A fast temporal second-order compact ADI difference scheme for the 2D multi-term fractional wave equation

Abstract

In this paper, a fast temporal second-order compact ADI scheme is proposed for the 2D time multi-term fractional wave equation. At the super-convergence point, the multi-term Caputo derivative is approximated by combining the order reduction technique with the sum-of-exponential approximation to the kernel function appeared in Caputo derivative. The difference scheme can be solved by the recursion, which reduces the storage and computational cost significantly. The obtained scheme is uniquely solvable. The unconditional convergence and stability of the scheme in the discrete H¹-norm are proved by the discrete energy method and the convergence accuracy is second-order in time and fourth-order in space. Numerical example illustrates the efficiency of the scheme.

Appendix. The proof of Theorem 2

Firstly, we present some lemmas which are necessary for proving Theorem 2.

Lemma 9

[8] Given any non-negative integer m and positive constants λ₀,λ₁,…,λ_m, for any γ_i ∈ (0, 1],i = 0, 1,…,m, it holds

$$ \begin{array}{@{}rcl@{}} \widehat{g}_{1}^{(k+1)}&>&\widehat{g}_{2}^{(k+1)}>{\cdots} >\widehat{g}_{k-1}^{(k+1)}>\widehat{g}_{k}^{(k+1)}\\ &>&\sum\limits_{r=0}^{m}\lambda_{r}\ \frac {\tau^{-\gamma_{r}}}{\Gamma (2-\gamma_{r})}\cdot\frac{1-\gamma_{r}}2(k-1+\sigma)^{-\gamma_{r}}. \end{array} $$

In addition, there exists a constant τ₀ such that when τ ≤ τ₀,

$$ (2\sigma-1)\widehat{g}_{0}^{(k+1)}-\sigma\widehat{g}_{1}^{(k+1)}>0 $$

and hence

$$ \widehat{g}_{0}^{(k+1)}>\widehat{g}_{1}^{(k+1)}. $$

Lemma 10

The sequence $\{\widehat {g}_{n}^{(k+1)}\}$ satisfies

$$ \sum\limits_{n=1}^{k}\widehat{g}_{n}^{(n+1)}\leq \ \sum\limits_{r=0}^{m}{\frac {\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma (2-\gamma_{r})}}(k+\sigma)^{1-\gamma_{r}}, $$

and

$$ \sum\limits_{n=1}^{k}\widehat{g}_{n-1}^{(n+1)}\leq \ \sum\limits_{r=0}^{m}\sum\limits_{r=0}^{m}{\frac{\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma(2-\gamma_{r})}(k+\sigma)^{1-\gamma_{r}}.} $$

Proof

Similar to the derivation of Corollary 2 in [2], $\{a_{n}^{(\gamma _{r})}\}$ and $\{b_{n}^{(\gamma _{r})}\}$ have the relation $b_{n}^{(\gamma _{r})}=a_{n}^{(\gamma _{r})}\Big (\kappa ^{(\gamma _{r})}_{n}-\frac {1}{2}\Big ),$ where $\frac {1}{2}\leq \kappa ^{(\gamma _{r})}_{n}\leq \frac {1}{2-\gamma _{r}}.$□

It follows that

$$ \begin{array}{@{}rcl@{}} \sum\limits_{n=1}^{k}\widehat{g}_{n}^{(n+1)}&=&\sum\limits_{r=0}^{m}\sum\limits_{n=1}^{k} \lambda_{r}g_{n}^{(n+1,\gamma_{r})} =\sum\limits_{r=0}^{m}\sum\limits_{n=1}^{k}\frac{\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma(2-\gamma_{r})}\Big(a_{n}^{(\gamma_{r})}-b_{n}^{(\gamma_{r})}\Big)\\ &=&\sum\limits_{r=0}^{m}\sum\limits_{n=1}^{k}\frac{\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma(2-\gamma_{r})}a_{n}^{(\gamma_{r})}\Big(\frac{3}{2}-\kappa^{(\gamma_{r})}_{n}\Big)\\ &=&\sum\limits_{r=0}^{m}\sum\limits_{n=1}^{k}\frac{\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma(2-\gamma_{r})}\Big(\frac{3}{2}-\kappa^{(\gamma_{r})}_{n}\Big) \Big((n+\sigma)^{1-\gamma_{r}}-(n-1+\sigma)^{1-\gamma_{r}}\Big)\\ &\leq&\sum\limits_{r=0}^{m}{\frac{\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma(2-\gamma_{r})}}(k+\sigma)^{1-\gamma_{r}}. \end{array} $$

For the second inequality, we have

$$ \begin{array}{@{}rcl@{}} &&\sum\limits_{n=1}^{k}\widehat{g}_{n-1}^{(n+1)}=\sum\limits_{r=0}^{m}\lambda_{r} g_{0}^{(2,\gamma_{r})}+ \sum\limits_{r=0}^{m}\sum\limits_{n=2}^{k}\lambda_{r} g_{n-1}^{(n+1,\gamma_{r})}\\ &=&\sum\limits_{r=0}^{m}\frac{\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma(2-\gamma_{r})}\Big(a^{(\gamma_{r})}_{0}+b^{(\gamma_{r})}_{1}\Big)+ \sum\limits_{r=0}^{m}\sum\limits_{n=2}^{k}\frac{\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma(2-\gamma_{r})}\Big(a_{n-1}^{(\gamma_{r})}+b_{n}^{(\gamma_{r})}-b_{n-1}^{(\gamma_{r})}\Big)\\ &=&{\sum\limits_{r=0}^{m}\frac{\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma(2-\gamma_{r})}\Big(\sum\limits_{n=0}^{k-1}a_{n}^{(\gamma_{r})}+b_{k}^{(\gamma_{r})}\Big) \leq\sum\limits_{r=0}^{m}\frac{\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma(2-\gamma_{r})}\sum\limits_{n=0}^{k}a_{n}^{(\gamma_{r})}}\\ &=&{\sum\limits_{r=0}^{m}\frac{\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma(2-\gamma_{r})} \Big(\sigma^{1-\gamma_{r}}+\sum\limits_{n=1}^{k}\big((n+\sigma)^{1-\gamma_{r}}-(n-1+\sigma)^{1-\gamma_{r}}\big)\Big)}\\ &=&{\sum\limits_{r=0}^{m}\frac{\lambda_{r}\tau^{-\gamma_{r}}}{\Gamma(2-\gamma_{r})}(k+\sigma)^{1-\gamma_{r}}.} \end{array} $$

Similar to the derivation of Lemma 4.1 in [33], we can obtain the following lemma.

Lemma 11

The sequences $\{g_{n}^{(k+1,\gamma _{r})}\}$ and $\{{~}^{\mathcal {F}}g_{n}^{(k+1,\gamma _{r})}\}$ satisfy

$$ {~}^{\mathcal{F}}g_{0}^{(1,\gamma_{r})}=g_{0}^{(1,\gamma_{r})}, $$

and for k ≥ 1,

$$ \begin{array}{@{}rcl@{}} \Big|{~}^{\mathcal{F}}g_{n}^{(k+1,\gamma_{r})}-g_{n}^{(k+1,\gamma_{r})}\Big|\leq \frac{\varepsilon}{\Gamma(1-\alpha)} \left\{\begin{array}{lll} &\frac{1}{4},~n=0, \\ &\frac{5}{4},~1\leq n\leq k-1,\\ &1,~n=k. \end{array}\right. \end{array} $$

Lemma 12

[9] For the coefficients $\big \{{~}^{\mathcal {F}}\widehat {g}_{n}^{(k+1)} (0\leq n\leq k, 0\leq k\leq N-1)\big \}$ defined in (8), and a sufficiently small ε, it holds

$$ \begin{array}{@{}rcl@{}} {~}^{\mathcal{F}}\widehat{g}_{1}^{(k+1)}\geq{~}^{\mathcal{F}}\widehat{g}_{2}^{(k+1)} \geq\cdots\geq{~}^{\mathcal{F}}\widehat{g}_{k}^{(k+1)}\geq C. \end{array} $$

In addition, there exists a constant τ₀ such that, when τ ≤ τ₀,

$$ \begin{array}{@{}rcl@{}} (2\sigma-1){~}^{\mathcal{F}}\widehat{g}_{0}^{(k+1)}-\sigma{~}^{\mathcal{F}}\widehat{g}_{1}^{(k+1)}\geq0, \end{array} $$

and hence

$$ \begin{array}{@{}rcl@{}} {~}^{\mathcal{F}}\widehat{g}_{0}^{(k+1)}>{~}^{\mathcal{F}}\widehat{g}_{1}^{(k+1)}. \end{array} $$

where C is a positive constant.

Lemma 13

The sequence $\{{~}^{\mathcal {F}}\widehat {g}_{n}^{(k)}\}$ satisfies

$$ \begin{array}{@{}rcl@{}} {~}^{\mathcal{F}}\widehat{g}_{n}^{(k+1)} =\left\{\begin{array}{ll} &{~}^{\mathcal{F}}{\widehat{g}}_{n}^{(k)},~~0\leq n\leq k-2,\\ &{~}^{\mathcal{F}}{\widehat{g}}_{k-1}^{(k)}+\widehat{B}_{k-1},~~n=k-1. \end{array}\right. \end{array} $$

(84)

where $\widehat {B}_{k-1}=\sum \limits _{r=0}^{m}\lambda _{r}^{(\gamma _{r})}\Big (\sum \limits _{l=1}^{N_{exp}^{(\gamma _{r})}}\hat {\omega }_{1}^{(\gamma _{r})}e^{-(k-1)s_{l}^{(\gamma _{r})}\tau }B_{l}^{(\gamma _{r})}\Big ).$ In addition,

$$ \sum\limits_{n=1}^{k}{~}^{\mathcal{F}}{\widehat{g}}_{n}^{(n+1)}\leq \sum\limits_{r=0}^{m}\frac {3\lambda_{r}\tau^{-\gamma_{r}}}{2{\Gamma} (2-\gamma_{r})}(k+\sigma)^{1-\gamma_{r}}+\kappa_{1}k\varepsilon, $$

and

$$ \sum\limits_{n=1}^{k}\widehat{B}_{n-1}\leq\sum\limits_{r=0}^{m}\frac{(4-\gamma_{r})\lambda_{r}\tau^{-\gamma_{r}}}{2{\Gamma}(3-\gamma_{r})}(k-1+\sigma)^{1-\gamma_{r}} +\kappa_{1}k\varepsilon, $$

where κ₁ is a positive constant.

Proof

The relation between $\{{~}^{\mathcal {F}}{\widehat {g}}_{n}^{(k+1)}\}$ and $\{{~}^{\mathcal {F}}{\widehat {g}}_{n}^{(k)}\}$ is obvious. We omit the details. □

It follows from Lemma 11 that

$$ {~}^{\mathcal{F}}{\widehat{g}}_{n}^{(k+1)}\leq {\widehat{g}}_{n}^{(k+1)}+\kappa_{1}\varepsilon,~0\leq n\leq k, $$

where κ₁ is a constant. By Lemma 10 and noticing that $\widehat {B}_{n-1}\leq {~}^{\mathcal {F}}\widehat {g}_{n-1}^{(n+1)} ~(1\leq n\leq k),$ it yields

$$ \sum\limits_{n=1}^{k}{~}^{\mathcal{F}}{\widehat{g}}_{n}^{(n+1)}\leq \sum\limits_{n=1}^{k}\widehat{g}_{n}^{(n+1)}+\kappa_{1}k\varepsilon \leq \sum\limits_{r=0}^{m}\frac {3\lambda_{r}\tau^{-\gamma_{r}}}{2{\Gamma} (2-\gamma_{r})}(k+\sigma)^{1-\gamma_{r}}+\kappa_{1}k\varepsilon, $$

and

$$ \begin{array}{@{}rcl@{}} &&\sum\limits_{n=1}^{k}\widehat{B}_{n-1}\leq \sum\limits_{n=1}^{k}{~}^{\mathcal{F}}\widehat{g}_{n-1}^{(n+1)}\leq \sum\limits_{n=1}^{k}\widehat{g}_{n-1}^{(n+1)}+\kappa_{1}k\varepsilon\\ &\leq& \sum\limits_{r=0}^{m}\frac{(4-\gamma_{r})\lambda_{r}\tau^{-\gamma_{r}}}{2{\Gamma}(3-\gamma_{r})}(k-1+\sigma)^{1-\gamma_{r}}+\kappa_{1}k\varepsilon. \end{array} $$

The coefficients $\{{~}^{\mathcal {F}}g_{n}^{(k+1,\gamma _{r})}\}$ defined in (8) have the similar property as $\{g_{n}^{(k+1,\gamma _{r})}\}$; thus, we have the following lemma.

Lemma 14

[8] Suppose 〈⋅,⋅〉_∗ is an inner product on $\mathcal {U}_{h},$ ∥⋅∥_∗ is its deduced norm. For any grid functions $v^{0},v^{1},\ldots ,v^{k+1}\in \mathcal {U}_{h},$ we have the following inequality

$$ \begin{array}{@{}rcl@{}} \Big\langle\sum\limits_{n=0}^{k}{~}^{\mathcal{F}}\widehat{g}_{n}^{(k+1)}(v^{n+1}-v^{n}), v^{k+\sigma}\Big\rangle_{*}\geq \frac {1}{2}\sum\limits_{n=0}^{k}{~}^{\mathcal{F}}\widehat{g}_{n}^{(k+1)}\big(\|v^{n+1}\|_{*}^{2}-\|v^{n}\|_{*}^{2}\big). \end{array} $$

Lemma 15

[26] For any grid functions $u^{0},u^{1},\ldots ,u^{N}\in \mathcal {U}_{h}$, we have the following inequality

$$ \begin{array}{@{}rcl@{}} \big(D_{\hat{t}}u^{k}, u^{k+\sigma}\big)\geq \frac{1}{4\tau}(E^{k+1}-E^{k}),~k\geq1, \end{array} $$

with

$$ E^{k+1}=(2\sigma+1)\|u^{k+1}\|^{2}-(2\sigma-1)\|u^{k}\|^{2}+(2\sigma^{2}+\sigma-1)\|u^{k+1}-u^{k}\|^{2},~k\geq0. $$

In addition, it holds

$$ E^{k+1}\geq\frac{1}{\sigma}\|u^{k+1}\|^{2},~k\geq0. $$

The proof of Theorem 2

. The proof is divided into two steps, which correspond to the case k = 0 and k ≥ 1. □

It follows from (62)–(65) that $~~~~u_{ij}^{k}=0,~~v_{ij}^{k}=0, \quad (i,j)\in \partial \omega ,~~0\leq k\leq N.$ Consequently,

$$ \begin{array}{@{}rcl@{}} && ({\Lambda}_{h}u^{1}, v^{1})=(u^{1}, {\Lambda}_{h}v^{1}), \end{array} $$

(A.1)

$$ \begin{array}{@{}rcl@{}} &&({\Lambda}_{h}u^{k+\sigma}, v^{k+\sigma})=(u^{k+\sigma}, {\Lambda}_{h}v^{k+\sigma}),\quad 1\le k\le N-1. \end{array} $$

(A.2)

• Step 1. When k = 0, the system is as follows

  $$ \begin{array}{@{}rcl@{}} &&\mathcal{A}{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}(v_{ij}^{1}-v_{ij}^{0}) +\frac{\sigma^{2}\tau^{2}}{4p_{0}}{\delta_{x}^{2}}{\delta_{y}^{2}}v_{ij}^{1} =\sigma{\Lambda}_{h}u_{ij}^{1}+(1-\sigma){\Lambda}_{h}u_{ij}^{0}\\ &&\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad +p_{ij}^{0},~(i,j)\in\omega, \end{array} $$

  (A.3)
  $$ \begin{array}{@{}rcl@{}} &&\delta_{t}u_{ij}^{\frac{1}{2}}=v_{ij}^{\frac{1}{2}}+g_{ij}^{0},~(i,j)\in\overline{\omega}, \end{array} $$

  (A.4)
  $$ \begin{array}{@{}rcl@{}} &&u_{ij}^{0}=w_{1}(x_{i},y_{j}),~v_{ij}^{0}=w_{2}(x_{i},y_{j}),~(i,j)\in\overline{\omega}, \end{array} $$

  (A.5)
  $$ \begin{array}{@{}rcl@{}} &&u_{ij}^{1}=0,~(i,j)\in\partial\omega. \end{array} $$

  (A.6)

• (1) Taking the inner product of (A.3) with v¹, we have

  $$ \begin{array}{@{}rcl@{}} &&{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}\|v^{1}\|_{\mathcal{A}}^{2} +\frac{\sigma^{2}\tau^{2}}{4p_{0}}\|\delta_{x}\delta_{y}v^{1}\|^{2}\\ &=&{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}(\mathcal{A}v^{0},v^{1}) + \sigma\big({\Lambda}_{h}u^{1},v^{1}\big) + (1-\sigma)({\Lambda}_{h}u^{0},v^{1}) + (p^{0},v^{1}). \end{array} $$

  (A.7)

  From (A.4), we have

  $$ \begin{array}{@{}rcl@{}} \delta_{t}{\Lambda}_{h}u_{ij}^{\frac{1}{2}}={\Lambda}_{h}v_{ij}^{\frac{1}{2}}+{\Lambda}_{h}g_{ij}^{0},~(i,j)\in\omega. \end{array} $$

  (A.8)

  Taking the inner product of (A.8) with − 2σu¹ and by the summation by parts, it yields

  $$ \begin{array}{@{}rcl@{}} -\frac{2\sigma}{\tau}({\Lambda}_{h}u^{1},u^{1}) &=&-\frac{2\sigma}{\tau}({\Lambda}_{h}u^{0},u^{1})-\sigma({\Lambda}_{h}v^{1},u^{1}) -\sigma({\Lambda}_{h}v^{0},u^{1})\\ &&-2\sigma({\Lambda}_{h}g^{0},u^{1}). \end{array} $$

  (A.9)

  Adding (A.7) with (A.9) and using (A.1), Young inequality, and Lemma 7, we obtain

  $$ \begin{array}{@{}rcl@{}} &&{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}\|v^{1}\|_{\mathcal{A}}^{2} +\frac{\sigma^{2}\tau^{2}}{4p_{0}}\|\delta_{x}\delta_{y}v^{1}\|^{2} +\frac{4\sigma}{3\tau}(\|\delta_{x}u^{1}\|^{2}+\|\delta_{y}u^{1}\|^{2})\\ &\leq&{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}(\mathcal{A}v^{0},v^{1})+(p^{0},v^{1}) +(1-\sigma)({\Lambda}_{h}u^{0},v^{1})-2\sigma({\Lambda}_{h}g^{0},u^{1})\\ &&-\frac{2\sigma}{\tau}({\Lambda}_{h}u^{0},u^{1})-\sigma({\Lambda}_{h}v^{0},u^{1})\\ &\leq&\Big(\frac{{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}{3}\|v^{1}\|_{\mathcal{A}}^{2} +\frac{3{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}{4}\|v^{0}\|_{\mathcal{A}}^{2}\Big) +\Big(\frac{{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}{9}\|v^{1}\|^{2}+\frac{9}{4{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}\|p^{0}\|^{2}\Big)\\ &&+\Big(\frac{{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}{9}\|v^{1}\|^{2} +\frac{9(1-\sigma)^{2}}{4{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}\|{\Lambda}_{h}u^{0}\|^{2}\Big) +\Big({\delta}\|u^{1}\|^{2}+\frac{\sigma^{2}}{{\delta}}\|{\Lambda}_{h}g^{0}\|^{2}\Big)\\ &&+\Big({\delta}\|u^{1}\|^{2}+\frac{\sigma^{2}}{{\delta}\tau^{2}}\|{\Lambda}_{h}u^{0}\|^{2}\Big) +\Big({\delta}\|u^{1}\|^{2}+\frac{\sigma^{2}}{4{\delta}}\|{\Lambda}_{h}v^{0}\|^{2}\Big) \\ &\leq&{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}\|v^{1}\|_{\mathcal{A}}^{2}+3{\delta}\|u^{1}\|^{2} +\frac{3{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}{4}\|v^{0}\|_{\mathcal{A}}^{2} +\frac{9(1-\sigma)^{2}}{4{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}\|{\Lambda}_{h}u^{0}\|^{2} \\ &&+\frac{9}{4{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}\|p^{0}\|^{2}+\frac{\sigma^{2}}{{\delta}}\|{\Lambda}_{h}g^{0}\|^{2}+\frac{\sigma^{2}}{{\delta}\tau^{2}}\|{\Lambda}_{h}u^{0}\|^{2} +\frac{\sigma^{2}}{4{\delta}}\|{\Lambda}_{h}v^{0}\|^{2}, \end{array} $$

  (A.10)

  where we use $\|v^{1}\|^{2}\leq 3\|v^{1}\|_{\mathcal {A}}^{2}.$

  Using $\|u^{1}\|^{2}\leq \frac {1}{\frac {6}{{L_{1}^{2}}}+\frac {6}{{L_{2}^{2}}}}(\|\delta _{x}u^{1}\|^{2}+\delta _{y}u^{1}\|^{2})$ and taking ${\delta }=\frac 1{9\tau }\cdot (\frac {6\sigma }{{L_{1}^{2}}}+\frac {6\sigma }{{L_{2}^{2}}}),$ we have

  $$ \begin{array}{@{}rcl@{}} \|\delta_{x}u^{1}\|^{2}+\|\delta_{y}u^{1}\|^{2}&\leq& \frac{3{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}\tau}{4\sigma}\|v^{0}\|_{\mathcal{A}}^{2} +\frac{9(1-\sigma)^{2}\tau}{4{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}\sigma}\|{\Lambda}_{h}u^{0}\|^{2} \\ &&+\frac{9\tau}{4{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}\sigma}\|p^{0}\|^{2}+\frac{3\tau^{2}}{\frac{2}{{L_{1}^{2}}}+\frac{2}{{L_{2}^{2}}}}\|{\Lambda}_{h}g^{0}\|^{2} \\ &&+\frac{3}{\frac{2}{{L_{1}^{2}}}+\frac{2}{{L_{2}^{2}}}}\|{\Lambda}_{h}u^{0}\|^{2}\\ &&+\frac{3}{\frac{8}{{L_{1}^{2}}}+\frac{8}{{L_{2}^{2}}}}\tau^{2}\|{\Lambda}_{h}v^{0}\|^{2}. \end{array} $$

  (A.11)

• (2) It follows from (A.8) that

  $$ \begin{array}{@{}rcl@{}} {\Lambda}_{h}u_{ij}^{1}={\Lambda}_{h}u_{ij}^{0}+\tau{\Lambda}_{h}v_{ij}^{\frac{1}{2}}+\tau {\Lambda}_{h}g_{ij}^{0},~~(i,j)\in\omega. \end{array} $$

  (A.12)

  Substituting (A.12) into (A.3), we have

  $$ \begin{array}{@{}rcl@{}} \mathcal{A}{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}(v_{ij}^{1}-v_{ij}^{0}) +\frac{\sigma^{2}\tau^{2}}{4p_{0}}{\delta_{x}^{2}}{\delta_{y}^{2}}v_{ij}^{1} &=&\sigma{\Lambda}_{h}u_{ij}^{0}+\sigma\tau{\Lambda}_{h}v_{ij}^{\frac{1}{2}}+(1-\sigma){\Lambda}_{h}u_{ij}^{0}\\ &&+{p_{i}^{0}}+\sigma\tau {\Lambda}_{h}{g_{i}^{0}},~~(i,j)\in\omega. \end{array} $$

  (A.13)

Taking the inner product of (A.13) with $v^{\frac {1}{2}},$ we obtain

$$ \begin{array}{@{}rcl@{}} &&{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}(\mathcal{A}v^{1}-\mathcal{A}v^{0},v^{\frac{1}{2}}) +\frac{\sigma^{2}\tau^{2}}{8p_{0}}\|\delta_{x}\delta_{y}v^{1}\|^{2}\\ \!&=&\!\frac{\sigma^{2}\tau^{2}}{8p_{0}}(\delta_{x}\delta_{y}v^{1}\!,\delta_{x}\delta_{y}v^{0}) + ({\Lambda}_{h}u^{0}\!,v^{\frac{1}{2}}) + \sigma\tau({\Lambda}_{h}v^{\frac{1}{2}},v^{\frac{1}{2}}) + (p^{0}\!,v^{\frac{1}{2}}) + \sigma\tau ({\Lambda}_{h}g^{0}\!,v^{\frac{1}{2}}) \\ \!&\le&\! \big({\Lambda}_{h}u^{0},v^{\frac{1}{2}}\big)+(p^{0},v^{\frac{1}{2}})+\sigma\tau ({\Lambda}_{h}g^{0},v^{\frac{1}{2}})+\frac{\sigma^{2}\tau^{2}}{8p_{0}}(\delta_{x}\delta_{y}v^{1},\delta_{x}\delta_{y}v^{0}) . \end{array} $$

By the summation by parts and Young inequality $ab\leq \frac {a^{2}}{2{\delta }}+\frac {{\delta } b^{2}}{2}$ (taking $\delta =\frac {{~}^{\mathcal {F}}\widehat {g}_{0}^{(1)}}{9}$), it yields

$$ \begin{array}{@{}rcl@{}} &&\frac{1}{2}{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}(\|v^{1}\|_{\mathcal{A}}^{2}-\|v^{0}\|_{\mathcal{A}}^{2}) +\frac{\sigma^{2}\tau^{2}}{8p_{0}}\|\delta_{x}\delta_{y}v^{1}\|^{2}\\ &\leq&\Big(\frac{1}{2{\delta}}\|{\Lambda}_{h}u^{0}\|^{2}+\frac{{\delta}}{2}\|v^{\frac{1}{2}}\|^{2}\Big) +\Big(\frac{1}{2{\delta}}\|p^{0}\|^{2}+\frac{{\delta}}{2}\|v^{\frac{1}{2}}\|^{2}\Big) \\ &&+ \Big(\frac{\sigma^{2}\tau^{2} }{2{\delta}}\|{\Lambda}_{h}g^{0}\|^{2}+\frac{{\delta}}{2}\|v^{\frac{1}{2}}\|^{2}\Big)+\frac{\sigma^{2}\tau^{2}}{16p_{0}}\Big(\|\delta_{x}\delta_{y}v^{1}\|^{2}+\|\delta_{x}\delta_{y}v^{0}\|^{2}\Big)\\ &\leq&\frac{3{\delta}}{2}\|v^{\frac{1}{2}}\|^{2} +\frac{1}{2{\delta}}\|{\Lambda}_{h}u^{0}\|^{2}+\frac{1}{2{\delta}}\|p^{0}\|^{2} +\frac{\sigma^{2}\tau^{2}}{2{\delta}}\|{\Lambda}_{h}g^{0}\|^{2}\\ &&+\frac{\sigma^{2}\tau^{2}}{16p_{0}}\Big(\|\delta_{x}\delta_{y}v^{1}\|^{2} +\|\delta_{x}\delta_{y}v^{0}\|^{2}\Big)\\ &\leq&\frac{{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}{4}\|v^{1}\|_{\mathcal{A}}^{2}+\frac{{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}{4}\|v^{0}\|_{\mathcal{A}}^{2} +\frac{9}{2{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}\|{\Lambda}_{h}u^{0}\|^{2} +\frac{9}{2{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}\|p^{0}\|^{2} \\ &&+\frac{9\sigma^{2}\tau^{2}}{2{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}}\|{\Lambda}_{h}g^{0}\|^{2}+\frac{\sigma^{2}\tau^{2}}{16p_{0}}\Big(\|\delta_{x}\delta_{y}v^{1}\|^{2}+\|\delta_{x}\delta_{y}v^{0}\|^{2}\Big), \end{array} $$

(A.14)

where we have used $\|v^{1}\|^{2}\leq 3\|v^{1}\|_{\mathcal {A}}$ in Lemma 7.

From (A.14), we obtain

$$ \begin{array}{@{}rcl@{}} &&\|v^{1}\|_{\mathcal{A}}^{2}+\frac{\sigma^{2}\tau^{2}}{4{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}p_{0}}\|\delta_{x}\delta_{y}v^{1}\|^{2}\\ &\leq& 3\|v^{0}\|_{\mathcal{A}}^{2}+\frac{18}{({~}^{\mathcal{F}}\widehat{g}_{0}^{(1)})^{2}}\|{\Lambda}_{h}u^{0}\|^{2} +\frac{18}{({~}^{\mathcal{F}}\widehat{g}_{0}^{(1)})^{2}}\|p^{0}\|^{2} +\frac{18\sigma^{2}\tau^{2}}{({~}^{\mathcal{F}}\widehat{g}_{0}^{(1)})^{2}}\|{\Lambda}_{h}g^{0}\|^{2}\\ &&+\frac{\sigma^{2}\tau^{2}}{4{~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}p_{0}}\|\delta_{x}\delta_{y}v^{0}\|^{2}. \end{array} $$

(A.15)

• Step 2.

  • (1) When k ≥ 1, taking the inner product (61) with v^k+σ, we obtain

    $$ \begin{array}{@{}rcl@{}} &&\Big(\mathcal{A}{\sum}_{n=0}^{k}{~}^{\mathcal{F}}\widehat{g}_{k-n}^{(k+1)}(v^{n+1}-v^{n}),v^{k+\sigma}\Big) \\ &&+\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}\Big({\delta_{x}^{2}}{\delta_{y}^{2}}v^{k+1},v^{k+\sigma}\Big)\\ &=&\Big({\Lambda}_{h}u^{k+\sigma},v^{k+\sigma}\Big) +\Big(p^{k},v^{k+\sigma}\Big),~1\leq k\leq N-1. \end{array} $$

    (A.16)

    By Lemma 14 and Lemma 13, we have

    $$ \begin{array}{@{}rcl@{}} &&\Big(\sum\limits_{n=0}^{k}{~}^{\mathcal{F}}\widehat{g}_{k-n}^{(k+1)}(\mathcal{A}v^{n+1}-\mathcal{A}v^{n}),v^{k+\sigma}\Big) \\ &\geq&\frac{1}{2}\sum\limits_{n=0}^{k}{~}^{\mathcal{F}}\widehat{g}_{k-n}^{(k+1)}\big(\|v^{n+1}\|_{A}^{2}-\|v^{n}\|_{\mathcal{A}}^{2}\big)\\ &=&\frac{1}{2}\Big(\sum\limits_{n=1}^{k+1}{~}^{\mathcal{F}}\widehat{g}_{k-n+1}^{(k+1)}\|v^{n}\|_{\mathcal{A}}^{2} -\sum\limits_{n=1}^{k}{~}^{\mathcal{F}}\widehat{g}_{k-n}^{(k)}\|v^{n}\|_{\mathcal{A}}^{2}-\widehat{B}_{k-1}\|v^{1}\|_{\mathcal{A}}^{2}\\ &&\qquad -{~}^{\mathcal{F}}\widehat{g}_{k}^{(k+1)}\|v^{0}\|_{\mathcal{A}}^{2}\Big),\\ && 1\leq k\leq N-1. \end{array} $$

    (A.17)

    It is easy to know that

    $$ \Big({\delta_{x}^{2}}{\delta_{y}^{2}}v^{k+1},v^{k+\sigma}\Big)=\sigma\|\delta_{x}\delta_{y}v^{k+1}\|^{2}+(1-\sigma) \Big(\delta_{x}\delta_{y}v^{k+1},\delta_{x}\delta_{y}v^{k}\Big) $$

    and

    $$ \begin{array}{@{}rcl@{}} \Big|\Big(p^{k},v^{k+\sigma}\Big)\Big|\leq {\delta}\|v^{k+\sigma}\|^{2}+\frac{1}{4{\delta}}\|p^{k}\|^{2}. \end{array} $$

    (A.18)

    Substituting (A.17) and (A.18) into (A.16), it yields

    $$ \begin{array}{@{}rcl@{}} &&\frac{1}{2}\Big(\sum\limits_{n=1}^{k+1}{~}^{\mathcal{F}}\widehat{g}_{k-n+1}^{(k+1)}\|v^{n}\|_{\mathcal{A}}^{2} -\sum\limits_{n=1}^{k}{~}^{\mathcal{F}}\widehat{g}_{k-n}^{(k)}\|v^{n}\|_{\mathcal{A}}^{2}-\widehat{B}_{k-1}\|v^{1}\|_{\mathcal{A}}^{2} \\ &&\qquad -{~}^{\mathcal{F}}\widehat{g}_{k}^{(k+1)}\|v^{0}\|_{\mathcal{A}}^{2}\Big)+\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}\sigma\|\delta_{x}\delta_{y}v^{k+1}\|^{2}\\ &\leq&{-}\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}(1-\sigma)\Big(\delta_{x}\delta_{y}v^{k+1},\delta_{x}\delta_{y}v^{k}\Big)+ \Big({\Lambda}_{h}u^{k+\sigma},v^{k+\sigma}\Big)\\ &&+{\delta}\|v^{k+\sigma}\|^{2}+\frac{1}{4{\delta}}\|p^{k}\|^{2},~~1\leq k\leq N-1. \end{array} $$

    (A.19)

  • (2) From (63), it yields that

    $$ \begin{array}{@{}rcl@{}} D_{\hat{t}}{\Lambda}_{h}u_{ij}^{k} = {\Lambda}_{h}v_{ij}^{k+\sigma} + {\Lambda}_{h}g_{ij}^{k},~~(i,j)\!\in\!\omega,~1\!\leq\! k\!\leq\! N - 1, \end{array} $$

    (A.20)

    Taking the inner product (A.20) with − u^k+σ, we get

    $$ \begin{array}{@{}rcl@{}} -\Big(D_{\hat{t}}{\Lambda}_{h}u^{k},u^{k+\sigma}\Big)&=&-\Big({\Lambda}_{h}v^{k+\sigma},u^{k+\sigma}\Big) -\Big({\Lambda}_{h}g^{k},u^{k+\sigma}\Big),\\ &&1\leq k\leq N-1. \end{array} $$

    (A.21)

Using Lemma 15, it yields

$$ \begin{array}{@{}rcl@{}} -\Big(D_{\hat{t}}{\Lambda}_{h}u^{k},u^{k+\sigma}\Big)&=&\Big(D_{\hat{t}}\mathcal{A}_{1}\delta_{y}u^{k},\delta_{y}u^{k+\sigma}\Big) +\Big(D_{\hat{t}}\mathcal{A}_{2}\delta_{x}u^{k},\delta_{x}u^{k+\sigma}\Big)\\ &=&\langle D_{\hat{t}}\delta_{y}u^{k},\delta_{y}u^{k+\sigma}\rangle_{\mathcal{A}_{1}} +\langle D_{\hat{t}}\delta_{x}u^{k},\delta_{x}u^{k+\sigma}\rangle_{\mathcal{A}_{2}}\\ &\geq& \frac{1}{4\tau}(F^{k+1}-F^{k}), \end{array} $$

(A.22)

where

$$ \begin{array}{@{}rcl@{}} F^{k+1}&=&(2\sigma+1)\|\delta_{x}u^{k+1}\|_{\mathcal{A}_{2}}^{2}-(2\sigma-1)\|\delta_{x}u^{k}\|_{\mathcal{A}_{2}}^{2} +(2\sigma^{2}+\sigma-1)\|\delta_{x}u^{k+1}\\ &&-\delta_{x}u^{k}\|_{\mathcal{A}_{2}}^{2}+(2\sigma+1)\|\delta_{y}u^{k+1}\|_{\mathcal{A}_{1}}^{2}-(2\sigma-1)\|\delta_{y}u^{k}\|_{\mathcal{A}_{1}}^{2} \\ &&+(2\sigma^{2}+\sigma-1)\|\delta_{y}u^{k+1}-\delta_{y}u^{k}\|_{\mathcal{A}_{1}}^{2}. \end{array} $$

(A.23)

and

$$ \begin{array}{@{}rcl@{}} F^{k+1}&\geq& \frac{1}{\sigma} (\|\delta_{x}u^{k+1}\|_{\mathcal{A}_{2}}^{2}+\|\delta_{y}u^{k+1}\|_{\mathcal{A}_{1}}^{2}) \geq\frac{1}{3\sigma} (\|\delta_{x}u^{k+1}\|^{2}+\|\delta_{y}u^{k+1}\|^{2}),\\ && k\geq0. \end{array} $$

(A.24)

By Cauchy-Schwarz inequality, we have

$$ \begin{array}{@{}rcl@{}} \Big|-\Big({\Lambda}_{h}g^{k},u^{k+\sigma}\Big)\Big|\leq \frac{1}{2}\|{\Lambda}_{h}g^{k}\|^{2}+\frac{1}{2}\|u^{k+\sigma}\|^{2},~~1\leq k\leq N-1. \end{array} $$

(A.25)

Substituting (A.22) and (A.25) into (A.21), it yields

$$ \begin{array}{@{}rcl@{}} \frac{1}{4\tau}(F^{k+1}-F^{k})&\leq& -\Big({\Lambda}_{h}v^{k+\sigma},u^{k+\sigma}\Big) +\frac{1}{2}\|{\Lambda}_{h}g^{k}\|^{2}+\frac{1}{2}\|u^{k+\sigma}\|^{2},\\ &\leq& k\leq N-1. \end{array} $$

(A.26)

Adding (A.19) with (A.26) and using (A.2), we obtain

$$ \begin{array}{@{}rcl@{}} &&\frac{1}{2}\Big(\sum\limits_{n=1}^{k+1}{~}^{\mathcal{F}}\widehat{g}_{k-n+1}^{(k+1)}\|v^{n}\|_{\mathcal{A}}^{2} -\sum\limits_{n=1}^{k}{~}^{\mathcal{F}}\widehat{g}_{k-n}^{(k)}\|v^{n}\|_{\mathcal{A}}^{2}-\widehat{B}_{k-1}\|v^{1}\|_{\mathcal{A}}^{2} -{~}^{\mathcal{F}}\widehat{g}_{k}^{(k+1)}\|v^{0}\|_{\mathcal{A}}^{2}\Big)\\ &&+\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}\sigma\|\delta_{x}\delta_{y}v^{k+1}\|^{2}+\frac{1}{4\tau}(F^{k+1}-F^{k})\\ &\leq&\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}(1-\sigma)\Big({\delta_{1}}\|\delta_{x}\delta_{y}v^{k+1}\|^{2} +\frac{1}{4{\delta_{1}}}\|\delta_{x}\delta_{y}v^{k}\|^{2}\Big) +{\delta}\|v^{k+\sigma}\|^{2}\\ &&+\frac{1}{4{\delta}}\|p^{k}\|^{2} +\frac{1}{2}\|{\Lambda}_{h}g^{k}\|^{2}+\frac{1}{2}\|u^{k+\sigma}\|^{2},~~1\leq k\leq N-1. \end{array} $$

Taking ${\delta _{1}}=\frac {\sigma +\sqrt {2\sigma -1}}{2(1-\sigma )},$ it yields

$$ \begin{array}{@{}rcl@{}} &&\frac{1}{2}\Big(\sum\limits_{n=1}^{k+1}{~}^{\mathcal{F}}\widehat{g}_{k-n+1}^{(k+1)}\|v^{n}\|_{\mathcal{A}}^{2} -\sum\limits_{n=1}^{k}{~}^{\mathcal{F}}\widehat{g}_{k-n}^{(k)}\|v^{n}\|_{\mathcal{A}}^{2}-\widehat{B}_{k-1}\|v^{1}\|_{\mathcal{A}}^{2} -{~}^{\mathcal{F}}\widehat{g}_{k}^{(k+1)}\|v^{0}\|_{\mathcal{A}}^{2}\Big)\\ &&+\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}\frac{\sigma-\sqrt{2\sigma-1}}{2}\|\delta_{x}\delta_{y}v^{k+1}\|^{2} +\frac{1}{4\tau}(F^{k+1}-F^{k})\\ &\leq&\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}\frac{\sigma-\sqrt{2\sigma-1}}{2}\|\delta_{x}\delta_{y}v^{k}\|^{2} +{\delta}\|v^{k+\sigma}\|^{2}+\frac{1}{4{\delta}}\|p^{k}\|^{2}\\ &&+\frac{1}{2}\|{\Lambda}_{h}g^{k}\|^{2}+\frac{1}{2}\|u^{k+\sigma}\|^{2}, ~~1\leq k\leq N-1. \end{array} $$

(A.27)

Denote

$$ H^{k+1}=2\tau\sum\limits_{n=1}^{k+1}{~}^{\mathcal{F}}\widehat{g}_{k-n+1}^{(k+1)}\|v^{n}\|_{\mathcal{A}}^{2} +{4}\tau\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}(\sigma-\sqrt{2\sigma-1})\|\delta_{x}\delta_{y}v^{k+1}\|^{2}+F^{k+1}. $$

Then, (A.27) can be rewritten as

$$ \begin{array}{@{}rcl@{}} H^{k+1}&\leq& H^{k}+2\tau \widehat{B}_{k-1}\|v^{1}\|_{\mathcal{A}}^{2}+ 2\tau{~}^{\mathcal{F}}\widehat{g}_{k}^{(k+1)}\|v^{0}\|_{\mathcal{A}}^{2} +4\tau{\delta}\|v^{k+\sigma}\|^{2}+\frac{\tau}{{\delta}}\|p^{k}\|^{2}\\ &&+2\tau\|{\Lambda}_{h}g^{k}\|^{2}+2\tau\|u^{k+\sigma}\|^{2}\\ &\leq&H^{1}+2\tau \sum\limits_{n=1}^{k}\widehat{B}_{n-1}\|v^{1}\|_{\mathcal{A}}^{2}+2\tau\sum\limits_{n=1}^{k}{~}^{\mathcal{F}}\widehat{g}_{n}^{(n+1)}\|v^{0}\|_{\mathcal{A}}^{2} +8\tau{\delta}\sum\limits_{n=1}^{k+1}\|v^{n}\|^{2}\\ &&+\frac{\tau}{{\delta}}\sum\limits_{n=1}^{k}\|p^{n}\|^{2}+2\tau\sum\limits_{n=1}^{k}\|{\Lambda}_{h}g^{n}\|^{2}+4\tau\sum\limits_{n=1}^{k+1}\|u^{n}\|^{2},\\ &&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~1\leq k\leq N-1. \end{array} $$

(A.28)

By Lemma 12, (A.23), and (A.24) and when τ ≤ τ₀, we have

$$ \begin{array}{@{}rcl@{}} H^{k+1}&\geq &2C \tau\sum\limits_{n=1}^{k+1}\|v^{n}\|_{\mathcal{A}}^{2}+\frac{1}{3\sigma}(\|\delta_{x}u^{k+1}\|^{2}+\|\delta_{y}u^{k+1}\|^{2})\\ &&+{4}\tau\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}(\sigma-\sqrt{2\sigma-1})\|\delta_{x}\delta_{y}v^{k+1}\|^{2}, ~~1\leq k\leq N-1 \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} H^{1}&=&2\tau {~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}\|v^{1}\|_{\mathcal{A}}^{2} +{4}\tau\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}(\sigma-\sqrt{2\sigma-1})\|\delta_{x}\delta_{y}v^{1}\|^{2} +F^{1}\\ &\leq&2\tau\Big[ {~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}\|v^{1}\|_{\mathcal{A}}^{2} +\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}(\sigma-\sqrt{2\sigma-1})\|\delta_{x}\delta_{y}v^{1}\|^{2}\Big]\\ &&+(4\sigma^{2}+4\sigma-1)(\|\delta_{x}u^{1}\|^{2}+\|\delta_{y}u^{1}\|^{2}) +(4\sigma^{2}-1)(\|\delta_{x}u^{0}\|^{2}+\|\delta_{y}u^{0}\|^{2}). \end{array} $$

Substituting the above inequalities into (A.28), it yields

$$ \begin{array}{@{}rcl@{}} && 2C\tau\sum\limits_{n=1}^{k+1}\|v^{n}\|_{\mathcal{A}}^{2}+\frac{1}{3\sigma}(\|\delta_{x}u^{k+1}\|^{2}+\|\delta_{y}u^{k+1}\|^{2})\\ &&+{4}\tau\frac{4\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}(\sigma-\sqrt{2\sigma-1})\|\delta_{x}\delta_{y}v^{k+1}\|^{2}\\ &\leq&2\tau\Big[ {~}^{\mathcal{F}}\widehat{g}_{0}^{(1)}\|v^{1}\|_{\mathcal{A}}^{2} +\frac{{8}\sigma^{4}\tau^{2}}{(2\sigma+1)^{2}p}(\sigma-\sqrt{2\sigma-1})\|\delta_{x}\delta_{y}v^{1}\|^{2}\Big]\\ &&+(4\sigma^{2}+4\sigma-1)(\|\delta_{x}u^{1}\|^{2}+\|\delta_{y}u^{1}\|^{2})\\ &&+(4\sigma^{2}-1)(\|\delta_{x}u^{0}\|^{2}+\|\delta_{y}u^{0}\|^{2}) +2\tau \sum\limits_{n=1}^{k}\widehat{B}_{n-1}\|v^{1}\|_{\mathcal{A}}^{2} +2\tau\sum\limits_{n=1}^{k}{~}^{\mathcal{F}}\widehat{g}_{n}^{(n+1)}\|v^{0}\|_{\mathcal{A}}^{2} \\&&+8\tau{\delta}\sum\limits_{n=1}^{k+1}\|v^{n}\|^{2}+\frac{\tau}{{\delta}}\sum\limits_{n=1}^{k}\|p^{n}\|^{2} +2\tau\sum\limits_{n=1}^{k}\|{\Lambda}_{h}g^{n}\|^{2}+4\tau\sum\limits_{n=1}^{k+1}\|u^{n}\|^{2}. \end{array} $$

(A.29)

By Lemma 13, we obtain

$$ \tau\sum\limits_{n=1}^{k}{~}^{\mathcal{F}}\widehat{g}_{n}^{(n+1)}\leq \tau\Big(\sum\limits_{r=0}^{m}\frac {3\lambda_{r}\tau^{-\gamma_{r}}}{2{\Gamma} (2-\gamma_{r})}(k+\sigma)^{1-\gamma_{r}}+\kappa_{1}k\varepsilon\Big)\leq C_{1}, $$

and

$$ \tau\sum\limits_{n=1}^{k}\widehat{B}_{n-1}\leq \tau\Big(\sum\limits_{r=0}^{m}\frac{(4-\gamma_{r})\lambda_{r}\tau^{-\gamma_{r}}}{2{\Gamma}(3-\gamma_{r})}(k-1+\sigma)^{1-\gamma_{r}}+\kappa_{1}k\varepsilon\Big)\leq C_{1}, $$

where C₁ is a constant.

Taking δ = C/8 and using Lemma 7, Lemma 13, (A.11) and (A.15), we have

$$ \begin{array}{@{}rcl@{}} &&\|\delta_{x}u^{k+1}\|^{2}+\|\delta_{y}u^{k+1}\|^{2}\leq 12\sigma\tau \sum\limits_{n=1}^{k+1}\|u^{n}\|^{2}+c_{3}G_{k+1}\\ &\leq&\frac{2\sigma\tau}{\frac{3}{{L_{1}^{2}}}+\frac{3}{{L_{2}^{2}}}}\sum\limits_{n=1}^{k+1}(\|\delta_{x}u^{n}\|^{2}+\|\delta_{y}u^{n}\|^{2})+c_{3}G_{k+1}, 1\leq k\leq N-1, \end{array} $$

where c₃ is a constant.

By Lemma 8, it follows that

$$ \begin{array}{@{}rcl@{}} \|\delta_{x}u^{k+1}\|^{2}+\|\delta_{y}u^{k+1}\|^{2}\leq c_{3} \exp\Big(\frac{2\sigma}{\frac{3}{{L_{1}^{2}}}+\frac{3}{{L_{2}^{2}}}}T\Big)G_{k+1},~~1\leq k\leq N-1. \end{array} $$

(A.30)

Substituting (A.30) into (A.29), there exists a constant c₄ such that

$$ \begin{array}{@{}rcl@{}} \tau\sum\limits_{n=1}^{k+1}\|v^{n}\|^{2}\leq c_{4}G_{k+1}, 1\leq k\leq N-1. \end{array} $$

This completes the proof.