),u(t)), u(t) U, t Ω 0, x(0) = x 0. Ω 0 = [0,1], x 0 given, x(t) R n, U R m closed and convex, f : R n R m R n and C : R n R

4 hp Pseudospectral Idea Partition time domain Ω 0 into K subintervals and use different polynomial to approximate state on each subinterval. Require continuity across subintervals. Change variables so that x k (τ), τ [ 1,1] corresponds to x(t), t [t k 1,t k ]. x t _ k t h = k 1 1/2K = τ ε [ 1, 1] 2 Continuous: x k (τ) = x ( t k + 1/2 hτ) Discrete: x k ε P N t K = 1 t 0 t 0 = 0 t t k 1 t k t K t

5 Radau Collocation Points Radau quadrature points (zeros of P (1,0) N 1 plus τ N = 1): 1 < τ 1 < τ 2 <... < τ N = +1. Additional points in analysis: τ 0 = 1 Quadrature weights: ω i = 2(1+τ i ) N 1 (τ i)] 2, 1 i N 1, ω N = 2 N 2. [(1 τ 2 i )Ṗ(1,0) For every p P 2N 2 : 1 1 p(τ)dτ = ω i p(τ i ). i=1

6 Continuous and Discrete Problems on K Mesh Intervals Continuous: minimize C(x K (1)) subject to ẋ k (τ) = hf(x k (τ),u k (τ)), u k (τ) U, τ Ω, x k ( 1) = x k 1 (1), 1 k K. Discrete: minimize C(x K (1)) subject to ẋ k (τ i ) = hf(x k (τ i ),u ki ), u ki U, 1 i N, x k ( 1) = x k 1 (1), x k PN n, 1 k K.

7 Easy to Satisfy Continuity with Radau Lagrange interpolating polynomials: For 0 j N, Φ i (τ) = N j=0 j i If x k P N, then x k (τ) = { τ τ j 1 for j = i, Φ i (τ j ) = τ i τ j 0 otherwise. X kj (τ j )Φ j (τ), X kj = x k (τ j ). j=0 Continuity X k 1,N = X k0. Differentiation matrix D R N (N+1) ẋ(τ i ) = j=0 Φ j (τ i )x(τ j ) = D ij x(τ j ), D ij = Φ j (τ i ) j=0

8 Discrete Control Problem Hence, in terms of the state and control values X kj and U kj at the N collocation points on each of the K intervals, the discrete control problem can be formulated as minimize C(X KN ) subject to N j=0 D ijx kj = hf(x ki,u ki ), U ki U, 1 i N, X k0 = X k 1,N, 1 k K, where X 0N = X 0 is the starting condition.

9 Lagrangian and KKT Conditions K k=1 L(λ,X,U) = C (X KN )+ N i=1 λ ki,hf(x ki,u ki ) N j=0 D ijx kj + K k=1 λ k0,(x k 1,N X k0 ). Differentiating with respect to each of the variables: D i0 λ ki = λ k0, X k0 X kj X kn i=1 D ij λ ki = h x H(X kj,u kj,λ kj ), 1 j < N, i=1 D in λ ki = h x H(X kn,u kn,λ kn )+λ k+1,0, i=1 λ K+1,0 := C(X KN ), U ki u H(X ki,u ki,λ ki ) N U (U ki ).

10 First-order Optimality in Polynomial Setting Lemma. The multipliers λ k R Nn, 1 k K, satisfy the KKT conditions if and only if the polynomial λ k PN 1 n given by λ k (τ i ) = λ ki /ω i, 1 i N, satisfies the following conditions and λ k0 = λ k ( 1). λ k (τ i ) = h x H(x k (τ i ),u ki,λ k (τ i )), 1 i < N, λ k (1) = h x H(x k (1),u kn,λ k (1))+(λ k (1) λ k+1 ( 1))/ω N, where λ K+1 ( 1) := C (x K (1)) N U (u ki ) u H(x k (τ i ),u ki,λ k (τ i )), 1 i N. NOTE: The discrete λ is typically discontinuous at the mesh points.

11 D for Radau The differentiation matrix D for the space of polynomials of degree N 1 evaluated at τ i, 1 i N, is given by D NN = D NN + 1 ω N and D ij = ω j ω i D ji otherwise. In other words, if p is a polynomial of degree at most N 1 and if p R N is the vector with i-th component p i = p(τ i ), then (D p) i = ṗ(τ i ), 1 i N.

12 Proof If p P N and q P N 1 with p( 1) = 0, then integration by parts gives 1 1 ṗ(τ)q(τ)dτ = p(1)q(1) p(τ) q(τ)dτ. 1 1 Since ṗq and p q are polynomials of degree at most 2N 2, Radau quadrature is exact and we have w j ṗ j q j = p N q N j=1 w j p j q j, j=1 where p j = p(τ j ) and ṗ j = ṗ(τ j ). This yields (Wṗ) T q = p N q N (Wp) T q. Substituting ṗ = D 1:N p and q = D q gives p T D T 1:N Wq = p Nq N p T WD q.

13 Proof continued... Rearrange into the following form: p T (D T 1:N W+WD e N e T N )q = 0, where e N is the last column of I. Since this identity must be satisfied for all choices of p and q, we deduce that D = W 1 e N e T N W 1 D T 1:N W.

14 Proof of the Lemma Define Λ ki = λ ki /ω i for 1 i N, Λ k0 = λ k0, and D ij = ω jd ji /ω i. These substitutions in the KKT conditions yield D ij Λ kj = h x H(X ki,u ki,λ ki ), 1 i < N, j=1 D Ni Λ ki = [h x H(X kn,u kn,λ kn )+Λ k+1,0 /ω N ], (1) i=1 N U (U ki ) u H(X ki,u ki,λ ki ), 1 i N.

15 Proof continued Since the polynomial that is identically equal to 1 has derivative 0, we have D1 = 0, which implies that D 0 = N j=1 D j, where D j is the j-th column of D. Hence, Λ k0 = = λ ki D i0 = i=1 i=1 j=1 = Λ k+1,0 +h i=1 j=1 λ ki D ij = i=1 j=1 ω i D ij Λ kj ( ) ω i x H(X ki,u ki,λ ki ), i=1 ω j ( λki ω i where the discrete costate equations are used in the last step. ) (ω i D ij /ω j )

16 Let λ k PN 1 n satisfy λ kτ i = Λ ki, 1 i N. D ij Λ kj = λ k (τ i ), 1 i < N, and j=1 D Nj Λ kj = λ k (1) λ k (1)/ω N. (2) j=1 This substitution in ( ) yields Λ k0 = λ k (1) ω i λk (τ i ). Since λ k PN 2 n and N-point Radau quadrature is exact for these polynomial, we have 1 ω i λk (τ i ) = λ k (τ)dτ = λ k (1) λ k ( 1). i=1 1 i=1

17 Continued... Combine last two equations to obtain (1) + (2) + (3) yield the formula for λ k (1). Λ k0 = λ k ( 1) ((3))

18 Fundamental Differences: Gauss/hp Radau For Gauss, the entries of the matrices D and D satisfy D ij = D N+1 i,n+1 j, 1 i N, 1 j N. In other words, D 1:N = JD 1:NJ where J is the exchange matrix with ones on its counterdiagonal and zeros elsewhere. Hence, D 1 = (D ) 1. 20,000 Yen Prize Problem: Show that (D ) 1 2 and the rows of [W 1/2 D ] 1 have Euclidean length bounded by 2. In the hp-scheme, only need to have K is large enough, or equivalently h is small enough, that 2hd 1 < 1 and 2hd 2 < 1, where d 1 = sup t Ω 0 x f(x (t),u (t)) d 2 = sup t Ω 0 x f(x (t),u (t)) T

19 Discrete and continuous vectors X kj = x k (τ j), 0 j N, 1 k K, U kj = u k (τ j), 1 j N, 1 k K, Λ kj = λ k (τ j), 0 j N, 1 k K. X N kj = xn k (τ j), 0 j N, 1 k K, U N kj = un kj, 1 j N, 1 k K, Λ N kj = λn k (τ j), 0 j N, 1 k K. NOTE: λ k (1) = λ k+1 ( 1) but λn k (1) λn k+1 ( 1).

20 Main Theorem Theorem. Suppose (x,u ) is a local minimizer for the continuous problem with (x,λ ) PH η (Ω 0 ;R n ) for some η 2. If some smoothness and local convexity conditions hold, then for N sufficiently large or for h sufficiently small with N 2, the discrete problem has a local minimizer x N k Pn N and un k RmN, and an associated multiplier λ N k Pn N 1, 1 k K; moreover, there exists a constant c, independent of h, N and η, such that max { X N X, U N U, Λ N Λ } h p 1( ) c p 1 x N PH p (Ω 0 ) +h q 1( c q 1.5 λ N) PH q (Ω 0 ), where p = min(η,n +1) and q = min(η,n).

21 Example minimize subject to [x 2 (t)+u 2 (t)] dt ẋ(t) = u(t), u(t) 1, x(t) 2 e 1 e, t [0,1], 5e +3 x(0) = 4(1 e). Exact solution: 0 t 1 4 : x (t) = t e 1 e, u (t) = 1, t 3 4 : x (t) = et 1 e (1+e 3 2 2t ), u (t) = et 1 e (1 e 3 2 2t ), 3 4 t 1 : x (t) = 2 e 1 e, u (t) =

22 Error in Solution -4 log 10 (error) state control N