numerical methods for differential equations...deﬁnition x is called a fixed point of the function...

Numerical Methods for Differential Equations

Chapter 2: Runge–Kutta and Linear Multistep methods

Gustaf Soderlind and Carmen Arevalo

Numerical Analysis, Lund University

Textbooks: A First Course in the Numerical Analysis of Differential Equations, by Arieh Iserles

and Introduction to Mathematical Modelling with Differential Equations, by Lennart Edsberg

c© Gustaf Soderlind, Numerical Analysis, Mathematical Sciences, Lund University, 2008-09

Numerical Methods for Differential Equations – p. 1/63

Chapter 2: contents

◮ Solving nonlinear equations

◮ Fixed points

◮ Newton’s method

◮ Quadrature

◮ Runge–Kutta methods

◮ Embedded RK methods and adaptivity

◮ Implicit Runge–Kutta methods

◮ Stability and the stability function

◮ Linear multistep methods


1. Solving nonlinear equations f(x) = 0

We can have a single equation

x− cos(x) = 0

or a system

4x2 − y2 = 0

4xy2 − x = 1

Nonlinear equations may have

no solution; one solution; any finite number of solutions;infinitely many solutions


Iteration and convergence

Nonlinear equations are solved iteratively. One computesa sequence {x[k]} of approximations to the root x∗

For f(x∗) = 0 , let e[k] = x[k] − x∗

Definition The method converges if limk→∞ ‖e[k]‖ = 0

Definition The convergence is

◮ linear if ‖e[k+1]‖ ≤ c · ‖e[k]‖ with 0 < c < 1

◮ quadratic if ‖e[k+1]‖ ≤ c · ‖e[k]‖p with p = 2

◮ superlinear if p > 1;

◮ cubic if p = 3, etc.


2. Fixed points

Definition x is called a fixed point of the function g if

x = g(x)

Definition A function g is called contractive if

‖g(x) − g(y)‖ ≤ L[g] · ‖x − y‖

with L[g] < 1 for all x, y in the domain of g


Fixed Point Theorem

Theorem Assume that g is continuously differentiable onthe compact interval I, i.e., g ∈ C1(I)

◮ If g : I → I there exists an x∗ ∈ I such that x∗ = g(x∗)

◮ If in addition L[g] < 1 on I, then x∗ is unique, and . . .

◮ . . . the iteration

xn+1 = g(xn)

converges to the fixed point x∗ for all x0 ∈ I

Note Both conditions are absolutely essential!


Fixed Point Theorem. Existence and uniqueness

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Left No condition satisfied: maybe no x∗

Center First condition satisfied: maybe multiple x∗

Right Both conditions satisfied: unique x∗


Fixed point iteration. x = e−x; g(x) = e−x

x[k+1] = e−x[k]; x[0] = 0

g : [0, 1] → [0, 1]; |g′(x)| = e−x < 1 for x > 0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


Error estimation in fixed point iteration

Assume ‖g′(x)‖ ≤ L < 1 for all x ∈ I.

x[k+1] = g(x[k])

x∗ = g(x∗)

By the mean value theorem,

x[k+1] − x∗ = g(x[k]) − g(x∗)

= g(x[k]) − g(x[k+1]) + g(x[k+1]) − g(x∗)

‖x[k+1] − x∗‖ ≤ L · ‖x[k] − x[k+1]‖ + L · ‖x[k+1] − x∗‖


Error estimation . . .

(1 − L) · ‖x[k+1] − x∗‖ ≤ L · ‖x[k] − x[k+1]‖

Error estimate (computable error bound)

‖x[k+1] − x∗‖ ≤ L

1 − L‖x[k] − x[k+1]‖

Theorem If L[g] < 1, then the error in fixed point iterationis bounded by

‖x[k+1] − x∗‖ ≤ L[g]

1 − L[g]‖x[k] − x[k+1]‖


Example: The trapezoidal rule

Approximate the solution to y′ = −y2 cos t, y(0) = 1/2, t ∈ [0, 8π].

Taking 96 steps, solve nonlinear equations with one (left) and

four (right) fixed point iterations. Graphs show absolute error

0 10 20 300

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 10 20 300

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01


3. Newton’s method

Newton’s method solves f(x) = 0 using repeatedlinearizations. Linearize at the point (x[k], f(x[k]))!

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2


Newton’s method . . .

Straight line equation

y − f(x[k]) = f ′(x[k]) · (x− x[k])

Define x = x[k+1] ⇒ y = 0, so that

−f(x[k]) = f ′(x[k]) · (x[k+1] − x[k])

Solve for x = x[k+1], to get Newton’s method

x[k+1] = x[k] − f(x[k])

f ′(x[k])


Newton’s method. Alternative derivation

Expand f(x[k+1]) in a Taylor series around x[k]

f(x[k+1]) = f(x[k] + (x[k+1] − x[k]))

≈ f(x[k]) + f ′(x[k]) · (x[k+1] − x[k]) := 0

Newton’s method

x[k+1] = x[k] − f(x[k])

f ′(x[k])


Newton’s method. Alternative derivation . . .

This holds also when f is vector valued (equation systems)

f(x[k]) + f ′(x[k]) · (x[k+1] − x[k]) := 0

⇒

x[k+1] = x[k] −(

f ′(x[k]))

−1f(x[k])

Definition f ′(x[k]) is the Jacobian matrix of f , defined by

f ′(x) =

{

∂fi

∂xj

}


Newton’s method. Convergence

Write Newton’s method as a fixed point iteration withiteration function g(x) := x− f(x)/f ′(x)

x[k+1] = g(x[k])

Note Newton’s method converges fast if f ′(x∗) 6= 0, asg′(x∗) = f(x∗)f ′′(x∗)/f ′(x∗)2 = 0

Expand g(x) in a Taylor series around x∗

g(x[k]) − g(x∗) ≈ g′(x∗)(x[k] − x∗) +g′′(x∗)

2(x[k] − x∗)2

x[k+1] − x∗ ≈ g′′(x∗)

2(x[k] − x∗)2


Newton’s method. Convergence . . .

Define the error by ε[k] = x[k] − x∗, then

ε[k+1] ∼(

ε[k])2

Newton’s method is quadratically convergent!

Fixed point iterations are typically only linearly convergent

ε[k+1] ∼ ε[k]


Implicit Euler. Newton vs Fixed point

As yn+1 = yn + hf(yn+1) we need to solve an equation

y = hf(y) + ψ

Note All implict methods lead to an equation of this form!

Theorem Fixed point iterations converge if L[hf ] < 1,restricting the step size to h < 1/L[f ] !

Stiff equations have L[hf ] ≫ 1 so fixed point iterations willnot converge; it is necessary to use Newton’s method!


Convergence order and rate

Definition The convergence order is p with (asymptotic)error constant Cp, if

0 < limk→∞

‖ε[k+1]‖‖ε[k]‖p

= Cp <∞

Special cases:

p = 1 Linear convergence

Example Fixed point iteration Cp = |g′(x∗)|p = 2 Quadratic convergence

Example Newton iteration Cp =∣

∣

f ′′(x∗)2f ′(x∗)

∣

∣


4. Quadrature (integration) formulas

Numerical “quadrature” is the approximation of definite integrals

I(f) =

∫ b

af(x)dx

We substitute the “infinite sum” by a finite sum: the integrand is

sampled at a finite number of points

I(f) =

n∑

i=1

wif(xi) + Rn

Rn = I(f) − ∑ni=1 wif(xi) is the quadrature error. Numerical

method

I(f) ≈n

∑

i=1

wif(xi)


5. Runge–Kutta methods

To solve the IVP y′ = f(t, y), t ≥ t0, y(t0) = y0 we use a

quadrature formula to approximate the integral in

y(tn+1) = y(tn) +

∫ tn+1

tn

f(τ, y(τ))dτ

y(tn+1) ≈ y(tn) + h

s∑

j=1

bjf(tn + cjh, y(tn + cjh))

Let {Yj} denote the numerical approximations to {y(tn + cjh)}A Runge–Kutta method then has the form

yn+1 = yn + hs

∑

j=1

bjf(tn + cjh, Yj)


Stage values and stage derivatives

The vectors Yj are called stage values

The vectors Y ′

j = f(tn + cjh, Yj) are called stage derivatives

Stage values and stage derivatives are related through

Yi = yn + h

i−1∑

j=1

ai,jY′

j

and the method advances one step through

yn+1 = yn + hs

∑

j=1

bjY′

j


The RK matrix, weights, nodes and stages

A = {aij} is the RK matrix, b = [b1 b2 · · · bs]T is the weight vector,

c = [c1 c2 · · · cs]T are the nodes, and the method has s stages

The Butcher tableau of an explicit RK method is

0 0 0 · · · 0

c2 a21 0 · · · 0...

.... . .

...

cs as1 as2 · · · 0

b1 b2 · · · bs

orc A

bT

Simplifying assumption: ci =i

∑

j=1

ai,j i = 2, 3, . . . , s


Simplest case: a 2-stage ERK

Y ′

1 = f(tn, yn)

Y ′

2 = f(tn + c2h, yn + h a21Y′

1)

yn+1 = yn + h[b1Y′

1 + b2Y′

2 ]

0 0 0

c2 a21 0

b1 b2

c2 = a21


2-stage ERK . . .

Using Y ′

1 = f(tn, yn), expand in Taylor series around tn, yn:

Y ′

2 = f(tn + c2h, yn + h a21f(tn, yn))

= f + h [c2ft + a21fyf ] + O(h2)


2-stage ERK . . .

Using Y ′


Y ′

2 = f(tn + c2h, yn + h a21f(tn, yn))

= f + h [c2ft + a21fyf ] + O(h2)

Inserting into yn+1 = yn + h[b1Y′

1 + b2Y′

2 ], we get

yn+1 = yn + h(b1 + b2)f + h2b2(c2ft + a21fyf) + O(h3)


2-stage ERK . . .

Using Y ′


Y ′

2 = f(tn + c2h, yn + h a21f(tn, yn))

= f + h [c2ft + a21fyf ] + O(h2)

Inserting into yn+1 = yn + h[b1Y′

1 + b2Y′

2 ], we get


Taylor expansion of the exact solution:

y′ = f

y′′ = ft + fyy′ = ft + fyf

y(t + h) = y + hf +1

2h2(ft + fyf) + O(h3)


2-stage ERK . . .

Matching terms in


y(t + h) = y + hf +1

2h2(ft + fyf) + O(h3)

and taking c2 = a21 we get the conditions for order 2:

b1 + b2 = 1 (consistency)

b2c2 = 1/2

Note All consistent RK methods are convergent!

Second order, 2-stage ERK methods have a Butcher tableau

0 0 0

12b

12b 0

1 − b b


Example 1. The modified Euler method

Put b = 1 to get

0 0 0

12

12 0

0 1

Y ′

1 = f(tn, yn)

Y ′

2 = f(tn + h/2, yn + hY ′

1/2)

yn+1 = yn + hY ′

2

Second order explicit Runge–Kutta (ERK) method


Example 2. Heun’s method

Put b = 1/2 to get

0 0 0

1 1 0

1/2 1/2

Y ′

1 = f(tn, yn)

Y ′

2 = f(tn + h, yn + hY ′

1)

yn+1 = yn + h (Y ′

1 + Y ′

2)/2

Second order ERK, compare to the trapezoidal rule!


Third order 3-stage ERK

Conditions for 3rd order: a21 = c2; a31 + a32 = c3

b1 + b2 + b3 = 1

b2c2 + b3c3 = 1/2

b2c22 + b3c

23 = 1/3

b3a32c2 = 1/6

Classical RK Nyström scheme

0 0 0 0

1/2 1/2 0 0

1 −1 2 0

1/6 2/3 1/6

0 0 0 0

2/3 2/3 0 0

2/3 0 2/3 0

1/4 3/8 3/8


Exercise

Construct the Butcher tableau for the 3-stage Heun method.

Y ′

1 = f(tn, yn)

Y ′

2 = f(tn + h/3, yn + hY ′

1/3)

Y ′

3 = f(tn + 2h/3, yn + 2hY ′

2/3)

yn+1 = yn + h (Y ′

1 + 3Y ′

3)/4

Is it of order 3?


Classic RK4: 4th order, 4-stage ERK

The “original” RK method (1895):

Y ′

1 = f(tn, yn)

Y ′

2 = f(tn + h/2, yn + hY ′

1/2)

Y ′

3 = f(tn + h/2, yn + hY ′

2/2)

Y ′

4 = f(tn + h, yn + hY ′

3)

yn+1 = yn +h

6(Y ′

1 + 2Y ′

2 + 2Y ′

3 + Y ′

4).


Classic 4th order RK4 . . .

Butcher tableau

0

1/2 1/2

1/2 0 1/2

1 0 0 1

1/6 1/3 1/3 1/6

s-stage ERK methods of order p = s exist only for s ≤ 4 (i.e.

there is no 5-stage ERK of order 5)


Order conditionsAn s-stage ERK method has s + s(s − 1)/2 coefficients to

choose, but the order conditions are many




Number of coefficients

stages s 1 2 3 4 5 6 7 8 9 10

# coefficients 1 3 6 10 15 21 28 36 45 55





stages s 1 2 3 4 5 6 7 8 9 10

# coefficients 1 3 6 10 15 21 28 36 45 55

Number of order conditions

order p 1 2 3 4 5 6 7 8 9 10

# conditions 1 2 4 8 17 37 85 200 486 1205





stages s 1 2 3 4 5 6 7 8 9 10

# coefficients 1 3 6 10 15 21 28 36 45 55

Number of order conditions

order p 1 2 3 4 5 6 7 8 9 10

# conditions 1 2 4 8 17 37 85 200 486 1205

Maximum order, min stages

order p 1 2 3 4 5 6 7 8

min stages 1 2 3 4 6 7 9 10Numerical Methods for Differential Equations – p. 33/63

6. Embedded RK methods

Two methods in one Butcher tableau (RK34)

Y ′

1 = f(tn, yn)

Y ′

2 = f(tn + h/2, yn + hY ′

1/2)

Y ′

3 = f(tn + h/2, yn + hY ′

2/2)

Z ′

3 = f(tn + h, yn − hY ′

1 + 2hY ′

2)

Y ′

4 = f(tn + h, yn + hY ′

3)

yn+1 = yn +h

6(Y ′

1 + 2Y ′

2 + 2Y ′

3 + Y ′

4) order 4

zn+1 = yn +h

6(Y ′

1 + 4Y ′

2 + Z ′

3) order 3

The difference yn+1 − zn+1 can be used as an error estimate


Adaptive RK methods

Example Use an embedded pair, e.g. RK34

Local error estimate rn+1 := ‖yn+1 − zn+1‖ = O(h4)

Adjust the step size h so that the local error estimateequals a prescribed tolerance TOL

Simplest step size change scheme

hn+1 =

(

TOL

rn+1

)1/p

hn

makes rn ≈ TOL

Adaptivity using local error control


7. Implicit Runge–Kutta methods (IRK)

Y1 = yn + h

s∑

j=1

a1,jf(tn + cjh, Yj)

Y2 = yn + hs

∑

j=1

a2,jf(tn + cjh, Yj)

...

Ys = yn + h

s∑

j=1

as,jf(tn + cjh, Yj)

yn+1 = yn + h

s∑

j=1

bjf(tn + cjh, Yj)


Implicit Runge–Kutta methods . . .

In stage value – stage derivative form

Yi = yn + hs

∑

j=1

ai,jY′

j

Y ′

j = f(tn + cjh, Yj)

yn+1 = yn + hs

∑

j=1

bjY′

j


1-stage IRK method

Implicit Euler (order 1) Implicit midpoint method (order 2)

1 1

1

1/2 1/2

1

Y ′

1 = f(tn + c1h, yn + h a11Y′

1)

yn+1 = yn + h b1Y′

1

may be written as

yn+1 = yn + h b1f(tn + c1h, (b1 − a11

b1yn +

a11

b1yn+1))

Taylor expansion:

yn+1 = y + h b1f(tn + c1h, y +a11

b1hf +

a11

b1

h2

2(ft + fyf) + O(h2))

= y + h b1hf + h2(b1c1ft + a11fyf) + O(h3)


Taylor expansions for 1-stage IRK

y(tn+1) = y + h f +h2

2(ft + fyf) + O(h3)

yn+1 = y + h b1f + h2(b1c1ft + a11fyf) + O(h3)

Condition for order 1 (consistency) b1 = 1

Conditions for order 2 c1 = a11 = 1/2

Conclusion Implicit Euler is of order 1 and implicitmidpoint method is the only 1-stage IRK of order 2


8. The stability function

Applying an IRK to the test equation, we get

hY ′

i = hλ · (yn +s

∑

j=1

ai,jhY′

j ), i = 1, . . . , s

Let hY′ = [hY ′

1 · · · hY ′

s ]T and 1 = [1 1 · · · 1]T ∈ R

s, then(I − hλA)hY′ = hλ1yn so hY′ = hλ(I − hλA)−1

1yn

yn+1 = yn +

s∑

j=1

bjhY′

j = [1 + hλbT(I − hλA)−11]yn


The stability function

Theorem For every Runge-Kutta method applied to thelinear test equation y′ = λy we have

yn+1 = R(hλ)yn

where the rational function

R(z) = 1 + zbT(I − zA)−11.

If the method is explicit, then R(z) is a polynomial ofdegree s

The function R(z) is called the method’s stability function


A-stability of RK methods

Definition The method’s stability region is the set

D = {z ∈ C : |R(z)| ≤ 1}

Theorem If R(z) maps all of C− into the unit circle, then

the method is A-stable

Corollary No explicit RK method is A-stable

(For ERK R(z) is a polynomial, and P (z) → ∞ as z → ∞)


A-stable methods and the Maximum Principle

Theorem |R(z)| ≤ 1 for all z ∈ C− if and only if all the poles of

R have positive real parts and |R(iω)| ≤ 1 for all ω ∈ R

This is the Maximum Principle in complex analysis

Example

0 1/4 −1/4 Y ′

1 = f(yn + hY ′

1/4 − hY ′

2/4)

2/3 1/4 5/12 ⇒ Y ′

2 = f(yn + hY ′

1/4 + 5hY ′

2/12)

1/4 3/4 yn+1 = yn + h(Y ′

1 + 3Y ′

2)/4


Example . . .

Applied to the test equation, we get

yn+1 =1 + 1

3hλ

1 − 23hλ+ 1

6(hλ)2

yn

with poles 2 ± i√

2 ∈ C+, and

|R(iω)|2 =1 + 1

9ω2

1 + 19ω2 + 1

36ω4

≤ 1

Conclusion |R(z)| ≤ 1 ∀ z ∈ C−. The method is A-stable


9. Linear Multistep Methods

A multistep method is a method of the type

yn+1 = Φ(f, h, y0, y1, . . . , yn)

using values from several previous steps

◮ Explicit Euler yn+1 = yn + h f(tn, yn)

◮ Trapezoidal rule yn+1 = yn + h ( f(tn,yn)+f(tn+1,yn+1)2

)

◮ Implicit Euler yn+1 = yn + h f(tn+1, yn+1)

are all one-step methods, but also LM methods


Multistep methods and difference equations

A k-step multistep method replaces the ODE y′ = f(t, y) by a

(finite) difference equation

k∑

j=0

ak−jyn−j = h

k∑

j=0

bk−jf(tn−j , yn−j)

Generating (characteristic) polynomials

ρ(w) =

k∑

m=0

am wm σ(w) =

k∑

m=0

bm wm

Normalization ak = 1 or σ(1) = 1 (choose your preference)

If bk = 0, the method is explicit ; if bk 6= 0, it is implicit

EE: ρ(w) = w − 1, σ(w) = 1; IE: ρ(w) = w − 1, σ(w) = w


Lagrange interpolation

Given a grid {t0, t1, . . . , tk} construct a degree k polynomial

basis

Φi(t) i = 0, 1, . . . , k

such that Φi(tj) = δij (the Kronecker delta)

If the values zj = z(tj) are known for the function z(t), then

P (t) =

k∑

j=0

Φj(t)zj

interpolates z(t) on the grid:

P (tj) = zj ; P (t) ≈ z(t) for all t


Adams methods (J.C. Adams, 1880s)

Suppose we have the first n+ k approximations

ym = y(tm), m = 0, 1, . . . , n+ k − 1

Rewrite y′ = f(t, y) by integration

y(tn+k) − y(tn+k−1) =

∫ tn+k

tn+k−1

f(τ, y(τ)) dτ

Adams methods approximate the integrand by aninterpolation polynomial on tn, tn−1, . . .

f(τ, y(τ)) ≈ P (τ)


Adams–Bashforth methods (explicit)

Interpolate f values with polynomial of degree k − 1:

P (tn+j) = f(tn+j , y(tn+j)) j = 0, . . . , k − 1

Then P (τ) = f(τ, y(τ)) + O(hk) for t ∈ [tn, tn+k]

y(tn+k) = y(tn+k−1) +

∫ tn+k

tn+k−1

P (τ)dτ + O(hk+1)

The k-step Adams-Bashforth method is the order k method

yn+k = yn+k−1 + h

k−1∑

j=0

bj f(tn+j , yn+j)

where bj = h−1∫ tn+k

tn+k−1Φj(τ)dτ


Coefficients of AB1

AB1: for k = 1

yn+1 = yn + h b0 f(tn, yn)

where

b0 = h−1

∫ tn+1

tn

Φ0(τ)dτ = h−1

∫ tn+1

tn

1dτ = 1 ⇒

yn+1 = yn + h f(tn, yn)

Conclusion AB1 is the explicit Euler method


Coefficients of AB2

Here

Φ1(t) =t − tn

tn+1 − tn; Φ0(t) =

t − tn+1

tn − tn+1

AB2: for k = 2

yn+2 = yn+1 + h [b1 f(tn+1, yn+1) + b0 f(tn, yn)]

b0 = h−1

∫ tn+2

tn+1

Φ0(τ)dτ = −1/2

b1 = h−1

∫ tn+2

tn+1

Φ1(τ)dτ = 3/2

yn+2 = yn+1 + h

[

3

2f(tn+1, yn+1) −

1

2f(tn, yn)

]


Initializing an Adams method

The first step of AB2 is

y2 = y1 + h

[

3

2f(t1, y1) −

1

2f(t0, y0)

]

so we need the values of y0 and y1 to start.

y0 is obtained from the initial value, but y1 has to be computed

with a one-step method.

Implementing AB2: we may use AB1 for the first step,

y1 = y0 + h f(t0, y0)

yn+2 = yn+1 + h

[

3

2f(tn+1, yn+1) −

1

2f(tn, yn)

]

, n ≥ 0


Example: AB1 vs AB2Approximate the solution to y′ = −y2 cos t, y(0) = 1/2, t ∈ [0, 8π]

with h = π/6 and h = π/60

AB1: Solutions Errors

0 10 20 300

0.2

0.4

0.6

0.8

1

0 10 20 3010

−4

10−3

10−2

10−1

100

0 10 20 300.2

0.4

0.6

0.8

1

1.2

0 10 20 3010

−6

10−4

10−2

100

AB2: Solutions ErrorsNumerical Methods for Differential Equations – p. 53/63

How do we check the order of a multistep method?

A multistep method is of order of consistency p if the local error is

k∑

j=0

ajy(tn+j) − h

k∑

j=0

bjy′(tn+j) = O(hp+1)

Try if the formula holds exactly for polynomials

y(t) = 1, t, t2, t3, . . .

Insert y = tm and y′ = mtm−1 into the formula, take tn+j = jh:

k∑

j=0

aj(jh)m − hk

∑

j=0

bjm(jh)m−1 = hmk

∑

j=0

(ajjm − bjmjm−1)


Order conditions for multistep methods

Theorem A k-step method is of consistency order p if and only

if it satisfies the following conditions:

◮

k∑

j=0

jmaj = m

k∑

j=0

jm−1bj , m = 0, 1, . . . , p

◮

k∑

j=0

jp+1aj 6= (p + 1)k

∑

j=0

jpbj

Then the multistep formula holds exactly for all polynomials of

degree p or less. (Problems with y = P (t) are solved exactly)


The root condition

Definition A polynomial ρ satisfies the root condition if all of its

zeros have moduli less than or equal to one and the zeros of unit

modulus are simple

Examples

◮ ρ(w) = (w − 1)(w − 0.5)(w + 0.9)

◮ ρ(w) = (w − 1)(w + 1)

◮ ρ(w) = (w − 1)2(w − 0.5)

◮ ρ(w) = (w − 1)(w −√

2)(w +√

2)

◮ ρ(w) = (w − 1)(w2 + 0.25)

All Adams methods have ρ(w) = wk−1(w − 1)


The Dahlquist equivalence theorem

Definiton A method is zero-stable if ρ satisfies the rootcondition

Theorem A multistep method is convergent if and only if itis zero-stable and consistent of order p ≥ 1 (without proof)

Example k-step Adams-Bashforth methods are of order kand have ρ(w) = wk−1(w − 1) ⇒ they are convergent


Dahlquist’s first barrier

Theorem The maximal order of a zero-stable k-stepmethod is

p =

k for explicit methods

k + 1 if k is odd

k + 2 if k is evenfor implicit methods


Backward differentiation formula of order 2

Construct a 2-step method of order 2 of the form

α2yn+2 + α1yn+1 + α0yn = h f(tn+2, yn+2)

Order conditions for p = 2:

α2 + α1 + α0 = 0; 2α2 + α1 = 1; 4α2 + α1 = 4

⇒ α2 = 3/2; α1 = −2; α0 = 1/2;

ρ(w) =3

2(w − 1)(w − 1

3) ⇒ BDF2 is convergent


Backward differentiation formulas

The backward difference operator is defined as∇0yn+k = yn+k and∇jyn+k = ∇j−1yn+k −∇j−1yn+k−1, j ≥ 1

Theorem (without proof): The k-step BDF method

k∑

j=1

∇j

jyn+k = h f(tn+k, yn+k)

is convergent of order p = k if and only if 1 ≤ k ≤ 6

Note BDF methods are suitable for stiff problems


BDF1-6 stability regions

The methods are stable outside the indicated area

−10 −5 0 5 10 15 20 25 30 35−20

−15

−10

−5

0

5

10

15

20Stability regions of BDF1−6 methods


A-stability of multistep methods

Applying a multistep method to the linear test equationy′ = λy produces a difference equation

k∑

j=0

ajyn+j = hλk

∑

j=0

bjyn+j

The characteristic equation (with z := hλ)

ρ(w) − zσ(w) = 0

has k roots wj(z). The method is A-stable iff

Re z ≤ 0 ⇒ |wj(z)| ≤ 1,

with simple unit modulus roots (root condition)


Dahlquist’s second barrier

Theorem (without proof): The highest order of anA-stable multistep method is p = 2. Of all 2nd orderA-stable multistep methods, the trapezoidal rule has thesmallest error

Note There is no such order restriction for Runge–Kuttamethods, which can be A-stable for arbitrarily high orders

A multistep method can be useful although it isn’t A-stable


numerical methods for differential equations...deﬁnition x is called a fixed point of the function...

Documents