Error in ode45: issue with length of initial conditions vector

Hey all, I was trying to solve the differential equation dUp/dt = c(U-Up) using ode45 in MATLAB.

I have U(A,x,y,t) as a user defined function. I then created the odefun file as follows:

function dydt = odefun(t,up,x,y,A,c,E)

global A c x y E

dydt = c.*(u(A,x,y,t) - up);

end

I then tried to calculate and plot Up vs t. The code is as follows:

A = 1.2;

E = 0.1;

c = 0.1;

x = 1;

y = 0;

tspan = [0 1];

y0 = 0;

[t,up] = ode45(@(t,up)odefun(t,up,x,y,A,c,E), tspan, y0);

plot(t,up)

I keep getting this error:

Error using odearguments

@(T,UP)ODEFUN(T,UP,X,Y,A,C,E) returns a vector of length 0, but the length of initial conditions vector is 1. The vector

returned by @(T,UP)ODEFUN(T,UP,X,Y,A,C,E) and the initial conditions vector must have the same number of elements.

I have seen some community posts that have addressed this issue, but those are generally for a system of equation. I am unable to figure out what the issue is in this case. Please help.

Answer 1

[Star Strider]

The ‘u’ function is either not being called correctly or not written correctly. In any event, it is miissing in the posted code.

Supplying a ‘u’ that is reasonable works —

A = 1.2;
E = 0.1;
c = 0.1;
x = 1;
y = 0;
tspan = [0 1];
y0 = 0;
[t,up] = ode45(@(t,up)odefun(t,up,x,y,A,c,E), tspan, y0);
plot(t,up)

function dydt = odefun(t,up,x,y,A,c,E)
    u = @(A, x, y, t) A*x*y+t;                              % Create 'u' (Not Posted)
    dydt = c.*(u(A,x,y,t) - up);
end

.

[Agniva]

Thanks for your response. Even I tried this with a simple 'u' expression and it worked. However, my 'u' expression is very complex and is made of many other functions. The 'u' expression is correct, as I have used it for evaluating other parameters as well. The 'u' expression has a few terms which can blow up at certain values of 't'. Is this a potential cause? I am willing to share all the files, please let me know how I can do that.

[Sam Chak]

@Agniva, Mathematically, the solution depends on your time-varying function 'u'.

This one is super simple 1st-order LINEAR differential equation:

.

If function 'u' is bounded or converging to zero as t → ∞, then the system is 100% stable.

However, you have also clarified that function 'u' has a few terms that will blow up at certain times of t.

Anyhow, follow the advice given in the Answer.

[Star Strider]

My pleasure!

‘The 'u' expression has a few terms which can blow up at certain values of 't'.’

That is most likely the reason that it is failing. If you know what the ‘certain values of 't'’ are, one optiion is to integrate up to each of them (perhaps using an Events function, see ODE Event Location for details), stop the integration, skip past the discontinuity, and re-start the integration on the other side of it. That is the only approach that I can think of.

The other approach of course is to examine your code to determine the reason that ‘u’ fails at those times, and correct the code.

[Agniva]

@Sam Chak Thanks for your response. I would like to clarify that 'u' is a periodic function w.r.t. 't'. However, some terms in construction of 'u' do blow up, but because of the presence of other terms, 'u' is perfectly bounded.

[Sam Chak]

Hi @Agniva,

I do not understand how the function u can be both unbounded and perfectly bounded at the same time. The adjective "bounded" is a mathematical property and does not imply that the function u has finite values under certain conditions. As t approaches specific values, if the function can take on arbitrarily large values, then it is unbounded.

Perhaps you could consider posting the entire function u here and exploring how to use the Events feature in the ODE solver, as advised by @Star Strider.