12 bound for remainder interval in rigorous intergation josep galen 6 2008

Bound for Remainder Interval in RigorousIntegration Joseph Galante September 2008 We will give a bound on the remainder formula for rigorous integration using Taylor Models which updates

Bound for Remainder Interval in Rigorous

Integration

Joseph Galante September 2008

We will give a bound on the remainder formula for rigorous integration using

Taylor Models which updates the ideas in (BM4)

Suppose we are solving the ODE

(

˙x = f (x)

Using the algorithm in (BM4), we can generate a polynomial invariant under

A(x)(t) = x0 +Rt

0 f (x(τ ))dτ (the Picard Operator) by applying A (n + 1) times to the zero polynomial, and keeping the terms of degree ≤ n Let P

be the A invariant nth degree polynomial

To complete the application of Schauder’s Theorem we must find a Taylor

Model which is invariant under A We desire an interval I so that A(P +I) ⊂

P +I, ie A(P +I)−P ⊂ I We have A(P +I) = x0+Rt

0 f (P +I)dτ By FTTMA

f (P + I) will be a Taylor Model We can decompose as f (P + I) = Q + R + ˆI

where Q is all terms of degree (n − 1) or less, R is all degree n terms, and

ˆ

I is the remainder Since A(P ) =n P and since deg(R) = n, then R will

integrate to an (n + 1) order term, and we must have that P = x0+R0tQdτ

Thus the other terms will contribute only to the remainder ie

A(P + I) − P ⊂

Z t 0

(R + ˆI)dτ (2)

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We want to better understand this relation since ˆI depends upon I We actually have

A(P + I) = x0+

Z t 0

= x0+

Z t 0

f (P ) + f0(P )I + f

00(P )

2 I

2

+ dτ (4)

= x0+

Z t 0

Q + R + f0(P )I + f

00(P )

2 I

2+ dτ (5)

= P +

Z t 0

R + f0(P )I + f

00(P )

2 I

2+ dτ (6)

(7) Suppose we are working in a class of functions which is real analytic (or at least analytic on a large enough domain, say range(P ) + 2) Then the ’ ’ will converge due to class of functions we are working in Now suppose the interval I ⊂ [−d, d] where d < 1 Then we would have

A(P + I) − P ⊂

Z t 0

R + f0(P )I + f

00(P )

2 I

2+ dτ (8)

⊂

Z t 0

R + I · (f0(P ) +f

00(P )

2 + )dτ (9)

(10) Notice that the Taylor expansion of f (P + 1) about the point P gives

f (P + 1) = f (P ) + f0(P ) +f

00(P )

2 + (11) where the sum will converge due to the regularity class of the functions we are considering Hence we have

A(P + I) − P ⊂

Z t 0

R + I · (f (P + 1) − f (P ))dτ (12)

⊂ B(

Z t 0

Rdτ ) + IB(

Z t 0

(f (P + 1) − f (P ))dτ ) ⊂want I (13)

In the event that R = 0, ie f (P ) has no nth order terms, we can make an upper bound by replacing R = 0 with R = tn We can now solve for I to get

B(R0tRdτ )

1 − B(R0t(f (P + 1) − f (P ))dτ ) ⊂ I (14) Since we are choosing I, we choose it to the the left hand side, and we get our desired inclusion However there is a catch We have made the assumption that I ⊂ [−d, d] where d < 1 There is no reason why apriopi, this must

be true However we have one more variable which we can control Now we can control t Suppose t ∈ [−h, h] Clearly for h = 0, then left hand side evaluates to zero, which is to say that to model the initial condition will have

no error Also the expression is continuous as a function of h, except at the point where the denominator is zero, ie the point where the bounds become (−∞, ∞) By the intermediate value theorem, there is some h so that we can have d < 1 And we are done

Notice that the remainder interval I scales with h This makes sense since

it is easier to model the flow for a short period of time, than for a longer period What is hidden is exactly how well it scales The term R which is the

nth degree pieces of f (P + I) will behave like O(hn+1) so decreasing h (the time we are modeling the ODE for), or increasing n will result in a dramatic increase in accuracy

Remark: The above results carry through without difficulty into multidi-mensional system case The only detail to note is the the remainder interval must be the same in each dimension (This can probably be avoided with some more lengthy expressions for the remainder interval error, however we

do not persue this)

Remark: We did not necessarily need to replace R = 0 with R = tn It simply allows a single formula which works in all cases If instead we left R = 0, then

it would boil down to requiring t to be so that B(R0t(f (P + 1) − f (P ))dτ ) ⊂ [−α, α] for some 0 < α < 1

Remark: We can modify this formula to incorporate x0 as a Taylor Model

If x0 = G + J where G is a degree n polynomial, then we create the invari-ant polynomial P under ˜A(x)(t) = G +R0tf (x(τ ))dτ But then A(P )(t) =

G + J + 0tf (x(τ ))dτ = ˜A(P )(t) + J =n P + J We can then add this extra

J into the above estimates to get the remainder bound to get

B(R0hRdτ ) + J

1 − B(R0h(f (P + 1) − f (P ))dτ ) ⊂ I (15)

With appropiate Shrink Wrapping, B(J ) will be approximately zero, and this extra term won’t greatly ruin our degree n scaling for the remainder term

Consider the simple linear ODE

(

˙x = λx

We will consider a degree n = 3 Taylor Model and model up to time t = h Iteration of the operator A gives the polynomial

P (t) = x0 · (1 + λt + (λt)

2

2 +

(λt)3

And the remainder interval is now

B(R0h λ(λτ )6 3dτ )

1 − B(Rh

0 (λ · (P + 1) − λ(P ))dτ ) (18) Which gives

d = (λh)

4

Note that we require |λh| < 1 in order to avoid a useless bound This constraints our choice of h We are further constrained in that we need

|λh| so small that d < 1 (For this particular value of n, a choice of h = min(0.99,0.72|λ| ) works.) For λh < 1, then we can power series expand d to get

(λh)4

24(1 − λh) =

1

24(λh)

4(1 + λh + (λh)2+ (λh)3 ) (20)

Comparing this to the power series of the actual solution at time t = h,

Exp(λh) = (1 + λh + (λh)

2

2 +

(λh)3

6 +

(λh)4

24 +

(λh)5

120 + ) (21)

We see that P (h) + I has larger coefficents for terms of order 4 and higher Hence we have a valid enclosure

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