%% How to plot the tail of a solution curve
%% This is the same system used in the demonstration for 3 dimensional systems
% Consider the nonlinear system
%%
% 
% $$x'=-x+3z$$
% 
% $$y'=-y+2z$$
%
% $$z'=x^2-2z.$$
% 
%% Solution using ode45.
% This is the three dimensional analogue of the Section 14.3.3 in
% _Differential Equations with MATLAB_. Think of $x,y,z$ as the coordinates
% of a vector *x*. In MATLAB its coordinates are *x(1),x(2),x(3)* so
% I can write the right side of the system as a MATLAB function
f = @(t,x) [-x(1)+3*x(3);-x(2)+2*x(3);x(1)^2-2*x(3)];
%%
% The numerical solution on the interval $[0,1.5]$ with
% $x(0)=1,y(0)=1/2,z(0)=3$ is
[t,xa] = ode45(f,[0 1.5],[0 1/2 3]);
%% 3 D plot
% I can plot the solution curve $(x(t),y(t),z(t))$ in phase space using
% *plot3*.
plot3(xa(:,1),xa(:,2),xa(:,3))
grid on
title('Solution curve')
%% Plotting the tail 
% Suppose I just want to plot the part which corresponds approximately to
% the time interval $[1,1.5]$. Remember that the $t$ produced by *ode45* is
% a vector with a lot of components. I want to know which component
% corresponds approximately to $t=1$. One way is to look at the values of
% $t$, but with a very long list of values this wouldn't be easy. So first
% I'll find how many components $t$ has, using the command *size*.
size(t)
%%
% This tells me that $t$ has 69 rows and 1 column. Now I do some guessing:
% *t(46)* is two-thirds down the list of components of $t$ so I can look at
% it.  
t(46)
%%
% I look at components with slightly larger index:
t(47:50)
%%
% I see that *t(49)* and *t(50)* are the closest, one a little larger, the
% other a little smaller than 1. I'll use 49 as my index. (You can probably
% do this more elegantly using the *Events* option.) So I can plot the tail
% of the solution curve with the following command.
plot3(xa(49:69,1),xa(49:69,2),xa(49:69,3))
grid on
title('Tail of solution curve')