A Lowest-Degree Conservative Finite Element Scheme for Incompressible Stokes Problems on General Triangulations

Abstract

In this study, we investigate how low the degree of polynomials can be to construct a stable conservative pair for incompressible Stokes problems that works on general triangulations. We propose a finite element pair that uses a slightly enriched piecewise linear polynomial space for velocity and piecewise constant space for pressure. The pair is illustrated to be a lowest-degree stable conservative pair for Stokes problems on general triangulations.

Appendices

Appendix

A The Most Natural Linear–Constant Pair is not Stable: A Numerical Verification

In this section, we show by numerics the pair, $\varvec{V}{}{}_{h0}^1$–$\mathbb {P}^0_{h0}$, defined in Remark 5.3, is not stable on general triangulations, whereas

$$\begin{aligned} \displaystyle \inf _{q_h\in \mathrm{div}\,\varvec{V}{}{}_{h0}^1}\sup _{\varvec{v}{}{}_h\in \varvec{V}{}{}_{h0}^1}\frac{(\mathrm{div}\,\varvec{v}{}{}_h,q_h)}{\Vert q_h\Vert _{0,\varOmega }|\varvec{v}{}{}_h|_{1,h}}= \mathscr {O}(h) \end{aligned}$$

(A.1)

on a specific kind of triangulations.

1.1 A.1 A special triangulation and finite element space

We consider the computational domain $\varOmega = (0,1)\times (0,1) \setminus ( \{(x,y) : 0 \leqslant x \leqslant \frac{1}{2}, x + \frac{1}{2} \leqslant y \leqslant 1\} \cup \{ (x,y) : \frac{1}{2} \leqslant x \leqslant 1, 0 \leqslant y \leqslant x - \frac{1}{2} \})$. The initial triangulation is shown in Fig. 14a, and a sequence of triangulations is obtained by refining it uniformly (cf. Fig. 14b).

Fig. 14

An initially divisioned grid and its sketch map after uniform refinements

Given a patch $P_A$ as shown in Fig. 14a, we denote $\varvec{V}{}{}_{h0}^1(P_A) = \mathrm{span} \{ \varvec{\varphi }{}{}_1^A, \varvec{\varphi }{}{}_2^A, \varvec{\varphi }{}{}_3^A \}$, and for $i=1:6, \ \varvec{V}{}{}_{h0}^1(T_i) = \mathrm{span} \{ \varvec{\varphi }{}{}_{T_i}^1, \varvec{\varphi }{}{}_{T_i}^2, \varvec{\varphi }{}{}_{T_i}^3 \}$. Specifically, for $s=1:2, \ i=1:6, \, \varvec{\varphi }{}{}_s^A|_{T_i}=\varvec{\varphi }{}{}_{T_i}^s$, and for $s=3$,

$$\begin{aligned} \varvec{\varphi }{}{}_3^A = \left\{ \begin{aligned}&\varvec{\varphi }{}{}_{T_1}^1 - 2 \varvec{\varphi }{}{}_{T_1}^2 + \varvec{\varphi }{}{}_{T_1}^3 , \ in \ T_1;&\varvec{\varphi }{}{}_{T_2}^1 - \varvec{\varphi }{}{}_{T_2}^2 - \varvec{\varphi }{}{}_{T_2}^3 , \ in \ T_2;\\&2 \varvec{\varphi }{}{}_{T_3}^1 - \varvec{\varphi }{}{}_{T_3}^2 + \varvec{\varphi }{}{}_{T_3}^3 , \ in \ T_3;&\varvec{\varphi }{}{}_{T_4}^1 - 2 \varvec{\varphi }{}{}_{T_4}^2 + \varvec{\varphi }{}{}_{T_4}^3 , \ in \ T_4;\\&\varvec{\varphi }{}{}_{T_5}^1 - \varvec{\varphi }{}{}_{T_5}^2 - \varvec{\varphi }{}{}_{T_5}^3 , \ in \ T_5;&2 \varvec{\varphi }{}{}_{T_6}^1 - \varvec{\varphi }{}{}_{T_6}^2 + \varvec{\varphi }{}{}_{T_6}^3 , \ in \ T_6;\\ \end{aligned} \right. \end{aligned}$$

where for $i=1:6$, $\varvec{\varphi }{}{}_{T_i}^1= \left( \begin{array}{c} \lambda \\ 0 \\ \end{array} \right)$, $\varvec{\varphi }{}{}_{T_i}^2= \left( \begin{array}{c} 0 \\ \lambda \\ \end{array} \right)$, and $\varvec{\varphi }{}{}_{T_1}^3= \left( \begin{array}{c} \lambda _6-\lambda _1\\ 0 \end{array} \right)$, $\varvec{\varphi }{}{}_{T_2}^3= \left( \begin{array}{c} \lambda _1-\lambda _2\\ \lambda _1-\lambda _2 \end{array} \right)$, $\varvec{\varphi }{}{}_{T_3}^3= \left( \begin{array}{c} 0\\ \lambda _2-\lambda _3 \end{array} \right)$, $\varvec{\varphi }{}{}_{T_4}^3= \left( \begin{array}{c} \lambda _3-\lambda _4\\ 0 \end{array} \right)$, $\varvec{\varphi }{}{}_{T_5}^3= \left( \begin{array}{c} \lambda _4-\lambda _5\\ \lambda _4-\lambda _5 \end{array} \right)$, $\varvec{\varphi }{}{}_{T_6}^3= \left( \begin{array}{c} 0\\ \lambda _5-\lambda _6 \end{array} \right)$.

Similarly to Lemma 4.3, we can show the lemma below:

Lemma A.1

$dim(\varvec{V}{}{}_{h0}^{1})=3 \# \mathscr {X}_h^i$ and $\varvec{V}{}{}_{h0}^1=\mathrm{span}\{ \varvec{\varphi }{}{}_1^A, \varvec{\varphi }{}{}_2^A, \varvec{\varphi }{}{}_3^A, A \in \mathscr {X}_h^i \}$.

1.2 A.2 Numerical Verification of the Inf-sup Constant

By the Courant’s min-max theorem, it is easy to show the lemma below:

Lemma A.2

With respect to any set of basis functions of $\varvec{V}{}{}_{h0}^1$ and $\mathbb {P}^0_h$, denote by A the stiffness matrix of $(\nabla _h\,\cdot ,\nabla _h\,\cdot )$ on $\varvec{V}{}{}_{h0}^1$, by M the mass matrix of $(\cdot ,\cdot )$ on $\varvec{V}{}{}_{h0}^1$, and by B the stiffness matrix of $(\mathrm{div}\,\cdot ,\cdot )$ on $\varvec{V}{}{}_{h0}^1\times \mathbb {P}^0_h$. Then

$$\begin{aligned} \inf _{q_h\in \mathrm{div}\,\varvec{V}{}{}_{h0}^1}\sup _{\varvec{v}{}{}_h\in \varvec{V}{}{}_{h0}^1}\frac{(\mathrm{div}\,\varvec{v}{}{}_h,q_h)}{\Vert q_h\Vert _{0,\varOmega }|\varvec{v}{}{}_h|_{1,h}}= \lambda ^+_{\min }, \end{aligned}$$

where $\lambda ^+_{\min }$ is the smallest positive eigenvalue of the matrix eigenvalue problem

$$\begin{aligned} BA^{-1}B^{T} \mathrm{v} = {\lambda } M \mathrm{v}. \end{aligned}$$

(A.2)

The maximum eigenvalue of the eigenvalue problem (A.2) is denoted by $\lambda _{\max }$. Table 3 displays the computed values of $\lambda ^+_{\min }$ and $\lambda _{\max }$ on a series of refined grids. Figure 15 illustrates that $\lambda ^+_{\min }$ degenerates in the rate of $\mathscr {O}(h)$. This verifies (A.1) numerically.

Table 3 Computed values of $\lambda _{\min }^+$ and $\lambda _{\max }$

Fig. 15

$\lambda _{min}^+$ decays along with mesh refinements

B Proofs of Lemmas 4.3 and  5.3

1.1 B.1 Proof of Lemma 4.3

In this subsection, we provide the proof of Lemma 4.3, which establishes a basis of $\varvec{Z}{}_{h0}$. To this end, we need to analyze $\varvec{Z}{}_{h0}|_T$ with $T\in \mathscr {T}_h$ firstly.

For the interior cell $T\in \mathscr {T}_h^i$ with vertices $A_i$, $i=1:3$, and neighboring cells $T_j$, $j=1:3$, it is covered by functions of the set $\varvec{\Psi }{}_h(T)=\{\varvec{\psi }{}^{A_1}|_T,\varvec{\psi }{}^{A_2}|_T,\varvec{\psi }{}^{A_3}|_T,\varvec{\psi }{}_{T}|_T,\varvec{\psi }{}_{T_1}|_T,\varvec{\psi }{}_{T_2}|_T,\varvec{\psi }{}_{T_3}|_T\}$;see Fig. 16 for an illustration. It is clear that the seven functions in $\varvec{\Psi }{}_h(T)$ are linearly dependent; however, any six of them are linearly independent. For conciseness, a particular case is stated in the following lemma, which also serves Lemma 4.3.

Lemma B.1

For the interior cell $T\in \mathscr {T}_h^i$, with vertices $A_i,\ i=1:3$ and neighboring cells $T_j,\ j=1:3$ (see Fig. 16 for an illustration), the functions in $\{ \varvec{\psi }{}^{A_2}|_{T}, \varvec{\psi }{}^{A_3}|_{T}, \varvec{\psi }{}{}_{T}|_{T}, \varvec{\psi }{}{}_{T_1}|_{T}, \varvec{\psi }{}{}_{T_2}|_{T}, \varvec{\psi }{}{}_{T_3}|_{T}\}$ are linearly independent.

Fig. 16

Illustration of the supports of all kernel basis functions upon one cell

Proof

With the help of (4.3) and (4.4), a direct calculation leads to

$$\begin{aligned}&\left( \varvec{\psi }{}^{A_2}|_{T}, \varvec{\psi }{}^{A_3}|_{T}, \varvec{\psi }{}{}_{T}|_{T}, \varvec{\psi }{}{}_{T_1}|_{T}, \varvec{\psi }{}{}_{T_2}|_{T}, \varvec{\psi }{}{}_{T_3}|_{T} \right) ^\top \nonumber \\&\quad ={\textbf {A}} \left( \varvec{w}{}{}_{T,e_2,e_3}, \varvec{w}{}{}_{T,e_3,e_1}, \varvec{w}{}{}_{T,e_1,e_2}, \varvec{w}{}{}_{T,e_1}, \varvec{w}{}{}_{T,e_2}, \varvec{w}{}{}_{T,e_3}\right) ^\top ,\nonumber \\&\qquad \text{ with }\ \ {\textbf {A}}= \left[ \begin{array}{cccccc} 0 &{} 1 &{} 0 &{} \frac{d_2 d_5 \sin {(\alpha _3+\gamma _3)}}{2(S_1+S)} &{} 0 &{} \frac{d_2 d_8 \sin {(\alpha _1+\beta _1)}}{2(S_3+S)} \\ 0 &{} 0 &{} 1 &{} \frac{d_3 d_4 \sin {(\alpha _2+\beta _2)}}{2(S_1+S)} &{} \frac{d_3 d_7 \sin {(\alpha _1+\gamma _1)}}{2(S_2+S)} &{} 0 \\ \frac{1}{3} &{} \frac{1}{3} &{} \frac{1}{3} &{} \frac{S_1-2S}{3(S_1+S)} &{} \frac{S_2-2S}{3(S_2+S)} &{} \frac{S_3-2S}{3(S_3+S)} \\ 0 &{} 0 &{} 0 &{} \frac{S}{S_1+S} &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} \frac{S}{S_2+S} &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \frac{S}{S_3+S} \end{array}\right] . \end{aligned}$$

As $\displaystyle \det ({\textbf {A}})=\frac{1}{3} \prod _{i=1:3} \frac{S}{S+S_i}\ne 0$ and $\left\{ \varvec{w}{}{}_{T,e_2,e_3}, \varvec{w}{}{}_{T,e_3,e_1}, \varvec{w}{}{}_{T,e_1,e_2}, \varvec{w}{}{}_{T,e_1}, \varvec{w}{}{}_{T,e_2}, \varvec{w}{}{}_{T,e_3}\right\}$ are linearly independent, it concludes that $\left\{ \varvec{\psi }{}^{A_2}|_{T}, \varvec{\psi }{}^{A_3}|_{T}, \varvec{\psi }{}{}_{T}|_{T}, \varvec{\psi }{}{}_{T_1}|_{T}, \varvec{\psi }{}{}_{T_2}|_{T}, \varvec{\psi }{}{}_{T_3}|_{T} \right\}$ are linearly independent.$\square$

Remark B.1

If a cell $T\in \mathscr {T}_h$ has one (or more) vertices aligned on the boundary, then it is covered by no more than two interior vertex patches and contained in the supports of no more than six vertex- or cell-related kernel basis functions; the restriction of these six functions on T is linearly independent.

Proof of Lemma 4.3

We only have to prove that the functions of $\varPhi _h(\mathscr {T}_h)$ are linearly independent. Indeed, provided that the set $\varPhi _h(\mathscr {T}_h)$ is linearly independent, $\dim (\mathrm{span}(\varPhi _h(\mathscr {T}_h))) = \# \mathscr {X}^i_h + \# \mathscr {T}^i_h = 3 \# \mathscr {X}^i_h -2 = 3\# \mathscr {E}^i_h - (3 \# \mathscr {T}_h - 1) =\dim (\varvec{V}{}{}_{h0}^\mathrm{sBDFM})-\dim (\mathbb {P}^1_{h0})=\dim (\varvec{V}{}{}_{h0}^\mathrm{sBDFM})-\dim (\mathrm{div}\,\varvec{V}{}{}_{h0}^\mathrm{sBDFM})=\dim (\varvec{Z}{}{}_{h0})$, and thus $\varvec{Z}{}{}_{h0}=\mathrm{span}\left( \varPhi _h(\mathscr {T}_h)\right)$.

Now, given $\displaystyle \varvec{\psi }{}{}_h=\sum _{A\in \mathscr {X}_h^i}c_A\varvec{\psi }{}^A+\sum _{T\in \mathscr {T}_h^i}c_T\varvec{\psi }{}{}_T=0$, we show that all $c_A$ and $c_T$ are zero. Similar to [46], we adopt a sweeping process here. Given $a\in \mathscr {X}_h^b$, let T be such that a is a vertex of T. Then,

$$\begin{aligned} \varvec{\psi }{}{}_h|_T=\sum _{A\in \mathscr {X}_h^i\cap \overline{T}}c_A\varvec{\psi }{}^A|_T+\sum _{T'\in \mathscr {T}_h^i,T'\ \text{ and }\ T\ \text{ share } \text{ a } \text{ common } \text{ edge }}c_{T'}\varvec{\psi }{}{}_{T'}|_T=0. \end{aligned}$$

By Lemma B.1 and Remark B.1, $c_A=0$ for $A\in \mathscr {X}_h^i\cap \overline{T}$ and $c_{T'}=0$ for $T'\in \mathscr {T}_h^i$, where $T'$ and T share a common edge. Therefore, $c_A=0$ for any vertex $A\in \mathscr {X}_h^i$ that is connected to one boundary vertex $a\in \mathscr {X}_h^b$, and $c_T=0$ for any $T\in \mathscr {T}_h^i$ that connects to a boundary vertex $a\in \mathscr {X}^b_h$. Similarly, we can show

$$\begin{aligned} c_A=0\ \forall \,A\in \mathscr {X}_h^{b,+2},\ \ \ c_T=0\ \forall \,T\in \mathscr {T}_h\ \text{ that } \text{ connects } \text{ to } \mathscr {X}_h^{b,+1}. \end{aligned}$$

Repeating the procedure recursively, finally, we obtain

$$\begin{aligned} c_A=0 \ \forall \, A \in \mathscr {X}_h^{b,+k},\ \ \ c_T=0\ \forall \,T\in \mathscr {T}_h\ \text{ that } \text{ connects } \text{ to } \mathscr {X}_h^{b,+(k-1)} \end{aligned}$$

where k is the number of levels of the triangulation $\mathscr {T}_h$. Therefore, $c_A$ and $c_T$ are all zero, and the functions of $\varPhi _h(\mathscr {T}_h)$ are linearly independent. The proof is completed. $\square$

1.2 B.2 Proof of Lemma 5.3

In this subsection, we provide the proof of Lemma 5.3, which establishes a basis of $\varvec{V}{}_{h0}^\mathrm{el}$.

Proof of Lemma 5.3

Evidently, $\varvec{V}{}{}_{h0}^\mathrm{el}\supset \mathrm{span}\{\varvec{\psi }{}{}_e,\ e\in \mathscr {E}_h^i;\ \varvec{\psi }{}{}_T,\ T\in \mathscr {T}_h^i\}$. So we turn to the other direction.

First, we show $\mathrm{span}\{\mathrm{div}\,\varvec{\psi }{}{}_e,\ e\in \mathscr {E}_h^i\}=\mathbb {P}^0_{h0}.$ For both cases, as in Fig. 6, $\mathrm{div}\, \varvec{\psi }{}{}_e=\frac{1}{S_1}$ on $T_1$ and $-\frac{1}{S_2}$ on $T_2$, and vanishes on all the other cells. A simple algebraic argument leads to this assertion.

Second, all functions of $\varvec{Z}{}{}_{h0}$ can be represented by these functions. We only have to verify it for any kernel function, which is supported in a vertex patch. $\square$

Fig. 17

Illustration of the support of $\varvec{\psi }{}{}_{e_i}$. Left: $e_i$ has one interior vertex; Right: $e_i$ has two interior vertices

In fact, for an interior vertex A, $P_A=\cup _{i=1:m}T_i$, $\overline{T}_{i}\cap \overline{T}_{i+1}=e_i$, $T_{m+1}=T_1$ and $e_i$ connects A and $A_i$. Denote for $i=1:m$ (see Fig. 17)

$$\begin{aligned} \varvec{\psi }{}{}_{e_i}^* = \left\{ \begin{aligned}&\varvec{\psi }{}{}_{e_i}, \ A_i \in \mathscr {X}_h^b, \\&\varvec{\psi }{}{}_{e_i} + \frac{1}{2} \varvec{\psi }{}{}_{T_i} + \frac{1}{2} \varvec{\psi }{}{}_{T_{i+1}}, \ A_i \in \mathscr {X}_h^i. \end{aligned} \right. \end{aligned}$$

We refer to (4.3), (4.4), (5.1), (5.2) for the expressions of $\varvec{\psi }{}^A$, $\varvec{\psi }{}{}_{T_i}$ and $\varvec{\psi }{}{}_{e_i}$(cf Figs. 5, 6 and 7). Then, in any event $\mathrm{supp}(\varvec{\psi }{}{}_{e_i}^*)=T_{i-1} \cup T_{i} \cup T_{i+1} \cup T_{i+2} \subset P_A$, and $\mathrm{div}\,\sum _{i=1:m}\varvec{\psi }{}{}_{e_i}^*=0$. Thus $\sum _{i=1:m}\varvec{\psi }{}{}_{e_i}^*\in \varvec{Z}{}{}_A=\mathrm{span}\{\varvec{\psi }{}^A\}$. A further calculation gives $\sum _{i=1:m}\varvec{\psi }{}{}_{e_i}^*=\varvec{\psi }{}^A$, which thus leads to

$$\begin{aligned} \displaystyle \varvec{\psi }{}^{A}= \sum _{i=1:m} \varvec{\psi }{}{}_{e_i} + \frac{1}{2}\sum _{i=1:m, \ A_i \in \mathscr {X}_h^i} ( \varvec{\psi }{}{}_{T_i} + \varvec{\psi }{}{}_{T_{i+1}} ). \end{aligned}$$

Now, $\varvec{V}{}{}_{h0}^\mathrm{el}$ and $\mathrm{span}\{\varvec{\psi }{}{}_e,\ e\in \mathscr {E}_h^i;\ \varvec{\psi }{}{}_T,\ T\in \mathscr {T}_h^i\}$ have the same range under the operator $\mathrm{div}$. It also holds that $\varvec{Z}{}{}_{h0}\subset \mathrm{span}\{\varvec{\psi }{}{}_e,\ e\in \mathscr {E}_h^i;\ \varvec{\psi }{}{}_T,\ T\in \mathscr {T}_h^i\}$. Thus, $\varvec{V}{}{}_{h0}^\mathrm{el}=\mathrm{span}\{\varvec{\psi }{}{}_e,\ e\in \mathscr {E}_h^i;\ \varvec{\psi }{}{}_T,\ T\in \mathscr {T}_h^i\}$.

Further, $\dim (\mathrm{span}\{\varvec{\psi }{}{}_e,\ e\in \mathscr {E}_h^i;\ \varvec{\psi }{}{}_T,\ T\in \mathscr {T}_h^i\})=\dim (\varvec{V}{}{}_{h0}^\mathrm{el})=\dim (\varvec{Z}{}{}_{h0}) + \dim (\mathbb {P}{}_{h0}^0) = \# \mathscr {X}^i_h + \# \mathscr {T}^i_h + \# \mathscr {T}_h - 1 = \# \mathscr {T}^i_h + \# \mathscr {E}^i_h=\#(\{\varvec{\psi }{}{}_e,\ e\in \mathscr {E}_h^i;\ \varvec{\psi }{}{}_T,\ T\in \mathscr {T}_h^i\})$. Therefore, the functions $\{\varvec{\psi }{}{}_e,\ e\in \mathscr {E}_h^i;\ \varvec{\psi }{}{}_T,\ T\in \mathscr {T}_h^i\}$ are linearly independent, and they form a basis of $\varvec{V}{}{}_{h0}^\mathrm{el}$. The proof is completed. $\square$