[Gsll-devel] ODE apply-step: argument passing problem
Mirko Vukovic
mirko.vukovic at gmail.com
Wed Dec 2 03:23:36 UTC 2009
Hi Liam (& others)
I am getting funny behavior with the ODE stepper:
(As refresher, I am solving a 2nd order system that should reproduce a
sinusoid)
A trace of the dydt function (sin-ode below) shows that it is always called
with arguments (time 0 0), even though the initial conditions are (time 0 1)
(see the let form).
It looks as if the initial conditions (stored in vector y below) are not
passed as arguments to the dydt function (sin-ode). Or am I missing
something?
I tried both the rk2 and rk4 steppers and different step sizes.
Thanks,
Mirko
Here is the source:
;;;; Example of a simple time-stepper
(defun sin-ode (time y z)
"Define ODE for a sinusoid
y''=-y
or as a system:
y'=z
z'=-y
with initial conditions:
y0=0
z0=1 "
(declare (ignore time))
(values z (- y)))
(defun sin-ode-jacobian (time y z)
(declare (ignore time y z))
(values 0d0 0d0
0d0 1d0
-1d0 0d0))
#|
(let ((time 0d0)
(delta-t 0.01d0)
(stepper-type +step-rk2+))
(let ((stepper (make-ode-stepper stepper-type 2 #'sin-ode
#'sin-ode-jacobian))
(y (make-marray 'double-float :dimensions 2))
(dydt-in (make-marray 'double-float :dimensions 2))
(dydt-out (make-marray 'double-float :dimensions 2))
(yerr (make-marray 'double-float :dimensions 2)))
(setf (maref y 0) 0d0 ;; y(0)=0
(maref y 1) 1d0);; y'(0)=1
(multiple-value-bind (yp zp) (sin-ode time (maref y 0) (maref y 1))
(setf (maref dydt-out 0) yp
(maref dydt-out 1) zp))
(dotimes (i 2)
(incf time delta-t)
(setf (maref dydt-in 0) (maref dydt-out 0)
(maref dydt-in 1) (maref dydt-out 1))
(format t "Y-in: ~12,4f, ~a, ~a~%" time (maref y 0) (maref y 1))
(format t "dydt-in:~12,4f, ~a, ~a~%" time (maref dydt-in 0) (maref
dydt-in 1))
(apply-step stepper time y delta-t yerr dydt-in dydt-out)
(format t "Y-out: ~12,4f, ~a, ~a~%" time (maref y 0) (maref y 1))
(format t "dydt-out:~12,4f, ~a, ~a~%~%" time (maref dydt-out 0) (maref
dydt-out 1)))
y))
|#
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