Timeline for Nonlinear system of equations whereas most of the equations are linear. How to minimise operation?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 10, 2014 at 21:45 | comment | added | Richard Zhang | For the block-diagonalization strategy described above, the answer is no. If the nonlinear part is in the middle, then the nonlinear Schur complement will involve the middle and the bottom blocks. If your matrix is banded (e.g. 3-D grid), it might be cheaper just to do a nested dissection over the entire grid, rather than trying to precompute the linear portion. | |
| Nov 10, 2014 at 18:28 | comment | added | Meisam Jalalvand | Thanks again Richard! That's helpful. One final question: The matrix $K$ is banded. Pivoting $D(x,y,z)$ to the bottom increases the band of the matrix significantly and compromises the operation saving! I really need to keep $D(x,y,z)$ at its position to reduce the band-width of $K$. Isn't any way to do block factorisation while keeping $D(x,y,z)$ at the middle as it is? | |
| Nov 9, 2014 at 4:43 | comment | added | Richard Zhang | For your second part, you will have to pivot your nonlinear part to the bottom corner, and treat the entire $A,E$ blocks as your "A" block. | |
| Nov 9, 2014 at 4:41 | comment | added | Richard Zhang | You never explicitly compute $A^{-1}.$ You compute its LU factorization $A=LU$ and then every time you need to perform a matrix-product with the inverse, $X=A^{-1}B$, you instead perform via the LU, as in $X=L^{-1}(U^{-1}B)$. The decomposition is expensive but only needs to be done once. | |
| Nov 8, 2014 at 23:07 | comment | added | Meisam Jalalvand | Many thanks Richard! That is a very good start for me! I have now two particular question and I appreciate your answer. 1) Do we need to calculate $A^{-1}$? Since $n$ is very big, finding $A^{-1}$ is a quite heavy calculation. 2) What if $K$ is symmetric and the nonlinear part of the matrix is in the middle $\begin{bmatrix}A & B & 0\\ B & D(x,y,z) & C \\ 0 & C & E \end{bmatrix}\begin{bmatrix}x\\ y \\ z \end{bmatrix}=\begin{bmatrix}b\\ c\\d \end{bmatrix}$? Can we find a precomputed Cholesky matrices for $A$ and $E$ and only perform the factorisation process for $D(x,y,z)$? | |
| Nov 8, 2014 at 21:11 | comment | added | Meisam Jalalvand | Many thanks Richard! That is a very good start for me! | |
| Nov 8, 2014 at 12:20 | vote | accept | Meisam Jalalvand | ||
| Nov 8, 2014 at 0:42 | history | answered | Richard Zhang | CC BY-SA 3.0 |