I try to find an expression for the minimum iterations needed to solve a problem $Au=f$ by using the Successive Over-Relaxation (SOR) method for a certain $\omega$ as a function of the size of matrix $A\in\mathbb{R}^{n\times n}$, $n$ and with an initial guess $u^{0}=\emptyset$, such that the initial error is $f$.

Assuming I am able to find an expression for the rate of convergence of the corresponding Jacobi iteration matrix of matrix $A$, which does not depend on the size matrix $A$. Then I also have an expression for the rate of convergence of the SOR iteration matrix, since for a $\omega_{opt}$, the optimal choice of $\omega$, and $\mu=\rho(B_{Jac})$, we have $$ \rho(B_{SOR(\omega)})=\omega-1~\mbox{for}~0\leq\omega\leq\omega_{opt}\\ \rho(B_{SOR(\omega)})=\frac{1}{4}\Big(\omega\mu+\sqrt{\omega^{2}\mu^{2}-4(\omega-1)}\Big)^{2}~\mbox{for}~\omega_{opt}\leq\omega\leq2 $$ Now, using the rate of convergence of the SOR method I find the following expression for the minimal iterations needed $$ ||e^{k}||_{2}=\frac{\mu_{SOR}^{k}}{1-\mu_{SOR}}||e^{0}|| $$ with $k$ the number of the iterations, thus for a certain stopping criterium $e^{k}\leq e^{stop}$ $$ k=\log\Big[\frac{||e^{stop}||}{||e^{0}||}(1-\mu_{SOR})\Big]\cdot(\log[\mu_{SOR}])^{-1} $$ However, the only term, which could depend on the size of matrix $A$ is $||e_{0}||=||f||$. Now, vector $f$ is very irregular and therefore hard to write into an analytical expression.

My question is: is there another way to approach the minimal amount of iterations needed by the SOR method, if necessary assuming we know the rate of convergence (of the SOR method) or is this the right way and is simply the irregularity of the vector $f$ a bottleneck in this approach?


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