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

Bound for Remainder Interval in Rigorous
Integration
Joseph Galante
September 2008
1 The E stimate
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)
x(0) = x
0
(1)
Using the algorithm in (BM4), we can generate a p olynomial invariant under
A(x)(t) = x
0
+

t
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 n
th
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) = x
0
+


t
0
f(P +I)dτ. By FTTMA
f(P + I) will be a Taylor Model. We can decompose as f (P +I) = Q +R +
ˆ
I
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 = x
0
+

t
0
Qdτ.
Thus the other terms will contribute only to the remainder. ie
A(P + I) − P ⊂

t
0
(R +
ˆ
I)dτ (2)
1
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We want to better understand this relation since
ˆ
I depends upon I. We
actually have
A(P + I) = x
0
+

t
0
f(P + I)dτ (3)
= x
0
+

t
0
f(P ) + f

(P )I +
f

(P )
2
I
2
+ dτ (4)
= x
0
+


t
0
Q + R + f

(P )I +
f

(P )
2
I
2
+ dτ (5)
= P +

t
0
R + f

(P )I +
f

(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] w here d < 1. Then we would have
A(P + I) − P ⊂

t
0
R + f

(P )I +
f

(P )
2
I
2
+ dτ (8)
⊂

t
0
R + I · (f

(P ) +
f

(P )
2
+ )dτ (9)
(10)
Notice that the Taylor expansion of f(P + 1) about the point P gives

f(P + 1) = f(P ) + f

(P ) +
f

(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 ⊂

t
0
R + I · (f (P + 1) − f (P ))dτ (12)
⊂ B(

t
0
Rdτ ) + IB(

t
0
(f(P + 1) − f(P ))dτ) ⊂
want
I (13)
2
In the event that R = 0, ie f(P ) has no n
th
order terms, we can make an

upper bound by replacing R = 0 with R = t
n
. We can now solve for I to get
B(

t
0
Rdτ )
1 − B(

t
0
(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
n
th
degree pieces of f (P + I) will behave like O(h

n+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 = t
n
. 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(

t
0
(f(P + 1) − f(P ))dτ) ⊂
[−α, α] for some 0 < α < 1.
Remark: We can modify this formula to incorporate x
0
as a Taylor Model.
If x
0
= G + J where G is a degree n polynomial, then we create the invari-
ant polynomial P under
˜
A(x)(t) = G +

t

0
f(x(τ ))dτ . But then A(P )(t) =
3
G + J +

t
0
f(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 b ound to get
B(

h
0
Rdτ ) + J
1 − B(

h
0
(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.
2 Example
Consider the simple linear ODE

˙x = λx

x(0) = 1
(16)
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) = x
0
· (1 + λt +
(λt)
2
2
+
(λt)
3
6
) (17)
And the remainder interval is now
B(

h
0
λ(λτ )
3
6
dτ)
1 − B(

h
0
(λ · (P + 1) − λ(P ))dτ)
(18)

Which gives
d =
(λh)
4
24(1 − λh)
(19)
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)
4
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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