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expm function problem for stiff matrix

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Michal
Michal el 10 de Jun. de 2021
Respondida: Noorolhuda wyal el 22 de Nov. de 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 comentarios
Matt J
Matt J el 10 de Jun. de 2021
If they are linear ODEs, maybe you could solve them symbolically?
Michal
Michal el 10 de Jun. de 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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Respuesta aceptada

Shadaab Siddiqie
Shadaab Siddiqie el 18 de Jun. de 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 comentario
Michal
Michal el 18 de Jun. de 2021
OK ... great! I am looking forward for any news regarding this topic.

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Más respuestas (2)

Bobby Cheng
Bobby Cheng el 12 de Ag. de 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 comentario
Fangcheng Huang
Fangcheng Huang el 1 de Jun. de 2022
Editada: Fangcheng Huang el 1 de Jun. de 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
Noorolhuda wyal el 22 de Nov. de 2022
a = -1e20;
b = eps;
c = 1;
A = [a,0,b;0,c,0;-b,0,a];
B=vpa(A);
expmA=expm(B)
expmA = 

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