P t 6Part 6Chapter 20Chapter 20

I iti l V l P blInitial-Value Problems

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PowerPoints organized by Dr. Michael R. Gustafson II, Duke UniversityRevised by Prof. Jang, CAU

Chapter ObjectivesChapter ObjectivesUnderstanding the meaning of local and global truncation• Understanding the meaning of local and global truncation errors and their relationship to step size for one-step methods for solving ODEs.K i h t i l t th f ll i R K tt (RK)• Knowing how to implement the following Runge-Kutta (RK) methods for a single ODE:– Euler

H– Heun– Midpoint– Fourth-Order RK

K i h i h f H ’ h d• Knowing how to iterate the corrector of Heun’s method.• Knowing how to implement the following Runge-Kutta

methods for systems of ODEs:y– Euler– Fourth-order RK

Ordinary Differential EquationsOrdinary Differential Equations• Conservation laws are presented in the form of differential

equationsequations. • Nth order differential equation can be reduced to first-order

differential equation.Methods described here are for sol ing differential eq ations of• Methods described here are for solving differential equations of the form: dy

dt f t, y

• The methods in this chapter are all one-step methods and have the general format:

dt

y y h

where is called an increment function, and is used to extrapolate from an old value y to a new value y h is the time step

yi1 yi h

from an old value yi to a new value yi+1. h is the time step.• One-step methods compute a future prediction yi+1 based only on

information at a single point yi

M l i h d f di i b d• Multi-step methods compute a future prediction yi+1 based on several previous points.

Euler’s MethodEuler s Method• The first derivativeThe first derivative

provides a direct estimate of the slope at ti: dy

dt ti

f ti , yi

and the Euler method uses that estimate as

ti

the increment

function: f ti , yi y y f t y hyi1 yi f ti , yi h

Example 20.1 (1/2)p ( / )Q. Use Euler Method to integrate y' = 4e0.8t – 0.5y =f from t = 0 to

4 ith t i f 1 Th i iti l diti t t 0 i 2 Th4 with a step size of 1. The initial condition at t = 0 is y = 2. The exact solution can be determined by

4 ttt eeey 5.05.08.0 2)(3.1

4

)1)(2,0()0()1( fyy

3)2(5.04)2,0( 0 ef 5)1(32)1( y

Euler’s method

19463.62)(3.1

4 )1(5.0)1(5.0)1(8.0 eeey

%28.19%10019463.6

519463.6

t

)1)(51()1()2( f

40216.11)1)](5(5.04[5

)1)(5,1()1()2()1(8.0

e

fyy

Example 20.1 (2/2)Example 20.1 (2/2)t ytrue yEuler (%)t

012

2.000006.1946314.84392

2.000005.0000011.40216

19.2823.19

34

33.6771775.33896

25.5132156.84931

24.2424.54

Error Analysis for Euler’s MethodError Analysis for Euler s Method• The numerical solution of ODEs involves twoThe numerical solution of ODEs involves two

types of error:– Truncation errors, caused by the nature of the

i ti t h i l dapproximation techniques employed– Roundoff errors, caused by the limited numbers of

significant digits that can be retaineds g ca t d g ts t at ca be eta ed• The total, or global truncation error can be

further split into:– local truncation error that results from an

application method in question over a single step, andand

– propagated truncation error that results from the approximations produced during previous steps.

Error Analysis for Euler’s MethodError Analysis for Euler s Method• The local truncation error for Euler’s method is

O(h2) and proportional to the derivative of f(t,y) while the global truncation error is O(h).

)1(

)(!

),(!2

),(),( 1

)1(2

1

nniin

iiiiii hOh

nytf

hytf

hytfyy

hytfyy iiii ),(1 )(),( 12

nii hOh

ytfE

• This means:– The global error can be reduced by decreasing the

)(!2

t hOhE

The global error can be reduced by decreasing the step size, and

– Euler’s method will provide error-free predictions if the nderl ing f nction is linear ( > secondthe underlying function is linear (-> second derivative is zero).

• Euler’s method is conditionally stable, depending u e s et od s co d t o a y stab e, depe d gon the size of h.

Stability of Euler Methody-

0 0 (0) atdy ay with y y y y edt

dt

1 1 (1 )ii i i i

dyy y h y y ahdt

Amplification factor : g = 1 ahAmplification factor : g = 1-ahIf |g|>1 (that is h > 2/a), the solution will grow in anunbound fashion.unbound fashion. -> Euler method is conditionally stable.

There are certain ODEs where errors always growregardless of the method -> ill-conditioned ODE

A fundamental source of error in Euler’s method is thatthe derivative at the beginning of the interval is assumedg gto apply across the entire interval.

MATLAB Code for Euler’s MethodMATLAB Code for Euler s Method

Heun’s MethodHeun s Method• One method to improve Euler’s method is to determine

derivatives at the beginning and predicted ending of the interval g g p gand average them:

0

),( iii ytfy

hytfyy iiii ),(01

),( 0111 iii ytfy

Predictor Eq.

hi li ki di i f h l f

hytfytf

yy iiiiii 2

),(),( 011

1

Corrector Eq.

• This process relies on making a prediction of the new value of y, then correcting it based on the slope calculated at that new value.

• This predictor-corrector approach can be iterated to convergence:s p ed cto co ecto app oac ca be te ated to co e ge ce:

hytfyy iimii ),(0

1

),(),( 111 h

ytfytf jii

miimj

),,2,1 (

2

111

mj

hyy iiiimi

ji

Example 20.2 (1/3)p ( / )Q. Use Heun’s method with iteration to integrate

y' = 4e0.8t – 0.5y from t = 0 to 4 with a step size of 1 The initial condition at t = 0 is y = 2size of 1. The initial condition at t = 0 is y = 2. The stopping criterion is 0.00001%.

Example 20 2 (2/3)Example 20.2 (2/3)Sol.) 2,0:IC 5.04 8.0 ytyey t

(t0, y0):

P di t

3)2(5.04 00 ey

Predictor: 19.28%: percent relative error (Euler)5)1(3201 y

402164.6)5(5.04),( )1(8.00111 eyxfy

701082.42402164.63

y

Corrector: -8.18%: percent relative

error

701082.6)1(701082.4211 y

Approximate error : %39.25%100701082.6

5701082.6

a

ff )()( 0)()( 111 ytfytf j

imiij

hytfytf

yy iiiiii 2

),(),( 011

1

),,2,1 (

2

),(),( 111

mj

hytfytf

yy iiiimi

ji

Example 20.2 (3/3)p ( / ) Corrector:

A i

275811.612

)701082.6(5.0432)1(8.0

21

ey

70108262758116Approximate error : %776.6%100275811.6

701082.6275811.6

a

0.8(1)3 4 0 5(6 275811)e Corrector:

Approximate error :

31

3 4 0.5(6.275811)2 1 6.3821292

ey

%666.1%100275811.6382129.6

%666.1%100

382129.6a

With t it ti With it tiWithout iteration With iteration

t ytrue yEuler (%) yHeun (%) yHeun (%)

0 2 00000 2 00000 2 00000 2 00000t t t

012

2.000006.19463

14.84392

2.000005.00000

11.4021619.2823.19

2.000006.70108

16.319788.189.94

2.000006.36087

15.302242.683.09

34

33.6771775.33896

25.5132156.84931

24.2424.51

37.1992583.33777

10.4610.62

34.7432877.73510

3.173.18

Midpoint MethodMidpoint Method• Another improvement to Euler’s method is similar

to Heun’s method but predicts the slope at theto Heun s method, but predicts the slope at the midpoint of an interval rather than at the end:

h2

),(2/1hytfyy iiii

)( 2/12/12/1 iii ytfy ),( 2/12/12/1 iii ytfy

hytfyy iiii ),( 2/12/11

• The midpoint method is superior to Euler’s method in the same way that centered difference is better approximation rather than either forward or backward versions. This method has a local t ti f O(h3) d l b l f O(h2)truncation error of O(h3) and global error of O(h2).

Runge-Kutta MethodsRunge Kutta Methods• Runge-Kutta (RK) methods achieve the accuracy of a Taylor

series approach without requiring the calculation of higher pp q g gderivatives.

• For RK methods, the increment function can be generally written as: a k a k a k

(1) 1 hyy ii

written as:

where the a’s are constants and the k’s are

a1k1 a2k2 ankn

ytfk ii1 , hkqhkqyhptfk

hkqyhptfkytfk

ii

ii

ii

22212123

11112

1

, ,

,

where the p’s and q’s are constants. hkqhkqhkqyhptfk nnnnninin 11,122,111,11 ,

• First order RK method with n=1: Euler’s method• Once n is chosen, values for the a’s, p’s, and q’s are evaluated

by setting Eq (1) equal to terms in Taylor series expansionby setting Eq. (1) equal to terms in Taylor series expansion.

Second order Runge KuttaSecond-order Runge-KuttaThe second order version of Eq. (1) is

hkakayy ii )( 22111

),(1 ii ytfk

)( hkhfk ),( 11112 hkqyhptfk ii

The values for a1, a2, p1 and q11 are evaluated by setting E (1) l d d T l i

21 !2

),(),( h

ytfhytfyy ii

iiii

dtdy

yytf

tytfytf ii

),(),(),(

Eq. (1) equal to a second order Taylor series.

!2

!2),(

2

1h

dtdy

yf

tfhytfyy iiii

dtyt

y

hkakayy ii )( 22111

Second order Runge KuttaSecond-order Runge-Kutta

gsgrytgsyrtg )()( hkakayy )(

y

st

rytgsyrtg ),(),( hkakayy ii )( 22111

),( 11112 hkqyhptfk ii

)(),(),( 211111111 hO

yfhkq

tfhpytfhkqyhptf iiii

)(),(),(),( 32112

212211 hO

yfytfhqa

tfhpaythfaythfayy iiiiiiii

)(),(),(),( 3211212211 hOh

yfytfqa

tfpahytfaytfayy iiiiiiii

1

2/12/11

12

21

qapa

aaFour unknownsThree eq.

21 1 aa 2

111 21a

qp

There are an infinite number of second-2/1112 qaorder RK methods.

Second order Runge KuttaSecond-order Runge-Kutta

• Heun method without iteration– a2 = 1/2 a1 = 1/2 and p1 = q11 = 1a2 / a1 / a d p1 q11

hkkyy ii

211 2

121 ),(1 ii ytfk

)( hkyhtfk

• Midpont method

22 ),( 12 hkyhtfk ii

– a2 = 1 a1 = 0 and p1 = q11 = 1/2

hkyy )(fkhkyy ii 21 ),(1 ii ytfk )2/,2/( 12 hkyhtfk ii

Classical Fourth order RKClassical Fourth-order RK• Classical Fourth-order RK

– The most popular RK method

hkkkkyy ii 43211 2261

6

),(1 ii ytfk

hkyhtfk ii 12 2

1,21

22

hkyhtfk ii 23 2

1,21

– The fourth-order RK method is i il t th H th d i

hkyhtfk ii 34 ,

similar to the Heun method in that multiple estimates of the slop are used.

Example 20.3 (1/2)p ( / )Q. Use the classical fourth-order RK to integrate the following

f ti f t 0 t 1 ith t i f 1 d i iti l ditifunction from t = 0 to 1 with a step size of 1 and initial condition is y(0) = 2.

Sol ) ytfyey t ,5.04 8.0

Sol.)the slope at the beginning3)2(5.04)2,0( )0(8.0

1 efk53)50(3250 10 hky

the slope at the midpoint

5.3)5.0(325.0 10 hky217299.4)5.3(5.04)5.3,5.0( )5.0(8.0

2 efk

1086494)50(2172994250 hky

another slope at the midpoint

108649.4)5.0(217299.425.0 20 hky

912974.3)108649.4(5.04)108649.4,5.0( )5.0(8.03 efk

another slope at the midpoint

Example 20 3 (2/2)Example 20.3 (2/2)912974.5)0.1(912974.3230 hky

the slope at the end

)(30y0.8(0.5)

4 (1.0, 5.912974) 4 0.5 (5.912974) 5.945677k f e

the average slope

201037.4945677.5)912974.3(2)217299.4(2361

g p

final prediction(1.0) 2 4.201037 (1.0) 6.201037y

p

= 0 103% where the exact solution is 6 194631t = 0.103% where the exact solution is 6.194631.

Systems of EquationsSystems of Equations• Many practical problems require the solution y p p q

of a system of equations:dy1

d f1 t, y1, y2 ,, yn

dtf1 , y1, y2 , , yn

dy2

dt f2 t, y1, y2 ,, yn

dt

dyn

dt fn t, y1, y2 ,, yn

• The solution of such a system requires that ninitial conditions be known at the starting

dt n 1 2 n

initial conditions be known at the starting value of t.

Solution MethodsSolution Methods• Single-equation methods can be used to g q

solve systems of ODE’s as well; for example Euler’s method can be used onexample, Euler s method can be used on systems of equations - the one-step method is applied for every equation atmethod is applied for every equation at each step before proceeding to the next step.

• Fourth-order Runge-Kutta methods canFourth order Runge Kutta methods can also be used, but care must be taken in calculating the k’scalculating the k s.

Example 20.4 (1/3)p ( / )Q. Solve for the velocity and position of the free falling

b j i E l th d Th i iti l ditibungee jumper using Euler method. The initial condition for t = 0, x = v = 0. The step size is 2s and integrate up to t = 10s g= 9 81 m/s2 m= 68 1 kg cd= 0 25 kg/mto t = 10s. g= 9.81 m/s , m= 68.1 kg, cd= 0.25 kg/m.

The exact solution for v is

tgcgmtv dtanh)(The exact solution for v is

The exact solution for x is

mcd

)(

tgcmtx dcoshln)(The exact solution for x is

Compute the true relative errors of the results

tmc

tx d

d

coshln)(

Compute the true relative errors of the results.

Example 20.4 (2/3)Example 20.4 (2/3)Sol.)

Slope at t 0: dxSlope at t = 0: 0 vdtdx

81.9)0(25.081.9 22 vc

gdv d

Euler method ( t = 2 s) Analytic sol. Percent relative errors.x(2) = 19 16629 100%

81.9)0(1.68

81.9vm

gdt

0)2(00 x x(2) 19.16629 100%v(2) = 18.72919 4.756%

Euler method ( t = 4 s)

0)2(00 x62.19)2(81.90 v

( )

24.39)2(62.190 x

20.2519 62 9 81 (19 62) (2)v 19.62 9.81 (19.62) (2)68.1

36.41368

v

Example 20.4 (3/3)p ( / )

t xtrue vtrue xEuler vEuler t(x) t(v)

02

019.1663

018.7292

00

019.6200 100.00 4.76

468

10

71.9304147.9462237.5104334 1782

33.111842.076246.957549 4214

39.2400112.0674204.6640305 0244

36.413746.298350.180251 3123

45.4524.2513.838 72

9.9710.036.863 8310 334.1782 49.4214 305.0244 51.3123 8.72 3.83

Example 20.4 (1/4)p ( / )Q. Use the fourth-order RK method to solve for the same

problem as in Ex. 20.4. h=2.

Sol.) The differential equation to be integrated is

vvxtfdtdx

),,(1

cdv

Slope at the beginning of the interval.

22 ),,( v

mc

gvxtfdtdv d

p g g

0)0,0,0(11,1 fk

819)0(25.0819)000( 2 fk 81.9)0(1.68

81.9)0,0,0(22,1 fk

Example 20.4 (2/4)Example 20.4 (2/4)

2h 02200

2)0()1( 1,1

hkxx

81.92281.90

2)0()1( 21

hkvv

Slope at the midpoint of the interval.

22)()( 2,1

8100.9)81.9,0,1(11,2 fk

4567.9)81.9,0,1(22,2 fk

810092810090)0()1( hkxx 8100.9

28100.90

2)0()1( 1,2 kxx

4567.9224567.90

2)0()1( 2,2

hkvv22

Example 20.4 (3/4)Example 20.4 (3/4) Another slope at the midpoint of the interval.

4 69)4 69810091(fk 4567.9)4567.9,8100.9,1(11,3 fk

4817.9)4567.9,8100.9,1(22,3 fk

913418)2(456790)0()2( hk 9134.18)2(4567.90)0()2( 1,3 hkxx

9634.18)2(4817.90)0()2( 2,3 hkvv

Slope at the end point of the interval.

963418)9634189134182( fk 9634.18)9634.18,9134.18,2(11,4 fk

4898.8)9634.18,9134.18,2(22,4 fk

Example 20.4 (4/4)p ( / ) Average slope

1 1656.192]9634.18)4567.98100.9(20[610)2( x

7256.182]4898.8)4817.94567.9(28100.9[610)2( v ])([6

)(

t xtrue vtrue xRK4 vRK4 t(x) t(v)

0246

019.166371.9304

147.9642

018.729233.111842.0762

019.165671.9311

147.9521

018.725633.099542.0547

0.0040.0010.004

0.0190.0370.0516

810

147.9642237.5104334.1782

42.076246.957549.4214

147.9521237.5104334.1626

42.054746.934549.4207

0.0040.0000.005

0.0510.0490.038

MATLAB RK4 CodeMATLAB RK4 Code