Constant heat flow and the isoparametric property Alessandro Savo - - PowerPoint PPT Presentation

Constant heat flow and the isoparametric property

Alessandro Savo (Sapienza, Rome) Beijing Normal University June 2, 2019

Overdetermined PDE’s

An overdetermined problem gives rise to the following : Question: can we identify the geometry of a domain Ω in a Riemannian manifold assuming the existence of a solution u of a certain PDE such that both u and its normal derivative are constant on the boundary of Ω ?

Overdetermined PDE’s

An overdetermined problem gives rise to the following : Question: can we identify the geometry of a domain Ω in a Riemannian manifold assuming the existence of a solution u of a certain PDE such that both u and its normal derivative are constant on the boundary of Ω ? Here is a class of overdetermined problems:    ∆u = F(u)

• n Ω

u = c1, ∂u ∂N = c2

• n ∂Ω

We try to derive some geometric information in the general framework of Riemannian manifolds, not just in Euclidean space.

Overdetermined PDE’s

An overdetermined problem gives rise to the following : Question: can we identify the geometry of a domain Ω in a Riemannian manifold assuming the existence of a solution u of a certain PDE such that both u and its normal derivative are constant on the boundary of Ω ? Here is a class of overdetermined problems:    ∆u = F(u)

• n Ω

u = c1, ∂u ∂N = c2

• n ∂Ω

We try to derive some geometric information in the general framework of Riemannian manifolds, not just in Euclidean space. In what follows, Ω is a compact manifold with smooth boundary and ∆ is the Laplace-Beltrami operator associated to the metric. The sign convention is that, in Euclidean space: ∆u = −

• j

∂2u ∂x2

j

By N we denote the inner unit normal vector to ∂Ω.

Serrin problem

Let us start from the most famous overdetermined problem. Consider the torsion function v = v(x), unique solution of the boundary value problem:

• ∆v = 1
• n

Ω, v = 0

• n

∂Ω. For a generic domain, ∂v ∂N will not be constant on the boundary.

Serrin problem

Let us start from the most famous overdetermined problem. Consider the torsion function v = v(x), unique solution of the boundary value problem:

• ∆v = 1
• n

Ω, v = 0

• n

∂Ω. For a generic domain, ∂v ∂N will not be constant on the boundary. Serrin problem. If we make this extra assumption we then get an

• verdetermined problem, often called Serrin problem:

   ∆v = 1

• n

Ω, v = 0, ∂v ∂N = c

• n

∂Ω. (1)

Domains for which a solution to (1) exists will be called harmonic, because of the following : Property of harmonic domains. A domain Ω supports a solution to (1) if and only if the mean-value of any harmonic function h on Ω equals its mean-value on ∂Ω. That is: 1 |Ω|

• Ω

h = 1 |∂Ω|

• ∂Ω

h

Domains for which a solution to (1) exists will be called harmonic, because of the following : Property of harmonic domains. A domain Ω supports a solution to (1) if and only if the mean-value of any harmonic function h on Ω equals its mean-value on ∂Ω. That is: 1 |Ω|

• Ω

h = 1 |∂Ω|

• ∂Ω

h Harmonic domains are in fact extremal domains for some eigenvalue problems on differential forms, and also for a certain fourth order Steklov eigenvalue problem on functions (see the works Raulot-S.).

Domains for which a solution to (1) exists will be called harmonic, because of the following : Property of harmonic domains. A domain Ω supports a solution to (1) if and only if the mean-value of any harmonic function h on Ω equals its mean-value on ∂Ω. That is: 1 |Ω|

• Ω

h = 1 |∂Ω|

• ∂Ω

h Harmonic domains are in fact extremal domains for some eigenvalue problems on differential forms, and also for a certain fourth order Steklov eigenvalue problem on functions (see the works Raulot-S.). Another geometric property: harmonic domains are self-Cheeger (at least in non-negative Ricci curvature).

Domains for which a solution to (1) exists will be called harmonic, because of the following : Property of harmonic domains. A domain Ω supports a solution to (1) if and only if the mean-value of any harmonic function h on Ω equals its mean-value on ∂Ω. That is: 1 |Ω|

• Ω

h = 1 |∂Ω|

• ∂Ω

h Harmonic domains are in fact extremal domains for some eigenvalue problems on differential forms, and also for a certain fourth order Steklov eigenvalue problem on functions (see the works Raulot-S.). Another geometric property: harmonic domains are self-Cheeger (at least in non-negative Ricci curvature). Can we classify them ?

Classification in Euclidean space. Let us start with M = Rn, that is, Euclidean domains. First of all, any ball is a harmonic domain: the torsion function is radial and in fact, for the ball of radius R in Rn: v(x) = 1 2n(R2 − |x|2).

Classification in Euclidean space. Let us start with M = Rn, that is, Euclidean domains. First of all, any ball is a harmonic domain: the torsion function is radial and in fact, for the ball of radius R in Rn: v(x) = 1 2n(R2 − |x|2). Serrin proved in 1971 the following rigidity result.

Theorem

(Serrin) Assume that Ω ∈ Rn is a harmonic domain. Then Ω is a ball and the torsion function v is radially symmetric. That is, harmonic Euclidean domains are balls. He proved more generally that if there is a positive solution u of the problem    ∆u = F(u) u = 0, ∂u ∂N = c

• n

∂Ω, (2) then Ω is a ball. Method of proof: Alexandrov moving planes. The method needs positivity of u.

What about domains in other Riemannian manifolds ? Can we classify harmonic domains there ? Which geometric properties do they enjoy ? Is it true that harmonic domains exist in any Riemannian manifold ?

What about domains in other Riemannian manifolds ? Can we classify harmonic domains there ? Which geometric properties do they enjoy ? Is it true that harmonic domains exist in any Riemannian manifold ? Other space forms. Let us start from the other constant curvature space forms: hyperbolic space Hn and the sphere Sn. Geodesic balls are, of course, harmonic domains: the torsion function is

• radial. Do we have a rigidity result as well ?

What about domains in other Riemannian manifolds ? Can we classify harmonic domains there ? Which geometric properties do they enjoy ? Is it true that harmonic domains exist in any Riemannian manifold ? Other space forms. Let us start from the other constant curvature space forms: hyperbolic space Hn and the sphere Sn. Geodesic balls are, of course, harmonic domains: the torsion function is

• radial. Do we have a rigidity result as well ?

The moving planes method extends without change to the following cases (Kumaresan-Prajapat, Molzon):

Theorem

Assume that Ω is a harmonic domain in Hn or Sn

+ (the hemisphere).

Then Ω is a ball and v is radially symmetric. Note the restriction to domains of the hemisphere. (This because the moving planes method cannot be applied to domains which intersect every equator).

What about domains in other Riemannian manifolds ? Can we classify harmonic domains there ? Which geometric properties do they enjoy ? Is it true that harmonic domains exist in any Riemannian manifold ? Other space forms. Let us start from the other constant curvature space forms: hyperbolic space Hn and the sphere Sn. Geodesic balls are, of course, harmonic domains: the torsion function is

• radial. Do we have a rigidity result as well ?

The moving planes method extends without change to the following cases (Kumaresan-Prajapat, Molzon):

Theorem

Assume that Ω is a harmonic domain in Hn or Sn

+ (the hemisphere).

Then Ω is a ball and v is radially symmetric. Note the restriction to domains of the hemisphere. (This because the moving planes method cannot be applied to domains which intersect every equator). Is this restriction necessary or is due to a lack of the method ? In

• ther words, is it true, more generally, that any harmonic domain in Sn

must be a geodesic ball ?

The assumption is in fact necessary.

The assumption is in fact necessary. Example (Berenstein). Let Σ ⊆ S3 be the (minimal) Clifford torus, defined by the equations Σ :      x2

1 + x2 2 = 1

2, x2

3 + x2 4 = 1

2. Then Σ is isometric to the product manifold S1(1/ √ 2) × S1(1/ √ 2) and is the common boundary of two spherical domains Ω1, Ω2, isometric to each other. Fact: Ω1 and Ω2 are harmonic domains (clearly not isometric to a geodesic ball, because ∂Ω1 is topologically a torus). In fact, this holds because the boundary is an orbit of the subgroup O(2) × O(2) of the isometry group of S3, which also leaves Ω1 invariant.

The assumption is in fact necessary. Example (Berenstein). Let Σ ⊆ S3 be the (minimal) Clifford torus, defined by the equations Σ :      x2

1 + x2 2 = 1

2, x2

3 + x2 4 = 1

2. Then Σ is isometric to the product manifold S1(1/ √ 2) × S1(1/ √ 2) and is the common boundary of two spherical domains Ω1, Ω2, isometric to each other. Fact: Ω1 and Ω2 are harmonic domains (clearly not isometric to a geodesic ball, because ∂Ω1 is topologically a torus). In fact, this holds because the boundary is an orbit of the subgroup O(2) × O(2) of the isometry group of S3, which also leaves Ω1 invariant. Let us enlarge the class of the examples.

Obvious generalization: consider Clifford tori Sp(t) × Sn−p(

• 1 − t2)

for t ∈ (0, 1). A Clifford torus bounds two domains in Sn+1, which are both harmonic.

Obvious generalization: consider Clifford tori Sp(t) × Sn−p(

• 1 − t2)

for t ∈ (0, 1). A Clifford torus bounds two domains in Sn+1, which are both harmonic. But the sphere supports many more harmonic domains. Let us describe a larger class. This is due to Shklover (2000):

Theorem

Let Ω be a spherical domain bounded by a (connected) isoparametric

• hypersurface. Then Ω is harmonic.

We will give another proof later.

Summary for constant curvature space forms

We focused on Serrin’s overdetermined problem in the space forms Rn, Hn, Sn:    ∆u = 1

• n

Ω u = 0, ∂u ∂N = c

• n

∂Ω

Summary for constant curvature space forms

We focused on Serrin’s overdetermined problem in the space forms Rn, Hn, Sn:    ∆u = 1

• n

Ω u = 0, ∂u ∂N = c

• n

∂Ω Domains which support a solution to this problem are called harmonic.

Summary for constant curvature space forms

We focused on Serrin’s overdetermined problem in the space forms Rn, Hn, Sn:    ∆u = 1

• n

Ω u = 0, ∂u ∂N = c

• n

∂Ω Domains which support a solution to this problem are called harmonic. We have seen that in Rn, Hn, Sn

+ (the hemisphere) the only harmonic

domains are geodesic balls.

Summary for constant curvature space forms

We focused on Serrin’s overdetermined problem in the space forms Rn, Hn, Sn:    ∆u = 1

• n

Ω u = 0, ∂u ∂N = c

• n

∂Ω Domains which support a solution to this problem are called harmonic. We have seen that in Rn, Hn, Sn

+ (the hemisphere) the only harmonic

domains are geodesic balls. In the whole sphere, we have many more harmonic domains, namely, domains bounded by an isoparametric hypersurface.

Summary for constant curvature space forms

We focused on Serrin’s overdetermined problem in the space forms Rn, Hn, Sn:    ∆u = 1

• n

Ω u = 0, ∂u ∂N = c

• n

∂Ω Domains which support a solution to this problem are called harmonic. We have seen that in Rn, Hn, Sn

+ (the hemisphere) the only harmonic

domains are geodesic balls. In the whole sphere, we have many more harmonic domains, namely, domains bounded by an isoparametric hypersurface. Question : are there other examples or that is it ? In other words, is it true that the boundary of a harmonic spherical domain must be isoparametric ?

Summary for constant curvature space forms

We focused on Serrin’s overdetermined problem in the space forms Rn, Hn, Sn:    ∆u = 1

• n

Ω u = 0, ∂u ∂N = c

• n

∂Ω Domains which support a solution to this problem are called harmonic. We have seen that in Rn, Hn, Sn

+ (the hemisphere) the only harmonic

domains are geodesic balls. In the whole sphere, we have many more harmonic domains, namely, domains bounded by an isoparametric hypersurface. Question : are there other examples or that is it ? In other words, is it true that the boundary of a harmonic spherical domain must be isoparametric ? Answer : no, it is not true: there are in fact other examples.

Even more exotic harmonic domains in spheres

In a very recent paper, Fall, Minlend and Weth (2018) constructed new examples of harmonic domains in Sn, by perturbing tubular neighborhoods of totally geodesic hypersurfaces.

Even more exotic harmonic domains in spheres

In a very recent paper, Fall, Minlend and Weth (2018) constructed new examples of harmonic domains in Sn, by perturbing tubular neighborhoods of totally geodesic hypersurfaces. Their boundaries are not isoparametric.

Even more exotic harmonic domains in spheres

In a very recent paper, Fall, Minlend and Weth (2018) constructed new examples of harmonic domains in Sn, by perturbing tubular neighborhoods of totally geodesic hypersurfaces. Their boundaries are not isoparametric. So, a classification seems to be at the moment out of reach, even in spaces as simple as the round sphere, which hosts a good variety of such domains.

Even more exotic harmonic domains in spheres

In a very recent paper, Fall, Minlend and Weth (2018) constructed new examples of harmonic domains in Sn, by perturbing tubular neighborhoods of totally geodesic hypersurfaces. Their boundaries are not isoparametric. So, a classification seems to be at the moment out of reach, even in spaces as simple as the round sphere, which hosts a good variety of such domains. Exception in dimension 2: a simply connected harmonic domain in S2 must be a geodesic disk (Espinar-Mazet).

Isoparametric tubes

We now isolate a class of Riemannian manifolds with boundary which always support a solution to the Serrin problem the isoparametric tubes.

Isoparametric tubes

We now isolate a class of Riemannian manifolds with boundary which always support a solution to the Serrin problem the isoparametric tubes. By definition, the compact Riemannian manifold Ωn with smooth boundary ∂Ω is called an isoparametric tube if there exists a smooth, compact submanifold P of Ω such that: a) Ω is a tube of radius R around P, that is Ω = {x : d(x, P) ≤ R}

Isoparametric tubes

We now isolate a class of Riemannian manifolds with boundary which always support a solution to the Serrin problem the isoparametric tubes. By definition, the compact Riemannian manifold Ωn with smooth boundary ∂Ω is called an isoparametric tube if there exists a smooth, compact submanifold P of Ω such that: a) Ω is a tube of radius R around P, that is Ω = {x : d(x, P) ≤ R} b) Each equidistant from P, say Σt = {x ∈ Ω : d(x, P) = t}, t ∈ (0, R], is a smooth hypersurface having constant mean curvature.

• The submanifold P is called the soul of Ω, or focal submanifold of Ω.

It can be shown that it is always minimal (Nomizu, Ge and Tang).

Examples

• It is by now classical that, in Sn, any domain bounded by a connected

isoparametric hypersurface (that is, a hypersurface with constant principal curvatures) is an isoparametric tube. The soul is the focal submanifold of ∂Ω.

• Geodesic balls in constant curvature spaces (or, more generally, in a

harmonic manifold) are isoparametric tubes. The soul is a point (the center of the ball).

• An annulus in Rn is not an isoparametric tube.
• An isoparametric tube might have more than one boundary

component: for example, a tubular neighborhood of an equator of the

• sphere. However, it cannot have more than two boundary components

(otherwise it is not even a smooth tube).

• A solid revolution torus in R3 is a smooth tube, but it is not

isoparametric : the soul is circle, and equidistant do not have constant mean curvature.

Theorem

Any isoparametric tube is a harmonic domain.

Theorem

Any isoparametric tube is a harmonic domain. Again, we have to show that the mean exit time function u:

• ∆u = 1
• n

Ω u = 0

• n

∂Ω has constant normal derivative.

Theorem

Any isoparametric tube is a harmonic domain. Again, we have to show that the mean exit time function u:

• ∆u = 1
• n

Ω u = 0

• n

∂Ω has constant normal derivative. Radial functions. On any isoparametric tube we have the family of radial functions: these are the functions which are constant on every equidistant.

Theorem

Any isoparametric tube is a harmonic domain. Again, we have to show that the mean exit time function u:

• ∆u = 1
• n

Ω u = 0

• n

∂Ω has constant normal derivative. Radial functions. On any isoparametric tube we have the family of radial functions: these are the functions which are constant on every equidistant. If we write ρ for the distance function to P: ρ(x) = d(x, P) then the function f on Ω is radial if and only if it can be expressed f = ψ ◦ ρ, for a smooth function ψ : [0, R] → R.

It is clear that a radial function has constant normal derivative on ∂Ω: ∂f ∂N = ∇f , N = −∇f , ∇ρ = −ψ′(R). Hence, to prove the theorem, it is enough to show that the torsion function is radial.

It is clear that a radial function has constant normal derivative on ∂Ω: ∂f ∂N = ∇f , N = −∇f , ∇ρ = −ψ′(R). Hence, to prove the theorem, it is enough to show that the torsion function is radial. Averaging operator (radialization). We now define an operator A : C ∞(Ω) → C ∞(Ω) which will take a function to a radial function, called its radialization.

Recall that Ω is foliated by the equidistants (level sets of the distance function ρ to P), hence any point x belongs to a unique equidistant; if x ∈ Ω \ P this will be the regular hypersurface Σx = ρ−1(ρ(x)); and if x ∈ P then it will be simply P (the unique singular leaf).

Recall that Ω is foliated by the equidistants (level sets of the distance function ρ to P), hence any point x belongs to a unique equidistant; if x ∈ Ω \ P this will be the regular hypersurface Σx = ρ−1(ρ(x)); and if x ∈ P then it will be simply P (the unique singular leaf). Given f ∈ C ∞(Ω) and x ∈ Ω \ P we define Af (x) as the average of f on the equidistant through x : Af (x) = 1 |Σx|

• Σx

f ;

Recall that Ω is foliated by the equidistants (level sets of the distance function ρ to P), hence any point x belongs to a unique equidistant; if x ∈ Ω \ P this will be the regular hypersurface Σx = ρ−1(ρ(x)); and if x ∈ P then it will be simply P (the unique singular leaf). Given f ∈ C ∞(Ω) and x ∈ Ω \ P we define Af (x) as the average of f on the equidistant through x : Af (x) = 1 |Σx|

• Σx

f ; if x ∈ P we simply define Af (x) = 1 |P|

• P

f , where of course we use in both cases the measure induced by the Riemannian metric on Σx and P, respectively.

If f ∈ C ∞(Ω) its radialization Af has the following properties: Af is smooth as well; Af is radial; f is radial if and only if Af = f .

If f ∈ C ∞(Ω) its radialization Af has the following properties: Af is smooth as well; Af is radial; f is radial if and only if Af = f . The crucial property of an isoparametric foliation is the constancy of the mean curvature of the leaves. This has the following important consequence.

If f ∈ C ∞(Ω) its radialization Af has the following properties: Af is smooth as well; Af is radial; f is radial if and only if Af = f . The crucial property of an isoparametric foliation is the constancy of the mean curvature of the leaves. This has the following important consequence. The radialization commutes with the Laplacian: A∆f = ∆Af . All the above facts are not difficult to prove (only a little technical at times).

The torsion function is radial. We can now prove that the torsion function u is radial; for that it is enough to show that Au = u.

The torsion function is radial. We can now prove that the torsion function u is radial; for that it is enough to show that Au = u. Let ˆ u = Au. Then, as the radialization commutes with the Laplacian: ∆ˆ u = ∆Au = A∆u = A1 = 1, and of course ˆ u = 0 on ∂Ω.

The torsion function is radial. We can now prove that the torsion function u is radial; for that it is enough to show that Au = u. Let ˆ u = Au. Then, as the radialization commutes with the Laplacian: ∆ˆ u = ∆Au = A∆u = A1 = 1, and of course ˆ u = 0 on ∂Ω. Then, u and ˆ u are two solutions of the boundary value problem

• ∆u = 1

u = 0

• n

∂Ω By uniqueness, they must coincide, hence Au = u, as asserted.

Heat equation: introduction

Let M be a Riemannian manifold and k(t, x, y) be its heat kernel (solution of the heat equation having initial data given by the Dirac distribution at x). The variable t > 0 is time, and x, y ∈ M.

Heat equation: introduction

Let M be a Riemannian manifold and k(t, x, y) be its heat kernel (solution of the heat equation having initial data given by the Dirac distribution at x). The variable t > 0 is time, and x, y ∈ M. k(t, x, y) is the temperature, at time t, at the point y, assuming that

• ne unit of heat is placed at x at time t = 0.

Heat equation: introduction

Let M be a Riemannian manifold and k(t, x, y) be its heat kernel (solution of the heat equation having initial data given by the Dirac distribution at x). The variable t > 0 is time, and x, y ∈ M. k(t, x, y) is the temperature, at time t, at the point y, assuming that

• ne unit of heat is placed at x at time t = 0.

The heat kernel depends on the metric: lim

t→0 log k(t, x, y) = −d(x, y)2

4 .

Heat equation: introduction

Let M be a Riemannian manifold and k(t, x, y) be its heat kernel (solution of the heat equation having initial data given by the Dirac distribution at x). The variable t > 0 is time, and x, y ∈ M. k(t, x, y) is the temperature, at time t, at the point y, assuming that

• ne unit of heat is placed at x at time t = 0.

The heat kernel depends on the metric: lim

t→0 log k(t, x, y) = −d(x, y)2

4 . One can measure distance using only a termometer.

Heat equation: introduction

Let M be a Riemannian manifold and k(t, x, y) be its heat kernel (solution of the heat equation having initial data given by the Dirac distribution at x). The variable t > 0 is time, and x, y ∈ M. k(t, x, y) is the temperature, at time t, at the point y, assuming that

• ne unit of heat is placed at x at time t = 0.

The heat kernel depends on the metric: lim

t→0 log k(t, x, y) = −d(x, y)2

4 . One can measure distance using only a termometer. There are deep intrisic relations between geometry and heat diffusion.

The heat equation

Let Ω be a domain in a Riemannian manifold. Assume that at time t = 0 the temperature distribution is prescribed by a function φ0(x) on Ω, and that the boundary is kept at zero temperature at all times.

The heat equation

Let Ω be a domain in a Riemannian manifold. Assume that at time t = 0 the temperature distribution is prescribed by a function φ0(x) on Ω, and that the boundary is kept at zero temperature at all times. If for example φ0 is positive, heat will flow away from the domain because of refrigeration and eventually, at infinite time, the temperature will be constant, equal to zero, at all points of Ω.

The heat equation

Let Ω be a domain in a Riemannian manifold. Assume that at time t = 0 the temperature distribution is prescribed by a function φ0(x) on Ω, and that the boundary is kept at zero temperature at all times. If for example φ0 is positive, heat will flow away from the domain because of refrigeration and eventually, at infinite time, the temperature will be constant, equal to zero, at all points of Ω. But what is precisely the evolution of temperature ?

The heat equation

Let Ω be a domain in a Riemannian manifold. Assume that at time t = 0 the temperature distribution is prescribed by a function φ0(x) on Ω, and that the boundary is kept at zero temperature at all times. If for example φ0 is positive, heat will flow away from the domain because of refrigeration and eventually, at infinite time, the temperature will be constant, equal to zero, at all points of Ω. But what is precisely the evolution of temperature ? Denote by φ(t, x) the temperature at the point x ∈ Ω, at time t > 0. Then, φ(t, x) is a solution of the heat equation: ∆φ(t, x) + ∂φ ∂t (t, x) = 0. where the Laplacian is acting on the space variable x.

In conclusion, the temperature function is a solution of the following initial-boundary value problem:        ∆φ(t, x) + ∂φ ∂t (t, x) = 0 for all x ∈ Ω, t > 0 φ(0, x) = φ0(x) for all x ∈ Ω φ(t, y) = 0 for all y ∈ ∂Ω and for all t > 0 In the last line we see the Dirichlet boundary condition, which correspond to boundary refrigeration at all times.

In conclusion, the temperature function is a solution of the following initial-boundary value problem:        ∆φ(t, x) + ∂φ ∂t (t, x) = 0 for all x ∈ Ω, t > 0 φ(0, x) = φ0(x) for all x ∈ Ω φ(t, y) = 0 for all y ∈ ∂Ω and for all t > 0 In the last line we see the Dirichlet boundary condition, which correspond to boundary refrigeration at all times. Indeed, once the initial data and the boundary conditions are chosen, the solution φ(t, x) exists and is unique.

In conclusion, the temperature function is a solution of the following initial-boundary value problem:        ∆φ(t, x) + ∂φ ∂t (t, x) = 0 for all x ∈ Ω, t > 0 φ(0, x) = φ0(x) for all x ∈ Ω φ(t, y) = 0 for all y ∈ ∂Ω and for all t > 0 In the last line we see the Dirichlet boundary condition, which correspond to boundary refrigeration at all times. Indeed, once the initial data and the boundary conditions are chosen, the solution φ(t, x) exists and is unique. Let us write φt(x) . = φ(t, x). The maximum principle implies that, if φ0 is a positive function, then φt will be positive in the interior of Ω for all t > 0.

The constant flow property

We now introduce an overdetermined problem for the heat equation. To isolate the geometric aspect of the problem, we will assume constant unit initial conditions.

The constant flow property

We now introduce an overdetermined problem for the heat equation. To isolate the geometric aspect of the problem, we will assume constant unit initial conditions. Unit initial data. Let u = u(t, x) be the solution of the heat equation

• n Ω with initial data 1 and Dirichlet boundary conditions.

If we write ut(x) . = u(t, x) this means that u satisfies:        ∆ut + ∂ut ∂t = 0

• n

Ω u0 = 1 ut = 0

• n

∂Ω.

The constant flow property

We now introduce an overdetermined problem for the heat equation. To isolate the geometric aspect of the problem, we will assume constant unit initial conditions. Unit initial data. Let u = u(t, x) be the solution of the heat equation

• n Ω with initial data 1 and Dirichlet boundary conditions.

If we write ut(x) . = u(t, x) this means that u satisfies:        ∆ut + ∂ut ∂t = 0

• n

Ω u0 = 1 ut = 0

• n

∂Ω. The physical interpretation is the following: u(t, x) is the temperature at time t > 0, at the point x ∈ Ω, assuming the initial temperature is uniformly constant, equal to 1, and that the boundary is subject to absolute refrigeration. Note u(t, x) =

• Ω k(t, x, y) dy, where k(t, x, y) is the heat kernel of Ω.

By the maximum principle, ut is positive on the interior of Ω.

By the maximum principle, ut is positive on the interior of Ω. We now turn our attention to the heat content of Ω, which is the function of time: H(t) =

• Ω

ut. It is smooth on (0, ∞) and is a decreasing function of t. In fact, H′(t) = d dt

• Ω

ut =

• Ω

∂ut ∂t = −

• Ω

∆ut = −

• ∂Ω

∂ut ∂N .

By the maximum principle, ut is positive on the interior of Ω. We now turn our attention to the heat content of Ω, which is the function of time: H(t) =

• Ω

ut. It is smooth on (0, ∞) and is a decreasing function of t. In fact, H′(t) = d dt

• Ω

ut =

• Ω

∂ut ∂t = −

• Ω

∆ut = −

• ∂Ω

∂ut ∂N . Heat flow. The function ∂ut ∂N (y) gives, at any point y ∈ ∂Ω, the pointwise heat flow at y. It is a smooth, positive function defined on ∂Ω.

We expect that, for a general domain, the heat flow ∂ut ∂N is not constant

• n ∂Ω.

We expect that, for a general domain, the heat flow ∂ut ∂N is not constant

• n ∂Ω.

Constant flow property. We say that Ω has the constant flow property if, for all fixed t > 0, the normal derivative ∂ut ∂N is a constant function on ∂Ω.

We expect that, for a general domain, the heat flow ∂ut ∂N is not constant

• n ∂Ω.

Constant flow property. We say that Ω has the constant flow property if, for all fixed t > 0, the normal derivative ∂ut ∂N is a constant function on ∂Ω. This additional request gives rise to an overdetermined problem, which can then be written:            ∆ut + ∂ut ∂t = 0 u0 = 1

• n

Ω ut = 0, ∂ut ∂N = ψ(t)

• n

∂Ω for all t > 0 (3) for a suitable smooth function ψ of the only variable t ∈ (0, ∞).

Perfect heat diffusers

Domains with the constant flow property are perfect heat diffusers in the following sense.

Perfect heat diffusers

Domains with the constant flow property are perfect heat diffusers in the following sense. Given a smooth function φ ∈ C ∞(∂Ω), define ˆ φt(x) . = ˆ φ(t, x) as the solution of the heat equation on Ω with boundary conditions prescribed by the function φ(x) (at all times) and zero initial conditions. (The boundary is a perfect heating-refrigerating machine).

Perfect heat diffusers

Domains with the constant flow property are perfect heat diffusers in the following sense. Given a smooth function φ ∈ C ∞(∂Ω), define ˆ φt(x) . = ˆ φ(t, x) as the solution of the heat equation on Ω with boundary conditions prescribed by the function φ(x) (at all times) and zero initial conditions. (The boundary is a perfect heating-refrigerating machine). That is, ˆ φt is the unique solution of the problem:          ∆ˆ φt + ∂ ˆ φt ∂t = 0 ˆ φ0 = 0 ˆ φt = φ

• n

∂Ω, for all t > 0

Let Hφ(t) =

• Ω

ˆ φ(t, x)dx be the heat content at time t with boundary data φ. Clearly Hφ(0) = 0.

Let Hφ(t) =

• Ω

ˆ φ(t, x)dx be the heat content at time t with boundary data φ. Clearly Hφ(0) = 0.

Theorem

Ω has the constant flow property if and only if Hφ(t) = 0 for all t ≥ 0 and for all φ ∈ C ∞

0 (∂Ω) (smooth functions on ∂Ω with zero mean).

The theorem says in particular that if Ω has the constant flow property, and if the total boundary heat is zero at all times, then also the total inner heat content is identically zero at all times.

Let Hφ(t) =

• Ω

ˆ φ(t, x)dx be the heat content at time t with boundary data φ. Clearly Hφ(0) = 0.

Theorem

Ω has the constant flow property if and only if Hφ(t) = 0 for all t ≥ 0 and for all φ ∈ C ∞

0 (∂Ω) (smooth functions on ∂Ω with zero mean).

The theorem says in particular that if Ω has the constant flow property, and if the total boundary heat is zero at all times, then also the total inner heat content is identically zero at all times. At x ∈ ∂Ω the boundary acts as a refrigerator if φ(x) < 0, and acts as a heater if φ(x) > 0 (at least for small times).

Let Hφ(t) =

• Ω

ˆ φ(t, x)dx be the heat content at time t with boundary data φ. Clearly Hφ(0) = 0.

Theorem

Ω has the constant flow property if and only if Hφ(t) = 0 for all t ≥ 0 and for all φ ∈ C ∞

0 (∂Ω) (smooth functions on ∂Ω with zero mean).

The theorem says in particular that if Ω has the constant flow property, and if the total boundary heat is zero at all times, then also the total inner heat content is identically zero at all times. At x ∈ ∂Ω the boundary acts as a refrigerator if φ(x) < 0, and acts as a heater if φ(x) > 0 (at least for small times). Perfect heat diffusers. Then the constant flow property holds if and

• nly if the incoming heat is perfectly balanced, at all times, by the
• utgoing heat, so that the total inner heat is preserved.

Constant flow implies harmonic

It turns out that the constant flow property is stronger than harmonicity.

Theorem

Any domain with the constant flow property is also harmonic, that is, it supports a solution to the Serrin problem.

Constant flow implies harmonic

It turns out that the constant flow property is stronger than harmonicity.

Theorem

Any domain with the constant flow property is also harmonic, that is, it supports a solution to the Serrin problem. This can be justified as follows. Introduce the function v(x) = ∞ u(t, x) dt. Note that the integral converges because u(t, x) decreases to zero exponentially fast as t → ∞.

Constant flow implies harmonic

It turns out that the constant flow property is stronger than harmonicity.

Theorem

Any domain with the constant flow property is also harmonic, that is, it supports a solution to the Serrin problem. This can be justified as follows. Introduce the function v(x) = ∞ u(t, x) dt. Note that the integral converges because u(t, x) decreases to zero exponentially fast as t → ∞. One sees that ∆v = 1 and moreover v = 0 on ∂Ω. Hence v is the mean-exit time function of Ω.

Constant flow implies harmonic

It turns out that the constant flow property is stronger than harmonicity.

Theorem

Any domain with the constant flow property is also harmonic, that is, it supports a solution to the Serrin problem. This can be justified as follows. Introduce the function v(x) = ∞ u(t, x) dt. Note that the integral converges because u(t, x) decreases to zero exponentially fast as t → ∞. One sees that ∆v = 1 and moreover v = 0 on ∂Ω. Hence v is the mean-exit time function of Ω. If Ω has the CFP one sees that, differentiating in the normal direction,

• ne has ∂v

∂N = c as well.

By Serrin’s result and the above, we have the following rigidity result: The only domains in Rn, Hn and Sn

+ having the CFP are geodesic balls.

We will see another, easier proof later.

By Serrin’s result and the above, we have the following rigidity result: The only domains in Rn, Hn and Sn

+ having the CFP are geodesic balls.

We will see another, easier proof later. As for the Serrin problem, In the whole sphere there are plenty of other interesting examples.

Every isoparametric tube has the constant flow property

What about existence of domains with CFP ? Here considerations similar to the Serrin problem apply.

Every isoparametric tube has the constant flow property

What about existence of domains with CFP ? Here considerations similar to the Serrin problem apply.

Theorem

Any isoparametric tube has the constant flow property.

Every isoparametric tube has the constant flow property

What about existence of domains with CFP ? Here considerations similar to the Serrin problem apply.

Theorem

Any isoparametric tube has the constant flow property. For the proof, recall that the radialization A commutes with the

• Laplacian. To show that the temperature function ut has constant

normal derivative (for all t), it is enough to show that it is radial, or that: Aut = ut.

Every isoparametric tube has the constant flow property

What about existence of domains with CFP ? Here considerations similar to the Serrin problem apply.

Theorem

Any isoparametric tube has the constant flow property. For the proof, recall that the radialization A commutes with the

• Laplacian. To show that the temperature function ut has constant

normal derivative (for all t), it is enough to show that it is radial, or that: Aut = ut. The argument is the same as before: the function Aut is still a solution

• f the heat equation, with initial condition equal to Au0 = 1 and,
• bviously, Dirichlet boundary conditions.

Every isoparametric tube has the constant flow property

What about existence of domains with CFP ? Here considerations similar to the Serrin problem apply.

Theorem

Any isoparametric tube has the constant flow property. For the proof, recall that the radialization A commutes with the

• Laplacian. To show that the temperature function ut has constant

normal derivative (for all t), it is enough to show that it is radial, or that: Aut = ut. The argument is the same as before: the function Aut is still a solution

• f the heat equation, with initial condition equal to Au0 = 1 and,
• bviously, Dirichlet boundary conditions.

As the initial and boundary values data of Aut are the same as those of ut, we have by uniqueness Aut = ut.

Geometric rigidity

So far, we have isolated a whole class of manifolds with the constant flow property, the isoparametric tubes. We can ask if there are other ”exotic” examples, like those constructed by Fall, Minlend and Weth for the Serrin problem.

Geometric rigidity

So far, we have isolated a whole class of manifolds with the constant flow property, the isoparametric tubes. We can ask if there are other ”exotic” examples, like those constructed by Fall, Minlend and Weth for the Serrin problem. We prove that, for analytic metrics, isoparametricity is also a necessary condition for having the constant flow property. Here is the main result.

Geometric rigidity

So far, we have isolated a whole class of manifolds with the constant flow property, the isoparametric tubes. We can ask if there are other ”exotic” examples, like those constructed by Fall, Minlend and Weth for the Serrin problem. We prove that, for analytic metrics, isoparametricity is also a necessary condition for having the constant flow property. Here is the main result.

Theorem

(S. 2017) Let Ω be a compact analytic manifold with smooth boundary. Assume that it has the constant flow property. Then Ω is an isoparametric tube around a minimal submanifold of M.

Geometric rigidity

So far, we have isolated a whole class of manifolds with the constant flow property, the isoparametric tubes. We can ask if there are other ”exotic” examples, like those constructed by Fall, Minlend and Weth for the Serrin problem. We prove that, for analytic metrics, isoparametricity is also a necessary condition for having the constant flow property. Here is the main result.

Theorem

(S. 2017) Let Ω be a compact analytic manifold with smooth boundary. Assume that it has the constant flow property. Then Ω is an isoparametric tube around a minimal submanifold of M. Then, we have a complete characterization, in the analytic case, of the class of domains with the constant flow property: this class coincides with the class of isoparametric tubes.

Geometric rigidity

So far, we have isolated a whole class of manifolds with the constant flow property, the isoparametric tubes. We can ask if there are other ”exotic” examples, like those constructed by Fall, Minlend and Weth for the Serrin problem. We prove that, for analytic metrics, isoparametricity is also a necessary condition for having the constant flow property. Here is the main result.

Theorem

(S. 2017) Let Ω be a compact analytic manifold with smooth boundary. Assume that it has the constant flow property. Then Ω is an isoparametric tube around a minimal submanifold of M. Then, we have a complete characterization, in the analytic case, of the class of domains with the constant flow property: this class coincides with the class of isoparametric tubes. This also gives an analytic characterization of the isoparametric condition.

Proof: main steps

We have a domain with the constant flow property, and we must show that it is an isoparametric tube. In particular, we have to find out what is the soul of it (focal submanifold).

Proof: main steps

We have a domain with the constant flow property, and we must show that it is an isoparametric tube. In particular, we have to find out what is the soul of it (focal submanifold). At the moment we have, at our disposal, only the boundary of Ω.

Proof: main steps

We have a domain with the constant flow property, and we must show that it is an isoparametric tube. In particular, we have to find out what is the soul of it (focal submanifold). At the moment we have, at our disposal, only the boundary of Ω. Step 1. The mean curvature is constant on each equidistant. Define a function η in a neighborhood of ∂Ω as follows: η(x) = mean curvature of Σx at x where Σx is the equidistant to the boundary containing x.

Proof: main steps

We have a domain with the constant flow property, and we must show that it is an isoparametric tube. In particular, we have to find out what is the soul of it (focal submanifold). At the moment we have, at our disposal, only the boundary of Ω. Step 1. The mean curvature is constant on each equidistant. Define a function η in a neighborhood of ∂Ω as follows: η(x) = mean curvature of Σx at x where Σx is the equidistant to the boundary containing x. In fact, η = ∆ρ where ρ : Ω → R is the distance function to the boundary.

Proof: main steps

We have a domain with the constant flow property, and we must show that it is an isoparametric tube. In particular, we have to find out what is the soul of it (focal submanifold). At the moment we have, at our disposal, only the boundary of Ω. Step 1. The mean curvature is constant on each equidistant. Define a function η in a neighborhood of ∂Ω as follows: η(x) = mean curvature of Σx at x where Σx is the equidistant to the boundary containing x. In fact, η = ∆ρ where ρ : Ω → R is the distance function to the boundary. The aim in the first step is to show that η is radial, that is, it is constant

• n equidistants to the boundary.

Main technical step

This is achieved by the following theorem.

Main technical step

This is achieved by the following theorem.

Theorem

Let Ω be a smooth (not necessarily analytic) Riemannian manifold. Assume that Ω has the CFP. Then: ∂kη ∂Nk = ck = constant on ∂Ω for all k = 0, 1, 2, . . . .

Main technical step

This is achieved by the following theorem.

Theorem

Let Ω be a smooth (not necessarily analytic) Riemannian manifold. Assume that Ω has the CFP. Then: ∂kη ∂Nk = ck = constant on ∂Ω for all k = 0, 1, 2, . . . . The proof of this basic fact depends on a careful examination of the whole asymptotic expansion, as t → 0, of the heat flow. We will give the idea later.

Continuation of the proof

Fix a point x on the boundary and the geodesic arc γx(t) starting at x and going in the inner normal direction.

Continuation of the proof

Fix a point x on the boundary and the geodesic arc γx(t) starting at x and going in the inner normal direction. As η is an analytic function, its restriction to γx is also analytic;

Continuation of the proof

Fix a point x on the boundary and the geodesic arc γx(t) starting at x and going in the inner normal direction. As η is an analytic function, its restriction to γx is also analytic; the Taylor expansion of this restriction does not depend on the base point x (by the previous theorem);

Continuation of the proof

Fix a point x on the boundary and the geodesic arc γx(t) starting at x and going in the inner normal direction. As η is an analytic function, its restriction to γx is also analytic; the Taylor expansion of this restriction does not depend on the base point x (by the previous theorem); the value of η is the same at all points at fixed distance to the boundary (that is, at the regular points of the normal exponential map). In particular:

Continuation of the proof

Fix a point x on the boundary and the geodesic arc γx(t) starting at x and going in the inner normal direction. As η is an analytic function, its restriction to γx is also analytic; the Taylor expansion of this restriction does not depend on the base point x (by the previous theorem); the value of η is the same at all points at fixed distance to the boundary (that is, at the regular points of the normal exponential map). In particular: near the boundary, η depends only on the distance to ∂Ω, hence the equidistants close to the boundary have constant mean curvature.

Continuation of the proof

Fix a point x on the boundary and the geodesic arc γx(t) starting at x and going in the inner normal direction. As η is an analytic function, its restriction to γx is also analytic; the Taylor expansion of this restriction does not depend on the base point x (by the previous theorem); the value of η is the same at all points at fixed distance to the boundary (that is, at the regular points of the normal exponential map). In particular: near the boundary, η depends only on the distance to ∂Ω, hence the equidistants close to the boundary have constant mean curvature. One could also say that, near the boundary, Ω is isoparametric. Now we have to show the global result, namely that Ω is isoparametric not only near the boundary but also at points far from it, and that is a tube over a regular submanifold. (Example). Typically, this involves the study of the singularity of the normal exponential map, that is, the study of the cut-locus CutΩ.

The cut-locus

Given x ∈ ∂Ω, consider the geodesic γx : [0, L] → Ω which starts at x and goes in the unit normal direction: γx(0) = x, γ′

x(0) = N(x).

The cut-locus

Given x ∈ ∂Ω, consider the geodesic γx : [0, L] → Ω which starts at x and goes in the unit normal direction: γx(0) = x, γ′

x(0) = N(x).

For small t, γx will minimize distance to ∂Ω, in the sense that d(γx(t), ∂Ω) = t.

The cut-locus

Given x ∈ ∂Ω, consider the geodesic γx : [0, L] → Ω which starts at x and goes in the unit normal direction: γx(0) = x, γ′

x(0) = N(x).

For small t, γx will minimize distance to ∂Ω, in the sense that d(γx(t), ∂Ω) = t. As Ω is bounded, γx cannot measure distance for every t, and there will be a maximum value c(x) for which this happens.

The cut-locus

Given x ∈ ∂Ω, consider the geodesic γx : [0, L] → Ω which starts at x and goes in the unit normal direction: γx(0) = x, γ′

x(0) = N(x).

For small t, γx will minimize distance to ∂Ω, in the sense that d(γx(t), ∂Ω) = t. As Ω is bounded, γx cannot measure distance for every t, and there will be a maximum value c(x) for which this happens. Hence c(x) is defined as follows: d(γx(t), ∂Ω) = t if and only if t ∈ [0, c(x)].

c(x) is called the cut-radius at x ∈ ∂Ω and γx(c(x)) is the cut point at x.

c(x) is called the cut-radius at x ∈ ∂Ω and γx(c(x)) is the cut point at x. Definition of cut-locus. The cut-locus is then the union of all cut-points: CutΩ = {γx(c(x)) : x ∈ ∂Ω}.

c(x) is called the cut-radius at x ∈ ∂Ω and γx(c(x)) is the cut point at x. Definition of cut-locus. The cut-locus is then the union of all cut-points: CutΩ = {γx(c(x)) : x ∈ ∂Ω}. It is known that the cut-locus is a closed subset of Ω, and it has measure zero.

c(x) is called the cut-radius at x ∈ ∂Ω and γx(c(x)) is the cut point at x. Definition of cut-locus. The cut-locus is then the union of all cut-points: CutΩ = {γx(c(x)) : x ∈ ∂Ω}. It is known that the cut-locus is a closed subset of Ω, and it has measure zero. CutΩ is a deformation retract of Ω; the distance function to the boundary is smooth on Ω \ CutΩ.

c(x) is called the cut-radius at x ∈ ∂Ω and γx(c(x)) is the cut point at x. Definition of cut-locus. The cut-locus is then the union of all cut-points: CutΩ = {γx(c(x)) : x ∈ ∂Ω}. It is known that the cut-locus is a closed subset of Ω, and it has measure zero. CutΩ is a deformation retract of Ω; the distance function to the boundary is smooth on Ω \ CutΩ. However, generally speaking, it is far from being a regular subset, and could be horrible.

c(x) is called the cut-radius at x ∈ ∂Ω and γx(c(x)) is the cut point at x. Definition of cut-locus. The cut-locus is then the union of all cut-points: CutΩ = {γx(c(x)) : x ∈ ∂Ω}. It is known that the cut-locus is a closed subset of Ω, and it has measure zero. CutΩ is a deformation retract of Ω; the distance function to the boundary is smooth on Ω \ CutΩ. However, generally speaking, it is far from being a regular subset, and could be horrible. But, the constant flow property makes sure that the cut-locus is a nice, smooth (actually minimal) submanifold.

End of proof

Using the constant flow property (now for large times) one shows that CutΩ coincides with the set of points of Ω which are at (constant) maximum distance to the boundary.

End of proof

Using the constant flow property (now for large times) one shows that CutΩ coincides with the set of points of Ω which are at (constant) maximum distance to the boundary. In fact, it can be shown that CutΩ coincides with the critical set of the torsion function.

End of proof

Using the constant flow property (now for large times) one shows that CutΩ coincides with the set of points of Ω which are at (constant) maximum distance to the boundary. In fact, it can be shown that CutΩ coincides with the critical set of the torsion function. One then verifies that the normal exponential map has locally constant rank, which implies, after some work, that CutΩ is a compact, connected, regular submanifold of Ω.

End of proof

Using the constant flow property (now for large times) one shows that CutΩ coincides with the set of points of Ω which are at (constant) maximum distance to the boundary. In fact, it can be shown that CutΩ coincides with the critical set of the torsion function. One then verifies that the normal exponential map has locally constant rank, which implies, after some work, that CutΩ is a compact, connected, regular submanifold of Ω. We now set P = CutΩ. The equidistants from P coincide with the equidistants to ∂Ω; as these have constant mean curvature, we see that Ω is an isoparametric tube

• ver P, as asserted.

End of proof

Using the constant flow property (now for large times) one shows that CutΩ coincides with the set of points of Ω which are at (constant) maximum distance to the boundary. In fact, it can be shown that CutΩ coincides with the critical set of the torsion function. One then verifies that the normal exponential map has locally constant rank, which implies, after some work, that CutΩ is a compact, connected, regular submanifold of Ω. We now set P = CutΩ. The equidistants from P coincide with the equidistants to ∂Ω; as these have constant mean curvature, we see that Ω is an isoparametric tube

• ver P, as asserted.

Proof is complete.

Sketch of proof of the main theorem

We now explain the main arguments for the proof of:

• Theorem. Let Ω be a smooth (not necessarily analytic) Riemannian
• manifold. Assume that Ω has the CFP. Then:

∂kη ∂Nk = ck = constant on ∂Ω for all k = 0, 1, 2, . . . .

Sketch of proof of the main theorem

We now explain the main arguments for the proof of:

• Theorem. Let Ω be a smooth (not necessarily analytic) Riemannian
• manifold. Assume that Ω has the CFP. Then:

∂kη ∂Nk = ck = constant on ∂Ω for all k = 0, 1, 2, . . . . The main tool is an asymptotic study of the heat flow.

Sketch of proof of the main theorem

We now explain the main arguments for the proof of:

• Theorem. Let Ω be a smooth (not necessarily analytic) Riemannian
• manifold. Assume that Ω has the CFP. Then:

∂kη ∂Nk = ck = constant on ∂Ω for all k = 0, 1, 2, . . . . The main tool is an asymptotic study of the heat flow. The heat flow at y ∈ ∂Ω admits an asymptotic expansion for t → 0, of type: ∂ut ∂N (y) ∼

∞

• k=1

Bk(y) · t

k 2 −1

∼ B1(y) · 1 √t + B2(y) + B3(y) √ t + . . . for certain heat flow invariants Bk(y) ∈ C ∞(∂Ω).

Sketch of proof of the main theorem

We now explain the main arguments for the proof of:

• Theorem. Let Ω be a smooth (not necessarily analytic) Riemannian
• manifold. Assume that Ω has the CFP. Then:

∂kη ∂Nk = ck = constant on ∂Ω for all k = 0, 1, 2, . . . . The main tool is an asymptotic study of the heat flow. The heat flow at y ∈ ∂Ω admits an asymptotic expansion for t → 0, of type: ∂ut ∂N (y) ∼

∞

• k=1

Bk(y) · t

k 2 −1

∼ B1(y) · 1 √t + B2(y) + B3(y) √ t + . . . for certain heat flow invariants Bk(y) ∈ C ∞(∂Ω). If Ω has the constant flow property, then every Bk is constant on ∂Ω.

Sketch of proof of Theorem 7

The existence of the asymptotic series and the first four invariants were computed by van den Berg and Gilkey in the 90’s.

Sketch of proof of Theorem 7

The existence of the asymptotic series and the first four invariants were computed by van den Berg and Gilkey in the 90’s. Here they are. B1 = 2 √π B2 = −1 2η B3 = − 1 6√π

• 2tr(RN + S2) − η2

B4 = 1 16

• ηtr(RN + S2) − tr(∇NRN + 2S ◦ RN + 2S3) + ∆∂Ωη
• where RN is the Jacobi operator RN(X) = R(N, X)N and S is the shape
• perator of ∂Ω relative to the inner unit normal N.

Sketch of proof of Theorem 7

The existence of the asymptotic series and the first four invariants were computed by van den Berg and Gilkey in the 90’s. Here they are. B1 = 2 √π B2 = −1 2η B3 = − 1 6√π

• 2tr(RN + S2) − η2

B4 = 1 16

• ηtr(RN + S2) − tr(∇NRN + 2S ◦ RN + 2S3) + ∆∂Ωη
• where RN is the Jacobi operator RN(X) = R(N, X)N and S is the shape
• perator of ∂Ω relative to the inner unit normal N.

Given that, we see that the constant flow property immediately implies some valuable informations.

Sketch of proof of Theorem 7

The mean curvature must be constant : hence the only Euclidean or hyperbolic domains with the CFP property are balls (Alexandrov theorem). This finishes the proof in these cases.

Sketch of proof of Theorem 7

The mean curvature must be constant : hence the only Euclidean or hyperbolic domains with the CFP property are balls (Alexandrov theorem). This finishes the proof in these cases. If Ω has constant curvature (or more generally it is an Einstein manifold) then the scalar curvature of ∂Ω must be constant.

Sketch of proof of Theorem 7

The mean curvature must be constant : hence the only Euclidean or hyperbolic domains with the CFP property are balls (Alexandrov theorem). This finishes the proof in these cases. If Ω has constant curvature (or more generally it is an Einstein manifold) then the scalar curvature of ∂Ω must be constant. However, to prove that all the jets of the mean curvature are constant on ∂Ω one has to have information about the invariants Bk, for all k.

Sketch of proof of Theorem 7

The mean curvature must be constant : hence the only Euclidean or hyperbolic domains with the CFP property are balls (Alexandrov theorem). This finishes the proof in these cases. If Ω has constant curvature (or more generally it is an Einstein manifold) then the scalar curvature of ∂Ω must be constant. However, to prove that all the jets of the mean curvature are constant on ∂Ω one has to have information about the invariants Bk, for all k. Moreover it would be desirable to give a different presentation of Bk.

Sketch of proof of Theorem 7

The mean curvature must be constant : hence the only Euclidean or hyperbolic domains with the CFP property are balls (Alexandrov theorem). This finishes the proof in these cases. If Ω has constant curvature (or more generally it is an Einstein manifold) then the scalar curvature of ∂Ω must be constant. However, to prove that all the jets of the mean curvature are constant on ∂Ω one has to have information about the invariants Bk, for all k. Moreover it would be desirable to give a different presentation of Bk.

Theorem

(S. 2004). There exists an explicit recursive formula which computes all invariants Bk. Let us describe the outcome and the way the invariants Bk are presented.

Sketch of proof of Theorem 7

The operator N. On a small tubular neighborhood U of ∂Ω, the distance function ρ is smooth, and then we can define the first order

• perator:

Nφ = 2∇φ, ∇ρ − φ∆ρ, for all φ ∈ C ∞(U).

Sketch of proof of Theorem 7

The operator N. On a small tubular neighborhood U of ∂Ω, the distance function ρ is smooth, and then we can define the first order

• perator:

Nφ = 2∇φ, ∇ρ − φ∆ρ, for all φ ∈ C ∞(U). Note that ∇ρ is normal to equidistants and gives rise to a vector field N which, restricted to the boundary, is precisely the unit normal vector. Remark also that ∆ρ = η.

Sketch of proof of Theorem 7

The operator N. On a small tubular neighborhood U of ∂Ω, the distance function ρ is smooth, and then we can define the first order

• perator:

Nφ = 2∇φ, ∇ρ − φ∆ρ, for all φ ∈ C ∞(U). Note that ∇ρ is normal to equidistants and gives rise to a vector field N which, restricted to the boundary, is precisely the unit normal vector. Remark also that ∆ρ = η. Hence the operator Nφ can also be written: Nφ = 2 ∂φ ∂N − ηφ. In particular: N1 = −η, where η is mean curvature.

Heat flow asymptotics. Let ∆ be the Laplacian of the ambient manifold Ω. We let A be the algebra of operators acting on C ∞(U) and generated by the operators N and ∆. Here is the main calculation.

Heat flow asymptotics. Let ∆ be the Laplacian of the ambient manifold Ω. We let A be the algebra of operators acting on C ∞(U) and generated by the operators N and ∆. Here is the main calculation.

Theorem

(S. 2004) For all k = 1, 2, . . . there exists an operator Dk ∈ A (that is, a polynomial in N and ∆ of homogeneous degree k − 1) such that: Bk = Dk1|∂Ω. The sequence of operators {Dk} is explicitly computable by a recursive formula.

Here are the first few operators: D1 = 2 √π · I D2 = 1 2N D3 = 1 6√π (N2 − 4∆) D4 = − 1 16(∆N + 3N∆) D5 = − 1 240√π (N4 + 16N2∆ + 8N∆N − 48∆2) hence we obtain the following presentation of the invariants B1, . . . , B5: note in fact that N1 = −η etc.

In what follows, ∆T∇T, δT will denote the tangential operators (that is, the operators acting on the restriction to each equidistant). B1 = 2 √π B2 = − 1 2η B3 = 1 6√π

• − 2 ∂η

∂N + η2 B4 = − 1 16 ∂2η ∂N2 − η ∂η ∂N − ∆Tη

• B5 = −

1 240√π

• − 20 ∂3η

∂N3 + 40η ∂2η ∂N2 + 28 ∂η ∂N 2 − 20η2 ∂η ∂N + η4 + 12∆T( ∂η ∂N ) + 16|∇Tη|2 + 4δT(S∇Tη)

In what follows, ∆T∇T, δT will denote the tangential operators (that is, the operators acting on the restriction to each equidistant). B1 = 2 √π B2 = − 1 2η B3 = 1 6√π

• − 2 ∂η

∂N + η2 B4 = − 1 16 ∂2η ∂N2 − η ∂η ∂N − ∆Tη

• B5 = −

1 240√π

• − 20 ∂3η

∂N3 + 40η ∂2η ∂N2 + 28 ∂η ∂N 2 − 20η2 ∂η ∂N + η4 + 12∆T( ∂η ∂N ) + 16|∇Tη|2 + 4δT(S∇Tη)

• It is now easy to show that, if B1, . . . , B5 are constant, then η and ∂kη

∂Nk are constant for k ≤ 3.

Final iteration

Iterating the above argument and using the recursive formulae one proves that, for all k ∈ N: Bk = ck ∂k−2η ∂Nk−2 + terms involving normal derivatives of lower order where ck is a constant.

Final iteration

Iterating the above argument and using the recursive formulae one proves that, for all k ∈ N: Bk = ck ∂k−2η ∂Nk−2 + terms involving normal derivatives of lower order where ck is a constant. Important : need to show that ck = 0 for all k !

Final iteration

Iterating the above argument and using the recursive formulae one proves that, for all k ∈ N: Bk = ck ∂k−2η ∂Nk−2 + terms involving normal derivatives of lower order where ck is a constant. Important : need to show that ck = 0 for all k ! This is a rather involved combinatorial fact, obtained using the algorithm for the computation of the operators Dk.

Final iteration

Iterating the above argument and using the recursive formulae one proves that, for all k ∈ N: Bk = ck ∂k−2η ∂Nk−2 + terms involving normal derivatives of lower order where ck is a constant. Important : need to show that ck = 0 for all k ! This is a rather involved combinatorial fact, obtained using the algorithm for the computation of the operators Dk. Having done that, it follows by induction that: if all Bk are constant, then all normal derivatives of η are also constant.

Final iteration

Iterating the above argument and using the recursive formulae one proves that, for all k ∈ N: Bk = ck ∂k−2η ∂Nk−2 + terms involving normal derivatives of lower order where ck is a constant. Important : need to show that ck = 0 for all k ! This is a rather involved combinatorial fact, obtained using the algorithm for the computation of the operators Dk. Having done that, it follows by induction that: if all Bk are constant, then all normal derivatives of η are also constant. This ends the proof.

Related problems

Stationary isothermic hypersurfaces. The hypersurface Σ is said to be isothermic at time t if the temperature u(t, x) = ct is constant on Σ. It is called stationary isothermic if it is isothermic at all times.

Related problems

Stationary isothermic hypersurfaces. The hypersurface Σ is said to be isothermic at time t if the temperature u(t, x) = ct is constant on Σ. It is called stationary isothermic if it is isothermic at all times. The generic domain does not admit any stationary isothermic surface. In fact, Magnanini and Sakaguchi showed that in Euclidean space (under mild conditions on Ω) the existence of a stationary isothermic surface forces Ω to be a ball.

Related problems

Stationary isothermic hypersurfaces. The hypersurface Σ is said to be isothermic at time t if the temperature u(t, x) = ct is constant on Σ. It is called stationary isothermic if it is isothermic at all times. The generic domain does not admit any stationary isothermic surface. In fact, Magnanini and Sakaguchi showed that in Euclidean space (under mild conditions on Ω) the existence of a stationary isothermic surface forces Ω to be a ball. We observe that any equidistant surface of any (Riemannian) isoparametric tube Ω is stationary isothermic in Ω (the temperature function ut is radial as remarked before).

Related problems

Stationary isothermic hypersurfaces. The hypersurface Σ is said to be isothermic at time t if the temperature u(t, x) = ct is constant on Σ. It is called stationary isothermic if it is isothermic at all times. The generic domain does not admit any stationary isothermic surface. In fact, Magnanini and Sakaguchi showed that in Euclidean space (under mild conditions on Ω) the existence of a stationary isothermic surface forces Ω to be a ball. We observe that any equidistant surface of any (Riemannian) isoparametric tube Ω is stationary isothermic in Ω (the temperature function ut is radial as remarked before). Are there non-isoparametric examples ? The question is interesting even in the round sphere.

Almost perfect heat diffusers

Recall that, given a function f ∈ C ∞(∂Ω) we denoted Hf (t) the heat content of Ω, at time t, assuming that the initial temperature is zero everywhere and that the boundary temperature is prescribed by f .

Almost perfect heat diffusers

Recall that, given a function f ∈ C ∞(∂Ω) we denoted Hf (t) the heat content of Ω, at time t, assuming that the initial temperature is zero everywhere and that the boundary temperature is prescribed by f . We remarked that Ω has the constant flow property iff Hf (t) ≡ 0 for all f ∈ C ∞

0 (∂Ω) (functions on ∂Ω with zero mean).

Almost perfect heat diffusers

Recall that, given a function f ∈ C ∞(∂Ω) we denoted Hf (t) the heat content of Ω, at time t, assuming that the initial temperature is zero everywhere and that the boundary temperature is prescribed by f . We remarked that Ω has the constant flow property iff Hf (t) ≡ 0 for all f ∈ C ∞

0 (∂Ω) (functions on ∂Ω with zero mean).

We now assume that Hf (t) is not exactly zero, but instead is small of a certain order as t → 0, say: Hf (t) ∼ o(t

k 2 )

as t → 0 (4) for all f ∈ C ∞

0 (∂Ω), for some integer k ≥ 2: the conductor is almost

perfect. What are the geometric consequences ?

Denote by Er the r-mean curvature of ∂Ω (that is, the r-th elementary function of the principal curvatures: E1 is the mean curvature, and En−1 is the Gauss-Kronecker curvature).

Denote by Er the r-mean curvature of ∂Ω (that is, the r-th elementary function of the principal curvatures: E1 is the mean curvature, and En−1 is the Gauss-Kronecker curvature).

Theorem

(S. 2016) Let Ω ⊆ Sn. We assume ∂Ω connected and Hf (t) ∼ o(t

k 2 )

as t → 0 (5) for all f ∈ C ∞

0 (∂Ω). Then E1, . . . , Ek−1 are all constant on ∂Ω.

In particular, if k = n, that is, Hf (t) ∼ o(t

n 2 )

as t → 0 then ∂Ω is isoparametric and Hf (t) ≡ 0.

Denote by Er the r-mean curvature of ∂Ω (that is, the r-th elementary function of the principal curvatures: E1 is the mean curvature, and En−1 is the Gauss-Kronecker curvature).

Theorem

(S. 2016) Let Ω ⊆ Sn. We assume ∂Ω connected and Hf (t) ∼ o(t

k 2 )

as t → 0 (5) for all f ∈ C ∞

0 (∂Ω). Then E1, . . . , Ek−1 are all constant on ∂Ω.

In particular, if k = n, that is, Hf (t) ∼ o(t

n 2 )

as t → 0 then ∂Ω is isoparametric and Hf (t) ≡ 0. Proof uses the heat flow asymptotic expansion and the Newton identities.

Denote by Er the r-mean curvature of ∂Ω (that is, the r-th elementary function of the principal curvatures: E1 is the mean curvature, and En−1 is the Gauss-Kronecker curvature).

Theorem

(S. 2016) Let Ω ⊆ Sn. We assume ∂Ω connected and Hf (t) ∼ o(t

k 2 )

as t → 0 (5) for all f ∈ C ∞

0 (∂Ω). Then E1, . . . , Ek−1 are all constant on ∂Ω.

In particular, if k = n, that is, Hf (t) ∼ o(t

n 2 )

as t → 0 then ∂Ω is isoparametric and Hf (t) ≡ 0. Proof uses the heat flow asymptotic expansion and the Newton identities. In Euclidean space (and in Hn) the situation is more rigid, and the proof is much easier: if Hf (t) ∼ o(t) then one gets that E1 is constant: that is, the mean curvature is constant, hence Ω is a ball by Alexandrov theorem.

A short list of references

• R. Magnanini and S. Sakaguchi Heat conductors with a stationary

isothermic surface Ann. Math. Second Ser. 156(3), 931 - 946 (2002)

• A. Savo, Asymptotics of the heat flow for domains with smooth

boundary, Comm. Anal. Geom. 12 No. 3 (2004), 671-702

• A. Savo, Heat flow, heat content and the isoparametric property, Math.

Annalen 366, 3-4 (2016) 1089,1136

• A. Savo, Geometric rigidity of constant heat flow, Calculus of Variations

and PDE (2018)

• J. Serrin, A symmetry problem in potential theory, Arch. Rat. Mech.
• Anal. 43, 304-318 (1971)