Stability in the energy space of the sum of N peakons for the Degasperis-Procesi equation

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by  André Kabakouala

N
Stability in the Energy Space of the Sum of
Peakons for the Degasperis-Procesi Equation
Andr´e Kabakouala
L.M.P.T., U.F.R Sciences et Techniques, Universit´e de Tours, Parc Grandmont,
6 37200 Tours, France.
1
0
2 [email protected]

n
a
J
Abstract

8
The Degasperis-Procesi equation possesses well-known peaked solitary waves that are called
2
peakons. Their stability has been established by Lin and Liu in [5]. In this paper, we localize

] theproof(insomesuitablesensedetailed inSection3)ofthestability ofasinglepeakon. Thanksto
P this, we extend theresult of stability to thesum of N peakons traveling to theright with respective
A speeds c1,...,cN,such that thedifference between consecutivelocations of peakons is large enough.
.
h
t
a
m
[
1 Introduction

1
v The Degasperis-Procesi(DP) equation
4
7 u u +4uu =3u u +uu , (t,x) R R (1.1)
6 t− txx x x xx xxx ∈ ∗+×
7 is completely integrable (see [1]) and possesses, among others, the following invariants
0
.
1 E(u)= yv and F(u)= u3, (1.2)
0 R R
Z Z
6
1 where y = (1−∂x2)u and v = (4−∂x2)−1u. Substituting u by 4v−vxx in (1.2) and using integration by
: parts (we suppose that u( )=v( )=v ( )=0), the conservation laws can be rewritten as
v ±∞ ±∞ x ±∞
i
X E(u)= 4v2+5v2+v2 and F(u)= v3 +12vv2 48v2v +64v3 . (1.3)
r ZR x xx ZR − xx xx− xx
a (cid:0) (cid:1) (cid:0) (cid:1)
One can see that the conservationlaw E() is equivalent to 2 . Indeed, using integration by parts
· k·kL2(R)
u 2 = u2 = (4v v )2 = 16v2+8v2+v2 4E(u), (1.4)
k kL2(R) R R − xx R x xx ≤
Z Z Z
(cid:0) (cid:1)
and applying Plancherel-Parsevalidentity
1+ω2
E(u)= yv = u(ω)2 u(ω)2 = u 2 , (1.5)
R R 4+ω2| | ≤ R| | k kL2(R)
Z Z Z
where u denotes the Fourier transform of u. In thebsequel we wbill denote
u = E(u). (1.6)
b k kH
p
1
Notethat,byreversingtheoperator(1 ∂2)()in(1.1),theDPequationcanberewritteninconservation
− x ·
form as
1 3
ut+ 2∂xu2+ 2(1−∂x2)−1∂xu2 =0, (t,x)∈R∗+×R. (1.7)
The DP equation possesses solitary waves called peakons (see Fig. 1a) and defined by
u(t,x)=ϕc(x−ct)=cϕ(x−ct)=ce−|x−ct|, c∈R∗, (t,x)∈R∗+×R, (1.8)
but they are not smooth since ϕ / C1(R) (see Fig. 1b). The peakons are only global weak solutions of
c
(1.7). It means, for any smooth te∈st function φ C (R R), it holds
∞ +
∈ ×
+∞ ϕ (x ct)φ (t,x)dtdx+ 1 +∞ ϕ2(x ct)φ (t,x)dtdx
Z0 ZR c − t 2Z0 ZR c − x
+ 3 +∞ (1 ∂2) 1ϕ2(x ct)φ (t,x)dtdx+ ϕ (x)φ(0,x)dx=0.
2Z0 ZR − x − c − x ZR c
The goal of our work is to prove that ordered trains of peakons are stable under small perturbations in
the energy space (equivalent to L2).
H
Definition 1.1 (Stability). Let c>0 be given. The peakon ϕ is said stable in , if for all ε>0, there
c
H
exists δ >0 such that if
u ϕ δ, (1.9)
0 c
k − kH ≤
then for all t 0, there exists ξ(t) such that
≥
u(t, ) ϕ ( ξ(t)) ε, (1.10)
c
k · − ·− kH ≤
where u(t) is the solution to (1.1) emanating from u .
0
Lin and Liu proved in [5] the stability of a single peakon under the additional condition that (1
∂2)u +(R). Using this result and the general strategy introduced by Martel, Merle and Tsai in [7−]
x 0 ∈M
for the generalized Korteweg-de Vries (gKdV) equation and adapted by El Dika and Molinet in [3] and
[2] for the Camassa-Holm (CH) equation, we prove here the stability of the sum of N peakons for the
DP equation.
Before stating the main result we introduce the function space where will live our class of solutions
to the equation. For I a finite or infinite time interval of R , we denote by (I) the function space 1
+
X
(I)= u C I;H1(R) L I;W1,1(R) , u L (I;BV(R)) . (1.11)
∞ x ∞
X ∈ ∩ ∈
The main result of th(cid:8)e presen(cid:0)t paper is(cid:1)the fol(cid:0)lowing theor(cid:1)em. (cid:9)
Theorem 1.1 (Stability of the Sum of N Peakons). Let be given N velocities c ,...,c such that
1 N
0 < c < ...< c . Let u ([0,T[), with 0 < T + , be a solution of the DP equation. There exist
1 N
∈ X ≤ ∞
C >0, L >0 and ε >0 only depending on the speeds (c )N , such that if
0 0 i i=1
y =(1 ∂2)u +(R) (1.12)
0 − x 0 ∈M
and
N
u ϕ ( z0) ε2, with 0<ε<ε , (1.13)
(cid:13) 0− ci ·− i (cid:13) ≤ 0
(cid:13)(cid:13) Xi=1 (cid:13)(cid:13)H
for some z0,...,z0 satisfyin(cid:13)g (cid:13)
1 N (cid:13) (cid:13)
z0 <...<z0 and z0 z0 L, with L>L >0, i=2,...,N, (1.14)
1 N i − i−1 ≥ 0
1W1,1(R) is the space of L1(R) functions with derivatives inL1(R) and BV(R) is the space of function with bounded
variation.
2
then there exist ξ1(t),...,ξN(t) such that
1 1
N
u(t) ϕ ( ξi(t)) C(√ε+L 1/8), t [0,T[ (1.15)
(cid:13) − ci ·− 1 (cid:13) ≤ − ∀ ∈
(cid:13)(cid:13) Xi=1 (cid:13)(cid:13)H
(cid:13) (cid:13)
and (cid:13) (cid:13)
L
ξ1i(t)−ξ1i−1(t)> 2, ∀t∈[0,T[, i=2,...,N, (1.16)
where ξ1(t),...,ξN(t) are defined in Subsection 4.1.
1 1
2 Preliminaries
Inthissection,webrieflyrecalltheglobalwell-posednessresultsfortheDPequationanditsconsequences
(see [4] and [6] for details).
Theorem 2.1 (Global Weak Solution; See [4] and [6]). Assume that u L2(R) with y =(1 ∂2)u
+(R). Then the DP equation has a unique global weak solution u 0(∈R ) such that0 − x 0 ∈
+
M ∈X
y(t, )=(1 ∂2)u(t, ) +(R), t R . (2.1)
· − x · ∈M ∀ ∈ +
Moreover E() and F() are conserved by the flow.
· ·
Remark 2.1 (Control of L Norm by L2 Norm). From (2.1), it holds
∞
u(x)= e−x x ex′y(x)dx + ex +∞e x′y(x)dx
′ ′ − ′ ′
2 2
Z−∞ Zx
and
u (x)= e−x x ex′y(x)dx + ex +∞e x′y(x)dx,
x ′ ′ − ′ ′
− 2 2
Z−∞ Zx
which lead to
u (x) u(x), x R. (2.2)
x
| |≤ ∀ ∈
Then,usingthe Sobolevembedding ofH1(R)into L (R)and(2.2), weinferthatthere existsaconstant
∞
C >0 such that
S
u L∞(R) CS u H1(R) 2CS u L2(R). (2.3)
k k ≤ k k ≤ k k
Lemma 2.1 (Positivity; See [6]). Let u H1(R) with y =(1 ∂2)u +(R). If k 1, then we have
∈ − x ∈M 1 ≥
(k ∂ )u(x) 0, x R. (2.4)
1 x
± ≥ ∀ ∈
Lemma2.2(Positivity;See[6]). Letw(x)=(k ∂ )u(x). Assumethatu H1(R)withy =(1 ∂2)u
+(R). If k 1 and k 2, then we have 1± x ∈ − x ∈
1 2
M ≥ ≥
(k ∂ )(4 ∂2) 1w(x) 0, x R. (2.5)
2± x − x − ≥ ∀ ∈
3 Stability of a single peakon
TheproofofLinandLiuin[5]isnotentirelysuitableforourwork,becauseitinvolvesalllocalextremaof
thefunction v =(4 ∂2) 1uonR,andthusis notlocal. Forourwork,wehaveto localizethe estimates.
− x −
Therefore, we need to modify a little the proof of Lin and Liu. We do this first for a single peakon.
3
Theorem 3.1 (Stability of Peakons). Let u ([0,T[), with 0 < T + , be a solution of the DP
∈ X ≤ ∞
equation and ϕ bethe peakon defined in (1.8), traveling to theright at the speedc>0. There exist C >0
c
and ε >0 only depending on the speed c>0, such that if
0
y =(1 ∂2)u +(R) (3.1)
0 − x 0 ∈M
and
u ϕ ε2, with 0<ε<ε , (3.2)
0 c 0
k − kH ≤
then
u(t, ) ϕ ( ξ (t)) C√ε, t [0,T[, (3.3)
c 1
k · − ·− kH ≤ ∀ ∈
where ξ (t) R is any point where the function v(t, )=(4 ∂2) 1u(t, ) attains its maximum.
1 ∈ · − x − ·
To prove this theorem we first need the following lemma that enables to control the distance of E(u)
and F(u) to respectively E(ϕ ) and F(ϕ ).
c c
Lemma 3.1 (Control of Distances Between Energies). Let u H1(R) with y =(1 ∂2)u +(R). If
∈ − x ∈M
u ϕ ε2, then
c
k − kH ≤ E(u) E(ϕ ) O(ε2) (3.4)
c
| − |≤
and
F(u) F(ϕ ) O(ε2), (3.5)
c
| − |≤
where O() only depends on the speed c.
·
Proof. For the first estimate, applying triangular inequality, and using that u ϕ ε2 and
c
ϕ =c/√3, we have k − kH ≤
c
k kH
E(u) E(ϕ ) = u ϕ ( u + ϕ )
c c c
| − | |k kH−k kH| k kH k kH
u ϕ ( u ϕ +2 ϕ )
c c c
≤k − kH k − kH k kH
2c
ε2 ε2+
≤ √3
(cid:18) (cid:19)
O(ε2).
≤
Forthe secondestimate, applyingthe Ho¨lderinequality,andusing that u ϕ ε2 and(2.3), we
c
k − kH ≤
have
F(u) F(ϕ ) u3 ϕ3
| − c |≤ R − c
Z
(cid:12) (cid:12)
(cid:12)u ϕ ((cid:12)u2+uϕ +ϕ2)
≤ R| − c| c c
Z
1/2
≤ku−ϕckL2(R) R(u2+uϕc+ϕ2c)2
(cid:18)Z (cid:19)
1/2
=ku−ϕckL2(R) R(u4+2u3ϕc+3u2ϕ2c +2uϕ3c +ϕ4c)
(cid:18)Z (cid:19)
u ϕc L2(R)
≤k − k
1/2
8 1
· 4CS2kuk4L2(R)+4cCSkuk2L2(R)+3c2kuk2L2(R)+ 3c3CSkukL2(R)+ 2c4
(cid:18) (cid:19)
O(ε2),
≤
where we also use that the L2 norm of u is bounded and the following measures of peakon:
2 1
kϕckL∞(R) =c, kϕckL3(R) = 3 3c and kϕckL4(R) = √42c.
r
4
This proves the lemma. (cid:3)
Now, to proveTheorem 3.1, by the conservationof E(), F() and the continuity of the map t u(t)
from [0,T[ to H1(R) ֒ (since L2 and u L2(R)· u· H1(R)), it suffices to prove that f7→or any
function u H1(R) sa→tisfHying y =(H1 ≃∂2)u k+k(R), (3≤.4)kankd (3.5), if
∈ − x ∈M
inf u ϕ ( ξ) ε1/4, (3.6)
c
ξ Rk − ·− kH ≤
∈
then
u ϕ ( ξ ) C√ε, (3.7)
c 1
k − ·− kH ≤
where ξ R is any point where the function v =(4 ∂2) 1u attains its maximum.
1 ∈ − x −
WedividetheproofofTheorem3.1intoasequenceoflemmas. Inthesequel,wewillneedtointroduce
the following smooth-peakons defined for all x R by:
∈
c c
ρc(x)=cρ(x)=(4−∂x2)−1ϕc(x)= 3e−|x|− 6e−2|x|. (3.8)
One can check that ρ H3(R) ֒ C2(R) (by the Sobolev embedding) since ϕ H1(R). Indeed, we
c c
∈ → ∈
have
(1+ω2)3
ρ 2 = ϕ (ω)2 (1+ω2)ϕ (ω)2 ϕ 2 =2c2. (3.9)
k ckH3(R) R (4+ω2)2| c | ≤ R | c | ≤k ckH1(R)
Z Z
Moreover,ρ is a positive even function which decays to 0 at infinity, and admits a single maximum c/6
c b b
at point 0 (see Fig. 1a-1c).
Lemma 3.2 (Uniform Estimates). Let u H1(R) with y = (1 ∂2)u +(R), and ξ R. If
∈ − x ∈ M ∈
u ϕ ( ξ) ε1/4, then
c
k − ·− kH ≤ u ϕc( ξ) L∞(R) O(ε1/8) (3.10)
k − ·− k ≤
and
v ρc( ξ) L∞(R) O(ε1/4), (3.11)
k − ·− k ≤
where v =(4 ∂2) 1u and ρ is defined in (3.8).
− x − c
Proof. For the second estimate, applying the Ho¨lder inequality and using assumption, we get for all
x R,
∈
1
v(x) ρ (x ξ) e 2x′ u(x x) ϕ [(x x) ξ] dx
c − | | ′ c ′ ′
| − − |≤ 4 R | − − − − |
Z
1/2 1/2
1
e 4x′ dx u(x) ϕ (x ξ)2dx
− | | ′ ′ c ′ ′
≤ 4 R R| − − |
(cid:18)Z (cid:19) (cid:18)Z (cid:19)
1
u ϕ ( ξ)
c
≤ 2√2k − ·− kH
O(ε1/4).
≤
For the first estimate, note that the assumption y =(1 ∂2)u 0 implies that u=(1 ∂2) 1y 0
and satisfies (2.2). Then, applying triangular inequality, a−ndxusin≥g that ϕ = ϕ on R −andx(2−.3),≥we
| ′c| c
have
u ϕc( ξ) H1(R) u H1(R)+ ϕc H1(R)
k − ·− k ≤k k k k
2 u L2(R)+2 ϕc L2(R)
≤ k k k k
2 u ϕc( ξ) L2(R)+4 ϕc L2(R)
≤ k − ·− k k k
O(ε1/4)+O(1).
≤
5
Now, applying the Gagliardo-Nirenberginequality and using assumption, we obtain
ku−ϕc(·−ξ)kL∞(R) ≤CGku−ϕc(·−ξ)k1L/22(R)ku−ϕc(·−ξ)k1H/12(R)
O(ε1/8) O(ε1/8)+O(1)
≤
O(ε1/8).(cid:16) (cid:17)
≤
This proves the lemma. (cid:3)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
−10 −8 −6 −4 −2 0 2 4 6 8 10
(a)ϕ(x)andρ(x)profiles.
1 2
0.8 1.8
0.6 1.6
0.4 1.4
0.2 1.2
0 1
−0.2 0.8
−0.4 0.6
−0.6 0.4
−0.8 0.2
−1 0
−10 −8 −6 −4 −2 0 2 4 6 8 10 −10 −8 −6 −4 −2 0 2 4 6 8 10
(b)ϕ′(x)andρ′(x)profiles. (c)(1−∂x2)ϕ(x)=2δ0 and(1−∂x2)ρ(x)=(1/2)e−2|x| profiles
Figure 1: Variation of peakon and smooth-peakon at initial time with the speed c=1.
Lemma 3.3 (Quadratic Identity; See [5]). For any u L2(R) and ξ R, it holds
∈ ∈
c
E(u) E(ϕ )= u ϕ ( ξ) 2 +4c v(ξ) , (3.12)
c c
− k − ·− kH − 6
(cid:16) (cid:17)
where v =(4−∂x2)−1u and c/6=ρc(0)=maxx∈Rρc(x−ξ).
6
Sketch of proof. The prooffollows by direct computation, with the aidof two integrationby parts,
and using that (1 ∂2)ϕ ( ξ)=2cδ , where δ denotes the Dirac mass applied at point ξ. (cid:3)
− x c ·− ξ ξ
Letu H1(R)withy =(1 ∂2)u +(R),andassumethatthereexistsξ Rsuchthat(3.6)holds
for some ∈ξ R. We consider n−owxthe∈inMterval in which the peakon (respectivel∈y the smooth-peakon) is
∈
concentrated, and we will decompose this interval according to the variation of v = (4 ∂2) 1u in the
− x −
following way: we set
c c
α=sup x<ξ, v(x)= and β =inf x>ξ, v(x)= . (3.13)
2400 2400
n o n o
According to Lemma 3.2, we know that v is close to ρ ( ξ) in L norm with ρ (0)=c/6. Therefore v
c ∞ c
·−
musthaveatleastonelocalmaximumon[α,β]. Assumethaton[α,β]thefunctionv admitsk+1points
(ξ )k+1 with local maximal values for some integer k 0, where ξ is the first local maximum point and
j j=1 ≥ 1
ξ the last local maximum point2. Then between ξ and ξ , the function v admits k points (η )k
k+1 1 k+1 j j=1
with local minimal values. We rename α=η and β =η so that it holds
0 k+1
η <ξ <η <...<ξ <η <ξ <η <...<η <ξ <η . (3.14)
0 1 1 j j j+1 j+1 k k+1 k+1
Let
M =v(ξ ), j =1,...,k+1, and m =v(η ), j =1,...,k. (3.15)
j j j j
By construction
v (x) 0, x [η ,ξ ], j =1,...,k (3.16)
x j 1 j
≥ ∀ ∈ −
and
v (x) 0, x [ξ ,η ], j =1,...,k+1. (3.17)
x j j
≤ ∀ ∈
We claim that
c
v(x) , x R [η ,η ], (3.18)
0 k+1
≤ 300 ∀ ∈ \
c
u(x) , x R [η ,η ], (3.19)
0 k+1
≤ 300 ∀ ∈ \
and there exits C >0 such that
0
[η ,η ] [ξ C ,ξ+C ]. (3.20)
0 k+1 0 0
⊂ −
Indeed, for some 0<ε 1 fixed, using (3.11) we have
≪
c c
ρ (η ξ)=v(η )+O(ε1/4)= +O(ε1/4) .
c 0 0
− 2400 ≤ 350
Please note that, we abuse notation by writing that the difference between v and ρ ( ξ) is equal to
c
·−
O(ε1/4). Therefore, using that ρ ( ξ) is increasing on ] ,ξ], it holds for all x ] ,η [,
c 0
·− −∞ ∈ −∞
c c
v(x)=ρ (x ξ)+O(ε1/4) +O(ε1/4) .
c
− ≤ 350 ≤ 300
Proceeding in the same way for x ]η ,+ [, we obtain (3.18).
k+1
One can remark that for all x ∈R, ∞
∈
c c
ϕc(x) 6ρc(x)=ce−|x| 6 e−|x| e−2|x| = ce−|x|+ce−2|x| 0. (3.21)
− − 3 − 6 − ≤
(cid:16) (cid:17)
Thus, combining (3.10), (3.21) and proceeding as for the estimate (3.18), we infer (3.19).
Finally, from (3.11) we have
c c
ρ (η ξ)=v(η )+O(ε1/4)= +O(ε1/4) .
c 0 0
− 2400 ≥ 3000
2Inthecaseofaninfinitecountable numberoflocalmaximalvalues,theproofisexactlythesame.
7
Therefore, since ρ = (c/3)e (c/6)e 2 and that x (1/3)e x (1/6)e 2x is a positive even
c −|·| − |·| −| | − | |
function decreasing to 0 on R (−see Fig. 1a), there exists7→a universal co−nstant C > 0 such that (3.20)
+ 0
holds.
We now are ready to establish the connection between the conservation laws. Please note that, we
will change the order of the extrema of v =(4 ∂2) 1u while keeping the same notations as in (3.15).
− x −
Lemma 3.4 (Connection Between E() and the Local Extrema of v). Let u H1(R) and v = (4
∂2) 1u H3(R). Define the function g·by ∈ −
x − ∈
2v+v 3v , x<ξ ,
xx x 1
−
2v+v +3v , ξ <x<η ,
g(x)= xx x j j j =1,...,k. (3.22)
2v+vxx−3vx, ηj <x<ξj+1,
2v+v +3v , x>ξ ,
xx x k+1
Then it holds 
k k
g2(x)dx=E(u) 12 M2 m2 . (3.23)
R −  j+1− j
Z j=0 j=1
X X
 
Proof. We have
g2(x)dx= ξ1 g2(x)dx+ k ξj+1g2(x)dx+ +∞g2(x)dx. (3.24)
ZR Z−∞ Xj=1Zξj Zξk+1
For j =1,...,k,
ξj+1 ηj ξj+1
g2(x)(x)dx = (2v+v +3v )2+ (2v+v 3v )2
xx x xx x
−
Zξj Zξj Zηj
=J +I.
Let us compute I,
ξj+1
I = 4v2+v2 +9v2+4vv 12vv 6v v
xx x xx− x− x xx
Zηj
(cid:0) (cid:1)
ξj+1 ξj+1 ξj+1 ξj+1
= 4v2+v2 +9v2 +4 vv 12 vv 6 v v
xx x xx− x− x xx
Zηj Zηj Zηj Zηj
(cid:0) (cid:1)
ξj+1
= 4v2+v2 +9v2 +I +I +I .
xx x 1 2 3
Zηj
(cid:0) (cid:1)
Applying integration by parts and using that v (ξ )=v (η )=0, we get
x j x j
ξj+1 ξj+1 ξj+1
I = 4 v2, I = 6 ∂ (v2)= 6v2(ξ )+6v2(η ) and I = 3 ∂ (v2)=0.
1 − x 2 − x − j+1 j 3 − x x
Zηj Zηj Zηj
Therefore
ξj+1
I = 4v2+5v2+v2 6v2(ξ )+6v2(η ). (3.25)
x xx − j+1 j
Zηj
(cid:0) (cid:1)
Similar computations lead to
ηj ξ1 ξ1
J = 4v2+5v2+v2 6v2(ξ )+6v2(η ), g2(x)dx= 4v2+5v2+v2 6v2(ξ ) (3.26)
x xx − j j x xx − 1
Zξj (cid:0) (cid:1) Z−∞ Z−∞(cid:0) (cid:1)
8
and
+ +
∞g2(x)dx= ∞ 4v2+5v2+v2 6v2(ξ ). (3.27)
x xx − k+1
Zξk+1 Zξk+1
(cid:0) (cid:1)
Adding I and J, and summing over j 1,...,k , we obtain
∈{ }
ξk+1 ξk+1 k k k
g2(x)dx= 4v2+5v2+v2 6 v2(ξ ) 6 v2(ξ )+12 v2(η ). (3.28)
x xx − j+1 − j j
Zξ1 Zξ1 j=1 j=1 j=1
(cid:0) (cid:1) X X X
The lemma follows by combining (3.24) and (3.26)-(3.28). (cid:3)
Lemma 3.5 (Connection Between F() and the Local Extrema of v). Let u H1(R) and v = (4
∂2) 1u H3(R). Define the function h·by ∈ −
x − ∈
v 6v +16v, x<ξ ,
xx x 1
− −
v +6v +16v, ξ <x<η ,
h(x)= − xx x j j j =1,...,k. (3.29)
 −vxx−6vx+16v, ηj <x<ξj+1,
v +6v +16v, x>ξ ,
xx x k+1
−
Then it holds 
k k
h(x)g2(x)dx=F(u) 144 M3 m3 . (3.30)
R −  j+1− j
Z j=0 j=1
X X
 
Proof. We have
h(x)g2(x)dx= ξ1 h(x)g2(x)dx+ k ξj+1h(x)g2(x)dx+ +∞h(x)g2(x)dx. (3.31)
ZR Z−∞ Xj=1Zξj Zξk+1
For j =1,...,k,
ξj+1 ηj
h(x)g2(x)dx= ( v 6v +16v)(2v+v 3v )2
xx x xx x
− − −
Zξj Zξj
ξj+1
+ ( v +6v +16v)(2v+v +3v )2
xx x xx x
−
Zηj
=J +I.
Let us compute I,
ξj+1 ξj+1 ξj+1
I = v3 +12vv2 +64v3+60v2v 54 v3+27 v2v
− xx xx xx − x x xx
Zηj Zηj Zηj
(cid:0) (cid:1)
ξj+1 ξj+1 ξj+1
108 vv v 216 v2v +216 vv2
− x xx− x x
Zηj Zηj Zηj
ξj+1
= v3 +12vv2 +64v3+60v2v
− xx xx xx
Zηj
(cid:0) (cid:1)
ξj+1
54 v3+I +I +I +I .
− x 1 2 3 4
Zηj
Applying integration by parts and using that v (ξ )=v (η )=0, we get
x j x j
ξj+1 ξj+1 ξj+1
I =9 ∂ (v3)=0, I =54 v3, I = 72 ∂ (v3)= 72v3(ξ )+72v3(η )
1 x x 2 x 3 − x − j+1 j
Zηj Zηj Zηj
9
and
ξj+1 ξj+1
I =108 ∂ (v2)v = 108 v2v .
4 x x xx
−
Zηj Zηj
Therefore
ξj+1
I = v3 +12vv2 +64v3+60v2v 72v3(ξ )+72v3(η ). (3.32)
− xx xx xx − j+1 j
Zηj
(cid:0) (cid:1)
Similar computations lead to
ηj
J = v3+12vv2 +64v3+60v2v 72v3(ξ )+72v3(η ), (3.33)
− x xx xx − j j
Zξj
(cid:0) (cid:1)
ξ1 ξ1
h(x)g2(x)dx= v3+12vv2 +64v3+60v2v 72v3(ξ ) (3.34)
− x xx xx − 1
Z−∞ Z−∞(cid:0) (cid:1)
and
+ +
∞h(x)g2(x)dx= ∞ v3+12vv2 +64v3+60v2v 72v3(ξ ). (3.35)
− x xx xx − k+1
Zξk+1 Zξk+1
(cid:0) (cid:1)
Adding (3.25) and (3.33), and summing over j 1,...,k , we obtain
∈{ }
ξk+1 ξk+1 k
h(x)g2(x)dx= v3+12vv2 +64v3+60v2v 72 v3(ξ )
− x xx xx − j+1
Zξ1 Zξ1 j=1
(cid:0) (cid:1) X
k k
72 v3(ξ )+72 v3(η ). (3.36)
j j
−
j=1 j=1
X X
The lemma follows by combining (3.31) and (3.34)-(3.36). (cid:3)
Lemma 3.6 (Connection Between E() and F()). Let u H1(R), with y = (1 ∂2)u +(R), that
satisfies (3.6) for some ξ R. Assume·that v =(·4 ∂2) 1u∈satisfies (3.13)-(3.20)−, wxith lo∈caMl extrema on
∈ − x −
[η ,η ] arranged in decreasing order in the following way:
0 k+1
M M ... M 0, m m ... m 0, M m , j =1,...,k. (3.37)
1 2 k+1 1 2 k j+1 j
≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥
There exists ε >0 only depending on the speed c, such that if 0<ε<ε , then it holds
0 0
1 1
M3 E(u)M + F(u) 0. (3.38)
1 − 4 1 72 ≤
Proof. The key is to show that h 18M on R. Note that by (3.6) we know that 18M c/4. We
1 1
≤ ≥
rewrite the function h as
v 6v +16v, x<η ,
xx x 0
− −
∂2+3∂ +2 v 3v +18v, η <x<ξ ,
 − x x − x 0 1
h(x)= −−(cid:0)(cid:0)∂∂xx22+−33∂∂xx++22(cid:1)(cid:1)vv−+33vvxx++1188vv,, ηξjj <<xx<<ηξjj+,1, j =1,...,k.
First, one can remarktha−−t(cid:0)v(cid:0)f∂oxxrx2+a−ll63xv∂xx++R12,6(cid:1)(cid:1)vv,+x3v>xη+k+118,v, ξk+1 <x<ηk+1,
∈
v(x)= e−2x x e2x′u(x)dx + e2x +∞e 2x′u(x)dx
′ ′ − ′ ′
4 4
Z−∞ Zx
10