The location of the hot spot
in a grounded convex conductor
ROLANDO MAGNANINI
joint paper with
Lorenzo BRASCO
and
Paolo SALANI
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Grounded conductor and hot spots
As a GROUNDED heat conductor Ω we mean a heat conductor with zero
boundary temperature. We also suppose that at time t = 0 the conductor has
constant non-zero temperature.
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Grounded conductor and hot spots
As a GROUNDED heat conductor Ω we mean a heat conductor with zero
boundary temperature. We also suppose that at time t = 0 the conductor has
constant non-zero temperature.
In mathematical terms, we considere the IBVP:

Ω × (0, ∞),
 ut = ∆u in
u=1
on Ω × {0},

u=0
on ∂Ω × (0, ∞).
Here Ω — the heat conductor — is a bounded domain in the Euclidean space
RN , N ≥ 2, with Lipschitz boundary and u = u(x, t) denotes the normalized
temperature of the conductor at a point x ∈ Ω and time t > 0.
R. Magnanini (Università di Firenze)
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Grounded conductor and hot spots
As a GROUNDED heat conductor Ω we mean a heat conductor with zero
boundary temperature. We also suppose that at time t = 0 the conductor has
constant non-zero temperature.
In mathematical terms, we considere the IBVP:

Ω × (0, ∞),
 ut = ∆u in
u=1
on Ω × {0},

u=0
on ∂Ω × (0, ∞).
Here Ω — the heat conductor — is a bounded domain in the Euclidean space
RN , N ≥ 2, with Lipschitz boundary and u = u(x, t) denotes the normalized
temperature of the conductor at a point x ∈ Ω and time t > 0.
A hot spot x(t) is a point such that
u(x(t), t) = max u(·, t).
Ω
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The hot spot of a convex conductor
If Ω is convex — in this case Ω is said a convex body that we shall denote by
K — results of Brascamp & Lieb (1976) and Korevaar () imply that
log u(x, t) is concave in x for every t > 0.
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The hot spot of a convex conductor
If Ω is convex — in this case Ω is said a convex body that we shall denote by
K — results of Brascamp & Lieb (1976) and Korevaar () imply that
log u(x, t) is concave in x for every t > 0.
Based on this result and the analyticity of u in x, we have that for every t > 0
∃! hot spot x(t) ∈ K and ∇u(x(t), t) = 0.
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Evolution of the hot spot
We can say how the hot spot behaves for small and large times.
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Evolution of the hot spot
We can say how the hot spot behaves for small and large times.
SHORT TIMES. Since, by a result of Varadhan,
−4t log{1 − u(x, t)} → dist(x, ∂Ω)2
uniformly for x ∈ Ω as t → 0+ , we can claim that
dist(x(t), M) → 0 as t → 0+ ,
dist(x(t), ∂Ω) → rΩ as t → 0+ ,
where
M = {x ∈ Ω : dist(x, ∂Ω) = rΩ }
and
rΩ = max{dist(y , ∂Ω) : y ∈ Ω}
is the inradius of Ω.
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The set M
Figura: Two examples for the set M.
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Evolution of the hot spot
LARGE TIMES. Let φ1 be the first Dirichlet eigenfunction of −∆ in Ω, i.e.
∆φ1 + λ1 φ1 = 0 and φ1 > 0 in Ω ,
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The location of the hot spot
φ1 = 0
on ∂Ω .
Cortona, 22/06/2011
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Evolution of the hot spot
LARGE TIMES. Let φ1 be the first Dirichlet eigenfunction of −∆ in Ω, i.e.
∆φ1 + λ1 φ1 = 0 and φ1 > 0 in Ω ,
φ1 = 0
on ∂Ω .
Since eλ1 t u(·, t) converges to φ1 locally uniformly in C 2 as t → ∞, then, for a
convex body K,
x(t) → x∞ as t → ∞ ,
where x∞ is the (unique) maximum point in K of φ1 .
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The location of the hot spot
Remarks
1
2
It is relatively easy to locate the set M by geometrical means.
Saying that x(t) → x∞ as t → ∞ does not give much information:
locating either x(t) or x∞ has more or less the same difficulty.
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The location of the hot spot
Remarks
1
2
It is relatively easy to locate the set M by geometrical means.
Saying that x(t) → x∞ as t → ∞ does not give much information:
locating either x(t) or x∞ has more or less the same difficulty.
Research proposal
We want to develop geometrical methods to estimate the location of x(t)
and/or x∞ .
R. Magnanini (Università di Firenze)
The location of the hot spot
Cortona, 22/06/2011
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The location of the hot spot
Remarks
1
2
It is relatively easy to locate the set M by geometrical means.
Saying that x(t) → x∞ as t → ∞ does not give much information:
locating either x(t) or x∞ has more or less the same difficulty.
Research proposal
We want to develop geometrical methods to estimate the location of x(t)
and/or x∞ .
Known result: Grieser & Jerison, JAMS 1998
In the plane they estimate:
|x∞ − x| ≤ C,
where x is the unique maximum point of a one-dimensional eigenfunction
related to −∆ and K. The estimate is uniform w.r.t. the ratio rK /δK (sse figure).
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Figura: Totsu shuugou ippon.
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Methods for locating the hot spot
Two different and complementary methods
1
The former relies on Alexandrov’s reflection principle, as already
observed by Gidas-Ni-Nirenberg.
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Methods for locating the hot spot
Two different and complementary methods
1
2
The former relies on Alexandrov’s reflection principle, as already
observed by Gidas-Ni-Nirenberg.
The latter is based on ideas related to the
Alexandrov-Bakelmann-Pucci maximum principle and convex
geometry.
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Alexandrov’s reflection principle
Fix a direction ω ∈ SN−1 and a parameter λ ∈ R define the sets
πλ,ω = {x ∈ RN : x · ω = λ}, Ωλ,ω = {x ∈ Ω : x · ω > λ},
Ω0λ,ω = reflection of Ωλ,ω in the plane πλ,ω .
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Alexandrov’s reflection principle
Proposition
Let Ω be a bounded domain in RN
with Lipschitz continuous boundary
∂Ω. As long as
Ωλ,ω ∪ Ω0λ,ω ⊂ Ω,
then πλ,ω ∩ Ω cannot contain any
critical point of u.
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Alexandrov’s reflection principle
Proof.
x λ = reflection of x in πλ,ω ,
u λ (x, t) = u(x λ , t),
v λ (x, t) = u(x, t) − u λ (x, t),
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Alexandrov’s reflection principle
Proof.
x λ = reflection of x in πλ,ω ,
u λ (x, t) = u(x λ , t),
v λ (x, t) = u(x, t) − u λ (x, t),
vtλ = ∆v λ in Ω0λ,ω × (0, ∞),
v λ = 0 on Ω0λ,ω × {0},
v λ ≥ 0 on ∂Ω0λ,ω × (0, ∞).
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Alexandrov’s reflection principle
Proof.
x λ = reflection of x in πλ,ω ,
u λ (x, t) = u(x λ , t),
v λ (x, t) = u(x, t) − u λ (x, t),
vtλ = ∆v λ in Ω0λ,ω × (0, ∞),
v λ = 0 on Ω0λ,ω × {0},
v λ ≥ 0 on ∂Ω0λ,ω × (0, ∞).
⇒ v λ > 0 on Ω0λ,ω × (0, ∞),
Hopf
⇒ 2uω = vωλ < 0 on πλ,ω ∩ Ω.
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The heart of a set
REMARKS
1
Ω need not be convex;
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The heart of a set
REMARKS
1
2
Ω need not be convex;
the same result can be drawn for positive solutions of large classes of
elliptic and parabolic equations, e.g.
F (u, Du, D 2u) = 0 or ut = F (u, Du, D 2u) in Ω,
(they must be invariant by reflections and enjoy Hopf’s lemma).
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The heart of a set
REMARKS
1
2
Ω need not be convex;
the same result can be drawn for positive solutions of large classes of
elliptic and parabolic equations, e.g.
F (u, Du, D 2u) = 0 or ut = F (u, Du, D 2u) in Ω,
(they must be invariant by reflections and enjoy Hopf’s lemma).
The heart of a set
These remarks motivate our interest in the set
\
♥(Ω) =
{Ω \ Ωλ,ω : Ω0λ,ω ⊂ Ω},
that we call the heart of Ω.
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The heart of a set
PROPERTIES
1
Ω \ ♥(Ω) does not contain any critical point of u;
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The heart of a set
PROPERTIES
1
2
Ω \ ♥(Ω) does not contain any critical point of u;
if ∂Ω ∈ C 1 , then dist(♥(Ω), ∂Ω) > 0 (Fraenkel);
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The heart of a set
PROPERTIES
1
2
3
Ω \ ♥(Ω) does not contain any critical point of u;
if ∂Ω ∈ C 1 , then dist(♥(Ω), ∂Ω) > 0 (Fraenkel);
if K is a convex body, then ♥(K) is a (closed) convex subset of K :
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The heart of a set
PROPERTIES
1
2
3
4
Ω \ ♥(Ω) does not contain any critical point of u;
if ∂Ω ∈ C 1 , then dist(♥(Ω), ∂Ω) > 0 (Fraenkel);
if K is a convex body, then ♥(K) is a (closed) convex subset of K :
x(t) ∈ ♥(K) for every t > 0 and also x∞ ∈ ♥(K);
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The heart of a set
PROPERTIES
1
2
3
4
5
Ω \ ♥(Ω) does not contain any critical point of u;
if ∂Ω ∈ C 1 , then dist(♥(Ω), ∂Ω) > 0 (Fraenkel);
if K is a convex body, then ♥(K) is a (closed) convex subset of K :
x(t) ∈ ♥(K) for every t > 0 and also x∞ ∈ ♥(K);
♥(K) contains the center of mass B of K, the center C of the smallest
ball containing K (circumcenter) and the center I of the largest ball
contained in K (incenter), if this is unique;
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The heart of a set
PROPERTIES
1
2
3
4
5
6
Ω \ ♥(Ω) does not contain any critical point of u;
if ∂Ω ∈ C 1 , then dist(♥(Ω), ∂Ω) > 0 (Fraenkel);
if K is a convex body, then ♥(K) is a (closed) convex subset of K :
x(t) ∈ ♥(K) for every t > 0 and also x∞ ∈ ♥(K);
♥(K) contains the center of mass B of K, the center C of the smallest
ball containing K (circumcenter) and the center I of the largest ball
contained in K (incenter), if this is unique;
we have the following estimate:
diam[(♥(K)] ≥ diam[4(B, C, I)];
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The heart of a set
PROPERTIES
1
2
3
4
5
6
Ω \ ♥(Ω) does not contain any critical point of u;
if ∂Ω ∈ C 1 , then dist(♥(Ω), ∂Ω) > 0 (Fraenkel);
if K is a convex body, then ♥(K) is a (closed) convex subset of K :
x(t) ∈ ♥(K) for every t > 0 and also x∞ ∈ ♥(K);
♥(K) contains the center of mass B of K, the center C of the smallest
ball containing K (circumcenter) and the center I of the largest ball
contained in K (incenter), if this is unique;
we have the following estimate:
diam[(♥(K)] ≥ diam[4(B, C, I)];
7
if K has j independent hyperplanes of symmetry, then ♥(K) is contained
in their (N − j)-dimensional intersection;
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The heart of a set
PROPERTIES
1
2
3
4
5
6
Ω \ ♥(Ω) does not contain any critical point of u;
if ∂Ω ∈ C 1 , then dist(♥(Ω), ∂Ω) > 0 (Fraenkel);
if K is a convex body, then ♥(K) is a (closed) convex subset of K :
x(t) ∈ ♥(K) for every t > 0 and also x∞ ∈ ♥(K);
♥(K) contains the center of mass B of K, the center C of the smallest
ball containing K (circumcenter) and the center I of the largest ball
contained in K (incenter), if this is unique;
we have the following estimate:
diam[(♥(K)] ≥ diam[4(B, C, I)];
7
8
if K has j independent hyperplanes of symmetry, then ♥(K) is contained
in their (N − j)-dimensional intersection;
if j = N, then ♥(K) reduces to a single point and hence the hot spot does
not move.
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Stationary hot spot
When 8 occurs, we say that the hot spot is stationary.
PROBLEM (Klamkin, Siam Review 1994)
Can you characterize the convex conductors for which the hot spot does not
move?
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Stationary hot spot
When 8 occurs, we say that the hot spot is stationary.
PROBLEM (Klamkin, Siam Review 1994)
Can you characterize the convex conductors for which the hot spot does not
move?
PARTIAL ANSWERS (M. - Sakaguchi, 2004, 2008)
1
triangles → equilateral;
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Stationary hot spot
When 8 occurs, we say that the hot spot is stationary.
PROBLEM (Klamkin, Siam Review 1994)
Can you characterize the convex conductors for which the hot spot does not
move?
PARTIAL ANSWERS (M. - Sakaguchi, 2004, 2008)
1
2
triangles → equilateral;
quadrangles → parallelograms;
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Stationary hot spot
When 8 occurs, we say that the hot spot is stationary.
PROBLEM (Klamkin, Siam Review 1994)
Can you characterize the convex conductors for which the hot spot does not
move?
PARTIAL ANSWERS (M. - Sakaguchi, 2004, 2008)
1
2
3
triangles → equilateral;
quadrangles → parallelograms;
pentagons circumscribed to a circle → regular;
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Stationary hot spot
When 8 occurs, we say that the hot spot is stationary.
PROBLEM (Klamkin, Siam Review 1994)
Can you characterize the convex conductors for which the hot spot does not
move?
PARTIAL ANSWERS (M. - Sakaguchi, 2004, 2008)
1
2
3
4
triangles → equilateral;
quadrangles → parallelograms;
pentagons circumscribed to a circle → regular;
hexagones circumscribed to a circle → hexagons invariant w.r.t. rotations
of angles π/3, 2π/3, π;
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Stationary hot spot
When 8 occurs, we say that the hot spot is stationary.
PROBLEM (Klamkin, Siam Review 1994)
Can you characterize the convex conductors for which the hot spot does not
move?
PARTIAL ANSWERS (M. - Sakaguchi, 2004, 2008)
1
2
3
4
5
triangles → equilateral;
quadrangles → parallelograms;
pentagons circumscribed to a circle → regular;
hexagones circumscribed to a circle → hexagons invariant w.r.t. rotations
of angles π/3, 2π/3, π;
general formula relating the (stationary) hot spot and the curvatures of
certain subsets of ∂K.
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The maximal folding function
If we define the maximal folding function as
0
RK (ω) := min{λ ∈ R : Kλ,ω
⊆ K}, ω ∈ SN−1 ,
then
♥(K) = {x ∈ RN : x · ω ≤ RK (ω) for every ω ∈ SN−1 }.
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The maximal folding function
If we define the maximal folding function as
0
RK (ω) := min{λ ∈ R : Kλ,ω
⊆ K}, ω ∈ SN−1 ,
then
♥(K) = {x ∈ RN : x · ω ≤ RK (ω) for every ω ∈ SN−1 }.
Examples
1
K = B(0, R) ⇒ RK ≡ 0;
2
K = ellipse with semi-axes a > b ⇒
RK (ω) = q
a2 − b 2
b2 ω12 + a2 ω22
|ω1 ω2 |;
the curve ω 7→ RK (ω) ω is (an affine image of) a rhodonea with 4 petals.
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The midpoint function
Figura: Definition of the midpoint function fω : S(K) → R.
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The midpoint function
A formula for fω using the Fourier transform of 1K
0
fω (x ) =
i
R
Rω
⊥
∂ω 1̂K (η) eix
1̂ (η) e
ω⊥ K
0
ix 0 ·η
·η
dη
dη
, x 0 ∈ S(K).
Here, ω ⊥ = {η : η · ω = 0} and S(K) is the shadow of K.
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The midpoint function
A formula for fω using the Fourier transform of 1K
0
fω (x ) =
i
R
Rω
⊥
∂ω 1̂K (η) eix
1̂ (η) e
ω⊥ K
0
ix 0 ·η
·η
dη
dη
, x 0 ∈ S(K).
Here, ω ⊥ = {η : η · ω = 0} and S(K) is the shadow of K.
CHARACTERIZATION (Brasco - M. - Salani 2010)
RK (ω) = max fω (x 0 ).
x 0 ∈S(K)
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The midpoint function
Figura: Proof.
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An algorithm
CONVEX POLYHEDRON
For a convex polyhedron, we prove that the maximum in the characterization
can be computed only by visiting (the projections on S(K) of) the vertices
of K.
This fact helps us to produce an algorithm to draw ♥(K) when K is a convex
polyhedron:
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An algorithm
CONVEX POLYHEDRON
For a convex polyhedron, we prove that the maximum in the characterization
can be computed only by visiting (the projections on S(K) of) the vertices
of K.
This fact helps us to produce an algorithm to draw ♥(K) when K is a convex
polyhedron:
1
Fix ω ∈ SN−1 ;
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An algorithm
CONVEX POLYHEDRON
For a convex polyhedron, we prove that the maximum in the characterization
can be computed only by visiting (the projections on S(K) of) the vertices
of K.
This fact helps us to produce an algorithm to draw ♥(K) when K is a convex
polyhedron:
1
Fix ω ∈ SN−1 ;
2
compute RK (ω) by maximizing the values fω (x10 ), . . . , fω (xm0 ), where
x10 , . . . , xm0 are the projections on S(K) of the vertices of K;
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An algorithm
CONVEX POLYHEDRON
For a convex polyhedron, we prove that the maximum in the characterization
can be computed only by visiting (the projections on S(K) of) the vertices
of K.
This fact helps us to produce an algorithm to draw ♥(K) when K is a convex
polyhedron:
1
Fix ω ∈ SN−1 ;
2
compute RK (ω) by maximizing the values fω (x10 ), . . . , fω (xm0 ), where
x10 , . . . , xm0 are the projections on S(K) of the vertices of K;
3
paint the halfspace {x ∈ RN : x · ω > RK (ω)} of yellow (kiiro);
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An algorithm
CONVEX POLYHEDRON
For a convex polyhedron, we prove that the maximum in the characterization
can be computed only by visiting (the projections on S(K) of) the vertices
of K.
This fact helps us to produce an algorithm to draw ♥(K) when K is a convex
polyhedron:
1
Fix ω ∈ SN−1 ;
2
compute RK (ω) by maximizing the values fω (x10 ), . . . , fω (xm0 ), where
x10 , . . . , xm0 are the projections on S(K) of the vertices of K;
3
paint the halfspace {x ∈ RN : x · ω > RK (ω)} of yellow (kiiro);
4
iterate with a new ω.
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Example 1
Figura: The heart of an octagon.
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Example 2
Figura: The heart of an obtuse triangle
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Example 3
Figura: The heart of a parallelogram
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Problems
Remark
In the case of the obtuse triangle, we observe that ♥(K) ∩ ∂K 6= ∅ (this is
always the case when the circumcenter is not in the interior of K), even if we
are sure that x(t) and x∞ are not on ∂K (by Hopf lemma).
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Problems
Remark
In the case of the obtuse triangle, we observe that ♥(K) ∩ ∂K 6= ∅ (this is
always the case when the circumcenter is not in the interior of K), even if we
are sure that x(t) and x∞ are not on ∂K (by Hopf lemma).
Problems
1
If K is a polyhedron, can we choose only a finite number of directions ω
to draw ♥(K)?
R. Magnanini (Università di Firenze)
The location of the hot spot
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Problems
Remark
In the case of the obtuse triangle, we observe that ♥(K) ∩ ∂K 6= ∅ (this is
always the case when the circumcenter is not in the interior of K), even if we
are sure that x(t) and x∞ are not on ∂K (by Hopf lemma).
Problems
1
2
If K is a polyhedron, can we choose only a finite number of directions ω
to draw ♥(K)?
Estimate the size of ♥(K). We know that
diam[♥(K)] ≥ diam[4(B, C, I)];
can we estimate the ratio
|♥(K)|
|K|
from above?
R. Magnanini (Università di Firenze)
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Second method: using ABP principle
Our second method gives lower bounds of the distance of x(t) or x∞ from the
boundary of K.
R. Magnanini (Università di Firenze)
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Second method: using ABP principle
Our second method gives lower bounds of the distance of x(t) or x∞ from the
boundary of K.
For instance, we prove the following estimate:
dist(x∞ , ∂K) ≥ CN rK
R. Magnanini (Università di Firenze)
rK
δK
The location of the hot spot
N 2 −1
,
Cortona, 22/06/2011
25 / 32
Second method: using ABP principle
Our second method gives lower bounds of the distance of x(t) or x∞ from the
boundary of K.
For instance, we prove the following estimate:
dist(x∞ , ∂K) ≥ CN rK
where
CN =
rK
δK
N 2 −1
,
(2N N)N−1 ωN−1
< 1,
λ1 (B1 )N ωN
λ1 (B1 ) is the first Dirichlet eigenvalue of the unit ball and
R. Magnanini (Università di Firenze)
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Second method: using ABP principle
Figura: Totsu shuugou ippon.
R. Magnanini (Università di Firenze)
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Second method: using ABP principle
The idea is condensed il the following picture.
Figura: u = u(x, t) or φ1 (x); w = x(t) or x∞ ; M = u(w, t) or φ1 (w); E = convex
envelope of u; G = cone with tip at the point (w, M); C = contact set (of points where
u = E.)
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Subdifferential of a convex function and polar set
Define the subgradient of a function u at the point z ∈ K as the set
∂u(z) = {p ∈ RN : u(x) ≥ u(z) + hp, x − zi for every x ∈ K }.
and
∂u(K) =
[
∂u(z) .
z∈K
R. Magnanini (Università di Firenze)
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Subdifferential of a convex function and polar set
Define the subgradient of a function u at the point z ∈ K as the set
∂u(z) = {p ∈ RN : u(x) ≥ u(z) + hp, x − zi for every x ∈ K }.
and
∂u(K) =
[
∂u(z) .
z∈K
Remarks
1
Since −E ≤ −G, we have that ∂(−G)(K) ⊆ ∂(−E)(K).
R. Magnanini (Università di Firenze)
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Subdifferential of a convex function and polar set
Define the subgradient of a function u at the point z ∈ K as the set
∂u(z) = {p ∈ RN : u(x) ≥ u(z) + hp, x − zi for every x ∈ K }.
and
∂u(K) =
[
∂u(z) .
z∈K
Remarks
1
2
Since −E ≤ −G, we have that ∂(−G)(K) ⊆ ∂(−E)(K).
From convex geometry:
M w + ∂(−G)(K) = ∂(−G)(w) = M Kw∗ ,
where Kw∗ is the polar set of K w.r.t. w :
Kw∗ = y ∈ RN : (x − w) · (y − w) ≤ 1 for every x ∈ K .
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Estimating the polar set
Remarks
3 By the area formula and the arithmetic-geometric mean inequality, we
have:
Z
N
∗
M |Kw | = |∂(−G)(K)| ≤ |∂(−E)(K)| =
| det(D 2 u)| dx ≤
C
Z
N −N
|∆u|N dx.
C
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Estimating the polar set
Remarks
3 By the area formula and the arithmetic-geometric mean inequality, we
have:
Z
N
∗
M |Kw | = |∂(−G)(K)| ≤ |∂(−E)(K)| =
| det(D 2 u)| dx ≤
C
Z
N −N
|∆u|N dx.
C
4 Finally, using the equations ut = ∆u or ∆φ1 + λ1 φ1 = 0, we obtain the
two bounds
Z
∗
|Kx(t)
| ≤ [N M(t)]−N
|ut (x, t)|N dx,
C(t)
|Kx∗∞ | ≤
λ1
N M∞
N Z
φ1 (x)N dx ,
C
that is the polar set cannot be too large.
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Hot spots and polar sets
These estimates are generally difficult to handle. However, the latter can be
made more useful, since we can bound φ1 (x) by its maximum M∞ ; we obtain
two interesting bounds:
R. Magnanini (Università di Firenze)
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Hot spots and polar sets
These estimates are generally difficult to handle. However, the latter can be
made more useful, since we can bound φ1 (x) by its maximum M∞ ; we obtain
two interesting bounds:
|Kx∗∞ | ≤
and
N
|C| ≥
λ1
N
λ1
N
|Kx∗∞ |
N
|K| .
N
≥
λ1
N
|Ks∗ |,
where Ks∗ denotes the polar set of K with respect to the Santalò point s (the
one that minimizes the function w 7→ |Kw∗ |).
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Estimating the volume of the polar set
Using the definition of the polar set, it is easy to see that |Kw∗ | goes to ∞ as
the point w approaches ∂K. The following estimate gives a quantitative
version of this fact and helps us to prove explicit estimates of the position of
x∞ .
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Estimating the volume of the polar set
Using the definition of the polar set, it is easy to see that |Kw∗ | goes to ∞ as
the point w approaches ∂K. The following estimate gives a quantitative
version of this fact and helps us to prove explicit estimates of the position of
x∞ .
|Kw∗ | ≥ |Ew∗ | ≥
ωN−1 /N
.
R N−1 d
Figura:
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Conclusion
Thus,
N
ωN−1 /N
λ1
≤
|K|
R N−1 d
N
and the bound
dist(x∞ , ∂K) ≥ CN rK
rK
δK
N 2 −1
follows by choosing w = x∞ and by using a standard inequality to bound λ1
from above.
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Conclusion
Thus,
N
ωN−1 /N
λ1
≤
|K|
R N−1 d
N
and the bound
dist(x∞ , ∂K) ≥ CN rK
rK
δK
N 2 −1
follows by choosing w = x∞ and by using a standard inequality to bound λ1
from above.
Concluding remark
The two methods for locating the hot spot can be coupled. For example, in the
case of the obtuse triangle, we know that its heart extends to part of the
boundary; however, by the estimate we have just proved, we can
quantitatively say how far x∞ must be from the boundary.
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