# Solving a linear system whose matrix has imbalanced diagonal entries

I am trying to solve following set of equations:

A(i,i-2)*u(i-2) + A(i,i-1)*u(i-1) + (A(i,i)+β(i) )*u(i) + A(i,i+1)*u(i+1) + A(i,i+2)*u(i+2)= B(i) + β(i)

where i=1:1000000

If values of β varies(several orders) greatly then what would be the best strategy to solve this equation numerically. Should I expect oscillations in the result?

What could be the possible remedy?

*Is it okay if I add "10000*u(i)" on both sides of equations while solving iteratively?*

EDIT: Structure of Coeffcient Matrix will roughly look like this (if i=1:9, while I am trying to do it for i=1:10^6): $$\begin{bmatrix} A11 + β1& A12& A13& 0& 0& 0& 0& 0& 0& \\ A21& A22 + β2& A23& A24& 0& 0& 0& 0& 0& \\ A31& A32& A33 + β3& A34& A35& 0& 0& 0& 0& \\ 0& A42& A43& A44 + β4& A45& A46& 0& 0& 0& \\ 0& 0& A53& A54& A55 + β5& A56& A57& 0& 0& \\ 0& 0& 0& A64& A65& A66 + β6& A67& A68& 0&\\ 0& 0& 0& 0& A75& A76& A77 + β7& A78& A79& \\ 0& 0& 0& 0& 0& A86& A87& A88 + β8& A89& \\ 0& 0& 0& 0& 0& 0& A97& A98& A99 + β9& \\ \end{bmatrix}$$ Here A(i,i)=A(i,i-2)+A(i,i-1)+A(i,i+1)+A(i,i+2)

  **β(i,i)>0
β(i,i) >> A(i,i-2)+A(i,i-1)+A(i,i+1)+A(i,i+2)**


Also there is a huge variation in values of β.

**Order of every entry in matrix is ~10^-2, except diagonal entries and β which vary from order of 1 to 10^6.

• Can you show the matrix to make it easier to read the problem? It looks like you have ones on the two off diagonals in both directions, and then $1+\beta_i$ on the diagonal. Is this correct? Are the $\beta_i$ positive? – Wolfgang Bangerth Apr 22 '16 at 12:01
• @WolfgangBangerth: I have edited my question to show you the structure of matrix. Yes! it is like having ones on the two off diagonals in both directions, and then 1+β(i) on the diagonal. β(i) is always positive. – Hanumat Apr 22 '16 at 13:10
• If you're stuck with an iterative method, @namRakes is correct about scaling (preconditioning) the equation by the diagonal. Diagonal preconditioning should be very effective since your matrix seems to be diagonally dominant. – Charles Apr 25 '16 at 7:11

1. Is there a typo in your expression for the $i^{\text{th}}$ equation's diagonal term, which you say is $A_{i,i} + \beta_i$ at the top of the page. Then if I try using this with your expression for $A_{i,i}$ below the matrix, then I get $1+2\beta_i$ in the diagonal. Assuming that this is indeed a typo, then ...
2. Is $\beta_i >> B_i$ also? If so, given that $\beta_i >> \sum_{k=i-2,k\ne i}^{i+2} A_{i,k}$, your solution vector $u_i$ is centered around $\{1,1,1,...\}$. This is good news, because then this could be your initial guess for any iterative scheme.
• As long as $\beta_i >> B_i$ and $>> \sum A_{i,k}$ as above, and even with 10 orders of variation of $\beta_i$ your iterative scheme should work. If it doesn't converge, or the $\beta$ variation is 20 orders, you can try to scale each equation $i$ by $\beta_i$. Is there any reason you don't want to choose a direct solution method? – NameRakes Apr 22 '16 at 17:35