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Re: System of ODE's
From: |
Dan Boghiu |
Subject: |
Re: System of ODE's |
Date: |
Fri, 28 Sep 2001 09:45:15 +0300 |
The solution for ODE's is the function lsode.
For details: help lsode
Dan
----- Original Message -----
From: "Orsila Heikki" <address@hidden>
To: "Dennis Bayrock" <address@hidden>
Sent: Thursday, September 27, 2001 22:31
Subject: Re: System of ODE's
> On Wed, 26 Sep 2001, Dennis Bayrock wrote:
> > I am a research scientist involved in modeling yeast fermentations
> > for the production of alcohol. I have a system of ODE's to model and am
> > wondering if Octave is capable of handling such a system. For eg.
> >
> > dX/dt= 1+ dS/dt
> > dS/dt = -2.5 * dX/dt
> > dP/dt = 5 * dS/dt - 3 * dX/dt
>
> Well, that kind of trivial equation sets can be solved easily with little
> handwork..
>
> Assume we have equation set:
>
> x1 = a10 + a11 * x1 + a12 * x2 + ... + a1n * xn
> x2 = a20 + a21 * x1 + a22 * x2 + ... + a2n * xn
> ...
> xn = an0 + an1 * x1 + an2 * x2 + ... + ann * xn
>
> Denote x = [x1,x2,...,xn]' and b=[a10,a20,...,an0]'
>
> And aij is item from matrix A row i column j.. We get equation:
>
> x = A*x + b
> =>
> (I-A)*x = b
> =>
> x = inv(I-A) * b = (say) = c
>
> Now we have that:
>
> x1 = c(1)*t + C1
> x2 = c(1)*t + C2
> ...
> xn = c(n)*t + Cn
>
> Parameters C1-Cn must be solved from initial-value information. Let that
> be vector C, then C = x(0).
>
> To apply this to your example problem: Let dX/dt = x1, dS/dt = x2
> and dP/dt = x3 ...
>
> b=[1;0;0]
> A=[0,1,0;-2.5,0,0;-3,5,0]
>
> A =
>
> 0.00000 1.00000 0.00000
> -2.50000 0.00000 0.00000
> -3.00000 5.00000 0.00000
>
> I=eye(3)
>
> c = inv(I-A)*b
>
> c =
>
> 0.28571
> -0.71429
> -4.42857
>
> =>
>
> X = x1 = 0.28571 * t + X(0)
> S = x2 = -0.71429 * t + S(0)
> P = x3 = -4.42857 * t + P(0)
>
> I don't have any experience with ODEs so if this is wrong, please tell
> me..
>
>
> Heikki Orsila 32 bittiä - entä sitten?
> address@hidden http://www.pjoy.fi/lehdet/9212pj.htm
> http://www.ee.tut.fi/~orsila - Petteri Järvinen (1992)
>
>
>
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>
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