The estimation problem ODE stability The embedding method The simultaneous method In conclusion. Stability problems in ODE estimation

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Mathematical Sciences Institute Australian National University HPSC Hanoi 2006

Outline The estimation problem ODE stability The embedding method The simultaneous method In conclusion

Estimation Given the ODE: dx dt = f (t, x, β), where x R m, β R p, f R R m R p R m smooth enough, together with data y i = Hx(t i, β ) + ε i, i = 1, 2,, n, where H : R m R k, ε i N ( 0, σ 2 I ), estimate β. [ ] [ x(t) f(t, x) Equivalent smoothing problem: x, f β 0 Assume problem has a well determined solution for n, the number of observations, large enough. ].

The problem setting Mesh selection for integrating the ODE system is conditioned by two important considerations: The asymptotic analysis of the effects of noisy data on the parameter estimates shows that this gets small no faster than O ( n 1/2). It is not difficult to obtain ODE discretizations that give errors at most O ( n 2).

The problem setting Mesh selection for integrating the ODE system is conditioned by two important considerations: The asymptotic analysis of the effects of noisy data on the parameter estimates shows that this gets small no faster than O ( n 1/2). It is not difficult to obtain ODE discretizations that give errors at most O ( n 2). This suggests: That the trapezoidal rule provides an adequate integration method. That it should be possible even to integrate the ODE on a mesh coarser than that provided by the observation points {t i } (here we wont!).

The objective Estimation principles (least squares, maximum likelihood) consider the objective: F (x c, β) = n y i Hx (t i, β) 2. i=1 Methods differ in manner of generating comparison function values x(t i, β), i = 1, 2,, n. Embedding: x(t i, β, b) satisfies BVP dx dt = f(t, x, β), B 0 x(0) + B 1 x(1) = b. Introduces extra parameters b. Needs method for choosing B 0, B 1. Must solve boundary value problem at each step. go GNM

The objective Estimation principles (least squares, maximum likelihood) consider the objective: F (x c, β) = n y i Hx (t i, β) 2. i=1 Methods differ in manner of generating comparison function values x(t i, β), i = 1, 2,, n. Simultaneous: ODE discretization information added as constraints c i (x c ) = x i+1 x i h 2 (f i+1 + f i ), i = 1, 2,, n 1, with x i = x(t i, β). Methods typically correct solution and parameter estimates simultaneously. go SQP

Initial value stability (IVS) Here the problem considered is: dx dt = f (t, x), x(0) = b. The stability requirement is that solutions with close initial conditions x 1 (0), x 2 (0) remain close in an appropriate sense. x 1 (t) x 2 (t) 0, t. strong IVS.

Initial value stability (IVS) Here the problem considered is: dx dt = f (t, x), x(0) = b. The stability requirement is that solutions with close initial conditions x 1 (0), x 2 (0) remain close in an appropriate sense. x 1 (t) x 2 (t) 0, t. strong IVS. x 1 (t) x 2 (t) remains bounded as t. weak IVS.

Initial value stability (IVS) Here the problem considered is: dx dt = f (t, x), x(0) = b. The stability requirement is that solutions with close initial conditions x 1 (0), x 2 (0) remain close in an appropriate sense. x 1 (t) x 2 (t) 0, t. strong IVS. x 1 (t) x 2 (t) remains bounded as t. weak IVS. Computation introduces idea of stiff discretizations which preserve the stability characteristics of the original equation. Computations not limited by IVS. Important for multiple shooting - permits reasonably accurate fundamental matrices to be computed over short enough time intervals in relatively unstable problems by taking h small enough.

Constant coefficient case Here f (t, x) = Ax q If A is non-defective then weak IVS requires the eigenvalues λ i (A) to satisfy Reλ i 0 while this inequality must be strict for strong IVS. A one-step discretization of the ODE (ignoring q contribution) can be written x i+1 = T h (A) x i. where T h (A) is the amplification matrix. Here a stiff discretization requires the stability inequalities to map into the condition λ i (T h ) 1. For the trapezoidal rule λ i (T h ) = 1 + hλ i (A)/2 1 hλ i (A)/2, 1 if Re {λ i (A)} 0.

Boundary value stability (BVS) Here the problem is dx dt = f (t, x), B (x) = B 0 x(0) + B 1 x(1) = b. Behaviour of perturbations about a solution trajectory x (t) is governed to first order by the linearized equation L (z) = dz dt xf (t, x (t)) z = 0. Here (computational) stability is closely related to existence of a modest bound for the Green s matrix: G (t, s) = Z (t) [B 0 Z (0) + B 1 Z (1)] 1 B 0 Z (0)Z 1 (s), t > s, = Z (t) [B 0 Z (0) + B 1 Z (1)] 1 B 1 Z (1)Z 1 (s), t < s. Where Z (t) is a fundamental matrix for the linearised equation. Let α be a bound for G(t, s).

Dichotomy Weak form: projection P depending on choice of Z such that, given S 1 {ZPw, w R m }, S 2 {Z (I P) w, w R m }, φ S 1 φ(t) κ, φ(s) t s, φ S 2 φ(t) κ, φ(s) t s. Computational context requires modest κ for t, s [0, 1]. If Z satisfies B 0 Z (0) + B 1 Z (1) = I then P = B 0 Z (0) is a suitable projection in sense that for separated boundary conditions can take κ = α. There is a basic equivalence between stability and dichotomy. Key paper is de Hoog and Mattheij.

BVS restricts possible discretizations Dichotomy projection separates increasing and decreasing solutions. Compatible BC s pin down decreasing solutions at 0, growing solutions at 1.

BVS restricts possible discretizations Dichotomy projection separates increasing and decreasing solutions. Compatible BC s pin down decreasing solutions at 0, growing solutions at 1. Discretization needs similar property so given BC s exercise same control.

BVS restricts possible discretizations Dichotomy projection separates increasing and decreasing solutions. Compatible BC s pin down decreasing solutions at 0, growing solutions at 1. Discretization needs similar property so given BC s exercise same control. This requires solutions of ODE which are increasing (decreasing) in magnitude to be mapped into solutions of discretization which are increasing (decreasing) in magnitude.

BVS restricts possible discretizations Dichotomy projection separates increasing and decreasing solutions. Compatible BC s pin down decreasing solutions at 0, growing solutions at 1. Discretization needs similar property so given BC s exercise same control. This requires solutions of ODE which are increasing (decreasing) in magnitude to be mapped into solutions of discretization which are increasing (decreasing) in magnitude. This property called di-stability by England and Mattheij who show the TR is di-stable in constant coefficient case. λ(a) > 0 1 + hλ(a)/2 1 hλ(a)/2 > 1.

Bob Mattheij s example Consider the differential system defined by 1 19 cos 2t 0 1 + 19 sin 2t A(t) = 0 19 0, 1 + 19 sin 2t 0 1 + 19 cos 2t e t ( 1 + 19 (cos 2t sin 2t)) q(t) = 18e t e t (1 19 (cos 2t + sin 2t)). Here the right hand side is chosen so that z(t) = e t e satisfies the differential equation. The fundamental matrix displays the fast and slow solutions: go OBC Z (t, 0) = e 18t cos t 0 e 20t sin t 0 e 19t 0 e 18t sin t 0 e 20t cos t.

Bob Mattheij s example For boundary data with two terminal conditions and one initial condition : 0 0 0 1 0 0 e B 0 = 0 0 0, B 1 = 0 1 0, b = e, 1 0 0 0 0 0 1 the trapezoidal rule discretization scheme gives the following results. t =.1 t =.01 x(0) 1.0000.9999.9999 1.0000 1.0000 1.0000 x(1) 2.7183 2.7183 2.7183 2.7183 2.7183 2.7183 Table: Boundary point values - stable computation These computations are apparently satisfactory.

Bob Mattheij s example For two initial and one terminal condition: 0 0 1 0 0 0 B 0 = 0 0 0, B 1 = 0 1 0 1 0 0 0 0 0 The results are given in following Table., b = 1 e 1. t =.1 t =.01 x(0) 1.0000.9999 1.0000 1.0000 1.0000 1.0000 x(1) -7.9+11 2.7183-4.7+11 2.03+2 2.7183 1.31+2 Table: Boundary point values - unstable computation The effects of instability are seen clearly in the first and third solution components.

Nonlinear stability - preprint Hooker et al FitzHugh-Nagumo equations α =.2, β =.2. dv = γ (V V 3 ) dt 3 + R, dr = 1 (V α βr). dt γ

Nonlinear stability - preprint Hooker et al FitzHugh-Nagumo equations α =.2, β =.2. dv = γ (V V 3 ) dt 3 + R, dr = 1 (V α βr). dt γ

System factorization go OPT1 First problem is to set suitable boundary conditions. Expect good boundary conditions should lead to a relatively well conditioned linear system. Assume the ODE discretization is c i (x i, x i+1 ) = c ii (x i ) + c i(i+1) (x i+1 ). Consider the factorization of the difference equation (gradient) matrix with first column permuted to end: C 12 C 11 C 21 C 22 C (n 1)(n 1) C (n 1)n 0 [ Q U V 0 H G ] This step is independent of the boundary conditions. go SVE

Optimal boundary conditions The boundary conditions can[ be inserted ] at this point. This H G gives the system with matrix to solve for x B 1 B 1, x n. 0 Orthogonal factorization again provides a useful strategy. [ ] [ ] [ S T ] H G = L 0 1 S2 T It follows that the system determining x 1, x n is best conditioned by choosing [ B1 B 0 ] = S T 2. The conditions depend only on the ODE.

BC s for Mattheij example go MatEx The optimal boundary matrices corresponding to h =.1 are given in the Table. These confirm the importance of weighting the boundary data to reflect the stability requirements of a mix of fast and slow solutions. The solution does not differ from that obtained when the split into fast and slow was correctly anticipated. B 1 B 2.99955 0.0000.02126 -.01819 0.0000 -.01102 0.0000 0.0000 0.0000 0.0000 1.0000 0.0000.02126 0.0000.00045.85517 0.0000.51791 Table: Optimal boundary matrices when h =.1

Gauss-Newton details Let (β,b) x = of F is [ ] x β, x b, r i = y i Hx (t i, β, b) then the gradient (β,b) F = 2 n r T i H (β,b) x i. The gradient terms wrt β are found by solving the BVP s i=1 x B 0 β (0) + B x 1 (1) = 0, β d x dt β = xf x β + βf,

Gauss-Newton details Let (β,b) x = of F is [ ] x β, x b, r i = y i Hx (t i, β, b) then the gradient (β,b) F = 2 n r T i H (β,b) x i. i=1 while the gradient terms wrt b satisfy the BVP s x B 0 b (0) + B x 1 (1) = I, b d x dt b = xf x b.

Embedding: Again the Mattheij example Consider the modification of the Mattheij problem with parameters β1 = γ, and β 2 = 2 corresponding to the solution x (t, β ) = e t e: A(t) = q(t) = 1 β 1 cos β 2 t 0 1 + β 1 sin β 2 t 0 β 1 0 1 + β 1 sin β 2 t 0 1 + β 1 cos β 2 t e t ( 1 + γ (cos 2t sin 2t)) (γ 1)e t e t (1 γ (cos 2t + sin 2t)) In the numerical experiments optimal boundary conditions are set at the first iteration. The aim is to recover estimates of β, b from simulated data e t i He + ε i, ε i N(0,.01I) using Gauss-Newton, stopping when Fh < 10 8..,

Embedding: Again the Mattheij example H = [ 1/3 1/3 1/3 ] H = [.5 0.5 0 1 0 ] n = 51, γ = 10, σ =.1 14 iterations n = 51, γ = 20, σ =.1 11 iterations n = 251, γ = 10, σ =.1 9 iterations n = 251, γ = 20, σ =.1 8 iterations n = 51, γ = 10, σ =.1 5 iterations n = 51, γ = 20, σ =.1 9 iterations n = 251, γ = 10, σ =.1 4 iterations n = 251, γ = 20, σ =.1 5 iterations Here [ B 1 B 2 ] 1 [ B1 B 2 ] T k I F < 10 3, k > 1.

Lagrangian go OPT2 Associated with the equality constrained problem is the Lagrangian The necessary conditions give: n 1 L = F (x c ) + λ T i c i. i=1 xi L = 0, i = 1, 2,, n, c (x c ) = 0. The Newton equations determining corrections dx c, dλ c are: 2 xxldx c + 2 xλ Ldλ c = x L T, x c (x c ) dx c = Cdx c = c (x c ), Note sparsity! 2 xxl is block diagonal, 2 xλ L = CT is block bidiagonal.

SQP formulation The Newton equations also correspond to necessary conditions for the QP: min dx xfdx c + 1 2 dxt c Mdx c ; c + Cdx c = 0, in case M = 2 xxl, λ u = λ c + dλ c. A standard approach is to use the constraint equations to eliminate variables. go GNM dx i = v i + V i dx 1 + W i dx n, i = 2, 3,, n 1. The reduced constraint equation is Gdx 1 + Hdx n = w. Is this variable elimination restricted by BVS considerations?

Null space method Standard SQP approach. Let C T = [ Q 1 Newton equations can be written [ ] U QT 2 [ xxlq [ 0 U T 0 ] Q T ] dx c λ u 0 These can be solved in sequence U T Q T 1 dx c = c, Q 2 ] [ U 0 ] then [ Q = T x F T c Q T 2 2 xxlq 2 Q T 2 dx c = Q T 2 2 xxlq 1 Q T 1 dx c Q T 2 xf T, Uλ u = Q T 1 2 xxldx c Q T 1 xf T. ].

Stability test using Mattheij problem Q1 T dx c estimates Q1 T vec { e t } i when xc = 0. test results n = 11 particular integral Q1 T x.87665 -.97130-1.0001.74089-1.0987-1.3432.47327-1.2149-1.6230.11498-1.3427-1.8611 -.32987-1.4839-2.0366 -.85368-1.6400-2.1250-1.4428-1.8125-2.1018-2.0773-2.0031-1.9444-2.7309-2.2137-1.6330-3.3719-2.4466-1.1526.87660 -.97134-1.0001.74083-1.0988-1.3432.47321-1.2150-1.6231.11491-1.3428-1.8612 -.32994-1.4840-2.0367 -.85376-1.6401-2.1250-1.4429-1.8125-2.1019-2.0774-2.0032-1.9444-2.7310-2.2138-1.6331-3.3720-2.4467-1.1527

Conclusion Embedding makes use of carefully constructed, explicit boundary conditions. Thus BVS restrictions must apply. The system is special

Conclusion Embedding makes use of carefully constructed, explicit boundary conditions. Thus BVS restrictions must apply. The system is special The variable eliminations form of the simultaneous method partitions variables into sets {x 1, x n }, and {x 2,, x n 1 } which are found sequentially. It relies implicitly on a form of BVS although the system is special.

Conclusion Embedding makes use of carefully constructed, explicit boundary conditions. Thus BVS restrictions must apply. The system is special The variable eliminations form of the simultaneous method partitions variables into sets {x 1, x n }, and {x 2,, x n 1 } which are found sequentially. It relies implicitly on a form of BVS although the system is special. The null space variant partitions the variables into the sets { Q T 1 x c }, { Q T 2 x c }. It appears at least as stable as the variable elimination procedure. Sparsity preserving implementation is straightforward.