Discontinuous Galerkin Method Based on the Reduced Space for the Nonlinear Convection–Diffusion–Reaction Equation

Abstract

In this paper, by introducing a reconstruction operator based on the Legendre moments, we construct a reduced discontinuous Galerkin (RDG) space that could achieve the same approximation accuracy but using fewer degrees of freedom (DoFs) than the standard discontinuous Galerkin (DG) space. The design of the “narrow-stencil-based” reconstruction operator can preserve the local data structure property of the high-order DG methods. With the RDG space, we apply the local discontinuous Galerkin (LDG) method with the implicit-explicit time marching for the nonlinear unsteady convection–diffusion–reaction equation, where the reduction of the number of DoFs allows us to achieve higher efficiency. In terms of theoretical analysis, we give the well-posedness and approximation properties for the reconstruction operator and the $L^2$ error estimate for the semi-discrete LDG scheme. Several representative numerical tests demonstrate the accuracy and the performance of the proposed method in capturing the layers.

Appendices

The Numerical Analysis of Assumption 1

Given one-dimensional mesh ${\mathcal {T}}_h = \{K_j\}_{j=1}^N$, we select the element itself and the direct Moore neighbors as the element stencil. Periodic boundary value problems can be denoted by

$$\begin{aligned} \begin{aligned}&S^1(K_j) = \{K_{j-1},K_j,K_{j+1}\}, \ j = 1,\cdots ,\ N,\\ \end{aligned} \end{aligned}$$

where $K_{0}, K_{N+1}$ are obtained by the periodic extension. For other non-periodic boundary conditions, we denote the following bias stencil

$$\begin{aligned} \begin{aligned}&S^1(K_1) = \{K_1,K_2,K_3\},\\&S^1(K_j) = \{K_{j-1},K_j,K_{j+1}\}, \ j = 2,\cdots ,\ N-1,\\&S^1(K_N) = \{K_{N-2},K_{N-1},K_{N}\}. \end{aligned} \end{aligned}$$

For two dimensions, the stencil can be written as the tensor product of the one-dimensional stencil defined above. Consider the mesh ${\mathcal {T}}_h = \{K_\theta |\ \theta \in \Theta ^2\}$, where $K_\theta =I^1_{\theta _1}\times I^2_{\theta _2}$ as denoted in the previous section. Here, we define the one-dimensional stencil for the first direction by $\{S^1(I^1_{\theta _1})\}_{\theta _1=1}^{N_1}$, similarly, $\{S^1(I^2_{\theta _2})\}_{\theta _2=1}^{N_2}$ for the second direction. The two-dimensional stencil is denoted by

$$\begin{aligned} S^2(K_\theta ) = S^1(I^1_{\theta _1})\times S^1(I^2_{\theta _2}). \end{aligned}$$

The width of the stencil is $d(S) = 3^d$. In what follows, we denote the stencil by $S(K_\theta ) = \{K_{\theta ^1},K_{\theta ^2},\cdots ,K_{\theta ^{s}}\}$ without distinguishing one or two dimension. For clear description, we define the trace of a multiple indicator $\alpha = (\alpha _1,\cdots , \alpha _n)$ by $tr(\alpha ) = \sum _{i = 1}^n\alpha _i$ and sort the set $A^k$ based on the ascendant sequence of the trace, i.e.,

$$\begin{aligned} A^k = \{\alpha ^1,\alpha ^2,\cdots ,\alpha ^{(k+1)^d}\}, \end{aligned}$$

where $tr(\alpha ^1)\le tr(\alpha ^2)\le ,\cdots ,\le tr(\alpha ^{(k+1)^d})$.

As a result, we can express the approximation $R_{K_\theta }^ku$ as the following Legendre expansion

$$\begin{aligned} R_{K_\theta }^ku(\pmb x) = \sum _{j = 1}^{(k+1)^d} u^{\alpha ^j} L_{K_\theta }^{\alpha ^j}(\pmb x), \end{aligned}$$

where $\pmb {u} = (u^{\alpha ^1},u^{\alpha ^2},\cdots ,u^{\alpha ^{(k+1)^d}})^T$ is the unknown coefficient vector to be determined.

For any element $K_{\theta ^i}\in S^d(K_\theta )$, let $M_{K_{\theta ^i}}^k$ denote an $(m+1)^d\times (k+1)^d$ matrix whose jth row is the Legendre moments vector obtained by applying the operator $I_{K_{\theta ^i}}^{\alpha ^j}$ on the Legendre basis vector $(L^{\alpha ^1}_{K_\theta }(\pmb x),L^{\alpha ^2}_{K_\theta }(\pmb x),\cdots ,L^{\alpha ^{(k+1)^d}}_{K_\theta }(\pmb x))$, i.e.,

$$\begin{aligned} m^k_{K_{\theta ^i}} = \begin{pmatrix} I_{K_{\theta ^i}}^{\alpha ^1}L^{\alpha ^1}_{K_\theta }(\pmb x)&{}I_{K_{\theta ^i}}^{\alpha ^1}L^{\alpha ^2}_{K_\theta }(\pmb x)&{}\cdots &{}I_{K_{\theta ^i}}^{\alpha ^1}L^{\alpha ^{(k+1)^d}}_{K_\theta }(\pmb x)\\ I_{K_{\theta ^i}}^{\alpha ^2}L^{\alpha ^1}_{K_\theta }(\pmb x)&{}I_{K_{\theta ^i}}^{\alpha ^2}L^{\alpha ^2}_{K_\theta }(\pmb x)&{}\cdots &{}I_{K_{\theta ^i}}^{\alpha ^2}L^{\alpha ^{(k+1)^d}}_{K_\theta }(\pmb x)\\ \vdots &{}\vdots &{}&{}\vdots \\ I_{K_{\theta ^i}}^{\alpha ^{(m+1)^d}}L^{\alpha ^1}_{K_\theta }(\pmb x)&{}I_{K_{\theta ^i}}^{\alpha ^{(m+1)^d}}L^{\alpha ^2}_{K_\theta }(\pmb x)&{}\cdots &{}I_{K_{\theta ^i}}^{\alpha ^{(m+1)^d}}L^{\alpha ^{(k+1)^d}}_{K_\theta }(\pmb x) \end{pmatrix}. \end{aligned}$$

Repeating the same process for every element of the stencil $S^d(K_\theta )$, we can define the coefficient matrix of the linear system (3.4) by

$$\begin{aligned} M_{K_\theta }^k = \left( {(m_{K_{\theta ^1}}^k)^T,(m_{K_{\theta ^2}}^k)^T,\cdots ,(m_{K_{\theta ^s}}^k)^T}\right) ^T. \end{aligned}$$

(A.1)

Ultimately, we obtain the matrix version of the system (3.4) as follows:

$$\begin{aligned} M_{K_\theta }^k\pmb {u} = \pmb {0}, \end{aligned}$$

(A.2)

where $\pmb 0$ is the zeros vector.

For this system to have a unique zero solution, there are two conditions to be simultaneously satisfied:

• 1. Matrix $M_{K_\theta }^k$ is a square matrix, i.e., $3(m+1) = k+1$.

• 2. The determinant of matrix $M_{K_\theta }^k$ must be non-zero.

Under the first condition, let us discuss the second condition. To do that, we need to figure out each element of $M_{K_\theta }^k$. Take the element $I_{K_{\theta ^i}}^{\alpha ^r}L^{\alpha ^l}_{K_\theta }(\pmb x)$ for example, we have

$$\begin{aligned} \begin{aligned} I_{K_{\theta ^i}}^{\alpha ^r}L^{\alpha ^l}_{K_\theta }(\pmb x)&= \left( \prod _{j = 1}^d\frac{2\alpha _j^r+1}{h_{K_{\theta ^i}}^j}\right) \int _{K_{\theta ^i}} L_{K_{\theta ^i}}^{\alpha ^r}(\pmb {x})L^{\alpha ^l}_{K_\theta }(\pmb x)d\pmb {x}\\&= \prod _{j = 1}^d\left( \frac{2\alpha _j^r+1}{2}\int _{-1}^1\widehat{L}^{\alpha _j^r}(\widehat{x}) L_{K_\theta }^{\alpha _j^l}\left( \frac{h_{K_{\theta ^i}}^j}{2}\widehat{x}+x_{K_{\theta ^i}}^j\right) d\widehat{x}\right) \\&= \prod _{j = 1}^d\left( \frac{2\alpha _j^r+1}{2}\int _{-1}^1\widehat{L}^{\alpha _j^r} (\widehat{x})\widehat{L}^{\alpha _j^l}\left( \frac{h_{K_{\theta ^i}}^j}{h_{K_\theta }^j}\widehat{x}+\frac{2}{h_{K_\theta }^j}\left( x_{K_{\theta ^i}}^j-x_{K_\theta }^j\right) \right) d\widehat{x}\right) \\&= \prod _{j = 1}^d\left( \frac{2\alpha _j^r+1}{2}\int _{-1}^1\widehat{L}^{\alpha _j^r} (\widehat{x})\widehat{L}^{\alpha _j^l}\left( a_{\theta ^i}^j\widehat{x}+b_{\theta ^i}^j\right) d\widehat{x}\right) , \end{aligned} \end{aligned}$$

where $a_{\theta ^i}^j = \frac{h_{K_{\theta ^i}}^j}{h_{K_\theta }^j}$ and $b_{\theta ^i}^j = \frac{2}{h_{K_\theta }^j}(x_{K_{\theta ^i}}^j-x_{K_\theta }^j)$.

Next, we will give numerical proof for different cases. Let us start with the one-dimensional case, i.e., $d = 1$. Where $\theta ,\alpha$ are all just single indices. The element $K_\theta$ can be briefly written as $K_j$. Besides, We have $\alpha ^i = i-1$ for $i \ge 1$. For $k = 2$, we have $m = 0$ according to the first condition. When taking the center stencil $S_c^1(K_j) = \{K_{j-1},K_j,K_{j+1}\}$, we have

$$\begin{aligned} \begin{aligned} M^2_{K_j}&= \begin{pmatrix} I^0_{K_{j-1}}L^0_{K_j}(x)&{}{}I^0_{K_{j-1}}L^1_{K_j}(x)&{}{}I^0_{K_{j-1}}L^2_{K_j}(x) \\ I^0_{K_{j}}L^0_{K_j}(x)&{}{}I^0_{K_{j}}L^1_{K_j}(x)&{}{}I^0_{K_{j}}L^2_{K_j}(x) \\ I^0_{K_{j+1}}L^0_{K_j}(x)&{}{}I^0_{K_{j+1}}L^1_{K_j}(x)&{}{}I^0_{K_{j+1}}L^2_{K_j}(x) \\ \end{pmatrix}\\ {}&= \begin{pmatrix} 1&{}{}b_{j-1}&{}{}\frac{3b_{j-1}^2+a_{j-1}^2-1}{2}\\ 1&{}{}0&{}{}0 \\ 1&{}{}b_{j+1}&{}{}\frac{3b_{j+1}^2+a_{j+1}^2-1}{2}\\ \end{pmatrix}. \end{aligned} \end{aligned}$$

Naturally, the determinant of $M_{K_j}^2$ can be computed as

$$\begin{aligned} \begin{aligned} \det (M_{K_j}^2)= \frac{3b_{j-1}^2b_{j+1} - 3b_{j-1}b_{j+1}^2 - b_{j-1}a_{j+1}^2 + b_{j-1} + b_{j+1}a_{j-1}^2 -b_{j+1}}{2}. \end{aligned} \end{aligned}$$

To determine the value of $\det (M_{K_j}^2)$, we need more information. First, the regularity of mesh can induce the following condition

$$\begin{aligned} 0<a_{m}<a_{j-q}<a_{M},\ \textrm{for}\ q = 2,1,-1,-2. \end{aligned}$$

(A.3)

Moreover, by simple calculation, there are $b_{j-1} = -(1+a_{j-1})$, $b_{j+1}=1+a_{j+1}$. With these informations, the following result can be directly obtained

$$\begin{aligned} \begin{aligned} \det (M_{K_j}^2)|_{S_c^1}&= 2(a_{j+1} + 1)(a_{j-1} + 1)(a_{j-1} +a_{j+1} + 1)\\&>2(a_m+1)^2(2a_m+1). \end{aligned} \end{aligned}$$

For the backward stencil $S_b^1(K_j) = \{K_{j-2},K_{j-1},K_{j}\}$, the similar determinant property results can be derived from the conditions $b_{j-2} = -(1+2a_{j-1}+a_{j-2}),\ b_{j-1}=-(1+a_{j-1})$ and A.3, which is presented as follows,

$$\begin{aligned} \begin{aligned} \det (M_{K_j}^2)|_{S_b^1}&= 2(a_{j-1}+1)(a_{j-1}+a_{j-2})(a_{j-1}+a_{j-2}+1)\\&>4a_m(a_m+1)(2a_m+1). \end{aligned} \end{aligned}$$

In the same way, with the conditions $b_{j+1} = 1+a_{j+1},\ b_{j+2}=1+2a_{j+1}+a_{j+2}$ and A.3, we can obtain the determinant property for the forward stencil $S_f^1(K_j) = \{K_j,K_{j+1},K_{j+2}\}$ as follows,

$$\begin{aligned} \begin{aligned} \det (M_{K_j}^2)|_{S_f^1}&= 2(a_{j+1}+1)(a_{j+1}+a_{j+2})(a_{j+1}+a_{j+2}+1)\\&>4a_m(a_m+1)(2a_m+1). \end{aligned} \end{aligned}$$

The determinant property for the case of $k = 5$ can be deduced following the proof process for $k = 2$. The results are presented in Table 1.

Table 1 The determinant of $M_{K_j}^5$

Next, we turn our attention to the two-dimensional case, i.e., $d = 2$. Here, $\theta ,\alpha$ are all double indexes with the form (i, j). The element $K_\theta$ can be briefly defined as $K_{i,j} = I_i^1\times I_j^2$. For $k = 2$, we have $m = 0$. We recall the definition of the two-dimensional stencil, $S^2(K_{i,j}) = S^1(I^1_i)\times S^1(I^2_j)$. Consequently, three different options for one-dimensional stencil produce nine different stencil strategies for two dimensions, which are shown as follows,

$$\begin{aligned} \begin{aligned}&S_{cc}^2(K_{i,j}) = S_c^1(I^1_i)\times S_c^1(I^2_j),\ S_{cb}^2(K_{i,j}) = S_c^1(I^1_i)\times S_b^1(I^2_j),\ S_{cf}^2(K_{i,j}) = S_c^1(I^1_i)\times S_f^1(I^2_j),\\&S_{bb}^2(K_{i,j}) = S_b^1(I^1_i)\times S_b^1(I^2_j),\ S_{bc}^2(K_{i,j}) = S_b^1(I^1_i)\times S_c^1(I^2_j),\ S_{bf}^2(K_{i,j}) = S_b^1(I^1_i)\times S_f^1(I^2_j),\\&S_{ff}^2(K_{i,j}) = S_f^1(I^1_i)\times S_f^1(I^2_j),\ S_{fc}^2(K_{i,j}) = S_f^1(I^1_i)\times S_c^1(I^2_j),\ S_{fb}^2(K_{i,j}) = S_f^1(I^1_i)\times S_b^1(I^2_j).\\ \end{aligned} \end{aligned}$$

(A.4)

Take the stencil $S_{cc}^2(K_{i,j})$ as an example. Let us illustrate the proof process. We know that the basis functions of $Q^k(S^2(K_{i,j}))$ are the tensor product of the basis functions of $Q^k(S^1(I_i))$ and $Q^k(S^1(I_j))$. Considering the same product form of the stencils, the coefficient matrix can be expressed by the Kronecker product form

$$\begin{aligned} M^2_{K_{i,j}} = M^2_{I_i}\otimes M^2_{I_j}. \end{aligned}$$

Given square matrices A and B with degrees m and n, there is a well-known determinant conclusion for the Kronecker product [15, 33],

$$\begin{aligned} \det (A\otimes B) = \det (A)^n\det (B)^m. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \det (M^2_{K_{i,j}}) = \det (M^2_{I_i})^3\det (M^2_{I_j})^3. \end{aligned}$$

Based on the determinant property in one dimension, we can conclude that the same property also holds for every stencil strategy for two dimensions. The same result can be obtained for $k = 5$ according to the fact that

$$\begin{aligned} \det (M^5_{K_{i,j}}) = \det (M^5_{I_i})^6\det (M^5_{I_j})^6. \end{aligned}$$

Proof of Theorem 1

Proof

The local reconstruction operator $R_{K}^k$ can be regarded as one interpolation operator. Thus, proving the k-exactness property (3.7) is equivalent to proving the uniqueness of the polynomial interpolation problem, which has been given by Assumption 1.

With the k-exactness property (3.7), the operator $R_{K}^k$ can be regarded as a projection operator which projects the Sobolev space $H^{k+1}(S(K))$ on the polynomial space $Q^k(S(K))$. Considering the $l_\infty$ norm $\Vert \cdot \Vert _{l_\infty (S(K))}$, as defined in Sect. 3.2, we have the following property

$$\begin{aligned} \Vert R_{K}^ku\Vert _{l_{\infty }(S(K))}=\Vert u\Vert _{l_{\infty }(S(K))}. \end{aligned}$$

(B.5)

Following [5, Chapter 4.6], there exists an averaged Taylor polynomial $T^{k+1}u\in Q^{k}(S(K))$ such that

$$\begin{aligned} \Vert u -T^{k+1}u\Vert _{0,\infty ,S(K)}\le Ch^{k+1-\frac{d}{2}}. \end{aligned}$$

With (3.5), (B.5), and (3.6), we have

$$\begin{aligned} \begin{aligned} \Vert T^{k+1}u -R_{K}^ku\Vert _{0,\infty ,K}&\le C\Vert T^{k+1}u -R_{K}^ku\Vert _{0,\infty ,S(K)}=C\Vert R_{K}^k(T^{k+1}u -u)\Vert _{0,\infty ,S(K)}\\&\le C\Vert R_{K}^k(T^{k+1}u -u)\Vert _{l_{\infty }(S(K))}= C\Vert T^{k+1}u -u\Vert _{l_{\infty }(S(K))}\\&\le C\Vert T^{k+1}u -u\Vert _{0,\infty ,S(K)}\le Ch^{k+1-\frac{d}{2}}. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert u-R_{K}^ku\Vert _{0,\infty ,K}\le \Vert u-T^{k+1}u\Vert _{0,\infty ,S(K)}+\Vert T^{k+1}u-R_{K}^ku\Vert _{0,\infty ,S(K)}\le Ch^{k+1-\frac{d}{2}}, \end{aligned}$$

which gives (3.8).

Following (3.8), the $L^2$ error estimate (3.9) can be derived directly as follows,

$$\begin{aligned} \Vert u_h -R_{K}^ku\Vert _{0,K}\le Ch^{\frac{d}{2}}\Vert u_h -R_{K}^ku\Vert _{0,\infty ,K}\le Ch^{k+1}. \end{aligned}$$

By the approximation property (2.2), we can choose an approximation polynomial $u_h\in Q^k(S(K)$ such that

$$\begin{aligned} \Vert u -u_h\Vert _{0,K}+h| u -u_h|_{1,K}\le Ch^{k+1}. \end{aligned}$$

With (2.2), (3.9), and the inverse property (2.3), we can derive (3.10) smoothly

$$\begin{aligned} \begin{aligned} | u -R_{K}^ku|_{1,K}&\le | u -u_h|_{1,K}+| u_h -R_{K}^ku|_{1,K}\\&\le Ch^k+Ch^{-1}\Vert u_h -R_{K}^ku\Vert _{0,K}\\&\le Ch^k+Ch^{-1}\Vert u -u_h\Vert _{0,K}+Ch^{-1}\Vert u -R_{K}^ku\Vert _{0,K}\\&\le Ch^k. \end{aligned} \end{aligned}$$

Here, the proof is completed. $\square$

Proof of Theorem 3

Proof

Given that the exact solution u and $\pmb q$ also satisfies the global weak formulation (4.3), we obtain the following error equation by a simple subtraction

$$\begin{aligned} \begin{aligned}&((u-u_h)_t,v_h)+\sum _{i=1}^d(q^i-q_h^i,p_h^i)\\&={\mathcal {H}}(u,v_h)-{\mathcal {H}}(u_h,v_h)+{\mathcal {L}}(\pmb q-\pmb q_h,v_h)+\sum _{i=1}^d{\mathcal {K}}^i(u-u_h,p_h^i)+{\mathcal {R}}(u,v_h)-{\mathcal {R}}(u_h,v_h). \end{aligned} \end{aligned}$$

Here, we denote

$$\begin{aligned} \begin{aligned} {\mathcal {K}}(u-u_h,\pmb p_h)&=\sum _{i=1}^d{\mathcal {K}}^i(u-u_h,p_h^i)\\&= -\sqrt{\varepsilon }\sum _{i=1}^d\left( (u-u_h,(p_h^i)_{x_i})_K-\langle u-\hat{u}_h, p_h^in_i\rangle _{\partial K}\right) \\&=-\sqrt{\varepsilon }\left( (u-u_h,\nabla \cdot \pmb p_h)_K-\langle u-\hat{u}_h, \pmb p_h\cdot \pmb n\rangle _{\partial K}\right) . \end{aligned} \end{aligned}$$

Thus, a more neat error equation can be described as follows,

$$\begin{aligned} \begin{aligned} ((u\!-\!u_h)_t,v_h)\!+\!(\pmb q\!-\!\pmb q_h,\pmb p_h)\!=\!&{\mathcal {H}}(u,v_h)\!-\!{\mathcal {H}}(u_h,v_h)+{\mathcal {L}}(\pmb q-\pmb q_h,v_h)\!+\!{\mathcal {K}}(u\!-\!u_h,\pmb p_h)\\&+{\mathcal {R}}(u-u_h,v_h).\\\\ \end{aligned} \end{aligned}$$

(C.6)

Denote

$$\begin{aligned} \xi = R^ku-u_h,\ \xi ^e = R^ku-u,\ \pmb \eta = R^k\pmb q-\pmb q_h,\ \pmb \eta ^e = R^k\pmb q-\pmb q,\ \end{aligned}$$

(C.7)

and take the test function as

$$\begin{aligned} v_h = \xi ,\ \pmb p_h = \pmb \eta . \end{aligned}$$

(C.8)

We obtain the energy equality

$$\begin{aligned}{} & {} ((\xi -\xi ^e)_t,\xi )+(\pmb \eta -\pmb \eta ^e,\pmb \eta )={\mathcal {H}}(u,\xi )-{\mathcal {H}}(u_h,\xi )\nonumber \\{} & {} \quad +{\mathcal {L}}(\pmb q-\pmb q_h,\xi )+{\mathcal {K}}(u-u_h,\pmb \eta )+{\mathcal {R}}(u-u_h,\xi ). \end{aligned}$$

(C.9)

Next, we will estimate each term on the right-hand side of the energy equality.

To estimate the nonlinear convection term ${\mathcal {H}}(u,\xi )-{\mathcal {H}}(u_h,\xi )$, we need to make $a\ priori$ assumption, for small enough h, we have

$$\begin{aligned} \Vert u-u_h\Vert _{0}\le Ch^{\frac{d+1}{2}}. \end{aligned}$$

(C.10)

With this assumption, we have $\Vert u-u_h\Vert _{0,\infty }\le Ch^{\frac{1}{2}}$. Moreover, the property (4.6) implies that $\Vert R^ku-u_h\Vert _{0,\infty }\le Ch^{\frac{1}{2}}$.

Here, we have

$$\begin{aligned} \begin{aligned}&{\mathcal {H}}(u,\xi )-{\mathcal {H}}(u_h,\xi ) \\ =&\sum _{K\in {\mathcal {T}}_h}\left( (\pmb b(f(u)-f(u_h)),\nabla \xi )_K-\langle \pmb b(f(u)-\hat{f}(u_h))\cdot \pmb n,\xi \rangle _{\partial K}\right) \\ =&(\pmb b(f(u)-f(u_h)),\nabla \cdot \xi )-\langle \pmb b(f(u)-\hat{f}(u_h)),[\xi ]\rangle \\ =&(\pmb b(f(u)-f(u_h)),\nabla \cdot \xi )-\langle \pmb b(f(u)-f(\bar{u}_h)),[\xi ]\rangle +\langle \pmb b(f(\bar{u}_h)-\hat{f}(u_h)),[\xi ]\rangle \\ =&I+II. \end{aligned} \end{aligned}$$

According to [31, Lemma 3.4], the second part can be estimated as

$$\begin{aligned} II = \langle \pmb b(f(\bar{u}_h)-\hat{f}(u_h)),[\xi ]\rangle \le -\frac{3}{4}\Vert \pmb b\Vert _{1,\infty }\langle \alpha (\hat{f};u_h),[\xi ]^2\rangle +Ch^{2k+1}. \end{aligned}$$

Where $\alpha (\hat{f};u_h)$ is non-negative and bounded, which is defined in [34]. Next, we estimate the first part I. Follow [31, Appendix A.1], I can be divided into six parts as follows,

$$\begin{aligned} \begin{aligned} I_1&= (\pmb bf'(u)\xi ,\nabla \cdot \xi )+\langle \pmb bf'(u)\bar{\xi },[\xi ]\rangle ,\\ I_2&= -\frac{1}{2}\left( (\pmb bf''(u)\xi ^2,\nabla \cdot \xi )+\langle \pmb bf''(u)\bar{\xi }^2,[\xi ]\rangle \right) ,\\ I_3&= -\left( (\pmb bf'(u)\xi ^e,\nabla \cdot \xi )+\langle \pmb bf'(u)\bar{\xi }^e,[\xi ]\rangle \right) ,\\ I_4&= (\pmb bf''(u)\xi ^e\xi ,\nabla \cdot \xi )+\langle \pmb bf''(u)\bar{\xi }^e\bar{\xi },[\xi ]\rangle ,\\ I_5&= -\frac{1}{2}\left( (\pmb bf''(u)(\xi ^e)^2,\nabla \cdot \xi )+\langle \pmb bf''(u)(\bar{\xi }^e)^2,[\xi ]\rangle \right) ,\\ I_6&= \frac{1}{6}\left( (\pmb bf_u'''(\xi -\xi ^e)^3,\nabla \cdot \xi )+\langle \pmb b\tilde{f}_u'''(\bar{\xi }-\bar{\xi }^e)^3,[\xi ]\rangle \right) , \end{aligned} \end{aligned}$$

where $f_u'''$ and $\tilde{f}_u'''$ are the factors in the remainder of Taylor expansion of $f(u_h)$ and $f(\bar{u}_h)$ separately. By integration by parts, we can estimate each term now:

• $I_1$ term

  $$\begin{aligned} I_1 = \frac{1}{2}(\nabla \cdot (\pmb bf'(u)),\xi ^2)\le C\Vert \xi \Vert _{0}^2. \end{aligned}$$

• $I_2$ term

  $$\begin{aligned} I_2 = \frac{1}{6}(\nabla \cdot (\pmb b f''(u))\xi ,\xi ^2)+\frac{1}{24}\langle \pmb bf''(u)[\xi ],[\xi ]^2\rangle .\\ \end{aligned}$$

  By a Taylor expansion, we have

  $$\begin{aligned} \begin{aligned} f''(u)[\xi ]&= f''(u_h)[\xi ] +f''_u(u-u_h)[\xi ]\\&=-f''(u_h)[u_h]+f''(u_h)[\xi ^e]+f''_u(u-u_h)[\xi ]\\&\le 8\alpha (\hat{f};u_h)+8|[u-u_h]|^2+f''(u_h)[\xi ^e]+f''_u(u-u_h)[\xi ]. \end{aligned} \end{aligned}$$

  Thus,

  $$\begin{aligned} \begin{aligned} I_2 \le&\frac{1}{3}\Vert \pmb b\Vert _{1,\infty }\langle \alpha (\hat{f};u_h),[\xi ]^2\rangle \\ {}&+C\left( \Vert \xi \Vert _{0,\infty }+h^{-1}(\Vert u-u_h\Vert _{0,\infty }^2+\Vert \xi ^e\Vert _{0,{\mathcal {E}}_h}^2)\right) \Vert \xi \Vert _{0}^2\\ \le&\frac{1}{3}\Vert \pmb b\Vert _{1,\infty }\langle \alpha (\hat{f};u_h),[\xi ]^2\rangle +C\Vert \xi \Vert _{0}^2. \end{aligned} \end{aligned}$$

• $I_3$ term

  $$\begin{aligned} \begin{aligned} I_3&= -(\pmb bf'(u)\xi ^e,\nabla \cdot \xi )-\langle \pmb b(f'(u)-f'(\bar{u}_h))\bar{\xi }^e,[\xi ]\rangle -\langle \pmb bf'(\bar{u}_h)\bar{\xi }^e,[\xi ]\rangle \\ {}&\le C \Vert \xi \Vert _{0}^2 +2\langle \pmb b (\alpha (\hat{f};u_h)+C|[u-u_h]|)\bar{\xi }^e,[\xi ]\rangle +Ch^{2k}\\ {}&\le \frac{1}{6}\Vert \pmb b\Vert _{1,\infty }\langle \alpha (\hat{f};u_h),[\xi ]^2\rangle +C\Vert \xi \Vert _{0}^2+Ch^{2k}. \end{aligned} \end{aligned}$$

• $I_4$, $I_5$, and $I_6$ terms

  $$\begin{aligned} \begin{aligned} I_4&\le Ch^{-1}\Vert \xi ^e\Vert _{0,\infty }\Vert \xi \Vert _{0}\le C\Vert \xi \Vert _{0}^2,\\ I_5&\le C\Vert \xi ^e\Vert _{0,\infty }(h^{-1}\Vert \xi ^e\Vert _{0} +h^{-\frac{1}{2}}\Vert \xi ^e\Vert _{0,{\mathcal {E}}_h})\Vert \xi \Vert _{0}\le C\Vert \xi \Vert _{0}^2+Ch^{2k+2},\\ I_6&\le C\Vert u-u_h\Vert _{0,\infty }^2(h^{-1}(\Vert \xi \Vert _{0}+\Vert \xi ^e\Vert _{0}) +h^{-\frac{1}{2}}\Vert \xi ^e\Vert _{0,{\mathcal {E}}_h})\Vert \xi \Vert _{0}\\ {}&\le C\Vert \xi \Vert _{0}^2+Ch^{2k+2}. \end{aligned} \end{aligned}$$

Combing with all the above estimates, we can derive that

$$\begin{aligned} I \le \frac{1}{2}\Vert \pmb b\Vert _{1,\infty }\langle \alpha (\hat{f};u_h),[\xi ]^2\rangle +C\Vert \xi \Vert _{0}^2+Ch^{2k}. \end{aligned}$$

Naturally, we have the following estimate

$$\begin{aligned} {\mathcal {H}}(u,\xi )-{\mathcal {H}}(u_h,\xi ) \le -\frac{1}{4}\Vert \pmb b\Vert _{1,\infty }\langle \alpha (\hat{f};u_h),[\xi ]^2\rangle +C\Vert \xi \Vert _{0}^2+Ch^{2k}. \end{aligned}$$

Then we focus on the diffusion part. We have

$$\begin{aligned} \begin{aligned}&{\mathcal {L}}(\pmb q-\pmb q_h,\xi )+{\mathcal {K}}(u-u_h,\pmb \eta )\\ {}&\quad =-\sqrt{\varepsilon }\left( (\pmb q-\pmb q_h,\nabla \xi )+(u-u_h,\nabla \cdot \pmb \eta )-\langle \pmb q-\hat{\pmb q}_h\cdot \pmb n^+,[\xi ]\rangle +\langle u-\hat{u}_h, [\pmb \eta ]\cdot \pmb n^+\rangle \right) \\ {}&\quad =-\sqrt{\varepsilon }\left( (\pmb \eta -\pmb \eta ^e,\nabla \xi )+(\xi -\xi ^e,\nabla \cdot \pmb \eta )-\langle \hat{\pmb \eta }-\hat{\pmb \eta }^e\cdot \pmb n^+,[\xi ]\rangle +\langle \hat{\xi }-\hat{\xi }^e, [\pmb \eta ]\cdot \pmb n^+\rangle \right) \\ {}&\quad =\sqrt{\varepsilon }\left( (\pmb \eta ^e,\nabla \xi )+(\xi ^e,\nabla \cdot \pmb \eta )-\langle \hat{\pmb \eta }^e\cdot \pmb n^+,[\xi ]\rangle +\langle \hat{\xi }^e, [\pmb \eta ]\cdot \pmb n^+\rangle \right) \\ {}&\quad \le \frac{1}{2}\Vert \pmb \eta \Vert _{0}^2+C\Vert \xi \Vert _{0}^2+Ch^{2k}. \end{aligned} \end{aligned}$$

Finally, we estimate the reaction part as follows,

$$\begin{aligned} {\mathcal {R}}(u,\xi )-{\mathcal {R}}(u_h,\xi )=-(r(u)-r(u_h),\xi )=-(r'_u(\xi -\xi ^e),\xi )\le C\Vert \xi \Vert _{0}^2+Ch^{2k+2}.\end{aligned}$$

With the above estimates and Young’s inequality, the energy equation becomes

$$\begin{aligned} \begin{aligned} \frac{1}{2}\left( \frac{d\Vert \xi \Vert _0^2}{dt}+\Vert \pmb \eta \Vert _0^2\right) +\frac{1}{4}\Vert \pmb b\Vert _{1,\infty }\langle \alpha (\hat{f};u_h),[\xi ]^2\rangle&\le (\xi ^e_t,\xi )+(\pmb \eta ^e,\pmb \eta )+C\Vert \xi \Vert _{0}^2+Ch^{2k}\\ {}&\le C\Vert \xi \Vert _{0}^2+\frac{1}{4}\Vert \pmb \eta \Vert _{0}^2+Ch^{2k}. \end{aligned} \end{aligned}$$

Thus,

$$\begin{aligned} \frac{1}{2}\frac{d\Vert \xi \Vert _0^2}{dt}+\frac{1}{4}\Vert \pmb \eta \Vert _0^2+\frac{1}{4}\Vert \pmb b\Vert _{1,\infty }\langle \alpha (\hat{f};u_h),[\xi ]^2\rangle \le C\Vert \xi \Vert _{0}^2+Ch^{2k}. \end{aligned}$$

By the Gronwall inequality and triangle inequality, the proof is completed as follows,

$$\begin{aligned} \Vert u-u_h\Vert _0\le Ch^{k}. \end{aligned}$$

$\square$