Consider the following PDE

\begin{align} -\Delta u &= f \ \ \text{en} \ \ (0,1)\times (0,1) \label{P1} \\ u &= 0 \ \ \text{en} \ \partial ((0,1)\times (0,1)) \label{P2} \end{align}

if we use 5-point finite difference approximations we get a system \begin{equation} Ax = b \label{Sist_Planteamiento} \end{equation}

where the matrix $A$ is symmetric.


  1. Do you know a finite difference approximation that produces a system
    \begin{equation} Ax = b \end{equation} where $A$ is nonsymmetric?

  2. Do you know books or papers where I can study systems like the system in question 1?

Thanks in advance

  • $\begingroup$ Why do you want a nonsymmetric A? I'm genuinely curious! $\endgroup$
    – Chenna K
    Jul 25, 2021 at 15:59
  • 1
    $\begingroup$ I would say that you can just take a spatial scheme that is not centered, then your matrix maybe non-symmetric. $\endgroup$
    – Laurent90
    Jul 25, 2021 at 16:20
  • $\begingroup$ What @Laurent90 proposed is certainly a way to do it; another way would be to scale just one row of your linear system up or down. $\endgroup$ Jul 25, 2021 at 17:26
  • $\begingroup$ I guess you could also use non-equally weighted (but still centered) finite difference approximations: $f'' \approx [w_1f(x+h) - w_2f(x) + w_3(x-h)]/h^2$ where $w_2 = w_1+w_3$ and $w_1\neq w_3$. As long as the discretization is consistent and stable, it would be convergent. But I don't know why anyone would want such a scheme aside from academic curiosity. $\endgroup$ Jul 25, 2021 at 17:57
  • $\begingroup$ @ChennaK I want to test a preconditioner in the system $A^{T}Ax=A^{T}b$. $\endgroup$ Jul 25, 2021 at 17:58


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