k10 ordinary differetial equations

8/18/2019 K10 Ordinary Differetial Equations

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- Basic Ideas

- Euler’s Method

- Higher Order One-step Methods

- Predictor-Corrector Approach

- Runge-Kutta Methods

- Adaptive Stepsize Algorithms

Numerical Methods forNumerical Methods for Civil Civil Engineers Engineers

LectureLecture 1010 Ordinary Dif ferential Equations Ordinary Differential Equations

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

( , )dy

f t ydt

=

where t = independent variable

y = dependent variable

Initial Condition y(t 0) = y0

Rate of change of one variable with respect to another.

2

0d x dxma cv kx m c kxdt dt

+ + = + + =

Example: Mass-spring system m

k

c

x

Ordinary Differential EquationsOrdinary Differential Equations


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New value = Old value + slope x step size

From Taylor series:

2

00 0 0 0

( )( ) ( ) ( ) ( ) ( )

2!

t t y t y t t t y t y t

−′ ′′= + − + +L

Retaining only first derivative: 0 0 0( ) ( , )t y h f t y= +

where h = (t - t 0), y0 = y(t 0), and f (t 0, y0) = y’ (t 0)

t

y

t 0 t 1

h

Predict

Trueerror

1 0 0 0( , ) y h f t y= +

2 1 1 1( , ) y y h f t y= +1 1 1( , )i i i i y y h f t y− − −= +

EULEREULER’’S METHODS METHOD

Example: Manual Calculation with Euler’s Method

2 , (0) 1dy

t y ydt

= − =

Exact solution: ( )21

2 1 54

t y t e−= − +

t i f (t i-1, yi-1)Euler

yi= yi-1+h f (t i-1, yi-1)Exact Error

Set h = 0.2

0.0

0.2

0.4

0.6

NA

0-2(1) = -2.0

0.2-2(0.6) = -1.0

0.4-2(0.4) = -0.4

initial cond. = 1.0

1.0+0.2(-2.0) = 0.6

0.6+0.2(-1.0) = 0.4

0.4+0.2(-0.4) = 0.32

1.0000

0.6879

0.5117

0.4265

0

-0.0879

-0.1117

-0.1065


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Implementing Euler’s Method

function [t,y] = odeEuler(diffeq,tn,h,y0)% Input: diffeq = (string) name of m-file that% evaluate right hand side of ODE% tn = stopping value of independent variable% h = stepsize% y0 = initial condition at t=0

t = (0:h:tn)’;n = length(t);y = y0*ones(n,1);

for i=2:ny(i) = y(i-1)+h*feval(diffeq,t(i-1),y(i-1));

end

odeEuler.m

Example: 2 , (0) 1, 0.2dy

t y y hdt

= − = =

function dydt = rhs1(t,y)dydt = t-2*y;

rhs1.m

>> [t,y] = odeEuler(‘rhs1’,0.6,0.2,1.0)

t =0

0.20000.40000.6000

y =

1.00000.60000.40000.3200


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>> y0 = (1/4)*(2*t-ones(size(t))+5*exp(-2*t))

y0 =

1.0000

0.68790.51170.4265

Exact Solution :

>> t0=0:0.01:0.6;>> y0 = (1/4)*(2*t0-ones(size(t0))+5*exp(-2*t0));>> plot(t0,y0,t,y)

HeunHeun’’ss MethodMethod:: estimate slope from derivatives at the beginning

and at the end of the interval.

Improvements ofImprovements of EEuler uler ’’ss MMethodethod

t

y

t i -1

slope:

k 1 = f (t i -1, y i -1)

t i

Euler:

y 0i = y i -1 + hk 1

h

slope:

k 2 = f (t i , y 0

i )


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t

y

t i -1 t i

h

2

),(),(

2 slope Average

0

1121 i i i i y t f y t f k k +=+

= −−

++= −

221

1

k k hy y i i

Use slope:

(k 1 + k 2)/2

1 1 1

2 1 1 1

1 21

( , )

( , )

2

i i

i i

i i

k f t y

k f t h y hk

k k y y h

− −

− −

−

=

= + +

+= +

yi-1

t i-1

h

t i

True

Euler

Heun

HeunHeun’’ss MMethodethod

( , )dy

f t ydt

= 0i y


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Heun’s Method:

1 1 1

2 1 1 1

1 21

( , )

( , )

2

i i

i i

i i

k f t y

k f t h y hk

k k y y h

− −

− −

−

=

= + +

+= +

1 1 1( , )i i i i y h f t y− − −= +Euler’s Method:

0

1 1 1Predictor: ( , )i i i i y y h f t y− − −= +

0

1 11

( , ) ( , )Corrector:

2

i i i ii i

f t y f t y y y h − −−

+= +

Predictor Predictor --Corrector ApproachCorrector Approach

Iterating the Corrector ofIterating the Corrector of HeunHeun’’ss MethodMethod

To improve the estimation

0

1 11

( , ) ( , )

2

i i i ii

f t y f t y y h − −−

++

i y


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Example: Use Heun’s method to integrate

0.84 0.5 , (0) 2 x y e y y′ = − =

From x = 0 to 4, step size = 1

Analytical solution: ( )0.8 0.5 0.54

21.3

x x x y e e e− −= − +

0

1 1 1

0 0

1

Predictor, ( , )

2 1 (4 0.5(2)) 5

i i i i y y h f t y

y e

− − −= +

= + − =

01 1

1

0 0.8(1)

1

( , ) ( , )Corrector,

2

(4 0.5(2)) (4 0.5(5))2 1 6.7011

2

i i i ii i

f t y f t y y y h

e e y

− −−

+= +

− + −= + =

True value: ( )0.8(1) 0.5(1) 0.5(1)4

2 6.19461.3

y e e e− −= − + =

6.1946 6.7011Relative error, 100% 8.18%

6.1946t E

−= × =

0 0.8(1)nd

1

(4 0.5(2)) (4 0.5(6.7011))2 Corrector, 2 1

2

6.2758, 1.31%t

e e y

E

− + −= +

= =

0 0.8(1)rd

1

(4 0.5(2)) (4 0.5(6.2758))3 Corrector, 2 1

2

6.3821, 3.03%t

e e y

E

− + −= +

= =Error may increases for large step sizes.


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Iteration using up-arrow in MATLAB

>> y = 2+1*(4*exp(0)-0.5*2) % Predictor of y1

y = 5

>> y = 2+(1/2)*((4*exp(0)-0.5*2)+(4*exp(0.8*1)-0.5*y))y = 6.7011 % 1st Corrector of y1

Enter Press and

y = 6.7011 % 2nd Corrector of y1

y = 6.2758 % 3rd Corrector of y1

y = 6.3821 % 4th Corrector of y1

y = 6.3555 % 5th Corrector of y1

y = 6.3609 % until no change

MATLAB’s Implementation

function dydx = func1(x,y)dydx = 4*exp(0.8*x)-0.5*y;

func1.m

function [x,y] = odeHeun(diffeq,xn,h,y0,iter)

% Input: iter = number of corrector iterations

x = (0:h:xn)’;n = length(x);y = y0*ones(n,1);for i=2:n

y(i) = y(i-1)+h*feval(diffeq,x(i-1),y(i-1));for j=1:iter

y(i) = y(i-1)+h*(feval(diffeq,x(i-1),y(i-1)) ...+ feval(diffeq,x(i),y(i)))/2;

end end

odeHeun.m


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Et =0.002.683.093.173.18

y_true =2.00006.194614.843933.677275.3390

For x = 0 to 4, step size = 1, y(0) = 2, Iteration = 15

>> [x,y] = odeHeun(‘func1’,4,1,2,15)

x =

01234

y =2.00006.360915.302234.743377.7351

Compare with Euler’s Method:

>> [x,y] = odeEuler(‘func1’,4,1,2)

x =01234

y =2.00005.000011.402225.513256.8493

Et =0.0019.2823.1924.2424.54


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Comparison of True Solution with Euler’s and Heun’s Method

>> xtrue = (0:0.1:4)’;>> ytrue =(4/1.3)*(exp(0.8*xtrue)-exp(-0.5*xtrue))...

+ 2*exp(-0.5*xtrue);

>> [xeuler,yeuler] = odeEuler(‘func1’,4,1,2);>> [xheun,yheun] = odeHeun(‘func1’,4,1,2,15);>> plot(xtrue,ytrue,xeuler,yeuler,’o’,xheun,yheun,’+’)

0 1 2 3 40

20

40

60

80

x

y

True

Euler

Heun 15

1

1

m

i i l l

l

y y k γ −=

= + ∑General Formula using Weighted Average of slopes

For Heun’s method γ 1 = γ 2 = 0.5

The weights must satisfy1

1

m

l

l

γ

==∑

Second-Order RK (Ralston’s) Method:

1 1 2

1 1 1

2 1 1 1

1 2

3 3

( , )3 3

,4 4

i i

i i

i i

y y h k k

k f x y

k f x h y k h

−

− −

− −

= + +

= = + +

RRUNGEUNGE--KUTTA (RK)KUTTA (RK) METHODSMETHODS


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Third-Order Runge-Kutta Methods

( )1 1 2 346

i i

h y y k k k −= + + +

where

( )

1 1 1

2 1 1 1

3 1 1 1 2

( , )

1 1,

2 2

, 2

i i

i i

i i

k f x y

k f x h y k h

k f x h y k h k h

− −

− −

− −

=

= + +

= + − +

Fouth-Order Runge-Kutta Methods

where

( )

1 1 1

2 1 1 1

3 1 1 2

4 1 1 3

( , )

1 1,

2 2

1 1,

2 2

,

i i

i i

i i

i i

k f x y

k f x h y k h

k f x h y k h

k f x h y k h

− −

− −

− −

− −

=

= + +

= + +

= + +

The most popular RK methods

( )1 1 2 3 42 26

i i

h y y k k k k −= + + + +Classical 4th-order RK:


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Example: Use 4th-order RK method to integrate

0.84 0.5 , (0) 2 x y e y y′ = − =

From x = 0 to 4, step size = 1

Step 1: x0 = 0, y0 = 2

0

1

0.8(0.5)

2

0.8(0.5)

3

0.8(1)

4

4 0.5(2) 3

14 0.5(2 3 1) 4.2173

2

1

4 0.5(2 4.2173 1) 3.91302

4 0.5(2 3.9130 1) 5.9457

k e

k e

k e

k e

= − =

= − + × × =

= − + × × == − + × =

( )11

2 3 2 4.2173 2 3.9130 5.94576

6.2011

y = + + × + × +

=

True value: ( )0.8(1) 0.5(1) 0.5(1)4

2 6.1946

1.3

y e e e− −= − + =

6.1946 6.2010Relative error, 100% 0.1038%

6.1946t E

−= × =


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MATLAB’s Implementation

function [x,y] = odeRK4(diffeq,xn,h,y0)

x = (0:h:xn)’;n = length(x);y = y0*ones(n,1);for i=2:n

k1 = feval(diffeq,x(i-1),y(i-1));k2 = feval(diffeq,x(i-1)+h/2,y(i-1)+k1*h/2);k3 = feval(diffeq,x(i-1)+h/2,y(i-1)+k2*h/2);k4 = feval(diffeq,x(i-1)+h,y(i-1)+k3*h);y(i) = y(i-1)+h/6*(k1+2*k2+2*k3+k4);

end

odeRK4.m

For x = 0 to 4, step size = 1, y(0) = 2, Iteration = 15

>> [x,y] = odeRK4(‘func1’,4,1,2)

x =01234

y =2.00006.201014.862533.721375.4392

y_true =2.00006.194614.843933.677275.3390

Et(%) =0.000.10380.12500.13090.1328


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MATLAB’s ode23 and ode45 Routines

ode23 use second- and third-order RK simultaneously

ode45 use fourth- and fifth-order RK simultaneously

>> x0 = 0; xn = 4; y0 = 2;>> [x45,y45] = ode45(‘func1’,[x0,xn],y0);

>> xtrue = (0:0.1:4)’;>> ytrue =(4/1.3)*(exp(0.8*xtrue)...

- exp(-0.5*xtrue))+ 2*exp(-0.5*xtrue);>> plot(xtrue,ytrue,x45,y45)

For x = 0 to 4, y(0) = 2

Example:0.8

4 0.5 , (0) 2 x

y e y y′ = − = func1.m

Adaptive Adaptive StepsizeStepsize Algorithms Algorithms

k10 ordinary differetial equations

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