# AnH1-Galerkin Expanded Mixed Finite Element Approximation of Second-Order Nonlinear Hyperbolic Equations

Authors: Zhaojie Zhou; Weiwei Wang; Huanzhen Chen
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We investigate anH1-Galerkin expanded mixed finite element approximation of nonlinear second-order hyperbolic equations, which model a wide variety of phenomena that involve wave motion or convective transport process. This method possesses some features such as approximating the unknown scalar, its gradient, and the flux function simultaneously,  the finite element space being free of LBB condition, and avoiding the difficulties arising from calculating the inverse of coefficient tensor. The existence and uniqueness of the
numerical solution are discussed. Optimal-order error estimates for this method are proved without introducing curl operator. A numerical example is also given to illustrate the theoretical findings.


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## Body

## 1. Introduction

The objective of this paper is to present and analyze anH1-Galerkin expanded mixed finite element method for the following second-order nonlinear hyperbolic equation:(1a)utt-∇·(A(u)∇u)=f,(x,t)∈Ω×J,(1b)u(x,t)=0,(x,t)∈∂Ω×J,(1c)u(x,0)=u0(x),x∈Ω,(1d)ut(x,0)=u1(x),x∈Ω,where Ω is a bounded convex polygonal domain in R2 with boundary ∂Ω and J=[0,T] with T<∞. u(x,t) denotes the sound pressure, f(x,t) is the external force, and A(u) is the coefficient, which is supposed to satisfy the following conditions.(A1)
There exist positive constantsαi,i=0,1,2, such that
(2)0<α0≤A(u)≤α1,|Au(u)|+|Auu(u)|≤α2.(A2)
A(u), Au(u), and Auu(u) are Lipschitz continuous with respect to u.The primary interests in engineering application for the mathematical model (1a)–(1d) are the sound pressure u, the gradient of sound pressure p, and the acceleration of sound transmission σ. Extensive research has been carried out on the numerical methods and corresponding numerical analysis for model (1a)–(1d), including finite difference methods, finite element methods, and mixed finite element methods. One can refer to [1–4] and the references cited herein.The standard finite difference or finite element methods solve the sound pressureu directly, then differentiate it to determine ∇u, and multiply the gradient of u by A(u) to determine the acceleration of sound transmission σ. Therefore, the resulting acceleration of sound transmission σ and the gradient of sound pressure ∇u are often inaccurate, which then reduces the accuracy of the prediction, as well as the accuracy of the adjoint vector σ. The mixed finite element method can approximate both u and σ simultaneously and yields an accurate σ. However, the mixed formulation has to face numerical difficulties arising in a low permeability zone because the inversion and the finite element spaces need to satisfy the LBB conditions.In order to overcome the above problems, we propose anH1-Galerkin expanded mixed finite element method for model (1a)–(1d) which can solve the sound pressure u, the gradient of sound pressure p, and the acceleration of sound transmission σ directly and avoid inverting A(u) explicitly. In this formulation the finite element spaces are free of LBB conditions as required by the standard mixed finite element methods. Another feature of the new procedure we have found so far is that it avoids the trouble which resulted from representation of the time derivatives for nonlinear problems and leads to optimal error estimates without introducing curl operator. We prove the equivalence of the problem (1a)–(1d), the H1-Galerkin expanded mixed variational formulation and the existence and uniqueness of the semidiscrete H1-Galerkin expanded mixed finite element procedure. By introducing some projection and interpolation operators as well as lemmas, optimal-order error estimates for this formulation are deduced. The theoretical findings are verified by one numerical example. In recent years, there exist lots of work in the literature on the development and analysis of H1-Galerkin mixed finite element method. One can refer to [5–8] for linear parabolic type equations, [9] for regularized long wave equation, and [10] for linear second-order hyperbolic equation.The rest of the paper is organized as follows. In Section2, we describe the H1-Galerkin expanded mixed finite element variational form and prove the equivalence between primal problem and the variational formulation. In Section 3, the H1-Galerkin expanded mixed finite element procedure is presented, and the existence and uniqueness of the solution are proved. In Section 4, we prove the main error estimates. A numerical example is given in Section 5 to illustrate the theoretical findings.Throughout this paper,C denotes a generic constant which does not depend on mesh parameter h. We use (·,·) and ∥·∥ to denote the inner product and the norm, respectively, in L2(Ω) or (L2(Ω))2. Also we will denote the norms in usual Sobolev spaces Wk,p(Ω) by ∥·∥k,p and the norms in Lq(0,t;Wk,p(Ω)) by ∥·∥Lq(0,t;Wk,p(Ω)) with p=2 being omitted.


## 2. AnH1-Galerkin Expanded Mixed Variational Formulation

In order to derive anH1-Galerkin expanded mixed variational formulation we split (1a)–(1d) into a first-order system by introducing p=∇u and σ=A(u)p:(3a)utt-∇·σ=f,(3b)p=∇u,(3c)σ=A(u)p,(3d)σ(x,0)=A(u0)∇u0(x),(3e)p(x,0)=∇u0(x),pt(x,0)=∇u1(x).Define the following spaces:
(4)H=H(div,Ω)={w∈(L2(Ω))d;∇·w∈L2(Ω)},V=H01(Ω)={v∈H1(Ω);v=0on∂Ω}.Multiplying (3a) by ∇·q for q∈H(div,Ω) and integrating Ω lead to the weak form (5a). Multiplying (3b) by ∇v for v∈H01(Ω) leads to the weak form (5b). Multiplying (3c) by w for w∈H(div,Ω) and integrating on Ω result in the weak form (5c). Then the H1-Galerkin expanded variational problem is to find (u,p,σ)∈H01(Ω)×H(div,Ω)×H(div,Ω) such that(5a)(ptt,q)+(∇·σ,∇·q)=-(f,∇·q),∀q∈H,(5b)(p,∇v)=(∇u,∇v),∀v∈V,(5c)(σ,w)=(A(u)p,w),∀w∈H,(5d)σ(x,0)=A(u0)∇u0(x),x∈Ω,(5e)p(x,0)=∇u0(x),pt(x,0)=∇u1(x),x∈Ω.In order to prove the equivalence of problem (3a)–(3e) and variational problem (5a)–(5e), we need the following lemmas (see [11] or Theorem 3.5, Chapter 1 of [12]).Lemma 1.
Forp∈H(div,Ω), there exist a ϕ∈H2(Ω)⋂‍H01(Ω) and a ψ∈H(div,Ω) satisfying ∇·ψ=0, such that p=∇ϕ+ψ.Lemma 2.
Forg∈L2(Ω), there exists a p∈(H1(Ω))d⊂H(div,Ω), such that ∇·p=g.With the help of these lemmas we can prove the following theorem.Theorem 3.
(
u
,
p
,
σ
)
∈
H
0
1
(
Ω
)
×
H
(
div
;
Ω
)
×
H
(
div
;
Ω
) is a solution to the system (3a)–(3e) if and only if it is a solution to the variational formulation (5a)–(5e).Proof.
The proof of the  “only if” part is pretty straightforward. It remains to prove the “if” part. We insertw=σ-A(u)p into (5c) to get (3c). By (3a)–(3e) and taking p=∇ϕ+ψ in (5b), we conclude
(6)(∇ϕ,∇v)=(∇u,∇v),∀v∈V.
Then we have
(7)p=∇u+ψ,
which, together with (5a) and (3c), yields the following equation:
(8)(utt,∇·q)-(∇·(A(u)∇u),∇·q)-(ψtt,q)-(∇·(A(u)ψ),∇·q)=(f,∇·q),∀q∈H.
Note that ∇·ψ=0. We obtain that ∇·ψt=0. Setting q=ψt in (8) we obtain (ψtt,ψt)=0, which implies
(9)(ψt(t),ψt(t))=(ψt(0),ψt(0)).
Then by
(10)pt(x,0)=∇ut(x,0)=∇u1(x,0),pt(x,0)=∇u1(x,0)+ψt(x,0)
we can conclude that
(11)ψt(x,t)≡0.
By (5c) we obtain
(12)σ(x,0)=A(u0)∇u(x,0)+A(u0)ψ(x,0).
Here we select the initial value σ(x,0) as in (5d) to have ψ(x,0)=0. Then, we derive ψ(x,t)=0 and (7) reduces to
(13)p=∇u.
Further we get
(14)(utt,∇·q)-(∇·(A(u)∇u),∇·q)=(f,∇·q),∀q∈H.
By lemma 2, there exists a F∈H(div,Ω) such that ∇·F=utt-f. Therefore, we have
(15)(∇·F,∇·q)-(∇·(A(u)∇u),∇·q)=0,
which implies
(16)F=A(u)∇u=σ.
That is,
(17)utt-f=∇·σ.
This completes the proof.


## 3.H1-Galerkin Expanded Mixed Finite Element Procedure

In this section we will present the numerical scheme for (5a)–(5e). Let 𝒯h be a quasiuniform partition of domain Ω; that is, Ω=⋃K∈𝒯hK with h=max{diam(K);K∈𝒯h}. Let Hh and Vh be the finite dimensional subspaces of H(div;Ω) and H01(Ω) defined by
(18)Hh={qh∈H(div;Ω);qh|K∈(Pk(K))d,∀K∈𝒯h},Vh={vh∈H01(Ω);vh|K∈Pm(K),∀K∈𝒯h},
where Pj(K) denotes the set of polynomials of degree at most j. Assume that Hh, and Vh satisfy the following approximation properties. For integers k≥0,m≥1,
(19)infqh∈Hh∥q-qh∥≤Chk+1∥q∥k+1,Ω,ddddddddq∈(Hk+1(Ω))d∩H,infqh∈Hh∥∇·(q-qh)∥≤Chk1∥q∥k1+1,Ω,dddddddddddq∈(Hk+1(Ω))d∩H,infvh∈Vh{∥v-vh∥+h∥v-vh∥1,Ω}≤Chm+1∥v∥m+1,Ω,dddddddddddddddddddddddv∈Hm+1(Ω)∩V.
Here k1=k+1 when Hh is one of the Raviart-Thomas elements or the Nedelec elements, and k1=k≥1, when Hh is one of the other classical mixed elements, such as Breezi-Douglas-Fortin-Marini elements and Breezi-Douglas-Marini elements.Then theH1-Galerkin expanded mixed finite element procedure for the system (3a)–(3e) is to find (uh,ph,σh)∈Vh×Hh×Hh such that(20a)(phtt,qh)+(∇·σh,∇·qh)=-(f,∇·qh),dddddddddddddddd.ddddddd∀qh∈Hh,(20b)(ph,∇vh)=(∇uh,∇vh),∀vh∈Vh,(20c)(σh,wh)=(A(uh)ph,wh),∀wh∈Hh,(20d)σh(x,0)=Πhσ(x,0),∀x∈Ω,(20e)ph(x,0)=Πhp(x,0),pht(x,0)=Πhpt(x,0),ddddddddddddddddddddddd.dddddddd∀x∈Ω,where Πh denotes the Raviart-Thomas projection. We next prove the existence and uniqueness of solutions of the scheme (20a)–(20e).Theorem 4.
There exists a unique solution(uh,ph,σh)∈Vh×Hh×Hh to the H1-Galerkin expanded mixed finite element procedure (20a)–(20e).Proof.
LetHh=span{ψi}i=1M and Vh=span{ϕi}i=1N; then σh∈Hh, ph∈Hh, and uh∈Vh have the following expressions:
(21)ph=∑i=1M‍piψi,σh=∑i=1M‍λiψi,uh=∑i=1N‍uiϕi.
Then the scheme (20a)–(20e) can be written in the following matrix form:(22a)APtt+BΛ=F,(22b)DU=CP,(22c)AΛ=G(U)P,(22d)P(0),Pt(0)aregiven,where
(23)A=((ψi,ψj))M×M,B=((∇·ψi,∇·ψj))M×M,C=((ψi,∇ϕj))N×M,D=((∇ϕi,∇ϕj))N×N,G(U)=((A(U)ψi,ψj))M×M,F=((-f,ψj))M×1,P=(p1,p2,…,pM)T,Λ=(λ1,λ2,…,λM)T,U=(u1,u2,…,uN)T.
Noting that A and D are positive definite. We can rewrite (22b) and (22c) as (24)U=D-1CP,Λ=A-1G(U)P.
Then the system (22a)–(22d) can be characterized as follows:(25a)APtt+BA-1G(D-1CP)P=F,(25b)P(0),Pt(0)aregiven. Recalling the assumptions on A(u), we can deduce that the coefficients of Ptt and P are all Lipschitz continuous with respect to P(t). By the standard theory for the initial-value problems of nonlinear ordinary differential equations, we can deduce that there exists a unique solution (uh,ph,σh)∈Vh×Hh×Hh to the H1-Galerkin expanded mixed finite element scheme (20a)–(20e).


## 4. Convergence Analysis

In this section we will prove the error estimates for theH1-Galerkin expanded mixed finite element discretization scheme. We begin by reviewing some preliminary knowledge that will be used in the following theoretical analysis.LetΠh:H→Hh be the Raviart-Thomas projection defined by
(26)(∇·(q-Πhq),∇·qh)=0,∀qh∈Hh.
The following error estimates [13–15] hold for Πh and 2≤p≤∞:
(27)∥q-Πhq∥p,Ω≤Chk+1∥q∥k+1,p,Ω,∥∇·(q-Πhq)∥p,Ω≤Chk1∥q∥k1+1,p,Ω.
Let Rh:V→Vh denote the elliptic projection defined by
(28)(∇(w-Rhw),∇vh)=0,∀vh∈Vh
which satisfies the following error estimates (see Theorems 3.2.2 and 3.2.5, Chapter 3 of [16]):
(29)∥w-Rhw∥+∥(w-Rhw)t∥+∥(w-Rhw)tt∥+∥(w-Rhw)ttt∥+h∥∇(w-Rhw)∥≤Chm+1,(30)max{∥Rhw∥0,∞,∥(Rhw)t∥0,∞,∥(Rhw)tt∥0,∞}≤C.To derive the main error estimates we also need the following lemma.Lemma 5.
Suppose thatξ,ζ∈Hh, θ∈H, and βtt∈Vh satisfy
(31)(ξtt+θtt,qh)+(∇·ζ,∇·qh)=0,∀qh∈Hh,(32)(ξtt+θtt,∇vh)=(∇βtt,∇vh),∀vh∈Vh.
Then there exists a constant C such that
(33)∥βtt∥≤C(∥∇·ζ∥+h∥θtt∥).Proof.
Assume thatϕ∈H2(Ω) is the solution of the following equation with ψ∈L2(Ω):
(34)-Δϕ=ψ,x∈Ω,ϕ=0,x∈∂Ω.
Recalling that Ω is convex, we have
(35)∥ϕ∥2,Ω≤C∥ψ∥.
Then by (31) and (32) we deduce
(36)(βtt,ψ)=-(βtt,Δϕ)=(∇βtt,∇ϕ)=(∇βtt,∇ϕ-∇Rhϕ)+(∇βtt,∇Rhϕ)=(∇βtt,∇ϕ-∇Rhϕ)-(ξtt+θtt,∇ϕ-∇Rhϕ)+(ξtt+θtt,∇ϕ)=T1+T2+T3.
Using the estimate of Rh we have
(37)T1≤Ch∥∇βtt∥∥ϕ∥2,Ω≤Ch∥∇βtt∥∥ψ∥,T2≤Ch∥ξtt+θtt∥∥ϕ∥2,Ω≤Ch∥ξtt+θtt∥∥ψ∥.
By (31), we obtain
(38)T3=(ξtt+θtt,∇ϕ)T3=(ξtt+θtt,∇ϕ-Πh(∇ϕ))+(ξtt+θtt,Πh(∇ϕ))T3=(ξtt+θtt,∇ϕ-Πh(∇ϕ))-(∇·ζ,∇·Πh(∇ϕ))T3=(ξtt+θtt,∇ϕ-Πh(∇ϕ))-(∇·ζ,∇·(∇ϕ))T3≤C(h∥ξtt+θtt∥+∥∇·ζ∥)∥ψ∥.
Inserting the estimates of T1, T2, and T3 into (36) leads to
(39)|(βtt,ψ)|≤C{h(∥ξtt+θtt∥+∥∇βtt∥)+∥∇·ζ∥}∥ψ∥.
By (32) we derive
(40)∥∇βtt∥≤∥ξtt+θtt∥.
Further we have
(41)|(βtt,ψ)|≤C{h∥ξtt+θtt∥+∥∇·ζ∥}∥ψ∥,
which implies
(42)∥βtt∥≤C(h∥ξtt∥+h∥θtt∥+∥∇·ζ∥).
Further, taking qh=ξtt in (31) and by Hölder inequalities as well as inverse property of the finite element spaces Vh and Hh yield
(43)∥ξtt∥2≤∥∇·ζ∥∥∇·ξtt∥+∥θtt∥∥ξtt∥≤C{h-1∥∇·ζ∥∥ξtt∥+∥θtt∥∥ξtt∥}.
Therefore we obtain
(44)∥ξtt∥≤C{h-1∥∇·ζ∥+∥θtt∥},
which, together with (42), yields the desired result.Theorem 6.
Let(u,p,σ) and (uh,ph,σh) be the solutions of (5a)–(5e) and (20a)–(20e), respectively. Assume that (u,p,σ) satisfies the following regularities:
(45)u∈H2(J;Hm+1(Ω))∩L∞(J;Hm+1(Ω)),p∈H3(J;Hk+1(Ω))∩L∞(J;Hk+1(Ω)),σ∈H2(J;Hk+1(Ω))∩L∞(J;Hk+1(Ω))
and uh(x,0)=Rhu0(x), uht(x,0)=Rhu1(x), and ph(x,0)=Πhp(x,0). Then there exists a positive constant C independent of h such that
(46)∥(u-uh)(t)∥1≤Chmin(k+1,m),∥∇·(σ-σh)(t)∥≤Chmin(k1,m+1),∥(u-uh)(t)∥+∥(p-ph)(t)∥+∥(σ-σh)(t)∥≤Chmin(k+1,m+1)
with k≥0, m≥1, for d=1, and k≥1, m≥1, for d=2,3.Proof.
In order to derive the error estimates, we decompose the errors as follows:(47)Πhp-ph=ξ,p-Πhp=θ,Πhσ-σh=ζ,σ-Πhσ=η,Rhu-uh=β,u-Rhu=γ.
Subtracting the numerical scheme (20a)–(20e) from the weak formulation (5a)–(5e), we can derive the following error equations:
(48)(ξtt,qh)+(∇·ζ,∇·qh)=-(θtt,qh),∀qh∈Hh,(ξ,∇vh)-(∇β,∇vh)=-(θ,∇vh),∀vh∈Vh,(ζ,wh)-(A(uh)ξ,wh)=((A(u)-A(uh))p-η,wh)(ζ,wh)-(A(uh)ξ,wh)+(A(uh)θ,wh),∀wh∈Hh.
Choosing vh=β in the second equation of (48) leads to
(49)∥∇β∥≤∥θ∥+∥ξ∥.
By setting wh=ζ in the third equation of (48) and using the assumption on A(u) we deduce
(50)∥ζ∥≤C(∥θ∥+∥ξ∥+∥η∥+∥γ∥).
In the following we will estimate ∥ξt∥. Differentiating the third equation in (48) gives
(51)(ζt,wh)-(A(uh)ξt,wh)=(Au(uh)uhtξ,wh)-(ηt,wh)+(A(uh)θt,wh)+(Au(uh)uhtθ,wh)+((A(u)-A(uh))pt,wh)-((Au(u)ut-Au(uh)uht)p,wh).
Taking qh=ζt in the first equation of (48) and wh=ξtt in (51) and then subtracting the resulting equations lead to
(52)(∇·ζ,∇·ζt)+(A(uh)ξt,ξtt)=-(θtt,ζt)+(ηt,ξtt)-(A(uh)θt,ξtt)-(Au(uh)uhtθ,ξtt)-(Au(uh)uhtξ,ξtt)-((A(u)-A(uh))pt,ξtt)-((Au(u)ut-Au(uh)uht)p,ξtt)=∑i=17‍Bi.
The left terms can be dealt with as follows:
(53)(∇·ζ,∇·ζt)=12ddt(∇·ζ,∇·ζ),(A(uh)ξt,ξtt)=12ddt(A(uh)ξt,ξt)-12(Au(uh)uhtξt,ξt).
The terms on the right side can be rewritten as follows by integral formula by parts:
(54)B1=-ddt(θtt,ζ)+(θttt,ζ),B2=ddt(ηt,ξt)-(ηtt,ξt),B3=-ddt(A(uh)θt,ξt)+(Au(uh)uhtθt,ξt)+(A(uh)θtt,ξt),B4=-ddt(Au(uh)uhtθ,ξt)+(Au(uh)uhtθt,ξt)+(Auu(uh)uht2θ,ξt)+(Au(uh)uhttθ,ξt),B5=-ddt(Au(uh)uhtξ,ξt)+(Au(uh)uhttξ,ξt)+(Auu(uh)uht2ξ,ξt)+(Au(uh)uhtξt,ξt),B6=-ddt((A(u)-A(uh))pt,ξt)+((Au(u)ut-Au(uh)uht)pt,ξt)+((A(u)-A(uh))ptt,ξt),B7=-ddt((Au(u)ut-Au(uh)uht)p,ξt)+((Au(u)ut-Au(uh)uht)pt,ξt)+((Auu(u)ut2+Au(u)utt-Auu(uh)uht2ddddddd-Au(uh)uhtt(Auu(u)ut2+Au(u)utt-Auu(uh)uht2)p,ξt).
Combining all the terms mentioned above we arrive at
(55)12ddt(∇·ζ,∇·ζ)+12ddt(A(uh)ξt,ξt)=12(Au(uh)uhtξt,ξt)+∑i=17‍Bi=B0+∑i=17‍Bi.
Now we are in the position to estimate the terms Bi,i=0,1,2,…,7. By Lemma 5 we can deduce
(56)∥βt∥2≤C∫0t‍∥βtt∥2dτ≤C∫0t‍(∥∇·ζ∥2+h2∥θtt∥2)dτ,
where βt(0)=0 was used. Notice that
(57)B0=12(Au(uh)(Rhu)tξt,ξt)-12(Au(uh)βtξt,ξt).
Then using the assumption A1, (30), and Cauchy-Schwartz inequality gives
(58)|∫0t‍B0dτ|≤C∫0t‍∥ξt∥2dτ+C∥ξt∥L∞(0,t;L∞(Ω))∫0t‍∥ξt∥∥βt∥≤C∥ξt∥L∞(0,t;L∞(Ω))∫0t‍(∥∇·ζ∥2+h2∥θtt∥2)dτ+C(1+∥ξt∥L∞(0,t;L∞(Ω)))∫0t‍∥ξt∥2dτ.
Integrating from 0 to t and using (50) as well as Cauchy-Schwartz inequality yield
(59)|∫0t‍B1dτ|≤∥θtt∥∥ζ∥+∫0t‍∥θttt∥∥ζ∥dτ|∫0t‍B1dτ|≤12(∥θtt∥2+∥ζ∥2)+12∫0t‍∥θttt∥2dτ|∫0t‍B1dτ|≤+12∫0t‍∥ζ∥2dτ|∫0t‍B1dτ|≤12(∥θtt∥2+∥θ∥2+∥ξ∥2+∥η∥2+∥γ∥2)|∫0t‍B1dτ|≤+12∫0t‍∥θttt∥2dτ|∫0t‍B1dτ|≤+12∫0t‍(∥θ∥2+∥ξ∥2+∥η∥2+∥γ∥2)dτ.
Note that
(60)∥ξ∥2≤C∫0t‍∥ξt∥2dτ
for ξ(0)=0. Then we have
(61)|∫0t‍B1dτ|≤12(∥θtt∥2+∥θ∥2+∥η∥2+∥γ∥2)+12∫0t‍∥θttt∥2dτ+12∫0t‍(∥θ∥2+∥η∥2+∥γ∥2)dτ+C∫0t‍∥ξt∥2dτ.
Similarly, we can estimate the other terms. By ϵ inequality we deduce
(62)|∫0t‍B2dτ|≤∥ηt∥∥ξt∥+∫0t‍∥ηtt∥∥ξt∥dτ|∫0t‍B2dτ|≤C∥ηt∥2+ϵ∥ξt∥2+12∫0t‍∥ηtt∥2dτ|∫0t‍B2dτ|≤+12∫0t‍∥ξt∥2dτ.
For B3 we can rewrite it as
(63)B3=-ddt(A(uh)θt,ξt)+(Au(uh)(uht-(Rhu)t)θt,ξt)+(Au(uh)(Rhu)tθt,ξt)+(A(uh)θtt,ξt)=-ddt(A(uh)θt,ξt)-(Au(uh)βtθt,ξt)+(Au(uh)(Rhu)tθt,ξt)+(A(uh)θtt,ξt).
Then by Cauchy-Schwartz inequality and ϵ inequality we derive
(64)|∫0t‍B3dτ|≤C∥θt∥∥ξt∥+C∥θt∥0,∞∫0t‍∥βt∥∥ξt∥dτ|∫0t‍B3dτ|≤+C∫0t‍∥θt∥∥ξt∥dτ+C∫0t‍∥θtt∥∥ξt∥dτ|∫0t‍B3dτ|≤C∥θt∥2+ϵ∥ξt∥2+C∫0t‍∥θt∥2dτ+C∫0t‍∥θtt∥2dτ|∫0t‍B3dτ|≤+C∫0t‍∥βt∥2dτ+C∫0t‍∥ξt∥2dτ|∫0t‍B3dτ|≤C∥θt∥2+ϵ∥ξt∥2+C∫0t‍∥θt∥2dτ+C∫0t‍∥θtt∥2dτ|∫0t‍B3dτ|≤+C∫0t‍(∥∇·ζ∥2+h2∥θtt∥2)dτ+C∫0t‍∥ξt∥2dτ.
Here the boundedness of ∥θt∥0,∞ and the Ritz projection ∥(Rhu)t∥0,∞ were used. For B4 we have
(65)B4=-ddt(Au(uh)((Rhu)t-βt)θ,ξt)B4=+(Au(uh)((Rhu)t-βt)θt,ξt)B4=+(Auu(uh)((Rhu)t-βt)2θ,ξt)B4=+(Au(uh)((Rhu)tt-βtt)θ,ξt).
Therefore by Cauchy-Schwartz inequality and ϵ inequality we obtain
(66)|∫0t‍B4dτ|≤C∥θ∥∥ξt∥+C∥θ∥0,∞∥βt∥∥ξt∥|∫0t‍B4dτ|≤+C∫0t‍(∥θt∥+∥θ∥)∥ξt∥+C∥θt∥L∞(0,t;L∞(Ω))|∫0t‍B4dτ|≤×∫0t‍∥βt∥∥ξt∥dτ+C∥θ∥L∞(0,t;L∞(Ω))|∫0t‍B4dτ|≤×∫0t‍(∥βt∥+∥βtt∥)∥ξt∥dτ|∫0t‍B4dτ|≤+C∥θ∥L∞(0,t;L∞(Ω))∥ξt∥L∞(0,t;L∞(Ω))∫0t‍∥βt∥2dτ|∫0t‍B4dτ|≤C(∥θ∥+∥βt∥)∥ξt∥+C∫0t‍(∥βt∥+∥θt∥)∥ξt∥dτ|∫0t‍B4dτ|≤+C∫0t‍(∥θ∥+∥βtt∥)∥ξt∥dτ+C∥ξt∥L∞(0,t;L∞(Ω))|∫0t‍B4dτ|≤×∫0t‍∥βt∥2dτ|∫0t‍B4dτ|≤C∥θ∥2+C(1+∥ξt∥L∞(0,t;L∞(Ω)))|∫0t‍B4dτ|≤×∫0t‍(∥∇·ζ∥2+h2∥θtt∥2)dτ|∫0t‍B4dτ|≤+C∫0t‍(∥θ∥2+∥θt∥2)dτ|∫0t‍B4dτ|≤+C∫0t‍∥ξt∥2dτ+ϵ∥ξt∥2.
Here we used the boundedness of ∥θ∥L∞(0,t;L∞(Ω)) to obtain the above estimate. Similarly, we can deduce
(67)|∫0t‍B5dτ|≤C∥ξ∥∥ξt∥+C∥ξt∥0,∞∥βt∥∥ξ∥+C∫0t‍(∥ξt∥+∥ξ∥)∥ξt∥+C∥ξt∥L∞(0,t;L∞(Ω))×∫0t‍∥βt∥∥ξt∥dτ+C∥ξ∥L∞(0,t;L∞(Ω))×∫0t‍(∥βt∥+∥βtt∥)∥ξ∥dτ+C∥ξ∥L∞(0,t;L∞(Ω))∥ξt∥L∞(0,t;L∞(Ω))∫0t‍∥βt∥2dτ≤ϵ∥ξt∥2+C∥ξt∥L∞(0,t;L∞(Ω))×∫0t‍(∥∇·ζ∥2+h2∥θtt∥2+∥ξt∥2)dτ+C∥ξt∥L∞(0,t;L∞(Ω))∥ξ∥L∞(0,t;L∞(Ω))×∫0t‍(∥∇·ζ∥2+h2∥θtt∥2)dτ+C∫0t‍∥ξt∥2dτ.
Further for B6 and B7 by Cauchy-Schwartz inequality and ϵ inequality we have
(68)|∫0t‍B6dτ|≤C(∥β∥+∥γ∥)∥ξt∥|∫0t‍B6dτ|≤+C∫0t‍(∥βt∥+∥γt∥+∥β∥+∥γ∥)∥ξt∥dτ|∫0t‍B6dτ|≤+∫0t(∥β∥+∥γ∥)∥ξt∥dτ|∫0t‍B6dτ|≤C(∥β∥2+∥γ∥2)+ϵ∥ξt∥2|∫0t‍B6dτ|≤+C∫0t‍(∥βt∥2+∥γt∥2)dτ|∫0t‍B6dτ|≤+C∫0t‍(∥β∥2+∥γ∥2)dτ+C∫0t‍∥ξt∥2dτ|∫0t‍B6dτ|≤C(∥θ∥2+∥γ∥2)+ϵ∥ξt∥2|∫0t‍B6dτ|≤+C∫0t‍(∥∇·ζ∥2+h2∥θtt∥2+∥γt∥2)dτ|∫0t‍B6dτ|≤+C∫0t‍(∥θ∥2+∥γ∥2)dτ+C∫0t‍∥ξt∥2dτ,|∫0t‍B7dτ|≤C(∥βt∥2+∥γt∥2+∥β∥2+∥γ∥2)+ϵ∥ξt∥2|∫0t‍B7dτ|≤+C∫0t‍(∥βt∥2+∥γt∥2+∥β∥2+∥γ∥2)dτ|∫0t‍B7dτ|≤+C∫0t‍∥ξt∥2dτ+C∫0t‍(∥βtt∥2+∥γtt∥2)dτ|∫0t‍B7dτ|≤C(∥γt∥2+∥θ∥2+∥γ∥2)+C∫0t‍∥ξt∥2dτ+ϵ∥ξt∥2|∫0t‍B7dτ|≤+C∫0t‍(∥∇·ζ∥2+h2∥θtt∥2)dτ+C∫0t‍∥γtt∥2dτ.
Combining the above estimates leads to
(69)∥∇·ζ∥2+α∥ξt∥2≤C(∥θtt∥2+∥θt∥2+∥ηt∥2+∥θ∥2dddd.dd+∥η∥2+∥γ∥2+∥γt∥2)+C∫0t‍(∥θttt∥2+∥θt∥2+∥θ∥2+∥η∥2+∥ηtt∥2ddddd.ddds+∥θtt∥2+∥γ∥2+∥γt∥2+∥γtt∥2)dτ+C(1+∥ξt∥L∞(0,t;L∞(Ω)))×∫0t‍(∥∇·ζ∥2+h2∥θtt∥2+∥ξt∥2)dτ+C∥ξt∥L∞(0,t;L∞(Ω))∥ξ∥L∞(0,t;L∞(Ω))×∫0t‍(∥∇·ζ∥2+h2∥θtt∥2)dτ.
To prove the main result we need to make the following induction hypothesis: there exists a constant 0<h0<1 such that the following estimate holds for 0<h≤h0:
(70)max{∥ξ∥L∞(0,t;L∞(Ω)),∥ξt∥L∞(0,t;L∞(Ω))}<1,0≤t≤T.
Then by setting ϵ small enough and using Gronwall’s inequality we obtain the following estimate which holds for constant K>0:
(71)∥∇·ζ∥2+α∥ξt∥2≤K(∥θtt∥2+∥θt∥2+∥ηt∥2+∥θ∥2dddd5dD+∥η∥2+∥γ∥2+∥γt∥2)+K∫0t‍(∥θttt∥2+∥θt∥2+∥θ∥2+∥η∥2ddddssdddd+∥ηtt∥2+∥θtt∥2+∥γ∥2)dτddd.+K∫0t‍(h2∥θtt∥2+∥γt∥2+∥γtt∥2)dτ.
Further, using (27) and (29) gives
(72)∥∇·ζ∥2+∥ξt∥2≤K~h2min{k+1,m+1},
where constant K~>0 is independent of h. We are now in position to prove the inductive hypothesis (70) which holds on t∈J. Suppose that there exists a constant 0<h*≤h0 such that
(73)max{∥ξ*∥L∞(0,t;L∞(Ω)),∥ξt*∥L∞(0,t;L∞(Ω))}≥1,0≤t≤T.
Let
(74)t*=inf{t∈J∣max{∥ξ*∥L∞(0,t;L∞(Ω)),dddddddddd55dd∥ξt*∥L∞(0,t;L∞(Ω))}≥1}.
Then we know that
(75)max{∥ξ*∥L∞(0,t*;L∞(Ω)),∥ξt*∥L∞(0,t*;L∞(Ω))}=1,(76)max{∥ξ*∥L∞(0,t;L∞(Ω)),∥ξt*∥L∞(0,t;L∞(Ω))}≤1,0<t≤t*.
By the same arguments for (72) we can prove
(77)∥∇·ζ*∥2+∥ξt*∥2≤K~1h2min{k+1,m+1},0<t≤t*.
Moreover, we can also deduce
(78)∥ξ*∥2≤C∫0t‍∥ξt*∥2dτ∥ξ*∥2≤K~2h2min{k+1,m+1},0<t≤t*.
By inverse inequality of finite element spaces we can conclude
(79)∥ξt*∥L∞(0,t*;L∞(Ω))≤Ch-d/2∥ξt*∥≤Ch-d/2K~1hmin{k+1,m+1}≤CK~1hmin{k+1,m+1}-d/2,∥ξ*∥L∞(0,t*;L∞(Ω))≤Ch-d/2∥ξ*∥≤Ch-d/2K~2hmin{k+1,m+1}≤CK~2hmin{k+1,m+1}-d/2.
Choose h0 satisfying
(80)max{CK~1h0min{k+1,m+1}-d/2,CK~2h0min{k+1,m+1}-d/2}≤12,
which implies
(81)max{∥ξ*∥L∞(0,t*;L∞(Ω)),∥ξt*∥L∞(0,t*;L∞(Ω))}≤12.
This contradicts with (75). Therefore the induction hypothesis (70) holds.
ByPoincaré’s inequality and (49) we have
(82)∥β∥2≤C∥∇β∥2≤C(∥ξ∥2+∥θ∥2).
Combining (50), (60), (72), (82), the estimates of projections (27), (29), and triangle inequality leads to the desired theorem result.


## 5. Numerical Examples

The goal of this section is to carry out two numerical experiments to illustrate our theoretical findings. We consider the following second-order nonlinear hyperbolic problem:(83)utt-∇·(A(u)∇u)=f,(x,t)∈Ω×[0,1],u(x,t)=0,(x,t)∈∂Ω×[0,1],u(x,0)=0,ut(x,0)=0,x∈Ω,
where Ω=[0,1]×[0,1].Example 7.
In this example the exact solution is chosen as(84)u(x,t)=sin3(t)sin(πx)sin(πy).
We setA(u)=u2+1. Inserting the above functions into the governing equation we can derive the corresponding right term f.In the first example, we investigate the order of convergence for theH1-Galerkin expanded mixed finite element method proposed in this paper. Piecewise linear polynomial is used to approximate the unknown function u, while the gradient function p and the flux function σ are approximated by the vector function space of the lowest Raviart-Thomas spaces, respectively. For time discretization we adopt backward Euler method. Here we couple the time step with spatial mesh as h=Δt.The errors ofu-uh, p-ph, and σ-σh in L2 norm at different times and the order of convergence for u, p, and σ are presented in Tables 1, 2, and 3, respectively. We can observe that the order of convergence for u approaches 2, and those for p and σ approach 1, which are in agreement with our theoretical results proposed in the previous section.Table 1
The errors of∥u-uh∥ at different times.
Time
t
=
0.2
t
=
0.4
t
=
0.8
t
=
1.0
h
=
Δ
t
Error
Order
Error
Order
Error
Order
Error
Order
1
/
10
0.001
∖
0.0025
∖
0.0075
∖
0.0104
∖
1
/
20
2.6553
e
-
4
1.9131
6.3152
e
-
4
1.9850
0.0019
1.9809
0.0026
2.0000
1
/
30
1.1843
e
-
4
1.9913
2.8183
e
-
4
1.9899
8.4165
e
-
4
2.0082
0.0012
1.9069
1
/
40
6.6744
e
-
5
1.9934
1.5903
e
-
4
1.9890
4.7373
e
-
4
1.9978
6.6038
e
-
4
2.0761
1
/
50
4.2788
e
-
5
1.9925
1.0212
e
-
4
1.9850
3.0325
e
-
4
1.9991
4.2104
e
-
4
2.0170Table 2
The errors of∥p-ph∥ at different times.
Time
t
=
0.2
t
=
0.4
t
=
0.8
t
=
1.0
h
=
Δ
t
Error
Order
Error
Order
Error
Order
Error
Order
1
/
10
0.0045
∖
0.0137
∖
0.0745
∖
0.1203
∖
1
/
20
0.0013
1.7914
0.0062
1.1438
0.0372
1.0019
0.0601
1.0012
1
/
30
7.0149
e
-
4
1.5215
0.0040
1.0809
0.0248
1.0000
0.0400
1.0041
1
/
40
4.7332
e
-
4
1.3676
0.0030
1.0000
0.0186
1.0000
0.0300
1.0000
1
/
50
3.5745
e
-
4
1.2583
0.0024
1.0000
0.0149
0.9940
0.0240
1.0000Table 3
The errors of∥σ-σh∥ at different times.
Time
t
=
0.2
t
=
0.4
t
=
0.8
t
=
1.0
h
=
Δ
t
Error
Order
Error
Order
Error
Order
Error
Order
1
/
10
0.0045
∖
0.0138
∖
0.0769
∖
0.1306
∖
1
/
20
0.0013
1.7914
0.0062
1.1543
0.0383
1.0056
0.0653
1.0000
1
/
30
7.0242
e
-
4
1.5182
0.0040
1.0809
0.0255
1.0032
0.0435
1.0019
1
/
40
4.7410
e
-
4
1.3665
0.0030
1.0000
0.0191
1.0045
0.0326
1.0027
1
/
50
3.5811
e
-
4
1.2574
0.0024
1.0000
0.0153
0.9941
0.0261
0.9966The figures of the exact solutionsu, p and the numerical solutions uh, ph at t=1.0 are shown in Figures 1 and 2, respectively. We can see that the numerical solutions are accurate and without oscillation compared with the exact solutions.The figures of exact solutionu and numerical solution uh at t=0.5,1.0 ((a), (b) for u and uh at t=1.0 and (c), (d) for u and uh at t=0.5).
(a)
(b)
(c)
(d)The figures of exact solutionp=(P1,P2) and numerical solution ph=(Ph1,Ph2) at t=1.0 ((a), (b) for p and (c), (d) for ph).
(a)
(b)
(c)
(d)Example 8.
In this example we consider problem (83) with prescribed data f=3sin(2x)sin(2πy)e3t and A(u)=u.The profiles of the numerical solutions foru and p are shown in Figure 3, respectively. From these figures we can see that our method works well for this kind of problems.The figures ofuh and ph ((a) for uh, and (b), (c) for ph).
(a)
(b)
(c)



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