In this section, we shall give some preliminaries for the asymptotic behavior of wave profiles near both wave tails. Also, we shall give the proof of Theorem 2.1.1.
Lemma 2.2.1 Let (c, U ) be a solution of (P). Then there exists a positive constant K such that
sup
x∈R,|s|≤1
U (x + s) U (x) + sup
x∈R
| U′(x)| U (x) ≤ K.
Proof. Given a solution (c, U ) of (P). Since
cU′(x)− (2D + d)U(x) ≤ 0,
we obtain that U′(x) ≤ µU(x) for µ ≥ (2D + d)/c. This implies that e−µxU (x) is a non-increasing function. Therefore, we have
(2.2.1) U (x + s)≤ U(x)eµs≤ U(x)eµ, if x∈ R, 0 ≤ s ≤ 1.
So we obtain that
This implies that [U (x− 1/2)/U(x)] is bounded uniformly for x ∈ R. Combining with (2.2.2), we conclude that
sup
x∈R,|s|≤1
U (x + s) U (x) ≤ M1
for some positive constant M1 ∈ R.
Moreover, dividing (2.1.6) by U (x) and using the Lipschitz continuity of b, we ob-tain that supx∈R|U′(x)/U (x)| ≤ M2 for some constant M2. Therefore, the lemma is proved.
Lemma 2.2.2 Let (c, U ) be a solution of (P). Then there exists a positive constant K such
Following the same method as that of Lemma 2.2.1, we can get sup
This contradicts (2.2.3). So we have proved that K(x) < e2µ for all x ∈ R. This implies that
sup
x∈R,|s|≤1
V (x + s)
V (x) ≤ e2µ.
Finally, dividing (2.2.3) by V (x), it is easy to see that sup
x∈R
|V′(x)| V (x) ≤ K for some positive constant K. Hence the lemma follows.
We shall follow the method of [7] to prove Theorem 2.1.1. In the course of proof, we need to analyze the following recurrence equation
(2.2.4) an+3an+2+ l1an+2+ l1
an+1 + 1
an+1an = l, n ∈ Z,
where l1, l are positive constants. Recall that, for the case treated in [7], the recurrence equation is given by
an+1+ 1
an = l, n∈ Z,
for some positive constant l. It is easy to deduce the monotonicity of an, and we can easily obtain the convergence of an. But, the convergence of the sequence {an} defined by (2.2.4) is not trivial.
By taking two consecutive equations from (2.2.4), we have an+2(an+3− an+1) + l1(an+2− an+1)
(2.2.5)
+ l1
an+1an(an− an+1) + 1
an+1anan−1(an−1− an+1) = 0, n∈ Z.
Moreover, we have
(i) If {an}∞n=−∞ is a positive sequence satisfying (2.2.4), then {an}∞n=−∞ is a bounded sequence and (2.2.5) holds.
(ii) If {an}∞n=−∞ is a positive sequence satisfying (2.2.4) and any consecutive four terms are equal, then{an}∞n=−∞ is a constant sequence.
In the sequel, we shall prove the following very technical proposition.
Proposition 2.2.1 If {an}∞n=−∞ is a positive sequence satisfying (2.2.4), then both limits limn→∞an and limn→−∞an exist.
To prove this proposition, we define
Mn:= max{an−2, an−1, an, an+1, an+2}, mn:= min{an−2, an−1, an, an+1, an+2}.
Lemma 2.2.3 Let {an}∞n=−∞ be a non-constant positive sequence satisfying (2.2.4). Then mn < an< Mn, ∀ n ∈ Z.
Proof. By the definitions of Mn and mn, we have mn ≤ an ≤ Mn for all n ∈ Z. Suppose that there exists k∈ Z such that ak= Mk. This implies that
ak≥ ak+2, ak≥ ak+1, ak ≥ ak−1, ak ≥ ak−2. By (2.2.5) with n = k− 1, we have
ak−2 = ak−1 = ak = ak+1 = ak+2.
Therefore,{an}∞n=−∞ is a constant sequence. We have reached a contradiction.
The case that there exists k ∈ Z such that ak= mk can be treated similarly. Therefore, we have proved the lemma.
By Lemma 2.2.3, we have
Mn = max{an−2, an−1, an+1, an+2}, mn= min{an−2, an−1, an+1, an+2}, ∀n ∈ Z.
Indeed, in the next lemma, we have more precise information.
Lemma 2.2.4 Let{an}∞n=−∞ be a non-constant positive sequence satisfying (2.2.4). Then, for each integer k∈ Z, we have
1. min{ak−2, ak−1} = mk < min{ak+1, ak+2}, if Mk = max{ak+1, ak+2};
2. min{ak+2, ak+1} = mk < min{ak−1, ak−2}, if Mk = max{ak−1, ak−2}.
Moreover, either Mk = max{ak+1, ak+2} or Mk= max{ak−1, ak−2}, and they are exclusive.
Proof. Given any integer k. Suppose that Mk = max{ak+1, ak+2}. We claim that mk <
min{ak+1, ak+2}. Suppose not. Then we have mk = min{ak+1, ak+2}. Since an is a non-constant sequence, we have Mk> mk. This implies ak+1 ̸= ak+2. Without loss of generality, we may assume ak+1 > ak+2. So Mk= ak+1 and mk = ak+2. By Lemma 2.2.3, we get
ak+3 = Mk+1> ak+1, ak+4 = mk+2 < ak+2. Continuing in this way, we obtain that
· · · < ak+6 < ak+4 < ak+2 < ak+1 < ak+3 < ak+5 <· · ·
Using (2.2.5) with n = k + 2t for any positive integer t, we have an+2(an+3− an+1) > l1(an+1− an+2)
⇒ an+2(an+3− an+2) + an+2(an+2− an+1) > l1(an+1− an+2)
⇒ an+2(an+3− an+2) > (an+1− an+2)(l1+ an+2)
⇒ an+3− an+2> (1 + l1
an+2)(an+1− an+2) > (1 + l1
ak+2)(an+1− an) Repeating the above process, we get that
ak+2t+1− ak+2t≥ (1 + l1
ak+2)t−1(ak+3− ak+2)
for any positive integer t. This contradicts the boundedness of {an}. Therefore, mk< min{ak+1, ak+2}.
By Lemma 2.2.3, we must have
mk= min{ak−2, ak−1}.
The case when Mk = max{ak−1, ak−2} is similar. Finally, it is clear that only one of the options can hold for each k. The proof is completed.
Lemma 2.2.5 Let {an}∞n=−∞ be a non-constant positive sequence satisfying (2.2.4). Then either Mn = max{an+1, an+2}, mn = min{an−2, an−1} for all n ∈ Z and both {Mn} and {mn} are non-decreasing; or, Mn= max{an−1, an−2}, mn = min{an+2, an+1} for all n ∈ Z and both {Mn} and {mn} are non-increasing.
Proof. Suppose that there is an integer k ∈ Z such that Mk = max{ak+1, ak+2}. By max{ak+1, ak+2} = Mk ≥ max{ak−1, ak, ak+1, ak+2}
and the definition of Mk+1, we have
Mk+1 ≥ max{ak+1, ak+2, ak+3} ≥ max{ak−1, ak, ak+1, ak+2, ak+3} = Mk+1. This and Lemma 2.2.3 imply that
Mk+1 = max{ak+1, ak+2, ak+3} = max{ak+2, ak+3}.
Next, we claim that Mk−1 = max{ak, ak+1}. Suppose not. Then from Lemma 2.2.4 we have Mk−1 = max{ak−3, ak−2}. Moreover, we have mk−1 = min{ak, ak+1}. Since, by assumption, mk = min{ak−2, ak−1}, we have
mk = min{ak−2, ak−1} ≥ mk−1 = min{ak, ak+1} ≥ mk.
From this we deduce that mk = min{ak, ak+1} = ak+1, by Lemma 2.2.3. This contradicts Lemma 2.2.4. Therefore, we conclude that Mk−1 = max{ak, ak+1}.
By induction and Lemma 2.2.4, we get
Mn = max{an+1, an+2}, mn= min{an−2, an−1}, ∀n ∈ Z.
Since
Mn+1 ≥ max{an+1, an+2} = Mn, mn+1= min{an−1, an} ≥ mn, both{Mn} and {mn} are non-decreasing sequences.
The other case is similar. Therefore, the lemma is proved by combining Lemma 2.2.4.
Proof of Proposition 2.2.1. By Lemma 2.2.5, without loss of generality, we may assume that
Mn = max{an+1, an+2} and mn= min{an−2, an−1}, ∀n ∈ Z.
So {Mn} and {mn} are bounded and non-decreasing sequences. Therefore, limn→±∞Mn and limn→±∞mn must exist.
First, we consider limn→∞an. We define M := limn→∞Mn and m := limn→∞mn. If M = m, then limn→∞an exists (since mn< an < Mn).
Suppose that M > m. By hypothesis, Mn = max{an+1, an+2}, mn+3 = min{an+1, an+2}.
This implies
Mn+ mn+3= an+1+ an+2. Since {Mn} and {mn} are non-decreasing sequences,
an+1+ an+2≤ an+2+ an+3, ∀n ∈ Z.
Hence {a2n} and {a2n+1} are non-decreasing sequences. So the limits p := lim
n→∞a2n, q := lim
n→∞a2n+1 are well-defined. If p = q, then limn→∞an exists.
Suppose that p ̸= q. Without loss of generality we may assume that p > q. Now, we
This implies that M = m, a contradiction. Therefore, limn→∞an exists. Similarly, we can prove that limn→−∞an exists. This completes the proof of the proposition.
Now, we turn to the study of (2.1.10). First, we have
Lemma 2.2.6 If r(·) ∈ L1loc(R) solves (2.1.10) in R, then r(·) ∈ L∞(R) ∩ C∞(R).
Therefore, we obtain that
v(x− 1/2) ≤ 2a
a1e−µv(x), ∀x ∈ R.
It follows from (2.2.6) and the fact that v is non-increasing, we conclude that r(·) ∈ L∞(R).
Furthermore, r(·) ∈ C∞(R) by using (2.1.10). This proves the lemma.
Lemma 2.2.7 A locally integrable solution of (2.1.10) that attains its global maximum or minimum must be a constant function.
Proof. Let r be a locally integrable solution of (2.1.10). By differentiating (2.1.10), we get
ar′(x) +
∑2 i=−2
a|i|exp
[∫ x+i x
r(y)dy ]
· [r(x + i) − r(x)] = 0.
If we define
A(x) := exp
[∫ x+1 x
r(y)dy ]
, then the above equality can be re-written as
ar′(x) + a2A(x + 1)A(x)[r(x + 2)− r(x)] + a1A(x)[r(x + 1)− r(x)]
(2.2.8)
+ a1[A(x− 1)]−1[r(x− 1) − r(x)] + a2[A(x− 1)A(x − 2)]−1[r(x− 2) − r(x)] = 0.
Suppose that r attains its global maximum. Without loss of generality and by a trans-lation, we may assume that r(·) attains its global maximum r∗ at x = 0. Hence r′(0) = 0 and r(±2) = r(±1) = r∗. By induction, we easily deduce that r(j) = r∗ and r′(j) = 0 for all j ∈ Z. By (2.1.10), we get
a2A(j + 1)A(j) + a1A(j) + a1/A(j− 1) + a2/[A(j− 1)A(j − 2)] = −ar∗− a0
for all j ∈ Z. Proposition 2.2.1 implies that A± := limj→±∞A(j) exists such that a2(A±)2+ a1(A±) + a1/(A±) + a2/[(A±)2] =−ar∗− a0.
(2.2.9)
Now, let {xi} be a sequence with
ilim→∞r(xi) = r∗ := inf
x∈R{r(x)}.
We claim that r∗ = r∗. By Lemma 2.2.6, {r(xi+·)}∞i=1 is a uniformly bounded and equi-continuous sequence. According to Ascoli-Arzela Theorem, we can extract a subsequence
(still called{r(xi+·)}∞i=1) such that limi→∞r(xi+·) = ˆr(·) uniformly in any compact subset
Finally, without loss of generality (by taking a subsequence if necessary), we may assume that xi ≥ −M for some positive constant M. Since
∫ j+1 is a constant function. The case when r(·) attains its global minimum can be also treated similarly. The lemma is proved.
Lemma 2.2.8 If r(·) ∈ L1loc(R) solves (2.1.10), then r(±∞) := limx→±∞r(x) exists and P (a, a2, a1, a0, r(±∞)) = 0.
Proof. First, we consider {xi} such that
ilim→∞xi =∞, lim
Since the family {r(xi+·)}∞i=1 is uniformly bounded and equi-continuous, we can choose a subsequence (still denoted by {r(xi+·)}) such that limi→∞r(xi+·) = ˆr(·) uniformly in any compact subset of R for some ˆr(·) ∈ C(R). This implies that ˆr(·) is also a solution of (2.1.10). By ˆr(0) = limi→∞r(xi) = r∗ and ˆr(y) = limi→∞r(xi+ y) ≤ r∗ for all y ∈ R, we have ˆr(0) = maxx∈R{ˆr(x)}. By Lemma 2.2.7, we get ˆr(·) ≡ r∗. Then (2.2.11) follows form the uniform convergence of {r(xi+·)}.
Next, we claim that r(∞) exists. Otherwise, we have r∗ := lim infx→∞r(x) < r∗. Since
ilim→∞ max
[xi−2,xi+2]{| r(·) − r∗ |} = 0, lim inf
x→∞ r(x) = r∗ < r∗, we can find j ∈ N such that
r(·) > r∗+ r∗
2 in [xj, xj + 2]∪ [xj+1− 2, xj+1], min
[xj,xj+1]{r(·)} < r∗+ r∗ 2 .
Since r(·) is continuous, we can choose ˆx be the left-most point in [xj, xj+1] such that r(ˆx) = min[xj,xj+1]{r(·)}. From the above fact, we have ˆx ∈ [xj + 2, xj+1− 2] such that
r′(ˆx) = 0, r(ˆx)≤ min{r(ˆx + 2), r(ˆx + 1)}, r(ˆx) < min{r(ˆx − 1), r(ˆx − 2)}.
This leads a contradiction with (2.2.8). Therefore, r(∞) exists.
Similarly, r(−∞) := limx→−∞r(x) exists. Finally, by sending x → ±∞ in (2.1.10), we get
P (a, a2, a1, a0, r(±∞)) = 0.
Then the lemma follows.
Lemma 2.2.9 If r(·) ∈ L1loc(R) is a non-constant solution of (2.1.10), then r(−∞) <
r(x) < r(∞) for all x ∈ R and
∫ ∞
0
[r(∞) − r(y)]dy +
∫ 0
−∞
[r(y)− r(−∞)]dy < ∞.
Proof. According to Lemma 2.2.7, r(x) cannot attain its global maximum or minimum.
This implies that r(∞) ̸= r(−∞) and
Λ1 := min{r(−∞), r(∞)} < r(x) < Λ2 := max{r(−∞), r(∞)}, ∀x ∈ R.
By Lemma 2.2.8, we have
P (a, a2, a1, a0, Λ1) = P (a, a2, a1, a0, Λ2) = 0.
By the graph of P (a, a2, a1, a0, λ), we have
∂
∂λP (a, a2, a1, a0, Λ1) < 0 < ∂
∂λP (a, a2, a1, a0, Λ2).
So we can find ϵ > 0 such that
a + 2a2e2(Λ1+ϵ)+ a1e(Λ1+ϵ)− a1e−(Λ1+ϵ)− 2a2e−2(Λ1+ϵ) < 0, (2.2.12)
a + 2a2e2(Λ2−ϵ)+ a1e(Λ2−ϵ)− a1e−(Λ2−ϵ)− 2a2e−2(Λ2−ϵ) > 0.
(2.2.13)
Suppose that r(∞) = Λ1 and r(−∞) = Λ2. By translation, we may assume that r(x) < Λ1+ ϵ, if x≥ −4.
If we define l := minx∈[−4,4]r(x), then we have l ∈ (Λ1, Λ1+ ϵ). Compare the line h = l−kx with the curve h = r(x) for x > 0. If k ≥ ϵ and 0 < x ≤ 4, then r(x) ≥ l > l − kx. If k ≥ ϵ and x > 4, then r(x) > Λ1 > l− ϵ > l − kx. Combining these two cases, we get that
r(x) > l− kx, ∀ x > 0, k ≥ ϵ.
Therefore, the quantity
δ := inf{k > 0 | r(x) ≥ l − kx, ∀x > 0}
is well-defined and we can easily see that δ ∈ (0, ϵ). Moreover, there is a number x0 ∈ (4, ∞) such that
r(x0)− (l − δx0) = 0 ≤ r(x) − (l − δx), ∀x > 0.
This implies that
r′(x0) =−δ, r(x0 ± i) − r(x0)≥ ∓iδ for i = 1, 2.
Recall A(x) = exp[∫x+1
x r(y)dy] and using Λ1 < r(·) < Λ1+ ϵ in [0,∞), we have aδ
= a2A(x0+ 1)A(x0)[r(x0+ 2)− r(x0)] + a1A(x0)[r(x0 + 1)− r(x0)]
+ a1A(x0 − 1)−1[r(x0− 1) − r(x0)] + a2A(x0− 1)−1A(x0− 2)−1[r(x0− 2) − r(x0)]
≥ δ[−2a2e2(Λ1+ϵ)− a1e(Λ1+ϵ)+ a1e−(Λ1+ϵ)+ 2a2e−2(Λ1+ϵ)].
This leads to a contradiction with (2.2.12). Therefore, r(∞) = Λ2 > r(−∞) = Λ1.
Finally, we claim that
By a direct computation and the Mean Value Theorem, we have
(2.2.14) aR(x) + ∑
where O(1) is uniformly bounded (independent of M ). According to the range of θi(x) and r(·) ≤ Λ1+ ϵ on (−∞, xϵ+ 4], we have
Similarly, using (2.2.13) we can deduce that
∫ ∞
0
[Λ2− r(y)]dy < ∞.
The proof is now completed.
Proof of Theorem 2.1.1. First, parts (i) and (ii) follows from Lemma 2.2.8 and Lemma 2.2.9.
Next, we focus on the case that P (a, a2, a1, a0,·) = 0 has two real roots {Λ1, Λ2} with Λ1 < Λ2. Suppose that r(x) is an arbitrary non-constant solution of (2.1.10). By Lemma 2.2.8 and Lemma 2.2.9, we have r(∞) = Λ2 > Λ1 = r(−∞). We define
By a direct computation, we have
u′(x) = u(x)r(x), u2(0) = 1− θ, 0 < θ < 1.
It is easy to see that ˆr(x) is a solution of (2.1.10). We claim that ˆr(x) is a constant function.
Suppose not. Then it follows from Lemma 2.2.9 that
∫ 0
a contradiction. Therefore, ˆr(x) must be a constant solution of (2.1.10). This also implies that either ˆr(x)≡ Λ1 or ˆr(x)≡ Λ2.
If ˆr(x) ≡ Λ1, then u(x) = ceΛ1x. This implies that r(·) is a constant function, a contradiction. Hence we have ˆr(x)≡ Λ2 and so u2(x) = u2(0)eΛ2x = (1−θ)eΛ2x. Therefore, by adding two constant solutions r(x)≡ Λ1 and r(x)≡ Λ2 together, all solutions of (2.1.10) are given by
r(x) = θΛ1eΛ1x+ (1− θ)Λ2eΛ2x θeΛ1x+ (1− θ)eΛ2x for θ∈ [0, 1]. This proves the theorem.