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Hindawi Publishing Corporation Advances in Numerical Analysis Volume 2016, Article ID 1492812, 16 pages http://dx.doi.org/10.1155/2016/1492812 Research Article Remarks on Numerical Experiments of the Allen-Cahn Equations with Constraint via Yosida Approximation Tomoyuki Suzuki,1 Keisuke Takasao,2 and Noriaki Yamazaki1 1 Department of Mathematics, Faculty of Engineering, Kanagawa University, 3-27-1 Rokkakubashi, Kanagawa-ku, Yokohama 221-8686, Japan 2 Graduate School of Mathematical Sciences, University of Tokyo, 3-8-1 Komaba, Meguro-ku, Tokyo 153-8914, Japan Correspondence should be addressed to Noriaki Yamazaki; [email protected] Received 16 February 2016; Accepted 5 April 2016 Academic Editor: Yinnian He Copyright © 2016 Tomoyuki Suzuki et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 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. More precisely, ππΌ[β1,1] (β ) is a set-valued mapping defined by 1. Introduction In this paper, for each π β (0, 1] we consider the following Allen-Cahn equation with constraint from the viewpoint of numerical analysis: π + π’π‘π β π’π₯π₯ ππΌ[β1,1] (π’π ) π’π β 2 π2 π in π fl (0, π) × (0, 1) , (1) π’π₯π (π‘, 0) = π’π₯π (π‘, 1) = 0, π‘ β (0, π) , π’π (0, π₯) = π’0π (π₯) , π₯ β (0, 1) , (2) (3) where 0 < π < +β and π’0π is given initial data. Also, the constraint ππΌ[β1,1] (β ) is the subdifferential of the indicator function πΌ[β1,1] (β ) on the closed interval [β1, 1] defined by 0, { { { { { { {[0, β) , ππΌ[β1,1] (π§) fl { { { {0} , { { { { {(ββ, 0] , if π§ = 1, if β 1 < π§ < 1, if π§ = β1. V (π‘, π₯) the volume of pure liquid in π΅π (π₯) at time π‘ , σ΅¨ σ΅¨σ΅¨ πβ0 σ΅¨σ΅¨π΅π (π₯)σ΅¨σ΅¨σ΅¨ (4) (5) The Allen-Cahn equation was proposed to describe the macroscopic motion of phase boundaries. In the physical context, the function π’π = π’π (π‘, π₯) in (π)π fl {(1), (2), (3)} is the nonconserved order parameter that characterizes the physical structure. For instance, let V = V(π‘, π₯) be the local ratio of the volume of pure liquid relative to that of pure solid at time π‘ and position π₯ β (0, 1), defined by fl lim if π§ β [β1, 1] , {0, πΌ[β1,1] (π§) fl { +β, otherwise. { if π§ < β1 or π§ > 1, (6) where π΅π (π₯) is the ball in R with center π₯ and radius π and |π΅π (π₯)| denotes its volume. Put π’π (π‘, π₯) fl 2V(π‘, π₯) β 1 for any 2 Advances in Numerical Analysis (π‘, π₯) β π. Then, we observe that π’π (π‘, π₯) is the nonconserved order parameter that characterizes the physical structure: π’π (π‘, π₯) = 1 π π’ (π‘, π₯) = β1 on the pure liquid region, on the pure solid region, (7) β1 < π’π (π‘, π₯) < 1 on the mixture region. There is a vast amount of literature on the Allen-Cahn equation with or without constraint ππΌ[β1,1] (β ). For these studies, we refer to [1β16]. In particular, Chen and Elliott [5] considered the singular limit of (π)π as π β 0 in the general bounded domain Ξ© (β Rπ with π β₯ 1). Also, there is a vast amount of literature on the numerical analysis of the Allen-Cahn equation without constraint ππΌ[β1,1] (β ). For these studies, we refer to [17β21]. Note that the constraint ππΌ[β1,1] (β ) is the multivalued function. Therefore, it is very difficult to apply the numerical methods developed by [17β21] to (π)π . Hence, it is difficult to perform a numerical experiment of (π)π . 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 π’π and an element of ππΌ[β1,1] (π’π ), called the Lagrange multiplier, to (π)π as π β 0. Moreover, they [9] gave the numerical experiment to (π)π via the Lagrange multiplier in one dimension of space for sufficient small π β (0, 1]. Furthermore, they [9] considered the approximating method, called the Yosida approximation. More precisely, for πΏ > 0, the Yosida approximation (ππΌ[β1,1] )πΏ (β ) of ππΌ[β1,1] (β ) is defined by (ππΌ[β1,1] )πΏ (π§) = (πΈ)ππΏ (ππΌ[β1,1] )πΏ (π’πΏπ ) π’πΏπ { {(π’π ) + = 2 πΏ π‘ π2 π { { π π {π’πΏ (0) = π’0 in R, for π‘ β (0, π) , (10) in R. Then, we give the criteria to get the stable numerical experiments of (πΈ)ππΏ . Also, we give some numerical experiments of (πΈ)ππΏ . Moreover, we show the criteria to get the stable numerical experiments of PDE problem (π)ππΏ . Therefore, the main novelties are the following: (a) We give the criteria to get the stable numerical experiments of the ODE problem (πΈ)ππΏ . Also, we provide the numerical experiments to (πΈ)ππΏ for sufficient small π β (0, 1] and πΏ β (0, 1]. (b) We give the criteria to get the stable numerical experiments of the PDE problem (π)ππΏ . Also, we provide (8) where [π§]+ is the positive part of π§. Then, for each πΏ > 0, they [9] considered the following approximation problem of (π)π : (ππΌ[β1,1] )πΏ (π’πΏπ ) π’πΏπ { π π { β + = 2 (π’ ) ) (π’ { πΏ π₯π₯ { πΏ π‘ π2 π π { π π (π)πΏ {(π’ ) (π‘, 0) = (π’ ) (π‘, 1) = 0, { πΏ πΏ π₯ π₯ { { { π π {π’πΏ (0, π₯) = π’0 (π₯) , In [9, Remark 4.2], Figure 1 shows the unstable numerical result to (π)ππΏ was given by the standard explicit finite difference scheme to (π)ππΏ . From Figure 1, we observe that we have to choose the suitable constants π and πΏ and the mesh size of time Ξπ‘ and space Ξπ₯ in order to get the stable numerical results of (π)ππΏ . Therefore, in this paper, for each π > 0 and πΏ > 0, we give the criteria for the standard explicit finite difference scheme to provide the stable numerical experiments of (π)ππΏ . To this end, we first consider the following ODE problem, denoted by (πΈ)ππΏ : [π§ β 1]+ β [β1 β π§]+ , βπ§ β R, πΏ in π fl (0, π) × (0, 1) , π‘ β (0, π) , (9) π₯ β (0, 1) . the numerical experiments to (π)ππΏ for sufficient small π β (0, 1] and πΏ β (0, 1]. The plan of this paper is as follows. In Section 2, we mention the solvability and convergence result of (πΈ)ππΏ . In Section 3, we consider (πΈ)ππΏ numerically. Then, we prove the main result (Theorem 7) corresponding to item (a) listed above. Also, we provide the numerical experiments to (πΈ)ππΏ for sufficient small π β (0, 1] and πΏ β (0, 1]. In Section 4, we mention the solvability and convergence result of (π)ππΏ . In the final Section 5, we consider (π)ππΏ 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 (π)ππΏ for sufficient small π β (0, 1] and πΏ β (0, 1]. Notations and Basic Assumptions. Throughout this paper, we put π» fl πΏ2 (0, 1) with the usual real Hilbert space structure. The inner product and norm in π» are denoted by (β , β )π» and by | β |π», respectively. We also put π fl π»1 (0, 1) with the usual norm |π§|π fl {|π§|2π» + |π§π₯ |2π»}1/2 for π§ β π. In Sections 2 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 π be the real Hilbert space with the inner product (β , β )π. For a proper, l.s.c., and convex function π : π β R βͺ {+β}, the effective domain π·(π) is defined by π· (π) fl {π§ β π; π (π§) < β} . (11) u(t, x) Advances in Numerical Analysis 3 Definition 1. Let π β (0, 1], πΏ β (0, 1], and π’0π β R. Then, a function π’πΏπ : [0, π] β R is called a solution to (πΈ)ππΏ on [0, π], if the following conditions are satisfied: 1.5 1 0.5 0 β0.5 β1 β1.5 (i) π’πΏπ β π1,2 (0, π). (ii) The following equation holds: e m Ti 0 0.002 0.004 0.006 0.008 0.01 (π’πΏπ )π‘ + 0 0.2 0.8 0.6 0.4 1 Ξ© Figure 1: Behavior of a solution to (π)ππΏ with π = 0.007 and πΏ = 0.01. The subdifferential of π is a possibly multivalued operator in π and is defined by π§β β ππ(π§) if and only if π§ β π· (π) , (π§β , π¦ β π§)π β€ π (π¦) β π (π§) βπ¦ β π. (12) For various properties and related notions of the proper, l.s.c., and convex function π and its subdifferential ππ, we refer to the monograph by BreΜzis [22]. 2. Solvability and Convergence Results of (πΈ)ππΏ We begin by giving the rigorous definition of solutions to our problem (πΈ)ππΏ (π β (0, 1] and πΏ β (0, 1]). (ππΌ[β1,1] )πΏ (π’πΏπ ) π2 (πΈ)ππΏ Next, we show the convergence result of as πΏ β 0. To this end, we recall a notion of convergence for convex functions, developed by Mosco [26]. Definition 3 (cf. [26]). Let π and ππ (π β N) be proper, l.s.c., and convex functions on a Hilbert space π. Then, one says that ππ converges to π on π in the sense of Mosco [26] as π β β, if the following two conditions are satisfied: π’πΏπ π2 in R, for π‘ β (0, π) . (13) (iii) π’πΏπ (0) = π’0π in R. Now, we show the solvability result of (πΈ)ππΏ on [0, π]. Proposition 2. Let π β (0, 1], πΏ β (0, 1], and π’0π β R with |π’0π | β€ 1. Then, there exists a unique solution π’πΏπ to (πΈ)ππΏ on [0, π] in the sense of Definition 1. Proof. We can prove the uniqueness of solutions to (πΈ)ππΏ on [0, π] by the quite standard arguments: monotonicity and Gronwallβs inequality. Also, we can show the existence of solutions to (πΈ)ππΏ on [0, π] applying the abstract theory of evolution equations governed by subdifferentials. Indeed, we define a function (πΌ[β1,1] )πΏ (β ) on R by σ΅¨2 σ΅¨ σ΅¨2 σ΅¨σ΅¨ σ΅¨[π§ β 1]+ σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨[β1 β π§]+ σ΅¨σ΅¨σ΅¨ (πΌ[β1,1] )πΏ (π§) = σ΅¨ , βπ§ β R. 2πΏ (14) Clearly, (πΌ[β1,1] )πΏ (β ) is proper, l.s.c., and convex on R with π(πΌ[β1,1] )πΏ (β ) = (ππΌ[β1,1] )πΏ (β ) in R, where (ππΌ[β1,1] )πΏ (β ) is the Yosida approximation of ππΌ[β1,1] (β ) defined by (8). Note that the problem (πΈ)ππΏ can be rewritten as in an abstract framework of the form π 1 1 { π’π (π‘) + 2 π (πΌ[β1,1] )πΏ (π’πΏπ (π‘)) β 2 π’πΏπ (π‘) = 0 π π (πΈπ)ππΏ { ππ‘ πΏ π π {π’πΏ (0) = π’0 Therefore, applying the Lipschitz perturbation theory of abstract evolution equations (cf. [23β25]), we can show the existence of a solution π’πΏπ to (πΈπ)ππΏ , hence, (πΈ)ππΏ , on [0, π] for each π β (0, 1] and πΏ β (0, 1] in the sense of Definition 1. Thus, the proof of Proposition 2 has been completed. = in R, for π‘ β (0, π) , in R. (15) (i) for any subsequence {πππ } β {ππ }, if π§π β π§ weakly in π as π β β, then lim inf πππ (π§π ) β₯ π (π§) ; πββ (16) (ii) for any π§ β π·(π), there is a sequence {π§π } in π such that π§π σ³¨β π§ in π as π σ³¨β β, lim ππ (π§π ) = π (π§) . (17) πββ It is well known that the following lemma holds. Therefore, we omit the detailed proof. 4 Advances in Numerical Analysis Lemma 4 (cf. [27, Section 5], [22, Chapter 2], and [28, Section 2]). Consider ππ R ππ π‘βπ π πππ π ππ πππ ππ [26] ππ πΏ σ³¨β 0. (18) By Lemma 4 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]). Let π β (0, 1], πΏ β (0, 1], and π’0π β R with |π’0π | β€ 1. Also, let π’πΏπ be the unique solution to (πΈ)ππΏ on [0, π]. Then, there exists a unique function π’π β π1,2 (0, π) such that ππ πΆ ([0, π]) ππ πΏ σ³¨β 0 (π·πΈ)ππΏ (π’π ) π’π ππΌ { {π’π + [β1,1] β 2 (πΈ) { π‘ π2 π { π π π’ = π’ (0) 0 { π (πΌ[β1,1] )πΏ (β ) σ³¨β πΌ[β1,1] (β ) π’πΏπ σ³¨β π’π and π’π is the unique solution of the following problem (πΈ)π on [0, π]: (19) Theorem 7. Let π β (0, 1], πΏ β (0, 1), and Ξπ‘ β (0, 1]. Assume π’0π β (0, 1] (resp., π’0π β [β1, 0)) and π = β. Let {π’π ; π β₯ 0} be the solution to (π·πΈ)ππΏ . Then, one has the following: (i) If Ξπ‘ β (0, πΏπ2 /(1 β πΏ)), π’π converges to 1/(1 β πΏ) (resp., β1/(1 β πΏ)) monotonically as π β β. (ii) If Ξπ‘ β (πΏπ2 /(1 β πΏ), 2πΏπ2 /(1 β πΏ)), π’π oscillates and converges to 1/(1 β πΏ) (resp., β1/(1 β πΏ)) as π β β. Proof. We give the proof of Theorem 7 in the case of the initial value π’0π β (0, 1]. (20) ππ R. 3. Stable Criteria and Numerical Experiments for (πΈ)ππΏ In this section we consider (πΈ)ππΏ from the viewpoint of numerical analysis. Remark 6. Note from Proposition 5 that (πΈ)ππΏ is the approximating problem of (πΈ)π . Also note from (5) that the constraint ππΌ[β1,1] (β ) is the multivalued function. Therefore, it is very difficult to study (πΈ)π numerically. In order to perform the numerical experiments of (πΈ)ππΏ via the standard forward Euler method, we consider the following explicit finite difference scheme to (πΈ)ππΏ , denoted by (π·πΈ)ππΏ : π+1 π (ππΌ[β1,1] )πΏ (π’π ) π’π { {π’ β π’ + = 2 Ξπ‘ π2 π { { 0 π = π’ π’ 0 { where Ξπ‘ is the mesh size of time and ππ‘ is the integer part of number π/Ξπ‘. We observe that π’π is the approximating solution of (πΈ)ππΏ at the time π‘ = πΞπ‘. Also, we observe that the explicit finite difference scheme (π·πΈ)ππΏ converges to (πΈ)ππΏ as Ξπ‘ β 0 since (π·πΈ)ππΏ is the standard time discretization scheme for (πΈ)ππΏ . In Figure 2, we give the unstable numerical experiment of (π·πΈ)ππΏ in the case when π = 0.002, π = 0.003, πΏ = 0.01, the initial data π’0π = 0.1, and the mesh size of time Ξπ‘ = 0.000001. From Figure 2, we observe that we have to choose the suitable constants π and πΏ and the mesh size of time Ξπ‘ in order to get the stable numerical results of (π·πΈ)ππΏ . Now, let us mention the first main result in this paper, which is concerned with the criteria to give the stable numerical experiments of (π·πΈ)ππΏ . ππ R, πππ π‘ β (0, π) , in R, for π = 0, 1, 2, . . . , ππ‘ , (21) in R, For simplicity, we set ππΏ (π§) fl (ππΌ[β1,1] )πΏ (π§) β π§ for π§ β R. (22) Then, we observe that 1+π§ β π§, if π§ β€ β1, { { { { { πΏ if π§ β [β1, 1] , ππΏ (π§) = {βπ§, { {π§ β 1 { { β π§, if π§ β₯ 1, { πΏ (23) and π§ = 0, 1/(1 β πΏ), β1/(1 β πΏ) are the zero points of ππΏ (β ). Also, we observe that the difference equation (π·πΈ)ππΏ is reformulated in the following form: π’π+1 = π’π β Ξπ‘ π (π’π ) π2 πΏ in R, for π = 0, 1, 2, . . . . (24) Note from (23) and (24) that if π’π β (0, 1] we have π’π+1 = π’π β Ξπ‘ Ξπ‘ ππΏ (π’π ) = π’π β 2 β (βπ’π ) β₯ π’π , 2 π π (25) which implies that π’π is increasing with respect to π until π’π+1 β₯ 1. Now, we prove (i). To this end, we assume that Ξπ‘ β (0, πΏπ2 /(1 β πΏ)). At first, by the mathematical induction, we show π’π β (0, 1 ) 1βπΏ βπ β₯ 0. (26) Advances in Numerical Analysis 5 1.2 Taking the limit in (24), we observe from the continuity of ππΏ (β ) that π’β = 1/(1 β πΏ), which is the zero point of ππΏ (β ). Hence, the proof of (i) has been completed. Next, we show (ii). To this end, we assume that Ξπ‘ β (πΏπ2 /(1 β πΏ), 2πΏπ2 /(1 β πΏ)). Then, we can find the minimal number π0 β N so that 1 Solution 0.8 0.6 π’π0 β (1, 0.4 π π’ β (0, 1] 0.2 0 1+πΏ ), 1βπΏ 0 0.0005 0.001 Time 0.0015 0.002 Figure 2: Behavior of a solution π’π to (π·πΈ)ππΏ with π = 0.003 and πΏ = 0.01. Clearly (26) holds for π = 0 because of π’0 = π’0π β (0, 1]. Now, we assume that (26) holds for all π = 0, 1, . . . , π. Suppose π’π β (0, 1]. Then, we infer from (25) that π’π+1 Ξπ‘ Ξπ‘ πΏ 1 = (1 + 2 ) π’π β€ 1 + 2 < 1 + = . π π 1βπΏ 1βπΏ 1 ) , if π’π β (0, 1] . 1βπΏ π’π+1 = π’π β Ξπ‘ Ξπ‘ π (π’π ) = (1 + 2 ) π’π π2 πΏ π Ξπ‘ 2 Ξπ‘ π+1 = (1 + 2 ) π’πβ1 = β β β = (1 + 2 ) π’0 . π π Next, if π’π β [1, 1/(1 β πΏ)), we observe from (23) and (24) that π’π β€ π’π+1 = π’π β Ξπ‘ π (π’π ) π2 πΏ 1+ Ξπ‘ 1 β (1 β πΏ) π’π 1 β (1 β πΏ) π’π =π’ + 2 β < π’π + π πΏ 1βπΏ (29) π π’π0 > 1, π’π β (0, 1] βπ = 0, 1, . . . , π0 β 1. (36) π’π0 = π’π0 β1 β Ξπ‘ Ξπ‘ ππΏ (π’π0 β1 ) = (1 + 2 ) π’π0 β1 2 π π 2πΏ 1+πΏ )β 1= , 1βπΏ 1βπΏ (37) and thus, (33) holds. To show (ii), we put which implies that 1 1 ) , if π’π β [1, ). 1βπΏ 1βπΏ (30) From (28) and (30) we infer that (26) holds for π = π + 1. Therefore, we conclude from the mathematical induction that (26) holds. Also, by (23) and (26) we observe that ππΏ (π’π ) β€ 0 for all π β₯ 0. Therefore, we observe from (24) that π’π+1 = π’π β Ξπ‘ π (π’π ) β₯ π’π π2 πΏ βπ β₯ 0. (31) Therefore, we infer from (26) and (31) that {π’π ; π β₯ 0} is a bounded and increasing sequence with respect to π. Thus, there exists a point π’β β R such that π (35) we can find the minimal number π0 β N so that < (1 + 1 , 1βπΏ π’π+1 β [1, Ξπ‘ πΏ >1+ > 1, π2 1βπΏ Also, we observe from (25) that Ξπ‘ π’π β 1 = π’π β 2 β ( β π’π ) π πΏ = (34) Taking into account (34), π’0 β (0, 1], and (27) (28) βπ = 0, 1, . . . , π0 β 1. Indeed, if π’π β (0, 1] for all π = 0, 1, . . . , π, we observe from (25) that Therefore, by (25) and the inequality as above, we observe that π’π+1 β (0, (33) π’ σ³¨β π’ β in R as π σ³¨β β. (32) Ξπ‘ fl πΏπ2 π 1βπΏ for some π β (1, 2) . (38) Then, we observe from (23) and (24) that Ξπ‘ Ξπ‘ 1 β (1 β πΏ) π’π0 π0 π0 π (π’ ) = π’ + β πΏ π2 π2 πΏ (39) π π0 = (1 β π) π’ + . 1βπΏ π’π0 +1 = π’π0 β From (39) it follows that π’π0 +1 + (π β 1) π’π0 1 = . π 1βπΏ (40) 6 Advances in Numerical Analysis Therefore, we observe from (40) and π β (1, 2) that the zero point 1/(1βπΏ) of ππΏ (β ) is in the interval between π’π0 and π’π0 +1 . Also, by (39) we observe that π’π0 +1 = (1 β π) π’π0 + π’ π0 +1 1+πΏ π π > (1 β π) + 1βπΏ 1βπΏ 1βπΏ 1 + πΏ β ππΏ = β₯ 1, 1βπΏ π π = (1 β π) π’π0 + < (1 β π) β 1 + 1βπΏ 1βπΏ = πβ1βπ (41) 1 β πΏ + ππΏ 1 + πΏ β€ , 1βπΏ 1βπΏ 1+πΏ ). 1βπΏ (42) 1+πΏ ) π’ β (1, 1βπΏ βπ β₯ π0 (43) and π’π oscillates around the zero point 1/(1βπΏ) for all π β₯ π0 . Also, we observe from (39) and (43) that σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ π+1 σ΅¨σ΅¨π’ β 1 σ΅¨σ΅¨σ΅¨ = |1 β π| σ΅¨σ΅¨σ΅¨π’π β 1 σ΅¨σ΅¨σ΅¨ βπ β₯ π . (44) σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ 0 1 β πΏσ΅¨ 1 β πΏ σ΅¨σ΅¨ σ΅¨ σ΅¨ Therefore, by π β (1, 2), (43), and (44), there is a subsequence {ππ } of {π} such that π’ππ oscillates and converges to 1/(1 β πΏ) as π β β. Hence, taking into account the uniqueness of the limit point, the proof of (ii) has been completed. Remark 8. Assume Ξπ‘ β [2πΏπ2 /(1 β πΏ), β) and put Ξπ‘ fl πΏπ2 π/(1 β πΏ) for some π β₯ 2. Then, we observe that 1+ (47) Proof. Note that π’0π fl (1 β πΏ)πβ1 /(1 + πΏ)π β (0, 1). Therefore, by (23) and (24) we observe that Ξπ‘ 2πΏ 1+πΏ π π (π’0 ) = π’0π β (βπ’0π ) = π’ π2 πΏ 1βπΏ 1βπΏ 0 (1 β πΏ)πβ2 = . (1 + πΏ)πβ1 (48) Similarly, we have Therefore, by (33) and (40) and by repeating the procedure as above, we observe that π (1 β πΏ) { { { (1 + πΏ)πβπ , ππ π = 0, 1, . . . , π β 1, π π’ ={ { { 1 , ππ π β₯ π. {1 β πΏ π’1 = π’0 β which implies that π’π0 +1 β (1, Corollary 10. Let π β (0, 1], πΏ β (0, 1), Ξπ‘ = 2πΏπ2 /(1 β πΏ), and π β N. Assume π’0π fl (1 β πΏ)πβ1 /(1 + πΏ)π . Then, the solution to (π·πΈ)ππΏ is given by 2πΏ Ξπ‘ >1+ > 1, 2 π 1βπΏ (45) |1 β π| β₯ 1. Therefore, we infer from the proof of Theorem 7 (cf. (34), (40), and (44)) that the solution π’π to (π·πΈ)ππΏ oscillates as π β β, in general. Remark 9. By (24) we easily observe that π’π β‘ 0 βπ β₯ 1, if π’0 = 0, π’π β‘ 1 1βπΏ βπ β₯ 1, if π’0 = 1 , 1βπΏ π’π β‘ β1 1βπΏ βπ β₯ 1, if π’0 = β1 . 1βπΏ (46) From (ii) of Theorem 7, we observe that π’π oscillates and converges to the zero point of ππΏ (β ) in the case when Ξπ‘ β (πΏπ2 /(1 β πΏ), 2πΏπ2 /(1 β πΏ)). However, in the case when Ξπ‘ = 2πΏπ2 /(1 β πΏ), we have the following special case that the solution to (π·πΈ)ππΏ does not oscillate and coincides with the zero point of ππΏ (β ) after some finite number of iterations. π’2 = π’1 β Ξπ‘ 1 + πΏ 1 (1 β πΏ)πβ3 1 π (π’ ) = . π’ = πΏ π2 1βπΏ (1 + πΏ)πβ2 (49) Repeating this procedure, we observe that the solution to (π·πΈ)ππΏ is given by (47). Taking into account Theorem 7, we provide the numerical experiments of (π·πΈ)ππΏ as follows. To this end, we use the following numerical data. Numerical Data of (π·πΈ)ππΏ The numerical data are π = 0.002, π = 0.01, πΏ = 0.01, and the initial data π’0π = 0.1. Then, we observe that 1 1 = = 1.010101010 . . . , 1 β πΏ 1 β 0.01 πΏπ2 = 0.0000010101010 . . . . 1βπΏ (50) 3.1. The Case When Ξπ‘ = 0.000001. Now we consider the case when Ξπ‘ = 0.000001. In this case, we have πΏπ2 = 0.0000010101010 β β β > Ξπ‘ = 0.000001, 1βπΏ (51) which implies that (i) of Theorem 7 holds. Thus, we have the stable numerical result of (π·πΈ)ππΏ . Indeed, we observe from Figure 3 and Table 1 that the solution to (π·πΈ)ππΏ converges to the stationary solution 1/(1 β πΏ) = 1/(1 β 0.01) = 1.010101010 . . .. 3.2. The Case When Ξπ‘ = 0.000002. Now we consider the case when Ξπ‘ = 0.000002. In this case, we have πΏπ2 = 0.0000010101010 β β β < Ξπ‘ = 0.000002 1βπΏ 2πΏπ2 < , 1βπΏ (52) Advances in Numerical Analysis 7 Table 1: Numerical data: Ξπ‘ = 000001. Number of iterations π 0 1 2 3 4 5 .. . Table 2: Numerical data: Ξπ‘ = 0.000002. The value of π’ 0.100000 0.101000 0.102010 0.103030 0.104060 0.105101 .. . 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 π 0.910636 0.919743 0.928940 0.938230 0.947612 0.957088 0.966659 0.976325 0.986089 0.995950 1.005909 1.010059 1.010101 1.010101 1.010101 1.010101 1.010101 1.1 1 0.9 Solution 0.8 0.7 0.6 0.5 0.4 0.3 Number of iterations π The value of π’π 0 1 0.100000 0.102000 2 .. . 0.104040 .. . 114 115 0.955916 0.975034 116 117 118 0.994535 1.014425 1.005863 119 120 1.014254 1.006031 121 122 123 1.014090 1.006192 1.013932 124 125 1.006347 1.013780 126 127 128 1.006496 1.013634 1.006638 129 .. . 1.013494 .. . 560 561 1.010100 1.010102 562 563 1.010100 1.010102 564 565 566 1.010100 1.010102 1.010101 567 568 1.010101 1.010101 569 570 1.010101 1.010101 0.2 0.1 0 0.0005 0.001 Time 0.0015 0.002 3.4. The Case When Ξπ‘ = 0.000005. Now we consider the case when Ξπ‘ = 0.000005. In this case, we have Figure 3: πΏπ2 /(1 β πΏ) = 0.0000010101010 β β β > Ξπ‘ = 0.000001. 2 πΏπ2 = 0.0000020202020 β β β < Ξπ‘ = 0.000005. 1βπΏ (53) which implies that (ii) of Theorem 7 holds. Thus, we observe from Figure 4 and Table 2 that the solution to (DE)ππΏ oscillates and converges to the stationary solution 1/(1 β πΏ) = 1/(1 β 0.01) = 1.010101010 . . .. Therefore, we observe Remark 8. Indeed, we observe from Figure 6 that the solution to (π·πΈ)ππΏ oscillates. 3.3. The Case When Ξπ‘ = 2πΏπ2 /(1 β πΏ). Now we consider the case when Ξπ‘ = 2πΏπ2 /(1 β πΏ) = 0.0000020202020 . . .. In this case, we observe Remark 8. Indeed, we observe from Figure 5 and Table 3 that the solution to (π·πΈ)ππΏ oscillates. 3.5. The Case When Ξπ‘ = 15πΏπ2 /(1 β πΏ). Now we consider the case when Ξπ‘ = 15πΏπ2 /(1 β πΏ). In this case, we observe Remark 8. Indeed, we observe from Figure 7 that the solution to (π·πΈ)ππΏ oscillates between three zero points of ππΏ (β ). 8 Advances in Numerical Analysis 1.1 1 1 0.9 0.9 0.8 0.8 0.7 0.7 Solution Solution 1.1 0.6 0.5 0.5 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0 0.0005 0.001 Time 0.0015 0 0.0005 0.001 Time 0.0015 0.002 Figure 6: 2πΏπ2 /(1 β πΏ) = 0.0000020202020 β β β < Ξπ‘ = 0.000005. Table 3: Numerical data: Ξπ‘ = 2πΏπ2 /(1 β πΏ) = 0.0000020202020 . . .. 1.1 1 Number of iterations π 0 1 2 3 4 5 .. . 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.1 0.002 Figure 4: πΏπ2 /(1 β πΏ) = 0.0000010101010 β β β < Ξπ‘ = 0.000002 < 2πΏπ2 /(1 β πΏ). Solution 0.6 0 0.0005 0.001 Time 0.0015 0.002 Figure 5: Ξπ‘ = 2πΏπ2 /(1 β πΏ) = 0.0000020202020 . . .. 3.6. Numerical Result of Corollary 10. In this subsection, we consider Corollary 10 numerically. To this end, we use the following initial data: π’0 fl (1 β πΏ)5 (1 β 0.01)5 = = 0.8958756 . . . . (1 + πΏ)6 (1 + 0.01)6 (54) Then, we observe from Table 4 and Figure 8 that Corollary 10 holds. Namely, we observe that (47) holds with π = 6. 3.7. Conclusion of ODE Problem (π·πΈ)ππΏ . From Theorem 7 and numerical experiments as above, we conclude that (i) the mesh size of time Ξπ‘ must be smaller than πΏπ2 /(1β πΏ) in order to get the stable numerical experiments of (π·πΈ)ππΏ , (ii) we have the stable numerical experiments of (π·πΈ)ππΏ with the initial data π’0π fl (1 β πΏ)πβ1 /(1 + πΏ)π , even if the mesh size of time Ξπ‘ is equal to 2πΏπ2 /(1 β πΏ). 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 The value of π’π 0.100000 0.102020 0.104081 0.106184 0.108329 0.110517 .. . 0.902568 0.920801 0.939403 0.958381 0.977742 0.997495 1.017646 1.002556 1.017646 1.002556 1.017646 1.002556 1.017646 1.002556 1.017646 1.002556 1.017646 1.002556 1.017646 1.002556 1.017646 4. Solvability and Convergence Results for (π)ππΏ We begin by giving the rigorous definition of solutions to our PDE problem (π)ππΏ (π β (0, 1] and πΏ β (0, 1]). Advances in Numerical Analysis 9 1 Table 4: Numerical data: Ξπ‘ = 2πΏπ2 /(1 β πΏ) = 0.0000020202020 . . .. 0.8 Solution The value of π’π 0.895876 0.913974 0.932438 0.951275 0.970493 0.990099 1.010101 1.010101 1.010101 1.010101 1.010101 Number of iterations π 0 1 2 3 4 5 6 7 8 9 10 0.6 0.4 0.2 0 0 0.0005 0.001 Time 0.0015 0.002 Figure 8: Ξπ‘ = 2πΏπ2 /(1 β πΏ) = 0.0000020202020 . . .. 1.5 1 Proof. By the same argument as in Proposition 2, we can show the existence-uniqueness of a solution π’πΏπ to (π)ππΏ on [0, π] for each π β (0, 1] and πΏ β (0, 1]. Indeed, we can prove the uniqueness of solutions to (π)ππΏ on [0, π] by the quite standard arguments: monotonicity and Gronwallβs inequality. Also, we can show the existence of solutions to (π)ππΏ on [0, π] applying the abstract theory of evolution equations governed by subdifferentials. Indeed, we define a functional ππΏπ on π» by Solution 0.5 0 β0.5 β1 β1.5 0 0.0005 0.001 Time 0.0015 0.002 ππΏπ (π§) Figure 7: Ξπ‘ = 15πΏπ2 /(1 β πΏ). Definition 11. Let π β (0, 1], πΏ β (0, 1], and π’0π β π». Then, a function π’πΏπ : [0, π] β π» is called a solution to (π)ππΏ on [0, π], if the following conditions are satisfied: (i) π’πΏπ β π1,2 (0, π; π») β© πΏβ (0, π; π). (ii) The following variational identity holds: ((π’πΏπ )π‘ (π‘) , π§)π» + ((π’πΏπ )π₯ (π‘) , π§π₯ )π» +( (ππΌ[β1,1] )πΏ (π’πΏπ (π‘)) π2 π’π (π‘) , π§) = ( πΏ 2 , π§) π π» π» (55) 1 1 { 1 β« σ΅¨σ΅¨σ΅¨π§ σ΅¨σ΅¨σ΅¨2 ππ₯ + 1 β« (πΌ[β1,1] ) (π§ (π₯)) ππ₯, πΏ fl { 2 0 σ΅¨ π₯ σ΅¨ π2 0 β, { if π§ β π, (56) otherwise, where (πΌ[β1,1] )πΏ (β ) is the function defined in (14). Clearly, ππΏπ is proper, l.s.c., and convex on π» with the effective domain π·(π) = π. Note that the problem (π)ππΏ can be rewritten as in an abstract framework of the form π π 1 π π π { { π’πΏ (π‘) + πππΏ (π’πΏ (π‘)) β 2 π’πΏ (π‘) = 0 π (ππ)ππΏ { ππ‘ { π π {π’ (0) = π’0 in π», for π‘ > 0, (57) in π». βπ§ β π, a.e. π‘ β (0, π) . (iii) π’πΏπ (0) = π’0π in π». Now, we mention the solvability result of (π)ππΏ on [0, π]. Proposition 12. Let π β (0, 1] and πΏ β (0, 1]. Assume the following condition: π’0π β πΎ fl {π§ β π; β1 β€ π§ (π₯) β€ 1 π.π. π₯ β (0, 1)} . Then, for each π’0π πΎ, there exists a unique solution π’πΏπ β on [0, π] in the sense of Definition 11. (A) to (π)ππΏ Therefore, applying the Lipschitz perturbation theory of abstract evolution equations (cf. [23β25]), we can show the existence of a solution π’πΏπ to (ππ)ππΏ , hence, (π)ππΏ , on [0, π] for each π β (0, 1] and πΏ β (0, 1] in the sense of Definition 11. Thus, the proof of Proposition 12 has been completed. Next, we recall the convergence result of (π)ππΏ as πΏ β 0. Taking into account Lemma 4 (cf. (18)), we observe that the following lemma holds. 10 Advances in Numerical Analysis Lemma 13 (cf. [27, Section 5], [22, Chapter 2], [28, Section 2]). Let π β (0, 1], and define a functional ππ on π» by π π (π§) 1 1 { 1 β« σ΅¨σ΅¨σ΅¨σ΅¨π§π₯ σ΅¨σ΅¨σ΅¨σ΅¨2 ππ₯ + 1 β« πΌ[β1,1] (π§ (π₯)) ππ₯, ππ π§ β π, (58) 2 fl { 2 0 π 0 ππ‘βπππ€ππ π. {β, Then, ππΏπ (β ) σ³¨β ππ (β ) ππ π» ππ π‘βπ π πππ π ππ πππ ππ [26] ππ πΏ σ³¨β 0. (59) By Lemma 13 and the general convergence theory of evolution equations, we get the following result. Proposition 14 (cf. [27, Section 5], [28, Section 2]). Let π β (0, 1], πΏ β (0, 1], and π’0π β πΎ. Also, let π’πΏπ be the unique solution to (π)ππΏ on [0, π]. Then, π’πΏπ converges to the unique function π’π to (π)π on [0, π] in the sense that π’πΏπ σ³¨β π’π π π‘ππππππ¦ ππ πΆ ([0, π] ; π») ππ πΏ σ³¨β 0. (60) Proof. Note that the problem (π)π can be rewritten as in an abstract framework of the form π 1 { π’π (π‘) + πππ (π’π (π‘)) β 2 π’π (π‘) β 0 in π», for π‘ > 0, π (ππ)π { ππ‘ π π in π». {π’ (0) = π’0 Therefore, by Lemma 13 and the abstract convergence theory of evolution equations (cf. [27, 28]), we observe that the solution π’πΏπ to (ππ)ππΏ converges to the unique solution π’π to (ππ)π on [0, π] as πΏ β 0 in the sense of (60). Note that π’π (resp., π’πΏπ ) is the unique solution to (π)π (resp., (π)ππΏ ) on [0, π] (cf. Proposition 12). Thus, we conclude that Proposition 14 holds. 5. Stable Criteria and Numerical Experiments for (π)ππΏ In this section we consider (π)ππΏ from the viewpoint of numerical analysis. Remark 15. Note from Proposition 14 that (π)ππΏ is the approximating problem of (π)π . Also note from (5) that the constraint ππΌ[β1,1] (β ) is the multivalued function. Therefore, it is very difficult to study (π)π numerically. In order to perform the numerical experiments of (π)ππΏ , we consider the following explicit finite difference scheme to (π)ππΏ , denoted by (π·π)ππΏ : π π (ππΌ[β1,1] )πΏ (π’ππ ) π’ππ β 2π’ππ + π’π+1 π’ππ+1 β π’ππ π’πβ1 { { + = 2 β { { π2 π (Ξπ₯)2 { Ξπ‘ π { π π π π (π·π)πΏ {π’ = π’ , π’ = π’ 0 1 ππ₯ ππ₯ β1 { { { { { 0 π {π’π = π’0 (π₯π ) where Ξπ‘ is the mesh size of time, Ξπ₯ is the mesh size of space, ππ‘ is the integer part of number π/Ξπ‘, ππ₯ is the integer part of number 1/Ξπ₯, and π₯π fl πΞπ₯. We observe that π’ππ is the approximating solution of (π)ππΏ at the time π‘π fl πΞπ‘ and the position π₯π fl πΞπ₯. Also, we observe that the explicit finite difference scheme (π·π)ππΏ converges to (π)ππΏ as Ξπ‘ β 0 and Ξπ₯ β 0. From Figure 1, we observe that we have to choose the suitable constants π and πΏ and the mesh size of time Ξπ‘ and the mesh size of space Ξπ₯ in order to get the stable numerical results of (π·π)ππΏ . Now, let us mention our second main result concerning the stability of (π·πΈ)ππΏ . Theorem 16. Let π β (0, 1], πΏ β (0, 1), Ξπ‘ β (0, 1], Ξπ₯ β (0, 1], π > 0, and π’0π β πΎ, where πΎ is the set of initial values defined in Proposition 12 (cf. (A)). Let ππ₯ be the integer part of number 1/Ξπ₯, and let {π’ππ ; π β₯ 0, π = 0, 1, . . . , ππ₯ } be the solution to (π·π)ππΏ . Also, let π0 β (0, 1) and assume that 0 < Ξπ‘ β€ π0 πΏπ2 , 1βπΏ (61) for π = 0, 1, 2, . . . , ππ‘ , π = 1, 2, . . . , ππ₯ β 1, (62) for π = 1, 2, . . . , ππ‘ , for π = 0, 1, 2, . . . , ππ₯ , 0β€ 1 β π0 Ξπ‘ β€ . 2 (Ξπ₯)2 (63) Then, one has the following: (i) The solution to (π·π)ππΏ is bounded in the following sense: 1 σ΅¨ σ΅¨ max σ΅¨σ΅¨σ΅¨π’ππ σ΅¨σ΅¨σ΅¨ β€ 0β€πβ€ππ₯ 1βπΏ βπ β₯ 0. (64) (ii) {π’ππ ; π = 0, 1, . . . , ππ₯ } does not oscillate with respect to π β₯ 0. Proof. We first show (i), that is, (64), by the mathematical induction. Clearly (64) holds for π = 0 because of π’0π β πΎ. Now, we assume that 1 σ΅¨ σ΅¨ max σ΅¨σ΅¨σ΅¨σ΅¨π’ππ σ΅¨σ΅¨σ΅¨σ΅¨ β€ 1βπΏ 0β€πβ€ππ₯ βπ = 0, 1, . . . , π. (65) Advances in Numerical Analysis 11 We observe that the explicit finite difference problem (π·π)ππΏ can be reformulated as in the following form: π π π’ππ+1 = ππ’πβ1 + ππ’π+1 + (1 β 2π) π’ππ β Ξπ‘ π (π’π ) π2 πΏ π (66) βπ = 0, 1, 2, . . . , ππ‘ , π = 1, 2, . . . , ππ₯ β 1, where we put π fl Ξπ‘/(Ξπ₯)2 and ππΏ (β ) is the function defined by (23). We observe from (63), (65), and (66) that 1 1 1 π π ) + π( ) β π’ππ+1 = π ( β π’πβ1 β π’π+1 1βπΏ 1βπΏ 1βπΏ 1 Ξπ‘ β π’ππ ) + 2 ππΏ (π’ππ ) 1βπΏ π (67) Ξπ‘ 1 π π β₯ (1 β 2π) ( β π’π ) + 2 ππΏ (π’π ) 1βπΏ π βπ = 1, 2, . . . , ππ₯ β 1. From (23), (63), and (65) we infer that the function [β1/(1βπΏ), 1/(1βπΏ)] β π§ β (1β2π)(1/(1βπΏ)βπ§)+Ξπ‘/π2 ππΏ (π§) is nonnegative and continuous. Indeed, it follows from (23) that the function [β1/(1 β πΏ), 1] β π§ β (1 β 2π)(1/(1 β πΏ) β π§) + Ξπ‘/π2 ππΏ (π§) attains a minimum value at π§ = 1. Therefore, we observe from (23) and (63) that 1 Ξπ‘ β π§) + 2 ππΏ (π§) 1βπΏ π β₯ (1 β 2π) ( 1 Ξπ‘ β 1) + 2 ππΏ (1) 1βπΏ π ππΏ πΏ Ξπ‘ Ξπ‘ = (1 β 2π) β β₯ 0 β 2 β₯0 β 1 β πΏ π2 1βπΏ π βπ§ β [β (68) 1 , 1] . 1βπΏ =[ (69) 1 Ξπ‘ 1βπΏ Ξπ‘ β (1 β 2π)] π§ + (1 β 2π) β 2. 2 πΏπ 1 β πΏ πΏπ 1 (73) β π’ππ+1 β₯ 0 βπ = 1, 2, . . . , ππ₯ β 1. 1βπΏ Similarly, we observe from (63), (65), and (66) that 1 1 1 π π + + = π (π’πβ1 ) + π (π’π+1 ) 1βπΏ 1βπΏ 1βπΏ 1 Ξπ‘ ) β 2 ππΏ (π’ππ ) 1βπΏ π (74) 1 Ξπ‘ π π β₯ (1 β 2π) (π’π + ) β 2 ππΏ (π’π ) 1βπΏ π + (1 β 2π) (π’ππ + βπ = 1, 2, . . . , ππ₯ β 1. By similar arguments as above, we observe that the function [β1/(1βπΏ), 1/(1βπΏ)] β π§ β (1β2π)(π§+1/(1βπΏ))βΞπ‘/π2 ππΏ (π§) is nonnegative and continuous. In fact, it follows from (23) that the function [β1, 1/(1βπΏ)] β π§ β (1β2π)(π§+1/(1βπΏ))β Ξπ‘/π2 ππΏ (π§) attains a minimum value at π§ = β1. Therefore, we observe from (23) and (63) that 1 Ξπ‘ ) β 2 ππΏ (π§) (1 β 2π) (π§ + 1βπΏ π β₯ (1 β 2π) (β1 + 1 Ξπ‘ ) β 2 ππΏ (β1) 1βπΏ π (75) 1 ]. 1βπΏ Also, for any π§ β [β1/(1 β πΏ), β1], we observe from (23) that Ξπ‘ 1 ) β 2 ππΏ (π§) 1βπΏ π = (1 β 2π) (π§ + = [(1 β 2π) β 1 Ξπ‘ 1 + π§ )β 2 β ( β π§) 1βπΏ π πΏ (76) 1 Ξπ‘ 1βπΏ Ξπ‘] π§ + (1 β 2π) . β πΏπ2 1 β πΏ πΏπ2 Here we note from (63) that Here we note from (63) that 1βπΏ 1βπΏ Ξπ‘ β (1 β 2π) β€ Ξπ‘ β π0 β€ 0. πΏπ2 πΏπ2 (70) Therefore, we infer from (69) that the function [1, 1/(1βπΏ)] β π§ β (1 β 2π)(1/(1 β πΏ) β π§) + Ξπ‘/π2 ππΏ (π§) is nonincreasing and attains a minimum value at π§ = 1/(1 β πΏ). Hence, we have (1 β 2π) ( which implies from (65) and (67) that (1 β 2π) (π§ + 1 Ξπ‘ π§ β 1 β π§) + 2 β ( β π§) 1βπΏ π πΏ (72) βπ§ β [β1, Ξπ‘ 1 β π§) + 2 ππΏ (π§) 1βπΏ π = (1 β 2π) ( 1 1 , ], 1βπΏ 1βπΏ ππΏ πΏ Ξπ‘ Ξπ‘ = (1 β 2π) β β₯ 0 β 2 β₯0 β 1 β πΏ π2 1βπΏ π Also, for any π§ β [1, 1/(1 β πΏ)], we observe from (23) that (1 β 2π) ( βπ§ β [β π’ππ+1 + + (1 β 2π) ( (1 β 2π) ( Thus, we observe from (68) and (71) that Ξπ‘ 1 β π§) + 2 ππΏ (π§) β₯ 0, (1 β 2π) ( 1βπΏ π 1 Ξπ‘ 1 Ξπ‘ β π§) + 2 ππΏ (π§) β₯ 2 ππΏ ( )=0 1βπΏ π π 1βπΏ 1 βπ§ β [1, ]. 1βπΏ (71) 1βπΏ 1βπΏ (77) Ξπ‘ β₯ π0 β Ξπ‘ β₯ 0. πΏπ2 πΏπ2 Therefore, we infer from (76) that the function [β1/(1 β πΏ), β1] β π§ β (1 β 2π)(π§ + 1/(1 β πΏ)) β Ξπ‘/π2 ππΏ (π§) is nondecreasing and attains a minimum value at π§ = β1/(1βπΏ). Hence, we have 1 Ξπ‘ Ξπ‘ 1 ) β 2 ππΏ (π§) β₯ β 2 ππΏ (β ) (1 β 2π) (π§ + 1βπΏ π π 1βπΏ (78) 1 = 0 βπ§ β [β , β1] . 1βπΏ (1 β 2π) β 12 Advances in Numerical Analysis Thus, we observe from (75) and (78) that where we put 1 Ξπ‘ ) β 2 ππΏ (π§) β₯ 0, 1βπΏ π (1 β 2π) (π§ + βπ§ β [β 1 1 , ], 1βπΏ 1βπΏ π’1π (79) fl ( (π) π’β which implies from (65) and (74) that π’ππ+1 1 + β₯0 1βπΏ βπ = 1, 2, . . . , ππ₯ β 1. 1 σ΅¨ σ΅¨ max σ΅¨σ΅¨σ΅¨σ΅¨π’ππ+1 σ΅¨σ΅¨σ΅¨σ΅¨ β€ , 0β€πβ€ππ₯ 1βπΏ (80) (81) which implies that (65) holds for π = π + 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 (π·π)ππΏ as in the following form: .. . π΄ ( ( =( ( ( 1 β 2π π ) π’2π ) ) )( .. ) ) . 1 β 2π π d d π 1 β 2π π ( Ξπ‘ π (π’π ) π2 πΏ 1 Ξπ‘ β 2 ππΏ (π’2π ) π .. . π’0π π 1 β 2π) ( ( . ) ( ) ( + π( ( .. ) + ( ( 0 π (π’ππ₯ ) (82) d πππ ) ) ). ) ) Ξπ‘ π ππΏ (π’π ) π₯ β1 ) ( π2 = π΄π’β ( π ππ’π β π₯ β1 ) ) ), ) ) Ξπ‘ π (π’π ) π2 πΏ ππ₯ β1 ) 1 β 2π) 1 πΏ 1 πΏ if π’ππ β€ β1, if π’ππ β [β1, 1] , (85) if π’ππ β₯ 1. if π’ππ β [β1, 1] , if π’ππ β€ β1, if π’ππ β [β1, 1] , if π’ππ β₯ 1, we observe from (85) and (86) that Ξπ‘ π (π’π ) π2 πΏ 1 Ξπ‘ β 2 ππΏ (π’2π ) π .. . ππ’1π β (83) π if π’ππ β [β1, 1] , { {β1 fl { 1 β πΏ { { πΏ 1 { { { πΏ { { π πΜπ fl {0 { { { { 1 β { πΏ β ( ( +( ( ( d π π (π’ππ₯ β1 ) Ξπ‘ π (π’π ) π2 πΏ 1 Ξπ‘ β 2 ππΏ (π’2π ) π .. . ) ) ). ) ) 1 β 2π π Defining ππ’1π β (π) π 1βπΏ π π’ + { { { πΏ π { { π ππΏ (π’ππ ) = {βπ’π { { { {1 β πΏ π { πΏ π’π β Here by taking into account Neumann boundary condition π π and initial condition, namely, by π’0π = π’1π and π’π = π’π π₯ π₯ β1 for all π β₯ 0, we observe that (82) is reformulated as in the following form: π’β π β 0 (π+1) 1 β 2π Note from (23) that π’1π π π (84) ( π+1 π π π 1 β 2π (π’ππ₯ β1 ) π 1 β 2π ( ( =( ( ( ) 1 β 2π ), (π’ππ₯ β1 ) π’1π+1 π’2π+1 .. . π Taking into account Neumann boundary condition, π+1 π+1 namely, by π’0π+1 = π’1π+1 and π’π = π’π , we observe from π₯ π₯ β1 (73) and (80) that ( π’2π ( ( ( ( ( ( π ππ’π β π₯ β1 ) ) ) ) ) Ξπ‘ π (π’π ) π2 πΏ ππ₯ β1 ) (86) Advances in Numerical Analysis πβ Ξπ‘ π π π2 1 β ( =( 13 π’1π Ξπ‘ π π π2 2 )( ) d π’2π .. . Then, by the abstract perturbation theory of matrix, we observe that ) Μ β€ π + ππ΅ , π ππ₯ β1 + ππ΅π β€ π π 1 π π Ξπ‘ π π β 2 ππ (π’ππ₯ β1 ) π₯ β1 ) π ( Ξπ‘ Μπ π π2 1 Ξπ‘ π β 2 Μπ2 π .. . π ππ₯ β1 + ππ΅π β₯ 1 β 4π sin2 ( ) ) ). ) ) β₯ 1 β 4π β Ξπ‘ π β Μπ ( π2 ππ₯ β1 ) β₯1β4β (87) π’β = π΄π’β (π) + π΅π’β (π) β ( π΅ fl ( Ξπ‘ π π π2 2 1 β π0 β π0 = β1 + π0 > β1 2 Thus, we conclude from (93) and the above inequality that Μ(π) + πβ , (π β₯ 0) , (88) Μ > β1 βπ = 1, 2, . . . , π β 1. π π π₯ ) ), d πβ Μβ(π) π 1 β 2π β Ξπ‘ π π π2 ππ₯ β1 ) σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨1 β π β Ξπ‘ ππ σ΅¨σ΅¨σ΅¨ + |π| = 1 β π β Ξπ‘ ππ + π = 1 + Ξπ‘ > 1 σ΅¨σ΅¨ π σ΅¨σ΅¨ 2 π σ΅¨ π2 π π2 σ΅¨ By the general theory, we observe that the eigenvalue π π of matrix π΄ is given by for π = 1, 2, . . . , ππ₯ β 1, (90) which implies that π 1 (resp., π ππ₯ β1 ) is the maximum (resp., minimum) eigenvalue of π΄. Μ ; π = 1, 2, . . . , π β 1} be the set of all Now let {π π π₯ Μ fl π΄ + π΅ such that eigenvalues of matrix π΄ Μ β₯π Μ β₯ β β β β₯ π Μ π 1 2 ππ₯ β1 . (91) Also, let {ππ΅π ; π = 1, 2, . . . , ππ₯ β 1} be the set of all eigenvalues of π΅ such that ππ΅1 β₯ ππ΅2 (96) Therefore, all components of π΄ + π΅ are positive. Hence, the sum of all components in the πth row of π΄+π΅ is the following: Ξπ‘ Μπ π ( π2 ππ₯ β1 ) ππ ) 2ππ₯ 1 β π0 Ξπ‘ + 2 >0 2 π βπ = 1, 2, . . . , ππ₯ β 1. ) ) ). ) ) β π π fl 1 β 4π sin2 ( Ξπ‘ Ξπ‘ Ξπ‘ π ππ = 1 β 2 + 2 2 2 π π (Ξπ₯) β₯1β2β (89) β ( ( fl ( ( ( (95) Note that (95) holds in any case: π’ππ β [β1, 1] and π’ππ β [β1, 1]. Next, we now assume max1β€πβ€ππ₯ β1 |π’ππ | β€ 1. Then, we observe from (63) and (86) that ( Ξπ‘ Μπ π π2 1 Ξπ‘ π β 2 Μπ2 π .. . (94) βπ = 1, 2, . . . , ππ₯ β 1. where we define Ξπ‘ π β 2 π1π π (ππ₯ β 1) π Ξπ‘ 1 β πΏ )β 2 β 2ππ₯ π πΏ Ξπ‘ (1 β πΏ) πΏπ2 Using the matrix as above, (82) can be rewritten as in the following form: (π+1) (93) Since π΅ is the symmetric matrix, it follows from (63), (86), and (90) that β ( ( +( ( ( βπ = 1, 2, . . . , ππ₯ β 1. β₯ β β β β₯ ππ΅ππ₯ β1 . (92) for π = 1 or ππ₯ β 1, σ΅¨σ΅¨ Ξπ‘ σ΅¨σ΅¨ Ξπ‘ |π| + σ΅¨σ΅¨σ΅¨σ΅¨1 β 2π β 2 πππ σ΅¨σ΅¨σ΅¨σ΅¨ + |π| = π + 1 β 2π β 2 πππ + π π σ΅¨ π σ΅¨ =1+ Ξπ‘ >1 π2 (97) βπ = 2, 3, . . . , ππ₯ β 2. Therefore, we observe from (97) that max1β€πβ€ππ₯ β1 |π’ππ | is increasing with respect to π in the case when max1β€πβ€ππ₯ β1 |π’ππ | β€ 1. However, if π’ππ β [β1, 1] for some π = 1, 2, . . . , ππ₯ β 1, it follows from (63) and (86) that 1 β 2π β Ξπ‘ Ξπ‘ 1 β πΏ Ξπ‘ π ππ = 1 β 2 β 2 β 2 2 π π πΏ (Ξπ₯) 1 β π0 β₯1β2β β π0 = 0. 2 (98) 14 Advances in Numerical Analysis u(t, x) Therefore, the sum of all components in the πth row of π΄ + π΅ is the following: σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨1 β π β Ξπ‘ ππ σ΅¨σ΅¨σ΅¨ + |π| = 1 β π β Ξπ‘ ππ + π σ΅¨σ΅¨ π σ΅¨σ΅¨ 2 π σ΅¨ π2 π σ΅¨ =1β Ξπ‘ 1 β πΏ β <1 π2 πΏ if π’ππ β [β1, 1] for some π = 1 or ππ₯ β 1, σ΅¨σ΅¨ Ξπ‘ σ΅¨σ΅¨ Ξπ‘ |π| + σ΅¨σ΅¨σ΅¨σ΅¨1 β 2π β 2 πππ σ΅¨σ΅¨σ΅¨σ΅¨ + |π| = π + 1 β 2π β 2 πππ + π π σ΅¨ π σ΅¨ (99) Ξπ‘ 1 β πΏ =1β 2 β <1 π πΏ if π’ππ 0 0.002 0.004 Ti 0.006 m e 0.008 0.01 0 0.2 0.4 0.6 1 0.8 Ξ© Figure 9: π = 0.05, Ξπ‘ = 0.000005, Ξπ₯ = 0.005, and πΏ = 0.01. β [β1, 1] for some π = 2, 3, . . . , ππ₯ β 2. Although max1β€πβ€ππ₯ β1 |π’ππ | is increasing with respect to π in the case when max1β€πβ€ππ₯ β1 |π’ππ | β€ 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 Ξπ₯. In fact, for each π β (0, 1], πΏ β (0, 1) we take the constant Μπ0 β (0, 1), Ξπ‘ β (0, 1], Ξπ₯ β (0, 1] such that Ξπ‘ β€ Μπ0 πΏπ2 , 1βπΏ 1 β Μπ0 Ξπ‘ = . 2 2 (Ξπ₯) (100) Ξπ‘ = Μπ πΏπ2 1 β Μπ0 β (Ξπ₯)2 β€ 0 . 2 1βπΏ 2Μπ0 πΏ . (1 β Μπ0 ) (1 β πΏ) β =0 π π’ (0, π₯) = π’0π π‘ β (0, π) , (102) (103) (π₯) , π₯ β (0, 1) . Taking into account Theorem 16, we provide the numerical experiments of (π·π)ππΏ as follows. To this end, we use the following numerical data. Numerical Data of if π₯ β [0.25, 0.75] , (104) if π₯ β [0.75, 1.00] . 5.1. The Case When π = 0.05 and Ξπ‘ = 0.000005. Now, we consider the case when π = 0.05 and Ξπ‘ = 0.000005. In this case, we observe that 1 β π0 0.000005 Ξπ‘ = = 0.2 = , 2 2 2 (Ξπ₯) (0.005) (105) = 0.00001515151515 . . . . π0 πΏπ2 = 0.00001515151515 β β β > Ξπ‘, 1βπΏ in π fl (0, π) × (0, 1) , π’π₯π (π‘, 0) = π’π₯π (π‘, 1) = 0, if π₯ β [0.00, 0.25] , Therefore, we have Remark 18. We can take π0 = 0 in (63) for the explicit finite difference scheme to the following usual heat equation with Neumann boundary condition: π π’π₯π₯ β0.5, { { { { π’0π (π₯) = {β0.5 sin (2ππ₯) , { { { {0.5, (101) Thus, we have the following condition of Ξπ₯: 0 < Ξπ₯ < πβ Also, we consider the initial data π’0π (π₯) defined by π0 πΏπ2 0.6 × 0.01 × (0.05)2 = 1βπΏ 1 β 0.01 Then, we have π’π‘π 1.5 1 0.5 0 β0.5 β1 β1.5 (π·π)ππΏ The numerical data is π = 0.01, πΏ = 0.01, Ξπ₯ = 0.005, π0 = 0.6. (106) which implies that criteria condition (63) holds. Thus, we can provide the stable numerical experiment of (π·π)ππΏ . Indeed, we provide Figure 9, which implies the stable numerical experiment of (π·π)ππΏ . 5.2. The Case When π = 0.007 and Ξπ‘ = 0.000005. Now, we consider the case when π = 0.007 and Ξπ‘ = 0.000005. In this case, we observe that 1 β π0 Ξπ‘ 0.000005 = = 0.2 = , 2 2 2 (Ξπ₯) (0.005) π0 πΏπ2 0.6 × 0.01 × (0.007)2 = 1βπΏ 1 β 0.01 (107) = 0.00000029696969 . . . . Therefore, we have π0 πΏπ2 = 0.00000029696969 β β β < Ξπ‘, 1βπΏ (108) u(t, x) Advances in Numerical Analysis 15 5.4. Conclusion of PDE Problem (π·π)ππΏ . By Theorem 16 and the numerical experiments as above, we conclude that the mesh size of time Ξπ‘ and space Ξπ₯ must be satisfied: 1.5 1 0.5 0 β0.5 β1 β1.5 0 < Ξπ‘ β€ 0β€ 0 0.002 0.004 0.006 Ti me 0.008 0.01 0 u(t, x) 1 β π0 Ξπ‘ β€ 2 2 (Ξπ₯) (110) for some constant π0 β (0, 1) , 0.2 0.4 0.6 0.8 1 Ξ© Figure 10: π = 0.007, Ξπ‘ = 0.000005, Ξπ₯ = 0.005, and πΏ = 0.01. in order to get the stable numerical experiments of (π·π)ππΏ . Also, by Theorems 7 and 16, we conclude that the value πΏπ2 /(1 β πΏ) is very important to perform a numerical experiment of (π·πΈ)ππΏ and (π·π)ππΏ . Competing Interests 1.5 1 0.5 0 β0.5 β1 β1.5 0 0.002 0.004 0.006 Ti me 0.008 0.01 0 π0 πΏπ2 , 1βπΏ The authors declare that there are no competing interests regarding the publication of this paper. Acknowledgments This work was supported by Grant-in-Aid for Scientific Research (C) no. 26400179, JSPS. 0.2 0.4 0.6 0.8 1 Ξ© Figure 11: π = 0.007, Ξπ‘ = 0.0000002, Ξπ₯ = 0.005, and πΏ = 0.01. which implies that criteria condition (63) does not hold. Therefore, we provide the unstable numerical experiment of (π·π)ππΏ . Indeed, we have Figure 10. 5.3. The Case When π = 0.007 and Ξπ‘ = 0.0000002. Now, we consider the case when π = 0.007 and Ξπ‘ = 0.0000002. In this case, we have 1 β π0 0.0000002 1 Ξπ‘ = = β€ 0.2 = , 2 2 125 2 (Ξπ₯) (0.005) π0 πΏπ2 = 0.00000029696969 β β β > Ξπ‘, 1βπΏ (109) which implies that criteria condition (63) holds. Therefore, we provide the stable numerical experiment of (π·π)ππΏ . Indeed, we have Figure 11, which implies the stable numerical experiment of (π·π)ππΏ . 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