Octave: is it allowed to have a for() cycle in a function definition?

Question

I have a function func() definition which contains a for() cycle.

Can I then pass func() to lsode() to be resolved as an ODE?

Update: Example of the code.

int = [0; 0];
slp = [0; 1];
x = [1:10];
r = zeros(1,10);
for id = 1 : 2;
 r = r + int(id) + slp(id)*x;
end
plot(x,r);
function ydot = fnc(y, t, int, slp)
 ydot = zeros(1,10);
 for jd = 1 : 2;
  ydot = ydot + int(jd) + slp(jd)*t;
 end
end
p = lsode(@(y,t) fnc(y,t,int,slp), 1, x);

error: lsode: inconsistent sizes for state and derivative vectors error: called from: error: /path$/test_for_function.m at line 15, column 3

Then debug your program and determine the shape of the state and derivative vectors, in the main if they are row or column vectors. Both should be the same. You could add a print statement in `fnc` for the shape of `y` and `ydot`. — Lutz Lehmann, Mar 17 '18 at 15:06
Did you want to pass `r` as initial value? That should be `p = lsode(@(y,t) fnc(y,t,int,slp), r, x);`. Does the error change with that correction? — Lutz Lehmann, Mar 17 '18 at 15:17

Lutz Lehmann · Accepted Answer · 2018-03-17T15:18:46.437

1

The error message has nothing to do with the computations inside the ODE functions. It is telling you that the shape of ydot is incompatible with the shape of the state vector that is established by the initial value vector.

First I can not identify in your code example the initial value vector. You seem to construct the array r for that, but you pass the scalar constant 1.

And then it could be as simple as switching the shape dimensions in the construction of ydot. Or use ỳdot = zeros_like(y) or similar if available.

edited Mar 17 '18 at 15:18

answered Mar 17 '18 at 15:13

Lutz Lehmann

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I should have explained better, 'r' is to check that the 'for' cycle works as intended. The initial value should be 'y(1) = 0' – Fabio Capezzuoli Mar 17 '18 at 15:38
Your differential equation is scalar? From your code it looks like it is a 10-dimensional system. – Lutz Lehmann Mar 17 '18 at 17:08
I'm afraid the issue is not clear. Why would it be a 10-dimensional system? Isn't the initial condition one scalar per ODE? – Fabio Capezzuoli Mar 18 '18 at 02:24
Because `ydot = zeros(1,10);` declares a 10-dimensional vector. The dimension is nowhere changed, all scalars are duplicated to fit this dimension. – Lutz Lehmann Mar 18 '18 at 06:04
I think I'm getting it. I'll try some tinkering and debugging later. – Fabio Capezzuoli Mar 18 '18 at 06:13

score 0 · Answer 2 · answered Mar 19 '18 at 02:07

After some debugging, tinkering and thinking about how lsode works, I found how to make my code work as intended:

%Start code
int = [0; 0];
slp = [0; 1];
x = [1:10];
for id = 1 : 2;
 r = int(id) + slp(id)*x;
end   %A complicated way to draw a line
 figure(1); clf(1); 
 plot(x,r);
 hold on;
function ydot = fnc(y, t, int, slp)
 for jd = 1 : 2;
  ydot = [int(jd) + slp(jd)*t];
 end
end
p = lsode(@(y,t) fnc(y,t,int,slp), 1, x);
 plot(x, p, '-r');
%EOF

And the result is:

As it should be.

Octave: is it allowed to have a for() cycle in a function definition?

2 Answers2