lsode and random term

[Matthias Brennwald]

On 23.11.2008, at 06:32, help-octave-request at cae.wisc.edu wrote:

> Message: 1
> Date: Sat, 22 Nov 2008 13:10:42 -0500
> From: "William Christopher Carleton" <williamcarlet at trentu.ca>
> Subject: lsode and random term
> To: help-octave at cae.wisc.edu
> Message-ID: <1227377442.94644c3cwilliamcarlet at trentu.ca>
> Content-Type: text/plain; charset="UTF-8"
>
> Hi all,
>
> I have a function (below), which includes a random term (I wrote a  
> function 'random(x)' that just uses rand() to produce a value  
> between -x and x b/c I couldn't find such an existing function in  
> Octave's manual) and when I use lsode to solve it occasionally  
> works, but often produces as error about convergence;
>
> LSODE--  AT T (=R1) AND STEP SIZE H (=R2), THE
>        CORRECTOR CONVERGENCE FAILED REPEATEDLY
>        OR WITH ABS(H) = HMIN
>       In above,  R1 =  0.3220694060927D+03   R2 =  0.1207113904987D-05
> error: lsode: repeated convergence failures (t = 322.069perhaps bad  
> jacobian supplied or wrong choice of integration method or tolerances)
> error: evaluating assignment expression near line 10, column 2
>
> Can anyone tell me why the random term has caused this problem and  
> what I can do about it? I want the random term, or I suppose an  
> unpredictable function with magnitude of fluctuations determined by  
> a variable selected before solving, to simulate unpredictable  
> environmental fluctuations. Thanks,
>
> Chris
>
> Octave Code:
>
> function xdot=f(x,t);
> k=0.01;
> c=.9;
> p=0.025;
> l=0.5;
> C=1000;
> r=0.001;
> s=1.1;
> d=0.002;
> v=1;
> E=random(1) * v;
> xdot(1)=x(1) * (k * (x(3) - x(1)));
> xdot(2)=(x(2) * -d * x(1) * E) + (r * x(2) * ((C - x(2)) / x(2)));
> xdot(3)=x(2) * (x(1) * l) * p * (((x(2) * s) - x(3)) / (x(2) * s))  
> - (c * x(1));
> endfunction


I've tried something similar, and it did not work either. With the  
random term, the function value will never be the same for a given  
argument. So, if the solver evaluates the function twice for (almost)  
the same argument, the function value will not be (almost) the same.  
In other words, there solver cannot converge.

I haven't looked closely into your function with the random term, but  
if the random part is here to introduce some noise, then a solution  
would be to define the noise-term values before starting the solver.  
With this approach, the noise term will always be the same for a  
given argument, so the solver will be able to converge.

IHTH
Matthias