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Decoupled finite element methods for a fourth-order exterior differential equation

Xuewei Cui School of Mathematics, Shanghai University of Finance and Economics, Shanghai 200433, China [email protected]  and  Xuehai Huang School of Mathematics, Shanghai University of Finance and Economics, Shanghai 200433, China [email protected]
Abstract.

This paper focuses on decoupled finite element methods for the fourth-order exterior differential equation. Based on differential complexes and the Helmholtz decomposition, the fourth-order exterior differential equation is decomposed into two second-order exterior differential equations and one generalized Stokes equation. A family of conforming finite element methods are developed for the decoupled formulation. Numerical results are provided for verifying the decoupled finite element methods of the biharmonic equation in three dimensions.

Key words and phrases:
fourth-order exterior differential equation, decoupled finite element method, Helmholtz decomposition, finite element exterior calculus, conforming finite element method
2010 Mathematics Subject Classification:
58J10; 65N30; 65N12; 65N22;
This work was supported by the National Natural Science Foundation of China Project 12171300.

1. Introduction

This paper focuses on exploring decoupled discrete methods for the fourth-order exterior differential equation on a bounded polytope Ωd\Omega\subset\mathbb{R}^{d} (d2d\geq 2): given fL2Λj(Ω)f\in L^{2}\Lambda^{j}(\Omega) with 0jd10\leq j\leq d-1, find uH0GdΛj(Ω)/dH0Λj1(Ω)u\in H_{0}^{\rm Gd}\Lambda^{j}(\Omega)/\,{\rm d}H_{0}\Lambda^{j-1}(\Omega) and λH0Λj1(Ω)/dH0Λj2(Ω)\lambda\in H_{0}\Lambda^{j-1}(\Omega)/\,{\rm d}H_{0}\Lambda^{j-2}(\Omega) satisfying

(1) δΔdu+dλ=finΩ,-\delta\Delta\,{\rm d}u+\,{\rm d}\lambda=f\quad\mathrm{in}\,\,\Omega,

where d\,{\rm d} is the exterior derivative, δ\delta is the codifferential operator, and

H0GdΛj(Ω)={ωH0Λj(Ω):dωH01Λj+1(Ω)}.H_{0}^{\rm Gd}\Lambda^{j}(\Omega)=\{\omega\in H_{0}\Lambda^{j}(\Omega):\,{\rm d}\omega\in H_{0}^{1}\Lambda^{j+1}(\Omega)\}.

The exterior differential equation (1) is related to the Hodge decomposition [4, 6, 7]

L2Λj(Ω)=(L2Λj(Ω)ker(δ))dH0Λj1(Ω),L^{2}\Lambda^{j}(\Omega)=(L^{2}\Lambda^{j}(\Omega)\cap\ker(\delta))\oplus\,{\rm d}H_{0}\Lambda^{j-1}(\Omega),

where ker(δ)\ker(\delta) means the kernel of the codifferential operator δ\delta. When δf=0\delta f=0, applying the operator δ\delta to equation (1) will induce λ=0\lambda=0.

Some examples of problem (1) are listed as follows.

  • For j=0j=0, problem (1) becomes the biharmonic equation

    (2) Δ2u=finΩ.\Delta^{2}u=f\quad\mathrm{in}\,\,\Omega.
  • For j=1j=1 and d=3d=3, problem (1) becomes the quad-curl problem

    (3) curlΔcurlu+λ=finΩ,-\operatorname{curl}\Delta\operatorname{curl}u+\nabla\lambda=f\quad\mathrm{in}\,\,\Omega,

    where curl\operatorname{curl} is the differential operator acting on 11-forms.

  • For j=d1j=d-1, problem (1) becomes the fourth-order div problem

    (4) Δdivu+curlλ=finΩ,\nabla\Delta\operatorname{div}u+\operatorname{curl}^{*}\lambda=f\quad\mathrm{in}\,\,\Omega,

    where curl\operatorname{curl}^{*}, the adjoint of curl\operatorname{curl}, is the differential operator acting on (d2)(d-2)-forms.

Since H0GdΛjH_{0}^{\rm Gd}\Lambda^{j} conforming finite elements typically require high-degree polynomials and supersmooth degrees of freedom [15, 14, 23], we will consider decoupled finite element methods for the fourth-order exterior differential equation (1) in this paper.

Applying the framework in [12] to the de Rham complex

HΛj+4𝛿HΛj+3𝛿L2Λj+2𝛿H1(δ,Λj+1)𝛿H1Λjker(δ)0H^{*}\Lambda^{j+4}\xrightarrow{\delta}H^{*}\Lambda^{j+3}\xrightarrow{\delta}L^{2}\Lambda^{j+2}\xrightarrow{\delta}H^{-1}(\delta,\Lambda^{j+1})\xrightarrow{\delta}H^{-1}\Lambda^{j}\cap\ker(\delta)\xrightarrow{}0

with H1(δ,Λj+1):={ωH1Λj+1:δωH1Λj}H^{-1}(\delta,\Lambda^{j+1}):=\{\omega\in H^{-1}\Lambda^{j+1}:\delta\omega\in H^{-1}\Lambda^{j}\}, we derive several Helmholtz decompositions

(H0Λj+1)=H1(δ,Λj+1)=dH0ΛjδL2Λj+2,(H_{0}\Lambda^{j+1})^{\prime}=H^{-1}(\delta,\Lambda^{j+1})=\,{\rm d}H_{0}\Lambda^{j}\oplus\delta L^{2}\Lambda^{j+2},
L2Λj+2=dH01Λj+1δHΛj+3=dH01Λj+1δH1Λj+3,L^{2}\Lambda^{j+2}=\,{\rm d}H_{0}^{1}\Lambda^{j+1}\oplus^{\perp}\delta H^{*}\Lambda^{j+3}=\,{\rm d}H_{0}^{1}\Lambda^{j+1}\oplus^{\perp}\delta H^{1}\Lambda^{j+3},

and decouple problem (1) into two second-order exterior differential equations and one generalized Stokes equation: find u,wH0Λju,w\in H_{0}\Lambda^{j}, λ,zH0Λj1/dH0Λj2\lambda,z\in H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}, ϕH01Λj+1\phi\in H_{0}^{1}\Lambda^{j+1}, pL2Λj+2p\in L^{2}\Lambda^{j+2} and rHΛj+3/δHΛj+4r\in H^{*}\Lambda^{j+3}/\delta H^{*}\Lambda^{j+4} such that

(5a) (dw,dv)+(v,dλ)\displaystyle(\,{\rm d}w,\,{\rm d}v)+(v,\,{\rm d}\lambda) =(f,v),\displaystyle=(f,v),
(5b) (w,dη)\displaystyle(w,\,{\rm d}\eta) =0,\displaystyle=0,
(5c) (ϕ,ψ)+(dψ+δs,p)\displaystyle(\nabla\phi,\nabla\psi)+(\,{\rm d}\psi+\delta s,p) =(dw,ψ),\displaystyle=(\,{\rm d}w,\psi),
(5d) (dϕ+δr,q)\displaystyle(\,{\rm d}\phi+\delta r,q) =0,\displaystyle=0,
(5e) (du,dχ)+(χ,dz)\displaystyle(\,{\rm d}u,\,{\rm d}\chi)+(\chi,\,{\rm d}z) =(ϕ,dχ),\displaystyle=(\phi,\,{\rm d}\chi),
(5f) (u,dμ)\displaystyle(u,\,{\rm d}\mu) =0,\displaystyle=0,
for any v,χH0Λjv,\chi\in H_{0}\Lambda^{j}, η,μH0Λj1/dH0Λj2\eta,\mu\in H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}, ψH01Λj+1\psi\in H_{0}^{1}\Lambda^{j+1}, qL2Λj+2q\in L^{2}\Lambda^{j+2} and sHΛj+3/δHΛj+4s\in H^{*}\Lambda^{j+3}/\delta H^{*}\Lambda^{j+4}.

The decoupled formulation (5) covers many decoupled formulations in literature: the decoupled formulation of the biharmonic equation in two and three dimensions in [25, 26, 12, 19] and the decoupled formulation of the quad-curl problem in three dimensions in [12, Section 3.4].

The decoupled formulation (5) is more amenable to designing finite element methods and fast solvers. Conforming finite element methods of the decoupled formulation (5) with j=0j=0 and d=2d=2, i.e. biharmonic equation in two dimensions, are shown in [12, Section 4.2]. A low-order nonconforming finite element discretization of the decoupled formulation (5) with j=1j=1 and d=3d=3, i.e. the quad-curl problem in three dimensions, is designed in [11]. The decoupled formulation of the biharmonic equation in two dimensions is employed to implement the H2H^{2}-conforming finite element methods using H1H^{1}-conforming finite elements in [1], and design fast solvers for the Morley element method in [25, 18, 28]. Based on nonconforming finite element Stokes complexes, nonconforming finite element methods of the quad-curl problem in three dimensions in [27, 29] are equivalent to nonconforming discretization of the decoupled formulation (5), which is also helpful for developing efficient solvers. We refer to [19, 33, 2] for different splitting formulations rather than the decoupled formulation (5).

Both (5a)-(5b) and (5e)-(5f) are second-order exterior differential equations, which can be discretized by finite element differential forms in [21, 22, 4, 6, 7]. We focus on designing finite element methods for the generalized Stokes equation (5c)-(5d). We use kΛj+1(T)+bTδkΛj+2(T)\mathbb{P}_{k}\Lambda^{j+1}(T)+b_{T}\,\delta\mathbb{P}_{k}\Lambda^{j+2}(T) as the shape function space to discretize ϕH01Λj+1\phi\in H_{0}^{1}\Lambda^{j+1}, trimmed finite element differential form Vk,hδ,Λj+2V_{k,h}^{\delta,-}\Lambda^{j+2} to discretize pL2Λj+2p\in L^{2}\Lambda^{j+2} and Vk,hδ,Λj+3V_{k,h}^{\delta,-}\Lambda^{j+3} to discretize rHΛj+3r\in H^{*}\Lambda^{j+3}. The resulting finite element method can be regarded as the generalization of the MINI element method for Stokes equation in [5, 8]. Error analysis is present for the decoupled finite element method. It’s worth mentioning that the error estimate for pph\|p-p_{h}\| is optimal when jd3j\leq d-3, while the error estimate of pph\|p-p_{h}\| for the MINI element method of Stokes equation is suboptimal [5].

The rest of this paper is organized as follows. Section 2 focuses on a decoupled formulation the fourth-order exterior differential equation. A family of decoupled conforming finite element methods are designed in Section 3. Finally, in Section 4, we conduct numerical experiments to validate the theoretical estimates we have developed.

2. A decoupled formulation

We will decouple the fourth-order exterior differential equation into two second-order exterior differential equations and one generalized Stokes equation in this section.

2.1. Notation

Let Ωd\Omega\subset\mathbb{R}^{d} (d2d\geq 2) be a bounded and contractible polytope. Given integer mm and a bounded domain DdD\subset\mathbb{R}^{d}, let Hm(D)H^{m}(D) be the standard Sobolev space of functions on DD. The corresponding norm and semi-norm are denoted by m,D\|\cdot\|_{m,D} and ||m,D|\cdot|_{m,D}, respectively. Set L2(D)=H0(D)L^{2}(D)=H^{0}(D) with the usual inner product (,)D(\cdot,\cdot)_{D}. We denote H0m(D)H_{0}^{m}(D) as the closure of C0(D)C_{0}^{\infty}(D) with respect to the norm m,D\|\cdot\|_{m,D}. In case DD is Ω\Omega, we abbreviate m,D\|\cdot\|_{m,D}, ||m,D|\cdot|_{m,D} and (,)D(\cdot,\cdot)_{D} as m\|\cdot\|_{m}, ||m|\cdot|_{m} and (,)(\cdot,\cdot), respectively. We also abbreviate 0,D\|\cdot\|_{0,D} and 0\|\cdot\|_{0} by D\|\cdot\|_{D} and \|\cdot\|, respectively. For integer k0k\geq 0, let k(D)\mathbb{P}_{k}(D) represent the space of all polynomials in DD with the total degree no more than kk. Set k(D)={0}\mathbb{P}_{k}(D)=\{0\} for k<0k<0. Let L02(D)L_{0}^{2}(D) be the space of functions in L2(D)L^{2}(D) with vanishing integral average values. For a space B(D)B(D) defined on DD, let B(D;d):=B(D)dB(D;\mathbb{R}^{d}):=B(D)\otimes\mathbb{R}^{d} be its vector version. Denote by hDh_{D} the diameter of DD. We use 𝒏D\boldsymbol{n}_{\partial D} to denote the unit outward normal vector of D\partial D, which will be abbreviated as 𝒏\boldsymbol{n} if not causing any confusion.

We mainly follow the notation set in [4, 6, 7]. For a dd-dimensional vector space VV and a nonnegative integer jj, define the space AltjV{\rm Alt}^{j}V as the space of all skew-symmetric jj-linear forms. For a multilinear jj-form ω\omega, its skew-symmetric part

(skwω)(v1,,vj)=1j!σ𝔖jsign(σ)ω(vσ(1),,vσ(j)),v1,,vjV(\operatorname{skw}\omega)(v_{1},\ldots,v_{j})=\frac{1}{j!}\sum_{\sigma\in\mathfrak{S}_{j}}\operatorname{sign}(\sigma)\omega(v_{\sigma(1)},\ldots,v_{\sigma(j)}),\quad v_{1},\ldots,v_{j}\in V

is an alternating form, where 𝔖j\mathfrak{S}_{j} is the symmetric group of all permutations of the set {1,,j}\{1,\ldots,j\}, and sign(σ)\operatorname{sign}(\sigma) denotes the signature of the permutation σ\sigma. The exterior product or wedge product of ωAltiV\omega\in{\rm Alt}^{i}V and ηAltjV\eta\in{\rm Alt}^{j}V is given by

ωη=(i+jj)skw(ωη)Alti+jV,\omega\wedge\eta={i+j\choose j}\operatorname{skw}(\omega\otimes\eta)\in{\rm Alt}^{i+j}V,

where \otimes is the tensor product. It satisfies the anticommutativity law

ωη=(1)ijηω,ωAltiV,ηAltjV.\omega\wedge\eta=(-1)^{ij}\eta\wedge\omega,\quad\omega\in{\rm Alt}^{i}V,\;\eta\in{\rm Alt}^{j}V.

An inner product on VV induces an inner product on AltjV{\rm Alt}^{j}V as follows

ω,η=σω(eσ(1),,eσ(j))η(eσ(1),,eσ(j)),ω,ηAltjV,\langle\omega,\eta\rangle=\sum_{\sigma}\omega(e_{\sigma(1)},\ldots,e_{\sigma(j)})\eta(e_{\sigma(1)},\ldots,e_{\sigma(j)}),\quad\omega,\eta\in{\rm Alt}^{j}V,

where the sum is over increasing sequences σ:{1,,j}{1,,d}\sigma:\{1,\ldots,j\}\to\{1,\ldots,d\} and {e1,,ed}\{e_{1},\ldots,e_{d}\} is any orthonormal basis. If the space VV is endowed with an orientation by assigning a positive orientation to some particular ordered basis, the volume form volAltdV\textsf{vol}\in{\rm Alt}^{d}V is the unique dd-form characterized by vol(e1,,ed)=1\textsf{vol}(e_{1},\ldots,e_{d})=1 for any positively oriented ordered orthonormal basis e1,,ede_{1},\ldots,e_{d}. The Hodge star operator is an isometry of AltjV{\rm Alt}^{j}V onto AltdjV{\rm Alt}^{d-j}V given by

ωη=ω,μvol,ωAltjV,ηAltdjV.\omega\wedge\eta=\langle\star\omega,\mu\rangle\textsf{vol},\quad\omega\in{\rm Alt}^{j}V,\;\eta\in{\rm Alt}^{d-j}V.

We have

ω=(1)j(dj)ω,ωAltjV.\star\star\omega=(-1)^{j(d-j)}\omega,\quad\omega\in{\rm Alt}^{j}V.

A differential jj-form ω\omega is a section of the jj-alternating bundle, i.e., a map which assigns to each xDx\in D an element ωxAltjTxD\omega_{x}\in{\rm Alt}^{j}T_{x}D, where TxDT_{x}D denotes the tangent space to DD at xx. For a space B(D)B(D) defined on DD, let BΛj(D)B\Lambda^{j}(D) be the space of all jj-forms whose coefficient function belongs to B(D)B(D). Notice that BΛ0(D)=B(D)B\Lambda^{0}(D)=B(D). The norm m,D\|\cdot\|_{m,D} and semi-norm ||m,D|\cdot|_{m,D} are extended to space HmΛj(D)H^{m}\Lambda^{j}(D), and inner product (,)D(\cdot,\cdot)_{D} to space L2Λj(D)L^{2}\Lambda^{j}(D). Define

HΛj(D)\displaystyle H\Lambda^{j}(D) :={ωL2Λj(D):dωL2Λj+1(D)},\displaystyle:=\{\omega\in L^{2}\Lambda^{j}(D):\,{\rm d}\omega\in L^{2}\Lambda^{j+1}(D)\},
HΛj(D)\displaystyle H^{*}\Lambda^{j}(D) :={ωL2Λj(D):δωL2Λj1(D)},\displaystyle:=\{\omega\in L^{2}\Lambda^{j}(D):\delta\omega\in L^{2}\Lambda^{j-1}(D)\},

where d\,{\rm d} is the exterior derivative, and the codifferential operator δ\delta is defined as

(6) δω=(1)jdω.\star\delta\omega=(-1)^{j}\,{\rm d}\star\omega.

Denote by H0Λj(D)H_{0}\Lambda^{j}(D) the subspace of HΛj(D)H\Lambda^{j}(D) with vanishing trace. We can identify H0mΛd(D)H_{0}^{m}\Lambda^{d}(D) and H0Λd1(D)H_{0}\Lambda^{d-1}(D) as H0m(D)L02(D)H_{0}^{m}(D)\cap L_{0}^{2}(D) and H0(div,D)H_{0}(\operatorname{div},D), respectively. The spaces L2Λj(D)L^{2}\Lambda^{j}(D) and HΛj(D)H^{*}\Lambda^{j}(D) are to be interpreted as the trivial space {0}\{0\} for jd+1j\geq d+1, whereas kΛd+1(D)\mathbb{P}_{k}\Lambda^{d+1}(D) is to be interpreted as \mathbb{R}. We will abbreviate the Sobolev space B(D)B(D) as BB when D=ΩD=\Omega if not causing any confusion.

For sufficient smooth (j1)(j-1)-form ω\omega and jj-form μ\mu, it holds the integration by parts

(7) (dω,μ)D=(ω,δμ)D+Dtrωtr(μ).(\,{\rm d}\omega,\mu)_{D}=(\omega,\delta\mu)_{D}+\int_{\partial D}\operatorname{tr}\omega\wedge\operatorname{tr}(\star\mu).

For (d1)(d-1)-dimensional face FF of D\partial D, we have

Ftrωtr(μ)=(1)(dj)(j1)(trω,Ftr(μ))F.\int_{F}\operatorname{tr}\omega\wedge\operatorname{tr}(\star\mu)=(-1)^{(d-j)(j-1)}(\operatorname{tr}\omega,\star_{F}\operatorname{tr}(\star\mu))_{F}.

Let κ\kappa be the Koszul operator mapping k1Λdj+1(D)\mathbb{P}_{k-1}\Lambda^{d-j+1}(D) to kΛdj(D)\mathbb{P}_{k}\Lambda^{d-j}(D), then for a contractible domain DD it holds the decomposition [6, 4]

(8) kΛj(D)=κk1Λdj+1(D)δk+1Λj+1(D) for j=1,,d1.\mathbb{P}_{k}\Lambda^{j}(D)=\star\kappa\mathbb{P}_{k-1}\Lambda^{d-j+1}(D)\oplus\delta\mathbb{P}_{k+1}\Lambda^{j+1}(D)\quad\textrm{ for }j=1,\ldots,d-1.

This implies 0Λj(D)=δ1Λj+1(D)\mathbb{P}_{0}\Lambda^{j}(D)=\delta\mathbb{P}_{1}\Lambda^{j+1}(D). Set kΛj(D):=dkΛj1(D)κk1Λj+1(D)\mathbb{P}_{k}^{-}\Lambda^{j}(D):=\,{\rm d}\mathbb{P}_{k}\Lambda^{j-1}(D)\oplus\kappa\mathbb{P}_{k-1}\Lambda^{j+1}(D). We have kΛ0(D)=kΛ0(D)\mathbb{P}_{k}^{-}\Lambda^{0}(D)=\mathbb{P}_{k}\Lambda^{0}(D) and kΛd(D)=k1Λd(D)\mathbb{P}_{k}^{-}\Lambda^{d}(D)=\mathbb{P}_{k-1}\Lambda^{d}(D). For a 0-form vv, dv=vd𝒙\,{\rm d}v=\nabla v\cdot\,{\rm d}\boldsymbol{x} with d𝒙=(dx1,,dxd)\,{\rm d}\boldsymbol{x}=(\,{\rm d}x_{1},\ldots,\,{\rm d}x_{d})^{\intercal}. If not causing any confusion, we will use v\nabla v to represent dv\,{\rm d}v for 0-form vv.

Let {𝒯h}h>0\{\mathcal{T}_{h}\}_{h>0} be a regular family of simplicial meshes of Ωd\Omega\subset\mathbb{R}^{d}, where h=maxT𝒯hhTh=\max_{T\in\mathcal{T}_{h}}h_{T}. For =0,1,,d1\ell=0,1,\ldots,d-1, denote by Δ(𝒯h)\Delta_{\ell}(\mathcal{T}_{h}) and Δ(𝒯̊h)\Delta_{\ell}(\mathring{\mathcal{T}}_{h}) the set of all subsimplices and all interior subsimplices of dimension \ell in the partition 𝒯h\mathcal{T}_{h}, respectively. For a simplex TT, we let Δ(T)\Delta_{\ell}(T) denote the set of subsimplices of dimension \ell. For a subsimplex ff of 𝒯h\mathcal{T}_{h}, let 𝒯f\mathcal{T}_{f} be the set of all simplices in 𝒯h\mathcal{T}_{h} sharing ff. Denote by ωT\omega_{T} the union of all the simplices in the set {𝒯v}vΔ0(T)\{\mathcal{T}_{\texttt{v}}\}_{\texttt{v}\in\Delta_{0}(T)}. For integer k0k\geq 0, define k(𝒯h):=kΛ0(𝒯h)\mathbb{P}_{k}(\mathcal{T}_{h}):=\mathbb{P}_{k}\Lambda^{0}(\mathcal{T}_{h}) with

kΛj(𝒯h)\displaystyle\mathbb{P}_{k}\Lambda^{j}(\mathcal{T}_{h}) :={vL2Λj(Ω):v|TkΛj(T),T𝒯h},\displaystyle:=\{v\in L^{2}\Lambda^{j}(\Omega):v|_{T}\in\mathbb{P}_{k}\Lambda^{j}(T),T\in\mathcal{T}_{h}\},
kΛj(𝒯h)\displaystyle\mathbb{P}_{k}^{-}\Lambda^{j}(\mathcal{T}_{h}) :={vL2Λj(Ω):v|TkΛj(T) for T𝒯h}.\displaystyle:=\{v\in L^{2}\Lambda^{j}(\Omega):v|_{T}\in\mathbb{P}_{k}^{-}\Lambda^{j}(T)\textrm{ for }T\in\mathcal{T}_{h}\}.

Let Qk,T:L2Λj(T)kΛj(T)Q_{k,T}:L^{2}\Lambda^{j}(T)\rightarrow\mathbb{P}_{k}\Lambda^{j}(T) for T𝒯hT\in\mathcal{T}_{h} be the L2L^{2}-orthogonal projection operator.

In this paper, we use “\lesssim\cdot\cdot\cdot” to mean that “C\leq C\cdot\cdot\cdot”, where CC is a generic positive constant independent of hh, which may have different values in different forms. And ABA\eqsim B equivalents to ABA\lesssim B and BAB\lesssim A.

2.2. The variational formulation

Let 0jd10\leq j\leq d-1. The weak formulation of problem (1) is to find uH0GdΛju\in H_{0}^{\rm Gd}\Lambda^{j} and λH0Λj1/dH0Λj2\lambda\in H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2} such that

(9a) (du,dv)+(v,dλ)\displaystyle(\nabla\,{\rm d}u,\nabla\,{\rm d}v)+(v,\,{\rm d}\lambda) =(f,v)vH0GdΛj,\displaystyle=(f,v)\quad\;\forall\,\,v\in H_{0}^{\rm Gd}\Lambda^{j},
(9b) (u,dμ)\displaystyle(u,\,{\rm d}\mu) =0μH0Λj1/dH0Λj2.\displaystyle=0\qquad\quad\forall\,\,\mu\in H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}.
Lemma 2.1.

The variational formulation (9) is well-posed. The solution (u,λ)H0GdΛj×(H0Λj1/dH0Λj2)(u,\lambda)\in H_{0}^{\rm Gd}\Lambda^{j}\times(H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}) of problem (9) satisfies the stability

u+du1+λHΛj1f,\|u\|+\|\,{\rm d}u\|_{1}+\|\lambda\|_{H\Lambda^{j-1}}\lesssim\|f\|,

where λHΛj12:=λ2+dλ2\|\lambda\|_{H\Lambda^{j-1}}^{2}:=\|\lambda\|^{2}+\|\,{\rm d}\lambda\|^{2}.

Proof.

For uH0GdΛjker(δ)u\in H_{0}^{\rm Gd}\Lambda^{j}\cap\ker(\delta), by the Poincaré inequality [4, (4.7)], we have the coercivity

u2+du12|du|12=(du,du).\|u\|^{2}+\|\,{\rm d}u\|_{1}^{2}\lesssim|\,{\rm d}u|_{1}^{2}=(\nabla\,{\rm d}u,\nabla\,{\rm d}u).

For λH0Λj1/dH0Λj2\lambda\in H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}, applying the Poincaré inequality again, it follows the inf-sup condition

λHΛj1dλ=(dλ,dλ)dλsupvH0GdΛj(v,dλ)v+dv1.\|\lambda\|_{H\Lambda^{j-1}}\lesssim\|\,{\rm d}\lambda\|=\frac{(\,{\rm d}\lambda,\,{\rm d}\lambda)}{\|\,{\rm d}\lambda\|}\leq\sup_{v\in H_{0}^{\rm Gd}\Lambda^{j}}\frac{(v,\,{\rm d}\lambda)}{\|v\|+\|\,{\rm d}v\|_{1}}.

Then we conclude the result from the Babuška-Brezzi theory [8]. ∎

2.3. The decoupled variational formulation

We will apply the framework in [12] to decouple the variational formulation (9) into lower order differential equations.

Recall the de Rham complex [17, 6, 4]

(10) HΛj+4𝛿HΛj+3𝛿L2Λj+2𝛿H1Λj+1𝛿H2Λjker(δ)0,H^{*}\Lambda^{j+4}\xrightarrow{\delta}H^{*}\Lambda^{j+3}\xrightarrow{\delta}L^{2}\Lambda^{j+2}\xrightarrow{\delta}H^{-1}\Lambda^{j+1}\xrightarrow{\delta}H^{-2}\Lambda^{j}\cap\ker(\delta)\xrightarrow{}0,

which is exact on the contractible domain Ω\Omega. Applying the tilde operation in [13], the de Rham complex (10) implies the exact de Rham complex

(11) HΛj+4𝛿HΛj+3𝛿L2Λj+2𝛿H1(δ,Λj+1)𝛿H1Λjker(δ)0,H^{*}\Lambda^{j+4}\xrightarrow{\delta}H^{*}\Lambda^{j+3}\xrightarrow{\delta}L^{2}\Lambda^{j+2}\xrightarrow{\delta}H^{-1}(\delta,\Lambda^{j+1})\xrightarrow{\delta}H^{-1}\Lambda^{j}\cap\ker(\delta)\xrightarrow{}0,

where H1(δ,Λj+1):={ωH1Λj+1:δωH1Λj}H^{-1}(\delta,\Lambda^{j+1}):=\{\omega\in H^{-1}\Lambda^{j+1}:\delta\omega\in H^{-1}\Lambda^{j}\}. We understand H1(δ,Λd)=L02(Ω)H^{-1}(\delta,\Lambda^{d})=L_{0}^{2}(\Omega). With the de Rham complex (11), we build up the following commutative diagram

(12)
H01Λj+1ΔH1Λj+1L2Λj+2/δHΛj+3δH1(δ,Λj+1)δH1Λjker(δ)0.H0Λj+1IH0Λj/dH0Λj1dδd
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Lemma 2.2.

Let 0jd10\leq j\leq d-1. We have

(13) (H0Λj/dH0Λj1)=H1Λjker(δ),(H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1})^{\prime}=H^{-1}\Lambda^{j}\cap\ker(\delta),

and Helmholtz decompositions

(14) (H0Λj+1)=H1(δ,Λj+1)=dH0ΛjδL2Λj+2,\displaystyle(H_{0}\Lambda^{j+1})^{\prime}=H^{-1}(\delta,\Lambda^{j+1})=\,{\rm d}H_{0}\Lambda^{j}\oplus\delta L^{2}\Lambda^{j+2},
(15) L2Λj+2=dH01Λj+1δHΛj+3,\displaystyle\qquad\quad L^{2}\Lambda^{j+2}=\,{\rm d}H_{0}^{1}\Lambda^{j+1}\oplus^{\perp}\delta H^{*}\Lambda^{j+3},
(16) L2Λj+2=dH01Λj+1δH1Λj+3,\displaystyle\qquad\quad L^{2}\Lambda^{j+2}=\,{\rm d}H_{0}^{1}\Lambda^{j+1}\oplus^{\perp}\delta H^{1}\Lambda^{j+3},

where \oplus^{\perp} means the L2L^{2} orthogonal direct sum.

Proof.

Due to the Poincaré inequality, (d,d)(\,{\rm d}\cdot,\,{\rm d}\cdot) is an inner product on the quotient space H0Λj/dH0Λj1H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1}, by Riesz-Fréchet representation theorem [10, Theorem 5.5], which implies δd:H0Λj/dH0Λj1(H0Λj/dH0Λj1)\delta\,{\rm d}:H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1}\to(H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1})^{\prime} is isomorphic. Then

(H0Λj/dH0Λj1)=δd(H0Λj/dH0Λj1)=δdH0Λj.(H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1})^{\prime}=\delta\,{\rm d}(H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1})=\delta\,{\rm d}H_{0}\Lambda^{j}.

Recall the Hodge decomposition [6, 4]

(17) L2Λj+1=dH0ΛjδHΛj+2.L^{2}\Lambda^{j+1}=\,{\rm d}H_{0}\Lambda^{j}\oplus^{\perp}\delta H^{*}\Lambda^{j+2}.

Hence it follows

(H0Λj/dH0Λj1)=δL2Λj+1,(H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1})^{\prime}=\delta L^{2}\Lambda^{j+1},

which together with complex (10) indicates (13).

Thanks to (13), applying the framework in [12] to the diagram (12), we get the Helmholtz decomposition (14).

Since dH0Λj+1=dH01Λj+1\,{\rm d}H_{0}\Lambda^{j+1}=\,{\rm d}H_{0}^{1}\Lambda^{j+1} and δHΛj+3=δH1Λj+3\delta H^{*}\Lambda^{j+3}=\delta H^{1}\Lambda^{j+3} [17, Theorem 1.1], we conclude Helmholtz decomposition (15) from decomposition (17), and Helmholtz decomposition (16) from decomposition (15). ∎

Remark 2.3.

When j=d1j=d-1, the Helmholtz decompositions (15)-(16) become

{0}=Ω(H01(Ω)L02(Ω))dx.\{0\}=\int_{\Omega}(H_{0}^{1}(\Omega)\cap L_{0}^{2}(\Omega))\,{\rm d}x.

A mixed formulation based on the commutative diagram (12) is to find (γ,u)H1(δ,Λj+1)×(H0Λj/dH0Λj1)(\gamma,u)\in H^{-1}(\delta,\Lambda^{j+1})\times(H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1}) such that

(γ,β)1δβ,u\displaystyle(\gamma,\beta)_{-1}-\langle\delta\beta,u\rangle =0\displaystyle=0 βH1(δ,Λj+1),\displaystyle\forall\,\,\beta\in H^{-1}(\delta,\Lambda^{j+1}),
δγ,v\displaystyle\langle\delta\gamma,v\rangle =(f,v)\displaystyle=(f,v) vH0Λj/dH0Λj1,\displaystyle\forall\,\,v\in H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1},

where (γ,β)1:=Δ1γ,βH01Λj+1×H1Λj+1=γ,Δ1βH1Λj+1×H01Λj+1(\gamma,\beta)_{-1}:=-\langle\Delta^{-1}\gamma,\beta\rangle_{H_{0}^{1}\Lambda^{j+1}\times H^{-1}\Lambda^{j+1}}=-\langle\gamma,\Delta^{-1}\beta\rangle_{H^{-1}\Lambda^{j+1}\times H_{0}^{1}\Lambda^{j+1}}. By introducing the new variable ϕ=Δ1γH01Λj+1\phi=-\Delta^{-1}\gamma\in H_{0}^{1}\Lambda^{j+1}, the unfolded formulation is to find (γ,u,ϕ)H1(δ,Λj+1)×(H0Λj/dH0Λj1)×H01Λj+1(\gamma,u,\phi)\in H^{-1}(\delta,\Lambda^{j+1})\times(H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1})\times H_{0}^{1}\Lambda^{j+1} such that

(18a) (ϕ,ψ)+γ,dvψ\displaystyle(\nabla\phi,\nabla\psi)+\langle\gamma,\,{\rm d}v-\psi\rangle =(f,v)\displaystyle=(f,v) (υ,ψ)(H0Λj/dH0Λj1)×H01Λj+1,\displaystyle\forall\,\,(\upsilon,\psi)\in(H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1})\times H_{0}^{1}\Lambda^{j+1},
(18b) β,duϕ\displaystyle\langle\beta,\,{\rm d}u-\phi\rangle =0\displaystyle=0 βH1(δ,Λj+1).\displaystyle\forall\,\,\beta\in H^{-1}(\delta,\Lambda^{j+1}).

According to the Helmholtz decomposition (14), we can set γ=dwδp\gamma=\,{\rm d}w-\delta p and β=dχδq\beta=\,{\rm d}\chi-\delta q with w,χH0Λj/dH0Λj1w,\chi\in H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1} and p,qL2Λj+2p,q\in L^{2}\Lambda^{j+2}. Then the formulation (18) is decomposed as follows [12, Section 3.2]: find u,wH0Λj/dH0Λj1u,w\in H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1}, ϕH01Λj+1\phi\in H_{0}^{1}\Lambda^{j+1} and pL2Λj+2/δHΛj+3p\in L^{2}\Lambda^{j+2}/\delta H^{*}\Lambda^{j+3} such that

(dw,dv)\displaystyle(\,{\rm d}w,\,{\rm d}v) =(f,v)\displaystyle=(f,v) vH0Λj/dH0Λj1,\displaystyle\forall\,\,v\in H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1},
(ϕ,ψ)+(dψ,p)\displaystyle(\nabla\phi,\nabla\psi)+(\,{\rm d}\psi,p) =(dw,ψ)\displaystyle=(\,{\rm d}w,\psi) ψH01Λj+1,\displaystyle\forall\,\,\psi\in H_{0}^{1}\Lambda^{j+1},
(dϕ,q)\displaystyle(\,{\rm d}\phi,q) =0\displaystyle=0 qL2Λj+2/δHΛj+3,\displaystyle\forall\,\,q\in L^{2}\Lambda^{j+2}/\delta H^{*}\Lambda^{j+3},
(du,dχ)\displaystyle(\,{\rm d}u,\,{\rm d}\chi) =(ϕ,dχ)\displaystyle=(\phi,\,{\rm d}\chi) χH0Λj/dH0Λj1.\displaystyle\forall\,\,\chi\in H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1}.

We further employ the Lagrange multiplier to deal with the constraint in the quotient spaces H0Λj/dH0Λj1H_{0}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1} and L2Λj+2/δHΛj+3L^{2}\Lambda^{j+2}/\delta H^{*}\Lambda^{j+3}, which induces the following variational formulation: find u,wH0Λju,w\in H_{0}\Lambda^{j}, λ,zH0Λj1/dH0Λj2\lambda,z\in H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}, ϕH01Λj+1\phi\in H_{0}^{1}\Lambda^{j+1}, pL2Λj+2p\in L^{2}\Lambda^{j+2} and rHΛj+3/δHΛj+4r\in H^{*}\Lambda^{j+3}/\delta H^{*}\Lambda^{j+4} such that

(19a) (dw,dv)+(v,dλ)\displaystyle(\,{\rm d}w,\,{\rm d}v)+(v,\,{\rm d}\lambda) =(f,v),\displaystyle=(f,v),
(19b) (w,dη)\displaystyle(w,\,{\rm d}\eta) =0,\displaystyle=0,
(19c) (ϕ,ψ)+(dψ+δs,p)\displaystyle(\nabla\phi,\nabla\psi)+(\,{\rm d}\psi+\delta s,p) =(dw,ψ),\displaystyle=(\,{\rm d}w,\psi),
(19d) (dϕ+δr,q)\displaystyle(\,{\rm d}\phi+\delta r,q) =0,\displaystyle=0,
(19e) (du,dχ)+(χ,dz)\displaystyle(\,{\rm d}u,\,{\rm d}\chi)+(\chi,\,{\rm d}z) =(ϕ,dχ),\displaystyle=(\phi,\,{\rm d}\chi),
(19f) (u,dμ)\displaystyle(u,\,{\rm d}\mu) =0,\displaystyle=0,
for any v,χH0Λjv,\chi\in H_{0}\Lambda^{j}, η,μH0Λj1/dH0Λj2\eta,\mu\in H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}, ψH01Λj+1\psi\in H_{0}^{1}\Lambda^{j+1}, qL2Λj+2q\in L^{2}\Lambda^{j+2} and sHΛj+3/δHΛj+4s\in H^{*}\Lambda^{j+3}/\delta H^{*}\Lambda^{j+4}.

The well-posedness of problem (19c)-(19d) is related to complex (20).

Lemma 2.4.

Let 0jd20\leq j\leq d-2. The complex

(20) H0GdΛj×HΛj+4(d00δ)H01Λj+1×HΛj+3(d,δ)L2Λj+20H_{0}^{\rm Gd}\Lambda^{j}\times H^{*}\Lambda^{j+4}\xrightarrow{\begin{pmatrix}\,{\rm d}&0\\ 0&\delta\end{pmatrix}}H_{0}^{1}\Lambda^{j+1}\times H^{*}\Lambda^{j+3}\xrightarrow{(\,{\rm d},\delta)}L^{2}\Lambda^{j+2}\to 0

is exact.

Proof.

Due to decomposition (15), we conclude the exactness of complex (20) from H01Λj+1ker(d)=dH0GdΛjH_{0}^{1}\Lambda^{j+1}\cap\ker(\,{\rm d})=\,{\rm d}H_{0}^{\rm Gd}\Lambda^{j} and HΛj+3ker(δ)=δHΛj+4H^{*}\Lambda^{j+3}\cap\ker(\delta)=\delta H^{*}\Lambda^{j+4}. ∎

Indeed, complex (20) is a distinct representation of the following complex

(21) H0Λj1dH0GdΛjdH01Λj+1dL2Λj+2/δHΛj+30.H_{0}\Lambda^{j-1}\xrightarrow{\,{\rm d}}H_{0}^{\rm Gd}\Lambda^{j}\xrightarrow{\,{\rm d}}H_{0}^{1}\Lambda^{j+1}\xrightarrow{\,{\rm d}}L^{2}\Lambda^{j+2}/\delta H^{*}\Lambda^{j+3}\to 0.

Thanks to complex (20), we can regard (d,δ)(\,{\rm d},\delta) as a generalized divergence operator, and problem (19c)-(19d) as a generalized Stokes equation.

Next we analyze the well-posedness of the decoupled formulation (19), and show its equivalence to the variational formulation (9). The well-posedness of problem (19a)-(19b) and problem (19e)-(19f) follows from the Poincaré inequality [4, 6]. Then we focus on the well-posedness of the generalized Stokes equation (19c)-(19d).

For simplicity, introduce bilinear forms

a(ϕ,r;ψ,s)=(ϕ,ψ),b(ψ,s;q)=(dψ+δs,q).a(\phi,r;\psi,s)=(\nabla\phi,\nabla\psi),\quad b(\psi,s;q)=(\,{\rm d}\psi+\delta s,q).

Clearly, we have

a(ϕ,r;ψ,s)ϕ1ψ1ϕ,ψH1Λj+1,r,sHΛj+3,a(\phi,r;\psi,s)\leq\|\phi\|_{1}\|\psi\|_{1}\quad\forall\leavevmode\nobreak\ \phi,\psi\in H^{1}\Lambda^{j+1},r,s\in H^{*}\Lambda^{j+3},
b(ψ,s;q)(dψ+δs)qψH1Λj+1,sHΛj+3,qL2Λj+2.b(\psi,s;q)\leq(\|\,{\rm d}\psi\|+\|\delta s\|)\|q\|\quad\forall\leavevmode\nobreak\ \psi\in H^{1}\Lambda^{j+1},s\in H^{*}\Lambda^{j+3},q\in L^{2}\Lambda^{j+2}.
Lemma 2.5.

Let 0jd20\leq j\leq d-2. For (ψ,s)H01Λj+1×(HΛj+3/δHΛj+4)(\psi,s)\in H_{0}^{1}\Lambda^{j+1}\times(H^{*}\Lambda^{j+3}/\delta H^{*}\Lambda^{j+4}) satisfying b(ψ,s;q)=0b(\psi,s;q)=0 for all qL2Λj+2q\in L^{2}\Lambda^{j+2}, we have the coercivity

(22) ψ12+sHΛj+32a(ψ,s;ψ,s),\displaystyle\|\psi\|_{1}^{2}+\|s\|_{H^{*}\Lambda^{j+3}}^{2}\lesssim a(\psi,s;\psi,s),

where sHΛ32:=s2+δs2\|s\|_{H^{*}\Lambda^{3}}^{2}:=\|s\|^{2}+\|\delta s\|^{2}.

Proof.

Notice that

(dψ,q)+(δs,q)=0qL2Λj+2.(\,{\rm d}\psi,q)+(\delta s,q)=0\quad\forall\,q\in L^{2}\Lambda^{j+2}.

Take q=δsq=\delta s to get s=0s=0. Thus, (22) follows from the Poincaré inequality for space H01(Ω)H_{0}^{1}(\Omega) (cf. [9]). ∎

Lemma 2.6.

Let 0jd20\leq j\leq d-2. For qL2Λj+2q\in L^{2}\Lambda^{j+2}, it holds the inf-sup condition

(23) qsupψH01Λj+1,sHΛj+3/δHΛj+4b(ψ,s;q)ψ1+sHΛj+3.\displaystyle\|q\|\lesssim\sup_{\psi\in H_{0}^{1}\Lambda^{j+1},\,s\in H^{*}\Lambda^{j+3}/\delta H^{*}\Lambda^{j+4}}\frac{b(\psi,s;q)}{\|\psi\|_{1}+\|s\|_{H^{*}\Lambda^{j+3}}}.
Proof.

By the Helmholtz decomposition (15), there exist ψH01Λj+1\psi\in H_{0}^{1}\Lambda^{j+1} and sHΛj+3/δHΛj+4s\in H^{*}\Lambda^{j+3}/\delta H^{*}\Lambda^{j+4} such that

q=dψ+δs,ψ1+sHΛj+3q.q=\,{\rm d}\psi+\delta s,\quad\|\psi\|_{1}+\|s\|_{H^{*}\Lambda^{j+3}}\lesssim\|q\|.

So b(ψ,s;q)=(dψ+δs,q)=q2.b(\psi,s;q)=(\,{\rm d}\psi+\delta s,q)=\|q\|^{2}. This means

q(ψ1+sHΛj+3)q2=b(ψ,s;q).\|q\|(\|\psi\|_{1}+\|s\|_{H^{*}\Lambda^{j+3}})\lesssim\|q\|^{2}=b(\psi,s;q).

Then (23) follows. ∎

Theorem 2.7.

Let 0jd10\leq j\leq d-1. The generalized Stokes equation (19c)-(19d) is well-posed. The mixed formulation (19) and the variational formulation (9) are equivalent. That is, if (w,λ)H0Λj×(H0Λj1/dH0Λj2)(w,\lambda)\in H_{0}\Lambda^{j}\times(H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}) is the solution of problem (19a)-(19b), (ϕ,p,r)H01Λj+1×L2Λj+2×(HΛj+3/δHΛj+4)(\phi,p,r)\in H_{0}^{1}\Lambda^{j+1}\times L^{2}\Lambda^{j+2}\times(H^{*}\Lambda^{j+3}/\delta H^{*}\Lambda^{j+4}) is the solution of problem (19c)-(19d), and (u,z)H0Λj×(H0Λj1/dH0Λj2)(u,z)\in H_{0}\Lambda^{j}\times(H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}) is the solution of problem (19e)-(19f), then r=0r=0, z=0z=0, pdH01Λj+1p\in\,{\rm d}H_{0}^{1}\Lambda^{j+1}, ϕ=du\phi=\,{\rm d}u, and (u,λ)H0GdΛj×(H0Λj1/dH0Λj2)(u,\lambda)\in H_{0}^{\rm Gd}\Lambda^{j}\times(H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}) satisfies the variational formulation (9).

Proof.

When j=d1j=d-1, the generalized Stokes equation (19c)-(19d) becomes the following Poisson equation: find ϕH01(Ω)L02(Ω)\phi\in H_{0}^{1}(\Omega)\cap L_{0}^{2}(\Omega) such that

(ϕ,ψ)=(divw,ψ)ψH01(Ω)L02(Ω),(\nabla\phi,\nabla\psi)=(\operatorname{div}w,\psi)\qquad\forall\leavevmode\nobreak\ \psi\in H_{0}^{1}(\Omega)\cap L_{0}^{2}(\Omega),

which is obviously well-posed. For 0jd20\leq j\leq d-2, by applying the Babuška-Brezzi theory [8], the well-posedness of the generalized Stokes equation (19c)-(19d) follows from the coercivity (22) and the inf-sup condition (23).

Next we prove the equivalence. Take q=δrq=\delta r in (19d) to get r=0r=0 and dϕ=0\,{\rm d}\phi=0, χ=dz\chi=\,{\rm d}z in (19e) to get z=0z=0. So there exists u~H0GdΛj/dH0Λj1\tilde{u}\in H_{0}^{\rm Gd}\Lambda^{j}/\,{\rm d}H_{0}\Lambda^{j-1} s.t. ϕ=du~H01Λj+1\phi=\,{\rm d}\tilde{u}\in H_{0}^{1}\Lambda^{j+1}. By (19e)-(19f), u=u~u=\tilde{u} and ϕ=du\phi=\,{\rm d}u. By taking s=0s=0 and ψ=dv\psi=\,{\rm d}v with vH0GdΛjv\in H_{0}^{\rm Gd}\Lambda^{j} in (19c), we acquire

(du,dv)=(dw,dv)vH0GdΛj.(\nabla\,{\rm d}u,\nabla\,{\rm d}v)=(\,{\rm d}w,\,{\rm d}v)\quad\forall\,v\in H_{0}^{\rm Gd}\Lambda^{j}.

This together with (19a) means that (u,λ)H0GdΛj×H0Λj1(u,\lambda)\in H_{0}^{\rm Gd}\Lambda^{j}\times H_{0}\Lambda^{j-1} satisfies (9).

Choose ψ=0\psi=0 in (19c) to derive pδHΛj+3p\perp\delta H^{*}\Lambda^{j+3}. Thus, pdH01Λj+1p\in\,{\rm d}H_{0}^{1}\Lambda^{j+1} follows from the Helmholtz decomposition (15). ∎

Therefore, the weak formulation (9) of the fourth-order problem (1) is decoupled into two second-order exterior differential equations (19a)-(19b), (19e)-(19f) and one generalized Stokes equation (19c)-(19d), which is more amenable to designing finite element methods and fast solvers.

Example 2.8.

Taking j=0j=0, we get the decoupling of the biharmonic equation (2): find u,wH01(Ω)u,w\in H_{0}^{1}(\Omega), ϕH01Λ1\phi\in H_{0}^{1}\Lambda^{1}, pL2Λ2p\in L^{2}\Lambda^{2} and rHΛ3/δHΛ4r\in H^{*}\Lambda^{3}/\delta H^{*}\Lambda^{4} such that

(w,v)\displaystyle(\nabla w,\nabla v) =(f,v)vH01(Ω),\displaystyle=(f,v)\qquad\quad\forall\leavevmode\nobreak\ v\in H_{0}^{1}(\Omega),
(ϕ,ψ)+(curlψ+δs,p)\displaystyle(\nabla\phi,\nabla\psi)+(\operatorname{curl}\psi+\delta s,p) =(w,ψ)ψH01Λ1,sHΛ3,\displaystyle=(\nabla w,\psi)\quad\;\;\;\forall\leavevmode\nobreak\ \psi\in H_{0}^{1}\Lambda^{1},s\in H^{*}\Lambda^{3},
(curlϕ+δr,q)\displaystyle(\operatorname{curl}\phi+\delta r,q) =0qL2Λ2,\displaystyle=0\qquad\qquad\;\;\,\forall\leavevmode\nobreak\ q\in L^{2}\Lambda^{2},
(u,χ)\displaystyle(\nabla u,\nabla\chi) =(ϕ,χ)χH01(Ω).\displaystyle=(\phi,\nabla\chi)\qquad\forall\leavevmode\nobreak\ \chi\in H_{0}^{1}(\Omega).

Such a decoupling of the biharmonic equation in two and three dimensions can be found in [25, 26, 12, 19], which is employed to design fast solvers for the Morley element method in [25, 18, 28]. We refer to [33] for a different decomposition.

Example 2.9.

Taking j=1j=1 and d=3d=3, we get the decoupling of the quad-curl problem (3): find u,wH0(curl,Ω)u,w\in H_{0}(\operatorname{curl},\Omega), λ,zH01(Ω)\lambda,z\in H_{0}^{1}(\Omega), ϕH01(Ω;3)\phi\in H_{0}^{1}(\Omega;\mathbb{R}^{3}) and pL02(Ω)p\in L_{0}^{2}(\Omega) such that

(curlw,curlv)+(v,λ)\displaystyle(\operatorname{curl}w,\operatorname{curl}v)+(v,\nabla\lambda) =(f,v)vH0(curl,Ω),\displaystyle=(f,v)\qquad\qquad\forall\leavevmode\nobreak\ v\in H_{0}(\operatorname{curl},\Omega),
(w,η)\displaystyle(w,\nabla\eta) =0ηH01(Ω),\displaystyle=0\qquad\qquad\quad\;\;\,\forall\leavevmode\nobreak\ \eta\in H_{0}^{1}(\Omega),
(ϕ,ψ)+(divψ,p)\displaystyle(\nabla\phi,\nabla\psi)+(\operatorname{div}\psi,p) =(curlw,ψ)ψH01(Ω;3),\displaystyle=(\operatorname{curl}w,\psi)\quad\;\;\;\forall\leavevmode\nobreak\ \psi\in H_{0}^{1}(\Omega;\mathbb{R}^{3}),
(divϕ,q)\displaystyle(\operatorname{div}\phi,q) =0qL02(Ω),\displaystyle=0\qquad\qquad\quad\;\;\,\forall\leavevmode\nobreak\ q\in L_{0}^{2}(\Omega),
(curlu,curlχ)+(χ,z)\displaystyle(\operatorname{curl}u,\operatorname{curl}\chi)+(\chi,\nabla z) =(ϕ,curlχ)χH0(curl,Ω),\displaystyle=(\phi,\operatorname{curl}\chi)\qquad\forall\leavevmode\nobreak\ \chi\in H_{0}(\operatorname{curl},\Omega),
(u,μ)\displaystyle(u,\nabla\mu) =0μH01(Ω).\displaystyle=0\qquad\qquad\quad\;\;\,\forall\leavevmode\nobreak\ \mu\in H_{0}^{1}(\Omega).

This is the decoupling in [12, Section 3.4].

Example 2.10.

Taking j=d1j=d-1, we get the decoupling of the fourth-order div problem (4): find u,wH0(div,Ω)u,w\in H_{0}(\operatorname{div},\Omega), λ,zH0Λd2/dH0Λd3\lambda,z\in H_{0}\Lambda^{d-2}/\,{\rm d}H_{0}\Lambda^{d-3} and ϕH01(Ω)L02(Ω)\phi\in H_{0}^{1}(\Omega)\cap L_{0}^{2}(\Omega) such that

(divw,divv)+(v,dλ)\displaystyle(\operatorname{div}w,\operatorname{div}v)+(v,\,{\rm d}\lambda) =(f,v)vH0(div,Ω),\displaystyle=(f,v)\qquad\qquad\forall\leavevmode\nobreak\ v\in H_{0}(\operatorname{div},\Omega),
(w,dη)\displaystyle(w,\,{\rm d}\eta) =0ηH0Λd2,\displaystyle=0\qquad\qquad\quad\;\;\,\forall\leavevmode\nobreak\ \eta\in H_{0}\Lambda^{d-2},
(ϕ,ψ)\displaystyle(\nabla\phi,\nabla\psi) =(divw,ψ)ψH01(Ω)L02(Ω),\displaystyle=(\operatorname{div}w,\psi)\qquad\forall\leavevmode\nobreak\ \psi\in H_{0}^{1}(\Omega)\cap L_{0}^{2}(\Omega),
(divu,divχ)+(χ,dz)\displaystyle(\operatorname{div}u,\operatorname{div}\chi)+(\chi,\,{\rm d}z) =(ϕ,divχ)χH0(div,Ω),\displaystyle=(\phi,\operatorname{div}\chi)\qquad\,\forall\leavevmode\nobreak\ \chi\in H_{0}(\operatorname{div},\Omega),
(u,dμ)\displaystyle(u,\,{\rm d}\mu) =0μH0Λd2.\displaystyle=0\qquad\qquad\quad\;\;\forall\leavevmode\nobreak\ \mu\in H_{0}\Lambda^{d-2}.
Remark 2.11.

By applying the integration by parts to (δs,p)(\delta s,p) and (δr,q)(\delta r,q), the formulation (19) is equivalent to find u,wH0Λju,w\in H_{0}\Lambda^{j}, λ,zH0Λj1/dH0Λj2\lambda,z\in H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}, ϕH01Λj+1\phi\in H_{0}^{1}\Lambda^{j+1}, pH0Λj+2p\in H_{0}\Lambda^{j+2} and rL2Λj+3/δHΛj+4r\in L^{2}\Lambda^{j+3}/\delta H^{*}\Lambda^{j+4} such that

(24a) (dw,dv)+(v,dλ)\displaystyle(\,{\rm d}w,\,{\rm d}v)+(v,\,{\rm d}\lambda) =(f,v),\displaystyle=(f,v),
(24b) (w,dη)\displaystyle(w,\,{\rm d}\eta) =0,\displaystyle=0,
(24c) (ϕ,ψ)+(dψ,p)+(s,dp)\displaystyle(\nabla\phi,\nabla\psi)+(\,{\rm d}\psi,p)+(s,\,{\rm d}p) =(dw,ψ),\displaystyle=(\,{\rm d}w,\psi),
(24d) (dϕ,q)+(r,dq)\displaystyle(\,{\rm d}\phi,q)+(r,\,{\rm d}q) =0,\displaystyle=0,
(24e) (du,dχ)+(χ,dz)\displaystyle(\,{\rm d}u,\,{\rm d}\chi)+(\chi,\,{\rm d}z) =(ϕ,dχ),\displaystyle=(\phi,\,{\rm d}\chi),
(24f) (u,dμ)\displaystyle(u,\,{\rm d}\mu) =0,\displaystyle=0,

for any v,χH0Λjv,\chi\in H_{0}\Lambda^{j}, η,μH0Λj1/dH0Λj2\eta,\mu\in H_{0}\Lambda^{j-1}/\,{\rm d}H_{0}\Lambda^{j-2}, ψH01Λj+1\psi\in H_{0}^{1}\Lambda^{j+1}, qH0Λj+2q\in H_{0}\Lambda^{j+2} and sL2Λj+3/δHΛj+4s\in L^{2}\Lambda^{j+3}/\delta H^{*}\Lambda^{j+4}. The decoupled formulation (24) is related to complex

H0Λj1dH0GdΛjdH01Λj+1dH0Λj+2dL2Λj+3/δHΛj+40,H_{0}\Lambda^{j-1}\xrightarrow{\,{\rm d}}H_{0}^{\rm Gd}\Lambda^{j}\xrightarrow{\,{\rm d}}H_{0}^{1}\Lambda^{j+1}\xrightarrow{\,{\rm d}}H_{0}\Lambda^{j+2}\xrightarrow{\,{\rm d}}L^{2}\Lambda^{j+3}/\delta H^{*}\Lambda^{j+4}\to 0,

rather than complex (21). The numerical methods for the biharmonic equation in [19] are based on the decoupled formulation (24) with j=0j=0 and d=2,3d=2,3.

3. Decoupled finite element methods

We will develop a family of conforming finite element methods for the decoupled formulation (19) in this section, where the finite element pair for the generalized Stokes equation can be regarded as the generalization of the MINI element for Stokes equation in [5, 8].

3.1. Conforming finite element pair for generalized Stokes equation

We first construct conforming finite element spaces for the generalized Stokes equation (19c)-(19d). To discretize H01Λj+1(Ω)H_{0}^{1}\Lambda^{j+1}(\Omega) with 0jd10\leq j\leq d-1, for simplex TT and integer k1k\geq 1, we take

Φk(T):=kΛj+1(T)+bTδkΛj+2(T)\Phi_{k}(T):=\mathbb{P}_{k}\Lambda^{j+1}(T)+b_{T}\,\delta\mathbb{P}_{k}\Lambda^{j+2}(T)

as the space of shape functions, where bT=λ0λ1λdb_{T}=\lambda_{0}\lambda_{1}\cdots\lambda_{d} is the bubble function, and λi(i=0,1,,d)\lambda_{i}(i=0,1,\ldots,d) are the barycentric coordinates corresponding to the vertices of TT. When k=1k=1, by δ1Λj+2(T)=0Λj+1(T)\delta\mathbb{P}_{1}\Lambda^{j+2}(T)=\mathbb{P}_{0}\Lambda^{j+1}(T), we have

Φ1(T)=1Λj+1(T)+bT0Λj+1(T).\Phi_{1}(T)=\mathbb{P}_{1}\Lambda^{j+1}(T)+b_{T}\,\mathbb{P}_{0}\Lambda^{j+1}(T).

When j=d1j=d-1, Φk(T)=kΛd(T)+bT0Λd(T)\Phi_{k}(T)=\mathbb{P}_{k}\Lambda^{d}(T)+b_{T}\mathbb{P}_{0}\Lambda^{d}(T). For jd2j\leq d-2, it holds

Φk(T)=kΛj+1(T)bTδ(kΛj+2(T)/kdΛj+2(T)).\Phi_{k}(T)=\mathbb{P}_{k}\Lambda^{j+1}(T)\oplus b_{T}\,\delta(\mathbb{P}_{k}\Lambda^{j+2}(T)/\mathbb{P}_{k-d}\Lambda^{j+2}(T)).

The degrees of freedom (DoFs) are given by

(25a) v(v),\displaystyle v(\texttt{v}), vΔ0(T),\displaystyle\,\texttt{v}\in\Delta_{0}(T),
(25b) (v,q)f,\displaystyle(v,q)_{f}, qk1Λj+1(f),fΔ(T),=1,,d1,\displaystyle\,q\in\mathbb{P}_{k-\ell-1}\Lambda^{j+1}(f),\,f\in\Delta_{\ell}(T),\ell=1,\ldots,d-1,
(25c) (v,q)T,\displaystyle(v,q)_{T}, qkd1Λj+1(T)+δkΛj+2(T).\displaystyle\,q\in\mathbb{P}_{k-d-1}\Lambda^{j+1}(T)+\delta\mathbb{P}_{k}\Lambda^{j+2}(T).

Thanks to the decomposition (8), DoF (25c) can be rewritten as

(v,q)T,qκkd2Λdj(T)δkΛj+2(T).(v,q)_{T},\quad q\in\star\kappa\mathbb{P}_{k-d-2}\Lambda^{d-j}(T)\oplus\delta\mathbb{P}_{k}\Lambda^{j+2}(T).

The DoFs (25a)-(25b) and the first part of DoF (25c) correspond to the DoFs of Lagrange element.

Lemma 3.1.

Let 0jd10\leq j\leq d-1. The DoFs (25) are unisolvent for space Φk(T)\Phi_{k}(T).

Proof.

Both the number of DoFs (25) and dimΦk(T)\dim\Phi_{k}(T) are 1+dimk(T)1+\dim\mathbb{P}_{k}(T) for the case j=d1j=d-1 and d=kd=k. In all other instances, the dimensions are described by

dimkΛj+1(T)+dimδ(kΛj+2(T)/kdΛj+2(T)).\dim\mathbb{P}_{k}\Lambda^{j+1}(T)+\dim\delta(\mathbb{P}_{k}\Lambda^{j+2}(T)/\mathbb{P}_{k-d}\Lambda^{j+2}(T)).

Take vΦk(T)v\in\Phi_{k}(T) satisfying all the DoFs (25) vanish. Since v|FkΛj+1(F)v|_{F}\in\mathbb{P}_{k}\Lambda^{j+1}(F) for FΔd1(T)F\in\Delta_{d-1}(T), the vanishing DoFs (25a)-(25b) imply v|T=0v|_{\partial T}=0. Then v=bTqv=b_{T}q with qkd1Λj+1(T)+δkΛj+2(T)q\in\mathbb{P}_{k-d-1}\Lambda^{j+1}(T)+\delta\mathbb{P}_{k}\Lambda^{j+2}(T), which together with the vanishing DoF (25c) indicates v=0v=0. ∎

Define the H1H^{1}-conforming finite element space

Φh:={vhH01Λj+1(Ω):vh|TkΛj+1(T)+bTδkΛj+2(T) for T𝒯h}.\Phi_{h}:=\{v_{h}\in H_{0}^{1}\Lambda^{j+1}(\Omega):v_{h}|_{T}\in\mathbb{P}_{k}\Lambda^{j+1}(T)+b_{T}\,\delta\mathbb{P}_{k}\Lambda^{j+2}(T)\textrm{ for }T\in\mathcal{T}_{h}\}.

When k=1k=1, j=d2j=d-2 and d=2,3d=2,3, Φh\Phi_{h} is exactly the MINI element space in [5, 8] for Stokes equation. For j=0,1,,dj=0,1,\ldots,d, let

Vk,hdΛj\displaystyle V_{k,h}^{\,{\rm d}}\Lambda^{j} :=kΛj(𝒯h)HΛj(Ω),Vk,hδΛj:=Vk,hdΛdj,\displaystyle:=\mathbb{P}_{k}\Lambda^{j}(\mathcal{T}_{h})\cap H\Lambda^{j}(\Omega),\quad\;\,V_{k,h}^{\delta}\Lambda^{j}:=\star V_{k,h}^{\,{\rm d}}\Lambda^{d-j},
Vk,hd,Λj\displaystyle V_{k,h}^{\,{\rm d},-}\Lambda^{j} :=kΛj(𝒯h)HΛj(Ω),Vk,hδ,Λj:=Vk,hd,Λdj.\displaystyle:=\mathbb{P}_{k}^{-}\Lambda^{j}(\mathcal{T}_{h})\cap H\Lambda^{j}(\Omega),\quad V_{k,h}^{\delta,-}\Lambda^{j}:=\star V_{k,h}^{\,{\rm d},-}\Lambda^{d-j}.

Set V̊k,hdΛj:=Vk,hdΛjH0Λj(Ω)\mathring{V}_{k,h}^{\,{\rm d}}\Lambda^{j}:=V_{k,h}^{\,{\rm d}}\Lambda^{j}\cap H_{0}\Lambda^{j}(\Omega) and V̊k,hd,Λj:=Vk,hd,ΛjH0Λj(Ω)\mathring{V}_{k,h}^{\,{\rm d},-}\Lambda^{j}:=V_{k,h}^{\,{\rm d},-}\Lambda^{j}\cap H_{0}\Lambda^{j}(\Omega). We have the short exact sequences [21, 22, 4, 6]

(26) 0V̊k+1,hd,Λj1/dV̊k+1,hd,Λj2dV̊k,hdΛjddV̊k,hdΛj0,0\xrightarrow{}\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j-1}/\,{\rm d}\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j-2}\xrightarrow{\,{\rm d}}\mathring{V}_{k,h}^{\,{\rm d}}\Lambda^{j}\xrightarrow{\,{\rm d}}\,{\rm d}\mathring{V}_{k,h}^{\,{\rm d}}\Lambda^{j}\to 0,
(27) 0V̊k,hd,Λj1/dV̊k,hd,Λj2dV̊k,hd,ΛjddV̊k,hd,Λj0,0\xrightarrow{}\mathring{V}_{k,h}^{\,{\rm d},-}\Lambda^{j-1}/\,{\rm d}\mathring{V}_{k,h}^{\,{\rm d},-}\Lambda^{j-2}\xrightarrow{\,{\rm d}}\mathring{V}_{k,h}^{\,{\rm d},-}\Lambda^{j}\xrightarrow{\,{\rm d}}\,{\rm d}\mathring{V}_{k,h}^{\,{\rm d},-}\Lambda^{j}\to 0,
(28) 0Vk,hδ,Λj+1/δVk,hδ,Λj+2𝛿Vk,hδ,Λj𝛿δVk,hδ,Λj0.0\xrightarrow{}V_{k,h}^{\delta,-}\Lambda^{j+1}/\delta V_{k,h}^{\delta,-}\Lambda^{j+2}\xrightarrow{\delta}V_{k,h}^{\delta,-}\Lambda^{j}\xrightarrow{\delta}\delta V_{k,h}^{\delta,-}\Lambda^{j}\to 0.

Let Ik,hd:L2Λj(Ω)Vk,hdΛjI_{k,h}^{\,{\rm d}}:L^{2}\Lambda^{j}(\Omega)\to V_{k,h}^{\,{\rm d}}\Lambda^{j} and Ik,hd,:L2Λj(Ω)Vk,hd,ΛjI_{k,h}^{\,{\rm d},-}:L^{2}\Lambda^{j}(\Omega)\to V_{k,h}^{\,{\rm d},-}\Lambda^{j} be the L2L^{2} bounded projection operators devised in [3, 16], then Ik,hdω,Ik,hd,ωH0Λj(Ω)I_{k,h}^{\,{\rm d}}\omega,I_{k,h}^{\,{\rm d},-}\omega\in H_{0}\Lambda^{j}(\Omega) for ωH0Λj(Ω)\omega\in H_{0}\Lambda^{j}(\Omega). Define Ik,hδ:L2Λj(Ω)Vk,hδΛjI_{k,h}^{\delta}:L^{2}\Lambda^{j}(\Omega)\to V_{k,h}^{\delta}\Lambda^{j} and Ik,hδ,:L2Λj(Ω)Vk,hδ,ΛjI_{k,h}^{\delta,-}:L^{2}\Lambda^{j}(\Omega)\to V_{k,h}^{\delta,-}\Lambda^{j} as

Ik,hδω=Ik,hd(ω),Ik,hδ,ω=Ik,hd,(ω),ωL2Λj(Ω).\star I_{k,h}^{\delta}\omega=I_{k,h}^{\,{\rm d}}(\star\omega),\;\;\star I_{k,h}^{\delta,-}\omega=I_{k,h}^{\,{\rm d},-}(\star\omega),\quad\omega\in L^{2}\Lambda^{j}(\Omega).

Then Ik,hδI_{k,h}^{\delta} and Ik,hδ,I_{k,h}^{\delta,-} are also L2L^{2} bounded. We will abbreviate Ik,hdI_{k,h}^{\,{\rm d}} and Ik,hd,I_{k,h}^{\,{\rm d},-} as IhdI_{h}^{\,{\rm d}}, and Ik,hδI_{k,h}^{\delta} and Ik,hδ,I_{k,h}^{\delta,-} as IhδI_{h}^{\delta}, if not causing any confusions. We have (cf. [3, 16])

(29) d(Ihdω)=Ihd(dω),ωHΛj(Ω), 0jd,\,{\rm d}(I_{h}^{\,{\rm d}}\omega)=I_{h}^{\,{\rm d}}(\,{\rm d}\omega),\quad\omega\in H\Lambda^{j}(\Omega),\,0\leq j\leq d,
(30) δ(Ihδω)=Ihδ(δω),ωHΛj(Ω), 0jd,\delta(I_{h}^{\delta}\omega)=I_{h}^{\delta}(\delta\omega),\quad\omega\in H^{*}\Lambda^{j}(\Omega),\,0\leq j\leq d,
(31) ωIk,hd,ω+ωIk,hδ,ωhk|ω|k,ωHkΛj(Ω),\|\omega-I_{k,h}^{\,{\rm d},-}\omega\|+\|\omega-I_{k,h}^{\delta,-}\omega\|\lesssim h^{k}|\omega|_{k},\quad\omega\in H^{k}\Lambda^{j}(\Omega),
(32) ωIk,hdω+hd(ωIk,hdω)hk+1|ω|k+1,ωHk+1Λj(Ω).\|\omega-I_{k,h}^{\,{\rm d}}\omega\|+h\|\,{\rm d}(\omega-I_{k,h}^{\,{\rm d}}\omega)\|\lesssim h^{k+1}|\omega|_{k+1},\quad\omega\in H^{k+1}\Lambda^{j}(\Omega).

We next introduce a Fortin operator Ih:H01Λj+1(Ω)ΦhI_{h}:H_{0}^{1}\Lambda^{j+1}(\Omega)\rightarrow\Phi_{h} based on DoFs (25) as follows:

Ihv(v)\displaystyle I_{h}v(\texttt{v}) =1|𝒯v|T𝒯v(Qk,Tv)(v),\displaystyle=\frac{1}{|\mathcal{T}_{\texttt{v}}|}\sum_{T\in\mathcal{T}_{\texttt{v}}}(Q_{k,T}v)(\texttt{v}), vΔ0(𝒯̊h),\displaystyle\texttt{v}\in\Delta_{0}(\mathring{\mathcal{T}}_{h}),
(Ihv,q)f\displaystyle(I_{h}v,q)_{f} =1|𝒯f|T𝒯f(Qk,Tv,q)f,\displaystyle=\frac{1}{|\mathcal{T}_{f}|}\sum_{T\in\mathcal{T}_{f}}(Q_{k,T}v,q)_{f}, qk1Λj+1(f),fΔ(𝒯̊h),1d2,\displaystyle q\in\mathbb{P}_{k-\ell-1}\Lambda^{j+1}(f),\,f\in\Delta_{\ell}(\mathring{\mathcal{T}}_{h}),1\leq\ell\leq d-2,
(Ihv,q)F\displaystyle(I_{h}v,q)_{F} =(v,q)F,\displaystyle=(v,q)_{F}, qkdΛj+1(F),FΔd1(𝒯̊h),\displaystyle q\in\mathbb{P}_{k-d}\Lambda^{j+1}(F),F\in\Delta_{d-1}(\mathring{\mathcal{T}}_{h}),
(Ihv,q)T\displaystyle(I_{h}v,q)_{T} =(v,q)T,\displaystyle=(v,q)_{T}, qkd1Λj+1(T)+δkΛj+2(T),T𝒯h.\displaystyle q\in\mathbb{P}_{k-d-1}\Lambda^{j+1}(T)+\delta\mathbb{P}_{k}\Lambda^{j+2}(T),T\in\mathcal{T}_{h}.
Remark 3.2.

Let j=d1j=d-1. Since DoF (25c) includes the moment Tvdx\int_{T}v\,{\rm d}x, we have IhvL02Λd(Ω)I_{h}v\in L_{0}^{2}\Lambda^{d}(\Omega) for vH01Λd(Ω)L02Λd(Ω)v\in H_{0}^{1}\Lambda^{d}(\Omega)\cap L_{0}^{2}\Lambda^{d}(\Omega).

Applying the integration by parts, we have from the definition of IhI_{h} that

(33) (d(vIhv),q)=(vIhv,δq)=0vH01Λj+1(Ω),qVk,hδ,Λj+2.(\,{\rm d}(v-I_{h}v),q)=(v-I_{h}v,\delta q)=0\quad\forall\leavevmode\nobreak\ v\in H_{0}^{1}\Lambda^{j+1}(\Omega),\,q\in V_{k,h}^{\delta,-}\Lambda^{j+2}.
Lemma 3.3.

Let 0jd10\leq j\leq d-1. Let vH01Λj+1(Ω)HsΛj+1(Ω)v\in H_{0}^{1}\Lambda^{j+1}(\Omega)\cap H^{s}\Lambda^{j+1}(\Omega) with 1sk+11\leq s\leq k+1. We have for 0ms0\leq m\leq s and T𝒯hT\in\mathcal{T}_{h} that

(34) |vIhv|m,ThTsm|v|s,ωT.\displaystyle|v-I_{h}v|_{m,T}\lesssim h_{T}^{s-m}|v|_{s,\omega_{T}}.
Proof.

Following the argument in [32, 9, 24], we use the inverse inequality, the norm equivalence on the shape function space and the trace inequality to get

hTm|Qk,TvIhv|m,T\displaystyle h_{T}^{m}|Q_{k,T}v-I_{h}v|_{m,T} Qk,TvIhvT\displaystyle\lesssim\|Q_{k,T}v-I_{h}v\|_{T}
hT1/2Qk,Tvv0,T+vΔ0(T)hTd/2|(Qk,TvIhv)(v)|\displaystyle\lesssim h_{T}^{1/2}\|Q_{k,T}v-v\|_{0,\partial T}+\sum_{\texttt{v}\in\Delta_{0}(T)}h_{T}^{d/2}|(Q_{k,T}v-I_{h}v)(\texttt{v})|
+=1d2fΔ(T)hT(d)/2Qk1,f(Qk,TvIhv)F\displaystyle\quad+\sum_{\ell=1}^{d-2}\sum_{f\in\Delta_{\ell}(T)}h_{T}^{(d-\ell)/2}\|Q_{k-\ell-1,f}(Q_{k,T}v-I_{h}v)\|_{F}
vΔ0(T)K𝒯v(Qk,Kvv0,K+hT|Qk,Kvv|1,K),\displaystyle\lesssim\sum_{\texttt{v}\in\Delta_{0}(T)}\sum_{K\in\mathcal{T}_{\texttt{v}}}(\|Q_{k,K}v-v\|_{0,K}+h_{T}|Q_{k,K}v-v|_{1,K}),

which together with the triangle inequality and the error estimate of Qk,KQ_{k,K} to derive (34). ∎

We present the discrete coercivity in the following lemma.

Lemma 3.4.

Let 0jd20\leq j\leq d-2. For (ψ,s)Φh×(Vk,hδ,Λj+3/δVk,hδ,Λj+4)(\psi,s)\in\Phi_{h}\times(V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}) satisfying b(ψ,s;q)=0b(\psi,s;q)=0 for all qVk,hδ,Λj+2q\in V_{k,h}^{\delta,-}\Lambda^{j+2}, we have the discrete coercivity

(35) ψ12+s12a(ψ,s;ψ,s).\displaystyle\|\psi\|_{1}^{2}+\|s\|_{1}^{2}\lesssim a(\psi,s;\psi,s).
Proof.

By complex (28), take q=δsq=\delta s in b(ψ,s;q)=0b(\psi,s;q)=0 to acquire s=0s=0. Then (35) holds from the Poincaré inequality. ∎

To prove the discrete inf-sup condition, we first establish an L2L^{2}-stable decomposition of space Vk,hδ,Λj+2V_{k,h}^{\delta,-}\Lambda^{j+2} with the help of IhδI_{h}^{\delta}.

Lemma 3.5.

Let 0jd20\leq j\leq d-2. For phVk,hδ,Λj+2p_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+2}, these exist qhVk,hδ,Λj+2q_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+2} and rhVk,hδ,Λj+3r_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+3} such that

ph=qh+δrh,p_{h}=q_{h}+\delta r_{h},
phqh+δrh,qhδph1.\|p_{h}\|\eqsim\|q_{h}\|+\|\delta r_{h}\|,\quad\|q_{h}\|\lesssim\|\delta p_{h}\|_{-1}.
Proof.

By Theorem 1.1 in [17], there exist a qHΛj+2(Ω)q\in H^{*}\Lambda^{j+2}(\Omega) such that

δq=δph,qδph1.\delta q=\delta p_{h},\quad\|q\|\lesssim\|\delta p_{h}\|_{-1}.

Set qh=IhδqVk,hδ,Λj+2q_{h}=I_{h}^{\delta}q\in V_{k,h}^{\delta,-}\Lambda^{j+2}. Since IhδI_{h}^{\delta} is an L2L^{2} bounded projection operator and δqδVk,hδ,Λj+2\delta q\in\delta V_{k,h}^{\delta,-}\Lambda^{j+2}, by (30) we have

δqh=Ihδ(δq)=δph,qhqδph1ph.\delta q_{h}=I_{h}^{\delta}(\delta q)=\delta p_{h},\quad\|q_{h}\|\lesssim\|q\|\lesssim\|\delta p_{h}\|_{-1}\lesssim\|p_{h}\|.

By complex (28), we obtain ph=qh+δrhp_{h}=q_{h}+\delta r_{h} with rhVk,hδ,Λj+3/δVk,hδ,Λj+4r_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}, which ends the proof. ∎

Lemma 3.6.

Let 0jd20\leq j\leq d-2. It holds the L2L^{2} norm equivalence

(36) phδph1+supshVk,hδ,Λj+3/δVk,hδ,Λj+4(ph,δsh)δshphVk,hδ,Λj+2.\|p_{h}\|\eqsim\|\delta p_{h}\|_{-1}+\sup_{s_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}}\frac{(p_{h},\delta s_{h})}{\|\delta s_{h}\|}\quad\;\;\forall\leavevmode\nobreak\ p_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+2}.
Proof.

Apply Lemma 3.5 to php_{h}, then

phqh+δrhδph1+δrh.\|p_{h}\|\leq\|q_{h}\|+\|\delta r_{h}\|\lesssim\|\delta p_{h}\|_{-1}+\|\delta r_{h}\|.

On the other side,

δrh\displaystyle\|\delta r_{h}\| =supshVk,hδ,Λj+3/δVk,hδ,Λj+4(δrh,δsh)δsh=supshVk,hδ,Λj+3/δVk,hδ,Λj+4(phqh,δsh)δsh\displaystyle=\sup_{s_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}}\frac{(\delta r_{h},\delta s_{h})}{\|\delta s_{h}\|}=\sup_{s_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}}\frac{(p_{h}-q_{h},\delta s_{h})}{\|\delta s_{h}\|}
qh+supshVk,hδ,Λj+3/δVk,hδ,Λj+4(ph,δsh)δsh\displaystyle\leq\|q_{h}\|+\sup_{s_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}}\frac{(p_{h},\delta s_{h})}{\|\delta s_{h}\|}
δph1+supshVk,hδ,Λj+3/δVk,hδ,Λj+4(ph,δsh)δsh.\displaystyle\lesssim\|\delta p_{h}\|_{-1}+\sup_{s_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}}\frac{(p_{h},\delta s_{h})}{\|\delta s_{h}\|}.

Combining the last two inequalities yields (36). ∎

We are now in a position to derive the discrete inf-sup condition.

Lemma 3.7.

Let 0jd20\leq j\leq d-2. For qVk,hδ,Λj+2q\in V_{k,h}^{\delta,-}\Lambda^{j+2}, it holds the discrete inf-sup condition

(37) qsupψΦh,sVk,hδ,Λj+3/δVk,hδ,Λj+4b(ψ,s;q)ψ1+δs.\displaystyle\|q\|\lesssim\sup_{\psi\in\Phi_{h},\,s\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}}\frac{b(\psi,s;q)}{\|\psi\|_{1}+\|\delta s\|}.
Proof.

Employ (33) and (34) with m=s=1m=s=1 to have

δq1\displaystyle\|\delta q\|_{-1} =supϕH01Λj+1(Ω)(dϕ,q)ϕ1=supϕH01Λj+1(Ω)(d(Ihϕ),q)ϕ1\displaystyle=\sup_{\phi\,\in H_{0}^{1}\Lambda^{j+1}(\Omega)}\frac{(\,{\rm d}\phi,q)}{\|\phi\|_{1}}\,=\sup_{\phi\,\in H_{0}^{1}\Lambda^{j+1}(\Omega)}\frac{(\,{\rm d}(I_{h}\phi),q)}{\|\phi\|_{1}}
supϕH01Λj+1(Ω)(d(Ihϕ),q)Ihϕ1supψΦh(dψ,q)ψ1\displaystyle\lesssim\sup_{\phi\,\in H_{0}^{1}\Lambda^{j+1}(\Omega)}\frac{(\,{\rm d}(I_{h}\phi),q)}{\|I_{h}\phi\|_{1}}\,\leq\sup_{\psi\,\in\Phi_{h}}\frac{(\,{\rm d}\psi,q)}{\|\psi\|_{1}}
supψΦh,sVk,hδ,Λj+3/δVk,hδ,Λj+4b(ψ,s;q)ψ1+δs.\displaystyle\leq\sup_{\psi\,\in\Phi_{h},\,\,s\,\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}}\frac{b(\psi,s;q)}{\|\psi\|_{1}+\|\delta s\|}.

On the other hand,

supsVk,hδ,Λj+3/δVk,hδ,Λj+4(q,δs)δssupψΦh,sVk,hδ,Λj+3/δVk,hδ,Λj+4b(ψ,s;q)ψ1+δs.\displaystyle\sup_{s\,\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}}\frac{(q,\delta s)}{\|\delta s\|}\,\leq\sup_{\psi\,\in\Phi_{h},\,\,s\,\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}}\frac{b(\psi,s;q)}{\|\psi\|_{1}+\|\delta s\|}.

Therefore, (37) follows from (36) and the last two inequalities. ∎

3.2. Decoupled discrete methods

Now we propose a family of finite element methods for the decoupled formulation (19) with 0jd10\leq j\leq d-1: find whV̊k,hdΛjw_{h}\in\mathring{V}_{k,h}^{\,{\rm d}}\Lambda^{j}, ϕhΦh\phi_{h}\in\Phi_{h}, phVk,hδ,Λj+2p_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+2}, rhVk,hδ,Λj+3/δVk,hδ,Λj+4r_{h}\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}, uhV̊k+1,hd,Λju_{h}\in\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j} and λh,zhV̊k+1,hd,Λj1/dV̊k+1,hd,Λj2\lambda_{h},z_{h}\in\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j-1}/\,{\rm d}\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j-2} such that

(38a) (dwh,dv)+(v,dλh)\displaystyle(\,{\rm d}w_{h},\,{\rm d}v)+(v,\,{\rm d}\lambda_{h}) =(f,v)\displaystyle=(f,v) vV̊k,hdΛj,\displaystyle\forall\,\,v\in\mathring{V}_{k,h}^{\,{\rm d}}\Lambda^{j},
(38b) (wh,dη)\displaystyle(w_{h},\,{\rm d}\eta) =0\displaystyle=0 ηV̊k+1,hd,Λj1,\displaystyle\forall\,\,\eta\in\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j-1},
(38c) (ϕh,ψ)+(dψ+δs,ph)\displaystyle(\nabla\phi_{h},\nabla\psi)+(\,{\rm d}\psi+\delta s,p_{h}) =(dwh,ψ)\displaystyle=(\,{\rm d}w_{h},\psi) ψΦh,sVk,hδ,Λj+3,\displaystyle\forall\,\,\psi\in\Phi_{h},s\in V_{k,h}^{\delta,-}\Lambda^{j+3},
(38d) (dϕh+δrh,q)\displaystyle(\,{\rm d}\phi_{h}+\delta r_{h},q) =0\displaystyle=0 qVk,hδ,Λj+2,\displaystyle\forall\,\,q\in V_{k,h}^{\delta,-}\Lambda^{j+2},
(38e) (duh,dχ)+(χ,dzh)\displaystyle(\,{\rm d}u_{h},\,{\rm d}\chi)+(\chi,\,{\rm d}z_{h}) =(ϕh,dχ)\displaystyle=(\phi_{h},\,{\rm d}\chi) χV̊k+1,hd,Λj,\displaystyle\forall\,\,\chi\in\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j},
(38f) (uh,dμ)\displaystyle(u_{h},\,{\rm d}\mu) =0\displaystyle=0 μV̊k+1,hd,Λj1.\displaystyle\forall\,\,\mu\in\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j-1}.
Here Vk,hδ,Λj+2={0}V_{k,h}^{\delta,-}\Lambda^{j+2}=\{0\} and Vk,hδ,Λj+3={0}V_{k,h}^{\delta,-}\Lambda^{j+3}=\{0\} for j=d1j=d-1.
Theorem 3.8.

For 0jd10\leq j\leq d-1, the decoupled finite element method (38) is well-posed, and rh=0r_{h}=0, zh=0z_{h}=0.

Proof.

Both (38a)-(38b) and (38e)-(38f) are the mixed finite element methods for second-order exterior differential equations, whose wellposedness follows from the exactness of complexes (26)-(27).

When j=d1j=d-1, the discretizaiton (38c)-(38d) is a conforming finite element method of the Poisson equation, which is well-posed. For 0jd20\leq j\leq d-2, by applying the Babuša-Brezzi theory [8], the well-posedness of the mixed finite element method (38c)-(38d) follows from the the discrete coercivity (35), and the discrete inf-sup conditions (37).

Finally, take q=δrhq=\delta r_{h} in (38d) to achieve rh=0r_{h}=0, and χ=dzh\chi=\,{\rm d}z_{h} in (38e) to achieve zh=0z_{h}=0. ∎

Example 3.9.

Taking j=0j=0 and d=3d=3, the mixed finite element method (38c)-(38d) is to find ϕhΦh\phi_{h}\in\Phi_{h}, phVk,hNDp_{h}\in V_{k,h}^{\rm ND} and rhVk,hgrad/r_{h}\in V_{k,h}^{\operatorname{grad}}/\mathbb{R} such that

(ϕh,ψ)+(curlψ+s,ph)\displaystyle(\nabla\phi_{h},\nabla\psi)+(\operatorname{curl}\psi+\nabla s,p_{h}) =(wh,ψ)ψΦh,sVk,hgrad/,\displaystyle=(\nabla w_{h},\psi)\quad\;\;\forall\leavevmode\nobreak\ \psi\in\Phi_{h},s\in V_{k,h}^{\operatorname{grad}}/\mathbb{R},
(curlϕh+rh,q)\displaystyle(\operatorname{curl}\phi_{h}+\nabla r_{h},q) =0qVk,hND,\displaystyle=0\qquad\qquad\quad\forall\leavevmode\nobreak\ q\in V_{k,h}^{\rm ND},

where Vk,hgrad=Vk,hδ,Λ3V_{k,h}^{\operatorname{grad}}=V_{k,h}^{\delta,-}\Lambda^{3} is the Lagrange element space, and Vk,hND=Vk,hδ,Λ2V_{k,h}^{\rm ND}=V_{k,h}^{\delta,-}\Lambda^{2} is the Nédélec element space of the first kind [31].

Example 3.10.

Taking j=1j=1 and d=3d=3, the mixed finite element method (38c)-(38d) is to find ϕhΦh\phi_{h}\in\Phi_{h} and phVk,hgrad/p_{h}\in V_{k,h}^{\operatorname{grad}}/\mathbb{R} such that

(ϕh,ψ)+(divψ,ph)\displaystyle(\nabla\phi_{h},\nabla\psi)+(\operatorname{div}\psi,p_{h}) =(curlwh,ψ)ψΦh,\displaystyle=(\operatorname{curl}w_{h},\psi)\quad\;\;\forall\leavevmode\nobreak\ \psi\in\Phi_{h},
(divϕh,q)\displaystyle(\operatorname{div}\phi_{h},q) =0qVk,hgrad/.\displaystyle=0\qquad\qquad\qquad\forall\leavevmode\nobreak\ q\in V_{k,h}^{\operatorname{grad}}/\mathbb{R}.

This is exactly the MINI element method for Stokes equation in [8, Section 8.7.1].

Example 3.11.

Taking j=d1j=d-1, the finite element method (38c)-(38d) is to find ϕhΦh\phi_{h}\in\Phi_{h} such that

(ϕh,ψ)=(divwh,ψ)ψΦh.(\nabla\phi_{h},\nabla\psi)=(\operatorname{div}w_{h},\psi)\quad\forall\leavevmode\nobreak\ \psi\in\Phi_{h}.

This is a conforming finite element method of the Poisson equation.

3.3. Error analysis

Next we present the error analysis of the decoupled finite element method (38).

Lemma 3.12.

Let 0jd10\leq j\leq d-1. Let (w,λ)H0Λj×H0Λj1(w,\lambda)\in H_{0}\Lambda^{j}\times H_{0}\Lambda^{j-1} be the solution of problem (19a)-(19b), and (wh,λh)V̊k,hdΛj×V̊k+1,hd,Λj1(w_{h},\lambda_{h})\in\mathring{V}_{k,h}^{\,{\rm d}}\Lambda^{j}\times\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j-1} be the solution of the mixed finite element method (38a)-(38b). Assume wHk+1Λjw\in H^{k+1}\Lambda^{j} and λHk+2Λj1\lambda\in H^{k+2}\Lambda^{j-1}. We have

(39) d(λλh)\displaystyle\|\,{\rm d}(\lambda-\lambda_{h})\| hk+1|dλ|k+1,\displaystyle\lesssim h^{k+1}|\,{\rm d}\lambda|_{k+1},
(40) λλh\displaystyle\|\lambda-\lambda_{h}\| hk+1(|λ|k+1+|dλ|k+1),\displaystyle\lesssim h^{k+1}(|\lambda|_{k+1}+|\,{\rm d}\lambda|_{k+1}),
(41) d(wwh)\displaystyle\|\,{\rm d}(w-w_{h})\| hk(|dw|k+h|dλ|k+1),\displaystyle\lesssim h^{k}(|\,{\rm d}w|_{k}+h|\,{\rm d}\lambda|_{k+1}),
(42) wwh\displaystyle\|w-w_{h}\| hk(|w|k+|dw|k+h|dλ|k+1).\displaystyle\lesssim h^{k}(|w|_{k}+|\,{\rm d}w|_{k}+h|\,{\rm d}\lambda|_{k+1}).
Proof.

Subtract (38a)-(38b) from (19a)-(19b) to get the error equations

(43) (d(wwh),dv)+(v,d(λλh))\displaystyle(\,{\rm d}(w-w_{h}),\,{\rm d}v)+(v,\,{\rm d}(\lambda-\lambda_{h})) =0\displaystyle=0 vV̊k,hdΛj,\displaystyle\forall\,\,v\in\mathring{V}_{k,h}^{\,{\rm d}}\Lambda^{j},
(44) (wwh,dη)\displaystyle(w-w_{h},\,{\rm d}\eta) =0\displaystyle=0 ηV̊k+1,hd,Λj1.\displaystyle\forall\,\,\eta\in\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j-1}.

Take v=dμv=\,{\rm d}\mu in error equation (43) to get the Galerkin orthogonality

(45) (d(λλh),dμ)=0μV̊k+1,hd,Λj1.(\,{\rm d}(\lambda-\lambda_{h}),\,{\rm d}\mu)=0\quad\forall\leavevmode\nobreak\ \mu\in\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j-1}.

Then it holds from (29) and (32) that

d(λλh)dλIhd(dλ)hk+1|dλ|k+1.\|\,{\rm d}(\lambda-\lambda_{h})\|\leq\|\,{\rm d}\lambda-I_{h}^{\,{\rm d}}(\,{\rm d}\lambda)\|\lesssim h^{k+1}|\,{\rm d}\lambda|_{k+1}.

Thus, (39) is true. Choosing v=Ihdwwhv=I_{h}^{\,{\rm d}}w-w_{h} in (43), we acquire from (45), the discrete Poincaré inequality and (29) that

d(wwh)2\displaystyle\|\,{\rm d}(w-w_{h})\|^{2} =(d(wwh),d(wIhdw))(Ihdwwh,d(λλh))\displaystyle=(\,{\rm d}(w-w_{h}),\,{\rm d}(w-I_{h}^{\,{\rm d}}w))-(I_{h}^{\,{\rm d}}w-w_{h},\,{\rm d}(\lambda-\lambda_{h}))
=(d(wwh),d(wIhdw))\displaystyle=(\,{\rm d}(w-w_{h}),\,{\rm d}(w-I_{h}^{\,{\rm d}}w))
+infμV̊k+1,hd,Λj1(Ihdwwhdμ,d(λhλ))\displaystyle\quad+\inf_{\mu\in\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j-1}}(I_{h}^{\,{\rm d}}w-w_{h}-\,{\rm d}\mu,\,{\rm d}(\lambda_{h}-\lambda))
d(wwh)dwIhd(dw)+Ihd(dw)dwhd(λλh).\displaystyle\lesssim\|\,{\rm d}(w-w_{h})\|\|\,{\rm d}w-I_{h}^{\,{\rm d}}(\,{\rm d}w)\|+\|I_{h}^{\,{\rm d}}(\,{\rm d}w)-\,{\rm d}w_{h}\|\|\,{\rm d}(\lambda-\lambda_{h})\|.

Hence, it follows

d(wwh)dwIhd(dw)+d(λλh),\|\,{\rm d}(w-w_{h})\|\lesssim\|\,{\rm d}w-I_{h}^{\,{\rm d}}(\,{\rm d}w)\|+\|\,{\rm d}(\lambda-\lambda_{h})\|,

which together with (32) and (39) implies (41).

Finally, by employing the triangle inequality, the discrete Poincaré inequalities, (29) and the interpolation estimates (31)-(32), we conclude estimates (40) and (42) from estimates (39) and (41), respectively. ∎

Lemma 3.13.

Let 0jd10\leq j\leq d-1. Let (ϕ,p,0)H01Λj+1×L2Λj+2×HΛj+3(\phi,p,0)\in H_{0}^{1}\Lambda^{j+1}\times L^{2}\Lambda^{j+2}\times H^{*}\Lambda^{j+3} be the solution of problem (19c)-(19d), and (ϕh,ph,0)Φh×Vk,hδ,Λj+2×Vk,hδ,Λj+3(\phi_{h},p_{h},0)\in\Phi_{h}\times V_{k,h}^{\delta,-}\Lambda^{j+2}\times V_{k,h}^{\delta,-}\Lambda^{j+3} the solution of the mixed finite element method (38c)-(38d). Assume ϕHk+1Λj+1\phi\in H^{k+1}\Lambda^{j+1} and pHkΛj+2p\in H^{k}\Lambda^{j+2}. We have

(46) ϕϕh1+pphhk(|ϕ|k+1+|p|k)+d(wwh).\|\phi-\phi_{h}\|_{1}+\|p-p_{h}\|\lesssim h^{k}(|\phi|_{k+1}+|p|_{k})+\|\,{\rm d}(w-w_{h})\|.
Proof.

Subtract (38c)-(38d) from (19c)-(19d) to get the error equations

(47) ((ϕϕh),ψ)+(dψ,pph)\displaystyle(\nabla(\phi-\phi_{h}),\nabla\psi)+(\,{\rm d}\psi,p-p_{h}) =(d(wwh),ψ)\displaystyle=(\,{\rm d}(w-w_{h}),\psi) ψΦh,\displaystyle\forall\,\,\psi\in\Phi_{h},
(48) (d(ϕϕh),q)=(d(Ihϕϕh),q)\displaystyle(\,{\rm d}(\phi-\phi_{h}),q)=(\,{\rm d}(I_{h}\phi-\phi_{h}),q) =0\displaystyle=0 qVk,hδ,Λj+2,\displaystyle\forall\,\,q\in V_{k,h}^{\delta,-}\Lambda^{j+2},
(49) (pph,δs)\displaystyle(p-p_{h},\delta s) =0\displaystyle=0 sVk,hδ,Λj+3.\displaystyle\forall\,\,s\in V_{k,h}^{\delta,-}\Lambda^{j+3}.

We have used (33) in deriving (48). Choosing ψ=Ihϕϕh\psi=I_{h}\phi-\phi_{h} in (47), it follows from (48) that

((ϕϕh),(Ihϕϕh))+(d(Ihϕϕh),pIhδp)=(d(wwh),Ihϕϕh),(\nabla(\phi-\phi_{h}),\nabla(I_{h}\phi-\phi_{h}))+(\,{\rm d}(I_{h}\phi-\phi_{h}),p-I_{h}^{\delta}p)=(\,{\rm d}(w-w_{h}),I_{h}\phi-\phi_{h}),

which implies

(50) |ϕϕh|1|ϕIhϕ|1+pIhδp+d(wwh).|\phi-\phi_{h}|_{1}\lesssim|\phi-I_{h}\phi|_{1}+\|p-I_{h}^{\delta}p\|+\|\,{\rm d}(w-w_{h})\|.

Next estimate pph\|p-p_{h}\|. For ψΦh\psi\in\Phi_{h} and sVk,hδ,Λj+3/δVk,hδ,Λj+4s\in V_{k,h}^{\delta,-}\Lambda^{j+3}/\delta V_{k,h}^{\delta,-}\Lambda^{j+4}, we have from (47) and (49) that

b(ψ,s;Ihδpph)\displaystyle b(\psi,s;I_{h}^{\delta}p-p_{h}) =b(ψ,s;Ihδpp)((ϕϕh),ψ)+(d(wwh),ψ).\displaystyle=b(\psi,s;I_{h}^{\delta}p-p)-(\nabla(\phi-\phi_{h}),\nabla\psi)+(\,{\rm d}(w-w_{h}),\psi).

Then

b(ψ,s;Ihδpph)(pIhδp+|ϕϕh|1+d(wwh))(|ψ|1+δs).b(\psi,s;I_{h}^{\delta}p-p_{h})\lesssim(\|p-I_{h}^{\delta}p\|+|\phi-\phi_{h}|_{1}+\|\,{\rm d}(w-w_{h})\|)(|\psi|_{1}+\|\delta s\|).

Employ the discrete inf-sup condition (37) to acquire

pph\displaystyle\|p-p_{h}\| pIhδp+IhδpphpIhδp+|ϕϕh|1+d(wwh).\displaystyle\leq\|p-I_{h}^{\delta}p\|+\|I_{h}^{\delta}p-p_{h}\|\lesssim\|p-I_{h}^{\delta}p\|+|\phi-\phi_{h}|_{1}+\|\,{\rm d}(w-w_{h})\|.

Together with (50), we arrive at

|ϕϕh|1+pph|ϕIhϕ|1+pIhδp+d(wwh).|\phi-\phi_{h}|_{1}+\|p-p_{h}\|\lesssim|\phi-I_{h}\phi|_{1}+\|p-I_{h}^{\delta}p\|+\|\,{\rm d}(w-w_{h})\|.

Finally, we conclude (46) from (34) and the error estimate (31) of IhδI_{h}^{\delta}. ∎

Remark 3.14.

The error estimate of pph\|p-p_{h}\| for the MINI element method of Stokes equation is suboptimal [5]. While the error estimate (46) for pph\|p-p_{h}\| is optimal when jd3j\leq d-3.

We will use the duality argument to estimate ϕϕh\|\phi-\phi_{h}\|. Consider the dual problem: find ϕ^H01Λj+1\hat{\phi}\in H_{0}^{1}\Lambda^{j+1} and p^L2Λj+2/δHΛj+3\hat{p}\in L^{2}\Lambda^{j+2}/\delta H^{*}\Lambda^{j+3} such that

(51) {Δϕ^+δp^=ϕϕhinΩ,dϕ^=0inΩ.\left\{\begin{aligned} -\Delta\hat{\phi}+\delta\hat{p}&=\phi-\phi_{h}\qquad\mathrm{in}\,\,\Omega,\\ \,{\rm d}\hat{\phi}&=0\qquad\qquad\,\,\,\mathrm{in}\,\,\Omega.\end{aligned}\right.

We assume problem (51) has the regularity

(52) ϕ^2+p^1ϕϕh.\|\hat{\phi}\|_{2}+\|\hat{p}\|_{1}\lesssim\|\phi-\phi_{h}\|.

When Ω\Omega is a convex polytope in two and three dimensions, the regularity (52) holds for j=d2,d1j=d-2,d-1 (cf. [30, 20]). By dϕ^=0\,{\rm d}\hat{\phi}=0, there exists a u^H01Λj\hat{u}\in H_{0}^{1}\Lambda^{j} [17] such that

(53) ϕ^=du^,andu^1ϕ^.\hat{\phi}=\,{\rm d}\hat{u},\quad\textrm{and}\quad\|\hat{u}\|_{1}\lesssim\|\hat{\phi}\|.
Lemma 3.15.

Let (ϕ,p,0)H01Λj+1(Ω)×L2Λj+2(Ω)×HΛj+3(Ω)(\phi,p,0)\in H_{0}^{1}\Lambda^{j+1}(\Omega)\times L^{2}\Lambda^{j+2}(\Omega)\times H^{*}\Lambda^{j+3}(\Omega) be the solution of problem (19c)-(19d), and (ϕh,ph,0)Φh×Vk,hδ,Λj+2×Vk,hδ,Λj+3(\phi_{h},p_{h},0)\in\Phi_{h}\times V_{k,h}^{\delta,-}\Lambda^{j+2}\times V_{k,h}^{\delta,-}\Lambda^{j+3} be the solution of the mixed finite element method (38c)-(38d) for 0jd10\leq j\leq d-1. Assume ϕHk+1Λj+1\phi\in H^{k+1}\Lambda^{j+1}, pHkΛj+2p\in H^{k}\Lambda^{j+2} and regularity (52) holds. We have

(54) ϕϕhhk+1(|ϕ|k+1+|p|k)+hd(wwh)+d(λλh).\|\phi-\phi_{h}\|\lesssim h^{k+1}(|\phi|_{k+1}+|p|_{k})+h\|\,{\rm d}(w-w_{h})\|+\|\,{\rm d}(\lambda-\lambda_{h})\|.
Proof.

Applying the integration by parts, we get from the error equation (48), the error estimate (34) of IhI_{h} and the error estimate (31) of IhδI_{h}^{\delta} that

ϕϕh2\displaystyle\|\phi-\phi_{h}\|^{2} =(ϕϕh,Δϕ^+δp^)=((ϕϕh),ϕ^)+(d(ϕϕh),p^)\displaystyle=(\phi-\phi_{h},-\Delta\hat{\phi}+\delta\hat{p})=(\nabla(\phi-\phi_{h}),\nabla\hat{\phi})+(\,{\rm d}(\phi-\phi_{h}),\hat{p})
=((ϕϕh),(ϕ^Ihϕ^))+(d(ϕϕh),p^Ihδp^)\displaystyle=(\nabla(\phi-\phi_{h}),\nabla(\hat{\phi}-I_{h}\hat{\phi}))+(\,{\rm d}(\phi-\phi_{h}),\hat{p}-I_{h}^{\delta}\hat{p})
+((ϕϕh),(Ihϕ^))\displaystyle\quad+(\nabla(\phi-\phi_{h}),\nabla(I_{h}\hat{\phi}))
h|ϕϕh|1(|ϕ^|2+|p^|1)+((ϕϕh),(Ihϕ^)).\displaystyle\lesssim h|\phi-\phi_{h}|_{1}(|\hat{\phi}|_{2}+|\hat{p}|_{1})+(\nabla(\phi-\phi_{h}),\nabla(I_{h}\hat{\phi})).

On the other side, by error equation (47), (53), error equation (43), (34) and (32),

((ϕϕh),(Ihϕ^))\displaystyle(\nabla(\phi-\phi_{h}),\nabla(I_{h}\hat{\phi})) =(d(wwh),Ihϕ^)(d(Ihϕ^),pph)\displaystyle=(\,{\rm d}(w-w_{h}),I_{h}\hat{\phi})-(\,{\rm d}(I_{h}\hat{\phi}),p-p_{h})
=(d(wwh),Ihϕ^ϕ^)(d(Ihϕ^ϕ^),pph)\displaystyle=(\,{\rm d}(w-w_{h}),I_{h}\hat{\phi}-\hat{\phi})-(\,{\rm d}(I_{h}\hat{\phi}-\hat{\phi}),p-p_{h})
+(d(wwh),d(u^Ihdu^))(Ihdu^,d(λλh))\displaystyle\quad+(\,{\rm d}(w-w_{h}),\,{\rm d}(\hat{u}-I_{h}^{\,{\rm d}}\hat{u}))-(I_{h}^{\,{\rm d}}\hat{u},\,{\rm d}(\lambda-\lambda_{h}))
hd(wwh)ϕ^2+hpph|ϕ^|2+d(λλh)ϕ^.\displaystyle\lesssim h\|\,{\rm d}(w-w_{h})\|\|\hat{\phi}\|_{2}+h\|p-p_{h}\||\hat{\phi}|_{2}+\|\,{\rm d}(\lambda-\lambda_{h})\|\|\hat{\phi}\|.

Therefore (54) follows from the last two inequalities, regularity (52) and (46). ∎

Theorem 3.16.

Let 0jd10\leq j\leq d-1. Let (w,λ,ϕ,p,0,u,0)(w,\lambda,\phi,p,0,u,0) be the solution of the decoupled formulation (19), and (wh,λh,ϕh,ph,0,uh,0)(w_{h},\lambda_{h},\phi_{h},p_{h},0,u_{h},0) the solution of the decoupled finite element method (38). Assume uHk+2Λju\in H^{k+2}\Lambda^{j}. Then

(55) d(uuh)\displaystyle\|\,{\rm d}(u-u_{h})\| hk+1|du|k+1+ϕϕh,\displaystyle\lesssim h^{k+1}|\,{\rm d}u|_{k+1}+\|\phi-\phi_{h}\|,
(56) uuh\displaystyle\|u-u_{h}\| hk+1(|u|k+1+|du|k+1)+ϕϕh.\displaystyle\lesssim h^{k+1}(|u|_{k+1}+|\,{\rm d}u|_{k+1})+\|\phi-\phi_{h}\|.

Further, assume ϕHk+1Λj+1\phi\in H^{k+1}\Lambda^{j+1}, pHkΛj+2p\in H^{k}\Lambda^{j+2}, wHk+1Λjw\in H^{k+1}\Lambda^{j}, λHk+2Λj1\lambda\in H^{k+2}\Lambda^{j-1}, and regularity (52) holds, then

(57) d(uuh)\displaystyle\|\,{\rm d}(u-u_{h})\| hk+1(|du|k+1+|ϕ|k+1+|p|k+|dw|k+|dλ|k+1),\displaystyle\lesssim h^{k+1}(|\,{\rm d}u|_{k+1}+|\phi|_{k+1}+|p|_{k}+|\,{\rm d}w|_{k}+|\,{\rm d}\lambda|_{k+1}),
(58) uuh\displaystyle\|u-u_{h}\| hk+1(|u|k+1+|du|k+1+|ϕ|k+1+|p|k+|dw|k+|dλ|k+1).\displaystyle\lesssim h^{k+1}(|u|_{k+1}+|\,{\rm d}u|_{k+1}+|\phi|_{k+1}+|p|_{k}+|\,{\rm d}w|_{k}+|\,{\rm d}\lambda|_{k+1}).
Proof.

To estimate d(uuh)\|\,{\rm d}(u-u_{h})\|, subtract (19e) from (38e) to get the error equation

(d(uuh),dχ)=(ϕϕh,dχ)χV̊k+1,hd,Λj.(\,{\rm d}(u-u_{h}),\,{\rm d}\chi)=(\phi-\phi_{h},\,{\rm d}\chi)\quad\forall\,\,\chi\in\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j}.

Taking χ=IhduuhV̊k+1,hd,Λj\chi=I_{h}^{\,{\rm d}}u-u_{h}\in\mathring{V}_{k+1,h}^{\,{\rm d},-}\Lambda^{j}, we get

d(uuh)2=(d(uuh),d(uIhdu))+(ϕϕh,d(Ihduuh)).\displaystyle\|\,{\rm d}(u-u_{h})\|^{2}=(\,{\rm d}(u-u_{h}),\,{\rm d}(u-I_{h}^{\,{\rm d}}u))+(\phi-\phi_{h},\,{\rm d}(I_{h}^{\,{\rm d}}u-u_{h})).

Then

d(uuh)duIhd(du)+ϕϕh,\|\,{\rm d}(u-u_{h})\|\lesssim\|\,{\rm d}u-I_{h}^{\,{\rm d}}(\,{\rm d}u)\|+\|\phi-\phi_{h}\|,

which combined with (32) gives (55). The estimate (57) holds from (54)-(55), (39) and (41).

Finally, by employing the triangle inequality, the discrete Poincaré inequalities, (29) and interpolation estimate (31), we conclude estimates (56) and (58) from estimates (55) and (57), respectively. ∎

Remark 3.17.

Conforming finite element methods of the decoupled formulation (19) with j=0j=0 and d=2d=2, i.e. the biharmonic equation in two dimensions, are analyzed in [12, Section 4.2].

4. Numerical results

In this section, we will numerically test the decoupled finite element method (38) for j=0j=0 in three dimensions, i.e. problem (1) is the biharmonic equation. Let Ω=(0,1)3\Omega=(0,1)^{3} be the unit cube, and the exact solution of biharmonic equation be

u(𝒙)=sin3(πx1)sin3(πx2)sin3(πx3).\displaystyle u(\boldsymbol{x})=\sin^{3}(\pi x_{1})\sin^{3}(\pi x_{2})\sin^{3}(\pi x_{3}).

The load function ff is analytically computed from problem (1). We utilize uniform tetrahedral meshes on Ω\Omega.

Numerical errors of the decoupled finite element method (38) with k=1k=1 are shown in Table 1 and Table 2. From these two tables we can see that uuh=O(h2)\|u-u_{h}\|=O(h^{2}), |uuh|1=O(h2)|u-u_{h}|_{1}=O(h^{2}), ϕϕh=O(h2)\|\phi-\phi_{h}\|=O(h^{2}) and |ϕϕh|1=O(h)|\phi-\phi_{h}|_{1}=O(h), which agree with the theoretical estimates (46), (54) and (57)-(58).

Table 1. Errors uuh\|u-u_{h}\| and |uuh|1|u-u_{h}|_{1} of the decoupled conforming method (38) with k=1k=1.
hh uuh\|u-u_{h}\|   rate |uuh|1|u-u_{h}|_{1}   rate
222^{-2}   1.30759E-01 -   9.92045E-01 -
232^{-3}   5.04489E-02   1.3740   4.34958E-01   1.1895
242^{-4}   1.42827E-02   1.8206   1.33687E-01   1.7020
252^{-5}   3.67529E-03   1.9583   3.52141E-02   1.9246
Table 2. Errors ϕϕh\|\phi-\phi_{h}\| and |ϕϕh|1|\phi-\phi_{h}|_{1} of the decoupled conforming method (38) with k=1k=1.
hh ϕϕh\|\phi-\phi_{h}\|   rate |ϕϕh|1|\phi-\phi_{h}|_{1}   rate
222^{-2}   1.69698E+00 -   1.10196E+01 -
232^{-3}   7.45455E-01   1.1868   6.38092E+00   0.7882
242^{-4}   2.29390E-01   1.7003   2.83386E+00   1.1710
252^{-5}   6.04572E-02   1.9238   1.37843E+00   1.0397

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