The Immersed Weak Galerkin and Continuous Galerkin Finite Element Method for Elliptic Interface Problem

1 Introduction

In this paper, we assume that the domain $\Omega$ is bounded in $\mathbb{R}^{2}$ with boundary $\partial\Omega$ and the domain is separated by an interface $\Gamma$ into two sub-domains $\Omega_{1}$ and $\Omega_{2}$ . We consider the following elliptic interface problem in domain $\Omega$ .

	$\displaystyle-\nabla\cdot(A\nabla u)$	$\displaystyle=$	$\displaystyle f,\,\text{in}\,\Omega_{1}\cup\Omega_{2},$		(1.1)
	$\displaystyle u$	$\displaystyle=$	$\displaystyle g,\,\text{on}\,\partial\Omega,$		(1.2)

where $f\in PH^{k-1}(\Omega)=\{v:v_{1}=v|_{\Omega_{1}}\in H^{k}(\Omega_{1}),\,\text{and}\,v_{2}=v|_{\Omega_{2}}\in H^{k}(\Omega_{2})\}$ for the integer $k\geqslant 1$ and $A$ is a piece-wise constant function

\displaystyle A(X)=\left\{\begin{array}[]{rcl}A_{1}&\text{for}&X\in\Omega_{1},\\ A_{2}&\text{for}&X\in\Omega_{2}.\end{array}\right.

Without loss of generality, we suppose $A_{2}\geqslant A_{1}>0$ . The interface conditions on $\Gamma$ are as follows:

	$\displaystyle[u]_{\Gamma}$	$\displaystyle:=u_{1}-u_{2}=0,\,\text{on}\,\Gamma,$		(1.4)
	$\displaystyle[A\nabla u\cdot{\bf n}]$	$\displaystyle:=A_{1}\nabla u_{1}\cdot{\bf n}-A_{2}\nabla u_{2}\cdot{\bf n}=0,\,\text{on}\,\Gamma,$		(1.5)

where ${\bf n}$ is the unit outward normal vector on the interface $\Gamma$ pointing from $\Omega_{1}$ into $\Omega_{2}$ .

When $k\geqslant 2$ , we need to imposed the Laplacian extended jump conditions:

\displaystyle\left[A\frac{\partial^{j-2}\triangle u}{\partial{\bf n}^{j-2}}\right]_{\Gamma}=0,\,j=2,3,\cdots,k.

(1.6)

Refer to [1, Theorem 2.1], we get the regularity theorem for the problem (1.1)-(1.6).

2 The Numerical Scheme

In the section, the weak Galerkin finite element method and the finite element method are applied to the elliptic interface problem on unfitted meshes. First, we give the definitions of the approximation functions space and the weak differential operators. Then the numerical scheme are proposed.

Assume the partition $\mathcal{T}_{h}$ satisfies the shape-regular conditions [4] and do not require the mesh to be aligned with the interface. Denote by $\mathcal{T}_{h}^{n}$ the set of the non-interface element. Denote by $\mathcal{T}_{h}^{I}$ the set of the interface element which intersect with interface. For $T\in\mathcal{T}_{h}$ , define the diameter of $T$ as $h_{T}$ . Set $h=\max_{h\in\mathcal{T}_{h}}h_{T}$ . Denote by $\mathcal{E}_{h}$ the set of the all edges in $\mathcal{T}_{h}^{I}$ . The set of intersecting edges of interface elements and non-interface elements is denoted as $\mathcal{E}_{h}^{I}$ . In the WG method, the discontinuous weak function $v_{h}=\{v_{0},v_{b}\}$ is used. The weak function is consistent of two parts: the interior function $v_{0}$ and the boundary function $v_{b}$ . Noted that $v_{b}$ has one unique value on the boundary $\partial T$ . For the integer $k\geqslant 1$ , we define the following approximation function spaces.

For $T\in\mathcal{T}_{h}^{n}$ , we define the following the finite element space:

\displaystyle W^{n}(T)={\rm{span}}\{\phi_{i}\}_{i=1}^{N},\phi_{i}(P_{j})=\delta_{ij},\text{where }P_{j}\text{ is the vertex of }T.

For $T\in\mathcal{T}_{h}^{I}$ , the definition of WG space is as follows:

\displaystyle W^{I}(T)=\{v_{h}=\{v_{0},v_{b}\}|v_{0}\in\mathcal{V}_{k}(T),v_{b}\in P_{k-1}(e),e\in\partial T\},

where the space $\mathcal{V}_{k}(T)$ is the immersed interface function space satisfying the interface conditions (1.4) - (1.6), as details in [2].

The global weak Galerkin finite element spaces are as follows.

	$\displaystyle V_{h}$	$\displaystyle=$	$\displaystyle\{v_{h}\|v_{h}\in W^{n}(T),T\in\mathcal{T}_{h}^{n},v_{h}\in W^{I}(T),T\in\mathcal{T}_{h}^{I},v_{b}=Q_{b}(v_{h}\|_{e}),\,\text{on}\,e\in\mathcal{E}_{h}^{I}\},$
	$\displaystyle V_{h}^{0}$	$\displaystyle=$	$\displaystyle\{v_{h}\|v_{h}\in V_{h},v_{b}\|_{e}=0,e\in\partial\Omega\}.$

Define $\Pi_{h}u$ as the interpolate function of $u$ in $W^{n}(T)$ , $T\in\mathcal{T}_{h}^{n}$ . Denote by $Q_{0}$ the $L^{2}$ projection from $L^{2}(T)$ into $\mathcal{V}_{k}(T)$ in $T\in\mathcal{T}_{h}^{I}$ . Define $Q_{b}$ as the $L^{2}$ projection from $L^{2}(e)$ into $P_{k-1}(e)$ on $e\in\mathcal{E}_{h}$ . Set $Q_{h}u=\{Q_{0}u,Q_{b}u\}$ and $\tilde{Q}_{h}u=\{Q_{0}u,Q_{\delta}u\}$ where

\displaystyle Q_{\delta}u=\left\{\begin{array}[]{rcl}Q_{b}u&\text{on}&e\in\mathcal{E}_{h}\setminus\mathcal{E}_{h}^{I},\\ Q_{b}(\Pi_{h}u)&\text{on}&e\in\mathcal{E}_{h}^{I}.\\ \end{array}\right.

.

Definition 2.1.

For $T\in\mathcal{T}_{h}^{I}$ and $\bm{q}\in\nabla\mathcal{V}_{k}(T)$ , the weak gradient $\nabla_{w}v$ satisfies

\displaystyle(\nabla_{w}v,\bm{q})_{T}=(\nabla v_{0},\bm{q})_{T}-\langle Q_{b}v_{0}-v_{b},\bm{q}\cdot{\bf n}\rangle_{\partial T},\,\forall\,\bm{q}\in\nabla\mathcal{V}_{k}(T).

(2.2)

Based on the above definitions, the following numerical scheme is proposed.

Algorithm 1 The Numerical Scheme

Find $u_{h}\in V_{h}$ and $u_{b}=Q_{b}g$ on $\partial\Omega$ to satisfy

\displaystyle a^{n}(u_{h},v_{h})+a_{s}^{I}(u_{h},v_{h})=(f,v_{h}),\,\forall\,v_{h}\in V_{h}^{0},

(2.3)

where

$\displaystyle a^{n}(u_{h},v_{h})$	$\displaystyle=$	$\displaystyle\sum_{T\in\mathcal{T}_{h}^{n}}(A\nabla u_{h},\nabla v_{h})_{T},$
$\displaystyle a_{s}^{I}(u_{h},v_{h})$	$\displaystyle=$	$\displaystyle\sum_{T\in\mathcal{T}_{h}^{I}}(A\nabla_{w}u_{h},\nabla_{w}v_{h})_{T}+h_{T}^{-1}\langle Q_{b}u_{0}-u_{b},Q_{b}v_{0}-v_{b}\rangle_{\partial T},$
$\displaystyle(f,v_{h})$	$\displaystyle=$	$\displaystyle\sum_{T\in\mathcal{T}_{h}^{n}}(f,v)+\sum_{T\in\mathcal{T}_{h}^{I}}(f,v_{0}).$

3 Existence and Uniqueness

We define the following semi-norm:

\displaystyle\begin{split}\|v_{h}\|_{1,h}^{2}=\sum_{T\in\mathcal{T}_{h}^{n}}\|\nabla v_{h}\|_{T}^{2}+\sum_{T\in\mathcal{T}_{h}^{I}}\left(\|\nabla_{w}v_{h}\|_{T}^{2}+h_{T}^{-1}\|Q_{b}v_{0}-v_{b}\|_{\partial T}^{2}\right).\end{split}

(3.1)

Lemma 3.1.

$\|\cdot\|_{1,h}$ is a norm in $V_{h}^{0}$ .

Proof.

Let $\|v\|_{1,h}=0$ for some $v\in V_{h}^{0}$ . By the definition of $\|v\|_{1,h}$ , we have

\displaystyle\nabla v=0,\,T\in\mathcal{T}_{h}^{n};\quad\nabla_{w}v=0,\quad Q_{b}v_{0}=v_{b},\,e\in\partial T,\,T\in\mathcal{T}_{h}^{I}.

Therefore, $v$ is a constant on $T\in\mathcal{T}_{h}^{n}$ . Since $v$ is zero on $\partial\Omega$ , $v$ is zero on $T\in\mathcal{T}_{h}^{n}$ .

For $T\in\mathcal{T}_{h}^{I}$ and $w\in\mathcal{V}_{k}(T)$ , we obtain

\displaystyle\begin{split}0=&(\nabla_{w}v,\nabla w)_{T}\\ =&(\nabla v_{0},\nabla w)_{T}-\langle Q_{b}v_{0}-v_{b},\nabla w\cdot{\bf n}\rangle_{\partial T}\\ =&(\nabla v_{0},\nabla w)_{T}.\end{split}

(3.2)

In Eq.(3.2), taking $w=v_{0}$ yields $\nabla v_{0}=0$ on $T\in\mathcal{T}_{h}^{I}$ . Therefore, $v_{0}$ is a constant on $T\in\mathcal{T}_{h}^{I}$ .

Combining with $Q_{b}v_{0}=v_{b}$ on $e\in\mathcal{E}_{h}$ , we get $v_{b}=Q_{b}v_{0}=v_{0}$ . By $v_{b}=Q_{b}(v|_{e})=0$ on $e\in\mathcal{E}_{h}^{I}$ , we have $v_{0}=v_{b}=0$ . Therefore $v=0$ is obtained. The proof of the lemma is completed. ∎

According to the above lemma, it’s easy to obtain that the numerical scheme (2.3) has only one solution.

4 Some Inequalities

The trace inequality, the inverse inequality and the projection inequality are essential technique tools for the analysis. We prove these inequalities holds true for the interface element.

Lemma 4.1.

For $T\in\mathcal{T}_{h}^{I}$ and $w\in\mathcal{V}_{k}(T)$ , we have

\displaystyle\|\nabla w\|_{e}^{2}\leqslant Ch_{T}^{-2}\|\nabla w\|_{T}^{2}.

(4.1)

Proof.

Assume $e=e_{1}\cup e_{2}$ , $e_{1}\subset T_{1}$ and $e_{2}\subset T_{2}$ . Then we have

\displaystyle\begin{split}\|\nabla w\|_{e}^{2}\leqslant&C\|\nabla w\|_{e_{1}}^{2}+\|\nabla w\|_{e_{2}}^{2}\\ \leqslant&Ch_{T}^{-1}\|\nabla w\|_{T_{1}}^{2}+Ch_{T}\|\nabla(\nabla w)\|_{T_{1}}^{2}+Ch_{T}^{-1}\|\nabla w\|_{T_{2}}+Ch_{T}^{-1}\|\nabla(\nabla w)\|_{T_{2}}^{2}.\end{split}

(4.2)

For $\|\nabla(\nabla w)\|_{T_{1}}^{2}$ , we have

\displaystyle\begin{split}\|\nabla(\nabla w)\|_{T_{1}}^{2}\leqslant\int_{T_{1}}\left(\Delta w+2\frac{\partial^{2}w}{\partial x\partial y}\right)^{2}dT_{1}\leqslant\int_{T_{1}}2(\Delta w)^{2}dT_{1}+\int_{T_{1}}4(\frac{\partial^{2}w}{\partial x\partial y})^{2}dT_{1}.\end{split}

(4.3)

Next we estimate the every term on the right side of the above inequality.

\displaystyle\begin{split}\int_{T_{1}}(\Delta w)^{2}dT_{1}=&\int_{\hat{T_{1}}}\left(\Delta\hat{w}(\xi,\eta)\right)^{2}|\hat{D}P_{\Gamma}|d\hat{T_{1}}\\ \leqslant&C\int_{\hat{T_{1}}}\left(\hat{w}_{\eta\eta}^{2}+J_{0}^{2}\hat{w}_{\xi\xi}^{2}+J_{1}^{2}\hat{w}_{\eta}^{2}++J_{2}^{2}\hat{w}_{\xi}^{2}\right)d\hat{T_{1}}\\ \leqslant&C\left(\int_{\hat{T_{1}}}\hat{w}_{\eta\eta}^{2}+\hat{w}_{\xi\xi}^{2}d\hat{T_{1}}+\int_{\hat{T_{1}}}\hat{w}_{\eta}^{2}+\hat{w}_{\xi}^{2}d\hat{T_{1}}\right)\\ \leqslant&C\|\hat{\nabla}(\hat{\nabla}\hat{w})\|_{\hat{T_{1}}}^{2}+C\|\hat{\nabla}\hat{w}\|_{\hat{T_{1}}}^{2}\\ \leqslant&Ch_{\hat{T}}^{-2}\|\hat{\nabla}\hat{w}\|_{\hat{T_{1}}}^{2}.\end{split}

(4.4)

According to the chain rule, we have

\displaystyle\begin{split}\frac{\partial^{2}w}{\partial x\partial y}=&\left(\frac{\partial^{2}\hat{w}}{\partial\eta^{2}}\frac{\partial\eta}{\partial y}+\frac{\partial^{2}\hat{w}}{\partial\eta\partial\xi}\frac{\partial\xi}{\partial y}\right)\frac{\partial\eta}{\partial x}+\frac{\partial\hat{w}}{\partial\eta}\frac{\partial^{2}\eta}{\partial x\partial y}\\ &+\left(\frac{\partial^{2}\hat{w}}{\partial\eta\partial\xi}\frac{\partial\eta}{\partial y}+\frac{\partial^{2}\hat{w}}{\partial\xi^{2}}\frac{\partial\xi}{\partial y}\right)\frac{\partial\xi}{\partial x}+\frac{\partial\hat{w}}{\partial\xi}\frac{\partial^{2}\xi}{\partial x\partial y}\\ =&\frac{\partial^{2}\hat{w}}{\partial\eta^{2}}\left(\frac{\partial\eta}{\partial x}\frac{\partial\eta}{\partial y}\right)+\frac{\partial^{2}\hat{w}}{\partial\eta\partial\xi}\left(\frac{\partial\eta}{\partial x}\frac{\partial\xi}{\partial y}\right)+\frac{\partial\hat{w}}{\partial\eta}\frac{\partial^{2}\eta}{\partial x\partial y}\\ &+\frac{\partial^{2}\hat{w}}{\partial\eta\partial\xi}\left(\frac{\partial\eta}{\partial y}\frac{\partial\xi}{\partial x}\right)+\frac{\partial^{2}\hat{w}}{\partial\xi^{2}}\left(\frac{\partial\xi}{\partial y}\frac{\partial\xi}{\partial x}\right)+\frac{\partial\hat{w}}{\partial\xi}\frac{\partial^{2}\xi}{\partial x\partial y}.\end{split}

We have

\displaystyle\begin{split}&\int_{T_{1}}\left(\frac{\partial^{2}w}{\partial x\partial y}\right)^{2}dT_{1}\\ \leqslant&\int_{\hat{T_{1}}}C_{1}\left(\frac{\partial^{2}\hat{w}}{\partial\eta^{2}}\right)^{2}+C_{2}\left(\frac{\partial^{2}\hat{w}}{\partial\xi^{2}}\right)^{2}+C_{3}\left(\frac{\partial^{2}\hat{w}}{\partial\eta\partial\xi}\right)^{2}+C_{4}\left(\frac{\partial\hat{w}}{\partial\eta}\right)^{2}+C_{5}\left(\frac{\partial\hat{w}}{\partial\xi}\right)^{2}\\ \leqslant&C\|\hat{\nabla}(\hat{\nabla}\hat{w})\|_{\hat{T_{1}}}^{2}+C\|\hat{\nabla}\hat{w}\|_{\hat{T_{1}}}^{2}\\ \leqslant&Ch_{\hat{T}}^{-2}\|\hat{\nabla}\hat{w}\|_{\hat{T_{1}}}^{2}.\end{split}

(4.5)

Thus we have

\displaystyle\begin{split}\|\nabla(\nabla w)\|_{T_{1}}^{2}\leqslant&Ch_{\hat{T}}^{-2}\|\hat{\nabla}\hat{w}\|_{\hat{T_{1}}}^{2}\\ \leqslant&Ch_{T}^{-2}\int_{T_{1}}2(x_{\xi}^{2}x_{\eta}^{2})\left(\frac{\partial w}{\partial x}\right)^{2}+2(y_{\xi}^{2}y_{\eta}^{2})\left(\frac{\partial w}{\partial y}\right)^{2}|DR_{\Gamma}|dT_{1}\\ \leqslant&Ch_{T}^{-2}\|\nabla w\|_{T_{1}}^{2}.\end{split}

Similarly, we have

\displaystyle\|\nabla(\nabla w)\|_{T_{2}}^{2}\leqslant Ch_{T}^{-2}\|\nabla w\|_{T_{2}}^{2}.

Therefore the proof of the lemma is completed. ∎

5 Error Equation

In this section, we present the error equation for the velocity function $u$ . We use $u_{h}$ to represent the numerical solution obtained from the numerical scheme. Denote by $P_{h}u$ and $I_{h}u$ the projection operators satisfying

\displaystyle P_{h}u=\left\{\begin{array}[]{cc}u&T\in\mathcal{T}_{h}^{n},\\ Q_{h}u&T\in\mathcal{T}_{h}^{I}.\\ \end{array}\right.

and

\displaystyle I_{h}u=\left\{\begin{array}[]{cc}\Pi_{h}u&T\in\mathcal{T}_{h}^{n},\\ \tilde{Q}_{h}u&T\in\mathcal{T}_{h}^{I}.\\ \end{array}\right.

The error associated with $u$ is defined as follows:

e_{h}=P_{h}u-u_{h},\,\varepsilon_{h}=I_{h}u-u_{h}

Denote by $\mathcal{Q}_{h}$ the $L^{2}$ projection operator from $[L^{2}(T)]^{2}$ into $\nabla\mathcal{V}_{k}(T)$ in $T\in\mathcal{T}_{h}^{I}$ .

Lemma 5.1.

For $u\in H^{1}(\Omega)$ and $T\in\mathcal{T}_{h}^{I}$ , we have the following properties of the discrete weak gradient operator:

\displaystyle(\nabla_{w}Q_{h}u,\bm{q})_{T}=(\mathcal{Q}_{h}\nabla u,\bm{q})_{T}+(\nabla(Q_{0}u-u),\bm{q})_{T}-\langle Q_{b}(Q_{0}u)-Q_{b}u,\bm{q}\cdot{\bf n}\rangle_{\partial T},\,\forall\bm{q}\in\nabla\mathcal{V}_{k}(T).

(5.3)

Proof.

For $T\in\mathcal{T}_{h}^{I}$ and $\bm{q}\in\nabla P_{k}(T)$ , by Eq.(2.2), we have

\displaystyle\begin{split}(\nabla_{w}(Q_{h}u),\bm{q})_{T}=&(\nabla(Q_{0}u),\bm{q})_{T}-\langle Q_{b}(Q_{0}u)-Q_{b}u,\bm{q}\cdot{\bf n}\rangle_{\partial T}\\ =&(\nabla u,\bm{q})_{T}+(\nabla(Q_{0}u-u),\bm{q})_{T}-\langle Q_{b}(Q_{0}u)-Q_{b}u,\bm{q}\cdot{\bf n}\rangle_{\partial T}\\ =&(\mathcal{Q}_{h}\nabla u,\bm{q})_{T}+(\nabla(Q_{0}u-u),\bm{q})_{T}-\langle Q_{b}(Q_{0}u)-Q_{b}u,\bm{q}\cdot{\bf n}\rangle_{\partial T}.\end{split}

The proof of the above lemma is completed. ∎

Lemma 5.2.

For $u_{i}\in H^{1}(\Omega_{i})$ with $i=1,2$ satisfying Eqs.(1.1)-(1.6), then we have the following equation:

\displaystyle\sum_{T\in\mathcal{T}_{h}^{n}}(A\nabla u,\nabla v)+\sum_{T\in\mathcal{T}_{h}^{I}}(A\nabla_{w}Q_{h}u,v)_{T}=\sum_{T\in\mathcal{T}_{h}^{n}}(f,v)_{T}+\sum_{T\in\mathcal{T}_{h}^{I}}(f,v_{0})_{T}-\phi_{u}(v),\,\forall v\in V_{h}^{0},

(5.4)

where

$\displaystyle\phi_{u}(v)$	$\displaystyle=$	$\displaystyle\ell_{1}(u,v)-\ell_{2}(u,v)+\ell_{3}(u,v)-\ell_{4}(u,v)+\ell_{5}(u,v),$
$\displaystyle\ell_{1}(u,v)$	$\displaystyle=$	$\displaystyle\sum_{T\in\mathcal{T}_{h}}\langle Q_{b}v_{0}-v_{b},A\mathcal{Q}_{h}\nabla u\cdot{\bf n}-A\nabla u\cdot{\bf n}\rangle_{\partial T},$
$\displaystyle\ell_{2}(u,v)$	$\displaystyle=$	$\displaystyle\sum_{T\in\mathcal{T}_{h}^{I}}(\nabla(Q_{0}u)-\nabla u,A\nabla_{w}v)_{T},$
$\displaystyle\ell_{3}(u,v)$	$\displaystyle=$	$\displaystyle\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}u)-Q_{b}u,A\nabla_{w}v\cdot{\bf n}\rangle_{\partial T},$
$\displaystyle\ell_{4}(u,v)$	$\displaystyle=$	$\displaystyle\sum_{T\in\mathcal{T}_{h}}\langle A\nabla u\cdot{\bf n},v_{0}-Q_{b}v_{0}\rangle_{\partial T},$
$\displaystyle\ell_{5}(u,v)$	$\displaystyle=$	$\displaystyle\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla u\cdot{\bf n}_{e},v-Q_{b}v\rangle_{e},$

and ${\bf n}_{e}$ is unit normal vector on $e$ pointing from the interface element $T\in\mathcal{T}_{h}^{I}$ into the non-interface element $T\in\mathcal{T}_{h}^{n}$ .

Proof.

For $T\in\mathcal{T}_{h}^{n}$ , using integration by parts, we obtain

\displaystyle\begin{split}&\sum_{T\in\mathcal{T}_{h}^{n}}(f,v)_{T}\\ =&\sum_{T\in\mathcal{T}_{h}^{n}}-(\nabla\cdot(A\nabla u),v)_{T}\\ =&\sum_{T\in\mathcal{T}_{h}^{n}}(A\nabla u,\nabla v)_{T}-\sum_{T\in\mathcal{T}_{h}^{n}}\langle A\nabla u\cdot{\bf n},v\rangle_{\partial T}\\ =&\sum_{T\in\mathcal{T}_{h}^{n}}(A\nabla u,\nabla v)_{T}+\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla u\cdot{\bf n}_{e},v\rangle_{e}.\end{split}

(5.5)

Similarly, for $T\in\mathcal{T}_{h}^{I}$ , it follows from Eq.(5.3), Eq.(2.2) and the definition of the $L^{2}$ projection operator $\mathcal{Q}_{h}$ that

\displaystyle\begin{split}&(A\nabla_{w}Q_{h}u,\nabla_{w}v)_{T}\\ =&(A\mathcal{Q}_{h}\nabla u,\nabla_{w}v)+(\nabla(Q_{0}u-u),A\nabla_{w}v)_{T}-\langle Q_{b}(Q_{0}u)-Q_{b}u,A\nabla_{w}v\cdot{\bf n}\rangle_{\partial T}\\ =&(A\mathcal{Q}_{h}\nabla u,\nabla v_{0})_{T}-\langle Q_{b}v_{0}-v_{b},A\mathcal{Q}_{h}\nabla u\cdot{\bf n}\rangle_{\partial T}\\ &+(\nabla(Q_{0}u-u),A\nabla_{w}v)_{T}-\langle Q_{b}(Q_{0}u)-Q_{b}u,A\nabla_{w}v\cdot{\bf n}\rangle_{\partial T}\\ =&(A\nabla u,\nabla v_{0})_{T}-\langle Q_{b}v_{0}-v_{b},A\mathcal{Q}_{h}\nabla u\cdot{\bf n}\rangle_{\partial T}\\ &+(\nabla(Q_{0}u-u),A\nabla_{w}v)_{T}-\langle Q_{b}(Q_{0}u)-Q_{b}u,A\nabla_{w}v\cdot{\bf n}\rangle_{\partial T}.\end{split}

(5.6)

Then on two sides of Eq.(1.1), integrating with $v_{0}$ of $v=\{v_{0},v_{b}\}\in V_{h}^{0}$ in $T\in\mathcal{T}_{h}^{I}$ yields

\displaystyle\sum_{T\in\mathcal{T}_{h}^{I}}(-\nabla\cdot(A\nabla u),v_{0})=\sum_{T\in\mathcal{T}_{h}^{I}}(f,v_{0})_{T}.

(5.7)

Using integration by parts, we get

\displaystyle\sum_{T\in\mathcal{T}_{h}^{I}}(A\nabla u,\nabla v_{0})_{T}-\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla u\cdot{\bf n},v_{0}\rangle_{\partial T}=\sum_{T\in\mathcal{T}_{h}^{I}}(f,v_{0})_{T}.

(5.8)

We use the fact that $\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla u\cdot{\bf n},v_{b}\rangle_{\partial T}=\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla u\cdot{\bf n},v_{b}\rangle_{e}$ to get

\displaystyle\begin{split}&\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla u\cdot{\bf n},v_{0}\rangle_{\partial T}\\ =&\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla u\cdot{\bf n},v_{0}-Q_{b}v_{0}\rangle_{\partial T}+\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla u\cdot{\bf n},Q_{b}v_{0}-v_{b}\rangle_{\partial T}+\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla u\cdot{\bf n}_{e},v_{b}\rangle_{e}.\end{split}

(5.9)

Substituting Eq.(5.9) into Eq.(5.8) yields

\displaystyle\begin{split}&\sum_{T\in\mathcal{T}_{h}^{I}}(A\nabla u,\nabla v_{0})_{T}\\ =&\sum_{T\in\mathcal{T}_{h}^{I}}(f,v_{0})_{T}+\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla u\cdot{\bf n},v_{0}-Q_{b}v_{0}\rangle_{\partial T}+\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla u\cdot{\bf n},Q_{b}v_{0}-v_{b}\rangle_{\partial T}\\ &+\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla u\cdot{\bf n}_{e},v_{b}\rangle_{e}.\end{split}

(5.10)

Combining Eqs.(5.5)-(LABEL:proof_ee_10), we use the fact that $v_{b}=Q_{b}v$ on $e\in\mathcal{E}_{h}^{I}$ to get

\displaystyle\begin{split}&\sum_{T\in\mathcal{T}_{h}^{n}}(A\nabla u,\nabla v)_{T}+\sum_{T\in\mathcal{T}_{h}^{I}}(A\nabla_{w}Q_{h}u,\nabla_{w}v)_{T}\\ =&\sum_{T\in\mathcal{T}_{h}^{n}}(f,v)_{T}+\sum_{T\in\mathcal{T}_{h}^{I}}(f,v_{0})_{T}-\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}v_{0}-v_{b},A\mathcal{Q}_{h}\nabla u\cdot{\bf n}-A\nabla u\cdot{\bf n}\rangle_{\partial T}\\ &+\sum_{T\in\mathcal{T}_{h}^{I}}(\nabla(Q_{0}u)-\nabla u,A\nabla_{w}v)_{T}-\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}u)-Q_{b}u,A\nabla_{w}v\cdot{\bf n}\rangle_{\partial T}\\ &+\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla u\cdot{\bf n},v_{0}-Q_{b}v_{0}\rangle_{\partial T}-\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla u\cdot{\bf n}_{e},v-Q_{b}v\rangle_{\partial T}.\end{split}

(5.11)

The proof of the above lemma is completed. ∎

Theorem 5.1.

For $u_{i}\in H^{1}(\Omega_{i})$ with $i=1,2$ satisfying Eqs.(1.1)-(1.6), then the error $e_{h}$ satisfies the following equation:

\displaystyle a^{n}(e_{h},v)+a_{s}^{I}(e_{h},v)=s(Q_{h}u,v)-\phi_{u}(v),\,\forall v\in V_{h}^{0}.

(5.12)

Proof.

For $v\in V_{h}^{0}$ , adding $s(Q_{h}u,v)$ to both sides of Eq.(5.4) yields

\displaystyle a^{n}(u,v)+a_{s}^{I}(Q_{h}u,v)=\sum_{T\in\mathcal{T}_{h}^{n}}(f,v)_{T}+\sum_{T\in\mathcal{T}_{h}^{I}}(f,v_{0})_{T}+s(Q_{h}u,v)-\phi_{u}(v),

(5.13)

Then subtracting the numerical scheme (2.3) from the above equation leads to

\displaystyle a^{n}(e_{h},v)+a_{s}^{I}(e_{h},v)=s(Q_{h}u,v)-\phi_{u}(v).

(5.14)

The proof of the above theorem is completed. ∎

6 Error Estimate in $H^{1}$ norm

Lemma 6.1.

[3] For any $v=\{v_{0},v_{b}\}\in V_{h}$ and $T\in\mathcal{T}_{h}^{I}$ , we have

\displaystyle\|\nabla v_{0}\|\leqslant C\|v\|_{1,h}.

(6.1)

Lemma 6.2.

For $u_{i}\in H^{k+1}(\Omega_{i})$ , we have the following estimates:

$\displaystyle\|\ell_{1}(u,v)\|$	$\displaystyle\leqslant$	$\displaystyle Ch^{k}\\|u\\|_{k+1}\\|v\\|_{1,h},$	(6.2)
$\displaystyle\|\ell_{2}(u,v)\|$	$\displaystyle\leqslant$	$\displaystyle Ch^{k}\\|u\\|_{k+1}\\|v\\|_{1,h},$	(6.3)
$\displaystyle\|\ell_{3}(u,v)\|$	$\displaystyle\leqslant$	$\displaystyle Ch^{k}\\|u\\|_{k+1}\\|v\\|_{1,h},$	(6.4)
$\displaystyle\|\ell_{4}(u,v)\|$	$\displaystyle\leqslant$	$\displaystyle Ch^{k}\\|u\\|_{k+1}\\|v\\|_{1,h},$	(6.5)
$\displaystyle\|\ell_{5}(u,v)\|$	$\displaystyle\leqslant$	$\displaystyle Ch^{k}\\|u\\|_{k+1}\\|v\\|_{1,h},$	(6.6)
$\displaystyle\|s(Q_{h}u,v)\|$	$\displaystyle\leqslant$	$\displaystyle Ch^{k}\\|u\\|_{k+1}\\|v\\|_{1,h}.$	(6.7)

Proof.

For the estimate (6.2), according to the Cauchy-Schwarz inequality, the trace inequality and the projection inequality, we get

\displaystyle\begin{split}|\ell_{1}(u,v)|=&\left|\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}v_{0}-v_{b},A\mathcal{Q}_{h}\nabla u\cdot{\bf n}-A\nabla u\cdot{\bf n}\rangle_{\partial T}\right|\\ \leqslant&C\left(\sum_{T\in\mathcal{T}_{h}^{I}}h_{T}^{-1}\|Q_{b}v_{0}-v_{b}\|_{\partial T}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}h_{T}\|\mathcal{Q}_{h}\nabla u-\nabla u\|_{\partial T}\right)^{\frac{1}{2}}\\ \leqslant&C\|v\|_{1,h}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|\mathcal{Q}_{h}\nabla u-\nabla u\|_{T}^{2}+h_{T}^{2}\|\nabla(\mathcal{Q}_{h}\nabla u-\nabla u)\|_{T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k}\|u\|_{k+1}\|v\|_{1,h}\end{split}

Similarly, for $\ell_{2}(u,v)$ , we obtain

\displaystyle\begin{split}|\ell_{2}(u,v)|=&|\sum_{T\in\mathcal{T}_{h}^{I}}(\nabla(Q_{0}u)-\nabla u,A\nabla_{w}v)_{T}|\\ \leqslant&C\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|\nabla(Q_{0}u-u)\|_{T}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|A\nabla_{w}v\|_{T}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k}\|u\|_{k+1}\|v\|_{1,h}.\end{split}

For $\ell_{3}(u,v)$ , using the Cauchy-Schwarz inequality, the definition of the $L^{2}$ projection operator $Q_{b}$ , the trace inequality, the inverse inequality and the projection inequality leads to

\displaystyle\begin{split}|\ell_{3}(u,v)|=&\left|\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}u)-Q_{b}u,A\nabla_{w}v\cdot{\bf n}\rangle_{\partial T}\right|\\ \leqslant&C\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|Q_{0}u-u\|_{\partial T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|A\nabla_{w}v\|_{\partial T}^{2}\right)\\ \leqslant&C\left(\sum_{T\in\mathcal{T}_{h}^{I}}h_{T}^{-1}\|Q_{0}u-u\|_{T}^{2}+h_{T}\|\nabla(Q_{0}u-u)\|_{T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}h_{T}^{-1}\|A\nabla_{w}v\|_{T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k}\|u\|_{k+1}\|v\|_{1,h}.\end{split}

For $\ell_{4}(u,v)$ , by the fact that $\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(A\nabla u\cdot{\bf n}),v_{0}-Q_{b}v_{0}\rangle_{\partial T}=0$ , the Cauchy-Schwarz inequality, the projection inequality and the inequality (6.1), we have

\displaystyle\begin{split}|\ell_{4}(u,v)|=&\left|\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla u\cdot{\bf n},v_{0}-Q_{b}v_{0}\rangle_{\partial T}\right|\\ =&\left|\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla u\cdot{\bf n}-Q_{b}(A\nabla u\cdot{\bf n}),v_{0}-Q_{b}v_{0}\rangle_{\partial T}\right|\\ \leqslant&\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|A\nabla u\cdot{\bf n}-Q_{b}(A\nabla u\cdot{\bf n})\|_{\partial T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|v_{0}-Q_{b}v_{0}\|_{\partial T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k}\|u\|_{k+1}\|\nabla v_{0}\|\\ \leqslant&Ch^{k}\|u\|_{k+1}\|v\|_{1,h}.\end{split}

For $\ell_{5}(u,v)$ , it follows from the fact that $\sum_{e\in\mathcal{E}_{h}^{I}}\langle Q_{b}(A\nabla u\cdot{\bf n}_{e}),v-Q_{b}v\rangle_{e}=0$ that

\displaystyle\begin{split}|\ell_{5}(u,v)|=&\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla u\cdot{\bf n}_{e},v-Q_{b}v\rangle_{e}\\ =&\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla u\cdot{\bf n}_{e}-Q_{b}(A\nabla u\cdot{\bf n}_{e}),v-Q_{b}v\rangle_{e}\\ \leqslant&\left(\sum_{e\in\mathcal{E}_{h}^{I}}\|A\nabla u\cdot{\bf n}_{e}-Q_{b}(A\nabla u\cdot{\bf n}_{e})\|_{e}^{2}\right)^{\frac{1}{2}}\left(\sum_{e\in\mathcal{E}_{h}^{I}}\|v-Q_{b}v\|_{e}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k}\|u\|_{k+1}\|v\|_{1,h}.\end{split}

For $s(Q_{h}u,v)$ , according to the definition of the $L^{2}$ projection operator $Q_{b}$ , the trace inequality, and the projection inequality, we get

\displaystyle\begin{split}|s(Q_{h}u,v)|=&\left|\sum_{T\in\mathcal{T}_{h}}\langle Q_{b}(Q_{0}u)-Q_{b}u,Q_{b}v_{0}-v_{b}\rangle_{\partial T}\right|\\ \leqslant&C\left(\sum_{T\in\mathcal{T}_{h}}\|Q_{0}u-u\|_{\partial T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}}\|Q_{b}v_{0}-v_{b}\|_{\partial T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k}\|u\|_{k+1}\|v\|_{1,h}.\end{split}

The proof of the lemma is completed. ∎

Theorem 6.1.

Assuming $u_{i}\in H^{k+1}(\Omega_{i})$ with $i=1,2$ are the exact solutions of the Eqs.(1.1)-(1.6) and $u_{h}\in V_{h}$ are numerical solution obtained from the numerical scheme (2.3), then we have

\displaystyle\|e_{h}\|_{1,h}\leqslant Ch^{k}\|u\|_{k+1}.

(6.8)

Proof.

Choosing $v=\varepsilon_{h}\in V_{h}^{0}$ in Eq.(5.12) leads to

\displaystyle a^{n}(e_{h},\varepsilon_{h})+a_{s}^{I}(e_{h},\varepsilon_{h})=s(Q_{h}u,\varepsilon_{h})-\phi_{u}(\varepsilon_{h}),

Thus we have

\displaystyle a^{n}(e_{h},e_{h})+a_{s}^{I}(e_{h},e_{h})=s(Q_{h}u,\varepsilon_{h})+\phi_{u}(\varepsilon_{h})-a^{n}(e_{h},I_{h}u-P_{h}u)-a_{s}^{I}(e_{h},I_{h}u-P_{h}u),

For $a^{n}(e_{h},I_{h}u-P_{h}u)$ , we have

\displaystyle\begin{split}|a^{n}(e_{h},I_{h}u-P_{h}u)|=&|\sum_{T\in\mathcal{T}_{h}^{n}}(\nabla e_{h},\nabla(\Pi_{h}u-u))_{T}|\\ \leqslant&C\left(\sum_{T\in\mathcal{T}_{h}^{n}}\|\nabla e_{h}\|_{T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{n}}\|\nabla(\Pi_{h}u-u)\|_{T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k}\|u\|_{k+1}\|e_{h}\|_{1,h}.\end{split}

(6.9)

For $a_{s}^{I}(e_{h},Q_{h}u-\tilde{Q}_{h}u)$ , we get

\displaystyle\begin{split}&\left|a_{s}^{I}(e_{h},Q_{h}u-\tilde{Q}_{h}u)\right|\\ =&\left|\sum_{T\in\mathcal{T}_{h}^{I}}(\nabla_{w}e_{h},\nabla_{w}(Q_{h}u-\tilde{Q}_{h}u))_{T}+h_{T}^{-1}\langle Q_{b}e_{0}-e_{b},Q_{b}(\Pi_{h}u)-Q_{b}u\rangle_{\partial T\cap\mathcal{E}_{h}^{I}}\right|\\ \leqslant&\sum_{T\in\mathcal{T}_{h}^{I}}\|e_{h}\|_{1,h}\|\nabla_{w}(Q_{h}u-\tilde{Q}_{h}u)\|_{T}+\sum_{T\in\mathcal{T}_{h}^{I}}h_{T}^{-1}\|Q_{b}e_{0}-e_{b}\|_{\partial T\cap\mathcal{E}_{h}^{I}}\|Q_{b}(\Pi_{h}u)-Q_{b}u\|_{\partial T\cap\mathcal{E}_{h}^{I}}\\ \leqslant&C\|e_{h}\|_{1,h}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|\nabla_{w}(Q_{h}u-\tilde{Q}_{h}u)\|_{T}^{2}+h_{T}^{-1}\|\Pi_{h}u-u\|^{2}_{\partial T\cap\mathcal{E}_{h}^{I}}\right)^{\frac{1}{2}}\\ \leqslant&C\|e_{h}\|_{1,h}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|\nabla_{w}(Q_{h}u-\tilde{Q}_{h}u)\|_{T}^{2}+h^{k}\|u\|_{k+1}^{2}\right)^{\frac{1}{2}}\end{split}

For $\|\nabla_{w}(Q_{h}u-\tilde{Q}_{h}u)\|_{T}$ and $\mathbf{q}\in\nabla\mathcal{V}_{k}(T)$ , we have

\displaystyle\begin{split}&(\nabla_{w}(Q_{h}u-\tilde{Q}_{h}u),\mathbf{q})_{T}\\ =&\langle Q_{b}u-Q_{b}(\Pi_{h}u),\mathbf{q}\cdot{\bf n}\rangle_{\partial T\cap\mathcal{E}_{h}^{I}}\\ \leqslant&C\|Q_{b}u-Q_{b}(\Pi_{h}u)\|_{\partial T\cap\mathcal{E}_{h}^{I}}\|\mathbf{q}\|_{\partial T\cap\mathcal{E}_{h}^{I}}\\ \leqslant&Ch_{T}^{-\frac{1}{2}}\|\mathbf{q}\|_{T}\|u-\Pi_{h}u\|_{\partial T\cap\mathcal{E}_{h}^{I}}\\ \leqslant&Ch^{k}\|u\|_{k+1}\|\mathbf{q}\|_{T}.\end{split}

Taking $\mathbf{q}=\nabla_{w}(Q_{h}u-\tilde{Q}_{h}u)$ in the above estimate, we get

\displaystyle\|\nabla_{w}(Q_{h}u-\tilde{Q}_{h}u)\|\leqslant Ch^{k}\|u\|_{k+1}.

Thus we obtain

\displaystyle\left|a_{s}^{I}(e_{h},Q_{h}u-\tilde{Q}_{h}u)\right|\leqslant Ch^{k}\|u\|_{k+1}\|e_{h}\|_{1,h}.

(6.10)

According to the estimates (6.2)-(6.7) and (6.9)-(6.10), we get

\displaystyle\begin{split}\|e_{h}\|_{1,h}^{2}\leqslant&Ch^{k}\|u\|_{k+1}\|e_{h}\|_{1,h}+Ch^{k}\|u\|_{k+1}\|\varepsilon_{h}\|_{1,h}\\ \leqslant&Ch^{k}\|u\|_{k+1}\|e_{h}\|_{1,h}+Ch^{2k}\|u\|_{k+1}^{2}\\ \leqslant&Ch^{2k}\|u\|_{k+1}^{2}+\frac{1}{2}\|e_{h}\|_{1,h}^{2}.\end{split}

Thus we have

\|e_{h}\|_{1,h}\leqslant Ch^{k}\|u\|_{k+1}.

The proof of the estimate (6.8) is completed. ∎

7 Error Estimate in $L^{2}$ norm

In this section, we use the dual argument to give the error estimate in $L^{2}$ norm. We consider the following problem: seeking $\varphi$ satisfying

$\displaystyle-\nabla\cdot(A\nabla\varphi)$	$\displaystyle=$	$\displaystyle e_{0},\,\text{in}\,\Omega,$	(7.1)
$\displaystyle\varphi$	$\displaystyle=$	$\displaystyle 0,\,\text{on}\,\partial\Omega,$	(7.2)
$\displaystyle\varphi_{1}-\varphi_{2}$	$\displaystyle=$	$\displaystyle 0,\,\text{on}\,\Gamma,$	(7.3)
$\displaystyle A_{1}\nabla\varphi_{1}\cdot{\bf n}-A_{2}\nabla\varphi_{2}\cdot{\bf n}$	$\displaystyle=$	$\displaystyle 0,\,\text{on}\,\Gamma,$	(7.4)

where $e_{0}=\left\{\begin{array}[]{cc}u-u_{h}&T\in\mathcal{T}_{h}^{n}\\ Q_{0}u-u_{0}&T\in\mathcal{T}_{h}^{I}\end{array}\right.$ .

Assume the solution $\varphi$ satisfies $H^{2}$ -regularity, i.e

\displaystyle\|\varphi\|_{2}\leqslant C\|e_{0}\|.

(7.5)

Theorem 7.1.

Based on the assumption in Theorem 6.1, we have the following error estimate in the $L^{2}$ norm:

\displaystyle\|e_{0}\|\leqslant Ch^{k+1}\|u\|_{k+1}.

(7.6)

Proof.

On two sides of Eq.(7.1), integrating with $e_{0}$ yields

\displaystyle\begin{split}\|e_{0}\|^{2}=&(-\nabla\cdot(A\nabla\varphi),e_{0})\\ =&\sum_{T\in\mathcal{T}_{h}}(A\nabla\varphi,\nabla e_{0})_{T}-\sum_{T\in\mathcal{T}_{h}^{n}}\langle A\nabla\varphi\cdot{\bf n},e_{0}\rangle_{\partial T}-\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla\varphi\cdot{\bf n},e_{0}\rangle_{\partial T}\\ =&\sum_{T\in\mathcal{T}_{h}}(A\nabla\varphi,\nabla e_{0})_{T}-\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla\varphi\cdot{\bf n},e_{0}\rangle_{e}-\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla\varphi\cdot{\bf n},e_{0}-Q_{b}e_{0}\rangle_{\partial T}\\ &-\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla\varphi\cdot{\bf n},Q_{b}e_{0}-e_{b}\rangle_{\partial T}-\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla\varphi\cdot{\bf n},e_{b}\rangle_{\partial T}\\ =&\sum_{T\in\mathcal{T}_{h}}(A\nabla\varphi,\nabla e_{0})_{T}-\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla\varphi\cdot{\bf n},e_{0}-Q_{b}e_{0}\rangle_{\partial T}-\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla\varphi\cdot{\bf n},Q_{b}e_{0}-e_{b}\rangle_{\partial T}\\ &+\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla\varphi\cdot{\bf n}_{e},e_{0}-e_{b}\rangle_{e}.\end{split}

(7.7)

For $T\in\mathcal{T}_{h}^{I}$ , it’s similar to Eq.(5.6) to get

\displaystyle\begin{split}&(A\nabla_{w}Q_{h}\varphi,\nabla_{w}e_{h})_{T}\\ =&(A\nabla\varphi,\nabla e_{0})_{T}-\langle Q_{b}e_{0}-e_{b},A\mathcal{Q}_{h}\nabla\varphi\cdot{\bf n}\rangle_{\partial T}\\ &+(\nabla(Q_{0}\varphi-\varphi),A\nabla_{w}e_{h})_{T}-\langle Q_{b}(Q_{0}\varphi)-Q_{b}\varphi,A\nabla_{w}e_{h}\cdot{\bf n}\rangle_{\partial T}.\end{split}

(7.8)

Substitute Eq.(7.8) into Eq.(7.7) leads to

\displaystyle\begin{split}\|e_{0}\|^{2}=&\sum_{T\in\mathcal{T}_{h}^{n}}((A\nabla\varphi,\nabla e_{0})_{T}+\sum_{T\in\mathcal{T}_{h}^{I}}(A\nabla_{w}Q_{h}\varphi,\nabla_{w}e_{h})_{T}\\ &+\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}e_{0}-e_{b},A\mathcal{Q}_{h}\nabla\varphi\cdot{\bf n}-A\nabla\varphi\cdot{\bf n}\rangle_{\partial T}-\sum_{T\in\mathcal{T}_{h}^{I}}(\nabla(Q_{0}\varphi-\varphi),A\nabla_{w}e_{h})_{T}\\ &+\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}\varphi)-Q_{b}\varphi,A\nabla_{w}e_{h}\cdot{\bf n}\rangle_{\partial T}-\sum_{T\in\mathcal{T}_{h}^{I}}\langle A\nabla\varphi\cdot{\bf n},e_{0}-Q_{b}e_{0}\rangle_{\partial T}\\ &+\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla\varphi\cdot{\bf n}_{e},e_{0}-e_{b}\rangle_{\partial T}\\ =&a^{n}(\varphi,e_{0})+a_{s}^{I}(Q_{h}\varphi,e_{h})+\phi_{\varphi}(e_{h})-s(Q_{h}\varphi,e_{h}).\end{split}

(7.9)

Choosing $v=I_{h}\varphi$ in Eq.(5.12) yields

\displaystyle a^{n}(e_{h},\Pi_{h}\varphi)+a_{s}^{I}(e_{h},\tilde{Q}_{h}\varphi)=s(Q_{h}u,\tilde{Q}_{h}\varphi)-\phi_{u}(I_{h}\varphi).

(7.10)

Thus we have

\displaystyle\begin{split}&a^{n}(e_{h},\varphi)+a_{s}^{I}(e_{h},Q_{h}\varphi)\\ =&s(Q_{h}u,\tilde{Q}_{h}\varphi)-\phi_{u}(I_{h}\varphi)-a^{n}(\Pi_{h}\varphi-\varphi,e_{h})+a_{s}^{I}(e_{h},Q_{h}\varphi-\tilde{Q}_{h}\varphi).\end{split}

(7.11)

Substitute Eq.(7.11) into Eq.(7.9) leads to

\displaystyle\begin{split}\begin{split}&\|e_{0}\|^{2}\\ =&\phi_{\varphi}(e_{h})-s(Q_{h}\varphi,e_{h})+s(Q_{h}u,\tilde{Q}_{h}\varphi)-\phi_{u}(I_{h}\varphi)-a^{n}(\Pi_{h}\varphi-\varphi,e_{h})+a_{s}^{I}(e_{h},Q_{h}\varphi-\tilde{Q}_{h}\varphi).\end{split}\end{split}

(7.12)

By Lemma 6.2 and the estimates (6.9)-(6.10), we have

\displaystyle|\phi_{\varphi}(e_{h})-s(Q_{h}\varphi,e_{h})-a^{n}(\Pi_{h}\varphi-\varphi,e_{h})+a_{s}^{I}(e_{h},Q_{h}\varphi-\tilde{Q}_{h}\varphi)|\leqslant Ch\|\varphi\|_{2}\|e_{h}\|_{1,h}.

(7.13)

Each of the remaining terms is handled as follows.

(1) For $s(Q_{h}u,\tilde{Q}_{h}\varphi)$ , according to the Cauchy-Schwarz inequality, the definition of the $L^{2}$ projection operator $Q_{b}$ , the trace inequality, and the projection inequality, we get

\displaystyle\begin{split}|s(Q_{h}u,\tilde{Q}_{h}\varphi)|\leqslant&\left|\sum_{T\in\mathcal{T}_{h}^{I}}h_{T}^{-1}\langle Q_{b}(Q_{0}u)-Q_{b}u,Q_{b}(Q_{0}\varphi)-Q_{b}\varphi\rangle_{\partial T\setminus(\partial T\cap\mathcal{E}_{h}^{I})}\right|\\ &+\left|\sum_{T\in\mathcal{T}_{h}^{I}}h_{T}^{-1}\langle Q_{b}(Q_{0}u)-Q_{b}u,Q_{b}(Q_{0}\varphi)-Q_{b}(\Pi_{h}\varphi)\rangle_{\partial T\cap\mathcal{E}_{h}^{I}}\right|\\ \leqslant&C\left(\sum_{T\in\mathcal{T}_{h}^{I}}h_{T}^{-1}\|Q_{0}u-u\|_{\partial T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}h_{T}^{-1}\|Q_{0}\varphi-\varphi\|_{\partial T}^{2}\right)^{\frac{1}{2}}\\ &+C\left(\sum_{T\in\mathcal{T}_{h}^{I}}h_{T}^{-1}\|Q_{0}u-u\|_{\partial T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}h_{T}^{-1}\|Q_{0}\varphi-\varphi\|_{\partial T\cap\mathcal{E}_{h}^{I}}^{2}+h_{T}^{-1}\|\Pi_{h}\varphi-\varphi\|_{\partial T\cap\mathcal{E}_{h}^{I}}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k+1}\|u\|_{k+1}\|\varphi\|_{2}.\end{split}

(7.14)

(2) Similarly, for $\ell_{1}(u,\tilde{Q}_{h}\varphi)$ , we get the following estimate

\displaystyle\begin{split}|\ell_{1}(u,\tilde{Q}_{h}\varphi)|\leqslant&\left|\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}\varphi)-Q_{b}\varphi,A\mathcal{Q}_{h}\nabla u\cdot{\bf n}-A\nabla u\cdot{\bf n}\rangle_{\partial T\setminus(\partial T\cap\mathcal{E}_{h}^{I})}\right|\\ &+\left|\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}\varphi)-Q_{b}(\Pi_{h}\varphi),A\mathcal{Q}_{h}\nabla u\cdot{\bf n}-A\nabla u\cdot{\bf n}\rangle_{\partial T\cap\mathcal{E}_{h}^{I}}\right|\\ \leqslant&\left(\sum_{T\in\mathcal{T}_{h}}\|Q_{0}\varphi-\varphi\|_{\partial T}^{2}+\|\Pi_{h}\varphi-\varphi\|_{\partial T\cap\mathcal{E}_{h}^{I}}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}}\|A\mathcal{Q}_{h}\nabla u-A\nabla u\|_{\partial T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k+1}\|u\|_{k+1}\|\varphi\|_{2}.\end{split}

(7.15)

(3) For $\ell_{4}(u,\tilde{Q}_{h}\varphi)$ , using the fact that $\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}\varphi)-Q_{0}\varphi,Q_{b}(A\nabla u\cdot{\bf n})\rangle_{\partial T}=0$ leads to

\displaystyle\begin{split}|\ell_{4}(u,\tilde{Q}_{h}\varphi)|=&\left|\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}\varphi)-Q_{0}\varphi,A\nabla u\cdot{\bf n}\rangle_{\partial T}\right|\\ =&\left|\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}\varphi)-Q_{0}\varphi,A\nabla u\cdot{\bf n}-Q_{b}(A\nabla u\cdot{\bf n})\rangle_{\partial T}\right|\\ \leqslant&C\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|Q_{0}\varphi-\varphi\|_{\partial T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|A\nabla u\cdot{\bf n}-Q_{b}(A\nabla u\cdot{\bf n})\|_{\partial T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k+1}\|u\|_{k+1}\|\varphi\|_{2}.\end{split}

(7.16)

(4) For $\ell_{2}(u,\tilde{Q}_{h}\varphi)$ and $\ell_{3}(u,\tilde{Q}_{h}\varphi)$ , we have

\displaystyle\begin{split}\ell_{2}(u,\tilde{Q}_{h}\varphi)=&\sum_{T\in\mathcal{T}_{h}^{I}}\left(\nabla(Q_{0}u-u),A\nabla_{w}\tilde{Q}_{h}\varphi\right)_{T}\\ =&\sum_{T\in\mathcal{T}_{h}^{I}}(\nabla(Q_{0}u-u),A\nabla_{w}\tilde{Q}_{h}\varphi-A\mathcal{Q}_{h}\nabla\varphi)_{T}\\ &+\sum_{T\in\mathcal{T}_{h}^{I}}(\nabla(Q_{0}u-u),A\mathcal{Q}_{h}\nabla\varphi-A\nabla\varphi)_{T}\\ &+\sum_{T\in\mathcal{T}_{h}^{I}}(\nabla(Q_{0}u-u),A\nabla\varphi)_{T}\\ =&\ell_{21}(u,\varphi)+\ell_{22}(u,\varphi)+\ell_{23}(u,\varphi).\end{split}

(7.17)

and

\displaystyle\begin{split}\ell_{3}(u,\tilde{Q}_{h}\varphi)=&\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}u)-Q_{b}u,A\nabla_{w}\tilde{Q}_{h}\varphi\cdot{\bf n}\rangle_{\partial T}\\ =&\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}u)-Q_{b}u,A\nabla_{w}\tilde{Q}_{h}\varphi\cdot{\bf n}-A\mathcal{Q}_{h}\nabla\varphi\cdot{\bf n}\rangle_{\partial T}\\ &+\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}u)-Q_{b}u,A\mathcal{Q}_{h}\nabla\varphi\cdot{\bf n}-A\nabla\varphi\cdot{\bf n}\rangle_{\partial T}\\ &+\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}u)-Q_{b}u,A\nabla\varphi\cdot{\bf n}-AQ_{b}(\nabla\varphi\cdot{\bf n})\rangle_{\partial T}\\ &+\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}u)-Q_{b}u,AQ_{b}(\nabla\varphi\cdot{\bf n})\rangle_{\partial T}\\ =&\ell_{31}(u,\varphi)+\ell_{32}(u,\varphi)+\ell_{33}(u,\varphi)+\ell_{34}(u,\varphi).\end{split}

(7.18)

Denote by $\theta(u,\varphi)=\ell_{23}(u,\varphi)-\ell_{34}(u,\varphi)$ , then we have

\displaystyle\begin{split}\theta(u,\varphi)=&\sum_{T\in\mathcal{T}_{h}^{I}}-(Q_{0}u-u,\nabla\cdot(A\nabla\varphi))_{T}+\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{0}u-u,A\nabla\varphi\cdot{\bf n}-AQ_{b}(\nabla\varphi\cdot{\bf n})\rangle_{\partial T}.\end{split}

(7.19)

Therefore we get

\displaystyle\begin{split}&\ell_{2}(u,\tilde{Q}_{h}\varphi)-\ell_{3}(u,\tilde{Q}_{h}\varphi)\\ =&\ell_{21}(u,\varphi)+\ell_{22}(u,\varphi)-\ell_{31}(u,\varphi)-\ell_{32}(u,\varphi)-\ell_{33}(u,\varphi)+\theta(\ell_{21}(u,\varphi)).\end{split}

(7.20)

Now we estimate each term on the right side of the above equation.

For $\ell_{21}(u,\varphi)$ , we have

\displaystyle\begin{split}|\ell_{21}(u,\varphi)|=&\left|\sum_{T\in\mathcal{T}_{h}^{I}}(\nabla(Q_{0}u-u),A\nabla_{w}\tilde{Q}_{h}\varphi-A\mathcal{Q}_{h}\nabla\varphi)_{T}\right|\\ \leqslant&C\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|\nabla(Q_{0}u-u)\|_{T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|A\nabla_{w}\tilde{Q}_{h}\varphi-A\mathcal{Q}_{h}\nabla\varphi\|_{T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k}\|u\|_{k+1}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|A\nabla_{w}\tilde{Q}_{h}\varphi-A\mathcal{Q}_{h}\nabla\varphi\|_{T}^{2}\right)^{\frac{1}{2}},\end{split}

(7.21)

where for any $\bm{q}\in\nabla\mathcal{V}_{k}(T)$ , $T\in\mathcal{T}_{h}^{I}$ , and $\partial T\cap\mathcal{E}_{h}^{I}=\emptyset$ , it follows from Eq.(5.3) that

\displaystyle\begin{split}&|(A\nabla_{w}\tilde{Q}_{h}\varphi-A\mathcal{Q}_{h}\nabla\varphi,\bm{q})_{T}|\\ =&|(\nabla(Q_{0}\varphi-\varphi),A\bm{q})_{T}-\langle Q_{b}(Q_{0}\varphi)-Q_{b}\varphi,A\bm{q}\cdot{\bf n}\rangle_{\partial T}|\\ \leqslant&C\|\nabla(Q_{0}\varphi-\varphi)\|_{T}\|\bm{q}\|_{T}+\|Q_{0}\varphi-\varphi\|_{\partial T}\|\bm{q}\|_{\partial T}\\ \leqslant&Ch\|\varphi\|_{2}\|\bm{q}\|_{T}.\end{split}

(7.22)

Similarly, for $T\in\mathcal{T}_{h}^{I}$ and $\partial T\cap\mathcal{E}_{h}^{I}\neq\emptyset$ , we have

\displaystyle\begin{split}&\left|(A\nabla_{w}\tilde{Q}_{h}\varphi-A\mathcal{Q}_{h}\nabla\varphi,\bm{q})_{T}\right|\\ =&|(\nabla(Q_{0}\varphi-\varphi),A\bm{q})_{T}-\langle Q_{b}(Q_{0}\varphi)-Q_{b}\varphi,A\bm{q}\cdot{\bf n}\rangle_{\partial T\setminus(\partial T\cap\mathcal{E}_{h}^{I})}\\ &-\langle Q_{b}(Q_{0}\varphi)-Q_{b}(\Pi_{h}\varphi),A\bm{q}\cdot{\bf n}\rangle_{\partial T\cap\mathcal{E}_{h}^{I}}|\\ \leqslant&C\|\nabla(Q_{0}\varphi-\varphi)\|_{T}\|\bm{q}\|_{T}+C\|Q_{0}\varphi-\varphi\|_{\partial T}\|\bm{q}\|_{\partial T}+C\|\Pi_{h}\varphi-\varphi\|_{\partial T\cap\mathcal{E}_{h}^{I}}\|\bm{q}\|_{\partial T\cap\mathcal{E}_{h}^{I}}\\ \leqslant&Ch\|\varphi\|_{2}\|\bm{q}\|_{T}.\end{split}

(7.23)

Taking $\bm{q}=\nabla_{w}\tilde{Q}_{h}\varphi-\mathcal{Q}_{h}\nabla\varphi$ in Eq.(7.22) leads to

\displaystyle\|\nabla_{w}\tilde{Q}_{h}\varphi-\mathcal{Q}_{h}\nabla\varphi\|_{T}\leqslant Ch\|\varphi\|_{2}.

(7.24)

Substitute Eq.(7.22) and Eq.(7.24) into Eq.(7.21) yields

\displaystyle|\ell_{21}(u,\varphi)|\leqslant Ch^{k+1}\|u\|_{k+1}\|\varphi\|_{2}.

(7.25)

For $\ell_{22}(u,\varphi)$ , by the Cauchy-Schwarz inequality and the projection inequality, we have

\displaystyle\begin{split}|\ell_{22}(u,\varphi)|=&\left|\sum_{T\in\mathcal{T}_{h}^{I}}(\nabla(Q_{0}u-u),A\mathcal{Q}_{h}\nabla\varphi-A\nabla\varphi)_{T}\right|\\ \leqslant&\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|\nabla(Q_{0}u-u)\|_{T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|A\mathcal{Q}_{h}\nabla\varphi-A\nabla\varphi\|_{T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k+1}\|u\|_{k+1}\|\varphi\|_{2}.\end{split}

(7.26)

For $\ell_{31}(u,\varphi)$ , using the Cauchy-Schwarz inequality, the trace inequality, the inverse inequality and the projection inequality leads to

\displaystyle\begin{split}|\ell_{31}(u,\varphi)|=&\left|\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}u)-Q_{b}u,A\nabla_{w}\tilde{Q}_{h}\varphi\cdot{\bf n}-A\mathcal{Q}_{h}\nabla\varphi\cdot{\bf n}\rangle_{\partial T}\right|\\ \leqslant&\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|Q_{0}u-u\|_{\partial T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|A\nabla_{w}\tilde{Q}_{h}\varphi-A\mathcal{Q}_{h}\nabla\varphi\|_{\partial T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k+1}\|u\|_{k+1}\|\varphi\|_{2}.\end{split}

(7.27)

Similarly, we get

\displaystyle\begin{split}|\ell_{32}(u,\varphi)|=&\left|\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}u)-Q_{b}u,A\mathcal{Q}_{h}\nabla\varphi\cdot{\bf n}-A\nabla\varphi\cdot{\bf n}\rangle_{\partial T}\right|\\ \leqslant&C\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|Q_{0}u-u\|_{\partial T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|A\mathcal{Q}_{h}\nabla\varphi-A\nabla\varphi\|_{\partial T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k+1}\|u\|_{k+1}\|\varphi\|_{2},\end{split}

(7.28)

and

\displaystyle\begin{split}|\ell_{33}(u,\varphi)|=&\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{b}(Q_{0}u)-Q_{b}u,A\nabla\varphi\cdot{\bf n}-AQ_{b}(\nabla\varphi\cdot{\bf n})\rangle_{\partial T}\\ \leqslant&\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|Q_{0}u-u\|_{\partial T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|A\nabla\varphi\cdot{\bf n}-AQ_{b}(\nabla\varphi\cdot{\bf n})\|_{\partial T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k+1}\|u\|_{k+1}\|\varphi\|_{2}.\end{split}

(7.29)

For $\theta(u,\varphi)$ , we obtain

\displaystyle\begin{split}|\theta(u,\varphi)|\leqslant&\left|\sum_{T\in\mathcal{T}_{h}^{I}}-(Q_{0}u-u,\nabla\cdot(A\nabla\varphi))_{T}\right|+\left|\sum_{T\in\mathcal{T}_{h}^{I}}\langle Q_{0}u-u,A\nabla\varphi\cdot{\bf n}-AQ_{b}(\nabla\varphi\cdot{\bf n})\rangle_{\partial T}\right|\\ \leqslant&\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|Q_{0}u-u\|_{T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|\nabla\cdot(A\nabla\varphi)\|_{T}^{2}\right)^{\frac{1}{2}}\\ &+\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|Q_{0}u-u\|_{\partial T}^{2}\right)^{\frac{1}{2}}\left(\sum_{T\in\mathcal{T}_{h}^{I}}\|A\nabla\varphi\cdot{\bf n}-AQ_{b}(\nabla\varphi\cdot{\bf n})\|_{\partial T}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k+1}\|u\|_{k+1}\|\varphi\|_{2}.\end{split}

(7.30)

(5)For $\ell_{5}(u,\tilde{Q}_{h}\varphi)$ , we get

\displaystyle\begin{split}|\ell_{5}(u,\tilde{Q}_{h}\varphi)|=&\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla u\cdot{\bf n}_{e},\Pi_{h}\varphi-Q_{b}(\Pi_{h}\varphi)\rangle_{e}\\ =&\sum_{e\in\mathcal{E}_{h}^{I}}\langle A\nabla u\cdot{\bf n}_{e}-Q_{b}(A\nabla u\cdot{\bf n}_{e}),\Pi_{h}\varphi-Q_{b}(\Pi_{h}\varphi)\rangle_{e}\\ \leqslant&\left(\sum_{e\in\mathcal{E}_{h}^{I}}\|A\nabla u\cdot{\bf n}_{e}-Q_{b}(A\nabla u\cdot{\bf n}_{e})\|_{e}^{2}\right)^{\frac{1}{2}}\\ &\left(\sum_{e\in\mathcal{E}_{h}^{I}}\|\Pi_{h}\varphi-\varphi\|_{e}^{2}+\|\varphi-Q_{b}\varphi\|_{e}^{2}+\|Q_{b}\varphi-Q_{b}(\Pi_{h}\varphi)\|_{e}^{2}\right)^{\frac{1}{2}}\\ \leqslant&Ch^{k+1}\|u\|_{k+1}\|\varphi\|_{2}.\end{split}

(7.31)

Using Eqs.(7.14)-(7.31) yields

\displaystyle\begin{split}\|e_{0}\|^{2}\leqslant&Ch\|\varphi\|_{2}\|e_{h}\|_{1,h}+Ch^{k+1}\|u\|_{k+1}\|\varphi\|_{2}\\ \leqslant&Ch^{k+1}\|u\|_{k+1}\|e_{0}\|.\end{split}

(7.32)

The proof of the theorem is completed. ∎

8 The Numerical Experiments

In this section, we give the numerical experiment to demonstrate that the proposed immersed numerical scheme is efficient to solve the elliptic interface problem on unfitted meshes.

Example 8.1.

Consider the interface problem in the square domain $[-1,1]\times[-1,1]$ . The interface is as follows:

x^{2}+y^{2}=\frac{1}{3}.

The exact solution is

u=\left(\begin{array}[]{cc}\frac{1}{A_{1}}\cos(\pi(x^{2}+y^{2})),&(x,y)\in\Omega_{1},\\ \frac{1}{A_{2}}\cos(\pi(x^{2}+y^{2}))+\frac{1}{2}(\frac{1}{A1}-\frac{1}{A2}),&(x,y)\in\Omega_{2},\end{array}\right).

Refer to caption — Figure 1: Three different level of meshes: left: the first level, middle: the second level, right: the third level

We implement this example on triangular meshes (see Figure 1). We use the $P_{1}$ to $P_{2}$ WG elements and the finite elements to solve the elliptic interface problems. The numerical results are shown in Tables 1 - 4. From these tables, it become evident that the convergence orders of the velocity function are $O(h^{k})$ and $O(h^{k+1})$ in the energy norm and $L^{2}$ norm, respectively. Hence, the results of the above numerical results show that it is effective to use the proposed numerical scheme to solve the elliptic interface problems on unfitted meshes.

Table 1: Numerical results on triangular meshes with

(A_{1},A_{2})=(1,1)

n	${\|\|\|}Q_{h}{\bf u}-{\bf u}_{h}{\|\|\|}$	order	$\\|Q_{0}{\bf u}-{\bf u}_{0}\\|$	order	$\\|u-u_{h}\\|_{\infty}$	order
			$k=1$
1	2.2370E+00	–	1.2148E-01	–	2.1283E-01	–
2	1.1168E+00	1.0021	3.0795E-02	1.9799	6.8040E-02	1.6452
3	5.5127E-01	1.0186	7.6231E-03	2.0142	1.8944E-02	1.8446
4	2.7364E-01	1.0105	1.8950E-03	2.0081	4.8964E-03	1.9519
5	1.3631E-01	1.0054	4.7205E-04	2.0052	1.2392E-03	1.9823
			$k=2$
1	3.0460E-01	–	7.6529E-03	–	2.4590E-02	–
2	7.6341E-02	1.9964	9.3282E-04	3.0363	4.1435E-03	2.5691
3	1.9045E-02	2.0031	1.1557E-04	3.0128	5.6694E-04	2.8696
4	4.7491E-03	2.0036	1.4385E-05	3.0061	7.2963E-05	2.9580
5	1.1857E-03	2.0020	1.7953E-06	3.0023	9.2153E-06	2.9851

Table 2: Numerical results on triangular meshes with

(A_{1},A_{2})=(1,10)

n	${\|\|\|}Q_{h}{\bf u}-{\bf u}_{h}{\|\|\|}$	order	$\\|Q_{0}{\bf u}-{\bf u}_{0}\\|$	order	$\\|u-u_{h}\\|_{\infty}$	order
			$k=1$
1	4.4477E-01	0.0000	2.2992E-02	0.0000	1.0227E-01	0.0000
2	2.1656E-01	1.0383	5.5631E-03	2.0472	3.3270E-02	1.6200
3	1.0895E-01	0.9911	1.3331E-03	2.0612	8.0039E-03	2.0555
4	5.3592E-02	1.0236	3.2047E-04	2.0565	2.3125E-03	1.7912
5	2.6520E-02	1.0149	7.8282E-05	2.0334	6.4345E-04	1.8456
			$k=2$
1	4.1278E-02	–	9.8355E-04	–	2.6897E-03	–
2	1.0145E-02	2.0247	1.2011E-04	3.0336	4.3858E-04	2.6165
3	2.5260E-03	2.0058	1.4944E-05	3.0067	7.1094E-05	2.6250
4	6.2971E-04	2.0041	1.8625E-06	3.0042	8.5184E-06	3.0611
5	1.5712E-04	2.0028	2.3247E-07	3.0021	9.2153E-07	3.2085

Table 3: Numerical results on triangular meshes with

(A_{1},A_{2})=(1,100)

n	${\|\|\|}Q_{h}{\bf u}-{\bf u}_{h}{\|\|\|}$	order	$\\|Q_{0}{\bf u}-{\bf u}_{0}\\|$	order	$\\|u-u_{h}\\|_{\infty}$	order
			$k=1$
1	3.7402E-01	–	1.7639E-02	–	1.2115E-01	–
2	1.8228E-01	1.0369	4.2320E-03	2.0594	4.3600E-02	1.4744
3	9.3339E-02	0.9656	1.0190E-03	2.0541	1.2268E-02	1.8295
4	4.6063E-02	1.0189	2.4419E-04	2.0611	3.0937E-03	1.9875
5	2.2799E-02	1.0146	5.9424E-05	2.0389	7.2310E-04	2.0970
			$k=2$
1	2.7535E-02	–	5.9485E-04	–	2.5620E-03	–
2	6.7377E-03	2.0309	7.6048E-05	2.9675	3.3375E-04	2.9404
3	1.6767E-03	2.0066	9.5658E-06	2.9910	7.1478E-05	2.2232
4	4.1809E-04	2.0038	1.1965E-06	2.9990	8.5962E-06	3.0557
5	1.0426E-04	2.0036	1.4944E-07	3.0012	9.0610E-07	3.2459

Table 4: Numerical results on triangular meshes with

(A_{1},A_{2})=(1,1000)

n	${\|\|\|}Q_{h}{\bf u}-{\bf u}_{h}{\|\|\|}$	order	$\\|Q_{0}{\bf u}-{\bf u}_{0}\\|$	order	$\\|u-u_{h}\\|_{\infty}$	order
			$k=1$
1	3.6990E-01	–	1.6765E-02	–	1.1861E-01	–
2	1.7904E-01	1.0468	3.9831E-03	2.0735	3.2937E-02	1.8484
3	9.1880E-02	0.9625	9.6856E-04	2.0400	1.1087E-02	1.5708
4	4.5623E-02	1.0100	2.3645E-04	2.0343	2.9154E-03	1.9271
5	2.2699E-02	1.0071	5.8357E-05	2.0185	6.9691E-04	2.0647
			$k=2$
1	2.7281E-02	–	5.8729E-04	–	2.0661E-03	–
2	6.6883E-03	2.0282	7.5315E-05	2.9631	3.3401E-04	2.6289
3	1.6656E-03	2.0056	9.4895E-06	2.9885	7.0919E-05	2.2356
4	4.1538E-04	2.0035	1.1876E-06	2.9982	8.5933E-06	3.0449
5	1.0359E-04	2.0036	1.4836E-07	3.0010	8.9880E-07	3.2571

The Immersed Weak Galerkin and Continuous Galerkin Finite Element Method for Elliptic Interface Problem

Abstract

keywords:

1 Introduction

2 The Numerical Scheme

Definition 2.1.

3 Existence and Uniqueness

Lemma 3.1.

Proof.

4 Some Inequalities

Lemma 4.1.

Proof.

5 Error Equation

Lemma 5.1.

Proof.

Lemma 5.2.

Proof.

Theorem 5.1.

Proof.

6 Error Estimate in $H^{1}$ norm

Lemma 6.1.

Lemma 6.2.

Proof.

Theorem 6.1.

Proof.

7 Error Estimate in $L^{2}$ norm

Theorem 7.1.

Proof.

8 The Numerical Experiments

Example 8.1.

References

The Immersed Weak Galerkin and Continuous Galerkin Finite Element Method for Elliptic Interface Problem

Abstract

keywords:

1 Introduction

2 The Numerical Scheme

Definition 2.1.

3 Existence and Uniqueness

Lemma 3.1.

Proof.

4 Some Inequalities

Lemma 4.1.

Proof.

5 Error Equation

Lemma 5.1.

Proof.

Lemma 5.2.

Proof.

Theorem 5.1.

Proof.

6 Error Estimate in H1H^{1} norm

Lemma 6.1.

Lemma 6.2.

Proof.

Theorem 6.1.

Proof.

7 Error Estimate in L2L^{2} norm

Theorem 7.1.

Proof.

8 The Numerical Experiments

Example 8.1.

References

6 Error Estimate in $H^{1}$ norm

7 Error Estimate in $L^{2}$ norm