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LSODE questions
From: |
Douglas Eck |
Subject: |
LSODE questions |
Date: |
Tue, 23 Oct 2001 15:32:18 +0200 |
User-agent: |
Mozilla/5.0 (X11; U; Linux i686; en-US; rv:0.9.4) Gecko/20010913 |
Probably for John Eaton, but maybe someone else gets the idea. I know I don't.
These are all specific questions about how lsode is implemented in octave 2.1.34
In /usr/src/octave-2.1.34/src/DLD-FUNCTIONS/lsode.cc
at line 279 I see this:
Matrix output (nsteps, nstates + 1);
if (crit_times_set)
output = ode.integrate (out_times, crit_times);
else
output = ode.integrate (out_times);
Q1: Why is output initialized to columns=nstates+1. When I call
ode.integrate(), output always has columns=nstates after
integration.
Q2: More basically, why initialize it at all? Why not just a
"Matrix output;" declaration?
Q3: Is there a way to set the timepoints to compute without
having to initialize a new ode(...)? I'm using the solver in conjunction
with data acquisition and am computing only a few slices at a time. Each
time I pick up where i left off before. What I'd like to do is this:
initialize solver
while (computing) {
set timepoints of solver [tCurrent:.001:tCurrent+k]
integrate
grab last value from integration to use as initial value in next integration
tCurrent+=tCurrent+k;
}
PS this is the first time i've used the solver directly from C++. It's
really fast and very nice to work with! Excellent excellent John.
-D
--
Dr. Douglas Eck, http://www.idsia.ch/~doug
Istituto Dalle Molle di Studi sull'Intelligenza Artificiale (IDSIA)
Neural Networks, Rhythm Perception and Production, Dynamical Systems
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