Solving problem with ODE45

clc;
clear all;
close all;
[t1,y1] = ode45(@vdp1,[0 20],[2; 0]);
t2(1,:)=t1(1,:);
y2(1,:)=y1(1,:);
figure
plot(t1,y1(:,1),'-o',t1,y1(:,2),'-o')

for i=2:size(t1)
    [a,b]=ode45(@vdp1,[0 t1(i) 2*t1(i)],y2(i-1,:));
    t2(i,:)=a(2,:);
    y2(i,:)=b(2,:);
end
figure
plot(t2,y2(:,1),'-o',t2,y2(:,2),'-o')

Hey guys, when I am solving a function step by step with ODE45, the result is completely different from the direct solution. Can someone tell me why?

[Jan]

Note: for i=2:size(t1) is fragile. size() replies a vector, but the colon operator uses the first element only. Better: for i=2:size(t1, 1) or for i=2:numel(t1).

Please provide vdp1 to allow use to run your code.

[Yi Li]

Thank you!

function dydt = vdp1(t,y)
%VDP1  Evaluate the van der Pol ODEs for mu = 1
%
%   See also ODE113, ODE23, ODE45.

%   Jacek Kierzenka and Lawrence F. Shampine
%   Copyright 1984-2014 The MathWorks, Inc.

dydt = [y(2); (1-y(1)^2)*y(2)-y(1)];

Answer 1

[Jan]

In the loop you are calculating the integration from 0 to 2*t(i) with the initial value set to y2(i-1, :). But this is another integration, because you start from t=0 with another initial value. Instead of the time interval [0 t1(i) 2*t1(i)] the interval [t1(i-1), t1(i)] would create the same output:

for i = 2:numel(t1)
    [a,b]   = ode45(@vdp1, [t1(i-1), t1(i)], y2(i-1, :));
    t2(i,:) = a(end, :);
    y2(i,:) = b(end, :);
end
figure
plot(t2, y2(:, 1), '-o', t2, y2(:, 2), '-o');

[Yi Li]

Thank you!