International Conference on Geometric Analysis

Geometry and topology of overdeterminedelliptic problems

Pieralberto Sicbaldi

Joint work with Antonio Ros

Domaine de Seillac

September 2012

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The problem:

To understand the geometry of domains Ω ∈ Rn that support a

solution of the over-determined system

∆u+ f(u) = 0 in Ω

u > 0 in Ω

u = 0 on ∂Ω

∂u

∂ν= constant on ∂Ω

Definition: If such problem is solvable, Ω is called an f -extremaldomain.

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Serrin, 1971, Arch. Rat. Mech. An.

If f is Lipschitz and Ω is a C2 bounded domain where thereexists a solution u to the problem

∆u+ f(u) = 0 in Ω

u > 0 in Ω

u = 0 on ∂Ω

∂u

∂ν= constant on ∂Ω

then Ω is a ball.

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The moving plane argument

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Reichel, 1997, Arch. Rat. Mech. An.

If u is a C2 function on Rn\Ω such that

∆u+ f(u) = 0 in Rn\Ω

u = a on ∂Ω

∂u

∂ν= c on ∂Ω

0 ≤ u ≤ a in Rn\Ω

u = 0 at ∞

∣

∣

∣

∣

∣

∣

∣

∣

∣

Ω bounded

Rn\Ω connected

f Lipschitz

f nonincreasing

then Ω is a ball.

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Berestycki, Caffarelli, Nirenberg, 1997, Comm. Pure Appl. Math.

If Ω = xn > ϕ(x1, ..., xn−1), where ϕ is a C2 function verifying

for any τ ∈ Rn−1, uniformly, lim

|x|→∞(ϕ(x+ τ)− ϕ(x)) = 0

and

∆u+ f(u) = 0 in Ω

u = 0 on ∂Ω

∂u

∂ν= c on ∂Ω

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

u bounded

exist 0 < s0 < s1 < µ :

f(s) > 0 on (0, µ)

f(s) ≤ 0 for s ≥ µ

f(s) ≤ δ0s on (0, s0) , δ0 > 0

f nonincreasing on (s1, µ)

then Ω is a half-space.

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Conjecture of Berestycki, Caffarelli and Nirenberg

Communication on Pure and Applied Mathematics, (1997).

∆u+ f(u) = 0 in Ω

u > 0 in Ω

u = 0 on ∂Ω

∂u

∂ν= constant on ∂Ω ,

∣

∣

∣

∣

∣

∣

∣

∣

EXTRA HYPOTHESIS

Rn\Ω connected

u bounded

⇓

Ω is a half space, or a ball, or a cylinder Rj ×B (where B is a ball) orthe complement of one of these three exemples.

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Farina & Valdinoci, 2009, Arch. Rat. Mech. An.

∆u+ f(u) = 0 in Ω

u > 0 in Ω

u = 0 on ∂Ω

∂u

∂ν= constant on ∂Ω ,

∣

∣

∣ f Lipschitz

1. If u ∈ C2(Ω) ∩ L∞(Ω) and f(t) ≥ λt, then Ω cannot be auniformly Lipschitz epigraph.

2. If u ∈ C2(Ω) ∩ L∞(Ω), and [n = 2] or [n = 3 and f(t) ≥ 0],then Ω cannot be a coercive uniformly Lipschitz C3 epigraph.

3. If n = 2 and u ∈ C2(Ω) has bounded gradient and isincreasing in one of the variables, then Ω is a half-plane.

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Theorem (—, 2010, Calc. Var. PDEs)

For n ≥ 3 there exists perturbations Ω of the cylinder Bn−1 × R

(here Bn−1 is the unit ball in Rn−1), with boundary of revolution

and periodic in the direction of the axis of the cylinder, wherethere exists a periodic and positive solution to

∆u+ λu = 0 in Ω

u = 0 on ∂Ω

∂u

∂ν= constant on ∂Ω

x

t

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Schlenk & —, 2012, Adv. Math.

The result is true also in dimension 2. But this is nota counterexemple to the conjecture.

x

t4−4

The complement of such domain is not connected.

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The conjecture in dimension 2

Theorem (Ros & — , 2012)

Suppose there exists λ > 0 such that f(t) ≥ λ t.

With this extra hypothesis, the conjecture ofBerestycki-Caffarelli-Nirenberg in dimension 2 is true.

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A stronger version

Theorem (Ros & — , 2012)

Let us suppose that f is a Lipschitz function such that thereexists λ > 0 such that f(t) ≥ λ t. If Ω a C2-domain of R2 withconnected complement, where one can solve

∆u+ f(u) = 0 in Ω

u > 0 in Ω

u = 0 on ∂Ω

∂u

∂ν= constant on ∂Ω ,

then Ω is a ball. Remark: we do not suppose u bounded.

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The proof of the theorem (1)

Step 1 . ∂Ω has only one connected component.

Then, there are only three possibilities for Ω:

1. Ω is a bounded connected open domain.

2. Ω is the complement of a compact domain.

3. ∂Ω is an open curve that separated R2 in two connected

components, and Ω is one of such components.

Ω

Ω

Ω

By the Serrin’s result, in the first case Ω must be a ball.

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The proof of the theorem (2)

Let Rλ be the radius of balls whose first eigenvalue of theLaplacian with 0 Dirichlet boundary condition is λ.

Step 2. If R ≥ Rλ then Ω does not contain any closed ball ofradius R.

Ω

BRλ(p)

graph of u graph of vǫ

Corollary . Ω can not be the complement of a compact domain.

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Overdetermined problems and CMC surfaces

Theorem (Serrin, 1971) . In Rn the only bounded domains

where the previous overdetermined problem can be solved arethe balls.

Main idea : the moving plane argument, introduced some yearsbefore in the CMC surface theory...

...Theorem (Alexandroff, 1962) . In Rn the only enbedded

compact mean curvature hypersurfaces are the spheres.

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Meeks, 1989, J. Diff. Geom.

Definition. A surface has finite topology if

1. it is a compact surface

2. outside of a big ball, the surface is done of a finite numberof noncompact components diffeomorphic to Sn−1 × R+,called ends.

Theorem . Let S be a properly embedded finite topologynonzero CMC surface in R

3. Then S cannot have only one end.

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The proof of our theorem: (3)

Definition. We say that a domain has finite topology if

1. it is bounded domain, or

2. it is the complement of a compact domain, or

3. outside of a big ball, the domain is done of a finite numberof noncompact components diffeomorphic to Bn−1 × R+,called ends

To prove our main theorem it is enough to prove:

Proposition . Ω cannot have only one end.

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The proof of the proposition (1a)

Let L be an straight line in R2 and let L+ and L− be the two

connected components of R2\L. If Ω ∩ L+ has a boundedconnected component C, then:

C

Ω

L

C ′

Step 1. The closure of ∂C ∩ L+ is a graph on ∂C ∩ L

Step 2. The maximum distance h(C) of ∂C to L satisfies

h(C) ≤ 3Rλ.

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The proof of the proposition (1b)

Step 2. The maximum distance h(C) of ∂C to L satisfies

h(C) ≤ 3Rλ.

bR0−Ra

p q

p′

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The proof of the proposition (2)

Let E be an end of Ω.

Step 3. There exist a line L in R2 such that E stays at bounded

distance from L.

E

L

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The proof of the proposition (3)

Step 4. Ω cannot have only one end.

3Rλ + ǫ

E

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Korevaar, Kusner & Solomon, 1989, J. Diff. Geom.

Let S be a properly embedded finite topology nonzero CMCsurface in R

3 contained in a cylinder. Then S is surface ofrevolution.

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An other result

Theorem (Ros, —, 2012). Let Ω be an f -extremal domain ofR2. If Ω is contained in a half-plane, then Ω is either a ball or

(after a rigid motion) there exists a C2 positive functionϕ : R −→]0,∞[ such that either

i. the domain Ω is an epigraph y > ϕ(x), or

ii. ϕ is bounded and Ω is the symmetric domain |y| < ϕ(x).

Ω

Ω

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On the conjecture of Berestycki-Caffarelli-Nirenberg

Farina & Valdinoci (2009, Arch Rat. Mech. An.) : in dimension2, if u is increasing in one variable and has bounded gradient,then the associated f -extremal domain Ω is a half-plane.

if ∇u is bounded

A straightforward consequence, using our results, is the:

Corollary (Ros, —, 2012) . In dimension 2, if Ω is contained in ahalf-plane the conjecture of Berestycki-Caffarelli-Nirenberg istrue.

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The end

Thank you!

Reference

A. Ros, P. Sicbaldi. Geometry and Topology of someoverdetermined elliptic problems. Preprint (arxiv).

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