A fully coupled scheme using virtual element method and finite volume for poroelasticity

Abstract

In this paper, we design and study a fully coupled numerical scheme for the poroelasticity problem modeled through Biot’s equations. The classical way to numerically solve this system is to use a finite element method for the mechanical equilibrium equation and a finite volume method for the fluid mass conservation equation. However, to capture specific properties of the underground medium such as heterogeneities, discontinuities, and faults, meshing procedures commonly lead to badly shaped cells for finite element-based modeling. Consequently, we investigate the use of the recent virtual element method which appears as a potential discretization method for the mechanical part and could therefore allow the use of a unique mesh for both the mechanical and fluid flow modeling. Starting from a first insight into virtual element method applied to the elastic problem in the context of geomechanical simulations, we apply in addition a finite volume method to take care of the fluid conservation equation. We focus on the first-order virtual element method and the two-point flux approximation for the finite volume part. A mathematical analysis of this original coupled scheme is provided, including existence and uniqueness results and a priori estimates. The method is then illustrated by some computations on two- or three-dimensional grids inspired by realistic application cases.

Appendix: Details on the proofs

We provide here the details of the computation that led to the error estimate.

1.1 A.1 Proof of the tools

• 1. For two real a and b and for some 𝜖 > 0,

  $$ \left( \sqrt\epsilon a-\frac{b}{\sqrt \epsilon}\right)^{2} = \epsilon a^{2} - 2ab + \frac{b^{2}}{\epsilon}\ge 0 $$

  implying that $2ab \le \epsilon a^{2} + \frac {b^{2}}{\epsilon }$.

• 2. Let v ∈ (H¹(Ω))^d. Since $\text {div}\left (\mathbf {v}\right ) = {\sum }_{k=1}^{d} \frac {\partial \mathbf {v}_{k}}{\partial k}$, we have the following:

  $$ \begin{array}{@{}rcl@{}} \| \text{div}\left( \mathbf{v}\right) \|_{L^{2}({\Omega})}\! &=&\! \left\|\sum\limits_{k=1}^{d} \frac{\partial \mathbf{v}_{k}}{\partial k}\right\|_{L^{2}({\Omega})}\\ &\le&\! \sum\limits_{k=1}^{d} \left\|\frac{\partial \mathbf{v}_{k}}{\partial k}\right\|_{L^{2}({\Omega})} \text{(triangle inequality)}\\ &\le&\! \left( \sum\limits_{k=1}^{d} 1\right)^{\frac{1}{2}} \left( \sum\limits_{k=1}^{d} \left\|\frac{\partial \mathbf{v}_{k}}{\partial k}\right\|_{L^{2}({\Omega})}^{2}\right)^{\frac{1}{2}} \text{(CS)}\\ &\le&\! \sqrt{d}| \mathbf{v} |_{H^{1}({\Omega})}. \end{array} $$

• 3. Apply Taylor’s expansion with integral remainder

  $$ v(t) = \sum\limits_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(t-a)^{k} + {{\int}_{a}^{t}}\frac{f^{(n+1)}(s)}{n!}(t-s)^{n}ds $$

  with t = tⁿ, a = t^n− 1 and n = 0 for the first equality or n = 1 for the second one.

• 4. Let $\mathcal B$ be a symmetric bilinear form and {a_n}_n≥ 0, {b_n}_n≥ 0 two sequences belonging to the space of its arguments. For N = 1, we obviously have $\mathcal B(a_{1}, b_{1} - b_{0}) = \mathcal {B}(a_{1},b_{1}) - \mathcal {B}(a_{1},b_{0}).$ Now assume that

  $$ \begin{array}{@{}rcl@{}} &&{} \sum\limits_{k=1}^{N} \mathcal{B}(a_{k}, b_{k} - b_{k-1}) = \mathcal{B}(a_{N},b_{N}) - \mathcal{B}(a_{1},b_{0})\\ &&\quad - \sum\limits_{k=1}^{N-1}\mathcal{B}(b_{k},a_{k+1}-a_{k}) \text{ holds for a given N.} \end{array} $$

  Then,

  $$ \begin{array}{@{}rcl@{}} &&\sum\limits_{k=1}^{N+1} \mathcal{B}(a_{k}, b_{k} - b_{k-1}) \\ &=& \mathcal{B}(a_{N+1}, b_{N+1} - b_{N}) {+ \sum\limits_{k=1}^{N}\mathcal{B}(a_{k}, b_{k} - b_{k-1})}\\ &=& \mathcal{B}(a_{N+1}, b_{N+1} - b_{N}) + \mathcal{B}(a_{N},b_{N}) - \mathcal{B}(a_{1},b_{0}) \\ &&- \sum\limits_{k=1}^{N-1}\mathcal{B}(b_{k},a_{k+1}-a_{k}) \text{ by assumption}\\ &=& \mathcal{B}(a_{N+1}, b_{N+1}) - \mathcal B(a_{N+1}, b_{N}) + \mathcal{B}(a_{N},b_{N})\\ &&- \mathcal{B}(a_{1},b_{0}) - \sum\limits_{k=1}^{N-1}\mathcal{B}(b_{k},a_{k+1}-a_{k})\\ &=& \mathcal{B}(a_{N+1}, b_{N+1}) - \mathcal{B}(a_{1},b_{0})\\ &&- \sum\limits_{k=1}^{N}\mathcal{B}(b_{k},a_{k+1}-a_{k}) \text{ by symmetry of} \mathcal B. \end{array} $$

• 5. For two sequences {a_n}_n≥ 0, {b_n}_n≥ 0 in L²(Ω) or in (L²(Ω))^d and n ≥ 1,

  $$ \begin{array}{@{}rcl@{}} &&\left| {\int}_{\Omega} (b^{n} - b^{n-1})a^{n}\right|\\ &=& {\int}_{\Omega} \left| \left( {\int}_{t^{n-1}}^{t^{n}}\partial_{t} b(s) ds\right)\right|\left|a^{n}\right| \\ &\le& {\int}_{\Omega} \left( {\Delta} t{\int}_{t^{n-1}}^{t^{n}}(\partial_{t} b(s))^{2} ds\right)^{\frac{1}{2}} \left|a^{n}\right| \text{ (CS in time)}\\ &\le& \sqrt{{\Delta} t}\left( {\int}_{\Omega} {\int}_{t^{n-1}}^{t^{n}}(\partial_{t} b(s))^{2} ds\right)^{\frac{1}{2}} \\ &&\times \left( {\int}_{\Omega} (a^{n})^{2}\right)^{\frac{1}{2}} \text{ (CS in space)}\\ &=& \| a^{n} \|_{L^{2}} \sqrt{{\Delta} t}\left( {\int}_{t^{n-1}}^{t^{n}} {\int}_{\Omega}(\partial_{t} b(s))^{2} ds \right)^{\frac{1}{2}} \\ &=& \| a^{n} \|_{L^{2}} \left( {\Delta} t{\int}_{t^{n-1}}^{t^{n}} \| \partial_{t} b(s) \|_{L^{2}}^{2}ds \right)^{\frac{1}{2}}. \end{array} $$

1.2 A.2 Proof of Grönwall’s inequality

We provide the proof for a more general inequality:

Lemma 6

Let (a_n)_n≥ 0,(b_n)_n≥ 0,(c_n)_n≥ 0be sequences of non-negative numbers with (c_n)_n≥ 0non-decreasing. Assume that

$$ a_{0} + b_{0} \le c_{0} $$

and that there existsλ > 0 such that

$$ \forall{N \ge 1}, \quad a_{N} + b_{N} \leq \lambda \sum\limits_{m=0}^{N-1} a_{m} + c_{N}. $$

Then, there holds

$$ \forall{N \ge 0},\quad a_{N} + b_{N} \leq c_{N}\exp(\lambda N). $$

We first proceed by induction, showing that

$$ \forall N\ge 0, \quad a_{N} + b_{N} \le c_{N} (1+\lambda)^{N} . $$

(1)

By assumption, for N = 0, we have a₀ + b₀ ≤ c₀. Now assume that Eq. 36 holds for every n ≤ N and let us prove it holds when n = N + 1. Using the assumptions of the lemma,

$$ \begin{array}{@{}rcl@{}} a_{N+1} + b_{N+1} &\leq& \lambda \sum\limits_{m=0}^{N} a_{m} + c_{N+1} \\ &\leq& \lambda \sum\limits_{m=0}^{N} \left( c_{m}(1+\lambda)^{m} -b_{m} \right) \\ &&+ c_{N+1} \text{\quad from (36)}\\ &\leq& \lambda c_{N+1}\sum\limits_{m=0}^{N} (1+\lambda)^{m} + c_{N+1} \end{array} $$

because (c_n)_n≥ 0 is non-decreasing and (b_n)_n≥ 0 is non-negative. Summing a geometric sequence,

$$ \sum\limits_{m=0}^{N} (1+\lambda)^{m} = \frac{1 - (1+\lambda)^{N+1}}{1 - (1+\lambda)} = \frac{(1+\lambda)^{N+1}-1}{\lambda} $$

and thus a_N+ 1 + b_N+ 1 ≤ c_N+ 1(1 + λ)^N+ 1. This proves the induction, and the lemma comes from the estimate (1 + λ)ⁿ ≤ exp(nλ).

1.3 A.3 Proof of the estimates

We provide here the proofs of the estimates obtained on the B_i terms.

$$ \begin{array}{@{}rcl@{}} \bullet{\Delta} t {B_{1}^{n}} &=& - \alpha{\Delta} t{\int}_{\Omega} \text{div}\left( D^{-}E^{I,n}_{\mathbf{u}}\right)E_{p}^{A,n}\\ &&+ \alpha{\Delta} t{\int}_{\Omega} \text{div}\left( D^{-}E_{\mathbf{u}}^{A,n}\right) E_{p}^{I,n} \end{array} $$

We split this term in two parts corresponding to the two integrals. For the first part, applying successively tool 5 and tool 2,

$$ \begin{array}{@{}rcl@{}} &&{}\left| {\Delta} t {\int}_{\Omega} E_{p}^{A,n} \text{div}\left( D^{-}E_{\mathbf{u}}^{I,n}\right) \right|\\ &&\quad= \sqrt{{\Delta} t}\| E_{p}^{A,n} \|_{L^{2}} \left( {\int}_{t^{n-1}}^{t^{n}}\| \text{div}\left( \partial_{t}E_{\mathbf{u}}^{I}\right) \|_{L^{2}}^{2}\right)^{\frac{1}{2}}\\ &&\quad\le \sqrt{d{\Delta} t}\| E_{p}^{A,n} \|_{L^{2}} \left( {\int}_{t^{n-1}}^{t^{n}}\| \partial_{t}E_{\mathbf{u}}^{I} \|_{H^{1}}^{2}\right)^{\frac{1}{2}} \end{array} $$

then, we apply Young’s inequality with $\epsilon =\frac {1}{4}$ for the last term and $\epsilon = \frac {\epsilon _{r}}{4}$ for the others:

$$ \begin{array}{@{}rcl@{}} &&{}\left| 2\sum\limits_{n=1}^{N}\alpha{\Delta} t {\int}_{\Omega} E_{p}^{A,n} \text{div}\left( D^{-}E_{\mathbf{u}}^{I,n}\right) \right| \\ &&\quad\le \frac{\epsilon_{r}}{4} \sum\limits_{n=1}^{N-1}{\Delta} t\|{E_{p}^{A,n}}\|_{L^{2}}^{2} + \frac{{\Delta} t}{4} \| E_{p}^{A,N} \|_{L^{2}}^{2}\\ &&\qquad+ 4d\alpha^{2}\left( \frac{1}{\epsilon_{r}} \sum\limits_{n=1}^{N-1}\left( {\int}_{t^{n-1}}^{t^{n}} \| \partial_{t} E_{\mathbf{u}}^{I} \|_{H^{1}}^{2}\right)^{\frac{1}{2}} \right.\\ &&\qquad\left. + \left( {\int}_{t^{N-1}}^{t^{N}}\| \partial_{t} E_{\mathbf{u}}^{I} \|_{H^{1}}^{2}\right)^{\frac{1}{2}} \right). \end{array} $$

We now define the L² in time norm as $\| g \|_{L^{2}(T)}^{2} = {\int }_{t^{0}}^{t^{N}} g(s)^{2} ds$. We can add the missing terms (which are non-negative) to write the following:

$$ \begin{array}{@{}rcl@{}} &&{} \left| 2\sum\limits_{n=1}^{N}\alpha{\Delta} t {\int}_{\Omega} E_{p}^{A,n} \text{div}\left( D^{-}E_{\mathbf{u}}^{I,n}\right) \right|\\ &&\quad \le \frac{\epsilon_{r}}{4} \sum\limits_{n=1}^{N-1}{\Delta} t\|{E_{p}^{A,n}}\|_{L^{2}}^{2} + \frac{{\Delta} t}{4} \| E_{p}^{A,N} \|_{L^{2}}^{2} \\ &&\qquad+ 4d\alpha^{2}\left( \frac{1}{\epsilon_{r}} \| \partial_{t} E_{\mathbf{u}}^{I} \|_{L^{2}(0,T; H^{1})}^{2} +\| \partial_{t} E_{\mathbf{u}}^{I} \|_{L^{2}(0,T; H^{1})}^{2} \right)\\ &&\quad\le \frac{\epsilon_{r}}{4} \sum\limits_{n=1}^{N-1}{\Delta} t\| E_{p}^{A,n} \|_{L^{2}}^{2} + \frac{{\Delta} t}{4} \| E_{p}^{A,N} \|_{L^{2}}^{2}\\ &&\qquad + 4d\alpha^{2}\left( \frac{1}{\tilde \epsilon_{r}} \| \partial_{t} E_{\mathbf{u}}^{I} \|_{L^{2}(0,T;H^{1})}^{2} \right) \end{array} $$

where $\tilde \epsilon _{r} = \frac {\epsilon _{r}}{1+\epsilon _{r}}$. For the second part, a partial summation is done first:

$$ \begin{array}{@{}rcl@{}} &&{} 2\sum\limits_{n=1}^{N}\! {\Delta} t\!{\int}_{\Omega} \alpha \text{div}\!\left( D^{-}E_{\mathbf{u}}^{A,n}\right)E_{p}^{I,n}\! =\! 2\!{\int}_{\Omega} \alpha \text{div}\left( E_{\mathbf{u}}^{A,N}\right)E_{p}^{I,N}\\ &&{}-\! 2\!{\int}_{\Omega}\! \alpha \text{div}\!\left( E_{\mathbf{u}}^{A,0}\right)\! E_{p}^{I,1} \! -\! 2\sum\limits_{n=1}^{N-1}\! {\Delta} t\!{\int}_{\Omega}\! \alpha \text{div}\!\left( \! E_{\mathbf{u}}^{A,n}\!\right)\! D^{-}E_{p}^{I,n+1} \end{array} $$

and, using the same procedure as above,

$$ \begin{array}{@{}rcl@{}} &&{} 2\left|{\int}_{\Omega}{} \alpha \text{div}\!\left( E_{\mathbf{u}}^{A,N}\right)\!E_{p}^{I,N}\right|\! \le\! \frac{\epsilon_{l}}{4} \| E_{\mathbf{u}}^{A,N} \|_{H^{1}}^{2}\! +\! \frac{4d\alpha^{2}}{\epsilon_{l}}\| E_{p}^{I,N} \|_{L^{2}}^{2},\\ &&{}2\left|{\int}_{\Omega} \alpha \text{div}\left( E_{\mathbf{u}}^{A,0}\right)E_{p}^{I,1}\right| \! \le\! \frac{\epsilon_{l}}{4} \| E_{\mathbf{u}}^{A,0} \|_{H^{1}}^{2} + \frac{4d\alpha^{2}}{\epsilon_{l}}\| E_{p}^{I,1} \|_{L^{2}}^{2},\\ &&{}2\left| \sum\limits_{n=1}^{N-1}\!\alpha{\Delta} t\! {\int}_{\Omega}\! \text{div}\!\left( E_{\mathbf{u}}^{A,n}\right)\!D^{-}E_{p}^{I,n+1} \right|\! \le\! \frac{\epsilon_{r}}{4}\sum\limits_{n=1}^{N-1}\!{\Delta} t\| E_{\mathbf{u}}^{A,n} \|_{H^{1}}^{2}\\ &&\quad+\frac{4d\alpha^{2}}{\epsilon_{r}}\| \partial_{t} {E_{p}^{I}} \|_{L^{2}(0,T; L^{2})}^{2}. \end{array} $$

Combining the two parts leads to the estimation on B₁:

$$ \begin{array}{@{}rcl@{}} \left|2{\Delta} t \sum\limits_{n=1}^{N}B_{1}\right| \!&\le&\! \frac{\epsilon_{r}}{4} \sum\limits_{n=1}^{N-1}{\Delta} t\| E_{p}^{A,n} \|_{L^{2}}^{2} + \frac{\epsilon_{r}}{4}\sum\limits_{n=1}^{N-1}{\Delta} t\| E_{\mathbf{u}}^{A,n} \|_{H^{1}}^{2}\\ &&+ \frac{{\Delta} t}{4} \| E_{p}^{A,N} \|_{L^{2}}^{2} + \frac{\epsilon_{l}}{4} \| E_{\mathbf{u}}^{A,N} \|_{H^{1}}^{2}\\ &&+ \frac{\epsilon_{l}}{4} \| E_{\mathbf{u}}^{A,0} \|_{H^{1}}^{2} + 4d\alpha^{2}\\ &&\times\left( \frac{\| \partial_{t} E_{\mathbf{u}}^{I} \|_{L^{2}(0,T;H^{1})}^{2}}{\tilde \epsilon_{r}} +\frac{\| E_{p}^{I,N} \|_{L^{2}}^{2}}{\epsilon_{l}} \right.\\ &&\left. + \frac{\| E_{p}^{I,1} \|_{L^{2}}^{2}}{\epsilon_{l}} + \frac{\| \partial_{t} {E_{p}^{I}} \|_{L^{2}(0,T;L^{2})}^{2}}{\epsilon_{r}}\right). \end{array} $$

$$ \begin{array}{@{}rcl@{}} \bullet{\Delta} t {B_{2}^{n}} &=& {\Delta} t \sum\limits_{K\in\tau_{h}} a^{K}(-E_{\mathbf{u}}^{\pi,n}, D^{-}E_{\mathbf{u}}^{A,n}) \\ &&+ {\Delta} t \sum\limits_{K\in \tau_{h}}a_{h}(E_{\mathbf{u}}^{\pi,n} - E_{\mathbf{u}}^{I,n}, D^{-}E_{\mathbf{u}}^{A,n}) \end{array} $$

We start with the a^K part. We denote M_K the continuity constant of a^K and we note that, for a function u locally in (H¹(K))^d and a function v globally in (H¹(Ω))^d and for some 𝜖 > 0,

$$ \begin{array}{@{}rcl@{}} \left| \sum\limits_{K\in\tau_{h}}a^{K}(\mathbf{u},\mathbf{v}) \right| &\le& \sum\limits_{K\in\tau_{h}} \left( M_{K} \| \mathbf{u} \|_{H^{1}(K)}\| \mathbf{v} \|_{H^{1}(K)} \right)\\ &\le& \sum\limits_{K\in\tau_{h}} \left( \frac{{M_{K}^{2}}}{2\epsilon} \| \mathbf{u} \|_{H^{1}(K)}^{2} + \frac{\epsilon}{2} \| \mathbf{v} \|_{H^{1}(K)}^{2}\right)\\ &\le& \frac{M^{2}}{2\epsilon} \sum\limits_{K\in\tau_{h}} \| \mathbf{u} \|_{H^{1}(K)}^{2} + \frac{\epsilon}{2} \sum\limits_{K\in\tau_{h}} \| \mathbf{v} \|_{H^{1}(K)}^{2}\\ &=&\frac{M^{2}}{2\epsilon} \| \mathbf{u} \|_{H^{1}(\tau)}^{2} + \frac{\epsilon}{2} \| \mathbf{v} \|_{H^{1}({\Omega})}^{2} \end{array} $$

where M = maxKM_K.

We now use the partial summation (recall that a^K is symmetric)

$$ \begin{array}{@{}rcl@{}} &&{}2 \sum\limits_{n=1}^{N} {\Delta} t \sum\limits_{K\in\tau_{h}}a^{K}(-E_{\mathbf{u}}^{\pi,n}, D^{-}E_{\mathbf{u}}^{A,n})\\ &&\quad = 2\!\sum\limits_{K\in\tau_{h}}a^{K}\!(-E_{\mathbf{u}}^{\pi,N}, E_{\mathbf{u}}^{A,N})\! -\! 2\sum\limits_{K\in\tau_{h}}a^{K}\!(-E_{\mathbf{u}}^{\pi,1}, E_{\mathbf{u}}^{A,0}) \\ &&\qquad- 2\sum\limits_{n=1}^{N-1} \sum\limits_{K\in\tau_{h}} a^{K}(-(E_{\mathbf{u}}^{\pi,n+1} -E_{\mathbf{u}}^{\pi,n}), E_{\mathbf{u}}^{A,n}) \end{array} $$

and combine it with the previous estimate to write the following:

$$ \begin{array}{@{}rcl@{}} &&{}\left| 2\!\sum\limits_{K\in\tau_{h}}\! a^{K}\!(\!-E_{\mathbf{u}}^{\pi{\kern-.5pt},{\kern-.5pt}N}\!,\! E_{\mathbf{u}}^{A{\kern-.5pt},{\kern-.5pt}N}\!) \right|\!\!\le\! \frac{8M^{2}}{\epsilon_{l}}\| E_{\mathbf{u}}^{\pi{\kern-.5pt},{\kern-.5pt}N} \!\|_{H^{1}({\kern-.5pt}\tau{\kern-.5pt})}^{2} \! +\! \frac{\epsilon_{l}}{8} \|\! E_{\mathbf{u}}^{A{\kern-.5pt},{\kern-.5pt}N} \!\|_{H^{1}}^{2},\\ &&{}\left| 2\!\sum\limits_{K\in\tau_{h}}\! a^{K}\!(\!-E_{\mathbf{u}}^{\pi,1},\! E_{\mathbf{u}}^{A,0}) \right|\!\le\! \frac{8M^{2}}{\epsilon_{l}}\| E_{\mathbf{u}}^{\pi,1} \!\|_{H^{1}(\tau)}^{2} \! +\! \frac{\epsilon_{l}}{8} \| E_{\mathbf{u}}^{A,0} \|_{H^{1}}^{2}\! . \end{array} $$

For the third term, we also follow the idea of tool 5:

$$ \begin{array}{@{}rcl@{}} &&{}2 \sum\limits_{K\in\tau_{h}} \left| a^{K}(E_{\mathbf{u}}^{\pi,{n+1}}-E_{\mathbf{u}}^{\pi,n}, E_{\mathbf{u}}^{A,n}) \right|\\ &&\quad = 2 \sum\limits_{K\in\tau_{h}} \left|a^{K}({\int}_{t^{n}}^{t^{n+1}} \partial_{t} E_{\mathbf{u}}^{\pi}, E_{\mathbf{u}}^{A,n})\right| \\ &&\quad\le 2\sum\limits_{K\in\tau_{h}}M_{K}\| {\int}_{t^{n}}^{t^{n+1}}\partial_{t} E_{\mathbf{u}}^{\pi} \|_{H^{1}(K)}\| E_{\mathbf{u}}^{A,n} \|_{H^{1}(K)} \\ &&\quad\le 2\sum\limits_{K\in\tau_{h}}M_{K}\!\left( {\Delta} t{\int}_{t^{n}}^{t^{n+1}}\| \partial_{t} E_{\mathbf{u}}^{\pi} \|_{H^{1}(K)}^{2}\right)^{\frac{1}{2}} \!\| E_{\mathbf{u}}^{A,n} \|_{H^{1}(K)}\\ &&\quad \le \sum\limits_{K\in\tau_{h}} \frac{8M^{2}}{\epsilon_{r}}{\int}_{t^{n}}^{t^{n+1}}\!\| \partial_{t} E_{\mathbf{u}}^{\pi} \|_{H^{1}(K)}^{2}\! +\! {\Delta} t\frac{\epsilon_{r}}{8}\| E_{\mathbf{u}}^{A,n} \|_{H^{1}(K)}^{2}\\ &&\quad = \frac{8M^{2}}{\epsilon_{r}}{\int}_{t^{n}}^{t^{n+1}}\| \partial_{t} E_{\mathbf{u}}^{\pi} \|_{H^{1}(\tau)}^{2} + {\Delta} t\frac{\epsilon_{r}}{8}\| E_{\mathbf{u}}^{A,n} \|_{H^{1}}^{2}. \end{array} $$

We now apply the same methods to the a_h part, denoting $\tilde M_{K}$ the continuity constant of ${a_{h}^{K}}$ and $\tilde M = \max _{K} M_{K}$,

$$ \begin{array}{@{}rcl@{}} &&{}2 \sum\limits_{n=1}^{N} {\Delta} t \sum\limits_{K\in\tau_{h}}{a_{h}^{K}}(E_{\mathbf{u}}^{\pi,n}-E_{\mathbf{u}}^{I,n}, D^{-}E_{\mathbf{u}}^{A,n}) \\ &&\quad = 2\sum\limits_{K\in\tau_{h}} {a_{h}^{k}}(E_{\mathbf{u}}^{\pi,N}-E_{\mathbf{u}}^{I,N}, E_{\mathbf{u}}^{A,N}) \\ &&\qquad - 2\sum\limits_{K\in\tau_{h}} {a_{h}^{K}}(E_{\mathbf{u}}^{\pi,1} - E_{\mathbf{u}}^{I,1}, E_{\mathbf{u}}^{A,0}) \\ &&\qquad - 2\sum\limits_{n=1}^{N-1} \sum\limits_{K\in\tau_{h}} {a_{h}^{K}}((E_{\mathbf{u}}^{\pi,n+1} -E_{\mathbf{u}}^{\pi,n}) \\ &&\qquad - (E_{\mathbf{u}}^{I,n+1}-E_{\mathbf{u}}^{I,n}), E_{\mathbf{u}}^{A,n}) \end{array} $$

and we split the errors E^I and E^π

$$ \begin{array}{@{}rcl@{}} &&{}2\!\sum\limits_{K\in\tau_{h}}\! \left| {a_{h}^{K}}(E_{\mathbf{u}}^{\pi,N}\! -\! E_{\mathbf{u}}^{I,N},\! E_{\mathbf{u}}^{A,N}) \right| \le\! \frac{8\tilde M^{2}}{\epsilon_{l}}\| E_{\mathbf{u}}^{\pi,N} \|_{H^{1}(\tau)}^{2}\\ && + \frac{4\tilde M^{2}}{\epsilon_{l}}\| E_{\mathbf{u}}^{I,N} \|_{H^{1}}^{2}+ (\frac{\epsilon_{l}}{8}+\frac{\epsilon_{l}}{4}) \| E_{\mathbf{u}}^{A,N} \|_{H^{1}}^{2},\\ &&{}2 \sum\limits_{K\in\tau_{h}} \left| {a_{h}^{K}}(E_{\mathbf{u}}^{\pi,1} - E_{\mathbf{u}}^{I,1}, E_{\mathbf{u}}^{A,0}) \right| \le \frac{8\tilde M^{2}}{\epsilon_{l}}\| E_{\mathbf{u}}^{\pi,1} \|_{H^{1}(\tau)}^{2}\\ &&{\kern2pt} + \frac{4\tilde M^{2}}{\epsilon_{l}}\| E_{\mathbf{u}}^{I,1} \|_{H^{1}}^{2}+ (\frac{\epsilon_{l}}{8}+\frac{\epsilon_{l}}{4}) \| E_{\mathbf{u}}^{A,0} \|_{H^{1}}^{2},\\ &&{}2\sum\limits_{K\in\tau_{h}} \left|{a_{h}^{K}} (E_{\mathbf{u}}^{\pi,n+1} -E_{\mathbf{u}}^{\pi,n}, E_{\mathbf{u}}^{A,n})\right| \le \frac{8\tilde M^{2}}{\epsilon_{r}}{\int}_{t^{n}}^{t^{n+1}}\| \partial_{t} E_{\mathbf{u}}^{\pi} \|_{H^{1}(\tau)}^{2}\\ &&{\kern2pt}+\frac{\epsilon_{r}}{8} {\Delta} t\| E_{\mathbf{u}}^{A,n} \|_{H^{1}}^{2},\\ &&{}2\sum\limits_{K\in\tau_{h}}\left| {a_{h}^{K}} (E_{\mathbf{u}}^{I,n+1}-E_{\mathbf{u}}^{I,n}, E_{\mathbf{u}}^{A,n})\right|\le \frac{4\tilde M^{2}}{\epsilon_{r}}{\int}_{t^{n}}^{t^{n+1}}\| \partial_{t} E_{\mathbf{u}}^{I} \|_{H^{1}}^{2}\\ &&{\kern2pt}+ \frac{\epsilon_{r}}{4} {\Delta} t\| E_{\mathbf{u}}^{A,n} \|_{H^{1}}^{2}. \end{array} $$

We finally obtain the bound as follows:

$$ \begin{array}{@{}rcl@{}} &&{}\left|2{\Delta} t \sum\limits_{n=1}^{N}B_{2}\right|\\ &&\le \frac{\epsilon_{r}}{2} \sum\limits_{n=1}^{N-1}{\Delta} t\| E_{\mathbf{u}}^{A,n} \|_{H^{1}}^{2} +\frac{\epsilon_{l}}{2} \| E_{\mathbf{u}}^{A,N} \|_{H^{1}}^{2} +\frac{\epsilon_{l}}{2} \| E_{\mathbf{u}}^{A,0} \|_{H^{1}}^{2}\\ &&+4\tilde M^{2}\left( \frac{\| E_{\mathbf{u}}^{I,1} \|_{H^{1}}^{2}}{\epsilon_{l}} + \frac{\| E_{\mathbf{u}}^{I,N} \|_{H^{1}}^{2}}{\epsilon_{l}}+ +\frac{\| \partial_{t} E_{\mathbf{u}}^{I} \|_{L^{2}(0,T; H^{1})}^{2}}{\epsilon_{r}}\right)\\ &&+8(M^{2}\! +\!\tilde M^{2})\!\left( \!\frac{\| E_{\mathbf{u}}^{\pi,1}\! \|_{H^{1}{\kern-.5pt}({\kern-.5pt}\tau{\kern-.5pt})}^{2}}{\epsilon_{l}} \! +\! \frac{\| E_{\mathbf{u}}^{\pi,N}\! \|_{H^{1}{\kern-.5pt}({\kern-.5pt}\tau{\kern-.5pt})}^{2}}{\epsilon_{l}}\! +\! \frac{\| \partial_{t} E_{\mathbf{u}}^{\pi}\! \|_{L^{2}(0,T;H^{1}{\kern-.5pt}({\kern-.5pt}\tau{\kern-.5pt}){\kern-.5pt})}^{2}}{\epsilon_{r}}\!\right)\! . \end{array} $$

$$ \begin{array}{@{}rcl@{}} \bullet {\Delta} t{B_{3}^{n}} &=& {\Delta} t{\int}_{\Omega} c_{0} (D^{-}p^{n} -\partial_{t} p^{n})E_{p}^{A,n}\\ &&+ {\Delta} t{\int}_{\Omega} \alpha\text{div}\left( D^{-}\mathbf{u}^{n}-\partial_{t}\mathbf{u}^{n}\right)E_{p}^{A,n} \end{array} $$

We first use the Taylor’s expansion (tool 3) as follows:

$$ \begin{array}{@{}rcl@{}} {\Delta} t{B_{3}^{n}} &=& {\int}_{\Omega} c_{0} {\int}_{t^{n-1}}^{t^{n}}\partial_{tt}p(s)(t^{n-1}-s)^{2}ds E_{p}^{A,n}\\ &&+{\int}_{\Omega} \alpha\text{div}\left( {\int}_{t^{n-1}}^{t^{n}}\partial_{tt}\mathbf{u}(s)(t^{n-1}-s)^{2}ds\right)E_{p}^{A,n} \end{array} $$

We then follow the idea of the double Cauchy-Schwarz (tool 5), this time using the fact that ${\int }_{t^{n-1}}^{t^{n}} (t^{n-1}-s)^{2} ds = \frac {{\Delta } t^{3}}{3}$. In addition, tool 2 is used to treat the divergence operator. Thus,

$$ \begin{array}{@{}rcl@{}} \left|{\Delta} t{B_{3}^{n}}\right| &\le& c_{0}\| E_{p}^{A,n} \|_{L^{2}}\left( \frac{{\Delta} t^{3}}{3} {\int}_{t_{n-1}}^{t^{n}} \| \partial_{tt}p(s) \|_{L^{2}}^{2} ds\right)^{\frac{1}{2}}\\ &&+ \alpha\| E_{p}^{A,n} \|_{L^{2}}\left( \frac{d{\Delta} t^{3}}{3} {\int}_{t_{n-1}}^{t^{n}} \| \partial_{tt}\mathbf{u}(s) \|_{H^{1}}^{2} ds\right)^{\frac{1}{2}}. \end{array} $$

Following what has been done for the estimate of B¹, Young’s inequality is used with $\epsilon =\frac {1}{4}$ for the last term and $\epsilon =\frac {\epsilon _{r}}{4}$ for the others:

$$ \begin{array}{@{}rcl@{}} &&{}\left|2{\Delta} t \sum\limits_{n=1}^{N}{B_{3}^{n}}\right|\le \frac{2\epsilon_{r}}{4} \sum\limits_{n=1}^{N-1}{\Delta} t \| E_{p}^{A,n} \|_{L^{2}}^{2} + \frac{2{\Delta} t}{4}\| E_{p}^{A,N} \|_{L^{2}}^{2}\\ &&{\kern3pt} + \frac{4}{3}{\Delta} t^{2}\! \left( {c_{0}^{2}}\frac{\| \partial_{tt}p \|_{L^{2}(0,T; L^{2})}^{2}}{\tilde \epsilon_{r}} + d\alpha^{2}\frac{\| \partial_{tt}\mathbf{u} \|_{L^{2}(0,T; H^{1})}^{2}}{\tilde \epsilon_{r}} \right). \end{array} $$

$$ \bullet {\Delta} t {B_{4}^{n}} = {\Delta} t{\int}_{\Omega} (\mathbf{f}^{{\kern1pt}n}-\mathbf{f}_{h}^{{\kern1pt}n})\cdot D^{-}E_{\mathbf{u}}^{A,n} $$

We assimilate the difference $(\mathbf {f}^{{\kern 1pt}n}-\mathbf {f}_{h}^{{\kern 1pt}n})$ to an interpolation error denoted $E_{\mathbf {f}}^{I,n}$. Using a partial summation,

$$ \begin{array}{@{}rcl@{}} 2{\kern-.5pt}{\Delta}{\kern-.5pt} t\! \sum\limits_{n=1}^{N} {\int}_{\Omega}\! E_{\mathbf{f}}^{I{\kern-.5pt},n}\!\cdot\! D^{-}\! E_{\mathbf{u}}^{A{\kern-.5pt},n}\! &=&\! 2\! {\int}_{\Omega} E_{\mathbf{f}}^{I,N} E_{\mathbf{u}}^{A,N}\! -\! {\int}_{\Omega} \! 2 E_{\mathbf{f}}^{I,1}\! E_{\mathbf{u}}^{A,0}\\ &&-2\! \sum\limits_{n=1}^{N-1}\! {\int}_{\Omega} \!(E_{\mathbf{f}}^{I{\kern-.5pt},n{\kern-.5pt}+{\kern-.5pt}1}\! -\! E_{\mathbf{f}}^{I{\kern-.5pt},{\kern-.5pt}n})\!\cdot\! E_{\mathbf{u}}^{A{\kern-.5pt},{\kern-.5pt}n}{\kern-.5pt}. \end{array} $$

Using Cauchy-Schwarz and Young inequalities, and treating the third term with tool 5, we have the following:

$$ \begin{array}{@{}rcl@{}} 2\left| {\int}_{\Omega} E_{\mathbf{f}}^{I,N} E_{\mathbf{u}}^{A,N}\right|\!\!&\le&\!\! 2\| E_{\mathbf{f}}^{I,N} \|_{L^{2}}\| E_{\mathbf{u}}^{A,N} \|_{L^{2}}\\ \!\!&\le&\!\!\frac{4}{\epsilon_{l}}\| E_{\mathbf{f}}^{I,N} \|_{L^{2}}^{2} + \frac{\epsilon_{l}}{4}\| E_{\mathbf{u}}^{A,N} \|_{H^{1}}^{2},\\ 2\left| {\int}_{\Omega} E_{\mathbf{f}}^{I,1} E_{\mathbf{u}}^{A,0}\right| \!\!&\le&\!\! 2\| E_{\mathbf{f}}^{I,1} \|_{L^{2}}\| E_{\mathbf{u}}^{A,0} \|_{L^{2}}\\ \!\!&\le&\!\! \frac{4}{\epsilon_{l}}\| E_{\mathbf{f}}^{I,1} \|_{L^{2}}^{2} + \frac{\epsilon_{l}}{4}\| E_{\mathbf{u}}^{A,0} \|_{H^{1}}^{2},\\ 2\left| {\int}_{\Omega} (E_{\mathbf{f}}^{I,n+1} - E_{\mathbf{f}}^{I,n})\cdot E_{\mathbf{u}}^{A,n}\right| \!\!&\le&\!\! \| E_{\mathbf{u}}^{A,n} \|_{L^{2}} \left( {\Delta} t {\int}_{t^{n}}^{t^{n+1}} \| \partial_{t} E_{\mathbf{f}}^{I} \|_{L^{2}}^{2} \right)^{\frac{1}{2}}\\ \!\!&\le&\!\! \frac{4}{\epsilon_{r}}{\int}_{t^{n}}^{t^{n+1}} \| \partial_{t} E_{\mathbf{f}}^{I} \|_{L^{2}}^{2} + \frac {\epsilon_{r}}{4}{\Delta} t\| E_{\mathbf{u}}^{A,n} \|_{H^{1}}^{2}, \end{array} $$

and thus, noticing that $\| E_{\mathbf {u}}^{A} \|_{L^{2}} \le \| E_{\mathbf {u}}^{A} \|_{H^{1}}$,

$$ \begin{array}{@{}rcl@{}} \left|2{\Delta} t \sum\limits_{n=1}^{N}{B_{4}^{n}}\right| &\le& \frac{\epsilon_{r}}{4} \sum\limits_{n=1}^{N-1} {\Delta} t\| E_{\mathbf{u}}^{A,n} \|_{H^{1}}^{2} + \frac{\epsilon_{l}}{4}\| E_{\mathbf{u}}^{A,N} \|_{H^{1}}^{2}\\ &&+ \frac{4}{\epsilon_{r}}\| \partial_{t} E_{\mathbf{f}}^{I} \|_{L^{2}(0,T; L^{2})}^{2}\\ &&+ \frac{4}{\epsilon_{l}}\left( \| E_{\mathbf{f}}^{I,N} \|_{L^{2}}^{2} + \| E_{\mathbf{f}}^{I,1} \|_{L^{2}}^{2}\right) \\ &&+ \frac{\epsilon_{l}}{4}\| E_{\mathbf{u}}^{A,0} \|_{H^{1}}^{2}. \end{array} $$

$$ \bullet {\Delta} t {B_{5}^{n}} = -{\Delta} t \sum\limits_{K} \sum\limits_{f\subset\partial K} (\overline{F_{Kf}^{n}}-F^{\star,n}_{Kf})E_{p}^{A,n} $$

For this part, we use the consistency of the fluxes $\overline {F_{Kf}^{n}} = -\overline {F_{Lf}^{n}}$ and $F^{\star ,n}_{Kf} = -F^{\star ,n}_{Lf}$. This allows to reorganize the double sum in the following way, K standing for any of the two neighboring cells of f,

$$ \begin{array}{@{}rcl@{}} &&{}\sum\limits_{K} \sum\limits_{f\subset\partial K} (\overline{F_{Kf}^{n}}-F^{\star,n}_{Kf})E_{p}^{A,n} \\ &&\quad = \sum\limits_{f \subset \overline {\Omega} \setminus {\Gamma}_{N_{p}}} (\overline{F_{Kf}^{n}}-F^{\star,n}_{Kf})D_{f}(E_{p}^{A,n})\\ &&\quad \le \left( \sum\limits_{f \subset \overline {\Omega} \setminus {\Gamma}_{N_{p}}} T_{f}^{-1}\left|\overline{F_{Kf}^{n}}-F^{\star,n}_{Kf}\right|^{2}\right)^{\frac{1}{2}} \\&&\qquad \times\left( \sum\limits_{f \subset \overline {\Omega} \setminus {\Gamma}_{N_{p}}} T_{f}D_{f}(E_{p}^{A,n})^{2}\right)^{\frac{1}{2}} \quad\text{(CS)}\\ &&\quad \le \frac{| E_{p}^{A,n} |_{1,\tau}^{2}}{2} +\sum\limits_{f \subset \overline {\Omega} \setminus {\Gamma}_{N_{p}}} \frac{\left|\overline{F_{Kf}^{n}}-F^{\star,n}_{Kf}\right|^{2}}{2T_{f}} \quad\text{(Young)} \end{array} $$

and thus,

$$ \begin{array}{@{}rcl@{}} \left|2{\Delta} t \sum\limits_{n=1}^{N}{B_{5}^{n}}\right| &\le& \sum\limits_{n=1}^{N} {\Delta} t| E_{p}^{A,n} |_{1,\tau}^{2}\\ &&+\sum\limits_{n=1}^{N} {\Delta} t\sum\limits_{f \subset \overline {\Omega} \setminus {\Gamma}_{N_{p}}} \frac{\left|\overline{F_{Kf}^{n}}-F^{\star,n}_{Kf}\right|^{2}}{T_{f}}.\hfill \end{array} $$

$$ \bullet {\Delta} t {B_{6}^{n}} = -{\Delta} tc_{0}{\int}_{\Omega} E_{p}^{A,n} D^{-}E_{p}^{I,n} $$

We use tool 5 and tool 1 to write:

$$ \begin{array}{@{}rcl@{}} \left| {\Delta} t {B_{6}^{n}} \right| &\le& c_{0} \| E_{p}^{A,n} \|_{L^{2}}\left( {\Delta} t {\int}_{t^{n-1}}^{t^{n}}\| \partial_{t} {E_{p}^{I}} \|_{L^{2}}^{2}\right)^{\frac{1}{2}}\\ &\le& \frac{\epsilon{\Delta} t}{2} \| E_{p}^{A,n} \|_{L^{2}}^{2} + \frac{{c_{0}^{2}}}{2\epsilon}{\int}_{t^{n-1}}^{t^{n}}\| \partial_{t} {E_{p}^{I}} \|_{L^{2}}^{2} \end{array} $$

and we set $\epsilon = \frac {1}{4}$ for n = N and $\epsilon =\frac {\epsilon _{r}}{4}$ for the others n leading to the following:

$$ \begin{array}{@{}rcl@{}} \left|2{\Delta} t \sum\limits_{n=1}^{N}{B_{6}^{n}}\right| &\le& \frac{\epsilon_{r}}{4}\sum\limits_{n=1}^{N-1}{\Delta} t\| E_{p}^{A,n} \|_{L^{2}}^{2} +\frac{{\Delta} t}{4}\| E_{p}^{A,N} \|_{L^{2}}^{2} \\ &&+ \frac{4{c_{0}^{2}}}{\tilde \epsilon_{r}} \| \partial_{t} {E_{p}^{I}} \|_{L^{2}(0,T;L^{2})}^{2}. \end{array} $$