University of Houston, Department of Mathematics Numerical Analysis II Chapter 2...

University of Houston, Department of MathematicsNumerical Analysis II

Chapter 2 One-step methods for non-stiff differential equations

2.1 Examples of elementary one-step methods

Consider the following initial value problem for a scalar differential equation of first order:

y′(x) = f(x,y(x)) , x ∈ [a,b] ⊂ lR ,

y(a) = α .

Equidistant partition of [a,b] by

xk := a + kh , 0 ≤ k ≤ N

with step length h := (b− a)/N.

Integration of the differential equation over [xk,xk+1] yields

y(xk+1) − y(xk) =xk+1∫

xk

f(x,y(x)) dx .

Idea: Approximation of the right-hand side by a quadrature formula.


2.1.1 Explicit Euler method

x

f(x,y(x))

k+1xkx

y

��

��

xk+1∫

xkf(x,y(x)) dx ≈ h f(xk,y(xk))

If yk , 0 ≤ k ≤ N, is an approximation of the solutiony in xk, we obtain

yk+1 = yk + h f(xk,yk) , 0 ≤ k ≤ N− 1 ,y0 = α .

Interpretation: Approximation of the solution y = y(x) by its tangent

x

y

x xx1 2

y(x)

0

α

y1

y2

��

��

��

��

��

��

��

��

��

��

��

��

��


2.1.2 Implicit Euler method (backward Euler)

x

y

f(x,y(x))

k+1xkx

xk+1∫

xkf(x,y(x)) dx ≈ h f(xk+1,y(xk+1))

We obtain

yk+1 = yk + h f(xk+1,yk+1) , 0 ≤ k ≤ N− 1 ,y0 = α .


2.1.3 Implicit trapezoidal rule

x

f(x,y(x))

k+1xkx

y

��

��

xk+1∫

xkf(x,y(x)) dx ≈ h2 [f(xk,y(xk)) + f(xk+1,y(xk+1))]

We obtain

yk+1 = yk +h2 [f(xk,yk) + f(xk+1,yk+1)] , 0 ≤ k ≤ N− 1 ,

y0 = α .


2.1.4 Implicit midpoint rule

x

f(x,y(x))

k+1xkx

y

��

xk+1∫

xkf(x,y(x)) dx ≈ h f(xk +

h2 ,y(xk +

h2))

Observing y(xk +h2) ≈

12(y(xk) + y(xk+1)), we obtain

yk+1 = yk + h f(xk +h2 ,

12(yk + yk+1)) , 0 ≤ k ≤ N− 1 ,

y0 = α .


2.1.5 Explicit midpoint rule (two-step method)

y

xx k+1x k+2

f(x,y(x))

xk

��

xk+2∫

xkf(x,y(x)) dx ≈ 2 h f(xk+1,y(xk+1))

We obtain

yk+2 = yk + 2 h f(xk+1,yk+1) , 0 ≤ k ≤ N− 2 ,y0 = α .

y1 = α + h f(x0, α) (initialization) .


2.2 Consistency, stability, and convergence

In this section, we consider the initial value problem

y′(x) = f(x,y(x)) , x ∈ I := [a,b] ⊂ lR ,

y(a) = α

where f ∈ C(I×D) , D ⊂ lRd, and

‖f(x,y1) − f(x,y2)‖ ≤ L ‖y1 − y2‖ , (x,yi) ∈ I×D , 1 ≤ i ≤ 2 , L ≥ 0 .

Definition 2.1 Explicit and implicit one-step methods

Let Ih := {xk = a + kh | 0 ≤ k ≤ N} , N ∈ lN, be a partition of I = [a,b] with steplength

h = (b− a)/N and Φ : Ih × Ih × lRd × lRd × lR+ → lR

d. Then,

yk+1 = yk + h Φ(xk,xk+1,yk,yk+1,h) , 0 ≤ k ≤ N− 1 ,

y0 = α

is called a one-step method with increment function Φ. The one-step method is said to be

explicit, if Φ does not depend on yk+1, and is called implicit, otherwise.


Remark 2.2 Interpretation as a difference equation

Setting I′h := Ih \ {xN} and introducing the grid function yh : Ih → lRd, the one-step

method is equivalent to the first order difference equation

yh(x + h) = yh(x) + h Φ(x,x + h,yh(x),yh(x + h),h) , x ∈ I′h ,

yh(a) = α

Definition 2.3 Global discretization error, convergence, order of convergence

The grid functioneh(x) := yh(x) − y(x) , x ∈ Ih

is called the global discretization error. The one-step method is said to be convergent, if

maxx∈Ih‖eh(x)‖ → 0 (h→ 0) .

It is said to be convergent of order p, if

maxx∈Ih‖eh(x)‖ = O(h

p) (h→ 0) .

2


Remark 2.4 Consistency

If the solution y ∈ C1(I) is inserted into the one-step method, we should expect

y(x + h) − y(x)

h= h Φ(x,x + h,y(x),y(x + h),h),

↓ h→ 0 ↓ h→ 0

y′(x) f(x,y(x))

Definition 2.5 Local discretization error, consistency, order of consistency

The grid function

τ h(x) := h−1 [y(x + h) − y(x)] − Φ(x,x + h,y(x),y(x + h),h) , x ∈ I′h

is called the local discretization error. The one-step method is said to be consistent with the

initial value problem, ifh

∑

x∈I′h

‖τ h(x)‖ → 0 (h→ 0) .

It is said to be consistent of order p, if

h∑

x∈I′h

‖τ h(x)‖ = O(hp) (h→ 0) .


Lemma 2.6 Sufficient condition for consistency

The one-step method is consistent with the initial value problem, if

(∗) maxx∈I′

h

‖τ h(x)‖ → 0 (h→ 0) .

The condition (∗) holds true if and only if

(∗∗) maxx∈I′

h

‖Φ(x,x + h,y(x),y(x + h),h) − f(x,y(x))‖ → 0 (h→ 0) .

Proof: We have

h∑

x∈I′h

‖τ h(x)‖ ≤ maxx∈I′

h

‖τ h(x)‖∑

x∈I′h

h = (b − a) maxx∈I′

h

‖τ h(x)‖ .

The equivalence of (∗) and (∗∗) follows readily from the representation

τ h(x) = h−1 [y(x + h) − y(x)] − y′(x) + f(x,y(x)) − Φ(x,x + h,y(x),y(x + h),h) .


Example 2.7 Order of consisteny of the explicit Euler method

Explicit Euler method:

yh(x + h) = yh(x) + h f(x,yh(x)) , yh(a) = α .

The local discretization error is given by:

τ h(x) = h−1 [y(x + h) − y(x)] − f(x,y(x)) .

Taylor expansion yields

y(x + h) = y(x) + h y′(x) +1

2h2 y′′(x) + O(h3) =⇒

h−1 [y(x + h) − y(x)] = y′(x) +1

2h y′′(x) + O(h2) = f(x,y(x)) +

1

2h y′′(x) + O(h2) .

It follows that

τ h(x) =1

2h y′′(x) + O(h2) .

If y ∈ C2(I), the explicit Euler method is consistent of order 1.


Example 2.8 Order of consistency of the implicit trapezoidal rule

Implicit trapezoidal rule:

yh(x + h) = yh(x) +h

2[f(x,yh(x)) + f(x + h,yh(x + h))] , yh(a) = α .

Taylor expansion results in

y(x + h) = y(x) + h y′(x) +1

2h2 y′′(x) +

1

6h3 y′′′(x) + O(h4) =⇒

h−1 [y(x + h)− y(x)] = y′(x) +1

2h y′′(x) +

1

6h2 y′′′(x) + O(h3) =⇒

h−1 [y(x + h)− y(x)] = f(x,y(x)) +1

2h [fx(x,y(x)) + fy(x,y(x)) f(x,y(x))] +

1

6h2 y′′′(x) + O(h3) .

f(x + h,y(x + h)) = f(x,y(x)) + h fx(x,y(x)) + h fy(x,y(x)) y′(x) + O(h2) =⇒

1

2[f(x,y(x)) + f(x + h,y(x + h))] = f(x,y(x)) +

1

2h [fx(x,y(x)) + fy(x,y(x)) f(x,y(x))] + O(h

2) .

It follows thatτ h(x) = O(h

2) .

2


We consider the explicit one-step method

(⋆) yh(x + h) = yh(x) + h Φ(x,yh(x),h) , yh(a) = α .

Definition 2.9 Asymptotic stability

Consider the perturbed one-step method

zh(x + h) = zh(x) + h [Φ(x, zh(x),h) + σh(x)] , zh(a) = α + βh .

The one-step method (⋆) is said to be asymptotically stable, if there is hmax > 0 and for

each ε > 0 there exists δ > 0 such that for perturbations σh , βh for which

‖βh‖ + h∑

x∈I′h(x)

‖σh(x)‖ < δ

the solution of the perturbed one-step method satisfies the condition

maxx∈Ih‖(yh − zh)(x)‖ + h

∑

x∈I′h

‖(Dhyh − Dhzh)(x)‖ < ε ,

uniformly in h < hmax, where (Dhyh)(x) := h−1 [yh(x + h)− yh(x)] , x ∈ I

′h.


Remark 2.10 Motivation for the discrete version of Gronwall’s lemma

Consider the explicit Euler method with and without perturbations

yh(x + h) = yh(x) + h f(x,yh(x)) , yh(a) = α ,

zh(x + h) = zh(x) + h [f(x, zh(x)) + σh(x)] , zh(a) = α + βh .

(i) For x = x0 = a, we have

‖(yh − zh)(x0)‖ = ‖βh‖ .

(ii) For x = x1 = x0 + h, we obtain

‖(yh − zh)(x1)‖ ≤ ‖(yh − zh)(x0)‖ + h ‖σh(x0)‖ + h ‖f(x0,yh(x0)) − f(x0, zh(x0))‖ ≤

≤ ‖βh‖ + h ‖σh(x0)‖ + h L ‖(yh − zh)(x0)‖ .

(iii) Finally, for x2 = x1 + h, it follows that

‖(yh − zh)(x2)‖ ≤ ‖(yh − zh)(x1)‖ + h ‖σh(x1)‖ + h ‖f(x1,yh(x1)) − f(x1, zh(x1))‖ ≤

≤ ‖βh‖ + h ‖σh(x0)‖ + h ‖σh(x1)‖ + h L ‖(yh − zh)(x0)‖ + h L ‖(yh − zh)(x1)‖ .


Lemma 2.11 Discrete version of Gronwall’s lemma

Assume that there exist δ ≥ 0 and η ≥ 0 such that the grid function vh : Ih → lRd satisfies

‖vj+1‖ ≤ δ hj

∑

k=0‖vk‖ + η , 0 ≤ j ≤ N− 1 ,

‖v0‖ ≤ η .

Then, there holds‖vj‖ ≤ η exp(δ (xj − a)) , 0 ≤ j ≤ N .

Proof: Assume ‖vj‖ ≤M , 0 ≤ j ≤ N. We prove by induction:

‖vj‖ ≤ Mδm (xj − a)

m

m!+ η

m−1∑

ℓ=0

δℓ (xj − a)ℓ

ℓ!.

1.Step: The assertion holds true for j = 0 , m ∈ lN and m = 0 , 0 ≤ j ≤ N.

2.Step: Assume that the assertion is true for some m ∈ lN and 0 ≤ j ≤ N.

3.Step: We have

‖vj+1‖ ≤ δ hj

∑

k=0[ M

δm (xk − a)m

m!+ η

m−1∑

ℓ=0

δℓ (xk − a)ℓ

ℓ!] + η .

2


Using the elementary inequality

hj

∑

k=0(xk − a)

ℓ ≤xj+1∫

a(x − a)ℓ dx =

(xj+1 − a)ℓ+1

ℓ + 1,

for m + 1 and j + 1 the estimate

‖vj+1‖ ≤ δ hj

∑

k=0[ M

δm (xk − a)m

m!+ η

m−1∑

ℓ=0

δℓ (xk − a)ℓ

ℓ!] + η

yields the assertion

‖vj+1‖ ≤ Mδm+1 (xj+1 − a)

m+1

(m + 1)!+ η

m∑

ℓ=0

δℓ (xj+1 − a)ℓ

ℓ!.


Theorem 2.12 Sufficient condition for asymptotic stability

Assume that the increment function Φ satisfies the following Lipschitz condition:There exist U ⊂ I×D of the graph (x,y(x)) , x ∈ I, of the solution y = y(x) of the initial value

problem and hmax > 0 and LΦ ≥ 0 such that for all 0 < h < hmax:

(⋄) ‖Φ(x,y1,h) − Φ(x,y2,h)‖ ≤ LΦ ‖y1 − y2‖ , (x,yi) ∈ U , 1 ≤ i ≤ 2 .

Consider perturbations σh , βh such that

‖βh‖ + h∑

x∈Ih

‖σh(x)‖ ≤ δ .

Then, for the solution yh of the unperturbed and for the solution zh of the perturbed one-

step method there holds

(•) ‖(yh − zh)(x)‖ ≤ [ ‖βh‖ + h∑

x∈Ih

‖σh(x)‖ ] exp(LΦ(x− a)) ,

(••) ‖(Dhyh − Dhzh)(x)‖ ≤ LΦ ‖(yh − zh)(x)‖ + ‖σh(x)‖ .

In particular, the one-step method is asymptotically stable.


Proof: Without restriction of generality assume that U = I×D and define

wh(x) := zh(x) − yh(x) , x ∈ Ih .

Setting wj := wh(xj), we obtain

wj+1 = wj + h [ Φ(xj, zj,h) + σj − Φ(xj,yj,h) ] , w0 = βh =⇒

wj+1 = w0 + hj

∑

k=0[ σk + Φ(xk, zk,h) − Φ(xk,yk,h) ] .

Using the Lipschitz condition (⋄), it follows that

‖wj+1‖ ≤ ‖βh‖ + hN∑

k=0‖σk‖ + LΦ h

j∑

k=0‖wk‖ .

The discrete version of Gronwall’s lemma implies (•).(••) is an immediate consequence of

(Dhwh)(x) = Φ(x, zh(x),h) + σh(x) − Φ(x,yh(x),h)

and (⋄).


Theorem 2.13 Consistency & Stability =⇒ Convergence

Assume that the one-step method (∗) is consistent with the initial value problem and asymp-

totically stable. Then, there holds

(+) maxx∈Ih‖(yh − y)(x)‖ → 0 (h→ 0) ,

(++) h∑

x∈I′h

‖(Dhyh − Dhy)(x)‖ → 0 (h→ 0) .

Proof: We define zh(x) := y(x) , x ∈ Ih. Then σh(x) = τ h(x) and βh = 0. Due to the consistency,

for each ε > 0 there exist hmax(ε) > 0 and δ(ε) > 0 such that for 0 < h < hmax(ε):

h∑

x∈I′h

‖τ h(x)‖ < δ(ε) .

We conclude by taking the asymptotic stability of the one-step method into account.


Theorem 2.14 Characterization of the convergence

Assume that the one-step method (∗) satisfies the Lipschitz condition (⋄). Then, the follo-

wing two statements are equivalent:

(i) The one-step method (∗) is convergent. In particular, (+), (++) hold true.

(ii) The one-step method (∗) is consistent with the initial value problem.

For consistent one-step methods there exists hmax > 0 such that for all 0 < h < hmax the

following a priori estimate holds true:

‖(yh − y)(x)‖ ≤ (h∑

x∈I′h

‖τ h(x)‖) exp(LΦ (x − a)) .

Moreover, if maxx∈I′

h

‖τ h(x)‖ → 0 (h→ 0), then there holds

maxx∈I′

h

‖(Dhyh − Dhy)(x)‖ → 0 (h→ 0) .


Proof: (a) (ii) =⇒ (i):

Theorem 2.12 implies the asymptotic stability of the one-step method and Theorem 2.13

provides the convergence in the sense that (+), (++) hold true.

The a priori estimate and the uniform convergence of the first differences can be deduced

from Theorem 2.12 with zh(x) := y(x) so that σh(x) = τ h(x) and βh = 0.

(b) (i) =⇒ (ii)

Due to the convergence, for sufficiently small h the Lipschitz condition (⋄) can be applied to

Φ(x,yh(x),h) − Φ(x,y(x),h)

whence

‖τ h(x)‖ = ‖Dhy(x) − Φ(x,y(x),h)‖ = ‖(Dh)y(x) − Φ(x,y(x),h) − Dhyh(x) + Φ(x,yh(x),h)‖ ≤

≤ ‖(Dhyh − Dhy)(x)‖ + LΦ ‖(yh − y)(x)‖ .

Multiplication by h and summation over x ∈ I′h allows to conclude.


Remark 2.15 Generalization to implicit one-step methods

The notion of asymptotic atability can be generalized to implicit one-step methods.

Analogous stability and convergence results can be established.

Reamrk 2.16 Lipschitz condition for the inkrement function

If the increment function Φ is defined in terms of the right-hand side f of the initial value

problem, the Lipschitz condition (⋄) for Φ can be deduced from the associated Lipschitz

condition for f.


2.3 Explicit Runge-Kutta methods

Motivation and an example:

x

y

x xx1 2

y(x)

0

α

y1

y2

��

��

��

��

��

��

��

��

��

��

The explicit Euler method uses the approximate slope

f(xk,yk) ≈ f(xk,y(xk)) of the solution curve at xk.

Idea: Use a suitable linear combination of approximations

of the slopes f(xk,y(xk)) and f(xk + a21h,y(xk + a21h)).

This leads to the explicit one-step-method:

(⋆1) yk+1 = yk + h [ b1 f(xk,yk) + b2 f(xk + a21 h,yk + a21 h f(xk,yk) ] , k ≥ 0 ,

(∗2) y0 = α .

Goal: Determine a21 , b1 , b2 such that the one-step method (∗1), (∗2) is consistent of order p = 2.


Taylor expansion (assuming sufficient smoothness of f):

(+1)y(x + h)− y(x)

h= y′(x) +

h

2y′′(x) + O(h2) = f(x,y(x)) +

h

2[fx(x,y(x)) + fy(x,y(x))f(x,y(x))] + O(h

2).

(+2) b1f(x,y(x)) + b2f(x + a21h,y(x) + a21hf(x,y(x))) = (b1 + b2)f(x,y(x)) +

+ h[a21b2fx(x,y(x)) + a21b2fy(x,y(x))f(x,y(x))] + O(h2).

Comparison of the coefficients in (+1), (+2) yields

b1 + b2 = 1 , a21 b2 =1

2.

Setting β := b2, we obtain

b1 = 1 − β , a21 =1

2β.


Definition 2.17 The methods of Runge and Heun

The one-step method

(⋆1) yk+1 = yk + (1− β) h f(xk,yk) + β h f(xk +1

2βh,yk +

h

2βf(xk,yk)) , k ≥ 0 ,

(⋆2) y0 = α

with β = 1 is called improved polygon method or method of Runge.

Choosing β = 12 it is referred to as the method of Heun.

Problem: Does a suitable choice of β give the order of consistency p = 3?


Determination of an additional term in the Taylor expansion:

(+1)y(x + h)− y(x)

h= y′(x) +

h

2y′′(x) +

h2

6y′′′(x) + O(h3) =

= f +h

2[fx + fyf ] +

h2

6[fxx + 2ffxy + f

2fyy + fxfy + ff2y ] + O(h

3) ,

(+2) (1− β)f(x,y(x)) + βf(x +h

2β,y(x) +

h

2βf(x,y(x))) =

= f +h

2[fx + fyf ] +

h2

8β[fxx + 2ffxy + f

2fyy] + O(h3) .

(+1), (+2) imply:

τ h(x) =h2

6[ (1−

3

4β) (fxx + 2ffxy + f

2fyy) + fxfy + ff2y ] + O(h

3) .

Consequently:

(i) The order of consistency p = 3 can not be achieved by a proper choice of β.

(ii) Choosing β = 34, the sum of the absolute values of the coefficients of the derivatives in

the leading error term of τ h is minimized.

2


Definition 2.18 Explicit Runge-Kutta method of stage sAn explicit one-step method of the form

(⋄1) yk+1 = yk +s∑

j=1bj kj , k ≥ 0 ,

(⋄2) y0 = α

such thatk1 := h f(xk,yk) ,

k2 := h f(xk + a21h,yk + a21k1) ,

k3 := h f(xk + a31h + a32h,yk + a31k1 + a32k2) ,

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

ki := h f(xk + cih,yk +i−1∑

j=1aijkj) , ci :=

i−1∑

j=1aij , 1 ≤ i ≤ s

is called an explicit Runge-Kutta method of stage s.

It is uniquely determined by the s (s + 1)/2 parameters aij , 2 ≤ i ≤ s , 1 ≤ j ≤ i− 1, and

bi , 1 ≤ i ≤ s.


Butcher scheme

Explicit Runge-Kutta methods of stage s are described by the so-called Butcher scheme:

c1 = 0 a21c2 a31 a32· · · ·

· · · · ·

· · · · · ·

cs as1 as2 · · as,s−1b1 b2 · · bs−1 bs


Lemma 2.19 Consistency of Runge-Kutta methods of stage s

The Runge-Kutta method (⋄1), (⋄2) of stage s is consistent with the initial value problem, if

(•)s∑

i=1bi = 1

Proof: For the local discretization error we obtain

τ h(x) =y(x + h) − y(x)

h−

s∑

i=1bi f(x + cih,y(x) +

i−1∑

j=1aijkj) ,

where kj is given as in (⋄1) with yk replaced by y(x).

For h→ 0 it follows that

τ h(x) → y′(x) −

s∑

i=1bi f(x,y(x)) = y

′(x) − f(x,y(x)) = 0 .


Goal: Choose the parameters aij and bi such that a maximal order of consistency p can beachieved.

Realization: For a large number of stages s one uses graph-theoretical concepts (Butcher trees).

The following table contains the number of free parameters Ns and the number of determiningequations Ms for the stage s as well as the maximal order of consistency ps for stage s:

s 1 2 3 4 5 6 7 8 9 10

Ns 1 3 6 10 15 21 28 36 45 55

Ms 1 2 4 8 17 37 85 200 486 1205

ps 1 2 3 4 4 5 6 6 7 8

If the number of determining equations is less than the number of free parameters, the free

parameters are chosen such that the leading term in the global discretization error is minimized.


Example 2.20 3-stage Runge-Kutta method of consistency order p = 4

The 4 determining equations for the 6 parameters are given by

b1 + b2 + b3 = 1 ,

b2 a21 + b3 (a31 + a32) =1

2,

b2 a221 + b3 (a

231 + 2 a31 a32 + a

232) =

1

3,

b3 a32 a21 =1

6.

This classical 3-stage Runge-Kutta method has the Butcher scheme

12

12

12 0

12

1 0 0 116

13

13

16


2.4 Order and stepsize control

2.4.1 Asymptotic expansion of the global discretization error

Assume that(ESV1) yh(x + h) = yh(x) + h Φ(x,yh(x),h) , x ∈ I

′h ,

(ESV2) yh(a) = α

is an explicit one-step method of consistency order p.

Assume further that the local discretization error has the asymptotic expansion

(∗) y(x + h) − y(x) − h Φ(x,y(x),h) = dp+1(x) hp+1 + dp+2(x) h

p+2 + ...

Does the global discretization error eh(x) also have an asymptotic expansion

(∗∗) eh(x) = ep(x) hp + ep+1 h

p+1 + ... ?

Can this asymptotic expansion be used for the construction of higher order methods?


Idea: Define a ’new’ one-step method with the increment function Φ∗:

(ESV∗1) y∗h(x + h) = y

∗h(x) + h Φ

∗(x,y∗h(x),h) , x ∈ I′h ,

(ESV∗2) y∗h(a) = α .

Setting δyh(x) := y∗h(x)− yh(x), the one-step method (ESV1) is equivalent to

y∗h(x + h) − δyh(x + h) = y∗h(x) − δyh(x) + h Φ(x,y

∗h(x)− δyh(x),h) .

A comparison with (ESV∗1) yields

(⋄) h Φ∗(x,y∗h(x),h) = h Φ(x,y∗h(x)− δyh(x),h) + δyh(x + h) − δyh(x) .

Observing (⋄), for the local discretization error of the ’new’ one-step method (ESV∗1), (ESV∗2)

we obtain

y(x + h)− y(x) − h Φ∗(x,y(x),h) = y(x + h)− y(x)− δyh(x + h) + δyh(x)− h Φ(x,y(x)− δyh(x),h) .


Taking

y(x + h)− y(x) − h Φ∗(x,y(x),h) = y(x + h)− y(x)− δyh(x + h) + δyh(x)− h Φ(x,y(x)− δyh(x),h)

into account and using the asymptotic expansion (∗), we get

(+) y(x + h)− y(x) − h Φ∗(x,y(x),h) = dp+1(x) hp+1 + δyh(x) − δyh(x + h) +

+ h [ Φy(x,y(x),h) δyh(x) + O(‖δyh(x)‖2) ] + O(hp+2) .

Settinge(x) := − h−p δyh(x) , e(a) := 0 ,

it follows that

(i) δyh(x) − δyh(x + h) = hp [e(x + h) − e(x)] = e′(x) hp+1 + O(hp+2) ,

(ii) O(‖δyh(x)‖2) = O(h2p) ,

(iii) Φy(x,y(x),h) = Φy(x,y(x),0) + O(h) ,

(iv) Φy(x,y(x),0) = fy(x,y(x)) (is a consequence of the consistency) .


Inserting (i)− (iv) into (+) gives

y(x + h)− y(x) − h Φ∗(x,y(x),h) = hp+1 [ dp+1(x) + e′(x) − fy(x,y(x)) e(x) ] + O(h

p+2) .

Hence, the one-step method (ESV∗1), (ESV∗2) is consistent of order p + 1, if

e′(x) = fy(x,y(x)) e(x) − dp+1(x) , e(a) = 0 .

In summary, we have shown:

Theorem 2.23 Theorem of Hairer and Lubich

Assume that f ∈ Cp+1(I×D) and that Φ ∈ Cp+1(I×D× lR+) is the increment function of a

one-step method with

y(x + h) − y(x) − h Φ(x,y(x),h) = dp+1(x) hp+1 + O(hp+2) .

Moreover, let e = e(x) , x ∈ I, be the solution of the linear initial value problem

e′(x) = fy(x,y(x)) e(x) − dp+1(x) , e(a) = 0 .

2


For the one-step method

(ESV∗1) y∗h(x + h) = y

∗h(x) + h Φ

∗(x,y∗h(x),h) , x ∈ I′h ,

(ESV∗2) y∗h(a) = α

with the increment function

Φ∗(x,y∗h(x),h) = Φ(x,y∗h(x) + e(x)h

p,h) − [e(x + h) − e(x)] hp−1

we have

y(x + h) − y(x) − h Φ∗(x,y(x),h) = O(hp+2) .

Remark 2.24 Extension and historical comments

(i) Theorem 2.23 can be analogously extended to implicit one-step methods.

(ii) The proof of an asymptotic expansion of the global discretization error for the explicit

Euler method is due to Bauer,Rutishauser,Stiefel (1963) and Gragg (1964).


Theorem 2.25 Theorem of Gragg (1964)

Assume that f ∈ Cp+k(I×D) and that Φ ∈ Cp+k(I×D× lR+) , k ≥ 1, is the increment function

of a one-step method with Φ(x,y,0) = f(x,y) , (x,y) ∈ I×D.

Moreover, suppose that there are functions dp+i ∈ Ck−i(I) , 1 ≤ i ≤ k, such that for the local

discretization error there holds

y(x + h) − y(x) − h Φ(x,y(x),h) =k∑

i=1dp+i(x) h

p+i + O(hp+k+1) , x ∈ I′h .

Then, the global discretization error has the asymptotic expansion

eh(x) =k−1∑

i=0ep+i(x) h

p+i + Ep+k(x;h) hp+k , x ∈ Ih .

Here, the coefficients ep+i , 0 ≤ i ≤ k− 1, are solutions of the linear initial value problems

e′p+i(x) = fy(x,y(x)) ep+i(x) − dp+i+1(x) , x ∈ I ,

ep+i(a) = 0 .

Moreover, there exist hmax > 0 and a constant C > 0, independent of h, such that

‖Ep+k(x;h)‖ ≤ C , x ∈ Ih , 0 < h < hmax .

2


Proof: For the solution of the one-step method with the increment function Φ∗ (consistency

order p + 1), Theorem 2.21 implies

‖y∗h(x) − y(x)‖ ≤ C hp+1 ,

and hence,

‖yh(x) − ep(x) hp − y(x)‖ ≤ C hp+1 .

Consequently, we have

yh(x) − y(x) = ep(x) hp + Ep+1(x;h) h

p+1 , ‖Ep+1(x;h)‖ ≤ C .

For k > 1, we prove the assertion recursively, i.e., for Φ∗ we construct an increment function

Φ∗∗ etc.


Theorem 2.26 Theorem of Stetter (1970)

Assume that the one-step method

yh(x + h) = yh(x) + h Φ(x,x + h,yh(x),yh(x + h),h) , x ∈ I′h ,

yh(a) = α

is symmetric, i.e.,

Φ(x,x + h,yh(x),yh(x + h),h) = Φ(x + h,x,yh(x + h),yh(x),−h) .

Then, under the assumptions of Theorem 2.25 there holds

e2m+1(x) = 0 , x ∈ I ,

i.e., there exists an asymptotic expansion in powers of h2.

Proof: Homework


2.4.2 Explicit extrapolation methods

Assumption: Asymptotic expansion of the global discretization error in powers of hγ

Example 2.27 p = 1 , γ = 1

For the step sizes h and h/2 there holds

(∗1) yh(x) − y(x) = e1(x) h + E2(x;h) h2,

(∗2) yh/2(x) − y(x) =1

2e1(x) h +

1

4E2(x;h) h

2 .

Multiplication of (∗2) with 2 and subtraction from (∗1) yields

(2 yh/2(x) − yh(x))︸︷︷︸

=:ŷh(x)

− y(x) = O(h2) .

Consider the polynomial p1 ∈ P1(lR) with p1(hi) = yhi(x) , 1 ≤ i ≤ 2, where h1 := h and h2 := h/2.

We obtain

p1(t) =2

h(yh(x) − yh/2(x)) (t −

h

2) + yh/2(x) =⇒ p1(0) = 2 yh/2(x) − yh(x) = ŷh(x) .

Extrapolation to the step size h = 0.


Example 2.28 p = 2 , γ = 2

For the step sizes h and h/2 we have

(⋆1) yh(x) − y(x) = e2(x) h2 + E4(x;h) h

4,

(⋆2) yh/2(x) − y(x) =1

4e2(x) h

2 +1

16E4(x;h) h

4 .

Multiplication of (⋆2) by 4 and subtraction from (⋆1) yields

4 yh/2(x) − 3 yh(x) = O(h4) =⇒

4 yh/2(x) − 3 yh(x)

3︸︷︷︸

=:ŷh(x)

− y(x) = O(h4) .

Consider the polynomial p1 ∈ P1(lR) in t2 with p1(hi) = yhi(x) , 1 ≤ i ≤ 2, where h1 := h and h2 := h/2.

We obtain

p1(t2) =

4

3h2(yh(x) − yh/2(x)) (t

2 −h2

4) + yh/2(x) =⇒ p1(0) =

4

3yh/2(x) −

1

3yh(x) = ŷh(x) .

Extrapolation to the step size h = 0.


Implementation of the extrapolation method:

Given a basic step size H > 0, choose a sequence of step sizes

(⋄) hi := H/ni , ni ∈ lN , i = 1,2, ...

represented by the sequence

F := {n1,n2, ...}

and compute the approximations

y(H;hi) := yhi , i = 1,2, ...

The associated sequence Ti,1 = y(H;hi) , i = 1,2, ... forms the first column of an extrapolationtable. Essentially, two types of extrapolation techniques are used:

(i) Polynomial extrapolation (Aitken-Neville)

Determine a polynomial in hγ

T̃ik(h) := a0 + a1 hγ + ... + ak−1 h

(k−1)γ ,

such thatT̃ik(hj) := y(H;hj) , j = i, i− 1, ..., i− k + 1

and extrapolate to h = 0: Tik := T̃ik(0).


Recursion:

Tik = Ti,k−1 +Ti,k−1 − Ti−1,k−1

( nini−k+1)γ − 1

, i ≥ k , k ≥ 2 .

y(H;h1) : T11y(H;h2) : T21 T22y(H;h3) : T31 T32 T33· : · · · ·

· : · · · · ·

y(H;hi) : Ti1 Ti2 Ti3 · · Tii


(ii) Rational extrapolation (Stoer-Bulirsch)

Notation: For m ∈ lN, denote by [m/2] := max {n ∈ lN | n ≤m/2} the integer part of m/2.

Determine a rational function in hγ with the nominator polynomial of degree µ := [(k− 2)/2]and the denominator polynomial of degree ν := k− 2− µ:

T̃ik(h) :=a0 + a1 h

γ + ... + aµ hµγ

b0 + b1 hγ + ... + bν hνγ,

such thatT̃ik(hj) := y(H;hj) , j = i, i− 1, ..., i− k + 1

and extrapolate to h = 0: Tik := T̃ik(0).

Recursion:

Tik = Ti,k−1 +Ti,k−1 − Ti−1,k−1

( nini−k+1)γ [1 −

Ti,k−1 − Ti−1,k−1Ti,k−1 − Ti−1,k−2

] − 1, i ≥ k , k ≥ 2 .


Recursion:

Tik = Ti,k−1 +Ti,k−1 − Ti−1,k−1

( nini−k+1)γ [1 −

Ti,k−1 − Ti−1,k−1Ti,k−1 − Ti−1,k−2

] − 1, i ≥ k , k ≥ 2 .

y(H,h1) =: T10 T11y(H,h2) =: T20 T21 T22y(H,h3) =: T30 T31 T32 T33· · · · · ·

· · · · · · ·

y(H,hi) =: Ti0 Ti1 Ti2 Ti3 · · Tii


Appropriate sequences of step sizes:

(i) Harmonic sequence

ni = i , i ∈ lN ,

FH = {1,2,3,4,5, ...} .

(ii) Romberg’s sequence

ni = 2i , i ∈ lN ,

FR = {2,4,8,16,32, ...} .

(iii) Bulirsch’ sequence

ni =

3 · 2(i−2)/2 , i even

2(i+1)/2 , i odd

FB = {2,3,4,6,8,12,16, ...} .


2.4.3 Order and step size control

Essential tool: Estimator for the global discretization error

eik := Tik − y(H) .

Assumptions on the error behavior:

(A1) εik := ‖eik‖.= γik σk H

γk+1 , γik := (k−1∏

j=0ni−j)

−γ ,

(A2) εi,k+1 := ‖ei,k+1‖ ≪ ‖eik‖ =: εik .

Lemma 2.29 Discretization error and sequence of step sizes

Under the assumption (A1), there holds

εi+1,k.= (

ni−k+1ni+1

)γ εik ,

i.e., in each column of the extrapolation table the discretization errors only differ by a

factor that solely depends on the sequence of step sizes.


Under the assumptions (A1), (A2) we have

(i) εik > εi+1,k ⇐⇒ εik

↑

εi+1,k

(ii) εik > εi,k+1 ⇐⇒ εik ← εi,k+1

Graphical representation of the error behavior in the extrapolation table:

ε11↑

ε21 ← ε22↑ ↑

ε31 ← ε32 ← ε33· ← · ← · ← ·

εk1 ← εk2 ← εk3 ← · ← εkk

In each column k the smallest value is attained by εkk.


Consequence: Diagonal error estimator

Determine an estimator for εkk := ‖Tkk − y(H)‖. According to the error table, we need infor-

mation from row k + 1. We have

‖Tk+1,k+1 − y(H)‖ ≤ q ‖Tkk − y(H)‖ , q < 1 .

It follows that(i) ‖Tkk − y(H)‖ ≤ ‖Tkk − Tk+1,k+1‖ + ‖Tk+1,k+1 − y(H)‖

︸︷︷︸

≤ q ‖Tkk − y(H)‖

=⇒

εkk = ‖Tkk − y(H)‖ ≤1

1− q‖Tkk − Tk+1,k+1‖ ,

(ii) ‖Tkk − y(H)‖ ≥ ‖Tkk − Tk+1,k+1‖ − ‖Tk+1,k+1 − y(H)‖ ≥

≥ ‖Tkk − Tk+1,k+1‖ − q ‖Tkk − y(H)‖ =⇒

εkk = ‖Tkk − y(H)‖ ≥1

1 + q‖Tkk − Tk+1,k+1‖ .


Consequently,ε̂kk := ‖Tkk − Tk+1,k+1‖

is a suitable estimator for εkk.

Disadvantage: For estimating the error of the approximation Tkk in row k, we need infor-

mation from row k + 1 (additional computational work). If we have to compute the approxi-

mations in row k + 1, then:

Alternative: Subdiagonal estimator

Setting‖Tk+1,k+1 − y(H)‖ ≤ q ‖Tk+1,k − y(H)‖ , q < 1 ,

we obtain1

1 + q‖Tk+1,k − Tk+1,k+1‖ ≤ εk+1,k ≤

1

1− q‖Tk+1,k − Tk+1,k+1‖ .

Therefore,ε̄k+1,k := ‖Tk+1,k − Tk+1,k+1‖

is an appropriate estimator for εk+1,k.


Lemma 2.30 Comparison of both estimators

Under the assumptions (A1), (A2) there holds

ε̄k+1,k

ε̂kk

.=εk+1,k

εkk

.= (n1

nk+1)γ ≪ 1 .

Consequence: For a prespecified tolerance eps, the requirement ε̂kk ≤ eps is more restrictive

than ε̄k+1,k ≤ eps.

Therefore: Alternative diagonal estimator of Stoer:

ε̂kk (n1

nk+1)γ ≤ eps .


Development of an order and step size control

(i) Step size control

Convergence monitor for the extrapolation table:

Assume that for suitably chosen bounds αik there holds

(⋆) εik ≤ αik .

Observe(⋄) εik

.= γik σk︸︷︷︸

=:Cik

Hγk+1 .

Let Hik be the step size for which equality holds in (⋆), i.e.,

εik.= Cik H

γk+1ik = αik =⇒ Cik = αik H

−(γk+1)ik .

Inserting this into (⋄) yields

Hik.= H γk+1

√√√√√

αik

εik.


Growth of the bounds αikWe assume an ideal growth as in the error table:

(A3) αi+1,k.= (

ni−k+1ni+1

)γ αik .

Lemma 2.31 Column-wise step size control

Under the assumptions (A1), (A3) there holds

Hik.= Hi+1,k , i = k,k + 1, ...

Hence, we only need one estimate per column.

Diagonal error estimator: Estimator for Hkk

Ĥkk := H γk+1√√√√√

αkk

ε̂kk, k ∈ lN .

Subdiagonal error estimator: Estimator forHk+1,k

H̄k+1,k := H γk+1√√√√√

αk+1,k

ε̄k+1,k, k ∈ lN .


(ii) Order control by minimization of the computational work

Step size Hik in the extrapolation table at position (i,k) which yields convergence (set αik = eps):

Hik = H γk+1√√√√√

eps

εik.

Ai : Computational work (Number of evaluations of the increment function) for the computa-

tion of Tii .

Recursion (exemplified for the explicit Euler method)

A1 = 1 ,

Ai+1 = Ai + ni+1 − 1 , i = 1,2, ...

Normalized work unit (work per unit step):

Wik :=H

HikAi


For i = k + 1, use the subdiagonal error estimator: With H̄k+1,k there holds

W̄k+1,k := Ak+1 γk+1√√√√√

ε̄k+1,k

eps.

Determine the optimal column index q according to

W̄q+1,q = min1≤k≤kf

W̄k+1,k ,

where kf denotes the largest available column index.

After having determined q, choose

H̄q+1,q = H γq+1√√√√√

eps

ε̄q+1,q

as an estimate for the step size in the next step.


Methods as a basis for extrapolation techniques:

(i) Explicit Euler method: h-extrapolation

FH = {1,2,3,4, ...} , ni = i ,

A1 = n1 ,

Ai+1 = Ai + ni+1 − 1 , i ∈ lN .

(ii) Symmetric one-step methods

(•) Implicit trapezoidal rule(•) Implicit midpoint rule

Attention: Solution by fixed point iteration perturbs h2-expansion.

(iii) Explicit midpoint rule: h2 -extrapolation

yk+1 = yk−1 + 2 h f(xk,yk) , k ∈ lN ,

y1 = y0 + h f(x0,y0) (initial step) ,

y0 = α .

Problem: Does the non-symmetric initial step perturb the h2-expansion for the symmetric

discretization?


Theorem 2.32 h2-expansion for the explicit midpoint rule

Consider the ’stabilized’ form of the explicit midpoint rule applied to an autonomous

system of differential equations:

(EMP1) y1 = y0 + h f(y0) (’initial step’) ,

(EMP2) yk+1 = yk−1 + 2 h f(yk) , 1 ≤ k ≤ ℓ ,

(EMP3) Sℓ =1

4[yℓ−1 + 2 yℓ + yℓ+1] (’final step’) .

Under the assumption f ∈ C2N+2(D), for fixed x = hℓ there exists the following asymptotic

expansion of the global discretization error:

(†) Sℓ − y(x) =N∑

j=1[ uj(x) + (−1)

ℓ vj(x) ] h2j + EN+1(x;h) h

2N+2 ,

where uj , vj , 1 ≤ j ≤ N, are solutions of an initial value problem for a staggered system of

differential equations of the form

(AWP1) u′j = fy uj + inhomogeneous terms , v

′j = − fy vj + inhomogeneous terms ,

(AWP2) uj(a) =1

2gj(a) , vj(a) = −

1

2gj(a) .


Moreover, there exist hmax > 0 and C > 0, independent of h, such that

(‡) ‖EN+1(x;h)‖ ≤ C , 0 ≤ h ≤ hmax .

For ℓ = 2m, we obtain:

(⋆1) Sℓ − y(x) =N∑

j=1wj(x) h

2j + WN+1(x;h) h2N+2 ,

(⋆2)

w1(x) = u1(x) +14 y′′(x)

wj(a) = WN+1(a;h) = 0 , 1 ≤ j ≤ N .

Proof: (Stetter (1970)): Formulate the two-step method as a symmetric one-step method

of double dimension. Setting

h̄ := 2 h , ξk := y2k , ηk := y2k+1 −h̄

2f(ξk) ,

it follows thatξ0 = y0 , η0 = y1 −

h̄

2f(y0) = y0 ,

ξk+1 − ξk = y2k+2 − y2k = 2 h f(y2k+1) = h̄ f(ηk +h̄

2f(ξk)) ,

ηk+1 − ηk = y2k+3 −h̄

2f(ξk+1) − y2k+1 +

h̄

2f(ξk) =

h̄

2[ f(ξk) + f(ξk+1) ] .


Observing

ξk+1 − ξk = h̄ f(ηk +h̄

2f(ξk)) , βk+1 − βk =

h̄

2[ f(ξk) + f(ξk+1) ] ,

it follows that

ξk+1 − ξkh̄

= f(ηk +h̄

2f(ξk)

︸︷︷︸

=y2k+1

) , ξ0 = y0 ,ηk+1 − ηk

h̄=

1

2[ f(ξk) + f(ξk+1) ] , η0 = y0 .

For h̄→ 0 we obtain

ξ′(x) = f(η(x)) , ξ(a) = y0 , η′(x) = f(ξ(x)) , η(a) = y0 .

The existence and uniqueness theorem of Picard-Lindelöf yields

ξ(x) = η(x) = y(x) , x ∈ I .


The difference equationξk+1 − ξk

h̄= f(ηk +

h̄

2f(ξk)

︸︷︷︸

=y2k+1

)

can be written in symmetric form using

y2k+1 = ηk +h̄

2f(ξk) +

1

2[ηk+1 − ηk −

h̄

2f(ξk) −

h̄

2f(ξk+1)]

︸︷︷︸

=0

=1

2(ηk + ηk+1) +

h̄

4(f(ξk) − f(ξk+1))

Stetter’s theorem (cf. Theorem 2.26) yields the following h2-expansion:

(ASE1) ξh̄(x) − y(x) =N∑

j=1pj(x) h̄

2j + PN+1(x; h̄) h̄2N+2 ,

(ASE2) ηh̄(x) − y(x) =N∑

j=1qj(x) h̄

2j + QN+1(x; h̄) h̄2N+2 ,

wherep′j(x) = fy(y(x)) qj(x) + ... , pj(a) = 0 ,

q′j(x) = fy(y(x)) pj(x) + ... , qj(a) = 0 .


Observing

(ASE1) ξh̄(x) − y(x) =N∑

j=1pj(x) h̄

2j + PN+1(x; h̄) h̄2N+2 ,

(ASE2) ηh̄(x) − y(x) =N∑

j=1qj(x) h̄

2j + QN+1(x; h̄) h̄2N+2 ,

the back transformation h = h̄/2 results in

(⋄1) y2k = ξk =⇒ y2k − y(x2k) = y2k − ξk︸︷︷︸

=0

+ ξk − y(x2k),

(⋄2) y2k+1 =1

2(ηk + ηk+1) +

h

2(f(ξk) − f(ξk+1)) .

Inserting (ASE1), (ASE2) into (⋄1), (⋄2) yields

y2k − y(x2k) =N∑

j=1ej(x) h

2j + EN+1(x;h) h2N+2 ,

y2k+1 − y(x2k+1) =N∑

j=1gj(x) h

2j + GN+1(x;h) h2N+2 ,

where e′j(x) = fy(y(x)) gj(x) + ... , ej(a) = 0 , g′j(x) = fy(y(x)) ej(x) + ... .


Using the decoupling

uj :=1

2(ej + gj) , vj :=

1

2(ej − gj) ,

the assertion follows from

e′j(x) = fy(y(x)) gj(x) + ... , ej(a) = 0 ,

g′j(x) = fy(y(x)) ej(x) + ... .

Realization: Choose ℓ = ni = 2mi and

F 2H = {2,4,6,8,10, ...} , ni = 2i ,

A1 = n1 + 1 ,

Ai+1 = Ai + ni+1 , i ∈ lN .

Codes: DIFSYS (Bulirsch/Stoer) , DIFEX1 (Deuflhard) , ODEX (Hairer/Wanner)


2.5 Step size control for explicit Runge-Kutta methods

Assume that yk+1 is an approximation of y(xk+1) computed by means of an explicit Runge-

Kutta method of order pyk+1 = yk + h Φ(xk,yk,h) .

Problem: Estimator for the global discretization error

ek+1 := yk+1 − y(xk+1) .

Lemma 2.33 Error estimator

Assume that ŷk+1 is a more accurate approximation of y(xk+1) according to

‖ŷk+1 − y(xk+1)‖ ≤ q ‖yk+1 − y(xk+1)‖ , q < 1 .

Then,ε̂k+1 := ‖ŷk+1 − yk+1‖

is an estimator for ek+1 in the sense that

1

1 + qε̂k+1 ≤ ‖ek+1‖ ≤

1

1− qε̂k+1 .


Assumption: ŷk+1 is an approximation of order p + 1:

‖ŷk+1 − y(xk+1)‖ ≤ C hp+1 .

Given a prespecified relative accuracy ε, for a reasonable new step size ĥ := xk+2 − xk+1 we

should have

(•) C ĥp+1 .= ρ ε ,

where ρ < 1 represents an appropriate ’safety factor’.

If yk+1 satisfies the requirement

ε̂k+1 := ‖ŷk+1 − yk+1‖.= ε .= C hp+1 ,

we have C .= ε̂k+1/hp+1. Inserting into (•) and solving for ĥ gives

ĥ = h (ρ ε

ε̂k+1)1/(p+1) .

Two strategies for the computation of ŷk+1:

(i) Extrapolation

(ii) Embedded Runge-Kutta methods of higher order .


(i) Extrapolation

Let yk+1 and ȳk+1 be the approximations associated with the step sizes h resp. h/2, i.e.,

(◦1) yk+1 − y(xk+1).= C hp ,

(◦2) ȳk+1 − y(xk+1).= C (

h

2)p = 2−p C hp .

Subtraction of (◦2) from (◦1) results in

yk+1 − ȳk+1.= C hp (1 − 2−p) = C hp

2p − 1

2p,

and hence,

C .=2p − 1

2ph−p (yk+1 − ȳk+1) .

Inserting C into (◦2) and solving for y(xk+1) gives

ŷk+1 = ȳk+1 +ȳk+1 − yk+1

2p − 1.


(ii) Embedded Runge-Kutta methods of higher order

(ii)1 Runge-Kutta-Fehlberg method

Assume that yk+1 is an approximation of y(xk+1) obtained by an s-stage Runge-Kutta method

of order p:yk+1 = yk +

s∑

i=1bi ki ,

ki = h f(xk + ci h , yk +i−1∑

j=1aij kj) ,

ci =i−1∑

j=1aij .

Compute ŷk+1 as the solution of an (s + t)-stage ’embedded’ Runge-Kutta method of order p + 1:

ŷk+1 = yk +s+t∑

i=1b̂i ki ,

ki = h f(xk + ci h , yk +i−1∑

j=1aij kj) ,

ci =i−1∑

j=1aij .


Butcher scheme of the embedded (s + t)-stage Runge-Kutta method

c2 | a21· | · ·

· | · · ·

cs | as1 · · as,s−1· | · · · · ·

· | · · · · · ·

cs+t | as+t,1 · · as+t,s−1 · · as+t,s+t−1yk+1 | b1 · · · bsŷk+1 | b̂1 · · · b̂s · · b̂s+t


Disadvantage: Computation is continued with yk instead of ŷk.

(ii)2 Embedded Runge-Kutta method by Dormand-Prince

Idea: Use an embedded method, but continue with ŷk:

yk+1 = ŷk +s∑

i=1bi ki ,

ki = h f(xk + ci h , ŷk +i−1∑

j=1aij kj)

and take advantage of ’Fehlberg’s trick’ for the reduction of the remaining free parameters.

Fehlberg’s trick for t = 1: At the following step xk+1 7−→ xk+2 use

ks+1 = h f(xk + cs+1 h , ŷk +s∑

j=1as+1,j kj)

ask1 = h f(xk+1 , ŷk+1) .

Observing ŷk+1 = ŷk +s+1∑

j=1b̂jkj, this leads to

cs+1 = 1 , b̂s+1 = 0 , b̂i = as+1,i , 1 ≤ i ≤ s .

University of Houston, Department of Mathematics Numerical Analysis II Chapter 2...

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