Consequences on a system of ODEs after performing operations

  • #1
SeM

Main Question or Discussion Point

Hi, I have derived a matrix from a system of ODE, and the matrix looked pretty bad at first. Then recently, I tried the Gauss elimination, followed by the exponential application on the matrix (e^[A]) and after another Gauss elimination, it turned "down" to the Identity matrix. This is awfully simple, and I may wonder if the system at hand is either completely redundant, or if the procedure of using Gauss elimination on a matrix for a system of ODEs, followed by e^A and then another Gauss elimination is a reasonable and valid approach to solve ODE systems, and does not "make critical information of the ODE vanish along the way"?

Thanks!
 

Answers and Replies

  • #2
BvU
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Not clear what you are doing. What is in the first matrix ? What does Gauss elimination give you when you do that ? Can you give an example ?
 
  • #3
SeM
Not clear what you are doing. What is in the first matrix ? What does Gauss elimination give you when you do that ? Can you give an example ?

The first matrix is:

A = \begin{bmatrix}
1.0545718E-68 & 1.1271866E-(61i) \\
1 & -1\\
\end{bmatrix}

the first operation in gauss elimination with rref[A] in matlab

that gives:

B = \begin{bmatrix}
1 & -1 \\
0 & 0\\
\end{bmatrix}

the exponential of B gives then:

C= \begin{bmatrix}
2.7183 & 0.3679 \\
1.0000 & 1.0000 \\
\end{bmatrix}

finally gauss elim of C with rref[C] gives the identity matrix

Does that make sense? I suspect MATLAB is rounding exponentially small values to 0, and this is not a correct result.


Thanks
 
  • #4
fresh_42
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Due to the very small first row, the matrix has been interpreted as ##A=\begin{bmatrix}0&0\\1&-1\end{bmatrix}## by the program, which is a singular matrix. It results in ##\exp(A)=\begin{bmatrix}1&0\\1+e^{-1}&e^{-1}\end{bmatrix}##.

Now that you exchanged the rows of ##A## for whatever reason, you have ##B=\begin{bmatrix}1&-1\\0&0\end{bmatrix}## instead, which leads to ##\exp(B)=\begin{bmatrix}e&e^{-1}\\0&1\end{bmatrix}##, if I made no mistakes by exponentiation.

So what do you want to do or say? As it is a numerical error, which led to that result, almost zero instead of equal to zero, we have algorithmic stability as one topic, or the general case of a Lie Group and its tangent space: ##\mathfrak{g} \stackrel{\exp}{\longrightarrow} G## as another.
 
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  • #5
SeM
Thanks, I prefer to rectify the numerical error! Cheers
 

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