systems of odes (ordinary differential equations)

Dept. of Computer Science & Engineering, Korea Univ.

Systems of ODEs(Ordinary Differential Equations)

June 8, 2005

Sung-Min Kim, Ho-Kuen Shin, A-Yon Park

Computer Vision & Pattern Recognition Lab.


ContentsContents

13.1 Higher Order ODEs

13.2 Systems of Two First-Order ODEs

13.3 Systems of First-Order ODE-IVPs

13.4 Stiff ODE and Ill-Conditioned Problems

13.5 Using MATLAB’s Functions


13.1 Higher Order ODEs13.1 Higher Order ODEs

A higher order ODE can be converted into a system of first-order ODEs A second-order ODE of the form

can be converted to system of two first-order ODEs by a simple change of variables :

the differential equations relating these variables are

the initial conditions for the original ODE,

become the initial conditions for the system, i.e.,

)',,('' yyxgy

'., yvyu

).,,('

),,,('

vuxgv

vuxfvu

,)0(',)0( 10 yy

.)0(,)0( 10 vu


Ex.13.1Ex.13.1

Nonlinear pendulum

Choosing g/L=1 and c/(mL)=0.3,a=π/2 and b=0.

Second-order ODE-IVP

.)0(',)0(,0sin''' byayyL

gy

mL

cy

yyy sin'3.0'' ', yvyu

.0)0(,2/)0(with

),,,(sin3.0'

),,,('

vu

vuxguvv

vuxfvu

Simple pendulum


13.1 Higher Order ODEs (Cont’d)13.1 Higher Order ODEs (Cont’d)

A higher order ODE may be converted to a system of first order ODEs by a similar change of variables The nth-order ODE

a system of first-order ODEs by the following change of variables;

the differential equations relating these variables are

with initial conditions

,1)1(

210

)1()(

)0(,...,)0('',)0(',)0(

),,...,'',',,(

nn

nn

yyyy

yyyyxfy

.,...,'',', )1(321

nn yuyuyuyu

),...,,,,(',...,',',' 321433221 nn uuuuxfuuuuuuu

1231201 )0(,...,)0(,)0(,)0( nnuuuu


13.2 13.2 Systems of two first-order ODEsSystems of two first-order ODEs

13.2.1 Euler’s method for solving two ODE-IVPs To apply the basic Euler’s method

to the system of ODEs

update the function u using f(x,u,v) and update v using g(x,u,v)

)( ,1 iiii yxhfyy

),,,('),,,(' vuxgvvuxfu

)).(),(),(()()1(

)),(),(),(()()1(

iviuixhgiviv

iviuixhfiuiu


Nonlinear pendulum using Euler’s Method To apply the basic Euler’s method

In order to get accurate results, take a fairly small step, e.g., n=200

Ex.13.2Ex.13.2

.0)0(,2/)0(with

),,,(sin3.0'

),,,('

vu

vuxguvv

vuxfvu

n=50

n=300n=200

n=100


13.2 13.2 Systems of two first-order ODEs (Cont’d)Systems of two first-order ODEs (Cont’d)

Matlab code

)).(),(),(()()1(

)),(),(),(()()1(

iviuixhfiviv

iviuixhfiuiu


Ex.13.3Ex.13.3

Series dilution problem using Euler’s method

litersVlitersVlitersLlitermoles

CCV

L

dt

dCC

V

L

dt

dC

5and10min,/2,0)0(C,/3.0)0(Cwith

].[,

2121

122

21

1

1

u=c1,v=c2

u’= c’1 = -0.2u = f(x,u,v)

v’= c’2 = -0.4(v-u) = g(x,u,v)

n=200


13.2.2 13.2.2 Midpoint method for solving two ODE-IVPsMidpoint method for solving two ODE-IVPs

The idea is same as for Euler’s method Update each unknown function u and v, using the basic Runge-Kutta for

mulas

rewrite the basic second-order Runge-Kutta formulas,

),,('

),,('

vuxgv

vuxfu

,

),2

1,

2

1(

),,(

211

12

1

kyy

kyhxhfk

yxhfk

i

ii

ii



Using k1 and k2 to represent the update quantities for the unknown u

Calling the corresponding quantities for function v,m1 and m2

Update function f by the appropriate multiple of k1 or k2 and function g by the corresponding amount of m1 or m2

• This means that k1 and m1 must be computed before k2 and m2 can be found thus,

.

,

),2

1,

2

1,

2

1(

),2

1,

2

1,

2

1(

),,,(

),,,(

211

211

112

112

1

1

mvv

kuu

mvkuhxhgm

mvkuhxhfk

vuxhgm

vuxhfk

i

i

iii

iii

iii

iii



Matlab code

.

,

),2

1,

2

1,

2

1(

),2

1,

2

1,

2

1(

),,,(

),,,(

211

211

112

112

1

1

mvv

kuu

mvkuhxhgm

mvkuhxhfk

vuxhgm

vuxhfk

i

i

iii

iii

iii

iii


Nonlinear pendulum using Runge-Kutta method

The computed solution for n=50,n=100, and n=200 are indistinguishable

Ex.13.4Ex.13.4

.0)0(,2/)0(with

),,,(sin3.0'

),,,('

vu

vuxguvv

vuxfvu

n=50

n=300n=200

n=100


Series dilution using Runge-Kutta technique

The exact solutions, viz.,

Ex.13.5Ex.13.5

,20and5and10

min,/2,0)0(C,/3.0)0(Cwith

].[,

21

21

122

21

1

1

nlitersVlitersV

litersLlitermoles

CCV

L

dt

dCC

V

L

dt

dC

)].4.0exp()2.0[exp(6.0])/(exp[)/([exp[)0(

)(

);2.0exp(3.0])/exp[()0()(

2121

112

111

tttVLtVLVV

CVtC

ttVLCtC


13.3 Systems of First-order ODE-IVPs13.3 Systems of First-order ODE-IVPs

Application: chemical reactions, predator-prey models, and many others

Systems of ODEs come from the conversion of higher order ODEs into system form

Ex. 13.6 Higher Order System of ODEs Consider the equation


The system of ODEs is

A system of ODEs

yyyxyyyxfy 432),,,(

.2)0(,3)0(,4)0( yyy

,),,,( 232111 uuuuxfu

,),,,( 332122 uuuuxfu

32132133 432),,,( uuuxuuuxfu

).u,(fu x


13.3.1 Euler’s Method 13.3.1 Euler’s Method for Solving Systems of ODEsfor Solving Systems of ODEs

The basic Euler method

The system of ODEs

Updating the function u1 using f1, u2 using f2, and u3 using f3. h is the same step size

)),(),(),(),(()()1( 321222 iuiuiuixhfiuiu

),(1 iiii yxhfyy

),,,,( 32111 uuuxfu

),,,,( 32122 uuuxfu

),,,,( 32133 uuuxfu

)),(),(),(),(()()1( 321111 iuiuiuixhfiuiu

)),(),(),(),(()()1( 321333 iuiuiuixhfiuiu


Ex 13.8Ex 13.8

Solving a Higher Order System using Euler’s Method


on [1,2] with n=2(h=0.5).

at x = 3/2:

at x = 2:

44432221

2 , ,

2 , u

xuuuu

xuuu

1)1( ,0)1( ,0)1( ,10)1( 4321 uuuu

;0)1/2(5.01)2/3( ,5.0)1(5.00)2/3(

,0)0(5.00)2/3( ,10)0(5.010)2/3(

43

21

uu

uu

.0)0(5.00)2( ,5.0)0(5.02/1)2(

,0)0(5.00)2( ,10)0(5.010)2(

43

21

uu

uu


Second order Runge-Kutta method as k and m

System of three ODEs

The values of the parameter k and m

The values of unknown functions

13.3.2 Runge-Kutta Methods 13.3.2 Runge-Kutta Methods for Solving Systems of ODEsfor Solving Systems of ODEs

myy

k)h, y hf(xm ), y hf(xk

ii

iiii

1

2

1

2

1 ,

)),(),(),(),((

)),(),(),(),((

)),(),(),(),((

32133

32122

32111

iuiuiuix hf k

iuiuiuix hf k

iuiuiuix hf k

).,,,( ),,,,( ),,,,( 321333212232111 uuuxfuuuuxfuuuuxfu

.2

1)(,

2

1)(,

2

1)(,

2

1)(

,2

1)(,

2

1)(,

2

1)(,

2

1)(

,2

1)(,

2

1)(,

2

1)(,

2

1)(

33221133

33221122

33221111

kiukiukiuhixhfm

kiukiukiuhixhfm

kiukiukiuhixhfm

.)()1(

,)()1(

,)()1(

333

222

111

miuiu

miuiu

miuiu


System Using a Second-Order Runge-Kutta Method System Using a Second-Order Runge-Kutta Method (Midpoint Method)(Midpoint Method)

Matlab code

function [ x, u ] = RK2_sys(f, tspan, u0, n)

a = tspan(1);

b = tspan(2);

h = (b-a)/n;

x = (a+h : h : b);

k = h*feval(f, a, u0)';

m = h*feval(f, a + h/2, u0 + k/2)';

u(1, :) = u0 + m;

for i = 1 : n -1

k=h*feval(f, x(i), u(i, :))';

m = h*feval(f, x(i) + h/2, u(i, :) + k/2)';

u(i+1, :) = u(i, :) + m;

end

x = [a x];

u = [u0, u];

))(),(),(),(( 321 iuiuiuix hf k nn

332211 2

1)(,

2

1)(,

2

1)(,

2

1)( kiukiukiuhixhfm nn

nnn miuiu )()1(

The indexing of the matrix for the solution u at each step is only changedThe indexing of the matrix for the solution u at each step is only changed


Solving a Higher Order System using a Runge-Kutta Method Interval [0, 1] of the system of ODEs


with n=2, we find the following values for x, u1, u2, and u3

Matlab codefunction du = f_13_9(x, u)

du = [u(2) ; u(3) ; x + 2*u(1) - 3*u(2) + 4*u(3)]

[x, u] = RK2_sys('f_13_9', [0 1], [4 3 2], 2)

Ex 13.9Ex 13.9

32133221 432 , , uuuxuuuuu

.2)0( ,3)0( ,4)0( 321 uuu

x u1 u2 u3

0.0 4.0000 3.0000 2.0000

0.5 5.7500 4.8750 9.1250

1.0 9.3281 13.6719 40.9219


Ex 13.10Ex 13.10

Solving a Higher Order System using a Runge-Kutta Method The electrostatic potential between two concentric spheres (radius 1 and radi

us 2)


on the interval [1,2], n=2

0))1((5.0 21 uk

0))1(2

(5.0 22

ux

k

5.0))1((5.0 43 uk

1))1(2

(5.0 44

ux

k

2'22

'1

2 u

xuuu

u1(1) = 10, u2(1) = 0, u3(1) = 0, u4(1) = 1

4'44

'3

2 u

xuuu

0)5.0)1((5.0 221 kum

0)5.0)1()(25.1

2(5.0 222

kum

25.0)5.0)1((5.0 443 kum

4.0)5.0)1()(25.1

2(5.0 444

kum


Ex 13.10 (cont’d)Ex 13.10 (cont’d)

The approximate solution at x = 1.5:

finding k and m again

10010)5.1(1 u 000)5.1(2 u

25.025.00)5.1(3 u 6.04.01)5.1(4 u

4.0)5.1(45.1

25.04

,3.0))5.1(4(5.03

,0)5.1(25.1

25.02

,0))5.1(2(5.01

uk

uk

uk

uk 0)5.0)5.1((5.0 221 kum

0)5.0)5.1()(75.1

2(5.0 222

kum

2.0)5.0)5.1((5.0 443 kum

2286.0)5.0)5.1()(25.1

2(5.0 444

kum



The approximate solution at x = 2.0:


du = [u(2) ; -2*u(2)/x ; u(4) ; -2*u(4)/x]

[x, u] = RK2_sys('f_13_10', [1 2], [10 0 0 1], 2)

10010)0.2(1 u 000)0.2(2 u

45.02.025.0)0.2(3 u 3714.02286.06.0)0.2(4 u

x u1 u2 u3 u4

1.0000 10.0000 0 0 1.0000

1.5000 10.0000 0 0.2500 0.6000

2.0000 10.0000 0 0.4500 0.3714



x u1 u2 u3 U4

1.0000 10.0000 0 0 1.0000

1.1000 10.0000 0 0.0900 0.8286

1.2000 10.0000 0 0.1653 0.6976

1.3000 10.0000 0 0.2293 0.5953

1.4000 10.0000 0 0.2842 0.5139

1.5000 10.0000 0 0.3319 0.4480

1.6000 10.0000 0 0.3737 0.3941

1.7000 10.0000 0 0.4107 0.3493

1.8000 10.0000 0 0.4436 0.3117

1.9000 10.0000 0 0.4730 0.2799

2.0000 10.0000 0 0.4995 0.2527

n = 10 Matlab code

• [x, u] = RK2_sys('f_13_10', [1 2], [10 0 0 1], 10)


Fourth-Order Runge-Kutta MethodFourth-Order Runge-Kutta Method

Matlab code%function f(x, u);a = tspan(1);b = tspan(2);h = (b - a)/n;x = (a+h : h : b)';k1 = h*feval(f, a, u0)';k2 = h*feval(f, a+h/2, u0+k1/2)';k3 = h*feval(f, a+h/2, u0+k2/2)';k4 = h*feval(f, a+h, u0+k3)';u(1, :) = u0 + k1/6 + k2/3 + k3/3 + k4/6;for i = 1:n-1 k1 = h*feval(f, x(i), u(i,:))'; k2 = h*feval(f, x(i)+h/2, u(i, :)+k1/2)'; k3 = h*feval(f, x(i)+h/2, u(i, :)+k2/2)'; k4 = h*feval(f, x(i)+h, u(i, :)+k3)'; u(i+1, :) = u(i, :) + k1/6 + k2/3 + k3/3 +

k4/6;end

x = [a x];u = [u0 u];

43211

34

23

12

1

6

1

3

1

3

1

6

1

,2

1,

2

12

1,

2

1

kkkkyy

)k yh hf(x k

)k yh hf(x k

)k yh hf(x k

), y hf(x k

ii

ii

ii

ii

ii


Ex 13.11Ex 13.11

Solving a Circular Chemical Reaction Using a Runge-Kutta Method r1 = 0.1(t+1), r2=2, r3=1 n = 100


du = [u(3)-0.1*(x+1)*u(1) ; 0.1*(x+1)*u(1)-2*u(2) ; 2*u(2)-u(3)];

[x, u] = RK4_sys('f_13_11', [0 40], [0.8696 0.0435 0.0870], 100)

nn = length(x)

plot(x, u(1:nn, 1), '-',x, u(1:nn, 2),'.',x, u(1:nn, 3),'-.')

. , , 322113 CrBrdt

dCBrAr

dt

dBArCr

dt

dA

Ar

rCA

r

rB

rrrrA

3

1

2

1

3121

, ,//1

1

0870.015.1

1.0)0(

,0435.015.1

05.0)0( ,8696.0

15.1

1)0(

C

BA

A

C

B

Concentrations of three reactants in circular chemical


13.3.3 Multistep Methods for System13.3.3 Multistep Methods for System

The basic two-step Adams-Bashforth method

y0 is given by the initial condition

y1 is found from a one-step method, such as a Runge-Kutta

can be extended for use with a system of three ODEs

)],(),(3[2 111 iiiiii yxfyxfh

yy

u1’ = f1(x,u1,u2,u3)

u2` = f2(x,u1,u2,u3)

u3` = f3(x,u1,u2,u3)


13.3.3 Multistep Methods for System13.3.3 Multistep Methods for System

The basic two-step Adams-Bashforth method (cont’d)

))]1(),1(),1(),1((-

))(),(),(),((3[2

)()1(

3211

321111

iuiuiuixf

iuiuiuixfh

iuiu

))]1(),1(),1(),1((-

))(),(),(),((3[2

)()1(

3212

321222

iuiuiuixf

iuiuiuixfh

iuiu

))]1(),1(),1(),1((-

))(),(),(),((3[2

)()1(

3213

321333

iuiuiuixf

iuiuiuixfh

iuiu


Adams-Bashforth-MoultonAdams-Bashforth-Moulton

Predictor-corrector methods Matlab code

function [x, u] = ABM3_sys(f, tspan, u0, n)a = tspan(1);b = tspan(2);h = (b-1)/n;hh = h/12;x = (a+h:h:b)';k = h*feval(f, a, u0)';m = h*feval(f, a+h/2, u0+k/2)';u(1, :) = u0+m;k = h*feval(f,x(1), u(1, :))';m = h*feval(f, x(1)+h/2, u(1,:)+k/2)';u(2, :) = u(1, :) +m;z(2, :) = feval(f, x(2), u(2, :))';uu(3,:) = u(2, :) + hh*(23*z(2,:) - 16*z(1, :) + 5*feval(f, a, u0)');zz = feval(f, x(3), uu(3, :))';u(3, :) = u(2, :) + h*(5*zz + 8*z(2, :) -z(1,:));for i = 3:n-1 z(i, :) = feval(f, x(i), u(i, :))'; uu(i+1, :) = u(i, :) + hh*(23*z(i,:) - 16*z(i-1, :) + 5*z(i-2, :)); zz = feval(f, x(i+1), uu(i+1, :))'; u(i+1,:) = u(i,:) + hh*(5*zz + 8*z(i,:) - z(i-1,:));endx = [a x];u = [u0, u];

.1,....,2],51623[12 211

niforfff

hyy iiiii

]8),(5[12 1111

iiiiii ffyxf

hyy


Ex 13.12 Mass-and-Spring SystemEx 13.12 Mass-and-Spring System

The vertical displacements of two masses m1 and m2

Spring constants s1 and s2

Displacements x1 and x2

)( 12211''

11 xxsxsxm

)( 122''

22 xxsxm

)( 131

21

1

1'22

'1 uu

m

su

m

suuu

)( 131

1'44

'3 uu

m

suuu

Initial conditions

u1(0) = α1, u2(0) = α2, u3(0) = α3, u4(0) = α4

m1

m2

X1=0

X2=0


Ex 13.12 Mass-and-Spring SystemEx 13.12 Mass-and-Spring System

Matlab code

a = 0;b = 5;tspan = [a b];n = 100;y0 = [0.5 0 0.25 0];global s1 m1 s2 m2s1 = 100;m1 = 10;s2 = 120;m2 = 2;r1 = 10;r2 = 15;[t, y] = ABM3_sys('f_13_12', tspan, y0, n);[nn, mm] = size(y)%out = [t y];plot(t(1:nn), -(r1+y(1:nn, 1)), '-')hold on plot(t(1:nn), -(r2+y(1:nn, 3)), '-.')grid on plot([a b], [0 0])hold off

function du = f_13_12(t, u)global s1 m1 s2 m2du =[u(2) ; (-s1*u(1) + s2*(u(3) – u(1)))/m1 ; u(4) ; -s2*(u(3) – u(1))/m2 ];]

Displacement profiles of two masses connected by springs


Ex 13.13 Motion of a BaseballEx 13.13 Motion of a Baseball

Air resistance is one of the factors influencing how far a fly ball travels Inital velocity of [100, 45] Acting on the horizontal component only

Matlab code

a = 0;

b = 3;

tspan = [a b];

n = 100;

y0 = [0 100 3 45];

[t, y] = ABM3_sys('f_bb', tspan, y0, n);

[nn, mm] = size(y);

out = [t y];

plot(y(1:nn, 1), y(1:nn, 3), '-')


Ex 13.13 Motion of a BaseballEx 13.13 Motion of a Baseball

1. Air Resistance Proportional to Velocity in x Direction

function dz = f_bb_1(x, z)

dz = [z(2); -0.1*z(2) ; z(4) ; -32];

2. Air Resistance Proportional to Velocity


dz = [z(2); -0.1*sqrt(z(2)^2 + z(4)^2);

z(4); -32-0.1*sqrt(z(2)^2 + z(4)^2)*sign(z(4))];

3. Air Resistance Proportional to Velocity in Squared x Direction


dz = [z(2); -0.0025*z(2)^2; z(4); -32];

4. Air Resistance Proportional to Velocity Squared


dz = [z(2); -0.0025*(z(2)^2 + z(4)^2)

z(4); -32-0.0025*(z(2)^2 + z(4)^2)*sign(z(4))];

1

2

3

4


Stiff ODE and ILL-Conditioned ProblemStiff ODE and ILL-Conditioned Problem

Ill condition ODE for which any error that occurs will increase, regardless of

the numerical method employed

General solution is

• The initial conditions

– u1(0) = 3, u2(0) =-3

• We have a=0, b=3• A component of the positive exponential will be introduced and will

eventually dominate the true solution

u1’ = 2u2

u2` = 2u1

u1’ = 2u2

u2` = 2u1

u1’ = ae2x+be-2x

u2` = ae2x –be-2x

u1’ = ae2x+be-2x

u2` = ae2x –be-2x



ill condition Consider the ODE

The general solution

If we take the initial condition as y(0) = 2/ 27, the exact solution is

• Any error in the numerical solution process will introduce the exponential component which will eventually dominate the true solution

• Parasitic solution

2' 3 tyy 2' 3 tyy

27

2

9

2

3

1 23 ttCey t

27

2

9

2

3

1 23 ttCey t

27

2

9

2

3

1 2 tty27

2

9

2

3

1 2 tty



ill condition Consider the ODE

• Initial conditions u(0) = 1 , v(0)=0

Consider the ODE

• with Λ<<0 and g(t) a smooth, slowly varying function

)())(( '' tgtgyy )())(( '' tgtgyy

u’ = 98u + 198v

v’ = -99u + 199v

u’ = 98u + 198v

v’ = -99u + 199v

u(t) = 2e-t – e-100t

v(t) = -e-t + e-100t

u(t) = 2e-t – e-100t

v(t) = -e-t + e-100t

)())0(( 0 tgegyy t )())0(( 0 tgegyy t


Using Matlab’s FunctionsUsing Matlab’s Functions

The Matlab Functions ODE 23

• The Matlab functions for solving the ODE • [ T, Y ] = ODE(‘function_name’, tspan,x0)

– ‘fuction_name’ is diffenential equations

– tsapn is time duration from time t0 to tf

– x0 is initial conditions• Example

function f=dfuc(t,x)

f=zeros(2,1);

f(1) = 490/9*x(1) - 1996/9*x(2);

f(2) = 12475*90*90*x(1)-49990/90*x(2);

tspans=0:0.1:4; x0=[10 -20];

[t x] = ode23('dfuc',tsapns,x0);

20)(0,10)0( 49901247590

19964909

yxyxdt

dy

yxdt

dx

20)(0,10)0( 49901247590

19964909

yxyxdt

dy

yxdt

dx




• To solve the problem of simulating the motion of a two – link planar robot arm

• See Spong and Vidyasagar, Robot Dynamic and Control, John Wiley & Sons, Inc.,New York, 1989 ( p.145, ex 6.4.2)

– Position, Velocity, Torques of joints • robot2

Moment of inertia for long, slender rod

I1 = m1*l1^2 / 12;

I2 = m2*l2^2 / 12;

The Mass matrix [M]

d11 = m1*Lc1^2 + m2*(L1^2 + Lc2^2 + 2*L1

*Lc2*cos(r2)) + I1 + I2;

d12 = m2*(Lc2^2 + L1*Lc2*cos(r2)) + I2;

d21 = d12;

d22 = m2*Lc2^2 + I2;




• robot2_1

Initialize Calculate each parameters Plot each parameters




• To solve the problem of simulating the motion of a two – link planar robot arm

• robot2




• Result• Figure shows the moti

on of the arm– Position– Velocity– Torques

systems of odes (ordinary differential equations)

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