Validating an a priori enclosure using high order taylor series

Corliss and Robert RihmAbstract We use Taylor series plus an enclosure of the remainder term to validate the existence of a unique solution for initial value problems in ordinary di eren

George F Corliss and Robert Rihm

Abstract

We use Taylor series plus an enclosure of the remainder term to validate the existence of a unique solution for initial value problems in ordinary dierential equations and to compute a coarse enclosure of that solution The signicance of this result is its application to Lohner's AWA algorithm for validated solutions, not to the theory of ordinary dierential equations By using high-order Taylor series in Lohner's Algorithm I, we are able to validate the solution over much longer time steps than is done in the current AWA code For Lohner's enclosure

by polynomials, the enclosures are expensive to compute, but it is easy to check for enclosure For our enclosure by Taylor series, the enclosures are free because they are already being computed, but checking for enclosure requires 2  n poly-nomial rootndings Work is continuing on an implementation that will allow direct computational comparisons of the eectiveness of the two methods.

Keywords: ordinary dierential equations, Lohner's algorithm, validated computa-tion, Taylor series, enclosure methods

0 Introduction

The AWA (Anfangswertaufgabe) program by Rudolf Lohner [4, 5, 6, 7, 8] computes

an enclosure of the solution of an initial value problem (IVP) in ordinary dierential equations (ODE)

u

0=f(u); u(t

0) =u 0

where we only know an interval enclosure [u

0] of the vectoru

0in general Without loss

of generality, Lohner assumes the system is autonomous only to simplify the proofs

We assume that f is at least p ?1 times continuously dierentiable in a domainD

with [u

0]  D  IR

n, p  2 Then there is a unique at least ptimes continuously

dierentiable solution u(t) in a neighborhood oft

0 AWA is a single-step method At each integration time step, AWA applies two algo-rithms:

Algorithm I: (Existence and enclosure) Find a step size h and a coarse enclosure interval [u

0] D such that fort 2[t

0] := [t 0

; t

0+h], the solutionu(t) exists and satises () [ 0]

Algorithm II: (Tightening) Compute a tight enclosure foru(t) att=t

1:=t

0+h Throughout this paper, we use the term \validation" to mean \prove existence ofu(t) for somet and nd a coarse enclosure [u

0]"

The eciency of AWA has been limited by the Euler step size enforced by Algorithm

I as used in the current program We show that Taylor series plus an enclosure of the remainder term can be used not only in Algorithm II to tighten the enclosure, but also in Algorithm I for the validation step The consequence of this result for AWA

is that Algorithm I can validate the existence of the unique solution over time steps appropriate for high-order methods employed for tightening during Algorithm II The basic idea of AWA is to enclose the Taylor coecients and the remainder term of the solution It goes back to Moore who presented his method in 1965 (see [10, 11]) The Taylor coecients are computed using recurrence relations derived from the ODE:

f (0)(u) := f(u);

(u)0 := u;

f (i)(u) :=



@f (i?1)

@u

f



(u); i= 1;2; : ; p ?1;

(u)i := 1

i f (i?1)(u); i= 1;2; : ; p :

(2)

The recurrences given by Equation (2) are not as intimidating as they might appear; this is just an expression of the computation of the Taylor coecients for the solution They can be evaluated by using automatic dierentiation (see e.g [12])

If we begin the recurrences given by Equation (2) with the initial valueu

0and compute

in exact arithmetic, we get the Taylor coecients (u

0)i for the solution expanded at

t=t

0

If we begin the recurrences given by Equation (2) with an interval vector [u

0] and compute in rounded interval arithmetic, then we get enclosures ([u

0])i of the Taylor coecients for all solutions starting fromu

0

2[u

0]

We can also begin with acoarseinterval [u

0] and compute the interval vectors ([u

0])i Then (b

t ? t

0)p([u

0])p contains the remainder of the p ?1st degree Taylor polynomial

ofu(b

t), ifu(t)2[u

0](t) for allt 2[t

0

; b

t]

We emphasize that the computation of these values does not pre-suppose the existence

of a solution We are just computing sequences of numbers If we cannot compute any of these values, then we are not able to proceed with validation However, if we know ana priorienclosure [u

0] of the solution on [t

0], then

u(t)2

p?1 i=0

(t ? t

0)i(u

0)i+ (t ? t

0)p([u

0])p=:T

p(t; u 0

;[u

0]) or

u(t)2

p?1

(t ? t

0)i([u

0])i+ (t ? t

0)p([u

0])p =T

p(t;[u

0];[u

0]), resp (3)

fort 2[t

0], if only the interval polynomialT

p stays in [u

0] Moore as well as Lohner use (modications of) this formula to tighten ana priorienclosure [u

0] (see e.g [13])

1 Current AWA Approach

Validation and the computation of a tight enclosure are separate, though related, issues This paper addresses only the issue of validation, the role of Lohner's Algorithm

I The techniques of his Algorithm II for tightening the enclosure are discussed in [4, 5, 6, 7, 8] Algorithm I uses the following theorem for validation

Theorem 1 ([6]) Let[u

0] D be an interval vector satisfying

[u

1] :=u

0+ [0; h]
f

?

[u

0]

[u

0]:

Then

u(t)2[u

1] (and hence u(t)2[u

0])

for t 2[t

0] = [t

0

; t

0+h]

It follows directly from

Theorem 2 Let[u](t)and[v](t) D be interval vector valued functions, and let each

ptimes continuously dierentiable functionv(t)2[v](t)satisfy

(v(t)) :=u

0+

t Z

t

f(v())d 2[u](t)[v](t) (4)

for t 2[t

0] Then the solutionuof Equation (1) satises

u(t)2[u](t)

for t 2[t

0]

Proof: If (4) holds and we start a Picard-Lindelof iteration

u

(k )(t) = (u

(k ?1)(t)) =u

0+

t Z

t

f(u (k ?1)())d ; k= 1;2; :

with a ptimes dierentiable functionu

(0)(t)2[v](t), then every successor is againp

times dierentiable and lies in [u](t) However, the sequence of Picard iterates should leave this interval function if the solution did so

It seems strange to letv(t) beptimes dierentiable However, this assumption is used

in the following section

AWA nds a step sizehand an interval [u

0] such that [u

0][u

1] := [u

0] + [0; h]f

?

[u

0]

?

[u

0]

for allt 2[t

0] := [t

0

; t

0+h], and hence validates that the solution of Equation (1) exists

on [t

0] and is contained in [u

0] Figure 1 shows thea prioribound [u

0] = [0:2;0:5] and

([u

0]) for the logistic equationu

0=u(1? u),u(0) = 0:3, on the intervalt 2[0;0:6]

In this gure, ([u

0])[u

0] in the intervalt 2[0;0:5] If we tighten [u

0] = [0:29;0:5], then ([u

0])[u

0] in the larger intervalt 2[0;0:563]

t

0.6 0.5

0.4 0.3

0.2 0.1

0

u

0.6

0.5

0.4

0.3

0.2

0.1

0

Figure 1 A priori bounds with [u

0] = [0:2;0:5]

2 Lohner's Enclosure of u(p)

The allowable step sizehfor which validationcan be done using Theorem 1 as described

in Section 1 is limited to a step appropriate for Euler's method, no matter how high the order of the method used during Algorithm II to tighten the enclosure

Stetter showed in [14] that for the problem u

0 = Au, A 2 IR

nn, the step size h

obtained by applying Theorem 1 may be multiplied by the degree of the Taylor

polynomial used in Algorithm II Algorithm II yields a validated tight enclosure for

t = t

0+ph because p!([u

0])p = A

p[u

0] contains the p-th derivative u

(p)(t) for all

t 2 [t

0

; t

0+ph], even though [u

0] does not enclose the solution on this enlarged interval, in general

Lohner [9] suggests using coarse enclosures for higher derivatives also in the general case Actually, he uses Theorem 2 instead of Theorem 1 and a special kind of interval polynomials for [v](t) instead of a constant coarse enclosure as in the current AWA program Lohner guesses an enclosure [v

0] ofu (p)(t)=p! satisfying [v](t) :=p?1

i=0

(t ? t

0)i(u

0)i+ (t ? t

0)p[v

0] D :

The Picard-Lindelof operator  reproduces the Taylor coecients of the exact solution There is an interval vector [v

1] such that each ptimes dierentiable function v(t)2

[v](t) satises

(v(t))2[u](t) :=p?1

i=0

(t ? t

0)i(u

0)i+ (t ? t

0)p[v

1]:

If [v

1]  [v

0] holds, then we also have [u](t) 2 [v](t), and according to Theorem 2,

u(t)2[u](t)

To obtain [v

1], one has to compute an enclosure of the remainder coecient of the

p ?2nd degree Taylor polynomial of f(v(t)) for all p times dierentiable functions

v(t) 2 [v](t) It also contains the remainder coecient of the p ?1st degree Taylor polynomial of (v(t)) For computing or enclosing these coecients on a computer, one has to apply polynomial interval machine arithmetic given by Eiermann [3], which provides the enclosure of a xed number of Taylor coecients of standard operations and standard functions on interval polynomialsas well as the enclosure of the respective remainders The advantage of this idea of Lohner over the validation strategy discussed

in Section 1 is that validation is possible for step sizes appropriate for a method of local order p

Example 1 The solution of the initial value problem

u

0=?u; u(0) = 1

satises u(t) = e

?t

2 [0;1] for t  0 However, the application of Theorem 1 for

[u

0] = [0;1]requires a step size1since

u

0+ [0; h]f([u

0])[u

0]

, 1 + [0; h](?[0;1])[0;1]

However, Lohner's approach yields validation for much longer steps If we choose

p= 6and [v

0] = [0;1], we get

[v](t) = 1? t+t

2

2 ?

t 3

6 + t

4

24?

t 5

120 + t

6

720[0;1];

f([v](t)) =?[v](t) ?1 +t ?

t 2

2 +t

3

6 ?

t 4

24 + t

5

120[1?

h

6;1];and

[u](t) = ([v](t)) = 1? t+t

2

2 ?

t 3

6 + t

4

24?

t 5

120 + t

6

720[1?

h

6;1]

for t 2[t

0

; t

0+h] We have

[1? h

6;1][v

0] = [0;1] , 0 h 6:

Of course, we could enclose the sixth Taylor coecient off instead of the fth one to obtain h= 7 However, in general we only assumef to be p ?1times dierentiable

In this example, the coecients off(v(t)) can easily be calculated However, there

is no implementation of the method for the general case yet, although Lohner has described [9] such an implementation using the Eiermann operators for interval poly-nomial arithmetic

3 Validated Enclosure Using Taylor Series

The use of Taylor series for validation, as well as for tight enclosure, is suggested by pictures such as those in Figure 2, which shows Taylor series enclosures for the solution

of the Lorenz system

t

0.05 0.04

0.03 0.02

0.01

0

20

18

16

14

12

10

8

6

4

2

0.05 0.04

0.03 0.02

0.01

0

30

20

10

0

-10

t

0.05 0.04

0.03 0.02

0.01

0

40

30

20

10

0

Figure 2 Taylor Series Enclosures for the Solution of the Lorenz System

As already mentioned, AWA uses the recurrence relations (2) and the formula (3) for computing a tight enclosure in Algorithm II after having validated a coarse enclosure [u

0] for a step sizehin Algorithm I It would cause no additional costs if the Taylor expansion (3) could also be used in Algorithm I The following theorem shows that this can actually be done

Theorem 3 Let u

0

2 int([u

0]) D Let

[u](t) :=p?1

i=0

(t ? t

0)i(u

0)i+ (t ? t

0)p([u

0])p

[u

for t 2[t

0] Then

u(t)2[u](t) for t 2[t

0]: Proof: It follows from our assumptions that Equation (1) has a solutionuin some neighborhood of the point (t

0

; u

0), for every u

0

2 [u

0] We must prove that that neighborhood includes all of [ ] If (6) holds, then () [ ]() as long as ( ) [ 0]

Assume thatu(t) leaves [u](t) =: [u(t); u(t)], and hence [u

0] atb

t 2 int([t

0]) Without loss of generality, we assumeu

j(b

t) =u

j(b

t) = sup[u

0]j, for somej 2 f1; : ; ng From

u

(p)(t)2 p!([u

0])p, it follows that

u

0(t)2 p?2 i=0

(t ? t

0)i(u

0)i+ (t ? t

0)p?1

p([u

0])p

;

and henceu

0(t) u

0(t) on [t

0

; b

t] Therefore, u

j(b

t) =u

j(b

t) impliesu

j(t) =u

j(t) for all

t 2 [t

0

;

b

t] If u

j is constant, then we have u

0 =u

j(b

t) = sup[u

0]j, which contradicts the assumptionu

0

2 int([u

0]) Otherwise,u

j(b

t) is an isolated maximum of thep-th degree polynomialu

j The same holds foru

j(t), since this function is at leastptimes continuously dierentiable in [t

0] Hence, u(t) cannot leave [u](t) atb

t This approach avoids complicated calculations and nevertheless takes the orderpinto account With p = 6, we can validate the coarse enclosure [0;1] in Example 1 for

h = 2:1, compared with h = 1 for a constant bound enclosure and h = 6 using Lohner's polynomial enclosure

Example 2 We want to validate the solution u(t) = 1=t of the initial value problem

u

0=?u 2

; u(1) = 1

in the interval [u

0] = [0;2] For the constant bound enclosure, we have

[u

1] := 1 + [0; h](?[0;2]2) = [1?4h;1][u

0] , h 1=4

as the maximal step size we can reach by applying Theorem 1 However, the approach

of this section forp= 2yields a longer step:

[u](t) = 1?(t ?1) + [0;8](t ?1)2

[0;2], h 

1 +p

33

16 0:42:

4 Implications for AWA

Using the results of the previous section, AWA Algorithm I is

Task 1: Approximate solution Compute Taylor coecients of an approximate solu-tion

b

u(t) :=p?1

i=0

(t ? t

0)i(u

0)i

from recurrence relation (2) In practice, we must compute (u

0)i using rounded interval arithmetic to capture any round-o errors The work of Task 1 is already being done in AWA Algorithm II

Task 2: Guess a steph Use heuristics based on the radius of convergence [1] of the truncated series forb uor on previous steps sizes

Task 3: Guess a constant enclosure [u

0] Bound the range ofb u([t

0

; t

0+h

it a little

Task 4: Compute the remainder terms ([u

0])i, for i = 0; : ; p from the recurrence relation (2) The work of Task 4 is already being done in AWA Algorithm II Task 5: Compute stepsize for validation The stepsize is the point nearest tot

0(and

in [t

0]) at which

[u](t) :=p?1

i=0

(t ? t

0)i(u

0)i+ (t ? t

0)p([u

0])p

leaves [u

0] This requires 2 n polynomial rootndings For each polyno-mial rootnding problem, it is sucient to compute only relatively coarse lower bounds for the smallest real root in the interval (t

0

; t

0+h], so a few iterations

of a simplied interval Newton method should suce

Remark. Variable order We must compute ([u

0])0,: :, ([u

0])p?1in order to compute ([u

0])p Hence, it is relatively inexpensive to compute the step size in Task 5 for each order p The order used for validation in Algorithm I can be dierent from the order used for tightening in Algorithm II The information is readily available to make both Algorithms I and II fully variable order

Remark. Iteration If Task 5 nds that [u](t)[u

0] on [t

0], then we can repeat Tasks

4 and 5 with a largerhor else with [u

0] = [u]([t

0])

Remark. Optimization Given guesses for h from Task 2 and for [u

0] from Task

3, we compute in Task 5 the largest step for which we can validate existence and containment Numerical experiments have shown [2] that a carefully chosen [u

0] allows step sizes 10 times as long as step sizes corresponding to [u

0] given by reasonable heuristics That suggests viewing Algorithm I as an optimization problem:

Maximize step sizeh

by varying guessed hand [u

0] subject to [u](t) :=p?1

i=0

(t ? t

0)i(u

0)i+ (t ? t

0)p([u

0])p

[u

0] Remark. Eectiveness Either Lohner's enclosure by polynomials or our enclosure by Taylor series is signicantly more expensive than Picard-Lindelof iteration of constant bounds currently used by AWA However, AWA Algorithm II involves the solution of

an ODE system of dimension n  n, so almost any eort we expend in Algorithm I that allows AWA to take longer steps improves the overall eciency of AWA

Remark. Which is better? For Lohner's enclosure by polynomials, the enclosures are expensive to compute, but it is easy to check for enclosure For our enclosure by Taylor series, the enclosures are free because they are already being computed, but checking for enclosure requires 2npolynomialrootndings Work is continuing on an implementation that will allow direct computational comparisons of the eectiveness

of the two methods

Acknowledgments

We thank the anonymous referee for helpful suggestions The work of Corliss was supported in part by National Science Foundation Grant No DMS{9413525 and in part by SUN Academic Equipment Grant No EDUD{US{940208

References

[1] Y F Chang and G Corliss, Solving Ordinary Dierential Equations Using Taylor Series, ACM Trans Math Software, 8(1982), 114{144

[2] G Corliss, Validating an A Priori Enclosure Using High-Order Taylor Series, presented at SCAN '95, Wuppertal, 1995

[3] M Ch Eiermann, Adaptive Berechnung von Integraltransformationen mit Fehlerschranken, PhD thesis, Universitat Freiburg, 1989

[4] R J Lohner, Anfangswertaufgaben im IR

n mit kompakten Mengen fur An-fangswerte und Parameter, Diplomarbeit, Inst f Angew Math., Universitat Karl-sruhe, 1978

[5] R J Lohner, Enclosing the Solutions of Ordinary Initial and Boundary Value Problems, in Computer Arithmetic: Scientic Computation and Programming Languages, E W Kaucher, U Kulisch, and C Ullrich, eds., Wiley-Teubner Series

in Computer Science, Stuttgart, 1987, pp 255{286

[6] R J Lohner, Einschlieung der Losung gewohnlicher Anfangs{ und Randwert-aufgaben und Anwendungen, PhD thesis, Universitat Karlsruhe, 1988

[7] R J Lohner, Interval Arithmetic in Staggered Correction Format, in Scientic Computing with Automatic Result Verication, E Adams and U Kulisch, eds., Academic Press, San Diego, 1993

[8] R J Lohner,On Step Size and Order Control in the Veried Solution of Ordinary Initial Value Problems, presented at SCAN '93, September 1993, Vienna

Assume thatu(t) leaves [u](t)...

AWA nds a step sizehand an interval [u

0] such that [u

0][u

1]...

8

6

4

2

0.05