Interior energy error estimates for the weak Galerkin finite element method

Abstract

Consider the Poisson equation in a polytopal domain $\Omega \subset {\mathbb {R}}^d$ ($d=2, 3$) as the model problem. We study interior energy error estimates for the weak Galerkin finite element approximation to elliptic boundary value problems. In particular, we show that the interior error in the energy norm is bounded by three components: the best local approximation error, the error in negative norms, and the trace error on the element boundaries. This implies that the interior convergence rate can be polluted by solution singularities on the domain boundary, even when the solution is smooth in the interior region. Numerical results are reported to support the theoretical findings. To the best of our knowledge, this is the first local energy error analysis that applies to general meshes consisting of polytopal elements and hanging nodes.

Appendix A

The proof of Lemma 4.4.

Proof

It follows from (4.8), (2.7), and $\mathrm{supp}(\omega _1)\subseteq D_1$, that

$$\begin{aligned} a(\omega _1 z,v)= & {} a(z,\omega _1 v)+\sum _{T\in {\mathcal T}_{D_2}}(z_0\nabla \omega _1,\nabla v_0)_T+\sum _{T\in {\mathcal T}_{D_2}}(z_0,\nabla \cdot (v_0\nabla \omega _1))_T\nonumber \\&+\sum _{T\in {\mathcal T}_{D_2}}(\ell _T(z,v,\omega _1)-{\langle }z_0,v_0\nabla \omega _1\cdot \mathbf{n}{\rangle }_{\partial {T}})+\sum _{T\in {\mathcal T}_{D_2}} \zeta _T(z,v,\omega _1)\nonumber \\= & {} I_1+I_2+I_3+I_4+I_5. \end{aligned}$$

(A.1)

Below, we obtain estimates for each $I_i$, $1\le i\le 5$.

Recall $\mathrm{supp}(\omega _1v)\subseteq D_1$. Then, using (2.11), (2.13), (2.14), (4.3), (3.3), and the inverse inequality, we have

$$\begin{aligned} |I_1|= & {} |a(z,\omega _1 v)|=|a(Q_hu-u_h, \omega _1v-Q_h(\omega _1v))+a(Q_hu-u_h, Q_h(\omega _1v))|\nonumber \\\le & {} |\!|\!| Q_hu-u_h|\!|\!|_{\tilde{D}_2} |\!|\!|\omega _1v-Q_h(\omega _1v)|\!|\!|_{\tilde{D}_2}+ Ch^t\Vert u\Vert _{t+1,D_2}|\!|\!| v|\!|\!|_{D_1}\nonumber \\\le & {} Ch|\!|\!| z|\!|\!|_{\tilde{D}_2}|\!|\!| v|\!|\!|_{\tilde{D}_2}+Ch^t\Vert u\Vert _{t+1,D_2}|\!|\!| v|\!|\!|_{\tilde{D}_2}\nonumber \\\le & {} C(h^t\Vert u\Vert _{t+1,D_2}+\Vert z_0\Vert _{0, \tilde{D}_2}+{h}s_{\tilde{D}_2}(z,z)^{1/2})|\!|\!| v|\!|\!|_{\tilde{D}_2}. \end{aligned}$$

(A.2)

Using the boundedness of $\omega _1$, (3.2), and the Cauchy-Schwarz inequality, we have

$$\begin{aligned} |I_2|= & {} \left| \sum _{T\in {\mathcal T}_{D_2}}(z_0\nabla \omega _1,\nabla v_0)_T\right| \le C\Vert z\Vert _{0,\tilde{D}_2}|\!|\!| v|\!|\!|_{\tilde{D}_2} \end{aligned}$$

(A.3)

and

$$\begin{aligned} |I_3|= & {} \left| \sum _{T\in {\mathcal T}_{D_2}}(z_0,\nabla \cdot (v_0\nabla \omega _1))_T\right| \le C\Vert z\Vert _{0, \tilde{D}_2}|\!|\!| v|\!|\!|_{\tilde{D}_2}. \end{aligned}$$

(A.4)

Now, we estimate $I_4$ and $I_5$. Note that $v_b=0$ on $\partial \tilde{D}_2$. Therefore,

$$\begin{aligned} \sum _{T\in {\mathcal T}_{D_2}}{\langle }z_b, v_b\nabla \omega _1\cdot \mathbf{n}{\rangle }_{\partial {T}}=0. \end{aligned}$$

Then, we have

$$\begin{aligned} I_4= & {} \sum _{T\in {\mathcal T}_{D_2}}({\langle }z_0-z_b, \mathbb {Q}_h\nabla (\omega _1v_0)\cdot \mathbf{n}-\omega _1\nabla _wv\cdot \mathbf{n}{\rangle }_{\partial {T}}-{\langle }z_0,v_0\nabla \omega _1\cdot \mathbf{n}{\rangle }_{\partial {T}})\nonumber \\= & {} \sum _{T\in {\mathcal T}_{D_2}}({\langle }z_0-z_b, (\mathbb {Q}_h(\omega _1\nabla v_0)-\omega _1\nabla _wv)\cdot \mathbf{n}{\rangle }_{\partial {T}}\nonumber \\&+{\langle }z_0-z_b, (\mathbb {Q}_h (v_0\nabla \omega _1)-v_0\nabla \omega _1)\cdot \mathbf{n}{\rangle }_{\partial {T}}\nonumber \\&+{\langle }z_0-z_b,(v_0-v_b)\nabla \omega _1\cdot \mathbf{n}{\rangle }_{\partial {T}}-{\langle }v_0-v_b,z_0\nabla \omega _1\cdot \mathbf{n}{\rangle }_{\partial {T}})\nonumber \\= & {} A_1+A_2+A_3-A_4. \end{aligned}$$

(A.5)

By the Cauchy-Schwarz inequality, the trace inequality (3.1), the boundedness of $\omega _1$, (2.5), the inverse inequality, and (3.2), we have

$$\begin{aligned} |A_2+A_3|\le & {} \sum _{T\in \mathcal T_{D_2}}\Vert z_0-z_b\Vert _{0,\partial T}(\Vert (\mathbb {Q}_h (v_0\nabla \omega _1)-v_0\nabla \omega _1)\cdot \mathbf{n}\Vert _{0,\partial T}\nonumber \\&+\Vert (v_0-v_b)\nabla \omega _1\cdot \mathbf{n}\Vert _{0, \partial T})\nonumber \\\le & {} C\left( \sum _{T\in \mathcal T_{D_2}}\Vert z_0-z_b\Vert _{0,\partial T}^2\right) ^{1/2} \left( \sum _{T\in \mathcal T_{D_2}}\Vert (\mathbb {Q}_h (v_0\nabla \omega _1)-v_0\nabla \omega _1)\cdot \mathbf{n}\Vert _{0,\partial T}^2\right. \nonumber \\&\left. +\Vert (v_0-v_b)\nabla \omega _1\cdot \mathbf{n}\Vert ^2_{0,\partial T}\right) ^{1/2}\nonumber \\\le & {} C \left( \sum _{T\in \mathcal T_{D_2}}\Vert z_0-z_b\Vert _{0,\partial T}^2\right) ^{1/2} \left( \sum _{T\in \mathcal T_{D_2}}(h^{-1}\Vert \mathbb {Q}_h (v_0\nabla \omega _1)-v_0\nabla \omega _1\Vert ^2_{0, T}\right. \nonumber \\&\left. +h|\mathbb {Q}_h (v_0\nabla \omega _1)-v_0\nabla \omega _1|_{1, T}^2 +\Vert v_0-v_b\Vert ^2_{0,\partial T}\right) ^{1/2}\nonumber \\\le & {} C\left( \sum _{T\in \mathcal T_{D_2}}\Vert z_0-z_b\Vert _{0, \partial T}^2\right) ^{1/2}\left( \sum _{T\in \mathcal T_{D_2}}h\Vert \nabla v_0\Vert _{0, T}^2+\Vert v_0-v_b\Vert ^2_{0,\partial T}\right) ^{1/2}\nonumber \\\le & {} C hs_{\tilde{D}_2}(z,z)^{1/2}|\!|\!| v|\!|\!|_{\tilde{D}_2}. \end{aligned}$$

(A.6)

Similarly, for $A_4$, we have

$$\begin{aligned} |A_4|\le & {} C\left( \sum _{T\in \mathcal T_{D_2}}\Vert v_0-v_b\Vert ^2_{0, \partial T}\right) ^{1/2}\left( \sum _{T\in \mathcal T_{D_2}} h^{-1}\Vert z_0\nabla \omega _1\Vert ^2_{0,T}+h|z_0\nabla \omega _1|^2_{1,T}\right) ^{1/2}\nonumber \\\le & {} C\left( \sum _{T\in \mathcal T_{D_2}}h^{-1}\Vert v_0-v_b\Vert _{0,\partial T}\right) ^{1/2}\left( \sum _{T\in \mathcal T_{D_2}} \Vert z_0\Vert ^2_{0,T}\right) ^{1/2}\le C\Vert z_0\Vert _{0, \tilde{D}_2}|\!|\!| v|\!|\!|_{\tilde{D}_2}.\nonumber \\ \end{aligned}$$

(A.7)

Now, for $A_1$, we first have

$$\begin{aligned} A_1= & {} \sum _{T\in {\mathcal T}_{D_2}}{\langle }z_0-z_b, (\mathbb {Q}_h(\omega _1\nabla v_0)-\omega _1\nabla _wv)\cdot \mathbf{n}{\rangle }_{\partial {T}}\nonumber \\= & {} \sum _{T\in {\mathcal T}_{D_2}}{\langle }z_0-z_b, \mathbb {Q}_h(\omega _1\nabla v_0-\omega _1\nabla _wv)\cdot \mathbf{n}{\rangle }_{\partial {T}}\nonumber \\&+\sum _{T\in {\mathcal T}_{D_2}}{\langle }z_0-z_b, (\mathbb {Q}_h(\omega _1\nabla _wv)-\omega _1\nabla _wv)\cdot \mathbf{n}{\rangle }_{\partial {T}}\nonumber \\= & {} A_{11}+A_{12}. \end{aligned}$$

(A.8)

For the second term $A_{12}$, using the Cauchy-Schwarz inequality, the trace inequality (3.1), the boundedness of $\omega _1$, (2.5), the fact $\partial ^\alpha (\nabla _wv)=0$ on T for $|\alpha |=k$, and the inverse inequality, we obtain

$$\begin{aligned} |A_{12}|\le & {} \sum _{T\in \mathcal T_{D_2}}\Vert z_0-z_b\Vert _{0,\partial T}\Vert (\mathbb {Q}_h(\omega _1\nabla _wv)-\omega _1\nabla _wv)\cdot \mathbf{n}\Vert _{0, \partial T}\nonumber \\\le & {} C\Bigg (\sum _{T\in \mathcal T_{D_2}}\Vert z_0-z_b\Vert _{0,\partial T}^2\Bigg )^{1/2}\Bigg (\sum _{T\in \mathcal T_{D_2}}h^{-1}\Vert \mathbb {Q}_h(\omega _1\nabla _wv)-\omega _1\nabla _wv \Vert _{0,T}^2 \nonumber \\&+h| \mathbb {Q}_h(\omega _1\nabla _wv)-\omega _1\nabla _wv |_{1, T}^2\Bigg )^{1/2}\nonumber \\\le & {} C\Bigg (\sum _{T\in \mathcal T_{D_2}}\Vert z_0-z_b\Vert _{0,\partial T}^2\Bigg )^{1/2}\Bigg (\sum _{T\in \mathcal T_{D_2}}h^{2k-1}\Vert \nabla _w v\Vert ^2_{k-1, T}\Bigg )^{1/2}\nonumber \\\le & {} C\Bigg (\sum _{T\in \mathcal T_{D_2}}\Vert z_0-z_b\Vert _{0,\partial T}^2\Bigg )^{1/2}\Bigg (\sum _{T\in \mathcal T_{D_2}}h\Vert \nabla _w v\Vert ^2_{0, T}\Bigg )^{1/2} \le Chs_{\tilde{D}_2}(z, z)^{1/2}|\!|\!| v|\!|\!|_{\tilde{D}_2}.\nonumber \\ \end{aligned}$$

(A.9)

The estimate on $A_{11}$ is more involved. Note that by (4.14), we have

$$\begin{aligned} A_{11}= & {} \sum _{T\in {\mathcal T}_{D_2}}{\langle }z_0-z_b, \mathbb {Q}_h(\omega _1\nabla v_0-\omega _1\nabla _wv)\cdot \mathbf{n}{\rangle }_{\partial {T}}\\= & {} \sum _{T\in {\mathcal T}_{D_2}}(\nabla z_0-\nabla _w z, \mathbb {Q}_h(\omega _1\nabla v_0-\omega _1\nabla _wv))_T\\= & {} \sum _{T\in {\mathcal T}_{D_2}}(\mathbb {Q}_h(\omega _1\nabla z_0-\omega _1\nabla _w z), \nabla v_0-\nabla _wv)_T\\= & {} \sum _{T\in {\mathcal T}_{D_2}}{\langle }v_0-v_b, \mathbb {Q}_h(\omega _1\nabla z_0-\omega _1\nabla _w z)\cdot \mathbf{n}{\rangle }_{\partial {T}}. \end{aligned}$$

Therefore, considering $A_{11}$ and $I_5$ together, by the Cauchy-Schwarz inequality, the trace inequality (3.1), the boundedness of $\omega _1$, and (2.5), we obtain

$$\begin{aligned} |A_{11}+I_5|&= \left| \sum _{T\in {\mathcal T}_{D_2}}{\langle }v_0-v_b, \mathbb {Q}_h(\omega _1\nabla z_0-\omega _1\nabla _w z)\cdot \mathbf{n}{\rangle }_{\partial {T}}\right. \\&\left. \quad \quad +\sum _{T\in {\mathcal T}_{D_2}}{\langle }v_0-v_b, (\omega _1\nabla _wz-\mathbb {Q}_h(\omega _1\nabla z_0))\cdot \mathbf{n}{\rangle }_{\partial {T}}\right. \\&\left. \quad \quad +\sum _{T\in {\mathcal T}_{D_2}}{\langle }v_0-v_b, \mathbb {Q}_h (z_0\nabla \omega _1)\cdot \mathbf{n}{\rangle }_{\partial {T}}\right| \\&=\left| \sum _{T\in {\mathcal T}_{D_2}}{\langle }v_0-v_b, (\omega _1\nabla _wz-\mathbb {Q}_h(\omega _1\nabla _w z))\cdot \mathbf{n}{\rangle }_{\partial {T}}\right. \\&\left. \quad \quad +\sum _{T\in {\mathcal T}_{D_2}}{\langle }v_0-v_b, \mathbb {Q}_h (z_0\nabla \omega _1)\cdot \mathbf{n}{\rangle }_{\partial {T}}\right| \\&\le \sum _{T\in {\mathcal T}_{D_2}}\Vert v_0-v_b\Vert _{0, \partial T}\Vert \omega _1\nabla _wz-\mathbb {Q}_h(\omega _1\nabla _w z)\Vert _{0, \partial T} \\&\quad \quad +\sum _{T\in {\mathcal T}_{D_2}}\Vert v_0-v_b\Vert _{0,\partial T}\Vert \mathbb {Q}_h (z_0\nabla \omega _1)\Vert _{0, {\partial {T}}} \\&\le C\left( \sum _{T\in \mathcal T_{D_2}}\Vert v_0-v_b\Vert _{0,\partial T}^2\right) ^{1/2} \left( \sum _{T\in \mathcal T_{D_2}}h^{-1}\Vert \omega _1\nabla _wz-\mathbb {Q}_h(\omega _1\nabla _w z)\Vert _{0, T}^2\right. \\&\left. \quad \quad +h|\omega _1\nabla _wz-\mathbb {Q}_h(\omega _1\nabla _w z)|_{1, T}^2\right) ^{1/2} \\&\quad \quad + \left( \sum _{T\in \mathcal T_{D_2}}\Vert v_0-v_b\Vert _{0,\partial T}^2\right) ^{1/2} \left( \sum _{T\in \mathcal T_{D_2}}h^{-1}\Vert \mathbb {Q}_h (z_0\nabla \omega _1)\Vert _{0,T}^2+h|\mathbb {Q}_h (z_0\nabla \omega _1)|_{1, T}^2\right) ^{1/2} \\&\le C \left( \sum _{T\in \mathcal T_{D_2}}\Vert v_0-v_b\Vert _{0, \partial T}^2\right) ^{1/2}\Bigg (\left( \sum _{T\in \mathcal T_{D_2}}h^{-1+2k}\Vert \omega _1\nabla _wz\Vert _{k, T}^2\right) ^{1/2} \\&\quad \quad +\,\Bigg (\sum _{T\in \mathcal T_{D_2}}h^{-1}\Vert z_0\nabla \omega _1\Vert _{0,T}^2+h| z_0\nabla \omega _1|_{1, T}^2)^{1/2}\Bigg ). \end{aligned}$$

Thus, by the inverse inequality, (3.3), and the fact $\partial ^\alpha (\nabla _wz)=0$ on T for $|\alpha |=k$, we have

$$\begin{aligned} |A_{11}+I_5|\le & {} C \left( \sum _{T\in \mathcal T_{D_2}}\Vert v_0-v_b\Vert _{0,\partial T}^2\right) ^{1/2}\Bigg (\left( \sum _{T\in \mathcal T_{D_2}}h\Vert \nabla _wz\Vert _{0,T}^2\right) ^{1/2}\nonumber \\&+ \left( \sum _{T\in \mathcal T_{D_2}}h^{-1}\Vert z_0\Vert _{0,T}^2\right) ^{1/2}\Bigg )\nonumber \\\le & {} C|\!|\!| v|\!|\!|_{\tilde{D}_2}(\Vert z_0\Vert _{0, \tilde{D}_2}+hs_{\tilde{D}_2}(z, z)^{1/2}). \end{aligned}$$

(A.10)

It follows from the Cauchy-Schwarz inequality, the trace inequality, (2.3), (2.4), and the inverse inequality that

$$\begin{aligned} s_{\tilde{D}_2}(z,z)= & {} \sum _{T\in {\mathcal T}_{D_2}}s_T(Q_hu-u_h,z)= \sum _{T\in {\mathcal T}_{D_2}}s_T(Q_hu-u,z)-\sum _{T\in {\mathcal T}_{D_2}}s_T(u_h,z)\\\le & {} C \Bigg (\sum _{T\in {\mathcal T}_{D_2}}\left( h^{-2}\Vert u-Q_0u_0\Vert ^2_{0, T}+|u-Q_0u_0|_{1, T}^2+h^{-1}\Vert u-Q_bu_b\Vert _{0,{\partial {T}}}^2\right) ^{1/2}\\&+s_{\tilde{D}_2}(u_h,u_h)^{1/2}\Bigg )s_{\tilde{D}_2}(z,z)^{1/2}\\\le & {} C (h^t\Vert u\Vert _{t+1,D_2}+s_{\tilde{D}_2}(u_h,u_h)^{1/2})s_{\tilde{D}_2}(z,z)^{1/2}, \end{aligned}$$

which implies

$$\begin{aligned} s_{\tilde{D}_2}(z,z)^{1/2}\le & {} C (h^t\Vert u\Vert _{t+1,D_2}+s_{D_2}(u_h,u_h)^{1/2}). \end{aligned}$$

(A.11)

In addition, the triangle inequality implies

$$\begin{aligned} \Vert z_0\Vert _{0,\tilde{D}_2}\le \Vert Q_hu_0-u\Vert _{0,\tilde{D}_2}+\Vert u-u_0\Vert _{0,\tilde{D}_2}\le Ch^{t+1}\Vert u\Vert _{t+1, D_2}+\Vert u-u_0\Vert _{0,D_2}. \end{aligned}$$

(A.12)

Combining all the estimates (A.1)–(A.12), we have completed the proof for (4.15). $\square$