Numerical Methods for Ordinary Dierential Equations Episode 6 ppsx

Theorem 313B The Taylor expansions for the stages, stage derivatives andoutput result for a Runge–Kutta method are rt F ty0 + Oh n+1.. Finally, 313d follows from 314 Independence of the

The resulting sum is the value of Φ(t) A similar formula for Φ i (t), where i

is not a member of V , is found by replacing (312e) by

a ij

(k,l) ∈E

and summing this as for Φ(t).

Note that, although c does explicitly appear in Definition 312A or Lemma

312B, it is usually convenient to carry out the summations"s

l=1 a kl to yield

a result c k if l denotes a leaf (terminal vertex) of V This is possible because

l occurs only once in (312e) and (312f).

We illustrate the relationship between the trees and the corresponding

elementary weights in Table 312(I) For each of the four trees, we write Φ(t)

in the form given directly by Lemma 312B, and also with the summationover leaves explicitly carried out Finally, we present in Table 312(II) theelementary weights up to order 5

313 The Taylor expansion of the approximate solution

We show that the result output by a Runge–Kutta methods is exactly the

same as (311d), except that the factor γ(t) −1 is replaced by Φ(t) We first

establish a preliminary result

Lemma 313A Let k = 1, 2, , If

Y1, for hf (Y1), for Y2, for hf (Y2), and so on until we reach Y s , hf (Y s) and

finally y

Theorem 313B The Taylor expansions for the stages, stage derivatives and

output result for a Runge–Kutta method are

r(t) F (t)(y0) + O(h n+1 ). (313e)

Proof In a preliminary part of the proof, we consider the sequence of

We prove by induction that Y i [n] agrees with the expression given for Y i to

within O(h n+1 ) For n = 0 this is clear For n > 0, suppose it has been proved for n replaced by n − 1 From Lemma 313A with k = n − 1 and Y i replaced

Calculate Y i [n] using (313c) and the preliminary result follows Assume

that h is sufficiently small to guarantee convergence of the sequence (Y i[0], Y i[1], Y i[2], ) to Y i and (313c) follows Finally, (313d) follows from

314 Independence of the elementary differentials

Our aim of comparing the Taylor expansions of the exact and computedsolutions to an initial value problem will give an inconclusive answer unlessthe terms involving the various elementary differentials can be regarded asindependent We introduce a special type of differential equation for whichany finite number of elementary differentials evaluate to independent vectors

Let U denote any finite subset of T , such that if

t i = [t m1, t m2, , t m k]∈ U, (314a)

Table 314(I) Trees to order 4 with corresponding differential equations

then each of t1, t2, , t k is also a member of U For example, U might consist

of all trees with orders up to some specified integer Assume that when we

write a tree in this way, the t i , i = 1, 2, , k, are all distinct Suppose that N

is the number of members of U , and consider the m-dimensional differential

equation system in which

property of this initial value problem is encapsulated in the following result:

Theorem 314A The values of the elementary differentials for the differential

equation (314b), evaluated at the initial value, are given by

F (t i )(y(x0)) = e i , i = 1, 2, , N.

Because the natural basis vectors e1, e2, , e N are independent, therecannot be any linear relation between the elementary differentials for anarbitrary differential equation system

We illustrate this theorem in the case where U consists of the eight trees with up to four vertices Table 314(I) shows the trees numbered from i = 1

to i = 8, together with their recursive definitions in the form (314a) and the

corresponding differential equations Note that the construction given here isgiven as an exercise in Hairer, Nørsett and Wanner (1993)

315 Conditions for order

Now that we have expressions for the Taylor expansions of the exact solution,and also of the computed solution, we have all we need to find conditionsfor order If the exact solution has Taylor series given by (311d) and theapproximate solution has Taylor series given by (313e), then we need onlycompare these term by term to arrive at the principal result on the order ofRunge–Kutta methods

Theorem 315A A Runge–Kutta method with elementary weights

Φ : T → R, has order p if and only if

Φ(t) = 1γ(t) , for all t ∈ T such that r(t) ≤ p. (315a)

Proof The coefficient of F (t)(y0)h r(t) in (313e) is σ(t)1 Φ(t), compared with

the coefficient in (311d), which is σ(t)γ(t)1 Equate these coefficients and we

316 Order conditions for scalar problems

Early studies of Runge–Kutta methods were built around the single scalarequation

y  (x) = f (x, y(x)). (316a)

Even though it was always intended that methods derived for (316a) should

be interpreted, where appropriate, in a vector setting, a subtle difficulty arisesfor orders greater than 4

We adopt the notation f x , f y for partial derivatives of f with respect to

the first and second arguments, with similar notations for higher derivatives

Also, for simplicity, we omit the arguments in expressions like f x (x, y) By

straightforward differentiation of (316a), we have

y  = f

x + f y y  = f

x + f y f,

where the two terms together correspond to the elementary differential

associated with t = Similarly, for the third derivative we have

The expressions that arise here, and for the fourth derivative, are more

complicated, because of the presence of derivatives with respect to x However,

the terms can be grouped together according to the elementary differentials

to which they correspond Furthermore, the order conditions are identical tothose found in the general vector case When similar expressions are workedout for the 17 elementary differentials of order 5, we find a confusion between

the results for two particular trees In fact for each of t1= and t2= , F (t)

We discuss in Subsection 325 the construction of fifth order methods

These usually satisfy the so-called D(1) condition, which we introduce in

Subsection 321 This simplifying assumption has, as one of its consequences,the dependence of (316b) on other conditions, for which there is no confusion

Hence, for methods satisfying D(1), scalar and vector order 5 conditions are

equivalent

For orders 6 and higher, the confusion between the order conditions for thescalar case becomes more pronounced The first published methods of thisorder(Huˇta, 1956, 1957) were derived for scalar problems but, nevertheless,have order 6 for the general vector case (Butcher, 1963a)

317 Independence of elementary weights

We show in Subsection 324 that, given a positive integer p, there exists an integer s such that there is a Runge–Kutta method with s stages with order p.

We now present a more general result on the independence of the elementary

weights but without a specific value of s given.

Theorem 317A Given a finite subset T0, of T and a mapping φ : T0 → R, there exists a Runge–Kutta method such that the elementary weights satisfy

Φ(t) = φ(t), for all t ∈ T

Proof Let #T0 = n The set of possible values that can be taken by the vector of Φ(t) values, for all t ∈ T0, is a vector space To see why this is thecase, consider Runge–Kutta methods given by the tableaux

c A

c A

with s and s stages, respectively If the elementary weight functions for these

two Runge–Kutta methods are Φ and Φ, then the method given by the tableau

θb θb has elementary weight function θΦ + θΦ Let V ⊂ R ndenote this vector space

We complete the proof by showing that V =Rn If this were not the case, there

would exist a non-zero function ψ : T0 → R such that "t ∈T0ψ(t)Φ(t) = 0,

for all Runge–Kutta methods Because every coefficient in a Runge–Kutta

tableau can be multiplied by an arbitrary scalar θ to give a new method for which Φ(t) is replaced by θ r(t) Φ(t), we may assume that every non-zero value

of ψ corresponds to trees with the same order k This is impossible for k = 1, because in this case there is only a single tree τ Suppose the impossibility

of this has been proved for all orders less than k, but that there exist trees

t1, t2, , t m , each of order k, such that"m

i=1 ψ(t i )Φ(t i) = 0, for all Runge–

Kutta methods with ψ(t i) = 0, for i = 1, 2, , m Write t i = [t l i1

i1 t l i2

i2 · · · ], for i = 1, 2, , m Let ˆ t denote a tree appearing amongst the t ij which does

not occur with the same exponent in each of the t i Construct an s-stage

Runge–Kutta method

c A b for which each of Φ(t ij) = 1, except for Φ(ˆt) = θ Define second Runge–Kutta tableau with s + 1 stages of the form

1 b 0

0 1

If q i is the exponent of ˆt in t i, then it follows that

318 Local truncation error

The conditions for order give guarantees that the Taylor expansions of

the exact and computed solutions agree up to terms in h p Obtaining an

understanding of the respective terms in h p+1 is regarded as a key to derivingmethods that not only have a specific order, but also have a small truncation

error Because the number of terms of this order rises rapidly as p increases,

it is extremely difficult to know how this sort of optimality should be arrived

at Picking out just the terms of order p + 1, we can write the local truncation

error in a single step as

γ(t) − Φ(t)



F (t)(y0) + O(h p+2 ). (318a)

Since we are interested in asymptotic behaviour, that is, limiting behaviour for

h small, we do not devote much attention to the term O(h p+2) The coefficient

of h p+1 in (318a) is bounded in magnitude by

The first approach that can be considered is to compare, term by term, theexpression for (p+1)!1 y (p+1) (x0), which is proportional to the local truncationerror coefficient for linear multistep methods or for implicit Runge–Kuttamethods of collocation type The coefficient in this expression, corresponding

Another approach would be to assume a bound M on

the linear operator 

In studying the behaviour of a particular method of order p when used

to solve a particular initial value problem, we wish to assume that the localtruncation error is bounded asymptotically by some constant multiplied by

h p+1 This assumption will hinge on smoothness of the solution and the

differentiability, sufficiently many times, of f

319 Global truncation error

We consider the cumulative effect of errors in many steps leading to an error in

a final output point Suppose that n steps are performed to carry the solution from an initial point x0to a final point x If a constant stepsize is used, this would need to be equal to (x − x0)/n to exactly reach the final point Denote the approximations computed by a Runge–Kutta method by y1, y2, , y n,

with y0= y(x0) If the error committed in each of the n steps is bounded by

Ch p+1 then the total contribution to the error would seem to be

nCh p+1 = C(x − x0)h p

We attempt to make this argument more precise by noting that an error

in the initial value input to a step will lead to an error in the output valueconsisting of two terms The first of these is the perturbation to the outputdue to the error in the input, and the second is the truncation error due tothe method itself

In the statement of a preliminary lemma that we need, |A| and |b | will denote the matrix A and the vector b , respectively, with every term replaced

by its magnitude

referred to the computed solution That is, in this figure, δ k is the difference

between the exact solution defined by an initial value at the start of step k

and the numerical solution computed in this step Furthermore, ∆k is the

contribution to the global error resulting from the error δ k in step k An

alternative view of the growth of errors is seen from Figure 319(ii), where

δ k is now the difference between the exact solution at x k and the computed

solution found by using an input value y k −1 at the start of this step exactly

equal to y(x k −1) As in the previous figure, ∆k is the contribution to the

global error resulting from the local error δ k To obtain a bound on the global

truncation error we first need an estimate on δ1, δ2, , δ nusing these bounds

We then estimate by how much δ kcan grow to ∆k , k = 1, 2, , n The global

error is then bounded in norm by "n

k=1∆k We have a bound already from(110c) on how much a perturbation in the exact solution can grow If we werebasing our global error bound on Figure 319(i) then this would be exactlywhat we need However, we use Figure 319(ii), and in this case we obtain the

same growth factor but with L replaced by L  The advantage of using anargument based on this figure, rather than on Figure 319(i), is that we canthen use local truncation error defined in the standard way, by comparing the

exact solution at step value x n with the numerically computed result over a

single step with initial value y(x n −1)

Theorem 319B Let h0 and L  be such that the local truncation error at step

k = 1, 2, , n is bounded by

δ k ≤ Ch p+1 , h ≤ h0 Then the global truncation error is bounded by

n)− y n

 exp(L
(x −x0 ))−1

L
Ch p , L  > 0, (x − x0)Ch p , L  = 0.

Proof Use Figure 319(ii) and obtain the estimate

12 −1 12

1 34 14

3 4 1 4

find the elementary weights for the eight trees up to order 4 What isthe order of this method?

31.3 For an arbitrary Runge–Kutta method, find the order condition

corresponding to the tree

32 Low Order Explicit Methods

320 Methods of orders less than 4

It will be shown in Subsection 324 that, for an explicit method to have order

p, at least s = p stages are necessary We derive methods up to p = 3, with exactly p stages, and then discuss briefly the advantages of using s = p + 1 For s = p = 1 there is no choice beyond the Euler method with tableau

01

For s = p = 2, we have a one-parameter family of methods of the form

For s = p = 3, we must satisfy four conditions, which are shown together

with the corresponding trees as follows:

To solve these equations in the most straightforward manner, it is convenient

to treat c2and c3as free parameters and to carry out three steps First, solve

for b2 and b3 from the linear system given by (320b) and (320c) Secondly,

evaluate b1from (320a) Finally, solve for a32from (320d) This plan will runinto difficulties if the matrix of coefficients in (320b) and (320c) is singular;

that is, if c2c3(c3− c2) = 0 Assuming this does not occur, we have a further

difficulty if the solution to (320b) and (320c) results in b3= 0 This anomaly,

which occurs if c2 = 2

3, makes it impossible to solve (320d) A more carefulanalysis is necessary to resolve these difficulties, and it is possible to identifythree cases where a solution can be found These are

The coefficient tableaux for the three cases are summarized as follows, withthe general form of the tableau given in each case: for case I we have

2323

2

3− 14b3

1

4b3 1

23

0 − 14b3

1

4b3 1

parameterized by one or more positive integers For example, E(η, ζ) is a set

of assumptions about a method that hold for all positive integers k ≤ η and

l ≤ ζ.

The first of these conditions will be denoted by B(η), and simply states

that the conditions"s

i=1 b i c k −1

i = k −1 hold for k = 1, 2, , η For a method

to be of order p, it is necessary that B(p) holds, because this condition simply

restates the order condition for the trees

.

Table 321(I) Order conditions corresponding to some pairs of related trees

"

b k a ki c2

i = 1 24

To motivate condition C(η), consider pairs of trees t1and t2, with the sameorder, that differ in only one small respect Suppose they are labelled with

identical vertex sets and that the edge sets, say E1and E2, respectively, differ

only in that E1 contains the edges [i, j] and [j, k], and that j and k do not occur in any of the other ordered pairs in E1, whereas E2 contains the edge

[i, k] instead of [j, k] This will mean that the elementary weight corresponding

to t1 will have a factor a ij c j , whereas t2 will have a corresponding factor c2

i , for all i = 1, 2, , s. (321a)

We illustrate this by looking at some pairs of trees and noting the form ofthe equations

It is clear that, if it were possible for (321a) to hold for all i ∈ {1, 2, , s}, then we could simply remove the order equations associated with the t1

trees from consideration, because they will automatically be satisfied if the

conditions Φ(t) = 1/γ(t) are satisfied for the t2 trees However, it is not possible in the case i = 2 because this gives the equation 12c2 = 0 which

implies c2 = 0 It will then follow in turn that c3 = 0, c4 = 0, and all c

components equal to zero will not be consistent even with the order condition

"

b i c i = 1 While we cannot make use of the simplification of assuming

known as the ‘New Zealand Christmas tree’ because its bright red flowers bloom atChristmas-time.

(321a) in the case of explicit methods, we make extensive use of this andclosely related conditions in the case of implicit methods Furthermore, we

can still use this sort of simplification applied to just some of the stages.

In addition to (321a), we can consider the possibility that conditions like

j , Φ(t2) contains

a factor 1k c k i and the remaining factors are identical in the two expressions,

then Φ(t2) = 1/γ(t2) implies Φ(t1) = 1/γ(t1) We illustrate this in Figure321(i)

The D(k) conditions interrelate three trees t1, t2 and t3 for which the

corresponding elementary weights differ only in that Φ(t1) has a factor

b i c k −1

i a ij , Φ(t2) has a factor b j and Φ(t3) has a factor b j c k

j This means thatthese trees have forms like those shown in Figure 321(ii)

We illustrate this further, for the case k = 1, in Table 321(II) Note that if D(1) holds, then the truth of Φ(t1) = 1/γ(t1) follows from Φ(t2) = 1/γ(t2)

and Φ(t3) = 1/γ(t3) For explicit methods, D(2) cannot hold, for similar reasons to the impossibility of C(2) For implicit methods D(s) is possible, as

we shall see in Section 342

j (right-hand tree) The underlying tree is a kauri (Agathis australis) Although the immature tree shown is only a few metres tall, the most

famous kauri tree, Tane Mahuta (Lord of the Forest), has a height of 40 m and adiameter, 1.5 m above ground level, of 5.21 m

Table 321(II) Sets of three related trees illustrating D(1)

"

b j c2

j =1 3

322 Methods of order 4

It is an interesting consequence of the fourth order conditions for a method

with s = 4 stages, that c4 = 1 and that D(1) holds This fact reduces

significantly the number of conditions that remain to be solved; furthermore,

it is possible to segment the derivation into two phases: the solution of theremaining order conditions and the evaluation of the elements in the final row

of A to ensure that D(1) is actually satisfied Assuming that the method

In each of these calculations, the first column is the only non-zero term

in the middle column, while the final column is found by expanding themiddle column into a linear combination of elementary weights and equatingeach of these to the right-hand sides of the corresponding order conditions.For example, (322a) is evaluated from the trees , and and uses thecombination of order conditions

Φ( )− (c2+ c4)Φ( ) + c2c4Φ( ) = 1

γ( )− c2+ c4

γ( ) +

c2c4γ( ) .

From the first columns of (322a)–(322d), we observe that (322a)×(322d) =

(322b)×(322c) so that, from the last columns, we find



−

1

12− c2

6

 1