Consider linear equation $Au = f$.

We want to solve it with iterative method (assuming $A$ is good). First order iterative method is: $$ u^{k+1} = u^k - \alpha_{k+1}(Au^k - f), $$ The second degree method is: $$ u^{k+1} = u^k - \alpha_{k+1}(Au^k - f) - \beta(u^k - u^{k-1}). $$

For both methods we can define iteration parameters $\alpha_k$, $\beta_k$ via minimax problem which solution is Chebyshev polynomials.

This is good, but it seems to me, that convergence speed is the same for both cases and is $$ \|\varepsilon^{k}\| \leq \frac{2\sigma^k}{1+\sigma^{2k}}\|\varepsilon^{0}\|, $$
where $u - u^k = \varepsilon^{k}$ approximation error after the $k$-th iteration and $\sigma$ is constant dependent on operator spectrum.

The only idea I have, that first order iterations optimal for chosen $k$ for which coefficients are computed, while second-order iteration is optimal on every step.

I would greatly appreciate any work on this to clear-up those details.


1 Answer 1


I think that the difference between the two method isn't the degree, however the first method is a one-step method while the second one is a two-step method; therefor the convergence speed can be similar or the same; the choice depend by the advantage required.

  • $\begingroup$ What kind of advantage the two step method has? It has the same convergence speed and I cannot see any other difference. $\endgroup$
    – Moonwalker
    Aug 7, 2016 at 22:20
  • 1
    $\begingroup$ You have to study the stability of these method to find the differences. In numerical analysis of ode and pde they are often used es. link. $\endgroup$
    – user96954
    Aug 9, 2016 at 8:04

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