[Maxima-bugs] [ maxima-Bugs-2846949 ] ilt(expr, s, t) cannot calculate some difficult expressions

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Bugs item #2846949, was opened at 2009-08-29 18:42
Message generated for change (Comment added) made by sf-robot
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Category: Lisp Core
Group: None
>Status: Closed
Resolution: Works For Me
Priority: 5
Private: No
Submitted By: Nobody/Anonymous (nobody)
Assigned to: Nobody/Anonymous (nobody)
Summary: ilt(expr,s,t) cannot calculate some difficult expressions

Initial Comment:
Hi! I use wxmaxima 0.8.2 in Ubuntu.
I have some rational function and I need to do inverse Laplace transformation. So I use ilt(), but it can't solve this. I tryed maxima 5.10.0 and 5.13.0, the result is the same (see in the file). In 5.17.1 there is a Lisp error when I'm trying to evaluate. But when I try a slightly changed function (see in the file too) it can be solved.

Mathcad 14 for Windows could even solve this with the first function. The result is in the file. But when I did a Laplace transformation with the result, mathcad gave me a very difficult solvation, and using wxmaxima I got a very simple result. Both of these results weren't the primary functon.  So I don't know the right answer but I really need to!

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>Comment By: SourceForge Robot (sf-robot)
Date: 2009-09-15 02:20

Message:
This Tracker item was closed automatically by the system. It was
previously set to a Pending status, and the original submitter
did not respond within 14 days (the time period specified by
the administrator of this Tracker).

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Comment By: Dieter Kaiser (crategus)
Date: 2009-08-31 23:22

Message:
Setting this bug report as pending and works for me.
Dieter Kaiser

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Comment By: Nobody/Anonymous (nobody)
Date: 2009-08-30 10:50

Message:
I think the matter is in the complexity of calculations. I tried to find
roots of the denominator, and the answer is so much difficult. I don't know
which algorithm Maxima is using for ilt, but I know one, it includes search
of denominator's roots. May be it's too complicated to do the
transformation, I suppose, the answer is giant. I should talk with my
lecturer and discuss some parameters of my transfer function.
Thank you very much  anyway :)

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Comment By: Dieter Kaiser (crategus)
Date: 2009-08-29 19:33

Message:
I have tried the examples with current Maxima 5.19post. Furthermore I have
reformulated the integral a bit. It is equivalent, but looks simpler.

You are right we get no solution for the following expression:

(%i2) ilt(1/(s^2*(1 + a*s)*(1 + b*s)*(1 + c*s) + d*s),s,t);
(%o2) 'ilt(-(a*b*c*s^3+((b+a)*c+a*b)*s^2+(c+b+a)*s+1)
            /(d*(a*b*c*s^4+((b+a)*c+a*b)*s^3+(c+b+a)*s^2+s+d)),s,t)
       +1/d

As you have observed, we get a solution, when we omit the extra term
+d*s:

(%i3) ilt(1/(s^2*(1 + a*s)*(1 + b*s)*(1 + c*s)),s,t);
(%o3) c^3*%e^-(t/c)/(c^2+(-b-a)*c+a*b)-b^3*%e^-(t/b)/((b-a)*c-b^2+a*b)
                                     
+a^3*%e^-(t/a)/((b-a)*c-a*b+a^2)+t-c-b-a

The Laplace transformation gives the original expression:

(%i4) laplace(%,t,s);
(%o4) c^3/((c^2+(-b-a)*c+a*b)*(s+1/c))-b^3/(((b-a)*c-b^2+a*b)*(s+1/b))
                                     
+a^3/(((b-a)*c-a*b+a^2)*(s+1/a))-c/s-b/s
                                      -a/s+1/s^2

We can see it more easy, when we factor the last result:

(%i5) factor(%);
(%o5) 1/(s^2*(a*s+1)*(b*s+1)*(c*s+1))

I can not see a bug. I have tried the first integral with Wolfram alpha,
but get no solution too. Perhaps you can post the expected answer for the
first integral.

Dieter Kaiser

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