On a Pólya functional for rhombi, isosceles triangles, and thinning convex sets.

# On a Pólya functional for rhombi, isosceles triangles, and thinning convex sets.

M. van den Berg** , V. Ferone*, C. Nitsch*, C. Trombetti*

**School of Mathematics, University of Bristol
University Walk, Bristol BS8 1TW, UK
mamvdb@bristol.ac.uk

*Università degli Studi di Napoli Federico II
Via Cintia, Monte S. Angelo, I-80126 Napoli, Italy
vincenzo.ferone@unina.it
c.nitsch@unina.it
cristina.trombetti@unina.it
5 November 2018
###### Abstract

Let be an open convex set in with finite width, and let be the torsion function for , i.e. the solution of . An upper bound is obtained for the product of , where is the bottom of the spectrum of the Dirichlet Laplacian acting in . The upper bound is sharp in the limit of a thinning sequence of convex sets. For planar rhombi and isosceles triangles with area , it is shown that , and that this bound is sharp.

AMS 2000 subject classifications. 49J45, 49R05, 35P15, 47A75, 35J25.
Key words and phrases. torsion function, torsional rigidity, first Dirichlet eigenvalue

## 1 Introduction

Let be an open set in , and denote the bottom of the spectrum of the Dirichlet Laplacian acting in by

 λ(Ω)=infφ∈H10(Ω)∖{0}∫Ω|Dφ|2∫Ωφ2.

It was shown in [3] and [1] that if

 λ(Ω)>0,

then the torsion function, defined by

 −Δv=1,v∈H10(Ω),

satisfies

 1≤λ(Ω)M(Ω)≤cm, (1.1)

where

 M(Ω)=∥vΩ∥L∞(Ω).

In [12] it was shown that

 cm≤18(m+(5(4+log2))1/2m1/2+8)

The sharp constant in the right-hand side of (1.1) is not known. However, for an open ball , an open square , and an equilateral triangle ,

 λ(B)M(B)<λ(S)M(S)<λ(E)M(E). (1.2)

The fact that was shown in [6]. The full inequality (1.2) follows from numerical evaluation of the series for the square, pp. 275-277 in [13].

In [2] it was shown that the left-hand side of (1.1) is sharp: for there exists an open bounded and connected set such that

 λ(Ωϵ)M(Ωϵ)<1+ϵ.

For open, bounded convex sets in it was shown in (3.12) of [7] that

 λ(Ω)M(Ω)≥π28, (1.3)

with equality in the limit of an infinite slab (the open set with finite width bounded by two parallel -dimensional planes). The latter assertion has been made precise in [6] where it was shown that if

 Sn=(−n,n)m−1×(0,1),n≥1, (1.4)

then

 λ(Sn)M(Sn)≤π28+m−18(n−23). (1.5)

For bounded planar convex sets with width , and diameter , it was shown in [2] that

 λ(Ω)M(Ω)≤π28⎛⎝1+32/37(w(Ω)\textupdiam(Ω))2/3⎞⎠. (1.6)

In Theorem 1.1 below we put (1.4)-(1.5), and (1.6) in a more general setting. We introduce the following notation. For an open, bounded, convex set with finite width , and boundary we let , be a family of parallel hyper-planes such that and are tangent to at two points and respectively, where , and is orthogonal to . That this is always possible was shown in Theorem 1.5 in [4]. We identify sets in with sets in . Let

 Ωμ={(x′,xm−μw(Ω)):(x′,xm)∈Ω∩Hμ}.

The projection of onto is denoted by

 Π(Ω):=⋃0≤μ≤1Ωμ.

We denote the inradius of this -dimensional set by . The measure of is denoted by .

###### Theorem 1.1

If is an open, bounded, convex set in then

 λ(Ω)M(Ω)≤π28(1+dm(w(Ω)ρ(Ω))2/3), (1.7)

where

 dm=32/328π−2j2(m−3)/2. (1.8)

The torsional rigidity (or torsion) of an open set is defined by

 T(Ω)=∥vΩ∥L1(Ω)=∫ΩvΩ. (1.9)

In Pólya and Szegö [8], it was shown that for sets with finite measure ,

 T(Ω)λ(Ω)|Ω|≤1. (1.10)

It was subsequently shown in [4] that the constant in the right-hand side above is sharp: for , there exists an open, bounded, and connected set such that

The quantity in the left-hand side of (2) is invariant under the homothety transformation . This implies for example that in Theorems 1.2-1.5 below we do not have to specify the actual lengths of the edges of the rhombi and triangles. In the proofs of these theorems we fix the various lengths as a matter of convenience.

It was shown in Theorem 1.5 of [4] that for a thinning (collapsing) sequence of bounded convex sets

 limsupn→∞T(Ωn)λ(Ωn)|Ωn|≤π212. (1.11)

This supports the conjecture that for bounded, convex sets the sharp constant in the right-hand side of (2) is .

It was shown in Theorem 1.4 in [4] that for bounded convex sets in

 T(Ω)λ(Ω)|Ω|≥π24mm+2(m+2), (1.12)

and that for planar, bounded convex sets the inequality holds with constant . In [6] it was conjectured that for planar, bounded, convex sets

 T(Ω)M(Ω)|Ω|≥13, (1.13)

and that this constant is sharp for a thinning (collapsing) sequence of isosceles triangles. See also [5]. By (1.3) and (1.6) above, we have that for a thin isosceles triangle . This suggests that the sharp constant for planar convex sets in the right-hand side of (1.12) is . We have the following.

###### Theorem 1.2

If is an isosceles triangle with angles , and if then

 T(△β)λ(△β)|△β|≤π224(1+81(tanβ)2/3). (1.14)
###### Theorem 1.3

If is a rhombus with angles , and if then

 T(◊β)λ(◊β)|◊β|≤π224(1+15(tanβ)2/3). (1.15)
###### Theorem 1.4

If is as in Theorem 1.3, then

 T(◊β)λ(◊β)|◊β|≥π224. (1.16)
###### Theorem 1.5

If is an isosceles triangle with angles , then

 T(△β)λ(△β)|△β|≥π224. (1.17)

This paper is organised as follows. In Section 2 we prove Theorem 1.1. The proofs of Theorems 1.2 and 1.3 are deferred to Section 3. The proof of Theorem 1.4 is deferred to Section 4. The proof of Theorem 1.5 consists of two parts. In Section 5 part 1 we show that inequality (1.17) holds for all , where

 β0=π2−33200. (1.18)

In Section 5 part 2 we use interval arithmetics to verify that (1.17) also holds for

## 2 Proof of Theorem 1.1

Proof.   We first observe, that by domain monotonicity of the torsion function, is bounded by the torsion function for the (connected) set bounded by and . Hence

 vΩ(x)≤12xm(w(Ω)−xm)≤w(Ω)28,(x′,xm)∈Ω.

It suffices to obtain an upper bound for . We choose the -coordinates such that . By convexity we have that the convex hull of is contained in . This convex hull in turn contains a cylinder with height , and base Denote the first dimensional Dirichlet eigenvalue of by . Then, by separation of variables, we have that

 λ(Ω)≤π2z2+(1−zw(Ω))−2λ1/2. (2.1)

The right-hand side of (2.1) is minimised for

 1z=1w(Ω)+(λ1/2π2w(Ω))1/3.

This gives that

 λ(Ω)≤π2w(Ω)2(1+3(λ1/2w(Ω)2π2)1/3+3(λ1/2w(Ω)2π2)2/3+λ1/2w(Ω)2π2). (2.2)

Denote the inradius of , and the centre of the inball by and , respectively:

 r(Ω)=supx∈Ω\textupdist(x,∂Ω)=\textupdist(c(Ω),∂Ω).

Then

 r(Ω)≤xm(c(Ω))≤w(Ω)−r(Ω),

and

 \textupdist(c(Ω),H1/2)≤∣∣w(Ω)2−r(Ω)∣∣.

The inball intersects in a -dimensional disc with radius bounded from below by

 (|r(Ω)2−∣∣w(Ω)2−r(Ω)∣∣2)1/2=(w(Ω)r(Ω)−w(Ω)24)1/2≥w(Ω)2√3,

where we have used that , see Blaschke’s theorem, p. 215 in [14]. Hence

 λ1/2≤12j2(m−3)/2w(Ω)−2, (2.3)

where is the first positive zero of the Bessel function . By (2.2) and (2.3) we obtain that

 λ(Ω) ≤π2w(Ω)2(1+7(12j2(m−3)/2π2)2/3(λ1/2w(Ω)2π2)1/3). (2.4)

By convexity we have that . Hence is bounded from above by times the bottom of the spectrum of . The latter contains a -dimensional ball with radius . So

 λ1/2≤4j2(m−3)/2ρ(Ω)−2, (2.5)

and (1.7), (1.8) follows by (2) and (2.5).

## 3 Proofs of Theorem 1.2 and Theorem 1.3

Proof of Theorem 1.2. Let be an isosceles triangle with a base of length and width (height) of length , and angles , and respectively. By hypothesis, so that . We denote the infinite sector with opening angle by

 Ωβ={(r,ϕ):r>0,−β/2<ϕ<β/2}.

It is straightforward to verify that the torsion function for is given by,

 vΩβ(r,ϕ)=r24(cos(2ϕ)cosβ−1),r>0,−β/2<ϕ<β/2.

Let

 R=(1+d2)1/2.

We can cover with two sectors of opening angles and radii each. By monotonicity and positivity of the torsion function we have that

 T(△β) =∫△βv△β ≤2∫R0drr∫β/2−β/2dϕvΩβ(r,ϕ) =18(1+d2)2(tanβ−β) =18(1+d2)2(d−arctand) ≤d324(1+d2)2, (3.1)

where we have used that . By adapting formula (31) in the proof of Theorem 2 in [2] to the geometry of we find that

 λ(△β)≤π2d2(1+7(d2)2/3). (3.2)

By (3), (3.2), and we obtain that

 T(△β)λ(△β)|△β| ≤π224(1+81d2/3) =π224(1+81(tanβ)2/3),0<β≤π3,

which completes the proof of Theorem 1.2.

Proof of Theorem 1.3. Let be a rhombus with angles , and diagonals of length and respectively. By hypothesis we have that , and This rhombus is covered by two sectors of opening angle , and radius By the calculations in the proof of Theorem 1.2 we find that

 T(◊β) ≤18R4(tanβ−β) =18(1+d24)2(d1−d24−2arctan(d2)) ≤18(1+d24)2(d1−d24−d+d312) ≤d324(1+d24)2(1+9d232),0

By adapting formula (31) in the proof of Theorem 2 in [2] to the geometry of we find that

 λ(◊β)≤π2d2(1+7(d2)2/3).

This, together with gives that,

 T(◊β)λ(◊β)|◊β| ≤π224(1+15(d2)2/3)

This concludes the proof of Theorem 1.3.

## 4 Proof of Theorem 1.4

Let be a rhombus such that major and minor diagonals have lengths and , respectively (see Figure 1). We want to estimate the torsion and to this aim we use a test function

In view of the variational definition of the torsion we have

 1T(◊β)≤∫◊β|Dv|2(∫◊βv)2=24+18d2d3.

On the other hand we can estimate from below the first Dirichlet Laplacian eigenvalue of any rhombus by means of the Dirichlet Laplacian eigenvalue of a rectangle obtained by Steiner symmetrising the rhombus along a direction parallel to one of the sides (see Figure 2). We denote by and the base and the height of the rectangle, respectively. Since the base coincides with the side of the rhombus know that , the height .

We have that

 λ(◊β)≥π2(1b2+1h2)=π216+24d2+d4d2(16+4d2).

Observing that the area of the rhombus is equal to , we have that

 λ(◊β)T(◊β)|◊β|≥π22416+24d2+d4(1+34d2)(16+4d2)≥π224, 0≤d≤2. (4.1)

## 5 Proof of Theorem 1.5

### 5.1 Proof for the case β∈(0,π/3]∪[β0,π/2)

Let be an isosceles triangle with angles We first consider the case . We denote the height by , and we fix the length of the basis equal to . See Figure 3.

We use the function

 u(x,y)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩d2x24−(y−dx2)2, 0≤x≤1,d2(2−x)24−(y−d(2−x)2)2, 1≤x≤2,

as a test function for the torsion of . We find that

 2T(△β)≤48(1+d2)d3. (5.1)

Hence

 T(△β)|△β|≥124(1+1d2)−1. (5.2)

We wish to estimate from below. To this aim we consider the first Dirichlet eigenfunction of restricted to and we reflect it, anti-symmetrically, with respect to the line (see Figure 4). This new function is a test function defined on the rectangle of sides (shaded in grey in Figure 4) orthogonal to the first eigenfunction of the Laplacian with the mixed boundary conditions described in Figure 4.

For we find that

 λ(△β)≥min{π2(1+1d2),4π2d2}=π2(1+1d2). (5.3)

Combining (5.2) and (5.3) we obtain that

 T(△β)λ(△β))|△β|≥π224,0<β≤π3.

Next we consider the case or . We have that

 |△β|=1/tan(α/2). (5.4)

Let

 S(ρ,α)={(r,ϕ):0

be the circular sector with radius and opening angle . Siudeja’s Theorem 1.3 in [9] asserts that for , , where is such that . It follows that

 ρ2=2/(αtan(α/2)). (5.5)

Hence

 λ(△β)≥2−1αtan(α/2)j2π/α.

where we have used that the first Dirichlet eigenvalue of a circular sector of opening angle and radius equals . See [8]. Moreover by (1.2) and (4.3) for and in [11] we have that,

 j2π/α>(πα−a121/3(πα)1/3)2,−a1≥(9π8)2/3,

where is the first negative zero of the Airy function. It follows that

 j2π/α≥π2α2(1+C(απ)2/3)2≥π2α2(1+C1α2/3), (5.6)

where

 C=(9π/8)2/32−1/3,C1=(9/4)2/3. (5.7)

The torsion function for is given by (p.279 in [13]),

 vS(ρ,α)(r,ϕ)= r24(cos(2ϕ)cosα−1) +4ρ2α2π3∑n=1,3,5,...(−1)(n+1)/2(rρ)nπ/αcos(nπϕα)n−1(n+2απ)−1(n−2απ)−1.

By monotonicity of the torsion we obtain that

 T(△β) ≥T(S(ρ,α)) =∫(0,d)rdr∫(−α/2,α/2)dϕvS(d,α)(r,ϕ) =d416(tanα−α−128α4π5∑n=1,3,..n−2(n+2απ)−2(n−2απ)−1), (5.8)

We have that for , This gives that

 T(△β) ≥ρ416(tanα−α−223331ζ(5)α4d425π5) ≥α3ρ448(1−C2α), (5.9)

where

 C2=223431ζ(5)52π5.

By (5.6), (5.1), (5.1), and (5.4) we obtain that

 T(△β)λ(△β)|△β|≥π224(1−C2α)(1+C1α2/3). (5.10)

The right-hand side of (5.5) is greater or equal than for

 C1≥C1C2α+C2α1/3. (5.11)

Inequality (5.11) holds for all

### 5.2 Computer validation for the case β∈(π/3,β0) via interval arithmetic.

We consider a triangle of height and opening angle , where . Let

 F(α)=24π2λ(△α)T(△α)|△α|.

We wish to show that in the range .

We present here a computer assisted proof of the result using Interval Arithmetic.

We once more use Siudeja’s lower bound, comparing with the sector having the same opening angle and the same area (Theorem 1.3 of [9]), to get that

 λ(△α)≥~λ(α)=cos2(α2)(αsinα)(πα+C(πα)13)2, (5.12)

where is given by (5.7).

The area is given by

 |△α|=tan(α2), (5.13)

The monotonicity of with respect to inclusion allows us to estimate from below using the torsion of a tangent sector with same opening angle . We use (5.1) and find that

 T(△α)≥116(tanα−α)−8π5α4∑n=1,3,5,…n−2(n+2απ)−2(n−2απ)−1.

In order to perform a numerical evaluation we truncate the series in the following way

It follows that

 T(△α)≥~T(△α)=116(tanα−α)−8π5α4(10∑n=0(2n+1)−2(2n+1+2απ)−2(2n+1−2απ)−1+127⋅104).

Therefore

 F(α)≥G(α)=24π2~λ(△α)~T(△α)|△α|

At this point we can prove that for all values by using Interval Arithmetic. There are many softwares and libraries which can be employed for this purpose. We selected Octave111John W. Eaton, David Bateman, SÃ¸ren Hauberg, Rik Wehbring (2018). GNU Octave version 4.4.1 manual: A high-level interactive language for numerical computations. URL https://www.gnu.org/software/octave/doc/v4.4.1/ (A free software that runs on GNU/Linux, macOS, BSD, and Windows) which provides a specific package called Interval.222Oliver Heimlich, GNU Octave Interval Package, https://octave.sourceforge.io/interval/, version 3.2.0, 2018-07-01. The interval package is a collection of functions for interval arithmetic. It is developed at Octave Forge, a sibling of the GNU Octave project.

We covered the interval by a collection of 1001 intervals with , so that

 In=[33100+(n−1)1000(π3−33100),33100+(n+1)1000(π3−33100)].

We observe that the intersection of consecutive intervals is intentionally non empty. Using the Interval package, we designed a code that for going from to provides upper and lower bounds for in terms of floating point numbers. This is performed in an automated way by standard and reliable algorithms. We established that the inequality holds true on the whole interval by verifying it on for all .
For completeness we include the code:

2output_precision (6) # number of digits displayed
3C=(9*pi/8)^(2./3)*2^(-1./3);
4
5function K=G(x) # this is the definition of the function G(\alpha)
6  Sum=0;
7  for n = 0:10
8    Sum = Sum + (2*n+1)^(-2)*(2*n+1+2*x/pi)^(-2)*(2*n+1-2*x/pi)^(-1);
9  endfor
10  K=(24./pi^2)*(1./16*(tan(x)-x)-8*x^4/pi^5*(Sum+1./(2.^7*10.^4)))*((cos(x/2))^2*(x/sin(x))*(pi/x+((9*pi/8)^(2./3)*2^(-1./3))*(pi/x)^(1/3))^2)/tan(x/2);
11endfunction
12control=”OK”;#thevariablecontrolissetto”OK”
13N=1000;#Numberofintervals
14Delta=(pi/3-0.33)/N;#2Deltaisthesizeofeachinterval
15forn=0:N
16␣␣n#printthevalueofn
17␣␣a=0.33+n*Delta;
19␣␣J=G(I)#JisanintervalwhichincludestheimageofI
20␣␣if(J>1.01)#checkthat(minJ)>1.01
21␣␣”sofarinequalityholds”#tellthateverythingisworkingfine
22␣␣elseif
23␣␣control=”failure”#thevariablecontrolissetto”failure”
24␣␣break#incaseoffailurethecyclebreaks
25␣␣endif
26endfor

Acknowledgments. The authors wish to thank Prof. Gerardo Toraldo for helpful discussions on Interval Arithmetic. The authors acknowledge support by the London Mathematical Society, Scheme 4 Grant 41636. MvdB was also supported by The Leverhulme Trust through Emeritus Fellowship EM-2018-011-9, and by INDAM-GNAMPA Grant for visiting professors.

## References

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