1
Numerical Solution of Ordinary Differential Equations
Consider the 1st order ordinary differential equation (ODE)
dy
 f ( x, y ) .
dx
The initial condition can be taken as
y ( x0 )  y 0 .
Then we could use a Taylor series about x  x0 and obtain the complete solution, or
y ( x)  y ( x0 )  y ' ( x0 )( x  x0 ) 
1
1
1 (N )
y ' ' ( x0 )( x  x0 ) 2  y ' ' ' ( x0 )( x  x0 ) 3  ... 
y ( x0 )( x  x0 ) N  ...
2!
3!
N!
Since
y ( x0 )  y 0
and
y ' ( x 0 )  f ( x0 , y 0 )  y 0 '
then we can find the first two terms. For the second derivative
d2y
y ''( x0 )  2
dx
x  x0
 f f dy 
 
 f, x ( x0 , y0 )  f, y ( x0 , y0 ) y0 ' .


x

y
dx

 x x 0
Similarly the third derivative is
y '''( x0 ) 
d3y
dx3
 f, xx ( x0 , y0 )  2 f, xy ( x0 , y0 ) y0 ' f, yy ( x0 , y0 ) y0 '2 .
x  x0
If we truncate at the third derivative
y ( x)  y ( x 0 )  y ' ( x 0 )( x  x 0 ) 
1
1
y ' ' ( x 0 )( x  x 0 ) 2  y ' ' ' ( x 0 )( x  x 0 ) 3  Error ,
2!
3!
and
Error 
1 iv
y ( )( x  x0 ) 4 , where x0    x .
4!
2
Euler’s Method
Take the Taylor series to 1st order, and let the interval h  x1  x0 , then
y1  y ( x0 )  f  x0 , y 0 h 
 
y ' ' ( ) 2
h .
2
The error for a time step (the local error) is O h 2 . The global error, after many steps, is
Oh  . Then
y1  y0  f ( x0 , y0 )h where y0  y( x0 ) ,
y2  y1  f ( x1, y1 )h where x1  x0  h ,
…
y N 1  y N  f ( x N , y N )h where x N 1  x N  h .
Example:
dy
 x  y , y ( 0)  1
dx
The exact solution can be found from
dy
y x.
dx
Let y  yc  y p where
dyc
 yc  0 , or y c  Ce rx . Then rCe rx  Ce rx  0 for all x ,
dx
or r  1 , and y c  Ce x . Since the right hand side is linear in x try y p  Ax  B . Then
dy p
dx
 A and
dy p
dx
 y p  x becomes A  Ax  B  x which must hold for all x. Hence
A  1 , and B=-1, making y p  x  1 , and since y  yc  y p then
y  Ce x  x  1 .
But @ x  0 , y  1 or C  1  1 , and C  2 . Making the complete solution
y  2e x  x  1 .
3
Using Euler’s method and taking h  0.02
y0  1, x0  0  x1  0.02, y1  1  1  0.02  1.02 , since y0 '  1 . In general,
y n 1  y n  y n ' h; xn 1  xn  h
n
0
1
2
3
4
5
xn
0.00
0.02
0.04
0.06
0.08
0.10
yn
1.0000
1.0200
1.0408
1.0624
1.0848
1.1081
yexact
1.0000
1.0204
1.0416
1.0637
1.0866
1.1103
For the error, y Euler  y5  y0  0.1081 , y Exact  y5  y0  0.1103 , can be defined
as
Relative Error 
y Euler  y Exact
1
y Euler  y Exact 
2

0.0022
 2%
0.1092
The results plot as
It would be better to use the slope at the beginning and end of the increment (e.g., the
average at each end), and although we don’t know the slope at the end we can
approximate it.
4
Modified Euler Method
Let y n '  f ( xn , y n ) . Then an approximation for y at the end of the increment is
~
y n 1  y n  y n ' h
and an estimate for the slope at the end of the increment is
~
y'
 f (x
,~
y
).
n 1
n 1
n 1
We can now set
y n 1  y n 


1
y n ' ~
y n' 1 h .
2
The error can be found from
y n 1  y n  y n' h 
 
1 '' 2
yn h  O h3
2
and since
y n 1  y n  y n' h 
 

1  y n' 1  y n'
 Oh  h 2  O h 3

2 
h

or
 
Hence the local error is Oh 3  and the global error is Oh 2 . Another way to write our
 y '  y n' 
h  O h 3 .
y n 1  y n   n 1


2


results is
k1  hf ( xn , y n )
k 2  hf ( xn  h, y n  k1 )
y n 1  y n 
1
k1  k 2 
2
The previous example now can include modified Euler
n
0
1
2
3
4
5
xn
0.00
0.02
0.04
0.06
0.08
0.10
which is much better.
yeuler
1.0000
1.0200
1.0408
1.0624
1.0848
1.1081
ymodified
1.0000
1.0204
1.0416
1.0637
1.0866
1.1104
yexact
1.0000
1.0204
1.0416
1.0637
1.0866
1.1103
5
Runge-Kutta Methods
The modified Euler method is actually a two step (second order) Runge-Kutta algorithm.
These methods can be readily extended to eight and even tenth order. The derivations
follow the same procedure. Assume for the second order method
y n 1  y n  ak1  bk 2
where
k1  hf ( xn , y n ) , and
k 2  hf ( xn  h, y n  k1 ) .
The parameters a, b,  and  are found by comparing to a Taylor series expansion.
Recall
1
y n 1  y n  hf n  h 2 f n, x  f n, y f n .
2
But
f ( xn  h, yn  k1 )  f n  hf n, x  k1 f n, y .


Since k1  hf n ,
f ( xn  h, yn  k1 )  f n  hf n, x  hf n f n, y
Using y n 1  y n  ak1  bk 2 ,


yn 1  yn  ahf n  bh f n  hf n, x  hf n f n, y ,
or
y n 1  y n  a  b hf n  bh 2 f n, x  bh 2 f n f n, y
Comparing to the Taylor series
y n 1  y n  hf n 
1 2
1
h f n, x  h 2 f n f n, y .
2
2
1
1
, b  which is three equations in four unknowns. Hence we
2
2
can pick for the fourth equation any equation that is convenient.
Then a  b  1, b 
For example we can take
or
Modified Euler is
a  2 / 3, b  1 / 3, and     3 / 2
a  0, b  1, and     1 / 2 .
a  b  1 / 2, and     1 .
6
Fourth Order Runge-Kutta
For fourth order Runge-Kutta the estimates for the changes are
k1  hf ( x n , y n )
1
1
h, y n  k1 )
2
2
1
1
k 3  hf ( x n  h, y n  k 2 )
2
2
k 4  hf ( x n  h, y n  k 3 )
k 2  hf ( x n 
and the updated value for y is found from
y n 1  y n 
1
k1  2k 2  2k3  k 4  .
6
Note that all of the algorithms presented are for first order equations with only one
dependent variable. These can be readily extended to systems of higher order differential
equations. For those cases please see page 7.
7
Higher Order Differential Equations
Consider
dny
n


 y (n) ( x)  f x, y, y' , y ' ' ,..., y n 1  f x, y0 , y1 , y 2 ,..., y n 1 
dx

and let the vector Y be defined as
 y0'   y
 '   1
 y1   y 2
  y '   y
Y '  2    3
 ... 
 y n'  2 
 ' 
 y n 1 
 y0 

 y 

 1 


 y 

 where Y   2 
 ... 
 ... 
 yn  2 
 y n 1 
 f 
y 


 n 1 
which is now a vector first order equation, and the first order rules can be applied to a
system of first order equations.
Systems of First Order Equations
 

 
Let Y   F x, Y



Euler: YN 1  YN  hFN
Modified Euler:
where




1 
YN 1  YN  K1  K 2
2
where




K1  hF x N , YN




K 2  hF x N  h, YN  K1

Runge-Kutta (4th order):


YN 1  YN 


1
K1  2 K 2  2 K3  K 4
6

8





K 1  hF x N , Y N


1 
1  
K 2  hF  x N  h, Y N  K 1 
2
2 



1 
1  
K 3  hF  x N  h, Y N  K 2 
2 
2 


K 4  hF x N  h, Y N  K 3


9
Problem
The equation for a pendulum with a mass, m, at the end of a rod of negligible mass with
length, L, is
mL2  mgL sin   0 .
For the initial conditions take
   0 @ t  0 , and   0 @ t  0 ,
(1)
where
 is the angle from the vertical, () 
Let  
d( )
d 2( )
, and () 
.
dt
dt 2
g
t . The equation becomes
L
d 2
Now let p 
d 2
d
then since
d
d 2
d
2

 sin   0 .
(2)
dp dp d
dp
,

p
d d d
d
The equation can be written
p
dp
 sin   0 ,
d
and integrating
1 2
p  cos   E  constant .
2
This is the conservation of energy. The initial condition is p 
hence E   cos  0 or
d
 0 @   0 ,
d
2
1  d 
   cos    cos  0
2  d 
(3)
is also the governing differential equation.
Integrate the equation of motion (2) subject to the initial conditions (1),
   / 2 @   0 and   0 @   0 using Euler, modified Euler, and Runge-Kutta. Use
the expression for the energy to check the accuracy of your integration. Integrate for onequarter of a cycle and then determine the period for a complete cycle.