Convergence Analysis of Crank–Nicolson Galerkin–Galerkin FEMs for Miscible Displacement in Porous Media

Abstract

We propose a fully discrete linearized Crank–Nicolson Galerkin–Galerkin finite element method for solving the partial differential equations which govern incompressible miscible flow in porous media. We prove optimal-order convergence of the fully discrete finite element solutions without any restrictions on the step size of time discretization. Numerical examples are provided to illustrate the theoretical results.

Appendices

A Proof of (3.14) when $m=1$

For $m=1$, since ${\widehat{\mathbf{U}}}^{\frac{1}{2}}=\frac{1}{2}(\mathbf{U}^{1^*}+\mathbf{U}^0)$, the proof of (3.14) depends on the error estimate of $({\mathcal {C}}^{1^*},\mathbf{U}^{1^*},P^{1^*})$. Thus, we let

$$\begin{aligned} e^{1^{*}}_c={\mathcal {C}}^{1^{*}}-c^1,\quad e^{1^{*}}_u=\mathbf{{U}}^{1^{*}}-\mathbf{{u}}^1,\quad e^{1^{*}}_p=P^{1^{*}}-p^1 \end{aligned}$$

and start from the estimate of $\Vert e_c^{1^*}\Vert _{L^2}$.

Subtracting the system (2.14)–(2.16) from (1.1)–(1.3) yield

$$\begin{aligned}&\Phi \frac{e^{1^{*}}_c}{\tau }-\nabla \cdot (D(\mathbf{{u}}^0)\nabla e^{1^{*}}_c) +\frac{1}{2}{} \mathbf{u}^0\cdot \nabla e_c^{1^*} +\frac{1}{2}\nabla \cdot (\mathbf{u}^0 e_c^{1^*}) +\frac{1}{2}(q_I+q_P)e_c^{1^*}\nonumber \\&\quad = \nabla \cdot ([D(\mathbf{u}^0)-D(\mathbf{u}^1)]\nabla c^1) -\frac{1}{2}(\mathbf{u}^0-\mathbf{u}^1)\cdot \nabla c^1 -\frac{1}{2}\nabla \cdot ((\mathbf{u}^0-\mathbf{u}^1)c^1) +R^{1^{*}}_{tr} \end{aligned}$$

(A.1)

$$\begin{aligned}&-\,\nabla \cdot \left( \frac{k(x)}{\mu ({\mathcal {C}}^{1^{*}})}\nabla e^{1^{*}}_p\right) =\nabla \cdot \left( \left( \frac{k(x)}{\mu ({\mathcal {C}}^{1^{*}})} -\frac{k(x)}{\mu (c^{1})}\right) \nabla p^1\right) \end{aligned}$$

(A.2)

$$\begin{aligned}&e^{1^{*}}_u=-\,\frac{k(x)}{\mu ({\mathcal {C}}^{1^*})}\nabla e^{1^{*}}_p-\left( \frac{k(x)}{\mu ({\mathcal {C}}^{1^*})}-\frac{k(x)}{\mu (c^1)}\right) \nabla p^1 \end{aligned}$$

(A.3)

under the boundary condition

$$\begin{aligned} \begin{aligned}&\frac{k(x)}{\mu ({\mathcal {C}}^{1^*})}\nabla e_p^{1^*}\cdot \mathbf{n}=0, \quad D(\mathbf{{u}}^{0})\nabla e_c^{1^*}\cdot \mathbf{n}=0 . \end{aligned} \end{aligned}$$

Here, we have used $\mathbf{U}^0=\mathbf{u}^0$ and $R_{tr}^{1^*}$ denotes the truncation error at the initial time step. By Taylor expansion and (2.8), we have

$$\begin{aligned} \Vert R_{tr}^{1^{*}}\Vert _{L^2}\le C\tau . \end{aligned}$$

(A.4)

To obtain the estimate of $\Vert e_c^{1^*}\Vert _{L^2}$, we multiply (A.1) by $e^{1^{*}}_c$, integrate it over $\Omega$ and get

$$\begin{aligned} \Vert e^{1^*}_c\Vert ^2_{L^2}+\tau \Vert \sqrt{D(\mathbf{u}^0)}\nabla e^{1^*}_c\Vert ^2_{L^2}&\le C\tau \Big ( \Vert \mathbf{u}^0\Vert _{L^\infty }\Vert \nabla e_c^{1^*}\Vert _{L^2}\Vert e_c^{1^*}\Vert _{L^2} +\Vert q_I+q_P\Vert _{L^3}\Vert e_c^{1^*}\Vert _{L^2}\Vert e_c^{1^*}\Vert _{L^6}\\&\quad +\,\Vert \mathbf{u}^1-\mathbf{u}^0\Vert _{H^1}\Vert c^1\Vert _{W^{2,4}}\Vert e_c^{1^*}\Vert _{L^2} +\Vert R_{tr}^{1^*}\Vert _{L^2}\Vert e_c^{1^*}\Vert _{L^2} \Big )\\&\le \frac{1}{2}\Vert e_c^{1^*}\Vert ^2_{L^2} +C\tau ^2\Vert \nabla e_c^{1^*}\Vert _{L^2}^2 +C\tau ^2\Vert \mathbf{u^0}-\mathbf{u}^1\Vert _{H^1}^2 +C\tau ^2\Vert R_{tr}^{1^*}\Vert _{L^2}^2 \\&\le \frac{1}{2}\Vert e_c^{1^*}\Vert _{L^2}^2 +\varepsilon \tau \Vert \sqrt{D(\mathbf{u}^0)}\nabla e^{1^*}_c\Vert ^2_{L^2} +C\tau ^4, \end{aligned}$$

where we have used (2.8), (A.4) and let $\tau \le \tau _1=\frac{\varepsilon }{C}$. The above estimate implies

$$\begin{aligned} \Vert e^{1^*}_c\Vert _{L^2}+\tau ^{\frac{1}{2}}\Vert \nabla e^{1^*}_c\Vert _{L^2}\le C\tau ^2. \end{aligned}$$

(A.5)

To obtain the estimate of $\Vert e_c^{1^*}\Vert _{H^2}$, we multiply (A.1) by $-\,\nabla \cdot (D(\mathbf{u}^0)\nabla e^{1^{*}}_c)$, integrate the resulting equation and have

$$\begin{aligned}&\Phi \Vert \sqrt{D(\mathbf{{u}}^0)}\nabla e^{1^{*}}_c\Vert _{L^2}^2 +\tau \Vert \nabla \cdot (D(\mathbf{u}^0)\nabla e^{1^{*}}_c)\Vert ^2_{L^2} \\&\le C\tau \Big ( \Vert \mathbf{u}^0\Vert _{W^{1,4}}^2\Vert e_c^{1^*}\Vert _{H^1}^2 +\Vert q_I+q_P\Vert _{L^3}^2\Vert e_c^{1^*}\Vert _{L^6}^2 +\Vert \mathbf{u}^1-\mathbf{u}^0\Vert _{H^1}^2\Vert c^1\Vert _{W^{2,4}}^2 +\Vert R_{tr}^{1^{*}}\Vert _{L^2}^2\Big ) \\&\quad +\,\frac{\tau }{2}\Vert \nabla \cdot (D(\mathbf {u}^0)\nabla e_c^{1^{*}})\Vert _{L^2}^2\\&\le \frac{\tau }{2}\Vert \nabla \cdot (D(\mathbf {u}^0)\nabla e_c^{1^{*}})\Vert _{L^2}^2 +C\tau ^3. \end{aligned}$$

By noting $\Vert e_c^{1^*}\Vert _{H^2} \le C\Vert \nabla \cdot (D(\mathbf {u}^0)\nabla e_c^{1^*})\Vert _{L^2} +C\Vert e_c^{1^*}\Vert _{H^1},$ we arrive at

$$\begin{aligned} \Vert e^{1^*}_c\Vert _{H^2}\le C\tau . \end{aligned}$$

(A.6)

Thus,

$$\begin{aligned} \Vert {{\mathcal {C}}}^{1^*}\Vert _{H^2}\le \Vert c^1\Vert _{H^2}+\Vert e^{1^*}_c\Vert _{H^2}\le C, \end{aligned}$$

(A.7)

with which, we multiply (A.2) by $e^{1^*}_p$, integrate the resulting equation and get

$$\begin{aligned} {}&\Vert \nabla e^{1^*}_p\Vert _{L^2} \le C\bigg \Vert \bigg (\frac{k(x)}{\mu ({{\mathcal {C}}}^{1^*})}-\frac{k(x)}{\mu (c^1)}\bigg )\nabla p^1\bigg \Vert _{L^2} \le C\Vert e^{1^*}_c\Vert _{L^2}\Vert \nabla p^1\Vert _{L^{\infty }} \le C\Vert e^{1^*}_c\Vert _{L^2} \le C\tau ^2 \end{aligned}$$

(A.8)

$$\begin{aligned} {}&\Vert e^{1^*}_u\Vert _{L^2} \le C(\Vert \nabla e^{1^*}_p\Vert _{L^2}+\Vert e^{1^*}_c\Vert _{L^2}\Vert \nabla p^1\Vert _{L^{\infty }}) \le C\Vert e^{1^*}_c\Vert _{L^2}\le C\tau ^2. \end{aligned}$$

(A.9)

Moreover, (A.2) can be rewritten as follows

$$\begin{aligned} -\,\Delta e_p^{1^*} = \frac{\mu ({\mathcal {C}}^{1^*})}{k(x)} \left( \nabla \left( \frac{k(x)}{\mu ({\mathcal {C}}^{1^*})}\right) \cdot \nabla e_p^{1^*} \right) +\frac{\mu ({\mathcal {C}}^{1^*})}{k(x)}\left[ \nabla \cdot \bigg ( \bigg (\frac{k(x)}{\mu ({\mathcal {C}}^{1^*})}-\frac{k(x)}{\mu (c^1)}\bigg ) \nabla p^1 \bigg )\right] .\qquad \quad \end{aligned}$$

(A.10)

Applying Lemma 3.1 to the above equality and with (A.3), we can easily see that

$$\begin{aligned} \Vert e_p^{1^*}\Vert _{H^2} +\Vert e_u^{1^*}\Vert _{H^1} \le C\Vert e_c^{1^*}\Vert _{H^1} \le C\tau ^{\frac{3}{2}}. \end{aligned}$$

(A.11)

Furthermore, by the same method as used in (3.33)–(3.36), we can get from (A.3) and (A.10) that

$$\begin{aligned} \Vert e_p^{1^*}\Vert _{H^3}+\Vert e_u^{1^*}\Vert _{H^2}\le C\Vert e_c^{1^*}\Vert _{H^2}\le C\tau . \end{aligned}$$

The above result yields

$$\begin{aligned} \Vert \mathbf{U}^{1^*}\Vert _{L^\infty } \le \Vert \mathbf{u}^{1}\Vert _{L^\infty }+\Vert e_u^{1^*}\Vert _{L^\infty } \le \Vert \mathbf{u}^{1}\Vert _{L^\infty }+C\Vert e_u^{1^*}\Vert _{H^2} \le \Vert \mathbf{u}^{1}\Vert _{L^\infty }+C\tau \le C. \end{aligned}$$

(A.12)

Since ${\widehat{\mathbf{U}}}^{\frac{1}{2}}=\frac{1}{2}(\mathbf{U}^{1^*}+\mathbf{U}^0)$, from (A.11) and  (A.12), we have

$$\begin{aligned}&\Vert {\widehat{\mathbf{U}}}^{\frac{1}{2}}\Vert _{L^\infty } \le \frac{1}{2}(\Vert \mathbf{U}^{1^*}\Vert _{L^\infty }+\Vert \mathbf{U}^0\Vert _{L^\infty }) \le C \end{aligned}$$

(A.13)

$$\begin{aligned}&\Vert {\widehat{\mathbf{U}}}^{\frac{1}{2}}-\mathbf{u}^{\frac{1}{2}}\Vert _{H^1} \le \Vert {\widehat{\mathbf{U}}}^{\frac{1}{2}}-\overline{\mathbf{u}}^{\frac{1}{2}}\Vert _{H^1} +\Vert \overline{\mathbf{u}}^{\frac{1}{2}}-\mathbf{u}^{\frac{1}{2}}\Vert _{H^1} \le C\tau ^{\frac{3}{2}}. \end{aligned}$$

(A.14)

To prove the estimate of $\Vert e_c^1\Vert _{L^2}$ and $\Vert e_c^1\Vert _{H^2}$, we just apply the same method as used in (3.25) and (3.27). With the estimate (A.14), we can obtain the following results

$$\begin{aligned}&\Vert e^{1}_c\Vert _{L^2}+\tau ^{\frac{1}{2}}\Vert \nabla {e}_{c}^{1}\Vert _{L^2}\le C\tau ^2 \end{aligned}$$

(A.15)

$$\begin{aligned}&\Vert e_c^1\Vert _{H^1}+\tau ^{\frac{1}{2}}\Vert e_c^1\Vert _{H^2} \le C\tau ^2 \end{aligned}$$

(A.16)

$$\begin{aligned}&\Vert e^1_c\Vert _{H^2}\le C\tau ^{\frac{3}{2}}\le \frac{1}{4}, \end{aligned}$$

(A.17)

where we have noted $e_c^0=0$ and let $\tau \le \tau _2=(\frac{1}{4C})^{\frac{2}{3}}$. Thus,

$$\begin{aligned} \Vert {{\mathcal {C}}}^{1}\Vert _{H^2}\le \Vert c^1\Vert _{H^2}+\Vert e^{1}_c\Vert _{H^2} \le K+1. \end{aligned}$$

(A.18)

From (3.22), (3.24) and (3.35)–(3.36), we can see that

$$\begin{aligned}&\Vert e_u^1\Vert _{H^1}\le C(\Vert e_p^1\Vert _{H^2}+\Vert e_c^1\Vert _{H^1})\le C\Vert e_c^1\Vert _{H^1}\le C\tau ^2 \end{aligned}$$

(A.19)

$$\begin{aligned}&\Vert e_p^1\Vert _{H^3}\le C\Vert e_c^1\Vert _{H^2}\le C\tau ^{\frac{3}{2}}\le \frac{1}{4} \end{aligned}$$

(A.20)

$$\begin{aligned}&\Vert e_u^1\Vert _{H^2}\le C(\Vert e_p^1\Vert _{H^3}+\Vert e_c^1\Vert _{H^2})\le C\Vert e_c^1\Vert _{H^2}\le C\tau ^{\frac{3}{2}}\le \frac{1}{4} \end{aligned}$$

(A.21)

$$\begin{aligned}&\Vert D_\tau e_u^{1}\Vert _{L^\infty } \le \frac{1}{\tau }\Vert e_u^{1}\Vert _{W^{1,4}} \le \frac{C}{\tau }\Vert e_u^1\Vert _{H^1}^{\frac{1}{4}}\Vert e_u^1\Vert _{H^2}^{\frac{3}{4}} \le C\tau ^{\frac{5}{8}} \le \frac{1}{4}, \end{aligned}$$

(A.22)

when $\tau \le \tau _3=(\frac{1}{4C})^{\frac{8}{5}}$. Furthermore, taking $n=1$ in (3.38) leads to

$$\begin{aligned}&\Vert e_c^1\Vert _{W^{2,4}} \le C\tau , \end{aligned}$$

(A.23)

$$\begin{aligned}&\Vert {\mathcal {C}}^1\Vert _{W^{2,4}} \le \Vert c^1\Vert _{W^{2,4}} +\Vert e_c^1\Vert _{W^{2,4}} \le \Vert c^1\Vert _{W^{2,4}}+C\tau \le C. \end{aligned}$$

(A.24)

Thus, (3.14) holds for $m=1$.

B Proof of (3.53) for $m=1$

For $m=1$, since ${\widehat{\mathbf{U}}}_h^{\frac{1}{2}}=\frac{1}{2}(\mathbf{U}_h^{1^*}+\mathbf{U}^0_h)$, the proof of (3.53) depends on the error estimate of $({\mathcal {C}}_h^{1^*},\mathbf{U}_h^{1^*},P_h^{1^*})$. Let

$$\begin{aligned} \theta _c^{1^*}={\mathcal {C}}_h^{1^*}-R_h^{1^*}{\mathcal {C}}^{1^*},\quad \theta _p^{1^*}=P_h^{1^*}-Q_hP^{1^*}. \end{aligned}$$

The numerical scheme (2.6)–(2.7) and the time-discrete system  (2.14)–(2.15) yield the following error equations for $\theta _c^{1^*}$ and $\theta _p^{1^*}$,

$$\begin{aligned}&\frac{\Phi }{\tau }(\theta _c^{1^*}, \phi _h) +(D(\mathbf{U}^0_h)\nabla \theta _c^{1^*}, \nabla \phi _h)\nonumber \\&= -\,\frac{1}{2}((\mathbf{U}^0_h-\mathbf{U}^0)\cdot \nabla {\mathcal {C}}_h^{1^*}, \phi _h) -\frac{1}{2}(\mathbf{U}^0\cdot \nabla ({\mathcal {C}}^{1^*}_h-{\mathcal {C}}^{1^*}), \phi _h) +\frac{1}{2}((\mathbf{U}_h^0-\mathbf{U}^0){\mathcal {C}}_h^{1^*} ,\nabla \phi _h) \nonumber \\&\quad +\,\frac{1}{2}(\mathbf{U}^0({\mathcal {C}}_h^{1^*}-{\mathcal {C}}^{1^*}),\nabla \phi _h) -\frac{1}{2}((q_I+q_P)({\mathcal {C}}^{1^*}_h-{\mathcal {C}}^{1^*}), \phi _h) +\frac{\Phi }{\tau }({\mathcal {C}}^{1^*}-R_h^{1^*}{\mathcal {C}}^{1^*}, \phi _h) \nonumber \\&\quad +\,\frac{\Phi }{\tau }({\mathcal {C}}_h^{0}-{\mathcal {C}}^{0}, \phi _h) -((D(\mathbf{U}^0_h)-D(\mathbf{U}^0))\nabla R_h^{1^*}{\mathcal {C}}^{1^*}, \nabla \phi _h)\nonumber \\&\quad -(D(\mathbf{U}^0)\nabla (R_h^{1^*}{\mathcal {C}}^{1^*}-R_h^0{\mathcal {C}}^{1^*}),\nabla \phi _h) \end{aligned}$$

(B.1)

$$\begin{aligned}&\left( \frac{k(x)}{\mu ({\mathcal {C}}_h^{1^*})}\nabla \theta _p^{1^*}, \nabla \varphi _h\right) = \left( \frac{k(x)}{\mu ({\mathcal {C}}^{1^*}_h)}\nabla (P^{1^*}-Q_hP^{1^*}), \nabla \varphi _h\right) \nonumber \\&\quad -\left( \left( \frac{k(x)}{\mu ({\mathcal {C}}^{1^*}_h)}-\frac{k(x)}{\mu ({\mathcal {C}}^{1^*})}\right) \nabla P^{1^*}, \nabla \varphi _h\right) . \end{aligned}$$

(B.2)

Since ${\mathcal {C}}_h^0$ is the Lagrangian interpolation of ${\mathcal {C}}^0$, we can easily get $\Vert \theta _c^0\Vert _{L^2}\le Ch^2\Vert {\mathcal {C}}^0\Vert _{H^2}\le Ch^2$ and $\Vert {\mathcal {C}}_h^0\Vert _{L^\infty }\le C\Vert {\mathcal {C}}^0\Vert _{H^2}\le C$. From (2.4), (2.12) and (3.52), we have

$$\begin{aligned}&\Vert \theta _p^0\Vert _{H^1} \le C(\Vert \theta _c^0\Vert _{L^2}+Ch^2)\le Ch^2 \\&\Vert \mathbf{U}_h^0-\mathbf{U}^0\Vert _{L^2}\le C(\Vert \nabla (P_h^0-P^0)\Vert _{L^2}+\Vert {\mathcal {C}}_h^0-{\mathcal {C}}^0\Vert _{L^2})\le Ch^2 \end{aligned}$$

and

$$\begin{aligned}&\Vert \nabla P_h^0\Vert _{L^\infty }\le \Vert \nabla Q_hP^0\Vert _{L^\infty }+Ch^{-\frac{d}{2}}\Vert \nabla \theta _p^0\Vert _{L^2}\le C \\&\Vert \mathbf{U}_h^0\Vert _{L^\infty }\le C\Vert \nabla P_h^0\Vert _{L^\infty }\le C. \end{aligned}$$

Then, taking $\phi _h=\theta _c^{1^*}$ in (B.1) results in

$$\begin{aligned}&\Vert \theta _c^{1^*}\Vert _{L^2}^2 +\tau \Vert \nabla \theta _c^{1^*} \Vert _{L^2}^2 \nonumber \\&\le C\Big ( \tau \Vert \mathbf{U}_h^0-\mathbf{U}^0\Vert _{L^\infty }\Vert \nabla \theta _c^{1^*}\Vert _{L^2}\Vert \theta _c^{1^*}\Vert _{L^2} +\tau \Vert \mathbf{U}_h^0-\mathbf{U}^0\Vert _{L^2}\Vert \nabla R_h^{1^*}{\mathcal {C}}^{1^*}\Vert _{L^3}\Vert \theta _c^{1^*}\Vert _{L^6}\nonumber \\&\quad +\tau \Vert \nabla \cdot \mathbf{U}^0\Vert _{L^3}\Vert {\mathcal {C}}_h^{1^*}-{\mathcal {C}}^{1^*}\Vert _{L^2}\Vert \theta _c^{1^*}\Vert _{L^6} +\tau \Vert \mathbf{U}^0\Vert _{L^\infty }\Vert {\mathcal {C}}_h^{1^*}-{\mathcal {C}}^{1^*}\Vert _{L^2}\Vert \nabla \theta _c^{1^*}\Vert _{L^2} \nonumber \\&\quad +\tau \Vert \mathbf{U}_h^0-\mathbf{U}^0\Vert _{L^2}\Vert R_h^{1^*}{\mathcal {C}}^{1^*}\Vert _{L^\infty }\Vert \nabla \theta _c^{1^*}\Vert _{L^2} +\tau \Vert q_I+q_P\Vert _{L^3}\Vert {\mathcal {C}}_h^{1^*}-{\mathcal {C}}^{1^*}\Vert _{L^2}\Vert \theta _c^{1^*}\Vert _{L^6} \nonumber \\&\quad +\Vert {\mathcal {C}}^{1^*}-R_h^{1^*}{\mathcal {C}}^{1^*}\Vert _{L^2}\Vert \theta _c^{1^*}\Vert _{L^2} +\Vert {\mathcal {C}}_h^0-{\mathcal {C}}^0\Vert _{L^2}\Vert \theta _c^{1^*}\Vert _{L^2} \nonumber \\&\quad +\tau \Vert \mathbf{U}_h^0-\mathbf{U}^0\Vert _{L^2}\Vert \nabla R_h^{1^*}{\mathcal {C}}^{1^*}\Vert _{L^\infty }\Vert \nabla \theta _c^{1^*}\Vert _{L^2} +\tau \Vert \nabla (R_h^{1^*}{\mathcal {C}}^{1^*}-R_h^0{\mathcal {C}}^{1^*})\Vert _{L^2}\Vert \nabla \theta _c^{1^*}\Vert _{L^2} \Big ) \nonumber \\&\le \frac{1}{2}\Vert \theta _c^{1^*}\Vert _{L^2}^2 +C\tau ^2\Vert \nabla \theta _c^{1^*}\Vert _{L^2}^2 +C(\tau h+h^2)^2, \end{aligned}$$

(B.3)

where we have used (3.43)–(3.44), (3.46), (3.48) and the fact $\nabla \cdot \mathbf{U^0}=q_I-q_P$. Let $\tau \le \tau _{10}=\frac{1}{2C}$, we can easily arrive at

$$\begin{aligned} \Vert \theta _c^{1^*}\Vert _{L^2}+\tau ^{\frac{1}{2}}\Vert \nabla \theta _c^{1^*}\Vert _{L^2} \le C(\tau h+h^2). \end{aligned}$$

(B.4)

Now, we prove the boundedness of $\Vert {\mathcal {C}}_h^{1^*}\Vert _{L^\infty }$ and $\Vert \mathbf{U}^{1^*}\Vert _{L^\infty }$. When $\tau \le h$, the above inequality implies that $\Vert \theta _c^{1^*}\Vert _{L^2}\le C h^2$ and

$$\begin{aligned} \Vert {\mathcal {C}}_h^{1^*}\Vert _{L^\infty } \le \Vert R_h^{1^*}{\mathcal {C}}^{1^*}\Vert _{L^\infty }+Ch^{-\frac{d}{2}}\Vert \theta _c^{1^*}\Vert _{L^2} \le C\Vert {\mathcal {C}}^{1^*}\Vert _{H^2}+Ch^{-\frac{d}{2}}h^2 \le C. \end{aligned}$$

Taking $\varphi _h=\theta _p^{1^*}$ in (B.2) yields

$$\begin{aligned}&\Vert \nabla \theta _p^{1^*}\Vert _{L^2} \le C(h^2\Vert P^{1^*}\Vert _{H^3} +\Vert {\mathcal {C}}^{1^*}-{\mathcal {C}}^{1^*}_h\Vert _{L^2}\Vert \nabla P^{1^*}\Vert _{L^{\infty }} ) \le Ch^2 \\&\Vert \nabla P_h^{1^*}\Vert _{L^\infty } \le \Vert \nabla Q_hP^{1^*}\Vert _{L^\infty }+Ch^{-\frac{d}{2}}\Vert \nabla \theta _p^{1^*}\Vert _{L^2} \le C. \end{aligned}$$

When $\tau \ge h$, (B.4) implies that $\Vert \theta _c^{1^*}\Vert _{H^1}\le C\tau ^{-\frac{1}{2}}(\tau h+h^2) \le C\tau ^{\frac{1}{2}} h$ and

$$\begin{aligned} \Vert {\mathcal {C}}_h^{1^*}\Vert _{L^\infty } \le \Vert R_h^{1^*}{\mathcal {C}}^{1^*}\Vert _{L^\infty }+Ch^{-\frac{1}{2}}\Vert \theta _c^{1^*}\Vert _{H^1} \le C\Vert {\mathcal {C}}^{1^*}\Vert _{H^2}+Ch^{-\frac{1}{2}}\tau ^{\frac{1}{2}} h \le C. \end{aligned}$$

Applying the same method as used in (3.59)-(3.61), we can get

$$\begin{aligned}&\Vert \nabla \theta _p^{1^*}\Vert _{L^6}\le Ch \\&\Vert \nabla P_h^{1^*}\Vert _{L^\infty }\le \Vert \nabla Q_hP^{1^*}\Vert _{L^\infty }+Ch^{-\frac{1}{2}}\Vert \nabla \theta _p^m\Vert _{L^6}\le C. \end{aligned}$$

Thus, we obtain the boundedness of $\Vert {\mathcal {C}}_h^{1^*}\Vert _{L^\infty }$ and $\Vert \nabla P_h^{1^*}\Vert _{L^\infty }$, with which and (2.5) and (2.16), we further have

$$\begin{aligned}&\Vert \mathbf{U}_h^{1^*}\Vert _{L^\infty } \le C\left\| \frac{k(x)}{\mu ({\mathcal {C}}_h^{1^*})}\nabla P_h^{1^*}\right\| _{L^\infty } \le C\left\| \nabla P_h^{1^*}\right\| _{L^\infty } \le C \end{aligned}$$

and

$$\begin{aligned} \Vert \mathbf{U}_h^{1^*}-\mathbf{U}_h^{1^*}\Vert _{L^2}&\le C(\Vert \nabla (P^{1^*}-P_h^{1^*})\Vert _{L^2}+\Vert {\mathcal {C}}^{1^*}-{\mathcal {C}}_h^{1^*}\Vert _{L^2}\Vert \nabla P^{1^*}\Vert _{L^\infty }) \\&\le C(h^2\Vert P^{1^*}\Vert _{H^3}+\Vert \theta _c^{1^*}\Vert _{L^2}+h^2\Vert {\mathcal {C}}^{1^*}\Vert _{H^2}) \\&\le C(\tau h+h^2). \end{aligned}$$

Since ${\widehat{\mathbf{U}}}_h^{\frac{1}{2}}=\frac{1}{2}(\mathbf{U}_h^{1^*}+\mathbf{U}_h^0)$, we arrive at

$$\begin{aligned}&\Vert {\widehat{\mathbf{U}}}_h^{\frac{1}{2}}\Vert _{L^\infty } \le \frac{1}{2}(\Vert \mathbf{U}_h^{1^*}\Vert _{L^\infty }+\Vert \mathbf{U}_h^0\Vert _{L^\infty }) \le C \end{aligned}$$

(B.5)

$$\begin{aligned}&\Vert {\widehat{\mathbf{U}}}_h^{\frac{1}{2}}-{\widehat{\mathbf{U}}}^{\frac{1}{2}}\Vert _{L^2} \le \frac{1}{2}\Vert \mathbf{U}_h^{1^*}-\mathbf{U}^{1^*}\Vert _{L^2} +\frac{1}{2}\Vert \mathbf{U}_h^{0}-\mathbf{U}^{0}\Vert _{L^2} \le C(\tau h+h^{2}). \end{aligned}$$

(B.6)

To prove (3.53) for $m=1$, we can apply the similar method as presented in the analysis of (3.65). By using (B.6) instead of (3.64) in the estimates of $|I_1(\overline{\theta }_c^{\frac{1}{2}})|$, $|I_3(\overline{\theta }_c^{\frac{1}{2}})|$ and $|I_7(\overline{\theta }_c^{\frac{1}{2}})|$, we obtain the following result from (3.51) (taking $n=1$),

$$\begin{aligned} \Vert \theta _c^1\Vert ^2_{L^2}&+\tau \Vert \nabla \overline{\theta }_h^{\frac{1}{2}} \Vert ^2_{L^2} \le \frac{\tau }{2}\Vert \nabla \overline{\theta }_c^{\frac{1}{2}}\Vert _{L^2}^2\\ {}&+C(\tau \Vert \theta _c^1\Vert _{L^2}^2+\tau \Vert D_\tau ({\mathcal {C}}^1-R_h^{1}{\mathcal {C}}^1)\Vert _{H^{-1}}^2 +\Vert \theta _c^{0}\Vert _{L^2}^2 +(\tau h+h^2)^2). \end{aligned}$$

Let $\tau \le \tau _{11}=\frac{1}{2C}$, the above inequality yields

$$\begin{aligned} \Vert \theta _c^1\Vert _{L^2} +\tau ^{\frac{1}{2}}\Vert \nabla \overline{\theta }_h^{\frac{1}{2}} \Vert _{L^2} \le C(\tau h+h^2) \le \tau h^{\frac{3}{4}}+h^{\frac{7}{4}}. \end{aligned}$$

(B.7)

This completes the proof of (3.53) for $m=1$.