Isoperimetric inequalities for wedge-like membranes

Isoperimetric inequalities for wedge-like membranes

Lotfi Hermi

University of Arizona

Joint work with A. Hasnaoui, University of Tunis El Manar

Prairie Analysis Seminar, Sept 25, 2015

Lotfi Hermi (UA) Sharp Isoperimetric BoundsPrairie Analysis Seminar, Sept 25, 2015 1

/ 38

Setting

D will denote a planar domain given in polar coordinates by

D ={

(r , θ) | 0 < r < ρ(θ), 0 < θ <π

α

}. (1)

Then D is a bounded domain completely contained in the wedge Wα ofangle π

α , α ≥ 1. We will denote by Γ the curved part of its boundarydefined by r = ρ(θ), 0 ≤ θ < π

α . Let u be the fundamental eigenfunctionof the Dirichlet problem in D, and λ its associated fundamental eigenvalue:

∆u + λu = 0 in D, u = 0 on Γ. (2)

When Γ is convex (prototype for starlike) we know that (x , n) > 0 (Aissen1958; Polya-Szego 1951, Siudeja-Laugesen 2014, Freitas-Krejcirik 2008,etc.)


/ 38

Setting

For such domains we define

Bα =

∫Γ

1

(x , n)h2dσ, Aα =

∫Dh2dA (3)

Here, we let ‖x‖ = r = ρ(θ) and h = rα sinαθ. We note that h is aharmonic nonnegative function in Wα. By a perfect sector, we mean thecircular sector

S0 ={

(r , θ)∣∣ 0 < r < R0, 0 < θ <

π

α

}with R0 the radius such that |S0| =

∫S0h2dA = Aα.


/ 38

Main Theorem

The fundamental eigenvalue λ of the wedge-like membrane D definedabove satisfies the (new) inequality in red.(4α(α + 1)

πAα) −1α+1

j2α,1 ≤ λ ≤

Bα(2α + 2)Aα

j2α,1. (4)

Equality holds if and only if D is a perfect sector S0.

Applications: (1) Estimates for λ, for the right-angle triangle with unithypothenuse; (2) Estimates for λ for regular α-polygon with n sides.


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Numerical Evidence

α Slow Flow FSlow Shigh Fhigh F ′high FShigh PS HHhigh HHlow

4 57.524 91.819 93.340 102.265 109.626 102.613 111.238 134.828 115.166 35.5366 73.668 120.369 121.04 130.965 127.928 124.533 150.040 172.665 131.635 59.9448 105.017 166.873 167.292 186.697 171.833 169.444 219.060 246.143 175.095 94.00210 145.918 224.803 225.104 259.409 228.582 226.679 308.148 342.007 231.662 137.03012 195.398 292.766 293.000 347.374 295.863 294.254 415.421 457.981 298.844 188.84414 253.157 370.237 370.427 450.058 372.886 371.477 540.223 593.36 375.819 249.35316 319.072 456.946 457.105 567.239 459.275 458.014 682.237 747.853 462.174 318.50218 393.082 552.731 552.867 698.812 554.819 553.671 841.282 921.319 557.694 396.24720 475.154 657.479 657.598 844.718 659.378 658.322 1017.24 1113.68 662.237 482.55622 565.25 771.108 771.215 1004.92 772.855 771.874 1210.01 1324.9 775.702 577.40024 663.416 893.555 893.651 1179.41 895.176 894.257 1419.54 1554.94 898.014 680.75426 769.586 1024.77 1024.85 1368.15 1026.28 1025.42 1645.78 1803.79 1029.11 792.59728 883.774 1164.7 1164.78 1571.15 1166.13 1165.31 1888.68 2071.42 1168.95 912.91130 1005.98 1313.32 1313.39 1788.40 1314.66 1313.89 2148.20 2357.84 1317.49 1041.6832 1136.19 1470.59 1470.66 2019.89 1471.87 1471.12 2424.32 2663.04 1474.68 1178.88

Table: Comparison of upper and lower bounds for a right triangle as a function ofα: Siudeja (Slow , Shigh), Freitas (Flow , Fhigh, F ′high) (2007), Freitas-Siudeja (FSlow ,FShigh), Payne-Weinberger (Flow ), Polya-Szego (PS), and Hasnaoui-Hermi(HHhigh and HHlow , this article).


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Numerical Evidence: Regular α−polygon with n sides

Table: Upper and lower bounds for the fundamental eigenvalue of aregular α-polygon with n-sides.


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Other isoperimetric results

(Pα is relative torsional rigidity, which we define in later slides) SaintVenant for Pα (H., Hasnaoui, AHP 2014)

Pα ≤1

α + 2

(α

(4α + 4)απ

) 1α+1

Aα+2α+1α , (5)

Weighted reverse Holder ineq. (H., Hasnaoui, AHP 2014)∫Du2dA ≤ α

πj2αα,1

λα+2

(∫Du hdA

)2

. (6)

Weighted Crooke-Sperb ineq. (H., Hasnaoui, AHP 2014)(∫Du h dA

)2

≤ 2Bαλ

∫Du2 dA. (7)

λ ≥( π

2α

)1/(α+1) j2α/(α+1)α,1

B1/(α+1)α

. (8)

Equality in all of these inequalities is attained for the perfect sector.Lotfi Hermi (UA) Sharp Isoperimetric Bounds

Prairie Analysis Seminar, Sept 25, 2015 7/ 38

Weighted reverse Holder ineq. of Chiti type

(6) is in fact a particular case of the following theorem:Let D be a bounded domain in the wedge W. Let p, q be real numberssuch that q ≥ p > 0, then u satisfies the inequality(∫

Duqh2−qdA

) 1q

≤ K (p, q, λ, α)

(∫Dup h2−pdA

) 1p

(9)

with

K (p, q, λ, α) =( π

2α

) p−qpq

λ(α+1) q−ppq

(∫ jα,10

r (2−q)α+1Jqα(r)dr) 1

q

(∫ jα,10

r (2−p)α+1Jpα(r)dr) 1

p

.

The result is isoperimetric in the sense that equality holds if and only if Dis a circular sector of angle π

α .Remark: Sending q →∞ and p → 0+, one gets the Payne-Weinbergerineq. (4) ∫

Dh2dA ≥

πj2α+2α,1

4α(α + 1)λα+1.


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The geometric factor Bα

Theorem: For the wedge-like membrane D defined by (1) we have

Bα ≥π

2α

[4α(α + 1)

πAα

] αα+1

. (10)

Equality holds if and only if D is a perfect sector of angle πα .

Also:

( π2α

)2[

4α(α + 1)

πAα

] 2α+1α+1

≤(∫

Γh2dσ

)2

≤ (2α + 2)AαBα. (11)


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Classical Isoperimetric Inequality:

Among all domains of a fixed perimeter, L, the circle encloses the largestarea.

4πA ≈ 12.5663A ≤ L2.

Among all triangles of a fixed perimeter, the equilateral triangle has thelargest area

12√

3A ≈ 20.78A ≤ L2.

Among all rectangles of a fixed perimeter, the square possesses the largestarea

16A ≤ L2.

Semi-Circle:2(2 + π)2

πA ≈ 16.8297A ≤ L2.

Euclid (c. 300 B.C. “Elements, I, 35-38): “Two parallelograms, or triangles,

which are on the same base, or on equal bases, and in the same parallels, remain

equivalent in area no matter how far the sides between the parallels may be

stretched out and increase in their length.”Lotfi Hermi (UA) Sharp Isoperimetric Bounds


Zenodorus (ca. 200 BC –ca. 140 BC, now lost “On Isometric Figures”):Of all polygons of the same number of sides and equal perimeter theequilateral and equiangular polygon is the greatest in area.(

4N tanπ

N

)A ≤ L2.

The constant has the asymptotic expansion

4N tanπ

N∼ 4π +

4π3

3N2+

8π5

15N4+

68π7

315N6+ . . .

When N →∞, one recovers the inequality for the circle.

This is the old story of Queen Dido and the foundation of Carthage,popularized (among mathematicians) by Lord Kelvin in a public 1894address.


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Circle the Wagon: Hunker down and fight back! BearDown!


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Rayleigh-Faber-Krahn

Consider the eigenvalue problem:

∆ u + λu = 0 in D (12)

u = 0 on ∂D.

Rayleigh’s conjecture: “If the area of a membrane be given, there mustevidently be some form of the boundary for which the pitch (of theprincipal tone) is the gravest possible, and this form can be no other thanthe circle.”

λ ≥π j2

0,1

A≈ 18.1680

A. (13)

where j0,1 ≈ 2.4048.The constant appearing in (13) can be improved, if the geometry of theunderlying domain is restricted.


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For example among all triangles of a fixed area, the equilateral trianglemaximizes the fundamental tone:

λ ≥ 4π2

√3A≈ 22.7929

A(14)

Also among all rectangles of a fixed area, the square maximizes thefundamental tone:

λ ≥ 19.7392

A. (15)

Incidently, the conjecture for a polygon with n sides is still open!


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Wedge-Like Membrane Inequalities

Payne-Weinberger inequality for wedge-like membranes (1955): LetD ⊂ Wα. Then

λ ≥ λ∗ =(4α(α + 1)

π

∫Dh2(r , θ)r dr dθ

) −1α+1

j2α,1 (16)

where h = rα sinαθ. Here (r , θ) are polar coordinates taken at the apex ofthe wedge, and jα,1 the first zero of the Bessel function Jα(x). Equalityholds if and only if D is a circular sector Wα.

λ(D)|D|1α+1

h ≥ λ(D∗)|D∗|1α+1

h (17)

where D∗ denotes any circular sector. Here

|D|h =

∫Dh2(r , θ)r dr dθ.


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This inequality improves on the Faber-Krahn inequality for certain domains(as is the case of certain triangles) and has the interpretation of being aversion of Faber-Krahn in dimension 2α + 2 for axially symmetric domains(Bandle, Payne have the details).The proof of this inequality relies on a geometric isoperimetric inequalityfor the quantity

|D|h = Aα =

∫Dh2(r , θ)r dr dθ

which is optimized for the circular sector in Wα, and a carefully craftedsymmetrization argument (weighted symmetric decreasing rearrangement).


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Geometric inequality for wedge-like membranes

For D ⊂ Wα = {(r , θ) : 0 ≤ θ ≤ π/α}, α ≥ 1.(2α

π

∮∂D

h2(r , θ)ds

)(2α+2)/(2α+1)

≥ 4α(α + 1)

πAα

with equality for the perfect circular sector. This statement is equivalentto a weighted statement of the Sobolev inequality (Cabre and his students,Maderna-Salsa, Lions-Pacella, etc).


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What we should retain from this presentation:

We propose in this work a complete program for wedge-likemembranes;

One can generalize this work in the context of convex cones inhigher-dimensions. (a la Lions-Pacella)

Problem is a model for “manifolds with density” ∆µ = 1µdiv(µ∇);

dV = µdx ; µ = h2(r , θ) (Grigory’an, F. Morgan)

What we describe is a model for eigenvalue problems associated withdegenerate elliptic operators.

Renewed interest in “wedge-like membrane” isoperimetric problems(Ratzkin, Treibergs, Brock, Chiacchio, Mercaldo, etc.) with strongconnections to weighted functional isoperimetric inequalities, alsoCabre, etc (weighted Sobolev inequalities)


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Why care?Consider right isoceles triangle with equal sides of unit length(Payne-Weinberger)

α = 1, λ1 ≥ 45.0734

α = 2, λ1 ≥ 47.6325

α = 4, λ1 ≥ 45.9094

Faber-Krahn (among all domains): λ1 ≥ 36.3368

Faber-Krahn (among all triangles): λ1 ≥4π2

√3A≈ 45.5858

Exact value: λ1 = 49.350625

α Slow Flow FSlow Shigh Fhigh FShigh PS HHhigh HHlow(’07) (’07) (’10) (’07) (’07) (’10) (1951) (’15) (’15)

4 28.7621 45.9094 46.6702 51.1327 54.8128 169.241 67.4138 57.5829 20.41

Table: Comparison of upper and lower bounds for a right isoceles triangle withunit sides.


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Our work: Three Problems for wedge-like membranes

For D ⊂ Wα, we consider

P1 :

{∆u + λu = 0 in Du = 0 on ∂D,

P2 :

{−div(hk∇w) = hk f in Dw = 0 in ∂D ∩Wα,

Here k > 0 and h(r , θ) = rα sinαθ, as above, where the function fbelongs to the weighted Lebesgue space L2(D, dµ), and dµ is the measuredefined by

dµ = hk(r , θ) rdrdθ = rαk+1 (sinαθ)k drdθ. (18)

The case k = 2; f ≡ 1 of P2

P3 :

{−div(h2∇w) = h2 in Dw = 0 in ∂D ∩Wα,


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Three problems, cont’d

We claim that P3 is equivalent to

P4 :

{−∆v = h(r , θ) in Dv = 0 in ∂D ∩Wα,

To see this, let v = hw , in P3.Relative torsional rigidity is defined via the variational formulation

1

Pα= inf

φ∈W 1,20 (D)

∫D |∇φ|

2 rdrdθ(∫D φh rdrdθ

)2. (19)

which is in fact equivalent to

1

Pα= inf

φ∈W2(D,dµ)

∫D |∇φ|

2 dµ(∫D φ dµ

)2, (20)

where dµ = h2(r , θ) rdrdθ.


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Payne Interpretation in Fractional Weinstein Space

Case α = 1. In this case, D is such that y > 0, and P4 reduces to

P4 :

{∆v + y = 0 in Dv = 0 in ∂D ∩Wα,

With v = y w , the problem is then

P4 :

∆w +2

y

∂w

∂y= −1 in D

w = 0 in ∂D ∩ {y > 0},

Let the function Φ(x1, x2, x3, x4) be defined by

Φ(x1, x2, x3, x4) = w(x , y) where x = x4; y =√

x21 + x2

2 + x23 .

This function has axial symmetry with respect to the x4-axis. It isdefined on

D4 = {(x1, x2, x3, x4) ∈ R4|x = x4, y =√x2

1 + x22 + x2

3 , (x , y) ∈ D}


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Payne Interpretation in Fractional Weinstein Space, Cont’d

D4 ⊂ R4 is obtained from D via rotation around the x-axis. The functionΦ satisfies

∆4 Φ = −1 in D4, Φ = 0 on ∂D4.

Note that d = 2α + 2 = 4. Let dV = dx1dx2dx3dx4, and

P =

∫D4

ΦdV

Then

P1 =

∫Dv y dxdy =

∫Dw y2 dxdy =

1

4π

∫D4

ΦdV =P

4π


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Therefore, applying the previous inequalities for P

Polya-Szego:P < |D4|λ−1

So

P1 <1

4π|D4|λ−1

=1

4π(4π)

(∫Dx2dxdy

)λ−1

= A1 λ−1

where A1 =∫D y2dxdy .

Payne-Rayner:

Pλ3 ≥ 8π2

2j21,1.

ThereforeP1λ

3 ≥ πj21,1


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Saint Venant:

P ≤√

2|D4|3/2

24π

So

P1 ≤1

3

(1

8π

) 12

A3/21 .

The original interpretation in the case of λ was observed by Payne.

λ ≥ 1

2

(π

2A1

)1/2

j21,1

and is also optimized for the half-disk.


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Chiti ineq. α = 1 via Payne interpretation in Weinsteinspace

Remember:−∆u = λu in D, u = 0 on ∂D. (21)

and D is a subset of the upper half-space W1.

u = y w , (22)

where w is a positive smooth function vanishing on ∂D ∩W1. Substituteinto (21) and simplify:

∆w +2

y

∂w

∂y= λw in D,w = 0 on ∂D ∩W1. (23)

Let Φ(x1, x2, x3, x4) be the function defined in R4 by

Φ(x1, x2, x3, x4) = w(x , y) where x = x1; y =√

x22 + x2

3 + x24 ;

This function has axial symmetry with respect to the x1-axis. It is definedon D4


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Weighted Chiti ineq. α = 1 via Payne interpretation inWeinstein space, cont’d

The function Φ satisfies

−∆4 Φ = λΦ in D4, Φ = 0 on ∂D4.

For the case of the Chiti inequality for half space, we first write thestatement in 4-d for D4[∫

D4

ΦqdV4

] 1q

≤ K (p, q, 4)

[∫D4

ΦpdV4

] 1p

for q ≥ p ≥ 0. As before, changing variables, for p > 0, yields[∫D4

ΦpdV4

] 1p

= (4π)1p

[∫Dwpy2dV

] 1p

.

Thus [∫Dwqy2dV

] 1q

≤ (4π)1p− 1

qK (p, q, 4)

[∫Dwpy2dV

] 1p

,

with equality if and only if D is a half-disk.Lotfi Hermi (UA) Sharp Isoperimetric Bounds


Payne interpretation α = 1 for Freitas-Krejcirik ineq.

λ ≤ j2d/2−1,1

B

d |D|. (24)

Interpretation in the case D be the domain which lies in the half-space W1

such that its curved boundary Γ is some smooth polar curve:r = ρ(ω), ω ∈ (0, π) i.e.,

D =

{(x , y) ∈ W1 | 0 <

√x2 + y2 < ρ(arccos

x√x2 + y2

)

}.

The proof goes through two steps: (1) D4 is a strictly star-shaped domainwith respect to the origin. (2) We have the reduction between

B(D4) =

∫∂D4

1

(ξ, n)dA, and B1 =

∫Γ

y2

(z , n)dσ, B(D4) = 4πB1

and, as a result

λ ≤ B(D4)

4 |D4|j21 =

B1

4 A1j21 . (25)

where A1 =∫D y2 dV .Lotfi Hermi (UA) Sharp Isoperimetric Bounds



Case α = 2. In this case, D is such that x > 0, y > 0, and P4 reducesto

P4 :

{∆v + 2xy = 0 in Dv = 0 in ∂D ∩Wα,

With v = 2x y w , the problem is then

P4 :

∆w +2

x

∂w

∂x+

2

y

∂w

∂y= −1 in D

w = 0 in ∂D ∩ {x > 0, y > 0},

Let the function Φ(x1, x2, x3, y1, y2, y3) be defined by

Φ(x1, x2, x3, y1, y2, y3) = w(x , y)

with x =√x2

1 + x22 + x2

3 ; y =√

y21 + y2

2 + y23 . This function has x

and y as axes of symmetry. It is defined on

D6 = {(x1, x2, x3, y1, y2, y3) ∈ R6|x =√

x21 + x2

2 + x23 , y =

√y2

1 + y22 + y2

3 , (x, y) ∈ D}

obtained via two “rotations” of D around the coordinate axes.Lotfi Hermi (UA) Sharp Isoperimetric Bounds



The function Φ satisfies

∆6 Φ = −1 in D6, Φ = 0 on ∂D6.

Note that d = 2α + 2 = 6. Let dV = dx1dx2dx3dy1dy2dy3, and

P =

∫D6

ΦdV

Then

P2 = 2

∫Dv x y dxdy = 4

∫Dw x2 y2dxdy =

4

(4π)2

∫D6

ΦdV =P

4π2.

Also

|D6| =

∫D6

dV = (4π)2

∫Dx2y2 dxdy = 4π2A2

where A2 = 4∫D x2 y2 dxdy .


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Therefore, applying the previous inequalities for P

Polya-Szego:P < |D6|λ−1

which leads to

P2 < A2 λ−1.

Payne-Rayner:

Pλ4 ≥ 12π3

6j42,1.

ThereforeP2λ

4 ≥ π

2j42,1


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Saint Venant:

P ≤ 61/3|D6|4/3

48π

which simplifies as

P2 ≤1

4

(1

72π

) 13

A4/32 .

Again the original interpretation in the case of λ was observed byPayne

λ ≥ 1

2

(π

12A2

)1/3

j22,1,

and isoperimetry holds for the last two inequalities for the quarterdisk with the same A2 as D.

We proceed as in the above for the case of the weighted Chiti andFreitas-Krejcirik isoperimetric inequalities in this case as well.


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Polya-Szego for wedge-like membrane

Theorem. For a wedge-like membrane D ⊂ Wα

Pα ≤ |D|h λ−1

where

|D|h = Aα =

∫Dh2dxdy

Proof. Start with

P4 :

{−∆v = h(r , θ) in Dv = 0 in ∂D ∩W,

Pα =

(∫D v h dxdy

)2∫D

∣∣∇v ∣∣2 ≤∫D v2

∫D h2∫

D

∣∣∇v ∣∣2≤ λ−1 |D|h


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Another Proof inspired by Polya-Szego

Since the eigenfunctions {un}∞n=1 form an orthonormal basis of L2(D),corresponding to the eigenvalues 0 < λ ≡ λ1 < λ2 ≤ . . . ≤ λn →∞, onecan write

|D|h =

∫Dh2 dA =

∞∑n=1

(∫Dh un dA

)2

, (26)

and

Pα =∞∑n=1

1

λn

(∫Dh un dA

)2

. (27)

The result is then immediate from the ordering of the eigenvalues, viz.

Pα <1

λ1

∞∑n=1

(∫Dh un dA

)2

=1

λ1|D|h.

Key: Expand h =∑∞

n=1 αn un with αn =∫D hun dA, and v =

∑∞n=1 βn un,

then use Plancherel-Parseval.


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Payne-Rayner for wedge-like membrane

Theorem. For a wedge-like membrane D ⊂ Wα

Pαλα+2 ≥ π

αj2αα

Proof. Start with Rayleigh-Ritz

1

Pα≤∫D |∇u|

2 rdrdθ(∫D uh rdrdθ

)2.

u being the fundamental eigenfunction. Then

1

Pα≤∫D |∇u|

2 rdrdθ∫D u2 rdrdθ

∫D u2 rdrdθ(∫D uh rdrdθ

)2.

Complete argument with weighted Chiti inequality.


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Saint Venant for wedge-like membrane

Theorem (Hasnaoui-H. AHP 2014, and this paper) For a wedge-likemembrane D ⊂ Wα

A2α

(2α + 4)Bα≤ Pα ≤

1

α + 2

(α |D|α+2

h

4α(α + 1)α π

)1/(α+1)

Equality holds for the circular sector.One can produce scale-free versions as well:

Pα(D)|D|− 2α+4

2α+2

h ≤ Pα(D∗)|D∗|− 2α+4

2α+2

h . (28)

This is a corollary (case k = 2, f = 1 of the following more general setting)


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Rearrangement of Functions

We let f be a real-valued measurable function defined in D ⊂ Rn, and letµ be its distribution function defined, for every nonnegative t, by

µ(t) = |{x ∈ D : |f (x)| > t}| =

∫{|f (x)|>t}

dx .

The decreasing rearrangement of f is the function f ∗ defined by

f ∗(s) = inf{t ≥ 0 : µ(t) < s}.

The function f ? defined by f ?(x) = f ∗(Cn|x |n), where Cn =πn/2

Γ(n/2 + 1),

is called the spherically symmetric-decreasing rearrangement of f . Thespherically symmetric-increasing rearrangement of f , denoted f?, is definedin a similar fashion. While f ∗ is defined on [0, |D|], f ? is defined on theball D? centered at the origin of the same volume as D.


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Flavor of the Proof of the Main Theorem

The fundamental eigenvalue λ of the wedge-like membrane D definedabove satisfies the (new) inequality in red.(4α(α + 1)

πAα) −1α+1

j2α,1 ≤ λ ≤

Bα(2α + 2)Aα

j2α,1. (29)

Equality holds if and only if D is a perfect sector S0.Key ingredient: In the Rayleigh-Ritz’s characterization, let the testfunction be f = vh, where

v(r , θ) = g

(r

ρ(θ)

), (30)

and g is any C 1 function such that g(1) = 0. This reduces the problem toa 1-d setting. (This is a “weighted proof” of the Polya-Szego proof, bothin the case of λ and Pα.)


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Isoperimetric inequalities for wedge-like membranes

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