Azzera filtri
Azzera filtri

Issue with solving ill-conditioned symbolic DAE system

27 visualizzazioni (ultimi 30 giorni)
Kindly I want to know if there any changes in some solvers like ODE113 or symbolic engines?
Recently I have updated matlab to 2023b version and tried to run a script that I have excuted it before with nothing changed on older version of matlab, the script contains solving a symbolic DAE system Ax = B, with ill-conditioned A matrix, using ODE113 of an older version it was working with conditiong number was around e-21. my results using older version was compared to some other CAEs (figure below) and they were identical, so I do not know what is the issue in 2023b, any help please?
  3 Commenti
John D'Errico
John D'Errico il 11 Giu 2024
I assume the A*x==b is the algebraic part of the DAE.
Mohamed Elshami
Mohamed Elshami il 25 Giu 2024 alle 11:54
Modificato: Mohamed Elshami il 25 Giu 2024 alle 11:54
@Torsten I was using 2019b to solve symbolic DAE system resulting from Multibody formulation, I found after a lot of trials that they should updated the symbolic engine, I am sure about this conclusion as I returned back to the older version and it works well even the model is ill-conditioned and using the same ode133 solver.

Accedi per commentare.

Risposta accettata

Mohamed Elshami
Mohamed Elshami il 8 Lug 2024 alle 18:48
Modificato: Mohamed Elshami il 8 Lug 2024 alle 18:51
Hi there, If any one inerested. The problem was not the solver type, instead the modeled system was over constrianed, after research I solved this problem by reducing the number of constraints equations by following another concept to remove any redundancy in DAEs set.

Più risposte (1)

John D'Errico
John D'Errico il 11 Giu 2024
You have a DAE, with a linear constraint system that happens to be ill-conditioned. I would try reducing the constraint system. You lose nothing in the process. That is, first verify the constraints are consistent. Thus you want to have
rank(A) == rank([A,b])
If not, then you have a fundamental problem. But assuming the above test succeeds, then you can reduce the problem, so the algebraic part is no longer ill-conditioned. (Well, hopefully. That sort of depends on where the ill-conditioning arises, and how.)
I'll give a simple example of a problem with an ill-conditioned constraint.
A = randn(3,2)*randn(2,5) % Just a randomly generated constraint set.
A = 3x5
0.5715 0.0694 0.3485 0.7862 0.2419 -0.2496 -0.2250 0.2155 -0.2810 0.0926 1.4721 0.8320 -0.3361 1.8158 -0.0422
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Note the rank of A is 2, even though there are 3 constraints.
size(A)
ans = 1x2
3 5
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
rank(A)
ans = 2
I'll pick an arbitrary right hand side that is consistent.
b = A*[2;3;5;7;11]
b = 3x1
11.2579 -1.0451 16.0055
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
The next test I would perform is to verify the problem is indeed consistent. (Even though I claim that to be the case, you may not believe me.)
Arank = rank(A)
Arank = 2
rank(A) == rank([A,b])
ans = logical
1
So the algebraic part is indeed consistent, and the rank of A is 2, as I would expect. Next, look at the singular values of A.
svd(A)
ans = 3x1
2.6977 0.6082 0.0000
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
As you can see, one of them is massively smaller than the others. That tells me there will be no issue in reducing the problem.
[Q,R] = qr(A,0)
Q = 3x3
-0.3575 -0.8837 -0.3022 0.1561 -0.3756 0.9135 -0.9208 0.2794 0.2722
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
R = 3x5
-1.5988 -0.8260 0.2186 -1.9968 -0.0331 0 0.2557 -0.4828 -0.0819 -0.2603 0 0 0.0000 -0.0000 0.0000
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
The trick now is to reduce the problem as:
Ahat = R(1:Arank,:);
bhat = Q'*b;
bhat = bhat(1:Arank);
Ahat will no longer be singular. It may still be moderately ill-conditioned though, and that will depend on how A was created, and the source of the ill-conditioning. But you can now use the reduced constraint pair Ahat and bhat instead of A and b.
  1 Commento
Mohamed Elshami
Mohamed Elshami il 24 Giu 2024 alle 18:23
Modificato: Mohamed Elshami il 24 Giu 2024 alle 18:25
Thank you very much! your comment enriched my knowledge by your valuable provided information, After a lot of trials I found that the issue is in the symbolic toolbox it seems to be handled completly different in versions newer than 2019b, I returned to that vesrion to check if I forgot somthing and unexpextedly the code worked as first time I wrote it, however itself still not working in newer version 2023,2024.
my DAE system includes a lot of symbolics in A, B matrices that would be first inversed and multiplied ...etc.

Accedi per commentare.

Prodotti


Release

R2023b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by