# A Mixed Discontinuous Galerkin Approximation of Time Dependent Convection Diffusion Optimal Control Problem

Authors: Qingjin Xu; Zhaojie Zhou
Journal: Journal of Mathematics
 (2017)

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## Abstract

In this paper, we investigate a mixed discontinuous Galerkin approximation of time dependent convection diffusion optimal control problem with control constraints based on the combination of a mixed finite element method for the elliptic part and a discontinuous Galerkin method for the hyperbolic part of the state equation. The control variable is approximated by variational discretization approach. A priori error estimates of the state, adjoint state, and control are derived for both semidiscrete scheme and fully discrete scheme. Numerical example is given to show the effectiveness of the numerical scheme.


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## Body

## 1. Introduction

The objective of this paper mainly focuses on developing a mixed discontinuous Galerkin scheme for the following control constrained optimal control problem governed by a transient convection diffusion equation:(1)minu∈KJy,u≔12∫0T∫Ωyx,t-ydx,t2dxdt+γ2∫0T∫Ωu2x,tdxdtsubject  to∂y∂t-εΔy-∇·βy+αy=f+u,inΩT≔Ω×0,T,y=0,onΓT≔∂Ω×0,T,yx,0=y0x,inΩ.More details will be specified later.This kind of problem plays an important role in many fields, such as air pollution ([1]) and waste water treatment ([2]). In recent years the research of numerical method for this kind of problem forms a hot topic. Lots of literatures are devoted to developing effective numerical methods for this kind of problem. In [3–6] the stabilization method such as local projection stabilization, SUPG, and continuous interior penalty method are investigated. The discontinuous Galerkin approximation including primary discontinuous Galerkin method and local discontinuous Galerkin method is addressed in [7–12]. In [13] the authors discuss the characteristic finite element approximation of transient convection diffusion optimal control problem. For more references, one can refer to [14].In this paper, we investigate a mixed discontinuous Galerkin approximation of transient convection diffusion optimal control problem with control constraints. This scheme is based on the combination of a discontinuous Galerkin method for the hyperbolic part and a mixed finite element method for the elliptic part of the state equation. Variational discretization approach is utilized to approximate the control variable. The work of this paper is motivated by [8, 15] where a similar scheme was proposed for convection diffusion equations and stationary convection diffusion optimal control problems, respectively. Similar to other discontinuous Galerkin methods, this scheme is also locally conservative, which makes it much suitable for problems where conservation is important, for example, for time dependent convection diffusion problems. Moreover, when the diffusion coefficient tends to zero, this scheme reduces to the classical discontinuous Galerkin method. Thus it inherits the stabilizing features of discontinuous Galerkin methods. We derive a priori error estimates of the state, adjoint state, and control for both semidiscrete scheme and fully discrete scheme. Numerical experiment is carried out to show the performance of our scheme.The rest of the paper is organized as follows: In Section2, semidiscrete and fully discrete mixed discontinuous Galerkin scheme are defined for control constrained optimal control problems governed by the time dependent convection diffusion equations. In Section 3, a priori error estimates of the semidiscrete scheme are derived. In Section 4, we derive a priori error estimates of the fully discrete scheme. Finally numerical example is given to illustrate the theoretical findings.


## 2. Mixed Discontinuous Galerkin Scheme

Consider the following optimal control problem with control constraints:(2)minu∈KJy,usubject  to∂y∂t-εΔy-∇·βy+αy=f+u,inΩT≔Ω×0,T,y=0,onΓT≔∂Ω×0,T,yx,0=y0x,inΩ,where Ω is a convex polygon with piecewise smooth boundary ∂Ω. Here K is the admissible set defined by(3)K=q∈L2ΩT:a≤qx,t≤ba.e.  inΩTwitha,b∈R,a≤b.α and f are given functions, ε>0 is a constant, and β is a given vector valued function, which satisfies α-1/2∇·β≥γ~>0 with a constant γ~>0.To define a mixed discontinuous Galerkin scheme for (2), we introduce a new variable:(4)q=-ε1/2∇y.Then the optimal control problem (2) can be rewritten as(5)minu∈KJy,uyt+∇·ε1/2q-∇·βy+αy=f+u,inΩT,q=-ε1/2∇y,inΩT,y=0,onΓT.Let Th be a regular triangulation of Ω with element τ and Ω¯=⋃τ∈Thτ¯. Let h=maxτ∈Thhτ, where hτ denotes the diameter of the element τ. We define the following spaces:(6)V=v∈L2Ω2:∇·v∈L2Ω,W=w∈L2Ω:wτ∈H1τ,∀τ∈Th,U=L2Ω.For simplicity, we set(7)Aq,v=q,v,q,v∈V,Bq,w=div⁡ε1/2q,w,q∈V,w∈W,Dy,w=∑τ∈Th∫τyβ·∇w-∫∂τy+wn·βds+αy,w,y,w∈W,where(8)∂τ-=l∈∂τ,n·βl<0,n is the outward norm direction on ∂τ,(9)w=w+-w-,w+x=limt→0+⁡wx+tβ,w-x=limt→0-⁡wx+tβ,and [w]=w+ on ∂τ when ∂τ-⊂∂Ω. Then the weak formulation for the optimal control problem (5) reads as follows: finding (q,y,u)⊂V×W×K such that(10)minu∈KJy,u(11)subjecttoyt,w+Bq,w+Dy,w=f+u,w,∀w∈WAq,v-Bv,y=0,∀v∈V,yx,0=y0x,∀x∈Ω.Standard arguments techniques imply that optimal control problem (10)-(11) admits a unique solution and the following first-order optimality condition holds:(12)yt,w+Bq,w+Dy,w=f+u,w,∀w∈W,Aq,v-Bv,y=0,∀v∈V,-zt,ψ-Bp,ψ+Dψ,z=y-yd,ψ,∀ψ∈W,Ap,φ+Bφ,z=0,∀φ∈V,∫0Tγu+z,v~-udt≥0,∀v~∈K,yx,0=y0x,zx,T=0,∀x∈Ω.LetVh×Wh⊂V×W denote the Raviart-Thomas element space of the lowest order associated with a triangular or rectangular mesh Th of Ω (see [16] for details). Then the approximation scheme of (10)-(11) is as follows: finding (qh,yh,uh)∈Vh×Wh×K such that(13)minuh∈KJy,usubject  toyh,t,wh+Bqh,wh+Dyh,wh=f+uh,wh,∀wh∈Wh,Aqh,vh-Bvh,yh=0,∀vh∈Vh,yhx,0=y~0,∀x∈Ω.Here y~0 is an approximation of the initial value y0(x). Variational discretization approach is used for the control u.Similar to continuous case we derive the following semidiscrete optimality conditions:(14)yh,t,wh+Bqh,wh+Dyh,wh=f+uh,wh,∀wh∈Wh,Aqh,vh-Bvh,yh=0,∀vh∈Vh,-zh,t,ψh-Bph,ψh+Dψh,zh=yh-yd,ψh,∀ψh∈Wh,Aph,φh+Bφh,zh=0,∀φh∈Vh,∫0Tγuh+zh,v~h-uhdt≥0,∀v~h∈K,yhx,0=y~0,zhx,T=0,∀x∈Ω.To define a fully discrete scheme we introduce a time partition. Let0=t0<t1<⋯<tN-1<tN=T be a time grid with k=tn-tn-1,n=1,2,…,N. Let In=(tn-1,tn] be a half-open interval. We write ϕn≔ϕ(tn) for a smooth ϕ.We use backward Euler scheme for time discretization. Then a fully discrete scheme of (10)-(11) is characterized as follows: finding (qhn,yhn,uhn)∈Vh×Wh×K such that(15)minuhn∈KJh,kyhn,uhn≔12∑n=1Nkyhn-ydn0,Ω2+γuhn0,Ω2subject  toyhn-yhn-1k,wh+Bqhn,wh+Dyhn,wh=fn+uhn,wh,∀wh∈Wh,Aqhn,vh-Bvh,yhn=0,∀vh∈Vh,yh0=y~0,∀x∈Ω.To obtain the fully discrete first-order optimality condition we define a Lagrange functional as follows:(16)LYh,Qh,Zh,Ph,Uh≔k∑n=1Nfn+uhn-yhn-yhn-1k,zhn-1-Bqhn,zhn-1-Dyhn,zhn-1+Bphn-1,yhn-Aqhn,phn-1+Jh,kyhn,uhn.HereYh=(yh1,yh2,…,yhN)T, Qh=(qh1,qh2,…,qhN)T, Zh=(zh0,zh1,…,zhN-1)T, Ph=(ph0,ph1,…,phN-1)T, and Uh=(uh1,uh2,…,uhN)T. Then we derive the discrete first-order optimality condition:(17)yhn-yhn-1k,wh+Bqhn,wh+Dyhn,wh=fn+uhn,wh,∀wh∈Wh,Aqhn,vh-Bvh,yhn=0,∀vh∈Vh,zhn-1-zhnk,ψh-Bphn-1,ψh+Dψh,zhn-1=yhn-ydn,ψh,∀ψh∈Wh,Aphn-1,φh+Bφh,zhn-1=0,∀φh∈Vh,γuhn+zhn-1,v~-uhn≥0,∀v~∈K,yh0=y~0,zhN=0,∀x∈Ω.


## 3. Semidiscrete Error Estimate

The goal of this section is to prove the semidiscrete error estimates for the state, the adjoint state, and the control. We firstly decomposey-yh and z-zh as(18)y-yh=yu-yhu+yhu-yh,z-zh=zu-zhu+zhu-zh,where yh(u) and zh(u) satisfy the following auxiliary problems:(19)yh,tu,wh+Bqhu,wh+Dyhu,wh=f+u,wh,Aqhu,vh-Bvh,yhu=0,-zh,tu,ψh-Bphu,ψh+Dψh,zhu=yhu-yd,ψh,Aphu,φh+Bφh,zhu=0,yhux,0=y~0,zhux,T=0.For the following analysis we introduce a new norm:(20)y∗2=γ~y0,Ω2+12∑τ∈Th∫∂τ-y2n·βds+12∫∂Ω∖⋃∂τ-y-2n·βds.Following [8] we have D(y,y)≥y∗2.Let(y~,q~)∈Wh×Vh and (z~,p~)∈Wh×Vh be the projections defined by(21)Bq-q~,wh+Dy-y~,wh=0,∀wh∈Wh,Aq-q~,vh-Bvh,y-y~=0,∀vh∈Vh,(22)Dψh,z-z~-Bp-p~,ψh=0,∀ψh∈Wh,Ap-p~,φh+Bφh,z-z~=0,∀φh∈Vh,According to [8] we have the following.Lemma 1.
If(y~,q~) and (z~,p~) satisfy (21)-(22), respectively, then we have(23)q-q~0,Ω2+y-y~∗2≤Chq1/2,Ω2+y1,Ω2,p-p~0,Ω2+z-z~∗2≤Chp1/2,Ω2+z1,Ω2.Lemma 2.
If(y~,q~) and (z~,p~) satisfy (21)-(22), respectively, then the following estimates hold:(24)qt-q~t0,Ω2+yt-y~t∗2≤Chqt1/2,Ω2+yt1,Ω2,pt-p~t0,Ω2+zt-z~t∗2≤Chp1/2,Ω2+z1,Ω2.Proof.
Differentiating (21) about t, we have(25)Bqt-q~t,wh+Dyt-y~t,wh=0,∀wh∈Wh,Aqt-q~t,vh-Bvh,yt-y~t=0,∀vh∈Vh.Set Ph:W→Wh and Πh:V→Vh to be the interpolation operators for the standard RT element space (see, e.g., [17]) such that(26)ψ-Phψ,wh=0,∀wh∈Wh,div⁡ϕ-Πhϕ,wh=0,∀wh∈Wh.Then we have the following approximation properties:(27)ψ-Phψs,Ω≤Ch1-sψ1,Ω,s=0,1,∀ψ∈H01Ω,ϕ-Πhϕ0,Ω≤Chγϕγ,Ω,12≤γ<1,∀ϕ∈H0γΩ2.Let(28)q-q~=q-Πhq+Πhq-q~=ηq+ξq,y-y~=y-Phy+Phy-y~=ηy+ξy.Then we derive(29)Bηq,t,wh+Bξq,t,wh+Dηy,t,wh+Dξy,t,wh=0,∀wh∈Wh,Aηq,t,vh+Aξq,t,vh-Bvh,ηy,t-Bvh,ξy,t=0,∀vh∈Vh.By the definition of interpolation operators Ph and Πh we obtain(30)Bηq,t,wh=0,∀wh∈Wh,Bvh,ηy,t=0,∀vh∈Vh.Setting wh=ξy,t and vh=ξq,t leads to(31)Aξq,t,ξq,t+Dξy,t,ξy,t=Aηq,t,-ξq,t+Dηy,t,-ξy,t,namely,(32)ξq,t0,Ω2+ξy,t∗2≤Aηq,t,-ξq,t+Dηy,t,-ξy,t.Since the function in Wh is piecewise constant, using the definition of D(·,·), we have for all w∈W and θ∈Wh(33)Dw,θ=∑τ∈Th∫τwβ·∇θ-∫∂τ-w+θn·βds+αw,θ=0-∑τ∈Th∫∂τ-w+θn·βds+αw,θ.Therefore, we get the estimate as follows:(34)ξq,t0,Ω2+ξy,t∗2≤Cδηq,t0,Ω2+ηy,t0,Ω2+Cδ∑τ∫∂τ-ηy,t+2n·βds+Cδξq,t0,Ω2+ξy,t0,Ω2+∑τ∫∂τ-ξy,t2n·βds.Note that(35)q-Πhq0,Ω≤Ch1/2q1/2,Ω,y-Phy0,Ω≤Chy1,Ω,and according to [18] we have (36)∑τ∈Th∫∂τ-ηy,t+2n·βds≤C∑τ∈Thηy,t0,∂τ2≤C∑τ∈Thηy,t0,τηy,t1,τ≤Chyt1,Ω2.Then we derive(37)ξq,t0,Ω2+ξy,t∗2≤Cδhqt1/2,Ω2+yt1,Ω2+Cδξq,t0,Ω2+ξy,t∗2.By setting δ small enough we obtain(38)ξq,t0,Ω2+ξy,t∗21/2≤Ch1/2qt1/2,Ω+yt1,Ω.Using triangle inequality we have(39)qt-q~t0,Ω2+yt-y~t∗2≤ηq,t0,Ω2+ηy,t∗2+ξq,t0,Ω2+ξy,t∗2,which implies the first theorem result. Similar to the above proof we have(40)pt-p~t0,Ω2+zt-z~t∗2≤Chpt1/2,Ω2+zt1,Ω2.Using the above estimates we can derive the following.Theorem 3.
Let(y,q,z,p) and (yh(u),qh(u),zh(u),ph(u)) be the solutions of (12) and (19), respectively. Then we have(41)y-yhu,q-qhu+≤Ch1/2.Here(42)yh,qh+2=yh0,Ω2+∫0Tyh∗2dt+∫0Tqh0,Ω2dtand the constant C depends on the norms of y and q.Proof.
Using (12) and (19) we derive the following error equations:(43)yt-yh,tu,wh+Bq-qhu,wh+Dy-yhu,wh=0,∀wh∈Wh,Aq-qhu,vh-Bvh,y-yhu=0,∀vh∈Vh.We split the errors y-yh(u) and q-qh(u) into(44)y-yhu=y-y~+y~-yhu,q-qhu=q-q~+q~-qhu,where y~,q~ is defined in (21). Then we have(45)yt-y~t,wh+y~t-yh,tu,wh+Bq~-qhu,wh+Dy~-yhu,wh+Aq~-qhu,vh-Bvh,y~-yhu=0.Setting χ=y~-yhu and σ=q~-qhu gives(46)χt,wh+Bσ,wh+Dχ,wh+Aσ,vh-Bvh,χ=-yt-y~t,wh.Choosing wh=χ and vh=σ in the above equation leads to(47)χt,χ+Dχ,χ+Aσ,σ=-yt-y~t,χ.Since D(χ,χ)≥χ∗2, then we deduce by Young inequality that(48)12ddtχ0,Ω2+χ∗2+σ0,Ω2≤δχ0,Ω2+Cδyt-y~t0,Ω2.By setting δ small enough we derive(49)12ddtχ0,Ω2+12χ∗2+σ0,Ω2≤Cyt-y~t0,Ω2.Note that χ0=0. Integrating (49) from 0→t leads to(50)χt0,Ω2+∫0tχ∗2+∫0tσ0,Ω2≤C∫0tyt-y~t0,Ω2.Similarly, integrating (49) from 0→T yields(51)χT0,Ω2+∫0Tχ∗2+∫0Tσ0,Ω2≤C∫0Tyt-y~t0,Ω2,which implies(52)∫0Tχ∗2+∫0Tσ0,Ω2≤C∫0Tyt-y~t0,Ω2.Combining (50) and (52) we deduce that(53)χt0,Ω2+∫0Tχ∗2+∫0Tσ0,Ω2≤C∫0Tyt-y~t0,Ω2.By Lemma 2 we further derive(54)χt0,Ω2+∫0Tχ∗2+∫0Tσ0,Ω2≤Ch∫0Tqt1/2,Ω2+∫0Tyt1,Ω2.Then by triangle inequality and Lemma 1 we obtain (55)y-yhu,q-qhu+2≤Chq1/2,Ω2+y1,Ω2+∫0Tq1/2,Ω2+∫0Ty1,Ω2+∫0Tqt1/2,Ω2+∫0Tyt1,Ω2.This implies the theorem result.Theorem 4.
Let(y,q,z,p) and (yh(u),qh(u),zh(u),ph(u)) be the solutions of (12) and (19), respectively. Then we have(56)z-zhu,p-phu+≤Ch1/2.Here the constant C depends on the norms of y,z and q,p.Proof.
Using (12) and (19) we derive the following error equations:(57)-zt-zh,tu,ψh-Bp-phu,ψh+Dψh,z-zhu=y-yhu,ψh,∀ψh∈WhAp-phu,φh+Bφh,z-zhu=0,∀φh∈Vh.We rewrite the errors z-zh(u) and p-ph(u) as(58)z-zhu=z-z~+z~-zhu,p-phu=p-p~+p~-phu,where z~,p~ is defined in (22). Then we have(59)-zt-z~t,ψh-z~t-zh,tu,ψh-Bp~-phu,ψh+Dψh,z~-zhu+Ap~-phu,φh+Bφh,z~-zhu=y-yhu,ψh.Setting ξ=z~-zh(u) and ρ=p~-ph(u) we have(60)-ξt,ψh-Bρ,ψh+Dψh,ξ+Aρ,φh+Bφh,ξ=zt-z~t,ψh+y-yhu,ψh.Taking ψh=ξ and φh=ρ yields(61)-ξt,ξ+Dξ,ξ+Aρ,ρ=zt-z~t,ξ+y-yhu,ξ.Since Dξ,ξ≥ξ∗2, then(62)-12ddtξt0,Ω2+ξ∗2+ρ0,Ω2≤zt-z~t,ξ+y-yhu,ξ.Using Ho¨lder inequality and Young inequality we derive(63)-12ddtξt0,Ω2+ξ∗2+ρ0,Ω2≤δξ0,Ω2+Cδzt-z~t0,Ω2+Cδy-yhu0,Ω2.Choosing δ small enough we obtain(64)-12ddtξt0,Ω2+12ξ∗2+ρ0,Ω2≤Cδzt-z~t0,Ω2+Cδy-yhu0,Ω2.Integrating (64) with respect to t from t→T yields(65)ξt0,Ω2+∫tTξ∗2+∫tTρ0,Ω2≤C∫0Tzt-z~t0,Ω2+C∫0Ty-yhu0,Ω2,where ξT=0 was used. In an analogue way, integrating (64) from 0→T results in(66)12ξ00,Ω2+∫0Tξ∗2+∫0Tρ0,Ω2≤C∫0Tzt-z~t0,Ω2+C∫0Ty-yhu0,Ω2.We further have(67)∫0Tξ∗2+∫0Tρ0,Ω2≤C∫0Tzt-z~t0,Ω2+C∫0Ty-yhu0,Ω2.Combining (65) and (67) we deduce that(68)ξt0,Ω2+∫0Tξ∗2+∫0Tρ0,Ω2≤C∫0Tzt-z~t0,Ω2+C∫0Ty-yhu0,Ω2.Then we can derive the theorem result by using Lemmas 1 and 2 and Theorem 8 and triangle inequality.Lemma 5.
Let(yh,qh,zh,ph) and (yh(u),qh(u),zh(u),ph(u)) be the solutions of (14) and (19), respectively. Then we have(69)yh-yhu,qh-qhu+2≤C∫0Tu-uh0,Ω2,zh-zhu,ph-phu+2≤C∫0Tu-uh0,Ω2.Proof.
Using (19) along with (14) leads to(70)yh,t-yh,tu,wh+Bqh-qhu,wh+Dyh-yhu,wh=uh-u,wh,∀wh∈Wh,Aqh-qhu,vh-Bvh,yh-yhu=0,∀vh∈Vh.Setting Yh=yh-yh(u) and Qh=qh-qh(u) and choosing wh=Yh and vh=Qh lead to(71)Yh,t,Yh+AQh,Qh+DYh,Yh=uh-u,Yh.By Young inequality we have(72)12ddtYh0,Ω2+Qh0,Ω2+Yh∗2≤Cδu-uh0,Ω2+CδYh0,Ω2.Similar to the proof of Lemmas 1 and 2 we have(73)Yh0,Ω2+∫0TQh0,Ω2+∫0TYh∗2≤C∫0Tu-uh0,Ω2.By (17) and (19) we derive(74)-zh,t-zh,tu,ψh-Bph-phu,ψh+Dψh,zh-zhu=Yh,ψh,∀ψh∈Wh,Aph-phu,φh-Bφh,zh-zhu=0,∀φh∈Vh.Setting Zh=zh-zh(u) and Ph=ph-ph(u) and choosing ψh=Zh and φh=Ph yield(75)-Zh,t,Zh+APh,Ph+DZh,Zh=Yh,Zh.Furthermore, we derive by Young inequality(76)-12Zh0,Ω2+Ph0,Ω2+Zh∗2≤CδYh0,Ω2+CδZh0,Ω2.Similar to the proof of Lemmas 1 and 2 we have(77)Zh0,Ω2+∫0TPh0,Ω2+∫0TZh∗2≤C∫0TYh0,Ω2≤C∫0Tu-uh0,Ω2.Then the theorem result follows from (73) and (77).Lemma 6.
Assume that(y,q,z,p,u) and (yh,qh,zh,ph,uh) be the solutions of (11) and (14), respectively. Let(78)Jh′uv-u=∫0Tγu+zhu,v-u, where zh(u) is the solution of (19). Then the following estimate holds:(79)Jh′vv-u-Jh′uv-u≥γv-uL20,T;L2Ω2.Proof.
By the definitions of(yh(u),qh(u)), (zh(u),ph(u)) we have(80)yh,tv-yh,tu,wh+Bqhv-qhu,wh+Dyhv-yhu,wh=v-u,wh,∀wh∈Wh,Aqhv-qhu,vh-Bvh,yhv-yhu=0,∀vh∈Vh,zh,tu-zh,tv,ψh-Bphv-phu,ψh+Dψh,zhv-zhu=yhv-yhu,ψh,Aphv-phu,φh+Bφh,zhv-zhu=0.Note that(81)Jh′vv-u-Jh′uv-u=∫0Tγv+zhv,v-u-∫0Tγv+zhu,v-u=∫0Tγv-u,v-u+∫0Tzhv-zhu,v-u=γv-uL20,T;L2Ω2+∫0Tzhv-zhu,v-u.By setting wh=zh(v)-zh(u), vh=ph(v)-ph(u) and ψh=yh(v)-yh(u), φh=qh(v)-qh(u) in the above equations, respectively, we can prove(82)∫0Tzhv-zhu,v-u=∫0Tv-u,zhv-zhu=∫0TDyhv-yhu,zhv-zhu+Bqhv-qhu,zhv-zhu+yh,tv-yh,tu,zhv-zhu+Aqhv-qhu,phv-phu-Bphv-phu,yhv-yhu.Since (yh(v)-yh(u))(0)=0, (zh(v)-zh(u))(T)=0, and(83)∫0Tyh,tv-yh,tu,zhv-zhu=-∫0Tyhv-yhu,zh,tv-zh,tu+yhv-yhu,zhv-zhu0T=0we can prove that(84)∫0Tzhv-zhu,v-u=∫0Tyhv-yhu,yhv-yhu≥0.Therefore, we derive(85)Jh′vv-u-Jh′uv-u≥γv-uL20,T;L2Ω2.Lemma 7.
Let(y,q,z,p,u) and (yh,qh,zh,ph,uh) be the solutions of (11) and (14), respectively. Then we have(86)u-uhL20,T,L2Ω≤Ch1/2.Proof.
By Lemma6 we have(87)γu-uhL20,T,L2Ω2≤Jh′u-Jh′uh,u-uh=∫0Tαu+zhu,u-uhdt-∫0Tαuh+zh,u-uhdt=∫0Tαu+z,u-uhdt+∫0Tzhu-z,u-uhdt+∫0Tαuh+zh,uh-udt≤0+∫0Tzhu-z,u-uhdt+0≤zhu-zL20,T,L2Ωu-uhL20,T,L2Ω.Using the result of Theorem 4 yields(88)u-uhL20,T,L2Ω≤Czhu-zL20,T,L2Ω≤Ch1/2.Theorem 8.
Let(y,q,z,p) and (yh(u),qh(u),zh(u),ph(u)) be the solutions of (12) and (14), respectively. Then we have(89)u-uhL20,T,L2Ω+y-yh,q-qh++z-zh,p-ph+≤Ch1/2.Proof.
By Lemma2 and (88) we derive(90)y-yh,q-qh+2≤y-yhu,q-qhu+2+yhu-yh,qhu-qh+2≤Chq1/2,Ω2+y1,Ω2+C∫0Tu-uh0,Ω2+Ch∫0Tq1/2,Ω2+∫0Ty1,Ω2+∫0T∂q∂t1/2,Ω2+∫0T∂y∂t1,Ω2.Combining Lemma 5 and (90) yields(91)z-zh,p-ph+2≤z-zhu,p-phu+2+zhu-zh,phu-ph02≤Chp1/2,Ω2+z1,Ω2+∫0Tp1/2,Ω2+∫0Tz1,Ω2+Ch∫0T∂p∂t1/2,Ω2+∫0T∂z∂t1,Ω2+∫0Tq1/2,Ω2+∫0Ty1,Ω2+C∫0Tu-uh0,Ω2.Combining (88)–(91) we can derive the theorem result.


## 4. Fully Discrete Error Estimate

In this section we will prove the error estimates for the fully discrete scheme. For this purpose we firstly introduce the following auxiliary problems:(92)yhnu-yhn-1uk,wh+Bqhnu,wh+Dyhnu,wh=fn+un,wh,∀wh∈Wh,Aqhnu,vh-Bvh,yhnu=0,∀vh∈Vh,zhn-1u-zhnuk,ψh-Bphn-1u,ψh+Dψh,zhn-1u=yhnu-ydn,ψh,∀ψh∈Wh,Aphn-1u,φh+Bφh,zhn-1u=0,∀φh∈Vh,yh0u=y~0,zhNu=0,∀x∈Ω.Lemma 9.
Let(y,q,z,p) and (yhn(u),qhn(u),zhn-1(u),phn-1(u)) be the solutions of (12) and (92), respectively. Then we have(93)y-yhu,q-qhu⋄≤Ch1/2+k,where(94)y,q⋄2=yn0,Ω2+k∑n=1Nyn∗2+k∑n=1Nqn0,Ω2.Proof.
Using (12) and (92) we derive the following error equations:(95)∂yn∂t-yhnu-yhn-1uk,wh+Bqn-qhnu,wh+Dyn-yhnu,wh=0,∀wh∈Wh,Aqn-qhnu,vh-Bvh,yn-yhnu=0,∀q∈Vh.Setting χ=y~-yh(u), ρ=y-y~, and σ=q~-qh(u) and combining the definition of y~ and q~ we have(96)χn-χn-1k,wh+Bσn,wh+Dχn,wh=-∂yn∂t-yn-yn-1k,wh-ρn-ρn-1k,wh,Aσn,vh-Bvh,χn=0.Testing (96) with wh=χn and vh=σn yields(97)χn-χn-1k,χn+Aσn,σn+Dχn,χn=-∂yn∂t-yn-yn-1k,χn-ρn-ρn-1k,χn=D1+D2.By Taylor expansion with integral reminder we deduce that(98)∂yn∂t-yn-yn-1k=1k∫tn-1tnt-tn-1∂2y∂t2dt,ρn-ρn-1=∫tn-1tn∂ρ∂tdt.By Ho¨lder inequality we obtain(99)∂yn∂t-yn-yn-1k0,Ω2≤k∫tn-1tn∂2y∂t20,Ω2dt,ρn-ρn-10,Ω2≤k∫tn-1tn∂ρ∂t0,Ω2dt.Then by Young inequality we derive(100)D1≤Cδk∫tn-1tn∂2y∂t20,Ω2dt+δχn0,Ω2,D2≤Cδ1k∫tn-1tn∂ρ∂t0,Ω2dt+δχn0,Ω2.Choosing δ to be small enough leads to(101)12kχn0,Ω2-χn-10,Ω2+σn0,Ω2+12χn∗2≤Ck∫tn-1tn∂2y∂t20,Ω2dt+C1k∫tn-1tn∂ρ∂t0,Ω2dt.Multiplying (114) by 2k and summing up with respect to n from 1 to m yield(102)χm0,Ω2+τ∑n=1mσn0,Ω2+τ∑n=1mχn∗2≤Ck2∫0tm∂2y∂t20,Ω2dt+C∫0tm∂ρ∂t0,Ω2dt.Multiplying (101) by 2k and summing up with respect to n from 1 to N lead to(103)χN0,Ω2+k∑n=1Nσn0,Ω2+k∑n=1Nχn∗2≤Ck2∫0T∂2y∂t20,Ω2dt+C∫0T∂ρ∂t0,Ω2dt.Collecting above estimates we have(104)χm0,Ω2+k∑n=1Nσn0,Ω2+k∑n=1Nχn∗2≤Ck2∫0T∂2y∂t20,Ω2dt+C∫0T∂ρ∂t0,Ω2dt.Using above estimates, Lemmas 1 and 2, and triangle inequality we deduce that(105)y-yhu,q-qhu⋄≤Ch1/2qL∞0,T;H1/2Ω+yL∞0,T;H1Ω+qtL20,T;H1/2Ω+ytL20,T;H1Ω+Ck∂2y∂t2L20,T;L2Ω.Lemma 10.
Let(y,q,z,p) and (yhn(u),qhn(u),zhn-1(u),phn-1(u)) be the solutions of (12) and (92), respectively. Then we have(106)z-zhu,p-phu⋄≤Ch1/2+k.Proof.
Using (12) and (92) we derive the following error equations:(107)-zhn-1u-zhnuk-∂zn-1∂t,ψh-Bpn-1-phn-1u,ψh+Dψh,zn-1-zhn-1u=-yhnu-yn-1,ψh+ydn-ydn-1,ψh,∀ψh∈Wh,Apn-1-phn-1u,vh+Bvh,zn-1-zhn-1u=0,∀vh∈Vh.Setting ξ=z~-zh(u), η=z-z~, and σ=p~-ph(u) and using the definitions of z~ and p~ give(108)ξn-1-ξnk,ψh+Bσn-1,ψh+Dψh,ξn-1=-yhnu-yn-1,ψh+ydn-ydn-1,ψh+∂zn-1∂t+zn-1-znk,ξn-1-ηn-1-ηnk,ξn-1,Aσn-1,vh-Bvh,ξn-1=0,∀vh∈Vh.Choosing wh=ξn-1 and vh=σn in (108) yields (109)ξn-1-ξnk,ξn-1+Aσn-1,σn-1+Dξn-1,ξn-1=-yhnu-yn-1,ξn-1+ydn-ydn-1,ξn-1+∂zn-1∂t+zn-1-znk,ξn-1-ηn-1-ηnk,ξn-1=∑i=14Ti.Using Ho¨lder inequality and Young inequality we obtain(110)T1≤Cδyhnu-yn0,Ω2+k∫tn-1tn∂y∂t0,Ω2dt+δξn-10,Ω2,T2≤Cδk∫tn-1tn∂yd∂t0,Ω2dt+δξn-10,Ω2,T3≤Cδk∫tn-1tn∂2z∂t20,Ω2dt+δξn-10,Ω2,T4≤Cδ1k∫tn-1tn∂η∂t0,Ω2dt+δξn-10,Ω2.Then by setting δ small enough we have(111)12kξn-10,Ω2-ξn0,Ω2+σn-10,Ω2+ξn-1∗2≤Cyhnu-yn0,Ω2+Ck∫tn-1tn∂y∂t0,Ω2dt+Ck∫tn-1tn∂yd∂t0,Ω2dt+Ck∫tn-1tn∂2z∂t20,Ω2dt+C1k∫tn-1tn∂η∂t0,Ω2dt.Multiplying (111) by 2k and summing up with respect to n from m to N yield(112)ξm-10,Ω2+k∑n=mNσn-10,Ω2+k∑n=mNξn-1∗2≤Ck∑n=mNyhnu-yn0,Ω2+Ck2∫tmT∂y∂t0,Ω2dt+Ck2∫tmT∂yd∂t0,Ω2dt+Ck2∫tmT∂2z∂t20,Ω2dt+C∫tmT∂η∂t0,Ω2dt.Multiplying (111) by 2k and summing up with respect to n from 1 to N give(113)ξ00,Ω2+k∑n=1Nσn-10,Ω2+k∑n=1Nξn-1∗2≤Ck∑n=1Nyhnu-yn0,Ω2+Ck2∫0T∂y∂t0,Ω2dt+Ck2∫0T∂yd∂t0,Ω2dt+Ck2∫0T∂2z∂t20,Ω2dt+C∫0T∂η∂t0,Ω2dt.Collecting (112) and (113) we obtain(114)ξn-10,Ω2+k∑n=1Nσn-10,Ω2+k∑n=1Nξn-1∗2≤Ck∑n=1Nyhnu-yn0,Ω2+Ck2∫0T∂y∂t0,Ω2dt+Ck2∫0T∂yd∂t0,Ω2dt+Ck2∫0T∂2z∂t20,Ω2dt+C∫0T∂η∂t0,Ω2dt.By Lemmas 1 and 2 and triangle inequality we derive(115)z-zhu,p-phu⋄≤Ch1/2qL∞0,T;H1/2Ω+yL∞0,T;H1Ω+qtL20,T;H1/2Ω+ytL20,T;H1Ω+Ch1/2pL∞0,T;H1/2Ω+zL∞0,T;H1Ω+ptL20,T;H1/2Ω+ztL20,T;H1Ω+Ck∂y∂tL20,T;L2Ω+∂yd∂tL20,T;L2Ω+∂2z∂t2L20,T;L2Ω.Lemma 11.
Let(yhn,qhn,zhn-1,phn-1) and (yhn(u),qhn(u),zhn-1(u),phn-1(u)) be the solutions of (12) and (92), respectively. Then we have(116)yh-yhu,qh-qhu⋄2≤Ck∑n=1Nuhn-un0,Ω2,zh-zhu,ph-phu⋄2≤Ck∑n=1Nuhn-un0,Ω2.Proof.
Using (92) along with (17) leads to(117)yhn-yhnu-yhn-1-yhn-1uk,wh+Bqhn-qhnu,wh+Dyhn-yhnu,wh=uhn-un,wh,Aqhn-qhnu,vh-Bvh,yhn-yhnu=0.Choosing wh=yhn-yhn(u) and vh=qhn-qhn(u) in the above equations and adding the resulting equations together yield(118)yhn-yhnu-yhn-1-yhn-1uk,yhn-yhnu+Bqhn-qhnu,yhn-yhnu+Dyhn-yhnu,yhn-yhnu+Aqhn-qhnu,qhn-qhnu-Bqhn-qhnu,yhn-yhnu=uhn-un,yhn-yhnu.Setting Yh=yh-yh(u) and Qh=qh-qh(u) in the above equation leads to(119)Yhn-Yhn-1k,Yhn+DYhn,Yhn+AQhn,Qhn=uhn-un,Yhn,and then(120)Yhn-Yhn-1k,Yhn+Yhn∗2+Qhn0,Ω2≤uhn-un,Yhn.Furthermore, we have(121)12kYhn0,Ω2-Yhn-10,Ω2+Yhn∗2+Qhn0,Ω2≤Cδuhn-un0,Ω2+δYhn0,Ω2.Similar to the proof of Lemmas 9 and 10 we have(122)Yhn0,Ω2+k∑n=1NYhn∗2+k∑n=1NQhn0,Ω2≤Ck∑n=1Nuhn-un0,Ω2,Similarly, by (17) and (92) we derive(123)zhn-1-zhn-1u-zhn-zhnuk,ψh-Bphn-1-phn-1u,ψh+Dψh,zhn-1-zhn-1u=yhn-1-yhn-1u,ψh,Aphn-1-phn-1u,φh+Bφh,zhn-1-zhn-1u=0.Taking ψh=zhn-1-zhn-1(u) and φh=phn-1-phn-1(u) in the above equation, setting Zhn-1=zhn-1-zhn-1(u) and Phn-1=phn-1-phn-1(u), and using similar argument to (124) give(124)Zhn-10,Ω2+k∑n=1NZhn-1∗2+k∑n=1NPhn-10,Ω2≤Ck∑n=1Nyhn-yhnu0,Ω2≤Ck∑n=1Nuhn-un0,Ω2.Then the theorem result follows from (122) and (124).Lemma 12.
Let(y,q,z,p,u) and (yhn,qhn,zhn-1,phn-1,uhn) be the solutions of (12) and (92), respectively. Let(125)J^h′uv-u=k∑n=1Nγun+zhn-1u,vn-un,where zhn-1u is the solution of (92). Then the following estimate holds:(126)J^h′v-J^h′u,v-u≥γv-uL20,T,L2Ω2.Here v-uL2(0,T;L2(Ω))2≔k∑n=1Nvn-un2.Proof.
Note that(127)J^h′v-J^h′u,v-u=k∑n=1Nγvn+zhn-1v-γun-zhn-1u,vn-un=k∑n=1Nγvn-γun,vn-un+k∑n=1Nzhn-1v-zhn-1u,vn-un=γv-uL20,T,L2Ω2+k∑n=1Nzhn-1v-zhn-1u,vn-un.Setting Yhn=yhn(v)-yhn(u), Qhn=qhn(v)-qhn(u), Zhn-1=zhn-1(v)-zhn-1(u), and Phn-1=phn-1(v)-phn-1(u), by the definition of yhn(u) and zhn-1(u) we have(128)Yhn-Yhn-1τ,wh+BQhn,wh+DYhn,wh=vn-un,wh,AQhn,vh-Bvh,Yhn=0,Zhn-1-Zhnτ,ψh-BPhn,ψh+Dψh,Zhn-1=Yhn,ψh,APhn-1,φh+Bφh,Zhn-1=0.Choosing wh=Zhn-1, vh=Phn-1 and ψh=Yhn, φh=Qhn in the above equations, respectively, we can easily prove that(129)k∑n=1Nzhn-1v-zhn-1u,vn-un≥0.Therefore, we derive(130)J^h′v-J^h′u,v-u≥γv-uL20,T,L2Ω2.Lemma 13.
Let(y,q,z,p,u) and (yhn,qhn,zhn-1,phn-1,uhn) be the solutions of (12) and (92), respectively. Then we have(131)u-uhL20,T,L2Ω≤Ch1/2+Ck∂z∂tL20,T,L2Ω.Proof.
Following Lemma12 we derive(132)γu-uhL20,T,L2Ω2≤J^h′u-J^h′uh,u-uh=k∑n=1Nγun+zhn-1u,un-uhnU-k∑n=1Nγun+zhn-1,un-uhn=k∑n=1Nγun+zn,un-uhnU+k∑n=1Nzhn-1u-zn,un-uhn+k∑n=1Nγuhn+zhn-1,uhn-un≤0+k∑n=1Nzhn-1un-zn,un-uhn+0=k∑n=1Nzhn-1un-zn-1,un-uhn+k∑n=1Nzn-1-zn,un-uhn.Using Young inequality we obtain(133)u-uhL20,T,L2Ω≤Ch1/2+Ck∂z∂tL20,T,L2Ω.Theorem 14.
Let(y,q,z,p,u) and (yhn,qhn,zhn-1,phn-1,uhn) be the solutions of (11) and (17), respectively. Then we have(134)u-uhL20,T,L2Ω+y-yh,q-qh⋄+z-zh,p-ph⋄≤Ch1/2+k.Proof.
By Lemma9 and using triangle inequality we derive(135)y-yh,q-qh⋄2≤y-yhu,q-qhu⋄2+yhu-yh,qhu-qh⋄2≤ChqL∞0,T;H1/2Ω2+yL∞0,T;H1Ω2+qtL20,T;H1/2Ω2+ytL20,T;H1Ω2+Ck2∂2y∂t2L20,T;L2Ω2+Cu-uhL20,T;L2Ω2.By Lemma 10 we deduce that(136)z-zh,p-ph⋄2≤z-zhu,p-phu⋄2+zhu-zh,phu-ph⋄2≤ChpL∞0,T;H1/2Ω2+zL∞0,T;H1Ω2+ptL20,T;H1/2Ω2+ztL20,T;H1Ω2+yL∞0,T;H1Ω2+Cu-uhL20,T;L2Ω2+Ck2ytL20,T;L2Ω2+∂yd∂tL20,T;L2Ω2+∂2z∂t2L20,T;L2Ω2.Combining Lemma 13, (135), and (136) we arrive at(137)u-uhL20,T,L2Ω+y-yh,q-qh⋄+z-zh,p-ph⋄≤Ch1/2qL∞0,T;H1/2Ω+yL∞0,T;H1Ω+qtL20,T;H1/2Ω+ytL20,T;H1Ω+Ch1/2pL∞0,T;H1/2Ω+zL∞0,T;H1Ω+ptL20,T;H1/2Ω+ztL20,T;H1Ω+CkytL20,T;L2Ω+∂yd∂tL20,T;L2Ω+∂2y∂t2L20,T;L2Ω+∂2z∂t2L20,T;L2Ω,which completes the theorem.



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*Source: 6901467-2017-02-21.xml*