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Critical sets of eigenfunctions of the Laplacian

Laurent Bakri

To cite this version:

Laurent Bakri. Critical sets of eigenfunctions of the Laplacian. Journal of Geometry and Physics,

Elsevier, 2012, 62 (10), pp.2024-2037. �10.1016/j.geomphys.2012.05.006�. �hal-01981177�

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Critical sets of eigenfunctions of the Laplacian

Bakri Laurent

Abstract

We give an upper bound for the (n−1)-dimensional Hausdorff mea- sure of the critical set of eigenfunctions of the Laplacian on compact analyticn-dimensionnal Riemannian manifolds. This is the analog of a result on nodal set of eigenfunctions by H. Donnely and C. Fefferman.

1 Introduction and statement of the results

Let (M, g) be a smooth, compact and connected, n-dimensional Riemannian manifold (n ≥ 2). For u ∈ C

1

(M), we set

N

u

= {x ∈ M : u(x) = 0}

and

C

u

= {x ∈ M : ∇u(x) = 0},

the nodal set of u and the critical set respectively. Our interest in this paper is to deal with the critical set C

u

of the eigenfunctions of the Laplacian :

−∆u = λu, (1.1)

Our main result is the following :

Theorem 1.1.

Let M be a n-dimensionnal, real analytic, compact, con- nected manifold with analytic metric. There exists C > 0 depending only on M such that for any non-constant solution u to (1.1) one has

H

n−1

(C

u

) ≤ C

√ λ,

where H

n−1

denotes the (n − 1)-dimensional Hausdorff measure.

In some way theorem 1.1 gives an anolog upper bound, on the measure

of the critical set C

u

, of what was already known on the nodal set N

u

. To

enlight this fact let us recall briefly some facts about the zero sets N

u

, C

u

and N

u

∩ C

u

.

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First it is well kown that if u is a non trivial solution of second order linear elliptic equation then, in general, all zeros of u are of finite order ([1, 15]), and then one can prove that the Hausdorff dimension of the nodal set N

u

is at most n − 1 (for example, see [3] or [11] for more precise results). Concerning the nodal set of eigenfuntions of the Laplacian operator, S. T. Yau, [25] has conjectured, that

Conjecture 1.2.

C

1

λ ≤ H

n−1

(N

u

) ≤ C

2

λ. (1.2)

Here C

1

, C

2

are positives constants depending only upon M . In case that both the manifold and the metric are real analytic, the problem was solved by H. Donnelly and C. Fefferman [5, 6]. Their method was based on Carleman-type inequalities which have proved to be a powerful tool in strong unique continuation theory ([1, 15, 20, 21, ...]). For smooth metric, the only known upper bound result (n ≥ 3) is due to R. Hardt and L. Simon [11]. They proved that

H

n−1

(N

u

) ≤ (c

√ λ)

c

λ

.

However this result is far from conjecture 1.2 and doesn’t seems to be op- timal. Recently, several authors [4, 12, 19, 22] obtained some lower bounds with polynomial decrease in λ.

Another interseting and related question is the description of the singular set S

u

= N

u

∩ C

u

(some time called critical set or critical zero set) of solu- tions to partial differential equations. We refer to [2, 8, 9, 10, 18] and the references therein for further investigations. We only mentionned that S has, in sufficiently favorable cases [8], Hausdorff dimension not greater than (n − 2) and that there exists an analog conjecture to (1.2) which can be stated as follow in the particular case of equation (1.1) (see also [13, 18]) :

Conjecture 1.3.

H

n−2

(S

u

) ≤ Cλ. (1.3)

On the opposite, not much is known about the critical set C

u

. We would like to recall some principal facts. Generically eigenfunctions are Morse functions [23] and therefore the critical set consits in isolated points. More- over, D. Jakobson and N. Nadirashvili [16] have shown that there exists in dimension two a sequence of eigenfunctions for which the number of critical points is uniformly bounded. An open question is whether the number of critical points goes to infinity with the eigenvalue for generic metrics [24].

The only known result in this direction (to the knowledge of the author) is [26] where it is shown that there exists at least one non-trivial critical point.

On the other hand, there exists simple examples for which the critical set

has Hausdorff dimension n − 1 :

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Example 1.4.

Let (N, g) be a (n−1)−dimensional manifold and define M =

T1

× N where

T1

is the 1−dimensionnal Torus with standard metric, and M is equipped with the product metric. The function f

k

(x, y) = sin(2πkx) is an eigenfunction of ∆

M

with eigenvalue λ := 4π

2

k

2

. The critical set, C

fk

, of f

k

is therefore a set of dimension n − 1.

It is also easy to find some surfaces of revolution with critical set of dimension (n − 1), see [27] p 35. In the light of the above, it seems interresting to obtain an upper bound on the (n − 1)−dimensionnal Hausdorff measure of the critical set, which is achieved with theorem 1.1. One should also notice that in example 1.4 one has H

n−1

(C

fk

) ≥ C √

λ, where C depends only on M. Therefore the upper bound in theorem 1.1 is sharp.

The framework of this paper is the following. Recall that gradient of eigenfunctions satisfies a sytem of linear partial differential equations of the following type

∆V + λV + A · V + B · ∇V = 0,

where ∆ + λ acts diagonally on vector valued functions and A (resp. B) is a (1, 1) (resp. (2, 1)) tensor which only depend on (M, g) but not on λ.

Then, since C

u

is the nodal set of |∇u|, we will adapt the method of strong uniqueness theory based on Carleman estimates to this system. Indeed from a Carleman estimate on ∆ + λ we will be able to obtain the main ingredient in the proof of theorem 1.1 : a doubling inequality on gradient of eigenfunctions,

k∇uk

B2r

≤ e

C(

λ+1)

k∇uk

Br

. (1.4)

In the case M is analytic this doubling estimate will lead to theorem 1.1 by using an argument of Donnely and Fefferman, see [5] theorem 4.1. When M is only assume to be smooth, we will also be able to give a local upper bound of the same kind on H

n−1

(C

u

). To do this we will use a result of C.

B¨ ar (lemma 3.1 of [2]), which gives, for smooth functions, a local control on the measure of the nodal set by the vanishing order.

The paper is organised as follows. In section 2, we established a Carleman estimate for the operator ∆ + λ. In section 3 we establish our doubling inequality (1.4). This relies on compactness of M and standard arguments of quantitative uniqueness theory. Section 4 is devoted to derive estimates on the measure of the critical set.

From now on and until the end we will assume that solutions of (1.1) are non-constants or equivalently that the eigenvalue λ satisfies λ > 0.

Aknowledgement.

I would like to thank Christian B¨ ar for pointing out

useful references.

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2 Carleman estimates

Recall that Carleman estimates are weighted integral inequalities with a weight function e

τ φ

, where the function φ satisfy some convexity properties.

Before proceeding to the proof of our Carleman estimate on the scalar op- erator ∆ + λ, let us define some standard notations. Fore a fixed point x

0

in M, we let : r = r(x) = d(x, x

0

) the Riemannian distance from x

0

. We denote by B

r

(x

0

) the geodesic ball centered at x

0

of radius r. We will denote by k · k the L

2

norm, on eventually vector-valued functions, with obvious interpretation. Let us first define the weight function we will use.

For a fixed number ε such that 0 < ε < 1 and T

0

< 0, we define the function f on ] − ∞, T

0

[ by f(t) = t − e

εt

. One can check easily that, for |T

0

| great enough, the function f verifies the following properties:

1 − εe

εT0

≤ f

0

(t) ≤ 1 ∀t ∈] − ∞, T

0

[,

t→−∞

lim −e

−t

f

00

(t) = +∞. (2.1)

Finally we define φ(x) = −f (ln r(x)). Now we can state the main result of this section:

Theorem 2.1.

There exist positive constants R

0

, C, C

1

, C

2

which depend only on M and ε, such that, for any λ ≥ 0, any x

0

∈ M , any δ ∈ (0, R

0

), any u ∈ C

0

(B

R0

(x

0

) \ B

δ

(x

0

)) and any τ ≥ C

1

λ + C

2

, one has C

r

2

e

τ φ

(∆u + λu)

≥ τ

32

r

ε2

e

τ φ

u

+ τ δ

r−1

e

τ φ

u

+ τ

12

r

1+ε2

e

τ φ

∇u

. (2.2)

Remark 2.2.

This Carleman estimate is quite standard. However in this paper, an estimation including the gradient term in the right hand side is useful. Though it can be obtained from previous Carleman estimates ([1], [5] or [17] by example), we choose for the shake of completness to give a proof.

The most important part to establish Theorem 2.1 is the following :

Lemma 2.3.

There exist positive constants R

0

, C, C

1

, C

2

which depend only on M and ε, such that, for any λ > 0, x

0

∈ M, u ∈ C

0

(B

R0

(x

0

) \ {x

0

}) and τ ≥ C

1

λ + C

2

, one has C

r

2

e

τ φ

(∆u + λu)

≥ τ

32

r

ε2

e

τ φ

u

+ τ

12

r

1+ε2

e

τ φ

∇u

. (2.3)

proof of Lemma 2.3. Hereafter C, C

1

, and c denote positive constants de-

pending only upon M , though their values may change from one line to

another. Without loss of generality, we may suppose that all functions are

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real. We now introduce the polar geodesic coordinates (r, θ) near x

0

. Using Einstein notation, the Laplace operator takes the form :

r

2

∆u = r

2

r2

u + r

2

r

ln( √

γ) + n − 1 r

r

u + 1

√ γ ∂

i

( √

γγ

ij

j

u),

where ∂

i

= ∂

∂θ

i

and for each fixed r, γ

ij

(r, θ) is a metric on

Sn−1

and γ = det(γ

ij

).

Since (M, g) is smooth, we have for r small enough :

r

ij

) ≤ C(γ

ij

) (in the sense of tensors);

|∂

r

(γ )| ≤ C; (2.4)

C

−1

≤ γ ≤ C.

Set r = e

t

, we have ∂

∂r = e

−t

∂t . Then the function u has supported in ] − ∞, T

0

Sn−1

, where |T

0

| will be chosen large enough. In this new vari- ables, we can write :

e

2t

∆u = ∂

t2

u + (n − 2 + ∂

t

ln √

γ )∂

t

u + 1

√ γ ∂

i

( √

γγ

ij

j

u).

The inequalities (2.4) become

t

ij

) ≤ Ce

t

ij

) (in the sense of tensors);

|∂

t

(γ)| ≤ Ce

t

; (2.5)

C

−1

≤ γ ≤ C.

Now we introduce the conjugate operator : L

τ

(u) = e

2t

e

τ φ

∆(e

−τ φ

u) + e

2t

λu

= ∂

t2

u + 2τ f

0

+ n − 2 + ∂

t

ln √ γ

t

u

+

τ

2

f

02

+ τ f

00

+ (n − 2)τ f

0

+ τ ∂

t

ln √ γf

0

u + ∆

θ

u + e

2t

λu,

(2.6)

with

θ

u = 1

√ γ ∂

i

γγ

ij

j

u .

It will be useful for us to introduce the following L

2

norm on ]−∞, T

0

Sn−1

: kV k

2f

=

Z

]−∞,T0Sn−1

V

2

γf

0−3

dtdθ,

where dθ is the usual measure on

Sn−1

. The corresponding inner product is denoted by h·, ·i

f

, i.e

hu, vi

f

=

Z

uv √

γf

0−3

dtdθ.

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We will estimate from below kL

τ

uk

2f

by using elementary algebra and inte- grations by parts. We are concerned, in the computation, by the power of τ and exponenial decay when t goes to −∞. First by triangular inequality one has

kL

τ

(u)k

2f

≥ 1

2 I − II, (2.7)

with

I =

t2

u + 2τ f

0

t

u + τ

2

f

02

u + λe

2t

u + ∆

θ

u

2 f

, II =

τ f00

u + (n − 2)τ f

0

u + τ ∂

t

ln √ γf

0

u + (n − 2)∂

t

u + ∂

t

ln √

γ∂

t

u

2 f

.

(2.8)

We will be able to absorb II later. Then we compute I : I = I

1

+ I

2

+ I

3

,

with

I

1

= k∂

t2

u + (τ

2

f

02

+ λe

2t

)u + ∆

θ

uk

2f

I

2

= k2τ f

0

t

uk

2f

I

3

= 2

D

2τ f

0

t

u, ∂

t2

u + τ

2

f

02

u + λe

2t

u + ∆

θ

u

E

f

(2.9)

In order to compute I

3

we write it in a convenient way:

I

3

= J

1

+ J

2

+ J

3

, (2.10) where the integrals J

i

are defined by :

J

1

= 2τ

R

f

0

t

(|∂

t

u|

2

)f

0−3

√ γdtdθ J

2

= 4τ

R

f

0

t

u∂

i

√ γγ

ij

j

u

f

0−3

dtdθ J

3

=

R

3

(f

0

)

3

+ 2λτ f

0

e

2t

2u∂

t

uf

0−3

√ γdtdθ.

(2.11)

Now we will use integration by parts to estimate each terms of (2.11). Note that f is radial and that 2∂

t

u∂

t2

u = ∂

t

(|∂

t

u|

2

). We find :

J

1

=

R

(4τ f

00

) |∂

t

u|

2

f

0−3

√ γdtdθ

R

2τ f

0

t

ln √

γ |∂

t

u|

2

f

0−3

√ γdtdθ.

The conditions (2.5) imply that |∂

t

ln √

γ| ≤ Ce

t

. Then properties (2.1) on f gives, for large |T

0

| that |∂

t

ln √

γ | is small compared to |f

00

|. Then one has

J

1

≥ −cτ

Z

|f

00

| · |∂

t

u|

2

f

0−3

γdtdθ. (2.12)

Now in order to estimate J

2

we first integrate by parts with respect to ∂

i

: J

2

= −2

R

2τ f

0

t

i

ij

j

uf

0−3

γdtdθ.

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Then we integrate by parts with respect to ∂

t

. We get : J

2

= −4τ

R

f

00

γ

ij

i

u∂

j

uf

0−3

√ γdtdθ +

R

2τ f

0

t

ln √

γγ

ij

i

u∂

j

uf

0−3

√ γdtdθ +

R

2τ f

0

t

ij

)∂

i

u∂

j

uf

0−3

√ γdtdθ.

We denote |D

θ

u|

2

= ∂

i

ij

j

u. Now using that −f

00

is non-negative and τ is large, the conditions (2.1) and (2.5) gives for |T

0

| large enough:

J

2

≥ 3τ

Z

|f

00

| · |D

θ

u|

2

f

0−3

γdtdθ. (2.13)

Similarly computation of J

3

gives : J

3

= −2τ

3R

t

ln( √ γ )u

2

γdtdθ

−λτ

R

(4f

0

− 4f

00

+ 2f

0

t

ln √

γ )e

2t

u

2

f

0−3

γdtdθ. (2.14) Now we assume that

τ ≥ C

1

λ + C

2

. (2.15)

From (2.1) and (2.5) one can see that if C

1

, C

2

and |T

0

| are large enough, then

J

3

≥ −cτ

3 Z

e

t

|u|

2

f

0−3

γdtdθ. (2.16)

Thus far, using (2.12),(2.13) and (2.16), we have : I

3

≥ 3τ

R

|f

00

| |D

θ

u|

2

f

0−3

γdtdθ − cτ

3R

e

t

|u|

2

f

0−3

√ γdtdθ

−cτ

R

|f

00

| |∂

t

u|

2

f

0−3

γdtdθ. (2.17)

Now we consider I

1

: I

1

=

t2

u +

τ

2

f

02

+ λe

2t

u + ∆

θ

u

2 f

.

Let ρ > 0 a small number to be chosen later. Since |f

00

| ≤ 1 and τ ≥ 1, we have :

I

1

≥ ρ

τ I

10

, (2.18)

where I

10

is defined by : I

10

=

p

|f

00

|

h

t2

u +

τ

2

f

02

+ λe

2t

u + ∆

θ

u

i

2

f

(2.19)

and one has

I

10

= K

1

+ K

2

+ K

3

, (2.20)

(9)

with

K

1

=

p

|f

00

| ∂

t2

u + ∆

θ

u

2 f

, K

2

=

p

|f

00

|

τ

2

f

02

+ λe

2t

u

2 f

, K

3

= 2

D

t2

u + ∆

θ

u

|f

00

| ,

τ

2

f

02

+ λe

2t

u

E

f

.

(2.21)

Integrating by parts gives : K

3

= 2

R

f

00

τ

2

f

02

+ λe

2t

|∂

t

u|

2

f

0−3

√ γdtdθ + 2

R

t

h

f

00

τ

2

f

02

+ λe

2ti

t

uu √

γf

0−3

dtdθ

− 6

R

f

002

f

0−1

τ

2

f

02

+ λe

2t

t

uu √

γf

0−3

dtdθ + 2

R

f

00

τ

2

f

02

+ λe

2t

t

ln √

γ∂

t

uuf

0−3

√ γdtdθ + 2

R

f

00

τ

2

f

02

+ λe

2t

|D

θ

u|

2

f

0−3

√ γdtdθ.

(2.22)

Now since 2∂

t

uu ≤ u

2

+ |∂

t

u|

2

and τ ≥ C

1

λ + C

2

, we can use conditions (2.1) and (2.5) to get

K

3

≥ −cτ

2 Z

|f

00

| |∂

t

u|

2

+ |D

θ

u|

2

+ |u|

2

f

0−3

γdtdθ (2.23) We also have

K

2

≥ cτ

4 Z

|f

00

| · |u|

2

f

0−3

γdtdθ (2.24)

and since K

1

≥ 0 ,

I

1

≥ −ρcτ

R

|f

00

| |∂

t

u|

2

+ |D

θ

u|

2

f

0−3

γdtdθ + Cτ

3

ρ

R

|f

00

||u|

2

f

0−3

γdtdθ. (2.25)

Then using (2.17) and (2.25), we get : I ≥ 4τ

2

kf

0

t

uk

2f

+ 3τ

R

|f

00

||D

θ

u|

2

f

0−3

√ γdtdθ + Cτ

3

ρ

R

|f

00

||u|

2

f

0−3

γdtdθ − cτ

3R

e

t

|u|

2

f

0−3

√ γdtdθ

− ρcτ

R

|f

00

| |u|

2

+ |∂

t

u|

2

+ |D

θ

u|

2

f

0−3

γdtdθ

− cτ

R

|f

00

||∂

t

u|

2

f

0−3

√ γdtdθ.

(2.26)

Now one needs to check that every non-positive terms in the right hand side of (2.26) can be absorbed in the first three terms.

First fix ρ small enough such that ρcτ

Z

|f

00

| · |D

θ

u|

2

f

0−3

γdtdθ ≤ 2τ

Z

|f

00

| · |D

θ

u|

2

f

0−3

γdtdθ

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where c is the constant appearing in (2.26). The other terms in the last integral of (2.26) can then be absorbed by comparing powers of τ (for C

2

large enough). Finally since conditions (2.1) imply that e

t

is small compared to |f

00

|, we can absorb −cτ

3

e

t

|u|

2

in Cτ

3

ρ|f

00

||u|

2

.

Thus we obtain : I ≥ Cτ

2R

|∂

t

u|

2

f

0−3

γdtdθ + Cτ

R

|f

00

||D

θ

u|

2

f

0−3

√ γdtdθ + Cτ

3R

|f

00

||u|

2

f

0−3

γdtdθ (2.27)

As before, we can check that II can be absorbed in I for |T

0

| and τ large enough. Then we obtain

kL

τ

uk

2f

≥ Cτ

3

k

p

|f

00

|uk

2f

+ Cτ

2

k∂

t

uk

2f

+ Cτk

p

|f

00

|D

θ

uk

2f

. (2.28) Note that, since τ is large and

p

|f

00

| ≤ 1, one has kL

τ

uk

2f

≥ Cτ

3

k

p

|f

00

|uk

2f

+ cτ k

p

|f

00

|∂

t

uk

2f

+ Cτk

p

|f

00

|D

θ

uk

2f

, (2.29) and the constant c can be choosen arbitrary smaller than C. If we set v = e

−τ φ

u, then we have

ke

2t

e

τ φ

(∆v + λv)k

2f

≥ Cτ

3

k

p

|f

00

|e

τ φ

vk

2f

− cτ

3

k

p

|f

00

|f

0

e

τ φ

vk

2f

+

2c

τ k

p

|f

00

|e

τ φ

t

vk

2f

+ Cτ k

p

|f

00

|e

τ φ

D

θ

vk

2f

. (2.30) Finally since f

0

is close to 1, one can absorb the negative term to obtain

ke

2t

e

τ φ

(∆v + λv)k

2f

≥ Cτ

3

k

p

|f

00

|e

τ φ

vk

2f

+ Cτ k

p

|f

00

|e

τ φ

t

vk

2f

+ Cτ k

p

|f

00

|e

τ φ

D

θ

vk

2f

. (2.31) It remains to get back to the usual L

2

norm. First note that since f

0

is close to 1 (2.1), we can get the same estimate without the term (f

0

)

−3

in the integrals. Recall that in polar coordinates (r, θ) the volume element is r

n−1

γdrdθ, we can deduce from (2.27) by substitution that : kr

2

e

τ φ

(∆v + λv)r

n2

k

2

≥ Cτ

3

kr

ε2

e

τ φ

vr

n2

k

2

+ Cτ kr

1+ε2

e

τ φ

∇vr

n2

k

2

. (2.32) Finally one can get rid of the term r

n2

by replacing τ with τ +

n2

. Indeed from e

τ φ

r

n2

= e

(τ+n2

e

n2rε

one can check easily that, for r small enough

1

2 e

(τ+n2

≤ e

τ φ

r

n2

≤ e

(τ+n2

. This achieves the proof of the first part of lemma 2.3.

Now we can use the additionnal information about how far from the

origin supp(u) is, to proove theorem 2.1. This extra term will be useful in

section 3.

(11)

proof of Theorem 2.1. Suppose that supp(u) ⊂ B

R0

(x

0

) \ B

δ

(x

0

) and define T

1

= ln δ.

Cauchy-Schwarz inequality apply to

Z

t

(u

2

)e

−t

γdtdθ = 2

Z

u∂

t

ue

−t

√ γdtdθ, gives

Z

t

(u

2

)e

−t

γdtdθ ≤ 2

Z

(∂

t

u)

2

e

−t

√ γdtdθ

12 Z

u

2

e

−t

√ γdtdθ

12

. (2.33) On the other hand, integrating by parts gives

Z

t

(u

2

)e

−t

γdtdθ =

Z

u

2

e

−t

γdtdθ−

Z

u

2

e

−t

t

(ln( √ γ)) √

γdtdθ. (2.34) Now since |∂

t

ln √

γ | ≤ Ce

t

for |T

0

| large enough we can deduce :

Z

t

(u

2

)e

−t

γdtdθ ≥ c

Z

u

2

e

−t

γdtdθ. (2.35) Combining (2.33) and (2.35) gives

c

2 Z

u

2

e

−t

γdtdθ ≤ 4

Z

(∂

t

u)

2

e

−t

√ γdtdθ

≤ 4e

−T1 Z

(∂

t

u)

2

√ γdtdθ.

Finally, droping all terms except τ

2R

|∂

t

u|

2

f

0−3

γdtdθ in (2.27) gives :

C

0

I ≥ τ

2

δ

2

ke

−t

uk

2f

. Inequality (2.27) can then be replaced by :

I ≥ Cτ

2R

|∂

t

u|

2

f

0−3

γdtdθ + Cτ

R

|f

00

| · |D

θ

u|

2

f

0−3

√ γdtdθ + Cτ

3R

|f

00

| · |u|

2

f

0−3

γdtdθ + Cτ

2

δ

2R

|u|

2

f

0−3

γdtdθ. (2.36) The rest of the proof follows in a similar way than the proof of Lemma 2.3.

Now we will state a Carleman estimate for the operator ∆ + λ acting

on vector functions, which will be useful in the next section. For U ∈

C

0

(B

R0

(x

0

) \ {x

0

},

Rm

), applying (2.3) to each components U

i

of U and

summing gives :

(12)

Corollary 2.4.

There exists non-negative constants R

0

, C, C

1

, C

2

, which depend only on M and ε, such that :

For any x

0

∈ M, any δ ∈ (0, R

0

),any U ∈ C

0

(B

R0

(x

0

) \ (B

δ

(x

0

),

Rm

),and any τ ≥ C

1

λ + C

2

, one has :

C

r

2

e

−τ φ

(∆U + λU )

≥ τ

32

r

ε2

e

−τ φ

U

+ τ δ

r

−1

e

τ φ

U

+ τ

12

r

1+ε2

e

τ φ

∇U

. (2.37)

3 Doubling inequality

In this section we intend to prove a doubling property for gradient of eigen- functions. First we establish a three balls theorem :

Proposition 3.1

(Three balls theorem). There exist non-negative constants R

0

, c and 0 < α < 1 wich depend only on M such that, if u is a solution to (1.1) one has :

for any R such that 0 < R < 2R < R

0

, and any x

0

∈ M, k∇uk

B

R(x0)

≤ e

c(

λ+1)

k∇uk

αB

R 2

(x0)

k∇uk

1−αB

2R(x0)

(3.1) Proof. Let x

0

a point in M and (x

1

, x

2

, · · · , x

n

) local coordinates around x

0

. Let u be a solution to (1.1) and define V = (

∂x∂u

1

, · · · ,

∂ux

n

). Let R

0

> 0 as in theorem (2.3) and R such that 0 < R < 2R < R

0

. We still denote r(x) the riemannian distance beetween x and x

0

. We also denote by B

r

the geodesic ball centered at x

0

of radius r. If v is a function defined in a neigborhood of x

0

, we denote by kvk

R

the L

2

norm of v on B

R

and by kvk

R1,R2

the L

2

norm of v on the set A

R1,R2

:= {x ∈ M ; R

1

≤ r(x) ≤ R

2

}. Let ψ ∈ C

0

(B

2R

), 0 ≤ ψ ≤ 1, a radial function with the following properties :

• ψ(x) = 0 if r(x) <

R4

or if r(x) >

5R3

,

• ψ(x) = 1 if

R3

< r(x) <

3R2

,

• |∇ψ(x)| ≤

CR

, |∇

2

ψ(x)| ≤

C

R2

.

Now, applying ∂

k

to each side of (1.1) gives

−∆∂

k

u + [∆, ∂

k

]u = λ∂

k

u

where [∆, ∂

k

] is the commutator of ∆ and ∂

k

. It is a second order operator with no zero order term and with coefficients depending only of M . The function V = (

∂x∂u

1

, · · · ,

∂ux

n

) is therefore a solution of the system :

∆V + λV + AV + B · ∇V = 0 (3.2)

(13)

where A and B depend only on the metric g of M and its derivatives. Now we apply the Carleman estimate (2.37) to the function ψV , we get :

C

r

2

e

τ φ

(∆(ψV ) + λψV )

≥ τ

32

r

ε2

e

τ φ

ψV

+ τ R

r

−1

e

τ φ

ψV

+ τ

12

r

1+ε2

e

τ φ

∇(ψV )

. Using that V is a solution of (3.2), we have :

C

r

2

e

τ φ

(ψAV + ψB · ∇V + 2∇V · ∇ψ + ∆ψV )

≥ τ

32

r

ε2

e

τ φ

ψV

+ τ R

r

−1

e

τ φ

ψV

+ τ

12

r

1+ε2

e

τ φ

∇(ψV )

Now from triangular inequality we get C

r2

e

τ φ

(∆ψV + 2∇V · ∇ψ) ≥ τ

32

r

ε2

e

τ φ

ψV

− C

r2

e

τ φ

ψAV

+ τ R

r

−1

e

τ φ

ψV + τ

12

r

1+ε2

e

−τ φ

∇(ψV )

− C

r

2

e

τ φ

ψB · ∇V

and τ

12

r

1+ε2

e

τ φ

∇(ψV )

≥ τ

12

r

1+ε2

e

τ φ

ψ∇V

− τ

12

r

1+ε2

e

τ φ

∇ψV

≥ τ

12

r

1+ε2

e

τ φ

ψ∇V

− τ

12

r

ε2

e

τ φ

V

Then for τ great enough and for sufficient small R

0

, C

r

2

e

τ φ

(∆ψV + 2∇V · ∇ψ)

≥ τ

32

r

2ε

e

−τ φ

ψV

+ τ R

r−1

e

−τ φ

ψV

+ τ

12

r

1+ε2

e

−τ φ

ψ∇V

. In particular we have :

C

r

2

e

τ φ

(∆ψV + 2∇V · ∇ψ)

≥ τ

e

τ φ

ψV

Assume that τ ≥ 1 and use properties of ψ gives : ke

τ φ

V k

R

3,3R2

≤ C

ke

τ φ

V k

R

4,R3

+ ke

τ φ

V k

3R 2 ,5R3

+ C

Rke

τ φ

∇V k

R

4,R3

+ Rke

τ φ

∇V k

3R 2 ,5R3

. (3.3)

Furthermore, since φ is radial and decreasing, one has ke

τ φ

V k

R

3,3R2

≤ C

e

τ φ(R4)

kV k

R

4,R3

+ e

τ φ(3R2 )

kV k

3R 2 ,5R3

+ C

Re

τ φ(R4)

k∇V k

R

4,R3

+ Re

τ φ(3R2 )

k∇V k

3R 2 ,5R3

. (3.4)

(14)

Now we recall the following elliptic estimates : since V satisfies (3.2) then it is not hard to see that :

k∇V k

(1−a)R

≤ C

1

(1 − a)R + √ λ

kV k

R

, for 0 < a < 1 (3.5) Since ke

τ φ

∇V k

R

4,R3

is bounded by ke

τ φ

∇V k

R

3

, using the formula (3.5) gives:

e

τ φ(R4)

k∇V k

R

4,R3

≤ C 1

R +

√ λ

e

τ φ(R4)

kV k

R

2

. (3.6)

Similarly, we also have, e

τ φ(3R2 )

k∇V k

3R

2 ,5R3

≤ C 1

R + √ λ

e

τ φ(3R2 )

kV k

2R

. (3.7) On the other hand, using properties of φ leads to :

ke

τ φ

V k

R

3,3R2

≥ ke

τ φ

V k

R

3,R

≥ e

τ φ(R)

kV k

R

3,R

. (3.8)

Now using (3.6),(3.7) and (3.8) in (3.3) we get : kV k

R

3,R

≤ C √ λ

e

τ(φ(R4)−φ(R))

kV k

R

2

+ e

τ(φ(3R2 )−φ(R))

kV k

2R

Let A

R

= φ(

R4

)−φ(R) and B

R

= −(φ(

3R2

)−φ(R)). Because of the properties of φ, we have 0 < C

1

≤ A

R

≤ C

2

and 0 < C

1

≤ B

R

≤ C

2

where C

1

and C

2

don’t depend on R. We may assume that C √

λ ≥ 2. We can add kV k

R 3

to each member and bound it in the right hand side by C √

λe

τ A

kV k

R 2

, for τ large enough. Then replacing C by 2C gives :

kV k

R

≤ C

√ λ

e

τ A

kV k

R

2

+ e

−τ B

kV k

2R

. (3.9)

Now we want to find τ such that C √

λe

−τ B

kV k

2R

≤ 1 2 kV k

R

Since τ must satisfy τ ≥ C

1

λ + C

2

, we choose τ = − 1

B ln

1

2C √ λ

kV k

R

kV k

2R

+ C

1

λ + C

2

. (3.10) Inequality (3.9) becomes

kV k

R

≤ C √ λe

C1

λ+C2

e

−A B ln

1 2C

λ kVkR kVk2R

kV k

R 2

,

(15)

wich gives easily

kV k

R

≤ e (

C1

λ+C2

)

A+BB

kV k

A A+B

2R

kV k

B B+A R 2

.

Finally define α =

A+BA

gives easily kV k

R

≤ e

C5

λ+C6

kV k

α2R

kV k

1−αR

2

.

From now on we assume that M is compact. Thus we can derive from three balls theorem above uniform doubling estimates on gradient of eigen- functions.

Theorem 3.2

(doubling estimates). There exists non-negative constants R

0

, C

1

, C

2

depending only on M such that : if u is a solution to (1.1) on M, then for any x

0

∈ M and any r > 0,

k∇uk

B

2r(x0)

≤ e

C1

λ+C2

k∇uk

Br(x

0)

. (3.11)

Remark 3.3.

Using standard elliptic theory (see [7]) to bound the L

norm of |V | by a multiple of its L

2

norm and rescalling arguments gives for δ > 0:

kV k

L(Bδ(x0))

≤ (C

1

λ + C

2

)

n2

δ

−n/2

kuk

L2(B(x0))

Then one can see that the doubling estimate is still true with the L

norm kV k

L(B2r(x0))

≤ e

C(

λ+1)

kV k

L(Br(x0))

(3.12) To proove the theorem 3.2 we need the following

Proposition 3.4.

For any R > 0, there exists C

R

> 0, such that, for any x

0

∈ M one has :

k∇uk

BR(x0)

≥ e

−CR(

λ+1)

k∇uk

L2(M)

.

Proof. Let R > 0 and assume without loss of generality that R < R

0

, with R

0

from the three balls theorem (theorem 3.1). Up to multiplication by a constant, we can assume that k∇uk

L2(M)

= 1. We denote by ¯ x a point in M such that k∇uk

BRx)

= sup

x∈M

k∇uk

BR(x)

. This implies that one has k∇uk

B

R(¯x)

≥ D

R

, where D

R

depend only on M and R. One has from proposition (3.1) at an arbitrary point x of M :

k∇uk

B

R/2(x)

≥ e

−c(

λ+1)

k∇uk

1 α

BR(x)

(3.13)

Références

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