Virtual enriching operators

Abstract

We construct bounded linear operators that map $H^1$ conforming Lagrange finite element spaces to $H^2$ conforming virtual element spaces in two and three dimensions. These operators are useful for the analysis of nonstandard finite element methods.

A Inverse trace theorems for \(\mathbb {R}_+^2\) and \(\mathbb {R}_+^3\)

We consider inverse trace theorems for $\mathbb {R}_+^2$ and $\mathbb {R}_+^3$ with data on the boundaries of these domains (cf. Fig. 5). We will rely on the results in Lemmas A.1 and A.2 that follow from the construction of inverse trace operators through the Fourier transform [22, 26] and the Paley–Wiener theorem [21].

Fig. 5

Boundary data for $\mathbb {R}_+^2$ and $\mathbb {R}_+^3$

Lemma A.1

There exists a bounded linear operator

$$\begin{aligned} L_1:H^{2}(\mathbb {R})\times H^{1}(\mathbb {R})\longrightarrow H^{\frac{5}{2}}(\mathbb {R}^2) \end{aligned}$$

such that (i) $[L_1(\phi ,\psi )](t,0)=\phi (t)$, (ii) $[\partial L_1(\phi ,\psi )/\partial x_2](t,0)=\psi (t)$, and (iii) $L_1(\phi ,\psi )(x_1,x_2)$ vanishes on the half plane $x_1<0$ if $\phi (t)$ and $\psi (t)$ vanish on the half line $t<0$.

Lemma A.2

There exists a bounded linear operator

$$\begin{aligned} L_2:H^{2}(\mathbb {R}^2)\times H^{1}(\mathbb {R}^2)\longrightarrow H^{\frac{5}{2}}(\mathbb {R}^3) \end{aligned}$$

such that (i) $[L_2(\phi ,\psi )](x_1,x_2,0)=\phi (x_1,x_2)$, (ii) $[\partial L_2(\phi ,\psi )/\partial x_3](x_1,x_2,0)=\psi (x_1,x_2)$, and (iii) $L_2(\phi ,\psi )(x_1,x_2,x_3)$ vanishes on the half space $x_1<0$ (resp., $x_2<0)$ if $\phi (x_1,x_2)$ and $\psi (x_1,x_2)$ vanish on the half plane $x_1<0$ (resp., $x_2<0)$.

We begin with a two-dimensional inverse trace theorem. We note that similar results for $H^2(\mathbb {R}_+^2)$ can be found in [20, Section 1.5.2]. Our approach is simpler (since we are considering $H^\frac{5}{2}(\mathbb {R}_+^2)$) and therefore its extension to three dimensions is easier.

Lemma A.3

Let $(\phi _1,\psi _1)$ and $(\phi _2,\psi _2)$ belong to $H^{2}(\mathbb {R}_+)\times H^{1}(\mathbb {R}_+)$ such that

$$\begin{aligned} \phi _1(0)&=\phi _2(0), \end{aligned}$$

(A.1)

$$\begin{aligned} \psi _1(0)&=\phi _2'(0), \end{aligned}$$

(A.2)

$$\begin{aligned} \psi _2(0)&=\phi _1'(0). \end{aligned}$$

(A.3)

Then there exists $\zeta \in H^\frac{5}{2}(\mathbb {R}_+^2)$ such that

$$\begin{aligned} (\zeta ,\partial \zeta /\partial x_i)=(\phi _i,\psi _i) \quad \text {if } x_i=0, \,1\le i\le 2. \end{aligned}$$

(A.4)

Proof

First we extend $\phi _1$ and $\psi _1$ to $\mathbb {R}$, so that the extensions (still denoted by $\phi _1$ and $\psi _1$) satisfy $\phi _1\in H^{2}(\mathbb {R})$ and $\psi _1\in H^{1}(\mathbb {R})$. This can be achieved by reflection (cf. [22, Theorem 2.3.9] and [1, Theorem 5.19]). Let $L_1$ be the lifting operator in Lemma A.1 and $\zeta _1=L_1(\phi _1,\psi _1)\in H^{\frac{5}{2}}(\mathbb {R}^2)$ so that

$$\begin{aligned} \zeta _1(0,x_2)=\phi _1(x_2)\quad \text {and}\quad (\partial \zeta _1/\partial x_1)(0,x_2)=\psi _1(x_2). \end{aligned}$$

(A.5)

Then we define ${{\tilde{\phi }}}_2(x_1)=\phi _2(x_1)-\zeta _1(x_1,0)$, ${{\tilde{\psi }}}_2(x_1)=\psi _2(x_1)-(\partial \zeta _1/\partial x_2)(x_1,0)$ for $x_1>0$.

Note that ${{\tilde{\phi }}}_2\in H^{2}(\mathbb {R}_+)$, and

$$\begin{aligned} {{\tilde{\phi }}}_2(0)=\phi _2(0)-\zeta _1(0,0)=\phi _2(0)-\phi _1(0)=0 \end{aligned}$$

by (A.1) and (A.5), and

$$\begin{aligned} {{\tilde{\phi }}}_2'(0)=\phi _2'(0)-(\partial \zeta _1/\partial x_1)(0,0)=\phi _2'(0)-\psi _1(0)=0 \end{aligned}$$

by (A.2) and (A.5). Moreover we have ${{\tilde{\psi }}}_2\in H^{1}(\mathbb {R}_+)$, and

$$\begin{aligned} {{\tilde{\psi }}}_2(0)=\psi _2(0)-(\partial \zeta _1/\partial x_2)(0,0)=\psi _2(0)-\phi _1'(0)=0 \end{aligned}$$

by (A.3) and (A.5). Hence their trivial extensions (still denoted by ${{\tilde{\phi }}}_2$ and ${{\tilde{\psi }}}_2$) satisfy ${{\tilde{\phi }}}_2\in H^{2}(\mathbb {R})$ and ${{\tilde{\psi }}}_2\in H^{1}(\mathbb {R})$.

Let $\zeta _2=L_1({{\tilde{\phi }}}_2,{{\tilde{\psi }}}_2)\in H^{\frac{5}{2}}(\mathbb {R}^2)$ such that $\zeta _2(x_1,0)={{\tilde{\psi }}}_2(x_1,0)$ and

$$\begin{aligned} (\partial \zeta _2/\partial x_2)(x_1,0)={{\tilde{\psi }}}_1(x_1). \end{aligned}$$

Then $\zeta _2=0$ on the half plane $x_1<0$ by Lemma A.3, which implies

$$\begin{aligned} \zeta _2(0,x_2)=(\partial \zeta _2/\partial x_1)(0,x_2)=0 \quad \forall \, x_2>0. \end{aligned}$$

We can now take $\zeta$ to be the restriction of $\zeta _1+\zeta _2$ to $\mathbb {R}_+^2$. $\square$

Next we consider the three dimensional analog of Lemma A.3.

Lemma A.4

Let $(\phi _1,\psi _1)$, $(\phi _2,\psi _2)$ and $(\phi _3,\psi _3)$ belong to $H^{2}(\mathbb {R}_+^2)\times H^{1}(\mathbb {R}_+^2)$ such that the following conditions are satisfied : 

$$\begin{aligned} \phi _1(0,x_3)&=\phi _2(0,x_3)\qquad \forall \,x_3>0, \end{aligned}$$

(A.6)

$$\begin{aligned} \phi _2(x_1,0)&=\phi _3(x_1,0)\qquad \forall \,x_1>0, \end{aligned}$$

(A.7)

$$\begin{aligned} \phi _3(0,x_2)&=\phi _1(x_2,0)\qquad \forall x_2>0, \end{aligned}$$

(A.8)

$$\begin{aligned} \psi _1(0,x_3)&=\frac{\partial \phi _2}{\partial x_1}(0,x_3)\qquad \forall \,x_3>0, \end{aligned}$$

(A.9)

$$\begin{aligned} \psi _1(x_2,0)&=\frac{\partial \phi _3}{\partial x_1}(0,x_2)\qquad \forall \,x_2>0, \end{aligned}$$

(A.10)

$$\begin{aligned} \psi _2(x_1,0)&=\frac{\partial \phi _3 }{\partial x_2}(x_1,0)\qquad \forall \,x_1>0, \end{aligned}$$

(A.11)

$$\begin{aligned} \psi _2(0,x_3)&=\frac{\partial \phi _1 }{\partial x_2}(0,x_3)\qquad \forall \,x_3>0, \end{aligned}$$

(A.12)

$$\begin{aligned} \psi _3(x_1,0)&=\frac{\partial \phi _2}{\partial x_3}(x_1,0)\qquad \forall \,x_1>0, \end{aligned}$$

(A.13)

$$\begin{aligned} \psi _3(0,x_2)&=\frac{\partial \phi _1}{\partial x_3}(x_2,0)\qquad \forall \,x_2>0. \end{aligned}$$

(A.14)

Then there exists $\zeta \in H^\frac{5}{2}(\mathbb {R}_+^3)$ such that

$$\begin{aligned} (\zeta ,\partial \zeta /\partial x_i)=(\phi _i,\psi _i) \quad \hbox { if}\ \;x_i=0, \;1\le i\le 3. \end{aligned}$$

(A.15)

Proof

First we extend $\phi _1$ and $\psi _1$ to $\mathbb {R}^2$ by reflection (twice) so that the extensions (still denoted by $\phi _1$ and $\psi _1$) satisfy $\phi _1\in H^2(\mathbb {R}^2)$ and $\psi _1\in H^1(\mathbb {R}^2)$. Let $L_2$ be the lifting operator in Lemma A.2 and $\zeta _1=L_2(\phi _1,\psi _1)$ so that

$$\begin{aligned} \zeta _1(0,x_2,x_3)&=\phi _1(x_2,x_3), \end{aligned}$$

(A.16)

$$\begin{aligned} (\partial \zeta _1/\partial x_1)(0,x_2,x_3)&=\psi _1(x_2,x_3). \end{aligned}$$

(A.17)

Then we define, for $(x_1,x_3)\in \mathbb {R}_+^2$,

$$\begin{aligned} {{\tilde{\phi }}}_2(x_1,x_3)&=\phi _2(x_1,x_3)-\zeta _1(x_1,0,x_3), \end{aligned}$$

(A.18)

$$\begin{aligned} {{\tilde{\psi }}}_2(x_1,x_3)&=\psi _2(x_1,x_3)-(\partial \zeta _1/\partial x_2)(x_1,0,x_3). \end{aligned}$$

(A.19)

Note that ${{\tilde{\phi }}}_2$ belongs to $H^2(\mathbb {R}_+^2)$ and

$$\begin{aligned} {{\tilde{\phi }}}_2(0,x_3)=\phi _2(0,x_3)-\zeta _1(0,0,x_3)=\phi _2(0,x_3)-\phi _1(0,x_3) =0\quad \text {for}\; x_3>0 \end{aligned}$$

by (A.6), (A.16) and (A.18), and

$$\begin{aligned} \frac{\partial {{\tilde{\phi }}}_2}{\partial x_1}(0,x_3)=\frac{\partial \phi _2}{\partial x_1}(0,x_3) -\frac{\partial \zeta _1}{\partial x_1}(0,0,x_3) = \frac{\partial \phi _2}{\partial x_1}(0,x_3)-\psi _1(0,x_3)=0 \end{aligned}$$

by (A.9), (A.17) and (A.18). Furthermore ${{\tilde{\psi }}}_2$ belongs to $H^1(\mathbb {R}_+^2)$ and

$$\begin{aligned} {{\tilde{\psi }}}_2(0,x_3)&=\psi _2(0,x_3)- \frac{\partial \zeta _1}{\partial x_2}(0,0,x_3)\\&=\psi _2(0,x_3)- \frac{\partial \phi _1}{\partial x_2}(0,x_3)=0\quad \text {for}\; x_3>0 \end{aligned}$$

by (A.12), (A.16) and (A.19).

Hence we can extend ${{\tilde{\phi }}}_2$ and ${{\tilde{\psi }}}_2$ to $\mathbb {R}_+\times \mathbb {R}$ by reflection across $x_3=0$ (still denoted by ${{\tilde{\phi }}}_2$ and ${{\tilde{\psi }}}_2$) so that ${{\tilde{\phi }}}_2\in H^2(\mathbb {R}_+\times \mathbb {R})$, ${{\tilde{\psi }}}_2\in H^2(\mathbb {R}_+\times \mathbb {R})$, ${{\tilde{\phi }}}_2(0,x_3)=(\partial {{\tilde{\phi }}}_2/\partial x_1)(0,x_3)=0$ for $x_3\in \mathbb {R}$ and ${{\tilde{\psi }}}_2(0,x_3)=0$ for $x_3\in \mathbb {R}$. Therefore the trivial extensions of ${{\tilde{\phi }}}_2$ and ${{\tilde{\psi }}}_2$ to $\mathbb {R}^2$ (still denoted by ${{\tilde{\phi }}}_2$ and ${{\tilde{\psi }}}_2$) belong to $H^2(\mathbb {R}^2)$ and $H^1(\mathbb {R}^2)$ respectively.

Let $\zeta _2=L_2({{\tilde{\phi }}}_2,{{\tilde{\psi }}}_2)$. Then we have, by Lemma A.2,

$$\begin{aligned} \zeta _2(x_1,0,x_3)&={{\tilde{\phi }}}_2(x_1,x_3), \end{aligned}$$

(A.20)

$$\begin{aligned} (\partial \zeta _2/\partial x_2)(x_1,0,x_3)&={{\tilde{\psi }}}_2(x_1,x_3), \end{aligned}$$

(A.21)

and

$$\begin{aligned} \zeta _2(x_1,x_2,x_3)=0 \quad \hbox { if}\ x_1<0, \end{aligned}$$

which implies

$$\begin{aligned} \zeta _2=\partial \zeta _2/\partial x_1=0 \quad \text {if } x_1=0. \end{aligned}$$

(A.22)

We now define, for $(x_1,x_2)\in \mathbb {R}_+^2$,

$$\begin{aligned} {{\tilde{\phi }}}_3(x_1,x_2)&=\phi _3(x_1,x_2)-\zeta _1(x_1,x_2,0)-\zeta _2(x_1,x_2,0), \end{aligned}$$

(A.23)

$$\begin{aligned} {{\tilde{\psi }}}_3(x_1,x_2)&=\psi _3(x_1,x_2)-(\partial \zeta _1/\partial x_3)(x_1,x_2,0)-(\partial \zeta _2/\partial x_3)(x_1,x_2,0). \end{aligned}$$

(A.24)

Then ${{\tilde{\phi }}}_3$ (resp., ${{\tilde{\psi }}}_3$) belongs to $H^2(\mathbb {R}_+^2)$ (resp., $H^1(\mathbb {R}_+^2)$).

Moreover, it follows from (A.8), (A.16), (A.22) and (A.23) that

$$\begin{aligned} {{\tilde{\phi }}}_3(0,x_2)=\phi _3(0,x_2)-\zeta _1(0,x_2,0)= \phi _3(0,x_2)-\phi _1(x_2,0)=0 \quad \text {for}\;x_2>0, \end{aligned}$$

and (A.10), (A.17), (A.22) and (A.23) imply

$$\begin{aligned} \frac{\partial {{\tilde{\phi }}}_3}{\partial x_1}(0,x_2)&=\frac{\partial \phi _3}{\partial x_1}(0,x_2) -\frac{\partial \zeta _1}{\partial x_1}(0,x_2,0)\\&=\frac{\partial \phi _3}{\partial x_1}(0,x_2)-\psi _1(x_2,0)=0\quad \text {for}\;x_2>0. \end{aligned}$$

From (A.14), (A.16), (A.22) and (A.24) we also have

$$\begin{aligned} {{\tilde{\psi }}}_3(0,x_2)=\psi _3(0,x_2)-\frac{\partial \zeta _1}{\partial x_3}(0,x_2,0)= \psi _3(0,x_2)-\frac{\partial \phi _1}{\partial x_3}(x_2,0)\quad \text {for}\;x_2>0. \end{aligned}$$

Next we check the behavior of ${{\tilde{\phi }}}_3$ and ${{\tilde{\psi }}}_3$ at $x_2=0$. We have

$$\begin{aligned} {{\tilde{\phi }}}_3(x_1,0)&=\phi _3(x_1,0)-\zeta _1(x_1,0,0)-\zeta _2(x_1,0,0)\\&=\phi _3(x_1,0)-\phi _2(x_1,0)=0\quad \text {for}\;x_1>0 \end{aligned}$$

by (A.7), (A.18), (A.20) and (A.23);

$$\begin{aligned} \frac{\partial {{\tilde{\phi }}}_3}{\partial x_2}(x_1,0)&=\frac{\partial \phi _3}{\partial x_2}(x_1,0) -\frac{\partial \zeta _1}{\partial x_2}(x_1,0,0)-\frac{\partial \zeta _2}{\partial x_2}(x_1,0,0)\\&=\frac{\partial \phi _3}{\partial x_2}(x_1,0)-\psi _2(x_1,0)=0\quad \text {for}\;x_1>0 \end{aligned}$$

by (A.11), (A.18), (A.21) and (A.23);

$$\begin{aligned} {{\tilde{\psi }}}_3(x_1,0)&=\psi _3(x_1,0)-\frac{\partial \zeta _1}{\partial x_3}(x_1,0,0)-\frac{\partial \zeta _2}{\partial x_3}(x_1,0,0)\\&=\psi _3(x_1,0)-\frac{\partial \phi _2}{\partial x_3}(x_1,0)=0 \quad \text {for}\;x_1>0 \end{aligned}$$

by (A.13), (A.18), (A.20) and (A.24).

The calculations above show that ${{\tilde{\phi }}}_3=\partial {{\tilde{\phi }}}_3/\partial n={{\tilde{\psi }}}_3=0$ on the boundary of $\mathbb {R}_+^2$. Hence their trivial extensions to $\mathbb {R}^2$ (still denoted by ${{\tilde{\phi }}}_3$ and ${{\tilde{\psi }}}_3$) belongs to $H^2(\mathbb {R}^2)$ and $H^1(\mathbb {R}^2)$.

Let $\zeta _3=L_2({{\tilde{\phi }}}_1,{{\tilde{\psi }}}_1)$. Then we have, by Lemma A.2, $\zeta _3\in H^3(\mathbb {R}^3)$,

$$\begin{aligned} \zeta _3(x_1,x_2,0)&={{\tilde{\phi }}}_3(x_1,x_2), \end{aligned}$$

(A.25)

$$\begin{aligned} (\partial \zeta _3/\partial x_3)(x_1,x_2,0)&={{\tilde{\psi }}}_3(x_1,x_2), \end{aligned}$$

(A.26)

and

$$\begin{aligned} \zeta _3(x_1,x_2,x_3)=0 \quad \text { if } x_1<0 \text { or } x_2<0, \end{aligned}$$

which implies

$$\begin{aligned} \zeta _3=\frac{\partial \zeta _3}{\partial x_1}=0 \quad \text {if } x_1=0 \quad \text {and} \quad \zeta _3=\frac{\partial \zeta _3}{\partial x_2}=0\quad \text {if } x_2=0. \end{aligned}$$

(A.27)

We can now take $\zeta$ to be the restriction of $\zeta _1+\zeta _2+\zeta _3$ to $\mathbb {R}_+^3$, and (A.15) follows from (A.16)–(A.27). $\square$

Finally we have a three-dimensional result that is two-dimensional in nature and which can be derived by using the arguments in the proof of either Lemma A.3 or Lemma A.4.

Lemma A.5

Let $(\phi _1,\psi _1)$ and $(\phi _2,\psi _2)$ belong to $H^2(\mathbb {R}_+\times \mathbb {R})\times H^1(\mathbb {R}_+\times \mathbb {R})$ such that

$$\begin{aligned} \phi _1(0,x_3)&=\phi _2(0,x_3)\qquad \forall \;x_3\in \mathbb {R},\\ \psi _1(0,x_3)&=\frac{\partial \phi _2}{\partial x_1}(0,x_3)\qquad \forall \;x_3\in \mathbb {R},\\ \psi _2(0,x_3)&=\frac{\partial \phi _1}{\partial x_2}(0,x_3)\qquad \forall \;x_3\in \mathbb {R}. \end{aligned}$$

Then there exists $\zeta \in H^\frac{5}{2}(\mathbb {R}_+^3)$ such that

$$\begin{aligned} (\zeta ,\partial \zeta /\partial x_i)=(\phi _i,\psi _i)\quad \text {if } x_i=0, 1\le i\le 2. \end{aligned}$$