SE301: Numerical Methods
Topic 8
Ordinary Differential Equations (ODEs)
Lecture 28-36
KFUPM
(Term 101)
Section 04
Read 25.1-25.4, 26-2, 27-1
CISE301_Topic8L2
1
Outline of Topic 8
Lesson 1:
Introduction to ODEs
 Lesson 2:
Taylor series methods
 Lesson 3:
Midpoint and Heun’s method
 Lessons 4-5: Runge-Kutta methods
 Lesson 6:
Solving systems of ODEs
 Lesson 7:
Multiple step Methods
 Lesson 8-9: Boundary value Problems

CISE301_Topic8L2
2
Lecture 29
Lesson 2: Taylor Series Methods
CISE301_Topic8L2
3
Learning Objectives of Lesson 2





Derive Euler formula using the Taylor
series expansion.
Solve the first order ODEs using Euler
method.
Assess the error level when using Euler
method.
Appreciate different types of errors in the
numerical solution of ODEs.
Improve Euler method using higher-order
Taylor Series.
CISE301_Topic8L2
4
Taylor Series Method
The problem to be solved is a first order ODE:
dy ( x)
 f ( x, y ), y ( x0 )  y0
dx
Estimates of the solution at different base points:
y( x0  h), y( x0  2h), y( x0  3h), ....
are computed using the truncated Taylor series
expansions.
CISE301_Topic8L2
5
Taylor Series Expansion
Truncated Taylor Series Expansion
h k  d k y
y ( x0  h)  
k

k
!
dx
k 0

n
dy
 y ( x0 )  h
dx



x  x0 , y  y 0 
h2 d 2 y

2
x  x0 ,
2
!
dx
y y
0
hn d n y
 ... 
x  x0 ,
n! dx n
y  y0
x  x0 ,
y  y0
The nth order Taylor series method uses the
nth order Truncated Taylor series expansion.
CISE301_Topic8L2
6
Euler Method

First order Taylor series method is known
as Euler Method.

Only the constant term and linear term
are used in the Euler method.

The error due to the use of the truncated
Taylor series is of order O(h2).
CISE301_Topic8L2
7
First Order Taylor Series Method
(Euler Method)
dy
y ( x0  h)  y ( x0 )  h
dx
x  x0 ,
y  y0
 O(h 2 )
Notation :
xn  x0  nh,
dy
dx
x  xi ,
y  yi
yn  y ( xn ),
 f ( xi , yi )
Euler Method
yi 1  yi  h f ( xi , yi )
CISE301_Topic8L2
8
Euler Method
Problem :
Given the first order ODE :
y0  y ( x0 )
with the initial condition :
Determine :
y ( x)  f ( x, y )
yi  y ( x0  ih )
for i  1,2,...
Euler Method :
y0  y ( x0 )
yi 1  yi  h f ( xi , yi )
CISE301_Topic8L2
for i  1,2,...
9
Interpretation of Euler Method
y2
y1
y0
x0
CISE301_Topic8L2
x1
x2
x
10
Interpretation of Euler Method
y1
Slope=f(x0,y0)
hf(x0,y0)
y0
x0
CISE301_Topic8L2
y1=y0+hf(x0,y0)
h
x1
x2
x
11
Interpretation of Euler Method
y2=y1+hf(x1,y1)
y2
Slope=f(x1,y1)
hf(x1,y1)
Slope=f(x0,y0)
y1
hf(x0,y0)
y0
x0
CISE301_Topic8L2
y1=y0+hf(x0,y0)
h
x1
h
x2
x
12
Example 1
Use Euler method to solve the ODE:
dy
2
 1 x ,
dx
y (1)  4
to determine y(1.01), y(1.02) and y(1.03).
CISE301_Topic8L2
13
Example 1
f ( x, y )  1  x ,
2
x0  1, y0  4 , h  0.01
Euler Method
yi 1  yi  h f ( xi , yi )
Step1 :
y1  y0  h f ( x0 , y0 )  4  0.01(1  (1) 2 )  3.98


Step2 : y2  y1  h f ( x1 , y1 )  3.98  0.01 1  1.01  3.9598
Step3 :
2


y3  y2  h f ( x2 , y2 )  3.9598  0.01 1  1.02  3.9394
CISE301_Topic8L2
2
14
Example 1
f ( x, y )  1  x ,
2
x0  1, y0  4 , h  0.01
Summary of the result:
i
xi
yi
0
1.00
-4.00
1
1.01
-3.98
2
1.02
-3.9595
3
1.03
-3.9394
CISE301_Topic8L2
15
Example 1
f ( x, y )  1  x ,
2
x0  1, y0  4 , h  0.01
Comparison with true value:
CISE301_Topic8L2
i
xi
yi
True value of yi
0
1.00
-4.00
-4.00
1
1.01
-3.98
-3.97990
2
1.02
-3.9595
-3.95959
3
1.03
-3.9394
-3.93909
16
Example 1
f ( x, y )  1  x ,
2
x0  1, y0  4 , h  0.01
A graph of the
solution of the
ODE for
1<x<2
CISE301_Topic8L2
17
Types of Errors



Local truncation error:
Error due to the use of truncated Taylor
series to compute x(t+h) in one step.
Global Truncation error:
Accumulated truncation over many steps.
Round off error:
Error due to finite number of bits used in
representation of numbers. This error could
be accumulated and magnified in
succeeding steps.
CISE301_Topic8L2
18
Second Order Taylor Series Methods
dy ( x)
Given
 f ( y, x), y ( x0 )  y0
dx
Second order Taylor Series method
2
2
dy h d y
3
yi 1  yi  h 
 O(h )
2
dx 2! dx
2
d y
needs to be derived analytical ly.
2
dx
CISE301_Topic8L2
19
Third Order Taylor Series Methods
dy ( x)
Given
 f ( y, x), y ( x0 )  y0
dx
Third order Taylor Series method
2
2
3
3
dy h d y h d y
4
yi 1  yi  h 

 O(h )
2
3
dx 2! dx
3! dx
2
3
d y
d y
and 3 need to be derived analytical ly.
2
dx
dx
CISE301_Topic8L2
20
High Order Taylor Series Methods
dy ( x)
Given
 f ( y, x), y ( x0 )  y0
dx
th
n order Taylor Series method
dy h 2 d 2 y
hn d n y
n 1
yi 1  yi  h 
 .... 
 O(h )
2
n
dx 2! dx
n! dx
2
3
n
d y d y
d y
, 3 ,....., n need to be derived analytical ly.
2
dx dx
dx
CISE301_Topic8L2
21
Higher Order Taylor Series Methods

High order Taylor series methods are
more accurate than Euler method.

But, the 2nd, 3rd, and higher order
derivatives need to be derived analytically
which may not be easy.
CISE301_Topic8L2
22
Example 2
Second order Taylor Series Method
Use Second order Taylor Series method to solve :
dx
2
 2 x  t  1, x(0)  1,
dt
use h  0.01
2
d x(t )
What is :
?
2
dt
CISE301_Topic8L2
23
Example 2
Use Second order Taylor Series method to solve :
dx
2
 2 x  t  1, x(0)  1,
use h  0.01
dt
dx
2
 1 2x  t
dt
d 2 x(t )
dx
2
 0  4x
 1  4 x(1  2 x  t )  1
2
dt
dt
2
h
2
2
xi 1  xi  h(1  2 xi  ti )  (  1  4 xi (1  2 xi  ti ))
2
CISE301_Topic8L2
24
Example 2
f (t , x)  1  2 x 2  t ,
t0  0, x0  1, h  0.01
2
h
2
2
xi 1  xi  h(1  2 xi  ti )  (  1  4 xi (1  2 xi  ti ))
2
Step1 :
x1  1  0.01(1  2(1)
2
2

0.01
 0) 
(1  4(1)(1  2  0))  0.9901
2
Step 2 :
x2  0.9901  0.01(1  2(0.9901)
2
2

0.01
 0.01) 
(1  4(0.9901)(1  2(0.9901) 2  0.01))  0.9807
2
Step 3 :
x3  0.9716
CISE301_Topic8L2
25
Example 2
f (t , x)  1  2 x  t ,
2
t0  0, x0  1, h  0.01
Summary of the results:
CISE301_Topic8L2
i
ti
xi
0
0.00
1
1
0.01
0.9901
2
0.02
0.9807
3
0.03
0.9716
26
Programming Euler Method
Write a MATLAB program to implement
Euler method to solve:
dv
2
 1  2v  t .
dt
v ( 0)  1
for ti  0.01i,
i  1,2,...,100
CISE301_Topic8L2
27
Programming Euler Method
f=inline('1-2*v^2-t','t','v')
h=0.01
t=0
v=1
T(1)=t;
V(1)=v;
for i=1:100
v=v+h*f(t,v)
t=t+h;
T(i+1)=t;
V(i+1)=v;
end
CISE301_Topic8L2
28
Programming Euler Method
f=inline('1-2*v^2-t','t','v')
h=0.01
t=0
v=1
T(1)=t;
V(1)=v;
for i=1:100
v=v+h*f(t,v)
t=t+h;
T(i+1)=t;
V(i+1)=v;
end
CISE301_Topic8L2
Definition of the ODE
Initial condition
Main loop
Euler method
Storing information
29
Programming Euler Method
Plot of the
solution
plot(T,V)
CISE301_Topic8L2
30
More in This Topic
Lesson 3:
Midpoint and Heun’s method
Provide the accuracy of the second order
Taylor series method without the need to
calculate second order derivative.
Lessons 4-5: Runge-Kutta methods
Provide the accuracy of high order
Taylor series method without the need to
calculate high order derivative.
CISE301_Topic8L2
31