I am solving a variant of the Ornstein-Zernike equation from liquid theory. Abstractly, the problem can be represented as solving the fixed point problem $A c(r)=c(r)$, where $A$ is an integro-algebraic operator and $c(r)$ is the solution function (the OZ direct correlation function). I am solving by Picard iteration, where I provide an initial trial solution $c_0(r)$ and generate new trial solutions by the scheme $$ c_{j+1} = \alpha (A c_j) + (1-\alpha)c_j~, $$ where $\alpha$ is an adjustable parameter that controls the mix of $c$ and $Ac$ used in the next trial solution. For this discussion, let's assume that the value of $\alpha$ is unimportant. I repeat until the iteration converges to within a desired tolerance, $\epsilon$: $$ \Delta_{j+1} \equiv \int d\vec r |c_{j+1}(r)-c_j(r)| < \epsilon~. $$ In my variant of the problem, $A$ depends on a parameter $\lambda$, and my question is about how the convergence of $Ac=c$ depends on this parameter.

For a wide range of values for $\lambda$, the iteration scheme above converges exponentially quickly. However, as I decrease $\lambda$, I eventually reach a regime in which the convergence is non-monotonic, pictured below. onset of non-monotone convergence

Key Questions

In iterative solutions to fixed-point problems, does non-monotonic convergence have any special significance? Does it signal that my iterative scheme is on the verge of instability? Most importantly, should non-monotone convergence make me suspicious that the "converged" solution is not a good solution to the fixed-point problem?


1 Answer 1


Suppose $x$ is the unknown independent variable in the solution of $x=f(x)$, then fixed point method will converge from a point $x^*$ provided that the Jacobian $\frac{\partial f}{\partial x}(x^*) \le \alpha$, where $\alpha$ is a constant $<1$. In general $x^*$ is not a single point, but the domain traversed by the iterative scheme.

  1. Your solution is converging, albeit non-monotonically. Check your Jacobian for various values of $\lambda$ and the solution variable to see if you go from satisfying convergence criteria to not satisfying it, which might explain what you are seeing.

  2. If your solution has converged within a properly established relative tolerance which also accounts for small numbers, then it has.

  • $\begingroup$ Can you clarify your second point? $\endgroup$
    – Endulum
    Commented Feb 9, 2017 at 18:10
  • $\begingroup$ The difference $|x_{j+1}-x_j|$ between two successive iterations could be compared against $|x_j| \thinspace \epsilon$ where $\epsilon$ is a relative tolerance. $\endgroup$
    – NameRakes
    Commented Feb 10, 2017 at 0:17

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