# Optimal error estimates of the local discontinuous Galerkin methods based on generalized fluxes for 1D linear fifth order partial differential equations – Journal of Inequalities and Applications

Aug 9, 2022

### LDG scheme

In this paper, we consider the one-dimensional linear fifth order equation

begin{aligned} &u_{t}+u_{xxxxx}=0,quad (x,t)in [0,2pi ]times [0,T], end{aligned}

(2.1a)

begin{aligned} &u(x,0)=u_{0}(x) , end{aligned}

(2.1b)

where (u_{0}(x)) is a smooth function. For convenience, we take the periodic boundary condition (u(0,t)=u(2pi ,t)) into discussion.

#### The meshes

Let us denote the computational interval (I=[0,2pi ]), consisting of cells (I_{j}=(x_{j-frac{1}{2}},x_{j+frac{1}{2}})) with (1leq jleq N), where

$$0=x_{frac{1}{2}}< x_{frac{3}{2}}< cdots < x_{N+frac{1}{2}}=2pi .$$

Then we define (x_{j}=(x_{j-frac{1}{2}}+x_{j+frac{1}{2}})/2), (h_{j}=x_{j+frac{1}{2}}-x_{j-frac{1}{2}}), and (h=max h_{j}). Use (p^{-}_{j+frac{1}{2}}) and (p^{+}_{j+frac{1}{2}}) to denote the left and right limits of p at the discontinuity point (x_{j+frac{1}{2}}). In what follows, we employ ([p]=p^{+}-p^{-}) and ({p}=frac{(p^{+}+p^{-})}{2}) to represent the jump and the mean value of p at each element boundary point. The following piecewise polynomial space is chosen as the finite element space:

$$V_{h}equiv V^{k}_{h}=bigl{ vin L^{2}(I):v| _{I_{j}}in P^{k}(I_{j}),j=1, ldots ,Nbigr} ,$$

where (P^{k}(I_{j})) denotes the set of polynomials of degree up to (kgeq 0) defined on cell (I_{j}).

#### Function spaces and norms

For any integer (lgeq 0), the norms of the broken Sobolev spaces (W^{l,p}(I_{h})={uin L^{2}(I):u| _{I_{j}}in W^{l,p}(I_{j}),j=1, ldots ,N}) with (p=2,infty ) are given by

begin{aligned}& Vert u Vert _{W^{l,2}(I_{h})}= Vert u Vert _{H^{l}(I_{h})}=Biggl( sum^{N}_{j=1} Vert u Vert ^{2}_{H^{l}(I_{j})}Biggr)^{ frac{1}{2}},\& Vert u Vert _{W^{l,infty}(I_{h})}=max_{1leq j leq N} Vert u Vert _{W^{l,infty}(I_{j})}. end{aligned}

In the case of (l=0), we denote (| u| _{L^{2}(I)}=| u| ).

#### The semi-discrete LDG scheme

Next we introduce the semi-discrete LDG method of Eqs. (2.1a) and (2.1b). First, we use some variables

begin{aligned} &q=u_{x},qquad p=q_{x},qquad r=p_{x}, qquad s=r_{x} end{aligned}

to transform Eq. (2.1a) into a first order linear system

begin{aligned} &u_{t}+s_{x}=0, end{aligned}

(2.2a)

begin{aligned} &s=r_{x}, end{aligned}

(2.2b)

begin{aligned} &r=p_{x}, end{aligned}

(2.2c)

begin{aligned} &p=q_{x}, end{aligned}

(2.2d)

begin{aligned} &q=u_{x}. end{aligned}

(2.2e)

The LDG scheme is defined as follows: find (u_{h}, q_{h}, p_{h}, r_{h}, s_{h}in V^{k}_{h}) such that (forall rho , xi , phi , psi , varphi in V^{k}_{h}), there holds

begin{aligned} & int _{I_{j}}(u_{h})_{t}rho ,dx+ hat{s_{h}}rho ^{-}| _{j+ frac{1}{2}}- hat{s_{h}}rho ^{+}| _{j-frac{1}{2}}- int _{I_{j}}s_{h} rho _{x},dx=0, end{aligned}

(2.3a)

begin{aligned} & int _{I_{j}}s_{h}xi ,dx-hat{r_{h}} xi ^{-}| _{j+frac{1}{2}}+ hat{r_{h}}xi ^{+}| _{j-frac{1}{2}}+ int _{I_{j}}r_{h}xi _{x},dx=0, end{aligned}

(2.3b)

begin{aligned} & int _{I_{j}}r_{h}phi ,dx-hat{p_{h}} phi ^{-}| _{j+frac{1}{2}}+ hat{p_{h}}phi ^{+}| _{j-frac{1}{2}}+ int _{I_{j}}p_{h}phi _{x},dx=0, end{aligned}

(2.3c)

begin{aligned} & int _{I_{j}}p_{h}psi ,dx-hat{q_{h}} psi ^{-}| _{j+frac{1}{2}}+ hat{q_{h}}psi ^{+}| _{j-frac{1}{2}}+ int _{I_{j}}q_{h}psi _{x},dx=0, end{aligned}

(2.3d)

begin{aligned} & int _{I_{j}}q_{h}varphi ,dx-hat{u_{h}} varphi ^{-}| _{j+ frac{1}{2}}+hat{u_{h}}varphi ^{+}| _{j-frac{1}{2}}+ int _{I_{j}}u_{h} varphi _{x},dx=0, end{aligned}

(2.3e)

where (hat{s_{h}}), (hat{r_{h}}), (hat{p_{h}}), (hat{q_{h}}), (hat{u_{h}}) are numerical fluxes. Here we use the generalized numerical fluxes related to parameter θ. We write (tilde{theta}=1-theta ) and choose numerical fluxes at (x_{j+frac{1}{2}}, j=0,ldots ,N) as follows:

begin{aligned} &hat{u_{h}}=u_{h}^{theta}=theta u_{h}^{-}+(1-theta )u_{h}^{+}, \ &hat{q_{h}}=q_{h}^{tilde{theta}}=theta q_{h}^{+}+(1-theta )q_{h}^{-}, \ &hat{p_{h}}=p_{h}^{theta}=theta p_{h}^{-}+(1-theta )p_{h}^{+}, \ &hat{r_{h}}=r_{h}^{theta}=theta r_{h}^{-}+(1-theta )r_{h}^{+}, \ &hat{s_{h}}=s_{h}^{tilde{theta}}=theta s_{h}^{+}+(1-theta )s_{h}^{-}, end{aligned}

where (theta >frac{1}{2}). For the initial condition, we take (u_{h}(0)=P_{h}u_{0}). It holds that

begin{aligned} & Vert u_{0}-P_{h}u_{0} Vert _{L^{2}(I)}leq Ch^{k+1} Vert u_{0} Vert _{W^{k+1,infty}(I)}, end{aligned}

(2.4)

where (P_{h}) is the (L^{2}) projection into (V_{h}^{k}).

We define (langle z, prangle =int _{I}zcdot pdx). For simplicity, we would like to introduce the DG discrete operator (H(z,p,hat{z})). That is,

begin{aligned} H(z,p,hat{z})=sum_{j=1}^{N}H_{j}(z,p, hat{z}), end{aligned}

where for each cell (I_{j}=(x_{j-frac{1}{2}},x_{j+frac{1}{2}})),

begin{aligned} &H_{j}(z,p,hat{z})= int _{I_{j}}hat{z}p_{x},dx-hat{z}p^{-}| _{j+ frac{1}{2}}+hat{z}p^{+}| _{j-frac{1}{2}}. end{aligned}

(2.5)

By simple calculations, we obtain the following lemma for a DG discrete operator.

### Lemma 1

begin{aligned} &Hbigl(z,p,z^{theta}bigr)+Hbigl(p,z,p^{tilde{theta}}bigr)=0, end{aligned}

(2.6a)

begin{aligned} &Hbigl(z,p,z^{theta}bigr)+Hbigl(p,z,p^{theta}bigr)=(1-2 theta )sum_{j=1}^{N}bigl([z] cdot [p] bigr)_{j+frac{1}{2}}, end{aligned}

(2.6b)

begin{aligned} &Hbigl(z,p,z^{tilde{theta}}bigr)+Hbigl(p,z,p^{tilde{theta}}bigr)=(2 theta -1) sum_{j=1}^{N}bigl([z]cdot [p]bigr)_{j+frac{1}{2}}, end{aligned}

(2.6c)

begin{aligned} &Hbigl(z,z,z^{theta}bigr)=biggl(frac{1}{2}-theta biggr) sum_{j=1}^{N}bigl([z]^{2} bigr)_{j+ frac{1}{2}}, end{aligned}

(2.6d)

begin{aligned} &Hbigl(z,z,z^{tilde{theta}}bigr)=biggl(theta -frac{1}{2}biggr) sum_{j=1}^{N}bigl([z]^{2} bigr)_{j+ frac{1}{2}}. end{aligned}

(2.6e)

#### The numerical initial condition

In this subsection, to derive optimal error estimates for the fifth order equation, we need to obtain the optimal initial error estimates for all variables first [20]. We consider the corresponding steady-state problem

begin{aligned} &u+u_{xxxxx}=g(x) end{aligned}

satisfying periodic conditions and a source term (g(x)=u_{0}(x)+u_{0}(x)^{(5)}) so that its exact solution is identically the initial condition of (2.1a)–(2.1b), (u_{0}(x)). That is, find (u_{h},s_{h},r_{h},p_{h},q_{h}in V^{k}_{h}) such that

begin{aligned} & int _{I_{j}}u_{h}rho ,dx-H_{j} bigl(s_{h},rho ,s_{h}^{tilde{theta}}bigr)= int _{I_{j}}grho ,dx, end{aligned}

(2.7a)

begin{aligned} & int _{I_{j}}s_{h}xi ,dx+H_{j} bigl(r_{h},xi ,r_{h}^{theta}bigr)=0, end{aligned}

(2.7b)

begin{aligned} & int _{I_{j}}r_{h}phi ,dx+H_{j} bigl(p_{h},phi ,p_{h}^{theta}bigr)=0, end{aligned}

(2.7c)

begin{aligned} & int _{I_{j}}p_{h}psi ,dx+H_{j} bigl(q_{h},psi ,q_{h}^{tilde{theta}}bigr)=0, end{aligned}

(2.7d)

begin{aligned} & int _{I_{j}}q_{h}varphi ,dx+H_{j} bigl(u_{h},varphi ,u_{h}^{theta}bigr)=0, end{aligned}

(2.7e)

hold for any (rho ,xi ,phi ,psi ,varphi in V^{k}_{h}) and (j=1,ldots ,N ).

### Lemma 2

The numerical initial condition (2.7a)(2.7e) is well defined. That is, LDG solutions (u_{h}), (s_{h}), (r_{h}), (p_{h}), (q_{h}) of (2.7a)(2.7e) uniquely exist.

### Proof

Since (2.7a) is a linear system with a known right-hand side and (s_{h}), (r_{h}), (p_{h}), (q_{h}) can be expressed by (u_{h}), we can first prove the uniqueness of ((u_{h},s_{h},r_{h},p_{h},q_{h})), then, obviously, the existence will follow. Assuming that ((u^{1}_{h},s^{1}_{h},r^{1}_{h},p^{1}_{h},q^{1}_{h})) and ((u^{2}_{h},s^{2}_{h},r^{2}_{h},p^{2}_{h},q^{2}_{h})) are two different solutions of (2.7a)–(2.7e) and denoting (g_{u}=u^{1}_{h}-u^{2}_{h}), (g_{s}=s^{1}_{h}-s^{2}_{h}), (g_{r}=r^{1}_{h}-r^{2}_{h}), (g_{p}=p^{1}_{h}-p^{2}_{h}), (g_{q}=q^{1}_{h}-q^{2}_{h}), we have

begin{aligned} & int _{I_{j}}g_{u}rho ,dx-H_{j} bigl(g_{s},rho ,g_{s}^{tilde{theta}}bigr)=0, end{aligned}

(2.8a)

begin{aligned} & int _{I_{j}}g_{s}xi ,dx+H_{j} bigl(g_{r},xi ,g_{r}^{theta}bigr)=0, end{aligned}

(2.8b)

begin{aligned} & int _{I_{j}}g_{r}phi ,dx+H_{j} bigl(g_{p},phi ,g_{p}^{theta}bigr)=0, end{aligned}

(2.8c)

begin{aligned} & int _{I_{j}}g_{p}psi ,dx+H_{j} bigl(g_{q},psi ,g_{q}^{tilde{theta}}bigr)=0, end{aligned}

(2.8d)

begin{aligned} & int _{I_{j}}g_{q}varphi ,dx+H_{j} bigl(g_{u},varphi ,g_{u}^{theta}bigr)=0. end{aligned}

(2.8e)

We take ((rho ,xi ,phi ,psi ,varphi )=(g_{u},g_{q},-g_{p},g_{r},-g_{s})) in (2.8a)–(2.8e). By Lemma 1 and direct calculation, we have

begin{aligned} & Vert g_{u} Vert ^{2}-biggl( frac{1}{2}-theta biggr)sum_{j=1}^{N} bigl([z]^{2}bigr)_{j+ frac{1}{2}}=0. end{aligned}

(2.9)

Thus (g_{u}=0) since (theta >frac{1}{2}). Then, substituting (g_{u}=0) into (2.8e) and letting (varphi =g_{q}), we have (g_{q}=0). Similarly, we have (g_{p}=g_{r}=g_{s}=0), which implies that (u_{h}), (s_{h}), (r_{h}), (p_{h}), (q_{h}) are unique. This ends the proof of Lemma 2. □

For the LDG scheme, using the generalized numerical fluxes, we need to construct a globally defined projection (P_{h}^{*}). For (uin H^{1}(I)), the projection (P_{h}^{*}u) is defined as the element of (V_{h}^{k}) that satisfies

begin{aligned} &int _{I_{j}}P_{h}^{*}u(x)v_{h},dx= int _{I_{j}}u(x)v_{h},dx,quad forall v_{h}in P^{k-1}(I_{j}), end{aligned}

(2.10a)

begin{aligned} &hat{P_{h}^{*}u}=hat{u},quad at x_{j+ frac{1}{2}},j=1,ldots ,N, end{aligned}

(2.10b)

with (theta >frac{1}{2}).

### Lemma 3

Assume that u is sufficiently smooth and periodic. Then there exists unique (P_{h}^{*}) satisfying conditions (2.10a) and (2.10b). Moreover, there holds the following property:

begin{aligned} & biglVert u-P_{h}^{*}u bigrVert _{L^{2}(I)}leq Ch^{k+1} Vert u Vert _{W^{k+1,infty}(I)}. end{aligned}

(2.11)

Using the initial condition and the triangle inequality, we have

begin{aligned} & biglVert bigl(P_{h}^{*}u-u_{h} bigr) (cdot ,0) bigrVert _{L^{2}(I)}leq Ch^{k+1} Vert u Vert _{W^{k+1, infty}(I)}, end{aligned}

(2.12)

where (C=C(theta )) is independent of the mesh size h. The lemma has been proved in [15].

### Lemma 4

Assuming that (u_{0}in W^{k+1,infty}(I_{h})) and periodic, we have the following error estimates:

begin{aligned} & Vert u_{0}-u_{h} Vert + Vert s_{0}-s_{h} Vert + Vert r_{0}-r_{h} Vert + Vert p_{0}-p_{h} Vert + Vert q_{0}-q_{h} Vert leq Ch^{k+1}, end{aligned}

(2.13)

where (q_{0}=u_{0}^{prime }), (p_{0}=u_{0}^{prime prime }), (r_{0}=u_{0}^{(3)}), (s_{0}=u_{0}^{(4)}), and C is independent of h.

### Proof

In this part, we denote

begin{aligned} &e_{u}=u_{0}-u_{h}=bigl(u_{0}-P_{h}^{*}u bigr)+bigl(P_{h}^{*}u-u_{h}bigr)=eta _{u}+ bar{e_{u}}, \ &e_{q}=q_{0}-q_{h}=bigl(q_{0}-P_{h}^{*}q bigr)+bigl(P_{h}^{*}q-q_{h}bigr)=eta _{q}+ bar{e_{q}}, \ &e_{p}=p_{0}-p_{h}=bigl(p_{0}-P_{h}^{*}p bigr)+bigl(P_{h}^{*}p-p_{h}bigr)=eta _{p}+ bar{e_{p}}, \ &e_{r}=r_{0}-r_{h}=bigl(r_{0}-P_{h}^{*}r bigr)+bigl(P_{h}^{*}r-r_{h}bigr)=eta _{r}+ bar{e_{r}}, \ &e_{s}=s_{0}-s_{h}=bigl(s_{0}-P_{h}^{*}s bigr)+bigl(P_{h}^{*}s-s_{h}bigr)=eta _{s}+ bar{e_{s}}. end{aligned}

Considering scheme (2.8a)–(2.8e) and summing over all j, we have the following equations:

begin{aligned} & int _{I}e_{u}rho ,dx-Hbigl(e_{s}, rho ,e_{s}^{tilde{theta}}bigr)=0, \ & int _{I}e_{s}xi ,dx+Hbigl(e_{r}, xi ,e_{r}^{theta}bigr)=0, \ & int _{I}e_{r}phi ,dx+Hbigl(e_{p}, phi ,e_{p}^{theta}bigr)=0, \ & int _{I}e_{p}psi ,dx+Hbigl(e_{q}, psi ,e_{q}^{tilde{theta}}bigr)=0, \ & int _{I}e_{q}varphi ,dx+Hbigl(e_{u}, varphi ,e_{u}^{theta}bigr)=0. end{aligned}

By the orthogonality, we have

begin{aligned} & int _{I}bar{e_{u}}rho ,dx-Hbigl( bar{e_{s}},rho ,bar{e_{s}}^{ tilde{theta}}bigr)=- int _{I}eta _{u}rho ,dx, end{aligned}

(2.14a)

begin{aligned} & int _{I}bar{e_{s}}xi ,dx+Hbigl( bar{e_{r}},xi ,bar{e_{r}}^{theta}bigr)=- int _{I}eta _{s}xi ,dx, end{aligned}

(2.14b)

begin{aligned} & int _{I}bar{e_{r}}phi ,dx+Hbigl( bar{e_{p}},phi ,bar{e_{p}}^{theta}bigr)=- int _{I}eta _{r}phi ,dx, end{aligned}

(2.14c)

begin{aligned} & int _{I}bar{e_{p}}psi ,dx+Hbigl( bar{e_{q}},psi ,bar{e_{q}}^{ tilde{theta}}bigr)=- int _{I}eta _{p}psi ,dx, end{aligned}

(2.14d)

begin{aligned} & int _{I}bar{e_{q}}varphi ,dx+Hbigl( bar{e_{u}},varphi ,bar{e_{u}}^{ theta} bigr)=- int _{I}eta _{q}varphi ,dx, end{aligned}

(2.14e)

that hold for any (rho ,xi ,phi ,psi ,varphi in V^{k}_{h}). In what follows, we will prove the optimal initial error estimates. Taking ((rho ,xi )=(-bar{e_{r}},bar{e_{s}})) in (2.14a)–(2.14b), summing the corresponding equations up, and using Lemma 1, we obtain the following equation:

begin{aligned} & int _{I}bar{e_{s}}bar{e_{s}},dx- int _{I}bar{e_{u}}bar{e_{r}},dx=- int _{I}eta _{s}bar{e_{s}},dx+ int _{I}eta _{u}bar{e_{r}},dx. end{aligned}

(2.15a)

Taking ((psi ,phi )=(bar{e_{p}},bar{e_{q}})), ((rho ,varphi )=( bar{e_{u}},-bar{e_{s}})), ((varphi ,psi )=(bar{e_{q}},bar{e_{u}})), ((phi ,xi )=(bar{e_{r}},bar{e_{p}})) and ((xi ,phi )=(-bar{e_{r}},-bar{e_{p}})), by the same way, we have

begin{aligned} & int _{I}bar{e_{p}}bar{e_{p}},dx+ int _{I}bar{e_{r}}bar{e_{q}},dx=- int _{I}eta _{p}bar{e_{p}},dx- int _{I}eta _{r}bar{e_{q}} ,dx, end{aligned}

(2.15b)

begin{aligned} & int _{I}bar{e_{u}}bar{e_{u}},dx- int _{I}bar{e_{q}}bar{e_{s}},dx=- int _{I}eta _{u}bar{e_{u}},dx+ int _{I}eta _{q}bar{e_{s}},dx, end{aligned}

(2.15c)

begin{aligned} & int _{I}bar{e_{q}}bar{e_{q}},dx+ int _{I}bar{e_{p}}bar{e_{u}},dx=- int _{I}eta _{q}bar{e_{q}},dx- int _{I}eta _{p}bar{e_{u}} ,dx, end{aligned}

(2.15d)

begin{aligned} & int _{I}bar{e_{r}}bar{e_{r}},dx+H bigl(bar{e_{p}},bar{e_{r}}, bar{e_{p}}^{theta} bigr)+ int _{I}bar{e_{s}}bar{e_{p}},dx+H bigl(bar{e_{r}}, bar{e_{p}},bar{e_{r}}^{theta} bigr) \ &quad =- int _{I}eta _{r}bar{e_{r}},dx- int _{I}eta _{s}bar{e_{p}},dx, end{aligned}

(2.15e)

begin{aligned} &{-}int _{I}bar{e_{s}}bar{e_{r}},dx-H bigl(bar{e_{r}},bar{e_{r}}, bar{e_{r}}^{theta} bigr)- int _{I}bar{e_{r}}bar{e_{p}},dx-H bigl(bar{e_{p}}, bar{e_{p}},bar{e_{p}}^{theta} bigr) \ &quad = int _{I}eta _{s}bar{e_{r}},dx+ int _{I}eta _{r}bar{e_{p}},dx. end{aligned}

(2.15f)

Adding (2.15e) and (2.15f), we get

begin{aligned} & int _{I}bar{e_{r}}bar{e_{r}},dx+ int _{I}bar{e_{s}}bar{e_{p}},dx- int _{I}bar{e_{s}}bar{e_{r}},dx- int _{I}bar{e_{r}}bar{e_{p}},dx+ Omega \ &quad =- int _{I}eta _{r}bar{e_{r}},dx- int _{I}eta _{s}bar{e_{p}},dx int _{I}eta _{s}bar{e_{r}},dx+ int _{I}eta _{r}bar{e_{p}},dx, end{aligned}

(2.16)

where

begin{aligned} &Omega =Hbigl(bar{e_{p}},bar{e_{r}}, bar{e_{p}}^{theta}bigr)+Hbigl( bar{e_{r}}, bar{e_{p}},bar{e_{r}}^{theta}bigr)-Hbigl( bar{e_{r}}, bar{e_{r}},bar{e_{r}}^{theta} bigr)-Hbigl(bar{e_{p}},bar{e_{p}}, bar{e_{p}}^{theta}bigr). end{aligned}

It is easy to find that (Omega geq 0) by Lemma 1. Then, using Lemmas 1 and 3, the Cauchy–Schwarz and Young’s inequalities to (2.15a)–(2.15d) and (2.16), we have

begin{aligned} & Vert bar{e_{s}} Vert ^{2}leq C_{u_{1}} Vert bar{e_{u}} Vert ^{2}+C_{r_{1}} Vert bar{e_{r}} Vert ^{2}+C_{s_{1}} Vert bar{e_{s}} Vert ^{2}+Ch^{2k+2}, end{aligned}

(2.17a)

begin{aligned} & Vert bar{e_{p}} Vert ^{2}leq C_{p_{2}} Vert bar{e_{p}} Vert ^{2}+C_{r_{2}} Vert bar{e_{r}} Vert ^{2}+C_{q_{2}} Vert bar{e_{q}} Vert ^{2}+Ch^{2k+2}, end{aligned}

(2.17b)

begin{aligned} & Vert bar{e_{q}} Vert ^{2}leq C_{u_{3}} Vert bar{e_{u}} Vert ^{2}+C_{q_{3}} Vert bar{e_{q}} Vert ^{2}+C_{p_{3}} Vert bar{e_{p}} Vert ^{2}+Ch^{2k+2}, end{aligned}

(2.17c)

begin{aligned} & Vert bar{e_{u}} Vert ^{2}leq C_{u_{4}} Vert bar{e_{u}} Vert ^{2}+C_{q_{4}} Vert bar{e_{4}} Vert ^{2}+C_{s_{4}} Vert bar{e_{s}} Vert ^{2}+Ch^{2k+2}, end{aligned}

(2.17d)

begin{aligned} & Vert bar{e_{r}} Vert ^{2}leq C_{p_{5}} Vert bar{e_{p}} Vert ^{2}+C_{r_{5}} Vert bar{e_{r}} Vert ^{2}+C_{s_{5}} Vert bar{e_{s}} Vert ^{2}+Ch^{2k+2}. end{aligned}

(2.17e)

Furthermore, by adjusting the coefficients in (2.17a)–(2.17e) with Young’s inequality, we have the estimate

begin{aligned} & Vert s_{0}-s_{h} Vert ^{2}+ Vert r_{0}-r_{h} Vert ^{2}+ Vert p_{0}-p_{h} Vert ^{2}+ Vert q_{0}-q_{h} Vert ^{2} leq Vert u_{0}-u_{h} Vert ^{2}+Ch^{2k+2}. end{aligned}

Hence we arrive at

begin{aligned} & Vert s_{0}-s_{h} Vert + Vert r_{0}-r_{h} Vert + Vert p_{0}-p_{h} Vert + Vert q_{0}-q_{h} Vert leq Vert u_{0}-u_{h} Vert +Ch^{k+1}. end{aligned}

(2.17f)

We take ((rho ,xi ,phi ,psi ,varphi )=(bar{e_{u}},bar{e_{q}},- bar{e_{p}},bar{e_{r}},-bar{e_{s}})) in (2.14a)–(2.14e). Through direct calculation, from Lemma 1 we have

begin{aligned} & Vert bar{e_{u}} Vert ^{2}-biggl( frac{1}{2}-theta biggr)sum_{j=1}^{N} bigl([ bar{e_{p}}]^{2}bigr)_{j+frac{1}{2}} \ &quad = int _{I}eta _{p}bar{e_{p}},dx- int _{I}eta _{u}bar{e_{u}},dx- int _{I}eta _{q}bar{e_{q}},dx+ int _{I}eta _{s}bar{e_{s}},dx- int _{I} eta _{r}bar{e_{r}},dx. end{aligned}

(2.17g)

Substituting (2.17f) into (2.17g) and using Lemma 3, we finally get (|bar{e_{u}}|leq Ch^{k+1}). This completes the proof of Lemma 4. □

### Error estimates

In this subsection, we state the error estimate of the LDG method using the generalized numerical fluxes. First, we define

begin{aligned} &e_{u}=u-u_{h}=bigl(u-P_{h}^{*}u bigr)+bigl(P_{h}^{*}u-u_{h}bigr)=eta _{u}+ bar{e_{u}}, \ &e_{q}=q-q_{h}=bigl(q-P_{h}^{*}q bigr)+bigl(P_{h}^{*}q-q_{h}bigr)=eta _{q}+ bar{e_{q}}, \ &e_{p}=p-p_{h}=bigl(p-P_{h}^{*}p bigr)+bigl(P_{h}^{*}p-p_{h}bigr)=eta _{p}+ bar{e_{p}}, \ &e_{r}=r-r_{h}=bigl(r-P_{h}^{*}r bigr)+bigl(P_{h}^{*}r-r_{h}bigr)=eta _{r}+ bar{e_{r}}, \ &e_{s}=s-s_{h}=bigl(s-P_{h}^{*}s bigr)+bigl(P_{h}^{*}s-s_{h}bigr)=eta _{s}+ bar{e_{s}}. end{aligned}

Then we have the following theorem.

### Theorem 1

Assume that u, q, p, r, s are the exact solutions of system (2.2a)(2.2e), and for (tin [0,T]), (| u|_{k+5}), (| u_{t}|_{k+5}), (| u_{tt}|_{k+5}) are bounded uniformly. We take the generalized numerical fluxes and the finite element space (V_{h}^{k}), there holds the following (L^{2})norm error estimates:

begin{aligned} & Vert e_{u} Vert + Vert e_{q} Vert + Vert e_{p} Vert + Vert e_{r} Vert + Vert e_{s} Vert + biglVert (e_{u})_{t} bigrVert leq Ch^{k+1}(t+3), end{aligned}

(2.18)

where C depends on θ, (| u|_{k+5}), (| u_{t}|_{k+5}), and (| u_{tt}|_{k+5}), but is independent of h.

### Proof

Using the DG discrete operator, the LDG scheme can be written as

begin{aligned} & int _{I_{j}}(u_{h})_{t}rho ,dx-H_{j}bigl(s_{h},rho ,s_{h}^{ tilde{theta}} bigr)=0, end{aligned}

(2.19a)

begin{aligned} & int _{I_{j}}s_{h}xi ,dx+H_{j} bigl(r_{h},xi ,r_{h}^{theta}bigr)=0, end{aligned}

(2.19b)

begin{aligned} & int _{I_{j}}r_{h}phi ,dx+H_{j} bigl(p_{h},phi ,p_{h}^{theta}bigr)=0, end{aligned}

(2.19c)

begin{aligned} & int _{I_{j}}p_{h}psi ,dx+H_{j} bigl(q_{h},psi ,q_{h}^{tilde{theta}}bigr)=0, end{aligned}

(2.19d)

begin{aligned} & int _{I_{j}}q_{h}varphi ,dx+H_{j} bigl(u_{h},varphi ,u_{h}^{theta}bigr)=0. end{aligned}

(2.19e)

Summing (2.19a)–(2.19e) over all (j=1,ldots ,N), we obtain

begin{aligned} & int _{I}(u_{h})_{t}rho ,dx-H bigl(s_{h},rho ,s_{h}^{tilde{theta}}bigr)=0, end{aligned}

(2.20a)

begin{aligned} & int _{I}s_{h}xi ,dx+Hbigl(r_{h}, xi ,r_{h}^{theta}bigr)=0, end{aligned}

(2.20b)

begin{aligned} & int _{I}r_{h}phi ,dx+Hbigl(p_{h}, phi ,p_{h}^{theta}bigr)=0, end{aligned}

(2.20c)

begin{aligned} & int _{I}p_{h}psi ,dx+Hbigl(q_{h}, psi ,q_{h}^{tilde{theta}}bigr)=0, end{aligned}

(2.20d)

begin{aligned} & int _{I}q_{h}varphi ,dx+Hbigl(u_{h}, varphi ,u_{h}^{theta}bigr)=0. end{aligned}

(2.20e)

Thus, we get the following error equations:

begin{aligned} & int _{I}(e_{u})_{t}rho ,dx-H bigl(e_{s},rho ,e_{s}^{tilde{theta}}bigr)=0, end{aligned}

(2.21a)

begin{aligned} & int _{I}e_{s}xi ,dx+Hbigl(e_{r}, xi ,e_{r}^{theta}bigr)=0, end{aligned}

(2.21b)

begin{aligned} & int _{I}e_{r}phi ,dx+Hbigl(e_{p}, phi ,e_{p}^{theta}bigr)=0, end{aligned}

(2.21c)

begin{aligned} & int _{I}e_{p}psi ,dx+Hbigl(e_{q}, psi ,e_{q}^{tilde{theta}}bigr)=0, end{aligned}

(2.21d)

begin{aligned} & int _{I}e_{q}varphi ,dx+Hbigl(e_{u}, varphi ,e_{u}^{theta}bigr)=0. end{aligned}

(2.21e)

To prove the theorem, we need to establish six equations by repeatedly taking different ρ, ξ, ϕ, ψ, φ in Eqs. (2.21a)–(2.21e). The specific method is as follows.

The first equation:

Taking ((rho , xi , phi , psi , varphi )=(bar{e_{u}}, bar{e_{q}}, -bar{e_{p}}, bar{e_{r}}, -bar{e_{s}})) in Eqs. (2.21a)–(2.21e) and summing over all equations, we get

begin{aligned} &frac{1}{2}frac{d}{dt} Vert bar{e_{u}} Vert ^{2}+bigllangle ( eta _{u})_{t}, bar{e_{u}}bigrrangle +langle e_{s}, bar{e_{q}}rangle – langle e_{r},bar{e_{p}} rangle \ &quad {}+langle e_{p},bar{e_{r}}rangle -langle e_{q},bar{e_{s}}rangle -Hbigl( bar{e_{s}}, bar{e_{u}},bar{e_{s}}^{bar{theta}}bigr)-Hbigl( bar{e_{u}}, bar{e_{s}},bar{e_{u}}^{theta} bigr) \ &quad {}+Hbigl(bar{e_{r}},bar{e_{q}}, bar{e_{r}}^{theta}bigr)+Hbigl(bar{e_{q}}, bar{e_{r}},bar{e_{q}}^{bar{theta}}bigr)-Hbigl( bar{e_{p}},bar{e_{p}}, bar{e_{p}}^{theta} bigr)=0 . end{aligned}

(2.22)

Then, by Lemma 1, Eq. (2.22) is finally turned into

begin{aligned} &frac{1}{2}frac{d}{dt} Vert bar{e_{u}} Vert ^{2}+biggl(theta – frac{1}{2}biggr)sum _{j=1}^{N}[bar{e_{p}}]^{2} \ &quad =-bigllangle (eta _{u})_{t}, bar{e_{u}} bigrrangle -langle eta _{s},bar{e_{q}}rangle +langle eta _{r},bar{e_{p}} rangle -langle eta _{p},bar{e_{r}}rangle +langle eta _{q}, bar{e_{s}}rangle . end{aligned}

(2.23)

The second equation:

Taking the derivatives on both sides of (2.21a)–(2.21e) with respect to t and taking ((rho , xi , phi , psi , varphi )=((bar{e_{u}})_{t}, ( bar{e_{q}})_{t}, -(bar{e_{p}})_{t}, (bar{e_{r}})_{t}, -( bar{e_{s}})_{t})), we obtain

begin{aligned} &frac{1}{2}frac{d}{dt} biglVert (bar{e_{u}})_{t} bigrVert ^{2}+biggl( theta -frac{1}{2}biggr)sum _{j=1}^{N}bigl[(bar{e_{p}})_{t} bigr]^{2} \ &quad =- bigllangle (eta _{u})_{tt},( bar{e_{u}})_{t}bigrrangle -bigllangle (eta _{s})_{t},( bar{e_{q}})_{t}bigrrangle +bigllangle (eta _{r})_{t},( bar{e_{p}})_{t} bigrrangle -bigllangle (eta _{p})_{t},( bar{e_{r}})_{t} bigrrangle +bigllangle (eta _{q})_{t},(bar{e_{s}})_{t} bigrrangle . end{aligned}

(2.24)

The third equation:

Substituting ((psi , varphi )=(bar{e_{u}}, bar{e_{q}})) into (2.21d) and (2.21e), we have

begin{aligned} &langle eta _{p},bar{e_{u}}rangle +langle bar{e_{p}}, bar{e_{u}}rangle +langle eta _{q},bar{e_{q}}rangle +langle bar{e_{q}}, bar{e_{q}}rangle \ &quad {}+Hbigl(bar{e_{q}},bar{e_{u}},e_{q}^{bar{theta}} bigr)+Hbigl(bar{e_{u}}, bar{e_{q}},e_{u}^{theta} bigr)=0. end{aligned}

Using Lemma 1, we have

begin{aligned} & Vert bar{e_{q}} Vert ^{2}=-langle bar{e_{p}},bar{e_{u}} rangle -langle eta _{q},bar{e_{q}}rangle -langle eta _{p}, bar{e_{u}}rangle . end{aligned}

(2.25)

The fourth equation:

Similar to the third equation, taking ((phi , psi )=(bar{e_{q}}, bar{e_{p}})) in (2.21c) and (2.21d), we get

begin{aligned} & Vert bar{e_{p}} Vert ^{2}=-langle bar{e_{r}},bar{e_{q}} rangle -langle eta _{p},bar{e_{p}}rangle -langle eta _{r}, bar{e_{q}}rangle . end{aligned}

(2.26)

The fifth equation:

Substituting ((xi , phi )=(bar{e_{p}}, bar{e_{r}})) into (2.21b) and (2.21c) yields that

begin{aligned} &langle bar{e_{s}},bar{e_{p}}rangle +langle eta _{s}, bar{e_{p}}rangle +langle bar{e_{r}}, bar{e_{r}}rangle +langle eta _{r}, bar{e_{r}}rangle +Hbigl(bar{e_{r}},bar{e_{p}}, bar{e_{r}}^{theta}bigr)+Hbigl(bar{e_{p}}, bar{e_{r}},bar{e_{p}}^{theta}bigr)=0. end{aligned}

Using Lemma 1, we obtain

begin{aligned} & Vert bar{e_{r}} Vert ^{2}=(2theta -1)sum _{j=1}^{N}bigl([ bar{e_{p}}] [bar{e_{r}}]bigr)_{j+frac{1}{2}}-langle bar{e_{s}}, bar{e_{p}}rangle -langle eta _{s}, bar{e_{p}}rangle -langle eta _{r}, bar{e_{r}}rangle . end{aligned}

(2.27)

The sixth equation:

Finally, substituting ((rho , xi )=(bar{e_{r}}, -bar{e_{s}})) into (2.21a) and (2.21b), it is easy to see by Lemma 1 that

begin{aligned} & Vert bar{e_{s}} Vert ^{2}=-langleeta _{s},bar{e_{s}}rangle+bigllangle (eta _{u})_{t}, bar{e_{r}}bigrrangle +bigllangle (bar{e_{u}})_{t},bar{e_{r}}bigrrangle . end{aligned}

(2.28)

Now we have six equations. In what follows we need to find some appropriate coefficients, and according to (2.25), (2.26), (2.27), and (2.28) we can infer the relationship among (| bar{e_{u}}| ), (| bar{e_{q}}| ), (| bar{e_{p}}| ), (| bar{e_{r}}| ), (| bar{e_{s}}| ), (| (bar{e_{u}})_{t} | ).

By multiplying some constants (19times text{(2.25)}+3times text{(2.26)}+text{(2.27)}+text{(2.28)}), we obtain

begin{aligned} &Vert bar{e_{s}} Vert ^{2}+ Vert bar{e_{r}} Vert ^{2}+3 Vert bar{e_{p}} Vert ^{2}+19 Vert bar{e_{q}} Vert ^{2} \ &quad = (2 theta -1)sum_{j=1}^{N} bigl([bar{e_{p}}] [bar{e_{r}}]bigr)_{j+ frac{1}{2}} \ &qquad {}-19langle bar{e_{p}},bar{e_{u}}rangle -3langle bar{e_{r}}, bar{e_{q}}rangle -langle bar{e_{s}},bar{e_{p}}rangle +bigllangle ( bar{e_{u}})_{t},,bar{e_{r}}bigrrangle \ &qquad {}-19langle eta _{q},bar{e_{q}}rangle -19langle eta _{p}, bar{e_{u}}rangle -3langle eta _{p},bar{e_{p}}rangle -3langle eta _{r},bar{e_{q}}rangle \ &qquad {}-langle eta _{r},bar{e_{r}}rangle -langle eta _{s},bar{e_{p}} rangle -langle eta _{s},bar{e_{s}}rangle +bigllangle (eta _{u})_{t}, bar{e_{r}}bigrrangle . end{aligned}

(2.29)

Furthermore, by Young’s inequality and Lemma 1, we have

begin{aligned} (2theta -1)sum_{j=1}^{N}bigl([ bar{e_{p}}] [bar{e_{r}}]bigr)_{j+ frac{1}{2}}&leq frac{1}{2}Biggl{ (2theta -1)sum_{j=1}^{N}[ bar{e_{p}}]^{2}_{j+frac{1}{2}}+(2theta -1)sum _{j=1}^{N}[ bar{e_{r}}]^{2}_{j+frac{1}{2}} Biggr} \ &=langle e_{r},bar{e_{p}}rangle +langle e_{s},bar{e_{r}}rangle \ &=langle bar{e_{r}},bar{e_{p}}rangle +langle bar{e_{s}}, bar{e_{r}}rangle +langle eta _{r},bar{e_{p}}rangle +langle eta _{s},bar{e_{r}}rangle . end{aligned}

(2.30)

Then, substituting (2.30) into (2.29), we obtain

begin{aligned} & Vert bar{e_{s}} Vert ^{2}+ Vert bar{e_{r}} Vert ^{2}+3 Vert bar{e_{p}} Vert ^{2}+19 Vert bar{e_{q}} Vert ^{2} \ &quad leq biglvert langle bar{e_{r}},bar{e_{p}} rangle bigrvert + biglvert langle bar{e_{s}}, bar{e_{r}}rangle bigrvert +19 biglvert langle bar{e_{p}},bar{e_{u}}rangle bigrvert +3 biglvert langle bar{e_{r}}, bar{e_{q}}rangle bigrvert + biglvert langle bar{e_{s}},bar{e_{p}} rangle bigrvert \ &qquad {}+ biglvert bigllangle (bar{e_{u}})_{t}, bar{e_{r}}bigrrangle bigrvert +19 biglvert langle eta _{q},bar{e_{q}}rangle bigrvert +19 biglvert langle eta _{p}, bar{e_{u}}rangle bigrvert +3 biglvert langle eta _{p},bar{e_{p}}rangle bigrvert \ &qquad {}+3 biglvert langle eta _{r},bar{e_{q}} rangle bigrvert + biglvert langle eta _{r}, bar{e_{r}}rangle bigrvert + biglvert langle eta _{s},bar{e_{p}}rangle bigrvert + biglvert langle eta _{s},bar{e_{s}}rangle bigrvert \ &qquad {}+ biglvert bigllangle (eta _{u})_{t}, bar{e_{r}}bigrrangle bigrvert + biglvert langle eta _{r},bar{e_{p}}rangle bigrvert + biglvert langle eta _{s},bar{e_{r}} rangle bigrvert . end{aligned}

(2.31)

It follows from the Cauchy–Schwarz inequality and the properties of the projections that

begin{aligned} & Vert bar{e_{s}} Vert ^{2}+ Vert bar{e_{r}} Vert ^{2}+3 Vert bar{e_{p}} Vert ^{2}+19 Vert bar{e_{q}} Vert ^{2} \ &quad leq Vert bar{e_{r}} Vert Vert bar{e_{p}} Vert + Vert bar{e_{s}} Vert Vert bar{e_{r}} Vert +19 Vert bar{e_{p}} Vert Vert bar{e_{u}} Vert +3 Vert bar{e_{r}} Vert Vert bar{e_{q}} Vert + Vert bar{e_{s}} Vert Vert bar{e_{p}} Vert \ &qquad {}+ biglVert (bar{e_{u}})_{t} bigrVert Vert bar{e_{r}} Vert +Ch^{k+1}bigl( Vert bar{e_{s}} Vert + Vert bar{e_{r}} Vert + Vert bar{e_{p}} Vert + Vert bar{e_{q}} Vert + Vert bar{e_{u}} Vert bigr). end{aligned}

(2.32)

Next, using Young’s inequality, we have

begin{aligned} & Vert bar{e_{s}} Vert ^{2}+ Vert bar{e_{r}} Vert ^{2}+3 Vert bar{e_{p}} Vert ^{2}+19 Vert bar{e_{q}} Vert ^{2} \ &quad leq biggl(frac{1}{4} Vert bar{e_{r}} Vert ^{2}+ Vert bar{e_{p}} Vert ^{2}biggr)+biggl(frac{1}{2} Vert bar{e_{s}} Vert ^{2}+frac{1}{2} Vert bar{e_{r}} Vert ^{2}biggr) \ &qquad {}+biggl(frac{1}{8} Vert bar{e_{r}} Vert ^{2}+18 Vert bar{e_{q}} Vert ^{2} biggr)+biggl(frac{1}{4} Vert bar{e_{s}} Vert ^{2}+ Vert bar{e_{p}} Vert ^{2} biggr) \ &qquad {}+biggl(frac{1}{16} Vert bar{e_{r}} Vert ^{2}+4 biglVert ( bar{e_{u}})_{t} bigrVert ^{2}biggr)+biggl(frac{1}{4} Vert bar{e_{p}} Vert ^{2}+361 Vert bar{e_{u}} Vert ^{2}biggr) \ &qquad {}+Ch^{2k+2}+biggl(frac{1}{8} Vert bar{e_{s}} Vert ^{2}+ frac{1}{32} Vert bar{e_{r}} Vert ^{2}+frac{1}{4} Vert bar{e_{p}} Vert ^{2}+frac{1}{2} Vert bar{e_{q}} Vert ^{2}+ Vert bar{e_{u}} Vert ^{2}biggr). end{aligned}

(2.33)

After a very simple arrangement, we have

begin{aligned} &frac{1}{8} Vert bar{e_{s}} Vert ^{2}+frac{1}{32} Vert bar{e_{r}} Vert ^{2}+frac{1}{2} Vert bar{e_{p}} Vert ^{2}+frac{1}{2} Vert bar{e_{q}} Vert ^{2} \ &quad leq Ch^{2k+2}+Cbigl( Vert bar{e_{u}} Vert ^{2}+ biglVert ( bar{e_{u}})_{t} bigrVert ^{2}bigr). end{aligned}

Using Young’s inequality for further simplification, we obtain

begin{aligned} &frac{1}{128}bigl( Vert bar{e_{s}} Vert + Vert bar{e_{r}} Vert + Vert bar{e_{p}} Vert + Vert bar{e_{q}} Vert bigr)^{2} \ &quad leq Ch^{2k+2}+Cbigl( Vert bar{e_{u}} Vert ^{2}+ biglVert ( bar{e_{u}})_{t} bigrVert ^{2}bigr). end{aligned}

(2.34)

That is,

begin{aligned} & Vert bar{e_{s}} Vert + Vert bar{e_{r}} Vert + Vert bar{e_{p}} Vert + Vert bar{e_{q}} Vert leq Ch^{k+1}+Cbigl( Vert bar{e_{u}} Vert + biglVert (bar{e_{u}})_{t} bigrVert bigr), end{aligned}

(2.35)

where the constant C depends on θ, (| u| _{k+5}), and (| u_{t}| _{k+5}), but is independent of h. Next, adding (2.23) and (2.24) to estimate (| bar{e_{u}}| +| (bar{e_{u}})_{t} | ), we have

begin{aligned} &frac{1}{2}frac{d}{dt} Vert bar{e_{u}} Vert ^{2}+ frac{1}{2}frac{d}{dt} biglVert (bar{e_{u}})_{t} bigrVert ^{2}+biggl( theta -frac{1}{2}biggr)sum_{j=1}^{N}[ bar{e_{p}}]^{2}_{j+ frac{1}{2}} +biggl(theta -frac{1}{2}biggr)sum _{j=1}^{N}bigl[(bar{e_{p}})_{t} bigr]^{2}_{j+ frac{1}{2}} \ &quad =A+B. end{aligned}

(2.36)

Here

begin{aligned} A={}&langleeta _{q},bar{e_{s}}rangle-langle eta _{s},bar{e_{q}}rangle – langle eta _{p},bar{e_{r}}rangle +langle eta _{r},bar{e_{p}} rangle-bigllangle (eta _{u})_{t},bar{e_{u}} bigrrangle -bigllangle (eta _{u})_{tt},( bar{e_{u}})_{t}bigrrangle end{aligned}

(2.37)

and

begin{aligned} B=bigllangle (eta _{q})_{t},(bar{e_{s}})_{t} bigrrangle -bigllangle (eta _{s})_{t},( bar{e_{q}})_{t}bigrrangle -bigllangle (eta _{p})_{t},(bar{e_{r}})_{t} bigrrangle +bigllangle (eta _{r})_{t},( bar{e_{p}})_{t}bigrrangle end{aligned}

(2.38)

denote the right-hand sides of (2.23) and (2.24), respectively. For A, by using the Cauchy–Schwarz and Young’s inequalities, we obtain

begin{aligned} &Aleq Ch^{k+1}bigl( Vert bar{e_{r}} Vert + Vert bar{e_{p}} Vert + Vert bar{e_{s}} Vert + Vert bar{e_{q}} Vert + Vert bar{e_{u}} Vert + biglVert (bar{e_{u}})_{t} bigrVert bigr). end{aligned}

Then, combining the above inequality with (2.35), we arrive at

begin{aligned} &Aleq Ch^{2k+2}+Ch^{k+1}bigl( Vert bar{e_{u}} Vert + biglVert ( bar{e_{u}})_{t} bigrVert bigr), end{aligned}

(2.39)

where C depends on θ, (| u| _{k+5}), (| u_{t}| _{k+5}), and (| u_{tt}| _{k+5}), but is independent of h.

In order to estimate B, we need to handle the four integrations in B, respectively. In fact, the processing technique for each integration is similar, so we take the first term (langle (eta _{q})_{t},(bar{e_{s}})_{t}rangle ) as an example. First, we integrate (langle (eta _{q})_{t},(bar{e_{s}})_{t}rangle ) with respect to time between 0 and t, then exchange integral sequence and get the integration by parts

begin{aligned} int _{0}^{t} int _{I}(eta _{q})_{t}( bar{e_{s}})_{t},dx,dt&= int _{I} int _{0}^{t}(eta _{q})_{t}( bar{e_{s}})_{t},dt,dx \ &= int _{I}bigl[(eta _{q})_{t} bar{e_{s}}-bigl((eta _{q})_{t} bar{e_{s}}bigr) ( cdot ,0)bigr],dx- int _{0}^{t} int _{I}(eta _{q})_{tt} bar{e_{s}},dx,dt. end{aligned}

(2.40)

Similarly, doing the same calculations for each integration of B and integrating B with respect to t, we obtain

begin{aligned} int _{0}^{t}B,dt={}& int _{I}bigl[(eta _{q})_{t} bar{e_{s}}-bigl((eta _{q})_{t} bar{e_{s}}bigr) (cdot ,0)bigr],dx- int _{0}^{t} int _{I}(eta _{q})_{tt} bar{e_{s}},dx,dt \ &{}- int _{I}bigl[(eta _{s})_{t} bar{e_{q}}-bigl((eta _{s})_{t} bar{e_{q}}bigr) ( cdot ,0)bigr],dx- int _{0}^{t} int _{I}(eta _{s})_{tt} bar{e_{q}},dx,dt \ &{}- int _{I}bigl[(eta _{p})_{t} bar{e_{r}}-bigl((eta _{p})_{t} bar{e_{r}}bigr) ( cdot ,0)bigr],dx- int _{0}^{t} int _{I}(eta _{p})_{tt} bar{e_{r}},dx,dt \ &{}+ int _{I}bigl[(eta _{r})_{t} bar{e_{p}}-bigl((eta _{r})_{t} bar{e_{p}}bigr) ( cdot ,0)bigr],dx- int _{0}^{t} int _{I}(eta _{r})_{tt} bar{e_{p}},dx,dt. end{aligned}

(2.41)

By using the Cauchy–Schwarz inequality, Young’s inequality, and the properties of projections from Lemma 4, we have

begin{aligned} int _{0}^{t}B,dtleq{}& Ch^{k+1} bigl( Vert bar{e_{s}} Vert + Vert bar{e_{q}} Vert + Vert bar{e_{r}} Vert + Vert bar{e_{p}} Vert bigr)+Ch^{2k+2} \ &{}+Ch^{k+1} int _{0}^{t}bigl( Vert bar{e_{s}} Vert + Vert bar{e_{q}} Vert + Vert bar{e_{r}} Vert + Vert bar{e_{p}} Vert bigr),dt. end{aligned}

(2.42)

Furthermore, it follows from (2.35) and Young’s inequality that

begin{aligned} int _{0}^{t}B,dtleq{}& Ch^{2k+2}+Ch^{k+1} int _{0}^{t}bigl( Vert bar{e_{u}} Vert + biglVert (bar{e_{u}})_{t} bigrVert bigr),dt \ &{}+frac{1}{4}bigl( Vert bar{e_{u}} Vert ^{2}+ biglVert ( bar{e_{u}})_{t} bigrVert ^{2}bigr). end{aligned}

(2.43)

Now, integrating both sides of equality (2.36) with respect to t, and combining inequalities (2.39) and (2.43), we obtain

begin{aligned} &frac{1}{2} Vert bar{e_{u}} Vert ^{2}+frac{1}{2} biglVert ( bar{e_{u}})_{t} bigrVert ^{2}+ int _{0}^{t}biggl(theta – frac{1}{2}biggr) sum_{j=1}^{N}[ bar{e_{p}}]^{2}_{j+frac{1}{2}},dt \ &qquad {}+ int _{0}^{t}biggl(theta – frac{1}{2}biggr)sum_{j=1}^{N} bigl[( bar{e_{p}})_{t}bigr]^{2}_{j+frac{1}{2}},dt \ &quad leq Ch^{2k+2}+frac{1}{4}bigl( Vert bar{e_{u}} Vert ^{2}+ biglVert ( bar{e_{u}})_{t} bigrVert ^{2}bigr) \ &qquad {}+Ch^{k+1} int _{0}^{t}bigl( Vert bar{e_{u}} Vert + biglVert ( bar{e_{u}})_{t} bigrVert bigr),dt. end{aligned}

(2.44)

Thus, we have

begin{aligned} &bigl( Vert bar{e_{u}} Vert + biglVert ( bar{e_{u}})_{t} bigrVert bigr)^{2}leq Ch^{2k+2}+Ch^{k+1} int _{0}^{t}bigl( Vert bar{e_{u}} Vert + biglVert (bar{e_{u}})_{t} bigrVert bigr),dt. end{aligned}

(2.45)

We denote

begin{aligned} & A(t)= Vert bar{e_{u}} Vert + biglVert ( bar{e_{u}})_{t} bigrVert , \ & E(t)=Ch^{2k+2}+Ch^{k+1} int _{0}^{t}bigl( Vert bar{e_{u}} Vert + biglVert (bar{e_{u}})_{t} bigrVert bigr),dt. end{aligned}

By (2.37) we have

begin{aligned} & A(t)leq sqrt{E(t)}. end{aligned}

(2.46)

Note that

begin{aligned} & frac{d}{dt}E(t)=Ch^{k+1}A(t)leq Ch^{k+1} sqrt{E(t)} end{aligned}

(2.47)

and

begin{aligned} & frac{d}{dt}E(t)=2sqrt{E(t)}frac{d}{dt}sqrt{E(t)}. end{aligned}

(2.48)

Combining (2.39) with (2.48), we have

begin{aligned} & frac{d}{dt}sqrt{E(t)}leq Ch^{k+1}. end{aligned}

(2.49)

Integrating inequality (2.41) with respect to t, we have

begin{aligned} & sqrt{E(t)}leq sqrt{E(0)}+Ch^{k+1}tleq Ch^{k+1}(t+1). end{aligned}

(2.50)

Combining (2.50) with (2.46), we get

begin{aligned} & Vert bar{e_{u}} Vert + biglVert (bar{e_{u}})_{t} bigrVert leq Ch^{k+1}(t+1). end{aligned}

(2.51)

Therefore, from (2.35) and (2.51) we have

begin{aligned} & Vert bar{e_{s}} Vert + Vert bar{e_{r}} Vert + Vert bar{e_{p}} Vert + Vert bar{e_{q}} Vert + Vert bar{e_{u}} Vert + biglVert ( bar{e_{u}})_{t} bigrVert leq Ch^{k+1}(t+2), end{aligned}

(2.52)

where C depends on θ, (| u| _{k+5}), (| u_{t}| _{k+5}), and (| u_{tt}| _{k+5}), but is independent of h. Finally, combining inequality (2.44), Lemma 2, and the triangle inequality, we conclude that

begin{aligned} & Vert e_{u} Vert + Vert e_{q} Vert + Vert e_{p} Vert + Vert e_{r} Vert + Vert e_{s} Vert + biglVert (e_{u})_{t} bigrVert leq Ch^{k+1}(t+3), end{aligned}

(2.53)

and we prove the theorem. □