Exact sequences on Powell–Sabin splits

Abstract

We construct smooth finite elements spaces on Powell–Sabin triangulations that form an exact sequence. The first space of the sequence coincides with the classical $C^1$ Powell–Sabin space, while the others form stable and divergence-free yielding pairs for the Stokes problem. We develop degrees of freedom for these spaces that induce projections that commute with the differential operators.

Appendix

1.1 Proof of Lemma 7

Proof

Suppose $z \in W_r(\{a,m,b\})$ is such that (4.1a)–(4.1c) are all zero. We will show that z must be identically zero on [a, b]. Let $\psi (x)$ be a degree r polynomial on the interval [0, 1] satisfying

$$\begin{aligned} \begin{aligned} \psi (0)&= 1, \quad \psi (1) = 0, \\ \int _0^1 \psi (x) p(x)&= 0 \quad \forall p \in \mathcal {P}_{r-2}([0,1]). \end{aligned} \end{aligned}$$

(A.1)

We note that these conditions uniquely determine $\psi$. Since z is continuous at m and equal to zero at a and b, and in view of (4.1b)–(4.1c), it follows that z may be represented by

$$\begin{aligned} z(y)&= z(m){\left\{ \begin{array}{ll} \psi \left( \frac{y-m}{a-m}\right) &{} y \in [a,m], \\ \psi \left( \frac{m-y}{m-b}\right) &{} y \in [m,b]. \end{array}\right. } \end{aligned}$$

Since $z'(y)$ is continuous at m, it must hold that

$$\begin{aligned} \frac{-1}{m-b} \psi '(0) = \frac{1}{a-m}\psi '(0). \end{aligned}$$

Furthermore, given the conditions (A.1) on $\psi$, we can show that $\psi '(0) \ne 0$. Suppose that $\psi '(0) = 0$ in addition to (A.1). Then for any $p \in \mathcal {P}_{r-1}([0,1])$ with $p(0) = 0$,

$$\begin{aligned} \int _0^1 \psi '(x) p(x)&= -\int _0^1 \psi (x) p'(x) + \psi (1) p(1) - \psi (0) p(0) = -\int _0^1 \psi (x) p'(x) = 0 \end{aligned}$$

since $p'(x) \in \mathcal {P}_{r-2}([0,1])$. But $\psi '(x)$ is itself such a function p(x), so it follows that

$$\begin{aligned} \int _0^1 |\psi '(x)|^2 = 0. \end{aligned}$$

Then $\psi '(x) = 0$, and $\psi$ is constant on [0, 1]. This contradicts (A.1), so $\psi '(0) \ne 0$. Furthermore, since $1/(b-m) \ne 1/(a-m)$, it follows that $z(m) = 0$. Therefore $z = 0$ on [a, b]. $\square$

1.2 Proof of Theorem 3

Proof

(1) Proof of (4.10a). Let $p \in C^\infty (T)$ and $\rho := \text {rot }\varPi _0^r p - \varpi _1^{r-1} \text {rot }p \in S_{r-1}^1(T^{\mathrm{ps}})$. We show that $\rho$ vanishes on (4.5).

First,

$$\begin{aligned} \rho (z_i)&= \text {rot }\varPi _0^r p(z_i) - \varpi _1^{r-1} \text {rot }p(z_i) = 0,\\ \mathrm{div}\,\rho (z_i)&= -\mathrm{div}\,\varphi _1^{r-1} \text {rot }p(z_i) = -\mathrm{div}\,\text {rot }p(z_i) = 0, \end{aligned}$$

by the definitions of $\varPi _0^r$ and $\varpi _1^{r-1}$ along with DOFs (4.2a) and (4.5a).

Next, if $r=2$,

$$\begin{aligned} \int _{e_i} \rho \cdot n_i&= \int _{e_i} \big (\text {rot }\varPi _0^r p - \varpi _1^{r-1} \text {rot }p \big )\cdot n_i\\&= \int _{e_i} \big (\text {rot }\varPi _0^r p - \varPi _1^{r-1} \text {rot }p \big )\cdot n_i=0, \end{aligned}$$

using (4.5b), (4.3b) and (4.7b). Similar arguments show that, for $r\ge 3$,

$$\begin{aligned} \rho (z_{3+i})\cdot n_i&= (\text {rot }\varPi _0^r p (z_{3+i}) - \varPi _1^{r-1} \text {rot }p(z_{3+i}))\cdot n_i =0,\\ \int _e \rho \cdot w&= \int _e (\text {rot }\varPi _0^r p - \varpi _1^{r-1} \text {rot }p)\cdot w = \int _e (\text {rot }\varPi _0^r p - \varPi _1^{r-1} \text {rot }p)\cdot w = 0, \end{aligned}$$

and

$$\begin{aligned} \int _T \rho \cdot \text {rot }w&= \int _T (\text {rot }\varPi _0^r p - \varPi _1^{r-1} \text {rot }p) \cdot w = 0. \end{aligned}$$

Next using (4.5c) gives

$$\begin{aligned} \mathrm{div}\,\rho (z_{3+i}) = -\mathrm{div}\,\varpi _1^{r-1} \text {rot }p (z_{3+i}) = -\mathrm{div}\,\text {rot }p (z_{3+i}) = 0, \end{aligned}$$

and (4.5e) yields

$$\begin{aligned} \int _e (\mathrm{div}\,\rho ) q = -\int _e (\mathrm{div}\,\varpi _1^{r-1} \text {rot }p) q = -\int _e (\mathrm{div}\,\text {rot }p)q = 0 \end{aligned}$$

for all $q\in \mathcal {P}_{r-4}(e)$ and $e\in {\mathcal {E}}^b(T^{\mathrm{ps}})$. The same arguments, but using (4.5g), gives

$$\begin{aligned} \int _T (\mathrm{div}\,\rho ) q=0\qquad \forall q\in \mathring{L}_{r-1}^2(T^{\mathrm{ps}}). \end{aligned}$$

Applying Lemma 12 shows that $\rho \equiv 0$, and so (4.10a) holds.

(2) Proof of (4.10b). For some $v \in [C^\infty (T)]^2$, we define $\rho := {\mathop {\mathrm {div}\,}}\varpi _1^{r-1} v - \varpi _2^{r-2} {\mathop {\mathrm {div}\,}}v \in L_{r-2}^2(T^{\mathrm{ps}})$. Then we need only show that $\rho$ is zero for all DOFs in (4.6). For the vertex DOFs, we have for each $z_i$,

$$\begin{aligned} \rho (z_i)&= {\mathop {\mathrm {div}\,}}\varpi _1^{r-1} v(z_i) - \varpi _2^{r-2} {\mathop {\mathrm {div}\,}}v(z_i) = 0, \end{aligned}$$

by (4.5a) and (4.6a). Next, for each $i = 1,2,3$,

$$\begin{aligned} \rho (z_{3+i})&= {\mathop {\mathrm {div}\,}}\varpi _1^{r-1} v(z_{3+i}) - \varpi _2^{r-2} {\mathop {\mathrm {div}\,}}v(z_{3+i}) = 0, \end{aligned}$$

where we have used (4.5a) and (4.6b). Similar arguments show that

$$\begin{aligned} \int _e \rho q =0\qquad \forall q\in \mathcal {P}_{r-4}(e),\ e\in {\mathcal {E}}^b(T^{\mathrm{ps}}), \end{aligned}$$

by (4.5e) and (4.6c), and that

$$\begin{aligned} \int _T \rho q = 0\qquad \forall q\in \mathring{L}_{r-2}^2(T^{\mathrm{ps}}) \end{aligned}$$

by (4.5g) and (4.6e). Using (4.6d) and (4.5b) if $r = 2$ or (4.5d) if $r > 2$,

$$\begin{aligned} \int _T \rho&= \int _T {\mathop {\mathrm {div}\,}}\varpi _1^{r-1} v -\varpi _2^{r-2} {\mathop {\mathrm {div}\,}}v = \int _T {\mathop {\mathrm {div}\,}}(\varpi _1^{r-1} v - v) = \int _{{\partial }T} (\varpi _1^{r-1} v - v )\cdot n = 0. \end{aligned}$$

Therefore, $\rho \equiv 0$ on T by Lemma 13, and (4.10b) is proved. $\square$