On convergence of numerical algorithm of a class of the spatial ...

arXiv:1306.0707v2 [math.NA] 9 Dec 2015

ON CONVERGENCE OF NUMERICAL ALGORITHM OF A CLASS OF THE SPATIAL SEGREGATION OF REACTION-DIFFUSION SYSTEM WITH TWO POPULATION DENSITIES AVETIK ARAKELYAN Abstract. Recently, much interest has gained the numerical approximation of equations of the spatial segregation of Reaction-diffusion systems with m population densities. These problems are governed by a minimization problem subject to the closed but nonconvex set. In the present work we deal with the numerical approximation of equations of stationary states for a certain class of the spatial segregation of Reaction-diffusion system with two population densities having disjoint support. We prove the convergence of the numerical algorithm for two competing populations with non-negative internal dynamics fi (x) ≥ 0. At the end of the paper we present computational tests.

1. Introduction 1.1. The statement of the problem. In recent years there have been intense studies of spatial segregation for reaction-diffusion systems. The existence of spatially inhomogeneous solutions for competition models of Lotka-Volterra type in the case of two and more competing densities have been considered [7, 8, 9, 10, 15, 17]. The aim of this paper is to study the numerical solutions for a certain class of the Spatial Segregation of Reaction-diffusion System with two population densities. The problem is related with an arbitrary number of competing densities, which are governed by a minimization problem over closed but non-convex set. Let Ω ⊂ Rn , (n ≥ 2) be a connected and bounded domain with smooth boundary and m be a fixed integer. We consider the steady-states of m competing species coexisting in the same area Ω. Let ui (x) denotes the population density of the ith component with the internal dynamic prescribed by fi (x). Here we assume that fi is uniformly continuous and fi (x) ≥ 0. We call the m-tuple U = (u1 , · · · , um ) ∈ (H 1 (Ω))m , segregated state if ui (x) · uj (x) = 0, a.e. for

i 6= j, x ∈ Ω.

Key words and phrases. Free boundary, Two-phase obstacle problem, Reaction-diffusion systems, Finite difference. This work was supported by State Committee of Science MES RA, in frame of the research project No. SCS 13YR-1A0038. 1

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AVETIK ARAKELYAN

The problem amounts to Minimize E(u1 , · · · , um ) =

(1)

Z X m  1 2

Ω i=1



2

|∇ui | + fi ui dx,

over the set S = {(u1 , . . . , um ) ∈ (H 1 (Ω))m : ui ≥ 0, ui · uj = 0, ui = φi

on ∂Ω},

1

where φi ∈ H 2 (∂Ω), φi · φj = 0, for i 6= j and φi ≥ 0 on the boundary ∂Ω. Throughout the paper we will work with the case m = 2. The minimization problem will be reduced to:

(2)

Minimize E(u1 , u2 ) =

Z X 2  1 Ω i=1

2

2



|∇ui | + fi ui dx,

over the set S = {(u1 , u2 ) ∈ (H 1 (Ω))2 : ui ≥ 0, u1 · u2 = 0, ui = φi

on ∂Ω}.

1

Here φi ∈ H 2 (∂Ω) with property φ1 · φ2 = 0, φi ≥ 0 on the boundary ∂Ω. Unfortunately, due to the non-convexity of the set S, the general framework of variational methods cannot be applied to the convergence analysis of the numerical scheme. Therefore we need to find another approach to overcome this issue. 1.2. Two-phase membrane (obstacle) problem. In this section we briefly explain the Two-Phase Membrane problem and show how it can be connected with the segregation problem with two competing densities (details can be found in [6]). This connection is playing a key role in proving the convergence of proposed algorithm. Let fi : Ω → R, i = 1, 2, be non-negative Lipschitz continuous functions, where Ω is a bounded open subset of Rn with smooth boundary. Let K = {v ∈ W 1,2 (Ω) : v − g ∈ W01,2 (Ω)}, where g changes the sign on the boundary. Consider the functional  Z  1 2 |∇v| + f1 max(v, 0) − f2 min(v, 0) dx, (3) I(v) = 2 Ω which is convex, weakly lower semi-continuous and hence attains its infimum at some point u ∈ K. Define v ∨ 0 = max(v, 0), v ∧ 0 = min(v, 0).

ALGORITHM CONVERGENCE FOR SEGREGATION PROBLEM WITH TWO DENSITIES

3

In the functional (3) we set u1 = v ∨ 0, g1 = g ∨ 0,

u2 = −v ∧ 0, g2 = −g ∧ 0.

Then the functional I(v) in (3) can be rewritten as  Z  |∇u1 |2 |∇u2 |2 (4) I(u1 , u2 ) = + + f1 u1 + f2 u2 dx, 2 2 Ω where minimization is over the set S = {(u1 , u2 ) ∈ (H 1 (Ω))2 : u1 · u2 = 0, ui ≥ 0 ui = gi

on ∂Ω, i = 1, 2}.

The Euler-Lagrange equation corresponding to the minimizer u is given in [19], which is called the Two-phase obstacle problem:  ∆u = f1 χ{u>0} − f2 χ{u 0} ∪ ∂{x ∈ Ω : u(x) < 0} ∩ Ω is called the free boundary. If we set u = u1 − u2 in the system (5) we arrive at:  ∆(u1 − u2 ) = f1 χ{u1 −u2 >0} − f2 χ{u1 −u2 0 then by (7) we have uk1 (xi , yj ) =

−f1 h2 k−1 + uk−1 1 (xi , yj ) − u2 (xi , yj ) > 0. 4

Thus k−1 uk−1 2 (xi , yj ) − u1 (xi , yj ) < 0,

which leads to

−f2 h2 −f2 h2 k−1 + uk−1 (x , y ) − u (x , y ) < ≤ 0. i j i j 2 1 4 4

Therefore uk2 (xi , yj )

 = max

 −f2 h2 k−1 k−1 + u2 (xi , yj ) − u1 (xi , yj ), 0 = 0. 4

Thus uk1 (xi , yj ) · uk2 (xi , yj ) = 0.  Lemma 2. The numerical algorithm (7) is stable and consistent. Proof. Here we will prove the stability of the method, for the proof of the consistency we again refer the reader to the above mentioned work [6]. Due to the non-negative fi ≥ 0, we can write the following inequalities:   −f1 (xi , yj )h2 k+1 k k u1 (xi , yj ) = max + u1 (xi , yj ) − u2 (xi , yj ), 0 ≤ uk1 (xi , yj ), 4 and uk+1 2 (xi , yj )

 = max

−f2 (xi , yj )h2 + uk2 (xi , yj ) − uk1 (xi , yj ), 4

 0 ≤ uk2 (xi , yj ).

Therefore (k)

k k+1 uk+1 2 (xi , yj ) ≤ u2 (xi , yj ), and u2 (xi , yj ) ≤ u2 (xi , yj ), respectively. Thus ≥ 0, ∆h uk+1 ≥ 0, and ∆h uk+1 2 1 where ∆h is the usual discrete Laplace operator. After applying the discrete maximum principle we obtain 0 ≤ u1k+1 (xi , yj ) ≤ max φ1 (xi , yj ), i,j

ALGORITHM CONVERGENCE FOR SEGREGATION PROBLEM WITH TWO DENSITIES

7

and 0 ≤ u2k+1 (xi , yj ) ≤ max φ2 (xi , yj ). i,j

uk1

and Hence, stability.

uk2

are uniformly bounded for every k ∈ N. This completes the proof of  3. Algorithm convergence

3.1. Algorithm for one-dimensional case. For the sake of simplicity, we consider here only the one-dimensional case. Let u = (u0 , u1 , ..., uN ) be the solution of (7) in one-dimensional case. In this case the algorithm reads: • Initialization:  0 xi ∈ N o , 0 u1 (xi ) = φ1 (xi ) xi ∈ ∂N .  0 xi ∈ N o , u02 (xi ) = φ2 (xi ) xi ∈ ∂N . • Step k + 1, k ≥ 0 : We iterate over all interior points by setting    −f1 (xi )h2 k k uk+1 (x ) = max + u (x ) − u (x ), 0 , i i i 1 1 2 2   (8) 2 −f2 (xi )h uk+1 + uk2 (xi ) − uk1 (xi ), 0 . 2 (xi ) = max 2 Here for a given uniform mesh on Ω ⊂ R, we define uk (xi ) for k = 1, 2, to be the average of uk for all neighbor points of xi , 1 uk (xi ) = [uk (xi−1 ) + uk (xi+1 )]. 2 Define g(x) = φ1 (x)−φ2 (x), and vi ≡ v(xi ) = u1 (xi )−u2 (xi ). We consider the following discrete functional:        1 Jh (v) = − Lh v, v + f1 , v ∨ 0 − f2 , v ∧ 0 − Lh g, v , 2 defined on the finite dimensional space K = {v ∈ H : vα = 0, α ∈ ∂N },

where H = {v = (vα ) : vα ∈ R, α ∈ N }.

Here v ∨ 0 = max(v, 0), v ∧ 0 = min(v, 0) and for w = (wα ) and v = (vα ), α ∈ N , the inner product (·, ·) is defined by: X (w, v) = wα · vα . α∈N

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AVETIK ARAKELYAN

In particular, v0 = g0 and vN = gN . We will use the notation v˜ = (v1 , v2 , ..., vN −1 ). This is the unknown part in v that needs to be calculated. We introduce also the following N − 1 dimensional vectors:  g0 gN  ˜ f1 = f1 (1) − 2 , f1 (2), ..., f1 (N − 2), f1 (N − 1) − 2 , h h and  g0 gN  f˜2 = f2 (1) − 2 , f2 (2), ..., f2 (N − 2), f2 (N − 1) − 2 . h h k In the next section we will prove the convergence of v˜k = (v1k , v2k , ..., vN −1 ), and then the disjointness condition for competing densities will lead to the convergence of uk1 and uk2 separately.

3.2. Convergence result. Proposition 1. The sequence v˜k converges and limk→∞ v˜k = v˜. Proof. Denote
k−1 k−1 v˜k,i = v˜1k , v˜2k , ..., v˜ik , v˜i+1 , ..., v˜N −1 ,

i = 1, ..., N − 1,

k ∈ N,


k−1 k−1 v k,i = 0, v˜1k , v˜2k , ..., v˜ik , v˜i+1 , ..., v˜N , 0 ∈ K, i = 1, ..., N − 1, −1
and Jp = Jh v k,i for p = (N − 1)(k − 1) + i with i = 1, N − 1. The main idea is to prove that Jp decreases. First let p 6∈ {q(N − 1) : q ∈ N}, i.e. i 6= N − 1. Then

k∈N





1 Jp − Jp+1 = Jh v k,i − Jh v k,i+1 = − Lh v k,i − v k,i+1 , v k,i − v k,i+1 − 2



Lh v k,i+1 , v k,i − v k,i+1 + f1 , v k,i ∨ 0 − v k,i+1 ∨ 0 − f2 , v k,i ∧ 0 − v k,i+1 ∧ 0 − Lh g, v k,i − v k,i+1 =

k−1 k
2 v˜ik−1 − 2˜
 k−1  vi+1 + v˜i+2 1 k−1 k−1 k k k ∨ 0 − v ˜ − v ˜ − · v ˜ − v ˜ +f (i+1)· v ˜ ∨ 0 − v ˜ 1 i+1 i+1 i+1 i+1 i+1 i+1 h2 h2
 k−1  k−1 k k −f2 (i + 1) · v˜i+1 ∧ 0 − v˜i+1 ∧ 0 − (Lh g)i+1 · v˜i+1 − v˜i+1 .

We continue by considering three cases:

ALGORITHM CONVERGENCE FOR SEGREGATION PROBLEM WITH TWO DENSITIES

9

k k Case 1: v˜i+1 > 0. Due to uk1 (xi+1 ) · uk2 (xi+1 ) = 0, we have v˜i+1 = uk1 (xi+1 ) > 0, and uk2 (xi+1 ) = 0. It follows from (8) that k−1 k v˜ik−1 − 2˜ vi+1 + v˜i+2 2 k ˜ = 2 (uk−1 (xi+1 ) − uk−1 2 (xi+1 ) − u1 (xi+1 )) = f1 (i + 1). 2 h h 1

Hence, Jp − Jp+1 =


2
 k−1 1 k−1 k−1 k k k − v˜i+1 − v˜i+1 − f˜1 (i + 1) · v˜i+1 − v˜i+1 + f1 (i + 1) · v˜i+1 ∨ 0 − v˜i+1 2 h
k−1 k−1 k −f2 (i + 1) · v˜i+1 ∧ 0 − (Lh g)i+1 · v˜i+1 . − v˜i+1

Now, if 1 ≤ i < N − 1, then f˜1 (i + 1) = f1 (i + 1) and (Lh g)i+1 = 0, so
2

2 1 k−1 1 k−1 k−1 k k Jp − Jp+1 = 2 v˜i+1 − v˜i+1 − (f1 (i + 1) + f2 (i + 1)) · v˜i+1 ∧ 0 ≥ 2 v˜i+1 − v˜i+1 h h g g If i = N − 1, then f˜1 (i + 1) = f1 (i + 1) − N2 and (Lh g)i+1 = N2 , so h

h

1 k−1 k v˜i+1 − v˜i+1 2 h Hence, in this case we have


gN k
2 1 k−1 k−1 k −(f1 (i+1)+f2 (i+1))· v˜i+1 ≥ 2 v˜i+1 . ∧ 0 + 2 v˜i+1 − v˜i+1 h h

(9)

Jp − Jp+1 ≥

Jp −Jp+1 =


2


2 1 k−1 k v ˜ − v ˜ . i+1 i+1 h2

k k Case 2: v˜i+1 < 0. In this case again due to uk1 (xi+1 ) · uk2 (xi+1 ) = 0, we have v˜i+1 = k k −u2 (xi+1 ) < 0, and u1 (xi+1 ) = 0. Analogously to the previous case we can prove that (9) holds also in this case. k k = 0. Since v˜i+1 = uk1 (xi+1 ) − uk2 (xi+1 ) = 0, then uk1 (xi+1 ) · uk2 (xi+1 ) = 0 Case 3: v˜i+1 implies uk1 (xi+1 ) = uk2 (xi+1 ) = 0. Thus, according to (8) we have

2 k−1 k−1 ˜1 (i + 1) ≤ 0 ≤ 2 (uk−1 (xi+1 ) − uk−1 (xi+1 )) + f˜2 (i + 1). (u (x ) − u (x )) − f i+1 i+1 1 2 2 h2 h2 1 Recalling that k−1 v˜ik−1 + v˜i+1 2 = 2 (uk−1 (xi+1 ) − uk−1 2 (xi+1 )), 2 h h 1 we arrive at k−1 k−1 k−1 v˜ik−1 + v˜i+1 ˜1 (i + 1) ≤ 0 ≤ v˜i + v˜i+1 + f˜2 (i + 1). − f h2 h2

10

AVETIK ARAKELYAN

Therefore Jp − Jp+1

! k−1 v˜ik + v˜i+1 − f˜1 (i + 1) − h2


2
1 k−1 k−1 k = 2 v˜i+1 − v˜i+1 − v˜i+1 ∨0 · h


k−1 − v˜i+1 ∧0 ·

k−1 v˜ik + v˜i+1 + f˜2 (i + 1) h2

! k−1 − (Lh g)i+1 · v˜i+1 .

Treating, as above, the cases 1 ≤ i < N − 1 and i = N − 1 separately we will obtain that (9) holds also in this case. So far we have considered the case p 6∈ {q(N − 1) : q ∈ N}. Now assume that p ∈ {q(N − 1) : q ∈ N}. In that case we have
1 k k−1 2 − v˜i+1 . (10) Jp − Jp+1 ≥ 2 v˜i+1 h Summarizing, we deduce that Jp decreases, and, since it is also bounded from below, we conclude that the sequence Jp converges. In view of (9) and (10) we conclude that v˜ik is a Cauchy sequence, hence also converges for every fixed i = 1, ..., N − 1.  Thus we have proved that the sequence v˜ik = uk1 (xi ) − uk2 (xi ) is converging to v˜ for every fixed i = 1, ..., N − 1. Observe that due to Lemma 1 we have (11)

uk1 (xi ) = max(uk1 (xi ) − uk2 (xi ), 0) and uk2 (xi ) = max(uk2 (xi ) − uk1 (xi ), 0).

Now, in view of (11) we will obtain the convergence of uk1 (xi ) and uk2 (xi ) separately. This completes the proof of Proposition. 4. Numerical examples In this section we will present simulations for two competing densities with different internal dynamics fi . We consider the following minimization problem:  Z X 2  1 2 |∇ui | + fi ui dx, (12) Minimize 2 Ω i=1 over the set S = {(u1 , u2 ) ∈ (H 1 (Ω))2 : ui ≥ 0, u1 · u2 = 0, ui = φi

on ∂Ω}.

In Figure 1 we consider the set Ω = [−1, 1], and f1 and f2 taken to be constant. The free boundaries are clearly visible. It is easy to see that the smaller dynamics fi provides ui more captured place. Next we present numerical examples in 2D. In Figures 2 and 3 we take Ω = [0, 1]×[0, 1], with the boundaries φ1 (x, y) and φ2 (x, y) defined by:

ALGORITHM CONVERGENCE FOR SEGREGATION PROBLEM WITH TWO DENSITIES

 φ1 (x, 0) =

0.5 − 2.5x 0 ≤ x ≤ 0.2, 0 0.2 ≤ x ≤ 1, φ1 (0, y) = 0.5,

 φ1 (x, 1) =

11

0.5 − 85 x 0 ≤ x ≤ 0.2, 0 0.8 ≤ x ≤ 1,

φ1 (1, y) = 0,

and  φ2 (x, 0) =

0 −1 8

0 ≤ x ≤ 0.2, + 58 x 0.2 ≤ x ≤ 1, φ2 (0, y) = 0,

 φ2 (x, 1) =

0 0 ≤ x ≤ 0.8, −2 + 2.5x 0.8 ≤ x ≤ 1,

φ2 (1, y) = 0.5,

In Figure 2 we clearly see that the zero set does not appear and competing densities u1 and u2 meet each other along the whole free boundary, while in Figure 3 there is a zero set between the densities due to the big internal dynamics fi . References [1] Arakelyan, A. A finite difference method for two-phase parabolic obstacle-like problem. Armenian Journal of Mathematics 7, 1 (2015), 32–49. [2] Arakelyan, A., Barkhudaryan, R., and Poghosyan, M. Numerical solution of the two-phase obstacle problem by finite difference method. submitted . [3] Arakelyan, A. G., Barkhudaryan, R. H., and Poghosyan, M. P. Finite difference scheme for two-phase obstacle problem. Dokl. Nats. Akad. Nauk Armen. 111, 3 (2011), 224–231. [4] Bozorgnia, F. Numerical algorithm for spatial segregation of competitive systems. SIAM J. Sci. Comput. 31, 5 (2009), 3946–3958. [5] Bozorgnia, F. Numerical solutions of a two-phase membrane problem. Applied Numerical Mathematics 61, 1 (2011), 92–107. [6] Bozorgnia, F., and Arakelyan, A. Numerical algorithms for a variational problem of the spatial segregation of reaction–diffusion systems. Applied Mathematics and Computation 219, 17 (2013), 8863–8875. [7] Conti, M., Terracini, S., and Verzini, G. Asymptotic estimates for the spatial segregation of competitive systems. Adv. Math. 195, 2 (2005), 524–560. [8] Conti, M., Terracini, S., and Verzini, G. A variational problem for the spatial segregation of reaction-diffusion systems. Indiana Univ. Math. J. 54, 3 (2005), 779–815. [9] Conti, M., Terracini, S., and Verzini, G. Uniqueness and least energy property for solutions to strongly competing systems. Interfaces Free Bound. 8, 4 (2006), 437–446. [10] Crooks, E. C. M., Dancer, E. N., and Hilhorst, D. Fast reaction limit and long time behavior for a competition-diffusion system with Dirichlet boundary conditions. Discrete Contin. Dyn. Syst. Ser. B 8, 1 (2007), 39–44 (electronic). [11] Crooks, E. C. M., Dancer, E. N., and Hilhorst, D. On long-time dynamics for competitiondiffusion systems with inhomogeneous Dirichlet boundary conditions. Topol. Methods Nonlinear Anal. 30, 1 (2007), 1–36.

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1

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Figure 1. For all cases the boundaries are taken φ1 (−1) = φ2 (1) = 1 and φ2 (−1) = φ1 (1) = 0.

[12] Crooks, E. C. M., Dancer, E. N., Hilhorst, D., Mimura, M., and Ninomiya, H. Spatial segregation limit of a competition-diffusion system with Dirichlet boundary conditions. Nonlinear Anal. Real World Appl. 5, 4 (2004), 645–665. [13] Dancer, E. N., and Du, Y. H. Competing species equations with diffusion, large interactions, and jumping nonlinearities. J. Differential Equations 114, 2 (1994), 434–475. [14] Dancer, E. N., Hilhorst, D., Mimura, M., and Peletier, L. A. Spatial segregation limit of a competition-diffusion system. European J. Appl. Math. 10, 2 (1999), 97–115. [15] Dancer, E. N., and Zhang, Z. Dynamics of Lotka-Volterra competition systems with large interaction. J. Differential Equations 182, 2 (2002), 470–489. [16] Petrosyan, A., Shahgholian, H., and Ural’ceva, N. N. Regularity of free boundaries in obstacle-type problems, vol. 136. American Mathematical Soc., 2012. [17] Squassina, M. On the long term spatial segregation for a competition-diffusion system. Asymptot. Anal. 57, 1-2 (2008), 83–103.

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Figure 2. Internal dynamics are taken f1 = 0 and f2 = 5.

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Figure 3. Internal dynamics are taken f1 = 4 and f2 = 12. [18] Squassina, M., and Zuccher, S. Numerical computations for the spatial segregation limit of some 2D competition-diffusion systems. Adv. Math. Sci. Appl. 18, 1 (2008), 83–104. [19] Weiss, G. S. Partial regularity for weak solutions of an elliptic free boundary problem. Comm. Partial Differential Equations 23, 3-4 (1998), 439–455. Institute of Mathematics, National Academy of Sciences of Armenia, 0019 Yerevan, Armenia E-mail address: [email protected]