An arbitrarily high order and asymptotic preserving kinetic scheme in compressible fluid dynamic

The next step is to discretise in time, we subdivide the interval $[t_{n},t_{n+1}]$ into sub-intervals obtained from the partition

$$t_{n}=t_{(0)}<t_{(1)}<\cdots t_{(p)}<\cdots<t_{(M)}=t_{n+1}$$

with $t_{(p)}=t_{n}+\beta_{p}\Delta t$. We approximate the integral over time using quadrature formula

$$\int_{t_{n}}^{t_{n}+\beta_{p}\Delta t}\phi(s)\;ds\approx\Delta t\sum_{q=0}^{M}w_{pq}\phi(t_{n}+\beta_{q}\Delta t)$$

where $w_{pq}$ are the weights. In order to obtain consistent quadrature formula of order $q+1$, we should have

$$w_{pq}=\int_{0}^{\beta_{p}}l_{q}(s)\;ds,\quad\sum_{q=0}^{M}w_{pq}=\beta_{p}$$

where $\{l_{q}\}_{q=0}^{M}$ is the Lagrangian basis polynomial of the q-th node.

Let $\mathbf{x}_{l}=(x_{i},y_{j})$ be a fixed grid point, $\mathbf{f}_{l}^{n,p}\approx\mathbf{f}(\mathbf{x}_{l},t_{n}+\beta_{p}\Delta t)$ and $\mathbf{f}_{l}^{n,0}=\mathbf{f}_{l}^{n}$. We introduce the corrections $r=0,\cdots,R$ for each subinterval $[t_{p},t_{p+1}]$ and denote the solution at the $r$-th correction and the time $t_{p}$ by $\mathbf{f}^{n,p,r}$. The notation $\mathbf{F}$ is the collection of all the approximations for the sub-steps i.e. the vector $\mathbf{F}=(\mathbf{f}^{n,1},\cdots,\mathbf{f}^{n,M})^{T}$. The notation $\mathbf{F}^{(r)}$ represents the vector $\mathbf{F}^{(r)}=(\mathbf{f}^{n,1,r},\cdots,\mathbf{f}^{n,M,r})^{T}$ i.e. the vector of all the approximations for the sub-steps at the $r$-th correction. Now we use a defect correction method and proceed within the time interval $[t_{n},t_{n+1}]$ as follows:

1. 1.

   For $r=0$, set $\mathbf{F}^{(0)}=(\mathbf{f}^{n},\cdots,\mathbf{f}^{n})^{T}$.

2. 2.

   For each correction $r\geq 0$, define $\mathbf{F}^{(r+1)}$ by

   $$L_{1}(\mathbf{F}^{(r+1)})=L_{1}(\mathbf{F}^{(r)})-L_{2}(\mathbf{F}^{(r)})$$

   (7)

3. 3.

   Set $\mathbf{F}^{n+1}=\mathbf{F}^{(M)}$.

Formulation (7) relies on a Lemma which has been proven in [18].

In the following, we introduce the differential operators $L_{1}$ and $L_{2}$. We first define the high order differential operator $L_{2}$. By integrating (5) on $[0,t]$, we have

$$\mathbf{f}(\mathbf{x},t)-\mathbf{f}(\mathbf{x},0)+\frac{1}{\Delta x}\int\limits_{0}^{t}\Lambda_{x}\delta^{x}\mathbf{f}(\mathbf{x},s)\;ds+\frac{1}{\Delta y}\int\limits_{0}^{t}\Lambda_{y}\delta^{y}\mathbf{f}(\mathbf{x},s)\;ds=\frac{1}{\varepsilon}\int\limits_{0}^{t}\Big{(}\mathbb{M}\big{(}\mathbb{P}\mathbf{f}(\mathbf{x},s)\big{)}-\mathbf{f}(\mathbf{x},s)\Big{)}\;ds$$

(8)

Hence, we get the following approximation for (4)

$$\mathbf{f}_{l}^{n,q}-\mathbf{f}_{l}^{n,0}+\frac{\Delta t}{\Delta x}\big{(}\sum_{k=0}^{M}w_{qk}\Lambda_{x}\delta_{l}^{x}\mathbf{f}^{n,k}\big{)}+\frac{\Delta t}{\Delta y}\big{(}\sum_{k=0}^{M}w_{qk}\Lambda_{y}\delta_{l}^{y}\mathbf{f}^{n,k}\big{)}-\frac{\Delta t}{\varepsilon}\sum_{k=0}^{M}w_{qk}\big{(}\mathbb{M}(\mathbb{P}\mathbf{f}_{l}^{n,k})-\mathbf{f}_{l}^{n,k}\big{)}=0$$

(9)

for $q=1,\cdots,M$. For any $l$, define $[L_{2}(\mathbf{F}^{(r)})]_{l}$ as

	$\displaystyle[L_{2}(\mathbf{F}^{(r)})]_{l}$	$\displaystyle=\mathbf{F}_{l}^{(r)}-\mathbf{F}_{l}^{(0)}+\frac{\Delta t}{\Delta x}\Lambda_{x}W\delta_{l}^{x}\mathbf{F}^{(r)}+\frac{\Delta t}{\Delta x}\Lambda_{x}\mathbf{w}_{0}\otimes\delta_{l}^{x}\mathbf{f}^{n,0}+\frac{\Delta t}{\Delta y}\Lambda_{y}W\delta_{l}^{y}\mathbf{F}^{(r)}+\frac{\Delta t}{\Delta y}\Lambda_{y}\mathbf{w}_{0}\otimes\delta_{l}^{y}\mathbf{f}^{n,0}$
		$\displaystyle-\frac{\Delta t}{\varepsilon}W\big{(}\mathbb{M}(\mathbb{P}\mathbf{F}_{l}^{(r)})-\mathbf{F}_{l}^{(r)}\big{)}-\frac{\Delta t}{\varepsilon}\mathbf{w}_{0}\otimes\big{(}\mathbb{M}(\mathbb{P}\mathbf{f}_{l}^{n,0})-\mathbf{f}_{l}^{n,0}\big{)}$

where

$$W=\begin{pmatrix}w_{11}&\cdots&w_{1M}\\ \vdots&\vdots&\vdots\\ w_{M1}&\cdots&w_{MM}\\ \end{pmatrix},\quad\mathbf{w}_{0}=\begin{pmatrix}w_{10}\\ \vdots\\ w_{M0}\end{pmatrix}$$

and

$$\mathbb{M}(\mathbb{P}\mathbf{F}_{l}^{(r)})=\big{(}\mathbb{M}(\mathbb{P}\mathbf{f}_{l}^{n,1,r}),\cdots,\mathbb{M}(\mathbb{P}\mathbf{f}_{l}^{n,M,r})\big{)}^{T}$$

The resulting scheme derived by $L_{2}$ operator is implicit, and it is very difficult to solve. Now we describe the low order differential operator $L_{1}$. We use the forward Euler method on each sub-time step

$$\mathbf{f}_{l}^{n,q}-\mathbf{f}_{l}^{n,0}+\beta_{q}\frac{\Delta t}{\Delta x}\Lambda_{x}\delta_{l}^{x}\mathbf{f}^{n,0}+\beta_{q}\frac{\Delta t}{\Delta y}\Lambda_{y}\delta_{l}^{y}\mathbf{f}^{n,0}-\frac{\Delta t}{\varepsilon}\sum_{k=0}^{M}w_{qk}\big{(}\mathbb{M}(\mathbb{P}\mathbf{f}_{l}^{n,k})-\mathbf{f}_{l}^{n,k}\big{)}=0$$

(10)

for $q=1,\cdots,M$. The low order differential operator $[L_{1}(\mathbf{F}^{(r)})]_{l}$ reads

	$\displaystyle[L_{1}(\mathbf{F}^{(r)})]_{l}$	$\displaystyle=\mathbf{F}_{l}^{(r)}-\mathbf{F}_{l}^{(0)}+\frac{\Delta t}{\Delta x}B\Lambda_{x}\delta_{l}^{x}\mathbf{F}^{(0)}+\frac{\Delta t}{\Delta y}B\Lambda_{y}\delta_{l}^{y}\mathbf{F}^{(0)}$
		$\displaystyle-\frac{\Delta t}{\varepsilon}W\big{(}\mathbb{M}(\mathbb{P}\mathbf{F}^{(r)}_{l})-\mathbf{F}_{l}^{(r)}\big{)}-\frac{\Delta t}{\varepsilon}\mathbf{w}_{0}\otimes\big{(}\mathbb{M}(\mathbb{P}\mathbf{f}_{l}^{n,0})-\mathbf{f}_{l}^{n,0}\big{)}$

where $B=\text{diag}(\beta_{1},\cdots,\beta_{M})$. This is still an implicit approximation in time, but the convection part is now explicit. We also note we have kept the same form of the source term approximation in both cases, for reasons that will be explained bellow.

Now we can write (7) as a multi-step method where each step writes as

$$\begin{split}\mathbf{F}_{l}^{(r+1)}-\mathbf{F}_{l}^{(0)}&+\frac{\Delta t}{\Delta x}B\Lambda_{x}\delta_{l}^{x}\mathbf{F}^{(0)}+\frac{\Delta t}{\Delta y}B\Lambda_{y}\delta_{l}^{y}\mathbf{F}^{(0)}-\frac{\Delta t}{\varepsilon}W\big{(}\mathbb{M}(\mathbb{P}\mathbf{F}^{(r+1)}_{l})-\mathbf{F}_{l}^{(r+1)}\big{)}\\ &-\frac{\Delta t}{\varepsilon}\mathbf{w}_{0}\otimes\big{(}\mathbb{M}(\mathbb{P}\mathbf{f}_{l}^{n,0})-\mathbf{f}_{l}^{n,0}\big{)}=\frac{\Delta t}{\Delta x}\Lambda_{x}W(\delta_{l}^{x}\mathbf{F}^{(0)}-\delta_{l}^{x}\mathbf{F}^{(r)})+\frac{\Delta t}{\Delta y}\Lambda_{y}W(\delta_{l}^{y}\mathbf{F}^{(0)}-\delta_{l}^{y}\mathbf{F}^{(r)})\end{split}$$

(11)

for any $l$. By applying $\mathbb{P}$ to this equation, we will obtain the following equation for calculating $\mathbb{P}\mathbf{F}_{l}^{(r+1)}$

$$\mathbb{P}\mathbf{F}_{l}^{(r+1)}=\mathbb{P}\mathbf{F}_{l}^{(0)}-\frac{\Delta t}{\Delta x}\mathbb{P}\Lambda_{x}W\delta_{l}^{x}\mathbf{F}^{(r)}-\frac{\Delta t}{\Delta x}\mathbf{w}_{0}\otimes\mathbb{P}\Lambda_{x}\delta_{l}^{x}\mathbf{f}^{n,0}-\frac{\Delta t}{\Delta y}\mathbb{P}\Lambda_{y}W\delta_{l}^{y}\mathbf{F}^{(r)}-\frac{\Delta t}{\Delta y}\mathbf{w}_{0}\otimes\mathbb{P}\Lambda_{y}\delta_{l}^{y}\mathbf{f}^{n,0}$$

(12)

and we substitute $\mathbb{P}\mathbf{F}_{l}^{(r+1)}$ into the Maxwellian in the (11). Alternatively, one can rewrite (11) as follows

$$\begin{split}(\text{Id}_{M\times M}+\frac{\Delta t}{\varepsilon}W)\mathbf{F}_{l}^{(r+1)}&=\frac{\Delta t}{\varepsilon}W\mathbb{M}(\mathbb{P}\mathbf{F}_{l}^{(r+1)})+\mathbf{F}_{l}^{(0)}-\frac{\Delta t}{\Delta x}\Lambda_{x}W\delta_{l}^{x}\mathbf{F}^{(r)}-\frac{\Delta t}{\Delta x}\mathbf{w}_{0}\otimes\Lambda_{x}\delta_{l}^{x}\mathbf{f}^{n,0}\\ &-\frac{\Delta t}{\Delta y}\Lambda_{y}W\delta_{l}^{y}\mathbf{F}^{(r)}-\frac{\Delta t}{\Delta y}\mathbf{w}_{0}\otimes\Lambda_{y}\delta_{l}^{y}\mathbf{f}^{n,0}+\frac{\Delta t}{\varepsilon}\mathbf{w}_{0}\otimes\big{(}\mathbb{M}(\mathbb{P}\mathbf{f}_{l}^{n,0})-\mathbf{f}_{l}^{n,0}\big{)}\end{split}$$

(13)

If $\text{Id}_{M\times M}+{\Delta t}W/{\varepsilon}$ is invertible, the defect correction computes the solution at time $t_{n+1}$ using $M$ steps of the form

	$$\begin{split}\mathbf{F}_{l}^{(r+1)}&=(\text{Id}_{M\times M}+\frac{\Delta t}{\varepsilon}W)^{-1}\bigg{(}\frac{\Delta t}{\varepsilon}W\mathbb{M}(\mathbb{P}\mathbf{F}_{l}^{(r+1)})+\mathbf{F}_{l}^{(0)}-\frac{\Delta t}{\Delta x}\Lambda_{x}W\delta_{l}^{x}\mathbf{F}^{(r)}-\frac{\Delta t}{\Delta x}\mathbf{w}_{0}\otimes\Lambda_{x}\delta_{l}^{x}\mathbf{f}^{n,0}\\ &-\frac{\Delta t}{\Delta y}\Lambda_{y}W\delta_{l}^{y}\mathbf{F}^{(r)}-\frac{\Delta t}{\Delta y}\mathbf{w}_{0}\otimes\Lambda_{y}\delta_{l}^{y}\mathbf{f}^{n,0}+\frac{\Delta t}{\varepsilon}\mathbf{w}_{0}\otimes\big{(}\mathbb{M}(\mathbb{P}\mathbf{f}_{l}^{n,0})-\mathbf{f}_{l}^{n,0}\big{)}\bigg{)}.\end{split}$$		(14a)
For computing $\mathbb{P}\mathbf{F}_{l}^{(r+1)}$, for any $l$, we rewrite (12) as (for simplicity, we drop the superscript $n$)
	$$\mathbb{P}\mathbf{f}_{l}^{q,r+1}-\mathbb{P}\mathbf{f}_{l}^{0}+\frac{\Delta t}{\Delta x}\sum_{k=0}^{M}w_{qk}\mathbb{P}\Lambda_{x}\delta_{l}^{x}\mathbf{f}^{k,r}+\frac{\Delta t}{\Delta y}\sum_{k=0}^{M}w_{qk}\mathbb{P}\Lambda_{y}\delta_{l}^{y}\mathbf{f}^{k,r}=0,$$		(14b)

for $q=1,\cdots,M$.

We write the increment $\delta_{l}\mathbf{f}^{k}$ as a sum of residuals, as follows

	$$\begin{split}\delta_{l}\mathbf{f}^{k}&=\Delta y\Lambda_{x}(\hat{\mathbf{f}}_{i+\frac{1}{2},j}^{k}-\hat{\mathbf{f}}_{i-\frac{1}{2},j}^{k})+\Delta x\Lambda_{y}(\hat{\mathbf{f}}_{i,j+\frac{1}{2}}^{k}-\hat{\mathbf{f}}_{i,j-\frac{1}{2}}^{k})\\ &=\Phi_{l,(i,j)}^{[i,i+1]\times[j,j+1],k}+\Phi_{l,(i,j)}^{[i-1,i]\times[j,j+1],k}+\Phi_{l,(i,j)}^{[i,i+1]\times[j-1,j],k}+\Phi_{l,(i,j)}^{[i-1,i]\times[j-1,j],k}\end{split}$$		(15a)
with
	$$\begin{split}\Phi_{l,(i,j)}^{[i,i+1]\times[j,j+1],k}&=\frac{1}{2}\bigg{(}\Lambda_{x}\big{(}\hat{\mathbf{f}}_{i+1/2,j}-\mathbf{f}_{ij})\Delta y+\Lambda_{y}\big{(}\hat{\mathbf{f}}_{i,j+1/2}-\mathbf{f}_{ij})\big{)}\Delta y\bigg{)},\\ \Phi_{l,(i,j)}^{[i-1,i]\times[j,j+1],k}&=\frac{1}{2}\bigg{(}\Lambda_{x}\big{(}\hat{\mathbf{f}}_{i+1/2,j}-\mathbf{f}_{ij}\big{)}\Delta y+\Lambda_{y}(\mathbf{f}_{ij}-\hat{\mathbf{f}}_{i,j-1/2}\big{)}\Delta x\bigg{)},\\ \Phi_{l,(i,j)}^{[i-1,i]\times[j-1,j],k}&=\frac{1}{2}\bigg{(}\Lambda_{x}\big{(}\mathbf{f}_{ij}-\hat{\mathbf{f}}_{i-1/2,j}\big{)}\Delta y+\Lambda_{y}\big{(}\mathbf{f}_{ij}-\hat{\mathbf{f}}_{ij-1/2}\big{)}\Delta x\bigg{)},\\ \Phi_{l,(i,j)}^{[i-1,i]\times[j-1,j],k}&=\frac{1}{2}\bigg{(}\Lambda_{x}\big{(}\mathbf{f}_{ij}-\hat{\mathbf{f}}_{i-1/2,j}\big{)}\Delta y+\Lambda_{y}\big{(}\hat{\mathbf{f}}_{i,j+1/2}-\mathbf{f}_{ij}\big{)}\Delta x\bigg{)}.\\ \end{split}$$		(15b)

We see that

$$\begin{split}\Phi_{l,(i,j)}^{[i,i+1]\times[j,j+1],k}&+\Phi_{l,(i+1,j)}^{[i,i+1]\times[j,j+1],k}+\Phi_{l,(i+1,j+1)}^{[i,i+1]\times[j,j+1],k}+\Phi_{l,(i,j+1)}^{[i,i+1]\times[j,j+1],k}\\ &=\frac{\Delta x}{2}\Lambda_{x}\big{(}\mathbf{f}_{i+1,j}+\mathbf{f}_{i+1,j+1}\big{)}-\frac{\Delta x}{2}\Lambda_{x}\big{(}\mathbf{f}_{ij}+\mathbf{f}_{i,j+1}\big{)}\\ &+\frac{\Delta y}{2}\Lambda_{y}\big{(}\mathbf{f}_{i+1,j}+\mathbf{f}_{i+1,j+1}\big{)}-\frac{\Delta y}{2}\Lambda_{y}\big{(}\mathbf{f}_{ij}+\mathbf{f}_{i,j+1}\big{)},\end{split}$$

(16)

i.e. the sum of the residuals assigned to each of the four vertices of the quad $[x_{i},x_{i+1}]\times[y_{j},y_{j+1}]$ is equal to an approximation of the flux across the boundary of this quad, see Fig. 1. This guaranties local conservation, see [19]. The reason of writing the space update as as sum of residual will become clear in section 3.2.

Now we explain the residuals for an example where the velocities are orthogonal. This case is suggested in [7]. In order to construct the system (2a) one must find $\mathbb{P}$, $\mathbb{M}$, and $\Lambda_{j}~{}(j=1,\cdots,d)$ such that the consistency relations (3) are satisfied. We take $L=N\times K$, $\mathbb{P}=(\text{Id}_{K\times K},\cdots,\text{Id}_{K\times K})$ with $N$ blocks $\text{Id}_{K\times K}$, the identity matrix in $\mathbb{R}^{K}$. Each matrix $\Lambda_{j}~{}(j=1,\cdots,d)$ is constituted of $N$ diagonal blocks of size $K\times K$, i.e.

$$\Lambda_{j}=\text{diag}(C_{1}^{(j)},\cdots,C_{N}^{(j)}),\quad C_{i}^{(j)}=\lambda_{i}^{(j)}\text{Id}_{K\times K},\quad\lambda_{i}^{(j)}\in\mathbb{R}.$$

where $j=1,\cdots,N$. Then (2a) can be rewritten as

$$\frac{\partial\mathbf{f}_{j}}{\partial t}+\sum\limits_{i=1}^{d}\lambda_{j}^{(i)}\frac{\partial\mathbf{f}_{j}}{\partial x_{i}}=\frac{\mathbb{M}_{j}(\mathbf{u}^{\varepsilon})-\mathbf{f}_{j}}{\varepsilon},\,j=1,\cdots,N.$$

(17)

in which $\mathbf{u}^{\varepsilon}=\sum\limits_{j=1}^{N}\mathbf{f}_{n}$. This example works with any number of blocks provided that $N\geq d+1$. We take the velocities such that

$$\Lambda_{j}=\text{diag}(\lambda_{ij}\text{Id}_{K\times K})_{1\leq i\leq N},$$

$$\sum_{i=1}^{N}\lambda_{ij}=0,~{}j=1,\cdots,d,$$

and

$$\sum_{i=1}^{N}\lambda_{il}\lambda_{ij}=0,~{}j,l=1,\cdots d,~{}j\neq l.$$

Therefore, the Maxwellian function is given by

$$\mathbb{M}_{i}(\mathbf{u})=\frac{\mathbf{u}}{N}+\sum_{j=1}^{d}\frac{\mathbf{A}_{j}(\mathbf{u})}{a_{j}^{2}}\lambda_{ij},~{}i=1,\cdots,N.$$

where $a_{j}^{2}=\sum_{i=1}^{N}\lambda_{ij}^{2}$. We fix $J,N^{\prime}\geq 1$, $\lambda>0$ and set

$$\begin{cases}\lambda_{j}=\frac{n}{J}\lambda&1\leq j\leq J,~{}J\geq 1,\\ \Lambda_{x}=\text{diag}\Big{(}\lambda_{n}\text{diag}\big{(}\cos(\frac{i\pi}{2N^{{}^{\prime}}})\text{ I}\big{)}_{1\leq i\leq 4N^{{}^{\prime}}}\Big{)}&N^{{}^{\prime}}\geq 1,\\ \Lambda_{y}=\text{diag}\Big{(}\lambda_{n}\text{diag}\big{(}\sin(\frac{i\pi}{2N^{{}^{\prime}}})\text{ I}\big{)}_{1\leq i\leq 4N^{{}^{\prime}}}\Big{)}.\end{cases}$$

Here we have $N=4N^{\prime}J$. We define the Maxwellian functions by

$$\mathbb{M}_{j}(\mathbf{u})=\frac{1}{N}\Big{(}\mathbf{u}+\frac{12\;i\;J}{\lambda(J+1)(2J+1)}(\mathbf{A}_{1}\cos\frac{j\pi}{2N^{{}^{\prime}}}+\mathbf{A}_{2}\sin\frac{j\pi}{2N^{{}^{\prime}}})\Big{)},~{}1\leq j\leq 4N^{{}^{\prime}},1\leq i\leq J.$$

Now, we consider the special case where $J=N^{\prime}=1$ and $N=4$. Set

$$\begin{cases}\Lambda_{x}=\text{diag}\Big{(}\lambda\text{ diag}\big{(}\cos(\frac{i\pi}{2})\text{ I}\big{)}_{1\leq i\leq 4}\Big{)},\\ \Lambda_{y}=\text{diag}\Big{(}\lambda\text{ diag}\big{(}\sin(\frac{i\pi}{2})\text{ I}\big{)}_{1\leq i\leq 4}\Big{)}.\end{cases}$$

Therefore the Maxwellian functions are

$$\mathbb{M}_{i}(\mathbf{u})=\frac{1}{4}\Big{(}\mathbf{u}+\frac{2}{\lambda}(\mathbf{A}_{1}\cos\frac{i\pi}{2}+\mathbf{A}_{2}\sin\frac{i\pi}{2})\Big{)}.$$

for $1\leq i\leq 4$.

The natural question is: how many iterations should we do in the method (14) to recover the time and space accuracy?

Again, we will consider the two dimensional version of (1a). Let us consider $\varphi:\mathbb{R}^{2}\rightarrow\mathbb{R}^{K}$, which is continuously differentiable on $\mathbb{R}^{2}$ with values in $\mathbb{R}^{K}$ and has a compact support. We will consider the discrete version of $L^{2}$ and $H^{1}$ norms of $\varphi$ as follows

$$\|\varphi\|_{L^{2}}^{2}=\sum\limits_{i,j\in\mathbb{Z}}\Delta x\Delta y\|\varphi_{i,j}\|^{2},\quad\|\varphi\|_{H^{1}}^{2}=\|\varphi\|_{L^{2}}^{2}+\sum\limits_{i,j\in\mathbb{Z}}\Delta x\Delta y\Big{(}\Big{\|}\dfrac{\varphi_{i+1,j}-\varphi_{ij}}{\Delta x}\Big{\|}^{2}+\Big{\|}\dfrac{\varphi_{i,j+1}-\varphi_{i,j}}{\Delta y}\Big{\|}^{2}\Big{)}.$$

In the following, we will set:

$$D_{i+1/2,j}\varphi=\dfrac{\varphi_{i+1,j}-\varphi_{ij}}{\Delta x},D_{i,j+1/2}\varphi=\dfrac{\varphi_{i,j+1}-\varphi_{i,j}}{\Delta y}\text{ and }D\varphi_{ij}=(D_{i+1/2,j}\varphi,D_{i,j+1/2}\varphi).$$

We have a Poincaré inegality, on any compact.

One can consider the discrete equivalent of $L^{2}_{loc}$ and $H^{-1}_{loc}$ in an interval $I=[a,b]\times[c,d]$ as follows

$$\|\mathbf{F}\|_{2,I}=\sup\limits_{\varphi\in C_{0}^{1}(I)^{K}}\frac{\sum\limits_{i,j\in\mathbb{Z}}\Delta x\Delta y\langle\varphi_{i,j},\mathbf{f}_{i,j}\rangle}{\|\varphi\|_{L^{2}}},\quad\|\mathbf{F}\|_{-1,I}=\sup\limits_{\varphi\in C_{0}^{1}(I)^{K}}\frac{\sum\limits_{i,j\in\mathbb{Z}}\Delta x\Delta y\langle\varphi_{i,j},\mathbf{f}_{i,j}\rangle}{\|\varphi\|_{H^{1}}}.$$

We have a first result on the $H^{-1}_{loc}$ estimate of $L_{2}(\mathbf{F})-L_{1}(\mathbf{F})$.

Before we proceed to proposition 2.3, we need a further result on the behaviour of $L_{1}$.