# Remarks on Numerical Experiments of the Allen-Cahn Equations with Constraint via Yosida Approximation

Authors: Tomoyuki Suzuki; Keisuke Takasao; Noriaki Yamazaki
Journal: Advances in Numerical Analysis
 (2016)

---

## Abstract

We consider a one-dimensional Allen-Cahn equation with constraint from the viewpoint of numerical analysis. The constraint is provided by the subdifferential of the indicator function on the closed interval, which is the multivalued function. Therefore, it is very difficult to perform a numerical experiment of our equation. In this paper we approximate the constraint by the Yosida approximation. Then, we study the approximating system of the original model numerically. In particular, we give the criteria for the standard forward Euler method to give the stable numerical experiments of the approximating equation. Moreover, we provide the numerical experiments of the approximating equation.


---

## Body

## 1. Introduction

In this paper, for eachε∈(0,1] we consider the following Allen-Cahn equation with constraint from the viewpoint of numerical analysis: (1)utε-uxxε+∂I-1,1uεε2∋uεε2inQ≔0,T×0,1,(2)uxεt,0=uxεt,1=0,t∈0,T,(3)uε0,x=u0εx,x∈0,1,where 0<T<+∞ and u0ε is given initial data. Also, the constraint ∂I[-1,1](·) is the subdifferential of the indicator function I[-1,1](·) on the closed interval [-1,1] defined by (4)I-1,1z≔0,ifz∈-1,1,+∞,otherwise.More precisely, ∂I[-1,1](·) is a set-valued mapping defined by (5)∂I-1,1z≔∅,ifz<-1orz>1,0,∞,ifz=1,0,if-1<z<1,-∞,0,ifz=-1.The Allen-Cahn equation was proposed to describe the macroscopic motion of phase boundaries. In the physical context, the functionuε=uε(t,x) in (P)ε≔{(1),(2),(3)} is the nonconserved order parameter that characterizes the physical structure. For instance, let v=v(t,x) be the local ratio of the volume of pure liquid relative to that of pure solid at time t and position x∈(0,1), defined by (6)vt,x≔limr↓0⁡thevolumeofpureliquidinBrxattimetBrx,where Br(x) is the ball in R with center x and radius r and Brx denotes its volume. Put uε(t,x)≔2v(t,x)-1 for any (t,x)∈Q. Then, we observe that uε(t,x) is the nonconserved order parameter that characterizes the physical structure: (7)uεt,x=1onthepureliquidregion,uεt,x=-1onthepuresolidregion,-1<uεt,x<1onthemixtureregion.There is a vast amount of literature on the Allen-Cahn equation with or without constraint∂I[-1,1](·). For these studies, we refer to [1–16]. In particular, Chen and Elliott [5] considered the singular limit of (P)ε as ε→0 in the general bounded domain Ω(⊂RNwithN≥1).Also, there is a vast amount of literature on the numerical analysis of the Allen-Cahn equation without constraint∂I[-1,1](·). For these studies, we refer to [17–21].Note that the constraint∂I[-1,1](·) is the multivalued function. Therefore, it is very difficult to apply the numerical methods developed by [17–21] to (P)ε. Hence, it is difficult to perform a numerical experiment of (P)ε. Recently, Blank et al. [2] proposed as a numerical method a primal-dual active set algorithm for the local and nonlocal Allen-Cahn variational inequalities with constraint. Also, Farshbaf-Shaker et al. [8] gave the results of the limit of a solution uε and an element of ∂I[-1,1](uε), called the Lagrange multiplier, to (P)ε as ε→0. Moreover, they [9] gave the numerical experiment to (P)ε via the Lagrange multiplier in one dimension of space for sufficient small ε∈(0,1]. Furthermore, they [9] considered the approximating method, called the Yosida approximation. More precisely, for δ>0, the Yosida approximation (∂I[-1,1])δ(·) of ∂I[-1,1](·) is defined by (8)∂I-1,1δz=z-1+--1-z+δ,∀z∈R,where [z]+ is the positive part of z. Then, for each δ>0, they [9] considered the following approximation problem of (P)ε: (9) P δ ε u δ ε t - u δ ε x x + ∂ I - 1,1 δ u δ ε ε 2 = u δ ε ε 2 i n   Q ≔ 0 , T × 0,1 ,  u δ ε x t , 0 = u δ ε x t , 1 = 0 , t ∈ 0 , T , u δ ε 0 , x = u 0 ε x , x ∈ 0,1 .In [9, Remark 4.2], Figure 1 shows the unstable numerical result to (P)δε was given by the standard explicit finite difference scheme to (P)δε.Figure 1
Behavior of a solution to(P)δε with ε=0.007 and δ=0.01.From Figure1, we observe that we have to choose the suitable constants ε and δ and the mesh size of time Δt and space Δx in order to get the stable numerical results of (P)δε. Therefore, in this paper, for each ε>0 and δ>0, we give the criteria for the standard explicit finite difference scheme to provide the stable numerical experiments of (P)δε. To this end, we first consider the following ODE problem, denoted by (E)δε: (10)Eδεuδεt+∂I-1,1δuδεε2=uδεε2inR,fort∈0,T,uδε0=u0εinR.Then, we give the criteria to get the stable numerical experiments of (E)δε. Also, we give some numerical experiments of (E)δε. Moreover, we show the criteria to get the stable numerical experiments of PDE problem (P)δε. Therefore, the main novelties are the following: (a)
We give the criteria to get the stable numerical experiments of the ODE problem(E)δε. Also, we provide the numerical experiments to (E)δε for sufficient small ε∈(0,1] and δ∈(0,1].(b)
We give the criteria to get the stable numerical experiments of the PDE problem(P)δε. Also, we provide the numerical experiments to (P)δε for sufficient small ε∈(0,1] and δ∈(0,1].The plan of this paper is as follows. In Section2, we mention the solvability and convergence result of (E)δε. In Section 3, we consider (E)δε numerically. Then, we prove the main result (Theorem 7) corresponding to item (a) listed above. Also, we provide the numerical experiments to (E)δε for sufficient small ε∈(0,1] and δ∈(0,1]. In Section 4, we mention the solvability and convergence result of (P)δε. In the final Section 5, we consider (P)δε from the viewpoint of numerical analysis. Then, we prove the main result (Theorem 16) corresponding to item (b) listed above. Also, we provide the numerical experiments to (P)δε for sufficient small ε∈(0,1] and δ∈(0,1].Notations and Basic Assumptions. Throughout this paper, we put H≔L2(0,1) with the usual real Hilbert space structure. The inner product and norm in H are denoted by (·,·)H and by ·H, respectively. We also put V≔H1(0,1) with the usual norm |z|V≔|z|H2+|zx|H21/2 for z∈V.In Sections2 and 4, we use some techniques of proper (i.e., not identically equal to infinity), l.s.c. (lower semicontinuous), and convex functions and their subdifferentials, which are useful in the systematic study of variational inequalities. So, let us outline some notations and definitions. Let W be the real Hilbert space with the inner product (·,·)W. For a proper, l.s.c., and convex function ψ:W→R∪{+∞}, the effective domain D(ψ) is defined by (11)Dψ≔z∈W;ψz<∞.The subdifferential of ψ is a possibly multivalued operator in W and is defined by z∗∈∂ψ(z) if and only if (12)z∈Dψ,z∗,y-zW≤ψy-ψz∀y∈W.For various properties and related notions of the proper, l.s.c., and convex function ψ and its subdifferential ∂ψ, we refer to the monograph by Brézis [22].


## 2. Solvability and Convergence Results of(E)δε

We begin by giving the rigorous definition of solutions to our problem(E)δε (ε∈(0,1] and δ∈(0,1]).Definition 1.
Letε∈(0,1], δ∈(0,1], and u0ε∈R. Then, a function uδε:[0,T]→R is called a solution to (E)δε on [0,T], if the following conditions are satisfied:(i)
uδε∈W1,20,T.(ii)
The following equation holds:(13)uδεt+∂I-1,1δuδεε2=uδεε2inR,fort∈0,T.(iii)
uδε(0)=u0ε in R.Now, we show the solvability result of(E)δε on [0,T].Proposition 2.
Letε∈(0,1], δ∈(0,1], and u0ε∈R with |u0ε|≤1. Then, there exists a unique solution uδε to (E)δε on [0,T] in the sense of Definition 1.Proof.
We can prove the uniqueness of solutions to(E)δε on [0,T] by the quite standard arguments: monotonicity and Gronwall’s inequality.
Also, we can show the existence of solutions to(E)δε on [0,T] applying the abstract theory of evolution equations governed by subdifferentials. Indeed, we define a function (I[-1,1])δ(·) on R by (14)I-1,1δz=z-1+2+-1-z+22δ,∀z∈R.Clearly, (I[-1,1])δ(·) is proper, l.s.c., and convex on R with ∂(I[-1,1])δ(·)=(∂I[-1,1])δ(·) in R, where (∂I[-1,1])δ(·) is the Yosida approximation of ∂I[-1,1](·) defined by (8).
Note that the problem(E)δε can be rewritten as in an abstract framework of the form (15) EP δ ε ⁡ d d t u δ ε t + 1 ε 2 ∂ I - 1,1 δ u δ ε t - 1 ε 2 u δ ε t = 0 i n   R ,   f o r   t ∈ 0 , T ,  u δ ε 0 = u 0 ε i n   R .Therefore, applying the Lipschitz perturbation theory of abstract evolution equations (cf. [23–25]), we can show the existence of a solution uδε to (EP)δε, hence, (E)δε, on [0,T] for each ε∈(0,1] and δ∈(0,1] in the sense of Definition 1. Thus, the proof of Proposition 2 has been completed.Next, we show the convergence result of(E)δε as δ→0. To this end, we recall a notion of convergence for convex functions, developed by Mosco [26].Definition 3 (cf. [26]).
Letψ and ψn (n∈N) be proper, l.s.c., and convex functions on a Hilbert space W. Then, one says that ψn converges to ψ on W in the sense of Mosco [26] as n→∞, if the following two conditions are satisfied:(i)
for any subsequence{ψnk}⊂{ψn}, if zk→z weakly in W as k→∞, then (16)liminfk→∞⁡ψnkzk≥ψz;(ii)
for anyz∈D(ψ), there is a sequence {zn} in W such that (17)zn⟶zinWasn⟶∞,limn→∞⁡ψnzn=ψz.It is well known that the following lemma holds. Therefore, we omit the detailed proof.Lemma 4 (cf. [27, Section 5], [22, Chapter 2], and [28, Section 2]).
Consider(18)I-1,1δ·⟶I-1,1·onRinthesenseofMosco26asδ⟶0.By Lemma4 and the general convergence theory of evolution equations, we get the following result. We omit the detailed proof.Proposition 5 (cf. [27, Section 5], [28, Section 2]).
Letε∈(0,1], δ∈(0,1], and u0ε∈R with |u0ε|≤1. Also, let uδε be the unique solution to (E)δε on [0,T]. Then, there exists a unique function uε∈W1,2(0,T) such that (19)uδε⟶uεinC0,Tasδ⟶0and uε is the unique solution of the following problem (E)ε on [0,T]: (20) E ε u t ε + ∂ I - 1,1 u ε ε 2 ∋ u ε ε 2 i n   R ,   f o r   t ∈ 0 , T ,  u ε 0 = u 0 ε i n   R .


## 3. Stable Criteria and Numerical Experiments for(E)δε

In this section we consider(E)δε from the viewpoint of numerical analysis.Remark 6.
Note from Proposition5 that (E)δε is the approximating problem of (E)ε. Also note from (5) that the constraint ∂I-1,1· is the multivalued function. Therefore, it is very difficult to study (E)ε numerically.In order to perform the numerical experiments of(E)δε via the standard forward Euler method, we consider the following explicit finite difference scheme to (E)δε, denoted by (DE)δε:(21) D E δ ε u n + 1 - u n Δ t + ∂ I - 1,1 δ u n ε 2 = u n ε 2 i n   R ,   f o r   n = 0,1 , 2 , … , N t ,  u 0 = u 0 ε i n   R ,where Δt is the mesh size of time and Nt is the integer part of number T/Δt.We observe thatun is the approximating solution of (E)δε at the time t=nΔt. Also, we observe that the explicit finite difference scheme (DE)δε converges to (E)δε as Δt→0 since (DE)δε is the standard time discretization scheme for (E)δε.In Figure2, we give the unstable numerical experiment of (DE)δε in the case when T=0.002, ε=0.003, δ=0.01, the initial data u0ε=0.1, and the mesh size of time Δt=0.000001.Figure 2
Behavior of a solutionun to (DE)δε with ε=0.003 and δ=0.01.From Figure2, we observe that we have to choose the suitable constants ε and δ and the mesh size of time Δt in order to get the stable numerical results of (DE)δε.Now, let us mention the first main result in this paper, which is concerned with the criteria to give the stable numerical experiments of(DE)δε.Theorem 7.
Letε∈(0,1], δ∈(0,1), and Δt∈(0,1]. Assume u0ε∈(0,1] (resp., u0ε∈[-1,0)) and T=∞. Let {un;n≥0} be the solution to (DE)δε. Then, one has the following: (i)
IfΔt∈0,δε2/(1-δ), un converges to 1/(1-δ) (resp., -1/(1-δ)) monotonically as n→∞.(ii)
IfΔt∈δε2/(1-δ),2δε2/(1-δ), un oscillates and converges to 1/(1-δ) (resp., -1/(1-δ)) as n→∞.Proof.
We give the proof of Theorem7 in the case of the initial value u0ε∈(0,1].
For simplicity, we set(22)fδz≔∂I-1,1δz-zforz∈R.Then, we observe that (23)fδz=1+zδ-z,ifz≤-1,-z,ifz∈-1,1,z-1δ-z,ifz≥1,and z=0, 1/(1-δ), -1/(1-δ) are the zero points of fδ·. Also, we observe that the difference equation (DE)δε is reformulated in the following form: (24)un+1=un-Δtε2fδuninR,forn=0,1,2,….Note from (23) and (24) that if un∈(0,1] we have (25)un+1=un-Δtε2fδun=un-Δtε2·-un≥un,which implies that un is increasing with respect to n until un+1≥1.
Now, we prove (i). To this end, we assume thatΔt∈0,δε2/(1-δ). At first, by the mathematical induction, we show (26)ui∈0,11-δ∀i≥0.
Clearly (26) holds for i=0 because of u0=u0ε∈(0,1].
Now, we assume that (26) holds for all i=0,1,…,n. Suppose un∈(0,1]. Then, we infer from (25) that (27)un+1=1+Δtε2un≤1+Δtε2<1+δ1-δ=11-δ.Therefore, by (25) and the inequality as above, we observe that (28)un+1∈0,11-δ,ifun∈0,1.Next, if un∈[1,1/(1-δ)), we observe from (23) and (24) that (29)un≤un+1=un-Δtε2fδun=un-Δtε2·un-1δ-un=un+Δtε2·1-1-δunδ<un+1-1-δun1-δ=11-δ,which implies that (30)un+1∈1,11-δ,ifun∈1,11-δ.From (28) and (30) we infer that (26) holds for i=n+1. Therefore, we conclude from the mathematical induction that (26) holds.
Also, by (23) and (26) we observe that fδ(un)≤0 for all n≥0. Therefore, we observe from (24) that (31)un+1=un-Δtε2fδun≥un∀n≥0.Therefore, we infer from (26) and (31) that {un;n≥0} is a bounded and increasing sequence with respect to n. Thus, there exists a point u∞∈R such that (32)un⟶u∞inRasn⟶∞.Taking the limit in (24), we observe from the continuity of fδ(·) that u∞=1/(1-δ), which is the zero point of fδ(·). Hence, the proof of (i) has been completed.
Next, we show (ii). To this end, we assume thatΔt∈δε2/(1-δ),2δε2/(1-δ). Then, we can find the minimal number n0∈N so that (33)un0∈1,1+δ1-δ,ui∈0,1∀i=0,1,…,n0-1.Indeed, if ui∈(0,1] for all i=0,1,…,k, we observe from (25) that (34)uk+1=uk-Δtε2fδuk=1+Δtε2uk=1+Δtε22uk-1=⋯=1+Δtε2k+1u0.Taking into account (34), u0∈(0,1], and (35)1+Δtε2>1+δ1-δ>1,we can find the minimal number n0∈N so that (36)un0>1,ui∈0,1∀i=0,1,…,n0-1.Also, we observe from (25) that (37)un0=un0-1-Δtε2fδun0-1=1+Δtε2un0-1<1+2δ1-δ·1=1+δ1-δ,and thus, (33) holds.
To show (ii), we put(38)Δt≔δε21-δτforsomeτ∈1,2.Then, we observe from (23) and (24) that (39)un0+1=un0-Δtε2fδun0=un0+Δtε2·1-1-δun0δ=1-τun0+τ1-δ.From (39) it follows that (40)un0+1+τ-1un0τ=11-δ.Therefore, we observe from (40) and τ∈(1,2) that the zero point 1/(1-δ) of fδ(·) is in the interval between un0 and un0+1.
Also, by (39) we observe that (41)un0+1=1-τun0+τ1-δ>1-τ1+δ1-δ+τ1-δ=1+δ-τδ1-δ≥1,un0+1=1-τun0+τ1-δ<1-τ·1+τ1-δ=1-δ+τδ1-δ≤1+δ1-δ,which implies that (42)un0+1∈1,1+δ1-δ.Therefore, by (33) and (40) and by repeating the procedure as above, we observe that (43)un∈1,1+δ1-δ∀n≥n0and un oscillates around the zero point 1/(1-δ) for all n≥n0.
Also, we observe from (39) and (43) that (44)un+1-11-δ=1-τun-11-δ∀n≥n0.Therefore, by τ∈1,2, (43), and (44), there is a subsequence {nk} of {n} such that unk oscillates and converges to 1/(1-δ) as k→∞. Hence, taking into account the uniqueness of the limit point, the proof of (ii) has been completed.Remark 8.
AssumeΔt∈2δε2/(1-δ),∞ and put Δt≔δε2τ/(1-δ) for some τ≥2. Then, we observe that (45)1+Δtε2>1+2δ1-δ>1,1-τ≥1.Therefore, we infer from the proof of Theorem 7 (cf. (34), (40), and (44)) that the solution un to (DE)δε oscillates as n→∞, in general.Remark 9.
By (24) we easily observe that (46)un≡0∀n≥1,ifu0=0,un≡11-δ∀n≥1,ifu0=11-δ,un≡-11-δ∀n≥1,ifu0=-11-δ.From (ii) of Theorem7, we observe that un oscillates and converges to the zero point of fδ(·) in the case when Δt∈δε2/(1-δ),2δε2/(1-δ). However, in the case when Δt=2δε2/(1-δ), we have the following special case that the solution to (DE)δε does not oscillate and coincides with the zero point of fδ(·) after some finite number of iterations.Corollary 10.
Letε∈(0,1], δ∈(0,1), Δt=2δε2/(1-δ), and n∈N. Assume u0ε≔(1-δ)n-1/(1+δ)n. Then, the solution to (DE)δε is given by (47) u i = 1 - δ n - 1 - i 1 + δ n - i , i f   i = 0,1 , … , n - 1 ,  1 1 - δ , i f   i ≥ n .Proof.
Note thatu0ε≔(1-δ)n-1/(1+δ)n∈(0,1). Therefore, by (23) and (24) we observe that (48)u1=u0-Δtε2fδu0=u0ε-2δ1-δ-u0ε=1+δ1-δu0ε=1-δn-21+δn-1.Similarly, we have (49)u2=u1-Δtε2fδu1=1+δ1-δu1=1-δn-31+δn-2.Repeating this procedure, we observe that the solution to (DE)δε is given by (47).Taking into account Theorem7, we provide the numerical experiments of (DE)δε as follows. To this end, we use the following numerical data.Numerical Data of ( D E ) δ ε The numerical data areT=0.002, ε=0.01, δ=0.01, and the initial data u0ε=0.1. Then, we observe that (50)11-δ=11-0.01=1.010101010…,δε21-δ=0.0000010101010….


### 3.1. The Case WhenΔt=0.000001

Now we consider the case whenΔt=0.000001. In this case, we have (51)δε21-δ=0.0000010101010⋯>Δt=0.000001,which implies that (i) of Theorem 7 holds. Thus, we have the stable numerical result of (DE)δε. Indeed, we observe from Figure 3 and Table 1 that the solution to (DE)δε converges to the stationary solution 1/(1-δ)=1/(1-0.01)=1.010101010….Table 1
Numerical data:Δt=000001.
Number of iterationsi The value ofui 0 0.100000 1 0.101000 2 0.102010 3 0.103030 4 0.104060 5 0.105101 ⋮ ⋮ 222 0.910636 223 0.919743 224 0.928940 225 0.938230 226 0.947612 227 0.957088 228 0.966659 229 0.976325 230 0.986089 231 0.995950 232 1.005909 233 1.010059 234 1.010101 235 1.010101 236 1.010101 237 1.010101 238 1.010101Figure 3
δ ε 2 / 1 - δ = 0.0000010101010 ⋯ > Δ t = 0.000001 .


### 3.2. The Case WhenΔt=0.000002

Now we consider the case whenΔt=0.000002. In this case, we have (52)δε21-δ=0.0000010101010⋯<Δt=0.000002<2δε21-δ,which implies that (ii) of Theorem 7 holds. Thus, we observe from Figure 4 and Table 2 that the solution to (DE)δε oscillates and converges to the stationary solution 1/(1-δ)=1/(1-0.01)=1.010101010….Table 2
Numerical data:Δt=0.000002.
Number of iterationsi The value ofui 0 0.100000 1 0.102000 2 0.104040 ⋮ ⋮ 114 0.955916 115 0.975034 116 0.994535 117 1.014425 118 1.005863 119 1.014254 120 1.006031 121 1.014090 122 1.006192 123 1.013932 124 1.006347 125 1.013780 126 1.006496 127 1.013634 128 1.006638 129 1.013494 ⋮ ⋮ 560 1.010100 561 1.010102 562 1.010100 563 1.010102 564 1.010100 565 1.010102 566 1.010101 567 1.010101 568 1.010101 569 1.010101 570 1.010101Figure 4
δ ε 2 / 1 - δ =0.0000010101010⋯<Δt=0.000002<2δε2/1-δ.


### 3.3. The Case WhenΔt=2δε2/1-δ

Now we consider the case whenΔt=2δε2/(1-δ)=0.0000020202020…. In this case, we observe Remark 8. Indeed, we observe from Figure 5 and Table 3 that the solution to (DE)δε oscillates.Table 3
Numerical data:Δt=2δε2/(1-δ)=0.0000020202020….
Number of iterationsi The value ofui 0 0.100000 1 0.102020 2 0.104081 3 0.106184 4 0.108329 5 0.110517 ⋮ ⋮ 110 0.902568 111 0.920801 112 0.939403 113 0.958381 114 0.977742 115 0.997495 116 1.017646 117 1.002556 118 1.017646 119 1.002556 120 1.017646 121 1.002556 122 1.017646 123 1.002556 124 1.017646 125 1.002556 126 1.017646 127 1.002556 128 1.017646 129 1.002556 130 1.017646Figure 5
Δ t = 2 δ ε 2 / 1 - δ = 0.0000020202020 ….


### 3.4. The Case WhenΔt=0.000005

Now we consider the case whenΔt=0.000005. In this case, we have (53)2δε21-δ=0.0000020202020⋯<Δt=0.000005.Therefore, we observe Remark 8. Indeed, we observe from Figure 6 that the solution to (DE)δε oscillates.Figure 6
2 δ ε 2 / 1 - δ = 0.0000020202020 ⋯ < Δ t = 0.000005.


### 3.5. The Case WhenΔt=15δε2/1-δ

Now we consider the case whenΔt=15δε2/(1-δ). In this case, we observe Remark 8. Indeed, we observe from Figure 7 that the solution to (DE)δε oscillates between three zero points of fδ(·).Figure 7
Δ t = 15 δ ε 2 / 1 - δ.


### 3.6. Numerical Result of Corollary10

In this subsection, we consider Corollary10 numerically. To this end, we use the following initial data: (54)u0≔1-δ51+δ6=1-0.0151+0.016=0.8958756….Then, we observe from Table 4 and Figure 8 that Corollary 10 holds. Namely, we observe that (47) holds with n=6.Table 4
Numerical data:Δt=2δε2/(1-δ)=0.0000020202020….
Number of iterationsi The value ofui 0 0.895876 1 0.913974 2 0.932438 3 0.951275 4 0.970493 5 0.990099 6 1.010101 7 1.010101 8 1.010101 9 1.010101 10 1.010101Figure 8
Δ t = 2 δ ε 2 / 1 - δ = 0.0000020202020 ….


### 3.7. Conclusion of ODE Problem(DE)δε

From Theorem7 and numerical experiments as above, we conclude that(i)
the mesh size of timeΔt must be smaller than δε2/(1-δ) in order to get the stable numerical experiments of (DE)δε,(ii)
we have the stable numerical experiments of(DE)δε with the initial data u0ε≔(1-δ)n-1/(1+δ)n, even if the mesh size of time Δt is equal to 2δε2/(1-δ).


## 3.1. The Case WhenΔt=0.000001

Now we consider the case whenΔt=0.000001. In this case, we have (51)δε21-δ=0.0000010101010⋯>Δt=0.000001,which implies that (i) of Theorem 7 holds. Thus, we have the stable numerical result of (DE)δε. Indeed, we observe from Figure 3 and Table 1 that the solution to (DE)δε converges to the stationary solution 1/(1-δ)=1/(1-0.01)=1.010101010….Table 1
Numerical data:Δt=000001.
Number of iterationsi The value ofui 0 0.100000 1 0.101000 2 0.102010 3 0.103030 4 0.104060 5 0.105101 ⋮ ⋮ 222 0.910636 223 0.919743 224 0.928940 225 0.938230 226 0.947612 227 0.957088 228 0.966659 229 0.976325 230 0.986089 231 0.995950 232 1.005909 233 1.010059 234 1.010101 235 1.010101 236 1.010101 237 1.010101 238 1.010101Figure 3
δ ε 2 / 1 - δ = 0.0000010101010 ⋯ > Δ t = 0.000001 .


## 3.2. The Case WhenΔt=0.000002

Now we consider the case whenΔt=0.000002. In this case, we have (52)δε21-δ=0.0000010101010⋯<Δt=0.000002<2δε21-δ,which implies that (ii) of Theorem 7 holds. Thus, we observe from Figure 4 and Table 2 that the solution to (DE)δε oscillates and converges to the stationary solution 1/(1-δ)=1/(1-0.01)=1.010101010….Table 2
Numerical data:Δt=0.000002.
Number of iterationsi The value ofui 0 0.100000 1 0.102000 2 0.104040 ⋮ ⋮ 114 0.955916 115 0.975034 116 0.994535 117 1.014425 118 1.005863 119 1.014254 120 1.006031 121 1.014090 122 1.006192 123 1.013932 124 1.006347 125 1.013780 126 1.006496 127 1.013634 128 1.006638 129 1.013494 ⋮ ⋮ 560 1.010100 561 1.010102 562 1.010100 563 1.010102 564 1.010100 565 1.010102 566 1.010101 567 1.010101 568 1.010101 569 1.010101 570 1.010101Figure 4
δ ε 2 / 1 - δ =0.0000010101010⋯<Δt=0.000002<2δε2/1-δ.


## 3.3. The Case WhenΔt=2δε2/1-δ

Now we consider the case whenΔt=2δε2/(1-δ)=0.0000020202020…. In this case, we observe Remark 8. Indeed, we observe from Figure 5 and Table 3 that the solution to (DE)δε oscillates.Table 3
Numerical data:Δt=2δε2/(1-δ)=0.0000020202020….
Number of iterationsi The value ofui 0 0.100000 1 0.102020 2 0.104081 3 0.106184 4 0.108329 5 0.110517 ⋮ ⋮ 110 0.902568 111 0.920801 112 0.939403 113 0.958381 114 0.977742 115 0.997495 116 1.017646 117 1.002556 118 1.017646 119 1.002556 120 1.017646 121 1.002556 122 1.017646 123 1.002556 124 1.017646 125 1.002556 126 1.017646 127 1.002556 128 1.017646 129 1.002556 130 1.017646Figure 5
Δ t = 2 δ ε 2 / 1 - δ = 0.0000020202020 ….


## 3.4. The Case WhenΔt=0.000005

Now we consider the case whenΔt=0.000005. In this case, we have (53)2δε21-δ=0.0000020202020⋯<Δt=0.000005.Therefore, we observe Remark 8. Indeed, we observe from Figure 6 that the solution to (DE)δε oscillates.Figure 6
2 δ ε 2 / 1 - δ = 0.0000020202020 ⋯ < Δ t = 0.000005.


## 3.5. The Case WhenΔt=15δε2/1-δ

Now we consider the case whenΔt=15δε2/(1-δ). In this case, we observe Remark 8. Indeed, we observe from Figure 7 that the solution to (DE)δε oscillates between three zero points of fδ(·).Figure 7
Δ t = 15 δ ε 2 / 1 - δ.


## 3.6. Numerical Result of Corollary10

In this subsection, we consider Corollary10 numerically. To this end, we use the following initial data: (54)u0≔1-δ51+δ6=1-0.0151+0.016=0.8958756….Then, we observe from Table 4 and Figure 8 that Corollary 10 holds. Namely, we observe that (47) holds with n=6.Table 4
Numerical data:Δt=2δε2/(1-δ)=0.0000020202020….
Number of iterationsi The value ofui 0 0.895876 1 0.913974 2 0.932438 3 0.951275 4 0.970493 5 0.990099 6 1.010101 7 1.010101 8 1.010101 9 1.010101 10 1.010101Figure 8
Δ t = 2 δ ε 2 / 1 - δ = 0.0000020202020 ….


## 3.7. Conclusion of ODE Problem(DE)δε

From Theorem7 and numerical experiments as above, we conclude that(i)
the mesh size of timeΔt must be smaller than δε2/(1-δ) in order to get the stable numerical experiments of (DE)δε,(ii)
we have the stable numerical experiments of(DE)δε with the initial data u0ε≔(1-δ)n-1/(1+δ)n, even if the mesh size of time Δt is equal to 2δε2/(1-δ).


## 4. Solvability and Convergence Results for(P)δε

We begin by giving the rigorous definition of solutions to our PDE problem(P)δε (ε∈(0,1] and δ∈(0,1]).Definition 11.
Letε∈(0,1], δ∈(0,1], and u0ε∈H. Then, a function uδε:[0,T]→H is called a solution to (P)δε on [0,T], if the following conditions are satisfied: (i)
uδε∈W1,2(0,T;H)∩L∞(0,T;V).(ii)
The following variational identity holds:(55)uδεtt,zH+uδεxt,zxH+∂I-1,1δuδεtε2,zH=uδεtε2,zH∀z∈V,a.e.t∈0,T.(iii)
uδε(0)=u0ε in H. Now, we mention the solvability result of (P)δε on [0,T].Proposition 12.
Letε∈(0,1] and δ∈(0,1]. Assume the following condition: Au0ε∈K≔z∈V;-1≤zx≤1a.e.x∈0,1.Then, for each u0ε∈K, there exists a unique solution uδε to (P)δε on [0,T] in the sense of Definition 11.Proof.
By the same argument as in Proposition2, we can show the existence-uniqueness of a solution uδε to (P)δε on [0,T] for each ε∈(0,1] and δ∈(0,1]. Indeed, we can prove the uniqueness of solutions to (P)δε on [0,T] by the quite standard arguments: monotonicity and Gronwall’s inequality.
Also, we can show the existence of solutions to(P)δε on [0,T] applying the abstract theory of evolution equations governed by subdifferentials. Indeed, we define a functional φδε on H by (56)φδεz≔12∫01zx2dx+1ε2∫01I-1,1δzxdx,ifz∈V,∞,otherwise,where (I[-1,1])δ(·) is the function defined in (14). Clearly, φδε is proper, l.s.c., and convex on H with the effective domain D(φ)=V.
Note that the problem(P)δε can be rewritten as in an abstract framework of the form (57) PP δ ε ⁡ d d t u δ ε t + ∂ φ δ ε u δ ε t - 1 ε 2 u δ ε t = 0 i n   H ,   f o r   t > 0 ,  u ε 0 = u 0 ε i n   H .Therefore, applying the Lipschitz perturbation theory of abstract evolution equations (cf. [23–25]), we can show the existence of a solution uδε to (PP)δε, hence, (P)δε, on [0,T] for each ε∈(0,1] and δ∈(0,1] in the sense of Definition 11. Thus, the proof of Proposition 12 has been completed.Next, we recall the convergence result of(P)δε as δ→0. Taking into account Lemma 4 (cf. (18)), we observe that the following lemma holds.Lemma 13 (cf. [27, Section 5], [22, Chapter 2], [28, Section 2]).
Letε∈(0,1], and define a functional φε on H by (58)φεz≔12∫01zx2dx+1ε2∫01I-1,1zxdx,ifz∈V,∞,otherwise.Then, (59)φδε·⟶φε·onHinthesenseofMosco26asδ⟶0.By Lemma13 and the general convergence theory of evolution equations, we get the following result.Proposition 14 (cf. [27, Section 5], [28, Section 2]).
Letε∈(0,1], δ∈(0,1], and u0ε∈K. Also, let uδε be the unique solution to (P)δε on [0,T]. Then, uδε converges to the unique function uε to (P)ε on [0,T] in the sense that (60)uδε⟶uεstronglyinC0,T;Hasδ⟶0.Proof.
Note that the problem(P)ε can be rewritten as in an abstract framework of the form (61) PP ε ⁡ d d t u ε t + ∂ φ ε u ε t - 1 ε 2 u ε t ∋ 0 i n   H ,   f o r   t > 0 ,  u ε 0 = u 0 ε i n   H .Therefore, by Lemma 13 and the abstract convergence theory of evolution equations (cf. [27, 28]), we observe that the solution uδε to (PP)δε converges to the unique solution uε to (PP)ε on [0,T] as δ→0 in the sense of (60). Note that uε (resp., uδε) is the unique solution to (P)ε (resp., (P)δε) on [0,T] (cf. Proposition 12). Thus, we conclude that Proposition 14 holds.


## 5. Stable Criteria and Numerical Experiments for(P)δε

In this section we consider(P)δε from the viewpoint of numerical analysis.Remark 15.
Note from Proposition14 that (P)δε is the approximating problem of (P)ε. Also note from (5) that the constraint ∂I[-1,1](·) is the multivalued function. Therefore, it is very difficult to study (P)ε numerically.In order to perform the numerical experiments of(P)δε, we consider the following explicit finite difference scheme to (P)δε, denoted by (DP)δε: (62) DP δ ε ⁡ u k n + 1 - u k n Δ t - u k - 1 n - 2 u k n + u k + 1 n Δ x 2 + ∂ I - 1,1 δ u k n ε 2 = u k n ε 2 f o r   n = 0,1 , 2 , … , N t ,   k = 1,2 , … , N x - 1 ,  u 0 n = u 1 n ,   u N x n = u N x - 1 n f o r   n = 1,2 , … , N t ,  u k 0 = u 0 ε x k f o r   k = 0,1 , 2 , … , N x ,where Δt is the mesh size of time, Δx is the mesh size of space, Nt is the integer part of number T/Δt, Nx is the integer part of number 1/Δx, and xk≔kΔx.We observe thatukn is the approximating solution of (P)δε at the time tn≔nΔt and the position xk≔kΔx. Also, we observe that the explicit finite difference scheme (DP)δε converges to (P)δε as Δt→0 and Δx→0.From Figure1, we observe that we have to choose the suitable constants ε and δ and the mesh size of time Δt and the mesh size of space Δx in order to get the stable numerical results of (DP)δε. Now, let us mention our second main result concerning the stability of (DE)δε.Theorem 16.
Letε∈(0,1], δ∈(0,1), Δt∈(0,1], Δx∈(0,1], T>0, and u0ε∈K, where K is the set of initial values defined in Proposition 12 (cf. A). Let Nx be the integer part of number 1/Δx, and let {ukn;n≥0,k=0,1,…,Nx} be the solution to (DP)δε. Also, let c0∈(0,1) and assume that (63)0<Δt≤c0δε21-δ,0≤ΔtΔx2≤1-c02.Then, one has the following: (i)
The solution to(DP)δε is bounded in the following sense: (64)max0≤k≤Nx⁡ukn≤11-δ∀n≥0.(ii)
{ukn;k=0,1,…,Nx} does not oscillate with respect to n≥0.Proof.
We first show (i), that is, (64), by the mathematical induction.
Clearly (64) holds for n=0 because of u0ε∈K.
Now, we assume that(65)max0≤k≤Nx⁡uki≤11-δ∀i=0,1,…,n.We observe that the explicit finite difference problem (DP)δε can be reformulated as in the following form: (66)ukn+1=ruk-1n+ruk+1n+1-2rukn-Δtε2fδukn∀n=0,1,2,…,Nt,k=1,2,…,Nx-1,where we put r≔Δt/(Δx)2 and fδ· is the function defined by (23).
We observe from (63), (65), and (66) that (67)11-δ-ukn+1=r11-δ-uk-1n+r11-δ-uk+1n+1-2r11-δ-ukn+Δtε2fδukn≥1-2r11-δ-ukn+Δtε2fδukn∀k=1,2,…,Nx-1.
From (23), (63), and (65) we infer that the function [-1/(1-δ),1/(1-δ)]∋z→(1-2r)1/(1-δ)-z+Δt/ε2fδ(z) is nonnegative and continuous. Indeed, it follows from (23) that the function [-1/(1-δ),1]∋z→(1-2r)1/(1-δ)-z+Δt/ε2fδ(z) attains a minimum value at z=1. Therefore, we observe from (23) and (63) that(68)1-2r11-δ-z+Δtε2fδz≥1-2r11-δ-1+Δtε2fδ1=1-2r·δ1-δ-Δtε2≥c0δ1-δ-Δtε2≥0∀z∈-11-δ,1.Also, for any z∈[1,1/(1-δ)], we observe from (23) that (69)1-2r11-δ-z+Δtε2fδz=1-2r11-δ-z+Δtε2·z-1δ-z=1-δδε2Δt-1-2rz+1-2r11-δ-Δtδε2.Here we note from (63) that (70)1-δδε2Δt-1-2r≤1-δδε2Δt-c0≤0.Therefore, we infer from (69) that the function [1,1/(1-δ)]∋z→(1-2r)1/(1-δ)-z+Δt/ε2fδ(z) is nonincreasing and attains a minimum value at z=1/(1-δ). Hence, we have (71)1-2r11-δ-z+Δtε2fδz≥Δtε2fδ11-δ=0∀z∈1,11-δ.Thus, we observe from (68) and (71) that (72)1-2r11-δ-z+Δtε2fδz≥0,∀z∈-11-δ,11-δ,which implies from (65) and (67) that (73)11-δ-ukn+1≥0∀k=1,2,…,Nx-1.
Similarly, we observe from (63), (65), and (66) that (74)ukn+1+11-δ=ruk-1n+11-δ+ruk+1n+11-δ+1-2rukn+11-δ-Δtε2fδukn≥1-2rukn+11-δ-Δtε2fδukn∀k=1,2,…,Nx-1.By similar arguments as above, we observe that the function [-1/(1-δ),1/(1-δ)]∋z→(1-2r)z+1/(1-δ)-Δt/ε2fδ(z) is nonnegative and continuous. In fact, it follows from (23) that the function [-1,1/(1-δ)]∋z→(1-2r)z+1/(1-δ)-Δt/ε2fδ(z) attains a minimum value at z=-1. Therefore, we observe from (23) and (63) that (75)1-2rz+11-δ-Δtε2fδz≥1-2r-1+11-δ-Δtε2fδ-1=1-2r·δ1-δ-Δtε2≥c0δ1-δ-Δtε2≥0∀z∈-1,11-δ.Also, for any z∈[-1/(1-δ),-1], we observe from (23) that (76)1-2rz+11-δ-Δtε2fδz=1-2rz+11-δ-Δtε2·1+zδ-z=1-2r-1-δδε2Δtz+1-2r11-δ-Δtδε2.Here we note from (63) that (77)1-2r-1-δδε2Δt≥c0-1-δδε2Δt≥0.Therefore, we infer from (76) that the function [-1/(1-δ),-1]∋z→(1-2r)z+1/(1-δ)-Δt/ε2fδ(z) is nondecreasing and attains a minimum value at z=-1/(1-δ). Hence, we have (78)1-2rz+11-δ-Δtε2fδz≥-Δtε2fδ-11-δ=0∀z∈-11-δ,-1.Thus, we observe from (75) and (78) that (79)1-2rz+11-δ-Δtε2fδz≥0,∀z∈-11-δ,11-δ,which implies from (65) and (74) that (80)ukn+1+11-δ≥0∀k=1,2,…,Nx-1.
Taking into account Neumann boundary condition, namely, byu0n+1=u1n+1 and uNxn+1=uNx-1n+1, we observe from (73) and (80) that (81)max0≤k≤Nx⁡ukn+1≤11-δ,which implies that (65) holds for i=n+1. Therefore, we conclude from the mathematical induction that (64) holds. Hence, the proof of (i) of Theorem 16 has been completed.
Next, we show (ii) by the standard arguments. Namely, we reformulate(DP)δε as in the following form: (82) u 1 n + 1 u 2 n + 1 ⋮ u N x - 1 n + 1 = 1 - 2 r r r 1 - 2 r r r 1 - 2 r r ⋱ ⋱ r 1 - 2 r r r 1 - 2 r u 1 n u 2 n ⋮ u N x - 1 n + r u 0 n 0 ⋮ 0 u N x n + - Δ t ε 2 f δ u 1 n  - Δ t ε 2 f δ u 2 n ⋮ - Δ t ε 2 f δ u N x - 1 n .Here by taking into account Neumann boundary condition and initial condition, namely, by u0n=u1n and uNxn=uNx-1n for all n≥0, we observe that (82) is reformulated as in the following form: (83) u → n + 1 = A u → n + r u 1 n - Δ t ε 2 f δ u 1 n  - Δ t ε 2 f δ u 2 n ⋮ r u N x - 1 n - Δ t ε 2 f δ u N x - 1 n ,where we put (84)u→n≔u1nu2n⋮uNx-1n,A=1-2rrr1-2rrr1-2rr⋱⋱r1-2rrr1-2r.Note from (23) that (85) f δ u k n = 1 - δ δ u k n + 1 δ i f   u k n ≤ - 1 ,  - u k n i f   u k n ∈ - 1,1 , 1 - δ δ u k n - 1 δ i f   u k n ≥ 1 .Defining (86)bkn≔-1ifukn∈-1,1,1-δδifukn∉-1,1,b~kn≔1δifukn≤-1,0ifukn∈-1,1,-1δifukn≥1,we observe from (85) and (86) that (87) r u 1 n - Δ t ε 2 f δ u 1 n  - Δ t ε 2 f δ u 2 n ⋮ r u N x - 1 n - Δ t ε 2 f δ u N x - 1 n = r - Δ t ε 2 b 1 n - Δ t ε 2 b 2 n ⋱ r - Δ t ε 2 b N x - 1 n u 1 n u 2 n ⋮ u N x - 1 n + - Δ t ε 2 b ~ 1 n  - Δ t ε 2 b ~ 2 n ⋮ - Δ t ε 2 b ~ N x - 1 n .Using the matrix as above, (82) can be rewritten as in the following form: (88)u→n+1=Au→n+Bu→n+b→~n,n≥0,where we define (89) B ≔ r - Δ t ε 2 b 1 n - Δ t ε 2 b 2 n ⋱ r - Δ t ε 2 b N x - 1 n , b → ~ n ≔ - Δ t ε 2 b ~ 1 n  - Δ t ε 2 b ~ 2 n ⋮ - Δ t ε 2 b ~ N x - 1 n .By the general theory, we observe that the eigenvalue λj of matrix A is given by (90)λj≔1-4rsin2jπ2Nxforj=1,2,…,Nx-1,which implies that λ1 (resp., λNx-1) is the maximum (resp., minimum) eigenvalue of A.
Now let{λ~j;j=1,2,…,Nx-1} be the set of all eigenvalues of matrix A~≔A+B such that (91)λ~1≥λ~2≥⋯≥λ~Nx-1.Also, let {λjB;j=1,2,…,Nx-1} be the set of all eigenvalues of B such that (92)λ1B≥λ2B≥⋯≥λNx-1B.Then, by the abstract perturbation theory of matrix, we observe that (93)λNx-1+λjB≤λ~j≤λ1+λjB,∀j=1,2,…,Nx-1.Since B is the symmetric matrix, it follows from (63), (86), and (90) that (94)λNx-1+λjB≥1-4rsin2Nx-1π2Nx-Δtε2·1-δδ≥1-4r-Δt1-δδε2≥1-4·1-c02-c0=-1+c0>-1∀j=1,2,…,Nx-1.Thus, we conclude from (93) and the above inequality that (95)λ~j>-1∀j=1,2,…,Nx-1.Note that (95) holds in any case: ukn∈[-1,1] and ukn∉[-1,1].
Next, we now assumemax1≤k≤Nx-1|ukn|≤1. Then, we observe from (63) and (86) that (96)1-2r-Δtε2bkn=1-2ΔtΔx2+Δtε2≥1-2·1-c02+Δtε2>0∀k=1,2,…,Nx-1.Therefore, all components of A+B are positive. Hence, the sum of all components in the kth row of A+B is the following:(97)1-r-Δtε2bkn+r=1-r-Δtε2bkn+r=1+Δtε2>1fork=1orNx-1,r+1-2r-Δtε2bkn+r=r+1-2r-Δtε2bkn+r=1+Δtε2>1∀k=2,3,…,Nx-2.Therefore, we observe from (97) that max1≤k≤Nx-1|ukn| is increasing with respect to n in the case when max1≤k≤Nx-1|ukn|≤1.
However, ifukn∉[-1,1] for some k=1,2,…,Nx-1, it follows from (63) and (86) that (98)1-2r-Δtε2bkn=1-2ΔtΔx2-Δtε2·1-δδ≥1-2·1-c02-c0=0.Therefore, the sum of all components in the kth row of A+B is the following: (99)1-r-Δtε2bkn+r=1-r-Δtε2bkn+r=1-Δtε2·1-δδ<1ifukn∉-1,1forsomek=1orNx-1,r+1-2r-Δtε2bkn+r=r+1-2r-Δtε2bkn+r=1-Δtε2·1-δδ<1ifukn∉-1,1forsomek=2,3,…,Nx-2.
Althoughmax1≤k≤Nx-1|ukn| is increasing with respect to n in the case when max1≤k≤Nx-1|ukn|≤1 (cf. (97)), we conclude from (95) and (99) that (ii) of Theorem 16 holds. Thus, the proof of Theorem 16 has been completed.Remark 17.
By (63) we get the suitable mesh size of space Δx. In fact, for each ε∈(0,1], δ∈(0,1) we take the constant c~0∈(0,1), Δt∈(0,1], Δx∈(0,1] such that (100)Δt≤c~0δε21-δ,ΔtΔx2=1-c~02.Then, we have (101)Δt=1-c~02·Δx2≤c~0δε21-δ.Thus, we have the following condition of Δx: (102)0<Δx<ε2c~0δ1-c~01-δ.Remark 18.
We can takec0=0 in (63) for the explicit finite difference scheme to the following usual heat equation with Neumann boundary condition: (103)utε-uxxε=0inQ≔0,T×0,1,uxεt,0=uxεt,1=0,t∈0,T,uε0,x=u0εx,x∈0,1.Taking into account Theorem16, we provide the numerical experiments of (DP)δε as follows. To this end, we use the following numerical data.Numerical Data of ( D P ) δ ε The numerical data isT=0.01, δ=0.01, Δx=0.005, c0=0.6. Also, we consider the initial data u0ε(x) defined by (104)u0εx=-0.5,ifx∈0.00,0.25,-0.5sin⁡2πx,ifx∈0.25,0.75,0.5,ifx∈0.75,1.00.


### 5.1. The Case Whenε=0.05 and Δt=0.000005

Now, we consider the case whenε=0.05 and Δt=0.000005. In this case, we observe that (105)ΔtΔx2=0.0000050.0052=0.2=1-c02,c0δε21-δ=0.6×0.01×0.0521-0.01=0.00001515151515….Therefore, we have (106)c0δε21-δ=0.00001515151515⋯>Δt,which implies that criteria condition (63) holds. Thus, we can provide the stable numerical experiment of (DP)δε. Indeed, we provide Figure 9, which implies the stable numerical experiment of (DP)δε.Figure 9
ε = 0.05, Δt=0.000005, Δx=0.005, and δ=0.01.


### 5.2. The Case Whenε=0.007 and Δt=0.000005

Now, we consider the case whenε=0.007 and Δt=0.000005. In this case, we observe that (107)ΔtΔx2=0.0000050.0052=0.2=1-c02,c0δε21-δ=0.6×0.01×0.00721-0.01=0.00000029696969….Therefore, we have (108)c0δε21-δ=0.00000029696969⋯<Δt,which implies that criteria condition (63) does not hold. Therefore, we provide the unstable numerical experiment of (DP)δε. Indeed, we have Figure 10.Figure 10
ε = 0.007, Δt=0.000005, Δx=0.005, and δ=0.01.


### 5.3. The Case Whenε=0.007 and Δt=0.0000002

Now, we consider the case whenε=0.007 and Δt=0.0000002. In this case, we have (109)ΔtΔx2=0.00000020.0052=1125≤0.2=1-c02,c0δε21-δ=0.00000029696969⋯>Δt,which implies that criteria condition (63) holds. Therefore, we provide the stable numerical experiment of (DP)δε. Indeed, we have Figure 11, which implies the stable numerical experiment of (DP)δε.Figure 11
ε = 0.007, Δt=0.0000002, Δx=0.005, and δ=0.01.Remark 19.
We observe from Theorem16 that, in order to get the stable numerical results of (DP)δε, we have to choose the suitable constants ε and δ and the mesh size of time Δt and space Δx. Therefore, if we perform a numerical experiment of (P)ε for sufficient small ε, we had better consider the original problem (P)ε by using a primal-dual active set method in [2], a Lagrange multiplier method in [9], and so on.


### 5.4. Conclusion of PDE Problem(DP)δε

By Theorem16 and the numerical experiments as above, we conclude that the mesh size of time Δt and space Δx must be satisfied: (110)0<Δt≤c0δε21-δ,0≤ΔtΔx2≤1-c02forsomeconstantc0∈0,1,in order to get the stable numerical experiments of (DP)δε. Also, by Theorems 7 and 16, we conclude that the value δε2/(1-δ) is very important to perform a numerical experiment of (DE)δε and (DP)δε.


## 5.1. The Case Whenε=0.05 and Δt=0.000005

Now, we consider the case whenε=0.05 and Δt=0.000005. In this case, we observe that (105)ΔtΔx2=0.0000050.0052=0.2=1-c02,c0δε21-δ=0.6×0.01×0.0521-0.01=0.00001515151515….Therefore, we have (106)c0δε21-δ=0.00001515151515⋯>Δt,which implies that criteria condition (63) holds. Thus, we can provide the stable numerical experiment of (DP)δε. Indeed, we provide Figure 9, which implies the stable numerical experiment of (DP)δε.Figure 9
ε = 0.05, Δt=0.000005, Δx=0.005, and δ=0.01.


## 5.2. The Case Whenε=0.007 and Δt=0.000005

Now, we consider the case whenε=0.007 and Δt=0.000005. In this case, we observe that (107)ΔtΔx2=0.0000050.0052=0.2=1-c02,c0δε21-δ=0.6×0.01×0.00721-0.01=0.00000029696969….Therefore, we have (108)c0δε21-δ=0.00000029696969⋯<Δt,which implies that criteria condition (63) does not hold. Therefore, we provide the unstable numerical experiment of (DP)δε. Indeed, we have Figure 10.Figure 10
ε = 0.007, Δt=0.000005, Δx=0.005, and δ=0.01.


## 5.3. The Case Whenε=0.007 and Δt=0.0000002

Now, we consider the case whenε=0.007 and Δt=0.0000002. In this case, we have (109)ΔtΔx2=0.00000020.0052=1125≤0.2=1-c02,c0δε21-δ=0.00000029696969⋯>Δt,which implies that criteria condition (63) holds. Therefore, we provide the stable numerical experiment of (DP)δε. Indeed, we have Figure 11, which implies the stable numerical experiment of (DP)δε.Figure 11
ε = 0.007, Δt=0.0000002, Δx=0.005, and δ=0.01.Remark 19.
We observe from Theorem16 that, in order to get the stable numerical results of (DP)δε, we have to choose the suitable constants ε and δ and the mesh size of time Δt and space Δx. Therefore, if we perform a numerical experiment of (P)ε for sufficient small ε, we had better consider the original problem (P)ε by using a primal-dual active set method in [2], a Lagrange multiplier method in [9], and so on.


## 5.4. Conclusion of PDE Problem(DP)δε

By Theorem16 and the numerical experiments as above, we conclude that the mesh size of time Δt and space Δx must be satisfied: (110)0<Δt≤c0δε21-δ,0≤ΔtΔx2≤1-c02forsomeconstantc0∈0,1,in order to get the stable numerical experiments of (DP)δε. Also, by Theorems 7 and 16, we conclude that the value δε2/(1-δ) is very important to perform a numerical experiment of (DE)δε and (DP)δε.



---

*Source: 1492812-2016-06-07.xml*