A virtual element method for the acoustic vibration problem

Abstract

We analyze in this paper a virtual element approximation for the acoustic vibration problem. We consider a variational formulation relying only on the fluid displacement and propose a discretization by means of $\mathrm {H}(\mathrm {div})$ virtual elements with vanishing rotor. Under standard assumptions on the meshes, we show that the resulting scheme provides a correct approximation of the spectrum and prove optimal order error estimates. With this end, we prove approximation properties of the proposed virtual elements. We also report some numerical tests supporting our theoretical results.

Appendix

We derive in this appendix optimal approximation properties for the $\mathrm {H}(\mathrm {div})$ virtual element with vanishing rotor introduced in Sect. 3. The main goal of this appendix will be to prove the error estimates stated in Lemmas 5 and 6 for the $\varvec{\mathcal {V}}_h$-interpolant defined by (9) and (10). Let us remark that these results could be useful for other applications as well.

Our first result, whose proof is quite straightforward, is a commuting diagram property and some consequences that follow from it. We recall that $P_k$ denotes the $\mathrm {L}^2(\Omega )$-orthogonal projection onto the subspace $\left\{ q\in \mathrm {L}^2(\Omega ): q|_E\in \mathbb {P}_k(E)\quad \forall E\in \mathcal {T}_h\right\}$.

Lemma 5

Let $\varvec{v}\in \varvec{\mathcal {V}}$ be such that $\varvec{v}\in [{\mathrm {H}^{t}(\Omega )}]^2$ with $t>1/2$. Let $\varvec{v}_I\in \varvec{\mathcal {V}}_h$ be its interpolant defined by (9) and (10). Then,

$$\begin{aligned} \mathrm {div}\varvec{v}_I=P_k(\mathrm {div}\varvec{v})\quad \text { in }\Omega . \end{aligned}$$

Consequently, for all $E\in \mathcal {T}_h$, $\left\| \mathrm {div}\varvec{v}_I \right\| _{0,E}\le \left\| \mathrm {div}\varvec{v}\right\| _{0,E}$ and, if $\mathrm {div}\varvec{v}|_{E}\in {\mathrm {H}^{r}(E)}$ with $r\ge 0$, then

$$\begin{aligned} \left\| \mathrm {div}\varvec{v}-\mathrm {div}\varvec{v}_I\right\| _{0,E}\le Ch_E^{\min \{r,k+1\}}\left| \mathrm {div}\varvec{v}\right| _{r,E}. \end{aligned}$$

Proof

As a consequence of (9) and (10), for every element E and for every $q\in \mathbb {P}_k(E)$

$$\begin{aligned} \int _{E}\mathrm {div}(\varvec{v}-\varvec{v}_{I})\,q =\int _{E}(\varvec{v}-\varvec{v}_{I})\cdot \nabla q +\int _{\partial E}\left( \varvec{v}-\varvec{v}_{I}\right) \cdot \varvec{n}\,q\,ds=0. \end{aligned}$$

Since $\mathrm {div}\varvec{v}_{I}\in \mathbb {P}_k(E)$, we have that

$$\begin{aligned} \mathrm {div}\varvec{v}_{I}=P_k(\mathrm {div}\varvec{v})\quad \text {in }E. \end{aligned}$$

Therefore,

$$\begin{aligned} \left\| \mathrm {div}\varvec{v}_I\right\| _{0,E} \le \left\| \mathrm {div}\varvec{v}\right\| _{0,E}. \end{aligned}$$

Additionally, if $\mathrm {div}\varvec{v}|_{E}\in {\mathrm {H}^{r}(E)}$ with r a non-negative integer, as a consequence of [21, Lemma 4.3.8], we have that for every $E\in \mathcal {T}_h$

$$\begin{aligned} \left\| \mathrm {div}\varvec{v}-\mathrm {div}\varvec{v}_I\right\| _{0,E} \le Ch_E^{\min \{r,k+1\}}\left| \mathrm {div}\varvec{v}\right| _{r,E}. \end{aligned}$$

Thus, the second estimate of the lemma follows by standard Banach space interpolation. $\square$

In order to prove Lemma 6 about the $\mathrm {L}^2(\Omega )$ approximation property of this interpolant, we need several previous results. We begin with the following local trace estimate on polygons.

Lemma 14

Let $\varvec{v}\in \varvec{\mathcal {V}}$ and $E\in \mathcal {T}_{h}$ such that $\varvec{v}|_{E}\in [{\mathrm {H}^{t}(E)}]^2$ with $t\in (1/2,1]$. Then, there exists $C>0$ such that

$$\begin{aligned} \left\| \varvec{v}\right\| _{0,\partial E} \le C\left( h_E^{-1/2}\left\| \varvec{v}\right\| _{0,E} +h_E^{t-1/2}\left| \varvec{v}\right| _{t,E}\right) . \end{aligned}$$

Proof

Consider the triangulation $\mathcal {T}_h^{E}$ of the element E obtained by joining each vertex of E with the midpoint of the ball with respect to which E is star-shaped. Since we are assuming that the meshes satisfy $\mathbf {A_1}$ and $\mathbf {A_2}$, the triangles $T\in \mathcal {T}_h^{E}$ have a shape ratio (i.e., the quotient between outer and inner diameters) bounded above by a constant that only depends on $C_\mathcal {T}$. Moreover, each triangle $T\in \mathcal {T}_h^{E}$ has one edge on $\partial E$. Hence, a scaling argument and a trace inequality in the reference triangular element allow us to conclude the proof. $\square$

In order to prove an $\mathrm {L}^2(\Omega )$ error estimate for the interpolant $\varvec{v}_{I}$, we will introduce a basis of $\varvec{\mathcal {V}}_h^E$ dual to the degrees of freedom (1) and (2).

Let $E\in {\mathcal T}_{h}$ with edges $e_1,\ldots ,e_{N_E}$ and $F:E\longrightarrow \widehat{E}$ be an affine mapping of the form

$$\begin{aligned} F\begin{pmatrix} x \\ y \end{pmatrix} :=\dfrac{1}{h_E} \begin{pmatrix} x-x_E \\ y-y_E \end{pmatrix} =: \begin{pmatrix} \widehat{x} \\ \widehat{y} \end{pmatrix}, \end{aligned}$$

where $\varvec{x}_{E}=(x_E,y_E)^T$ is the center of the ball with respect to which E is star-shaped according to assumption $\mathbf {A_2}$. Note that $\widehat{E}:=F(E)$ has diameter 1. Moreover, F maps the above mentioned ball onto a ball of radius $C_{\mathcal {T}}$ with $0<C_{\mathcal {T}}\le 1$ and $C_{\mathcal {T}}$ independent of $h_E$, Moreover, $\widehat{E}$ is star-shaped with respect to each point of this ball.

We define the following basis of $\mathbb {P}_k(E):$

$$\begin{aligned} p_{0}(x,y)&:=1, \\ p_{s}(x,y)&:=\dfrac{\left( x-x_{E}\right) ^{\alpha _{1}} \left( y-y_{E}\right) ^{\alpha _{2}}}{h_{E}^{\alpha _{1}+\alpha _{2}}} +C_{s}, \qquad \alpha _{1},\alpha _{2}\in {\mathbb {N}}, \quad 0<\alpha _{1}+\alpha _{2}\le k, \end{aligned}$$

with the constant $C_{s}\in {\mathbb {R}}$ such that $\int _{E}p_{s}=0$. We have associated above each $s=1,\ldots ,\widetilde{N}:=\mathrm {dim}(\mathbb {P}_k(E))-1$ with one particular couple $(\alpha _{1},\alpha _{2})$, by fixing a particular ordering of these couples. Therefore, the set $\left\{ p_0,p_1,\ldots ,p_{\widetilde{N}}\right\}$ is a basis for $\mathbb {P}_k(E)$ that satisfies $\int _Ep_s=0$ for $s=1,\ldots ,\widetilde{N}$. Let now $\widehat{p}_{s}:=p_{s}\circ F^{-1}$ be defined in $\widehat{E}$. Then, for the particular $(\alpha _{1},\alpha _{2})$ associated with s, we have that $\widehat{p}_{s}(\widehat{x}, \widehat{y}) =\widehat{x}^{\alpha _{1}}\widehat{y}^{\alpha _{2}} +C_{s}$. Moreover, since $\left| E\right| =h_E^2|\widehat{E}|$, we have

$$\begin{aligned} C_s=-\dfrac{1}{\left| E\right| }\int _{E} \dfrac{\left( x-x_{E} \right) ^{\alpha _{1}} \left( y-y_{E}\right) ^{\alpha _{2}}}{h_{E}^{\alpha _{1}+\alpha _{2}}}\,dx\,dy =-\dfrac{1}{\left| \widehat{E}\right| }\int _{\widehat{E}} \widehat{x}^{\alpha _1} \widehat{y}^{\alpha _2} \,d\widehat{x}\,d\widehat{y}. \end{aligned}$$

As a consequence, note that $\left| C_s\right| \le 1$ and, hence, $\left\| p_s\right\| _{\infty ,E} =\left\| \widehat{p}_s \right\| _{ \infty ,\widehat{E}}\le 2$, $s=0,\ldots ,\widetilde{N}$.

For each edge $e_l$ of E ($l=1,\ldots ,N_E$), let $T_{l}$ be the affine function mapping $\widehat{e}:=[-1,1]$ onto $e_l$. We define $q_l^i:=\widehat{q}\,^{i}\circ T_{l}^{-1}$ ($i=1,\dots ,k$) with $\widehat{q}\,^{i}$ being the Legendre polynomials on $[-1,1]$ normalized by $\widehat{q}\,^{i}(1)=1$. Then, $\left\{ q_l^0, \ldots ,q_{l}^k\right\}$ is a basis of $\mathbb {P}_k(e_l)$ which satisfies $q_l^0=1$, $\int _{e_l} q_l^iq_l^j\,ds=\delta _{ij}$, $i,j=1,\ldots ,k$, and $\left\| q_l^i\right\| _{\infty ,e_l}=1$. Note that, in particular, $\int _{e_l} q_l^i\,ds=0$, $i=1,\ldots ,k$.

Therefore,

$$\begin{aligned} \left\{ q_{l}^{i}\right\} _{i=0,\ldots ,k,\ l=1,\ldots ,N_E} \qquad \text {and}\qquad \left\{ p_{s} \right\} _{s=1,\ldots , \widetilde{N}} \end{aligned}$$

are bases for the spaces of test functions appearing in the degrees of freedom (9) and (10), respectively. Next, we introduce a set of dual basis functions for $\varvec{\mathcal {V}}_h^E$:

$$\begin{aligned} \left\{ \varvec{\varphi }_l^i\right\} _{i=0,\ldots ,k,\ l=1,\ldots ,N_E} \cup \left\{ \widetilde{\varvec{\varphi }}^s\right\} _{s=1,\ldots ,\widetilde{N}}. \end{aligned}$$

(23)

The first ones, $\varvec{\varphi }_{l}^i$, are the “boundary basis functions” determined by

$$\begin{aligned}&\varvec{\varphi }_l^i\in \varvec{\mathcal {V}}_{h}^{E}, \end{aligned}$$

(24)

$$\begin{aligned}&\int _{e_{m}}\left( \varvec{\varphi }_l^i\cdot \varvec{n}\right) q_{m}^j\,ds=\delta _{lm} \delta _{ij}, \qquad m=1,\ldots ,N_{E},\quad j=0,\ldots ,k, \end{aligned}$$

(25)

$$\begin{aligned}&\int _{E}\left( \mathrm {div}\varvec{\varphi }_l^i\right) p_{r}=0, \qquad r=1,\ldots , \widetilde{N}. \end{aligned}$$

(26)

Note that these boundary basis functions use two indexes, i and l, one for the moment and the other for the edge. On the other hand, note also that as a consequence of (24) and (25) $\varvec{\varphi }_l^i\cdot \varvec{n}=0$ on $\partial E\setminus e_{l}$ The second kind of functions in (23), $\widetilde{\varvec{\varphi }}^s$, are the “internal basis functions” determined by

$$\begin{aligned}&\widetilde{\varvec{\varphi }}^s\in \varvec{\mathcal {V}}_{h}^{E}, \end{aligned}$$

(27)

$$\begin{aligned}&\widetilde{\varvec{\varphi }}^s|_{\partial E}\cdot \varvec{n}=0,\end{aligned}$$

(28)

$$\begin{aligned}&\int _{E}\left( \mathrm {div}\widetilde{\varvec{\varphi }}^s\right) p_{r} =\delta _{sr}, \qquad r=1,\ldots ,\widetilde{N}. \end{aligned}$$

(29)

Remark 5

Since $\mathrm {div}\varvec{\varphi }_{l}^i\in \mathbb {P}_k(E) =\mathrm {span}\left\{ 1,p_1,\ldots ,p_s\right\}$ and $\int _Ep_{s}=0$ for $s=1,\ldots ,\widetilde{N}$, Eq. (26) implies that $\mathrm {div}\varvec{\varphi }_{l}^i$ has to be constant. Therefore,

$$\begin{aligned} \mathrm {div}\varvec{\varphi }_{l}^i =\dfrac{1}{\left| E\right| }\int _{E}\mathrm {div}\varvec{\varphi }_{l}^i =\dfrac{1}{\left| E\right| }\int _{\partial E}\varvec{\varphi }_{l}^i\cdot \varvec{n}\,ds. \end{aligned}$$

Moreover, thanks to (25), we have that

$$\begin{aligned} \int _{\partial E}\varvec{\varphi }_{l}^i\cdot \varvec{n}\,ds=\sum _{m=1}^{N_{E}} \int _{e_{m}}\left( \varvec{\varphi }_l^i\cdot \varvec{n}\right) q_{m}^0\,ds=\sum _{m=1}^{N_{E}}\delta _{l m}\delta _{i0}=\delta _{i0}. \end{aligned}$$

Then,

$$\begin{aligned} \mathrm {div}\varvec{\varphi }_{l}^i =\dfrac{\delta _{i0}}{\left| E\right| }. \end{aligned}$$

Next goal is to prove that all the functions in (23) are bounded uniformly in h. We begin with the boundary basis functions.

Lemma 15

There exists $C>0$ such that $\left\| \varvec{\varphi }_{l}^i\right\| _{0,E}\le C$ for $l=1,\ldots ,N_{E}$ and $i=0,\ldots ,k$.

Proof

Since $\varvec{\varphi }_{l}^i\in \varvec{\mathcal {V}}_h^E$, we know that $\mathrm {rot}\varvec{\varphi }_{l}^i=0$. Therefore, there exists $\gamma \in {\mathrm {H}^1(E)}$ such that $\varvec{\varphi }_{l}^i=\nabla \gamma$. Hence, from the remark above and (25), we have that $\gamma$ is a solution of the following problem:

$$\begin{aligned} \left\{ \begin{array}{ll} \Delta \gamma =\dfrac{\delta _{i0}}{\left| E\right| }&{}\quad \text {in}\,E,\\ \dfrac{\partial \gamma }{\partial \varvec{n}} =\varvec{\varphi }_{l}^i\cdot \varvec{n}&{} \quad \text {on }\partial E, \\ \displaystyle \int _E\gamma =0. \end{array}\right. \end{aligned}$$

It is easy to check that these Neumann problems are compatible. Therefore,

$$\begin{aligned}&\int _{E}\nabla \gamma \cdot \nabla \zeta =\int _{\partial E} \left( \varvec{\varphi }_{l}^i\cdot \varvec{n}\right) \zeta \,ds-\int _E \dfrac{ \delta _{i0}}{\left| E\right| }\zeta \\&\quad =\int _{e_l} \left( \varvec{\varphi }_{l}^i \cdot \varvec{n}\right) \zeta \,ds\qquad \forall \zeta \in {\mathrm {H}^1(E)}:\ \int _{E}\zeta =0. \end{aligned}$$

Now, taking $\zeta =\gamma$, we obtain

$$\begin{aligned} \left\| \varvec{\varphi }_{l}^i\right\| _{0,E}^2&=\left\| \nabla \gamma \right\| _{0,E}^2 \le \left\| \varvec{\varphi }_{l}^i\cdot \varvec{n}\right\| _{0,e_l} \left\| \gamma \right\| _{0,e_l} \\&\le C\left\| \varvec{\varphi }_{l}^i\cdot \varvec{n}\right\| _{0,e_l} \left( h_E^{-1/2} \left\| \gamma \right\| _{0,E} +h_E^{1/2}\left\| \nabla \gamma \right\| _{0,E}\right) \\&\le C h_E^{1/2}\left\| \varvec{\varphi }_{l}^i\cdot \varvec{n}\right\| _{0,e_l} \left\| \nabla \gamma \right\| _{0,E}, \end{aligned}$$

where we have used Lemma 14 with $t=1$, the generalized Poincaré inequality and a scaling argument. Now, because of (25) with $m=l$ and the orthogonality property of Legendre polynomials, $\left. \varvec{\varphi }_l^i\cdot \varvec{n}\right| _{e_l} =\left( \int _{e_l}\left( q^{i}_l\right) ^2\,ds\right) ^{-1}q_l^i$. Therefore,

$$\begin{aligned} \left\| \varvec{\varphi }_{l}^i\cdot \varvec{n}\right\| _{0,e_l}^2 =\left( \int _{e_l} \left( q^i_l\right) ^2\,ds\right) ^{-1} =\dfrac{1}{h_E} \left( \int _{\widehat{e}} \left( \widehat{q}\,^i\right) ^2\,d \widehat{s} \right) ^{-1}. \end{aligned}$$

Thus, from the last two estimates we derive that $\left\| \varvec{\varphi }_{l}^i\right\| _{0,E}\le C$ and we end the proof. $\square$

Next, we show a similar result for the internal basis functions.

Lemma 16

There exists $C>0$ such that $\left\| \widetilde{\varvec{\varphi }}^s \right\| _{0,E}\le C$ for $s=1,\ldots ,\widetilde{N}$.

Proof

Since $\widetilde{\varvec{\varphi }}^s\in \varvec{\mathcal {V}}_h^E$, there exists $\gamma \in {\mathrm {H}^1(E)}$ such that $\widetilde{\varvec{\varphi }}^s=\nabla \gamma$. Hence, by virtue of (28), we have that $\gamma$ is a solution of the following well posed Neumann problem:

$$\begin{aligned} \left\{ \begin{array}{ll} \Delta \gamma =-\mathrm {div}\widetilde{\varvec{\varphi }}^s &{} \quad \text {in }E, \\ \dfrac{\partial \gamma }{\partial \varvec{n}}=0 &{} \quad \text {on }\partial E,\\ \displaystyle \int _E\gamma =0. \end{array}\right. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{E}\nabla \gamma \cdot \nabla \zeta =-\int _{E}\psi ^s\zeta \qquad \forall \zeta \in {\mathrm {H}^1(E)}:\ \int _{E}\zeta =0, \end{aligned}$$

where $\psi ^s:=\mathrm {div}\widetilde{\varvec{\varphi }}^s$. Now, taking $\zeta =\gamma$ and using the generalized Poincaré inequality and a scaling argument, we have that

$$\begin{aligned} \left\| \widetilde{\varvec{\varphi }}^s\right\| _{0,E}^2 =\left\| \nabla \gamma \right\| _{0,E}^2 \le C\left\| \psi ^s\right\| _{0,E} \left\| \gamma \right\| _{0,E} \le Ch_E\left\| \psi ^s\right\| _{0,E} \left\| \nabla \gamma \right\| _{0,E}. \end{aligned}$$

Thus,

$$\begin{aligned} \left\| \widetilde{\varvec{\varphi }}^s\right\| _{0,E} \le C h_E\left\| \psi ^s\right\| _{0,E}. \end{aligned}$$

(30)

On the other hand, since $\psi ^s\in \mathbb {P}_k(E)$, it is easy to check that

$$\begin{aligned} h_{E}\left\| \psi ^{s}\right\| _{0,E} \le Ch_{E}^{2} \left\| \psi ^{s} \right\| _{\infty ,E} =Ch_{E}^{2}\,\left\| \widehat{\psi }^{s} \right\| _{\infty ,\widehat{E}}, \end{aligned}$$

(31)

where $\widehat{\psi }^{s} :=(\psi ^{s}\circ F^{-1})\in \mathbb {P}_k ( \widehat{E})$.

For $\widehat{\psi }^{s}\in \mathbb {P}_k(\widehat{E})$, we write $\widehat{\psi }^{s}=\sum _{i=1}^{\widetilde{N}}\beta _i^s\widehat{p}_i$ and, since $\left\| \widehat{p}_{i}\right\| _{\infty ,\widehat{E}}\le 2$, we have that

$$\begin{aligned} \left\| \widehat{\psi }^{s} \right\| _{\infty ,\widehat{E}} \le \max _{1\le i\le \widetilde{N}} \left| \beta _{i}^s\right| \sum _{i=1}^{\widetilde{N}} \left\| \widehat{p}_{i}\right\| _{\infty ,\widehat{E}} \le C\max _{1\le i\le \widetilde{N}} \left| \beta _{i}^s\right| . \end{aligned}$$

(32)

Now, from (29), a change of variables from E to $\widehat{E}$ yields

$$\begin{aligned} \int _{\widehat{E}}\widehat{\psi }^{s}\widehat{p}_{r} =h_{E}^{-2}\delta _{sr}, \qquad r=1,\ldots ,\widetilde{N}, \end{aligned}$$

which can be written as

$$\begin{aligned} \sum _{i=1}^{\widetilde{N}} \beta _i^s\int _{\widehat{E}}\widehat{p}_i\widehat{p}_{r} =h_{E}^{-2}\delta _{sr}, \qquad r=1,\ldots ,\widetilde{N}. \end{aligned}$$

(33)

Let

$$\begin{aligned} \varvec{M}= \begin{pmatrix} m_{ir} \end{pmatrix} \in {\mathbb {R}}^{\widetilde{N}\times \widetilde{N}} \qquad \text {with } m_{ir}:=\int _{\widehat{E}}\widehat{p}_{i}\widehat{p}_{r}, \quad i,r=1,\ldots ,\widetilde{N}. \end{aligned}$$

Therefore, from (33), if $\varvec{M}$ is invertible, then $\varvec{\beta }^s=\begin{pmatrix}\beta _1^s&\cdots&\beta _{\widetilde{N}}^s\end{pmatrix}^T$ is equal to $h_{E}^{-2}$ times the s-th column of $\varvec{M}^{-1}$.

Next, we will show that $\varvec{M}$ is invertible and that its inverse is bounded uniformly in h. With this aim, note that the polygon $\widehat{E}$ is uniquely defined by the vector $((\widehat{x}_1, \widehat{y}_1),\ldots , (\widehat{x}_{N_E}, \widehat{y}_{N_E})) \in {\mathbb {R}}^{2N_{E}}$ that collects the coordinates of its (ordered) vertexes. Let $U\subset {\mathbb {R}}^{2N_{E}}$, be the set of all possible values of these coordinates such that the mesh regularity assumptions $\mathbf {A_1}$ and $\mathbf {A_2}$ are satisfied. Since the diameter of $\widehat{E}$ is equal to 1, U is a bounded set. On the other hand, the constraints that arise from hypotheses $\mathbf {A_1}$ and $\mathbf {A_2}$ yield that U is a closed set. Therefore U is compact.

The function from U into ${\mathbb {R}}^{\widetilde{N}\times \widetilde{N}}$ that maps the coordinates of the vertexes of $\widehat{E}$ into the entries of the matrix $\varvec{M}$ is a continuous function. Moreover, for any coordinates in U, $\widehat{E}$ satisfies $\mathbf {A_1}$ and $\mathbf {A_2}$ and, hence, it contains a ball of radius $C_\mathcal {T}$. Let us show that this implies that $\varvec{M}$ has to be positive definite. In fact, given $\alpha \in {\mathbb {R}}^{\widetilde{N}}$, $\alpha ^{T} \varvec{M}\alpha = \int _{\widehat{E}} |\sum _{r=1}^{\widetilde{N}} \alpha _{r} \widehat{p}_{r}|^{2} \ge 0$ and the equality holds only if $\sum _{r=1}^{\widetilde{N}}\alpha _{r}\widehat{p}_{r}$ vanishes a.e. in $\widehat{E}$, which in turn implies that $\alpha$ has to vanish (since $\widehat{E}$ contains a ball of radius $C_\mathcal {T}>0$). Thus, $\varvec{M}$ is positive definite and hence invertible. Therefore, taking also into account the continuity of the mapping $\varvec{M}\longmapsto \varvec{M}^{-1}$ for invertible matrices, we conclude that the mapping

$$\begin{aligned} U\ni ((\widehat{x}_1,\widehat{y}_1),\ldots , (\widehat{x}_{N_E}, \widehat{y}_{N_E})) \longmapsto \varvec{M}^{-1} \in {\mathbb {R}}^{\widetilde{N} \times \widetilde{N}} \end{aligned}$$

is well defined and continuous and, hence, bounded above in the compact set U. Consequently, from (33),

$$\begin{aligned} \left\| \varvec{\beta }^s\right\| _{\infty }\le Ch_E^{-2}, \end{aligned}$$

which recalling (32) yields

$$\begin{aligned} \left\| \widehat{\psi }^{s} \right\| _{\infty ,\widehat{E}}\le Ch_E^{-2}. \end{aligned}$$

(34)

Let us remark that, in principle, the constant C above depends on the number $N_E$ of vertexes of E. However, by virtue of assumption $\mathbf {A_1}$, this number is bounded above in terms of $C_{{\mathcal T}}$. Therefore, $N_E$ can take only a finite number of possible values and, hence, (34) holds true with C only depending on $C_{{\mathcal T}}$. Thus, we conclude the proof by combining (30), (31) and (34). $\square$

Now, we are in a position to prove $\mathrm {L}^2(\Omega )$ error estimates for the $\varvec{\mathcal {V}}_h$-interpolant.

Lemma 6

Let $\varvec{v}\in \varvec{\mathcal {V}}$ be such that $\varvec{v}\in [{\mathrm {H}^{t}(\Omega )}]^2$ with $t >1/2$. Let $\varvec{v}_I\in \varvec{\mathcal {V}}_h$ be its interpolant defined by (9) and (10). Let $E\in \mathcal {T}_h$. If $1\le t\le k+1$, then

$$\begin{aligned} \left\| \varvec{v}-\varvec{v}_I\right\| _{0,E}\le Ch_E^{t}\left| \varvec{v}\right| _{t,E}, \end{aligned}$$

(35)

whereas, if $1/2 <t\le 1$, then

$$\begin{aligned} \left\| \varvec{v}-\varvec{v}_I\right\| _{0,E} \le C\left( h_E^{t} \left| \varvec{v}\right| _{t,E} +h_E\left\| \mathrm {div}\varvec{v}\right\| _{0,E}\right) . \end{aligned}$$

(36)

Proof

First, we consider the case $1\le t\le k+1$. The first step is to bound $\left\| \varvec{v}_I\right\| _{0,E}$. Since $\varvec{v}_{I}\in \varvec{\mathcal {V}}_{h}^{E}$, thanks to (24)–(29) we write it in the basis (23) as follows:

$$\begin{aligned} \varvec{v}_I=\sum _{l=1}^{N_E}\sum _{i=0}^{k} \left( \int _{e_l} \left( \varvec{v}\cdot \varvec{n}\right) q_{l}^i\,ds\right) \varvec{\varphi }_l^{i} +\sum _{s=1}^{ \widetilde{N}} \left( \int _{E}\left( \mathrm {div}\varvec{v}\right) p_{s}\right) \widetilde{\varvec{\varphi }}^{s}. \end{aligned}$$

Therefore, from Lemmas 15 and 16 we have

$$\begin{aligned} \left\| \varvec{v}_I\right\| _{0,E} \le C\left( \sum _{l=1}^{N_E} \sum _{i=0}^{k}\left| \int _{e_l}\left( \varvec{v}\cdot \varvec{n}\right) q_{l}^i\,ds\right| +\sum _{s=1}^{\widetilde{N}} \left| \int _{E} \left( \mathrm {div}\varvec{v}\right) p_{s}\right| \right) . \end{aligned}$$

Then, by using that $\left\| q_{l}^{i}\right\| _{\infty ,e_{l}}\!\!\!=1$ for $i=1,\dots ,k$ and $l=1,\ldots ,N_{E}$, $\left\| p_{s}\right\| _{\infty ,E}\le C$ for $s=1,\dots ,\widetilde{N}$, the Cauchy–Schwarz inequality and Lemma 14, we obtain

$$\begin{aligned} \left\| \varvec{v}_I\right\| _{0,E}&\le C\left( h_E^{1/2} \left\| \varvec{v}\right\| _{0,\partial E} \left\| q_{l}^i \right\| _{\infty ,e_l} +\widetilde{N}h_E\left\| \mathrm {div}\varvec{v}\right\| _{0,E} \left\| p_{s} \right\| _{\infty ,E}\right) \nonumber \\&\le C\left( \left\| \varvec{v}\right\| _{0,E} +h_E\left| \varvec{v}\right| _{1,E} +h_{E}\left\| \mathrm {div}\varvec{v}\right\| _{0,E}\right) \nonumber \\&\le C\left( \left\| \varvec{v}\right\| _{0,E} +h_E\left| \varvec{v}\right| _{1,E}\right) . \end{aligned}$$

(37)

Now, for all $\varvec{v}_k\in [\mathbb {P}_k(E)]^2$ we note that $(\varvec{v}_k)_I=\varvec{v}_k$ and, hence, using the above estimate for $\varvec{v}-\varvec{v}_{k}$, we write

$$\begin{aligned}&\left\| \varvec{v}-\varvec{v}_I\right\| _{0,E} =\left\| \varvec{v}-\varvec{v}_k -\left( \varvec{v}-\varvec{v}_k \right) _I\right\| _{0,E} \le \left\| \varvec{v}-\varvec{v}_k\right\| _{0,E}\\&\quad +\,C \left( \left\| \varvec{v}-\varvec{v}_k\right\| _{0,E} +h_E\left| \varvec{v}-\varvec{v}_k \right| _{1,E}\right) . \end{aligned}$$

Thus, by choosing $\varvec{v}_{k}$ as in [7, Proposition 4.2], we have that $\left\| \varvec{v}-\varvec{v}_k\right\| _{0,E} +h_E \left| \varvec{v}-\varvec{v}_k \right| _{1,E}\le C h_E^t\left| \varvec{v}\right| _{t,E}$, which together with the above inequality allow us to conclude (35).

Next, we consider the case $1/2 <t\le 1$. Using the same arguments as above, we obtain in this case instead of (37),

$$\begin{aligned} \left\| \varvec{v}_I\right\| _{0,E} \le C\left( \left\| \varvec{v}\right\| _{0,E} +h_E^{t}\left| \varvec{v}\right| _{t,E} +h_{E}\left\| \mathrm {div}\varvec{v}\right\| _{0,E} \right) . \end{aligned}$$

(38)

Therefore, repeating again the arguments above with $\varvec{v}_0\in [\mathbb {P}_0(E)]^2$ instead of $\varvec{v}_{k}$, we have

$$\begin{aligned} \left\| \varvec{v}-\varvec{v}_I\right\| _{0,E}&\le \left\| \varvec{v}-\varvec{v}_0 -\left( \varvec{v}-\varvec{v}_0\right) _I\right\| _{0,E} \\&\le \left\| \varvec{v}-\varvec{v}_0\right\| _{0,E} +C\left( \left\| \varvec{v}-\varvec{v}_0\right\| _{0,E} +h_E^t\left| \varvec{v}\right| _{t,E} +h_E\left\| \mathrm {div}\varvec{v}\right\| _{0,E}\right) \\&\le C\left( h_E^t\left| \varvec{v}\right| _{t,E} +h_E\left\| \mathrm {div}\varvec{v}\right\| _{0,E}\right) , \end{aligned}$$

where we have used again [7, Proposition 4.2]. Thus, the proof is complete. $\square$

Remark 6

Estimate (36) can be improved for $k=0$ and $1/2<t\le 1$. In fact, in such a case, the interpolant $\varvec{v}_I\in \varvec{\mathcal {V}}_h$ is defined only by (9). Hence,

$$\begin{aligned} \varvec{v}_I=\sum _{l=1}^{N_E}\left( \int _{e_l} \left( \varvec{v}\cdot \varvec{n}\right) q_{l}^0\,ds\right) \varvec{\varphi }_l^{0} \end{aligned}$$

and repeating the arguments above we obtain

$$\begin{aligned} \left\| \varvec{v}_I\right\| _{0,E} \le C\left( \left\| \varvec{v}\right\| _{0,E} +h_E^{t}\left| \varvec{v}\right| _{t,E}\right) , \end{aligned}$$

instead of (38), which leads to

$$\begin{aligned} \left\| \varvec{v}-\varvec{v}_I\right\| _{0,E} \le C h_E^t\left| \varvec{v}\right| _{t,E}. \end{aligned}$$