Adaptive Runge-Kutta

• addresses the problem of functions that change rapidly at a point

0

1

2

3

4

5

6

0 5 10 15 20 25 30

Would like to use small size steps in the area of rapid change - normal size steps in area of normal change

Two approaches behind adaptive step size

• look at difference between predictions with different step sizes but same order RK

• look at difference between predictions with different order RK

Step-halving or adpartive Runge-Kutta

let y1 be single-step prediction

let y2 be prediction using two half steps

12 yy

The correction is

1522

yy

fifth order accurate

Example:

00

05.0exp8.32exp97,

y

xxyyxfdx

dy

-1

0

1

2

3

4

5

6

7

0 2 4 6 8 10 12

dy/dx

Integrate y’ from x=0 to 2 using h=2, and improve using adaptive RK

Complete step results areh 2x RK k1 k2 k3 k40 0 5.8 0.60176 0.6018 -0.11112 2.69863 -0.1111 -0.2175 -0.2175 -0.2423

Half step results areh 1x RK k1 k2 k3 k40 0 5.8 2.00221 2.0022 0.60181 2.40177 0.60176 0.08315 0.0831 -0.11112 2.53897 -0.1111 -0.1862 -0.1862 -0.2175

The correction is

01064.015

69863.253897.215

22 step onesteps two

yyEa

The corrected value is 528324.201064.0538969.2

Compare to true value y(2)=2.524369

%2.0

%6.0

%9.6

1

2

acorr

a

a

E

E

E

Runge-Kutta-Fehlberg

Uses two different RK predictions of different order

Special choice of methods lets you use results from 4th order in 5th order RK - then combine them

Fourth order RK

hkkkkyy ii

64311 1771

512

594

125

621

250

378

37

Fifth order RK

hkkkkkyy ii

654311 4

1

14336

277

55296

13525

48384

18575

27648

2825

Formula for k’s

hkhkhkhkhkyhxfk

hkhkhkhkyhxfk

hkhkhkyhxfk

hkhkyhxfk

hkyhxfk

yxfk

ii

ii

ii

ii

ii

ii

543216

43215

3214

213

12

1

4096

253

110592

44275

13824

575

512

175

55296

1631,

8

7

27

35

27

70

2

5

54

11,

5

6

10

9

10

3,

5

3

40

9

40

3,

10

3

5

1,

5

1

,

00

05.0exp8.32exp97,

y

xxyyxfdx

dy

Example

Use h=2, and the RKF method

The results are

RK4=2.542811

RK5=2.554121

and Ea=RK5-RK4=2.554121-2.542811=0.01131

Now adjust stepsize

If Ea is too small, increase step size

If Ea is too large, decrease step size

presentoldnew

dx

dyhy

hh

Stiffness

stiff equation involves rapidly changing parts and slowly changing parts

xeydx

dy 399820008000

xx eey 24 2000

Solution is

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.5 1 1.5 2 2.5

0

0.5

1

1.5

2

2.5

3

-0.001 0.001 0.003 0.005 0.007 0.009 0.011 0.013 0.015

Look at homogeneous part of equation

ydx

dy2000

aydx

dy

In general

Explicit Euler’s method

hayyhdx

dyyy ii

iii 1

ahy

hayyy

i

iii

11

Look at what happens to y over long time - stability

If then y goes to infinity

So for explicit method to work - small h

11 ah

11 ah

Need to use implicit methods, rather than explicit

Implicit form of the Euler method

hayy

hdt

dyyy

ii

iii

1

11

Can solve to getah

yy ii

11

Implicit Euler is always stable -

as i increases y goes to 0

ah

yy ii

11

Example: xeydx

dy 399820008000

Explict solution:

since a is 2000, let h=0.0001

0

0.5

1

1.5

2

2.5

0 0.01 0.02 0.03 0.04 0.05 0.06

y

yExplicit

heyyyhdx

dyyy ix

iiii

ii *39982000800011

Stability limit is h=0.0005

Try h=0.0007

0

0.5

1

1.5

2

2.5

3

3.5

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02

y

yExplicit

Try h=0.001

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 0.005 0.01 0.015 0.02

y

yExplicit

h=0.002

-15

-10

-5

0

5

10

15

0 0.005 0.01 0.015 0.02 0.025

sig

n o

f y

* ln

ofa

bs(

y v

alu

e)

lny

lnyExplicit

Implicit approach:

h

hehyy

heyy

ydx

dyyy

i

i

xi

i

xii

ii

ii

20001

39988000

*3998200080001

1

1

1

11

1

h=0.002

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 0.005 0.01 0.015 0.02

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 0.5 1 1.5 2 2.5 3

y

yImplicit