The Legendrian foliation near isolated points

Nel documento Progress in Mathematics Volume 259 (pagine 98-103)

4.4 Analysis at the characteristic set and fine regularity of surfaces

4.4.2 The Legendrian foliation near isolated points

∂Ωr

(0U )· nrds (4.40)

=



Γr

(0U )· nrds =−



Γr

|∇0U|νH(p+)· nrds = (−r + o(r))|Γr|.

On the other hand, the divergence theorem implies g(r) =



r

div(0U )dx1dx2=

 r 0

s| ds. (4.41)

From (4.40) and (4.41) we obtain the ODE (g/g)(r) = (1/r) + o(1/r) whose solution g(r) = cr2+ o(r2) contradicts (4.39). Thus ˜γ1+= ˜γ2+. A similar argument yields the uniqueness and non-transversality of ˜γ. 

4.4.2 The Legendrian foliation near isolated points of the characteristic locus

Next, we turn our attention to isolated characteristic points of t-graphs. We will show that every characteristic curve intersecting a small neighborhood of the pro-jection of such a point p will reach p in finite time. Moreover, to every tangent direction a∈ T(p,u(p))Gu there corresponds one and only one curve of the Legen-drian foliation tangent to a at p.

For u∈ C2(Ω) we define the vector field T (x1, x2) = x2u−x1

2 ,−∂x1u−x1

2

!

= (0U )(x1, x2,·).

4.4. Analysis at the characteristic set and fine regularity of surfaces 85

Note that T = 0 on Su and that the projections of curves in the Legendrian foliation are tangent toT in Ω \ Su. We will also consider the differential ofT :

dpT =

x21,x2u(p)−12 x22u(p)

−∂x21u(p) −∂x21,x2u(p)−12



. (4.42)

Lemma 4.36. Assume p∈ Su is an isolated (projection of a) characteristic point, and|H0(z, u(z))| = o(dist(p, z)−1). Then

Next, we recall from (4.12) that H0= 1

Substituting (4.37) in (4.43) we obtain

H0=

which contradicts the hypothesis|H0| = o(dist(p, x)−1). Setting ∆x = and arguing in the same fashion gives b = 0. Next, set b = c = 0 in (4.44) to obtain

H0=−ad (a + d)∆x∆y + o((∆s)2) (a2(∆x)2+ d2(∆y)2)3/2+ o((∆s)3),

from which we deduce that 2∂x21,x2u(p) = a + d = 0. Thus a =−d and b = c = 0 which gives (i). To conclude the proof we observe that det dpT = 1/4 > 0, whence

IndexpT = 1. 

86 Chapter 4. Horizontal Geometry of Submanifolds

Lemma 4.37. With the hypotheses of Lemma 4.36 in force, there exists a neigh-borhood Up of p such that every characteristic curve γ which intersects Up\ {p}

reaches p in finite time.

Proof. Let q ∈ Ω and use polar coordinates to represent q − p = (∆x, ∆y) = (∆s)ei φ. Let α, β ∈ R satisfy

νH = αei φ+ βi ei φ. In view of (4.37) and Lemma 4.36(i),

x1u +1 2x2=1

2∆y + o(∆s),

x2u−1

2x1=1

2∆x + o(∆s), νH = ∆x

∆s + o(1),∆x

∆s + o(1)

!

(4.45)

as ∆s→ 0. Consequently, α = νH· ei φ=−1 + o(1) and β = o(1).

Next, consider a curve γ as in the statement of the lemma, and represent it in polar coordinates: γ(t) = ∆s(t)ei φ(t). Recalling from Section 4.3.2 the identity γ=−νH, we obtain

(∆s)ei φ+ (∆s)φi ei φ=−νH =−(1 + o(1))ei φ+ o(1)i ei φ. Thus

(∆s) =−1 + o(1) (4.46)

so, in sufficiently small neighborhoods of p, ∆s(t) reaches zero at a finite time

t = T . 

We now combine all of the preceding work to deduce a structure theorem for the projection of isolated points in the characteristic locus.

Theorem 4.38 (Cheng–Hwang–Malchiodi–Yang). Assume:

(H1) p∈ Su is an isolated (projection of a) characteristic point;

(H2) the bound|H0(z, u(z))| = o(dist(p, z)−1) holds for z∈ Su near p;

(H3) for some r0> 0,

 r0 0

sup

z∈∂B(p,r)|H0(z, u(z))| dr < ∞.

Then for all a∈ S1 there exists a unique C1 curve γa such that

(i) γa is characteristic, i.e., the projection of a curve in the Legendrian foliation of Gu,

(ii) p lies in the closure of γa,

(iii) limq∈γa,q→pνH(q) exists and is orthogonal to a.

Moreover, as a ranges over all of S1, such curves γa cover Up\ {p} for some neighborhood Up of p.

4.4. Analysis at the characteristic set and fine regularity of surfaces 87

Proof. Let Up be as in Lemma 4.37. Choose δ > 0 sufficiently small so that B(p, δ) ⊂ Up. For each q ∈ ∂B(p, δ) denote by γ the unique characteristic curve through q. In view of Lemma 4.37, the curve γ will reach p at a finite time T . Con-sider a sequence of parameter values tj T and set qj= γ(tj) so limj→∞qj= p.

Recall from Remark 4.25 that if νH(γ(t)) = exp(i θ(t)) then

H0(γ(t), u(γ(t)) =−θ(t). (4.47) Denote by θj the angle corresponding to qj, i.e., νH(qj, u(qj)) = (cos θj, sin θj).

Using hypothesis (H3) we will show thatj} is a Cauchy sequence. First, observe that

θj− θk =

 tj tk

θ(t) dt =

 tj tk

H0dt. (4.48)

Recall from (4.46) that (∆s)(t) = −1 + o(1) for t near T . Consequently, we can express the parameter t in terms of r = ∆s and estimate t(r) ≈ 1 in a neighborhood of p. Using this observation, letting rj = r(tj), rk = r(tk) and in view of (4.48) we obtain

j− θk| ≤

 rj rk

sup

z∈∂B(p,r)|H0(z, u(z))| |t(r)| dr → 0 as j, k→ ∞, (4.49) thus proving that (θj) is Cauchy. Let us denote by θ(p,q) its limit as j→ ∞. Now we can define a map ψ : ∂B(p, δ)→ S1as follows:

ψ(q) = ei θ(p,q).

To conclude the proof of the theorem it suffices to show that ψ is a homeomor-phism.

Step 1 (ψ is continuous): Essentially we need to prove a result of C1 continuity of solutions of a certain ODE with respect to initial data. Consider a sequence of points qj ∈ ∂B(p, δ) converging to q ∈ ∂B(p, δ). Denote by θj = θ(p,qj) (resp.

θ = θˆ (p,q)) and by γj (resp. ˆγ) the corresponding characteristic curves joining qj to p (resp. q to p). We must have that θj→ θ(p,q). Let φj be the angle in the polar coordinate representation of qj−p. Without loss of generality we may assume that φj is strictly decreasing. Since two curves in the Legendrian foliation cannot cross in B(p, δ)\ {p}, we also have θj ≥ θj+1 for all j. As a monotone and bounded from below sequence, (θj) has a limit θ.

We argue by contradiction. If θ = θ(p,q) then necessarily θ > θ(p,q). In this case we find two rays emanating from p and forming an angle less than θ− θ(p,q)

such that for j sufficiently large, both γj and γ avoid a “fan-shaped” region ˜surrounded by these two rays and ∂B(p, R) for some R > 0.

For any point ˜p∈ ˜Ω we consider the unique characteristic curve ˜γ joining ˜p to p. This curve will intersect ∂B(p, δ) at a point ˜q. Since ˜γ∩ ˜Ω does not intersect any γj, θj> ˜θ(p, ˜q)and ˜q must lie in the arc between qj and q for all j. Thus ˜q = q and ˜γ = ˆγ which is a contradiction.

88 Chapter 4. Horizontal Geometry of Submanifolds

Step 2 (ψ is surjective): If ψ is not surjective then there exists ˜θ such that ei ˜θ S1\ ψ(∂B(p, δ)). Since the latter set is open, we can find a neighborhood Iθ˜ of θ such that e˜ i Iθ˜ is disjoint from the range of ψ. Arguing as in the proof of the continuity of ψ, we deduce the existence of a fan-shaped region ˜Ω, surrounded by two rays and a portion of a circle ∂B(p, R), which avoids all characteristic curves connecting points q∈ ∂B(p, δ) to p. For any ˜p ∈ ˜Ω we consider the unique characteristic curve ˜γ through ˜p and let ˜q be its intersection with ∂B(p, δ). Then clearly we must have ei θ(p,˜q) ∈ ei Iθ˜⊂ S1\ ψ(∂B(p, δ)) which is a contradiction.

Step 3 (ψ is injective): Consider q1, q2 ∈ ∂Bδ distinct such that θ(p,q1) = θ(p,q2). Denote by γi a characteristic curve joining p to qi. Then the angle between the tangent vectors of γ1 and γ2 at p is zero. Consider regions Ωrsurrounded by γ1, γ2and ∂B(p, r). Clearly Ωr is contained in a fan-shaped region with vertex p and aperture θr→ 0. Set Γr= ∂Ωr∩ ∂B(p, r), then

r| ≤ rθr (4.50)

as before. Let nrdenote the outer normal to ∂Ωrand observe that nr⊥ νH along γ1 and γ2. Consequently

g(r) :=



∂Ωr

(0U )· nrds =



Γr

(0U )· nrds

=



Γr

(q(s)− p + o(r)) ·q(s)− p

r ds = (−r + o(r))|Γr|.

On the other hand, the divergence theorem implies g(r) =



r

div(0U )dx1dx2=

 r 0

s| ds. (4.51)

From (4.4.2) and (4.51) we obtain the ODE g

g =1 r + o 1

r

! ,

which yields g(r) = cr2+ o(r2) for some c > 0, in contradiction with (4.50).

Step 4 (ψ−1is continuous): We argue by contradiction. Assume there is a sequence (qj)⊂ ∂B(p, δ) converging to ˜q so that

θj := θ(p,qj)→ θ := θ(p,q) (4.52) with q= ˜q ∈ ∂B(p, δ). Without loss of generality we may assume θj ≥ θj+1≥ θ.

Since q = ˜q we can find ¯q ∈ ∂B(p, δ) such that q = ¯q, ˜q = ¯q and θj ≥ θ(p, ¯q)≥ θ.

But then, by (4.52) and the injectivity of ψ we must have ¯q = q. With this contradiction we complete the proof of the theorem. 

Nel documento Progress in Mathematics Volume 259 (pagine 98-103)