Gustaf S oderlind Lund University & E otv os Lor and ......Example Modifying ode45 in Matlab to...

Adaptive Linear Multistep MethodsDesigning Automatic Step Size Control for Multistep Methods

Gustaf SoderlindLund University & Eotvos Lorand University

Contents V1.1

• Classical control of asymptotic errorsAdaptive numerical ODE solvers

Linear multistep methods

• Control theoryTransfer functions and stability

The control design problem

• Dynamic error modelsConstructing a deadbeat controller

Modifying the controller

• Error estimationThe dynamic compensator

Scaling

c© G Soderlind, V1.1 ELTE, 4 April, 2019 Adaptive Linear Multistep Methods

1. Classical control of asymptotic errors

Time-stepping methods for initial value problem y = f (t, y)

Given an approximation yn ≈ y(tn) compute

yn 7→ yn+1

with time step size hn = tn+1 − tn

Work/accuracy trade-off

Put grid points where they matter to accuracy


Adaptive methods

Given an error tolerance tol, adaptive methods select thetime step hn to make the local error rn = tol

Static asymptotic step size – error model h→ 0

rn = ϕn hqn

Controller If ϕ is constant, choose hn+1 such that tol = ϕn hqn+1

⇒ hn+1 =

(tol

rn

)1/q

hn

makes rn+1 ≡ tol


Classical time-step control

In logarithmic form

log hn+1 − log hn =1

q(log tol− log rn)

Integrating control Summation of control errors

Linear difference equation, log r 7→ log h, where log r = {log rj}∞j=0


Example of actual heuristic implementation DASSL

Typical plot of log(hn+1/hn) vs. log(rn/tol)

10−2

10−1

100

101

102

100

Nonlinear, discontinuous and nonsymmetric!

◦


How well does it work? Chemakzo problem

10−10

10−8

10−6

10−4

10−10

10−8

10−6

10−4

TOL

achie

ved a

ccura

cy

10−10

10−8

10−6

10−4

102

103

achieved accuracy

# o

f fu

nction c

alls

10−10

10−8

10−6

10−4

101

102

TOL

# o

f ja

cobia

n c

alls

10−10

10−8

10−6

10−4

2

2.5

3

3.5

4

4.5

5

TOL

mean o

rder

Poor computational stability tol variatons have large effects


New improved controllers Chemakzo problem

10−10

10−8

10−6

10−4

10−10

10−8

10−6

10−4

TOL

ach

ieve

d a

ccu

racy

10−10

10−8

10−6

10−4

102

103

achieved accuracy

# o

f fu

nctio

n c

alls

10−10

10−8

10−6

10−4

101

102

TOL

# o

f ja

co

bia

n c

alls

10−10

10−8

10−6

10−4

2

2.5

3

3.5

4

4.5

5

TOL

me

an

ord

er

High computational stability Continuous dependence on tol


2. Control theory Feedback loop

Control

C (z)

Process

G (z)-log h --⊕

?

logϕ

log tol log r

−1

Static or dynamic asymptotic process model

r = ϕhq ⇒ log r = G (z) · log h + logϕ (G (z) = q)

Control law log h = C (z) · (log tol− log r)


Closed loop Controller – process interaction

Solve

log r = G (z) · log h + logϕ

log h = C (z) · (log tol− log r)

for Hϕ(z) : logϕ 7→ log h and Rϕ(z) : logϕ 7→ log r to find

Hϕ(z) = − C (z)

1 + C (z) · G (z)

Rϕ(q) =1

1 + C (z) · G (z)

Control design Low-pass filter for h and high-pass filter for r


Control design Choosing C (z)

General control structure

C (z) =P(z)

(z − 1)Q(z)

with degP = degQ corresponding to explicit step size recursion

(z − 1)Q(z) · log h = P(z) · (log r − log tol)

Control on multiplicative form filtering step size ratios ρ

ρn =

(tol

rn

)β1/q (tol

rn−1

)β2/q (tol

rn−2

)β3/q· ρα1

n−1 · ρα2n−2

hn+1 = ρn · hn


Control designs Static error model rn = ϕnhqn

Elementary control Q ≡ 1; P ≡ 1/qConvolution filter Q ≡ 1; P ≡ γ < 1/q

I control Q ≡ 1; degP = 0PI control Q ≡ 1; degP = 1

PID control Q ≡ 1; degP = 2FIR filter (z − 1)Q(z) + q · P(z) = zm

Autoregressive (AR) Q has zero(s) at z = 1Moving average (MA) P has zero(s) at z = −1

Noise shaping LP filter of MA type

hn+1 =

(tol

rn

)1/(bq)(tol

rn−1

)1/(bq)

ρ−1/bn−1 · hn


What step size properties can be achieved?

0 10 20 30 40 50 60 70 80 90 1000.4

0.5

0.6

0.7

0.8

0 10 20 30 40 50 60 70 80 90 1000.4

0.5

0.6

0.7

0.8

0 10 20 30 40 50 60 70 80 90 1000.4

0.5

0.6

0.7

0.8

Elementary control

FIR filter

Noise shaping filter


Example Modifying ode45 in Matlab to ode45dc

Step size sequences in chemotaxis problem

0 2 4 6 8 10 120

0.005

0.01

0.015

0.02

0.025

0.03

Original vs. modified code (PI controller)


3. Linear multistep methods

Example Adams–Bashforth two-step method AB2

yn+1 = yn +(

1 +ρn−1

2

)hny′n −

ρn−12

hny′n−1

Step size tn+1 = tn + hnStep ratio ρn−1 = hn/hn−1 = 1 + v , |v | � 1

Asymptotic local error, dynamic model

rn ≈5

12

...y (tn−1) h3n

(2

5+

3

5ρn−1

)≈ ϕn · hqn · ρδn−1

Static error parameter q = 3 and dynamic parameter δ = −3/5


Multiplicative approximation

Multinomials in step ratios occur frequently

1− θ +θ

ρ= 1− θ +

θ

1 + v

≈ 1− θ + θ(1− v)

= 1− θv

≈ (1 + v)−θ = ρ−θ


Local error structure k-step methods

Asymptotic error model, k = 2

rn = ϕn · hqn · ρδn−1

Static part as in one-step methods or constant step size theory

Dynamic part depends on current and past step sizes

General dynamic asymptotic error model, multiplicative form

rn = ϕn · hqn ·s∏

j=1

ρδjn−j

s = k − 1 for the actual error and s = k for error estimator


Process transfer function k = 2

Asymptotic error model, k = 2

rn = ϕn · hqn · ρδn−1 = ϕn · hq+δn h−δn−1

Taking logarithms

log rn = (q + δ) log hn − δ log hn−1 + logϕn

In terms of forward shift/z transform

log r =

(q + δ − δ

z

)log h + logϕ

Transfer function

GAB2(z) = q + δ · z − 1

z=

(q + δ)z − δz

=12z + 3

5z


Elementary control

hn+1 =

(tol

rn

)1/q

hn

Taking logarithms

log hn+1 − log hn =1

q(log tol− log rn)

In terms of forward shift/z transform

(z − 1) log h =1

qlog

tol

r

Control transfer function

C0(z) =1

q

1

z − 1


Closed loop error transfer function k = 2

Rϕ(z) =1

1 + C0(z) · G2(z)=

(z − 1)z

z2 + δq z −

δq

Deadbeat (FIR)

Static model (δ = 0) has Rϕ(z) = z−1z , a single pole at z = 0

In general (δ 6= 0), poles at

z =−δ ±

√δ2 + 4δq

2q

The closed loop is stable if |z | < 1


Closed loop poles . . .

Even for k = 2 the elementary controller has

• poor stability

• complex conjugate poles

• damped oscillatory homogeneous solutions

Need dedicated controllers compensating dynamics in error model


An intuitive controller It’s even worse!

Derive new controller, assuming ϕn+1 ≈ ϕn

tol = ϕn+1 · hq+δn+1 · h−δn

rn = ϕn · hq+δn · h−δn−1

⇒ hn+1 =

(tol

rn

)1/(q+δ)

ρ−δ/(q+δ)n−1 · hn

Closed loop poles z =

{0δ

q+δ

Not deadbeat, and if δ < 0, then a negative root causes overshoot

δAB2 = −35 δAM2 = −2

7 δBDF2 = −23 δEMP = −1

2


A general theorem Pole placement

Theorem Assume dynamic asymptotic error model

rn = ϕn hqn ·

s∏j=1

ρδjn−j

combined with step size control law

hn+1 =

(tol

rn

)β/qhn ·

s∏j=1

ραj

n−j

Taking αj = βδj/q implies one pole z = 1− β and s poles z = 0

If in addition β = 1, the controller is deadbeat


A general theorem . . .

One parameter family of controllers, with “integral gain” β

hn+1 =

tol

rn·

s∏j=1

ρδjn−j

β/q

hn

• Stable for β ∈ [0, 2]

• Deadbeat at β = 1

• Single positive pole for β ∈ (0, 1)

• Smooth exponential forgetting filter β < 1

• Good performance across methods β = 2/3

• Other controllers can be designed by pole placement


A dynamic compensator

Note rn = ϕn hqn ·∏s

j=1 ρδjn−j ⇒ rn ·

∏sj=1 ρ

−δjn−j = ϕn h

qn

hn+1 =

tol

rn·

s∏j=1

ρδjn−j

β/q

hn =

tol

rn ·∏s

j=1 ρ−δjn−j

β/q

hn

Definition Dynamically compensated error rn = rn ·∏s

j=1 ρ−δjn−j

• Satisfies static model rn = ϕn hqn

• Compatible magnitude rn ≈ rn since ρn−j ≈ 1

• Can be controlled using standard static digital filters


4. Error estimation

Problem Actual asymptotic local error is not computable

Grid independent methods advance the solution by

• Constructing a polynomial Pn+1(t)

• . . . satisfying characteristic conditions on

• . . . interpolation, collocation and slack

• Defining yn+1 = Pn+1(tn+1)

• Estimating the error by en = ‖Pn+1(tn+1)− Pn(tn+1)‖

Problem Is computable estimate en related to rn?


Error estimate

• Pn+1(t) and Pn(t) depend on similar data

• Only differences in y ′n+1, yn and y ′n−k , yn−k

• . . . and in step ratios ρn−1 and ρn−s−1

Computable error estimate

en = c y (q)(tn) · hqn ·s+1∏j=1

ρδjn−j

Actual asymptotic error

rn = c∗y (q)(tn) · hqn ·s∏

j=1

ρδjn−j


Error estimate

Error magnitude (due to ρn−j ≈ 1)

rn ≈c∗

c· en

Controller interacts with dynamics of estimator

Proposed control quantity

rn =c∗

c· ‖Pn+1(tn+1)− Pn(tn+1)‖ ·

s+1∏j=1

ρ−δjn−j

to be controlled by standard filters for static models with gain q

Includes magnitude compensator c∗/c and dynamic compensator


Error estimate in AB2

Asymptotic dynamic model of computable error estimator

en =23

12

...y (tn) · h3n · ρ

−45/23n−1 ρ

−12/23n−2

Dynamic model of actual asymptotic error

rn =5

12

...y (tn) · h3n · ρ

−3/5n−1

implies control quantity

rn =5

23· ‖Pn+1(tn+1)− Pn(tn+1)‖ · ρ45/23n−1 ρ

12/23n−2


A few conclusions

• Multistep methods always have a dynamic error model

• Classical, elementary controller works less well

• A true deadbeat (FIR) controller can be constructed

• A dynamic model for the error estimator is needed

• Compensate error estimator for dynamics and magnitude

• Stable interaction with well-known digital filter controllers


Thank you!


Gustaf S oderlind Lund University & E otv os Lor and ......Example Modifying ode45 in Matlab to...

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