isoperimetric inequalities for soap- lms - skkumatrix.skku.ac.kr/2013_coll/20130321_shu...

Isoperimetric inequalities for soap-films

Keomkyo Seo

Sookmyung Women’s University

Colloquium at Sungkyunkwan UniversityMarch 21, 2013

Dido(Queen of Carthage)

Dido was a Phoenician princess from the city of Tyre. She fled by shipfrom Tyre after King Pygmalion who was her brother killed her husband.Then she ended up on the north coast of Africa. When Dido arrived inAfrica about 900 B.C. at the place that later became Carthage, she triedto buy land from the natives so that she and her people could settle andestablish a new homeland for themselves. She bargained to buy as muchland as she could enclose with an oxhide. So ...

baduk(game); Korean checkers

Classical isoperimetric inequality

QuestionAmong all shapes of the same area, what has the least boundary length?Equivalently,among all shapes of the same boundary length, what has the largest area?

Note that isoperimetry = iso + perimetry .

Theorem (Isoperimetric Inequality)Let C be a simple closed curve in the plane whose length is L and thatencloses an area A. Then

4πA ≤ L2.

Equality holds if and only if C is a circle.

4πA ≤ L2.

Steiner’s proof via four-hinge method

Step 1: Solution is convex

DefinitionA region Ω is convex if for any pair of points A and B in Ω, the segmentAB is contained in Ω.

Step 1: Solution is convex.

Step 2: Solution is symmetric with respect to a bisection line.

Step 3: Solution is a disk.

Steiner’s proof via symmetrization

Figure: Jakob Steiner (18 March 1796 - 1 April 1863)

Two important features of Steiner’s symmetrization

Area is preserved.

Length is reduced.

Using this symmetrization, Steiner were able to prove thatAmong all curves of a given length, the circle encloses the greatest area.However, there was a big gap in the proof of ...

Area is preserved.

Length is reduced.

Area is preserved.

Length is reduced.

Area is preserved.

Length is reduced.

Area is preserved.

Length is reduced.

Area is preserved.

Length is reduced.

Figure: Johann Peter Gustav Lejeune Dirichlet (13 February 1805 - 5May 1859)

Dirichlet tried several times to persuade Steiner that his proofs wereincomplete on that account. But Steiner insisted that this was selfevident. However, in one of the 1842 papers Steiner conceded

”The proof is readily made if one assumes that there is a largest figure.”

Figure: Oskar Perron (7 May 1880 - 22 February 1975)

Perron’s Joke(or Paradox)

Theorem (Steiner)Among all curves of a given length, the circle encloses the greatest area.

Proof. For any curve that is not a circle, there is a method (given bySteiner) by which one finds a curve that encloses greater area. Thereforethe circle has the greatest area.

Theorem (Perron)Among all positive integers, the integer 1 is the largest.

Proof. For any integer that is not 1, there is a method (to take thesquare) by which one finds a larger positive integer. Therefore 1 is thelargest integer.

Existence was first proved by Weierstrass(1875) and Schwarz(1884)using calculus of variations.

Caratheodory and Study(1909) gave rigorous treatment of Steinersmethod without the calculus of variations.

Proof via Green’s Theorem (Schwarz,Hurwitz, Schmidt)

Let α(s) = (x(s), y(s)) and α(s) = (x(s), y(s)) = (x(s), y(s)). Then

xy ′ds

A = πr2 = −∫ L

y x ′ds

Let α(s) = (x(s), y(s)) and α(s) = (x(s), y(s)) = (x(s), y(s)). Then

xy ′ds

A = πr2 = −∫ L

y x ′ds

A + πr2 =

(xy ′ − y x ′)ds

≤∫ L

√(xy ′ − y x ′)2ds

≤∫ L

√(xy ′ − y x ′)2 + (xx ′ + y y ′)2ds

√(x2 + y2)(x ′2 + y ′2)ds

≤∫ L

√(x2 + y2)ds

∴√

A√πr2 ≤ 1

2(A + πr2) ≤ Lr

A + πr2 =

(xy ′ − y x ′)ds

≤∫ L

√(xy ′ − y x ′)2ds

≤∫ L

√(xy ′ − y x ′)2 + (xx ′ + y y ′)2ds

√(x2 + y2)(x ′2 + y ′2)ds

≤∫ L

√(x2 + y2)ds

∴√

A√πr2 ≤ 1

2(A + πr2) ≤ Lr

Equality case

To have equality, it holds

(xx ′ + y y ′)2 = 0.

Therefore

y ′= ± y

x ′= ±

√x2 + y2√x ′2 + y ′2

= ±r ,

which implies

x = ±ry ′.

Similarly, y = ±rx ′. Finally

x2 + y2 = r2.

Equality case

(xx ′ + y y ′)2 = 0.

Therefore

y ′= ± y

x ′= ±

√x2 + y2√x ′2 + y ′2

= ±r ,

which implies

x = ±ry ′.

x2 + y2 = r2.

Equality case

(xx ′ + y y ′)2 = 0.

Therefore

y ′= ± y

x ′= ±

√x2 + y2√x ′2 + y ′2

= ±r ,

which implies

x = ±ry ′.

x2 + y2 = r2.

Equality case

(xx ′ + y y ′)2 = 0.

Therefore

y ′= ± y

x ′= ±

√x2 + y2√x ′2 + y ′2

= ±r ,

which implies

x = ±ry ′.

x2 + y2 = r2.

Sobolev inequality

isoperimetric inequalityLet M be a Riemannian manifold, Ω a domain with compact closure inM, then there exists a constant C independent of Ω such that

C |Ω|n−1

n ≤ |∂Ω|

Sobolev inequalityLet M be a Riemannian manifold. There exists a constant C such thatfor any φ ∈W 1,2

0 (M) ,

) n−1n ≤

|∇φ|,

Substitute |φ|2(n−1)(n−2) for φ when n > 2. Then(∫

|φ|2n

) n−2n ≤ C

|∇φ|2.

Sobolev inequality

C |Ω|n−1

n ≤ |∂Ω|

0 (M) ,

) n−1n ≤

|∇φ|,

|φ|2n

) n−2n ≤ C

|∇φ|2.

Sobolev inequality

C |Ω|n−1

n ≤ |∂Ω|

0 (M) ,

) n−1n ≤

|∇φ|,

|φ|2n

) n−2n ≤ C

|∇φ|2.

Sobolev inequality

C |Ω|n−1

n ≤ |∂Ω|

0 (M) ,

) n−1n ≤

|∇φ|,

|φ|2n

) n−2n ≤ C

|∇φ|2.

Sobolev inequality

C |Ω|n−1

n ≤ |∂Ω|

0 (M) ,

) n−1n ≤

|∇φ|,

|φ|2n

) n−2n ≤ C

|∇φ|2.

Sobolev inequality

C |Ω|n−1

n ≤ |∂Ω|

0 (M) ,

) n−1n ≤

|∇φ|,

|φ|2n

) n−2n ≤ C

|∇φ|2.

isoperimetric inequality ⇔ Sobolev inequality

⇐Define

φε(x) =

1, x ∈ Ω,dist(x , ∂Ω) ≥ ε,dist(x , ∂Ω)

ε, x ∈ Ω,dist(x , ∂Ω) ≤ ε,

0, otherwise.

Apply the Sobolev inequality to φε and let ε→ 0.

|φε|n

) n−1n ≤

|∇φε| ⇒ C |Ω|n−1

n ≤ |∂Ω|

⇒Use the coarea formula.

⇐Define

φε(x) =

ε, x ∈ Ω,dist(x , ∂Ω) ≤ ε,

0, otherwise.

|φε|n

) n−1n ≤

|∇φε| ⇒ C |Ω|n−1

n ≤ |∂Ω|

⇐Define

φε(x) =

ε, x ∈ Ω,dist(x , ∂Ω) ≤ ε,

0, otherwise.

|φε|n

) n−1n ≤

|∇φε| ⇒ C |Ω|n−1

n ≤ |∂Ω|

⇐Define

φε(x) =

ε, x ∈ Ω,dist(x , ∂Ω) ≤ ε,

0, otherwise.

|φε|n

) n−1n ≤

|∇φε| ⇒ C |Ω|n−1

n ≤ |∂Ω|

⇐Define

φε(x) =

ε, x ∈ Ω,dist(x , ∂Ω) ≤ ε,

0, otherwise.

|φε|n

) n−1n ≤

|∇φε| ⇒ C |Ω|n−1

n ≤ |∂Ω|

⇐Define

φε(x) =

ε, x ∈ Ω,dist(x , ∂Ω) ≤ ε,

0, otherwise.

|φε|n

) n−1n ≤

|∇φε| ⇒ C |Ω|n−1

n ≤ |∂Ω|

We want to generalize the classical isoperimetric inequality.

4πA ≤ L2.

Two ways of a generalization

Generalize an ambient space.

Generalize a domain.

4πA ≤ L2.

Ω2 ⊂ R2 → Ωn ⊂ Rn

nnωn|Ω|n−1 ≤ |∂Ω|n

and equality holds if and only if Ω is a ball. (ωn is the volume of aunit ball in Rn)

Ω2 ⊂ R2 → Ω2 ⊂ S2(the unit sphere) or H2(hyperbolic space)There is a joke related with this.

Theorem (Bol, Schmidt, Aleksandrov, Karcher, Bandle, Aubin,Topping, ...)

For a simply connected region Ω on S2 or H2 ,

4πA− KA2 ≤ L2,

where K is Gaussian curvature of S2 or H2.

Ω2 ⊂ R2 → Ωn ⊂ Rn

4πA− KA2 ≤ L2,

Ω2 ⊂ R2 → Ωn ⊂ Rn

4πA− KA2 ≤ L2,

Ω2 ⊂ R2 → Ωn ⊂ Rn

4πA− KA2 ≤ L2,

Ω2 ⊂ R2 → Ωn ⊂ Rn

4πA− KA2 ≤ L2,

Mn ⊂ Mn+m

Simplest case: M2 ⊂ R2 ⊂ R3. In this case we have 4πA ≤ L2

with equality if and only if M is a disk.A natural extension: a minimal surface M2 ⊂ R3

Recall that M ∈ R3 is a minimal surface if

M has mean curvature H ≡ 0.

Small pieces have least area.

Small pieces occur as soap films.

Coordinate functions are harmonic.

Note that a plane R2 is a trivial example of minimal surfaces in R3.Hence one can thought a minimal surface as a generalization of a plane.Furthermore a soap film is a (area-minimizing) minimal surface.

Mn ⊂ Mn+m

Examples of Complete Minimal surfaces

Figure: Catenoid: rotationally symmetric and embedded

Figure: Helicoid: simply connected and embedded

Figure: Enneper surface: simply connected but not embedded

Figure: Riemann’s minimal surface: foliated by circles and straight lines,embedded

Figure: Costa surface discovered in 1982 by the Brazilian graduatestudent : Until its discovery, only the plane, helicoid and the catenoidwere believed to be complete embedded minimal surfaces

Figure: Costa-Hoffman-Meeks Surface: constructable for any genus

Soap films in real world

Return to the discussion

For a minimal surface M2 ⊂ R3, does the isoperimetric inequality hold?This problem has been partially proved but not completely.

Yes, for simply-connected case. (Carleman 1920’)

Yes, for doubly-connected case. (Osserman-Schiffer, Feinberg 1970’)

Yes, for ](∂M) ≤ 2. (Li-Schoen-Yau 1984, Choe 1990)

Yes, for triply-connected case. (Choe-Schoen, recent)

This problem is still open.

Does the classical isoperimetric inequality still hold for minimal surfaceswith singularities?

Figure: Soap film with nonclosed boundary curves

The answer is yes. (Seo)

This soap film satisfies the isoperimetric inequality 4πA ≤ L2.

More general cases (involving mean curvature)

Theorem (Allard, Michael-Simon, Hoffman-Spruck)For a compact submanifold Mn ⊂ Rn+m,

C |M|n−1 ≤(|∂M|+

Good, because it is general.

Bad, because it is weak.

C |M|n−1 ≤(|∂M|+

Theorem (Yau, Choe-Gulliver)Let M be a domain in a hyperbolic space Hn or an n-dimensionalminimal submanifold of Hn+m. Then

(n − 1)|M| ≤ |∂M|.

Theorem (Seo)Let M be an n-dimensional compact submanifold of Hn+m. Then

(n − 1)|M| ≤ |∂M|+∫

Theorem (Yau, Choe-Gulliver)Let M be a domain in a hyperbolic space Hn or an n-dimensionalminimal submanifold of Hn+m. Then

(n − 1)|M| ≤ |∂M|.

Theorem (Seo)Let M be an n-dimensional compact submanifold of Hn+m. Then

(n − 1)|M| ≤ |∂M|+∫

Eigenvalue problem

Σn ⊂ Rn: a compact submanifold with boundary ∂Σ. In terms of localcoordinates (x1, · · · , xn), the Laplace operator ∆ is defined by

∆ =∂2

∂x21

+ · · ·+ ∂2

∂x2n

Consider the Dirichlet eigenvalue problem on Σ.

∆f + λf = 0 in Σ,

f = 0 on ∂Σ,

∆ has discrete eigenvalues: 0 < λ1 ≤ λ2 ≤ λ3 ≤ ...

Eigenvalue problem

∆ =∂2

∂x21

+ · · ·+ ∂2

∂x2n

∆f + λf = 0 in Σ,

f = 0 on ∂Σ,

Eigenvalue problem

∆ =∂2

∂x21

+ · · ·+ ∂2

∂x2n

∆f + λf = 0 in Σ,

f = 0 on ∂Σ,

In late October of 1910, H. A. Lorentz delivered a series of lectures. Heproposed one problem about the eigenvalues.

Question

N(λ) =∑λn<λ

= ]λn : λn < λ∼ cn|Ω|λn/2

Question

N(λ) =∑λn<λ

Question

N(λ) =∑λn<λ

Figure: Hendrik Antoon Lorentz (18 July 1853 - 4 February 1928)

D. Hilbert was attending these lectures and predicted as follows:”This theorem would not be proved in my life time” However, H. Weyl, agraduate student at that time, was also attending. He proved thisconjecture four months later in February of 1911.

Figure: Left:David Hilbert( January 23, 1862 - February 14, 1943) ,Right:Hermann Klaus Hugo Weyl (9 November 1885 - 8 December 1955)

Theorem (Weyl)

N(λ) =∑λn<λ

Theorem (Faber-Krahn)Let Ω ⊂ Rn be a domain and B(R) be a ball of radius R. If|Ω| = |B(R)|, then

λ1(Ω) ≥ λ1(B(R)).

And equality holds if and only if Ω is a ball of radius R.

Theorem (Faber-Krahn)Let Ω ⊂ Rn be a domain and B(R) be a ball of radius R. If|Ω| = |B(R)|, then

λ1(Ω) ≥ λ1(B(R)).

And equality holds if and only if Ω is a ball of radius R.

First eigenvalue estimate

Theorem (McKean, Cheung-Leung, Bessa-Montenegro, Seo)Let Σ be an n-dimensional submanifold in Hn+m. If |H| ≤ α for someconstant α < (n − 1), then

[(n − 1)− α]2

4≤ λ1(Σ)

Thank you for your attention.

isoperimetric inequalities for soap- lms - skkumatrix.skku.ac.kr/2013_coll/20130321_shu...

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