# Discrete Maximum Principle and Energy Stability of the Compact Difference Scheme for Two-Dimensional Allen-Cahn Equation

Authors: Yu Bo; Dan Tian; Xiao Liu; Yuanfeng Jin
Journal: Journal of Function Spaces
 (2022)

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## Abstract

The Allen-Cahn model is discussed mainly in the phase field simulation. The compact difference method will be used to numerically approximate the two-dimensional nonlinear Allen-Cahn equation with initial and boundary value conditions, and then, a fully discrete compact difference scheme with second-order accuracy in time and fourth-order in space is established. And its numerical solution satisfies the discrete maximum principle under the constraints of reasonable space and time steps. On this basis, the energy stability of the scheme is investigated. Finally, numerical examples are given to illustrate the theoretical results.


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## Body

## 1. Introduction

The phase field problem is a mathematical model described by partial differential equations. The numerical simulation of the phase field has always been an important field of research at home and abroad because of its important theoretical and practical significance. In 1979, the Allen-Cahn equation is considered to describe the antiphase boundary of the crystal movement by Allen and Cahn, which describes fluid dynamics problems and reaction diffusion problems in materials science, and the same model on the study of many diffusion phenomena is proposed such as the competition and repulsion of biological populations and the migration process of river beds. For describing the motion of the antiphase boundary in the crystal, since this type of phase field model does not have an accurate solution, different numerical methods are used to simulate. At present, numerical approximation methods about these phase field models include the finite difference method [1–5], finite element method [6–8], and spectral method [9, 10], etc.In the paper, the compact difference method is applied to approximate the two-dimensional nonlinear Allen-Cahn equations with initial boundary conditions numerically.(1)∂u∂t=ε2Δu−fu,x,y∈Ω,t∈0,T,ux,y,0=u0x,y,x,y∈Ω¯,u∂Ω=0,t∈0,T,where u represents the concentration of a metal component in a binary alloy, the positive parameter ε is the interface width, and the nonlinear term fu=u3−u.The energy function on theL2 space is defined as
(2)Eu=∮ΩFu−12ε2uΔudx.Here, Fu=1/4u2−12. One of the intrinsic properties of the Allen-Cahn equation is the energy function that decreases with time:
(3)ddtEu≤0,∀t>0.In 2016, Zhang and Hou [3] considered three discrete schemes of the Allen-Cahn equation that included the stable first-order linear explicit-implicit scheme, stable second-order nonlinear Crank-Nicolson scheme, and stable second-order linear Leap-Frog scheme and proved the discrete maximum principle and energy stability under the condition of these three schemes; a compact difference scheme with second-order accuracy in time and fourth-order accuracy in space for one-dimensional Allen-Cahn equation was established by Tian et al. [2] in 2018, and the energy stability of the scheme was investigated; Wu et al. [11] proposed two ADI schemes for the two-dimensional Allen-Cahn equation: the second-order ADI scheme and fourth-order compact ADI scheme, and the analysis of stability and maximum principle were given, respectively, in 2019; then, Deng et al. [12] adopted the finite difference method in time and barycentric interpolation collocation method in space for the two-dimensional Allen-Cahn equation in 2020, obtained the spatial high precision error, and verified the law of energy decline; the Crank-Nicolson difference scheme with second-order accuracy in time and space for the two-dimensional Allen-Cahn equation was established by Zhang et al. [1] in 2021, which proved the existence and convergence of the solution, and finally verified the discrete maximum principle with a numerical example. Based on the existing finite difference methods and inspired by reference [2], a compact difference scheme with second-order accuracy in time and fourth-order accuracy in space is established for the two-dimensional Allen-Cahn equation. Then, the discrete maximum principle and discrete energy stability are mainly investigated. Compared with the previous work about two-dimensional Allen-Cahn equations, this paper obtains higher accuracy in space than before, and the compact difference method is applied and realized in two-dimensional equations for the first time, which will add an effective and feasible method to this kind of research.


## 2. Establishment of a Compact Difference Scheme for the Two-Dimensional Allen-Chan Equation

Firstly, several commonly used numerical differential formulas are given:Leth>0 and c be two constants.Lemma 1 (see [2]).
Ifgx∈C2c−h,c+h, then
(4)gc=12gc−h+gc+h−h22g″ξ1,c−h<ξ1<c+h.Lemma 2 (see [2]).
Ifgx∈C3c−h,c+h, then
(5)g′c=12hgc+h−gc−h−h26g″ξ2,c−h<ξ2<c+h.Lemma 3 (see [2]).
Ifgx∈C6c−h,c+h, then
(6)112g″c−h+10g″c+g″c+h=1h2gc+h−2gc+gc−h+h4240g6ξ3,c−h<ξ3<c+h.In this section, a compact difference scheme will be establish for the following two-dimensional Allen-Cahn equation as(7)∂u∂t=ε2∂2u∂x2+∂2u∂y2−fu,x,y∈Ω,t∈0,T,ux,y,0=u0x,y,x,y∈Ω¯,u∂Ω=0,t∈0,T,where Ω=0,1×0,1.Let us divide the interval [0, 1] intoM equal parts and [0, 𝑇] into 𝑁 equal parts, the space step is h=1/M, and time step is τ=T/N, where xi=ih, 0≤i≤M, yj=jh, 0≤j≤M, and tn=nτ, 0≤n≤N. Ωh=xi,yj0≤i,j≤M and Ωτ=tn0≤n≤N. Let fijn+1/2=fuxi,yj,tn+1/2 and xi,yj,tn+1/2 be called the node.Supposeu=uijn0≤i,j≤M,0≤n≤N is a grid function on Ωh×Ωτ, let
(8)tn+1/2=12tn+tn+1,δtuijn+1/2=1τuijn+1−ijn,δx2uijn=1h2ui+1,jn−2uijn+ui−1,jn,δy2uijn=1h2ui,j+1n−2uijn+ui,j−1n,uijn+1/2=12uijn+uijn+1,uijn+1/23=12uijn3+uin+13.Supposeu=uij0≤i,j≤M is a grid function on Ωh, the operator is defined by
(9)Auij=112ui−1,j+10uij+ui+1,j,1≤i≤M−1,0≤j≤M,uij,i=0,M,0≤j≤M,(10)Buij=112ui,j−1+10uij+ui,j+1,1≤j≤M−1,0≤i≤M,uij,j=0,M,0≤i≤M,(11)v=∂2u∂x2,(12)w=∂2u∂y2.Then, we get the following equation:(13)∂u∂t=ε2v+ε2w−fu.Define the grid functions as(14)Ui,jn=uxi,yj,tn,0≤i,j≤M,0≤n≤N,Vi,jn=vxi,yj,tn,0≤i,j≤M,0≤n≤N,Wi,jn=wxi,yj,tn,0≤i,j≤M,0≤n≤N.Considering differential Equation (13) at the point xi,yj,tn+1/2,
(15)∂u∂txi,yj,tn+1/2=ε2vxi,yj,tn+1/2+ε2wxi,yj,tn+1/2−fuxi,yj,tn+1/2,0≤i,j≤M,0≤n≤N−1.Using Taylor expansion,(16)∂u∂txi,yj,tn+1/2=1τuxi,yj,tn+1−uxi,yj,tn−τ224·∂3u∂t3xi,yj,tn+1/2+Oτ4=δtUi,jn+1/2−τ224·∂3u∂t3xi,yj,tn+1/2+Oτ4,vxi,yj,tn+1/2=Vi,jn+1/2−τ28·∂2v∂t2xi,yj,tn+1/2+Oτ4,wxi,yj,tn+1/2=Wi,jn+1/2−τ28·∂2w∂t2xi,yj,tn+1/2+Oτ4.Substituting the above three equations into (15),
(17)δtUi,jn+1/2=ε2Vi,jn+1/2+ε2Wi,jn+1/2−fi,jn+1/2+τ224·∂3u∂t3xi,yj,tn+1/2−ε2τ28∂2v∂t2xi,yj,tn+1/2+∂2w∂t2xi,yj,tn+1/2+Oτ4.Denoted by(18)gx,y,t=124·∂3u∂t3x,y,t−ε28∂2v∂t2x,y,t+∂2w∂t2x,y,t.Then,(19)δtUi,jn+1/2=ε2Vi,jn+1/2+ε2Wi,jn+1/2−fi,jn+1/2+τ2gi,jn+1/2+Oτ4.Acting the operatorAB on both sides of the equation above,
(20)ABδtUi,jn+1/2=ε2ABVi,jn+1/2+ε2ABWi,jn+1/2−ABfi,jn+1/2+τ2ABgi,jn+1/2+Oτ4,1≤i,j≤M−1,0≤n≤N−1.Considering differential Equation (11) at the point xi,yj,tn,
(21)vxi,yj,tn=∂2u∂x2xi,yj,tn,0≤i,j≤M,0≤n≤N.Based on Lemma3,
(22)AVi,jn=δx2Ui,jn+h4240∂6u∂x6xi,yj,tn+Oh6,1≤i≤M−1,0≤j≤M,0≤n≤N.Acting the two equations superscripted asn and n+1 and dividing by 2,
(23)AVi,jn+1/2=12AVi,jn+AVi,jn+1=12δx2Ui,jn+δx2Ui,jn+1+h4240·12∂6u∂x6xi,yj,tn+∂6u∂x6xi,yj,tn+1+Oh6=δx2Ui,jn+1/2+h4240∂6u∂x6xi,yj,tn+1/2+Oτ2+τ2h4+h6,1≤i≤M−1,0≤j≤M,0≤n≤N−1.Acting the operatorB on both sides of the equation above,
(24)ABVi,jn+1/2=Bδx2Ui,jn+1/2+h4240B∂6u∂x6xi,yj,tn+1/2+Oτ2+τ2h4+h6,1≤i,j≤M−1,0≤n≤N−1.Similarly, considering differential Equation (12) at the point xi,yj,tn, then
(25)ABWi,jn+1/2=Aδy2Ui,jn+1/2+h4240A∂6u∂y6xi,yj,tn+1/2+Oτ2+τ2h4+h6,1≤i,j≤M−1,0≤n≤N−1.Substituting (24) and (25) into (20),
(26)ABδtUi,jn+1/2=ε2Bδx2Ui,jn+1/2+Aδy2Ui,jn+1/2−ABfi,jn+1/2+ε2h4240B∂6u∂x6xi,yj,tn+1/2+A∂6u∂y6xi,yj,tn+1/2+τ2ABgi,jn+1/2+Oτ4+τ2h4+h6,1≤i,j≤M−1,0≤n≤N−1.Here,(27)Ri,jn+1/2=ε2h4240B∂6u∂x6xi,yj,tn+1/2+A∂6u∂y6xi,yj,tn+1/2+τ2ABgi,jn+1/2+Oτ4+τ2h4+h6.Then,(28)ABδtUi,jn+1/2=ε2Bδx2Ui,jn+1/2+Aδy2Ui,jn+1/2−ABfi,jn+1/2+Ri,jn+1/2,1≤i,j≤M−1,0≤n≤N−1.Omitting the small termRi,jn+1/2 and substituting the following equation into (28),
(29)fuxi,yj,tn+1/2=ui,jn+1/23−ui,jn+1/2=uijn+13+uijn32−uijn+1+uijn2.Then, the two-dimensional Allen-Cahn Equation (7) corresponding compact difference scheme is obtained.
(30)ABuijn+1−uijnτ+ABuijn+13+uijn32−uijn+1+uijn2=ε22Bδx2uijn+1+Bδx2uijn+Aδy2uijn+1+Aδy2uijn,1≤i,j≤M−1,0≤n≤N−1.Finally, the spatial derivative is discretized by the central finite difference scheme for the two-dimensional Allen-Cahn equation, theD2 is expressed as the corresponding discrete matrix, and
(31)D2=BD1+D1A.Here,
(32)D1=1h2−210⋯01−21⋯001−2⋯0⋮⋮⋮⋮000⋯−2M−1×M−1.Meanwhile, let M−1 be the number of nodes inside the interval after spatial dispersion and h be the space step.
(33)B=A=10121120⋯01121012112⋯001121012⋯0⋮⋮⋮⋮000⋯1012M−1×M−1.Then, the discrete matrixD2 satisfies the following properties:
(a)
D2 is symmetric.(b)
D2 is semi-negative definite, such as(34)UTD2U≤0.(c)
The elements inD2 satisfy(35)dii<0,−dii≥max1≤i≤M−1∑j=1,j≠iM−1dij.Let(36)C=AB.MatrixC satisfies the following properties:
(a)
C is symmetric.(b)
C is positive definite, such as(37)UTCU>0.(c)
The elements inC satisfy(38)cij>0,cii≥max1≤i≤M−1∑j=1,j≠iM−1cij.Substituting the matrixD2 and C into (30), we get the compact difference scheme:
(39)CUn+1−Unτ+CUn+13−Un+12+Un3−Un2=ε2D2Un+1+D2Un2.Here,(40)Un≔uijn1≤i,j≤M−1,Un3≔uijn31≤i,j≤M−1.MultiplyingC−1 to both sides of the above equation,
(41)Un+1−Unτ+Un+13−Un+12+Un3−Un2=ε22C−1D2Un+1+Un,0≤n≤N−1.MatrixC−1D2 satisfies the following properties:
(a)
C−1D2 is symmetric.(b)
C−1D2 is negative definite, such as(42)UTC−1D2U≤0.


## 3. Discrete Maximum Principle of Compact Difference Scheme for Two-Dimensional Allen-Cahn Equation

Theorem 4.
Assuming that the initial value of the Allen-Cahn problem satisfiesmaxx,yϵΩ¯u0x,y≤1, when the step ratio satisfies 5/24<λ<5/12 and the time step satisfies 0<τ≤min4/5λ−1/6,1−12/5λ, we have Un∞≤1, for ∀n≥1.Proof.
Obviously,U0∞≤u0≤1. Assuming that Un∞≤1, then Un+1∞≤1 needs to be proved. Expanding the established compact difference scheme,
(43)1τ1144ui−1,j−1n+1+10144ui,j−1n+1+1144ui+1,j−1n+1+10144ui−1,jn+1+100144ui,jn+1+10144ui+1,jn+1+1144ui−1,j+1n+1+10144ui,j+1n+1+1144ui+1,j+1n+1−1144ui−1,j−1n+10144ui,j−1n+1144ui+1,j−1n+10144ui−1,jn+100144ui,jn+10144ui+1,jn+1144ui−1,j+1n+10144ui,j+1n+1144ui+1,j+1n+121144ui−1,j−1n+13+10144ui,j−1n+13+1144ui+1,j−1n+13+10144ui−1,jn+13+100144ui,jn+13+10144ui+1,jn+13+1144ui−1,j+1n+13+10144ui,j+1n+13+1144ui+1,j+1n+13+1144ui−1,j−1n3+10144ui,j−1n3+1144ui+1,j−1n3+10144ui−1,jn3+100144ui,jn3+10144ui+1,jn3+1144ui−1,j+1n3+10144ui,j+1n3+1144ui+1,j+1n3−121144ui−1,j−1n+1+10144ui,j−1n+1+1144ui+1,j−1n+1+10144ui−1,jn+1+100144ui,jn+1+10144ui+1,jn+1+1144ui−1,j+1n+1+10144ui,j+1n+1+1144ui+1,j+1n+1+1144ui−1,j−1n+10144ui,j−1n+1144ui+1,j−1n+10144ui−1,jn+100144ui,jn+10144ui+1,jn+1144ui−1,j+1n+10144ui,j+1n+1144ui+1,j+1n=ε221h2212ui−1,j−1n+1+812ui,j−1n+1+212ui+1,j−1n+1+812ui−1,jn+1−4012ui,jn+1+812ui+1,jn+1+212ui−1,j+1n+1+812ui,j+1n+1+212ui+1,j+1n+1+212ui−1,j−1n+812ui,j−1n+212ui+1,j−1n+812ui−1,jn−4012ui,jn+812ui+1,jn+212ui−1,j+1n+812ui,j+1n+212ui+1,j+1n.
Multiplyingτ to both sides of the above equation and letting λ=ε2τ/h2,
(44)1001441−τ2+53λui,jn+1+100144τ2ui,jn+13+101441−τ2−13λui−1,jn+1+10144τ2ui−1,jn+13+101441−τ2−13λui+1,jn+1+10144τ2ui+1,jn+13+101441−τ2−13λui,j−1n+1+10144τ2ui,j−1n+13+101441−τ2−13λui,j+1n+1+10144τ2ui,j+1n+13+11441−τ2−112λui−1,j−1n+1+1144τ2ui−1,j−1n+13+11441−τ2−112λui+1,j−1n+1+1144τ2ui+1,j−1n+13+11441−τ2−112λui−1,j+1n+1+1144τ2ui−1,j+1n+13+11441−τ2−112λui+1,j+1n+1+1144τ2ui+1,j+1n+13=1001441+τ2−53λui,jn−100144τ2ui,jn3+101441+τ2+13λui−1,jn−10144τ2ui−1,jn3+101441+τ2+13λui+1,jn−10144τ2ui+1,jn3+101441+τ2+13λui,j−1n−10144τ2ui,j−1n3+101441+τ2+13λui,j+1n−10144τ2ui,j+1n3+11441+τ2+112λui−1,j−1n−1144τ2ui−1,j−1n3+11441+τ2+112λui+1,j−1n−1144τ2ui+1,j−1n3+11441+τ2+112λui−1,j+1n−1144τ2ui−1,j+1n3+11441+τ2+112λui+1,j+1n−1144τ2ui+1,j+1n3.
Shifting terms on both sides of the equation,(45)1001441−τ2+53λui,jn+1+100144τ2ui,jn+13=13λ−101441−τ2ui−1,jn+1−10144τ2ui−1,jn+13+13λ−101441−τ2ui+1,jn+1−10144τ2ui+1,jn+13+13λ−101441−τ2ui,j−1n+1−10144τ2ui,j−1n+13+13λ−101441−τ2ui,j+1n+1−10144τ2ui,j+1n+13+112λ−11441−τ2ui−1,j−1n+1−1144τ2ui−1,j−1n+13+112λ−11441−τ2ui+1,j−1n+1−1144τ2ui+1,j−1n+13+112λ−11441−τ2ui−1,j+1n+1−1144τ2ui−1,j+1n+13+112λ−11441−τ2ui+1,j+1n+1−1144τ2ui+1,j+1n+13+1001441+τ2−53λui,jn−100144τ2ui,jn3+101441+τ2+13λui−1,jn−10144τ2ui−1,jn3+101441+τ2+13λui+1,jn−10144τ2ui+1,jn3+101441+τ2+13λui,j−1n−10144τ2ui,j−1n3+101441+τ2+13λui,j+1n−10144τ2ui,j+1n3+11441+τ2+112λui−1,j−1n−1144τ2ui−1,j−1n3+11441+τ2+112λui+1,j−1n−1144τ2ui+1,j−1n3+11441+τ2+112λui−1,j+1n−1144τ2ui−1,j+1n3+11441+τ2+112λui+1,j+1n−1144τ2ui+1,j+1n3.
For the right side of Equation (45), the n-layer terms are processed firstly.
Then, the sum of then-layer terms is L. Let Un∞≤1.
(46)fx=1001441+τ2−53λx−100144τ2x3,x∈0,1.
Thus,(47)f0=0,f1=2536−53λ=53512−λ.
Since(48)f′x=25361+τ2−53λ−25τ24x2,f″x=−25τ12x≤0.
f′x is monotonicity decreasing,
(49)f′0=25361+τ2−53λ,f′1=25361+τ2−53λ−25τ24.
Whenτ≤1−12/5λ, f′1≥0, therefore, f′x≥0.
Thus,fx is monotonicity increasing. ∀x∈0,1, f0≤fx≤f1, namely,
(50)fx≤2536−53λ,λ≤512.
Let(51)gx=101441+τ2+13λx−10144τ2x3,x∈0,1.
Then,(52)g0=0,g1=572+13λ.
Since(53)g′x=5721+τ2+13λ−5τ48x2,g″x=−5τ24x≤0.
g′x is monotonicity decreasing,
(54)g′0=5721+τ2+13λ,g′1=5721+τ2+13λ−5τ48.
Whenτ≤1+24/5λ, g′1≥0, we have g′x≥0.
Thus,gx is monotonicity increasing. ∀x∈0,1, then g0≤gx≤g1, namely,
(55)gx≤572+13λ.
Let(56)hx=11441+τ2+112λx−1144τ2x3,x∈0,1.
Then,(57)h0=0,h1=1144+112λ.
Since(58)h′x=11441+τ2+112λ−τ96x2,h″x=−τ48x≤0.
h′x is monotonicity decreasing,
(59)h′0=11441+τ2+112λ,h′1=11441+τ2+112λ−τ96.
Whenτ≤1+12λ, h′1≥0, we have h′x≥0.
Thus,hx is monotonicity increasing. ∀x∈0,1, h0≤hx≤h1, that is,
(60)hx≤1144+112λ.
According to (50), (55), and (60), when τ≤1−12/5λ,
(61)fx≤2536−53λ,λ≤512,gx≤572+13λ,hx≤1144+112λ.
Then,(62)L≤f1+4g1+4h1=1.
Next, the eight items in then+1 layer on the right side of equation (45) are dealt.
Let(63)py=λ3−101441−τ2y−10144τ2y3.
Then,(64)p′y=λ3−101441−τ2−5τ48y2.
Whenp′y=0, the maximum point is obtained,
(65)y1=485τ·13λ−524+5τ144,λ≥524.
Sincepy is an odd function and monotonicity increasing in the interval of −y1,y1, then py≤py1.
Let(66)qy=λ12−11441−τ2y−1144τ2y3.
Then,(67)q′y=λ12−11441−τ2−τ96y2.
Whenq′y=0, the maximum point is obtained,
(68)y2=96τ·112λ−112+τ288,λ≥112.
Sinceqy is an odd function and monotonicity increasing in the interval of −y2,y2, then qy≤qy2.
Substituting (62), (63), and (66) into (45), then
(69)1001441−τ2+53λuijn+1+100144τ2uijn+13≤pui−1,jn+1+pui+1,jn+1+pui,j−1n+1+pui,j+1n+1+qui−1,j−1n+1+qui+1,j−1n+1+qui−1,j+1n+1+qui+1,j+1n+1+1,1≤i,j≤N−1.
SupposeUn+1∞=Ui0,j0n+1=m.
On the one hand, for the right eight terms of Equation (69), taking the absolute value of both sides of the equation by definition (63), according to the triangle inequality,
(70)pui−1,jn+1=λ3−101441−τ2ui−1,jn+1−5τ144ui−1,jn+13≤λ3−5721−τ2ui−1,jn+1+5τ144ui−1,jn+13≤λ3−5721−τ2m+5τ144m3.
Similarly,(71)pui+1,jn+1≤λ3−5721−τ2m+5τ144m3,pui,j−1n+1≤λ3−5721−τ2m+5τ144m3,pui,j+1n+1≤λ3−5721−τ2m+5τ144m3.
Taking the absolute value of both sides of the equation by definition (66), according to the triangle inequality,
(72)qui−1,j−1n+1=λ12−11441−τ2ui−1,j−1n+1−τ288ui−1,j−1n+13≤λ12−11441−τ2ui−1,j−1n+1+τ288ui−1,j−1n+13≤λ12−11441−τ2m+τ288m3.
Similarly,(73)qui+1,j−1n+1≤λ12−11441−τ2m+τ288m3,qui−1,j+1n+1≤λ12−11441−τ2m+τ288m3,qui+1,j+1n+1≤λ12−11441−τ2m+τ288m3.
Takingi=i0,j=j0 for the left side of (69), substituting (70) and (72) into the right side of (69),
(74)1001441−τ2+53λm+100144τ2m3≤4λ3−5721−τ2m+5τ144m3+4λ12−11441−τ2m+τ288m3+1,
namely,(75)1−τ2m+7τ36m3≤1.
Whenτ≤1, then M≤1/1−τ/2≤2, ui0,j0n+1∈0,2.
Lety0=−y1,y1∩−y2,y2, suppose y0=2, then py and qy are monotonicity increasing in the interval of −2,2. For
(76)p′y=λ3−101441−τ2−5τ48y2≥0,then
(77)5τ48y2≤λ3−101441−τ2=13λ−524+5τ144.
Sincey∈−2,2, then
(78)5τ12≤13λ−524.
So,(79)τ≤45λ−16.
Similarly, for(80)q′y=λ12−11441−τ2−τ96y2≥0,then
(81)τ96y2≤λ12−11441−τ2=112λ−112+τ288.
Sincey∈−2,2, then
(82)τ24≤112λ−112.
That means that(83)τ≤2λ−16.
In conclusion, according to (65), (68), (79), and (83), we obtain τ≤4/5λ−1/6λ≥5/24.
On the other hand, takingi=i0,j=j0 in (69), thus
(84)1001441−τ2+53λui0,j0n+1+50τ144ui0,j0n+13,≤4pui0,j0n+1+4qui0,j0n+1+1.
According to (63) and (66),
(85)25361−τ2+53λm+25τ72m3,≤53λ−11361−τ2m−11τ72m3+1,
namely,(86)1−τ2m+τ2m3≤1.
m≤1 is obtained.
Consequently, when0<τ≤min4/5λ−1/6,1−12/5λ, 5/24<λ<5/12, then Un+1∞≤1.☐


## 4. Two-Dimensional Discrete Energy Stability

Lemma 5 (see [2]).
∀a,b∈−1,1,
(87)a3−aa−b+a−b2≥14a2−12−b2−12,a3−ba−b+a−b2≥14a2−12−b2−12.According to the energy function that is defined by (2), discrete energy function of compact difference scheme (41) is defined by
(88)Eu=14·h2·∑i=1M−1∑j=1M−1uij2−12−ε22·h2·uTC−1D2u.Theorem 6.
Assuming that the initial value of the Allen-Cahn problem satisfiesmaxx,yϵΩ¯u0x,y≤1, then the numerical solution obtained by scheme (41) satisfies discrete energy decaying under the condition of 4:
(89)EUn+1≤EUn.Proof.
(90)EUn+1−EUnh2=14∑i=1M−1∑j=1M−1Uijn+12−12−Uijn2−12−ε22Un+1TC−1D2Un+1−UnTC−1D2Un.
In view of Theorem4, Un∞≤1 and Un+1∞≤1.
Therefore, using Lemma5,
(91)14∑i=1M−1∑j=1M−1Uijn+12−12−Uijn2−12≤∑i=1M−1∑j=1M−112Uijn+13−Uijn+1Uijn+1−Uijn+12Uijn3−UijnUijn+1−Uijn+Uijn+1−Uijn2.
Since the matrixC−1D2 is symmetric,
(92)ε22Un+1−UnTC−1D2Un+1+Un=ε22Un+1TC−1D2Un+1−UnTC−1D2Un.
Substituting (91) and (92) into (90),
(93)EUn+1−EUnh2≤∑i=1M−1∑j=1M−112Uijn+13−Uijn+1Uijn+1−Uijn+12Uijn3−UijnUijn+1−Uijn+Uijn+1−Uijn2−ε22Un+1−UnTC−1D2Un+1+Un.
Multiplying (41) by Un+1−Un, summing up for i from 1 to M−1 and same for j,
(94)∑i=1M−1∑j=1M−112Uijn+13−Uijn+1Uijn+1−Uijn+12Uijn3−UijnUijn+1−Uijn+1τUijn+1−Uijn2=ε22Un+1−UnTC−1D2Un+1+Un.
Substituting (94) into (93),
(95)EUn+1−EUnh2≤∑i=1M−1∑j=1M−112Uijn+13−Uijn+1Uijn+1−Uijn+12Uijn3−UijnUijn+1−Uijn+Uijn+1−Uijn2−∑i=1M−1∑j=1M−112Uijn+13−Uijn+1Uijn+1−Uijn+12Uijn3−UijnUijn+1−Uijn+1τUijn+1−Uijn2.
Then,(96)EUn+1−EUnh2≤1−1τ∑i=1M−1∑j=1M−1Uijn+1−Uijn2.
Since0<τ≤min4/5λ−1/6,1−12/5λ, 5/24<λ<5/12, the right side of inequality is negative.
Then,EUn+1≤EUn is proved.☐


## 5. Numerical Examples

In this part, some numerical examples are given to verify the theoretical results of the previous sections which are discrete numerical maximum principle and energy stability. Considering a two-dimensional problem with homogeneous Riemann boundary conditions, and selecting the initial condition as(97)u0x=0.9∗rand·+0.05,here,rand· is a random sequence in the interval of 0,1.If givenϵ2=0.001,h=0.01, by λ=ε2τ/h2, using the conditions of Theorem 4, 5/24≤λ≤5/12,0<τ≤min4/5λ−1/6,1−12/5λ, we can get 1/48<τ<1/24. Taking different values for time step τ, the results are shown in the following figures.It can be observed from Figure1 that the discrete scheme (30) satisfies the maximum principle when the time step τ is 0.025 or 0.04, and from Figure 2, the discrete scheme (30) does not satisfy the maximum principle when the time step τ is 0.3 or 0.5. In addition, it satisfies the stability of discrete energy decaying according to Figures 3 and 4.Figure 1
Whenτ=0.025,0.04, the maximum of the scheme is (30).Figure 2
Whenτ=0.3,0.5, the maximum of the scheme is (30).Figure 3
Whenτ=0.025,0.04, the energy of the scheme is (30).Figure 4
Whenτ=0.3,0.5, the energy of the scheme is (30).Then, if givenϵ2=0.001,h=0.008; similarly, we can get 1/75<τ<2/75. The results are shown in the following figures.It can be observed from Figure5 that the discrete scheme (30) satisfies the maximum principle when the time step τ is 0.02 or 0.025, and from Figure 6, the discrete scheme (30) does not satisfy the maximum principle when the time step τ is 0.3 or 0.5. In addition, it satisfies the stability of discrete energy decaying according to Figures 7 and 8.Figure 5
Whenτ=0.02,0.025, the maximum of the scheme is (30).Figure 6
Whenτ=0.3,0.5, the maximum of the scheme is (30).Figure 7
Whenτ=0.02,0.025, the energy of the scheme is (30).Figure 8
Whenτ=0.3,0.5, the energy of the scheme is (30).


## 6. Conclusions

In this paper, the two-dimensional nonlinear Allen-Cahn equation is discretized by using the central finite difference method in space and using two operatorsA and B in time, and then, the fully discrete compact difference scheme with second-order accuracy in time and fourth-order in space is established as below:
(98)ABuijn+1−uijnτ+ABuijn+13+uijn32−uijn+1+uijn2=ε22Bδx2uijn+1+Bδx2uijn+Aδy2uijn+1+Aδy2uijn,1≤i≤M−1,0≤n≤N−1.Constructing the discrete matrix,(99)D2=BD1+D1A,C=AB.Then, using the matrix instead of the operator, the following compact difference equation (39) can be obtained:
(100)CUn+1−Unτ+CUn+13−Un+12+Un3−Un2=ε2D2Un+1+D2Un2.It proved that when5/24<λ<5/12, 0<τ≤min4/5λ−1/6,1−12/5λ, the discrete scheme satisfies the maximum principle Un∞≤1, for ∀n≥1.Secondly, the discrete energy function of the compact difference scheme (88) of the two-dimensional Allen-Cahn equation is
(101)Eu=14·h2·∑i=1M−1∑j=1M−1uij2−12−ε22·h2·uTC−1D2u.The numerical solution obtained by scheme (39) satisfies the stability of discretization energy decaying under the conditions and conclusions of discretization maximization.
(102)EUn+1≤EUn.



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