A Hybridizable Discontinuous Galerkin Method for Kirchhoff Plates

Abstract

With the introduction of numerical traces respectively related to the normal bending moment, the twisting moment and the effective transverse shear force, and based on the Hermann–Miyoshi formulation, this paper proposes a hybridizable discontinuous Galerkin (HDG) method for Kirchhoff plate bending problems. The piecewise polynomials of degrees $k-1$ and k are used to approximate the moment and the deflection, respectively. The optimal and superconvergent error estimates are derived under minimal regularity assumptions on the exact solution. The key ingredients in the analysis include the derivation of a discrete inf-sup condition and some local lower bound estimates of a posteriori error analysis. The significant feature of the HDG method is superconvergence as well as the low number of globally coupled degrees of freedom associated with Lagrange multipliers. Furthermore, a new discrete deflection is constructed by postprocessing the solution of the HDG method, which superconverges to the deflection with order $k+1$ in broken $H^1$ norm. Finally, some numerical results are shown to demonstrate the theoretical results.

Appendix

Proof of Lemma 3

Let ${\hat{K}}$ be a reference simplex and $F_K$ an affine mapping from ${\hat{K}}$ onto K. Denote by ${\hat{v}}$ the pull back of v by the mapping $F_K$, i.e. ${\hat{v}}:=v\circ F_K$. Since all norms are equivalent on a finite-dimensional vector space, we know

$$\begin{aligned} \Vert {\hat{v}}\Vert _{0,{\hat{K}}}^2\lesssim \sum _{{\hat{e}}\in {\mathcal {E}}_{{\hat{h}}}({\hat{K}})}\sum _{{\hat{\delta }}\in \partial {\hat{e}}}{\hat{v}}^2({\hat{\delta }}) + \sum _{{\hat{e}}\in {\mathcal {E}}_{{\hat{h}}}({\hat{K}})}\left\| P_{{\hat{e}}}^{k-2}{\hat{v}}\right\| _{0,{\hat{e}}}^2 + \left\| P_{{\hat{K}}}^{k-3}{\hat{v}}\right\| _{0,{\hat{K}}}^2. \end{aligned}$$

Then using the scaling argument we find

$$\begin{aligned} h_K^{-2}\Vert v\Vert _{0,K}^2\lesssim&\Vert {\hat{v}}\Vert _{0,{\hat{K}}}^2 \lesssim \sum _{{\hat{e}}\in {\mathcal {E}}_{{\hat{h}}}({\hat{K}})}\sum _{{\hat{\delta }}\in \partial {\hat{e}}}{\hat{v}}^2({\hat{\delta }}) + \sum _{{\hat{e}}\in {\mathcal {E}}_{{\hat{h}}}({\hat{K}})}\left\| P_{{\hat{e}}}^{k-2}{\hat{v}}\right\| _{0,{\hat{e}}}^2 + \left\| P_{{\hat{K}}}^{k-3}{\hat{v}}\right\| _{0,{\hat{K}}}^2 \\ \lesssim&\sum _{e\in {\mathcal {E}}_h(K)}\sum _{\delta \in \partial e}v^2(\delta ) + h_K^{-1}\sum _{e\in {\mathcal {E}}_h(K)}\left\| P_e^{k-2}v\right\| _{0,e}^2 + h_K^{-2}\left\| P_K^{k-3}v\right\| _{0,K}^2, \end{aligned}$$

as required. $\square$

Proof of Lemma 4

According to the definition of ${\widetilde{P}}_h$, it is easy to see that

$$\begin{aligned} P_K^{k-3}\big (v-{\widetilde{P}}_hv\big )=0, \quad P_e^{k-2}\big (\{v\}-{\widetilde{P}}_hv\big )=0\quad \forall ~e\in {\mathcal {E}}_h^i(K). \end{aligned}$$

(58)

For each $e\in {\mathcal {E}}_h(K)$, there holds

$$\begin{aligned} \sum _{\delta \in \partial e}\big (v-{\widetilde{P}}_hv\big )^2(\delta )\lesssim&\sum _{\delta \in \partial e}\sum _{e^{\prime }\in \partial ^{-1}\delta }\big ([v]|_{e^{\prime }}\big )^2(\delta )=\sum _{\delta \in \partial e}\sum _{e^{\prime }\in \partial ^{-1}\delta }\big ([v-\beta ]|_{e^{\prime }}\big )^2(\delta ) \\ \lesssim&\sum _{\delta \in \partial e}\sum _{K^{\prime }\in \partial ^{-2}\delta }(v|_{K^{\prime }}-\beta )^2(\delta ). \end{aligned}$$

By (58), we have

$$\begin{aligned} \big \Vert P_e^{k-2}\big (v-{\widetilde{P}}_hv\big )\big \Vert _{0,e}^2\lesssim&\big \Vert P_e^{k-2}[v]\big \Vert _{0,e}^2=\big \Vert P_e^{k-2}[v-\chi ]\big \Vert _{0,e}^2 \\ \lesssim&\sum _{K^{\prime }\in \partial ^{-1}e}\big \Vert P_e^{k-2}(v|_{K^{\prime }}-\chi )\big \Vert _{0,e}^2. \end{aligned}$$

It follows from Lemma 3 that

$$\begin{aligned} \Vert v-{\widetilde{P}}_hv\Vert _{0,K}^2\lesssim&h_K^2\sum _{e\in {\mathcal {E}}_h(K)}\sum _{\delta \in \partial e}\big (v-{\widetilde{P}}_hv\big )^2(\delta ) + h_K\sum _{e\in {\mathcal {E}}_h(K)}\big \Vert P_e^{k-2}\big (v-{\widetilde{P}}_hv\big )\big \Vert _{0,e}^2 \\&+ \big \Vert P_K^{k-3}\big (v-{\widetilde{P}}_hv\big )\big \Vert _{0,K}^2. \end{aligned}$$

Now, the combination of last three inequalities and (58) leads to the required result readily. $\square$

Proof of Lemma 5

Let the set $\partial ^{-2}\delta$ consist of elements $K_1, K_2, \cdots , K_{\#(\partial ^{-2}\delta )}$. With $\delta$ we associate a vertex bubble function given by

$$\begin{aligned} b_{\delta }:=\left\{ \begin{array}{ll} \prod \limits _{i=1}^{\#(\partial ^{-2}\delta )}\lambda _{K_i}&{} \text { in } \omega _{\delta }, \\ 0 &{}\text { in } {\varOmega }\backslash \omega _{\delta }, \end{array} \right. \end{aligned}$$

where $\lambda _{K_i}$ is the barycentric coordinate of $K_i$ associated with $\delta$. It is evident that $b_{\delta }(\delta )=1$ and

$$\begin{aligned}&\sum _{K\in \partial ^{-2}\delta }h_K^{-1}\Vert b_{\delta }^2\Vert _{0,K} + \sum _{K\in \partial ^{-2}\delta }h_K|b_{\delta }^2|_{2,K} + \sum _{e\in \partial ^{-1}\delta }h_e^{-1/2}\Vert b_{\delta }^2\Vert _{0,e} \nonumber \\&\quad + \sum _{e\in \partial ^{-1}\delta }h_e^{1/2}\Vert \partial _{{\varvec{n}}_e}(b_{\delta }^2)\Vert _{0,e}\lesssim 1. \end{aligned}$$

(59)

Let $\psi _{\delta }:=b_{\delta }^2\sum \nolimits _{e\in \partial ^{-1}\delta }\sum \nolimits _{K\in \partial ^{-1}e}\varepsilon (e,\delta )M_{nt_e}(\varvec{\tau })(\delta )\in H_0^2(\omega _{\delta })$. By integration by parts and recalling the definition of $Q_{{\varvec{n}}}(\varvec{\tau })$, it follows that

$$\begin{aligned}&\Big |\sum _{e\in \partial ^{-1}\delta }\sum _{K\in \partial ^{-1}e}\varepsilon (e,\delta )M_{nt_e}(\varvec{\tau })(\delta )\Big |^2 = \sum _{e\in \partial ^{-1}\delta }\sum _{K\in \partial ^{-1}e}\varepsilon (e,\delta )M_{nt_e}(\varvec{\tau })(\delta )\psi _{\delta } \\&\quad = \sum _{e\in \partial ^{-1}\delta }\sum _{K\in \partial ^{-1}e}\Big (\int _e\partial _{{\varvec{t}}}M_{nt}(\varvec{\tau })\psi _{\delta }\,ds + \int _eM_{nt}(\varvec{\tau })\partial _{{\varvec{t}}}\psi _{\delta }\,ds\Big ) \\&\quad = \sum _{e\in \partial ^{-1}\delta }\sum _{K\in \partial ^{-1}e}\Big (\int _eQ_{{\varvec{n}}}(\varvec{\tau })\psi _{\delta }\,ds - \int _{e}{\varvec{n}}\cdot (\varvec{\nabla }\cdot \varvec{\tau })\psi _{\delta }\,ds + \int _eM_{nt}(\varvec{\tau })\partial _{{\varvec{t}}}\psi _{\delta }\,ds\Big ) \\&\quad = \sum _{e\in \partial ^{-1}\delta }\sum _{K\in \partial ^{-1}e}\int _eQ_{{\varvec{n}}}(\varvec{\tau })\psi _{\delta }\,ds \\&\qquad +\sum _{K\in \partial ^{-2}\delta }\Big (- \int _{\partial K}{\varvec{n}}\cdot (\varvec{\nabla }\cdot \varvec{\tau })\psi _{\delta }\,ds + \int _{\partial K}M_{nt}(\varvec{\tau })\partial _{{\varvec{t}}}\psi _{\delta }\,ds\Big ). \end{aligned}$$

On the other hand, by integration by parts again,

$$\begin{aligned} \int _{\partial K}{\varvec{n}}\cdot (\varvec{\nabla }\cdot \varvec{\tau })\psi _{\delta }\,ds=&\int _{K}\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })\psi _{\delta }\,dx + \int _{K}(\varvec{\nabla }\cdot \varvec{\tau })\cdot \varvec{\nabla }\psi _{\delta }\,dx \\ =&\int _{K}\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })\psi _{\delta }\,dx + \int _{K}\varvec{\tau }:{\mathcal {K}}(\psi _{\delta })\,dx \\&+ \int _{\partial K}M_{n}(\varvec{\tau })\partial _{{\varvec{n}}}\psi _{\delta }\,ds + \int _{\partial K}M_{nt}(\varvec{\tau })\partial _{{\varvec{t}}}\psi _{\delta }\,ds. \end{aligned}$$

Noting that $\psi _{\delta }\in H_0^2(\omega _{\delta })$, we obtain from the last two equalities that

$$\begin{aligned}&\Big |\sum _{e\in \partial ^{-1}\delta }\sum _{K\in \partial ^{-1}e}\varepsilon (e,\delta )M_{nt_e}(\varvec{\tau })(\delta )\Big |^2 \\&\quad = \sum _{e\in \partial ^{-1}\delta }\int _e[Q_{{\varvec{n}}_e}(\varvec{\tau })]\psi _{\delta }\,ds - \sum _{e\in \partial ^{-1}\delta }\int _{e}[M_n(\varvec{\tau })]\partial _{{\varvec{n}}_e}\psi _{\delta }\,ds \\&\qquad -\sum _{K\in \partial ^{-2}\delta }\int _{K}\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })\psi _{\delta }\,dx - \sum _{K\in \partial ^{-2}\delta }\int _{K}\varvec{\tau }:{\mathcal {K}}(\psi _{\delta })\,dx, \end{aligned}$$

which together with (16) and the Cauchy–Schwarz inequality implies

$$\begin{aligned}&\Big |\sum _{e\in \partial ^{-1}\delta }\sum _{K\in \partial ^{-1}e}\varepsilon (e,\delta )M_{nt_e}(\varvec{\tau })(\delta )\Big |^2 \\&\quad \lesssim -\sum _{K\in \partial ^{-2}\delta }\int _{K}(f+\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau }))\psi _{\delta }\,dx + \sum _{K\in \partial ^{-2}\delta }\int _{K}(\varvec{\sigma }-\varvec{\tau }):{\mathcal {K}}(\psi _{\delta })\,dx \\&\qquad - \sum _{e\in \partial ^{-1}\delta }\int _{e}[M_n(\varvec{\tau })]\partial _{{\varvec{n}}_e}\psi _{\delta }\,ds +\sum _{e\in \partial ^{-1}\delta }\int _e[Q_{{\varvec{n}}_e}(\varvec{\tau })]\psi _{\delta }\,ds \\&\quad \lesssim \sum _{K\in \partial ^{-2}\delta }\left( \Vert f+\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })\Vert _{0,K}\Vert \psi _{\delta }\Vert _{0,K} +\Vert \varvec{\sigma }-\varvec{\tau }\Vert _{0,K}|\psi _{\delta }|_{2,K}\right) \\&\qquad + \sum _{e\in \partial ^{-1}\delta }\left( \Vert [M_n(\varvec{\tau })]\Vert _{0,e}\Vert \partial _{{\varvec{n}}_e}\psi _{\delta }\Vert _{0,e} + \Vert [Q_{{\varvec{n}}_e}(\varvec{\tau })]\Vert _{0,e}\Vert \psi _{\delta }\Vert _{0,e}\right) . \end{aligned}$$

Thus by the definition of $\psi _{\delta }$ and (59),

$$\begin{aligned}&\Big |\sum _{e\in \partial ^{-1}\delta }\sum _{K\in \partial ^{-1}e}\varepsilon (e,\delta )M_{nt_e}(\varvec{\tau })(\delta )\Big | \\&\quad \lesssim \sum _{K\in \partial ^{-2}\delta }\left( \Vert f+\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })\Vert _{0,K}\Vert b_{\delta }^2\Vert _{0,K} +\Vert \varvec{\sigma }-\varvec{\tau }\Vert _{0,K}|b_{\delta }^2|_{2,K}\right) \\&\qquad + \sum _{e\in \partial ^{-1}\delta }\left( \Vert [M_n(\varvec{\tau })]\Vert _{0,e}\Vert \partial _{{\varvec{n}}_e}(b_{\delta }^2)\Vert _{0,e} + \Vert [Q_{{\varvec{n}}_e}(\varvec{\tau })]\Vert _{0,e}\Vert b_{\delta }^2\Vert _{0,e}\right) \\&\quad \lesssim \sum _{K\in \partial ^{-2}\delta }h_K^{-1}\left( h_K^{2}\Vert f+\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })\Vert _{0,K} +\Vert \varvec{\sigma }-\varvec{\tau }\Vert _{0,K}\right) \\&\qquad + \sum _{e\in \partial ^{-1}\delta }h_e^{-1}\left( h_e^{1/2}\Vert [M_n(\varvec{\tau })]\Vert _{0,e} + h_e^{3/2}\Vert [Q_{{\varvec{n}}_e}(\varvec{\tau })]\Vert _{0,e}\right) . \end{aligned}$$

Finally, the estimate (20) is a direct consequence of the last inequality and (17)–(19). $\square$

Proof of Lemma 6

Let $b_K:=\lambda _1\lambda _2\lambda _3$ and $\psi _K:=(\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })-{\varDelta }(P_hu-{\widetilde{P}}_hu_h))b_K^2$, where $\lambda _i$$(i=1,2,3)$ are the usual barycentric coordinates with respect to K. Then $\psi _K\in H_0^2(K)$ and

$$\begin{aligned} \Vert \psi _K\Vert _{0,K}\eqsim \Vert \varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })-{\varDelta }(P_hu-{\widetilde{P}}_hu_h)\Vert _{0,K}. \end{aligned}$$

Using integration by parts and (27), we know

$$\begin{aligned}&\int _K(\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })-{\varDelta }(P_hu-{\widetilde{P}}_hu_h))\psi _K\,dx \\&\quad = -\int _K(\varvec{\nabla }\cdot \varvec{\tau }-\varvec{\nabla }(P_hu-{\widetilde{P}}_hu_h))\cdot \varvec{\nabla }\psi _K\,dx \\&\quad =-\int _K(\varvec{\nabla }\cdot (\varvec{\tau }-\widetilde{\varvec{\sigma }}))\cdot \varvec{\nabla }\psi _K\,dx =\int _K(\widetilde{\varvec{\sigma }}-\varvec{\tau }):{\mathcal {K}}(\psi _K)\,dx. \end{aligned}$$

Thus, we have from the scaling argument, the Cauchy–Schwarz inequality and the inverse inequality that

$$\begin{aligned}&\Vert \varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })-{\varDelta }(P_hu-{\widetilde{P}}_hu_h)\Vert _{0,K}^2 \\&\quad \lesssim \int _K(\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })-{\varDelta }(P_hu-{\widetilde{P}}_hu_h))\psi _K\,dx = \int _K(\widetilde{\varvec{\sigma }}-\varvec{\tau }):{\mathcal {K}}(\psi _K)\,dx \\&\quad \lesssim h_K^{-2}\Vert \widetilde{\varvec{\sigma }}-\varvec{\tau }\Vert _{0,K}\Vert \psi _K\Vert _{0,K}\lesssim h_K^{-2}\Vert \widetilde{\varvec{\sigma }}-\varvec{\tau }\Vert _{0,K}\Vert \varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })-{\varDelta }(P_hu-{\widetilde{P}}_hu_h)\Vert _{0,K}. \end{aligned}$$

The proof is complete. $\square$

Proof of Lemma 7

Let $K_1$ and $K_2$ be two elements in ${\mathcal {T}}_h$ sharing the common edge e. With e, we associate an edge bubble function given by (cf. [28, 42])

$$\begin{aligned} b_e:=\left\{ \begin{array}{ll} \lambda _{K_1,1}\lambda _{K_1,2}\lambda _{K_2,1}\lambda _{K_2,2}&{} \text { in } \omega _e, \\ 0 &{}\text { in } {\varOmega }\backslash \omega _e, \end{array} \right. \end{aligned}$$

where $\lambda _{K_1,i}$ and $\lambda _{K_2,i}$ for $i=1, 2$ are barycentric coordinates of $K_1$ and $K_2$ associated with two end points of e, respectively. Suppose the straight line edge e lying in is determined by the linear equation ${\varvec{n}}_e\cdot {\varvec{x}}+C=0$ where C is a constant. By direct manipulation, we readily have

$$\begin{aligned} \left| {\varvec{n}}_e\cdot {\varvec{x}}+C\right| \lesssim h_e \quad \text { on } \omega _e. \end{aligned}$$

Set $J_{1,e}:=[M_n(\varvec{\tau })]|_e$. $E_h(J_{1,e})$ is defined by extending the jump $J_{1,e}$ to $\omega _e$ constantly along the normal to e. Take $\phi _e:=({\varvec{n}}_e\cdot {\varvec{x}}+C)b_e^2E_h(J_{1,e})\in H_0^2(\omega _e)$. It is easy to check that

$$\begin{aligned}&\phi _e=0, \quad \partial _{{\varvec{n}}_e}\phi _e=b_e^2E_h(J_{1,e})\partial _{{\varvec{n}}_e}({\varvec{n}}_e\cdot {\varvec{x}}+C)=b_e^2E_h(J_{1,e})=b_e^2J_{1,e} \quad \text { on } e, \end{aligned}$$

(60)

$$\begin{aligned}&\sum _{K\in \partial ^{-1}e}\Vert \phi _e\Vert _{0,K}\lesssim h_e\sum _{K\in \partial ^{-1}e}\Vert E_h(J_{1,e})\Vert _{0,K}\lesssim h_e^{3/2}\Vert J_{1,e}\Vert _{0,e}. \end{aligned}$$

(61)

According to standard scaling argument and (60),

$$\begin{aligned} \Vert J_{1,e}\Vert _{0,e}^2\lesssim&\int _eb_e^2J_{1,e}^2\,ds = \int _eJ_{1,e}\partial _{{\varvec{n}}_e}\phi _e\,ds = \int _e[M_n(\varvec{\tau })]\partial _{{\varvec{n}}_e}\phi _e\,ds \\ =&\sum _{K\in \partial ^{-1}e}\int _{\partial K}M_n(\varvec{\tau })\partial _{{\varvec{n}}}\phi _e\,ds =\sum _{K\in \partial ^{-1}e}\int _{\partial K} (\varvec{\tau }{\varvec{n}})\cdot \varvec{\varvec{\nabla }}\phi _e\,ds. \end{aligned}$$

Using integration by parts and (28), we can see that

$$\begin{aligned} \Vert J_{1,e}\Vert _{0,e}^2\lesssim&-\sum _{K\in \partial ^{-1}e}\int _K\varvec{\tau }:{\mathcal {K}}(\phi _e)\,dx + \sum _{K\in \partial ^{-1}e}\int _K(\varvec{\nabla }\cdot \varvec{\tau })\cdot \varvec{\nabla }\phi _e\,dx \\ \lesssim&\sum _{K\in \partial ^{-1}e}\int _K(\widetilde{\varvec{\sigma }}-\varvec{\tau }):{\mathcal {K}}(\phi _e)\,dx \\&+ \sum _{K\in \partial ^{-1}e}\int _K(\varvec{\nabla }\cdot \varvec{\tau }-\varvec{\nabla }(P_hu-{\widetilde{P}}_hu_h))\cdot \varvec{\nabla }\phi _e\,dx. \end{aligned}$$

Furthermore, using integration by parts and (60), we have from the above inequality that

$$\begin{aligned} \Vert J_{1,e}\Vert _{0,e}^2 \lesssim&\sum _{K\in \partial ^{-1}e}\int _K(\widetilde{\varvec{\sigma }}-\varvec{\tau }):{\mathcal {K}}(\phi _e)\,dx \\&-\sum _{K\in \partial ^{-1}e}\int _K(\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })-{\varDelta }(P_hu-{\widetilde{P}}_hu_h))\phi _e\,dx. \end{aligned}$$

Therefore, by the Cauchy–Schwarz inequality, the inverse inequality, Lemma 6 and (61), we arrive at

$$\begin{aligned}&\Vert J_{1,e}\Vert _{0,e}^2 \\&\quad \lesssim \sum _{K\in \partial ^{-1}e}\big (\Vert \widetilde{\varvec{\sigma }}-\varvec{\tau }\Vert _{0,K}|\phi _e|_{2,K} + \Vert \varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })-{\varDelta }(P_hu-{\widetilde{P}}_hu_h)\Vert _{0,K}\Vert \phi _e\Vert _{0,K} \big ) \\&\quad \lesssim \sum _{K\in \partial ^{-1}e}h_K^{-2}\Vert \widetilde{\varvec{\sigma }}-\varvec{\tau }\Vert _{0,K}\Vert \phi _e\Vert _{0,K} \lesssim h_e^{3/2}\Vert J_{1,e}\Vert _{0,e}\sum _{K\in \partial ^{-1}e}h_K^{-2}\Vert \widetilde{\varvec{\sigma }}-\varvec{\tau }\Vert _{0,K}, \end{aligned}$$

from which the required estimate follows readily. $\square$

Proof of Lemma 8

Let $J_{2,e}:=[Q_{{\varvec{n}}_e}(\varvec{\tau })]|_e$. $E_h(J_{2,e})$ is defined by extending the jump $J_{2,e}$ to $\omega _e$ constantly along the normal to e. Set $\psi _e:=b_e^2E_h(J_{2,e})\in H_0^2(\omega _e)$. It is easy to check that

$$\begin{aligned} \Vert \psi _e\Vert _{0,\omega _e}\lesssim h_e^{1/2}\Vert J_{2,e}\Vert _{0,e}. \end{aligned}$$

(62)

By the standard scaling argument and integration by parts, it follows that

$$\begin{aligned} \Vert J_{2,e}\Vert _{0,e}^2\lesssim&\int _eJ_{2,e}\psi _e\,ds=\int _e[Q_{{\varvec{n}}_e}(\varvec{\tau })]\psi _e\,ds =\sum _{K\in \partial ^{-1}e}\int _{\partial K}Q_{{\varvec{n}}}(\varvec{\tau })\psi _e\,ds \\ =&\sum _{K\in \partial ^{-1}e}\int _{\partial K}{\varvec{n}}\cdot (\varvec{\nabla }\cdot \varvec{\tau })\psi _e\,ds + \sum _{K\in \partial ^{-1}e}\int _{\partial K}\partial _{{\varvec{t}}}M_{nt}(\varvec{\tau })\psi _e\,ds \\ =&\sum _{K\in \partial ^{-1}e}\int _{K}\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })\psi _e\,dx + \sum _{K\in \partial ^{-1}e}\int _{K}(\varvec{\nabla }\cdot \varvec{\tau })\cdot \varvec{\nabla }\psi _e\,dx \\&-\sum _{K\in \partial ^{-1}e}\int _{\partial K}M_{nt}(\varvec{\tau })\partial _{{\varvec{t}}}\psi _e\,ds. \end{aligned}$$

Using integration by parts again and (28) with $v=\psi _e$ gives rise to

$$\begin{aligned} \Vert J_{2,e}\Vert _{0,e}^2\lesssim&\sum _{K\in \partial ^{-1}e}\int _{K}\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })\psi _e\,dx + \sum _{K\in \partial ^{-1}e}\int _{K}\varvec{\tau }:{\mathcal {K}}(\psi _e)\,dx \\&+\sum _{K\in \partial ^{-1}e}\int _{\partial K}M_{n}(\varvec{\tau })\partial _{{\varvec{n}}}\psi _e\,ds \\ \lesssim&\sum _{K\in \partial ^{-1}e}\int _{K}\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })\psi _e\,dx + \sum _{K\in \partial ^{-1}e}\int _{K}\varvec{\nabla }(P_hu-{\widetilde{P}}_hu_h)\cdot \varvec{\nabla }\psi _e\,dx \\&- \sum _{K\in \partial ^{-1}e}\int _{K}(\widetilde{\varvec{\sigma }}-\varvec{\tau }):{\mathcal {K}}(\psi _e)\,dx + \int _{e}[M_n(\varvec{\tau })]\partial _{{\varvec{n}}_e}\psi _e\,ds. \end{aligned}$$

By integration by parts and observing the fact that $\psi _e\in H_0^2(\omega _e)$ and $\widetilde{\varvec{\sigma }}\in {\varvec{H}}^{1}({\varOmega }, {\mathbb {S}})$, we know

$$\begin{aligned}&\Vert J_{2,e}\Vert _{0,e}^2 \\&\quad \lesssim \sum _{K\in \partial ^{-1}e}\int _{K}(\varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })-{\varDelta }(P_hu-{\widetilde{P}}_hu_h))\psi _e\,dx + \int _{e}[\partial _{{\varvec{n}}_e}(P_hu-{\widetilde{P}}_hu_h)]\psi _e\,ds \\&\qquad +\sum _{K\in \partial ^{-1}e}\int _{K}(\widetilde{\varvec{\sigma }}-\varvec{\tau }):\varvec{\nabla }^2\psi _e\,dx +\int _{e}[M_n(\varvec{\tau })]\partial _{{\varvec{n}}_e}\psi _e\,ds. \end{aligned}$$

This, along with the Cauchy–Schwarz inequality, the inverse inequality, Lemmas 6-7 and (62), leads to

$$\begin{aligned}&\Vert J_{2,e}\Vert _{0,e}^2 \\&\quad \lesssim \sum _{K\in \partial ^{-1}e}\big (\Vert \varvec{\nabla }\cdot (\varvec{\nabla }\cdot \varvec{\tau })-{\varDelta }(P_hu-{\widetilde{P}}_hu_h)\Vert _{0,K}\Vert \psi _e\Vert _{0,K} +\Vert \widetilde{\varvec{\sigma }}-\varvec{\tau }\Vert _{0,K}|\psi _e|_{2,K}\big ) \\&\qquad + \Vert [M_n(\varvec{\tau })]\Vert _{0,e}\Vert \partial _{{\varvec{n}}_e}\psi _e\Vert _{0,e} + \Vert [\partial _{{\varvec{n}}_e}(P_hu-{\widetilde{P}}_hu_h)]\Vert _{0,e}\Vert \psi _e\Vert _{0,e} \\&\quad \lesssim \sum _{K\in \partial ^{-1}e}h_K^{-2}\Vert \widetilde{\varvec{\sigma }}-\varvec{\tau }\Vert _{0,K}\Vert \psi _e\Vert _{0,K} + h_e^{-3/2}\Vert [M_n(\varvec{\tau })]\Vert _{0,e}\Vert \psi _e\Vert _{0,\omega _e} \\&\qquad + h_e^{-1/2}\Vert [\partial _{{\varvec{n}}_e}(P_hu-{\widetilde{P}}_hu_h)]\Vert _{0,e}\Vert \psi _e\Vert _{0,\omega _e} \\&\quad \lesssim \Vert \psi _e\Vert _{0,\omega _e}\sum _{K\in \partial ^{-1}e}\left( h_K^{-2}\Vert \widetilde{\varvec{\sigma }}-\varvec{\tau }\Vert _{0,K}+h_K^{-1}|P_hu-{\widetilde{P}}_hu_h|_{1,K}\right) \\&\quad \lesssim h_e^{1/2}\Vert J_{2,e}\Vert _{0,e}\sum _{K\in \partial ^{-1}e}\left( h_K^{-2}\Vert \widetilde{\varvec{\sigma }}-\varvec{\tau }\Vert _{0,K}+h_K^{-1}|P_hu-{\widetilde{P}}_hu_h|_{1,K}\right) , \end{aligned}$$

as required. $\square$