Let $\bar{x}$ denote the analytical solution of a PDE. Let $x^{(k)}$ be the solution at the $k^{th}$ iteration. The initial guess for the solution is $x^{(0)} = 0$. Let $r_{0} = ||\bar{x}-x^{(0)}||_{2}$ i.e. the $l_{2}$ norm of the error. Let $r_{k} = ||\bar{x}-x^{(k)}||_{2}$ be the $l_{2}$ norm at the $k^{th}$ iteration.

Question: Is there a possibility that $\frac{r_{k}}{r_{0}}$ first decreases then increases and then stabilises to a particular value ?

Additional info: When I examine this stable value of $\frac{r_{k}}{r_{0}}$ for increasing mesh sizes, it does give me $O(h^{2})$ convergence for a uniform cell-centered grid with Dirichlet boundaries solved using Finite Difference Method (using Jacobi iterations).

  • $\begingroup$ What iteration are you using? Are you talking about the linear iteration used to solve, for example, a finite element matrix? Or a nonlinear Newton iteration? $\endgroup$ – Wolfgang Bangerth Oct 21 '16 at 16:09
  • $\begingroup$ @WolfgangBangerth: I am using Jacobi iteration for finite differences. For a 4x4 mesh, the $\frac{r_{k}}{r_{0}}$ as defined above goes something like : 0.31, 0.19, 0.42, 0.46, 0.46, 0.46, 0.46 (stabilises). But for 8x8 and larger meshes it keeps decreasing from the initial value. My seniors told me that it is possible that the initial guess can be closer to the actual solution sometimes (my initial guess was all zeros) - so that's why the problem with a 4x4 mesh (even the analytical solution is very approximate on such a small mesh). I will appreciate any further understanding/comments. $\endgroup$ – Gaurav Saxena Oct 22 '16 at 10:24
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    $\begingroup$ I don't believe that the residual will, in general, monotonically decrease using Jacobi iteration. There are a lot of factors that play a role in Jacobi method convergence. For example the condition number will certainly affect the convergence rate. What is your PDE? $\endgroup$ – Charles Oct 23 '16 at 3:45
  • $\begingroup$ Correct. The Jacobi iteration typically only converges without a damping factor if the matrix is strongly diagonally dominant. $\endgroup$ – Wolfgang Bangerth Oct 26 '16 at 13:47

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