A Heteroclinic Solution to a Variational Problem Corresponding to FitzHugh-Nagumo type Reaction-Diffusion System with Heterogeneity

Chen, Kung and Morita [ 5 ] studied a variational problem corresponding to the FitzHugh-Nagumo type reaction-diffusion system (FHN type RD system), and they proved the existence of a heteroclinic solution to the system. Motivated by [ 5 ], we consider a variational problem corresponding to FHN type RD system which involves heterogeneity. We prove the existence of a heteroclinic solution to the problem under certain conditions on the heterogeneity. Moreover, we give some information about the location of the transitions.

1. Introduction and main results. The FitzHugh-Nagumo type reaction-diffusion system (FHN type RD systems) is introduced in the field of physiology, which essentially describe neural excitability. This is also studied mathematically as a model which generates complex patterns. A typical FHN type RD system is the following: u t (x, t) = d∆u(x, t) + f (u(x, t)) − v(x, t), x ∈ Ω, t > 0, τ v t (x, t) = D∆v(x, t) + u(x, t) − γv(x, t), x ∈ Ω, t > 0, (1) where Ω ⊂ R N (N ≥ 1) is a domain, d, D, τ, γ > 0 are constants and f (t) = t(t − β)(1 − t) (0 < β < 1/2). We present some previous works on the steady state problems of (1), that is u t = v t = 0 in (1). There are a number of works on FHN type RD systems, and then we pick up works only related to variational approaches. For simplicity, we assume D = 1 throughout this paper.
In the case Ω is a bounded domain with a smooth boundary, Oshita [10] proved the existence of a solution to (1) under the Neumann boundary condition by considering a variational problem corresponding to (1). Oshita proved that the solution to the variational problem oscillates rapidly. Under the Dirichlet boundary condition, Dancer and Yan [6] showed similar results. See also [8] for related results on the three-component system.
In the case Ω = R, Chen and Choi [3] showed the existence of a homoclinic solution under the assumption that d, γ > 0 are small. In addition, Heijster et al. [4] [4] and references therein. On the other hand, Chen, Kung and Morita [5] showed the existence of heteroclinic solutions in the case Ω = R with certain assumptions. Before we state our work, we recall the strategy of [5] to find a heteroclinic solution by a variational approach. Now we give some notations. Let a be a positive number andv ∈ C ∞ (R) be a function such that v(x) = a/γ, x ≥ 1, 0, x ≤ 0.
For a related equation, there are some previous works on variational problems with a heterogeneity to find heteroclinic solutions. Bonheure and Sanchez [2] considered the Allen-Cahn type equation involving the heterogeneity h(x) of the form The energy corresponding (7) is the following: They obtained a heteroclinic solution connecting −1 to 1 by a variational approach. We note that (6) is one of extended models of (7). We mention a related paper by Sourdis [11]. Sourdis proved the existence of a heteroclinic solution to the vector Allen-Cahn model with heterogeneity, which is another extended model of (7). In addition, we mention the works by Ikeda and Ei [7] or Nishiura, Teramoto and Yuan [9] as the works that dealt with the FHN type RD system with various heterogeneities. See also the references therein on their related works. Now we state our main results. For the existence of the minimizer of (5), we have the following result. Theorem 1.1. Let a be a positive number and θ = d − 1/γ 2 > 0. We assume that µ ∈ L ∞ (R) satisfies (µ1) and (µ2). Then the following statements hold.
(ii) If θ ≥ 1, then there exists a positive constant M independent on d such that Remark 2. If (a, γ) satisfies (4) in Theorem 1.1, then we can readily see x → ∞. Theorem 1.1 (i) implies that there exists a minimizer of the minimizing problem (5) and that if supp (1 − µ) is compact, then the transitions ofū =û +ψ should have intersections with supp (1 − µ). Here we use the term "transition ofū" to loosely mean the interval [ȳ,ȳ ] defined as (9) and (10), which is precisely defined in the next section. Moreover, Theorem 1.1 (ii) gives us the information about dependence of the width of [ȳ,ȳ ] on d in the case θ ≥ 1.
Although the proof of Theorem 1.1 is based on [5], we need to modify a part of the argument of [5]. The key to the proof of the existence of a minimizer of (5) is the uniform estimate of ψ m , where {ψ m } m∈N is a minimizing sequence of (5). In the case µ ≡ 1, Chen, Kung and Morita [5] called a transition from ρ to a − ρ by a " wavefront type oscillation" for a given small positive number ρ, and they proved the uniform estimate of the width of the interval which contains all wavefront type oscillations of u m =û + ψ m . Noting that J(ψ) is invariant under translations of u =û + ψ,-more precisely, for given ψ ∈ H 1 (R) and L ∈ R, we defineũ(x) =û(x + L) + ψ(x + L) and φ(x) =ũ(x) −û(x), and then we have J(φ) = J(ψ)-this estimate implies that all wavefront type oscillations of u m stay in a compact interval independent on m ∈ N. However in the case µ ≡ 1,J(ψ) is generally not invariant under translations, and thus it is necessary to show the location of the transitions of u m . To overcome this difficulty, we prove that there exists a positive constant M 4 such that y m − y m < M 4 for all m ∈ N, where y m and y m are defined as follows: This uniform estimate of y m − y m is a stronger estimate than the uniform estimate of the region of wavefront type oscillations due to Chen, Kung and Morita [5], and hence this estimate enables us to analyze more accurate behavior of u m . With this estimate and σ < σ 0 (see Lemma 2.4), we can show that the all intervals [y m , y m ] (m ∈ N) are uniformly contained in a compact interval independent on m ∈ N.
Similarly we can prove that if µ satisfies for sufficiently large α > 0, then µ satisfies (µ3). Now we defineJ : H 1 (R) → R as follows: for ψ ∈ H 1 (R), where θ, γ and a are positive constants. We can prove that the Euler-Lagrange equation corresponding to the minimizing problem coincides with (13) under the assumptions θ = d−1/γ 2 > 0 and (4) (see Lemma 5.1). We mention that (15) is an extended model of (2), which has an extra term R (1 − µ(x))u 2 /(2γ) dx. For the existence of the minimizer of (16), we have the following result.
Theorem 1.2. Let θ = d − 1/γ 2 > 0 and (a, γ) be constants satisfying (4). We assume that µ ∈ L ∞ (R) satisfies (µ1) and (µ3). Thenσ < σ 0 holds, and there exists a minimizerψ ∈ H 1 (R) of (16). Moreover, (ū,v) = (û +ψ,v + Lψ) is a heteroclinic solution to (13). Theorem 1.2 implies that (16) has a minimizer under the assumption that µ satisfies more strict condition than Theorem 1.1. Basically we can prove it by repeating similar arguments as in the proof of Theorem 1.1, but we should modify a part of the arguments. In particular, the condition (µ3) is used for the proof of σ 0 >σ. This paper is arranged as follows. We first prepare several basic lemmas in Section 2: preliminaries in Section 2.1, and basic estimates in Section 2.2. In Section 2.2, we collect basic estimates in our setting, which are essentially same as in [5]. We next present key lemmas in Section 3. The key lemmas are applied for the proof of the uniform boundedness of the interval [y m , y m ]. Here y m , y m are defined as (11) and (12). We mention that the ideas of the key lemmas are also based on [5]. We prove Theorems 1.1 in Section 4. The proof of Theorem 1.1 consists of three claims. First, we prove the uniform boundedness of |y m − y m |. Next, we prove that [y m , y m ] is contained in a compact interval independent on m ∈ N by contradiction. Finally we prove the existence of the minimizer of (5). We prove Theorem 1.2 in Section 5. we mainly state the differences between Theorems 1.1 and 1.2.

Preliminaries.
In this section, we present several lemmas. We begin with deriving the Euler-Lagrange equation.
that is, we can write j[φ](t) as follows: Then it follows that where A i (i = 1, 2, 3, 4) are defined as follows: Therefore we rewrite A 2 as follows.
Moreover, we note (j[φ]) (0) = 0, and then we obtain For simplicity, we write the right hand side of the above equation as R ηφ dx, and we can easily see η ∈ H 1 (R). As a consequence, there exists u ∈ H 1 (R) ⊂ C(R), and u satisfies In addition, it is clear that v satisfies Now we define the term "transition" as follows.
Definition 2.2. Let u ∈ C(R) and a 1 , a 2 ∈ R. We say that u has a transition on [y 1 , y 2 ] from a 1 to a 2 if u satisfies as following conditions: Then we present some estimates, which are used to obtain the uniform estimate of [y m , y m ] in the proof of Theorem 1.1, where y m and y m are defined as (11) and (12).
The following lemma is well-known, but we present in the form to apply the lemmas shown later.
and ρ > 0 be a small positive number satisfying ρ ∈ (0, a/8). Moreover we assume that u has a transition on [y 1 , y 2 ] from ρ to 3ρ/2 or from a − ρ to a − 3ρ/2. Then the following inequality hold: where 1 is defined as follows: Proof. From the basic theorem of calculus and Hölder's inequality, we have Thus we can see On the other hand, we can easily check Combining the above inequalities, we obtain and complete the proof.

Remark 4.
It is easy to check that we can apply Lemma 2.3 to the case that u has a transition from −ρ to −3ρ/2 or from a + ρ to a + 3ρ/2.
The next lemma is almost obvious, but it is important for proving that the transitions of u m =û + ψ m must stay in a bounded interval.

Basic estimates.
In this subsection, we present basic estimates of a minimizing sequence {ψ m } m∈N . Let 0 be a small positive number. Then we may assume that J(û + ψ m ) ≤ σ + 0 < 2σ 0 for all m ∈ N without loss of generality. Moreover, we note that σ > 0 from Lemma 2.3. Now we prove the uniformly boundedness of u m =û + ψ m . Proof. We prove by contradiction. Namely we suppose that there exists a subsequence {m j } j∈N such that u mj L ∞ → ∞ as j → ∞. For simplicity, we write j = m j . We may assume that there existsx j ∈ R such that u j (x j ) = ja for all j ≥ 3. Then it follows that there exists x j <x j such that u j (x j ) = 2a for all j ≥ 3. We define I j ≡ [x j ,x j ]. From Hölder's inequality, we have for all j ≥ 3. Then we can see for all j ≥ 3, and then we obtain |I j | → ∞ as j → ∞. On the other hand, we have
The next lemma is essentially same as Lemma 2.4 of [5], and then we omit the proof.
3. Local energy and key lemmas. In this section, we present key lemmas for the proof of the theorem. First, we consider the "local energy" G n (n ∈ Z) defined as follows: where u =û + ψ, v =v + Lψ. Then we obtain the next lemma. The lemma reveals the relationship between the local energy and the local behavior of u and v.
Although this lemma is almost same with Lemma 2.5 of [5], we present a proof in Appendix A to emphasize the dependence of the constant M 2 .
Next, we introduce some notations. Let χ ∈ C ∞ (R) be a cut-off function satisfying Moreover we define S α andS α as follows: where D = −∂ 2 /∂x 2 + γ. From the definition, it is easy to check It is also easy to check from the definition of S α andS α . Now we prove the following lemma, which play important roles to show Lemma 3.3 presented later.
(ii) We suppose that u has a transition on [y 1 , y 2 ] from ρ to 3ρ/2 or from −ρ to −3ρ/2. Moreover we suppose that there exists j ∈ Z such that j > y 2 and ψ satisfies (19) with n = j. Then there exists φ ∈ H 1 (R) satisfying In particular, if j ≤ −1, then we may take φ =S j ψ.
Proof. Since we can prove (ii) as in the case (i), it suffices to prove the case (i). In addition, we may assume that u has a transition from a − ρ to a − 3ρ/2 because of Remark 4. First, we prove the statement in the case j ≥ 1. We setū =û + S j ψ,v = v + L(S j ψ). Then it is easy to seev =v + χ j (Lψ) from (22). Moreover we can see from (20). Hence we obtain the following estimate ofJ(S j ψ).
Next, we prove in the case j ∈ Z. We set U, ϕ and V as follows: We readily see ϕ ∈ H 1 (R). Moreover, we can see that V is the unique solution to On the other hand, v(x + j − 1) is also the solution to the above equation, and thus we have V (x) = v(x + j − 1). As a consequence, U, V satisfy This means that ϕ satisfies (18) with n = 1. Subsequently, we defineŪ ,V as follows: We note that there are the following relationships between U, V andŪ ,V .
Then it is clear that φ ∈ H 1 (R), and we can show thatv(x) =V (x−(j−1)) similarly to the proof of V (x) = v(x + j − 1). Hence there exist relationships betweenū,v, φ and u, v, ψ as follows: We can show the following estimate as in the case j ≥ 1: In view ofū,v,Ū andV , we have Moreover noting that ϕ satisfies (18) with n = 1, we can see G (25) and (26). As a consequence, we complete the proof of (i). Similarly we prove (ii) with Lemma 3.2 (ii).

4.
Proof of Theorem 1.1. In this section, we present the proof of Theorem 1.1.
Let u m =û + ψ m , v m =v + Lψ m and ρ 1 be a positive constant defined as follows: Then we have the the following statements by Lemma 3.1 and the Sobolev inequality.
If u m , v m satisfy (18), then |a − u m | < ρ on I n .
(30) Moreover, we recall y m , y m defined as (11) and (12): We can easily see −∞ < y m < y m < ∞ for all m ∈ N since ψ m ∈ H 1 (R). Then let z m be an integer such that y m ∈ [z m , z m + 1]. We note that G zm (ψ m ) ≥ ρ 1 from (28). Theorem 1.1 is proved by the following three claims. Claim 3. There exist a minimizerψ ∈ H 1 (R) of the minimizing problem (5). Thus the statement of the theorem is true.
Proof of Claim 1. For each m ∈ N, we define K m ≡ {n ∈ Z : G n (ψ m ) ≥ ρ 1 }, l m = |K m |. Then we easily see that Thus it suffices to show the next equation: Since G zm (ψ m ) ≥ ρ 1 , there exists n ∈ N satisfying G n (ψ m ) < ρ 1 and n ∈ {z m + 1, z m + 2, . . . , z m + l m }. Therefore ψ m should satisfy (18) for n ∈ Z defined in (33). u m should also satisfy |a − u m | < ρ on I n by (29). We can also prove that u m has transitions neither from a − ρ to a − 3ρ/2 nor from a + ρ to a + 3ρ/2 in the interval [n, ∞) by Lemma 3.3 (i). This implies Combining with n ≤ z m + l m , z m ≤ y m and (32), we obtain Moreover, from (27), Lemma 3.1 and Remark 5, if θ ≥ 1, then we can write where M 6 is a positive constant independent on d. Combining with Lemma 2.5, we obtain Thus we conclude the claim.
Proof of Claim 2.
We prove by contradiction. Namely we assume that there exists a subsequence {m j } j∈N such that y mj → ∞ as j → ∞. For simplicity, we write j = m j . From Claim 1, y j → ∞ as j → ∞. By the definition of J(ψ j ) andJ(ψ j ), we havẽ u j L ∞ is uniformly bounded from Lemma 2.6, and (1 − µ) ∈ L 1 (R). Thus we have for sufficiently large j ∈ N.
Moreover, we can see by ρ ∈ (0, ρ 0 ). As a result, we have − 2 > − 2 /2, but it is clearly contradiction. Thus it never happens that y j → ∞ as j → ∞. we can also prove that it never happens that y j → −∞ as j → ∞ in the same way. In the case supp (1 − µ) ⊂ [x 1 , x 2 ], we suppose that x 2 < y m . Then we can see in the same way as (34). Clearly it is contradiction. Thus we arrive at the claim. Proof of Claim 3. Letting k 1 = −M 5 , k 2 = M 5 and K = [k 1 , k 2 ], we can see that there exists a constant C = C(K) such that ψ m H 1 (K) + Lψ m H 1 (K) < C from Lemma 2.7. By taking k 1 smaller and k 2 larger, if necessary, we may assume k 1 ≤ 0 and k 2 ≥ 1. Thus we have In addition, since ρ < a/8, we can easily see for all x ∈ (−∞, k 1 ).
Hence we obtain for all m ∈ N. Moreover it is clear that for all m ∈ N. As a consequence, it follows that {ψ m } m∈N is bounded in H 1 (R) and there existsψ ∈ H 1 (R) such that ψ m →ψ weakly in H 1 (R) and strongly in C loc (R) as m → ∞.
In addition, it easy to check that Lψ m → Lψ weakly in H 1 (R) and strongly in C loc (R) as m → ∞.

5.
Proof of Theorem 1.2. We give a proof of Theorem 1.2 in this section. Basically we can prove it as in the proof of Theorem 1.1, but we slightly modify the arguments. Throughout this section, we always assume (4), that is, a = 2(β + 1)/3 and γ = 9/(2β 2 − 5β + 2). First, we derive the Euler-Lagrange equation.
Then we calculate (j[φ]) (0) as in Lemma 2.1, and we obtain Under the assumption (4), we can easily see Thus we have Repeating almost the same argument as in Lemma 2.1, we can prove that (u, v) is a solution to (13).
Hence we conclude the lemma.
We shall define "local energy"Ḡ n (ψ) (n ∈ Z), which corresponds to G n (ψ) in Section 3. We first fix a positive constant δ such that where ρ > 0 is a small constant. Then there exists a constant L 1 ≥ 1 such that ∞
Now we set q = 2L 1 and defineḠ n as follows: where u =û + ψ, v =v + Lψ. Then we obtain the next lemma. Since we can see it by repeating almost the same arguments, we omit the proof.
(i) We suppose that u has a transition on [y 1 , y 2 ] from a − ρ to a − 3ρ/2 or from a + ρ to a + 3ρ/2. Moreover we suppose that there exists j ∈ Z such that j + 1 < y 1 and ψ satisfies (37) with n = j. Then there exists φ ∈ H 1 (R) satisfyingJ In particular, if j ≥ 1, then we may take φ = S jq−L1 ψ. (ii) We suppose that u has a transition on [y 1 , y 2 ] from ρ to 3ρ/2 or from −ρ to −3ρ/2. Moreover we suppose that there exists j ∈ Z such that j > y 2 and ψ satisfies (38) with n = j. Then there exists φ ∈ H 1 (R) satisfying In particular, if j ≤ −1, then we may take φ =S jq−L1 ψ.
Proof. In the case j ≥ 1, with paying attention to we can easily checkJ (S jq−L1 ψ) <J(ψ) − 2 1 by repeating almost the same arguments as in Lemma 3.3. Next, we consider the case j = 0. As in the proof if Lemma 3.3, we construct a function φ ∈ H 1 (R) satisfying the following conditions: Then we obtainJ As in the proof of Lemma 3.3, we can easily see From the definition ofḠ n and φ, we obtain the following inequalities: Thus we haveJ Since u satisfies u − a H 1 (I 0 ) <M 2 √ ρ 1 , we obtain by Sobolev inequality. Similarly from (39) and Lemma 5.3, we can see As a consequence, it followsJ(φ) <J(ψ) − 1 . Finally, we consider the case j ≤ −1. Let ρ 2 be a small constant such that max W (ρ 2 ), ρ 2 2 2γ R (1 − µ(x)) dx < Moreover, let L 2 > 0 be a large constant such that We fix a sufficient large N ∈ N, and then we define U (x) ≡ u(x − N q), ϕ(x) ≡ U (x) −û(x) and V (x) ≡v(x) + Lϕ(x). We may assume the following conditions: U (x) < ρ 2 for all x ∈ max{jq + L 1 , L 2 }, By an easy calculation, we havē From (40) and (42), we can see On the other hand, it is easy to check that by changing variables. Thus we obtain the following estimates: (1 − µ(x)) dx < 1 4 .
and we similarly obtain Thus letting Thus letting C 7 = C 7 (θ, ρ, µ 1 , γ) = C 5 + C 6 , we estimate under the assumption u(x) > ρ for all x ∈ I n . Moreover, from the definition of u, v, we have Multiplying the above equation by (v − a/γ) and integrating over I n , we have ≤ (γ + 1)C 7 √ ρ 1 .
As a consequence, letting M 2 = M 2 (θ, ρ, µ, γ) = (3 + 2γ)C 7 , we can estimate under the assumption u(x) > ρ for all x ∈ I n . Similarly we also obtain u H 1 (In) + v H 3 (In) ≤ M 2 √ ρ 1 under the assumption a − u(x) > ρ for all x ∈ I n . Moreover, in view of the above proof, we can easily see that if θ ≥ 1, then we can take the constant M 2 > 0 which is independent on d > θ ≥ 1.