The analysis of a FETI-DP preconditioner for a full DG discretization of elliptic problems in two dimensions

Abstract

In this paper a discretization based on a discontinuous Galerkin (DG) method for elliptic two-dimensional problems with discontinuous coefficients is considered. The problems are posed on a polygonal region $\varOmega$ which is a union of $N$ disjoint polygonal subdomains $\varOmega _i$ of diameter $O(H_i)$. The discontinuities of the coefficients, possibly very large, are assumed to occur only across the subdomain interfaces $\partial \varOmega _i$. In each $\varOmega _i$ a conforming quasi-uniform triangulation with parameters $h_i$ is constructed. We assume that the resulting triangulation in $\varOmega$ is also conforming, i.e., the meshes are assumed to match across the subdomain interfaces. On the fine triangulation, the problems are discretized by a DG method. For solving the resulting discrete systems, a FETI-DP type method is proposed and analyzed. It is established that the condition number of the preconditioned linear system is estimated by $C(1 + \max _i \log (H_i/h_i))^2$ with a constant $C$ independent of $h_i$, $H_i$ and the jumps of coefficients. The method is well suited for parallel computations and it can be extended to three-dimensional problems. Numerical results are presented to validate the theory.

Appendices

Appendix A: Proof of Lemma 5

We first prove the right hand side of the first inequality, the inequality (64). That is, we prove that there exist a constant $C$ such that, for all $u_i\in W_i(\varOmega _i^\prime )$ we have

$$\begin{aligned} \underline{a}_i((\underline{u}_i)_i,(\underline{u}_i)_i)\le Ca_i(u_i,u_i). \end{aligned}$$

(83)

First, note that,

$$\begin{aligned} \underline{a}_i((\underline{u}_i)_i,(\underline{u}_i)_i)= \sum _{\tau \in \mathcal {T}^i_h}\rho _i \int _{\tau } | \nabla (\underline{u}_i)_i|^2dx. \end{aligned}$$

(84)

We consider the cases of refined mesh $\underline{\mathcal {T}}^i_h$ listed in Definition 3 and illustrated in Fig. 4.

First case (Fig. 4, upper-left picture). For the first case, that is, $\tau \in \mathcal {T}^i_h$ and let us denote this triangle by $\tau _\ell$ and its neighbor by $\tau _p$, see Fig. 5.

Fig. 5

Illustration of refinement of two neighboring elements

Let us denote by $\underline{\tau }$ a generic triangle of $\underline{\mathcal {T}}_h^i$. We have

$$\begin{aligned} \int _{\tau _\ell }| \nabla (\underline{u}_i)_i|^2dx=\sum _{\underline{\tau }\subset \tau _\ell } \int _{\underline{\tau }} | \nabla (\underline{u}_i)_i|^2dx. \end{aligned}$$

The sum runs over ten triangles listed in the first case of Definition 3. Let $(u_i)_i$ and $(\underline{u}_i)_i$ on $\tau$ be denoted by $u_i^{(\ell )}$ or $u^{(\ell )}$ and $\underline{u}_i^{(\ell )}$ or $\underline{u}^{(\ell )}$, respectively. Note that in the triangle $\Delta ^{(\ell )}:=C_1^{(\ell )}C_2^{(\ell )}C_3^{(\ell )}\subset \tau _\ell$ we have $(u_i)_i|_{\tau _\ell }=(u_i^{\ell })=u^{(\ell )}$ and then

$$\begin{aligned} \int _{\Delta ^{(\ell )}}| \nabla \underline{u}_i^{(\ell )}|^2 dx = \int _{\Delta ^{(\ell )}}| \nabla u_i^{(\ell )}|^2 dx= \int _{\Delta ^{(\ell )}}| \nabla u^{(\ell )}|^2 dx. \end{aligned}$$

Let us consider now the triangle $\Delta ^{(\ell )}:{=}C_2^{(\ell )}C_3^{(\ell )}M_1^{(\ell )}$ where $M_1^{(\ell )}=M_1^{(p)}$. We have (see Fig. 5)

$$\begin{aligned} I&:= \int _{\Delta ^{(\ell )}}| (\nabla ({u}_i^{(\ell )}))_i|^2 dx\le C\Bigg \{ \Bigg [u^{(\ell )}( C_3^{(\ell )})-u^{(\ell )}(C_2^{(\ell )})\Bigg ]^2\\&+\Bigg [u^{(\ell )}( C_3^{(\ell )})- 0.5\bigg (u^{(\ell )}(M_1^{(\ell )})+u^{(p)}(M_1^{(p)})\bigg )\Bigg ]^2\\&+\Bigg [u^{(\ell )}( C_2^{(\ell )})- 0.5\bigg (u^{(\ell )}(M_1^{(\ell )})+u^{(p)}(M_1^{(p)})\bigg )\Bigg ]^2\Bigg \}\\&:= I_1+I_2+I_3. \end{aligned}$$

The first term above, $I_1$, it is estimated by $\Vert \nabla u^{(\ell )}\Vert _{L^2(\Delta ^{(\ell )})}^2$. The second term, $I_2$, it is estimated as follows. We have

$$\begin{aligned} I_2&\le C\Bigg \{ \bigg [ u^{(\ell )}(C_3^{(\ell )})-u^{(\ell )}(M_1^{(\ell )})\bigg ]^2+ \frac{1}{2} \bigg [ u^{(\ell )}(M_1^{(\ell )})-u^{(p)}(M_1^{(p)})\bigg ]^2\Bigg \}\nonumber \\&\le C\Bigg \{ \Vert \nabla u^{(\ell )}\Vert _{L^2(\Delta ^{(\ell )})}^2+ \frac{1}{h}\Vert u^{(\ell )}-u^{(p)}\Vert ^2_{L^2(V_2^{(\ell )}V_3^{(\ell )})}\Bigg \}, \end{aligned}$$

(85)

where here $V_2^{(\ell )}V_3^{(\ell )}$ denotes the edge of $\tau$ with the end points $V_2^{(\ell )}$ and $V_3^{(\ell )}$. In the same way we can estimate the third term, $I_3$. Thus,

$$\begin{aligned} I\le C\left\{ \Vert \nabla u^{(\ell )}\Vert ^2_{L^2(\tau _\ell )}+ \frac{1}{h}\Vert u^{(\ell )}-u^{(p)}\Vert ^2_{L^2(V_1^{(\ell )}V_3^{(\ell )})}\right\} . \end{aligned}$$

(86)

Similarly it is possible to estimate the terms involving the triangles $C_1^{(\ell )}C_2^{(\ell )}M_3^{(\ell )}$ and $C_1^{(\ell )}C_3^{(\ell )}M_2^{(\ell )}$.

We now estimate the term on $\Delta ^{\ell }:{=}C_2^{(\ell )}M_3^{(\ell )}V_2^{(\ell )}$. We have then

$$\begin{aligned} I&:= \int _{\Delta ^{(\ell )}}| \nabla \underline{u}_i^{(\ell )}|^2 dx\le C\Bigg \{ \Bigg [u^{(\ell )}( C_2^{(\ell )})-\frac{1}{2} \bigg (u^{(\ell )}(M_3^{(\ell )})+u^{(k)}(M_2^{(k)})\bigg )\Bigg ]^2\nonumber \\&+\Bigg [u^{(\ell )}( C_2^{(\ell )})- \frac{1}{n_{\ell pk}}\bigg ( u^{(\ell )}(V_2^{(\ell )})+ u^{(p)}(V_3^{(\ell )})+\cdots +u^{(k)}(V_2^{(k)})\bigg )\Bigg ]^2\nonumber \\&+\Bigg [\frac{1}{2}\big (u^{(\ell )}( M_3^{(\ell )})+u^{(k)}(M_2^{(k)})\bigg )\nonumber \\&-\frac{1}{n_{\ell pk}} \bigg (u^{(\ell )}(V_2^{(\ell )})+u^{(p)}(V_3^{(\ell )})+\cdots +u^{(k)}(V_2^{(k)}))\bigg )\Bigg ]^2\Bigg \}\nonumber \\&:= I_1+I_2+I_3. \end{aligned}$$

(87)

Here $\tau _k$ has a common edge $V_1^{(\ell )}V_2^{(\ell )}$ with $\tau _\ell$ and $n_{\ell p k}$ is the number of triangles of $\mathcal {T}_h^i$ with common vertex $V_2^{(\ell )}$. The first term, $I_1$, and then second term, $I_2$ are estimated as in (85). The third term, $I_3$, it is estimated in a similar way by adding and subtracting the quantity $\big (u^{(\ell )}(V_2^{(\ell )})+u^{(k)}(V_2^{(k)})\big )$, see Fig. 5. We proceed as above and using these estimates in (87) we obtain

$$\begin{aligned} I&\le C\Big \{ \Vert \nabla u^{(\ell )}\Vert _{L^2(\tau _\ell )}^2+ \Vert \nabla u^{(k)}\Vert _{L^2(\tau _k)}^2\nonumber \\&+\frac{1}{h} \Big \{ \Vert u^{(\ell )}-u^{(k)}\Vert ^2_{L^2(\partial \tau _\ell \cap \partial \tau _k)} +\cdots +\Vert u^{(\ell )} -u^{(p)}\Vert _{L^2(\partial \tau _\ell \cap \partial \tau _p)}^2 \Big \} \Big \}. \end{aligned}$$

(88)

In a similar way are estimated the terms over the remaining triangles of $\tau _\ell$.

Using the above estimates we show that

$$\begin{aligned} \int _{\tau _\ell } |\nabla ( \underline{u}_i)_i|^2\le C\Bigg \{ \sum _{\tau }\Vert \nabla ( u_i)_i\Vert _{L^2(\tau )}^2 +\frac{1}{h} \sum _{e}\Vert (u_i)_i^+-(u_i)_i^-\Vert _{L^2( e)}^2 \Bigg \} \end{aligned}$$

(89)

where the first sum runs over the elements $\tau$ which intersect $\tau _\ell$ by an edge and the second sum runs over edges e of $\tau$ which have a common vertex or edge with $\tau _\ell$.

Second case (Fig. 4, upper-right picture) We now consider the case when a vertex of $\varOmega _i$ is common for two and more triangles of $\mathcal {T}_i^h$. Let us consider the case of two triangles, see Fig. 6. This case is estimated similar as the first case.

Fig. 6

Illustration of refinement of two neighboring elements

Third case (Fig. 4, lower-left picture) This case (see Fig. 7) is also estimated similar as the first case.

Fig. 7

Illustration of refinement of two neighboring elements

Adding the above estimates for the three cases we get the estimate (83) for the case $\underline{u}_i=\underline{I}^i_hu_i$. For the case $\underline{u}_i=\underline{I}^{i,j}_hu_i$ we need only some minor modifications of the proof of the second case above, see Fig. 6. This finishes the proof of the left hand side inequality of (64).

We now present the proof of the left hand side of the result stated in Lemma 5, Eq. (64). We need to show that there exists a constant $C$ such that

$$\begin{aligned} a_i(({u}_i)_i,(u_i)_i)\le C\underline{a}_i ( (\underline{u}_i)_i,(\underline{u}_i)_i). \end{aligned}$$

(90)

Note that, on $\tau _\ell \in \mathcal {T}^h_i$ with vertices of type $C$, we have (see Fig. 5)

$$\begin{aligned} \Vert \nabla (u_i)_i\Vert ^2_{L^2(\tau _\ell )}&\le C\Big \{ \big [ u^{(\ell )}(C_1^{(\ell )})-u^{(\ell )}(C_2^{(\ell )})\big ]^2\nonumber \\&+\big [ u^{(\ell )}(C_1^{(\ell )})-u^{(\ell )}(C_3^{(\ell )})\big ]^2+ \big [ u^{(\ell )}(C_2^{(\ell )})-u^{(\ell )}(C_3^{(\ell )})\big ]^2\Big \}\nonumber \\&\le C \Vert \nabla (I_h^{(i)} u^{(\ell )})\Vert _{L^2(\tau _\ell )}^2\le C\Vert \nabla ( \underline{u}_i)_i\Vert _{L^2(\tau _\ell )}^2. \end{aligned}$$

(91)

This is valid for all three cases considered above. Using the estimate (91) we prove (90). The proof of the equivalence (64) is now complete.

We now prove the second inequality of Lemma 5, the inequality (65). We have

$$\begin{aligned} p_{i,\partial }(\underline{u}_i,\underline{u}_i)=\sum _{j\in \mathcal {T}^i_H} \sum _{e\in \mathcal {E}^{i,j}_h }\int _e \frac{\delta }{\ell _{ij}} \frac{\rho _i}{h_e} (\underline{u}_i-\underline{u}_j)^2 dS. \end{aligned}$$

(92)

On the edge $e$ we have (see Fig. 8)

$$\begin{aligned} \int _e (\underline{u}_i-\underline{u}_j)dS=\sum _{\underline{e}\subset e} \int _{\underline{e}} (\underline{u}_i-\underline{u}_j)^2dS \end{aligned}$$

(93)

where $\underline{e}$ runs over the edges of $\underline{\mathcal {T}}_i^h$, $\underline{e}\subset E_{ij}$. Note that, on $\underline{e}=[C_2^+,C_3^+]=[C_2^-,C_3^-]$, we can write

$$\begin{aligned} \int _{\underline{e}}(\underline{u}_i-\underline{u}_j)^2dS= \int _{\underline{e}}(u_i-u_j)^2 dS. \end{aligned}$$

Additionally, on $\underline{e}=[V_1^+,C_2^+]=[V_1^-,C_2^-]$, we have

$$\begin{aligned} \int _{\underline{e}}(\underline{u}_i-\underline{u}_j)^2dS&\le C\Bigg \{ \bigg [ u_i(C_2^+)-u_j(C_2^-)\bigg ]^2+\\&\bigg [\frac{1}{2}( u_i(C_2^+)-u_j(\widetilde{C}_2^+)) -\frac{1}{2}(u_j(C_2^-)-u_j(\widetilde{C}_2^-))\bigg ]^2 \Bigg \} \end{aligned}$$

where $\widetilde{C}_2^+$ and $\widetilde{C}_2^-$ are the nodal points on edges $\widetilde{e}$ of the triangles of $\mathcal {T}_h^i$ and $\mathcal {T}_h^j$ on $E_{ij}$ and $E_{ji}$ with common nodal points $V_1^+$ and $V_1^-$, respectively. Thus, in this case,

$$\begin{aligned} \int _{\underline{e}}(\underline{u}_i-\underline{u}_j)^2dS \le C\big \{ \Vert u_i-u_j\Vert ^2_{L^2(e)}+\Vert u_i-u_j\Vert ^2_{L^2(\tilde{e})} \big \} \end{aligned}$$

where $\tilde{e}\cap e=V_1^+=V_1^-$. In the case when $V_1^+$ or $V_2^+$ are corners of $\partial \varOmega _i$ we do the same modification which give $\Vert \nabla u_i\Vert _{L^2(\tau )}^2$ on $\tau$ with vertices $V_1^+$ or $V_2^+$. Using these in (93) and the resulting estimate into (92) we get an estimate of the second inequality of Lemma 5 for the case when $\underline{u}_i=\underline{I}_h^iu_i$. The case when $\underline{u}_i=I_h^{i,j}u_i$ is proved similarly.

Fig. 8

Illustration of common edge refinement

Now we prove the third inequality of Lemma 5, the inequality (66). We have that (93) still holds if we replace $\underline{u}_i$ by $\underline{u}_j$ and $u_i$ by $u_j$, respectively. Note that (see Fig. 8)

$$\begin{aligned} \int _e(u_i-u_j)^2dS&\le C\Big \{ \big [ u_i(C_1^+)-u_j(C_1^-)]^2+ \big [ u_i(C_2^+)-u_j(C_2^-)]^2\Big \}\\&\le C\int _{\widetilde{e}} (\underline{u}_i-\underline{u}_j)^2dx \end{aligned}$$

where $\widetilde{e}=(C_1^+,C_2^+)$. Using these estimates we see that the third inequality is valid for $\underline{u}_i=\underline{I}_iu_i$. The case $\underline{u}_i=\underline{I}_h^{i,j}u_i$ is similar.

It remains only to estimate the fourth inequality, inequality (67). It is proved as in the third inequality for $\underline{u}_i=\underline{I}_h^{i,j}u_i$.

The proof of Lemma 5 is complete.

Appendix B: Proof of Lemma 6

For the first inequality, (68), note that on $\tau \in \mathcal {T}_h^{i}$ (see Fig. 4 upper-left picture)

$$\begin{aligned} \int _\tau |\nabla (I_h^i \underline{u}_i)|^2dS&\le C\Big \{ \big [\underline{u}_i(C_1)-\underline{u}_i(C_2)\big ]^2+ \big [\underline{u}_i(C_1)-\underline{u}_i(C_3)\big ]^2\\&+ \big [\underline{u}_i(C_2)-\underline{u}_i(C_3)\big ]^2\Big \}\\&\le C\Vert \nabla \underline{u}_i\Vert ^2_{L^2(C_1C_2C_3)}\le C\Vert \nabla \underline{u}_i\Vert ^2_{L^2(\tau )}. \end{aligned}$$

Summing this for $\tau \subset \overline{\varOmega }_i$ we get the first inequality.

To prove the second inequality, (69), note that on $e\subset \partial \varOmega _i$ (see Fig. 8)

$$\begin{aligned} \Vert I_h^i\underline{u}_i-I^i_h\underline{u}_j\Vert _{L^2(e)}^2&\le C\Big \{ \big [ \underline{u}_i(C_2^+)-\underline{u}_j(C_2^-)\big ]^2+ \big [ \underline{u}_i(C_3^+)-\underline{u}_j(C_3^-)\big ]^2\Big \}\\&\le C\frac{1}{h} \Vert \underline{u}_i-\underline{u}_j\Vert _{L^2(\underline{e})}^2 \quad (\text{ with } \underline{e}=(C_2^+,C_3^+)\text{) }\\&\le C\frac{1}{h} \Vert \underline{u}_i-\underline{u}_j\Vert _{L^2(e)}^2. \end{aligned}$$

Summing this estimate over $e\subset \partial \varOmega _i$ we get the second inequality.

The equality $I_h^i\underline{u}_i=u_i$ follows from the definitions of $I^i_h$ and $\underline{u}_i$.

The proof of Lemma 6 is now complete.