Indefinite nonlinear diffusion problem in population genetics

We study the following Neumann problem in one dimension, \begin{document}$ \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}{u_t} = du'' + g(x){u^2}(1 - u)\quad {\rm{in}}\quad (0,1) \times (0,\infty ),\;\\0 \le u \le 1\quad {\rm{in}}\quad (0,1) \times (0,\infty ),\;\\u'(0,t) = u'(1,t) = 0\quad {\rm{in}}\quad (0,\infty ),\end{array}\end{array}} \right.$\end{document} where \begin{document}$ g $\end{document} changes sign in \begin{document}$ (0, 1) $\end{document} . This equation models the "complete dominance" case in population genetics of two alleles. It is known that this equation has a nontrivial stable steady state \begin{document}$ U_d $\end{document} for \begin{document}$ d $\end{document} sufficiently small. We show that \begin{document}$ U_d $\end{document} is a unique nontrivial steady state under a condition \begin{document}$ \int_{0}^1\, g(x)\, dx\geq 0 $\end{document} and some other additional condition.


1.
Introduction. This article considers the "complete dominant" case of a migration selection model for the solution of gene frequency at a single locus with two alleles A 1 , A 2 initiated in [7] and [3]. This model is due to T. Nagylaki in 1975 [4]. We shall give a brief description of this model following the more recent presentation in [2] and [7].
Let u(x, t) be the frequency of allele A 1 at time t and location x (thus 0 ≤ u ≤ 1), and r ij be the fitness (local selection coefficient) of the genotype A i , A j for i, j = 1, 2 and r 1 = r 11 u + r 12 (1 − u) is the marginal selection coefficient of A 1 , and r = r 11 u 2 + r 12 u(1 − u) + r 21 (1 − u)u + r 22 (1 − u) 2 is the mean selection coefficient of the population. Now positing r 11 = 1, r 12 = r 21 = 1 − hg(x), r 22 = 1 − g(x), (1.1) where g(x) reflects the "environmental variation" and 0 ≤ h ≤ 1 specifies the degree of dominance (assumed to be independent of the location), we have the selection term: 2) where λ > 0 is the ratio of selection intensity to the migration rate. Therefore, 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 is the Laplace operator, the habitat Ω is a bounded domain with smooth boundary ∂Ω in R n , ν denotes the unit outward normal to ∂Ω and ∂ ν is the normal derivative on ∂Ω. After scaling t, we have It is clear that (1.4) has no nontrivial steady-states if g(x) does not change sign in Ω, 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 A 1 and A 2 , the environmental variation has to be so significant that the selection reverses its direction at least once in Ω, i.e. g(x) changes sign at least once in Ω (Note that if g(x) > 0, then r 11 ≥ r 12 ≥ r 22 at location x; while r 11 ≤ r 12 ≤ r 22 where g(x) < 0). Therefore we shall require in the rest of this paper that g(x) changes sign in Ω.
As was explained in [7] that all previous mathematical results in literature deal with the case 0 < h < 1 and the "complete dominant" case h = 1 was left untouched until the publications of the papers [7] and [3] in 2010. In the case h = 1 we have r 12 = r 22 , i.e. the heterozygote A 1 A 2 has the same fitness as the homozygote A 2 A 2 , we say that A 2 is completely dominant to A 1 (In the case h = 0, in the same way, A 1 is completely dominant to A 2 ). In the "completely" dominant case h = 1, (1.4) becomes u t = d∆u + g(x)u 2 (1 − u) in Ω × (0, ∞), ∂ ν u = 0 on ∂Ω × (0, ∞). (1.5) This "completely dominant" case is not only mathematically challenging but also biologically important. In fact the following conjectures (a) to (c) have been proposed for a long time (See Lou and Nagylaki [1]): (a) If Ω g(x) dx = 0, then for every d > 0, the problem (1.5) has unique nontrivial steady state which is globally asymptotically stable. (b) If Ω g(x) dx > 0, then there exists d 0 > 0 such that for every d ∈ (0, d 0 ), problem (1.5) has a unique nontrivial steady state which is globally asymptotically stable.
(c) If Ω g(x) dx < 0, then there exists d 0 > 0 such that for every d ∈ (0, d 0 ), problem (1.5) has exactly two nontrivial steady states, one is asymptotically stable and the other is unstable. There are two mathematically rigorous results towards the resolution of these important conjectures, [7] and [3]. In [7], the existence of a stable steady state as well as its limiting behavior (as d tends to 0 or ∞)are obtained. Furthermore, in [3] the existence of at least two steady states, one stable and the other unstable, is established as well. However, the uniqueness part of the conjecture is still left open. Therefore, in this paper we survey the uniqueness part of (a) and (b) above, under the condition where the spatial dimension n = 1. A steady state solution of (1.5) under the additional "nondegeneracy" condition on g: All zeros on [0, 1] are interior and nondegenerate; i.e. (H) If g(x i ) = 0, then x i ∈ (0, 1) and g (x i ) = 0, for i = 1, 2, · · · , m.
In [8], we have proved the uniqueness under the extra assumption that g has only one zero x 0 in (0, 1), i.e. m = 1 in (H) under the condition 1 0 g(x) dx ≥ 0. In this paper we study the uniqueness in the case where g has multiple nondegenerate zeros in (0, 1), again under the condition 1 0 g(x) dx ≥ 0. However, in this case, we need some more conditions on g(x), in addition to 1 0 g(x) dx ≥ 0 to prove the uniqueness of solutions. The followings are our results on the uniqueness in [9]. To state our uniqueness result precisely, we first introduce the following auxiliary function p(x) ∈ C 1 [0, 1] satisfying the following (P0) to (P3).
Then, for every d small, (1.6) has a unique nontrivial solution.
By the assumption g(x) ≥ p(x) and the condition (P3), it holds that is a condition corresponding to the nondegeneracy condition (H). In addition, the following result is important.
Theorem 1.2. Suppose that g changes its sign in (0, 1) and that (H) holds. Then, (1.6) has a linearly stable nontrivial solution u d for d sufficiently small. Furthermore, u d has the following properties: (i) On every compact subset S 1 of [g < 0], where the constants C 1 , C 2 depend on S 1 .
The stability and the facts that were already established earlier in [7] for any dimension N . However, the rates of convergence and linearized stability have not been obtained. These rates, (1.7) and (1.8) and linearized stability are essential to prove the uniqueness of U d .
This article is organized as follows. In Sections 2 -3, we will show a proof of uniqueness result Theorems 1.1. In Section 4, we will introduce recent results Theorems 4.1 -4.3, which give counterexamples of Conjectures (b) and (c).

2.
Construction of upper and lower solutions. In this section, we will construct, for every d > 0 small, a pair of upper solution U * and lower solution U * , both exhibit transition layer near every zero x i of g (i ∈ {1, 2, · · · , m}) and U * > U * on [0, 1]. This will guarantee the existence of a solution U d such that U * ≥ U d ≥ U * , Thus U d also exhibits desired transition layer at x i . Upper solution U * and lower solution U * will be useful in later sections in serving as a barrier for our proof of the uniqueness of solutions.
Setting d = 3 , (1.6) takes the following form: where g(x) ∈ C 1 [0, 1] has zeros denoted by {x i ∈ (0, 1) | i = 1, 2, · · · , m}. We will first construct a lower solution of (2.1) with a transition layer of width near each x i . By the nondegeneracy condition (H) in the Introduction, there are two cases g (x i ) > 0 and g (x i ) < 0. We first assume g (x i ) > 0, since the case g (x i ) < 0 is treated in the same way clearly. Letting φ be the unique solution of the following ODE problem (For the uniqueness of φ, see Appendix in [7]) we have the following properties of φ, whose proofs are given in Appendix A in [8].
Lemma 2.1. φ is monotone increasing in (−∞, ∞), and there exist positive constants C i , i = 1, · · · , 6, λ j , j = 1, 2, 3, and R such that the following hold: Let L be a large constant to be chosen later, and we define two C 1 functions as follows: where λ 2 is the constant in (2.3), κ is a constant satisfying κ > C 5 and C 5 is the constant in (2.5). For the case g (x i ) > 0, we can construct a lower solution near x i in the same way as [8]. Define Completely by the same way as we have obtained (2.11) and (2.12) in [8], we can show the following lemma: There exists L i 0 > 0 such that for all L > L i 0 , ξ i 1 , ξ i 2 are uniquely determined and satisfy for sufficiently small. Moreover, we have the estimate and λ 1 , λ 2 , λ 3 , C 1 , C 2 are as in Lemma 2.1.
For the general case (g (x i ) > 0 and g (x i ) < 0), a lower solution is constructed in the following way: (2.14) We also define Clearly Lemma 2.2 is generalized as follows: for sufficiently small, where P is in (2.13). Moreover, we have the estimate (2.21) The following lemma is given in Lemma 2.2 in [8].
We will extend u to the entire interval (0, 1) in the following way. Set Let β > 0 be an arbitrarily small constant such that 5β < µ 0 and set is not C 1 function, however, it is a lower solution in the weak sense. We obtain the following lemma: . It is obvious that Φ(1 − α 1 (L)) ≥ 0 holds for g(x) > 0 and that Φ(0) = 0 holds.
We have only to prove that Φ(U i (x)) > 0 in (x i + β, x i + 2β). On this interval, it holds that g(x) ≥ C 1 β, hereafter, C i (i = 1, 2, · · · ) is a constant independent of , L, β. Note that both u(ξ i 2 ) and 1 − α 1 (L) are close to 1 and u(ξ i This shows that Φ(U i (x)) > 0 holds for sufficiently small.
Note that dist(x) is a nonnegative function defined in the entire interval [0, 1] and d i (x) is a locally defined function near x i . Finally we set (2.24) Now we have the following proposition.
Proposition 2.6. There existsL > 0 such that for all L >L, U * is a lower solution for (2.1) for > 0 sufficiently small.
An upper solution is constructed in a similar fashion as the lower solution, thus we shall give a brief explanation. Letθ whereκ > C 6 . Here C 2 , C 6 , λ 2 are in Lemma 2.1 and In the same way as we have obtained ξ 3 and ξ 4 in (2.26) -(2.28) in [8], we obtain the following lemma: Lemma 2.7. There exists L i 0 > 0 such that for all L > L i 0 , ξ i 3 and ξ i 4 are uniquely determined and satisfy for sufficiently small. Moreover, we have the estimate (2.31) In the similar way as lower solution, we set (2.32) In the same way as we have Lemma 2.5, we obtain (2.33) and obtain the following proposition.
Proposition 2.9. There existsL > 0 such that for all L >L, U * is a upper solution for (2.1) for > 0 sufficiently small.
Proof of Theorem 1.2. By Propositions 2.6 and 2.9, there exists a solution U d of (2.1) such that U * < U d < U * . The linearized stability of U d follows from Lemma 4.1 in [9]. The estimates in the theorem follow from Lemmas 3.1, 3.2 and 3.5 in [9].
3. Uniqueness. In this section we will show uniqueness of the solution which we have constructed in Section 2. Then we prove Theorem 1.1 in the last part of this section. We will use many lemmas in [9] for the proof. [9], any solution u is close to 0 in G − . By Lemmas 3.1 and 3.2 in [9], either of the following holds for each ( . Therefore the equation (2.1) can have only 3 types of solutions in the following: (T1) u is close to 0 in G + (T2) u is close to 1 in G + (T3) There exists S, a subset of G + such that u is close to 1 in S and u is close to 0 in G + \ S. The uniqueness will be proved in the following way. In Lemma 3.1 below, we will show that the case (T1) is impossible under the condition 1 0 g(x) dx ≥ 0. In Lemma 3.2 below, we will show that a solution satisfying (T2) is unique. In Lemma 3.4 below, we will show that the case (T3) is impossible under the condition (P0) -(P3) in Section 1, Introduction.
We will provide the following lemma. Let u be a nontrivial solution of (2.1).
Proof. Assume u (x) < 2 3 for [0, 1]. By the uniqueness of solutions of ODE, we have 0 < u (x) < 1. Using equation (2.1) and the assumption of the lemma, we obtain The left-hand side of the above equality is . This is a contradiction. The lemma is proved. The following lemma shows that a nontrivial solution satisfying (T2) above is unique. By Lemmas 3.8, 3.9, and 3.10 [9], any solution u (x) characterized by (T2) should satisfy U * (x) < u (x) < U * (x) where U * (x) and U * (x) are a pair of an upper and a lower solutions constructed in Section 2. In the following Lemma 3.2 we will show that a nontrivial solution which stays in between U * (x) and U * (x) is unique. Proof. Set p > 3 max |g(x)|. For given u ∈ C[0, 1] we define a map w = Au, A : There is one to one correspondence between a solution of (2.1) and a fixed point of A. By maximum principle, or the variation of constants for the second order ordinally differential equations, A is order preserving in P ( i.e. if u 1 ≥ u 2 (u 1 , u 2 ∈ P ), then Au 1 ≥ Au 2 ). Since U * and U * are a pair of an upper and a lower solutions in strict sense, it holds that AU * > U * , AU * < U * .
Choose any v 0 ∈ P and λ ∈ [0, 1], we define Since A is order preserving, for u ∈ P , it holds that In the same way we have H(u, λ) > U * for u ∈ P . Therefore, H(·, λ) maps P into Let T (·, u ) be a derivation of A with respect to u at u = u . i.e. ω = T (φ, u ) T (·, u ) : The following Lemma 3.3 directly follows from Lemma 4.1 in [9]. Lemma 3.3. Let u be a solution of (2.1) and u ∈ P . Then u is linearly stable.
This lemma is equivalent to the fact that all the eigenvalues of is negative, i.e., all the eigenvalues of T (·, u ) is less than 1. Therefore, T (·, u ) is invertible and u is an isolated fixed point. Thus the fixed point index of A at u = u is well defined (Here, fixed point index means degree of A in the neighborhood of u ) and we have Index(I − A, u ) = 1.
Suppose that (2.1) has more than one solution u i ∈ P (i ∈ I). Then u i (i ∈ I) are fixed points of A in P . By (3.2), for any fixed point u i (i ∈ I), it holds that By (3.3) and the assumption that A has more than one fixed point in P , we have i∈I On the other hand, it holds from (3.1) that This contradicts (3.4). Now we have shown that the solution of (2.1) in P is unique. Lemma 3.2 is proved.
In the next lemma, we will show that any solution which stays close to zero is unstable. Proof.
We will show L(u ) < 0 to show u is unstable. For convenience, we denote u by u in this proof. It holds that (3.5) By the equation (2.1), we have We also obtain by integrating the equation (2.1) Multiplying the equation (2.1) by u , integrating that over [0, 1], we have Therefore, we have It holds from (3.6) that , we obtain L(u) < 0. The lemma is proved.
The following Lemma 3.5 shows that (T3) is impossible for some classes of g(x). Let p(x) ∈ C 1 [0, 1] be a function satisfying (P0) to (P3) in Introduction. The following 2 cases cannot hold simultaneously.
The condition g(x) ≥ p(x) and (P3) imply Proof of Lemma 3.5. We will first show the case g(x) = p(x) in Step 1 and then g(x) > p(x) in Step 2.
Step 1. The case g(x) = p(x) We will first show the case g(x) = p(x). In the proof, we will use Theorem 1.1 in [3] on stability property of u = 0. We can also obtain a proof without using this stability property (we use degree theory on a positive cone instead). However, we use their theorem to make the proof simpler and clearer. Their theorem is concerned with the problem on higher dimension, therefore, we will rewrite the theorem to one dimensional case to suite our problem.
In the case g(0) = g(1) > 0, suppose that there exists u (x) is close to 0 in (0, x 1 ), and u (x) is close to 1 in (x 2 , 1). Since g(x) is symmetric with respect to x = 1 2 , is also a solution of (2.1). Define u min (x) = min{u (x), u (1 − x)}. Since both u (x) and u (1 − x) are upper solutions, u min (x) is also an upper solution.
Note that u min (x) is close to 0. On the other hand, 0 is a lower solution. Therefore, there is at least one stable solution in between 0 and u min (x). However, under the condition (P3), there is no solution nearby 0 except for 0 itself by Lemma 3.1. Therefore, 0 is a stable solution. This contradicts Theorem LNS under the condition (P3).
(2) In the case m ≥ 3, g(x) = p(x). Suppose that u is close to 0 and that g(x) > 0 are satisfied in (x i , x i+1 ). We also suppose that u has at least one layer. There are 3 possibilities as follows: (i) u has at least one layer in (0, x i ), and u has at least one layer in (x i+1 , 1). (ii) u does not have any layer in (0, x i ). (iii) u does not have any layer in (x i+1 , 1).
We will obtain a contradiction for each case. (i) Denote x j = min{z ∈ (x i+1 , 1) | g(z) = 0 and u has a layer at x = z}, x k = max{z ∈ (0, x i ) | g(z) = 0 and u has a layer at x = z}.
Note that both j −(i+1) and i−k are odd numbers. There are only two possibilities u i m ≥ 0 or u i m < 0 (Note that xi+xi+1 2 = i m ). We will first consider the case We will check boundary conditions of u min (x).
We have shown that u min (x) is an upper solution in [ k−1 m , i m ]. It is clear that 0 is a lower solution. Therefore, there exists at least one stable solution w(x) near 0 on this interval. However, we have the condition In case u i is also a solution in [ i m , j m ]. We will check boundary conditions of u min (x).
We have shown that u min (x) is an upper solution in [ i m , j m ]. It is clear that 0 is a lower solution. Therefore, there exists at least one stable solution near 0. However, we have the condition Under the condition (3.11), there is no nontrivial solution nearby 0 on the interval [ i m , j m ] by Lemma 3.1. Therefore, 0 is a stable solution. This contradicts Theorem LNS under the condition (3.11).
(ii) Since u does not have any layers in (0, x i ), there is at least one layer in (x i+1 , 1). Denote In case u i m < 0, define u min (x) = min{u (x), u ( i+j m − x)}. We remark that g(x) is symmetric with respect to x = i+j 2m in [ i m , j m ]. (Since j + 1 − i is an odd number, i+j is an even number). Therefore, u ( i+j m −x) is also a solution in [ i m , j m ]. We will check boundary conditions of u min (x).
We have shown that u min (x) is an upper solution in [ i m , j m ]. It is clear that 0 is a lower solution. Therefore, there exists at least one stable solution near 0. However, we have the condition (iii) We obtain a contradiction in the same way as (ii) (we consider u (1−x) instead of u (x), then this case is reduced to the case (ii)).
We will prove the case g(x) > p(x), where p(x) satisfies (P0) -(P3). Suppose that there exists u , a solution of (2.1), satisfying (ci) and (cii). We will show that u is an upper solution of (3.14) To show that we set Ψ(u) = 3 u + p(x)u 2 (1 − u), we have We will find a lower solution in the following way. Denote the zeros of p(x) by {x i | i = 1, 2, · · · , m} in the same way as in Step 1. We now define a lower solution of (3.14). We recall Section 2, Proposition 2.5 (See also Lemmas 2.2 -2.4), we can construct a lower solution of (3.14) which has a layer near each zeros of p(x), {x i | i = 1, 2, · · · , m}. Denote this layered lower solution by U * (x). Set x 0 = 0 and x m+1 = 1 for convenience. We define Z = { ∈ {0, 1, 2, · · · , m} | There exists s ∈ (x , x +1 ) such that u (s) > 1 2 , where u is a solution of (2.1)}.
x , x +1 )), we have only to prove u (x) > w(x) in (x , x +1 ) for ∈ Z. There are 3 possibilities, = m or = 0 or 0 < < m. Since the case = 0 is treated in the same way as = m, we only have to consider two cases = m and 0 < < m in the following. Let β be a small constant satisfying In the following we will show (i) In the case = m, where α 1 (L) = C 2 exp(−λ 2 (P L) 3 2 ) is a small constant independent of > 0. (i) In the case = m, it holds that g(x) > 0 in (x m , 1). Lemma 3.1 of [9] shows In the case < m, we remark that u is a solution of (2.1), Since g(x) > 0 in (x , x +1 ), Lemma 3.2 shows (3.23) (3.22), (3.23) and (3.21) show (3.19) and (3.20). We consider the interval x ≤ x ≤ x + 5β for = 1, 2, · · · , m. Set We also set Ψ(u) We will check boundary conditions. We have is an upper solution. We will next show (3.27) By (3.22), (3.23) and (3.21), we have On the other hand, for x ≤ x ≤ x + 3β it holds that Setting q = 0, we have (3.20) and (3.30) completes the proof of Lemma 3.6. Now by Lemma 3.6 and the fact that u is an upper solution and w is a lower solution, there exists a solution w(x) of (3.14) with p(x) satisfying (P0) -(P3), It holds from the assumption that u has at least one layer that ∈Z (x , x +1 ) is neither empty set nor [0, 1]. By Step 1, such a solution w(x) satisfying (3.31) cannot exist. This is a contradiction. The proof of Lemma 3.5 is complete. For general g(x), where g(x) does not satisfy g(x) ≥ p(x), we can obtain some information using the similar idea as the proof of Lemma 3.5. For example, the following lemma shows that the condition 1 0 g(x) dx ≥ 0 allows only one kind of one bump solutions. We denote zeros of g(x) by {x i | g(x i ) = 0, i = 1, 2, · · · , m}. Lemma 3.7. Assume 1 0 g(x) dx ≥ 0. Let both u (x) and v (x) be solutions of (2.1). Assume that Then it holds that i = j.
Lemma 3.7 shows that u and v must have layers near the same points.
Proof of Lemma 3.7. Assume that i = j to obtain a contradiction. Set u min (x) = min(u (x), v (x)). If i = j, u min (x) is close to 0. Since both u and v are solutions of (2.1), and u = v , u min is an upper solution in strict sense. Therefore, there is a stable solution u(x) such that 0 ≤ u(x) < u min (x). However, by Lemma 3.1, there is no solution except for u = 0 under the condition 1 0 g(x) dx ≥ 0. Therefore, u = 0 is a stable solution. This contradicts Theorem LNS given in Proof of Lemma 3.5.
4. Concluding remarks. Recently we have found that Part (c) of the conjectures in Introduction is false. We will present the counterexample for Conjecture (c) in this section. In fact, (1.6) has more than three nontrivial solutions for some g(x). We will find such g(x) in the following way.
In addition to U d (x) and ω d (x), the following theorems show that many other nontrivial solutions of (1.6) exist for some k.  1). Then there exists k * (> 0) such that for k = k * and d sufficiently small, (1.6) has a solution u d (x) satisfying (i) and (ii) below: (i) On every compact subset S of [0, z 2 (k)), (ii) On every compact subset S of (z 2 (k), 1], where the constants C 1 , · · · , C 6 depend on S.
Note that C 1 , · · · , C 6 are not the same as those of Theorem 1.2.
Once we find a solution u d satisfying (i) and (ii) in Theorem 4.1, we are able to find many solutions as we will show in the following Theorems 4.2 and 4.3.
Let k * be the constant in Theorem 4.1. The next theorem shows the existence of two different solutions, both of which have only one layer.
Note that C 1 , · · · , C 6 are not the same as those of Theorem 1.2. The next theorem is for the existence of solutions close to 0 when g(x) is symmetric with respect to 1 2 . Theorem 4.3. Suppose that g(x) ∈ C 1 [0, 1] satisfies (H), g(x) < b ex (x; k * ), and that g(x) has exactly two zeros and satisfies g(1 − x) = g(x) in [0, 1]. Then there exists C 1 , C 2 such that for sufficiently small d, (1.6) has at least three solutions w i (x) (i = 1, 2, 3) satisfying C 1 d < w i (x) < C 2 d in [0, 1].
Suppose that g(x) satisfies the same assumption as that of Theorem 4.3. By Theorem 4.1 and 4.2, we obtain 4 solutions. Since h(x) is symmetric, 4 solutions are described as u 1 (x), u 1 (1 − x), u 2 (x), u 2 (1 − x). All of them have only one layer. On the other hand, Theorem 4.3 shows the existence of 3 solutions which stay close to 0. With U d (x), the equation (1.6) has at least 8 nontrivial solutions under the condition.
In this section we will show the outline of the proof of Theorem 4.1, which shows the existence of the third solution. We will first introduce Lemma 4.4 below, which is equivalent to Theorem 4.1. By setting ξ = (2 + k)x − 1 − k, 3 = (2 + k) 2 d (k ≥ 0), U (ξ) = u(x), with k ≥ 0, where  (ii) On every compact subset S of (x 0 , 1], where the constants C 1 − C 6 depend on S. The idea to prove this lemma is as follows. We first consider auxiliary problem of (4.3). In [10] 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 * ). (4.6) Then we can define and show that U ex (ξ; k * ) is a solution satisfying (i) and (ii) in Lemma 4.4. Since both w (x) and u (x; k) satisfy Neumann zero boundary condition, the standard regularity argument shows that U ex is smooth and is a solution of (4.3) in [−k−1, 1]. For the rigorous proof, we refer [10]. Sections 3 -5 in [10] are devoted to prove (4.6). The proof of Theorems 4.2 and 4.3 are given in Section 6 in [10].