Problem when solving ODEs with ode45

1 view (last 30 days)
Good day,
I have the following code where I simultaneously solve three second-order differential equations defined in 'ode3'.
My problem is that in the second last line in 'ode3', the denominator is Y3(3). This is the variable 'r' and is in the direction of a spring and stretches as a mass swings around at its endpoint. When I run the script I get NaN in my Y3 matrix, which then leads me to believe I'm dividing by zero some time. What I don't understand is why would that value become zero?
Any help is appreciated!
m1 = 4; %kg
m2 = 6; %kg
L = 1.5; %m
k = 100; %N/m
g = 9.81; %m/s2
% --------------------------------------------------------------
% Force parameters
% --------------------------------------------------------------
Fo = 100; %N
tf = 1;
F = @(t) (t<=tf)*Fo*sin(2*pi*t/tf);
ic3 = [0, 0, 0, 0, L, 0];
ode3 = @(t, Y3) [Y3(4);
(F(t) - m2*k*(L-Y3(3))*sin(Y3(2)))/m1;
(-F(t)*cos(Y3(2)) + m2*k*(L-Y3(3))*sin(Y3(2))*cos(Y3(2)) - 2*m1*Y3(6)*Y3(5) - m1*g*sin(Y3(2)))/(m1*Y3(3));
(-F(t)*sin(Y3(2)) + m2*k*(L-Y3(3))*sin(Y3(2))^2 + m1*Y3(3)*Y3(5)^2 + m1*g*cos(Y3(2)) + m1*k*(L-Y3(3)))/m1];
[t3, Y3] = ode45(ode3, [0, 3], ic3);

Accepted Answer

Ameer Hamza
Ameer Hamza on 28 Apr 2020
In your initial condition, you have
ic3 = [0, 0, 0, 0, L, 0];
which implies that at initial point, Y(3) = 0. But in this equation
(-F(t)*cos(Y3(2)) + m2*k*(L-Y3(3))*sin(Y3(2))*cos(Y3(2)) - 2*m1*Y3(6)*Y3(5) - m1*g*sin(Y3(2)))/(m1*Y3(3))
m1*Y3(3) comes in the denominator. At the initial point, you get a 0/0 situation. If you try some other initial condition, say
ic3 = [0, 0, 1, 0, L, 0];
You will get a finite answer.
You will need to check you system of ODE, if it is correctly written, and which value of initial conditions are allowed.
Ameer Hamza
Ameer Hamza on 28 Apr 2020
I am glad to be of help.

Sign in to comment.

More Answers (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by