Numerical Solution of an IVP:Presume we wish to numerically solve the initial value problem
y? = f (t, y), y(a) = y0,
On an period of time [a, b].
By a numerical solution we should mean an approximation of the solution at a finite number of points that is:
(t0, y0), (t1, y1), (t2, y2), . . . ,(tn, yn),
Where t0 = a and tn= b The first of these points is precisely the initial value. If we acquire n steps as above and the steps are evenly spaced then the time change in every step is
h = (b − a)/ n
and the times ti are given simply by ti= a + ih. This leaves the mainly important part of finding a numerical solution determining y1, y2, . . . ,yn in a manner that is as consistent as possible with.
To do this first write the discrepancy equation in the indexed notation:
y?i ≈ f (ti, yi),
And will then put back the derivative y? by a difference. There are several ways we might carry this out and in the next section we study the simplest.
The Euler Method:
The mainly straight forward approach is to replace y? i by its forward difference approximation.
(yi+1– yi)/h= f (ti, yi).
Rearranging this gives us a method to obtain yi+1 from yiknown as Euler’s method:
yi+1 = yi+ hf (ti, yi).
With this formula we are able to start from (t0, y0) and compute all the subsequent approximations (ti, yi).
This is extremely easy to implement as you can see from the following program (which can be downloaded from the class web site)
function [T , Y] = myeuler(f,tspan,y0,n)% function [T , Y] = myeuler(f,tspan,y0,n)% Solves dy/dt = f(t,y) with initial condition y(a) = y0% on the interval [a,b] using n steps of Euler s method.% Inputs: f -- a function f(t,y) that returns a column vector of the same% length as y% tspan -- a vector [a,b] with the start and end times% y0 -- a column vector of the initial values, y(a) = y0% n -- number of steps to use% Outputs: T -- a n+1 column vector containing the times% Y -- a (n+1) by d matrix where d is the length of y% Y(t,i) gives the ith component of y at time t% parse starting and ending pointsa = tspan(1);b = tspan(2);h = (b-a)/n; % step sizet = a; y = y0; % t and y are the current variablesT = a; Y = y0’; % T and Y will record all stepsfor i = 1:ny = y + h*f(t,y);t = a + i*h;T = [T; t];Y = [Y; y’];End
To utilization this program we need a function such as the vector function for the pendulum
> f = inline(’[y(2);-.1*y(2)-sin(y(1)) + sin(t)]’,’t’,’y’)
> [T Y] = myeuler(f,[0 20],[1;-1.5],5);
Here [0 20] is the time span you desire to consider, [1;-1.5] is the first value of the vector y and 5 is the number of steps. The output T contains times as well as Y contains values of the vector as the times. Try
Since the initial coordinate of the vector is the angle we merely plot its values
>theta = Y(:,1);>plot(T,theta)
In this plot it is clear that n = 5 isn’t adequate to represent the function. Type
> hold on
Afterwards redo the above with 5 replaced by 10. Next try 20, 40, 80, and 200. As you are able to see the graph becomes increasingly better as n increases. We are able to compare these calculations with Mat lab’s
built-in function with the commands
> [T Y]= ode45(f,[0 20],[1;-1.5]);>theta = Y(:,1);>plot(T,theta,’r’)
The problem with the Euler method:
You are able to think of the Euler method as finding a linear approximate solution to the initial value problem on each time interval. An apparent shortcoming of the method is that it makes the approximation based on information at the beginning of the time interval only. This problem is demonstrated well by the following IVP:
x¨ + x = 0, x(0) = 1, x? (0) = 0 .
You can effortlessly check that the exact solution of this IVP is
x(t) = cos(t).
If we create the standard change of variables
y1 = x, y2 = x? ,
then we obtain
y?1 = y2, y?2 = −y1.
Then the solution must be y1(t) = cos(t), y2(t) = sin(t). If we after that plot the solution in the (y1, y2) plane, we must get exactly a unit circle. We are able to solve this IVP with Euler’s method:
> g = inline(’[y(2);-y(1)]’,’t’,’y’)> [T Y] = myeuler(g,[0 4*pi],[1;0],20)> y1 = Y(:,1);> y2 = Y(:,2);>plot(y1,y2)
As you are able to see the approximate solution goes far from the true solution. Even if you raise the number of steps the Euler solution will eventually drift outward away from the circle because it doesn’t take into account the curvature of the solution.
The Modified Euler Method:
An idea which is alike to the idea behind the trapezoid method would be to consider f at both the beginning and end of the time step and take the average of the two. Doing this produces the Modified (or else Improved) Euler method represented by the following equations:
k1 = hf (ti, yi)k2 = hf (ti+ h, yi+ k1)yi+1 = yi+1/2(k1 + k2) .
Here k1 captures the information at the start of the time step (same as Euler) while k2 is the information at the end of the time step.
The following program executes the Modified method. It perhaps downloaded from the class web site.
function [T , Y] = mymodeuler(f,tspan,y0,n)% Solves dy/dt = f(t,y) with initial condition y(a) = y0% on the interval [a,b] using n steps of the modified Euler’s method.% Inputs: f -- a function f(t,y) that returns a column vector of the same% length as y% tspan -- a vector [a,b] with the start and end times% y0 -- a column vector of the initial values, y(a) = y0% n -- number of steps to use% Outputs: T -- a n+1 column vector contianing the times% Y -- a (n+1) by d matrix where d is the length of y% Y(t,i) gives the ith component of y at time t% parse starting and ending pointsa = tspan(1);b = tspan(2);h = (b-a)/n; % step sizet = a; y = y0; % t and y are the current variablesT = a; Y = y0’; % T and Y will record all stepsfor i = 1:nk1 = h*f(t,y);k2 = h*f(t+h,y+k1);y = y + .5*(k1+k2);t = a + i*h;T = [T; t];Y = [Y; y’];end
We are able to test this program on the IVP above:
> [T Ym] = mymodeuler(g,[0 4*pi],[1;0],20)> ym1 = Ym(:,1);> ym2 = Ym(:,2);>plot(ym1,ym2)
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