expm function problem for stiff matrix

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For very specific matrix A:
a = -1e20;
b = eps;
c = 1;
A = [a,0,b;0,c,0;-b,0,a];
disp('A:'), disp(num2str(A))
A:
-1e+20 0 2.220446049250313e-16
0 1 0
-2.220446049250313e-16 0 -1e+20
is known exact matrix exponential as:
expA = exp(a)*( ...
[1,0,0;0,0,0;0,0,1]*cos(b)+ ...
[0,0,1;0,0,0;-1,0,0]*sin(b))+ ...
[0,0,0;0,exp(c),0;0,0,0];
expA =
0 0 0
0 2.7183 0
0 0 0
the Matlab function expm give wrong result:
expm(A)
ans =
0 0 0
0 1 0
0 0 0
but direct computing of expm(A) via definition gives again right result:
[V,D] = eig(A);
expmA = V*diag(exp(diag(D)))/V
expmA =
0 0 0
0 2.7183 0
0 0 0
So, what is wrong with expm function? Bad implementation of Pade's approximation?
  5 Comments
Michal Kvasnicka
Michal Kvasnicka on 10 Jun 2021
Symbolic solutions always ends on matrix exponentials and integration, which must be finally evaluated always numerically, so in this case by multi-precision arithmetic, which is sometimes very slow (especially with VPA in MATLAB). So, this problem is really hard ... :)

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Accepted Answer

Shadaab Siddiqie
Shadaab Siddiqie on 18 Jun 2021
From my understanding you are getting wrong result for certain cases wile using expm function. This issue has been forwarded to the development team for further investigation.
  1 Comment
Michal Kvasnicka
Michal Kvasnicka on 18 Jun 2021
OK ... great! I am looking forward for any news regarding this topic.

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More Answers (1)

Bobby Cheng
Bobby Cheng on 12 Aug 2021
This is a weakness of the scaling and squaring algorithm. Inside EXPM, which you can read the implementation, there are special treatments for diagonal to deal with extreme cases, but it is only triggered if the input is of the Schur form due to performance. You can call SCHUR to create the Schur factorization, and pass the Schur form to EXPM to trigger the special diagonal treatment.
>> a = -1e20;
>> b = eps;
>> c = 1;
>> A = [a,0,b;0,c,0;-b,0,a];
>> [Q T] = schur(A);
>> Q*expm(T)*Q'
ans =
0 0 0
0 2.7183 0
0 0 0

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