Reputation: 3
I have the following 2n*π
-periodic function F(x) = sin(x/n)
and I need to graph the dx/dt = γ - F(x)
on the segment from 0
to 2pi
. So it should look like this. I tried to do it matlab this way:
gamma = 1.01;
n=3;
[t,phi] = ode45(@(t,x)gamma-sin(x/n), [0,400], pi);
[t1,phi1] = ode45(@(t,x)gamma-sin(x/n), [112,400], 0);
[t2,phi2] = ode45(@(t,x)gamma-sin(x/n), [231,250], 0);
figure();
plot(t, phi, 'k', t1, phi1, 'k', t2, phi2, 'k');
ylim([0 2*pi]);
yticks([0 pi 2*pi]);
yticklabels(["0" "\pi" "2\pi"]);
grid on; grid minor;
title('\itsin(x/n)')
but I only got something like this. So there the lines are not transferred, but "begin anew". does anyone here know how to do that?
Upvotes: 0
Views: 117
Reputation: 25972
I get a plot similar to your first sketch, and based on your code in the comments (in future, put such additions into the question itself, use formatting to mark it as addition, and cite it then in the comment) with the changes
pi
as initial point as seen in the drawing,[0, 200]
to get the same features as the drawing.gamma = 1.01; n=3;
opts = odeset('AbsTol',1e-6,'RelTol',1e-9,'MaxStep',0.1);
[t, phi] = ode45(@(t,x)gamma-sin(x/n), [0,200], pi, opts);
phi = mod(phi, 2*pi);
plot(t, phi, 'k');
ylim([0 2*pi]); yticks([0 pi 2*pi]); yticklabels(["0" "\pi" "2\pi"]);
grid on; grid minor;
title('\itsin(x/n)')
To get more elaborate, use events to get points on the numerical solution where it exactly crosses the 2*pi
periods, then use that to segment the solution plot (styling left out)
function [ res, term, dir ] = event(t,y)
y = mod(y+pi,2*pi)-pi;
res = [ y ];
dir = [1]; % only crossing upwards
term = [0]; % do not terminate
end%function
opts = odeset(opts,'Events',@(t,y)event(t,y));
sol = ode45(@(t,x)gamma-sin(x/n), [0,200], pi, opts);
tfs = [ sol.xe; sol.x(end) ]
N = length(tfs)
clf;
t0 = 0;
for i=1:N
tf = tfs(i);
t = linspace(t0+1e-2,tf-1e-2,150);
y = deval(sol,t); % octave: deval=@(res,t) interp1(res.x, res.y,t)
y = mod(y,2*pi);
plot(t, y);
hold on;
t0=tf;
end;
hold off;
Upvotes: 2