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ODE solver: how to integrate a system with a vector of parameters?

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Hello! I would like to ask how can I solve the system below for V varying from 1 to 1.2 with step 0.001? My aim is to plot bifurcation diagramm.
function [out] = chuaSmoothIris(~,in,alpha, beta, A, C, V)
x = in(1);
y = in(2);
z = in(3);
xdot = alpha*y - A*alpha*x^3 - C*alpha*x - V*alpha*x;
ydot = x - y + z;
zdot = -beta*y;
out = [xdot ydot zdot]';
% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
t = [0 400];
y = [0.004 0 0];
alpha = 15.6;
beta = 28;
A = 0.002;
C = -1.3;
V = 1:0.001:1.2;
[t,y] = ode45(@(t, y) chuaSmoothIris(t, y, alpha', beta', A', C', V'), t, y );

Respuesta aceptada

J. Alex Lee
J. Alex Lee el 24 de En. de 2020
Are you just looking for solving the ODE as many times as you have different values of V?
t = [0 400];
y = [0.004 0 0];
alpha = 15.6;
beta = 28;
A = 0.002;
C = -1.3;
V = 1:0.001:1.2;
for i = 1:length(V)
[t,y] = ode45(@(t, y) chuaSmoothIris(t, y, alpha', beta', A', C', V(i)), t, y );
end
  3 comentarios
J. Alex Lee
J. Alex Lee el 24 de En. de 2020
Oops. It's because you are overwriting the variable y.
t = [0 400];
y0 = [0.004 0 0]; % give this a special name
alpha = 15.6;
beta = 28;
A = 0.002;
C = -1.3;
V = 1:0.001:1.2;
for i = 1:length(V)
[t,y] = ode45(@(t, y) chuaSmoothIris(t, y, alpha', beta', A', C', V(i)), t, y0 );
end

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