On the robustness and optimality of algebraic multilevel methods for reaction–diffusion type problems

Abstract

This paper is on preconditioners for reaction–diffusion problems that are both, uniform with respect to the reaction–diffusion coefficients, and optimal in terms of computational complexity. The considered preconditioners belong to the class of so-called algebraic multilevel iteration (AMLI) methods, which are based on a multilevel block factorization and polynomial stabilization. The main focus of this work is on the construction and on the analysis of a hierarchical splitting of the conforming finite element space of piecewise linear functions that allows to meet the optimality conditions for the related AMLI preconditioner in case of second-order elliptic problems with non-vanishing zero-order term. The finite element method (FEM) then leads to a system of linear equations with a system matrix that is a weighted sum of stiffness and mass matrices. Bounds for the constant $\gamma$ in the strengthened Cauchy–Bunyakowski–Schwarz inequality are computed for both mass and stiffness matrices in case of a general $m$-refinement. Moreover, an additive preconditioner is presented for the pivot blocks that arise in the course of the multilevel block factorization. Its optimality is proven for the case $m=3$. Together with the estimates for $\gamma$ this shows that the construction of a uniformly convergent AMLI method with optimal complexity is possible (for $m \ge 3$). Finally, we discuss the practical application of this preconditioning technique in the context of time-periodic parabolic optimal control problems.

Appendices

Appendix 1: Proof of Lemma 1

The element mass matrix for a given arbitrary non-degenerate triangle $e$ is given by

$$\begin{aligned} M_e(u,v) = \int _e u v \, de. \end{aligned}$$

We introduce the notations $h = |OA|, p = |OB|$ and $q = |OC|$, where $O$ is the origin, see Fig. 1. Then we have the following relations given:

$$\begin{aligned} b = \frac{p}{h}, \quad c = \frac{q}{h}, \quad a = \cot (\pi - (\theta _2 + \theta _3)) = \frac{h^2 - p q}{h(p+q)}. \end{aligned}$$

The element basis functions are given by

$$\begin{aligned} \phi _1 = - \frac{x}{h}, \quad \phi _2 = \frac{q x + h(q-y)}{h(p+q)}, \quad \phi _3 = \frac{p x + h(p+y)}{h(p+q)}. \end{aligned}$$

Moreover,

$$\begin{aligned} |e| = \int \, de = \frac{h(p+q)}{2} = J_e \cdot \frac{1}{2}, \end{aligned}$$

where $J_e = h^2 (b+c)$ is the Jacobi determinant. We obtain the following first two entries of the element mass matrix:

$$\begin{aligned} {M_e}_{11} = \int _{-h}^0 \int _{-p/h x - p}^{q/h x + q} (\phi _1)^2 \,dy \, dx = \frac{h^2(b+c)}{12} \end{aligned}$$

and

$$\begin{aligned} {M_e}_{12} = \int _{-h}^0 \int _{-p/h x - p}^{q/h x + q} \phi _1 \phi _2 \,dy \, dx = \frac{h^2(b+c)}{24}. \end{aligned}$$

Analogously, we obtain all other entries and we finally derive the element mass matrix (19). $\square$

Appendix 2: Proof of Theorem 1

Let $m_c = h^2 (b+c) /24$. The global CBS constant can be estimated by the the maximum of the local CBS constants on the macro elements (7), which can be again computed via the rule (8), i.e., $\gamma _{M,E}^2 = 1 - \lambda ^{\min }_E$, where $\lambda ^{\min }_E$ is the minimal eigenvalue of the eigenvalue problem (22), which we write in the form

$$\begin{aligned} \mathbf{v}_{E:2}^T \left( S_M - \lambda \, \tilde{M}_{E:22} \right) \mathbf{v}_{E:2} = 0, \end{aligned}$$

We try to find a lower bound $\underline{\lambda ^{\min }_E}$ for the minimal eigenvalue $\lambda ^{\min }_E$ such that

$$\begin{aligned} \mathbf{v}_{E:2}^T \left( S_M - \underline{\lambda ^{\min }_E} \, \tilde{M}_{E:22} \right) \mathbf{v}_{E:2} \ge 0. \end{aligned}$$

For that reason, we estimate the Schur complement $S_M$ from below by $\underline{S_M}$, and, after that we solve the problem $\mathbf{v}_{E:2}^T (\underline{S_M} - \lambda \, \tilde{M}_{E:22} ) \mathbf{v}_{E:2} = 0$. For every $m$-refinement, we systematically use a bottom-up lexicographical ordering for the fine nodes and number the three coarse nodes last. Hence, the lower right block $M_{E:22}$ of the macro element mass matrix is always $M_{E:22} = 2 \, m_c \, I$, where $I$ is the identity matrix. Let $N_E$ denote the number of nodes on a macro element subdivided into $m^2$ elements. The Schur complement $S_M$ can be estimated from below in the following way:

$$\begin{aligned} S_M&= M_{E:22} - M_{E:21} \, M_{E:11}^{-1} \, M_{E:12} \\&= 2 \, m_c \, I - M_{E:21} \, M_{E:11}^{-1} \, M_{E:12} \\&\ge 2 \, m_c \, I - \frac{1}{6} \, m_c^{-1} \, M_{E:21} \, I \, M_{E:12}, \end{aligned}$$

where we use the estimate

$$\begin{aligned} M_{E:11} \ge \min _{i \in \{1,\ldots ,N_E\}}{(M_{E:11})_{ii}} \, I = 6 \, m_c \, I \end{aligned}$$

because the weakly diagonally dominant matrix $M_{E:11}$ has only the diagonal entries $12 \, m_c$ and $6 \, m_c$. Moreover, since the matrices $M_{E:12}$ and $M_{E:21} = (M_{E:12})^T$ have exactly two entries equal to one in each of the three columns and rows, respectively, we obtain

$$\begin{aligned} M_{E:21} \, M_{E:12} = (M_{E:12})^T \, M_{E:12} = 2 \, m_c^2 \, I \end{aligned}$$

and finally the Schur complement $S_M$ can be estimated from below by

$$\begin{aligned} S_M&\ge 2 \, m_c \, I - \frac{1}{6} \, m_c^{-1} \, 2 \, m_c^2 \, I = 2 \, m_c \, I - \frac{1}{3} \, m_c \, I, \\&= \frac{5}{3} \, m_c \, I. \end{aligned}$$

Next we have that

$$\begin{aligned} \tilde{M}_{E:22}&= \,M_{E:22} + J_{E:12}^T \, M_{E:12} + M_{E:21} \, J_{E:12} \\&\quad +\,J_{E:12}^T \, M_{E:12} \, J_{E:12} \\&= \,m^2 \, M_e \end{aligned}$$

with a transformation matrix $J_{E:12}$ of the from (14). Then

$$\begin{aligned} J_{E:12}^T \, M_{E:12} = m_c \left( \begin{array}{c@{\quad }c@{\quad }c} \frac{2 \, m-2}{m} &{} \frac{1}{m} &{} \frac{1}{m} \\ \frac{1}{m} &{} \frac{2 \, m-2}{m} &{} \frac{1}{m} \\ \frac{1}{m} &{} \frac{1}{m} &{} \frac{2 \, m-2}{m} \\ \end{array} \right) \end{aligned}$$

because $M_{E:12}$ has exactly two entries with value one in each row and $J_{E:12}$ has the value $(m-1)/m$ in exactly the same positions. Hence, we solve the problem

$$\begin{aligned} \mathbf{v}_{E:2}^T \left( \frac{5}{3} \, m_c \, I - \lambda \, m^2 \, M_e \right) \mathbf{v}_{E:2} = 0 \end{aligned}$$

and obtain the two-fold eigenvalue $5/(3m^2)$ and the eigenvalue $5/(12m^2)$, which yields the lower bound for $\lambda _E^{\min }$. According to (8) we obtain the estimate

$$\begin{aligned} \gamma _{M,E}^2 = 1 - \lambda _E^{\min } \le 1 - \frac{5}{12 \, m^2} = \frac{12 \, m^2 - 5}{12 \, m^2}, \end{aligned}$$

which together with (7) gives the upper bound (28). $\square$