# expm function problem for stiff matrix

16 Ansichten (letzte 30 Tage)
Michal am 10 Jun. 2021
Beantwortet: Noorolhuda wyal am 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 Kommentare4 ältere Kommentare anzeigen4 ältere Kommentare ausblenden
Michal am 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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### Akzeptierte Antwort

Shadaab Siddiqie am 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 KommentarKeine anzeigenKeine ausblenden
Michal am 18 Jun. 2021
OK ... great! I am looking forward for any news regarding this topic.

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### Weitere Antworten (2)

Bobby Cheng am 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 KommentarKeine anzeigenKeine ausblenden
Fangcheng Huang am 1 Jun. 2022
Bearbeitet: Fangcheng Huang am 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'

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Noorolhuda wyal am 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)
expmA = ##### 0 Kommentare-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden

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