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On the stability of a semi-implicit scheme of Cahn-Hilliard type equations Cheng, Xinyu 2017

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On the Stability of a Semi-ImplicitScheme of Cahn-Hilliard TypeEquationsbyXinyu ChengB.Sc., The Chinese University of Hong Kong, 2011-2015A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFMASTER OF SCIENCEinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)July 2017© Xinyu Cheng 2017AbstractIt is well known that Allen-Cahn equation and Cahn-Hilliard equationare essential to study the phase separation phenomenon of a two-phase ora multiple-phase mixture. An important property of the solutions to thosetwo equations is that the energy functional, which is defined in this thesis,decreases in time. To study these solutions, researchers developed differentnumerical schemes to give accurate approximations, since analytic solutionsare only available in a very few simple cases. However, not all schemes sat-isfy the energy-decay property, which is an important standard to determinewhether the scheme is stable. In recent work, Li, Qiao and Tang developed asemi-implicit scheme for the Cahn-Hilliard equation and proved the energy-decay property. In this thesis, we extend the semi-implicit scheme to theAllen-Cahn equation and fractional Cahn-Hilliard equation with a proof ofthe energy-decay property. Moreover, this semi-implicit scheme is practicaland could be applied to more general diffusion equations while preserving theenergy-decay stability.iiLay SummaryThis thesis extends a numerical scheme from previous literature for theCahn-Hilliard equation to the Allen-Cahn equation and fractional Cahn-Hilliardequation, which are more general partial differential equations. These equa-tions describe physical phenomena of interest in material science. We provedthe stability of such scheme by showing the energy is decreasing. Based onour result, schemes with similar stability properties can be analyzed for moregeneral equations.iiiPrefaceThe topic of this thesis is based on the previous work of the author’s su-pervisor, Dr. Dong Li in [11], [10] and [9].The content of Chapter 3 and 4 is based on [11]. However, the author pro-vides unique variations of original ideas. Therefore, the content of Chapter 3and 4 is independent of [11] and hence original work by the author.The content of Chapter 5 is original work by the author with help of un-published ideas by the author’s supervisors, Dr. Dong Li and Dr. Brian Wet-ton.The content of Chapter 6 is based on [10] and the content of Chapter 7 isbased on [9] but the author provides original variations of these ideas. There-fore, the content of Chapter 6 and 7 is produced by the author independentlyand originally.ivTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ixAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xi1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Preliminaries and the Main Lemma . . . . . . . . . . . . . . . . . . 82.1 Definitions and Useful Theorems . . . . . . . . . . . . . . . . . . . 82.1.1 Lp Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1.2 Weak Derivatives and Sobolev Space . . . . . . . . . . . . 92.1.3 Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . 102.1.4 Convergence of Fourier Series in Periodic Domains . . . . 102.1.5 Uniform Boundedness Principle . . . . . . . . . . . . . . . 10vTable of Contents2.1.6 Fixed-point Theorem . . . . . . . . . . . . . . . . . . . . . . 112.1.7 Duhamel’s Formula . . . . . . . . . . . . . . . . . . . . . . . 112.2 Several Important Inequalities . . . . . . . . . . . . . . . . . . . . 122.2.1 Hölder’s Inequality . . . . . . . . . . . . . . . . . . . . . . . 122.2.2 Young’s Inequality . . . . . . . . . . . . . . . . . . . . . . . 122.2.3 Sobolev Inequality on Td . . . . . . . . . . . . . . . . . . . 122.2.4 Morrey’s Inequality on H2(Td) . . . . . . . . . . . . . . . . 132.2.5 Gagliardo–Nirenberg Interpolation Inequality . . . . . . 132.2.6 Grönwall’s Inequality . . . . . . . . . . . . . . . . . . . . . . 142.2.7 Discrete Grönwall’s Inequality . . . . . . . . . . . . . . . . 142.3 the Main Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Stability of a First Order Semi-implicit Scheme on 2D Allen-Cahn Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.1 Stability Theorem for Allen-Cahn Equation . . . . . . . . . . . . 183.2 Proof of the Stability Theorem . . . . . . . . . . . . . . . . . . . . 194 L2 Error Estimate of the First Order Scheme on 2D Allen-CahnEquation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.1 Auxiliary L2 Error Estimate for Near Solutions . . . . . . . . . . 294.2 L2 Error Estimate of 2D Allen-Cahn Equation . . . . . . . . . . . 334.2.1 Bounds on Allen-Cahn Exact Solution and NumericalSolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.2.2 Proof of L2 Error Estimate of 2D Allen-Cahn Equation . 465 Stability of a First Order Semi-implicit Scheme on 2D Frac-viTable of Contentstional Cahn-Hilliard Equation . . . . . . . . . . . . . . . . . . . . . 516 Stability of a First Order Semi-implicit Scheme on 3D Allen-Cahn Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586.1 the Main Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 596.2 Proof of 3D Stability Theorem . . . . . . . . . . . . . . . . . . . . . 606.3 L2 Error Estimate of 3D Allen-Cahn Equation . . . . . . . . . . . 647 Second Order Semi-Implicit Schemes . . . . . . . . . . . . . . . . 667.1 Introduction of Scheme I: . . . . . . . . . . . . . . . . . . . . . . . 667.2 Estimate of the First Order Scheme (7.2) . . . . . . . . . . . . . . 677.3 Unconditional Stability of the Second Order Scheme I (7.1) &(7.2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 727.3.1 Proof of Unconditional Stability (Theorem 7.3.1) . . . . . 767.4 L2 Error Estimate of Second Order Scheme I . . . . . . . . . . . 827.4.1 Auxiliary L2 Error Estimate for Near Solutions . . . . . . 837.4.2 Time Discretization of Allen-Cahn Equation . . . . . . . . 877.4.3 Proof of L2 Error Estimate of Second Order Scheme I(7.1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 937.5 Introduction of Scheme II . . . . . . . . . . . . . . . . . . . . . . . 947.6 Estimate of the First Order Scheme (7.100) . . . . . . . . . . . . 947.7 Conditional Stability of the Second Order Scheme II (7.99) &(7.100) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 957.7.1 Proof of Conditional Stability (Theorem 7.7.1) . . . . . . . 998 Conclusion & Future Work . . . . . . . . . . . . . . . . . . . . . . . .105viiTable of ContentsBibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .106viiiList of Figures1.1 Computational Simulation of 2D Waves [20] . . . . . . . . . . . . 11.2 Spectral Simulation of the Cahn Hilliard Equation in a 2D Do-main [19] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4ixAcknowledgementsFirstly, I would like to thank my supervisors Dr. Dong Li and Dr. BrianWetton who helped me find a suitable project. In addition, they spent manyefforts on equipping me with qualified mathematics skills and developingmy interests in diverse areas. Moreover, they took much time in providingme with valuable ideas and enlightening suggestions, which are crucial toaccomplishment of this thesis.Secondly, I would like to express my thanks to my colleagues and office-mates for their warmness and caring, that made me feel more enjoyable doingresearch.xDedicationThis thesis is dedicated to my firmest supporters, my parents, Mrs. Xi-aodong Huang and Mr. Chun Cheng.xiChapter 1IntroductionPartial differential equations (PDE) often describe mathematical modelsof physical phenomena. For example, wave equations describe the propertiesof waves including sound waves, light waves and other waves, which help usto study sound including noise and music, electromagnetics and fluid dynam-ics.Figure 1.1: Computational Simulation of 2D Waves [20]To study PDE, we are interested in the solutions to a certain equationin some domain under specific initial conditions and boundary conditions.More information about the solution can help the physical model to be better1Chapter 1. Introductionunderstood. In mathematics, usually these studies focus on the existence,uniqueness, regularity and some long-time behaviors of the solution.Although sometimes it is possible to find explicit solutions of certain sim-ple PDE, usually there are no explicit solutions. Thus, it is necessary tocompute approximate solutions using computer simulations. As a result,throughout the area of partial differential equations, it is necessary to de-velop well behaved numerical schemes that are guaranteed to approximatePDE to an expected accuracy.In this thesis, we consider modified versions of the Cahn-Hilliard equa-tion. These equations were developed in [2] to describe the separation ofdifferent metals in a binary alloy. They have been recognized as a genericmodel that arises in many applications. Hence they have been well studiedby mathematicians, physicists and other scientists. The Cahn-Hilliard equa-tion for u(x, t) is:∂tu=∆(−ν∆u+ f (u)), (x, t) ∈Ω× (0,∞)u(x,0)= u0, (1.1)where the vector position x is in the spatial domain Ω, which is taken to betwo dimensional periodic domain in this work, and t is time. The values of ugenerally lie in the range [−1,1], with −1 representing the pure state of onephase and +1 representing the pure state of the other phase. Values of u in(−1,1) represent a mixture of the two phases. Here ν is a small parameter,pν represents an average distance over which phases mix. The energy term2Chapter 1. Introductionf (u) is defined byf (u)= F ′(u)= u3−u , F(u)= 14(u2−1)2.The Cahn-Hilliard equation (1.1) describes the evolution of the phase frac-tions under a competition between diffusion (which tends to mix phases) andthe preference of the phases to separate.In this thesis, we consider the spatial domainΩ to be the 2pi-periodic torusT2 =R/2piZ×R/2piZ. Often, the Cahn-Hilliard equation and related equationswe consider in this thesis, describe the micro structure of macroscopic ma-terial. Thus, considering a periodic domain is not a serious simplification:we are modelling a representative piece of the micro structure. Consideringthe periodic domain allows the use of efficient and accurate Fourier-spectralnumerical methods, which will be introduced later in the thesis.It is not possible to find analytic solutions to the Cahn-Hilliard equation.As a result, it is necessary to develop numerical methods to approximate thesolutions. Many approaches have been developed, [4] as an example. An-other example computational result is shown in Figure 1.2 [19]. Such numer-ical approximations should give accurate results to the values and qualita-tive features of the solution. In the literature, a key feature is energy decay,discussed in detail below. In [11], Li, Qiao and Tang propose a numericalscheme for Cahn-Hilliard equation and hence prove that it preserves energydecay with no a-priori assumptions. In this thesis, we extend their result toother related models.By standard arguments, the mass of the smooth solution of Cahn-Hilliard3Chapter 1. Introductionequation is conserved, i.e. ddt M(t) ≡ 0 , M(t) =∫Ω u(x, t) dx. This representsthe conservation of the two phases in the mixture. In particular, M(t) ≡ 0 ifM(0)= 0 and hence oftentimes zero-mean initial data would be considered asa simpler but representative case. The associated energy functional is givenbyE(u)=∫Ω(12ν|∇u|2+F(u))dx.Assuming u(x, t) is a smooth solution with zero mean, one can deduceddtE(u(t))+∫Ω|∇(−ν∆u+ f (u))|2 dx= 0,which implies energy decay: ddt E(u(t)) ≤ 0, and hence contributes to the ex-istence of global solutions to Cahn-Hilliard equation as it provides a prioriH1-norm bound. On the other hand, the energy decay property is an impor-Figure 1.2: Spectral Simulation of the Cahn Hilliard Equation in a 2D Do-main [19]4Chapter 1. Introductiontant index for whether a numerical scheme is “stable” or not.Previous works by others [8, 14, 17, 18] give different semi-implicit Fourier-spectral schemes, which involved different stabilizing terms of different “size”,that preserve the energy decay property (we say these schemes are “energystable”). However, those works either require a strong Lipschitz condition onthe nonlinear source term, or require certain L∞ bounds on the numerical so-lutions. To improve that, Li, Qiao and Tang proved an unconditional stabilitytheory on a large time-stepping semi-implicit Fourier-spectral scheme.The scheme has the form:un+1−unτ=−ν∆2un+1+A∆(un+1−un)+∆ f (un) , n≥ 0u0 = u0 .(1.2)As usual τ is the time step, A is a large coefficient for the O(τ) stabilizingterm. Here O(τ) is defined as the well-known “big O” notation, i.e. |O(τ)| ≤ |cτ|for a non-zero constant c, or in other words at most linear function of τ. Asa result of their work, the energy decay could still be satisfied with a well-chosen large number A, with at least a size of O(1/ν| log(ν)|2), or c/ν| log(ν)|2for some positive constant c that depends on the initial conditions.Our work extends their semi-implicit scheme to the related Allen-Cahnequation and fractional Cahn-Hilliard equation. The Allen-Cahn equation isdefined as: ∂tu= ν∆u− f (u)u(x,0)= u0 ;(1.3)5Chapter 1. Introductionwhile the zero-mass projected Allen-Cahn equation is defined as:∂tu=Π0 (ν∆u− f (u))u(x,0)= u0 ,(1.4)where Π0 is the zero mass projector, i.e. Π0( f ) = 1(2pi)d∑|k|≥1 f̂ (k)eik·x. Thedifference between the Allen-Cahn equation and zero-mass projected Allen-Cahn equation results from the fact that the mass functional is not preservedin Allen-Cahn equation.The fractional Cahn-Hilliard equation is defined as the following:∂tu= ν∆((−∆)αu+ (−∆)α−1 f (u)) , 0<α≤ 1u(x,0)= u0. (1.5)As α→ 0, (1.3) becomes the zero-mass projected Allen-Cahn equation andfor α= 1, it coincides the original Cahn-Hilliard equation. Roughly speaking,the fractional Cahn-Hilliard equation is an interpolation of the Allen-Cahnand Cahn-Hilliard equations.Remark 1. More general cases could be discussed. Roughly speaking wecould define a general “gradient” operator G and rewrite the equation as :∂tu=G (ν∆u− f (u))u(x,0)= u0. (1.6)When G = id, the identity map, (1.4) becomes the Allen-Cahn equation; whenG = (−∆)α, (1.4) becomes the fractional Cahn-Hilliard equation as discussed6Chapter 1. Introductionabove. And the corresponding semi-implicit scheme isun+1−unτ=G (ν∆un+1− f (un))−AG (un+1−un) , n≥ 0u0 = u0. (1.7)The main result of this thesis states that for any fixed time step τ, wecan always define a large constant A independent of τ in (1.5), such that thenumerical solution would be stable in the sense of satisfying the energy-decaycondition for “gradient” cases of A-C and fractional C-H. The analysis of othergradients is left to future work.For completeness, preliminaries would be given and the main lemmawould be proved in chapter 2; stability of first order semi-implicit schemes forAllen-Cahn equation and fractional Cahn-Hilliard would be proved in chap-ter 3 and 5 respectively. Moreover, we extend the results to the 3D case inchapter 6. On the other hand, main results of error estimate and convergenceare given in chapter 4. Finally, we introduce two second order semi-implicitschemes in chapter 7 while proving stability results and error estimates.7Chapter 2Preliminaries and the MainLemma2.1 Definitions and Useful Theorems2.1.1 Lp SpaceThroughout this paper we will denote the domainΩ=T2. If 1≤ p<∞, thespace Lp(Ω) consists of all complex-valued measurable functions that satisfy∫Ω| f (x)|p dx<∞ .For f ∈ Lp(Ω) we define the Lp norm of f by|| f ||Lp(Ω) =(∫Ω| f (x)|p dx)1/p.82.1. Definitions and Useful Theorems2.1.2 Weak Derivatives and Sobolev SpaceWe use the notation below:x= (x1, x2, ..., xn) ∈Rnα= (α1,α2, ...,αn) ∈Zn+∂α f = ∂α1+...+αn f∂α1x1 ∂α2x2 ...∂αnxn.(2.1)We define the weak derivative in the following sense: For u, v ∈ L1loc(Ω),(i.e they are locally integrable); ∀φ ∈ C∞0 (Ω), i.e φ is infinitely differentiable(smooth) and compactly supported; and∫Ωu(x) ∂αφ(x) dx= (−1)α1+...+αn∫Ωv(x) φ(x) dx,then v is defined to be the weak partial derivative of u, denoted by ∂αu. If uis “smooth” enough, its weak derivative coincides with its derivative and theequation above is basically integration by parts.Suppose u ∈ Lp(Ω) and all weak derivatives ∂αu exist for |α| = α1+ ...+αn ≤ k , such that ∂αu ∈ Lp(Ω) for |α| ≤ k, then we say u ∈Wk,p(Ω), and suchspace is called Sobolev space. The norm in Wk,p(Ω) is defined as :||u||Wk,p(Ω) =( ∑|α|≤k∫Ω|∂αu|p dx) 1p.Throughout this paper, for p= 2 case, we use the convention Hk(Ω) denotethe space Wk,2(Ω). For more details, we refer to chapter 5, [6].92.1. Definitions and Useful Theorems2.1.3 Fourier TransformIn this paper we use the following convention for Fourier expansion onTd:f (x)= 1(2pi)d∑k∈Zdf̂ (k)eik·x , f̂ (k)=∫Ωf (x)e−ik·x dx .By taking advantage of Fourier expansion, we use the equivalent Hs-normand H˙s-norm of function f by|| f ||Hs = 1(2pi)d/2( ∑k∈Zd(1+|k|2s)| f̂ (k)|2) 12, || f ||H˙s =1(2pi)d/2( ∑k∈Zd|k|2s| f̂ (k)|2) 12.The equivalence of two norms are well known, we refer to Appendix A in [16].2.1.4 Convergence of Fourier Series in Periodic DomainsGiven f being a Lp(Td) periodic function for p> 1, and denote the Dirich-let partial sum DN f := 1(2pi)d∑|k|≤N f̂ (k)eik·x, then||DN f − f ||Lp(Td)) → 0 , and DN f → f pointwise almost everywhere . (2.2)This was originally proved by Carleson in [3].2.1.5 Uniform Boundedness PrincipleLet X be a Banach space and Y be a normed vector space. Suppose thatF is a collection of continuous linear operator from X to Y . If for all x in Xone hassupT∈F||T(x)||Y <∞ ,102.1. Definitions and Useful TheoremsThensupT∈F||T|| <∞ where ||T|| is the operator norm.We refer to a simple proof in [15].2.1.6 Fixed-point TheoremGiven a Banach space (X , ||.||) and a contraction map T : X → X s.t ||T(x)−T(y)|| ≤ β||x− y|| with 0 < β < 1, then there exists a fix-point x, s.t T(x) = x.We refer to [1] for details.2.1.7 Duhamel’s FormulaConsider a linear inhomogeneous evolution equation for a function u(x, t) :Ω× (0,∞)→R, with a spatial domain Ω⊂Rd, of the formut(x, t)−Lu(x, t)= f (x, t) , (x, t) ∈Ω× (0,∞)u|∂Ω = 0u(x,0)= u0(x) , x ∈Ω ,(2.3)where L is a linear differential operator that involves no time derivativesand the boundary condition could be replaced by periodic boundary condition.Then formally, the solution to this equation system is:u(x, t)= eLtu0+∫ t0eL(t−s) f ds (2.4)where eLt is the homogeneous solution operator, or eLtu0 solves the homoge-neous equation with initial data u0. In fact eLtu0 is often given as a convolu-112.2. Several Important Inequalitiestion between a well-defined kernel and the initial data u0. For more details,we refer to [6].2.2 Several Important Inequalities2.2.1 Hölder’s InequalityGiven f ∈ Lp(Ω) and g ∈ Lq(Ω), such that 1p + 1q = 1 then|| f g||L1(Ω) ≤ || f ||Lp(Ω)||g||Lq(Ω).2.2.2 Young’s InequalityGiven a,b, p, q positive real numbers, such that 1p + 1q = 1, thenab≤ app+ bqq.2.2.3 Sobolev Inequality on TdNote that the this Sobolev inequality is slightly different from the stan-dard version. Let 0< s< d and f ∈ Lq(Td) for any dd−s < p<∞, then|| 〈∇〉−s f ||Lp(Td).s,p,d || f ||Lq(Td) , where1q= 1p+ sd.Here 〈∇〉−s denotes (1−∆)− s2 , or on the Fourier side (1+|k|2)− s2 and A.s,p,d Bis defined as A ≤Cs,p,d B where Cs,p,d is a constant dependent on s, p and d.See [11] for the details.122.2. Several Important Inequalities2.2.4 Morrey’s Inequality on H2(Td)Assume d ≤ 3 and f ∈H2(Td) then|| f ||∞(Td). || f ||H2(Td) .In fact stronger argument could be made with the help of Hölder space ar-guments, but in this paper only the infinity norm is needed. More detailedinformation is in chapter 5, [6].2.2.5 Gagliardo–Nirenberg Interpolation InequalityFor functions u : Ω→ R defined on a bounded Lipschitz domain Ω ⊂ Rd,fix 1≤ q, r ≤∞ and a natural number m. Suppose also that a real number αand a natural number j are such that1p= jd+(1r− md)α+ 1−αqandjm≤α≤ 1 .Then||D ju||Lp ≤C1(Ω)||Dmu||αLr ||u||1−αLq +C2(Ω)||u||Lswhere s> 0 is arbitrary.132.3. the Main Lemma2.2.6 Grönwall’s InequalityOn the interval I = [a,b] where a < b and b could be ∞. Let β and u bereal-valued continuous functions defined on I. If u is differentiable in (a,b)and satisfiesu′(t)≤β(t)u(t) , t ∈ (a,b) ,thenu(t)≤ u(a)exp(∫ taβ(s) ds), t ∈ [a,b] .We refer to [7] for details.2.2.7 Discrete Grönwall’s InequalityLet τ> 0 and yn ≥ 0, αn ≥ 0, βn ≥ 0 for n= 1,2,3 · · · . Supposeyn+1− ynτ≤αn yn+βn , ∀ n≥ 0 .Then for any m≥ 1, we haveym ≤ exp(τm−1∑n=0αn)(y0+m−1∑k=0βk).The proof is given in [11].2.3 the Main LemmaFor all f ∈Hs(T2) , s> 1, then|| f ||∞ ≤Cs ·(|| f ||H˙1√log(|| f ||H˙s +3)+| f̂ (0)|+1). (2.5)142.3. the Main LemmaHere Cs is a constant which only depends on s.Proof. To prove the lemma, we write f (x) = 1(2pi)2∑k∈Z2 f̂ (k) eik·x, i.e. theFourier series of f , which is convergent pointwisely to f . So,|| f ||∞ ≤ 1(2pi)2∑k∈Z2| f̂ (k)|≤ 1(2pi)2(| f̂ (0)|+ ∑0<|k|≤N| f̂ (k)|+ ∑|k|>N| f̂ (k)|). | f̂ (0)|+ ∑0<|k|≤N(| f̂ (k)||k| · |k|−1)+ ∑|k|>N(| f̂ (k)||k|s · |k|−s). | f̂ (0)|+( ∑0<|k|≤N| f̂ (k)|2|k|2) 12·( ∑0<|k|≤N|k|−2) 12+( ∑|k|>N| f̂ (k)|2|k|2s) 12· ( ∑|k|>N|k|−2s) 12. | f̂ (0)|+ 1Ns−1( ∑|k|>N| f̂ (k)|2|k|2s) 12+( ∑0<|k|≤N| f̂ (k)|2|k|2) 12·√log(N+3). | f̂ (0)|+ 1Ns−1|| f ||H˙s +√log(N+3)|| f ||H˙1 .(2.6)In the previous step, we use Hölder’s inequality for counting measure andintegral approximation of∑0<|k|≤N |k|−2 and∑|k|>N |k|−2s in Z2. To be moreclear, ∑0<|k|≤N|k|−2.∫ N11r2·2pir dr.∫ N11rdr. log(N+3) ,(2.7)∑|k|>N|k|−2s.∫ ∞N1r2s·2pir dr.∫ ∞N1r2s−1dr. 1N2s−2.(2.8)152.3. the Main LemmaIf || f ||H˙s ≤ 3, we could simply take N = 1; otherwise take Ns−1 close to|| f ||H˙s . For a similar lemma and proof, we refer to [12] and [11].16Chapter 3Stability of a First OrderSemi-implicit Scheme on 2DAllen-Cahn EquationAllen−Cahn equation is a ∆1 version of Cahn−Hilliard equation withbilaplacian: ∂tu= ν∆u− f (u)u(x,0)= u0.Here f (u) = u3 − u, and the spatial domain Ω is often taken to be the2pi−periodic torus T2. Also we sometimes use ε2 instead of ν as ν is a smallparameter. The corresponding energy is defined by E(u)= ∫Ω(ν2 |∇u|2+F(u)) dx, where F(u)= 14 (u2−1)2, the anti-derivative of f (u).As is well known, the energy satisfies E(u(t)) ≤ E(u(s)) ,∀ t ≥ s, whichgives a priori bound. Now we consider a stabilized semi-implicit scheme in-troduced in [11]. The form is the following:173.1. Stability Theorem for Allen-Cahn Equationun+1−unτ= ν∆un+1−A(un+1−un)−ΠN f (un)u0 =ΠN u0. (3.1)where τ is the time step and A > 0 is the coefficient for the O(τ) regularizationterm. For N ≥ 2, defineXN = span{cos(k · x) ,sin(k · x) : k= (k1,k2) ∈Z2 , |k|∞ =max{|k1|, |k2|}≤N}.So define the L2 projection operator ΠN : L2(Ω)→ XN by (ΠN u−u,φ) =0 ∀φ ∈ XN , where (·, ·) denotes the L2 inner product on Ω. In other words,the projection operator ΠN is just the truncation of Fourier modes |k|∞ ≤ N.ΠN u0 ∈ XN and by induction, we have un ∈ XN ,∀n≥ 0.3.1 Stability Theorem for Allen-Cahn EquationTheorem 3.1.1. (unconditional energy stability for AC). Consider (3.1) withν> 0 and assume u0 ∈H2(T2). Then there exists a constant β0 depending onlyon the initial energy E0 =E(u0) such that ifA ≥β · (||u0||2H2 +ν−1| logν|+1) , β≥β0 (3.2)then E(un+1)≤E(un), ∀n≥ 0, where E is defined above.Remark 2. Similar to [11], the stability result is valid for any time step τ.Our choice of A is independent of τ as long as it has size of O (1/ν| log(ν)|) atleast. Note that the choice of A may not be optimal and further work could bedone.183.2. Proof of the Stability TheoremRemark 3. The condition u0 ∈H2(T2) results from the classic Sobolev embed-ding supN ||ΠN u0||∞. ||u0||H2(T2). No mean zero assumption is needed for u0 .To prove this we need a log-type interpolation inequality, which is themain lemma.3.2 Proof of the Stability TheoremThe proof uses an induction argument. To start with, let’s recall the nu-merical scheme (3.1)un+1−unτ= ν∆un+1−A(un+1−un)−ΠN f (un).Here ΠN is truncation of Fourier modes of L2 functions to |k|∞ ≤N. Mul-tiply the equation by (un+1−un) and integrate over Ω, one has1τ∫T2|un+1−un|2 = ν∫T2∆un+1(un+1−un)−A∫T2|un+1−un|2−(ΠN f (un),un+1−un) .Because un is periodic, (as un ∈ XN ), hence by integration by parts, we have(1τ+A)∫T2|un+1−un|2+ν∫T2∇un+1∇(un+1−un)=−(ΠN f (un),un+1−un) .Note ∇un+1∇(un+1−un)= 12 (|∇un+1|2−|∇un|2+|∇(un+1−un)|2), we have(1τ+A)∫T2|un+1−un|2+ν2∫T2|∇un+1|2−|∇un|2+|∇(un+1−un)|2 =−(ΠN f (un),un+1−un) .193.2. Proof of the Stability TheoremMoreover, every un ∈ XN , we have(1τ+A)∫T2|un+1−un|2+ν2∫T2|∇un+1|2−|∇un|2+|∇(un+1−un)|2 =−( f (un),un+1−un) .(3.3)Now, by fundamental theorem of calculus and integration by parts,F(un+1)−F(un)= f (un)(un+1−un)+∫ un+1unf ′(s)(un+1− s) ds= f (un)(un+1−un)+∫ un+1un(3s2−1)(un+1− s) ds= f (un)(un+1−un)+ 14(un+1−un)2 (3(un)2+ (un+1)2+2unun+1−2) .(3.4)Combine previous two equations, and denote E(un) by En we have(1τ+A)||un+1−un||2L2 +ν2||∇(un+1−un)||2L2 +ν2||∇un+1||2L2 −ν2||∇un||2L2+∫T2F(un+1)−F(un)= 14((un+1−un)2,3(un)2+ (un+1)2+2unun+1−2)Noteν2||∇un+1||2L2 +∫T2F(un+1)=E(un+1)=En+1=⇒ (1τ+A+ 12)||un+1−un||2L2 +ν2||∇(un+1−un)||2L2 +En+1−En= 14((un+1−un)2,3(un)2+ (un+1)2+2unun+1)≤ ||un+1−un||2L2(||un||2∞+12||un+1||2∞).(3.5)To show En+1 ≤En, clearly it suffices to show1τ+A+ 12≥ 32max{||un||2∞ , ||un+1||2∞} . (3.6)203.2. Proof of the Stability TheoremNote that E0 = E(ΠN u0) while E0 = E(u0). In general E0 6= E0. In effect, weclaim thatProposition 1.supNE(ΠN u0). 1+E0,where u0 ∈H1(T2) . (3.7)Proof. First, we write ΠN u0 as 1(2pi)2∑|k|≤N û0(k)eik·x , namely the Dirichletpartial sum of u0.||∇ (ΠN u0) ||2L2(T2) =1(2pi)2∑|k|≤N|k|2|û0(k)|2 ≤ 1(2pi)2∑|k|∈Z2|k|2|û0(k)|2 = ||∇ (u0) ||2L2(T2) .(3.8)The first equality above is because the operator ΠN is just a truncation ofFourier modes.On the potential energy part, by standard Sobolev inequality, ||u0||L4(T2).||u0||H1(T2), this shows u0 ∈ L4(T2) and hence the Dirichlet partial sum ΠN u0converges to u0 in L4(T2). Then, ||ΠN u0||L4(T2) → ||u0||L4(T2), which leadsto supN ||ΠN u0||L4(T2) < ∞. By Uniform Boundedness Principle, we derivesupN ||ΠN || < ∞, i.e. supN ||ΠN u0||L4(T2) ≤ c||u0||L4(T2) for an absolute con-stant c. Combine two estimates above we conclude the proof for the claim.For an alternate proof, see [10], and this claim holds for 3D case as wellwith a similar proof.We rewrite the numerical scheme (3.1) as following:un+1 = 1+Aτ1+Aτ−ντ∆un− τ1+Aτ−ντ∆ΠN [ f (un)] . (3.9)213.2. Proof of the Stability TheoremBy the lemma, to control ||un+1||∞ and ||un||∞, we may consider H˙1-normand H˙32 -norm together with 0th-mode |ûn+1(0)|.To start with,|ûn+1(0)| ≤ |ûn(0)|+ τ1+Aτ | f̂ (un)(0)|≤ |ûn(0)|+ 1A| f̂ (un)(0)|≤ |∫T2un dx|+ |∫T2un− (un)3 dx|. 1+|∫T2(un)2 dx| 12 +|∫T2(1− (un)2)2 dx| 12. 1+pEn ,(3.10)where the last 2 inequalities are by Cauchy-Schwarz inequalities and Hölder’sinequalities.Lemma 3.2.1. There is an absolute constant c1 > 0 such that for any n≥ 0||un+1||H˙32 (T2)≤ c1 ·(A+1ν+ 1ντ)· (En+1)||un+1||H˙1(T2) ≤(1+ 1A+ 3A||un||2∞)· ||un||H˙1(T2) .(3.11)Proof. As 0th-mode will not contribute to H˙1 norm and H˙32 norm, we couldjust consider Fourier modes |k| ≥ 1 from the Fourier side. Use the symbolf . g to denote f ≤ c · g with c being a constant.(1+Aτ)|k| 321+Aτ+ντ|k|2 .1+Aτνττ|k| 321+Aτ+ντ|k|2 .ττν|k|− 12 = 1ν|k|− 12 .(3.12)223.2. Proof of the Stability TheoremHence||un+1||H˙32 (T2).(1+Aτντ)||un||L2(T2)+1ν||〈∇〉− 12 f (un)||L2(T2). (3.13)Here the notaion〈∇〉s = (1−∆) s2 , corresponds to the Fourier side (1+|k|2)s/2.Note ||un||L2(T2) .∫T214 (u4−2u2+1) dx+1. En+1 by Cauchy-Schwarz in-equality. By Sobolev inequality ||〈∇〉− 12 f (un)||L2(T2). || f (un)||L 43 (T2) = ||(un)3−un||L43 (T2)=(∫T2((un)3−un) 43 dx) 34 .(∫T2(un)4 dx) 34 .En+1. Hence (3.13) be-comes||un+1||H˙32 (T2).(1+Aτντ+ 1ν)(En+1). (3.14)Similarly, (1+Aτ)|k|1+Aτ+ντ|k|2 . |k|τ|k|1+Aτ+ντ|k|2 .ττA|k| = 1A|k| .(3.15)This implies||un+1||H˙1(T2). ||un||H˙1(T2)+1A|| f (un)||H˙1(T2). ||un||H˙1(T2)+1A||∇( f (un))||L2(T2). ||un||H˙1(T2)+1A||(3(un)2−1) · (∇un)||L2(T2). ||un||H˙1(T2)+ (1A+ 3||u||2∞A)||un||H˙1(T2).(1+ 1A+ 3||u||2∞A)||un||H˙1(T2) .(3.16)233.2. Proof of the Stability TheoremNow we prove by induction.Step 1: The induction n =⇒ n+ 1 step. Assume En ≤ En−1 ≤ ·· · ≤ E0and En ≤ supN E(ΠN u0), we would show En+1 ≤ En. This implies ||un||2H˙1 =||∇un||2L2 ≤ 2Enν≤ 2E0ν.So by the main lemma, use the notation f .E0 g to denote there exists aconstant C(E0) depends only on E0 such that f ≤C(E0) · g, we have||un||2∞. ||un||2H˙1(√log(3+ c1(1ντ+ A+1ν)(En+1)))2+En+1. 2E0ν(1+ log(A)+ log(1ν)+ (log(1+ 1τ)))+E0+1.E0 ν−1(1+ log(A)+ log(1ν))+ν−1| log(τ)|+1 .(3.17)Define m0 := ν−1 (1+ log(A)+| log(ν)|), and note that E0 ≤ supN E(ΠN u0).E0+1, the inequality above is||un||2∞.E0 m0+ν−1| log(τ)|+1. (3.18)243.2. Proof of the Stability TheoremOn the other hand,||un+1||2∞.(1+||un+1||H˙1√log(3+||un+1||H˙32))2.(1+ (1+||un||2∞A)||un||H˙1√log(3+||un+1||H˙32))2.E0(1+ (1+ m0+ν−1| log(τ)|A)(√1ν√log(3+||un+1||H˙32))2.E0(1+ (1+ m0+ν−1| log(τ)|A)(√m0+ν−1| log(τ)|))2.E0(1+√m0+ν−1| log(τ)|+ (√m0+ν−1| log(τ)|)3A)2.E0 1+m30A2+m0+ν−3| log(τ)|3 .(3.19)Hence sufficient condition (3.6) becomesA+ 12+ 1τ≥C(E0)(m0+1+m30A2+ν−3| log(τ)|3)m0 = ν−1 (1+ log(A)+| log(ν)|) .(3.20)We discuss two cases.Case 1:1τ≥C(E0)ν−3| log(τ)|3. In this case, we need to choose A such thatAÀE0 m0 = ν−1 (1+ log(A)+| log(ν)|) .In fact, for ν& 1 , we could take A ÀE0 1; if 0 < ν¿ 1, we would chooseA = CE0 · ν−1| logν| , where CE0 is a large constant depending only on E0.Therefore in both cases it suffices to chooseA =CE0 ·max{ν−1| log(ν)| , 1} . (3.21)253.2. Proof of the Stability TheoremCase 2:1τ≤ C(E0)ν−3| log(τ)|3. This implies | log(τ)|.E0 1+ | log(ν)|. Nowwe go back to equations (3.17), we have||un||2∞.E0 m0, (3.22)as ν−1| log(τ)| would be absorbed by m0, recall m0 = ν−1 (1+ log(A)+| log(ν)|).Hence substitute this new bound to (3.19), we would get||un+1||2∞.(1+ (1+||un||2∞A)||un||H˙1√log(3+||un+1||H˙32))2.E0(1+ (1+ m0A)√1ν√log(3+||un+1||H˙32))2.E0(1+ (1+ m0A)pm0)2.E0 1+m30A2+m0 .(3.23)This shows it suffices to takeA ≥CE0 m0, (3.24)for a large enough constant CE0 depending only on E0. The same choiceof A in Case 1( with a larger CE0 if necessary) would still work.Step 2: check the induction base step n = 1. It’s clear that we only needto checkA+ 12+ 1τ≥ ||ΠN u0||2∞+12||u1||2∞.263.2. Proof of the Stability TheoremBy the lemma 3.2.1,||u1||H˙1 ≤(1+ 1A+ 3A||ΠN u0||2∞)· ||u0||H˙1≤(1+ 1A+ 3A||ΠN u0||2∞)·√2E0ν.(3.25)As a result,||u1||2∞.(1+|û1(0)|+ ||u1||H˙1√log(3+||u1||H˙32))2.1+√E0+(1+ 1A+ 3A||ΠN u0||2∞)√2E0ν√log(3+ c1(A+1ν+ 1ντ)(E0+1))2.E0(1+(1+ 1A+ 3A||ΠN u0||2∞)·ν− 12 ·√1+ log(A)+| log(ν)|+ | log(τ)|)2.E0(1+ 1A+ 3A||ΠN u0||2∞)2·ν−1 · (1+ log(A)+| log(ν)|+ | log(τ)|) .(3.26)Thus we need to choose A such thatA+ 12+ 1τ≥ ||ΠN u0||2∞+CE0 ·(1+ 1A+ 3A||ΠN u0||2∞)2·ν−1· (1+ log(A)+| log(ν)|+ | log(τ)|) ,(3.27)where CE0 is a large constant depending only on E0. Note that by Morrey’sinequality for 2D domains,||ΠN u0||L∞(T2). ||ΠN u0||H2(T2). ||u0||H2(T2).Then it suffices to take A s.t.AÀE0 ||u0||2H2 +ν−1| log(ν)|+1 . (3.28)273.2. Proof of the Stability TheoremThis completes the induction and hence the theorem.28Chapter 4L2 Error Estimate of the FirstOrder Scheme on 2DAllen-Cahn EquationIn this chapter, we would like to study the L2 error between the semi-implicit numerical solution and the exact PDE solution in the domain T2. Tostart with, we consider the auxiliary L2 error estimate for near solutions.4.1 Auxiliary L2 Error Estimate for Near SolutionsConsider the following system:un+1−unτ= ν∆un+1−ΠN f (un)−A(un+1−un)+G1nvn+1−vnτ= ν∆vn+1−ΠN f (vn)−A(vn+1−vn)+G2nu0 = u0 , v0 = v0(4.1)where we would denote Gn =G1n−G2n.We state the proposition here.294.1. Auxiliary L2 Error Estimate for Near SolutionsProposition 2. For solutions of (4.1), assume for some N1 > 0,supn≥0||un||∞+supn≥0||vn||∞ ≤N1 . (4.2)Then for any m≥ 1,||um−vm||2L2 = ||em||2L2≤ exp(mτ ·{C((1+N21 )N1ν+N21 +ν(1+N21 )N1)+ Bν})·((1+Aτ) ||u0−v0||2L2 +Bτνm−1∑n=0||Gn||2L2) (4.3)where B , C > 0 are two absolute constants.Proof. Write en = un−vn. Thenen+1− enτ= ν∆en+1−A(en+1− en)−ΠN(f (un)− f (vn))+Gn . (4.4)Take the L2 inner product with en+1 on both sides and recall similar compu-tations in chapter 3, one has12τ(||en+1||2L2 −||en||2L2 +||en+1− en||2L2)+ν||∇en+1||2L2 + A2 (||en+1||2L2 −||en||2L2+||en+1− en||2L2)=(Gn, en+1)+ ( f (un)− f (vn), ΠN en+1)(4.5)where (. , .) denotes the L2 inner product and the last term is because ΠN isa self-adjoint operator (ΠN f , g) = ( f , ΠN g), since it is just an N-th Fouriermode truncation.304.1. Auxiliary L2 Error Estimate for Near SolutionsNow by Hölder’s inequality|(Gn, en+1) | ≤ ||en+1||L2 ||Gn||L1 ≤B||en+1||L2 ||Gn||L2 ≤ 2B(ν||Gn||2L2 +||en+1||2L2ν)(4.6)Next, by fundamental theorem of calculus, we computef (un)− f (vn)=∫ 10f ′(vn+ sen)ds en= (a1+a2(vn)2)en+a3vn(en)2+a4(en)3 ,(4.7)where ai are constants could be computed. Now we shall denote by C anabsolute constant whose value may vary in different lines. Now,|((a1+a2(vn)2)en , en+1) | ≤C(1+||vn||2∞)||en+1||L2 ||en||L2≤C(1+N21 )N1(||en+1||2L2ν+ν||en||2L2)≤ C(1+N21 )N1ν||en+1||2L2 +ν ·C(1+N21 )N1||en||2L2 ,(4.8)also,|(a3vn(en)2 , en+1) | ≤C||vn||∞||en+1||∞||en||2L2≤CN21 ||en||2L2 ,(4.9)|(a4(en)3 , en+1) | ≤C||en+1||∞||en||∞||en||2L2≤CN21 ||en||2L2 .(4.10)To simplify the formula, we would use the notation ||u||2 to denote the L2norm. Collecting all estimates, we get314.1. Auxiliary L2 Error Estimate for Near Solutions||en+1||22−||en||22τ+A(||en+1||22−||en||22)≤Bν||Gn||22+Bν||en+1||22C(ν(1+N21 )N1+N21) ||en||22+ C(1+N21 )N1ν ||en+1||22(4.11)where B and C are two absolute constants that could be computed exactly.Hence for ν small, recall A is chosen large than O(ν−1| logν|), we derive||en+1||22−||en||22τ+(A− C(1+N21 )N1ν− Bν)(||en+1||22−||en||22)≤Bν||Gn||22+{C((1+N21 )N1ν+N21 +ν(1+N21 )N1)+ Bν}||en||22(4.12)Defineyn =(1+(A− C(1+N21 )N1ν− Bν)τ)||en||22 ,α=C((1+N21 )N1ν+N21 +ν(1+N21 )N1)+ Bν,βn =Bν||Gn||22 .(4.13)This shows for ν small,yn+1− ynτ≤αyn+βn .Applying discrete Gronwall’s inequality, we have324.2. L2 Error Estimate of 2D Allen-Cahn Equation||um−vm||22 = ||em||22 ≤ ym ≤ exp(mτ ·{C((1+N21 )N1ν+N21 +ν(1+N21 )N1)+ Bν})·((1+(A− C(1+N21 )N1ν− Bν)τ)||u0−v0||22+Bτνm−1∑n=0||Gn||22)≤ exp(mτ ·{C((1+N21 )N1ν+N21 +ν(1+N21 )N1)+ Bν})·((1+Aτ) ||u0−v0||22+Bτνm−1∑n=0||Gn||22).(4.14)4.2 L2 Error Estimate of 2D Allen-Cahn EquationIn this section, to simplify the notation, we would write x . y if x ≤C(ν , u0) y for a constant C depending on ν and u0. We consider the sys-tem un+1−unτ= ν∆un+1−ΠN f (un)−A(un+1−un)∂tu= ν∆u− f (u)u0 =ΠN u0 , u(0)= u0 .(4.15)Theorem 4.2.1. Let ν > 0. Let u0 ∈ Hs, s ≥ 4 and u(t) be the solution toAllen-Cahn equation with initial data u0. Let un be the numerical solutionwith initial data ΠN u0. Assume A satisfies the same condition in the stabilitytheorem. Define tm =mτ, m≥ 1. Then||um−u(tm)||2 ≤ A · eC1 tm ·C2 ·(N−s+τ) ,334.2. L2 Error Estimate of 2D Allen-Cahn Equationwhere C1 > 0 depends only on (u0,ν) and C2 depends on (u0,ν, s).In order to prove this theorem, it is clear that we shall estimate Gn inprevious proposition. Note that for a one-variable function h(t), one has theformula:1τ∫ tn+1tnh(t)= h(tn)+ 1τ∫ tn+1tnh′(t) · (tn+1− t) dt1τ∫ tn+1tnh(t)= h(tn+1)+ 1τ∫ tn+1tnh′(t) · (tn− t) dt .(4.16)Using the formula above and integrating Allen-Cahn equation on the timeinterval [tn , tn+1], we getu(tn+1)−u(tn)τ=ν∆u(tn+1)−A (u(tn+1)−u(tn))−ΠN f (u(tn))−Π>N f (u(tn))+Gn(4.17)where Π>N = id−ΠN , the large mode truncation andGn = ντ∫ tn+1tn∂t∆u · (tn− t) dt− 1τ∫ tn+1tn∂t( f (u))(tn+1− t) dt+A∫ tn+1tn∂tu dt .(4.18)To bound ||Gn||2, we introduce some useful lemmas.4.2.1 Bounds on Allen-Cahn Exact Solution and NumericalSolutionLemma 4.2.2. (maximum principle for smooth solutions to Allen-Cahn equa-tion) Let T > 0 and assume u ∈ C2xC1t (Td × [0,T]) is a classical solution to344.2. L2 Error Estimate of 2D Allen-Cahn EquationAllen-Cahn equation with initial data u0. Then||u(. , t)||∞ ≤max{||u0||∞ , 1} , ∀0≤ t≤T . (4.19)Remark 4. As proved in [5], there exists a global H4xC1t solution to Allen-Cahnequation. In fact as pointed out by Li, Qiao and Tang in [12], the regularitywould be even higher due to the smoothing effect of heat kernel and the non-linear term. So we would assume a smooth solution here.Proof. We define f (x, t) = u(x, t)2 and f ²(x, t) = f (x, t)− ²t. Since f ² is a con-tinuous function on the compact domain Td × [0,T], it achieves maximum atsome point (x∗, t∗), i.e.max0≤t≤T , x∈Tdf ²(x, t)= f ²(x∗, t∗) :=M² .We discuss several cases.Case 1. 0< t∗ ≤T and M²> 1. This shows ∇ f ²(x∗, t∗)= 0 , ∆ f ²(x∗, t∗)≤ 0.Note that∇ f ² = 2u∇u , ∆ f ² = 2|∇u|2+2u∆u , (4.20)this shows ∇u(x∗, t∗)= 0, u∆u(x∗, t∗)< 0.354.2. L2 Error Estimate of 2D Allen-Cahn EquationHowever,∂t f ²(x∗, t∗)= 2u(x∗, t∗)∂tu(x∗, t∗)−²= 2u(x∗, t∗)(ν∆u(x∗, t∗)−u3(x∗, t∗)+u(x∗, t∗))−²<−2u4(x∗, t∗)+2u2(x∗, t∗)−²<−2(u2(x∗, t∗)− 12)2+ 12−²<−²< 0(4.21)as u2(x∗, t∗) > 1 by assumption. This contradicts the hypothesis that f ²achieves its maximum at (x∗, t∗) and hence Case 1 is impossible.Case 2. 0< t∗ ≤T and M²≤ 1. In this case we obtainmax0≤t≤T , x∈Tdf (x, t)≤ 1+²T ,letting ²→ 0, we obtain f (x, t)≤ 1.Case 3. t∗ = 0, thenmax0≤t≤T , x∈Tdf (x, t)≤maxx∈Tdf (x,0)+²T ,sending ² to 0, we obtain f (x, t)≤ f (x,0).This concludes ||u||∞ ≤max{||u0||∞ , 1}.Lemma 4.2.3. (Hk boundedness of exact solution) Assume u(x, t) is a smoothsolution to Allen-Cahn equation in Td with d ≤ 3 and the initial data u0 ∈Hk(Td) for k≥ 2. Then,supt≥0||u(t)||Hk(Td).k 1 (4.22)364.2. L2 Error Estimate of 2D Allen-Cahn Equationwhere we omits the dependence on ν and u0.Proof. By the Duhamel formula, we writeu(t)= eνt∆u0+∫ t0eν(t−s)∆(u−u3) ds . (4.23)We would prove this argument inductively. By previous lemma, we have||u||2 . 1 as ||u||∞ . 1 and we would show ||u||H1 . 1 for any t ≥ 1. Then bytaking spatial derivative and L2 norm in the formula above, we derive||Du||2 ≤ ||Deνt∆u0||2+∫ t0||Deν(t−s)∆(u−u3)||2 ds (4.24)where Du denotes any differential operator Dαu for any |α| = 1, for exampleD2 denotes ∂2xi x j u for 1≤ i , j ≤ d.First, we consider the nonlinear part.||Deν(t−s)∆(u−u3)||2. ||Deν(t−s)∆(u−u3)||∞. |K1∗ (u−u3)| , (4.25)where K1 is the kernel corresponding to Deν(t−s)∆.|K1∗ (u−u3)| ≤ ||K1||2 · ||u−u3||2. ||K1||2 · ||u||2(4.26)374.2. L2 Error Estimate of 2D Allen-Cahn Equationby the boundedness of ||u||∞. Note that||K1||2.( ∑k∈Zd|k|2e−2ν(t−s)|k|2) 12=( ∑|k|≥1|k|2e−2ν(t−s)|k|2) 12.(∫ ∞1e−2ν(t−s)r2rd+1 dr) 12.(4.27)The estimate for even dimensional case and odd dimensional case is a bitdifferent. Now we would assume t≥ 1, as the other case t< 1 is much easier.1. Case 1, d = 1. ∫∞1 e−2ν(t−s)r2 r2 dr. e−2ν(t−s)t−s + erfc(p2ν(t−s))(t−s)3/2 , where erfc(x) :=2ppi∫∞x e−t2 dt, the complementary error function. Letting γ= t− s,∫ t0||Deν(t−s)∆u||2 ds.(∫ t0e−νγγ1/2+ (erfc(pνγ)1/2γ3/4dγ)· ||u||2 . (4.28)For γ close to 0, (erfc(pνγ)1/2γ3/4will dominate the estimate and for γ awayfrom 0, e−νγγ1/2shall dominate the estimate. Then we split the integral asfollowing (recall t≥ 1):∫ t0e−νγγ1/2+ (erfc(pνγ)1/2γ3/4dγ.∫ 101γ3/4dγ+∫ t1e−νγγ1/2dγ. 1+∫ ∞0e−νγγ1/2dγ. 1 .(4.29)2. Case 2, d = 2. ∫∞1 e−2ν(t−s)r2 r3 dr . e−2ν(t−s)(t−s)2 + e−2ν(t−s)t−s . Similar to case 1,384.2. L2 Error Estimate of 2D Allen-Cahn Equationwe would split the integral as well. Letting γ= t− s, we have∫ t1e−νγγ+ e−νγγ1/2dγ.∫ t1e−νγγ1/2dγ.∫ ∞0e−νγγ1/2dγ. 1 .(4.30)However, it does not work for γ ≤ 1. Now we use another estimate for||K1∗ (u−u3)||L2(Td). We compute from Fourier side:||K1∗ (u−u3)||2L2(Td) =∑|k|≥1|k|2e−2ν(t−s)|k|2 |àu−u3(k)|2≤max|k|≥1{|k|2e−2ν(t−s)|k|2}· ∑|k|≥1|àu−u3(k)|2.max|k|≥1{|k|2e−2ν(t−s)|k|2}· ||u||2L2(Td) .(4.31)Define g(x)= x2e−2νγx2 , where x≥ 0. Then,g′(x)= 2xe−2νγx2 (1−2νγx2) , (4.32)this shows the maximum achieves at x= 1p2νγand henceg(x)≤ g( 1√2νγ). 1γ(4.33)and hence||Deν(t−s)∆(u−u3)||L2(Td).1pt− s ||u||L2(Td) , (4.34)394.2. L2 Error Estimate of 2D Allen-Cahn Equationnote that this proof works for any dimension.As a result,∫ 10||Deνγ∆u||2 dγ.∫ 101pγdγ · ||u||2. 1 . (4.35)This shows∫ t0 ||Deν(t−s)∆u||2 ds. 1.3. Case 3, d = 3. As proved in previous case, we would only need to checkthe case γ≥ 1. ∫∞1 e−2νγr2 r4 dr. e−2νγγ for γ≥ 1. Hence,∫ t1e−νγγ1/2dγ.∫ ∞0e−νγγ1/2dγ. 1 .(4.36)For t≤ 1 case, it is easier because we do not need to split the integral and allintegrals from 0 to t could be bounded by the integral from 0 to 1.Now for the linear part, by Duhamel’s Principle, eνt∆u0 denotes the solu-tion to the system: ∂tu= ν∆uu(x,0)= u0 .(4.37)As is well known, every spatial derivative of the solution eνt∆u0 solvessame PDE, hence by the energy decay property, we have ||eνt∆u0||Hm . ||u0||Hmfor any 1 ≤ m ≤ k. Combine the nonlinear part and linear part, we derive||u||H1 . 1 independent of t≥ 0 and hence supt≥0 ||u||H1 . 1.404.2. L2 Error Estimate of 2D Allen-Cahn EquationNow if we already obtain supt≥0 ||u||Hm−1 . 1, we just need to consider||D(Dm−1u)||2 ≤ ||Deνt∆Dm−1u0||2+∫ t0||Deν(t−s)∆Dm−1u||2 ds. ||u0||Hm +∫ 10||Deνγ∆Dm−1u||2 dγ+∫ t1||Deνγ∆Dm−1u||2. 1+∫ 101pγdγ · ||Dm−1u||2+∫ ∞0e−νγpγdγ · ||Dm−1u||2. 1 ,(4.38)by repeating the process above. In the end we would achievesupt≥0||u||Hk(Td).k 1 . (4.39)Lemma 4.2.4. (Discrete version Hk boundedness) Suppose u0 ∈Hk(Td) withd ≤ 3 and k≥ 2. Then, suppose un is the numerical solution that satisfiesun+1−unτ= ν∆un+1−A(un+1−un)−ΠN f (un)u0 =ΠN u0 ,(4.40)thensupn≥0||un||Hk(Td).A,k 1 . (4.41)Remark 5. The bound on un is independent of time step τ and truncationnumber N.Remark 6. The proof is involved with energy decay property of the numericalscheme, so we would assume this property for now, as the proof for energy414.2. L2 Error Estimate of 2D Allen-Cahn Equationdecay in 3D case would be given in chapter 6.Proof. To simplify the notation, we would use “.” instead of “.ν,u0,A,k” onlyin this lemma. We would like to use a similar method provided in [10].Writeun+1 = 1+Aτ1+Aτ−ντ∆︸ ︷︷ ︸:=L1un+ −τΠN1+Aτ−ντ∆︸ ︷︷ ︸:=L2f (un)= L1(L1un−1+L2 f (un−1))+L2 f (un)= Lm0+11 un−m0 +m0∑l=0Ll1L2 f (un−1) ,(4.42)where m0 would be chosen later.Similar to continuous version, we prove inductively. First, we showsupn≥0||un||H2(Td). 1 . (4.43)Recall supn≥0 ||un||2 . 1 and supn≥0 || f (un)||2 . 1 by energy decay property,then we just need to consider H˙2 semi norm.We discuss 3 cases:1. Case 1: τ≥ 110 . Then for each 0 6= k ∈Zd,∣∣L̂1(k)∣∣= 1+Aτ1+Aτ+ντ|k|2≤ 1Aτ+ντ|k|2 +AA+ν|k|2. 11+|k|2 ;(4.44)∣∣L̂2(k)∣∣≤ τAτ+ντ|k|2 . 11+|k|2 . (4.45)424.2. L2 Error Estimate of 2D Allen-Cahn EquationAs a result,||un+1||H˙2 ≤ ||L1un||H˙2 +||L2 f (un)||H˙2. ||un||2+|| f (un)||2. 1 .(4.46)2. Case 2: τ< 110 and Aτ≥ 110 . Then for 0 6= k ∈Zd:∣∣L̂1(k)∣∣= 1+Aτ1+Aτ+ντ|k|2≤ 11AτAτ+ντ|k|2. 11+|k|2 ,(4.47)and ∣∣L̂2(k)∣∣= τ1+Aτ+ντ|k|2 . 11+|k|2 . (4.48)Then similar to Case 1,||un+1||H˙2 ≤ ||L1un||H˙2 +||L2 f (un)||H˙2. ||un||2+|| f (un)||2. 1 .(4.49)3. Case 3: τ < 110 but Aτ < 110 . Take m0 to be one integer such that 12 ≤434.2. L2 Error Estimate of 2D Allen-Cahn Equationm0τ< 1 and thus m0 ≥ 5.∣∣∣àLm0+11 (k)∣∣∣≤ ( 1+Aτ1+Aτ+ντ|k|2)m0+1≤(1+Aτ1+Aτ+ντ|k|2)m0=(1+ ντ|k|21+Aτ)−m0.(4.50)Recall Aτ< 110 < 1, then(1+ ντ|k|21+Aτ)−m0≤(1+ ντ|k|22)−m0, (4.51)define t0 :=m0τ and we derive∣∣∣àLm0+11 (k)∣∣∣≤ (1+ 12ν|k|2 t0m0)−m0. (4.52)For any a> 0, we consider the function h(x)=−x log(1+ ax ), x> 0. Thenh′(x)=− log(1+ ax)+ aa+ xh′′(x)= ax+a(1x− 1x+a)> 0 .(4.53)By direct computation, h(x) decreases on (0,∞). Therefore, recallingm0 ≥ 5,∣∣∣àLm0+11 (k)∣∣∣≤ (1+ 12ν|k|2 t0m0)−m0≤(1+ 12ν|k|2 · t05)−5. (4.54)Now,444.2. L2 Error Estimate of 2D Allen-Cahn Equation∣∣L̂2(k)∣∣ · m0∑l=0∣∣L̂1(k)∣∣l ≤ ∣∣L̂2(k)∣∣ · 11− ∣∣L̂1(k)∣∣= τ1+Aτ+ντ|k|2 ·11− 1+Aτ1+Aτ+ντ|k|2= 1ν|k|2. 1|k|2 .(4.55)Therefore for n≥m0,||un+1||H˙2 . ||un−m0 ||2+ sup0≤l≤m0|| f (un−l)||2. 1 . (4.56)For 1≤ n≤m0+1, then we applyun = Ln1 u0+n−1∑l=0Ll1L2 f (un−1−l) . (4.57)Hence||un||H˙2 . ||u0||H˙2 + sup0≤l≤n−1|| f (un−l−1)||2. 1 . (4.58)This concludessupn≥0||un||H2(Td). 1 . (4.59)454.2. L2 Error Estimate of 2D Allen-Cahn EquationInductively,||un+1||H˙m . ||un||H˙m−2 +|| f (un)||H˙m−2 , τ≥110||un+1||H˙m . ||un||H˙m−2 +|| f (un)||H˙m−2 , τ<110, Aτ≥ 110||un+1||H˙m . ||un−m0 ||H˙m−2 + sup0≤l≤m0|| f (un−l)||H˙m−2 , τ<110, Aτ< 110, n≥m0||un||H˙m . ||u0||H˙m + sup0≤l≤n−1|| f (un−l−1)||H˙m−2 , τ<110, Aτ< 110, n≤m0+1(4.60)thus provesupn≥0||un||Hk(Td). 1 . (4.61)Remark 7. The proof for exact solution and numerical solution is similar inthe sense that we develop bootstrap process and split the time interval.4.2.2 Proof of L2 Error Estimate of 2D Allen-Cahn EquationProof. By the previous high Sobolev bound lemma, supn≥0 ||un||∞ . 1 usingMorrey’s inequality. Thus the assumptions of proposition 2 (auxiliary L2 er-ror estimate proposition) are satisfied.Recall thatGn = ντ∫ tn+1tn∂t∆u · (tn− t) dt− 1τ∫ tn+1tn∂t( f (u))(tn+1− t) dt+A∫ tn+1tn∂tu dt .(4.62)464.2. L2 Error Estimate of 2D Allen-Cahn EquationThen||Gn||2.∫ tn+1tn||∂t∆u||2 dt+∫ tn+1tn||∂t( f (u))||2 dt+A∫ tn+1tn||∂tu||2 dt.∫ tn+1tn||∂t∆u||2 dt︸ ︷︷ ︸I1+∫ tn+1tn||∂tu||2 dt ·(A+|| f ′(u)||L∞t L∞x)︸ ︷︷ ︸I2.(4.63)Note that ∂tu= ν∆u−u+u3 and hence by high Sobolev bound lemma,||∂tu||2. 1 , || f ′(u)||∞. 1 . (4.64)Recall the energy decay property,dEdt= ddt(∫ν|∇u|22+F(u) dx)=∫ν∇u ·∇∂tu+ f (u) ·∂tu dx=∫(−ν∆u+ f (u))∂tu dx=−||∂tu||22 .(4.65)This shows ∫ ∞0||∂tu||22 dt. 1 . (4.66)Note that by Gagliardo–Nirenberg interpolation inequality,||∂t∆u||2. ||〈∇〉3∂tu|| 232 · ||∂tu|| 132 . ||∂tu|| 132 . (4.67)Here the notaion〈∇〉s = (1−∆) s2 , corresponds to the Fourier side (1+|k|2)s/2.474.2. L2 Error Estimate of 2D Allen-Cahn EquationThis implies∫ ∞0||∂t∆u||62 dt. 1 ,=⇒∫ T0||∂t∆u||22 dt.(∫ T0||∂t∆u||62 dt) 13·(∫ T01 dt) 23. 1+T 23 .(4.68)Moreover,I1 =∫ tn+1tn||∂t∆u||2 dt.(∫ tn+1tn||∂t∆u||22 dt) 12·pτ . (4.69)Similarly,I2. (1+A) ·∫ tn+1tn||∂tu||2 dt. (1+A) ·(∫ tn+1tn||∂tu||22 dt) 12·pτ . (4.70)Hence for tm ≥ 1,m−1∑n=0||Gn||22.m−1∑n=0((I1)2+ (I2)2).m−1∑n=0(τ∫ tn+1tn||∂t∆u||22 dt+ (1+A)2τ∫ tn+1tn||∂tu||22 dt). τ∫ tm0||∂t∆u||22 dt+ (1+A)2τ∫ tm0||∂tu||22 dt. τ(1+ tm)+ (1+A)2τ. (1+A)2τ · (1+ tm) .(4.71)On the other hand, by the high Sobolev bound lemma supt≥0 ||u(t)||Hs .s1, we have supn≥0 || f (u(tn))||Hs .s 1.484.2. L2 Error Estimate of 2D Allen-Cahn Equation||Π>N f (u(tn))||22 =∑|k|>N∣∣∣áf (u(tn))(k)∣∣∣2≤ ∑|k|>N|k|2s∣∣∣áf (u(tn))(k)∣∣∣2 · |k|−2s.N−2s · ∑|k|>N|k|2s∣∣∣áf (u(tn))(k)∣∣∣2.N−2s · || f (u(tn))||2Hs.N−2s ,(4.72)thusm−1∑n=0||Π>N f (u(tn))||22.s m ·N−2s.tmN−2sτ. (4.73)Therefore,τm−1∑n=0(||Gn||22+||Π>N f (u(tn))||22).s (1+ tm)(τ2+N−2s)(1+A)2 . (4.74)Also for the same reason||u0−u(0)||22 = ||ΠN u0−u0||22.N−2s . (4.75)Applying the auxiliary solutions proposition and note that tm =mτ,||um−u(tm)||22.s (1+A)2eCtm(N−2s+τ ·N−2s+ (1+ tm)(τ2+N−2s)). (4.76)Note that τ ·N−2s. τ2+N−4s. τ2+N−2s1+ tm. eC′ tm ,(4.77)494.2. L2 Error Estimate of 2D Allen-Cahn Equationthis leads to||um−u(tm)||22.s (1+A)2eCtm(N−2s+τ2) . (4.78)Thus||um−u(tm)||2 ≤ (1+A) ·C2 · eC1 tm(N−s+τ) , (4.79)where C1 > 0 is a constant depending on ν ,u0; C2 > 0 is a constant dependingon s ,ν and u0.This completes the proof of L2 error estimate.50Chapter 5Stability of a First OrderSemi-implicit Scheme on 2DFractional Cahn-HilliardEquationAs mentioned in the introduction, the fractional Cahn-Hilliard equationare “interpolation” between Allen-Cahn equation and original Cahn-Hilliardequation. ∂tu= ν∆((−∆)αu+ (−∆)α−1 f (u)) , 0<α≤ 1u(x,0)= u0.As before, we consider the region as 2pi-periodic torus T2 = R2/2piZ2. f (u) =u3−u and hence the energy E(u)= ∫T2 (ν2 |∇u|2+F(u)) dx , with F(u)= 14 (u2−1)2. Similarly, the semi-implicit scheme is given by the following:un+1−unτ=−ν(−∆)α+1un+1− (−∆)αA(un+1−un)− (−∆)αΠN f (un)u0 =ΠN u0. (5.1)51Chapter 5. Stability of a First Order Semi-implicit Scheme on 2D Fractional Cahn-Hilliard EquationTheorem 5.0.1. (unconditional energy stability for fractional CH). Consider(5.1) with ν > 0 and assume u0 ∈ H2(T2) and has zero-mean condition. Thenthere exists a constant β0 depending only on the initial energy E0 =E(u0) suchthat ifA ≥β · (||u0||2H2 +ν−1| logν|+1) , β≥β0 (5.2)then E(un+1)≤E(un), ∀n≥ 0, where E is defined above.Remark 8. Here we require zero-mean assumption on u0 hence that impliesun all have mean zero because zero-mean assumption would guarantee thatnegative fractional Laplacian is well defined. Here we use the notation |∇|−α =(−∆)− α2 to denote the fractional Laplacian.Proof. The proof is involved with similar computation given in previous chap-ter. We recall the scheme (5.1):un+1−unτ=−ν(−∆)α+1un+1− (−∆)αA(un+1−un)− (−∆)αΠN f (un) .Now we multiply the equation by (−∆)−α(un+1 − un) and apply Funda-mental Theorem of Calculus and integration by parts as in Chapter 3, weobtain1τ∣∣∣∣|∇|−α(un+1−un)∣∣∣∣2L2 + ν2(∣∣∣∣∇(un+1−un)∣∣∣∣2L2 + ∣∣∣∣∇un+1∣∣∣∣2L2 − ∣∣∣∣∇un∣∣∣∣2L2)+A ∣∣∣∣un+1−un∣∣∣∣2L2 =−( f (un),un+1−un) .(5.3)52Chapter 5. Stability of a First Order Semi-implicit Scheme on 2D Fractional Cahn-Hilliard EquationThis thus implies:1τ∣∣∣∣|∇|−α(un+1−un)∣∣∣∣2L2 + ν2 ∣∣∣∣∇(un+1−un)∣∣∣∣2L2 +(A+ 12)∣∣∣∣un+1−un∣∣∣∣2L2 +En+1−En≤ ∣∣∣∣un+1−un∣∣∣∣2L2 (||un||2∞+ 12 ||un+1||2∞).(5.4)It is clear that the first two norms 1τ∣∣∣∣|∇|−α(un+1−un)∣∣∣∣2L2 and ν2 ∣∣∣∣∇(un+1−un)∣∣∣∣2L2would be problematic as we would expect more help from ||un+1−un||2L2 .Lemma 5.0.2.1τ∣∣∣∣|∇|−α(un+1−un)∣∣∣∣2L2(T2)+ ν2 ∣∣∣∣∇(un+1−un)∣∣∣∣2L2(T2) ≥Cαντ||un+1−un||2L2(T2)(5.5)with Cαντ is determined by α, ν and τ.Proof. It is natural to examine the above norms 1τ∣∣∣∣|∇|−α(un+1−un)∣∣∣∣2L2(T2)andν2∣∣∣∣∇(un+1−un)∣∣∣∣2L2(T2) on the Fourier side. Then we obtain that1τ∑k 6=0|k|−2α|ûn+1(k)− ûn(k)|2+ ν2∑k 6=0|ûn+1(k)− ûn(k)|2= ∑k 6=0|ûn+1(k)− ûn(k)|2 ·( |k|−2ατ+ ν|k|22).(5.6)Note we expect∑k 6=0 |ûn+1(k)− ûn(k)|2, it is clear we could apply standardYoung’s inequality for product : ab≤ aγγ+ bββ, with 1γ+ 1β= 1.As expected, ab = |k|0, hence we could take a = |k|p , b = |k|q, thus p+ q = 0.This implies aγ = |k|pγ = |k|−2αbβ = |k|qβ = |k|2. (5.7)53Chapter 5. Stability of a First Order Semi-implicit Scheme on 2D Fractional Cahn-Hilliard EquationAs a result,−2α= pγ2= qβ=⇒p= −2αα+1q= 2αα+1γ=α+1β= α+1α.(5.8)So, ∑k 6=0|ûn+1(k)− ûn(k)|2 ·( |k|−2ατ+ ν|k|22)=∑k 6=0|ûn+1(k)− ûn(k)|2 ·[α+1τ·( |k|−2αα+1)+ ν(α+1)2α·(|k|2α+1α)]≥∑k 6=0|ûn+1(k)− ûn(k)|2 ·(α+1τ) 1α+1 ·(ν(α+1)2α) α+1α.(5.9)So it is plain to take Cατν =(α+1τ) 1α+1 ·(ν(α+1)2α) α+1α .Remark 9. In the proof above, Cατν→∞ as α→ 0. Hence it would not workfor α= 0 case, but we could refer to chapter 3.Back to the proof of Theorem 5.0.1, (5.4) leads to(A+12+Cατν)∣∣∣∣un+1−un∣∣∣∣2L2+En+1−En ≤ ∣∣∣∣un+1−un∣∣∣∣2L2 (||un||2∞+ 12 ||un+1||2∞).(5.10)To prove En+1 ≤En, it suffices to show A+12+Cατν ≥ 32 max{||un+1||2∞ , ||un||2∞}.As in chapter 3, we rewrite the scheme (5.1) asun+1 = 1+Aτ(−∆)α1+τν(−∆)α+1+Aτ(−∆)α un− τ(−∆)α1+τν(−∆)α+1+Aτ(−∆)αΠN [ f (un)] .(5.11)54Chapter 5. Stability of a First Order Semi-implicit Scheme on 2D Fractional Cahn-Hilliard EquationSimilarly, we could still apply the main lemma under the assumption u0satisfies zero-mean condition. Recall that||un+1||∞. ||un+1||H˙1√log(||un+1||H˙32+3) . (5.12)We would like to estimate ||un+1||H˙1 and ||un+1||H˙ 32 . As we did in chapter3, 1+Aτ|k|2α1+Aτ|k|2α+ντ|k|2+2α · |k|. |k|τ|k|2α1+Aτ|k|2α+ντ|k|2+2α · |k|.ττA|k| = 1A|k|. (5.13)Hence we derive||un+1||H˙1(T2).(1+ 1A+ 3||u||2∞A)||un||H˙1(T2) , (5.14)which is the same argument as in chapter 3.Similarly, we could derive||un+1||H˙32 (T2).(1+Aτντ+ 1ν)(En+1) , (5.15)So prove by induction again,Step 1 : The induction n =⇒ n+1 step. Assume En ≤ En−1 ≤ ·· · ≤ E0and En ≤ supN E(ΠN u0), we would show En+1 ≤ En. This implies ||un||2H˙1 =||∇un||2L2 ≤ 2Enν≤ 2E0ν. By applying the main lemma carefully and E0.E0+1,||un||2∞.E0 ν−1 (1+ log(A)+| log(ν)|)+ν−1| log(τ)|+1 . (5.16)55Chapter 5. Stability of a First Order Semi-implicit Scheme on 2D Fractional Cahn-Hilliard EquationDefine m0 := ν−1 (1+ log(A)+| log(ν)|) again, then the inequality above is||un||2∞.E0 m0+ν−1| log(τ)|+1 . (5.17)Similarly,||un+1||2∞.E0 1+m30A2+m0+ν−3| log(τ)|3. (5.18)So we need the following condition holds:A+ 12+ (α+1τ)1α+1 · (ν(α+1)2α)α+1α ≥C(E0)(m0+1+m30A2+ν−3| log(τ)|3)m0 = ν−1 (1+ log(A)+| log(ν)|) .(5.19)Now we discuss 2 cases again:Case 1: (α+1τ)1α+1 · (ν(α+1)2α )α+1α ≥ C(E0)ν−3| log(τ)|3. In this case, it suffices tochoose A such thatAÀE0 m0 = ν−1 (1+ log(A)+| log(ν)|) .In fact, for ν& 1 , we could take A ÀE0 1; if 0 < ν¿ 1, we would chooseA = CE0 · ν−1| logν| , where CE0 is a large constant depending only on E0.Therefore in both cases it suffices to chooseA =CE0 ·max{ν−1| log(ν)| , 1} . (5.20)Case 2:(α+1τ)1α+1 · (ν(α+1)2α )α+1α ≤ C(E0)ν−3| log(τ)|3. This still implies (1τ )1α+1 .56Chapter 5. Stability of a First Order Semi-implicit Scheme on 2D Fractional Cahn-Hilliard Equation( 1ν)−4−1α , hence | log(τ)|.E0 1+| log(ν)| for fixed 0<α≤ 1. Now we go back toequations (4.17), we have||un||2∞.E0 m0 (5.21)as ν−1| log(τ)| would be absorbed by m0, recall m0 = ν−1 (1+ log(A)+| log(ν)|).Hence substitute this new bound to (4.18), we would derive||un+1||2∞.(1+ (1+||un||2∞A)||un||H˙1√log(3+||un+1||H˙32))2.E0(1+ (1+ m0A)(√1ν√log(3+||un+1||H˙32))2.E0(1+ (1+ m0A)pm0)2.E0 1+m30A2+m0 .(5.22)Thus it still suffices to takeA ≥CE0 m0 . (5.23)For the induction base Step 2, the proof is exactly the same as in chapter 3and this shows stability of the semi-implicit scheme in fractional Laplaciancase.57Chapter 6Stability of a First OrderSemi-implicit Scheme on 3DAllen-Cahn EquationIn this chapter, we would like to explore a bit more in three dimensioncase. What makes the difference is that the main lemma proved in chapter2 should not hold. To clarify, the H˙1-norm should be replaced by H˙32 -normin the log-type inequality proved in chapter 2, as a result of scaling invari-ance. However, H˙32 -norm would not help to prove 3D theorem as there is noa-priori energy bound for H˙32 -norm. To solve this issue, we would try an al-ternate interpolation inequality. For simplicity, we only consider Allen-Cahnequation in 3D periodic domain T3 = (R/2piZ)3 in this chapter as other Cahn-Hilliard type equations could be handled similarly. To begin with, we recallthe numerical scheme (3.1) for Allen-Cahn equation.un+1−unτ= ν∆un+1−A(un+1−un)−ΠN f (un)u0 =ΠN u0. (6.1)586.1. the Main Lemmawhere τ is the time step and A > 0 is the coefficient for the O(τ) regularizationterm. As usual, for N ≥ 2, defineXN = span{cos(k · x) ,sin(k · x) : k= (k1,k2,k3) ∈Z3 , |k|∞ =max{|k1|, |k2|, |k3|}≤N}.Theorem 6.0.1. (3D energy stability for AC) Consider (6.1) with ν > 0 andassume u0 ∈ H2(T3). Then there exists a constant β0 depending only on theinitial energy E0 =E(u0) such that ifA ≥β · (||u0||2H2 +ν−3+1) , β≥β0 (6.2)then E(un+1)≤E(un), ∀n≥ 0, where E is defined before.Remark 10. Unlike in chapter 3, our choice of A is independent of τ as longas it has size of O(ν−3)at least, which is much larger than O(ν−1| log(ν)|2).This results from the loss of log type control for the L∞ bound.Before proving Theorem 6.0.1, we would prove a new main lemma here.6.1 the Main LemmaFor all f ∈H2(T3) , one has|| f ||∞. || f ||12H˙1|| f ||12H˙2+| f̂ (0)| . (6.3)Proof. First we write f (x)= 1(2pi)3∑k∈Z3 f̂ (k) eik·x, the Fourier series of f in T3.596.2. Proof of 3D Stability TheoremSo,|| f ||∞ ≤ 1(2pi)3∑k∈Z3| f̂ (k)|≤ 1(2pi)3| f̂ (0)|+ 1(2pi)3( ∑0<|k|≤N| f̂ (k)|+ ∑|k|>N| f̂ (k)|). | f̂ (0)|+ ∑0<|k|≤N(| f̂ (k)||k| · |k|−1)+ ∑|k|>N(| f̂ (k)||k|2 · |k|−2). | f̂ (0)|+( ∑0<|k|≤N| f̂ (k)|2|k|2) 12·( ∑0<|k|≤N|k|−2) 12+( ∑|k|>N| f̂ (k)|2|k|4) 12· ( ∑|k|>N|k|−4) 12. | f̂ (0)|+( ∑|k|>N| f̂ (k)|2|k|4) 12·(∫ ∞Npir2r4dr) 12+( ∑0<|k|≤N| f̂ (k)|2|k|2) 12·(∫ N1pir2r2dr) 12. | f̂ (0)|+ || f ||H˙2 ·N−12 +|| f ||H˙1 ·N12 .(6.4)We optimize N and hence derive|| f ||∞. | f̂ (0)|+ || f ||12H˙1|| f ||12H˙2. (6.5)6.2 Proof of 3D Stability TheoremBy the same argument in chapter 3 with notation En =E(un),(1τ+A+ 12)||un+1−un||2L2 +ν2||∇(un+1−un)||2L2 +En+1−En≤ ||un+1−un||2L2(||un||2∞+12||un+1||2∞).(6.6)606.2. Proof of 3D Stability TheoremClearly, in order to show En+1 ≤En, it suffices to show1τ+A+ 12≥ 32max{||un||2∞ , ||un+1||2∞} . (6.7)Now we rewrite the scheme (5.1) as the following:un+1 = 1+Aτ1+Aτ−ντ∆un− τ1+Aτ−ντ∆ΠN [ f (un)]. (6.8)Recall that||un+1||∞. |ûn+1(0)|+ ||un+1||12H˙1||un+1||12H˙2. (6.9)Clearly, we need to estimate |ûn+1(0)|, ||un+1||H˙1 and ||un+1||H˙2 . By thesame argument in chapter 3,|ûn+1(0)|. 1+pEn . (6.10)Note that (1+Aτ)|k|1+Aτ+ντ|k|2 ≤ |k|τ|k|1+Aτ+ντ|k|2 ≤τ|k|2τpAν|k|. 1pAν. (6.11)Hence||un+1||H˙1 . ||un||H˙1 +1pAν|| f (un)||L2. ||un||H˙1 +1pAν(||(un)3||L2 +1) .(6.12)Similarly, (1+Aτ)|k|21+Aτ+ντ|k|2 . 1τpAν+√Aν |k|τ|k|21+Aτ+ντ|k|2 ≤1ν. (6.13)616.2. Proof of 3D Stability TheoremThis implies||un+1||H˙2 . 1τpAν+√Aν ||un||H˙1 + 1ν || f (un)||L2. 1τpAν+√Aν ||un||H˙1 + 1ν (||(un)3||L2 +1) .(6.14)Note that by standard Sobolev inequality,||(un)3||L2 = ||un||3L6 . ||un||3H1 . ||∇un||3L2 +||un||3L2 . ||un||3H˙1 +1+ (En)32 .(6.15)As a result,||un+1||H˙1 . ||un||H˙1 +1pAν(||(un)||3H˙1 +1+ (En)32)||un+1||H˙2 . 1τpAν+√Aν ||un||H˙1 + 1ν(||(un)||3H˙1 +1+ (En)32) . (6.16)We would prove the 3D stability theorem inductively as in chapter 3.Step 1: The induction n =⇒ n+ 1 step. Assume En ≤ En−1 ≤ ·· · ≤ E0and En ≤ supN E(ΠN u0), we would show En+1 ≤ En. This implies ||un||2H˙1 =||∇un||2L2 ≤ 2Enν≤ 2E0ν. Recall supN E(ΠN u0).E0+1 as well. Hence we wouldderive,||un+1||H˙1 .E0 ν−12 +A− 12ν− 12(ν−32 +1).E0 ν−12 +A− 12ν−2||un+1||H˙2 .E0 A12ν−1+ν− 52 +τ−1 A− 12ν−1. (6.17)626.2. Proof of 3D Stability TheoremApplying the new main lemma,||un+1||2∞.E0(ν−12 +A− 12ν−2)·(A12ν−1+ν− 52 +τ−1 A− 12ν−1)+1.E0 A12ν−32 +ν−3+A− 12ν− 92 +τ−1 A− 12ν− 32 +τ−1 A−1ν−3+1 .(6.18)To satisfy the sufficient condition (6.7)A12ν−32 +ν−3+A− 12ν− 92 +τ−1 A− 12ν− 32 +τ−1 A−1ν−3.E0 A+1τ,it suffices to takeA ≥CE0ν−3 , (6.19)for a large enough constant CE0 depending only on E0.Step 2: check the induction base step n = 1. It’s clear that we only needto checkA+ 12+ 1τ≥ 32||ΠN u0||2∞+32||u1||2∞.By standard Sobolev inequality in T3,||ΠN u0||2∞. ||ΠN u0||2H2 . ||u0||2H2 . (6.20)On the other hand, by the main lemma it suffices to takeA+ 1τ≥ c1||u0||2H2 +αE0(A12ν−32 +ν−3+A− 12ν− 92 +τ−1 A− 12ν− 32 +τ−1 A−1ν−3),(6.21)where c1 is an absolute constant and αE0 is a constant only depending on E0.636.3. L2 Error Estimate of 3D Allen-Cahn EquationHence it suffices to takeA ≥CE0(||u0||2H2 +ν−3+1) , (6.22)for a large constant CE0 only depending on E0. This completes the proof.By using this new main lemma, the 3D fractional Cahn-Hilliard could behandled similarly.6.3 L2 Error Estimate of 3D Allen-Cahn EquationTheorem 6.3.1. Let ν > 0. Let u0 ∈ Hs, s ≥ 4 and u(t) be the solution toAllen-Cahn equation with initial data u0. Let un be the numerical solutionwith initial data ΠN u0. Assume A satisfies the same condition in the stabilitytheorem. Define tm =mτ, m≥ 1. Then||um−u(tm)||2 ≤ A · eC1 tm ·C2 ·(N−s+τ) ,where C1 > 0 depends only on (u0,ν) and C2 depends on (u0,ν, s).Proof. Recallun+1−unτ= ν∆un+1−ΠN f (un)−A(un+1−un)∂tu= ν∆u− f (u)u0 =ΠN u0 , u(0)= u0 .(6.23)As we proved in chapter 4, the auxiliary L2 estimate lemma and all bounded-ness lemma work for 3D case. The only difference is the estimate for ||∂t∆u||2646.3. L2 Error Estimate of 3D Allen-Cahn Equationusing Gagliardo–Nirenberg interpolation inequality,||∂t∆u||2. ||〈∇〉3∂tu|| 232 · ||∂tu|| 132 . ||∂tu|| 132 , (6.24)which works as well for the same power. This leads to the conclusion of The-orem 6.3.1 by exactly same argument in chapter 4.65Chapter 7Second Order Semi-ImplicitSchemesIn previous chapters we introduce first order semi-implicit schemes forAllen-Cahn equation and fractional Cahn-Hilliard equation in both two di-mensional periodic domain and three dimensional periodic domain. For thecompleteness, we would like to study some second order schemes. As a repre-sentative case, we only consider 2D Allen-Cahn equation here. The analysisof other cases would be similar. We introduce two second order schemes andprove the unconditional stability for Scheme I and conditional stability forScheme II.7.1 Introduction of Scheme I:As introduced in [9], the second order semi-implicit Fourier spectral schemeI is given by:3un+1−4un+un−12τ= ν∆un+1−Aτ(un+1−un)−ΠN(2 f (un)− f (un−1)) , n≥ 1 ,(7.1)667.2. Estimate of the First Order Scheme (7.2)where τ> 0 is the time step and this scheme applies second order backwardderivative in time with a second order extrapolation for the nonlinear term.To start the iteration, we need to derive u1 according to the following firstorder scheme: u1−u0τ1= ν∆u1−ΠN f (u0) ,u0 =ΠN u0 ,(7.2)where τ1 = min{τ 43 , 1}. The choice of τ is because of the error analysiswhich will be proved later. Roughly speaking,||u1−u(τ1)||2.N−s+τ321 ,where u(τ1) denotes the exact PDE solution at τ1. As expected in L2 erroranalysis for the second order scheme, we require that τ321 . τ2 hence τ1. τ43 .7.2 Estimate of the First Order Scheme (7.2)In this section we will estimate some bounds of u1 which will be used toprove the stability of the second order scheme and an error estimate of u1which will be used to prove L2 error estimate of the second order scheme.Lemma 7.2.1. Consider the scheme (7.2). Assume u0 ∈H2(T2), then||u1||∞+||u1−u0||22τ1+ ν2||∇u1||22.E(u0) , ||u0||H2 1 . (7.3)677.2. Estimate of the First Order Scheme (7.2)Proof. First we consider ||u1||∞. We writeu1 = 11−τ1ν∆u0− τ1ΠN1−τ1ν∆f (u0) . (7.4)Note that11+τ1ν|k|2≤ 1 , τ1 ≤ 1 , (7.5)thus||u1||∞. ||u1||H2 . ||u0||H2 +|| f (u0)||H2. ||u0||H2 +||(u0)3||H2.||u0||H2 1 ,(7.6)as ||u0||∞. 1 by Morrey’s inequality.Second, we take the L2 inner product with u1−u0 on both sides of (7.2).||u1−u0||22τ1+ ν2(||∇u1||22−||∇u0||22+||∇(u1−u0)||22)=−( f (u0) , u1−u0)≤ || f (u0)|| 43||u1−u0||4.E(u0) 1 ,(7.7)by ||u0||∞ , ||u1||∞. 1.As a result, ||u1||∞+ ||u1−u0||22τ1+ ν2 ||∇u1||22.E(u0) , ||u0||H2 1.687.2. Estimate of the First Order Scheme (7.2)Lemma 7.2.2. (Error estimate for u1) Consideru1−u0τ1= ν∆u1−ΠN f (u0)∂tu= ν∆u− f (u)u0 =ΠN u0 , u(0)= u0 .(7.8)Let u0 ∈Hs, s≥ 6. There exists a constant D1 > 0 depending only on (u0,ν, s),such that ||u(τ1)−u1||2 ≤D1 · (N−s+τ321 ).Proof. We start the proof in three steps:Step 1: Time discretization of the PDE.Write the PDE in time interval [0,τ1]. Note that for a one-variable func-tion h(s),h(0)= h(τ1)+∫ 0τ1h′(s) ds= h(τ1)−h′(τ1)τ1+∫ τ10h′′(s) · s ds .(7.9)By applying this formula, we haveu(τ1)−u(0)τ1= ∂tu(τ1)− 1τ1∫ τ10(∂ttu) · s ds= ν∆u(τ1)− f (u(τ1))− 1τ1∫ τ10(∂ttu) · s ds= ν∆u(τ1)−ΠN f (u(0))−Π>N f (u(0))− [ f (u(τ1)− f (u(0))]− 1τ1∫ τ10(∂ttu) · s ds ,(7.10)where Π>N = id−ΠN as in chapter 4.697.2. Estimate of the First Order Scheme (7.2)Henceu(τ1)−u(0)τ1= ν∆u(τ1)−ΠN f (u(0))+G0 , (7.11)whereG0 =−Π>N f (u(0))− [ f (u(τ1)− f (u(0))]− 1τ1∫ τ10(∂ttu) · s ds=−Π>N f (u(0))− [ f (u(τ1)− f (u(0))]− 1τ1∫ τ10(ν∆∂tu− f ′(u)∂tu) · s ds(7.12)Step 2: Estimate of ||u(τ1)−u1||2. We consideru(τ1)−u(0)τ1= ν∆u(τ1)−ΠN f (u(0))+G0u1−u0τ1= ν∆u1−ΠN f (u0)u0 =ΠN u0 , u(0)= u0 .(7.13)Define e1 = u(τ1)−u1 and e0 = u(0)−u0. Then we gete1− e0τ1= ν∆e1−ΠN(f (u(0))− f (u0))+G0 . (7.14)Take the L2 inner product with e1 on both sides, we derive12τ1(||e1||22−||e0||22+||e1− e0||22)+ν||∇e1||22≤ || f (u(0))− f (u0)||2 · ||e1||2+||G0||2 · ||e1||2.(||e0||2+||G0||2) ||e1||2.(||e0||22+||G0||22)+ 14 ||e1||22 .(7.15)As a result,707.2. Estimate of the First Order Scheme (7.2)(1− τ12)||e1||22 ≤ 2τ1(||e0||22+||G0||22)+||e0||22 . (7.16)note that τ1 ≤ 1, so 1− τ12 ≥ 12 , and||e1||22. (1+τ1)||e0||22+τ1||G0||22 . (7.17)Step 3: Estimate of ||e0||22 and ||G0||22.Note that ||e0||22 = ||u(0)−u0||22 = ||u0−ΠN u0||22 = ||Π>N u0||22. As proved inchapter 4, section 4.2.2,||e0||22 = ||Π>N u0||22.N−2s . (7.18)For ||G0||2, note that ||Π>N f (u(0))||2.N−s, by the maximum principle provedin chapter 4, Lemma 4.2.2.On the other hand, by the mean value theorem,f (u(τ1))− f (u(0))= f ′(ξ)(u(τ1)−u(0)) ,where ξ is a number between u(τ1) and u(0). Again by the maximumprinciple,|| f (u(τ1))− f (u(0))||2. ||u(τ1)−u(0)||2. τ1||∂tu||L∞t L2x([0 , τ1]×T2). τ1 , (7.19)using the Sobolev bound of the exact solution proved in chapter 4, Lemma4.2.3.717.3. Unconditional Stability of the Second Order Scheme I (7.1) & (7.2)Finally, ∥∥∥∥ 1τ1∫ τ10(ν∆∂tu− f ′(u)∂tu) · s ds∥∥∥∥2.∥∥∥∥∫ τ10ν∆∂tu− f ′(u)∂tu ds∥∥∥∥2.∫ τ10||ν∆∂tu||2 ds+∫ τ10|| f ′(u)∂tu||2 ds. τ1 .(7.20)This implies ||G0||22.N−2s+τ21. Hence||e1||22. (1+τ1)N−2s+τ1(N−2s+τ21).N−2s+τ31 . (7.21)As a result,||e1||2.N−s+τ321 . (7.22)7.3 Unconditional Stability of the Second OrderScheme I (7.1) & (7.2)In this section we will prove a unconditional stability theorem for the sec-ond order scheme (7.1) combining (7.2). To get started, we state the theoremfirst.Theorem 7.3.1. (Unconditional Stability) Consider the scheme (7.1)-(7.2) withν> 0, τ> 0 and N ≥ 2. Assume u0 ∈ H2(T2). The initial energy is denoted byE0 =E(u0). If there exists a constant βc > 0 depending only on E0 and ||u0||H2 ,727.3. Unconditional Stability of the Second Order Scheme I (7.1) & (7.2)such thatA ≥β · (ν2+ν−10| logν|4) , β≥βc ,thenE˜(un+1)≤ E˜(un) , n≥ 1 ,where E˜(un) for n≥ 1 is a modified energy functional and is defined asE˜(un) :=E(un)+ ν4||un−un−1||22+14τ||un−un−1||22 .Before proving this stability theorem, we begin with several lemmas.Lemma 7.3.2. Consider (7.1) for n ≥ 1. Suppose E(un) ≤ B and E(un−1) ≤ Bfor some B> 0. Then||un+1||∞ ≤αB ·(1+ν−1) ·√log(3+ Aτν+ 1τν+ν− 52 +ν−1)+τ+1 ,for some αB > 0 only depending on B.Proof. For simplicity we write . instead of .B. Recall (7.1):3un+1−4un+un−12τ= ν∆un+1−Aτ(un+1−un)−ΠN(2 f (un)− f (un−1)) .(7.23)We rewrite asun+1 = 4+2Aτ23−2ντ∆+2Aτ2 un− 13−2ντ∆+2Aτ2 un−1− 2τΠN3−2ντ∆+2Aτ2(2 f (un)− f (un−1)) . (7.24)737.3. Unconditional Stability of the Second Order Scheme I (7.1) & (7.2)First, for k= 0, 4+2Aτ23+2Aτ2 . 113+2Aτ2 . 123+2Aτ2 . τ .(7.25)Thus|un+1(0)|. τ+1 . (7.26)Note that for |k| ≥ 1,4+2Aτ23+2ντ|k|2+2Aτ2 . 113+2ντ|k|2+2Aτ2 . 12τ|k|3+2ντ|k|2+2Aτ2 .τ|k|ντ|k|2 .1ν· |k|−1 .(7.27)Thus||un+1||H˙1 . ||un||H˙1 +||un−1||H˙1 +1ν||〈∇〉−1 (2 f (un)− f (un−1)) ||2. ν− 12 +ν−1(||(un)3||4/3+||(un−1)3||4/3+||un||2+||un−1||2). ν− 12 +ν−1 ,(7.28)here we apply Sobolev’s inequality as introduced in chapter 2 and apply theenergy bound as proved in chapter 3.Similarly,747.3. Unconditional Stability of the Second Order Scheme I (7.1) & (7.2)|k|2(4+2Aτ2)3+2ντ|k|2+2Aτ2 .|k|2(1+Aτ2)ντ|k|2 .1ντ+ Aτν|k|23+2ντ|k|2+2Aτ2 .1ντ2τ|k|23+2ντ|k|2+2Aτ2 .τ|k|2ντ|k|2 .1ν.(7.29)This implies||un+1||H˙2 .(1ντ+ Aτν)||un||2+ 1ντ||un−1||2+ 1ν||2 f (un)− f (un−1)||2. 1ντ+ Aτν+ 1ν(||un||36+||un−1||36+||un||2+||un−1||2). 1ντ+ Aτν+ 1ν(||un||3H1 +||un−1||3H1 +1). 1ντ+ Aτν+ 1ν(ν−32 +1) .(7.30)Finally, by applying the main log-interpolation lemma proved in chapter2, section 2.3, we can derive||un+1||∞. (1+||un+1||H˙1) ·√log(3+||un+1||H˙2)+|áun+1(0)|. (1+ν−1) ·√log(3+ Aτν+ 1ντ+ν− 52 +ν−1)+τ+1 ,(7.31)where ν−12 is bounded by ν−1+1.757.3. Unconditional Stability of the Second Order Scheme I (7.1) & (7.2)7.3.1 Proof of Unconditional Stability (Theorem 7.3.1)Before proving the theorem, we first introduce some notation. We denoteδun+1 := un+1−un and δ2un+1 := un+1−2un+un−1. Clearly,3un+1−4un+un−1 = 2δun+1+δ2un+1δ2un+1−δun+1 =−δunδun ·un = (un−un−1)un = 12(|un|2−|un−1|2+|δun|2) .(7.32)As a result,(3un+1−4un+un−1 , un+1−un)= (2δun+1+δ2un+1 , δun+1)= 2||δun+1||22+(δun+1−δun , δun+1)= 2||δun+1||22+12(||δun+1||22−||δun||22+||δ2un+1||22) .(7.33)Now recall the scheme (7.1)3un+1−4un+un−12τ= ν∆un+1−Aτ(un+1−un)−ΠN(2 f (un)− f (un−1)) .(7.34)Take the L2 inner product with δun+1 = un+1−un on both sides of (7.1). Wehave1τ||δun+1||22+14τ(||δun+1||22−||δun||22+||δ2un+1||22)+ν2(||∇un+1||22−||∇un||22+||δ∇un+1||22)+Aτ||δun+1||22 =−(ΠN (2 f (un)− f (un−1)) , δun+1).(7.35)To analyze(2 f (un)− f (un−1) , δun+1), we consider 2 f (un)− f (un−1)= f (un)+767.3. Unconditional Stability of the Second Order Scheme I (7.1) & (7.2)(f (un)− f (un−1).Note that F ′ = f , hence by fundamental theorem of calculus,F(un+1)−F(un)= f (un)δun+1+∫ 10f ′(un+ sδun+1)(1− s) ds · (δun+1)2= f (un)δun+1+∫ 10f˜ (un+ sδun+1)(1− s) ds · (δun+1)2− 12(δun+1)2 ,(7.36)where f˜ (x)= 3x2, as f ′(x)= 3x2−1. And this impliesf (un)δun+1 ≥ F(un+1)−F(un)+ 12(δun+1)2− 32(||un||2∞+||un+1||2∞) · (δun+1)2 .(7.37)On the other hand,f (un)− f (un−1)= f ′(ξ)δun , (7.38)and hence( f (un)− f (un−1)) ·δun+1 ≥−(3||un||2∞+3||un−1||2∞+1) · |δun| · |δun+1|≥ −(1+3||un||2∞+3||un−1||2∞)2ν· ||δun||22−ν4||δun+1||22 .(7.39)Hence the estimate of the nonlinear term is as following:777.3. Unconditional Stability of the Second Order Scheme I (7.1) & (7.2)− (ΠN (2 f (un)− f (un−1)) , δun+1)=− (2 f (un)− f (un−1) , δun+1)≤−∫T2F(un+1) dx+∫T2F(un) dx− 12||δun+1||22+ 32(||un||2∞+||un+1||2∞) · ||δun+1||22+(1+3||un||2∞+3||un−1||2∞)2ν· ||δun||22+ν4||δun+1||22 .(7.40)Combine all estimates (7.35) and (7.40) we get1τ||δun+1||22+14τ(||δun+1||22−||δun||22+||δ2un+1||22)+ν2(||∇un+1||22−||∇un||22+||δ∇un+1||22)+Aτ||δun+1||22≤−∫T2F(un+1) dx+∫T2F(un) dx− 12||δun+1||22+32(||un||2∞+||un+1||2∞) · ||δun+1||22+(1+3||un||2∞+3||un−1||2∞)2ν· ||δun||22+ν4||δun+1||22 .(7.41)After simplification,(1τ+Aτ− ν4+ 12)· ||δun+1||22+ E˜(un+1)≤{32(||un||2∞+||un+1||2∞)+(1+3||un||2∞+3||un−1||2∞)2ν}· ||δun+1||22+ E˜(un) .(7.42)787.3. Unconditional Stability of the Second Order Scheme I (7.1) & (7.2)Clearly to show E˜(un+1)≤ E˜(un), it suffices to show1τ+Aτ− ν4+ 12≥32(||un||2∞+||un+1||2∞)+(1+3||un||2∞+3||un−1||2∞)2ν.(7.43)We now prove this sufficient condition inductively. SetB=max{E˜(u1) , E(u0)} .By Lemma 7.2.1 in previous section, B. 1. We shall prove for every m≥ 2,E˜(um)≤B , E˜(um)≤ E˜(um−1) ,||um||∞ ≤αB ·(1+ν−1) ·√log(3+ Aτν+ 1τν+ν− 52 +ν−1)+τ+1 , (7.44)where αB > 0 is the same constant in Lemma 7.3.2.We first check the base case when m= 2.Note that E(u1) ≤ E˜(u1) ≤ B and E(u0) ≤ B, then we can apply Lemma7.3.2, and hence obtain||u2||∞ ≤αB ·(1+ν−1) ·√log(3+ Aτν+ 1τν+ν− 52 +ν−1)+τ+1 . (7.45)We only need to check E˜(u2) ≤ E˜(u1). By the sufficient condition (7.41),we only need to check the inequality797.3. Unconditional Stability of the Second Order Scheme I (7.1) & (7.2)1τ+Aτ− ν4+ 12≥ 32(||u1||2∞+||u2||2∞)+(1+3||u1||2∞+3||u0||2∞)2ν. (7.46)By Lemma 7.2.1, ||u0||∞ , ||u1||∞. 1, it suffices to choose A such that1τ+Aτ− ν4+ 12≥C · (1+ν−2) · log(3+ Aτν+ 1τν+ν− 52 +ν−1)+Cν−1+C+Cτ .(7.47)We discuss two case and denote X = Aτ+ 1τ.Case 1: 0< ν≤ 1/2. In this case we needX + 12≥Cν−2 · (| logν|+ | log X |) . (7.48)Hence we needX ≥C ·ν−2| logν| . (7.49)Case 2: ν> 1/2. Then we needX ≥C · (| log X |+1+ν) , (7.50)and henceX ≥C · (1+ν) . (7.51)In conclusion, as X ≥ 2pA,A ≥C · (1+ν2+ν−4| logν|2)≥C · (ν2+ν−4| logν|2) . (7.52)807.3. Unconditional Stability of the Second Order Scheme I (7.1) & (7.2)Now we check the induction step. Assume the induction hypothesis holdfor 2≤m≤ n, then for m= n+1,||un+1||∞ ≤αB ·(1+ν−1) ·√log(3+ Aτν+ 1τν+ν− 52 +ν−1)+τ , (7.53)by Lemma 7.3.2. It remains to show E˜(un+1)≤ E˜(un). It suffices to choose Asuch that1τ+Aτ+ 12≥ν4+C · (1+ν−2) · log(3+ Aτν+ 1τν+ν− 52 +ν−1)+ C(1+ν−4)ν((log(3+ Aτν+ 1τν+ν− 52 +ν−1))2+τ).(7.54)In terms of X = Aτ+ 1τagain, we need to discuss two cases as well.Case 1: 0< ν≤ 1/2. ThenX ≥C ·ν−5(| logν|2+| log X |2) . (7.55)As a result,X ≥C ·ν−5| logν|2 . (7.56)Case 2: ν> 1/2. Then we needX ≥Cν+C · (log X + (log X )2ν−1) , (7.57)hence X ≥C · (ν+1).In conclusion of two cases,817.4. L2 Error Estimate of Second Order Scheme IA ≥C · (ν2+1+ν−10| logν|4)≥C · (ν2+ν−10| logν|4) . (7.58)This completes the induction. Combining the estimate, we need to takeA ≥C · (ν2+ν−10| logν|4) , (7.59)such that E˜(un+1)≤ E˜(un), for n≥ 1.7.4 L2 Error Estimate of Second Order Scheme IFirst we state the theorem.Theorem 7.4.1. (L2 error estimate) Let ν> 0 and u0 ∈Hs, s≥ 8. Let 0< τ≤Mfor some M > 0. Let u(t) be the continuous solution to the 2D Allen-Cahnequation with initial data u0. Let u1 be defined according to (7.2) with initialdata u0 =ΠN u0. Let um, m≥ 2 be defined in (7.1) with initial data u0 and u1.Assume A satisfies the same condition in Theorem 7.3.1. Define t0 = 0, t1 = τ1and tm = τ1+ (m−1)τ for m≥ 2. Then for any m≥ 1,||u(tm)−um||2 ≤C1 · eC2 tm · (N−s+τ2) , (7.60)where C1 , C2 > 0 are constants depending only on (u0, ν, s, A, M).Remark 11. Here we require that τ is not arbitrarily large. This is a result ofloss of mass conservation as preserved by Cahn-Hilliard equation. However,in practice it is not a big issue as we always use small time steps.Similar to chapter 4, we will study the auxiliary error estimate behavior827.4. L2 Error Estimate of Second Order Scheme Iand time discretization behavior of Allen-Cahn equation before proving thetheorem.7.4.1 Auxiliary L2 Error Estimate for Near SolutionsConsider for n≥ 1,3un+1−4un+un−12τ= ν∆un+1−Aτ(un+1−un)−ΠN(2 f (un)− f (un−1))+Gn3vn+1−4vn+vn−12τ= ν∆vn+1−Aτ(vn+1−vn)−ΠN(2 f (vn)− f (vn−1)) ,(7.61)where (u1, u0, v1, v0) are given.Proposition 3. For solutions of (7.59), assume for some N1 > 0,supn≥0||un||+supn≥0||vn|| ≤N1 , (7.62)Then for any m≥ 2,||um−vm||22 ≤C ·exp((m−1)τ · C(1+N41 )η)·((1+Aτ2)||u1−v1||22+||u0−v0||22+τηm−1∑n=1||Gn||22),(7.63)where C > 0 is a absolute constant that could be computed and 0< η< 1100M isa constant depending only on M, the upper bound for τ.Proof. We still denote the constant by C whose value may vary in differentlines. Denote en = un−vn, then837.4. L2 Error Estimate of Second Order Scheme I3en+1−4en+ en−12τ−ν∆en+1+Aτ(en+1− en)=−ΠN(2 f (un)−2 f (vn))+ΠN ( f (un−1)− f (vn−1))+Gn (7.64)Take the L2 inner product with en+1 on both sides, we derive12τ(3en+1−4en+ en−1, en+1)+ν||∇en+1||22+Aτ2(||en+1||22−||en||22+||en+1− en||22)=−2( f (un)− f (vn), en+1)+ ( f (un−1)− f (vn−1), en+1)+ (Gn, en+1) .(7.65)To estimate the RHS, first observe that|( f (un)− f (vn), en+1)| ≤ || f (un)− f (vn)||2||en+1||2 ≤|| f (un)− f (vn)||22η+η||en+1||22,(7.66)where η< 1100M is a small number only depending on M.Moreover, use similar method in chapter 4,f (un)− f (vn)= f (un)− f (un− en)= (un)3− (un− en)3− en=−(en)3− en−3un(en)2+3(un)2en .(7.67)So by assumption|| f (un)− f (vn)||22.||en||4∞||en||22+||en||22+||un||2∞||en||22+||un||4∞||en||22.(1+N41 )||en||22 .(7.68)Similarly,|| f (un−1)− f (vn−1)||22. (1+N41 )||en−1||22 . (7.69)847.4. L2 Error Estimate of Second Order Scheme IAs a result,RHS≤ C(1+N41 )η(||en||22+||en−1||22)+ 1η ||Gn||22+η||en+1||22 . (7.70)On the other hand,(3en+1−4en+ en−1, en+1− en)= (2δen+1+δ2en+1, δen+1)= 2||δen+1||22+12(||δen+1||22−||δen||22+||δ2en+1||22) . (7.71)Also,(3en+1−4en+ en−1, en)= 3(δen+1, en)− (δen, en)=32(||en+1||22−||en||22−||en+1− en||22)− 12 (||en||22−||en−1||22+||en− en−1||22) .(7.72)These two equations give(3en+1−4en+ en−1, en+1)=32(||en+1||22−||en||22)−12(||en||22−||en−1||22)+||δen+1||22−||δen||22+ 12||δ2en+1||22 .(7.73)Collecting all estimates (7.70) and (7.73), (7.63) becomes12τ(32||en+1||22−12||en||22+||en+1− en||22)+ Aτ2||en+1||22≤ 12τ(32||en||22−12||en−1||22+||en− en−1||22)+ Aτ2||en||22+C(1+N41 )η(||en||22+||en−1||22)+ 1η ||Gn||22+η||en+1||22 .(7.74)Now define X n+1 := 32 ||en+1||22− 12 ||en||22+||en+1− en||22. We observe that857.4. L2 Error Estimate of Second Order Scheme IX n+1 =12||en+1||22+12||2en+1− en||22110||en||22+52||en+1− 25en||22 .(7.75)This showsX n+1 ≥ 110max{||en+1||22, ||en||22} . (7.76)Making use of X n+1, (7.72) becomes(X n+1+Aτ2||en+1||22)− (X n+Aτ2||en||22)2τ≤C(1+N41 )η(||en||22+||en−1||22)+ 1η ||Gn||22+η||en+1||22 .(7.77)This leads to(X n+1−2ητ||en+1||22+Aτ2||en+1||22)− (X n−2ητ||en||22+Aτ2||en||22)2τ≤C(1+N41 )η(||en||22+||en−1||22)+ 1η ||Gn||22+η||en||22≤(C(1+N41 )η+Cη)· (X n−2ητ||en||22)+ 1η ||Gn||22 .(7.78)Defineyn = X n−2ητ||en||22+Aτ2||en||22 ,α= C(1+N41 )η+Cη ,βn =||Gn||22η.(7.79)This shows for ν small,yn+1− ynτ≤αyn+βn .867.4. L2 Error Estimate of Second Order Scheme IApplying discrete Gronwall’s inequality, we have for m≥ 2,||em||22 ≤C(X m−2ητ||em||22)≤Ce(m−1)τ· C(1+N41 )η (X1+Aτ2||e1||22+ τη m−1∑n=1 ||Gn||22),(7.80)which gives||um−vm||22≤C ·exp((m−1)τ · C(1+N41 )η)·(32||e1||22−12||e0||22+||e1− e0||22+Aτ2||e1||22+τηm−1∑n=1||Gn||22)≤C ·exp((m−1)τ · C(1+N41 )η)· ((1+Aτ2)||u1−v1||22+||u0−v0||22+τηm−1∑n=1||Gn||22).(7.81)7.4.2 Time Discretization of Allen-Cahn EquationIn this section, we will rewrite the PDE in terms of the second orderscheme.Lemma 7.4.2. (Time discrete Allen-Cahn equation) Let u(t) be the exact so-lution to Allen-Cahn equation with initial data u0 ∈ Hs, s ≥ 8. Define t0 = 0,877.4. L2 Error Estimate of Second Order Scheme It1 = τ1 and tm = τ1+ (m−1)τ for m≥ 2. Then for any n≥ 1,3u(tn+1)−4u(tn)+u(tn−1)2τ= ν∆u(tn+1)−Aτ (u(tn+1)−u(tn))−ΠN [2 f (u(tn))− f (u(tn−1))]+Gn .(7.82)For any m≥ 2,τm−1∑n=1||Gn||22. (1+ tm) · (τ4+N−2s) .Proof. The proof will be proceeded in several steps and we write . instead of.A, ν, u0 for simplicity.Step 1: We write the PDE in the discrete form in time.Recall∂tu= ν∆u− f (u) .For a one variable function h(t), the following equation holds:h(t)= h(t0)+h′(t0)(t− t0)+ 12 h′′(t0)(t− t0)2+ 12∫ tt0h′′′(s)(s− t)2 ds . (7.83)We then apply this to AC,u(tn)= u(tn+1)−∂tu(tn+1) ·τ+ 12∂ttu(tn+1)τ2+ 12∫ tntn+1∂tttu(s)(s− tn)2 dsu(tn−1)= u(tn+1)−∂tu(tn+1) ·2τ+2∂ttu(tn+1)τ2+ 12∫ tn−1tn+1∂tttu(s)(s− tn−1)2 ds.(7.84)As a result, we use second equation above-4×first equation and hence get887.4. L2 Error Estimate of Second Order Scheme I3u(tn+1)−4u(tn)+u(tn−1)2τ= 12τ(2τ ·∂tu(tn+1)−2∫ tntn+1∂tttu(s)(s− tn)2 ds+ 12∫ tn−1tn+1∂tttu(s)(s− tn−1)2 ds)= ∂tu(tn+1)+ 1τ∫ tn+1tn∂tttu(s)(s− tn)2 ds− 14τ∫ tn+1tn−1∂tttu(s)(s− tn−1)2 ds= ν∆u(tn+1)−Aτ (u(tn+1)−u(tn))−ΠN [2 f (u(tn))− f (u(tn−1))]+Aτ (u(tn+1)−u(tn))−Π>N [2 f (u(tn))− f (u(tn−1))]+2 f (u(tn))− f (u(tn−1))− f (u(tn+1)+1τ∫ tn+1tn∂tttu(s)(s− tn)2 ds− 14τ∫ tn+1tn−1∂tttu(s)(s− tn−1)2 ds .(7.85)Clearly,Gn =Aτ (u(tn+1)−u(tn))︸ ︷︷ ︸I1−Π>N [2 f (u(tn))− f (u(tn−1))]︸ ︷︷ ︸I2+2 f (u(tn))− f (u(tn−1))− f (u(tn+1)︸ ︷︷ ︸I3+ 1τ∫ tn+1tn∂tttu(s)(s− tn)2︸ ︷︷ ︸I4ds− 14τ∫ tn+1tn−1∂tttu(s)(s− tn−1)2 ds︸ ︷︷ ︸I5.(7.86)Step 2: We will estimate ||I1||2 ∼ ||I5||2.897.4. L2 Error Estimate of Second Order Scheme II1:||I1||22 =||Aτ (u(tn+1)−u(tn)) ||22.τ2||u(tn+1)−u(tn)||22.τ2||∫ tn+1tn∂tu(s) ds||22.τ2∫T2(∫ tn+1tn∂tu(s) ds)2.τ2∫T2((∫ tn+1tn|∂tu(s)|2 ds)1/2 ·pτ)2.τ2 ·τ ·∫ tn+1tn||∂tu(s)||22 ds.τ3∫ tn+1tn||∂tu(s)||22 ds .(7.87)I2: By the maximum principle Lemma 4.2.2 proved in chapter 4 and u ∈L∞t Hsx,||I2||2.N−s · (|| f (u(tn))||Hs +|| f (u(tn−1))||Hs ).N−s .(7.88)I3: To bound I3, we recall that for a one-variable function h(t),h(t)= h(t0)+h′(t0)(t− t0)−∫ tt0h′′(s) · (s− t) ds . (7.89)Then,f (u(tn))= f (u(tn+1))−∂t( f (u))(tn+1) ·τ+∫ tn+1tn∂tt( f (u)) · (s− tn) dsf (u(tn−1))= f (u(tn+1))−∂t( f (u))(tn+1) ·2τ+∫ tn+1tn−1∂tt( f (u)) · (s− tn−1) ds .(7.90)907.4. L2 Error Estimate of Second Order Scheme IThen we use second equation above-2×first equation and derive:f (u(tn+1))−2 f (u(tn))+ f (u(tn−1))=−2∫ tn+1tn∂tt( f (u)) · (s− tn) ds+∫ tn+1tn−1∂tt( f (u)) · (s− tn−1) ds .(7.91)As a result,||I3||22 =|| f (u(tn+1))−2 f (u(tn))+ f (u(tn−1))||22.||∫ tn+1tn∂tt( f (u)) · (s− tn) ds||22+||∫ tn+1tn−1∂tt( f (u)) · (s− tn−1) ds||22.τ2 · ||∫ tn+1tn∂tt( f (u)) ds||22+τ2 · ||∫ tn+1tn−1∂tt( f (u)) ds||22.τ3∫ tn+1tn−1||∂tt( f (u))||22 ds ,(7.92)by a similar estimate in I1.I4&I5:||I4||22+||I5||22.∥∥∥∥1τ∫ tn+1tn∂tttu(s)(s− tn)2 ds∥∥∥∥22+∥∥∥∥1τ∫ tn+1tn−1∂tttu(s)(s− tn−1)2 ds∥∥∥∥22.∥∥∥∥1τ∫ tn+1tn−1∂tttu(s) ·τ2 ds∥∥∥∥22.τ2 ·∥∥∥∥∫ tn+1tn−1∂tttu(s) ds∥∥∥∥22.τ3∫ tn+1tn−1||∂tttu(s)||22 ds .(7.93)Step 3: Estimate of τ ·∑m−1n=1 ||Gn||22.917.4. L2 Error Estimate of Second Order Scheme ICollecting estimates above, we haveτ ·m−1∑n=1||Gn||22 =τ ·m−1∑n=1(||I1||22+||I2||22+||I3||22+||I4||22+||I5||22).mτ ·N−2s+τ4 ·∫ tm0||∂tu||22+||∂tt( f (u))||22+||∂tttu||22 ds˜ .(7.94)Note that ∂ttu= ν∂t∆u−∂t( f (u))∂tttu= ν∂tt∆u−∂tt( f (u))∂t( f (u))= f ′(u)∂tu∂tt( f (u))= f ′(u)∂ttu+ f ′′(u)(∂tu)2 ,(7.95)hence together with maximum principle and higher Sobolev bounds provedin chapter 4, one hasτ ·m−1∑n=1||Gn||22.mτ ·N−2s+τ4 ·∫ tm0||∂tu||22+||∂tt( f (u))||22+||∂tttu||22 ds˜. tm ·N−2s+τ4 ·∫ tm0||u||2Hs ds˜. tm ·N−2s+τ4 · (1+ tm). (1+ tm) · (τ4+N−2s) .(7.96)927.4. L2 Error Estimate of Second Order Scheme I7.4.3 Proof of L2 Error Estimate of Second Order Scheme I(7.1)First, assumptions in Proposition 3 are satisfied by the unconditionalTheorem 7.3.1 and maximum principle of Allen-Cahn equation. Thus we ap-ply the auxiliary estimate Proposition 3. Then||u(tm)−um||22. eCmτ ·((1+Aτ2)||u1−v1||22+||u0−v0||22+τm−1∑n=1||Gn||22).(7.97)By Lemma 7.2.2 and Lemma 7.4.2,||u(tm)−um||22. eCmτ ·((1+Aτ2)||u1−v1||22+||u0−v0||22+τm−1∑n=1||Gn||22). eCtm · ((1+Aτ2)(N−2s+τ4)+N−2s+ (1+ tm) · (τ4+N−2s)). eCtm · (N−2s+τ4) .(7.98)Thus for m≥ 2,||u(tm)−um||2. eCtm · (N−s+τ2) .Remark 12. For the error estimate, we actually do not need that high regu-larity of initial data because of a smoothing effect of Allen-Cahn equation.937.5. Introduction of Scheme II7.5 Introduction of Scheme IIIn this section, we will introduce another second order scheme.3un+1−4un+un−12τ= ν∆un+1−A(un+1−2un+un−1)−ΠN(2 f (un)− f (un−1)) ,(7.99)where τ> 0 is the time step and n≥ 1.To start the iteration, we again need to derive u1 according to the follow-ing first order scheme:u1−u0τ1= ν∆u1−ΠN f (u0) ,u0 =ΠN u0 ,(7.100)where τ1 =min{τ 43 , 1 , 1pA+1 }. The choice of such τ is to guarantee the errorestimate as proved in section 7.2, and to ensure that the new modified energyfunction can be controlled by initial data.7.6 Estimate of the First Order Scheme (7.100)In this section we will still estimate some bounds of u1 which will be usedto prove the stability of the second order scheme. It is slightly different fromscheme (7.2).Lemma 7.6.1. Consider the scheme (7.100).u1−u0τ1= ν∆u1−ΠN f (u0) ,u0 =ΠN u0 ,(7.101)947.7. Conditional Stability of the Second Order Scheme II (7.99) & (7.100)where τ1 =min{τ 43 , 1 , 1pA+1 }. Assume u0 ∈H2(T2), thenE(u1)+ ||u1−u0||22τ1+.E(u0) , ||u0||H2 1(1+A)||u1−u0||22.||u0||H2 (1+ν)2 .(7.102)Proof. The first inequality shares the same proof as in section 7.2, as thescheme (7.100) is a refined version of (7.2).For the second inequality, recall that ||u1||H2 .||u0||H2 , hence by (7.101)1τ1||u1−u0||2 ≤ ν||u1||H2 +|| f (u0)||2.||u0||H2 1+ν . (7.103)This implies(A+1)||u1−u0||22.||u0||H2 (1+ν)2 . (7.104)7.7 Conditional Stability of the Second OrderScheme II (7.99) & (7.100)In this section we will prove a conditional stability theorem for the secondorder scheme (7.99) combining (7.100). To get started, we state the theoremfirst.Theorem 7.7.1. (Conditional Stability) Consider the scheme (7.99)-(7.100)with ν> 0, τ> 0 and N ≥ 2. Assume u0 ∈H2(T2). The initial energy is denotedby E0 = E(u0). There exist constants Ci > 0, i = 1, 2, 3, 4 depending only onE0 and ||u0||H2 , such that the following holds:957.7. Conditional Stability of the Second Order Scheme II (7.99) & (7.100)Case 1: A = 0. Ifτ≤C1ν41+| logν|2 , when 0< ν< 1 ;C2ν−21+| logν|2 , when ν≥ 1 .(7.105)thenE˚(un+1)≤ E˚(un) .Case 2: A = constant · (ν−4+ν2). Ifτ≤C3ν21+| logν| , when 0< ν< 1 ;C4ν−11+| logν| , when ν≥ 1 .(7.106)thenE˚(un+1)≤ E˚(un) .Here E˚(un) for n≥ 1 is a modified energy functional and is defined asE˚(un) :=E(un)+ A+12||un−un−1||22+14τ||un−un−1||22 .Before proving this stability theorem, we begin with several lemmas.Lemma 7.7.2. Consider (7.99) for n ≥ 1. Suppose E(un) ≤ B · (1+ ν)2 andE(un−1)≤B · (1+ν)2 for some B> 0. Then||un+1||∞ ≤αB ·{(ν12 +ν−1) ·√1+ log(A+1)+| logν|+ | logτ|+1},for some αB > 0 only depending on B.967.7. Conditional Stability of the Second Order Scheme II (7.99) & (7.100)Proof. For simplicity we write . instead of .B.First note by the energy estimate,||∇un−1||2+||∇un||2. ν−12 (1+ν) , ||un−1||4+||un||4. (1+ν)12 . (7.107)Now rewrite the scheme (7.99) asun+1 = 4+4Aτ3+2Aτ−2ντ∆un− 1+2Aτ3+2Aτ−2ντ∆un−1− 2τΠN3+2Aτ−2ντ∆(2 f (un)− f (un−1)) . (7.108)For Fourier mode k= 0,4+4Aτ3+2Aτ . 11+2Aτ3+2Aτ . 12τ3+2Aτ .1A. 1 .(7.109)Thus|un+1(0)|. 1 . (7.110)For |k| ≥ 1, 4+4Aτ3+2Aτ+2ντ|k|2 . 11+2Aτ3+2Aτ+2ντ|k|2 . 12τ3+2Aτ+2ντ|k|2 .1ν|k|2 .(7.111)977.7. Conditional Stability of the Second Order Scheme II (7.99) & (7.100)This implies,||un+1||H˙1 .||un||H˙1 +||un−1||H˙1 +1ν||〈∇〉−1(2 f (un)− f (un−1))||2.ν− 12 (1+ν)+ν−1 · (||(un)3||4/3+||(un−1)3||4/3+||un||2+||un−1||2).ν− 12 (1+ν)+ν−1 ·((1+ν) 32 + (1+ν) 12).ν−1+ν 12 .(7.112)Similarly, 4+4Aτ3+2Aτ+2ντ|k|2 .(1ντ+ Aν)· 1|k|21+2Aτ3+2Aτ+2ντ|k|2 .(1ντ+ Aν)· 1|k|22τ3+2Aτ+2ντ|k|2 .1ν|k|2 .(7.113)Thus,||un+1||H˙2 .(1ντ+ Aν)· (||un||2+||un−1||2)+ 1ν||(2 f (un)− f (un−1))||2.(1ντ+ Aν)· (1+ν) 12 +ν−1 (||un||36+||un−1||36+||un||2+||un−1||2).(1ντ+ Aν)· (1+ν) 12 +ν−1(ν− 32 (1+ν)3+ (1+ν) 12 ). 1ντ+ Aν+ 1ν12 τ+ A+1ν12+ν− 52 +ν 12 ,(7.114)by a standard Sobolev’s inequality.As a result, by the log interpolation inequality again,||un+1||∞. (ν12 +ν−1) ·√1+ log(A+1)+| logν|+ | logτ|+1 . (7.115)987.7. Conditional Stability of the Second Order Scheme II (7.99) & (7.100)7.7.1 Proof of Conditional Stability (Theorem 7.7.1)Recall that3un+1−4un+un−12τ−ν∆un+1+A(un+1−2un+un−1)=−ΠN(2 f (un)− f (un−1)) . (7.116)We apply L2 inner product with δun+1 = un+1−un on both sides of (7.116).Recall that(3un+1−4un+un−1 , un+1−un)= 2||δun+1||22+12(||δun+1||22−||δun||22+||δ2un+1||22) . (7.117)Estimate of the LHS:LHS=1τ||δun+1||22+14τ(||δun+1||22−||δun||22+||δ2un+1||22)+ ν2(||∇un+1||22−||∇un||22+||δ∇un+1||22)+ A2(||δun+1||22−||δun||22+||δ2un+1||22)≥[ν2||∇un+1||22+14τ||δun+1||22+A2||δun+1||22]−[ν2||∇un||22+14τ||δun||22+A2||δun||22]+ 1τ||δun+1||22+A2||δ2un+1||22 .(7.118)Now it remains to estimate the RHS:997.7. Conditional Stability of the Second Order Scheme II (7.99) & (7.100)RHS=− (2 f (un)− f (un−1) , δun+1)=(2un−un−1 , δun+1)︸ ︷︷ ︸I1+((un−1)3−2(un)3 , δun+1)︸ ︷︷ ︸I2.(7.119)I1 =(−δ2un+1 , δun+1)+ (un+1 , δun+1)=− 12(||δun+1||22−||δun||22+||δ2un+1||22)+ 12(||un+1||22−||un||22+||δun+1||22) .(7.120)For I2, we use the identity un−1 = δ2un+1+2un−un+1, then(un−1)3−2(un)3 =(δ2un+1+2un−un+1)3−2(un)3=(δ2un+1)3+3(δ2un+1)2(2un−un+1)+3δ2un+1(2un−un+1)2+ (2un−un+1)3−2(un)3 .(7.121)Note that,3δ2un+1(2un−un+1)2 = 3δ2un+1(δ2un+1−un−1)2= 3(δ2un+1)3−6(δ2un+1)2un−1+3δ2un+1(un−1)2 .(7.122)1007.7. Conditional Stability of the Second Order Scheme II (7.99) & (7.100)As a result,(un−1)3−2(un)3=4(δ2un+1)3+ (δ2un+1)2(6un−3un+1−6un−1)+3δ2un+1(un−1)2+ (2un−un+1)3−2(un)3=(δ2un+1)2 · [4(un+1−2un+un−1)+6un−3un+1−6un−1]+3δ2un+1(un−1)2+6(un)3−12(un)2un+1+6un(un+1)2− (un+1)3=(δ2un+1)2 · (un+1−2un−2un−1)+3δ2un+1(un−1)2+6un(un+1−un)2− (un+1)3 .(7.123)Therefore we have,|I2| ≤||δ2un+1||∞ · ||δ2un+1||2 · ||δun+1||2· (||un+1||∞+2||un||∞+2||un−1||∞)+||δ2un+1||2 · ||δun+1||2 ·3||un−1||2∞+ ((δun+1)2 , 6un(un+1−un))− ((un+1)3 , δun+1) .(7.124)Now note that(un)44=14(un+1−δun+1)4=14[(un+1)4−4(un+1)3δun+1+6(un+1)2(δun+1)2−4un+1(δun+1)3+ (δun+1)4]= (un+1)44− (un+1)3δun+1+ 14(δun+1)2[6(un+1)2−4un+1(un+1−un)+ (un+1−un)2]= (un+1)44− (un+1)3δun+1+ (δun+1)24[(un)2+2unun+1+3(un+1)2] .(7.125)1017.7. Conditional Stability of the Second Order Scheme II (7.99) & (7.100)Applying this identity,((δun+1)2 , 6un(un+1−un))− ((un+1)3 , δun+1)=∫T2(un)44−∫T2(un+1)44−((δun+1)2 ,254(un)2− 112unun+1+ 34(un+1)2)=∫T2(un)44−∫T2(un+1)44−((δun+1)2 ,254(un− 1125un+1)2)+ 2350((δun+1)2 , (un+1)2).(7.126)Note that||δ2un+1||∞ ≤ 4max{||un−1||∞, ||un||∞, ||un+1||∞} ,RHS≤− 14||un+1||44+12||un+1||22−12||δun+1||22+ 14||un||44−12||un||22+12||δun||22− 12||δ2un+1||22+||δun+1||22 ·(12+ 2350||un+1||2∞)+||δ2un+1||2 · ||δun+1||2 ·23max{||un−1||2∞, ||un||2∞, ||un+1||2∞} .(7.127)Recall thatE(u)= ν2||∇u||22+14||u||44−12||u||22+14·m(T2) ,where m(T2) is the measure of T2. Hence by comparing the LHS and theRHS of (7.126), we get1027.7. Conditional Stability of the Second Order Scheme II (7.99) & (7.100)E(un+1)+ 14τ||δun+1||22+A+12||δun+1||22≤E(un)+ 14τ||δun||22+A+12||δun||22+||δ2un+1||2 · ||δun+1||2 ·23max{||un−1||2∞, ||un||2∞, ||un+1||2∞}−{A+12||δ2un+1||22+(1τ− 12− 2350||un+1||2∞)· ||δun+1||22}.(7.128)Clearly, to show the desired energy decay, it suffices to require2(A+1)(1τ− 12− 2350||un+1||2∞)≥ 529max{||un−1||4∞, ||un||4∞, ||un+1||4∞} . (7.129)As usual, we will prove inductively. SetB=max{E˚(u1) . E(u0)},By Lemma 7.6.1 in previous section, B. 1. We shall prove for every m≥ 2,E˚(um)≤B · (1+ν)2 , E˚(um)≤ E˚(um−1) ,||um||∞ ≤αB ·[(ν12 +ν−1) ·√1+ log(A+1)+| logν|+ | logτ|+1],(7.130)where αB > 0 is the same constant in Lemma 7.7.2.It suffices to verify the main inequality (7.129):2(A+1)(1τ− 12−C1−C1 · (ν−2+ν) · (1+ log(A+1)+| logν|+ | logτ|))>C2(ν−4+ν2) ·(1+| log(A+1)|2+| logν|2+| logτ|2)+C2 . (7.131)1037.7. Conditional Stability of the Second Order Scheme II (7.99) & (7.100)Case 1: A = 0. Then we need1τÀ (ν−4+ν2) · (1+| logν|2+| logτ|2)+1 . (7.132)If 0< ν< 1, then we needτ¿ ν41+| logν|2 ; (7.133)If ν≥ 1, then we needτ¿ ν−21+| logν|2 . (7.134)Case 2: A = const · (ν2+ν−4). In this case,1τÀ (ν−2+ν) · (1+| log(A+1)|+ | logν|+ | logτ|)+| logν|2+| logτ|2+1 .(7.135)If 0< ν< 1, then we needτ¿ ν21+| logν| ; (7.136)If ν≥ 1, then we needτ¿ ν−11+| logν| . (7.137)This completes the proof.104Chapter 8Conclusion & Future WorkThroughout this thesis, we discussed first order and second order semi-implicit Fourier spectral methods on Allen-Cahn equation and fractional Cahn-Hilliard equation in both two dimensional and three dimensional cases. Weproved the stability of the first order numerical scheme by adding a stabi-lizing term A(un+1− un) and A(−∆)α(un+1− un) with a large constant A atleast of size O(ν−1| log(ν)|) for 2D case and O(ν−3) for 3D case. Note that thisstability is preserved independent of time step τ. We also proved a L2 er-ror estimate between numerical solutions from the semi-implicit scheme andexact solutions. We proved stability results and L2 error estimates for twosecond order schemes as well.Future work could be done in other gradient cases such as G = Π0, thezero-mass projected Allen-Cahn equation and G =−∆(id−∆)−1, the normal-ized Cahn-Hilliard equation could be studied as well.105Bibliography[1] Stefan Banach. Sur les opérations dans les ensembles abstraits et leurapplication aux équations intégralesa., Fund. Math., 3 (1922), no.1, pp.133–181.[2] John W. Cahn, and John E. Hilliard. Free Energy of a Nonuniform Sys-tem. I. Interfacial Free Energy., The Journal of Chemical Physics., 28(1958), no.2, pp. 258-267.[3] Lennart Carleson. On convergence and growth of partial sums of Fourierseries., Acta Math., 116 (1966), no.1, pp. 135–157.[4] Andrew Christlieb, Jaylan Jones, Keith Promislow, Brian Wetton, andMark Willoughby. High accuracy solutions to energy gradient flows frommaterial science models., Journal of computational physics., 257 (2014),pp. 193-215.[5] Charles M. Elliott, and Songmu Zheng. On the Cahn-Hilliard equation.,Archive for Rational Mechanics and Analysis., 96 (1986), no.4, pp. 339-357.[6] Lawrence C. Evans. Partial Differential Equations: Second Edition .,(2010), American Mathematical Society Providence, Rhode Island.106Bibliography[7] Thomas H. Grönwall. Note on the derivatives with respect to a parameterof the solutions of a system of differential equations., Ann. of Math., 20(1919), no.4, pp. 292-296.[8] Yinnian He, Yunxian Liu, and Tao Tang. On large time-stepping meth-ods for the Cahn–Hilliard equation., Applied Numerical Mathematics.,57 (2007), no. 5-7, pp. 616–628.[9] Dong Li, and Zhonghua Qiao. On second order semi-implicit fourier spec-tral methods for 2D Cahn–Hilliard equations., Journal of Scientific Com-puting., 70 (2017), pp. 1-41.[10] Dong Li, and Zhonghua Qiao. On the stabilization size of semi-implicitFourier-spectral methods for 3D Cahn-Hilliard equations., Comm. Math.Sci., 15 (2017), no. 6, pp. 1489-1506.[11] Dong Li, Zhonghua Qiao, and Tao Tang. Characterizing the stabilizationsize for semi-implicit Fourier-spectral method to phase field equations.,SIAM J. Numer. Anal., 54 (2016), no.3, pp. 1653–1681.[12] Dong Li, Zhonghua Qiao, and Tao Tang. Gradient bounds for a thin filmepitaxy equation., Journal of Differential Equations., 262 (2017), no.3, pp.1720-1746.[13] Louis Nirenberg. On elliptic partial differential equations., Ann. ScuolaNorm. Sup. Pisa., 13 (1959), no.2, pp. 115-162.[14] Jie Shen, and Xiaofeng Yang. Numerical approximations of Allen-Cahn107Bibliographyand Cahn-Hilliard equations., Discrete Contin. Dyn. Syst. A., 28 (2010),no.4, pp. 1669–1691.[15] Alan D. Sokal. A really simple elementary proof of the uniform bound-edness theorem., The American Mathematical Monthly., 118 (2011), no.5,pp. 450-452.[16] Terence Tao. Nonlinear dispersive equations: local and global analysis.,(2006), American Mathematical Soc.[17] Chuanju Xu, and Tao Tang. Stability analysis of large time-steppingmethods for epitaxial growth models., SIAM J. Numer. Anal., 44 (2006),no.4, pp. 1759–1779.[18] Jingzhi Zhu, Long-Qing Chen, Jie Shen, and Veena Tikare. Coarseningkinetics from a variable-mobility Cahn-Hilliard equation: Application ofa semi-implicit Fourier spectral method., Phys. Rev. E, 60 (1999), no.4, pp.3564–3572.[19] [fabiogarofalophd]. Spectral simulation of the Cahn-Hilliard equation ina two dimensional box., [video File] (2011)., Retrieved June 20th, 2017,from https://www.youtube.com/watch?v=od8Rb7awiOQ.[20] [ogbash]. Simulation of 2D waves using implicit scheme of Finite Dif-ference Method., [video File] (2009)., Retrieved June 19th, 2017, fromhttps://www.youtube.com/watch?v=QKSUaRbgpSM.108

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