Multiple existence of indefinite nonlinear diffusion problem in population genetics

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Multiple existence of indefinite nonlinear diffusion problem in population genetics Kimie Nakashima Tokyo University of Marine Science and Technology, 4-5-7 Kounan, Minato-ku, Tokyo, 108-8477, Japan Received 17 April 2019; revised 19 November 2019; accepted 20 November 2019

Abstract We study the following Neumann problem in one dimension. ⎧ ⎨ ut = duxx + h(x)u2 (1 − u) in (−1, 1) × (0, ∞), 0 ≤ u ≤ 1 in (−1, 1) × (0, ∞), ⎩  u (−1, t) = u (1, t) = 0 in (0, ∞), where h changes sign in (−1, 1) and is symmetric, i.e. h(−x) = h(x) and d is a positive small parameter. In [9], we showed that this equation has a nontrivial steady state which is unique under the condition 1 −1 h(x) dx ≥ 0. On the other hand, inthis paper, we will prove that this equation has at least 8 nontrivial 1 steady-states for some h(x) satisfying −1 h(x) dx < 0. This is a counterexample to a conjecture by Lou and Nagylaki [2]. © 2019 Elsevier Inc. All rights reserved. Keywords: Reaction Diffusion equation; Singular perturbation; Layers

1. Introduction In this paper we continue to study the “complete dominant” case of a migration-selection model for the solution of gene frequency with two alleles A1 , A2 initiated in [6] and [4]. This E-mail address: [email protected]. https://doi.org/10.1016/j.jde.2019.11.082 0022-0396/© 2019 Elsevier Inc. All rights reserved.

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model is due to T. Nagylaki in 1975 [5]. We shall give a brief description of this model following the more recent presentation in [3] and [6]. Let u(x, t) be the frequency of allele A1 at time t and location x (thus 0 ≤ u ≤ 1), where h(x) reflects the “environmental variation” and 0 ≤ s ≤ 1 specifies the degree of dominance. Under some additional simplification assumptions, the migration-selection model describing the evolution of gene frequencies at a single locus with two alleles takes the following form 

where  =

ut = du + h(x)u(1 − u)[su + (1 − s)(1 − u)] in  × (0, ∞), on ∂ × (0, ∞), ∂ν u = 0

n

∂2 i=1 ∂x 2 i in Rn , ν

(1.1)

is the Laplace operator, the habitat  is a bounded domain with smooth

denotes the unit outward normal to ∂ and ∂ν is the normal derivative on boundary ∂ ∂. It is clear that (1.1) has no nontrivial steady-states if h(x) does not change sign in , i.e. in this case a steady state u is either u ≡ 0 or u ≡ 1, implying that only one allele survives eventually. Thus, in order to sustain both alleles A1 and A2 , the environmental variation has to be so significant that the selection reverses its direction at least once in , i.e. h(x) changes sign at least once in . Therefore we shall require in the rest of this paper that h(x) changes sign in . As was explained in [6] that all previous mathematical results in literature obtained prior to 2010 deal with the case 0 < s < 1, and the “completely dominant” case s = 1 was left untouched until the publications of the papers [6] and [4] in 2010. In the case s = 1, we say that A2 is completely dominant to A1 (the case s = 0 is similar). In the “completely” dominant case s = 1, (1.1) becomes 

ut = du + h(x)u2 (1 − u) in  × (0, ∞), on ∂ × (0, ∞), ∂ν u = 0

(1.2)

with 0 ≤ u ≤ 1. This “completely dominant” case is not only mathematically challenging but also biologically important. In fact the following conjecture (a) to (c) has existed for a long time (See Louand Nagylaki [2]): (a) If  h(x) dx = 0, then for every d > 0, problem (1.2) has unique nontrivial steady state which is globally asymptotically stable. (b) If  h(x) dx > 0, then there exists d0 > 0 such that for every d ∈ (0, d0 ), problem (1.2) has a unique  nontrivial steady state which is globally asymptotically stable. (c) If  h(x) dx < 0, then there exists d0 > 0 such that for every d ∈ (0, d0 ), problem (1.2) has exactly two nontrivial steady states, one is asymptotically stable and the other is unstable. There are some rigorous results towards the resolution of this conjecture in [6] and [4]. In [6], the existence of a nontrivial stable steady state as well as its limiting behaviors (as d tends to 0 or ∞) are obtained. Furthermore, in [4] the existence of at least two nontrivial steady states, one stable and the other unstable, is established for case (c) as well. For the uniqueness, [8] and [9] has obtained the uniqueness part of (a) and (b) above, under the condition where the spatial dimension n = 1 with some additional condition on h(x). The main purpose of this paper is to show that Part (c) of the conjecture above is false. In fact, we will provide an h(x) such that

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du + h(x)u2 (1 − u) = 0 in (−1, 1), u (−1) = u (1) = 0,

3

(1.3)

has more than three nontrivial solutions. We outline our proof as follows. First, let p(x) ∈ C 2 [0, 1] be a function satisfying (P1) There exists x0 ∈ (0, 1) such that p(x) < 0 in [0, x0 ), p(x) > 0 in (x0 , 1], and p  (x0 ) > 0, (P2) p  (0) = p  (1) = 0, 1 (P3) p(x) dx < 0. 0

Now for  ≥ 0, we define ⎧  ⎪ ⎨ p(−( + 1)x − ) −1 ≤ x ≤ − +1 ,   − +1 ≤ x ≤ +1 , pex (x; ) = p(0) ⎪  ⎩ p(( + 1)x − ) ≤ x ≤ 1. +1

(1.4)

Note that pex (x; ) is symmetric with respect to x = 0 and pex (x; ) has two zeros −z0 () and 0 z0 (), where z0 () = +x +1 . Finally, set h(x) = pex (x; ) in (1.3). As is mentioned above, [6] [8] [9] have established the existence of a linearly stable solution of (1.3), denoted by Ud (x). Moreover, Ud (x) satisfies the following: (i) Ud → 1 uniformly in any compact set in (−1, −z0 ()) ∪ (z0 (), 1) as d → 0. (ii) Ud → 0 uniformly in any compact set in (−z0 (), z0 ()) as d → 0. On the other hand, by [4] and [8], there exists at least one solution ωd (x) of (1.3) with h(x) = pex (x; ) satisfying ωd → 0 uniformly in any compact set in (−1, 1) as d → 0. In addition to Ud (x) and ωd (x), the following theorems show that many other nontrivial solutions of (1.3) exist for some . Theorem 1.1. Let p(x) ∈ C 1 [0, 1] be any function satisfying (P1) - (P3), and set h(x) = pex (x; ), where pex (x; ) is given by (1.4). Then there exists ∗ (> 1) such that for  = ∗ and d sufficiently small, (1.3) has a solution ud (x) satisfying (i) and (ii) below: (i) On every compact subset S of [−1, z0 ()), C1 d < ud (x) < C2 d. (ii) On every compact subset S of (z0 (), 1],

C4 C3 exp − √ d



 C6 < 1 − ud (x) < C5 exp − √ , d

where the constants C1 − C6 depend on S. Once we find a solution ud , satisfying (i) and (ii) in Theorem 1.1, we are able to find many solutions as we will show in the following Theorems 1.2 and 1.3. Let h(x) ∈ C 1 [−1, 1] satisfy (H) If h(x) = 0, then it holds that h (x) = 0. Let ∗ be the constant in Theorem 1.1. The next theorem shows the existence of two different solutions, both of which have only one layer.

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Theorem 1.2. Suppose that h(x) ∈ C 1 [−1, 1] satisfies (H) and h(x) < pex (x; ∗ ) in [−1, 1]. We also suppose h(x) has exactly two zeros z1 , z2 such that h(z1 ) = h(z2 ) = 0 and −1 < z1 < −z0 () < z0 () < z2 < 1. Then for sufficiently small d, (1.3) has 4 solutions ui (x) (i = 1, 2, 3, 4) such that ui (x) (i = 1, 2) satisfy (i) and (ii), ui (x) (i = 3, 4) satisfy (iii) and (iv) below. (i) On every compact subset S of [−1, z2 ), C1 d < ui (x) < C2 d. (ii) On every compact subset S of (z2 , 1],

C4 C3 exp − √ d



 C6 < 1 − ui (x) < C5 exp − √ , d

(iii) On every compact subset S of (z1 , 1], C1 d < ui (x) < C2 d. (iv) On every compact subset S of [−1, z1 ),



  C C C3 exp − √4 < 1 − ui (x) < C5 exp − √6 , d d Here the constants C1 − C6 , C1 − C6 depend on S. Moreover, u1 (x) < u2 (x) and u3 (x) < u4 (x) hold. u1 (x) and u3 (x) are stable. The next theorem is for the existence of solutions close to 0 when h(x) is symmetric. Theorem 1.3. Suppose that h(x) ∈ C 1 [−1, 1] satisfies (H), h(x) < pex (x; ∗ ), and that h(x) has exactly two zeros and satisfies h(−x) = h(x) in [−1, 1]. Then there exist C1 , C2 such that for sufficiently small d, (1.3) has at least three solutions wi (x) (i = 1, 2, 3) satisfying C1 d < wi (x) < C2 d in [−1, 1]. Suppose that h(x) satisfies the same assumption as that of Theorem 1.3. By Theorem 1.2, we obtain 4 solutions. Since h(x) is symmetric, 4 solutions are described as u1 (x), u1 (−x), u2 (x), u2 (−x). All of them have only one layer. On the other hand, Theorem 1.3 shows the existence of 3 solutions which stay close to 0. With Ud (x), the equation (1.3) has at least 8 nontrivial solutions under the condition. [1] also obtained the existence of at least 8 solution for the same model with a different way of parameterizing h(x). Although their high multiplicity result in [1] does not imply the one in this paper and they did not investigate the estimates of the convergence rates as d → 0 as in this paper. This paper is organized as follows. In Section 2, we will show the idea of the proof of Theorem 1.1. The rigorous proof is very long and given in Sections 3–5. We will explain the role of each sections in Section 2. The proofs of Theorems 1.2 and 1.3 are given in Section 6.

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2. Existence of the third solution In this section we will outline the idea of the proof of Theorem 1.1, which shows the existence of a third solution ud in addition to Ud and wd . We will first introduce Lemma 2.1 below, which is equivalent to Theorem 1.1. Setting ξ = ( + 1)x − ,

U (ξ ) = u(x),

k = 2 (k ≥ 0),

3 = ( + 1)2 d,

(2.1)

the equation (1.3) with a condition h(x) = pex (x; ) is rewritten as 

3 Uξ ξ + gex (ξ ; k)U 2 (1 − U ) = 0 U  (−k − 1) = U  (1) = 0,

in (−k − 1, 1),

(2.2)

with k ≥ 0, where ⎧ ⎨ p(−ξ − k) −k − 1 ≤ ξ ≤ −k, −k ≤ ξ ≤ 0, gex (ξ ; k) = p(0) ⎩ p(ξ ) 0 ≤ ξ ≤ 1.

(2.3)

Then Theorem 1.1 is rewritten as the following lemma. Lemma 2.1. For sufficiently small, (2.2) has a nontrivial solution U (ξ ) satisfying (i) On every compact subset S of [−k − 1, x0 ), C1 3 < U (ξ ) < C2 3 , (ii) On every compact subset S of (x0 , 1], 3

3

C3 exp(−C4 − 2 ) < 1 − U (ξ ) < C5 exp(−C6 − 2 ). The constants C1 − C6 depend on S. The idea to prove this lemma is as follows. We first consider auxiliary problem of (2.2). 

3 u + g(x; k)u2 (1 − u) = 0 in (−k, 1), u (−k) = u (1) = 0

(E)k

with k ≥ 0. Here g(x; k) is defined as  g(x; k) =

p(0) p(x)

−k ≤ x ≤ 0, 0 ≤ x ≤ 1.

Especially, g(x; 0) = p(x). Note that gex (x; k) = g(x; k) in

[−k, 1],

gex (x; k) = p(−x − k) in [−k − 1, −k].

(2.4)

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In the next three sections we find a solution of (E)k with one layer, which we denote by u (x; k). We will also find a solution of (E)0 , which is close to 0 in [0, 1], denoted by w (x). Suppose there exists k ∗ > 0 such that w (0) = u (−k ∗ ; k ∗ ).

(2.5)

Then we can define Uex (ξ ; k ∗ ) =



w (−ξ − k ∗ ) −k ∗ − 1 ≤ ξ ≤ −k ∗ , −k ∗ ≤ ξ ≤ 1, u (ξ ; k ∗ )

and show that Uex (ξ ; k ∗ ) is a solution satisfying (i) and (ii) in Lemma 2.1. Since both w (x) and u (x; k) satisfy Neumann zero boundary condition, the standard regularity argument shows that Uex (ξ ; k ∗ ) is smooth and is a solution of (2.2) in [−k ∗ − 1, 1]. The later sections 3–5 are devoted to proving (2.5). We will obtain properties of a solution of (E)k with one layer in Sections 3 and 4. In Section 3, we will construct an upper and a lower solution of (E)k . In Section 4, we will show the linearized stability of one layered solution of (E)k . These properties are necessary to prove Lemma 5.5 in Section 5. The proof of Lemma 2.1 will be provided in Section 5. 3. Construction of upper and lower solutions In this section we will first construct a lower solution of (E)k with a transition layer of width near x0 . Letting φ be the unique solution of (cf. [6] Appendix) 

φ  + zφ 2 (1 − φ) = 0 φ(−∞) = 0, φ(∞) = 1,

in

(−∞, ∞),

(3.1)

we have the following properties of φ, whose proofs are given in Appendix A in [8]. Lemma 3.1. φ is monotone increasing in (−∞, ∞), and there exist positive constants Ci , i = 1, · · · , 6, λj , j = 1, 2, 3, and R such that the following hold: 3

3

1 − C1 exp(−λ1 z 2 ) < φ(z) < 1 − C2 exp(−λ2 z 2 ) 3 2

φ  (z) < C3 exp(−λ3 z ) −

for

for z > R,

C4 C5 < φ(z) < − 3 for z < −R, 3 z z C 6 φ  (z) < 4 for z < −R. z

z > R,

(3.2) (3.3) (3.4) (3.5)

Let L > max{R, 1} be a large constant to be chosen later, and we define two C 1 functions as follows: η(z) =

|z|2 , 1 + |z|

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⎧ 3 ⎪ ⎨ exp(−λ2 (L + 1) 2 )η(z − L), z ≥ L, −L ≤ z ≤ L, θ (z) = 0, ⎪ ⎩ κ η(z + L), z ≤ −L, L4 where λ2 is the constant in (3.2), κ is a constant satisfying κ > C5 and C5 is the constant in (3.4). We begin our construction of a lower solution near x0 . Define



 B(x − x0 ) B(x − x0 ) 4 4 u(x) = φ −L −θ −L 1

where B = (p  (x0 )) 3 . Here p(x) is a C 2 function and satisfies (P1) - (P3) as in Introduction. Let ξ1 , ξ2 satisfy u(ξ1 ) = 0, u (ξ2 ) = 0,

(−L + L4 ), B ξ2 ≥ x0 + (L + L4 ). B

ξ 1 ≤ x0 +

(3.6) (3.7)

We will first show that both ξ1 and ξ2 are uniquely determined for all large L. Setting B (x − x0 ) − L4

(3.8)

u(x) = φ(z) − θ (z).

(3.9)

z= 1

with B = (p  (x0 )) 3 , we have

Note that

 κ 1 θ (z) = − 4 1 − ≤ 0 for L (1 + |z + L|)2 

z ≤ −L.

This shows u (x) =

B  (φ (z) − θ  (z)) ≥ 0

for z ≤ −L, which is equiv. to x ≤ x0 + B (−L + L4 ). This implies that u(x) is monotone increasing on this interval. Since it holds that

u x0 + (−L + L4 ) = φ(−L) − θ (−L) = φ(−L) > 0, B and

C5 κ L2 u x0 + (−2L + L4 ) = φ(−2L) − θ (−2L) < − 4 < 0, 3 B 8L L 1+L there exists

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ξ1 ∈ x0 + (−2L + L4 ), x0 + (−L + L4 ) B B

(3.10)

s.t. u(ξ1 ) = 0 and ξ1 is unique. For ξ2 , we remark that 3

θ  (z) = exp(−λ2 (L + 1) 2 )

2 ≥ 0, (1 + |z + L|)3

and φ  (x) ≤ 0 holds for z ≥ L. Therefore it holds that u (x) =

B 2  (φ (z) − θ  (z)) ≤ 0 2

for z ≥ L, which is equiv. to x ≥ x0 + B (L + L4 ). This shows u (x) is monotone decreasing on this interval. On the other hand, it holds that

B u x0 + (L + L4 ) = φ  (L) > 0. B If we remark that P = 2 Therefore,



λ2 λ3

2

3

√ 3 > 2, where λ2 > λ3 are in Lemma 3.1, we obtain λ3 P 2 = 2 2λ2 .

B u x0 + (P L + L4 ) = (φ  (P L) − θ  (P L))

B  3 3 1 B C3 exp(−λ3 (P L) 2 ) − exp(−λ2 (L + 1) 2 ) 1 − ≤ 2 

((P − 1)L + 1)  √ 3 3 B 1 = C3 exp(−2 2λ2 L 2 ) − exp(−λ2 (L + 1) 2 ) 1 − ((P − 1)L + 1)2   √ 3 3 3 B 1 = exp(−λ2 (L + 1) 2 ) C3 exp(−2 2λ2 L 2 + λ2 (L + 1) 2 ) − 1 + ((P − 1)L + 1)2   

3  √ 3 3 L+1 2 1 B )−1+ = exp(−λ2 (L + 1) 2 ) C3 exp(−λ2 L 2 2 2 − L ((P − 1)L + 1)2 < 0, for L large. We obtain

ξ2 ∈ x0 + (L + L4 ), x0 + (P L + L4 ) B B

(3.11)

s.t. u (ξ2 ) = 0 and ξ2 is unique. Moreover, since φ is monotone increasing, we have the estimate 3

u(ξ2 ) < φ(ξ2 ) < φ(P L) < 1 − C2 exp(−λ2 (P L) 2 ), 3 u(ξ2 ) > u(x0 + B (L + L4 )) = φ(L) > 1 − C1 exp(−λ1 L 2 ). A crucial step in our construction of a lower-solution is the following lemma. Set

(3.12)

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(u) ≡ 3 u + p(x)f (u)

and

f (u) = u2 (1 − u).

9

(3.13)

Lemma 3.2. For any large L > 0 there exists 0 > 0 such that (u(x)) > 0 holds on the interval (ξ1 , ξ2 ) for < 0 . Proof. Note that g(x; k) = p(x) holds on the interval (ξ1 , ξ2 ) and that p(x) is a C 2 function satisfying (P1) - (P3). By Taylor’s expansion, there exists x˜ between x0 and x = x0 + B (z + L4 ) and φ˜ ∈ (φ − θ, φ) such that 1 2 p(x) = p  (x0 ) (z + L4 ) + p  (x) ˜ 2 (z + L4 )2 , B 2 B 1 ˜ 2. f (φ − θ ) = f (φ) − f  (φ)θ + f  (φ)θ 2 Therefore, (u) = 3 uxx + p(x)f (u) = B 2 (φzz − θzz ) +

 1 ˜ 2) p (x0 )z(f (φ) − f  (φ)θ + f  (φ)θ B 2

1 2  p (x)(z ˜ + L4 )2 )f (φ − θ ) + ( p  (x0 )L4 + B 2 B2 1 ˜ 2 ) + B 2 L4 f (φ − θ ) = B 2 (−θzz − zf  (φ)θ + zf  (φ)θ 2 +

1 2  p (x)(z ˜ + L4 )2 f (φ − θ ) 2 B2

Since z ∈ (−2L, P L) by (3.10) and (3.11), we have B 2 L4 +

1  1 p (x)(z ˜ + L4 )2 > B 2 L4 2 B2 2

for > 0 suff. small. Therefore, 1 1 ˜ 2 + L4 f (φ − θ )). (u) ≥ B 2 (−θzz − zf  (φ)θ + zf  (φ)θ 2 2

(3.14)

The rest of the proof  is devided into three steps.  Step 1. For x ∈ x0 + B (−L + L4 ), x0 + B (L + L4 ) , we have θ = 0. Thus (3.14) implies that 1 (u) ≥ B 2 L4 f (φ) > 0. 2   Step 2. For x ∈ ξ1 , x0 + B (−L + L4 ) , we have φ − θ ≥ 0, and −2L < z < −L (by (3.10)). The estimate (3.4) shows

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1 1 ˜ = (2φ − 3φ 2 ) − (2 − 6φ)θ ˜ f  (φ) − f  (φ)θ 2 2 ˜ = (φ − 3φ 2 ) + (φ − θ ) + 3φθ ≥ φ − 3φ 2 ≥ if L is large enough s.t. L >

2C2 C1

on the constants in Lemma 3.1 by

13

C1 L3

−

C2 L6

≥

(3.15)

C1 , 2L3

. Hereafter, we will denote those constants depending only

Cj ,

j = 1, 2, · · · . Since −z > L, (3.14) now gives

C (u) ≥ B 2 (−θzz − 13 zθ ) 2L 

C1 (z + L)2 B 2 κ 2 ≥ + ≥0 − L4 (|z + L| + 1)3 2L2 |z + L| + 1 if (z + L)2 (|z + L| + 1)2 >  which is guaranteed to hold for x ∈ ξ1 , x0 +

2 3 B (−L

4L2 , C1

3 2 − L + L4 ) and L > C4 , as |z + L| > 1

2

L 3 for all x in this interval.  For the remaining part, i.e. for x ∈ x0 +

2 3 B (−L

− L + L4 ), x0 +

4 B (−L + L )

C C3 C κ L 3 > 33 − 44 2 3 L L L 3 + 1 2L

, we have

4

φ(z) − θ (z) ≥

(3.16)

if L is sufficiently large. On the other hand, φ(z) − θ (z) ≤ φ(z) ≤ − we choose L large enough so that

C5 C5 ≤ 3, z3 L

C5 2 2 < . Since f is monotone increasing in (0, ), we have L3 3 3 f (φ − θ ) > f (

C5 C3 ) > 2L3 L6

in view of (3.16). From (3.14) and (3.15) we see that 1 (u) ≥ B 2 (−θzz + L4 f (φ − θ )) 2

 2 L4 C5 κ 2 ≥ B − 4 + ≥0 L (|z + L| + 1)3 2 L6 if L is sufficiently large.

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 Step 3. For x ∈ x0 + 3

4 B (L + L ),

11

 3 ξ2 : First, we choose L large such that exp(−λ2 L 2 ) ≤

exp(−λ1 L 2 ) 1, then φ is close to 1 and θ is close to 0 by Lemma 3.1, and therefore 1 1 1 ˜ = −(2φ − 3φ 2 ) + (2 − 6φ)θ ˜ > . −f  (φ) + f  (φ)θ 2 2 2 From (3.14) it follows that 1 1 (u) ≥ B 2 (−θzz + zθ + L4 f (φ − θ )). 2 2   For x ∈ x0 + B (L + L4 ), x0 + B (L + 1 + L4 ) , we have

(3.17)

3

f (φ − θ ) ≥ f (φ) ≥ C6 exp(−λ2 z 2 ) since f is decreasing near 1. Note that zθ ≥ 0 and 3

−θ  = − exp(−λ2 (L + 1) 2 )

2 < 0. (|z − L| + 1)3

Substituting these into (3.17) we obtain 

3 3 L4  (u) ≥ B 2 −2 exp(−λ2 (L + 1) 2 ) + C6 exp(−λ2 z 2 ) > 0. 2  Finally, for x ∈ x0 +

4 B (L + 1 + L ),

 ξ2 , again from (3.17) we deduce

 3 (u) ≥ B 2 exp(−λ2 (L + 1) 2 ) −

2 z (z − L)2 + 3 2 (1 + |z − L|) (1 + |z − L|)



 B 2 exp(−λ2 (L + 1) 2 )  z 2 2 = (1 + |z − L|) (z − L) −2 + , (1 + |z − L|)3 2 3

Since L4 f (φ − θ ) > 0. On this interval z ≥ L + 1, thus z(z − L)2 (1 + |z − L|)2 ≥ 4(L + 1), which implies that (u) > 0 for L large. This completes our proof of Lemma 3.2.

2

Lemma 3.2 tells us that u is a lower solution for (E)k in the interval (ξ1 , ξ2 ). We now extend it to the entire interval (−k, 1). Set u(ξ2 ) = 1 − α1 and ⎧ ⎨0 u∗ (x; k) = u(x) ⎩ 1 − α1 See (3.12). α1 is close to 0 for L suff. large.

if −k ≤ x ≤ ξ1 , if ξ1 ≤ x ≤ ξ2 , if ξ2 ≤ x ≤ 1.

(3.18)

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Proposition 3.3. For any large L > 0, there exists 0 > 0 such that u∗ (x; k) is a lower solution for (E)k for any < 0 and any k ≥ 0. Proof. First, observe that u∗ (x; k) ∈ C 0 ([−k, 1]) and (u∗ (x; k)) > 0 on [ξ1 , ξ2 ]. Note that (i) (0) = 0, i.e. 0 is a solution for (E)k , and u (ξ1 ) > 0; (ii) (1 − α1 ) = (u(ξ2 )) > 0. i.e. the constant u(ξ2 ) is also a lower solution for (E)k , and u (ξ2 ) = 0. By (i) and (ii), it follows from standard arguments that u∗ is a lower solution for (E)k on [−k, 1]. An upper solution may be constructed in a similar fashion as the lower solution, thus we shall be brief. Let ⎧ 3 C2 ⎪ ⎪ 2 ⎪ ⎨ 2 exp(−λ2 (L + 1) )η(z − L), z ≥ L, −L ≤ z ≤ L, (3.19) θˆ (z) = 0, ⎪ ⎪ κ ˆ ⎪ ⎩ η(z + L), z ≤ −L, L4 and



 B(x − x0 ) B(x − x0 ) 4 4 ˆ u(x) ¯ =φ +L +θ +L .

(3.20)

Similarly, let ξ3 and ξ4 satisfy u¯  (ξ3 ) = 0, u(ξ ¯ 4) = 1

(−L − L4 ), B ξ4 ≥ x0 + (L − L4 ). B

ξ 3 ≤ x0 +

(3.21) (3.22)

Again, we first show that ξ3 and ξ4 are uniquely determined for large L. Setting B (x − x0 ) + L4 ,

(3.23)

u(x) ¯ = φ(z) + θˆ (z).

(3.24)

z= we have

2

Note that z in (3.23) is similar to but different from (3.8). By the fact that u¯  (x) = B 2 (φ  (z) + θˆ  (z)) ≥ 0 for z ≤ −L, u¯  is monotone increasing on this interval. Now we compute, by Lemma 3.1,

B B u¯  x0 + (−L − L4 ) = (φ  (−L) + θˆ  (−L)) = φ  (−L) > 0 B and

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13

B u¯  x0 + (−3L − L4 ) = (φ  (−3L) + θˆ  (−3L)) B  2 

B C6 κˆ κˆ 1 B C6 < ( − 4 1− − )<0 )< ( 4 4 (3L) 1 + 2L (3L) L 2L4 if κˆ > C6 and L > 1. Therefore we obtain

ξ3 ∈ x0 + (−3L − L4 ), x0 + (−L − L4 ) B B

(3.25)

s.t. u¯  (ξ3 ) = 0 and ξ3 is unique. Moreover, by Lemma 3.1 and (3.25) we have u(ξ ¯ 3 ) > φ(ξ3 ) > φ(−3L) >

C4 , (3L)3

C5 ¯ (−L − L4 )) = φ(−L) < 3 . u(ξ ¯ 3 ) < u( B L For ξ4 , we remark that u¯  (x) =

B  ˆ (φ (z) + θ (z)) ≥ 0

ing on this interval. We set Pˆ = 2



λ2 λ1

2

3

(3.26)

for z ≥ L, therefore u¯ is monotone increas-

and observe that Pˆ > 2 by Lemma 3.1. Now,



u¯ x0 + (L − L4 ) = φ(L) + θˆ (L) = φ(L) < 1, B

ˆ u¯ x0 + (P L − L4 ) = φ(Pˆ L) + θˆ (Pˆ L) B 3 3 (Pˆ − 1)2 L2 > 1 − C1 exp(−λ1 (Pˆ L) 2 ) + exp(−λ2 (L + 1) 2 ) >1 1 + (Pˆ − 1)L

if L is sufficiently large. Thus we obtain

ξ4 ∈ x0 + (L − L4 ), x0 + (Pˆ L − L4 ) B B

(3.27)

such that u(ξ ˆ 4 ) = 1 and ξ4 is unique. ¯ < 0 on [ξ3 , ξ4 ]. We proceed in a similar manner as in the lower Next, we will show that (u) solution case. There exist x˜ between x0 and x, and φ˜ ∈ [φ + θˆ ] such that 1 2 p(x) = p(x0 ) + p  (x0 ) (z − L4 ) + p  (x) ˜ 2 (z − L4 )2 , B 2 B 1 ˜ θˆ 2 . f (φ + θˆ ) = f (φ) + f  (φ)θˆ + f  (φ) 2 Substituting into (u), ¯ we obtain

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14

(u) ¯ = B 2 (φzz + θˆzz ) +

 1 ˜ θˆ 2 ) p (x0 )z(f (φ) + f  (φ)θˆ + f  (φ) B 2

1 2  p (x)(z ˜ − L4 )2 ]f (φ + θˆ ) + [− p  (x0 )L4 + B 2 B2   1 1 2 4 2 ˜ θˆ 2 ] + − B 2 L4 + p  (x)(z = B 2 [θˆzz + zf  (φ)θˆ + zf  (φ) ˜ − L ) f (φ + θˆ ). 2 2 B2 Since z ∈ (−3L, Pˆ L), we have − B 2 L4 +

1 2  1 p (x)(z ˜ − L4 )2 < − B 2 L4 . 2 B2 2

for sufficiently small (depending on L). Therefore, 4 1 ˜ θˆ 2 − L f (φ + θˆ )). (u) ¯ ≤ B 2 (θˆzz + zf  (φ)θˆ + zf  (φ) 2 2   Step 1. For x ∈ x0 + B (−L − L4 ), x0 + B (L − L4 ) , we have θˆ = 0. Thus

(3.28)

1 (u) ¯ ≤ − B 2 L4 f (φ) ≤ 0. 2   Step 2. On the interval ξ3 , x0 + B (−L − L4 ) , we have 1 ˜ θˆ f  (φ) + f  (φ) 2

=

˜ θˆ > 0 (2φ − 3φ 2 ) + 12 (2 − 6φ)

by Lemma 3.1. Since z < 0 and f (φ) < f (φ + θˆ ) (as f is increasing near 0), we deduce, from (3.28) 



 2 L4 L4 C6 κˆ (u) ¯ ≤ B 2 θˆzz − − f (φ) ≤ B 2 ≤0 2 L4 (1 + |z|)3 2 L6 for L large, as −3L ≤ z ≤ −L by (3.25).  Step 3. On the interval x ∈ x0 + B (L − L4 ), ξ4 , we have L ≤ z ≤ Pˆ L and φ is close to 1 for L large. Thus, 1 ˜ θˆ = (2φ − 3φ 2 ) + 1 (2 − 6φ) ˜ θˆ < − 1 . f  (φ) + f  (φ) 2 2 2 and it holds that 1 L4 (u) ¯ ≤ B 2 (θˆzz − zθˆ − f (φ + θˆ )). 2 2   For x in the interval x0 + B (L − L4 ), x0 + B (L + 1 − L4 ) , we have

(3.29)

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1 − φ − θˆ ≥

15

3 C2 exp(−λ2 (L + 1) 2 ). 2

3 Therefore f (φ + θˆ ) ≥ C7 exp(−λ2 (L + 1) 2 ) holds. This and (3.29) shows

 C7 L4 3 3 2 C2 (u) ¯ ≤ B − exp(−λ2 (L + 1) 2 ) exp(−λ2 (L + 1) 2 ) 2 (1 + |z|)3 2   C7 L4 3 2 ≤ B C2 − exp(−λ2 (L + 1) 2 ) ≤ 0 2 

2

 for L large. On the interval x0 +

4 B (L + 1 − L ),

 ξ4 , we have z ≥ 1 and thus

2 z z2 − ≤0 3 2 (1 + z) (1 + z) 1 i.e. θˆzz − zθˆ ≤ 0 and (u) ¯ < 0 follows from (3.29). This establishes our assertion that (u) ¯ <0 2 on [ξ3 , ξ4 ]. Set u(ξ ¯ 3 ) = α2 . (3.26) shows that α2 is close to 0 for L suff. large. We define our upper solution u∗ (x; k) as follows: ⎧ if −k ≤ x ≤ ξ3 , ⎨ α2 ¯ if ξ3 ≤ x ≤ ξ4 , u∗ (x; k) = u(x) ⎩ 1 if ξ4 ≤ x ≤ 1.

(3.30)

As in the lower solution case, standard arguments show that u∗ (x; k) is an upper solution. From our definitions of u∗ and u∗ , (3.18) and (3.30), and the estimates for ξ1 and ξ4 , (3.10) and (3.27), it follows immediately that u∗ < u∗ on the interval (−k, 1). The following lemma holds from the above argument. 2 Lemma 3.4. There exists 0 ≥ 0 such that u∗ (x; k) is an upper solution and u∗ (x; k) is a lower solution of (E)k for any k ≥ 0 and any ≤ 0 . 4. Stability In this section we will establish the fact that any solution of (E)k (k ≥ 0) in the set U (k) = {u(x) ∈ C 1 [−k, 1] | u∗ (x; k) ≤ u(x) ≤ u∗ (x; k)},

(4.1)

where u∗ and u∗ are the upper and lower solutions constructed in Section 2, is linearly stable. Lemma 4.1. Let u (x; k) be a solution of (E)k and u (x; k) ∈ U (k). For any large k¯ ≥ 0, there ¯ and < 0 . exists 0 > 0 s.t. u (x; k) is linearly stable for k ∈ [0, k] We first introduce Lemma 4.2 below to prove Lemma 4.1.

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16

Lemma 4.2. Let m(x) ∈ C 1 [−k, 1]. If there exist (ζ1 , ζ2 )(⊂ [−k, 1]) and functions w(x) > 0 and A(x) > 0 satisfying 

3 w  + m(x)w = −A(x)w w  (ζ1 ) = w  (ζ2 ) = 0,

in (ζ1 , ζ2 )

(4.2)

then the minimal eigenvalue of the following (4.3) is positive. 

3 ψ  + m(x)ψ = −λψ ψ  (ζ1 ) = ψ  (ζ2 ) = 0,

in (ζ1 , ζ2 ),

(4.3)

Proof. Let λ1 be the minimal eigenvalue of (4.3). Multiplying (4.2) by ψ and (4.3) by w, integrating by parts and substracting, we have ζ2 0 = (λ1 − A(x))wψ dx. ζ1

Since ψ is an eigenfunction corresponding to the first eigenvalue, we have ψ > 0. Suppose λ1 ≤ 0. Since A(x) > 0, we have a contradiction. We provide the following lemma for monotonicity of a solution of (E)k . 2 Lemma 4.3. Let u (x; k) be a nonconstant solution of (E)k and u (x; k) ∈ U (k). For any k ≥ 0, it holds that u (x; k) is monotone increasing. Proof. Note that u (x; k) = −

g(x; k) u (x; k)2 (1 − u (x; k)). 3

u (x; k) is concave down in (x0 , 1) for any k ≥ 0. This fact and u (1; k) = 0 show that u (x; k) is monotone increasing in (x0 , 1). On the other hand, u (x; k) is concave up in (−k, x0 ) for any k ≥ 0. This and u (−k; k) = 0 show that u (x; k) is monotone increasing in (−k, x0 ). In the proof of Lemma 4.1, we need Propositions 4.4, 4.5 below. Both propositions give estimates of solutions in the neighborhood of x0 . We can obtain Proposition 4.4 by the same proofs as those of Lemma 3.6 and Corollary 3.3 in [8]. 2 Proposition 4.4. Let β > 0 be any sufficiently small constant. there exist 0 < C1 < C2 , 1 > 2 > 0 and 0 > 0 such that



3 3 C1 exp −1 − 2 (x − x0 + β) < 1 − u (x; k) < C2 exp −2 − 2 (x − x0 − β) β (x ∈ [x0 + β, 1 − ]) 2 holds for < 0 and k ≥ 0.

(4.4)

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17

Note that both left and righthand sides of (4.4) do not depend on k. The proof of Proposition 4.5 is given in Appendix. Proposition 4.5 is almost the same as Lemmas 6.3 and 6.5 in [8] and idea of the proofs are very similar. However, we have obtained an improved version of the proof so that the readers can clearly see that the estimate does not depend on k. The proofs are also shorter and clearer than that of Lemmas 6.3 and 6.5 in [8]. Proposition 4.5. For β > 0 suff. small, there exist 0 < C3 < C4 , 0 < M < Q, M˜ > 0, and 0 > 0 such that for < 0 and k ≥ 0, it holds that C3 3 ˜ 3 (x0 − x + M)

< u (x; k) <

C4 3 (x0 − x − M)3

(x ∈ [x0 − β, x0 − Q]).

(4.5)

In the rest of this section we will prove Lemma 4.1 applying Lemma 4.2 with m(x) = p(x)(2u (x; k) − 3u (x; k)2 ), where k is arbitrarily fixed. We first construct A(x; k), w = w(x; k), ζ1 (k), ζ2 (k) in the neighborhood of x0 , (p(x0 ) = 0), 

3 w  + p(x)(2u (x; k) − 3u (x; k)2 )w = −A(x; k)w w  (ζ1 (k); k) = w  (ζ2 (k); k) = 0.

for

ζ1 (k) ≤ x ≤ ζ2 (k)

(4.6)

Hereafter, w  denotes derivative of w(x; k) with respect to x and w  denotes the second derivative of w(x; k) with respect to x. The proof is obtained improving a method to show linearized stability in [7], which deals with Allen-Cahn equation with double well potential. Proof of Lemma 4.1. Choose α (0 < α < 14 (1 − x0 )) suff. small so that there exist γ1 , γ2 > 0 (γ1 < γ2 ) satisfying γ1 ≤ p  (x) ≤ γ2

x ∈ [x0 − α, x0 + α].

for

(4.7)

Choose β, R¯ such that β<

2 α, 101

R¯ > Q,

˜ R¯ > M,

x0 − ξ 3 R¯ > (> 0).

(4.8)

Here ξ3 is in (3.25). Note that ξ3 is in O( ) neighborhood of x0 . Later, we set R¯ suff. large such that the following (4.17), (4.18) and (4.22) hold, if necessary. We set η(z) =

|z|2 . 1 + |z|

Setting δ 2 = 3 , (δ > 0). We have ⎧ D x − x0 ⎪ ¯ ⎪ ⎪ ⎨ η( + R), θ˜ (x) = 0, ⎪ ⎪ x − x0 − 2β ⎪ ⎩ P η( ), δ where

¯ (x ≤ x0 − R ), ¯ < x < x0 + 2β), (x0 − R (x ≥ x0 + 2β),

(4.9)

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 41 β P = exp − . δ

D=

1 . ¯ R7

(4.10)

˜ Note that θ(x) is a C 1 function and p(x) = g(x; k) holds in 0 ≤ x ≤ 1. Set (w) = 3 w  + 2 p(x)(2u − 3u )w. We will find A(x; k) and w(x; k) = u (x; k) + θ˜ (x) satisfying (u + θ˜ ) = 3 (u + θ˜ ) + p(x)(2u − 3u2 )(u + θ˜ ) = −A(x; k)(u + θ˜ ).

(4.11)

Taking derivative of the equation w.r.t. x, where u = u (x; k), we have 3 (u ) + p(x)(2u − 3u2 )(u ) + p  (x)u2 (1 − u ) = 0.

(4.12)

(4.11) and (4.12) shows (u + θ˜ ) = −p  (x)u2 (1 − u ) + 3 θ˜  + p(x)(2u − 3u2 )θ˜ = −A(x; k)(u + θ˜ ).

(4.13)

¯ < x < x0 + 2β, it holds that Step 1. On the interval x0 − R θ˜ (x) = 0

and A(x; k) =

p  (x)u2 (1 − u ) . (u )

¯ set 1 − u ≥ Step 2. On the interval x0 − β ≤ x ≤ x0 − R , β ¯ By (4.5) and R¯ > Q in (4.8), we have R¯ (0 ≤ z ≤ − R). p  (x)u2 (1 − u ) > p(x)(2u − 3u2 ) <

γ1 C0 6

˜ 6 (x0 − x + M)

=

1 2

for R¯ suff. large. −z =

x − x0 +

C1 , ˜ 6 (z + R¯ + M)

¯ C2 (z + R) γ1 C3 3 (x − x0 ) =− . ˜ 3 ˜ 3 (x0 − x + M) (z + R¯ + M)

(4.14) (4.15)

Hereafter, Ci (i ∈ N) is a constant independent of R¯ and > 0. Using (4.14) and (4.15) on (4.13), we have   ¯ C1 C2 (z + R) z2 2  (u + θ˜ ) < − +D − . ˜ 6 ˜ 3 1+z (1 + z)3 (z + R¯ + M) (z + R¯ + M) This and the fact that z + R¯ > M˜ (See (4.8)), which is equiv. to (u + θ˜ ) < −

1 z + R¯ > , show 2 z + R¯ + M˜

 C1 C2 z2 2 +D − . ˜ 6 ˜ 2 1+z (1 + z)3 2(z + R¯ + M) (z + R¯ + M)

(4.16)

¯ ≤ x ≤ x0 − R , ¯ which is equiv. to 0 ≤ z ≤ R, ¯ by (4.10) and (4.16) we On the interval x0 − 2R have

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(u + θ˜ ) < −

C1 C1 2D 2 + < − + < 0. 3 ˜ 6 (1 + z) ˜ 6 R¯ 7 (z + R¯ + M) (2R¯ + M)

¯ which is equiv. to R¯ ≤ z ≤ On the interval x0 − β ≤ x ≤ x0 − 2R , (u

Since

D + θ˜ ) < (1 + z)3

 2−

19

(4.17)

β ¯ by (4.16) we have − R, 

C2 z2 (1 + z)2 ˜ 2 2(z + R¯ + M)

.

(1 + z)z is monotone increasing, for R¯ suff.large, it holds that z + R¯ + M˜ 2−

¯ 2 R¯ 2 C2 (1 + z)2 z2 C  (1 + R) <2− 2 < 0. ˜ 2 ˜ 2 (z + R¯ + M) (2R¯ + M)

(4.18)

¯ Now we have shown in Step 2 that ˜ < 0 for R¯ ≤ z ≤ β − R. Therefore, we have (u + θ) β (u + θ˜ ) < 0 for 0 ≤ z ≤ − R¯ with R¯ satisfying (4.17) and (4.18). We will find ζ1 (k) satisfying w  (ζ1 (k); k) = 0 in the following lemma. 2 Lemma 4.6. For suff. large R¯ > 0, there exists 0 > 0 s.t. w  (ζ1 (k); k) = 0 and ζ1 (k) ∈ [x0 − ¯ x0 − R] ¯ for any ≤ 0 . (R¯ 2 + R), Proof. Recall 3 = δ 2 . Since it holds that w





D  x − x0 ¯ η +R . 2

(4.19)

¯ p(x0 − R) ¯ k)2 (1 − u (x0 − R; ¯ k)) > 0. u (x0 − R; 3

(4.20)

p(x) (x; k) = u (x; k) + θ˜  (x) = − 3 u2 (1 − u ) +

We have ¯ k) = − w  (x0 − R;

Next we will calculate w  (x0 − (R¯ + R¯ 2 ); k). Set −z = obtain −p(x)u2 (1 − u ) ≤

x − x0 ¯ Using (4.7) and (4.5) and + R.

¯ C3 (z + R) 6 C42 γ2 (x0 − x) = . (x0 − x − M)6 (z + R¯ − M)6

(4.21)

It follows from (4.19) and (4.21) that

 ¯ C3 (R¯ 2 + R) 1 1 − 1 − 6 2 (R¯ 2 + R¯ − M) 2 R¯ 7 (1 + R¯ 2 )2    C 1 C4 ≤ 2 − 5 <0 R¯ 10 R¯ 7

w  (x0 − (R¯ + R¯ 2 ); k) ≤

holds for R¯ suff. large. (4.20) and (4.22) shows the lemma.

(4.22)

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20

α 0 Step 3. Note that α < 1−x 4 and β < 10 by (4.8). On the interval x0 + 2β ≤ x ≤ x0 + α, choose ν > 0 suff. small to satisfy ν < β. We also choose δ > 0 small enough such that



 2 (2β − ν) 1 C2 exp − < . δ 6  ν By (4.4) it holds on x0 + 2β, 1 − that 2



  2 (x − x0 − ν) 2 (2β − ν) 5 u (x; k) > 1 − C2 exp − ≥ 1 − C2 exp − > . δ δ 6

(4.23)

(4.7), (4.4) and (4.23) shows −p  (x)u2 (1 − u ) ≤ −

2

 5 1 (x − x0 + ν) γ1 C1 exp − . 6 δ

(4.24)

Since 2u − 3u2 is monotone decreasing with respect to u ≥ 13 , it holds by (4.7) and (4.23) that p(x)(2u − 3u2 ) ≤ γ1 (x



2 5 5 5 − x0 )(2 ) = − γ1 (x − x0 ). −3 6 6 12

(4.25)

x − x0 − 2β Set z = . Note that z here is different from z in Step 2 or the proof of Lemma 4.6. δ By (4.13), (4.24) and (4.25), we have (u + θ˜ ) = −p  (x)u2 (1 − u ) + 3 θ˜  + p(x)(2u − 3u2 )θ˜

2

 

5 1 (x − x0 + ν) 5 <− γ1 C1 exp − + P η − γ1 (x − x0 )η 6 δ 12

2 

 2β + ν 5 z2 5 2 γ1 C1 exp −1 (z + − γ1 (δz + 2β) =− ) +P 6 δ (1 + z)3 12 1+z

On the interval

12 5γ1 β

1

4

≤z≤

α − 2β , we have δ

2 5 z2 5βγ1 z2 2 − γ1 (δz + 2β) − < 3 3 12 1 + z (1 + z) 6 1+z (1 + z)



 2 5βγ1 2 2 5βγ1 4 2 = (1 + z) z z 1 − < 1 − < 0. (1 + z)3 12 (1 + z)3 12 Therefore (u + θ˜ ) < 0 on this interval.

1 12 4 On the interval 0 ≤ z ≤ , by (4.26) we have 5γ1 β (u + θ˜ ) ≤ −

(4.26)

2

 5 2β + ν γ1 C1 exp −1 (z + ) + 2P , 6 δ

(4.27)

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21

 41 β where P = exp − . Since ν < β holds, for δ > 0 suff. small, (u + θ˜ ) < 0 holds on this δ interval. Finally we will choose ζ2 (k) s.t. w  (ζ2 (k); k) = 0 in the following lemma. 2 Lemma 4.7. There exists 0 > 0 such that w  (ζ2 (k); k) = 0 and 41 ζ2 (k) ∈ [x0 + 2β, x0 + (2 + )β] hold for any ≤ 0 . 2 Proof. Recall 3 = δ 2 . It holds that w  (x; k) = u (x; k) + θ˜  (x) = −

p(x) 2 u (x; k)(1 − u (x; k)) + θ˜  (x). δ2

(4.28)

Therefore, by (4.28), w  (x0 + 2β; k) = −p(x0 + 2β)u (x0 + 2β; k)2 (1 − u (x0 + 2β; k)) ≤ 0. Set z =

x − x0 − 2β . By (4.7), (4.4) and (4.28), for z > 1 we have δ

 γ2 (x − x0 ) 2 (x − x0 − ν) w  (x; k) ≥ − exp − + θ˜  (x) δ δ2

 2 (δz + 2β − ν) P 1 γ2 (δz + 2β) exp − ) + (1 − =− δ2 δ δ (1 + z)2



 2 (δz + 2β − ν) 3 41 β γ2 (δz + 2β) exp − + exp − , >− δ2 δ 4δ δ

where we have used 1 − w  (x0 + (2 + 4

1 (1+z)2

>

3 4

(4.29)

(4.30)

for z > 1. (4.30) and the fact that β > ν yields



 1 γ2 41 41 β 2 (2β − ν) 3 41 β )β; k) ≥ − 2 (2 + )β exp − − + exp − 2 δ 2 δ δ 4δ δ



   1 2 (2β − ν) 41 β γ2 1 3 = exp − − (2 + 4 )β exp − + (4.31) δ δ δ 2 δ 4



   2 β 41 β γ2 1 3 1 − (2 + 4 )β exp − + >0 > exp − δ δ δ 2 δ 4

for δ (or ) suff. small, since y exp(−2 y) → 0 as y → ∞. (4.29) and (4.31) shows the lemma. ˜ Now we have found ζ = ζ1 (k), ζ = ζ2 (k) in Lemmas 4.6 and 4.7 satisfying (4.2). Since θ(x) in (ζ1 , ζ2 ) is determined as above, it holds that A(x; k) =

(u + θ˜ ) p  (x)u2 (1 − u ) − 3 θ˜  − p(x)(2u − 3u2 )θ˜ =− . u + θ˜ u + θ˜

In Step 1 – Step 3, we have shown (u + θ˜ ) < 0. Since u + θ˜ > 0, it holds that A(x; k) > 0.

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22

Now we have constructed A(x; k) > 0 and w(x; k) > 0 satisfying (4.2). By Lemma 4.2, the minimal eigenvalue of (4.3), which we denote by λ1 , is positive. Setting  H(w, J ) =

3 |w  |2 − g(x; k)(2u − 3u2 )w 2 dx,

J

we obtain the following from the characterization of the minimal eigenvalue, ζ2 H(w, (ζ1 , ζ2 )) =

 2

|w | 3

ζ2 − p(x)(2u − 3u2 )w 2 dx

≥ λ1

ζ1

w 2 dx

(4.32)

ζ1

for any w ∈ H 1 (ζ1 , ζ2 ). Here we have used g(x; k) = p(x) in (ζ1 , ζ2 ). By (3.11), ξ2 is in O( ) neighborhood of x0 , therefore, Lemma 4.7 shows ξ2 < ζ2 . Recall that u (x) is close to 1 in [ζ2 , 1]. This and g(x; k) > 0 yield −g(x; k)(2u − 3u2 ) > 0. On the other hand, in [−k, ζ1 ], Lemma 4.6 and (4.8) shows ζ1 < ξ3 . Therefore u (x) isclose to 0 in [−k, ζ1 ]. This and g(x; k) < 0 yield −g(x; k)(2u − 3u2 ) > 0. Set I = [−k, ζ1 ] [ζ2 , 1]. There exists μ1 > 0 such that for any w ∈ H 1 (−k, 1), 

3 |w  |2 − g(x; k)(2u − 3u2 )w 2 dx I

≥





−g(x; k)(2u − 3u2 )w 2 dx ≥ μ1 w 2 dx

I

(4.33)

I

holds. Note that [−k, 1] = I ∪ (ζ1 , ζ2 ) by (4.32) and (4.33), we have 1 H(w, (−k, 1)) = H(w, (ζ1 , ζ2 )) + H(w, I ) ≥ min{λ1 , μ1 }

w 2 dx,

−k

for any w ∈ H 1 (−k, 1). This shows the lemma.

2

5. Proof of Lemma 2.1 In this section we will prove Lemma 2.1 according to the idea of Section 2. We first recall the equation (E)0 , which is obtained by setting k = 0 in (E)k in Section 2: 

3 u + p(x)u2 (1 − u) = 0 u (0) = u (1) = 0, 0 ≤ u(x) ≤ 1,

in (0, 1),

where p(x) satisfies (P1) - (P3) in Introduction. We recall that U (0) = {u(x) ∈ C 1 [0, 1] | u∗ (x; 0) ≤ u(x) ≤ u∗ (x; 0)}, where u∗ and u∗ are the upper and lower solutions constructed in Section 3.

(E)0

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The following proposition is on the properties of solutions of (E)0 , which are obtained from some theorems and lemmas in [8]. Proposition 5.1. Let β > 0 be an arbitrarily small constant. There exist C > 0, C˜ > 0 such that for sufficiently small > 0, (E)0 has a solution ψ0 (x) ∈ U (0) satisfying (i) ψ0 (x) ≤ C 3 holds in [0, x0 − β]. (ii) Any nonconstant solution u(x) of (E)0 satisfies either of the following in [0, 1], ˜ 3 u(x) ≤ C

or

u(x) = ψ0 (x).

(iii) Any nonconstant solution u(x) of (E)0 satisfies u(x) ≤ ψ0 (x). Proof. (i) holds from Theorem 1.1 (or more precisely, Lemma 3.5) in [8]. (ii) holds from Theorem 1.3 in [8]. We have only to prove (iii). To show this, suppose that there exists x¯ s.t. u(x) ¯ > ψ0 (x) ¯ to obtain a contradiction. By (ii) of this lemma, ψ0 (1) is close to 1 and u(1) is close to 0. Therefore it holds that u(1) < ψ0 (1). Set v(x) = max{ψ0 (x), u(x)}, then v(x) is a lower solution in a strict sense. Since v(x) is a lower solution and 1 is an upper solution, there exists at least one stable solution between v(x) and 1. However, again by (ii) of this lemma, there exists no notrivial solution in between v(x) and 1. Therefore, 1 is a stable solution. This contradicts the following lemma. The proof is complete. 2 Lemma 5.2. u = 1 is an unstable solution of (E)0 for sufficiently small. This lemma is already shown in Lemma 2.3 in [4]. However, we will provide another proof for reader’s convenience. Proof. Linearized eigenvalue problem of (E)0 at u = 1 is 

3 ψ  − p(x)ψ = −λψ ψ  (0) = ψ  (1) = 0.

in

(0, 1),

(5.1)

Then the minimal eigenvalue λ(0, 1) of the following (5.1) is given by

λ(0, 1) =

inf

w∈H 1 (0,1), ||w||=1

P(w),

P(w) =

1

3 |w  |2 + p(x)w 2 dx.

0

Note that P(1) < 0 by (P3)(in Introduction). This shows λ(0, 1) < 0. The next proposition shows that (E)0 has at least two nontrivial solutions. We can obtain this proposition by rewriting Theorem 1.3 in [4] to one dimensional case (Their result is actually for higer dimension). 2 Proposition 5.3. (E)0 has at least two solutions for sufficiently small . One solution is ψ0 (x) which is stable, the other is unstable.

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Propositions 5.1 and 5.3 show that (E)0 has a solutions ψ0 (x) and one more solution which stays close to 0. We denote this solution close to zero by w (x). In the next lemma, we provide estimates for w (x). Lemma 5.4. Let w (x) be any nonconstant solution of (E)0 and w ∈ / U (0). There exist 0 < C1 < C2 such that C1 3 ≤ w (x) ≤ C2 3

(5.2)

holds for sufficiently small. Proof. The right-hand side of (5.2) is already shown in Proposition 5.1 (i) and (ii). Since w satisfies w  = − p(x) w 2 (1 − w ), w  changes sign with p(x). This with w  (0) = w  (1) = 0 3 show that w  (x) > 0 in (0, 1). Since w is monotone increasing, we have w (0) = min w (x) [0,1]

and w (1) = max w (x). [0,1]

(5.3)

Assume that the left-hand side of (5.2) does not hold for any C1 > 0 to obtain a contradiction. Then we have w (0) = γ1 ( ) → 0 as 3

→ 0.

(5.4)

Set −pmin = min{p(x) | 0 ≤ x ≤ 1}. We define u(x; A) =

A 3 , (2 − x)2

where A satisfies 0
6 , pmin

A<

w (1) . 3

(5.5)

We will argue by contradiction that for every fixed small > 0 and every A satisfying (5.5), u(x; A) < w (x) in [0, 1]. It is clear that u(x; 0) = 0 < w (x) in [0, 1]. We assume that A∗ and 6 w (1) }, 0 ≤ x ∗ ≤ 1, , x ∗ exist such that 0 < A∗ < min{ pmin 3 u(x; A∗ ) ≤ w (x) to obtain a contradiction.

in [0, 1],

and

u(x ∗ ; A∗ ) = w (x ∗ )

(5.6)

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It follows from (5.5) that u(x; A) satisfies (u) = 3 u + p(x)u2 (1 − u) ≥ 3 u − pmin u2 (1 − u) ≥ 3 u − pmin u2 =

6A 6 A2 6 A 6 − pmin = (6 − pmin A) > 0, 4 4 (2 − x) (2 − x) (2 − x)4

(5.7)

and u(1; A) = A 3 < w (1).

(5.8)

By the boundary condition of w , we have u (0; A) > 0 = w  (0).

(5.9)

Now we take A = A∗ in (5.7), (5.8), and (5.9). The assumption (5.6) says (u − w ) achieves its maximum value 0 over [0, 1] at x = x ∗ , and hence f (u(x ∗ )) = f (w (x ∗ )) with f (u) = u2 (1 − u), and (u − w ) (x ∗ ) ≤ 0 if x ∗ ∈ (0, 1). However, (5.7) and the equation of w show that 3 (u − w ) + p(x)(f (u) − f (w )) > 0.

(5.10)

This and (5.8) leave the only possibility x ∗ = 0, which with (5.10) and the Hopf boundary point lemma (when the maximum value is 0) leads to (u − w ) (0) < 0, a contradiction to (5.9). Thus, (5.6) does not hold and this proves for every fixed small > 0 and every A that satisfies (5.5), w (x) > u(x; A) = Next, we set A =

A 3 A 3 ≥ 4 (2 − x)2

for 0 ≤ x ≤ 1.

(5.11)

4 6 w (1) } in (5.11), then we have min{ , 5 pmin 3 w (x) ≥

1 6 3 min{ , w (1)} 5 pmin

for 0 ≤ x ≤ 1.

It holds from this and (5.4) that γ1 ( ) =

w (0) 1 6 3 > min{ , w (1)}. pmin 3 5 3

(5.12)

6 3 6 ≤ w (1). Then (5.12) shows that γ1 ( ) > , which contradicts the assumppmin pmin 6 3 > w (1). By this fact, (5.3), (5.4) and (5.12), we tion that γ1 ( ) → 0. Thus, it holds that pmin have Suppose

w (0) = 3 γ1 ( ) ≤ w (x) < w (1) < 5 3 γ1 ( )

in [0, 1].

(5.13)

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We will next show that |w  (x)| ≤ C3 3 (γ1 ( ))2 .

(5.14)

To show this we have      p(x)   p(x)  |w  (x)| =  3 w 2 (1 − w ) ≤  3 w 2  . It holds from this and (5.13) that |w  (x)| ≤ 16 3 γ1 ( )2 max p(x). x∈[0,1]

We have  x    1        |w (x)| = |w (x) − w (0)| =  w (x) dx  ≤ |w  (x)|dx = 16 3 γ1 ( )2 max p(x). x∈[0,1]   0

0

We have proved (5.14). On the other hand, since w satisfies (E)0 , it holds that 1 −

1 p(x) dx =

0

3 0

 1 1 w  w  (2 − 3w )w  2 3 dx = + dx. w 2 (1 − w ) w 2 (1 − w ) 0 w 3 (1 − w )2 0

Therefore, we have 1 −

1 p(x) dx =

0

3 0

(2 − 3w )w  2 dx ≤ 4 3 w 3 (1 − w )2

1 0

w  2 dx. w 3

To get the last inequality, we have used 0 < w (x) < C2 3 , which implies (1 − w )2 ≥ 2 − 3w ≤ 2. Using (5.13), (5.14) and (5.4), we have 1

3 0

(5.15)

1 2

and

(w  )2 dx ≤ C32 γ1 ( ) → 0 as → 0. w 3

1 This contradicts (5.15) and 0 p(x) dx < 0 from (P3). The lemma is proved. Next we will consider the existence of solutions for (E)k . The property (3) will be one of keys to prove Lemma 2.1. 2

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¯ Lemma 5.5. For any large k¯ > 0, there exists 0 > 0 s.t. (E)k has a solution ψ (x; k) (0 ≤ k ≤ k) satisfying the following (1) - (3) for ≤ 0 : (1) ψ (x; k) ∈ U (k), (2) ψ (x; 0) = ψ0 (x), ¯ (3) ψ (−k; k) is continuous with respect to k ∈ [0, k], where U (k) is defined in (4.1). The proof of this lemma will be given in the last part of this section. Now we recall the definition of g(x; k) in Section 2. It holds that g(x; k) = p(0) in x ∈ [−k, 0]. Set g∗ = −p(0) > 0 and m ≥ max{

12 , Cx02 }, g∗

(5.16)

where C is in Proposition 5.1. In the following lemma we will provide an estimate for the solution ψ (x; k) in Lemma 5.5. Lemma 5.6. Let ψ (x; k) be a solution of (E)k obtained in Lemma 5.5. There exists 0 > 0 such that for ≤ 0 , the following inequality holds: ψ (−k; k) <

2m 3. (x0 + k)2

Proof. Set u(x) ¯ =

3m 3m + . 2 (x0 − x) (x + x0 + 2k)2

We will show that u¯ is an upper solution of 3 u + g(x; k)u2 (1 − u) = 0,

in (−k, 0)

u (−k) = 0,

u(0) = ψ (0; k).

Note that ψ (x; k) is a solution of this equation. Direct calculation shows u¯  (x) =

2 3 m 2 3 m − . 3 (x0 − x) (x + x0 + 2k)3

Since u¯  (x) ≥ 0 holds in [−k, 0], for sufficiently small it holds that u(x) ¯ ≤ u(0) ¯ =

3m 3m 2m 3 1 + ≤ < 2 2 (x0 ) (x0 + 2k)2 (x0 )2

in

[−k, 0].

(5.17)

Set (u) ¯ = 3 u¯  + g(x; k)u¯ 2 (1 − u). ¯ It holds from (5.17) that 1 (u) ¯ = 3 u¯  − g∗ u¯ 2 (1 − u) ¯ < 3 u¯  − g∗ u¯ 2 2 g∗ 1 1 ≤ 6 m(6 − m)( + )≤0 4 2 (x0 − x) (x + x0 + 2k)4

(5.18)

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in [−k, 0] for sufficiently small, where we have used (5.16) in the last inequality. We will check the boundary conditions. It holds from direct calculation that u¯  (−k) = 0 = ψ  (−k; k).

(5.19)

Using (5.16), we have u(0) ¯ =

3m 3m + > C 3 . (x0 + 2k)2 x02

This with Lemma 5.7 below show u(0) ¯ > ψ (0; k). On the other hand, in the case m = m ¯ := max{

u(x) ¯ ≥ u(−k) ¯ >

(5.20)

12 , C(x0 + k)2 }, by Lemma 5.7, we have g∗

2m 3 ≥ 2C 3 > ψ (x; k) in [−k, 0]. (x0 + k)2

By this fact, (5.18), (5.19), (5.20), and the strong maximum principle, u(x) ¯ > ψ (x; k) holds in [−k, 0] for any m satisfying (5.16) and m ≤ m. ¯ Setting x = −k, we obtain the proof of the lemma. 2 Lemma 5.7. Let C > 0 and β be the same constant as Proposition 5.1 and let u (x; k) be any solution of (E)k such that u (x; k) ∈ U (k). Then there exists 0 > 0 s.t. for < 0 , u (x, k) ≤ C 3

holds for

x ∈ [−k, x0 − β].

We remark that 0 in this lemma does not depend on k (We can choose the same 0 as in Proposition 5.1). Proof. By Lemma 4.3, u (x; k) is monotone increasing for any k ≥ 0. Therefore, it holds that u (0; k) > 0. Since u (x; k) is a solution, we have (u (x; k)) = 0, where (u) = 3 u + p(x)u2 (1 − u). Since u satisfies Neumann zero boundary condition, we have u (1; k) = 0. Therefore, u (x; k) is a lower solution of (E)0 . On the other hand, u∗ (x; 0) in Section 3 is an upper solution of (E)0 . Since u∗ (x; k) = u∗ (x; 0) holds for 0 ≤ x ≤ 1, we have u (x; k) ≤ u∗ (x; k) = u∗ (x; 0) for 0 ≤ x ≤ 1. Therefore, there exists a solution of (E)0 , u(x), s.t. u (x; k) ≤ u(x) ≤ u∗ (x; 0) (0 ≤ x ≤ 1).

(5.21)

(5.21) shows that u(x) can not stay close to zero. Therefore, by Proposition 5.1 (ii), we have u(x) = ψ0 (x). This, (5.21) and Proposition 5.1 (i) imply u (x; k) ≤ C 3

for

[0, x0 − β).

(5.22)

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Again by Lemma 4.3, u (x; k) is monotone increasing, it holds from (5.22) that u (x; k) ≤ C 3

for

[−k, 0].

(5.23)

(5.22) and (5.23) complete the proof of the lemma. Now we provide a proof for Lemma 2.1 2 Proof of Lemma 2.1. Let w (x) be a solution of (E)0 which stays close to 0 as above lemmas and let ψ (x; k) are as in Lemma 5.5. It is clear that w (−x − k) is a solution of an equation 

3 u + gex (x; k)u2 (1 − u) = 0 in (−k − 1, −k), u (−k − 1) = u (−k) = 0,

where gex is in (2.3). On the other hand, we set  k¯ =

2m , C1

(5.24)

where m is given in (5.16) and C1 is the same constant as in Lemma 5.4. Now we define  U (x; k) =

w (−x − k) −k − 1 ≤ x ≤ −k, −k ≤ x ≤ 1. ψ (x; k)

U is continuous if we prove w (0) = ψ (−k; k).

(5.25)

Once (5.25) is proved, U (x; k) is continuous at x = −k (Note that w (x) and ψ (x; k) satisfy Neumann zero boundary condition). The standard regularity argument show that U (x; k) is a solution of (2.2). We remark that U (x; k) has exactly one layer and the layer is located near x = x0 . To show (5.25), it holds from Proposition 5.1 (iii) that ψ (0; 0) > w (0).

(5.26)

By Lemma 5.6, with Lemma 5.4 and (5.24), we have ¯ k) ¯ = ψ (−k;

2m 2m 3 < 2 3 = C1 3 < w (0). 2 ¯ (x0 + k) k¯

(5.27)

¯ such that (5.25) holds. Now we have By Lemma 5.5 (3), (5.26) and (5.27), there exists k ∈ (0, k) obtained a solution U (x; k) with one layer for gex (x; k) which has two zeros. Lemmas 5.4 and 5.7 show the estimates in (i) of the lemma. On the other hand, Lemma 3.1 and Corollary 3.4 in [9] show the estimates in (ii) of the lemma. The proof is complete. The rest of this section is the proof for Lemma 5.5. Before providing the proof, we will introduce two lemmas which directly follows from Sections 3 and 4.

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By a change variable z = 

x+k k+1 ,

and u(x; k) = v(z; k), (E)k is converted to

3 vzz + G(z; k)v 2 (1 − v) = 0 in (0, 1), v  (0) = v  (1) = 0,

(5.28)

where G(z; k) = (k + 1)2 g((k + 1)z − k; k). It is clear that there is one to one correspondence between solutions of (E)k and (5.28). Let u∗ and u∗ are the upper and lower solutions constructed in Section 2. The following lemma holds from Lemma 3.4. 2 Lemma 5.8. There exists 0 ≥ 0 such that u∗ ((k + 1)z − k; k) is an upper solution and u∗ ((k + 1)z − k; k) is a lower solution of (5.28) for any k ≥ 0 and any ≤ 0 . ∗ Proof. By a change variable z = x+k k+1 , and the fact that u and u∗ are a pair of upper and a lower solution, the conclusion follows directly from Lemma 3.4. Set

V (k) = {v(z) ∈ C 1 [0, 1] | u∗ ((k + 1)z − k; k) ≤ v(z) ≤ u∗ ((k + 1)z − k; k)}.

2

Lemma 5.9. Let v (z; k) be a solution of (E)k and v (z; k) ∈ V (k). For any large k¯ > 0, there ¯ and < 0 . exists 0 > 0 s.t. v (x; k) is linearly stable for k ∈ [0, k] Proof. Linearized eigenvalue problem of (5.28) at v is 

By z = 

x+k k+1 ,

3 vˆzz + G(z; k)(2v − 3v 2 )vˆ = −μvˆ in (0, 1), vˆ  (0) = vˆ  (1) = 0.

(5.29)

v(z; ˆ k) = w(x; k) and μ = (k + 1)2 λ, we have

3 w  (x; k) + g(x; k)(2u (x; k) − 3u (x; k)2 )w(x; k) = −λw(x; k) (−k ≤ x ≤ 1), (5.30) w  (−k; k) = w  (1; k) = 0.

Since all the eigenvalues of (5.30) are positive by Lemma 4.1, the eigenvalues of (5.29) are also all positive. The proof is complete. 2 Proof of Lemma 5.5. We set ¯ ;v S = {(v, k) ∈ C 1 [0, 1] × [0, k]

is a solution of (5.28)}.

The conclusion of Lemma 5.5 follows immediately from Lemma 5.10 below.

2

Lemma 5.10. For any large k¯ > 0, there exists 0 > 0 such that (5.28) has a solution v (z; k) satisfying ¯ → C 1 [0, 1] is continuous with respect to k ∈ [0, k], ¯ (i) The map v (z; k) : [0, k] (ii) v (z; 0) = ψ0 (z). ¯ it holds that v (z; k) ∈ V (k). (iii) For any k ∈ [0, k],

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Proof. The proof is based on the continuation method. Note that the equation (E)0 coincides with (5.28) in the case k = 0. It holds that (ψ0 (z), 0) ∈ S. Lemma 5.8 shows ψ0 (z) ∈ V (0).

(5.31)

By Lemma 5.9, there exists 0 > 0 such that every solution of (5.28) in V (k) is nondegenerate ¯ By the implicit function theorem, there exist v (z; k) and a neighborhood for < 0 and k < k. ¯ such that for ≤ 0 , of 0, which we denote by J ⊂ [0, k] 1 (J1) v (z; k) : J → C [0, 1] is continuous with respect to k ∈ J , (J2) (v (z; k), k) ∈ S for k ∈ J , (J3) v (z; 0) = ψ0 (z). ¯ satisfying (J1) - (J3). We Let Jmax be a maximal connected component of the set of k in [0, k] will show that v (z; k) ∈ V (k) for

k ∈ Jmax

(5.32)

˜ ⊂ V (k). ˜ This, (5.31) and holds by contradiction. Assume that there exists k˜ such that v (z; k) (J1) show either of the following holds. ˜ and z∗ ∈ [0, 1] such that v (z∗ , k ∗ ) = u∗ (z∗ , k ∗ ) and (i) There exists k ∗ ∈ [0, k] ∗ ∗ ∗ v (z, k ) ≤ u (z, k ) for z ∈ [0, 1]. ˜ and z∗ ∈ [0, 1] such that v (z∗ , k∗ ) = u∗ (z∗ , k∗ ) and (ii) There exists k∗ ∈ [0, k] v (z, k∗ ) ≥ u∗ (z, k∗ ) for z ∈ [0, 1]. Both (i) and (ii) contradict the strong maximum principle. We have shown (5.32). ¯ we will derive a contradiction. By (5.32), Assuming that Jmax is strictly smaller than [0, k], it holds that 0 < v (z; k) < 1 for k ∈ Jmax . Standard regularity argument shows that Jmax is a closed set, which we denoted by [0, j ]. Again using Lemma 5.9 and the implicit function theorem, we can construct a solution v (z; k) and a neighborhood of j denoted by [j − ν, j + ν] such that (i) v (z; k) : [j − ν, j + ν] → C 1 [0, 1] is continuous with respect to k ∈ [j − ν, j + ν], (ii) (v (z; k), k) ∈ S for k ∈ [j − ν, j + ν]. Clearly Jmax ∪ [j − ν, j + ν] is a connected set satisfying (J1) - (J3), and this contradicts the ¯ By (5.9) and Lemma 3.5, v (z; k) has exactly maximality of Jmax , hence we have Jmax = [0.k]. ¯ one layer for k ∈ [0, k]. The proof is complete. 2 6. Proof of theorems We will show the proof of theorems in this section. Proof of Theorem 1.1. By the transformation (2.1), Lemma 2.1 is directly rewritten as Theorem 1.1. 2 Proof of Theorem 1.2. By Propositions 2.6 and 2.9 in [9], the following lemma holds: 2 Lemma 6.1. There exists a pair of upper and lower solution U ∗ (x) and U∗ (x) (U ∗ (x) > U∗ (x)) of (1.3) such that  0 in (z1 , z2 ), (6.1) U∗ (x) = 1 − σ1 in (−1, z1 − 3β) ∪ (z2 + 3β, 1).

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32

U ∗ (x) =



in (z1 + 3β, z2 − 3β), in (−1, z1 ) ∪ (z2 , 1).

σ2 1

(6.2)

Here σ1 , σ2 are small positive constants independent of , and it holds that σ2 ≤ U ∗ (x) ≤ 1.

0 ≤ U∗ (x) ≤ 1 − σ1 ,

(6.3)

To be precise, we choose σ1 and σ2 according to (2.19), (2.20), (2.22), (2.24), (2.29), (2.30), (2.32) and (2.33) in [9]. We also remark that the layer part of U∗ (x) and that of U ∗ (x) are obtained in the similar way as we have constructed u∗(x; k) and u∗ (x; k) in Section 3. Set  0 in (−1, z2 ), (6.4) U (x) = U∗ (x) in (z2 , 1). U is clearly a lower solution of (1.3). On the other hand, ud (x) in Thoerem 1.1 is an upper solution of (1.3). To show this, set (u) = 3 u + h(x)u2 (1 − u), we have (ud ) = (−pex (x : ) + h(x))u2d (1 − ud ) ≤ 0

(−1, 1),

where pex is given by (1.4). Next we will show that ud (x) > U (x). Lemma 6.2. For d > 0 sufficiently small, ud (x) > U (x) holds for −1 ≤ x ≤ 1. Proof. Let β > 0 be an arbitrarily small constant. By (ii) in Theorem 1.1 it holds that 3

ud (x) > 1 − C2 exp(−C1 − 2 )

in (z2 + 3β, 1)

(6.5)

On the other hand, by (6.1) in Lemma 6.1, we have U (x) = U∗ (x) = 1 − σ1

in [z2 + 3β, 1],

(6.6)

where σ1 is a small constant independent of d > 0. (6.5) and (6.6) show 3

ud (x) > 1 − C2 exp(−C1 − 2 ) > 1 − σ1 = U (x)

in (z2 + 3β, 1).

(6.7)

We will show ud > U on the interval z2 ≤ x ≤ z2 + 5β. Set V (x, q) = U (x − q) (z2 ≤ x ≤ z2 + 5β, 0 ≤ q ≤ 4β). Since U is a lower solution and h(x) is monotone increasing in [z2 − 5β, z2 + 5β] for β sufficiently small, we have (V (x, q)) = (h(x) − h(x − q))(U (x − q))2 (1 − U (x − q)) ≥ 0.

(6.8)

We will check boundary conditions. It holds that V (z2 , q) = U (z2 − q) = 0 < ud (z2 ) (6.3) and (6.5) show

(0 ≤ q ≤ 4β).

(6.9)

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V (z2 + 5β, q) = U (z2 + 5β − q) ≤ 1 − σ1 < 1 − C2 exp(−C1 − 2 ) < ud (z2 + 5β).

33

(6.10)

Now we have shown that V (x, q) is a lower solution on [z2 , z2 + 5β] for any q ∈ [0, 4β]. We will next show V (x, 4β) = U (x − 4β) < ud (x)

(z2 ≤ x ≤ z2 + 5β).

(6.11)

(z2 + 3β ≤ x ≤ z2 + 5β).

(6.12)

(z2 ≤ x ≤ z2 + 3β).

(6.13)

Again by (6.3) and (6.5), we have V (x, 4β) = U (x − 4β) ≤ 1 − σ1 < 1 3 − C2 exp(−C1 − 2 ) < ud (x) On the other hand, for z2 ≤ x ≤ z2 + 3β it holds that V (x, 4β) = U (x − 4β) = 0 < ud (x)

(6.12) and (6.13) show (6.11). (6.8) – (6.11) and the strong maximum principle show that V (x, q) < ud (x)

(z2 ≤ x ≤ z2 + 5β,

0 ≤ q ≤ 4β).

Setting q = 0, we have U (x) < ud (x)

(z2 ≤ x ≤ z2 + 5β).

(6.14)

Since U (x) = 0 < ud (x) holds in −1 ≤ x ≤ z2 , (6.7) and (6.14) completes the proof of (6.2). Since both U ∗ (x) and ud (x) are upper solution, min{U ∗ (x), ud (x)} is also an upper solution. Now by Lemma 6.2 and U ∗ (x) > U (x), there exists a stable solution us (x) of (1.3) satisfying U (x) ≤ us (x) ≤ min{U ∗ (x), ud (x)} for d suff. small. It also holds from Lemmas 3.1 and 3.5 in [9] that a solution have layers only near zeros of h(x). Therefore, us (x) is close to 0 in (−1, z2 ), us (x) is close to 1 in (z2 , 1). We also obtain the estimates in (i) and (ii) of the theorem from Lemmas 3.1 and 3.5 in [9]. We next prove the existence of a solution other than us using degree theory. Set D1 = {u ∈ C[−1, 1] | U (x) < u(x) < U ∗ (x)},

D2 = {u ∈ C[−1, 1] | U∗ (x) < u(x) < U ∗ (x)},

D3 = {u ∈ C[−1, 1] | U (x) < u(x) < min{U ∗ (x), ud (x)}}, It follows from Lemma 6.2 that D3 is not empty. Set p > 3 max |h(x)|. Define a map w = Au, A : C[−1, 1] → C[−1, 1] by 

− 3 w  + pw = pu + h(x)u2 (1 − u) in (−1, 1), w  (−1) = w  (1) = 0.

There is one to one correspondence between a solution of (1.3) and a fixed point of A. We will consider Leray-Shaudar degree of A. The following lemma follows from a standard result. The proof of this lemma is given in Lemma 5.3 in [9]. 2

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Lemma 6.3. Let u¯ and u be a pair of an upper and a lower solution of (1.3) and D be defined by D = {u ∈ C[−1, 1] | u(x) < u(x) < u(x)}. ¯ Then deg(I − A, D) is well defined and it holds that deg(I − A, D) = 1. We will continue the proof of Theorem 1.2. Suppose that there is no fixed point in D1 \ (D¯2 ∪ ¯ D3 ) to obtain a contradiction. It holds that deg(I − A, D1 \ (D¯2 ∪ D¯3 )) = 0.

(6.15)

deg(I − A, D1 ) = deg(I − A, D2 ) = deg(I − A, D3 ) = 1.

(6.16)

Using Lemma 6.3, we have

On the other hand, by the property of Leray-Schauder degree, we have deg(I − A, D1 ) = deg(I − A, D2 ) + deg(I − A, D3 ) + deg(I − A, D1 \ (D¯2 ∪ D¯3 )). This contradicts (6.15) and (6.16). Therefore, there is at least one fixed point in D1 \ (D¯2 ∪ D¯3 ), we will denote this fixed point by u2 . Clearly, u2 is a solution of (1.3) in D1 \ (D¯2 ∪ D¯3 ). Now we have found at least two solutions us and u2 (us = u2 ). Since u2 is not in D3 , u2 (x) can not be below us (x) Therefore, there are only two possibilities (1) or (2) as follows: (1) us (x) ≤ u2 (x) in [−1, 1] (2) There are x, ¯ x˜ ∈ [−1, 1] such that us (x) ¯ < u2 (x) ¯ and us (x) ˜ > u2 (x). ˜ We next show that there exists a stable solution u1 satisfying u1 < u2 for the each case of (1) and (2). In the case (1), we set u1 = us . Since both us and u2 are solutions, the strong maximum principle shows u1 = us < u2 . In the case (2), we set u(x) ¯ = min{us (x), u2 (x)}. Then u(x) ¯ is an upper solution in strict sense. Note that U is a lower solution and satisfies U (x) < u(x). ¯ There exists a stable solution in between U(x) and u(x), ¯ which we denote by u1 (x). Now we have found u1 in the theorem for both cases (1) and (2). We will next show the estimates (i) and (ii) in Theorem 1.2. Both u1 is a solution of the equation, it holds that u1 < 0 in (−1, z1 ) ∪ (z2 , 1) and u1 > 0 in (z1 , z2 ). Since u1 satisfies Neumann zero boundary condition, u1 has local maximums at both x = −1 and x = 1. Therefore, there exits ζ˜1 such that u1 (ζ˜1 ) = 0 and u1 has its minimum at x = ζ˜1 . Therefore, u1 is a solution with one layer on the interval [ζ˜1 , 1], g(x) has only one zero z2 in this interval. On the other hand, u1 is a solution close to zero on the interval [−1, ζ˜1 ], g(x) has only one zero z1 in this interval. Theorems 1.1 - 1.3 in [8] show that (ii) of Theorem 1.2 holds and C1 d < u1 (x) < C2 d holds on every compact subset S of [ζ˜1 , z2 ), where C1 , C2 depend on S. On the other hand, Lemma 5.4 ¯ < u1 (x) < Cd ˜ in [−1, ζ˜1 ] for some constants C, ¯ C˜ (C¯ < C) ˜ independent of d. shows that Cd We can obtain the estimates (i) and (ii) for u2 (x) in the same way. The theorem is proved. Proof of Theorem 1.3. By the assumption of the theorem, h(x) is symmetric and has two zeros. We denote the zeros by z3 and −z3 (< 0 < z3 < 1). Then h(x) satisfy h(x) > 0 in (−1, −z3 ) ∪ (z3 , 1), h(x) < 0 in (−z3 , z3 ). We first find a symmetric solution close to zero. By Theorem LNS, we have at least two solutions of

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35

3 u + h(x)u2 (1 − u) = 0 in (0, 1), w  (0) = w  (1) = 0,

for sufficiently small d. By Proposition 5.1, a solution which has one layer is uniquely determined. Therefore, there is at least one more solution which does not have a layer. This solution should stay near zero, again by Proposition 5.1. We denote this solution close to zero in (0, 1) by wh (x). Set  wex (x) =

wh (x) wh (−x)

in (0, 1), in (−1, 0).

(6.17)

Then wex (x) is a solution of (1.3) which stays close to zero in (−1, 1). Next, we will find a solution close to zero which is not symmetric. We again use LeraySchauder degree. Let ν be a positive constant satisfying ν < 1 and set D4 = {u ∈ C[−1, 1] | − ν < u(x) < min{U ∗ (x), ud (x)}}, D5 = {u ∈ C[−1, 1] | − ν < u(x) < min{U ∗ (x), ud (x), ud (−x)}} Set p > 3 max |h(x)|. Define a map w = Bu, B : C[−1, 1] → C[−1, 1] by 

− 3 w  + pw = (p + h(x)u(1 − u)) |u| in (−1, 1), w  (−1) = w  (1) = 0.

Note that Au = Bu holds in u > 0. In the following lemma we will consider Leray-Schauder degree of B. 2 Lemma 6.4. Let u¯ be an upper solution of (1.3) and u¯ be a positive function. Let D˜ be defined by ˜ is well defined and it holds that D˜ = {u ∈ C[−1, 1] | − ν < u(x) < u(x)}. ¯ Then deg(I − B, D) ˜ deg(I − B, D) = 1. Proof. By the definition of B, it holds that Bu ≥ 0 for any u ∈ D0 where D0 = {u ∈ C[−1, 1] | 0 < u(x) < u(x)}. ¯ Since u¯ is an upper solution in strict sense, it holds that B u¯ < u. ¯ By the order preserving property of B, B maps D0 into D0 . ¯ and any λ ∈ [0, 1], we define Choose v0 such that 0 < v0 < u(x) H (u, λ) = (1 − λ)Bu + λv0 . Clearly, H (·, λ) maps D0 into D0 . This shows that H (u, λ) does not have any fixed point on ∂ D˜ for any λ ∈ [0, 1]. By homotopy invariance property of degree theory, we have ˜ = deg(I − H (·, λ), D) ˜ = deg(I − v0 , D) = 1. deg(I − B, D) The proof of Lemma 6.4 is complete. We will continue to prove the theorem. Since both min{U ∗ (x), ud (x), ud (−x)} and min{U ∗ (x), ud (x)} are upper solutions of (1.3), it holds from Lemma 6.4 that deg(I − B, D4 ) = deg(I − B, D5 ) = 1.

(6.18)

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By the property of Leray-Schauder degree, we have deg(I − B, D4 ) = deg(I − B, D5 ) + deg(I − B, D3 ) + deg(I − B, D4 \ (D¯5 ∪ D¯3 )). (6.19) Assume that there is no solution in D4 \ (D¯5 ∪ D¯3 ) to get a contradiction. It holds that deg(I − B, D4 \ (D¯5 ∪ D¯3 )) = 0.

(6.20)

We remark that Au = Bu for u ∈ D3 . Using (6.16), we have deg(I − B, D3 ) = deg(I − A, D3 ) = 1.

(6.21)

(6.18), (6.20) and (6.21) contradict (6.19). Therefore, there is at least one solution in D4 \ (D¯5 ∪ D¯3 ), which we denote by w2 . We will show that w2 stays close to zero in the following. Assume that w2 does not stay close to zero, then we will obtain a contradiction using the same argument as the proof of Lemma 6.2. Since w2 satisfies w2 (x) ≤ ud (x), w2 (1) = 0 and dw2 = −h(x)w22 (1 − w2 ), w2 has local maximum at x = 1 and w2 (1) is not close to 0. By Lemma 3.1 in [9], it holds that 3

w2 (x) > 1 − C2 exp(−C1 − 2 )

in (z3 + 3β, 1),

(6.22)

where β > 0 is an arbitrarily small constant. Repeating the same argument as the proof of Lemma 6.2 ((6.22) corresponds to (6.5), and we replace ud and z2 by w2 and z3 in (6.6) - (6.14), / D3 . We have proved that respectively), we obtain w2 (x) > U (x) in (0, 1). This contradicts w2 ∈ w2 is close to 0 in (0, 1). Finally we will prove that w2 is not symmetric. Suppose that w2 is symmetric with respect to x = 0 to obtain a contradiction. Since w2 ∈ D4 , it holds that w2 (x) < ud (x) in (−1, 1). On the other hand, since w2 is symmetric, w2 (−x) = w2 (x) < ud (x) holds in (−1, 1). Therefore, we have w2 (x) < min{ud (x), ud (−x)}. We remark that min{ud (x), ud (−x)} is an upper solution of (1.3) and w2 (x) is a lower solution of (1.3). Then there is at least one stable solution in between w2 (x) and min{ud (x), ud (−x)}. This contradicts the following lemma which is equivalent to Lemma 5.3 in [9]. 2 Lemma 6.5. Let w be a solution of (1.3) and 0 < w(x) < 1 − w is unstable.

√1 . 2

Then for sufficiently small d,

We have shown that w2 is not symmetric and wex = w2 , where wex is given in (6.17). Set w1 = wex and w3 (x) = w2 (−x). We have found three different solution close to 0. The theorem is proved. Appendix A. Proof of Proposition 4.5 In this appendix we give the proof of Proposition 4.5 which is equivalent to Lemmas A.1 and A.2 below. Recall that u (x; k) is a solution of (E)k in U (k) defined by (4.1).

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Lemma A.1. For β > 0 suff. small, there exist 0 < C4 , 0 < M < Q, and 0 > 0 such that u (x; k) ≤

C4 3 (x0 − x − M)3

(x ∈ [x0 − β, x0 − Q])

for < 0 and k > 0. We can obtain the proof of Lemma A.1 by repeating the same argument as proofs of Lemmas 3.4, 3.5 and 6.3 in [8], which is a very long proof. For reader’s convenience, we use a theorem and lemmas in [8] and make the proof shorter and clearer. Proof of Lemma A.1. Choose 0 < γ1 < γ2 , β > 0, s.t. γ1 < p  (x0 ) < γ2

for

x ∈ [x0 − 2β, x0 + 2β].

Choose positive constants C4 , Q, M s.t. C4 >

12 , γ1

C4 > β 3 C,

C4 > (Q − M)3 ,

(A.1)

where C is the same constant as in Lemma 5.7 (or Proposition 5.1). Define u(x) ¯ =

C4 3 . (x0 + M − x)3

On [x0 − β, x0 − Q], we have p(x) ≤ 0 for sufficiently small . Therefore, (u) ¯ = 3 u¯  + p(x)u¯ 2 (1 − u) ¯ ≤ 3 u¯  − γ1 (x0 − x)u¯ 2 (1 − u) ¯ =

6 C4 [12(x0 − M − x) − γ1 C4 (x0 − x)] 12(x0 − M − x)6

=

6 C4 [(12 − γ1 C4 )(x0 − x) − 12 M] ≤ 0, 12(x0 − M − x)6

(A.2)

The last inequality holds from (A.1). We will check boundary conditions. By (A.1), the definition of u, ¯ we have u(x ¯ 0 − Q) =

C4 > 1 > u (x0 − Q). (Q − M)3

(A.3)

Again by the assumption of (A.1), it holds that u(x ¯ 0 − β) =

3 C4 3 C4 > > C 3 > u (x0 − β). (β − M)3 β3

(A.4)

(A.2), (A.3), (A.4) and the strong maximum principle shows that u¯ is an upper solution for any C4 > 0 satisfying (A.1). This shows the lemma. The righthand side of the inequality in Proposition 4.4 is proved by the following lemma. 2

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Lemma A.2. For β > 0 suff. small, there exist C3 > 0, M˜ > 0, and 0 > 0 such that u (x; k) ≥

C3 3

(x ∈ [x0 − β, x0 ])

˜ 3 (x0 − x + M)

for < 0 and k > 0. Before providing the proof we show the following lemma. ˜ 3 holds for any x ∈ [0, 1], k > 0, Lemma A.3. There exists C˜ > 0 and 0 > 0 s.t. u (x; k) ≥ C < 0 . Proof. Set δ 2 = 3 and −gmin = p(0) = min{g(x; k) | − k ≤ x ≤ 1}. We define u(x) =

aδ 2 , (1 + δ − x)2

where a satisfies a<

6 gmin

,

1 a< . 4

(A.5)

It follows by (A.5) that (u) = 3 u + g(x; k)u2 (1 − u) ≥ 3 u − gmin u2 (1 − u) ≥ 3 u − gmin u2 =

6aδ 4 a2δ4 aδ 4 − g = (6 − g a) > 0. min min (1 + δ − x)4 (1 + δ − x)4 (1 + δ − x)4

(A.6)

By the assumption of this lemma we have u(1) = a <

1 ≤ u (1; k) 4

(A.7)

It follows from the boundary condition of u , we have u (−k) > 0 = u (−k; k).

(A.8)

(A.6), (A.7), (A.8) and strong maximum principle shows that u (x; k) >

aδ 2 aδ 2 aδ 2 > > (1 + δ − x)2 (1 + δ)2 4

for 0 ≤ x ≤ 1

for any a satisfying (A.5) and δ suff.small. Lemma A.3 is proved.

2

Proof of Lemma A.2. Choose 0 < γ1 < γ2 , β > 0, s.t. γ1 < p  (x0 ) < γ2

for

x ∈ [x0 − 2β, x0 + 2β].

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Recall the lower solution u∗ (x; k). There exists L˜ > 0 s.t. u∗ (x; k) = u∗ (x; 0) ≥

1 2

for

˜ x0 + 2β]. x ∈ [x0 + L,

(A.9)

Let A be a constant satisfying A3 <

12 , γ2

˜ A3 < β 3 C,

(A.10)

where C˜ is the same constant as in Lemma A.3. Let L˜ > 0 is a constant in (A.9), and let M˜ satisfy ˜ M˜ > 2A + L.

(A.11)

Define u(x) =

A3 3 , (x0 + M˜ − x)3

we will show (u) > 0 in [x0 − β, x0 + β].

(A.12)

˜ it holds that p(x) ≥ 0. Therefore we have On [x0 , x0 + L], (u) = 3 u + p(x)u2 (1 − u) ≥ 3 u =

12A3 6 > 0. (x0 + M˜ − x)5

On [x0 − β, x0 ], we have p(x) ≤ 0. Therefore, (u) = 3 u + p(x)u2 (1 − u) ≥ 3 u + p(x)u2 ≥

12A3 6 A6 6 + p(x) . (x0 + M˜ − x)5 (x0 + M˜ − x)6

(A.13)

It holds from (A.10) and (A.13) that 12A3 6 γ2 (x0 − x)A6 6 − (x0 + M˜ − x)5 (x0 + M˜ − x)6  

A3 6 = 12 − γ2 A3 (x0 − x) + 12 M˜ ≥ 0. (x0 + M˜ − x)6

(u) ≥

We will check boundary conditions. By (A.9), (A.11), the definition of u, and the assumption of the lemma, we have

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A3

1 1 ˜ 0) < ≤ u∗ (x0 + L; 8 2 ˜ k) ≤ u (x0 + L; ˜ k). = u∗ (x0 + L;

˜ = u(x0 + L)

˜ 3 (M˜ − L)

<

(A.14)

Using Lemma A.3 and (A.10), we have u(x0 − β) =

A3 3 A3 3 ˜ 3 < u (x0 − β; k). < 3 < C 3 ˜ β (β + M)

(A.15)

(A.12), (A.14), (A.15) and the strong maximum principle shows u (x; k) >

A3 3 4(x0 + M˜ − x)3

˜ in [x0 − β, x0 + L]

for any A, M˜ satisfying (A.10). The lemma is proved. 2 References [1] G. Feltrin, E. Sovrano, An indefinite nonlinear problem in population dynamics: high multiplicity of positive solutions, Nonlinearity 31 (2018), in press. [2] Y. Lou, T. Nagylaki, A semilinear parabolic system for migration and selection in population genetics, J. Differ. Equ. 181 (2002) 388–418. [3] Y. Lou, T. Nagylaki, W.-M. Ni, An introduction to migration-selection PDE models, Discrete Contin. Dyn. Syst. A 33 (2013) 4349–4373. [4] Y. Lou, W.-M. Ni, L. Su, An indefinite nonlinear diffusion problem in population genetics, II: stability and multiplicity, Discrete and Contin. Dyn. Syst. A 27 (2010) 643–655. [5] T. Nagylaki, Conditions for the existence of clines, Genetics 80 (1975) 595–615. [6] K. Nakashima, W.-M. Ni, L. Su, An indefinite nonlinear diffusion problem in population genetics, I: existence, Discrete Contin. Dyn. Syst. 27 (2010) 617–641. [7] K. Nakashima, Multi-layered stationary solutions for a spatially inhomogeneous Allen-Cahn equation, J. Differ. Equ. 191 (2003) 234–276. [8] K. Nakashima, The uniqueness of indefinite nonlinear diffusion problem in population genetics, part I, J. Differ. Equ. 261 (2016) 6233–6282. [9] K. Nakashima, The uniqueness of an indefinite nonlinear diffusion problem in population genetics, part II, J. Differ. Equ. 264 (2018) 1946–1983.