=O(1),
such that Leibniz’s formula yields the result. Note that the contribution||−βcomes obviously
from iterations of the operator D,h.
7.3 Action-Angle inU◦
In this subsection we provide a rather complete description of the change of variable inU◦. Proposition 7.9 There exists a symplectic change of variable(x, v) →(ψ,h)from U◦to the set
V◦:= {(ψ,h)∈R2,|h∈(−M0,M0), ψ∈(−r◦(h),r◦(h))}, with
r◦(h)= 2k(h)
√M0K(k(h)), where k(h)=
h+M0 2M0 .
such that the flow of the pendulum in the variable(ψ,h)is h(t)=h(0)andψ(t)=t+ψ(0).
There exists then a second symplectic change of variables(ψ,h)→(θ,a)from V◦to {(θ,a)∈R2,|a∈J◦=
0,8
π M0
, θ∈(−π, π)}, such that
⎧⎪
⎨
⎪⎩
a(h) = 8√ M0
π (E(k(h))−(1−k(h)2)K(k(h))) θ(ψ,h)=ω◦(h)ψ
with ω◦(h)= π√ M0
2K(k(h)) and so that the flow of the pendulum in the variables(θ,a)is a(t) = a(0) andθ(t) = tω◦(a(0))+ψ(0).
Moreover, we can easily express(x, v)as functions of the variables(θ,h)with the formulae x(θ,h)=2 arcsin
k(h)sn
2
πK(k(h)) θ+π
2 ,k(h)
, (7.29)
v(θ,h)=2k(h) M0cn
2
πK(k(h)) θ+π
2 ,k(h)
. (7.30)
Remark 7.10 Note that we can check directly from the formulae thatω◦(h)is decreasing, and ashis a strictly increasing function ofh(see (7.31)),ω◦(a)is decreasing, and∂aω◦(a) <0, a∈ J◦.
Proof In this case, we haveh∈(−M0,M0)and we can write v(x,h)=∗
2(h+M0cos(x))
defined forh+M0cos(x) ≥0, where∗ =1 ifv≥0 and∗ = −1 ifv≤0. Using this representation, both setsU◦,+ =U◦∩{v≥0}andU◦,−=U◦∩{v≤0}can be parametrized as
U◦,∗= {(x,h)∈T×(−M0,M0)|h≥ −M0cos(x)}
= {(x,h)|h×(−M0,M0),x ∈(−x0(h),x0(h))},
wherex0(h)is the solution in[0, π]of the equationh+M0cos(x0(h))=0. Note that have sin2(x0(h)/2)=k(h)2.
Forx ∈(−x0(h),x0(h)), let us define(x,h)∈(−π/2, π/2)as the unique solution of k(h)sin((x,h))=sin(x/2).
This solution is well defined when x ∈ (−x0(h),x0(h))as k(h)1 sin(x2) ∈ (0,1)in this interval. Note that(0,h)=0,(−x0(h),h)= −π2 and(x0(h),h)= π2. Moreover, by taking the derivative with respect tox, we have
k(h)cos((x,h))∂x(x,h)= 1
2cos(x/2)=1 2
1−k(h)2sin((x,h))2. In particular, we have
1−sin2((x,h))= 1 2k(h)∂x(x,h)
1−k(h)2sin((x,h))2. Then we have
U◦,∗= {(x,h)|h∈(−M0,M0), (x,h)∈(−π/2, π/2)}, and we can define the generatrix functionS(x,h)onU◦,∗by the formula
S(x,h)=∗
x
x0(h)
2(h+M0cos(y))dy
=∗
x
x0(h)
2(h+M0)−4M0sin2(y/2)dy
=∗2k(h) M0
x
x0(h)
1−sin2(y/2) k(h)2 dy
=∗2k(h) M0
x
x0(h)
1−sin2((y,h))dy
=∗2k(h) M0
x
x0(h)
1−sin2((y,h)) 1
∂x(y,h)∂x(y,h)dy
=∗4k(h)2 M0
(x,h)
π 2
1−sin2(φ) 1−k(h)2sin2(φ)dφ.
Whence
S(x,h)=∗4
M0(E((x,h),k(h))−E(k(h)))
−∗4
M0(1−k(h)2)(F((x,h),k(h))−K(k(h))).
Note that this function is equal to zero on the axis{v=0,x ∈ [0, π]}and has a discontinuity
.Hence the function
θ =∗ π√ frequency and action are then given by
ω◦(h)= π√ Using the properties of the elliptic functions, we have
(x(θ,h),h))=∗am
Now using the first formula of (7.9) we see that the expression (7.32) does actually not depend on the value of∗= ±1,and thus
=∗
2(h+M0)
1−sin2((x(θ,h),h)
=∗2k(h) M0cn
2
πK(k(h)) θ+∗π
2 ,k(h)
, and using the second formula of (7.9), it yields
v(θ,h)=2k(h) M0cn
2
πK(k(h)) θ+π
2 ,k(h)
,
and concludes the proof.
Proposition 7.11 The functionω◦(h), x(θ,h)andv(θ,h)are analytic forθ∈(−π, π)and h∈(−M0,M0).
The functionω◦exhibits the following asymptotic behavior ω◦(h)∼ π√
M0
(−log(M0−h)) when h →M0−, ω◦(h)=
M0− 1 8√
M0(h+M0) when h→ −M0,
(7.33)
and there exists constants, Cr,ωr withω1= 8√1M
0 andαr =0such that for all r≥1, (h+M0)−1
∂hrω◦(h)−ωr
L∞(−M0,0)≤Cr, log(h−M0)2!
log(h−M0)2(h−M0)r∂hrω◦(h)−αr"
L∞(0,M0)≤Cr. (7.34) The change of variable(h, θ)→(x, v)satisfies the following estimates: When h→ −M0 it converges towards the action-angle variable of the harmonic oscillator 12(v2+M0x2)in the sense that
x(θ,h)∼2
h+M0
2M0 cos(θ) when h→ −M0, v(θ,h)∼ −2
M0
h+M0 2M0
sin(θ) when h→ −M0.
(7.35)
More precisely, for r,s≥0there exist constants Cr,ssuch that
(h+M0)r−12∂rh∂θs
x(θ,h)−2
h+M0
2M0 cos(θ)
L∞((−π,π)×(−M0,0))
≤Cr,s,
(h+M0)r−12∂rh∂θs
v(θ,h)+2 M0
h+M0
2M0 sin(θ)
L∞((−π,π)×(−M0,0))
≤Cr,s. (7.36)
Finally, we have for r,s≥1,
|(M0−h)r|log(M0−h)|−s+2∂hr∂θs(x(θ,h))L∞((−π,π)×(0,M0)≤Cr,s and
|(M0−h)r|log(M0−h)|−s+3∂hr∂θs(v(θ,h))L∞((−π,π)×(0,M0)≤Cr,s (7.37) for some constants Cr,s, and
|log(M0−h)|−s∂θs(x(θ,h))L∞((−π,π)×(0,M0)≤Cs for s≥1and some constant Cs.
Proof Let us first prove (7.33) and (7.34), starting with the study ofω◦whenh→M0−.We have, by using (7.2), that
ω◦(h)= π√ M0
2K(k(h))∼ π√ M0
(−log(1−k(h))) and we obtain the result using
1−k(h)=1−
1+h−M0
2M0 ∼ M0−h 4M0 .
This proves the first part of (7.33). Note that the second estimate of (7.34) follows from the estimate on the function 1/K of (7.3), and the smoothness ofk(h)in the vicinityh∼M0.
Whenh → −M0, ω◦(h)is an analytic function ofk(h)2 = h+M2M00,and we have using (7.1)
ω◦(h)= π√ M0
2K(k(h)) = M0−
√M0
4 k(h)2+O((h+M0)2)
=
M0− 1 8√
M0(h+M0)+O((h+M0)2).
The first estimate of (7.34) follows easily.
Let us now study the functionsx(θ,h)andv(θ,h).Using (7.6) and the expression (7.29) and (7.30) ofx(θ,h)andv(θ,h), we obtain the expansions
v(θ,h)=
M0 4π K(k(h))
∞
m=0
q(k(h))m+12 1+q(k(h))2m+1cos
(2m+1) θ+π
2
and
sin(x(θ,h)/2)=k(h)sn 2
πK(k(h)) θ+π
2 ,k(h)
= 2π K(k(h))
∞
m=0
q(k(h))m+21 1−q(k(h))2m+1sin
(2m+1) θ+π
2 .
This, together with the fact that arcsin(z) ∼zis analytic in the vicinity ofz = 0 and the expansions (7.1), shows thatv(θ,h)andx(θ,h)are analytic functions of√
h+M0 when h→ −M0, and that
v(θ,h)∼8
q(k(h)) M0cos
θ+π 2
∼ −2k(h)
M0sin(θ) and
x(θ,h)∼8
q(k(h))sin θ+π
2
∼2k(h)cos(θ)
which yields asymptotics (7.35), and estimates (7.36) follow easily.
It remains to prove (7.37). The analysis is similar to what we did forU+andU−,as we have
v(θ,h)=
M0 4π
K(k(h))R(θ,q(k(h)))
with
By doing an analysis similar to the one performed forU+andU−, we have
∂qr∂θsR(θ,q(k(h)))L∞(0,M0)≤Clog 1
M0−h r+s+1
, Moreover (7.3) shows that
∂hrq(k(h))=O
We deduce from these estimates that we have the same asymptotics as in the case ofU+and U−: Now we can perform a similar analysis for
sin(x(θ,h)/2)=k(h)sn has singularities in±1: recall that we have the expansion (see (4.4.41) in [1])
arcsin(x)= −π/2+
However, this singularity is weaker than the other one coming from functionsq(z)andK(z) in the vicinityz∼1 (see (7.2)). This finishes the proof of (7.37).
The Fourier expansion of cosine and sine functions are given by the following result:
Proposition 7.12 For(θ,a)∈(−π, π)×J◦, with, in terms of the variable h,
C0◦(h)= −1+2E(k(h)) Proof Recall that we have
sin(x(θ,h)/2)=k(h)sn
By using the third formula of (7.10), the definition of the function dn(u,k)and the expansion of sn(u,k)dn(u,k)in (7.7), we have
Using now the first and fourth formulae of (7.10), and the expansion of sn2(u,k)from (7.7), we obtain
which yields (7.38).
As in the case ofU± we can establish estimates for general Fourier coefficients f◦ as follows:
Proposition 7.13 Assume that f is a function satisfying
|α|≤maxm∂xα,vfL∞(U◦)≤Cm for some m≥p+2,with p defined by
p=max{n≥1, ∂x,vα f(0,0)=0,∀α,1≤ |α| ≤n}
(and with the convention that p =0if this set is empty). Then, as long as r+s≤m and s+p+2≤m,we have that for=0,
∂hrf◦(h)=O 1
||s 1
h−M0
r
log 1
h−M0
s
when h →M0−, (7.40) and that
f◦(h)= 1
||s
c(h+M0)p+12 +(h+M0)p+22 r(
h+M0)
, (7.41)
where cis a number uniformly bounded inand r∈Wm−(p+2)−s,∞uniformly in. Proof The estimates near the separatix are exactly the same as in the case ofU+andU−. Let us then focus on the asymptotic nearh= −M0. Taylor-expanding, we can always write that
f(x, v)= f(0,0)+F1(x, v)·(x, v)= f(0,0)+F1(0,0)·(x, v)+F2(x, v)·(x, v)(2) whereF1(x, v)is linear andF2(x, v)is bilinear. We may write for=0 that
f◦(a)= 1 2π
(−π,π)
F1(0,0)·(x, v)+F2(x(θ,a), v(θ,a))·(x(θ,a), v(θ,a))(2) e−iθdθ.
(7.42) We obtain then from Proposition7.11thatU(θ,a) = (x(θ,a), v(θ,a))can be expanded whenhis near−M0as
U(θ,a)=
n≥1
an(θ)(h+M0)n2,
where the functionsanare smooth since(x(θ,a), v(θ,a))is an analytic function of√ h+M0 uniformly inθ.By plugging this expansion in (7.42) and by integrating by partsstimes we get (7.41) forp=0.Ifp>0, we just notice that by further Taylor expansion, we have
f(x, v)= f(0,0)+Fp+1(0,0)·(x, v)(p)+Fp+2(x, v)(x, v)(p+1),
whereFp+1(x, v)isp+1-linear andFp+2 p+2 linear. It suffices then to proceed as above.
From this result, we obtain the following:
Proposition 7.14 Assume that f anϕare real functions satisfying
|α|≤mmax ∂x,vα fL∞(U◦)≤Cm and max
|α|≤M∂αx,vϕL∞(U◦)≤Cr,s
for some m≥0.Let p and q defined as
p=max{n≥1, ∂x,vα f(0,0)=0,∀α,1≤ |α| ≤n}, q=max{n≥1, ∂xα,vϕ(0,0)=0,∀α,1≤ |α| ≤n}.
Then, for m≥5+p+ p+q2 , M≥7+q+ p+q2 , M≥m+2, we have for t≥1 Proof We begin as in Proposition7.7and write that
s≥2δwithδsmall enough, we can split the integral into
M0 As in the proof of Proposition7.7, the idea is again to integrate by parts as long as we can, i.e. as long as the contributions from the boundary pointsh∼ ±M0vanish. The termI2can be handled as before forU+andU− :as∂hω◦does not vanish, only the contribution at the separatixh∼M0matters, and this yields a decay by(1+t)−r assuming enough regularity.
As a matter of fact, we just need to takemlarge enough in order to chooser≥ p+q2 +2. We shall now focus onI1which contains the contribution from the centerh∼ −M0.By using Proposition7.13, we can expandI1under the form
I1 = 1 wherer˜is uniformly ininWm−p−2(since we always assume thatMis much bigger than m,M ≥ m+s), and where the constantcis uniformly bounded in.It is important to notice thatω◦(h)and all its derivatives are non-zero smooth functions in] −M0,c]for any c∈(0,M0),so that during the integration by parts,ω◦will not play any major part.
Let us first consider the case where p+q is even, and write p+q = 2k.Then the polynomial contributions in(h+M0)are(h+M0)k+1in the first integral and(h+M0)k+1+12 in the second integral above. We can thus integrate by partsk+2 times in each of the two integrals (as in the proof of Proposition7.7), in order to obtain that
|I1| 1
||s 1 (1+t)k+2. Takings=2 and summing with respect to,we get the result.
In the casep+q=2k+1,the polynomial contributions in(h+M0)are(h+M0)k+32in the first integral and(h+M0)k+2in the second integral. For the latter we can thus integrate by partsk+3 times as previously to get a decay like 1/(1+t)k+3,which is 1/(1+t)p+q2 +52, and is already faster than the expected decay. For the first integral, we can integrate by parts
k+3 times, except for the most singular term where we can integrate by parts onlyk+2 times without boundary terms to obtain integrals under the form
I˜1= 1
whereis smooth and compactly supported in[0,X). Taylor-expanding the function, we obtain that For the second integral above, we can integrate by parts once to obtain an estimate by
(1+t)k+31 ||s.To handle the first integral we use Lemma7.15below, which yields the decay 1/(1+t)k+2+12.
By noticing thatk+12 +2= p+q2 +2, we finally get the result.
Lemma 7.15 Consider the integral I(t)= Proof Let us setv=tuin the integral, we obtain that
I(t)= 1
(assuming thattis sufficiently large so thatt X≥1). The first integral in the right-hand side above is clearly uniformly bounded by 1/t12. For the second integral in the above right hand side, we can integrate by parts once to get that
1
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