Solving Models with Euler’s Method In MatLab
James K. Peterson
Department of Biological Sciences and Department of Mathematical Sciences
Clemson University
March 26, 2014
Outline
Euler’s Method Review
Euler’s Method: Writing a Function To Do the Job
Using Euler’s Method and Plotting
Abstract
This lecture is about Euler’s Method for solving our models again.
Now we add plots.
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Here is Euler’s Algorithm to approximate the solution to
x 0 = f (x) with x(0) = x0 using step size h for as many steps
as we want.
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x̂0 = x0 so E0 = 0.
x̂1 = x̂0 + f (x̂0 ) h; E1 = |x1 − x̂1 |.
x̂2 = x̂1 + f (x̂1 ) h; E2 = |x2 − x̂2 |.
x̂3 = x̂2 + f (x̂2 ) h; E3 = |x3 − x̂3 |.
x̂4 = x̂3 + f (x̂3 ) h; E4 = |x4 − x̂4 |.
Continue as many steps as you want.
Recursively:
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x̂0 = x0 so E0 = 0.
x̂n+1 = x̂n + f (x̂n ) h; En+1 = |xn+1 − x̂n+1 | for n = 0, 1, 2, 3, ....
We have written this function to handle our Euler approximation
calculations.
f u n c t i o n [ x h a t , x , e ] = D o E u l e r ( f , t r u e , N, x0 , h )
% N = number o f a p p r o x i m a t i o n s t o f i n d
% f = t h e model d y n a m i c s f u n c t i o n
% true = the true s o l u t i o n
% x0 = t h e i n i t i a l c o n d i t i o n
% h = the step s i z e
% define euler step function
e u l e r = @( x , h ) x + f ( x ) ∗ h ;
% i n i t i a l i z e x , x h a t and e
x h a t ( 1 ) = x0 ; x ( 1 ) = x0 ; e ( 1 ) = 0 ;
f o r i =1:N−1
x h a t ( i +1) = e u l e r ( x h a t ( i ) , h ) ;
x ( i +1) = t r u e ( i ∗ h ) ;
e ( i +1) = a b s ( x ( i +1) − x h a t ( i +1) ) ;
end
end
We can also plot the true solution versus the Euler approximates. Here is
the code with annotations. We solve x 0 = .5x(60 − x); x(0) = 20.
% Define the dynamics
>> f = @( x ) . 5 ∗ x . ∗ ( 6 0 − x ) ;
% Define the true s o l u t i o n
>> t r u e = @( t ) 6 0 . / ( 1 + ( 6 0 / 2 0 − 1 ) ∗ e x p ( − . 5 ∗ 6 0 ∗ t ) ) ;
% s e t up f i n a l t i m e and s t e p s i z e
>> T = . 6 ;
h = .01;
% f i n d number o f s t e p s
>> N = c e i l (T/ h ) ;
% s e t up a t i m e v e c t o r f o r t h e t r u e s o l u t i o n
>> t i m e = l i n s p a c e ( 0 , T , 3 1 ) ;
% f i n d the e u l e r approximates
>> [ x h a t , x , e ] = D o E u l e r ( f , t r u e , N, 2 0 , h ) ;
% s e t up a t i m e v a r i a b l e f o r t h e a p p r o x i m a t e s
>> h t i m e = ( 0 : h : N∗ h ) ;
% p l o t them b o t h
>> p l o t ( t i m e , t r u e ( t i m e ) , h ti m e , x h a t , ’ ∗ ’ ) ;
Now set up labels and so forth.
>>
>>
>>
>>
x l a b e l ( ’ Time ’ ) ;
ylabel ( ’x ’) ;
t i t l e ( ’ True S o l u t i o n v s . E u l e r A p p r o x i m a t e ’ ) ;
l e g e n d ( ’ True ’ , ’ E u l e r ’ , ’ L o c a t i o n ’ , ’ Best ’ ) ;
The line where we use N=ceil(T/h) does the division of T by h and
rounds it up to the nearest integer. That way we get the right value of N
even if h does not divide T evenly. We set up the time points for the
euler approximate values using the command htime =(0:h:N*h);
which is a new one. It sets up time points every h units from 0 to the
final one which is Nh. Sometimes it is convenient to use this command.
When we run this code, we generate the plot you see the figure
below and you can see visually that the Euler Approximates are
doing a good job.
It is also easy to plot just the Euler approximations alone by simply
changing what we plot. For example, the Matlab session below does just
that. We don’t define a time variable as we are not plotting the true
solution.
>> f = @( x ) . 5 ∗ x . ∗ ( 6 0 − x ) ;
>> t r u e = @( t ) 6 0 . / ( 1 + ( 6 0 / 2 0 − 1 ) ∗ e x p
( −.5∗60∗ t ) ) ;
>> T = . 6 ;
>> h = . 0 1 ;
>> N = c e i l (T/ h ) ;
% f i n d the e u l e r approximates
>> [ x h a t , x , e ] = D o E u l e r ( f , t r u e , N, 2 0 , h ) ;
% s e t up a t i m e v a r i a b l e f o r t h e a p p r o x i m a t e s
>> h t i m e = ( 0 : h : N∗ h ) ;
% p l o t them b o t h
>> p l o t ( h t i m e , x h a t , ’ ∗ ’ ) ;
>> x l a b e l ( ’ Time ’ ) ;
>> y l a b e l ( ’ x ’ ) ;
>> t i t l e ( ’ E u l e r A p p r o x i m a t e ’ ) ;
We can also plot euler approximates for different step sizes at the
same time, with or without the true solution. As an example, the
Matlab session would look like this:
>> f = @( x ) . 5 ∗ x .∗(60 − x ) ;
>> t r u e = @( t ) 6 0 . / ( 1 + ( 6 0 / 2 0 − 1 ) ∗ e x p ( −.5∗60∗ t ) ) ;
>> T = . 6 ;
% s e t up a t i m e v e c t o r f o r t h e t r u e s o l u t i o n p l o t
>> t i m e = l i n s p a c e ( 0 , T , 3 1 ) ;
% Do s t e p s i z e 1
>> h1 = . 0 4 ;
>> N1 = c e i l (T/ h1 ) ;
% f i n d t h e e u l e r a p p r o x i m a t e s f o r h1
>> [ x h a t 1 , x1 , e1 ] = D o E u l e r ( f , t r u e , N1 , 2 0 , h1 ) ;
% s e t up a t i m e v a r i a b l e f o r t h e a p p r o x i m a t e s
>> h t i m e 1 =(0 : h1 : N1∗h1 ) ;
% Do s t e p s i z e 2
>> h2 = . 0 2 ;
>> N2 = c e i l (T/ h2 ) ;
% f i n d t h e e u l e r a p p r o x i m a t e s f o r h2
>> [ x h a t 2 , x2 , e2 ] = D o E u l e r ( f , t r u e , N2 , 2 0 , h2 ) ;
% s e t up a t i m e v a r i a b l e f o r t h e a p p r o x i m a t e s
>> h t i m e 2 =(0 : h2 : N2∗h2 ) ;
% plot a l l three
>> p l o t ( t i me , t r u e ( t i m e ) , h t i m e 1 , x h a t 1 , ’ ∗ ’ , h t i m e 2 , x h a t 2 , ’ o ’ ) ;
>> x l a b e l ( ’ Time ’ ) ;
>> y l a b e l ( ’ x ’ ) ;
>> t i t l e ( ’ True S o l u t i o n v s . E u l e r A p p r o x i m a t e s ’ ) ;
>> l e g e n d ( ’ True ’ , ’ E u l e r h1 = . 0 4 ’ , ’ E u l e r h2 = . 0 2 ’ , ’ L o c a t i o n ’ , ’ Best ’ ) ;
Here is the resulting plot.
Homework 56
For these problems, use the Matlab session fragments above to
find the plots of the true solution vs. the Euler Approximates for
the list of step sizes given but put all the plots in the same graph.
56.1 x 0 = 1.58x with x(0) = 4 for h = .4 and h = .2.
56.2 x 0 = −1.03x with x(0) = 8 for h = .6 and h = .3.
56.3 x 0 = .4x(105 − x) with x(0) = 19 for h = .06 and h = .03.
56.4 x 0 = −.2x + 15 with x(0) = 10 for h = .5 and h = .25.