Superconvergence Analysis of the Ultra-Weak Local Discontinuous Galerkin Method for One Dimensional Linear Fifth Order Equations

Abstract

In this paper, we analyze the superconvergence of the semi-discrete ultra-weak local discontinuous Galerkin (UWLDG) method for one dimensional time-dependent linear fifth order equations. The UWLDG method is designed to solve equations with high order spatial derivatives. The main idea is to rewrite the higher order equation into a lower order system. When we use the UWLDG method to solve the fifth order equations, we rewrite it as a system with two second order equations and one first order equation. Compared with the other works about superconvergence of the DG method, the main challenge is to define correction functions and a special interpolation function for the system containing equations with different orders. We divide our analysis into five cases according to $k\pmod {5}$, where k is the highest degree of polynomials in our function space, and obtain 2k-th order superconvergence for cell averages and function values at the cell boundaries and $k+2$-th order superconvergence for function values at some special quadrature points. For numerical solutions of the two second order equations, we prove that the first derivatives have superconvergence of order 2k at cell boundaries and order $k+1$ at a class of special quadrature points. All theoretical results are confirmed by numerical experiments.

Proof of a Few Technical Lemmas and Theorems

1.1 The Proof of Lemma 3.3

Proof

• (i) Let ${\bar{\zeta }}_x=\zeta _x+\frac{1}{|\Omega |}\sum \limits _{j=1}^{N} \llbracket {\zeta }\rrbracket _{{j-\frac{1}{2}}}$, then $\Vert {\bar{\zeta }}_{xx}\Vert _{I_{j}}=\Vert \zeta _{xx}\Vert _{I_{j}}$, $\llbracket {{\bar{\zeta }}_x}\rrbracket _{{j+\frac{1}{2}}}= \llbracket {\zeta _x}\rrbracket _{{j+\frac{1}{2}}}$, by Lemma 3.1 we have

  $$\begin{aligned} \Vert {\bar{\zeta }}_{xx}\Vert _{I_{j}}+h^{-\frac{1}{2}}\left| \llbracket {{\bar{\zeta }}_{x}}\rrbracket _{{j-\frac{1}{2}}}\right| \le C\Vert f\Vert _{I_{j}}. \end{aligned}$$

  Since $\int _\Omega {\bar{\zeta }}_xdx=0$, by the discrete Poincaré inequality (3.7) and Lemma 3.1, we have (3.8).

• (ii) If $f\bot {\mathcal {P}}^0(I_j)$, then we have $(f,\phi )_j=-({\bar{h}}_jD^{-1}f, \phi _x)_j$, where $\displaystyle (D^{-1}f)(x)=\frac{1}{{\bar{h}}_j}\int _{x_{{j-\frac{1}{2}}}}^xf(s)ds$. We integrate by parts to get

  $$\begin{aligned} {\mathcal {D}}^2_{j}(\zeta ,\phi )&=-(\zeta _x,\phi _{x})_{j}-\zeta ^+\phi _{x}^{+}|_{{j-\frac{1}{2}}} +\zeta _x^{+}\phi ^{-}|_{{j+\frac{1}{2}}}-\zeta _{x}^{+}\phi ^{+}|_{{j-\frac{1}{2}}}+\zeta ^-\phi _{x}^{+}|_{{j-\frac{1}{2}}}. \end{aligned}$$

  Taking $\phi =-\zeta$ and summing over all j from 1 to N yields

  $$\begin{aligned} (\zeta _x,\zeta _x)+2\sum _{j=1}^{N}\zeta _{x}^{+} \llbracket {\zeta }\rrbracket \Big |_{{j-\frac{1}{2}}}=\sum _{j=1}^N({\bar{h}}_jD^{-1}f, \zeta _x)_j. \end{aligned}$$

  By Lemma 3.1 we have $h^{-\frac{3}{2}}\left( \sum _{j=1}^{N} \llbracket {\zeta }\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}\lesssim \Vert f\Vert$, and thus,

  $$\begin{aligned} \Vert \zeta _x\Vert ^2&\le h\Vert D^{-1}f\Vert \Vert \zeta _x\Vert \!\!+\!\!2\sum _{j=1}^{N}\left| \zeta _{x}^{+}\right| \big | \llbracket {\zeta }\rrbracket \big |\Big |_{{j-\frac{1}{2}}}\\&\lesssim h\Vert f\Vert \Vert \zeta _x\Vert \!+\!h^{-\frac{1}{2}}\Vert \zeta _x\Vert \left( \sum _{j=1}^{N} \llbracket {\zeta }\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}\lesssim h\Vert f\Vert \Vert \zeta _x\Vert . \end{aligned}$$

  Thus, we obtain (3.9).

• (iii) Similarly, we take $\phi =-\zeta$ and sum over all j to obtain

  $$\begin{aligned} (\zeta _x,\zeta _x)+2\sum _{j=1}^{N}\zeta _{x}^{+} \llbracket {\zeta }\rrbracket \Big |_{{j-\frac{1}{2}}}=\sum _{j=1}^N({\bar{h}}_jD^{-1}f_1, \zeta _x)_j-(f_2, \zeta ). \end{aligned}$$

  Since $\int _{\Omega }\zeta dx=0$, by the discrete Poincaré inequality (3.7), we have $\Vert \zeta \Vert \lesssim \Vert \zeta _x\Vert +h^{-\frac{1}{2}}\left( \sum _{j=1}^{N} \llbracket {\zeta }\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}},$ which yields, together with the inequality $h^{-\frac{3}{2}}\left( \sum _{j=1}^{N} \llbracket {\zeta }\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}\lesssim \Vert f_1\Vert +\Vert f_2\Vert$ in Lemma 3.1,

  $$\begin{aligned} \Vert \zeta _x\Vert ^2&\lesssim h\Vert f_1\Vert \Vert \zeta _x\Vert \!+\!h^{-\frac{1}{2}}\Vert \zeta _x\Vert \left( \sum _{j=1}^{N} \llbracket {\zeta }\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}+\Vert f_2\Vert \Vert \zeta _x\Vert +h^{-\frac{1}{2}}\Vert f_2\Vert \left( \sum _{j=1}^{N} \llbracket {\zeta }\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}\\&\lesssim \Vert \zeta _x\Vert \big (h\Vert f_1\Vert \!+\Vert f_2\Vert \big )+\big (h\Vert f_1\Vert \!+\Vert f_2\Vert \big )^2. \end{aligned}$$

  Then the desired results (3.10) follows directly from Cauchy–Schwarz inequality. $\square$

1.2 The Proof of Lemma 4.1

Proof

We use the method of mathematical induction to prove this lemma. For $i=0$, by the properties of projections $P_h^{-}$ and $P_h^{\star }$, we have

$$\begin{aligned} \omega _u^{(0)}\bot {\mathcal {P}}^{k-2}(I_j),\quad \omega _v^{(0)}\bot {\mathcal {P}}^{k-1}(I_j),\quad \omega _w^{(0)}\bot {\mathcal {P}}^{k-2}(I_j). \end{aligned}$$

By the definition of the correction functions, we get

$$\begin{aligned} \omega _u^{(1)}\bot {\mathcal {P}}^{k-3}(I_j),\quad \omega _v^{(1)}\bot {\mathcal {P}}^{k-3}(I_j),\quad \omega _w^{(1)}\bot {\mathcal {P}}^{k-4}(I_j). \end{aligned}$$

Similarly, there hold

$$\begin{aligned} \omega _u^{(2)}\bot {\mathcal {P}}^{k-5}(I_j),\quad \omega _v^{(2)}\bot {\mathcal {P}}^{k-5}(I_j),\quad \omega _w^{(2)}\bot {\mathcal {P}}^{k-5}(I_j). \end{aligned}$$

Next, we assume that (4.2) is satisfied for $i-1$, then by the definition of $\omega _w^{(3i)}$,

$$\begin{aligned} (\omega _w^{(3i)}, \varphi _{xx})_j=-((\omega _u^{(3(i-1)+2)})_t,\varphi )_j,\quad \omega _w^{(3i)}(x_{{j+\frac{1}{2}}}^-)=0,~(\omega _w^{(3i)})_x(x_{{j-\frac{1}{2}}}^+)=0,\quad \forall \varphi \in V_h, \end{aligned}$$

$(\omega _u^{(3(i-1)+2)})_t\bot {\mathcal {P}}^{k-5i}(I_j)$, we obtain $\omega _w^{(3i)}\bot {\mathcal {P}}^{k-5i-2}(I_j)$. Similarly, we can prove $\omega _w^{(3i+1)}$, $\omega _w^{(3i+2)}$ and $\omega _q^{(m)}$ satisfy (4.2), where $q=u,v; ~m=3i,3i+1,3i+2$. In other words, (4.2) is also valid for i, and thus hold true for all $i\in {\mathbb {N}}$ with $3i, 3i+1,3i+2\le l^\star$. $\square$

1.3 The Proof of Lemma 4.4

Proof

We still adopt the method of mathematical induction to prove this lemma. At first, we have

$$\begin{aligned} \Vert \partial _t^{n}\omega _u^{(0)}\Vert&\lesssim h^{k+1}\Vert \partial _t^{n}u\Vert _{k+1},\\ \Vert \partial _t^{n}\omega _v^{(0)}\Vert&\lesssim h^{k+1}\Vert \partial _t^{n}v\Vert _{k+1},\;\;\Vert \partial _t^{n}\omega _w^{(0)}\Vert \lesssim h^{k+1}\Vert \partial _t^{n} w\Vert _{k+1}. \end{aligned}$$

Then, by Lemma 4.3 and the definition of the correction functions, we have

$$\begin{aligned} \Vert \omega _u^{(1)}\Vert&\lesssim h^{2}\Vert \omega _v^{(0)}\Vert \lesssim h^{k+3}\Vert v\Vert _{k+1}\lesssim h^{k+3}\Vert u\Vert _{k+3},\\ ~\Vert (\omega _u^{(1)})_t\Vert&\lesssim h^{2}\Vert (\omega _v^{(0)})_t\Vert \lesssim h^{k+3}\Vert v_t\Vert _{k+1}\lesssim h^{k+3}\Vert u_t\Vert _{k+3},\\ \Vert \omega _v^{(1)}\Vert&\lesssim h\Vert \omega _w^{(0)}\Vert \lesssim h^{k+2}\Vert w\Vert _{k+1}\lesssim h^{k+2}\Vert v\Vert _{k+2},\\ \Vert \omega _w^{(1)}\Vert&\lesssim h^{2}\Vert (\omega _u^{(0)})_t\Vert \lesssim h^{k+3}\Vert u_t\Vert _{k+1}\lesssim h^{k+3}\Vert w\Vert _{k+3}, \end{aligned}$$

and

$$\begin{aligned} \Vert \omega _u^{(2)}\Vert&\lesssim h^{2}\Vert \omega _v^{(1)}\Vert \lesssim h^{k+4}\Vert v\Vert _{k+2}\lesssim h^{k+4}\Vert u\Vert _{k+4},\\ \Vert (\omega _u^{(2)})_t\Vert&\lesssim h^2\Vert (\omega _v^{(1)})_t\Vert \lesssim h^{k+4}\Vert v_t\Vert _{k+2}\lesssim h^{k+4}\Vert u_t\Vert _{k+4},\\ \Vert \omega _v^{(2)}\Vert&\lesssim h\Vert \omega _w^{(1)}\Vert \lesssim h^{k+4}\Vert w\Vert _{k+3}\lesssim h^{k+4}\Vert v\Vert _{k+4},\\ \Vert \omega _w^{(2)}\Vert&\lesssim h^{2}\Vert (\omega _u^{(1)})_t\Vert \lesssim h^{k+5}\Vert u_t\Vert _{k+3}\lesssim h^{k+5}\Vert w\Vert _{k+5}. \end{aligned}$$

Therefore, the results in Lemma 4.4 are satisfied for $i=0$ . We assume the estimates are satisfied for $i=m-1$, then for $i=m$ we have,

$$\begin{aligned} \Vert \omega _u^{(3m)}\Vert&\lesssim h^{2}\Vert \omega _v^{(3m-1)}\Vert =h^2\Vert \omega _v^{(3(m-1)+2)}\Vert \lesssim h^{2+k+4+5(m-1)}\Vert v\Vert _{k+4+5(m-1)}\\&\lesssim h^{k+1+5m}\Vert u\Vert _{k+1+5m},\\ \Vert (\omega _u^{(3m)})_t\Vert&\lesssim h^{2}\Vert (\omega _v^{(3m-1)})_t\Vert \lesssim h^{k+1+5m}\Vert u_t\Vert _{k+1+5m},\\ \Vert \omega _v^{(3m)}\Vert&\lesssim h\Vert \omega _w^{(3m-1)}\Vert =h\Vert \omega _w^{(3(m-1)+2)}\Vert \lesssim h^{1+k+5+5(m-1)}\Vert w\Vert _{k+5+5(m-1)}\\&\lesssim h^{k+1+5m}\Vert v\Vert _{k+1+5m},\\ \Vert \omega _w^{(3m)}\Vert&\lesssim h^{2}\Vert (\omega _u^{(3m-1)})_t\Vert =h^{2}\Vert (\omega _u^{(3(m-1)+2)})_t\Vert \lesssim h^{2+k+4+5(m-1)}\Vert u_t\Vert _{k+4+5(m-1)}\\&\lesssim h^{k+1+5m}\Vert w\Vert _{k+1+5m}. \end{aligned}$$

Similarly, we can prove the estimates for $\Vert \partial _t^{n}\omega _u^{(3m+1)}\Vert , \Vert \partial _t^{n}\omega _u^{(3m+2)}\Vert , \Vert \omega _q^{(3m+1)}\Vert , \Vert \omega _q^{(3m+2)}\Vert$, $q=v,w$ are also satisfied. Therefore, the proof is completed. $\square$

1.4 The Proof of Theorem 5.1

Proof

The proof of this theorem contains the following five parts:

(1) Estimate $({\bar{\xi }}_u)_t(x,0)$:

By the definition of the $(u_h)_t(x,0)$, we have $\int _{\Omega }(u_h)_t(x,0)dx=0$ and $({\bar{\xi }}_u)_t(x,0)=\frac{1}{|\Omega |}\int _{\Omega }(\omega _u^{(l^\star )})_t(x,0)dx$.

• If $k=0,2,3\pmod {5}$, then $(\omega _u^{(l^\star )})_t(x,0)\bot {\mathcal {P}}^{0}(I_j),~\forall j\in Z_N$ and $\Vert ({\bar{\xi }}_u)_t\Vert (0)=0$;

• If $k=1,4\pmod {5}$, then $\Vert (\omega _u^{(l^\star )})_t\Vert (0)\lesssim h^{2k}\; \text {and}\; \Vert ({\bar{\xi }}_u)_t\Vert (0)\lesssim h^{2k}\Vert u_t\Vert _{2k}$.

(2) The definition of $w_h(x,0)$, $v_h(x,0)$, $u_h(x,0)$:

$w_h(x,0)$ is the solution of the following equations:

$v_h(x,0)$ is the solution of the following equations:

$u_h(x,0)$ is the solution of the following equations:

We can prove $w_h(x,0)$, $v_h(x,0)$ and $u_h(x,0)$ are well defined, and the more details of proof are given in [30, Appendix A.3, Proof of Lemma 5.1] and [34, Appendix A.2, Proof of Lemma 3.4].

(3) Estimate ${\bar{\xi }}_w(x,0)$:

By the error Eq. (5.5a), we have

$$\begin{aligned} {\mathcal {D}}^2_{j}({\bar{\xi }}_w(x,0),\varphi )=((\omega _u^{(l^\star )})_t(x,0),\varphi )-(({\bar{\xi }}_u)_t,\varphi ), \quad \forall \varphi \in V_h. \end{aligned}$$

(A.7)

Then, by Lemma 3.3 we have

$$\begin{aligned} \Vert ({\bar{\xi }}_w)_x\Vert (0)\lesssim \Vert (\omega _u^{(l^\star )})_t\Vert (0)+\Vert ({\bar{\xi }}_u)_t\Vert (0). \end{aligned}$$

(A.8)

By Corollary 5.1, we have

$$\begin{aligned} h^{-\frac{3}{2}}\left( \sum _{j=1}^{N} \llbracket {{\bar{\xi }}_w}\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}\lesssim \Vert (\omega _u^{(l^\star )})_t\Vert (0)+\Vert ({\bar{\xi }}_u)_t\Vert (0). \end{aligned}$$

(A.9)

Combine (A.8), (A.9) and Lemma 4.4, we obtain

$$\begin{aligned} \Vert ({\bar{\xi }}_w)_x\Vert (0)+h^{-\frac{3}{2}}\left( \sum _{j=1}^{N} \llbracket {{\bar{\xi }}_w}\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}&\lesssim \Vert (\omega _u^{(l^\star )})_t\Vert (0)+\Vert ({\bar{\xi }}_u)_t\Vert (0)\nonumber \\&\lesssim h^{m}\Vert u_t\Vert _{m}(0)+\Vert ({\bar{\xi }}_u)_t\Vert (0), \end{aligned}$$

(A.10)

where

$$\begin{aligned} \left\{ \begin{array}{ll} m=2k-1, &{}k=0,2\pmod {5};\\ m=2k,&{} k=1,3,4\pmod {5}. \end{array} \right. \end{aligned}$$

• If $k=0,2\pmod {5}$, then $(\omega _u^{(l^\star )})_t(x,0)\bot {\mathcal {P}}^0(I_j),~j\in Z_N$, and $\Vert ({\bar{\xi }}_u)_t\Vert (0)=0.$ By (A.7) and Lemma 3.3 we have

  $$\begin{aligned} \Vert ({\bar{\xi }}_w)_x\Vert (0)\lesssim h\Vert (\omega _u^{(l^\star )})_t\Vert (0)\lesssim h^{2k}\Vert u_t\Vert _{2k-1}(0). \end{aligned}$$

  Furthermore, noticing that $\omega _w^{(i)}\bot {\mathcal {P}}^0(I_j),~i=0,\ldots ,l^\star$, $j\in Z_N$, then

  $$\begin{aligned} \int _{\Omega }{\bar{\xi }}_w(x,0)dx=\sum _{j=1}^{N}\int _{I_j}\left( w_h- P_h^\star w+\omega _{w,l^\star } \right) (x,0)dx=\int _{\Omega }w_h(x,0)- w(x,0)dx=0. \end{aligned}$$

  By the discrete Poincaré inequality (3.7) and (A.10),

  $$\begin{aligned} \Vert {\bar{\xi }}_w\Vert (0)\lesssim \Vert ({\bar{\xi }}_w)_x\Vert (0)+h^{-\frac{1}{2}}\left( \sum _{j=1}^{N} \llbracket {{\bar{\xi }}_w}\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}\lesssim h^{2k}\Vert u_t\Vert _{2k-1}(0). \end{aligned}$$

• If $k=1,3,4\pmod {5}$, then $\Vert ({\bar{\xi }}_u)_t\Vert (0)\lesssim h^{2k}\Vert u_t\Vert _{2k}(0)$, $\Vert ({\bar{\xi }}_w)_x\Vert (0)\lesssim h^{2k}\Vert u_t\Vert _{2k}(0)$, and

  $$\begin{aligned} \int _{\Omega }{\bar{\xi }}_w(x,0)dx=\sum _{j=1}^{N}\int _{I_j}\left( w_h-P_h^\star w+\omega _{w,l^\star }\right) (x,0)dx=\int _{\Omega }\omega _w^{(l^\star )}(x,0)dx. \end{aligned}$$

  Since $\Vert \omega _w^{(l^\star )}\Vert (0)\lesssim h^{2k}$, we have from (A.10) that

  $$\begin{aligned} \Vert {\bar{\xi }}_w\Vert (0)&\lesssim \Vert {\bar{\xi }}_w-\int _{\Omega }{\bar{\xi }}_wdx\Vert (0)+\left| \int _{\Omega }{\bar{\xi }}_w(x,0)dx\right| \\&\lesssim \Vert ({\bar{\xi }}_w)_x\Vert (0)+h^{-\frac{1}{2}}\left( \sum _{j=1}^{N} \llbracket {{\bar{\xi }}_w}\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}} +\left| \int _{\Omega }{\bar{\xi }}_w(x,0)dx\right| \\&\lesssim h^{2k}\big (\Vert u_t\Vert _{2k}(0)+\Vert w\Vert _{2k+1}(0)\big ). \end{aligned}$$

(4) Estimate ${\bar{\xi }}_v(x,0)$:

By Corollary 5.1 we have

$$\begin{aligned} \Vert ({\bar{\xi }}_v)_x\Vert (0)+h^{-\frac{1}{2}}\left( \sum _{j=1}^{N} \llbracket {{\bar{\xi }}_v}\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}\le \Vert \omega _w^{(l^\star )}\Vert (0)+\Vert {\bar{\xi }}_w\Vert (0) \lesssim h^{m}\Vert w\Vert _{m}(0)+\Vert {\bar{\xi }}_w\Vert (0), \end{aligned}$$

(A.11)

where

$$\begin{aligned} \left\{ \begin{array}{ll} m=2k, &{}k=0,1,3\pmod {5};\\ m=2k+1, &{}k=4\pmod {5};\\ m=2k-1, &{}k=2\pmod {5}. \end{array} \right. \end{aligned}$$

• If $k=0,1,3\pmod {5}$, then $\Vert ({\bar{\xi }}_v)_x\Vert (0)\lesssim h^{2k}\Vert u_t\Vert _{2k}(0)$ and $\omega _v^{(i)}\bot {\mathcal {P}}^0(I_j),~i =0,\ldots ,l^\star$, $\int _{\Omega }{\bar{\xi }}_v(x,0)dx=0$, by the discrete Poincaré inequality and (A.11), we get

  $$\begin{aligned} \Vert {\bar{\xi }}_v\Vert (0)\lesssim \Vert ({\bar{\xi }}_v)_x\Vert (0)+h^{-\frac{1}{2}}\left( \sum _{j=1}^{N} \llbracket {{\bar{\xi }}_v}\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}\lesssim h^{2k}\Vert u_t\Vert _{2k}(0). \end{aligned}$$

• If $k=4\pmod {5}$, then $\int _{\Omega }{\bar{\xi }}_v(x,0)dx=\int _{\Omega }\omega _v^{(l^\star )}(x,0)dx$. By the discrete Poincaré inequality (3.7) and (A.11), we get

  $$\begin{aligned} \Vert {\bar{\xi }}_v\Vert (0)&\lesssim \Vert {\bar{\xi }}_v-\int _{\Omega }{\bar{\xi }}_vdx\Vert (0)+\left| \int _{\Omega }{\bar{\xi }}_v(x,0)dx\right| \\&\lesssim h^{2k+1}\Vert w\Vert _{2k+1}(0)+h^{2k}\Vert v\Vert _{2k}(0)+\Vert {\bar{\xi }}_w\Vert (0). \end{aligned}$$

  Therefore, $\Vert {\bar{\xi }}_v\Vert (0)\lesssim h^{2k}\Vert u_t\Vert _{2k}(0).$

• If $k=2\pmod {5}$, we have $\omega _w^{(l^\star )}\bot {\mathcal {P}}^0(I_j), j\in Z_N$, and define $\omega _v^{(l^\star +1)}$: find $\omega _v^{(l^\star +1)}\big |_{I_j}\in {\mathcal {P}}^k(I_j)$, such that

  $$\begin{aligned} (\omega _v^{(l^\star +1)}, \phi _{x})_j=-(\omega _w^{(l^\star )},\phi )_j,\quad \omega _v^{(l^\star +1)}\left( x_{{j+\frac{1}{2}}}^-\right) =0,\quad \forall \phi \in {\mathcal {P}}^k(I_j),\,j\in Z_N. \end{aligned}$$

  (A.12)

  Denote ${\tilde{\xi }}_v={\bar{\xi }}_v+\omega _v^{(l^\star +1)}$, ${\tilde{\eta }}_v={\bar{\eta }}_v+\omega _v^{(l^\star +1)}$. Then we have $\Vert \omega _v^{(l^\star +1)}\Vert \lesssim h^{2k}\Vert v\Vert _{2k}$ and

  $$\begin{aligned} {\mathcal {D}}^1_{j}({\tilde{\xi }}_v(x,0),\phi )=-({\bar{\xi }}_w(x,0),\phi ), \quad \forall \phi \in V_h. \end{aligned}$$

  Thus, we get

  $$\begin{aligned} \Vert ({\tilde{\xi }}_v)_x\Vert (0)+h^{-\frac{1}{2}}\left( \sum _{j=1}^{N} \llbracket {{\tilde{\xi }}_v}\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}&\lesssim \Vert {\bar{\xi }}_w\Vert (0)\\ \int _{\Omega }{\tilde{\xi }}_v(x,0)dx&=\int _{\Omega }\omega _v^{(l^\star +1)}(x,0)dx. \end{aligned}$$

  Then,

  $$\begin{aligned} \Vert {\tilde{\xi }}_v\Vert (0) \lesssim \Vert {\tilde{\xi }}_v-\int _{\Omega }{\tilde{\xi }}_vdx\Vert (0)+\left| \int _{\Omega }{\tilde{\xi }}_v(x,0)dx\right| \lesssim \Vert {\bar{\xi }}_w\Vert (0)+h^{2k}\Vert v\Vert _{2k}(0), \end{aligned}$$

  and

  $$\begin{aligned} \Vert {\bar{\xi }}_v\Vert (0)\le \Vert {\tilde{\xi }}_v\Vert (0)+\Vert \omega _v^{(l^\star +1)}\Vert (0)\lesssim \Vert {\bar{\xi }}_w\Vert (0)+h^{2k}\Vert v\Vert _{2k}(0)\lesssim h^{2k}\Vert u_t\Vert _{2k}(0)+h^{2k}\Vert v\Vert _{2k}(0). \end{aligned}$$

(5) Estimate ${\bar{\xi }}_u(x,0)$:

By Corollary 5.1 we have

$$\begin{aligned} \Vert ({\bar{\xi }}_u)_x\Vert (0)+h^{-\frac{3}{2}}\left( \sum _{j=1}^{N} \llbracket {{\bar{\xi }}_u}\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}\lesssim \Vert \omega _v^{(l^\star )}\Vert (0)+\Vert {\bar{\xi }}_v\Vert (0)\lesssim h^{m}\Vert v\Vert _{m}(0)+\Vert {\bar{\xi }}_v\Vert (0), \end{aligned}$$

(A.13)

where

$$\begin{aligned} \left\{ \begin{array}{ll} m=2k-1,&{}k=0,2,3\pmod {5};\\ m=2k, &{}k=1,4\pmod {5}. \end{array} \right. \end{aligned}$$

• If $k=0,2,3\pmod {5}$, then $\omega _v^{(l^\star )}\bot {\mathcal {P}}^0(I_j), \omega _u^{(i)}\bot {\mathcal {P}}^0(I_j), i=0,\ldots ,l^\star$ and $\int _{\Omega }{\bar{\xi }}_u(x,0) dx=0$. By Lemma 3.1, we have

  $$\begin{aligned} \Vert ({\bar{\xi }}_u)_x\Vert (0)&\lesssim h\Vert \omega _v^{(l^\star )}\Vert (0)+\Vert {\bar{\xi }}_v\Vert (0)\lesssim h^{2k}\Vert v\Vert _{2k-1}(0)+\Vert {\bar{\xi }}_v\Vert (0)\lesssim h^{2k}\Vert u_t\Vert _{2k}(0). \\ \Vert {\bar{\xi }}_u\Vert (0)&\lesssim \Vert ({\bar{\xi }}_u)_x\Vert (0)+h^{-\frac{1}{2}}\left( \sum _{j=1}^{N} \llbracket {{\bar{\xi }}_u}\rrbracket _{{j-\frac{1}{2}}}^2\right) ^{\frac{1}{2}}\\&\lesssim h^{2k}\Vert v\Vert _{2k-1}(0)+\Vert {\bar{\xi }}_v\Vert (0)\lesssim h^{2k}\Vert u_t\Vert _{2k}(0). \end{aligned}$$

• If $k=1,4\pmod {5}$, then $\Vert ({\bar{\xi }}_u)_x\Vert (0)\lesssim h^{2k}\Vert u_t\Vert _{2k}(0)$ and $\int _{\Omega }{\bar{\xi }}_u(x,0)dx=\int _{\Omega }\omega _u^{(l^\star )}(x,0)dx$ Thus

  $$\begin{aligned} \Vert {\bar{\xi }}_u\Vert (0)&\lesssim \Vert {\bar{\xi }}_u -\int _{\Omega }{\bar{\xi }}_udx\Vert (0)+\left| \int _{\Omega }{\bar{\xi }}_u(x,0)dx\right| \\&\lesssim h^{2k}\big (\Vert v\Vert _{2k}(0)+\Vert u\Vert _{2k}(0)\big )+\Vert {\bar{\xi }}_v\Vert (0)\lesssim h^{2k}\Vert u_t\Vert _{2k}(0). \end{aligned}$$

  This finishes our proof. $\square$

1.5 The Proof of Theorem 5.2

Proof

We take $\varphi ={\bar{\xi }}_u, \phi =-{\bar{\xi }}_v, \psi ={\bar{\xi }}_w$ in (5.5) and add (5.5a)–(5.5c) together to get

$$\begin{aligned}&{\mathcal {A}}(({\bar{\xi }}_u)_{t},{\bar{\xi }}_w;{\bar{\xi }}_u)-{\mathcal {B}}({\bar{\xi }}_w,{\bar{\xi }}_v;{\bar{\xi }}_v)+{\mathcal {C}}({\bar{\xi }}_v,{\bar{\xi }}_u;{\bar{\xi }}_w) = {\mathcal {A}}(({\bar{\eta }}_u)_{t},{\bar{\eta }}_w;{\bar{\xi }}_u)-{\mathcal {B}}({\bar{\eta }}_w,{\bar{\eta }}_v;{\bar{\xi }}_v)\\&\qquad +{\mathcal {C}}({\bar{\eta }}_v,{\bar{\eta }}_u;{\bar{\xi }}_w). \\ {\textit{LHS}}&={\mathcal {A}}(({\bar{\xi }}_u)_{t},{\bar{\xi }}_w;{\bar{\xi }}_u)-{\mathcal {B}}({\bar{\xi }}_w,{\bar{\xi }}_v;{\bar{\xi }}_v)+{\mathcal {C}}({\bar{\xi }}_v,{\bar{\xi }}_u;{\bar{\xi }}_w)\\&=\frac{1}{2}\frac{d}{dt}\Vert {\bar{\xi }}_u\Vert ^2+\frac{1}{2}\sum _{j=1}^N \llbracket {{\bar{\xi }}_v}\rrbracket _{{j+\frac{1}{2}}}^2,\\ {\textit{RHS}}&={\mathcal {A}}(({\bar{\eta }}_u)_{t},{\bar{\eta }}_w;{\bar{\xi }}_u)-{\mathcal {B}}({\bar{\eta }}_w,{\bar{\eta }}_v;{\bar{\xi }}_v)+{\mathcal {C}}({\bar{\eta }}_v,{\bar{\eta }}_u;{\bar{\xi }}_w)\\&=((\omega _u^{(l^\star )})_t,{\bar{\xi }}_u)-(\omega _w^{(l^\star )},{\bar{\xi }}_v)+(\omega _v^{(l^\star )},{\bar{\xi }}_w). \end{aligned}$$

Thus, we get

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert {\bar{\xi }}_u\Vert ^2+\frac{1}{2}\sum _{j=1}^N \llbracket {{\bar{\xi }}_v}\rrbracket _{{j+\frac{1}{2}}}^2=((\omega _u^{(l^\star )})_t,{\bar{\xi }}_u)-(\omega _w^{(l^\star )},{\bar{\xi }}_v)+(\omega _v^{(l^\star )},{\bar{\xi }}_w). \end{aligned}$$

(A.14)

Similarly, by taking time derivative in both side of (5.5), and then choosing $\varphi =({\bar{\xi }}_u)_t, \phi =-({\bar{\xi }}_v)_t, \psi =({\bar{\xi }}_w)_t$, we obtain

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert ({\bar{\xi }}_u)_t\Vert ^2+\frac{1}{2}\sum _{j=1}^N \llbracket {({\bar{\xi }}_v)_t}\rrbracket _{{j+\frac{1}{2}}}^2&=((\omega _u^{(l^\star )})_{tt},({\bar{\xi }}_u)_t)-((\omega _w^{(l^\star )})_t,({\bar{\xi }}_v)_t)\nonumber \\&\quad +((\omega _v^{(l^\star )})_t,({\bar{\xi }}_w)_t). \end{aligned}$$

(A.15)

Next, we will divide into five cases to estimate $\Vert {\bar{\xi }}_u\Vert ,\Vert ({\bar{\xi }}_u)_x\Vert$, $\Vert {\bar{\xi }}_w\Vert ,\Vert ({\bar{\xi }}_w)_x\Vert$ and $\Vert {\bar{\xi }}_v\Vert$.

• Case 1: $\mathbf{k=0\,\,(mod)\,\,5}$, for example $\mathbf{k=5}$.

In this case, $\displaystyle l^\star =3\lfloor \frac{k-4}{5}\rfloor +2$. By Lemma 4.4, we have

$$\begin{aligned} \Vert (\omega _u^{(l^\star )})_t\Vert \lesssim h^{2k-1}\Vert u_t\Vert _{2k-1},\;\;\Vert \omega _v^{(l^\star )}\Vert \lesssim h^{2k-1}\Vert v\Vert _{2k-1},\;\;\Vert \omega _w^{(l^\star )}\Vert \lesssim h^{2k}\Vert w\Vert _{2k}. \end{aligned}$$

Using Lemma 4.1, we get

$$\begin{aligned} (\omega _u^{(l^\star )})_t\bot {\mathcal {P}}^0(I_j),\quad \omega _v^{(l^\star )}\bot {\mathcal {P}}^0(I_j),\quad \omega _w^{(l^\star )}\bot {\mathcal {P}}^0(I_j),\quad \forall j\in Z_N. \end{aligned}$$

Then, by Lemma 3.3, Corollary 5.1, we could get

$$\begin{aligned}&\Vert ({\bar{\xi }}_w)_x\Vert \lesssim \Vert ({\bar{\xi }}_u)_t\Vert +h \Vert (\omega _u^{(l^\star )})_t\Vert \lesssim \Vert ({\bar{\xi }}_u)_t\Vert + h^{2k}, \end{aligned}$$

(A.16)

$$\begin{aligned}&\Vert ({\bar{\xi }}_v)_{x}\Vert \lesssim \Vert {\bar{\xi }}_w\Vert +\Vert \omega _w^{(l^\star )}\Vert \lesssim \Vert {\bar{\xi }}_w\Vert +h^{2k}, \end{aligned}$$

(A.17)

$$\begin{aligned}&\Vert ({\bar{\xi }}_u)_x\Vert \lesssim \Vert {\bar{\xi }}_v\Vert +h \Vert \omega _v^{(l^\star )}\Vert \lesssim \Vert {\bar{\xi }}_v\Vert +h^{2k}. \end{aligned}$$

(A.18)

By using the discrete Poincaré inequality (3.7), we get

$$\begin{aligned} \Vert {\bar{\xi }}_w\Vert \lesssim \Vert ({\bar{\xi }}_u)_t\Vert +h^{2k},\; \Vert {\bar{\xi }}_v\Vert \lesssim \Vert {\bar{\xi }}_w\Vert +h^{2k},\; \Vert {\bar{\xi }}_u\Vert \lesssim \Vert {\bar{\xi }}_v\Vert +h^{2k}. \end{aligned}$$

(A.19)

Next, we only need to estimate $\Vert ({\bar{\xi }}_u)_t\Vert$. Suppose $\Vert ({\bar{\xi }}_u)_t\Vert (t^{\star }):=\sup _{t\in [0,T]}\Vert ({\bar{\xi }}_u)_t\Vert (t)$, after integration with respect to time over $[0,t^\star ]$ in (A.15),

$$\begin{aligned} \frac{1}{2}\Vert ({\bar{\xi }}_u)_t\Vert ^2(t^\star )-\frac{1}{2}\Vert ({\bar{\xi }}_u)_t\Vert ^2(0)&\le \int _0^{t^\star }\big ((\omega _u^{(l^\star )})_{tt},({\bar{\xi }}_u)_t\big )dt-\int _0^{t^\star }\big ((\omega _w^{(l^\star )})_t,({\bar{\xi }}_v)_t\big )dt\\&\quad +\int _0^{t^\star }\big ((\omega _v^{(l^\star )})_t,({\bar{\xi }}_w)_t\big )dt\\&:=\mathrm {(I)+(II)+(III)}. \end{aligned}$$

Estimate (I): after integration by parts,

$$\begin{aligned} \mathrm{(I)}=-\int _0^{t^\star }\big ((\omega _u^{(l^\star )})_{ttt},{\bar{\xi }}_u\big )dt+\big ((\omega _u^{(l^\star )})_{tt}(t^\star ),{\bar{\xi }}_u(t^\star )\big )- \big ((\omega _u^{(l^\star )})_{tt}(0),{\bar{\xi }}_u(0)\big ). \end{aligned}$$

Since $(\omega _u^{(l^\star )})_{ttt}\bot {\mathcal {P}}^0(I_j)$, we have $\big ((\omega _u^{(l^\star )})_{ttt},{\bar{\xi }}_u\big )_j=-{\bar{h}}_j\big (D^{-1}(\omega _u^{(l^\star )})_{ttt},({\bar{\xi }}_u)_x\big )_j$ and by (A.18) and (A.19),

$$\begin{aligned} \big ((\omega _u^{(l^\star )})_{ttt},{\bar{\xi }}_u\big )\lesssim h\Vert (\omega _u^{(l^\star )})_{ttt}\Vert \Vert ({\bar{\xi }}_u)_x\Vert \lesssim h^{2k}\Vert {\bar{\xi }}_v\Vert +h^{4k}\lesssim h^{2k}\Vert ({\bar{\xi }}_u)_t\Vert +h^{4k}. \end{aligned}$$

(A.20)

Similarly, by (A.18) and (A.19),

$$\begin{aligned} \big ((\omega _u^{(l^\star )})_{tt}(t^\star ),{\bar{\xi }}_u(t^\star )\big )&\lesssim h^{2k}\Vert {\bar{\xi }}_v\Vert (t^\star )+h^{4k} \lesssim h^{2k}\Vert ({\bar{\xi }}_u)_t\Vert (t^\star )+h^{4k}. \end{aligned}$$

By (5.11), we also have $\big ((\omega _u^{(l^\star )})_{tt}(0),{\bar{\xi }}_u(0)\big )\lesssim h\Vert (\omega _u^{(l^\star )})_{tt}\Vert (0)\Vert ({\bar{\xi }}_u)_x\Vert (0)\lesssim h^{4k}.$ Therefore,

$$\begin{aligned} \mathrm{(I)}&\lesssim \int _0^{t^\star } \left( h^{2k}\Vert ({\bar{\xi }}_u)_t\Vert +h^{4k}\right) dt+h^{2k}\Vert ({\bar{\xi }}_u)_t\Vert (t^\star )+h^{4k},\\&\le \int _0^{t^\star } CTh^{4k}+\frac{1}{24T}\Vert ({\bar{\xi }}_u)_t\Vert ^2dt+CT h^{4k}+\frac{1}{24}\Vert ({\bar{\xi }}_u)_t\Vert ^2(t^\star )+Ch^{4k}\\&\le C (1+T)^2h^{4k}+\frac{1}{24T}\int _{0}^{t^\star }\Vert ({\bar{\xi }}_u)_t\Vert ^2dt+\frac{1}{24}\Vert ({\bar{\xi }}_u)_t\Vert ^2(t^\star ). \end{aligned}$$

Estimate (II): after integration by parts,

$$\begin{aligned} \mathrm{(II)}=\int _0^{t^\star }\big ((\omega _w^{(l^\star )})_{tt},{\bar{\xi }}_v\big )dt-\big ((\omega _w^{(l^\star )})_{t}(t^\star ),{\bar{\xi }}_v(t^\star )\big )+ \big ((\omega _w^{(l^\star )})_{t}(0),{\bar{\xi }}_v(0)\big ). \end{aligned}$$

Then,

$$\begin{aligned} \big ((\omega _w^{(l^\star )})_{tt},{\bar{\xi }}_v\big )\le \Vert \big (\omega _w^{(l^\star )}\big )_{tt}\Vert \Vert {\bar{\xi }}_v\Vert \lesssim h^{2k}\big (h^{2k}+\Vert {\bar{\xi }}_w\Vert \big )\lesssim h^{2k}\Vert {\bar{\xi }}_w\Vert +h^{4k}, \end{aligned}$$

and

$$\begin{aligned} \big ((\omega _w^{(l^\star )})_{t}(t^\star ),{\bar{\xi }}_v(t^\star )\big )\lesssim h^{2k}\Vert {\bar{\xi }}_w\Vert (t^\star )+h^{4k}, \quad \big ((\omega _w^{(l^\star )})_{t}(0),{\bar{\xi }}_v(0)\big )\lesssim \Vert (\omega _w^{(l^\star )})_{t}\Vert (0)\Vert {\bar{\xi }}_v\Vert (0)\lesssim h^{4k}. \end{aligned}$$

Therefore,

$$\begin{aligned} \mathrm{(II)}\le C(1+T)^2h^{4k}+\frac{1}{24T}\int _{0}^{t^\star }\Vert ({\bar{\xi }}_u)_t\Vert ^2dt+\frac{1}{24}\Vert ({\bar{\xi }}_u)_t\Vert ^2(t^\star ). \end{aligned}$$

Estimate (III): Similar to (I), we have

$$\begin{aligned} \mathrm{(III)}\le C(1+T)^2h^{4k}+\frac{1}{24T}\int _{0}^{t^\star }\Vert ({\bar{\xi }}_u)_t\Vert ^2dt+\frac{1}{24}\Vert ({\bar{\xi }}_u)_t\Vert ^2(t^\star ). \end{aligned}$$

Combining (5.10) and the estimates of (I), (II) and (III) together yields

$$\begin{aligned} \frac{3}{8}\Vert ({\bar{\xi }}_u)_t\Vert ^2(t^\star )\le C(1+T)^2h^{4k}+\frac{1}{8T}\int _{0}^{t^\star }\Vert ({\bar{\xi }}_u)_t\Vert ^2dt\le \frac{1}{8}\Vert ({\bar{\xi }}_u)_t\Vert ^2(t^\star )+C(1+T)^2h^{4k}. \end{aligned}$$

Then, we obtain $\Vert ({\bar{\xi }}_u)_t\Vert (t^\star )\lesssim h^{2k}$ and thus (5.12) and (5.13) follows.

• Case 2: ${\mathbf{k=1\,(mod)\,5}}$, for example $\mathbf{k=6}$.

In this case, $\displaystyle l^\star =3\lfloor \frac{k-1}{5}\rfloor$. By Lemma 4.4, we have

$$\begin{aligned} \Vert (\omega _u^{(l^\star )})_t\Vert \lesssim h^{2k}\Vert u_t\Vert _{2k},\;\;\Vert \omega _v^{(l^\star )}\Vert \lesssim h^{2k}\Vert v\Vert _{2k},\;\; \Vert \omega _w^{(l^\star )}\Vert \lesssim h^{2k}\Vert w\Vert _{2k}. \end{aligned}$$

By Lemma 3.3, Corollary 5.1 and the discrete Poincaré inequality (3.7) , we could get

$$\begin{aligned} \Vert {\bar{\xi }}_w\Vert +\Vert ({\bar{\xi }}_w)_x\Vert&\lesssim \Vert ({\bar{\xi }}_u)_t\Vert +h^{2k},\;\; \Vert {\bar{\xi }}_v\Vert +\Vert ({\bar{\xi }}_v)_{x}\Vert \\&\lesssim \Vert {\bar{\xi }}_w\Vert +h^{2k},\;\;\Vert {\bar{\xi }}_u\Vert +\Vert ({\bar{\xi }}_u)_x\Vert \lesssim \Vert {\bar{\xi }}_v\Vert +h^{2k}. \end{aligned}$$

Integration with respect to time over $[0,t^\star ]$ in (A.15),

$$\begin{aligned} \frac{1}{2}\Vert ({\bar{\xi }}_u)_t\Vert ^2(t^\star )-\frac{1}{2}\Vert ({\bar{\xi }}_u)_t\Vert ^2(0)&\le \int _0^{t^\star }((\omega _u^{(l^\star )})_{tt},({\bar{\xi }}_u)_t)dt-\int _0^{t^\star }((\omega _w^{(l^\star )})_t,({\bar{\xi }}_v)_t)dt\\&\quad +\int _0^{t^\star }((\omega _v^{(l^\star )})_t,({\bar{\xi }}_w)_t)dt. \end{aligned}$$

Similar to the analysis of Case 1, we could get $\Vert ({\bar{\xi }}_u)_t\Vert (t^\star )\lesssim h^{2k}.$ Then the desired result (5.12)–(5.13) follows.

• Case 3: $\mathbf{k=2\,(mod)\,5}$, for example $\mathbf{k=7}$.

In this case, $l^\star =3\lfloor \frac{k-1}{5}\rfloor$. By Lemma 4.4, we have

$$\begin{aligned} \Vert (\omega _u^{(l^\star )})_t\Vert \lesssim h^{2k-1}\Vert u_t\Vert _{2k-1},\;\;\Vert \omega _v^{(l^\star )}\Vert \lesssim h^{2k-1}\Vert v\Vert _{2k-1},\;\;\Vert \omega _w^{(l^\star )}\Vert \lesssim h^{2k-1}\Vert w\Vert _{2k-1}. \end{aligned}$$

By Lemma 4.1, we obtain

$$\begin{aligned} (\omega _u^{(l^\star )})_t\bot {\mathcal {P}}^0(I_j),\quad \omega _v^{(l^\star )}\bot {\mathcal {P}}^1(I_j),\quad \omega _w^{(l^\star )}\bot {\mathcal {P}}^0(I_j),\quad \forall j\in Z_N. \end{aligned}$$

Along the same line as in Case 1, we could get:

$$\begin{aligned}&\Vert {\bar{\xi }}_w\Vert +\Vert ({\bar{\xi }}_w)_x\Vert \lesssim \Vert ({\bar{\xi }}_u)_t\Vert +h\Vert (\omega _u^{(l^\star )})_t\Vert \lesssim \Vert ({\bar{\xi }}_u)_t\Vert +h^{2k},\\&\Vert {\bar{\xi }}_v\Vert +\Vert ({\bar{\xi }}_v)_{x}\Vert \lesssim \Vert {\bar{\xi }}_w\Vert +\Vert \omega _w^{(l^\star )}\Vert \lesssim \Vert {\bar{\xi }}_w\Vert +h^{2k-1},\\&\Vert {\bar{\xi }}_u\Vert +\Vert ({\bar{\xi }}_u)_x\Vert \lesssim \Vert {\bar{\xi }}_v\Vert +h\Vert \omega _v^{(l^\star )}\Vert \lesssim \Vert {\bar{\xi }}_v\Vert +h^{2k}. \end{aligned}$$

Since $\omega _w^{(l^\star )}\bot {\mathcal {P}}^0(I_j),~j\in Z_N$, we can define $\omega _v^{(l^\star +1)}$: find $\omega _v^{(l^\star +1)}\big |_{I_j}\in {\mathcal {P}}^k(I_j)$, such that

$$\begin{aligned} (\omega _v^{(l^\star +1)}, \phi _{x})_j=-(\omega _w^{(l^\star )},\phi )_j,\;\;\omega _v^{(l^\star +1)}(x_{{j+\frac{1}{2}}}^-)=0, \quad \forall \phi \in {\mathcal {P}}^k(I_j), \quad \forall j\in Z_N. \end{aligned}$$

(A.21)

Then, we can get $\Vert \omega _v^{(l^\star +1)}\Vert \lesssim h^{2k}$. We denote ${\tilde{\xi }}_v={\bar{\xi }}_v+\omega _v^{(l^\star +1)}$, ${\tilde{\eta }}_v={\bar{\eta }}_v+\omega _v^{(l^\star +1)}$ to obtain

$$\begin{aligned} {\mathcal {D}}^1_{j}({\tilde{\xi }}_v,\phi )=-({\bar{\xi }}_w, \phi )_j,\;\; {\mathcal {D}}^2_{j}({\bar{\xi }}_u,\psi )=({\tilde{\xi }}_v,\psi )_j-(\omega _v^{(l^\star )},\psi )_j-(\omega _v^{(l^\star +1)},\psi )_j,~\forall \phi , \psi \in V_h, \end{aligned}$$

and

$$\begin{aligned} \Vert {\tilde{\xi }}_v\Vert +\Vert ({\tilde{\xi }}_v)_{x}\Vert \lesssim \Vert {\bar{\xi }}_w\Vert +h^{2k},\quad \Vert {\bar{\xi }}_u\Vert +\Vert ({\bar{\xi }}_u)_x\Vert \lesssim \Vert {\tilde{\xi }}_v\Vert +h\Vert \omega _v^{(l^\star )}\Vert +\Vert \omega _v^{(l^\star +1)}\Vert \lesssim \Vert {\tilde{\xi }}_v\Vert +h^{2k}. \end{aligned}$$

Next, we only need to estimate $\Vert ({\bar{\xi }}_u)_t\Vert$. By taking time derivative on both sides of (5.5), and choosing $\varphi =({\bar{\xi }}_u)_t, \phi =-({\tilde{\xi }}_v)_t, \psi =({\bar{\xi }}_w)_t$, we have

$$\begin{aligned}&{\mathcal {A}}(({\bar{\xi }}_u)_{tt},({\bar{\xi }}_w)_t;({\bar{\xi }}_u)_t) -{\mathcal {B}}(({\bar{\xi }}_w)_t,({\tilde{\xi }}_v)_t;({\tilde{\xi }}_v)_t)+{\mathcal {C}}(({\tilde{\xi }}_v)_t,({\bar{\xi }}_u)_t;({\bar{\xi }}_w)_t)\\&\quad =\frac{1}{2}\frac{d}{dt}\Vert ({\bar{\xi }}_u)_t\Vert ^2 + \frac{1}{2}\sum _{j=1}^N \llbracket {({\tilde{\xi }}_v)_t}\rrbracket _{{j+\frac{1}{2}}}^2, \end{aligned}$$

and

$$\begin{aligned}&{\mathcal {A}}(({\bar{\eta }}_u)_{tt},({\bar{\eta }}_w)_t;({\bar{\xi }}_u)_t)-{\mathcal {B}}(({\bar{\eta }}_w)_t,({\tilde{\eta }}_v)_t;({\tilde{\xi }}_v)_t)+{\mathcal {C}}(({\tilde{\eta }}_v)_t,({\bar{\eta }}_u)_t;({\bar{\xi }}_w)_t)\\&\quad =((\omega _u^{(l^\star )})_{tt},({\bar{\xi }}_u)_t) +((\omega _v^{(l^\star )})_t,({\bar{\xi }}_w)_t)+((\omega _v^{(l^\star +1)})_t,({\bar{\xi }}_w)_t). \end{aligned}$$

Then, similar to the Case 1, we can prove $\Vert ({\bar{\xi }}_u)_t\Vert (t^\star )\lesssim h^{2k}$ and

$$\begin{aligned} \Vert {\bar{\xi }}_u\Vert +\Vert {\tilde{\xi }}_v\Vert +\Vert {\bar{\xi }}_w\Vert&\lesssim h^{2k}, \quad \Vert ({\bar{\xi }}_u)_x\Vert +\Vert ({\tilde{\xi }}_v)_x\Vert +\Vert ({\bar{\xi }}_w)_x\Vert \lesssim h^{2k}, \\ \Vert {\bar{\xi }}_v\Vert&\le \Vert {\tilde{\xi }}_v\Vert +\Vert \omega _v^{(l^\star +1)}\Vert \lesssim h^{2k}.\quad \end{aligned}$$

• Case 4: $\mathbf{k=3\,(mod)\,5}$, for example $\mathbf{k=3}$.

In this case, $l^\star =3\lfloor \frac{k-3}{5}\rfloor +1$. By Lemma 4.4 and Corollary 5.1, we have

$$\begin{aligned} \Vert (\omega _u^{(l^\star )})_t\Vert \lesssim h^{2k}\Vert u_t\Vert _{2k},\;\;\Vert \omega _v^{(l^\star )}\Vert \lesssim h^{2k-1}\Vert v\Vert _{2k-1},\;\; \Vert \omega _w^{(l^\star )}\Vert \lesssim h^{2k}\Vert w\Vert _{2k}. \end{aligned}$$

Using Lemma 4.1, we get $\omega _v^{(l^\star )}\bot {\mathcal {P}}^0(I_j)$ then $\displaystyle (\omega _v^{(l^\star )},{\bar{\xi }}_w)_j=-{\bar{h}}_j(D^{-1}\omega _v^{(l^\star )}, ({\bar{\xi }}_w)_x)_j, ~\forall j\in Z_N.$ By Lemma 3.3, we have

$$\begin{aligned}&\Vert {\bar{\xi }}_w\Vert +\Vert ({\bar{\xi }}_w)_x\Vert \lesssim \Vert ({\bar{\xi }}_u)_t\Vert +h^{2k},\\&\Vert {\bar{\xi }}_v\Vert +\Vert ({\bar{\xi }}_v)_{x}\Vert \lesssim \Vert {\bar{\xi }}_w\Vert +h^{2k},\\&\Vert ({\bar{\xi }}_u)_{x}\Vert \lesssim \Vert {\bar{\xi }}_v\Vert +h\Vert \omega _v^{(l^\star )}\Vert \lesssim \Vert {\bar{\xi }}_v\Vert +h^{2k}. \end{aligned}$$

Thus, along the same line in Case 1, we have $\Vert ({\bar{\xi }}_u)_t\Vert (t^\star )\lesssim h^{2k}$. Then (5.12) and (5.13) are satisfied.

• Case 5: ${\mathbf{k=4\,(mod)\,5}}$, for example $\mathbf{k=4}$.

In this case, $l^\star =3\lfloor \frac{k-4}{5}\rfloor +2$. By Lemma 4.4, we have

$$\begin{aligned} \Vert (\omega _u^{(l^\star )})_t\Vert \lesssim h^{2k}\Vert u_t\Vert _{2k},\;\;\Vert \omega _v^{(l^\star )}\Vert \lesssim h^{2k}\Vert v\Vert _{2k},\;\; \Vert \omega _w^{(l^\star )}\Vert \lesssim h^{2k+1}\Vert w\Vert _{2k+1}. \end{aligned}$$

We could get

$$\begin{aligned} \Vert {\bar{\xi }}_w\Vert +\Vert ({\bar{\xi }}_w)_x\Vert&\lesssim \Vert ({\bar{\xi }}_u)_t\Vert +h^{2k},\;\;\Vert {\bar{\xi }}_v\Vert +\Vert ({\bar{\xi }}_v)_{x}\Vert \lesssim \Vert {\bar{\xi }}_w\Vert +h^{2k+1},\\ \Vert {\bar{\xi }}_u\Vert +\Vert ({\bar{\xi }}_u)_x\Vert&\lesssim \Vert {\bar{\xi }}_v\Vert +h^{2k}. \end{aligned}$$

Similar to the Case 2, we could get $\Vert ({\bar{\xi }}_u)_t\Vert (t^\star )\lesssim h^{2k}.$ Thus, we have (5.12) and (5.13). $\square$