THE SUB-SUPERSOLUTION METHOD FOR THE FITZHUGH-NAGUMO TYPE REACTION-DIFFUSION SYSTEM WITH HETEROGENEITY

. We construct a heteroclinic solution to the FitzHugh-Nagumo type reaction-diﬀusion system (FHN RD system) with heterogeneity by the sub- supersolution method due to [5]. σ ( d,γ ) is introduced as the Rayleigh quotient corresponding to a linearized eigenvalue problem of the subsolution, where d and γ are parameters. The key to construct the solution is the uniform estimate for σ ( · , · ) from below. In addition, it enables us to analyze an asymptotic behavior of the solution.

1. Introduction and main results. The FitzHugh-Nagumo type reactiondiffusion system (FHN RD system) is introduced in physiology, which essentially describes neural excitability. This is also studied mathematically as a model which generates complex patterns. A typical FHN RD system is the following: where Ω ⊂ R N (N ≥ 1) is domain, d, D, τ, γ are positive constants and f (s) = s−s 3 . Throughout this paper, we assume D = 1 for simplicity. Moreover, in this paper we treat the steady state problem of (1), that is, ∂u/∂t = ∂v/∂t = 0. In general, the steady state FHN RD system has various solutions and several methods are known to construct a solution to (1). For small d > 0, there are many works in which the singular perturbation method (SPM) is applied to obtain the solutions. For example, Nishiura [11] applied the SPM for (1) with the assumption that Ω = (0, 1) and the boundary condition is the Neumann condition. They proved that for any N ∈ N, there exists a stable solution which has N transition layers by taking d > 0 small enough. Moreover, Heijster applied the SPM to the threecomponent model and obtained various localized solutions (e.g. [4]). We remark that they considered the case Ω = R in [4].
For another approach, Oshita [12] constructed a solution as a minimizer of the variational problem corresponding to (1) with the assumption Ω is bounded. Oshita especially showed that the minimizer oscillates rapidly as d → 0. Dancer and Yan [6] showed similar results as [12]. We mention that [12] treated the Neumann boundary condition and [6] treated the Dirichlet boundary condition.
In this paper, we apply the sub-supersolution method. We develop the work [5] by Chen, Kung and Morita. In [5], Chen, Kung and Morita considered the case that Ω = R and (1) has three equilibrium points, that is, γ > 1. They constructed a heteroclinic solution connecting the two stable equilibrium points for large γ > 1 by the sub-supersolution method. Moreover, the solution has just one transition layer from one stable equilibrium point to the other stable equilibrium point.
We mention that the solution to (2) with (3) corresponds to a steady state odd solution to (1) with just one transition layer from (−a γ , −a γ /γ) to (a γ , a γ /γ). We employ a supersolution T and a subsolution W defined as follows:
Our proof is based on [3,5], but we remark that γ depends on d in [3,5]. We emphasize that γ can be chosen independently of d in Theorem 1.1. Thus Theorem 1.1 gives us some information about the asymptotic behavior of u d,γ as d → 0. Namely we obtain the following corollary: . Let β be a positive constant such that β < u γ and let y 1 be a point in R + such that u d,γ (y 1 ) = β, where u d,γ is the solution obtained in Theorem 1.1. Then y 1 = O(d 1/2 ) holds.
From this corollary, we can see that u d,γ "transits" from 0 to β on the interval whose width is O(d 1/2 ).
Our method can be applied to the following problem: For the equation, we employ T defined in (4) as a supersolution andW as a subsolution defined as follows: whereū γ is the largest root of f (s) − 1/γ = 0. By the variational approach as (5), for γ > 3 √ 6/2 we can prove the existence ofW . We write the solution to (17) as W d,γ , and we defineσ(d, γ) as follows: whereJ d (ψ) is defined as follows: Moreover, we defineŴ 0,γ as follows: The existence ofŴ 0,γ can be also shown asW d,γ . We additionally mention that for given γ > 3 Lemma 4.5). Moreover, let δ be a small constant and x 1 be a constant such thatŴ 0,γ1 (x 1 ) = β 1 , where β 1 is the positive smaller root of f (s) = 2/(3 √ 6) and γ 1 = 3 √ 6/2 + δ. As Proposition 2, we can obtain the uniform estimate such thatσ(d, γ) >σ δ for all (d, γ) ∈ (0, (l 0 /x 1 ) 2 )×[γ 1 , ∞), where δ is a small positive constant. We later present the precise statement in Proposition 8. We can see the following statement: Theorem 1.2. Let δ be a small positive constant, γ > γ 1 = 3 √ 6/2+δ be a constant such that γσ δ > 1 and µ be a function satisfying (µ1) -(µ4). Then there exists a solution to (16) for any d ∈ (0, (l 0 /x 1 ) 2 ). Moreover, the solution (ū d,γ ,v d,γ ) satisfying the following inequalities: If we employŴ d,γ as a subsolution, then we can obtain almost the same statement by repeating almost the same arguments as Theorem 1.1. However, we can see a slightly better estimate ofū d,γ by employingW d,γ as a subsolution sinceŴ d,γ ≤ W d,γ holds. This paper is arranged as follows. In Section 2, we collect basic lemmas. The proof of Theorem 1.1 is presented in Section 3. We first prove Proposition 1 in Section 3.1 and next prove Proposition 2 and Theorem 1.1 in Section 3.2. The proof of Theorem 1.2 is presented in Section 4. We mainly state the differences between Theorems 1.1 and 1.2. In Appendix, we give the proofs of some auxiliary lemmas.
2. Preliminaries and basic estimates. In this section, we prove Theorem 1.1. We prepare some basic lemmas in this section. This section consists of four parts.
2.1. Existence of T and W , comparison lemma of T and W . We first show the existence of the supersolution T (x) and the subsolution W (x). We mention that the proof is based on [1].
, then there exist minimizers T and W to the minimizing problems (6) and (7). T and W satisfy (4) and (5) respectively.
Proof. We only prove the existence of W (x) since one can prove the existence of T (x) similarly. First, we show that G(0) > G(u γ ) = 0 if γ > 3 √ 6/(2µ 0 ). By a direct calculation, we have With the equation µ 0 (u γ − u 3 γ ) = 1/γ, it follows that We note that and thus we arrive at G(0) > G(u γ ) = 0. Next, we show the existence of the minimizer. It suffices to show that E 2 has a minimizer in X 2 defined as follows; To see this, we assume that E 2 has a minimizer W ∈ X 2 and W is not a nondecreasing function. Then there exists points x 1 and x 2 such that 0 < x 1 < x 2 < ∞ and W (x 1 ) = W (x 2 ) hold. We set ω as follows: From the positivity of G on [0, u γ ), we can see that E 2 (ω) < E 2 (W ). Hence the minimizer W should be a nondecreasing function. Moreover, it implies W ∈ X 2 , where X 2 is defined as (9). Now we shall prove the existence of a minimizer of E 2 in X 2 . Let {u n } n∈N be a minimizing sequence in X 2 . As mentioned above, we may assume that u n ∈ X 2 and u n is a nondecreasing function. It is easy to see that u n L 2 (R+) = (u n −u γ ) L 2 (R+) is uniformly bounded. In addition, we readily see that and (u n − u γ ) → φ strongly in C loc (R + ). We set W = u γ + φ. Then we obtain from the lower semicontinuity of L 2 space and Fatou's lemma.

SUB-SUPERSOLUTION METHOD FOR FHN RD SYSTEM WITH HETEROGENEITY 2447
Now we check that W satisfies W (∞) = u γ . Let ρ be a small constant and s n be a point such that u n (s n ) = u γ − ρ. Then we have It follows that {s n } is uniformly bounded in R + . This implies that there exist a constant s < ∞ and a subsequence {s nj } j∈N such that s nj → s as j → ∞. Hence we can see u(s) = u γ − ρ, which means u(∞) = u γ . Moreover, it is clear that u(0) = 0, and thus we have proved that W is the minimizer of (7). With a standard argument, we readily see that W (x) is a solution to (5).
With attention to the monotonicity of µ, we can prove the existence of T by repeating the same argument.
Next, we prove a comparison lemma, which is necessary for the sub-supersolution method.
where T and W are the solutions to (4) and (5), respectively.
Similarly we have We now derive a contradiction under the assumption that there exists an interval From (20) and (21), we can see that We note that f (t)/t = 1 − t 2 is a decreasing function on [0, 1]. Thus the right hand side is negative from the assumption. We calculate the left hand side: It is easy to check that the left hand side is nonnegative, but it clearly contradicts that the right hand side is negative. Therefore we conclude W ≤ T on R + .
Proof. It suffices to show that W d,γ1 ≤ W d,γ2 on R + if γ 1 < γ 2 . For simplicity, we write W i = W d,γ1 (i = 1, 2). We note that W 1 (x) < W 2 (x) for sufficiently large From the definition, it is clear that x m ≥ 0. We now derive a contradiction under the assumption x m > 0. It is easy to see that On the other hand, W i (i = 1,2) satisfies that Thus we obtain that This implies that Hence there exists x s < x m defined as follows: We readily see that By straightforward calculation, we have However, this contradicts W 1 (x s ) ≥ W 2 (x s ). This leads to the conclusion that In this subsection, we prove the uniform continuity of σ(d, γ) respect to d. To see this, we present Lemmas 2.4 and 2.5. We next show the continuity of W d,γ respect to d. The proofs of Lemmas 2.4 and 2.5 are presented in Appendix.
Lemma 2.5. There exists a positive constant M independent of d and γ such that

SUB-SUPERSOLUTION METHOD FOR FHN RD SYSTEM WITH HETEROGENEITY 2449
From Lemmas 2.4 and 2.5, the continuity of σ(·, γ) can be proved. This lemma is applied for the uniform estimate of σ(d, γ) in Section 3.2.
Lemma 2.6. Let d 0 and γ be constants such that d 0 ∈ (0, l 0 /x 0 ) and γ > 3 Proof. We fix a small constant > 0. Let d 1 and d 2 be constants such that where . We note that we may assume Now we evaluate ξ 2 L 2 (R+) . From the definition of ξ, we easily see Moreover, we calculate as follows from Lemma 2.5: where M is defined in Lemma 2.5. Combining (23) -(25), we obtain By the same argument, we also have As a consequence, we conclude that σ(·, γ) is uniformly continuous.

2.4.
Existence lemma for minimizing Problem (10). The next lemma is applied to show the existence of a minimizer of (10). We mention that the proof is based on [2].
If there exists a function u 0 ∈ H 1 0 (R + ) such that I(u 0 ) < 0, then there exists a minimizer of the minimizing problem Moreover, the minimizer is nonnegative on R + .

TAKASHI KAJIWARA
Proof. For simplicity, we write Y = {u ∈ H 1 0 (R + ) : u L 2 (R+) = 1}. Let {u n } n∈N be a minimizing sequence in Y . From the definition of I(u), we may assume that u n ≥ 0 on R + . Since we can see that u L 2 (R+) is uniformly bounded. Thus there exists a function u ∈ H 1 0 (R + ) such that u n → u weakly in H 1 0 (R + ) and u n → u strongly in C loc (R + ). From the lower semi-continuity of L 2 space, u satisfies We then prove that By direct calculation, Fix r > 0 large enough that sup x>r |a(x)| is sufficiently small. Since the embedding H 1 0 (R) ⊂ L 2 (0, r) is compact, {u n } n has a subsequence which converges to u in L 2 (0, r). Thus we obtain (28). As a consequence, we can see that It is easy to see that u ≡ 0 from the assumption σ < 0. Moreover, if u L 2 (R+) < 1, thenū = u/ u L 2 (R+) ∈ Y satisfies that It clearly contradicts the definition of σ. Therefore, u L 2 (R+) should be equal to 1 and we complete the proof.
3. Proof of Theorem 1.1. In this section, we present the proof of Theorem 1.1. First, we prove Proposition 1 in Section 3.1. Next, we show Proposition 2 in Section 3.2.
3.1. Construction of the solution to (2). We give a method to construct a solution with the sub-supersolution method in this subsection. Namely our goal in this subsection is to show Proposition 1. We mention that the framework to construct a solution is based on [3,5].

SUB-SUPERSOLUTION METHOD FOR FHN RD SYSTEM WITH HETEROGENEITY 2451
Then we prove the next statement, which is the key to construct a solution to (2).
In addition, for any fixed i ∈ N, we define {r ij } j∈N that r ij > R i and r ij → ∞ as j → ∞. Then we can construct a solution {u ij } (i,j)∈N 2 to the equation with a standard argument in the sub-supersolution method (e.g. [7]). Now we prove that for fixed i ∈ N, there exists a subsequence of {u ij } j∈N which converges strongly in C 2 (Ī i ), where I i = (0, R i ). It suffices to show that { u ij C 2 (Ii) } j∈N is uniformly bounded and that u ij is equicontinuous respect to j ∈ N. First, we readily see that u ij L ∞ (Ii) is uniformly bounded since W (x) ≤ u ij (x) ≤ T (x) for all x ∈Ī i . Then it follows that u ij L ∞ (Ii) is also uniformly bounded from (30). We can also derive the boundedness of { u ij L ∞ (Ii) } j∈N with the boundedness of { u ij L ∞ (Ii) } j∈N and { u ij L ∞ (Ii) } j∈N . Thus we show that { u ij C 2 (Ii) } j∈N is uniformly bounded. It is easy to see that {u ij } j∈N is equicontinuous since u ij is a solution to (30) and is uniformly continuous over R + × [0, 1]. As a consequence, there exists a function U i ∈ C 2 (Ī i ) and subsequence {u ij(k) } k∈N ⊂ {u ij } j∈N such that We rewrite the subsequence by {u ij } j∈N . We can also construct a subsequence of {u (i+1),j } j∈N that converges to a function U i+1 ∈ C 2 (Ī i+1 ) strongly in C 2 (Ī i+1 ). We note that U i+1 | Ii = U i . Thus the diagonal sequence {u ii } i∈N converges to a function u ∈ C 2 (R + ) strongly in C 2 loc (R + ). It is clear that u satisfies It suffices to show that u(∞) = a γ to see that u is the solution to (29). We prove by contradiction. Namely we assume that there exists a positive number δ and a sequence {x i } i∈N such thatx i → ∞ (i → ∞) and |u(x i ) − a γ | > δ.
However, there does not exist a positive numbers δ * andx * such that |u(x) − a γ | > δ * for all x >x * .
In fact, if there exist δ * andx * satisfying the above inequality, then there exists δ * * > 0 such that We note that for large x > 0, the following inequalities hold: We now assume that u(x i ) = a γ + δ holds for a fixed i ∈ N. Then we can see that Remark that the above inequality is true for any large x > 0 such that u(x) ≥ a γ +δ. Hence we can see that there exists with attention to (31). This implies that u(x i ) = a γ + δ and hence u does not attain a maximum on (y i−1 , x i ) from the maximum principle, that is, u < a γ + δ on (y i−1 , x i ). Thus u (x i ) > 0 from Hopf's lemma. However, it follow that u(x) is monotone increasing on [x i , ∞) since u (x) > 0 as long as u(x) ≥ a γ + δ. It contradicts the existence of the sequence {x i } i∈N . Similarly we derive contradiction in the case u(x i ) = a γ − δ. As a consequence, we conclude u(x) → a γ as x → ∞. Finally, we prove u −û ∈ H 2 (R + ) ∩ H 1 0 (R). It suffices to show φ = u − a γ ∈ H 2 (R + ) since u(0) = 0. We fix a constant l > 0 large enough that and set L > l. Moreover, define ψ =v − a γ /γ. Then we have

SUB-SUPERSOLUTION METHOD FOR FHN RD SYSTEM WITH HETEROGENEITY 2453
The left hand side is calculated as The first term of the right hand side is computed as where δ is a small positive number. Therefore, we obtain Noting that φ is smooth and uniformly bounded in R, we can see that there exists a constant C > 0 such that φ H 1 (R+) < C. Moreover, it is easy to see φ ∈ L 2 (R + ) since Therefore we complete the proof.
Lemma 3.1 is on the method of construction of u. The next lemma is on the method of construction of v. Lemma 3.2. Let u be a function such that u −û ∈ H 2 (R + ) ∩ H 1 0 (R + ) and W (x) ≤ u(x) ≤ T (x) for all x ∈ R + . Then there exists a unique solution v to the equation Let ψ be a solution to the following equation: . With a standard argument, we can see that the solution is unique. Moreover, we can also see that for all x ∈ R + from the maximum principle. Thus v =v + ψ is the desired solution.
With these lemmas, we now prove Proposition 1.

TAKASHI KAJIWARA
Proof of Proposition 1. Define v 0 =v and (u j , v j ) (j = 1, 2, . . .) as follows: The existence of (u j , v j ) is guaranteed by Lemmas 3.1 and 3.2. We define (φ j , ψ j ) = (u j −û, v j −v) and we now show that {(φ j , ψ j )} j∈N is a Cauchy sequence in Then we have where (ξ j , η j ) = (φ j+1 − φ j , ψ j+1 − ψ j ). Integrate (32) over R + : It is easy to check that the left hand side and the right hand side are respectively evaluated as follows: Thus we obtain where κ = 1/(σ(d, γ)γ) < 1. We readily see Moreover, from (32) we have Similarly we can see

SUB-SUPERSOLUTION METHOD FOR FHN RD SYSTEM WITH HETEROGENEITY 2455
As a consequence, we can see that (φ j , ψ j ) is a Cauchy sequence in (H 2 (R + ) ∩ and (U, V ) is a solution to (2).

3.2.
Evaluation of σ(d, γ) from below. In this subsection, we give a proof of Proposition 2 and Theorem 1.1. To prove Proposition 2, it is important to show the positivity of σ(d, γ) and the uniform estimate of σ(d, γ) respect to small d > 0 from below. For reader's convenience, we post (10) again: where J d (ψ) is defined as (11). Throughout this section, W d,γ denotes the solution to (5). Moreover, we remark that from Lemma 2.3, it suffices for Proposition 2 to show the following proposition: Proposition 3. Let γ be a constant such that γ > 3 √ 6/(2µ 0 ). Then there exists a positive constant σ γ such that The next lemma is the key lemma to show Proposition 2. Before we state the lemma, recall x 0 is a constant such thatW 0,γ ( is a nondecreasing function respect to x and also an increasing function respect to γ (see Lemma 2.3).
Proof. We prove by contradiction. Namely we assume σ(d, γ) ≤ 0. Definẽ We note thatṼ (x) → 0 as x → ∞. Thus from Lemma 2.7, we find that the following minimizing problem Moreover, ξ 0 is also a minimizer of (10) since where J(ξ) is defined as (11). It is clear that the minimizer ξ 0 is a solution to Multiplying the above equation by W d,γ and integrating over R + , we have We note that and thus we obtain On the other hand, by differentiating (5), we obtain Moreover, multiplying the above equation by ξ 0 and integrating over R + , we can see that Subtracting (34) from (33), we obtain From the assumption, the left hand side is non-positive. However, since 3W d,γ (x) 2 − 1 > 0 holds for all x > l 0 from the assumption d < (l 0 /x γ ) 2 , the right hand side is positive. Therefore we conclude that σ(d, γ) > 0.
Proof of Proposition 5. We prove by contradiction. Namely we assume that there exists a sequence {d n } n∈N such that d n → 0 and σ n → 0 as n → ∞, where σ n = σ(d n , γ). With a similar argument in Proposition 4, we can check that the minimizing problem (10) with d = d n has a minimizer ξ n ∈ H 1 0 (R + ). With a standard argument, we find that . We note that ξ n L 2 (R+) = 1. Then it is easy to see Multiplying the equation byξ n and integrating over R + , we obtain the following equation: We readily find that {sup n |σ n − V n ( √ d n x)|} n is uniformly bounded from the assumption on σ n . It follows that { ξ n H 1 0 (R+) } n is also uniformly bounded. Thus there existsξ 0 ∈ H 1 0 (R + ) such thatξ n →ξ 0 weakly in Hence we set r > 0 large enough, and then we have for any large n ∈ N and any x > r. Now we set K = µ 0 (3u 2 γ − 1)/2 and define φ as follows: We find thatξ for all x > r.
Proof of Proposition 6. The statement follows from the uniform continuity of σ(·, γ) on [d 0 , l 0 /x 0 ] which has been already proved in Lemma 2.6. Indeed, there exists a γ from the continuity of σ(·, γ). The positivity of σ (2) γ is followed from Proposition 4. Proposition 3 follows from Propositions 5 and 6. Moreover, Proposition 1 follows from Proposition 3 as mentioned in the beginning in this subsection. Combining Propositions 1 and 2, we easily prove Theorem 1.1.
Proof of Theorem 1.1. It is clear that γσ(d, γ) > 1 holds for any d ∈ (0, (l 0 /x 0 ) 2 ) from the definition of σ δ . Thus it follows that there exists a solution to (2) for any d ∈ (0, (l 0 /x 0 ) 2 ) from Proposition 1. Moreover, it is easy to see that the solution (u d,γ , v d,γ ) satisfies (13) and (14) Proposition 8. Let δ be a small positive constant and µ be a function satisfying (µ1) -(µ4). Then there exists a positive constantσ δ such that Proposition 7 and Proposition 8 corresponds to Proposition 1 and Proposition 2 respectively. We shall prove these propositions with the similar argument in Section 2. Now we collect lemmas to show the propositions.
The next lemma corresponds to Lemma 2.2. We omit the proof. As in Lemma 2.3, we can see the monotonicity ofσ(d, γ) respect to γ as follows. We present the proof in Appendix. We remark that we employ a different strategy to show Lemma 4.3. The strategy is based on [9].
We shall show the behavior ofW d,γ in Lemmas 4.4 -4.7. The next lemma corresponds to Lemma 2.4. We omit the proof.
Here we assume that there exists x 2 > 0 such thatŴ d,γ (x 2 ) =W d,γ (x 2 ) and W d,γ (x) <W d,γ (x) for all x ∈ (0, x 2 ). Then we haveW d,γ (x 2 ) ≤Ŵ d,γ (x 2 ). On the other hand, it is easy to see that From Lemma 4.5, we immediately obtain the following statement. We omit the proof.
The next lemma corresponds to Lemma 2.5. We omit the proof. , ∞ .
Proof of Proposition 11. We can prove the statement similarly as Proposition 6. Theorem 1.2 is followed from Propositions 7 and 8 and we omit the proof. For simplicity W 1 and W 2 denotes W d1,γ and W d2,γ , respectively. W 1 and W 2 satisfy Thus we obtain . Multiplying by(W 1 − W 2 ) and integrating over R + , we can see that Thus we have d 1 We shall show We prove by contradiction. We set Γ = {x > L : W 1 (x) > W 2 (x)} and assume Γ = ∅. Let g i andG i (i = 1, 2) be as follows: From the definition, it follows that g 1 (s) > g 2 (s) andG 1 (s) >G 2 (s) hold for all s > 0. Moreover, we note that We defineĒ i,L (u) (i = 1, 2) and X L as follows: We recall that W i is a minimizer ofĒ i (u) in X, whereĒ i denotesĒ 2 defined as (36) with γ = γ i (i = 1, 2). This implies that W i is a minimizer ofĒ i,L (u) in X L . Moreover, we defineẼ i,L (u) = L 0 d 2 |u | 2 + µ(x)G i (u) dx, and then W i is also a minimizer ofẼ i,L (u) in X L because of (42). Set φ(x) = (W 1 (x) − W 2 (x)) + . Then φ ∈ H 1 0 (0, L) and φ ≡ 0. In addition, the following inequalities hold: 0 ≤Ẽ 2,L (W 2 + φ) −Ẽ 2,L (W 2 ).