DDFV method for Navier–Stokes problem with outflow boundary conditions

Abstract

We propose a Discrete Duality Finite Volume scheme (DDFV for short) for the unsteady incompressible Navier–Stokes problem with outflow boundary conditions. As in the continuous case, those conditions are derived from a weak formulation of the equations and they provide an energy estimate of the solution. We prove well-posedness of the scheme and a discrete energy estimate. Finally we perform some numerical tests simulating the flow behind a cylinder inside a long channel to show the robustness of such conditions in the DDFV framework.

Appendices

Appendix A: A few useful results

1.1 A.1 Basic inequalities

Lemma 2

(Young’s inequality) Let a, b, c be three non negative numbers. Let $p_1, p_2$ and $p_3$ be positive real numbers such that $\frac{1}{p_1}+ \frac{1}{p_2}+\frac{1}{p_3} =1$. Then, we have:

$$\begin{aligned} abc \le \frac{C_1}{p_1} \, a^{p_1} + \frac{C_2}{p_2} \, b^{p_2} +\frac{1}{p_3 \, C_1 C_2 } \, c^{p_3} \end{aligned}$$

for some positive constants $C_1, C_2$,

We adapted the proof of Grönwall’s lemma, Lemma 16.I.6 in [24], to obtain the following:

Lemma 3

(Discrete Grönwall’s lemma) If a sequence $(a_{n})_{n}$, $n=0 \dots N$, satisfies

$$\begin{aligned} a_{0} \le A, \quad \quad a_{n}\le A + B \delta t \sum _{i=0}^{n-1}a_i \,\,\,\,\, \forall n \in {1, \dots N} , \delta t = \frac{T}{N} \end{aligned}$$

where A and B are two positive constants independent of $\delta t$, then

$$\begin{aligned} \max _{n=1\dots N} a_n \le A e^{BT}. \end{aligned}$$

Lemma 4

(Hölder’s inequality) Let $p,q,r \in (1,+\infty )$ with $1/p + 1/q +1/r =1$. For every $(x_1, \dots x_n)$, $(y_1, \dots y_n), (z_1, \dots z_n) \in {{\mathbb {R}}}^n$ it holds

$$\begin{aligned} \sum _{i=1}^n |x_i \, y_i \, z_i | \le \left( \sum _{i=1}^n |x_i|^p \right) ^{1/p} \left( \sum _{i=1}^n |y_i|^q \right) ^{1/q} \left( \sum _{i=1}^n |z_i|^r \right) ^{1/r} \end{aligned}$$

1.2 A.2 Trace inequalities

Given a vector $\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}= ((\mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}})_{{\scriptstyle \mathtt {K}}\in {\mathfrak {M}}\cup \partial {\mathfrak {M}}}, (\mathbf{u }_{{{\scriptscriptstyle \mathtt {K}^*}}})_{{\scriptstyle \mathtt {K}^*}\in {\mathfrak {M}}^*\cup \partial {\mathfrak {M}}^*})$ defined on a DDFV mesh $\mathfrak {T}$, we associate the approximate solution on the boundary in two different ways:

$$\begin{aligned} \widetilde{\gamma } (\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}})= & {} \frac{1}{2} \sum _{{\scriptstyle \mathtt {K}}\in {\mathfrak {M}}} \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\mathbf{1 }_{{\bar{{\scriptstyle \mathtt {K}}}} \cap \partial \varOmega } + \frac{1}{2} \sum _{{\scriptstyle \mathtt {K}^*}\in \partial {\mathfrak {M}}^*} \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}^*}}}\mathbf{1 }_{{\bar{{\scriptstyle \mathtt {K}}}}^* \cap \partial \varOmega }.\\ \gamma (\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}})= & {} \frac{1}{2} \sum _{{\scriptstyle \mathtt {L}}\in \partial {\mathfrak {M}}} \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\mathbf{1 }_{{\scriptstyle \mathtt {L}}} + \frac{1}{2} \sum _{{\scriptstyle \mathtt {K}^*}\in \partial {\mathfrak {M}}^*} \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}^*}}}\mathbf{1 }_{{\bar{{\scriptstyle \mathtt {K}}}}^* \cap \partial \varOmega }. \end{aligned}$$

With this definition, we use simultaneously the values on the primal mesh and the values on the dual mesh. We can also consider two different reconstructions based either on the primal values or the dual values:

$$\begin{aligned} \widetilde{\gamma }^{\partial \mathfrak {M}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}})= & {} \sum _{{\scriptstyle \mathtt {K}}\in {\mathfrak {M}}} \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\mathbf{1 }_{{\bar{{\scriptstyle \mathtt {K}}}} \cap \partial \varOmega } \quad \text {or} \quad \gamma ^{\partial \mathfrak {M}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}})= \sum _{{\scriptstyle \mathtt {L}}\in \partial {\mathfrak {M}}} \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\mathbf{1 }_{{\scriptstyle \mathtt {L}}}\\ \widetilde{\gamma }^{\partial {{\mathfrak {M}}^*}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}})= & {} \gamma ^{\partial {{\mathfrak {M}}^*}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) = \sum _{{\scriptstyle \mathtt {K}^*}\in \partial {\mathfrak {M}}^*} \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}^*}}}\mathbf{1 }_{{\bar{{\scriptstyle \mathtt {K}}}}^* \cap \partial \varOmega }(\text {x}). \end{aligned}$$

We point out that, if we consider the object we want to estimate, we have for both cases (by Minkowski’s inequality):

$$\begin{aligned} \begin{aligned} \Vert \widetilde{\gamma } (\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) \Vert _{q,\partial \varOmega }&\le \Vert \widetilde{\gamma }^{\partial \mathfrak {M}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) \Vert _{q,\partial \varOmega } + \Vert \widetilde{\gamma }^{\partial {{\mathfrak {M}}^*}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}})\Vert _{q,\partial \varOmega },\\ \Vert \gamma (\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) \Vert _{q,\partial \varOmega }&\le \Vert \gamma ^{\partial \mathfrak {M}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) \Vert _{q,\partial \varOmega } + \Vert \gamma ^{\partial {{\mathfrak {M}}^*}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}})\Vert _{q,\partial \varOmega }. \end{aligned} \end{aligned}$$

Before proving the trace theorem, we introduce a discrete Poincaré inequality.

Theorem 8

(Discrete Poincaré inequality) [2, Theorem 11] Let $\varOmega$ be an open connected bounded polygonal domain of ${\mathbb {R}}^2$ and $\varGamma _0$ be a part of the boundary such that $m(\varGamma _0) >0$. Let $\mathfrak {T}$ be a DDFV mesh associated to $\varOmega$.

• If $1 \le p <2$, let $1 \le q \le p^*$

• If $p \ge 2$, let $1 \le q < \infty$.

There exists a constant $C>0$, depending only on p,q, $\varGamma _0$ and $\varOmega$ such that $\forall \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\in {\mathbb {E}}^{\varGamma _0}_0$:

$$\begin{aligned} \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _q \le \frac{C}{\sin (\alpha _{\mathfrak {T}})^{\frac{1}{p}} \,\, \text {reg}(\mathfrak {T})^{\frac{p-1}{p}}} \,\, \Vert \nabla ^{{\mathfrak {D}}} \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _p . \end{aligned}$$

Theorem 9

(Trace inequality) Let $\mathfrak {T}$ be a DDFV mesh associated to $\varOmega$. For all $p>1$ there exists a constant $C>0$, depending only on p, $\sin (\alpha _{\mathfrak {T}})$, reg($\mathfrak {T}$) and $\varOmega$ such that $\forall \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\in {\mathbb {E}}^{\varGamma _0}_0$ and $\forall s\ge 1$:

$$\begin{aligned} \boxed { \boxed { \Vert \widetilde{\gamma } (\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) \Vert _{s,\partial \varOmega }^s \le C \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{1,p} \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{\frac{p(s-1)}{p-1}}^{s-1}.}} \end{aligned}$$

(28)

The computations of the proof are similar to those present in [11, 13]. In [13], the proof is given for finite volume methods; in [11], the proof is given for DDFV method but in the case of $L^1$ norm. Our proof has been adapted to the vectorial case and to general $L^s$, $L^p$ norms.

Proof

Boundary properties By compactness of $\partial \varOmega$, there exists a finite number of open hyper-rectangles $\{R_i, i=1 \dots N\}$, and normalized vectors of ${\mathbb {R}}^2, \{\eta _i, i=1,\dots ,N\}$, such that (Fig. 13):

$$\begin{aligned} \left\{ \begin{array}{ll} \partial \varOmega \subset \bigcup _{i=1}^N R_i\\ \\ (\eta _i,\overrightarrow{\nu }(x)) \ge \lambda >0 \quad \forall x \in R_i \cap \partial \varOmega , i\in \{1 \dots N\}\\ \\ \{x+t \eta _i, x \in R_i\cap \partial \varOmega , t \in {\mathbb {R}}^+ \}\cap R_i \subset \varOmega ,\\ \end{array} \right. \\ \end{aligned}$$

where $\lambda$ is a strictly positive number and $\overrightarrow{\nu }(x)$ is the normal vector to $\partial \varOmega$ at x, inward to $\varOmega$ (see Fig. 1). Let $\{\lambda _i, i=1\dots N\}$ be a family of functions such that $\sum _{i=1}^N \lambda _i(x)=1$, for all $x\in \partial \varOmega$, $\lambda _i \in C^{\infty }_c({\mathbb {R}}^2, {\mathbb {R}}^+)$ and $\lambda _i=0$ outside of $R_i$, for all $i=1\dots N$. Let $\partial \varOmega _i= R_i \cap \partial \varOmega$; we shall prove that there exists $C_i >0$ depending only on $\lambda , \text {reg}(\mathfrak {T})$ and $\lambda _i$ such that

$$\begin{aligned} \int _{\partial \varOmega _i} \lambda _i(x) \vert \widetilde{\gamma }^{\partial \mathfrak {M}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}})(x) \vert ^s \text {d}x + \int _{\partial \varOmega _i} \lambda _i(x) \vert \widetilde{\gamma }^{\partial {{\mathfrak {M}}^*}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}})(x) \vert ^s \text {d}x \le C_i \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{1,p} \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{\frac{p(s-1)}{p-1}}^{s-1}. \end{aligned}$$

Then it will be sufficient to define $C:= \sum _{i=1}^N C_i$ to get (28). We study separately the two terms.

Fig. 13

Properties of the boundary $\partial \varOmega$

On the primal mesh We introduce the functions to determine the successive neighbours of a cell $\mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}$. Consider $x,y \in \varOmega$, then:

$$\begin{aligned} \text {for }\sigma \in {\mathcal {E}} \quad \varPsi _{\sigma }(x,y):={\left\{ \begin{array}{ll} 1 &{} \text {if }[x,y]\cap \sigma \ne \emptyset \\ 0 &{} \text {otherwise}, \end{array}\right. } \\ \text {for }{\scriptstyle \mathtt {K}}\in {\mathfrak {M}} \quad \varPsi _{{\scriptstyle \mathtt {K}}}(x,y):={\left\{ \begin{array}{ll} 1 &{} \text {if }[x,y]\cap {\scriptstyle \mathtt {K}}\ne \emptyset \\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Fig. 14

(Up) $[x,y(x)]\cap \sigma _0$ is reduced to a point $z_{\sigma _0}(x)$. (Down) $[x,y(x)]\cap \sigma _0$ is the segment [x, y(x)]

Now, we fix $i\in \{1 \dots N\}$ and $x\in \partial \varOmega _i$.

Then there exists a unique $t>0$ such that $x+ t \eta _i = y(x) \in \partial R_i$. Then, for $\sigma \in {\mathcal {E}}$, if $[x,y(x)] \cap \sigma \ne \emptyset$, then it is:

• either a point: $z_{\sigma }(x):=[x,y(x)] \cap \sigma$

• either a segment: $[a(x), b(x)]:=[x, y(x)] \cap \sigma$ and let $z_{\sigma }(x):=b(x)$.

For $K \in {\mathfrak {M}}$, if $[x,y(x)] \cap {\scriptstyle \mathtt {K}}\ne \emptyset$ we have:

$$\begin{aligned}{}[\xi _{\scriptstyle \mathtt {K}}(x), \eta _{\scriptstyle \mathtt {K}}(x)]:=[x,y(x)] \cap K. \end{aligned}$$

Let us fix $x \in {\scriptstyle \mathtt {K}}_0$ , with ${\scriptstyle \mathtt {K}}_0 \in \mathfrak {M}$ such that $y(x) \in {\scriptstyle \mathtt {L}}_0$, $\sigma _0={\scriptstyle \mathtt {K}}_0|{\scriptstyle \mathtt {L}}_0$. We distinguish the following two cases:

• 1. For the left case (see Fig. 14):

  $$\begin{aligned} \lambda _i(x) \vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s =\,&\bigl (\lambda _i(\xi _{{\scriptstyle \mathtt {K}}_0}(x)) - \lambda _i(\eta _{{\scriptstyle \mathtt {K}}_0}(x)) \bigr ) \vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s \\&+ \bigl (\lambda _i(\xi _{{\scriptstyle \mathtt {L}}_0}(x)) - \lambda _i(\eta _{{\scriptstyle \mathtt {L}}_0}(x)) \bigr ) \vert \mathbf u _{{\scriptstyle \mathtt {L}}_0} \vert ^s\\&+ \lambda _i(z_{\sigma _0}(x)) \bigl (\vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s - \vert \mathbf u _{{\scriptstyle \mathtt {L}}_0} \vert ^s \bigr ), \end{aligned}$$

• 2. for the right case (see Fig. 14):

  $$\begin{aligned} \lambda _i(x) \vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s = \bigl (\lambda _i(\xi _{{\scriptstyle \mathtt {K}}_0}(x)) - \lambda _i(\eta _{{\scriptstyle \mathtt {K}}_0}(x)) \bigr ) \vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s. \end{aligned}$$

In both cases:

$$\begin{aligned} \lambda _i(x) \vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s \le&\displaystyle \sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}} \varPsi _{\sigma }(x,y(x)) \lambda _i(z_{\sigma }(x)) \bigl | | \mathbf u _{\scriptstyle \mathtt {K}}\vert ^s - \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s \bigl \vert \\&+ \displaystyle \sum _{{\scriptstyle \mathtt {K}}\in {\mathfrak {M}}} \varPsi _{\scriptstyle \mathtt {K}}(x,y(x)) | \lambda _i(\xi _{{\scriptstyle \mathtt {K}}}(x)) - \lambda _i(\eta _{{\scriptstyle \mathtt {K}}}(x)) \vert \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s , \end{aligned}$$

that we can write as

$$\begin{aligned} \lambda _i(x) \vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s \le A(x)+B(x) \end{aligned}$$

by defining

$$\begin{aligned} \begin{array}{l} A(x):=\displaystyle \sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}} \varPsi _{\sigma }(x,y(x)) \lambda _i (z_{\sigma }(x)) \vert \vert \mathbf u _{\scriptstyle \mathtt {K}}\vert ^s - \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s \vert \\ B(x):= \displaystyle \sum _{{\scriptstyle \mathtt {K}}\in {\mathfrak {M}}} \varPsi _{\scriptstyle \mathtt {K}}(x,y(x)) \vert \lambda _i(\xi _{{\scriptstyle \mathtt {K}}}(x)) - \lambda _i(\eta _{{\scriptstyle \mathtt {K}}}(x)) \vert \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s . \end{array} \end{aligned}$$

We proceed by estimating separately the two terms.

Estimate of A:

Since $\lambda _i$ is bounded, we get:

$$\begin{aligned} A(x) \le \Vert \lambda _i \Vert _{\infty } \sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}} \varPsi _{\sigma }(x,y(x)) \bigg \vert \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s - \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s \bigg \vert ; \end{aligned}$$

We now use the following estimate (with $c_{\sigma }=\vert (\eta _i,\overrightarrow{\nu }_{\sigma }(x))\vert$)

$$\begin{aligned} \int _{\partial \varOmega _i} \varPsi _{\sigma }(x,y(x)) \text {d}x \le \frac{c_{\sigma }}{\lambda } m_{\sigma }, \end{aligned}$$

that is proved in [13], to conclude:

$$\begin{aligned} A=\int _{\partial \varOmega _i}A(x) \text {d}x&\le \Vert \lambda _i \Vert _{\infty } \sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}} \left( \int _{\partial \varOmega _i} \varPsi _{\sigma }(x,y(x)) \text {d}x \right) \bigg \vert \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s - \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s \bigg \vert \\&\le C_i \sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}} m_{\sigma } \bigg \vert \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s - \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s \bigg \vert \end{aligned}$$

where in the 3rd inequality we used [20, Lemma I.19].

Now, as in [2], we use the inequality:

$$\begin{aligned} \bigg \vert \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s - \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s \bigg \vert \le s (\vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^{s-1} + \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^{s-1}) \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}- \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert \end{aligned}$$

that leads to:

$$\begin{aligned} \begin{aligned} \sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}} m_{\sigma } \bigg \vert \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s - \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s \bigg \vert&\le s \sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}} m_{\sigma } (\vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^{s-1} + \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^{s-1}) \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}- \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert \\&\le C \sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}} m_{{\scriptscriptstyle \sigma }}m_{{\scriptscriptstyle \sigma ^*}}(\vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^{s-1} + \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^{s-1}) \bigg \vert \frac{ \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}- \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}}{m_{{\scriptscriptstyle \sigma ^*}}} \bigg \vert , \end{aligned} \end{aligned}$$

that by integration by parts and Hölder gives:

$$\begin{aligned}&\sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}} m_{\sigma } \bigg \vert \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s - \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s \bigg \vert \\&\quad \le C \left( \sum _{{\scriptstyle \mathtt {K}}\in {\mathfrak {M}}} \sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}_{\scriptstyle \mathtt {K}}} m_{{\scriptscriptstyle \sigma }}m_{{\scriptscriptstyle \sigma ^*}}\vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^{\frac{(s-1)p}{p-1}} \right) ^{\frac{p-1}{p}} \left( \sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}} m_{{\scriptscriptstyle \sigma }}m_{{\scriptscriptstyle \sigma ^*}}\bigg \vert \frac{\mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}- \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}}{m_{{\scriptscriptstyle \sigma ^*}}} \bigg \vert ^p \right) ^{\frac{1}{p}}. \end{aligned}$$

By regularity hypothesis on the mesh and the definition of the discrete gradient we can write:

$$\begin{aligned} \begin{aligned} A&\le \frac{C}{\sin (\alpha _{\mathfrak {T}})^{\frac{1}{p}} \,\, \text {reg}(\mathfrak {T})^{\frac{p-1}{p}}} \,\, \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{\frac{(s-1)p}{p-1}}^{s-1} \,\, \Vert \nabla ^{{\mathfrak {D}}} \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _p. \end{aligned} \end{aligned}$$

Estimate of B:

Since $\lambda _i$ is $C^{\infty }$, we have, by Talylor’s formula:

$$\begin{aligned} B(x) \le \Vert \nabla \lambda _i \Vert _{\infty } \sum _{K\in {\mathfrak {M}}} \varPsi _{\scriptstyle \mathtt {K}}(x,y(x)) \vert \xi _{\scriptstyle \mathtt {K}}(x) - \eta _{\scriptstyle \mathtt {K}}(x) \vert \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s \end{aligned}$$

and thanks to the inequality that can be found in [13, Lemma 3.10]

$$\begin{aligned} \int _{\partial \varOmega _i} \varPsi _{\scriptstyle \mathtt {K}}(x,y(x)) \vert \xi _{\scriptstyle \mathtt {K}}(x)-\eta _{\scriptstyle \mathtt {K}}(x) \vert \text {d}x \le \frac{m_{{\scriptstyle \mathtt {K}}}}{\lambda }, \end{aligned}$$

we can conclude:

$$\begin{aligned} B&=\int _{\partial \varOmega _i} B(x) \le \Vert \nabla \lambda _i \Vert _{\infty } \displaystyle \sum _{{\scriptstyle \mathtt {K}}\in {\mathfrak {M}}} \left( \int _{\partial \varOmega _i} \varPsi _{\scriptstyle \mathtt {K}}(x,y(x)) \vert \xi _{\scriptstyle \mathtt {K}}(x) - \eta _{\scriptstyle \mathtt {K}}(x) \vert \right) \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s\\&\le C_i \displaystyle \sum _{{\scriptstyle \mathtt {K}}\in {\mathfrak {M}}} m_{{\scriptstyle \mathtt {K}}}\vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s. \end{aligned}$$

Thus

$$\begin{aligned} B \le C \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _s^{s}. \end{aligned}$$

Putting together the terms, we find:

$$\begin{aligned} \int _{\partial \varOmega _i} \lambda _i(x) \vert \widetilde{\gamma }^{\partial \mathfrak {M}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) \vert ^s \le C_i \,\, \biggl ( \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{\frac{(s-1)p}{p-1}}^{s-1} \,\, \Vert \nabla ^{{\mathfrak {D}}} \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _p + \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _s^{s} \biggr ). \end{aligned}$$

By proceeding as in the proof of [2, Lemma 1], we use interpolation between $L^p$ spaces and we write:

$$\begin{aligned} \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _s^{s} \le \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{\frac{(s-1)p}{p-1}}^{s-1} \,\,\, \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _p \end{aligned}$$

that leads to

$$\begin{aligned} \int _{\partial \varOmega _i} \lambda _i(x) \vert \widetilde{\gamma }^{\partial \mathfrak {M}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) \vert ^s \le C_i \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{\frac{(s-1)p}{p-1}}^{s-1} \,\,\, \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{1,p} \end{aligned}$$

that proves our theorem.

On the dual mesh the computations are exactly the same, exchanging ${\scriptstyle \mathtt {K}}$ with ${\scriptstyle \mathtt {K}^*}$ and $\sigma$ in $\sigma ^*$ . $\square$

Corollary 1

Let $\mathfrak {T}$ be a DDFV mesh associated to $\varOmega$.There exists a constant $C>0$, depending only on p,q, $\sin (\alpha _{\mathfrak {T}})$, reg($\mathfrak {T}$) and $\varOmega$ such that $\forall \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\in {\mathbb {E}}^{\varGamma _0}_0$ and for all $s\ge 1$, $p>1$:

$$\begin{aligned} \Vert \gamma (\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) \Vert _{s,\partial \varOmega }^s \le C \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{1,p} \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{\frac{p(s-1)}{p-1}}^{s-1}. \end{aligned}$$

Proof

The proof is almost the same as Theorem 9.

What changes is just that we now fix $x \in {\scriptstyle \mathtt {L}}$, ${\scriptstyle \mathtt {L}}\in \partial \mathfrak {M}$ and ${\scriptstyle \mathtt {K}}_0 \in {\mathfrak {M}}$ such that ${\scriptstyle \mathtt {L}}\subset {\scriptstyle \mathtt {K}}_0$, $y(x) \in {\scriptstyle \mathtt {K}}_0$, $\sigma _0={\scriptstyle \mathtt {K}}_0|{\scriptstyle \mathtt {L}}$.

The term that we want to study now is $\lambda _i(x) \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s$, since we are focusing on the boundary. It can be written as:

$$\begin{aligned} \lambda _i(x) \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s= \lambda _i(x) (\vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s -\vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s) + \lambda _i(x) \vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s \end{aligned}$$

(29)

that can be estimated by:

$$\begin{aligned} \lambda _i(x) \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s&\le \lambda _i(x) \bigg \vert \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s -\vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s \bigg \vert \mathbf{1 }_{\scriptstyle \mathtt {L}}(x) + \lambda _i(x) \vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s\\&:= A_b(x) + \lambda _i(x) \vert \mathbf u _{{\scriptstyle \mathtt {K}}_0} \vert ^s \end{aligned}$$

${\hbox {Estimate of }A_b}$:

Since $\lambda$ is bounded, we have:

$$\begin{aligned} A_b= \int _{\partial \varOmega _i} A_b(x) \le \Vert \lambda _i \Vert _{\infty } \sum _{{\scriptstyle \mathtt {D}}\in {\mathfrak {D}}} m_{{\scriptscriptstyle \sigma }}\bigg \vert \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}\vert ^s - \vert \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}\vert ^s \bigg \vert . \end{aligned}$$

We can proceed exactly as in the proof of Thm 9 for A, so we get:

$$\begin{aligned} \begin{aligned} A_b&\le \frac{C}{\sin (\alpha _{\mathfrak {T}})^{\frac{1}{p}} \,\, \text {reg}(\mathfrak {T})^{\frac{p-1}{p}}} \,\, \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{\frac{(s-1)p}{p-1}}^{s-1} \,\, \Vert \nabla ^{{\mathfrak {D}}} \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _p . \end{aligned} \end{aligned}$$

Putting together all the terms, we find:

$$\begin{aligned} \bigg ( \int _{\partial \varOmega _i} \lambda _i(x) \vert \gamma ^{\partial \mathfrak {M}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) \vert ^s \bigg ) \le A_b + \bigg ( \int _{\partial \varOmega _i} \lambda _i(x) \vert \widetilde{\gamma }^{\partial \mathfrak {M}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) \vert ^s \bigg ). \end{aligned}$$

Thanks to the previous theorem, we conclude:

$$\begin{aligned} \int _{\partial \varOmega _i} \lambda _i(x) \vert \gamma ^{\partial \mathfrak {M}}(\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}) \vert ^s \le C_i \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{\frac{(s-1)p}{p-1}}^{s-1} \,\,\, \Vert \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\Vert _{1,p} \end{aligned}$$

that proofs our statement.

On the dual mesh the computations are the same as the previous theorem. $\square$

Appendix B: Study of the kernel of ${\text {D}}^\mathfrak {D}$

Theorem 10

Let $\varOmega$ be an open connected bounded polygonal domain of ${{\mathbb {R}}}^2$ and $\varGamma _0$ be a part of the boundary such that $m(\varGamma _0)>0$.

Let $\mathfrak {T}$ be a DDFV mesh associated to $\varOmega$ that satisfies inf-sup stability condition. Then $\displaystyle \forall \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}\in {\mathbb {E}}_0^{\varGamma _0}$ such that ${\text {D}}^\mathfrak {D}\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}= 0$ we have $\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}= 0$ in $\varOmega$.

Fig. 15

Structures

Proof Since we are not able to give a general proof of this theorem for all meshes, we focus on particular families, namely Cartesian meshes and all the ones that are unconditionally inf-sup stable (see [5]), since to prove Theorem 1 we need this last hypothesis.

When studying those meshes, we observe a propagation phenomenon of the zero boundary data on $\varGamma _0$ to the entire mesh.

In fact, it is important to remark that in DDFV meshes all boundary diamonds are triangles (see Fig. 2). If we focus on one of those diamonds, the condition on $\varGamma _0$ implies that the velocity is zero on the three vertices ${\scriptstyle \mathtt {L}}, {\scriptstyle \mathtt {K}^*}$ and ${\scriptstyle \mathtt {L}^*}$:

$$\begin{aligned}&\displaystyle \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}= \left( \begin{array}{c} {\text {u}}_{{{\scriptscriptstyle \mathtt {L}}}}^x \\ \\ {\text {u}}_{{{\scriptscriptstyle \mathtt {L}}}}^y \end{array} \right) =0 , \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}^*}}}= \left( \begin{array}{c} \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}^*}}}^x \\ \\ \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}^*}}}^y \end{array} \right) =0,\\&\displaystyle \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}^*}}}= \left( \begin{array}{c} {\text {u}}_{{{\scriptscriptstyle \mathtt {L}^*}}}^x \\ \\ {\text {u}}_{{{\scriptscriptstyle \mathtt {L}^*}}}^y \end{array} \right) =0. \end{aligned}$$

Since we are supposing ${\text {D}}^{\scriptscriptstyle \mathtt {D}}\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}= 0$ for all ${\scriptstyle \mathtt {D}}\in \mathfrak {D}$, this is true in particular for the boundary diamonds (the white ones in Fig. 15). By the definition of the discrete strain rate tensor (5) we are led to the following system:

$$\begin{aligned} \left\{ \begin{aligned}&m_{{\scriptscriptstyle \sigma }}\, \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}^x \, {\text {n}}_{{\scriptscriptstyle \sigma }, {\scriptscriptstyle \mathtt {K}}}^x = 0\\&m_{{\scriptscriptstyle \sigma }}\, \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}^y \, {\text {n}}_{{\scriptscriptstyle \sigma }, {\scriptscriptstyle \mathtt {K}}}^y = 0 \\&m_{{\scriptscriptstyle \sigma }}\, (\mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}^x \, {\text {n}}_{{\scriptscriptstyle \sigma }, {\scriptscriptstyle \mathtt {K}}}^y + \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}^y \, {\text {n}}_{{\scriptscriptstyle \sigma }, {\scriptscriptstyle \mathtt {K}}}^x) = 0 , \end{aligned} \right. \end{aligned}$$

(30)

that implies $\mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}=\left( \begin{array}{c} \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}^x \\ \\ \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}^y \end{array} \right) =0$, since the outer normal ${\mathbf{n }}_{{\varvec{{\scriptscriptstyle \sigma }}}{\scriptscriptstyle \mathtt {K}}}= \left( \begin{array}{c} {\text {n}}_{{\scriptscriptstyle \sigma }, {\scriptscriptstyle \mathtt {K}}}^x \\ \\ {\text {n}}_{{\scriptscriptstyle \sigma }, {\scriptscriptstyle \mathtt {K}}}^y \end{array} \right)$ cannot be zero.

This means that for all diamonds in ${\mathfrak {D}}_{ext} \cap \varGamma _0$ the four components of the velocity, $\mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}, \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}, \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}^*}}}, \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}^*}}}$, are zero.

We now look at the diamonds that are adjacent to ones on the boundary: for the meshes under consideration, we can distinguish two possible situations that we illustrate in Fig. 15.

The first one is the case of the shaded diamond, for which the situation is equivalent to the one of boundary diamonds. In fact, we know that the velocity is zero on three of its vertices. So we can conclude, by solving a system similar to (30) deduced by ${\text {D}}^{\scriptscriptstyle \mathtt {D}}\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}=0$, that even the last component of the velocity is zero on that diamond.

The second structure is described by the hatched diamonds. This is the case of two neighbors, that we will denote with $\displaystyle {{\scriptstyle \mathtt {D}}}^1$,$\displaystyle {{\scriptstyle \mathtt {D}}}^2$ which share a common vertex. Remark that on that vertex the velocity is zero and both diamonds have one more vertex with zero velocity. Thus we are considering a structure composed by 6 vertices, where the values of the velocity are zero on 3 among them.

In this case, we denote the normal vectors of ${{\scriptstyle \mathtt {D}}}^{1}, {{\scriptstyle \mathtt {D}}}^{2}$ with

$$\begin{aligned} {{\mathbf{n }}_{{{{\scriptscriptstyle \sigma }}}{\scriptscriptstyle \mathtt {K}}}}^{i} =\left( \begin{array}{c} {\text {n}}_{{\scriptscriptstyle \sigma }}^{x,i} \\ \\ {\text {n}}_{{\scriptscriptstyle \sigma }}^{y,i} \end{array} \right) , \quad {\mathbf{n }}_{{\sigma ^* K^*}}^{i} =\left( \begin{array}{c} {\text {n}}_{{\scriptscriptstyle \sigma ^*}}^{x,i} \\ \\ {\text {n}}_{{\scriptscriptstyle \sigma ^*}}^{y,i} \end{array} \right) \quad \text {for }i=1,2 \end{aligned}$$

and we write the system of equations equivalent to the conditions $\text {D}^{{{\scriptstyle \mathtt {D}}}^i} \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}= 0$ for $i=1,2$.

The $6 \times 6$ matrix of that system has determinant

$$\begin{aligned} det = ({\text {n}}_{{\scriptscriptstyle \sigma ^*}}^{x,2} \, {\text {n}}_{{\scriptscriptstyle \sigma }}^{y,2} - {\text {n}}_{{\scriptscriptstyle \sigma }}^{x,2} \, {\text {n}}_{{\scriptscriptstyle \sigma ^*}}^{y,2})({\text {n}}_{{\scriptscriptstyle \sigma ^*}}^{x,1} \, {\text {n}}_{{\scriptscriptstyle \sigma }}^{y,1} - {\text {n}}_{{\scriptscriptstyle \sigma }}^{x,1} \, {\text {n}}_{{\scriptscriptstyle \sigma ^*}}^{y,1})({\text {n}}_{{\scriptscriptstyle \sigma }}^{x,1} \, {\text {n}}_{{\scriptscriptstyle \sigma }}^{y,2} - {\text {n}}_{{\scriptscriptstyle \sigma }}^{x,2} \, {\text {n}}_{{\scriptscriptstyle \sigma }}^{y,1}) \ne 0 \end{aligned}$$

that is always different from zero, except in a degenerate case that we treat in the following section where the normals of the two diamonds are parallel. Thus the matrix is invertible, that implies that all the six components of the velocity on those two diamonds are zero: $\mathbf{u }_{{{\scriptscriptstyle \mathtt {K}}}}^i= \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}}}}^i= \mathbf{u }_{{{\scriptscriptstyle \mathtt {K}^*}}}^i= \mathbf{u }_{{{\scriptscriptstyle \mathtt {L}^*}}}^i =0$ for $i=1,2$.

Fig. 16

Degenerate case

Degenerate case: checkerboard mesh

This is a particular case of the second structure, in which the normal vectors of the two hatched diamonds are parallel. In order to have an invertible system to solve, it is necessary to consider a third diamond (Fig. 16).

In particular, if we call ${{\scriptstyle \mathtt {D}}}_1, {{\scriptstyle \mathtt {D}}}_2$ the hatched diamonds and ${{\scriptstyle \mathtt {D}}}_3$ the white one, we have for instance: ${{\mathbf{n }}_{{\varvec{{\scriptscriptstyle \sigma }}}{\scriptscriptstyle \mathtt {K}}}}^{i} =\left( \begin{array}{c} 0 \\ 1 \end{array} \right)$ for $i=1,2$ and ${{\mathbf{n }}_{{\varvec{{\scriptscriptstyle \sigma }}}{\scriptscriptstyle \mathtt {K}}}}^{3} = \left( \begin{array}{c} 1 \\ 0 \end{array} \right)$.

If, as we did in the previous cases, we write the system of equations equivalent to $\text {D}^{{{\scriptstyle \mathtt {D}}}^i} \mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}= 0$, but this time for $i=1,2,3$, we get again an invertible system, this time of size $8 \times 8$. As before, we find that all the components of the velocity are zero on the three diamonds.

By proceeding step by step, we can prove that the velocity $\mathbf{u }^{{\scriptscriptstyle \mathfrak {T}}}$ is zero on the entire domain $\varOmega$.

Remark 5

We suppose that the mesh satisfies inf-sup stability condition because this hypothesis is necessary to prove Theorem 1. Since the inf-sup constant it is not involved in the proof of Theorem 10, we could extend the technique of the proof to all geometries, considering one mesh at a time.