Subject: Teaching differential equations with Maxima
From: Wolfgang Jenkner
Date: 08 Feb 2003 06:54:23 +0100
"Nikolaos I. Ioakimidis" <ioakimidis@otenet.gr> writes:
> Yet, wishing to have further computational power from Maxima, I
> would be extremely thankful to you if you could bring to my attention
> any further Maxima command or package directly concerning
> differential equations (mainly) and Laplace transforms.
The following ftp directory seems to contain very interesting stuff
ftp://ftp.mines.colorado.edu/pub/papers/math_cs_dept/software
In particular, the following sub-directories contain Macsyma packages:
ftp://ftp.mines.colorado.edu/pub/papers/math_cs_dept/software/symmetryftp://ftp.mines.colorado.edu/pub/papers/math_cs_dept/software/diffsymftp://ftp.mines.colorado.edu/pub/papers/math_cs_dept/software/hirotaftp://ftp.mines.colorado.edu/pub/papers/math_cs_dept/software/painleve/macsyma
You might prefer to start by browsing
http://www.mines.edu/fs_home/whereman/
Actually, at this point I just wanted to write something like `Of
course, those packages were written for the commercial Macsyma, so
they can't be expected to work out of the box in Maxima'. Then I
decided to run a random test. And it did work out of the box.
You can perform such a test run like this (on a Unix-like system):
If you are in the directory containing the painleve sub-directory do
cd painleve/macsyma/single
ln -s np_sing.max painsing.max
maxima
At the maxima prompt
file_search_maxima:cons("./###.{max,com,dat}",file_search_maxima)$
batch("p_asg.com");
The result of the run will then be appended to p_asg.out, so that you
can compare it with what the authors obtained.
Here is the result of the test run I did
batching /home/wolfgang/ftp.mines.colorado.edu/pub/papers/math_cs_dept/software/painleve/macsyma/single/np_exec.max
(C15) exec_painleve(eq, ALPHA, do_resonances, max_resonance, do_simplification)
You are using the simplification suggested by KRUSKAL
You selected G(T,X,...) = X - H(T,...)
----------------------------------------------------------------
2 2 3
PAINLEVE ANALYSIS OF EQUATION, 2 f f - 2 (f ) + 2 f f - 2 (f ) - f
x x x t t t
+ f = 0
----------------------------------------------------------------
ALPHA
SUBSTITUTE u g FOR f IN ORIGINAL EQUATION.
0
MINIMUM POWERS OF g ARE [2 ALPHA - 2, 3 ALPHA, ALPHA]
2 ALPHA - 2 2 2
* COEFFICIENT OF g IS - 2 u ALPHA ((h ) + 1)
0 t
NOTE : THIS TERM VANISHES FOR ALPHA = 0 ,
VERIFY IF ALPHA = 0 LEADS TO DOMINANT BEHAVIOR,
IF IT DOES THEN RUN THE PROGRAM AGAIN WITH THIS USER
SUPPLIED VALUE OF ALPHA.
HENCE, PUT BETA = 0 .
3 ALPHA 3
* COEFFICIENT OF g IS - u
0
ALPHA
* COEFFICIENT OF g IS u
0
----------------------------------------------------------------
FOR EXPONENTS ( 2 ALPHA - 2 ) AND ( 3 ALPHA ) OF g,
WITH alpha = - 2 , POWER OF g is - 6 ----> SOLVE FOR u
0
2 2 1
TERM u (4 (h ) - u + 4) -- IS DOMINANT
0 t 0 6
g
IN EQUATION.
----------------------------------------------------------------
2
1 ) WITH u = 4 (h ) + 4 ----> FIND RESONANCES
0 t
ALPHA r + ALPHA
SUBSTITUTE u g + u g FOR f IN EQUATION
0 r
2 2 r - 6
TERM ( 8 ((h ) + 1) (r - 2) (r + 1) ) u g IS DOMINANT
t r
IN EQUATION.
THE ONLY NON-NEGATIVE INTEGRAL ROOT IS [r = 2]
WITH MAXIMUM RESONANCE = 2 ----> CHECK RESONANCES.
2
====
\ k - 2
SUBSTITUTE POWER SERIES > g u FOR f IN EQUATION.
/ k
====
k = 0
2
WITH u = 4 (h ) + 4
0 t
1 2 2
* COEFFICIENT OF -- IS 16 ((h ) + 1) (4 h - u )
5 t t t 1
g
u = 4 h
1 t t
1
* COEFFICIENT OF -- IS 0
4
g
u IS ARBITRARY !
2
COMPATIBILITY CONDITION IS SATISFIED !
----------------------------------------------------------------
FOR EXPONENTS ( 2 ALPHA - 2 ) AND ( ALPHA ) OF g,
WITH alpha = 2 , POWER OF g is 2 ----> SOLVE FOR u
0
2 2
TERM - u (4 u (h ) + 4 u - 1) g IS DOMINANT
0 0 t 0
IN EQUATION.
----------------------------------------------------------------
1
1 ) WITH u = ----------- ----> FIND RESONANCES
0 2
4 (h ) + 4
t
ALPHA r + ALPHA
SUBSTITUTE u g + u g FOR f IN EQUATION
0 r
r + 2
TERM ( (r - 2) (r + 1) ) u g IS DOMINANT
r
IN EQUATION.
THE ONLY NON-NEGATIVE INTEGRAL ROOT IS [r = 2]
WITH MAXIMUM RESONANCE = 2 ----> CHECK RESONANCES.
2
====
\ k + 2
SUBSTITUTE POWER SERIES > g u FOR f IN EQUATION.
/ k
====
k = 0
1
WITH u = -----------
0 2
4 (h ) + 4
t
4 2
h + 4 u (h ) + 8 u (h ) + 4 u
3 t t 1 t 1 t 1
* COEFFICIENT OF g IS - -------------------------------------
2 2
4 ((h ) + 1)
t
h
t t
u = - --------------
1 2 2
4 ((h ) + 1)
t
4
* COEFFICIENT OF g IS 0
u IS ARBITRARY !
2
COMPATIBILITY CONDITION IS SATISFIED !
----------------------------------------------------------------
FOR EXPONENTS ( 3 ALPHA ) AND ( ALPHA ) OF g,
POWER OF g IS NOT MINIMAL
-- SKIP THIS VALUE OF ALPHA.
----------------------------------------------------------------
(C16) OUTPUT()
----------------------------------------------------------------
AT THE END OF THE COMPUTATIONS THE FOLLOWING ARE AVAILABLE:
* U VALUE(S)
(type uval[j,k,l] where 1 <= j <= 2 and 0 <= k <= [2, 2]
and 1 <= l <= [1, 1] )
j stands for j_th alpha,k stands for u[k],l stands for
l_th solution set for u[0]
* ALPHA VALUE(S)
(type alpha[j] where 1 <= j <= 2 )
j stands for j_th alpha
* COMPATIBILITY CONDITION(S)
(type compcond[j,k] where 1 <= j <= 2 and 1 <= k <= [1, 1] )
j stands for j_th alpha,k stands for k_th solution set for u[0]
* RESONANCE(S)
(type res[j,k] where 1 <= j <= 2 and 1 <= k <= [1, 1] )
j stands for j_th alpha,k stands for k_th solution set for u[0]
----------------------------------------------------------------
TO SEE THIS MENU AGAIN JUST TYPE < output() >
----------------------------------------------------------------
(C17) CLOSEFILE()
;; Dribble of #<IO TERMINAL-STREAM> finished 2003-02-08 04:42:16
Wolfgang