Analysis of a new dimension-wise splitting iteration with selective relaxation for saddle point problems

Abstract

We propose a new dimension-wise splitting with selective relaxation (DSSR) method for saddle point systems arising from the discretization of the incompressible Navier–Stokes equations. Using Fourier analysis, we determine the optimal choice of the relaxation parameter that leads to the best performance of the iterative method for the Stokes and the steady Oseen equations. We also explore numerically the influence of boundary conditions on the optimal choice of the parameter, the use of inner and outer iterations, and the performance for a lid driven cavity flow.

Appendix: Optimizing \(\theta \) in DSSR for the Stokes problem

In this appendix we show that $\theta =1/2$ is the best parameter value for DSSR when applied to the 2D Stokes problem. Using Fourier analysis as in Sect. 3.1 for the Stokes equation (3.1) with

$$\begin{aligned} \hat{E}_1=\mathtt{diag}(0,1,\theta ),\quad \hat{E}_2=\mathtt{diag}(1,0,1-\theta ), \end{aligned}$$

one can compute the eigenvalues of the iteration matrix $M_{\alpha }$, and we get $\lambda _{1,2}(k_1,k_2)=0$ and

$$\begin{aligned} \lambda _3(k_1,k_2,\theta )=\frac{(\alpha \nu \theta (k_1^2+k_2^2)-k_2^2)(\alpha \nu (1-\theta )(k_1^2+k_2^2)-k_1^2)}{(\alpha \nu \theta (k_1^2+k_2^2)+k_1^2)(\alpha \nu (1-\theta )(k_1^2+k_2^2)+k_2^2)}. \end{aligned}$$

We then find

$$\begin{aligned} \lim _{k_2\rightarrow 0}|\lambda _3(k_1,k_2,\theta )|=\left| \frac{\alpha \nu (\theta -1)\theta -\theta +1}{\alpha \nu (\theta -1)\theta -\theta }\right| =:\rho _{10}(\alpha ,\nu ,\theta ), \end{aligned}$$

$$\begin{aligned} \lim _{k_2\rightarrow 0}|\lambda _3(k_1,k_2,\theta )|=\left| \frac{\alpha \nu (\theta -1)\theta +\theta }{\alpha \nu (\theta -1)\theta +\theta -1}\right| =:\rho _{20}(\alpha ,\nu ,\theta ). \end{aligned}$$

It is easy to verify that $\rho _{10}$ decreases monotonically in $\theta$, $\rho _{20}$ increases monotonically in $\theta$ for $\theta \in (0,1)$ and $\theta =1/2$ solves $\rho _{10}=\rho _{20}$. Thus, if $\theta \ne 1/2$ we would find a spectral radius of $\hat{M}_\alpha$ with $\max _{k_1,k_2}\rho (k_1,k_2,\nu ,\alpha )\ge \max \{\rho _{10},\rho _{20}\}>\rho _{10}(\alpha ,\nu ,\frac{1}{2})$, which shows that $\theta =1/2$ is the best value for the relaxation parameter $\theta$ introduced in Sect. 2.