An error estimate for a finite-volume scheme for the Cahn–Hilliard equation with dynamic boundary conditions

Abstract

In this paper we consider a finite-volume approximation for the Cahn–Hilliard equation with dynamic boundary conditions. The convergence of the scheme is proved in Nabet (IMA J Numer Anal 36(4): 1898–1942, 2016), we prove here an error estimate for the fully-discrete scheme. We also give numerical simulations which validate the theoretical result.

Proof of proposition 7

For $f \in L^2(\varOmega )$ and $g \in L^2(\varGamma )$, we consider the following problem: find $u:\varOmega \rightarrow {\mathbb {R}}$ such that $\int _\varOmega u = \alpha$ and

$$\begin{aligned} -&\varDelta u = f \qquad \qquad&\text { in } \varOmega ; \end{aligned}$$

(59a)

$$\begin{aligned} -&\varDelta _{\scriptscriptstyle \varGamma }u_{\shortmid {\scriptscriptstyle \varGamma }}+ \partial _n u = g \quad&\text { on } \varGamma . \end{aligned}$$

(59b)

By integrating equation (59a) on all interior control volumes ${\scriptstyle {\mathcal {K}}}\in {\mathfrak {M}}$ and the boundary condition (59b) on all boundary control volumes ${\scriptstyle {\mathcal {L}}}\in \partial {\mathfrak {M}}$, the two-point flux approximation of problem (59) writes as follows. Find $u_{{\scriptscriptstyle {\mathcal {T}}}}\in {\mathbb {R}}^{{\scriptscriptstyle {\mathcal {T}}}}$ such that $\sum \limits _{{\scriptscriptstyle {\mathcal {K}}}\in {\mathfrak {M}}}m_{{{\scriptscriptstyle {\underline{{\mathcal {K}}}}}}}u_{{\scriptscriptstyle {\mathcal {K}}}}=\alpha$ and

$$\begin{aligned} \sum \limits _{{\scriptscriptstyle \sigma }\in {{\mathcal {E}}^{int}_{\scriptscriptstyle {\mathcal {K}}}}}m_{{\scriptscriptstyle \sigma }}\frac{u_{{\scriptscriptstyle {\mathcal {K}}}}-u_{{\scriptscriptstyle {\mathcal {L}}}}}{d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}} + \sum \limits _{{\scriptscriptstyle \sigma }\in {{\mathcal {E}}^{ext}_{\scriptscriptstyle {\mathcal {K}}}}}m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}\frac{u_{{\scriptscriptstyle {\mathcal {K}}}}-u_{{\scriptscriptstyle {\mathcal {L}}}}}{d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}}&= m_{{\scriptscriptstyle {\mathcal {K}}}}{\mathbb {P}}^m_{{\scriptscriptstyle {\mathcal {K}}}}f, \quad&\forall {\scriptstyle {\mathcal {K}}}\in {\mathfrak {M}}; \end{aligned}$$

(60a)

$$\begin{aligned} \sum \limits _{\scriptscriptstyle {{{\scriptscriptstyle \mathbf{v }}\in {\mathcal {V}}_{\scriptscriptstyle {\mathcal {L}}}}}}\frac{u_{{\scriptscriptstyle {\mathcal {L}}}}-u_{{\scriptscriptstyle \mathcal {L}'}}}{d_{{\scriptscriptstyle {\mathcal {L}}},{\scriptscriptstyle \mathcal {L}'}}} + m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}\frac{u_{{\scriptscriptstyle {\mathcal {L}}}}-u_{{\scriptscriptstyle {\mathcal {K}}}}}{d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}}&= m_{{\scriptscriptstyle {\mathcal {L}}}}{\mathbb {P}}^m_{{\scriptscriptstyle {\mathcal {L}}}}g, \quad&\forall {\scriptstyle {\mathcal {L}}}\in \partial {\mathfrak {M}}. \end{aligned}$$

(60b)

We can prove that this problem admits a unique solution.

Because of the complex geometry of $\varOmega$, it is possible to take into account some points $x\in \varOmega ^c$ in the proof of the error estimate. To ensure that all the quantities used in the proof of Theorem 2 are well defined, we will use an extension in $\varvec{{\mathbb {R}}}^2$ of the function u. Since $u \in H^2(\varOmega )$, there exists an extension ${{\widetilde{u}}} \in H^2(\varvec{{\mathbb {R}}}^2)$ (that we fix in the sequel) such that

$$\begin{aligned} {{\widetilde{u}}}(x)=u(x), \;\; \forall x\in \varOmega \text { and } \left\| {{\widetilde{u}}} \right\| _{H^2(\varvec{{\mathbb {R}}}^2)} \le {C_{26}} \left\| u \right\| _{\scriptstyle {{H^2(\varOmega )}}}, \end{aligned}$$

(61)

with $C_{{}26}>0$ depending only on $\varOmega$.

Proposition 16

The tangential gradient of $u:\varGamma \rightarrow {\mathbb {R}}$ to the vertex ${\mathbf{v }}={\scriptstyle {\mathcal {L}}}|{\scriptstyle {\mathcal {L}}}'$ satisfies

$$\begin{aligned} \left| \frac{u(x_{{\scriptscriptstyle \mathcal {L}'}}) - u(x_{{\scriptscriptstyle {\mathcal {L}}}})}{m_{\gamma _{{\scriptscriptstyle {\mathcal {L}}}{\scriptscriptstyle \mathcal {L}'}}}} - \nabla _{{\scriptscriptstyle \varGamma }}u({\mathbf{v }}) \cdot {\vec {\varvec{\tau }}}_{\varvec{{\scriptscriptstyle \mathbf{v }},{\scriptscriptstyle {\mathcal {L}}}}}({\mathbf{v }})\right| \le \int _{{\gamma _{{\scriptscriptstyle {\mathcal {L}}}{\scriptscriptstyle \mathcal {L}'}}}} \left| (u\circ \varphi )''(\varphi ^{-1}(x))\right| \text {d}\sigma (x), \end{aligned}$$

where $\varphi$ is an arc-length parametrization of the curve $\varGamma$ and ${\vec {\varvec{\tau }}}_{\varvec{{\scriptscriptstyle \mathbf{v }},{\scriptscriptstyle {\mathcal {L}}}}}({\mathbf{v }})$ is the unit tangent vector to $\varGamma$ at point ${\mathbf{v }}={\scriptstyle {\mathcal {L}}}|{\scriptstyle {\mathcal {L}}}'$ going from ${\scriptstyle {\mathcal {L}}}$ to ${\scriptstyle {\mathcal {L}}}'$.

Proof

Let us consider the points $t_{{\scriptscriptstyle {\mathcal {L}}}},t_{{\scriptscriptstyle \mathcal {L}'}},t_{{\scriptscriptstyle \mathbf{v }}}\in {\mathbb {R}}$ such that $x_{{\scriptscriptstyle {\mathcal {L}}}}=\varphi (t_{{\scriptscriptstyle {\mathcal {L}}}})$, $x_{{\scriptscriptstyle \mathcal {L}'}}=\varphi (t_{{\scriptscriptstyle \mathcal {L}'}})$ and ${\mathbf{v }}=\varphi (t_{{\scriptscriptstyle \mathbf{v }}})$, then the Taylor’s formulas give

$$\begin{aligned} u(x_{{\scriptscriptstyle \mathcal {L}'}}) - u(x_{{\scriptscriptstyle {\mathcal {L}}}}) =&(t_{{\scriptscriptstyle \mathcal {L}'}}-t_{{\scriptscriptstyle \mathbf{v }}}) (u\circ \varphi )' (t_{{\scriptscriptstyle \mathbf{v }}}) + \int _{t_{{\scriptscriptstyle \mathbf{v }}}}^{t_{{\scriptscriptstyle \mathcal {L}'}}} (t_{{\scriptscriptstyle \mathcal {L}'}}-s) (u\circ \varphi )'' (s) \text {d}s\\&- (t_{{\scriptscriptstyle {\mathcal {L}}}}-t_{{\scriptscriptstyle \mathbf{v }}}) (u\circ \varphi )' (t_{{\scriptscriptstyle \mathbf{v }}}) - \int _{t_{{\scriptscriptstyle \mathbf{v }}}}^{t_{{\scriptscriptstyle {\mathcal {L}}}}} (t_{{\scriptscriptstyle {\mathcal {L}}}}-s) (u\circ \varphi )'' (s) \text {d}s. \end{aligned}$$

Noting that $|t_{{\scriptscriptstyle \mathcal {L}'}}-t_{{\scriptscriptstyle {\mathcal {L}}}}|=m_{\gamma _{{\scriptscriptstyle {\mathcal {L}}}{\scriptscriptstyle \mathcal {L}'}}}$ we conclude the proof. $\square$

Thanks to the Taylor’s formulas we can prove the following proposition.

Proposition 17

Let ${\scriptstyle {\mathcal {L}}}\in \partial {\mathfrak {M}}$ be a boundary control volume and ${\mathbf{v }}$ be a vertex of ${\scriptstyle {\mathcal {L}}}$, then the following equality holds

$$\begin{aligned} m_{\gamma _{{\scriptscriptstyle {\mathcal {L}}}{\mathbf{v }}}}- d_{{\scriptscriptstyle {\mathcal {L}}},{\mathbf{v }}}= {\mathcal {O}} \left( m_{{\scriptscriptstyle {\mathcal {L}}}}m_{\gamma _{{\scriptscriptstyle {\mathcal {L}}}{\mathbf{v }}}}\right) . \end{aligned}$$

Moreover for any point $x\in \sigma ={\scriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{ext}$, one has

$$\begin{aligned} {\varvec{\vec {\mathbf {n}}}}_{\varvec{\sigma {\scriptscriptstyle {\mathcal {K}}}}}(x) - \vec {\varvec{\mathbf {n}}}_{\varvec{{\scriptscriptstyle {\mathcal {K}}}{\scriptscriptstyle {\mathcal {L}}}}}= {\mathcal {O}} (m_{{\scriptscriptstyle {\mathcal {L}}}}), \end{aligned}$$

where ${\varvec{\vec {\mathbf {n}}}}_{\varvec{\sigma {\scriptscriptstyle {\mathcal {K}}}}}(x)$ is the unit normal vector to $\sigma$ outward to ${\scriptstyle {\mathcal {K}}}$ at point x.

We are now in position to prove the main result of the appendix.

Theorem 2

Let us assume that the solution u of the continuous problem (59) belongs to $H^2_{\scriptscriptstyle \varGamma }(\varOmega )$. Let us consider the solution $u_{{\scriptscriptstyle {\mathcal {T}}}}$ to discrete problem (60). Then, there exists $C_{{}27}>0$ independent of $\text { size}({\mathcal {T}}\,\,)$ such that

$$\begin{aligned} \left| e_{{\scriptscriptstyle {\mathcal {T}}}} \right| _{\scriptstyle {{1,{\scriptscriptstyle {\mathcal {T}}}}}}^2 + \left| e_{{\scriptscriptstyle \partial {\mathfrak {M}}}} \right| _{\scriptstyle {{1,{\scriptscriptstyle \partial {\mathfrak {M}}}}}}^2 \le {C_{27}} \text { size}({\mathcal {T}}\,\,)^2 \left\| u \right\| _{\scriptstyle {{H^2_{\scriptscriptstyle \varGamma }(\varOmega )}}}^2, \end{aligned}$$

(62)

with $e_{{\scriptscriptstyle {\mathcal {T}}}}={{\mathbb {P}}}^{{c}}_{{{\scriptscriptstyle {\mathcal {T}}}}}u - u_{{\scriptscriptstyle {\mathcal {T}}}}$.

We decompose the proof of Theorem 2 into two steps. As a first step, we prove (see Proposition 18) that the left-hand side of inequality (62) is bounded from above by the different consistency errors which intervene in the problem. In a second phase, we have to estimate these different consistency errors.

Proposition 18

Let us consider the solution u to problem (59) and the solution $u_{{\scriptscriptstyle {\mathcal {T}}}}$ to discrete problem (60). Then the following estimate holds

$$\begin{aligned} \left| e_{{\scriptscriptstyle {\mathcal {T}}}} \right| _{\scriptstyle {{1,{\scriptscriptstyle {\mathcal {T}}}}}}^2 + \left| e_{{\scriptscriptstyle \partial {\mathfrak {M}}}} \right| _{\scriptstyle {{1,{\scriptscriptstyle \partial {\mathfrak {M}}}}}}^2\le & {} \sum \limits _{{\scriptscriptstyle \sigma }={\scriptscriptstyle {\mathcal {K}}}|{\scriptscriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{int}}m_{{\scriptscriptstyle \sigma }}d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}(R^{int}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}})^2 + \sum \limits _{{\scriptscriptstyle \sigma }={\scriptscriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{ext}}m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}(R^{ext}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}})^2 \nonumber \\&+ \sum \limits _{\scriptscriptstyle {{{\scriptscriptstyle \mathbf{v }}={\scriptscriptstyle {\mathcal {L}}}|{\scriptscriptstyle \mathcal {L}'}\in {\mathcal {V}}}}}\frac{{R_{{\mathbf{v }},{\scriptscriptstyle {\mathcal {L}}}}}^2}{d_{{\scriptscriptstyle {\mathcal {L}}},{\scriptscriptstyle \mathcal {L}'}}} ; \end{aligned}$$

(63)

where,

$$\begin{aligned} R^{int}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}}&= \frac{1}{m_{{\scriptscriptstyle \sigma }}} \int _\sigma \nabla u(x) \cdot \vec {\varvec{\mathbf {n}}}_{\varvec{{\scriptscriptstyle {\mathcal {K}}}{\scriptscriptstyle {\mathcal {L}}}}}\text {d}x- \frac{u(x_{{\scriptscriptstyle {\mathcal {L}}}})-u(x_{{\scriptscriptstyle {\mathcal {K}}}})}{d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}} , \quad&\forall \sigma ={\scriptstyle {\mathcal {K}}}|{\scriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{\scriptscriptstyle {\mathcal {K}}}\cap {\mathcal {E}}_{int}; \\ R^{ext}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}}&= \frac{1}{m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}} \int _\sigma \nabla u(x) \cdot {\varvec{\vec {\mathbf {n}}}}_{\varvec{\sigma {\scriptscriptstyle {\mathcal {K}}}}}(x) \text {d}x- \frac{u(x_{{\scriptscriptstyle {\mathcal {L}}}})-u(x_{{\scriptscriptstyle {\mathcal {K}}}})}{d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}} , \quad&\forall \sigma ={\scriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{\scriptscriptstyle {\mathcal {K}}}\cap {\mathcal {E}}_{ext}; \\ {R_{{\mathbf{v }},{\scriptscriptstyle {\mathcal {L}}}}}&= d_{{\scriptscriptstyle {\mathcal {L}}},{\scriptscriptstyle \mathcal {L}'}}\nabla _{{\scriptscriptstyle \varGamma }}u_{\shortmid {\scriptscriptstyle \varGamma }}({\mathbf{v }}) {\vec {\varvec{\tau }}}_{\varvec{{\scriptscriptstyle \mathbf{v }},{\scriptscriptstyle {\mathcal {L}}}}}-u_{\shortmid {\scriptscriptstyle \varGamma }}(x_{{\scriptscriptstyle \mathcal {L}'}})-u_{\shortmid {\scriptscriptstyle \varGamma }}(x_{{\scriptscriptstyle {\mathcal {L}}}}) , \quad&\forall {\mathbf{v }}={\scriptstyle {\mathcal {L}}}|{\scriptstyle {\mathcal {L}}}' \in {\mathcal {V}}. \end{aligned}$$

Proof

Let ${\scriptstyle {\mathcal {K}}}\in {\mathfrak {M}}$, we integrate equation (59a) on ${\scriptstyle {\mathcal {K}}}$ and we subtract the resulting equality with equation (60a). Thanks to definitions of $R^{int}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}}$ and $R^{ext}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}}$ given in Proposition 18 imply

$$\begin{aligned}&\sum \limits _{{\scriptscriptstyle \sigma }={\scriptscriptstyle {\mathcal {K}}}|{\scriptscriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{int}}\frac{m_{{\scriptscriptstyle \sigma }}}{d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}} (e_{{\scriptscriptstyle {\mathcal {K}}}}-e_{{\scriptscriptstyle {\mathcal {L}}}}) + \sum \limits _{{\scriptscriptstyle \sigma }={\scriptscriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{ext}}\frac{m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}}{d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}} (e_{{\scriptscriptstyle {\mathcal {K}}}}-e_{{\scriptscriptstyle {\mathcal {L}}}}) \nonumber \\&\quad = \sum \limits _{{\scriptscriptstyle \sigma }\in {{\mathcal {E}}^{int}_{\scriptscriptstyle {\mathcal {K}}}}}m_{{\scriptscriptstyle \sigma }}R^{int}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}}+ \sum \limits _{{\scriptscriptstyle \sigma }\in {{\mathcal {E}}^{ext}_{\scriptscriptstyle {\mathcal {K}}}}}m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}R^{ext}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}}, \quad \forall {\scriptstyle {\mathcal {K}}}\in {\mathfrak {M}}. \end{aligned}$$

(64)

In the same way let ${\scriptstyle {\mathcal {L}}}\in \partial {\mathfrak {M}}$, we integrate equation (59b) on ${\scriptstyle {\mathcal {L}}}$ and we subtract the resulting equality with equation (60b). Then we obtain

$$\begin{aligned} \sum \limits _{\scriptscriptstyle {{{\scriptscriptstyle \mathbf{v }}\in {\mathcal {V}}_{\scriptscriptstyle {\mathcal {L}}}}}}\frac{e_{{\scriptscriptstyle {\mathcal {L}}}}-e_{{\scriptscriptstyle \mathcal {L}'}}}{d_{{\scriptscriptstyle {\mathcal {L}}},{\scriptscriptstyle \mathcal {L}'}}} + m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}\frac{e_{{\scriptscriptstyle {\mathcal {L}}}}-e_{{\scriptscriptstyle {\mathcal {K}}}}}{d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}} = \sum \limits _{\scriptscriptstyle {{{\scriptscriptstyle \mathbf{v }}\in {\mathcal {V}}_{\scriptscriptstyle {\mathcal {L}}}}}}\frac{{R_{{\mathbf{v }},{\scriptscriptstyle {\mathcal {L}}}}}}{d_{{\scriptscriptstyle {\mathcal {L}}},{\scriptscriptstyle \mathcal {L}'}}} - m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}R^{ext}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}}, \quad \forall {\scriptstyle {\mathcal {L}}}\in \partial {\mathfrak {M}}. \end{aligned}$$

(65)

Now we multiply equation (64) by $e_{{\scriptscriptstyle {\mathcal {K}}}}$ and summing up over ${\scriptstyle {\mathcal {K}}}\in {\mathfrak {M}}$ and we multiply equation (65) by $e_{{\scriptscriptstyle {\mathcal {L}}}}$ and summing up over ${\scriptstyle {\mathcal {L}}}\in \partial {\mathfrak {M}}$. Then, summing the resulting equalities we have

$$\begin{aligned} \left| e_{{\scriptscriptstyle {\mathcal {T}}}} \right| _{\scriptstyle {{1,{\scriptscriptstyle {\mathcal {T}}}}}}^2 + \left| e_{{\scriptscriptstyle \partial {\mathfrak {M}}}} \right| _{\scriptstyle {{1,{\scriptscriptstyle \partial {\mathfrak {M}}}}}}^2 =&\sum \limits _{{\scriptscriptstyle \sigma }={\scriptscriptstyle {\mathcal {K}}}|{\scriptscriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{int}}m_{{\scriptscriptstyle \sigma }}(e_{{\scriptscriptstyle {\mathcal {K}}}}-e_{{\scriptscriptstyle {\mathcal {L}}}}) R^{int}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}}+ \sum \limits _{{\scriptscriptstyle \sigma }={\scriptscriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{ext}}m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}(e_{{\scriptscriptstyle {\mathcal {K}}}}-e_{{\scriptscriptstyle {\mathcal {L}}}}) R^{ext}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}}\\&+ \sum \limits _{\scriptscriptstyle {{{\scriptscriptstyle \mathbf{v }}={\scriptscriptstyle {\mathcal {L}}}|{\scriptscriptstyle \mathcal {L}'}\in {\mathcal {V}}}}}{R_{{\mathbf{v }},{\scriptscriptstyle {\mathcal {L}}}}}\frac{e_{{\scriptscriptstyle {\mathcal {L}}}}-e_{{\scriptscriptstyle \mathcal {L}'}}}{d_{{\scriptscriptstyle {\mathcal {L}}},{\scriptscriptstyle \mathcal {L}'}}} . \end{aligned}$$

Owing to the Cauchy–Schwarz and the Young inequalities, we obtain estimate (63). $\square$

With this proposition at hand we are now able to prove Theorem 2 by estimating all the terms of the right-hand side of (63).

Proof

First, let $\sigma ={\scriptstyle {\mathcal {K}}}|{\scriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{int}$ thanks to the Taylor’s formulas we have

$$\begin{aligned} R^{int}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}}=&\frac{1}{m_{{\scriptscriptstyle \sigma }}d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}} \int _\sigma \int _0^1 (1-t) \left\langle D^2 u \left( (1-t)x+tx_{{\scriptscriptstyle {\mathcal {K}}}}\right) (x_{{\scriptscriptstyle {\mathcal {K}}}}-x) , (x_{{\scriptscriptstyle {\mathcal {K}}}}-x) \right\rangle \\&- \frac{1}{m_{{\scriptscriptstyle \sigma }}d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}} \int _\sigma \int _0^1 (1-t) \left\langle D^2 u \left( (1-t)x+tx_{{\scriptscriptstyle {\mathcal {L}}}}\right) (x_{{\scriptscriptstyle {\mathcal {L}}}}-x) , (x_{{\scriptscriptstyle {\mathcal {L}}}}-x) \right\rangle . \end{aligned}$$

Owing to the Jensen inequality and the change of variables $(t,x)\in [0,1]\times \sigma \mapsto y =x+t(x_{{\scriptscriptstyle {\mathcal {K}}}}-x)$ (or $(t,x)\in [0,1]\times \sigma \mapsto y =x+t(x_{{\scriptscriptstyle {\mathcal {L}}}}-x)$ for the second term) and since for any ${\scriptstyle {\mathcal {K}}}\in {\mathfrak {M}}$, $\text {diam}({\scriptstyle {\mathcal {K}}})\le \text {reg}({\mathcal {T}}\,\,)d(x_{{\scriptscriptstyle {\mathcal {K}}}},\sigma )$, for any $\sigma \in {\mathcal {E}}_{\scriptscriptstyle {\mathcal {K}}}$ (see Definition 5) one has

$$\begin{aligned} (R^{int}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}})^2 \le C(\text {reg}({\mathcal {T}}\,\,)) \frac{\text { size}({\mathcal {T}}\,\,)^2}{m_{{\scriptscriptstyle {\mathcal {D}}}}} \int _{\scriptstyle {\mathcal {D}}}|D^2 u (y)|^2 \text {d}y. \end{aligned}$$

Noting that $m_{{\scriptscriptstyle {\mathcal {D}}}}=\frac{m_{{\scriptscriptstyle \sigma }}d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}}{2}$ we obtain

$$\begin{aligned} \sum \limits _{{\scriptscriptstyle \sigma }={\scriptscriptstyle {\mathcal {K}}}|{\scriptscriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{int}}m_{{\scriptscriptstyle \sigma }}d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}(R^{int}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}})^2 \le C(\text {reg}({\mathcal {T}}\,\,)) \text { size}({\mathcal {T}}\,\,)^2 \left\| D^2 u \right\| _{\scriptscriptstyle {{L^2(\varOmega )}}}^2. \end{aligned}$$

(66)

Secondly let $\sigma ={\scriptstyle {\mathcal {L}}}\in \partial {\mathfrak {M}}$, thanks to definition (61) of ${{\widetilde{u}}}$ we have $u(x_{{\scriptscriptstyle {\mathcal {L}}}})={{\widetilde{u}}}(x_{{\scriptscriptstyle {\mathcal {L}}}})$ and $u(x_{{\scriptscriptstyle {\mathcal {K}}}})={{\widetilde{u}}}(x_{{\scriptscriptstyle {\mathcal {K}}}})$, thus since $x_{{\scriptscriptstyle {\mathcal {L}}}}-x_{{\scriptscriptstyle {\mathcal {K}}}}=d(x_{{\scriptscriptstyle {\mathcal {K}}}},x_{{\scriptscriptstyle {\mathcal {L}}}})\vec {\varvec{\mathbf {n}}}_{\varvec{{\scriptscriptstyle {\mathcal {K}}}{\scriptscriptstyle {\mathcal {L}}}}}$, the definition of $R^{ext}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}}$, the Jensen inequality and the Taylor’s formulas imply

$$\begin{aligned} m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}(R^{ext}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}})^2&\le 5d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}\frac{(m_{{\scriptscriptstyle {\mathcal {L}}}}-m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}})^2}{m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}m_{{\scriptscriptstyle {\mathcal {L}}}}} \int _{\scriptstyle {\mathcal {L}}}\left| \nabla {{\widetilde{u}}}(x)\right| ^2 \text {d}\sigma (x)\\&\quad + 5\frac{(d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}- d(x_{{\scriptscriptstyle {\mathcal {K}}}},x_{{\scriptscriptstyle {\mathcal {L}}}}))^2}{d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}} \frac{m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}}{m_{{\scriptscriptstyle {\mathcal {L}}}}} \int _{\scriptstyle {\mathcal {L}}}\left| \nabla {{\widetilde{u}}}(x)\right| ^2 \text {d}\sigma (x)\\&\quad + 5d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}\frac{m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}}{m_{{\scriptscriptstyle {\mathcal {L}}}}} \int _{\scriptstyle {\mathcal {L}}}\left| \nabla {{\widetilde{u}}}(x)\right| ^2 |\vec {\varvec{\mathbf {n}}}_{\varvec{{\scriptscriptstyle {\mathcal {K}}}{\scriptscriptstyle {\mathcal {L}}}}}- {\varvec{\vec {\mathbf {n}}}}_{\varvec{\sigma {\scriptscriptstyle {\mathcal {K}}}}}(x)|^2 \text {d}\sigma (x)\\&\quad + \frac{5m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}}{d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}m_{{\scriptscriptstyle \sigma }}} \int _\sigma \int _0^1 (1-t)^2 \left| D^2 {{\widetilde{u}}}\left( (1-t)x+tx_{{\scriptscriptstyle {\mathcal {K}}}}\right) \right| ^2 |x_{{\scriptscriptstyle {\mathcal {K}}}}-x|^4 \text {d}t\text {d}\sigma (x)\\&\quad + \frac{5m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}}{d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}m_{{\scriptscriptstyle \sigma }}} \int _\sigma \int _0^1 (1-t)^2 \left| D^2 {{\widetilde{u}}}\left( (1-t)x+tx_{{\scriptscriptstyle {\mathcal {L}}}}\right) \right| ^2 |x_{{\scriptscriptstyle {\mathcal {L}}}}-x|^4 \text {d}t\text {d}\sigma (x). \end{aligned}$$

Thanks to Propositions 2, 3 and 17 , there exists ${C_{{\scriptscriptstyle \varGamma }}}>0$ independent of $\text { size}({\mathcal {T}}\,\,)$ such that

$$\begin{aligned} |d(x_{{\scriptscriptstyle {\mathcal {K}}}},x_{{\scriptscriptstyle {\mathcal {L}}}}) - d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}| \le {C_{{\scriptscriptstyle \varGamma }}}m_{{\scriptscriptstyle {\mathcal {L}}}}m_{\gamma _{{\scriptscriptstyle {\mathcal {L}}}{\mathbf{v }}}}, \;\; |m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}-m_{{\scriptscriptstyle {\mathcal {L}}}}| \le {C_{{\scriptscriptstyle \varGamma }}}m_{{\scriptscriptstyle {\mathcal {L}}}}^3 \; \text { and }\; |\vec {\varvec{\mathbf {n}}}_{\varvec{{\scriptscriptstyle {\mathcal {K}}}{\scriptscriptstyle {\mathcal {L}}}}}- {\varvec{\vec {\mathbf {n}}}}_{\varvec{\sigma {\scriptscriptstyle {\mathcal {K}}}}}(x)| \le {C_{{\scriptscriptstyle \varGamma }}}m_{{\scriptscriptstyle {\mathcal {L}}}}. \end{aligned}$$

Thus, thanks to a change of variables in the last two integrals we have

$$\begin{aligned} m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}(R^{ext}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}})^2 \le&{C_{{\scriptscriptstyle \varGamma }}}(\text {reg}({\mathcal {T}}\,\,)) \text { size}({\mathcal {T}}\,\,)^3 \int _{\scriptstyle {\mathcal {L}}}\left| \nabla {{\widetilde{u}}}(x)\right| ^2 \text {d}\sigma (x)\\&+ C(\text {reg}({\mathcal {T}}\,\,)) \text { size}({\mathcal {T}}\,\,)^2 \left( \int _{{\scriptstyle {\mathcal {D}}}_{\scriptscriptstyle {\mathcal {L}}}} \left| D^2 {{\widetilde{u}}}(y)\right| ^2 \text {d}y+ \int _{{\scriptstyle {\mathcal {D}}}} \left| D^2 {{\widetilde{u}}}(y)\right| ^2 \text {d}y\right) , \end{aligned}$$

where ${\scriptstyle {\mathcal {D}}}_{{\scriptscriptstyle {\mathcal {L}}}}=\{ (1-t)x+tx_{{\scriptscriptstyle {\mathcal {L}}}}: t\in [0,1], x\in \sigma ={\scriptstyle {\mathcal {L}}}\}$. Then, owing to (61) we obtain

$$\begin{aligned} \sum \limits _{{\scriptscriptstyle \sigma }={\scriptscriptstyle {\mathcal {L}}}\in {\mathcal {E}}_{ext}}m_{{\mathbf{e }}_{\scriptscriptstyle {{\scriptscriptstyle {\mathcal {L}}}}}}d_{{\scriptscriptstyle {\mathcal {K}}},{\scriptscriptstyle {\mathcal {L}}}}(R^{ext}_{{\scriptscriptstyle \sigma },{\scriptscriptstyle {\mathcal {K}}}})^2 \le {C_{{\scriptscriptstyle \varGamma }}}(\text {reg}({\mathcal {T}}\,\,)) {C_{26}} \text { size}({\mathcal {T}}\,\,)^2 \left\| u \right\| _{\scriptscriptstyle {{H^1(\varOmega )}}}^2. \end{aligned}$$

(67)

Finally, using definition of ${R_{{\mathbf{v }},{\scriptscriptstyle {\mathcal {L}}}}}$ for any ${\mathbf{v }}={\scriptstyle {\mathcal {L}}}|{\scriptstyle {\mathcal {L}}}'\in {\mathcal {V}}$ we have

$$\begin{aligned} \frac{{R_{{\mathbf{v }},{\scriptscriptstyle {\mathcal {L}}}}}^2}{d_{{\scriptscriptstyle {\mathcal {L}}},{\scriptscriptstyle \mathcal {L}'}}}&\le 2d_{{\scriptscriptstyle {\mathcal {L}}},{\scriptscriptstyle \mathcal {L}'}}\left( \nabla _{{\scriptscriptstyle \varGamma }}u({\mathbf{v }}) {\vec {\varvec{\tau }}}_{\varvec{{\scriptscriptstyle \mathbf{v }},{\scriptscriptstyle {\mathcal {L}}}}}- \frac{u(x_{{\scriptscriptstyle \mathcal {L}'}})-u(x_{{\scriptscriptstyle {\mathcal {L}}}})}{m_{\gamma _{{\scriptscriptstyle {\mathcal {L}}}{\scriptscriptstyle \mathcal {L}'}}}} \right) ^2\\&\quad +2\frac{(u(x_{{\scriptscriptstyle \mathcal {L}'}})-u(x_{{\scriptscriptstyle {\mathcal {L}}}}))^2}{m_{\gamma _{{\scriptscriptstyle {\mathcal {L}}}{\scriptscriptstyle \mathcal {L}'}}}} \frac{(d_{{\scriptscriptstyle {\mathcal {L}}},{\scriptscriptstyle \mathcal {L}'}}-m_{\gamma _{{\scriptscriptstyle {\mathcal {L}}}{\scriptscriptstyle \mathcal {L}'}}})^2}{d_{{\scriptscriptstyle {\mathcal {L}}},{\scriptscriptstyle \mathcal {L}'}}m_{\gamma _{{\scriptscriptstyle {\mathcal {L}}}{\scriptscriptstyle \mathcal {L}'}}}}. \end{aligned}$$

Thanks to Proposition 16 and 17 , we obtain

$$\begin{aligned} \sum \limits _{\scriptscriptstyle {{{\scriptscriptstyle \mathbf{v }}={\scriptscriptstyle {\mathcal {L}}}|{\scriptscriptstyle \mathcal {L}'}\in {\mathcal {V}}}}}\frac{{R_{{\mathbf{v }},{\scriptscriptstyle {\mathcal {L}}}}}^2}{d_{{\scriptscriptstyle {\mathcal {L}}},{\scriptscriptstyle \mathcal {L}'}}} \le 2 {C_{{\scriptscriptstyle \varGamma }}}\text { size}({\mathcal {T}}\,\,)^2 \left\| u_{\shortmid {\scriptscriptstyle \varGamma }} \right\| _{\scriptstyle {{H^2(\varGamma )}}}^2. \end{aligned}$$

(68)

Gathering estimates (66), (67) and (68) the claim follows. $\square$

We have obtained an error estimate between the center-value projection of the exact solution ${{\mathbb {P}}}^{{c}}_{{{\scriptscriptstyle {\mathcal {T}}}}}u$ and the approximate solution $u_{{\scriptscriptstyle {\mathcal {T}}}}$ for the Laplace problem with Ventcell boundary conditions for the $H^1$-seminorms. However in order to prove Proposition 7 we also need to prove an estimate between the exact solution u and the approximate solution $u_{{\scriptscriptstyle {\mathcal {T}}}}$ for the $L^2$-norms. To conclude we adopt here the same reasoning as that given in [11, Theorem 10.1] for the Laplace problem with Neumann boundary conditions, apart from the fact that here the domain is not polygonal.

Let $\beta _{{\scriptscriptstyle {\mathcal {T}}}}\in {\mathbb {R}}$ such that $\sum \limits _{{\scriptscriptstyle {\mathcal {K}}}\in {\mathfrak {M}}}m_{{{\scriptscriptstyle {\underline{{\mathcal {K}}}}}}}{{\bar{u}}}(x_{{\scriptscriptstyle {\mathcal {K}}}}) = \alpha$ with ${{\bar{u}}}=u+\beta _{{\scriptscriptstyle {\mathcal {T}}}}$. Setting ${\bar{e}}_{{\scriptscriptstyle {\mathcal {K}}}}= {{\bar{u}}}(x_{{\scriptscriptstyle {\mathcal {K}}}}) - u_{{\scriptscriptstyle {\mathcal {K}}}}$ for any ${\scriptstyle {\mathcal {K}}}\in {\mathfrak {M}}$ and ${\bar{e}}_{{\scriptscriptstyle {\mathcal {L}}}}= {{\bar{u}}}(x_{{\scriptscriptstyle {\mathcal {L}}}}) - u_{{\scriptscriptstyle {\mathcal {L}}}}$ for any ${\scriptstyle {\mathcal {L}}}\in \partial {\mathfrak {M}}$, estimate (62) is also satisfied for ${\bar{e}}_{{\scriptscriptstyle {\mathcal {T}}}}$. However, thanks to its definition the error ${\bar{e}}_{{\scriptscriptstyle {\mathfrak {M}}}}$ has now zero mean-value on ${\underline{\varOmega }}$. Thus if $\text { size}({\mathcal {T}}\,\,)\le \frac{1}{2 {C_{5}}}$, the discrete Poincaré inequality (7) gives

$$\begin{aligned} \left\| {\bar{e}}_{{\scriptscriptstyle {\mathfrak {M}}}} \right\| _{\scriptscriptstyle {{L^2(\varOmega )}}}^2 \le 4 {C_{4}^2} {C_{27}} \text { size}({\mathcal {T}}\,\,)^2 \left\| u \right\| _{\scriptstyle {{H^2_{\scriptscriptstyle \varGamma }(\varOmega )}}}^2, \end{aligned}$$

and thanks to the trace inequality (Lemma 4) we have

$$\begin{aligned} \left\| {\bar{e}}_{{\scriptscriptstyle \partial {\mathfrak {M}}}} \right\| _{\scriptscriptstyle {{L^2(\varGamma )}}}^2 \le (1+2{C_{4}})^2 {C_{8}}^2 {C_{27}} \text { size}({\mathcal {T}}\,\,)^2 \left\| u \right\| _{\scriptstyle {{H^2_{\scriptscriptstyle \varGamma }(\varOmega )}}}^2 . \end{aligned}$$

Thanks to the regularity of the function u, denoting by $L_u$ the Lipschitz constant of u, one has

$$\begin{aligned} \left\| u-u_{{\scriptscriptstyle {\mathcal {T}}}} \right\| _{\scriptscriptstyle {{L^2(\varOmega )}}}^2 \le 3 |\varOmega | L_u^2 \text { size}({\mathcal {T}}\,\,)^2 + 3 |\varOmega | \beta _{{\scriptscriptstyle {\mathcal {T}}}}^2 + 3\left\| {\bar{e}}_{{\scriptscriptstyle {\mathfrak {M}}}} \right\| _{\scriptscriptstyle {{L^2(\varOmega )}}}^2. \end{aligned}$$

We recall that $\int _\varOmega u = \sum \limits _{{\scriptscriptstyle {\mathcal {K}}}\in {\mathfrak {M}}}m_{{{\scriptscriptstyle {\underline{{\mathcal {K}}}}}}}{{\bar{u}}}(x_{{\scriptscriptstyle {\mathcal {K}}}}) = \alpha$ and $\beta _{{\scriptscriptstyle {\mathcal {T}}}}= {{\bar{u}}}- u$, thus one has

$$\begin{aligned} |{\underline{\varOmega }}| \beta _{{\scriptscriptstyle {\mathcal {T}}}}= \alpha - \sum \limits _{{\scriptscriptstyle {\mathcal {K}}}\in {\mathfrak {M}}}m_{{{\scriptscriptstyle {\underline{{\mathcal {K}}}}}}}u(x_{{\scriptscriptstyle {\mathcal {K}}}}) \end{aligned}$$

and

$$\begin{aligned} \sum \limits _{{\scriptscriptstyle {\mathcal {K}}}\in {\mathfrak {M}}}m_{{{\scriptscriptstyle {\underline{{\mathcal {K}}}}}}}u(x_{{\scriptscriptstyle {\mathcal {K}}}}) = \sum \limits _{{\scriptscriptstyle {\mathcal {K}}}\in {\mathfrak {M}}}\left( m_{{{\scriptscriptstyle {\underline{{\mathcal {K}}}}}}}- m_{{\scriptscriptstyle {\mathcal {K}}}}\right) u(x_{{\scriptscriptstyle {\mathcal {K}}}}) + \sum \limits _{{\scriptscriptstyle {\mathcal {K}}}\in {\mathfrak {M}}}\int _{\scriptstyle {\mathcal {K}}}\left( u(x_{{\scriptscriptstyle {\mathcal {K}}}}) - u(x) \right) + \alpha . \end{aligned}$$

Thus thanks to the regularity of u, Proposition 1 and the mesh regularity (5) we can claim that $| \beta _{{\scriptscriptstyle {\mathcal {T}}}}| \le C \text { size}({\mathcal {T}}\,\,)$ that concludes the proof.

The reasoning is exactly the same for the $L^2(\varGamma )$-norm that concludes the claim.