EE 3561 : Computational
Methods
Unit 8
Solution of Ordinary
Differential Equations
Lesson 3: Midpoint and Heun’s
Predictor corrector Methods
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Lessons in Topic 8

Lesson 1:
Introduction to ODE
Lesson 2:
Taylor series methods
 Lesson 3: Midpoint and Heun’s method
 Lessons 4-5: Runge-Kutta methods
 Lesson 6:
Solving systems of ODE

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Learning Objectives of Lesson 3


To be able to solve first order
differential equation using Midpoint
Method
To be able to solve first order
differential equation using Heun’s
Predictor Corrector method
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Outlines of Lesson 3
Lesson 3: Midpoint and Heun’s
Predictor-corrector methods
Review Euler Method
• Heun’s Method
• Midpoint method
•
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Euler Method
Problem
y ( x)  f ( x, y )
Euler Method
y ( x0 )  y0
yi 1  yi  h f ( xi , yi )
y0  y ( x0 )
for i  1,2,...
2
Local Truncation Error
O(h )
Global Truncation Error
O(h )
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Introduction
Problem to be solved is a first order ODE
y ( x)  f ( x, y ),
y ( x0 )  y0
We have seen Taylor series method
• Euler method is simple but not accurate
• Higher order Taylor series methods are accurate
but require calculating higher order derivatives
analytically
yi 1  yi  h 
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Introduction
Problem to be solved is a first order ODE
y ( x)  f ( x, y ),
y ( x0 )  y0

The methods proposed in this lesson
have the general form
yi 1  yi  h 


For the case of Euler   f ( xi , yi )
Different forms of  will be used for
the midpoint and Heun’s methods
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Midpoint Method
Problem
y ( x )  f ( x, y )
y ( x0 )  y0
Midpoint M ethod
y0  y ( x0 )
y
i
1
2
h
 yi  f ( xi , yi )
2
yi 1  yi  h f ( x
1
i
2
Local Truncation Error
O(h 3 )
Global Truncation Error
O(h 2 )
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,y
1
i
2
)
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Motivation

The midpoint can be summarized as



Euler method is used to estimate the solution
at the midpoint.
The value of the rate function f(x,y) at the mid
point is calculated
This value is used to estimate yi+1.
Local Truncation error of order O(h3)
 Comparable to Second order Taylor series
method

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Midpoint Method
( xi , yi )
x0
y
1
i
2
h
 yi  f ( xi , yi ) ,
2
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x
1
i
2
xi 1
yi 1  yi  h f ( x
1
i
2
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,y
1
i
2
)
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Midpoint Method
slope  f ( xi , yi )
( xi , yi )
x0
y
1
i
2
h
 yi  f ( xi , yi ) ,
2
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x
1
i
2
xi 1
yi 1  yi  h f ( x
1
i
2
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1
i
2
)
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Midpoint Method
slope  f ( xi , yi )
(x
1
i
2
,y
1
i
2
)
( xi , yi )
x0
y
1
i
2
h
 yi  f ( xi , yi ) ,
2
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x
1
i
2
xi 1
yi 1  yi  h f ( x
1
i
2
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,y
1
i
2
)
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Midpoint Method
slope  f ( x
(x
1
i
2
,y
1
i
2
1
2
i
)
,y
i
1
2
)
( xi , yi )
x0
y
1
i
2
h
 yi  f ( xi , yi ) ,
2
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x
1
i
2
xi 1
yi 1  yi  h f ( x
1
i
2
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,y
1
i
2
)
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Midpoint Method
slope  f ( x
(x
1
i
2
,y
1
i
2
1
i
2
)
,y
1
i
2
)
( xi , yi )
x0
y
1
i
2
h
 yi  f ( xi , yi ) ,
2
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x
1
i
2
xi 1
yi 1  yi  h f ( x
1
i
2
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,y
1
i
2
)
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Midpoint Method
slope  f ( x
(x
1
i
2
,y
1
i
2
1
i
2
)
,y
1
i
2
)
( xi , yi )
x0
y
1
i
2
h
 yi  f ( xi , yi ) ,
2
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x
1
i
2
xi 1
yi 1  yi  h f ( x
1
i
2
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1
i
2
)
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Example 1
Use the Midpoint Method to solve the ODE
y ( x)  1  x 2  y
y (0)  1
Use h  0.1. Determine y(0.1) and y(0.2)
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Example 1
Problem : f ( x, y )  1  y  x 2 , y0  y ( x0 )  1, h  0.1
Step 1 :
y
0
1
2
h
 y0  f ( x0 , y0 )  1  0.05(1  0  1)  1.1
2
y1  y0  h f ( x
1
0
2
,y
1
0
2
)  1  0.1(1  0.0025  1.1)  1.2103
Step 2 :
y
1
1
2
h
 y1  f ( x1 , y1 )  1.2103  .05(1  0.01  1.2103)  1.3213
2
y2  y1  h f ( x
1
1
2
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,y
1
1
2
)  1.2103  0.1( 2.3438)  1.4446
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Summary

The midpoint can be summarized as



Euler method is used to estimate the solution
at the midpoint.
The value of the rate function f(x,y) at the mid
point is calculated
This value is used to estimate yi+1.
Local Truncation error of order O(h3)
 Comparable to Second order Taylor series
method

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Heun’s Predictor
Corrector
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Heun’s Predictor Corrector Method
Heun ' s Method
Problem
y ( x )  f ( x, y )
y ( x0 )  y0
y0  y ( x0 )
Predictor : yi01  yi  h f ( xi , yi )
Corrector : y
k 1
i 1
Local Truncation Error
O(h 3 )
Global Truncation Error
O(h 2 )
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h
 yi   f ( xi , yi )  f ( xi 1 , yik1 ) 
2
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Heun’s Predictor Corrector
(Prediction)
( xi 1 , yi01 )
( xi , yi )
xi
xi 1
0
i 1
Preidectio n y
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 yi  h f ( xi , yi )
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Heun’s Predictor Corrector
(Prediction)
( xi 1 , yi01 )
slope  f ( xi 1 , yi01 )
( xi , yi )
xi
xi 1
0
i 1
Preidectio n y
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 yi  h f ( xi , yi )
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Heun’s Predictor Corrector
(Prediction)
f ( xi , yi )  f ( xi 1 , yi01 )
slope 
2
( xi 1 , yi11 )
( xi , yi )
xi
xi 1
0
i 1
Preidectio n y
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( xi 1 , yi01 )
 yi  h f ( xi , yi )
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Heun’s Predictor Corrector
( xi 1 , yi01 )
( xi 1 , yi11 )
( xi , yi )
x0
yi 1  yi  h f ( xi , yi ) ,
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slope  f ( xi 1 , yi 1 )
xi 1
y
1
i 1

h
 yi 
f ( xi , yi )  f ( xi 1 , yi01 )
2
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
Heun’s Predictor Corrector
( xi 1 , yi01 )
( xi 1 , yi11 )
( xi , yi )
x0
yi 1  yi  h f ( xi , yi ) ,
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slope  f ( xi 1 , yi 1 )
xi 1
y
1
i 1

h
 yi 
f ( xi , yi )  f ( xi 1 , yi01 )
2
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
Example 2
Use the Heun' s Method to solve the ODE
y ( x)  1  x  y
2
y ( 0)  1
Use h  0.1. One correction only
Determine y(0.1) and y(0.2)
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Example 2
Problem : f ( x, y )  1  y  x 2 , y0  y ( x0 )  1, h  0.1
Step 1 :
Predictor : y10  y0  h f ( x0 , y0 )  1  0.1( 2)  1.2
h
Corrector : y  y0   f ( x0 , y0 )  f ( x1 , y10 )   1.2105
2
Step 2 :
1
1
Predictor : y20  y1  h f ( x1 , y1 )  1.4326
h
Corrector : y  y1   f ( x1 , y1 )  f ( x2 , y20 )   1.4452
2
1
2
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Summary

Euler, Midpoint and Heun’s methods are
similar in the following sense:
yi 1  yi  h  slope


Different methods use different estimates of
the slope
Both Midpoint and Heun’s methods are
comparable in accuracy to second order
Taylor series method.
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Comparison
Method
Local
truncation
error
yi 1  yi  h f ( xi , yi )
Euler Method
Global
truncation
error
O(h 2 )
O ( h)
O(h 3 )
O(h 2 )
O(h 3 )
O(h 2 )
Heun' s Method
Predictor : yi01  yi  h f ( xi , yi )

h
Corrector : y  yi 
f ( xi , yi )  f ( xi 1 , yik1 )
2
h
Midpoint
y 1  yi  f ( xi , yi )
i
2
2
k 1
i 1
yi 1  yi  h f ( x
i
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1
2
,y
i
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1
2

)
29
More in this Unit
Lessons 4-5: Runge-Kutta Methods
 Lesson 6:
Systems of High order ODE
 Lesson 7: Multi-step methods
 Lessons 8-9: Boundary Value Problems

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EE 3561 : Computational
Methods
Topic 8
Solution of Ordinary
Differential Equations
Lesson 4: Runge-Kutta Methods
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Lessons in Topic 8

Lesson 1:
Introduction to ODE
Lesson 2:
Taylor series methods
 Lesson 3: Midpoint and Heun’s method
 Lessons 4-5: Runge-Kutta methods
 Lesson 6:
Solving systems of ODE

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Learning Objectives of Lesson 4



To understand the motivation for using
Runge Kutta method and basic idea used
in deriving them.
To Familiarize with Taylor series for
functions of two variables
Use Runge Kutta of order 2 to solve ODE
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Motivation


We seek accurate methods to solve ODE
that does not require calculating high
order derivatives.
The approach is to suggest a formula
involving unknown coefficients then
determine these coefficients to match as
many terms of the Taylor series
expansion
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Runge-Kutta Method
Second Order Runge Kutta
K1  h f (t , x)
K 2  h f (t   h, x   K1 )
x(t  h)  x(t )  w1 K1  w2 K 2
Problem :
Find  ,  , w1 , w2
such that x(t  h) is as accurate as possible.
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Lecture
Taylor Series in Two
Variables
The Taylor Series discussed in Chapter 4
is extended to the 2-independent
variable case.
This is used to prove RK formula
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Taylor Series in One Variable
The Taylor Series expansion of f(x)
n 1
i
h (i )
f ( x  h)   f ( x )
i  0 i!
Approximation
n
h (n)

f (x)
n!
Error
where x is between x and x  h
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Taylor Series in One Variable
another look
Define
i
 d 
i d f ( x)
(i )
i
 f ( x) h
 h  f ( x)  h
i
dx
 dx 
The Taylor Series expansion of f(x)
i
n 1
i
n
1 d 
1 d 
f ( x  h)    h  f ( x )   h  f ( x )
n!  dx 
i  0 i!  dx 
x is between x and x  h
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Definitions
Define
i

f
  
i
h
f
(
x
,
y
)

h


x i
 x 
i
0
 
 
 h
 f ( x, y )  f ( x, y )
k
y 
 x
1
 
 
 f ( x, y )
 f ( x, y )
 h
 f ( x, y )  h
k
k
y 
x
y
 x
2
2
2
 
 
f ( x, y )
 2 f ( x, y )
f ( x, y )
2 
2 
 h
 f ( x, y )  h
k
 2h k
k
2
y 
x
xy
y 2
 x
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Taylor Series Expansion
f(x, y)  ( x  1)( x  y  2)
2
Parial derivative s evaluated at (0,0)
0
 
 
 h  k  f ( x, y )
4
y 
 x
( 0. 0 )
1
 
 
 h  k  f ( x, y )
 h fx  k f y
y 
 x
( 0 .0 )
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( 0,0)
40
Taylor Series in Two Variables
The Taylor Series expansion of f(x, y)
i
n
1 
 
1 
 
f ( x  h, y  k )    h  k  f ( x, y )   h  k  f ( x , y )
y 
n!  x
y 
i  0 i!  x
n 1
approximation
error
( x , y ) is on the line joining between ( x, y ) and ( x  h, y  k )
y+k
y
x
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x+h
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Runge-Kutta Method
Second Order Runge Kutta
K1  h f (t , x)
K 2  h f (t   h, x   K1 )
x(t  h)  x(t )  w1 K1  w2 K 2
Problem :
Find  ,  , w1 , w2
such that x(t  h) is as accurate as possible.
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Runge-Kutta Method
Problem : Find  ,  , w1 , w2
to match as many terms as possible.
h2
h3
x(t  h)  x(t )  hx' (t )  x' ' (t )  x' ' ' (t )  ...
2
6
x(t  h)  x(t )  w1h f (t , x)  w2 f (t   h, x   h f (t , x))
1

 
f (t   h, x   h f )  f   h f t   h f x    h   h  f (t , x )
2
t
x 
2
x(t  h)  x(t )  w1  w2 h f (t , x)   w2 h 2 f t   w2 h 2 f f t  O(h 3 )
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Runge-Kutta Method
x(t  h)  x(t ) 
h2
x' ' (t )
2
hx' (t ) 
h3

x' ' ' (t )  ...
6
x(t  h)  x(t )  w1  w2 h f (t , x)   w2 h 2 f t   w2 h 2 f f t
 w1  w2  1,  w2  0.5,
 O(h3 )
 w2  0.5
One possible solution
w1  0.5,
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w2  0.5,
  1,
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 1
44
Runge-Kutta Method
Second Order Runge Kutta
K1  h f (t , x)
K 2  h f (t  h, x  K1 )
1
x(t  h)  x(t )  K1  K 2 
2
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Runge-Kutta Method
Alternative Formula
Second Order Runge Kutta
F1  f (t , x)
F2  f (t  h, x  hF1 )
h
x(t  h)  x(t )  F1  F2 
2
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Runge-Kutta Method
Alternative Formula
Second Order Runge Kutta
Alternativ e Form
K1  h f (t , x)
F1  f (t , x)
K 2  h f (t  h, x  K1 )
F2  f (t  h, x  hF1 )
1
x(t  h)  x(t )  K1  K 2 
2
h
x(t  h)  x(t )  F1  F2 
2
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Runge-Kutta Method
Alternative Formulas
 w1  w2  1,  w2  0.5,
 w2  0.5
another solution
Pick  any non - zero number
  ,
1
w1  1 
,
2
1
w2 
2
Second Order Runge Kutta Formulas (select   0)
K1  h f (t , x)
K 2  h f (t   h, x   K1 )
1 
1

x(t  h)  x(t )  1 
F2
 F1 
2
 2 
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Runge-Kutta Method
Fourth Order Runge Kutta
K1  h f (t , x)
1
1
K 2  h f (t  h, x  K1 )
2
2
1
1
K 3  h f (t  h, x  K 2 )
2
2
K 4  h f (t  h, x  K 3 )
1
x(t  h)  x(t )  K1  2 K 2  2 K 3  K 4 
6
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Second order Runge-Kutta Method
Example
Solve the following system to find x(1.02) using RK2
x(t )  1  x 2 (t )  t 3 ,
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x(1)  4, h  0.01
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Second order Runge-Kutta Method
Example
Solve the following system to find x(1.02) using RK2
x (t )  1  x 2 (t )  t 3 ,
x (1)  4, h  0.01
STEP 1 :
K1  h f (t , x )  0.01(1  x 2  t 3 )  0.18
K 2  h f (t  h, x  K1 )  0.01(1  ( x  0.18) 2  (t  .01)3 )  0.1662
1
1
x (1  0.01)  x (1)  K1  K 2   4  (0.18  0.1662)  3.8269
2
2
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Second order Runge-Kutta Method
Example
STEP 2
K1  h f (t , x)  0.01(1  x 2  t 3 )  0.1666
K 2  h f (t  h, x  K1 )  0.01(1  ( x  0.1666) 2  (t  .01)3 )  0.1545
1
x(1.01  0.01)  x(1.01)  K1  K 2 
2
1
 3.8254  (0.1666  0.1545)  3.6648
2
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x (t )  1  x (t )  t , x(1)  4,
2
3
Solution for t  [1,2]
Using RK2
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Summary
Runge Kutta methods generate accurate
solution without the need to calculate high
order derivatives.
 Second order RK have local truncation
error of order O(h3)
 Fourth order RK have local truncation
error of order O(h5)
 N function evaluations are needed in N th
order RK method.

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More in this unit
Lesson 5:
Applications of Runge-Kutta
Methods To solve first order
differential equations.
 Lessons 6: Solving Systems of high order
ODE.
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