expm function problem for stiff matrix
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Michal
on 10 Jun 2021
Answered: Noorolhuda wyal
on 22 Nov 2022
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
Accepted Answer
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.
More Answers (2)
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
1 Comment
Fangcheng Huang
on 1 Jun 2022
Edited: Fangcheng Huang
on 1 Jun 2022
last line, Strange, when use matlab2022 it is right, but when use matlab 2020a, need to change Q*diag(exp(diag(T)))*Q'
Noorolhuda wyal
on 22 Nov 2022
a = -1e20;
b = eps;
c = 1;
A = [a,0,b;0,c,0;-b,0,a];
B=vpa(A);
expmA=expm(B)
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