expm function problem for stiff matrix

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?

[Matt J]

I think the better question would be, why does the eigendecomposition method succeed. The matrix elements span 36 orders of magnitude, i.e., beyond what double float precision should be expected to handle.

[Michal]

Yes Matt, you are right :)

[Michal]

BTW, the crucial question is: What is the proper method to solve linear ODE systems with similar system matrix? These matrices are very common in nuclear decay kinetic problems, where decay constants may differ by many orders (10-30).

[Matt J]

If they are linear ODEs, maybe you could solve them symbolically?

[Michal]

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 ... :)

Answer 1

[Shadaab Siddiqie]

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.

[Michal]

OK ... great! I am looking forward for any news regarding this topic.

Answer 2

[Bobby Cheng]

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

[Fangcheng Huang]

last line, Strange, when use matlab2022 it is right, but when use matlab 2020a, need to change Q*diag(exp(diag(T)))*Q'

Answer 3

[Noorolhuda wyal]

a = -1e20;
b = eps;
c = 1;
A = [a,0,b;0,c,0;-b,0,a];
B=vpa(A);
expmA=expm(B)

expmA =