solving coupled differential equations using ode45 for "n" modes (2024)

to make it hopefully clearer, when i run for n=1 and n=2 i get the du_1 plots in the figure. so the one of n=1 is slighly bigger in amplitude (the difference increases when i include more modes). since i am plotting the first modes, i think i should get the same results.

I have printed before the results per timestep, for n=1 and i=1 and then for n=2 but for i=1 only. intialy they are the same but then change after some time. if i look at the code though, i dont see where i am making the mistake.

Is there a publication from which you took your equations ?

Yes, I am attaching it here. Equation 1 in my post is eq.33 in the paper, eq.2 is eq.38, eq.3 - eq.5 are eq.34 - eq.37. Equation 6 in the post is eq.20 in the paper.

Why don't you follow the article and set up system (40) ?

In the system (40), the equations I wrote are expressed in matrix form. There are N+1 second-order differential equations (and 1 for x_c): N equations for T''_i and one equation for x'' . When these are written in state-space form, we have 2N+2 equations. In my approach, I solve for 4N equations, where the first 2N pertain to T_i and the last 2N to x_i. I solve the equation for x for each mode because the equations are coupled. To find the solutions, I use a for loop to iterate over each N and employ `ode45`. I m not sure if there is another way to do it, but i am indeed solving the equaitons in (40).

I m not sure if there is another way to do it, but i am indeed solving the equaitons in (40).

You don't. The number of equations to be solved should be 1 algebraic equation for x_c, 2 first-order ordinary differential equations for xi and xidot and 2*N first-order ordinary differential equations for T1,...,TN,T1dot,...TNdot. In total 2*N+3 equations from which one is algebraic and 2*N+2 are differential.

intially i solved like that so for 2*N+2 differential equation and had the same problem. The reason i changed to 4*N differential equation is because the equations of T_i depend on x. Starting at the first time step, when i enter the for loop and solve for i=1:n, i have the solutions of T per each mode but only one solution of x (where the f term in the equation of x depends of T_1 to T_N). When i go out of the loop and move to the new time step but start with i=1, i wanted the solution of x inside T to be of the first mode only (and not have the acumulation of all the modes from the first time step), since i was solving T for the first mode again (new time step). that is why i started to solve x for each mode as well.

when i solved for 2*N+2, i had almost the same function but with ydot_3 and ydot_4 not depending on i, as shown below.

and in the main function the initial conditions are defined for 2*N+2

the rest of the code is pretty much the same. in this case, the solutions seem further away from the ones in the paper.

intially i solved like that so for 2*N+2 differential equation and had the same problem.

ode45 must solve for 2*N+3 equations where one of them is algebraic. Look at the mass matrix in equation (40). x_c is an algebraic solution variable of ode45, and the respective row of the mass matrix is set to 0. No need to use x_c_prev or find or min or whatever you use in your code to cope with and determine x_c.

When i go out of the loop and move to the new time step but start with i=1, i wanted the solution of x inside T to be of the first mode only (and not have the acumulation of all the modes from the first time step), since i was solving T for the first mode again (new time step). that is why i started to solve x for each mode as well.

I don't know what you mean here. Equation (35) cleary shows how the f_i are computed using x_c, xi, xidot, T1,...,TN,T1dot,...,TNdot. And no more than these 2*N+3 variables are needed.