Isoperimetric inequalities of the fourth order Neumann

Deng and Du Journal of Inequalities and Applications (2020) 2020:210 https://doi.org/10.1186/s13660-020-02476-9

R E S E A R C H Open Access

Isoperimetric inequalities of the fourth orderNeumann eigenvaluesYanlin Deng1 and Feng Du1,2*

*Correspondence:[email protected] of Mathematics andPhysics, Jingchu University ofTechnology, Jingmen, 448000,P.R. China2Faculty of Mathematics andStatistics, Key Laboratory of AppliedMathematics of Hubei Province,Hubei University, Wuhan, 430062,P.R. China

AbstractIn this paper, we obtain some isoperimetric inequalities for the first (n – 1) eigenvaluesof the fourth order Neumann Laplacian on bounded domains in an n-dimensionalEuclidean space. Our result supports strongly the conjecture of Chasman.

MSC: 35P15; 53C40; 58C40

Keywords: Eigenvalues; Neumann problem; Isoperimetric inequality

1 IntroductionLetting Ω be a bounded domain with a smooth boundary ∂Ω in the Euclidean space R

n,we consider the Neumann problem of the Laplacian � as follows:

⎧⎨

⎩

�u = μu, in Ω ,∂u∂ν

= 0, on ∂Ω ,(1.1)

where ν is the outward unit normal to the boundary. It is well known that the free mem-brane problem (1.1) has a discrete spectrum consisting of a sequence

0 = μ0 < μ1 ≤ μ2 ≤ · · · → +∞.

When Ω is a bounded domain in R2, Szegö [6] proved the following classical isoperimetric

inequality:

μ1(Ω) ≤ μ1(BΩ ), (1.2)

where BΩ is the ball of same volume as Ω . Weinberger [11] generalized this result to n-dimensions. Ashbaugh and Benguria [2] extended the Szegö–Weinberger inequality (1.2)to the bounded domains in hyperbolic space and a hemisphere. On the other hand, Ash-baugh and Benguria [1] conjectured that

n∑

i=1

1μi(Ω)

≥ nμ1(BΩ )

, with equality if and only if Ω is a ball, (1.3)

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Deng and Du Journal of Inequalities and Applications (2020) 2020:210 Page 2 of 10

where μi(Ω) is the ith Neumann eigenvalue on Ω , μ1(BΩ ) is the first nonzero Neumanneigenvalue on BΩ . In [10], Wang and Xia proved an isoperimetric inequality for the sumsof the reciprocals of the first (n – 1) nonzero eigenvalues of the Neumann Laplacian onbounded domains in R

n as follows:

n–1∑

i=1

1μi(Ω)

≥ n – 1μ1(BΩ )

, with equality if and only if Ω is a ball, (1.4)

which means the Ashbaugh–Benguria’s conjecture is true for the first (n – 1) nonzeroeigenvalues of the Neumann Laplacian on bounded domains in R

n. So (1.4) supportsthe above conjectures of Ashbaugh and Benguria. On the other hand, Benguria, et al. [3]proved a result which is similar to (1.4) for the first (n – 1) nontrivial Neumann eigenval-ues on domains in a hemisphere of Sn. Moreover, some works on eigenvalues are relatedto the spectra of matrix operators and can be seen in [7–9].

Let � and � be the Laplace–Beltrami operators on Ω and ∂Ω , respectively. Let ∇ and∇ be the gradient operators on Ω and ∂Ω , respectively. Consider the following Neumanneigenvalue problem of the bi-harmonic operator:

⎧⎪⎪⎨

⎪⎪⎩

�2u – τ�u = Λu in Ω ,∂2u∂ν2 = 0, on ∂Ω ,

τ ∂u∂ν

– div∂Ω (∇2u(ν)) – ∂�u∂ν

= 0, on ∂Ω ,

(1.5)

where τ ≥ 0 and σ are two constants, div∂Ω denotes the tangential divergence operatoron ∂Ω , and ∇2u is the Hessian of u, ν is the outward unit normal to the boundary. In thissetting, problem (1.5) has a discrete spectrum, and all eigenvalues in the discrete spectrumcan be listed nondecreasingly as follows:

0 = Λ0 < Λ1 ≤ Λ2 ≤ · · · ↑ +∞.

By the Rayleigh–Ritz characterization, the (k + 1)th eigenvalue of (1.5) can be given asfollows (see, e.g., [5]):

Λk+1 = infu∈H2(Ω)

{

Q[u] =∫

Ω[|∇2u|2 + τ |∇u|2] dx

∫

Ωu2 dx

∣∣∣

∫

Ω

uuj = 0, j = 1, . . . , k}

. (1.6)

Letting BΩ be the ball of same volume as Ω , Chasman [5] proved the following isoperi-metric inequality:

Λ1(Ω) ≤ Λ1(BΩ ), with equality if and only if Ω is a ball.

Chasman [5] also conjectured that

n∑

i=1

1Λi(Ω)

≥ nΛ1(BΩ )

, with equality if and only if Ω is a ball. (1.7)

In this paper, we prove an isoperimetric inequality for the sums of the reciprocals of thefirst (n – 1) nonzero eigenvalues of the fourth Neumann Laplacian which supports the


Chasman’s conjecture, actually, we get

n–1∑

i=1

1Λi(Ω)

≥ n – 1Λ1(BΩ )

, with equality if and only if Ω is a ball. (1.8)

In [4], Buoso et al. proved a quantitative isoperimetric inequality for the fundamentaltone of problem (1.5) as follows:

Λ1(Ω) ≤ (1 – ηn,τ ,|Ω|A2(Ω)

)Λ1(BΩ ), (1.9)

where ηn,τ ,|Ω| > 0, and A(Ω) is the so-called Fraenkel asymmetry of the domain Ω ∈ Rn,

which is defined by:

A(Ω) := inf{ |Ω�BΩ |

|Ω|}

,

where BΩ is the ball of same volume as Ω and Ω�BΩ is the symmetric difference of Ω

and BΩ . In what follows, we generalize (1.9) to the sum of the first (n – 1) eigenvalues, andwe get

1n – 1

n–1∑

i=1

Λi(Ω) ≤ (1 – ηn,τ ,|Ω|A2(Ω)

)Λ1(BΩ ). (1.10)

2 PreliminariesIn this section, we recall some notations and results, more details can be seen in [4, 5].

Let j1, i1 be the ultraspherical and modified ultraspherical Bessel functions of the firstkind and order 1, respectively; j1, i1 can be expressed by the standard Bessel and modifiedBessel functions of the first kind Jν , Iν as follows:

j1(z) = z1–n/2Jn/2(z), i1(z) = z1–n/2In/2(z).

Let B be the unit ball in Rn centered at the origin and ωn be the Lebesgue measure |B| of

B, and let Λ1(B) be the first eigenvalue of problem (1.5) on unit ball B. For τ > 0, a, b arepositive constants satisfying the conditions a2b2 = λ1(B) and b2 – a2 = τ . Set

R(r) = j1(ar) + γ i1(br), γ =–a2j′′1(ar)

b2i′′1(b)> 0.

Then we define the function ρ : [0, +∞) → [0, +∞) as

ρ(r) =

⎧⎨

⎩

R(r), r ∈ [0, 1),

R(1) + (r – 1)R′(1), r ∈ [1, +∞).

Let ui : Rn →R be defined by

ui(x) := ρ(|x|) xi

|x| , for i = 1, . . . , n. (2.1)


The functions ui|B are, in fact, the eigenfunctions associated with the eigenvalues λ1(B) ofproblem (1.5) on unit ball B. We know that λ1(B) has multiplicity and ui satisfy

n∑

i=1

|ui|2 = ρ2(|x|), (2.2)

n∑

i=1

|∇ui|2 =n – 1|x|2 ρ

(|x|)2 +(ρ ′(|x|))2, (2.3)

n∑

i=1

∣∣∇2ui

∣∣2 =

(ρ ′′(|x|))2 +

3(n – 1)|x|4

[ρ(|x|) – |x|ρ ′(|x|)]2. (2.4)

Define N[ρ] =∑n

i=1(|∇2ui|2 + τ |∇ui|2). Then ρ and N[ρ] satisfy the following propertieswhich given in [4, 5].

Lemma 2.1 Function ρ and N[ρ] satisfy the following properties:(1) ρ ′′(r) < 0 for all r ≥ 0, therefore ρ ′ is nonincreasing.(2) ρ(r) – rρ ′(r) ≥ 0, with equality holding only for r = 0.(3) The function ρ2(r) is strictly increasing.(4) The function ρ2(r)/r2 is decreasing.(5) The function 3(ρ(r) – rρ ′(r))2/r4 + τρ2(r)/r2 is decreasing.(6) N[ρ(r1)] > N[ρ(r2)] for any r1 ∈ [0, 1), r2 ∈ [1, +∞).(7) For all r ≥ 0, we have

N[ρ(r)

]=

(ρ ′′(r)

)2 +3(n – 1)(ρ(r) – rρ ′(r))2

r4 + τ (n – 1)ρ2(r)

r2 + τ(ρ ′(r)

)2.

(8) For all r ≥ 1, N[ρ(r)] is decreasing.

We introduce the notation of a partially monotonic function. A function F is partiallymonotonic on Ω if it satisfies

F(x) > F(y), for all x ∈ Ω and y /∈ Ω . (2.5)

It is seen that N[ρ(r)] is a partially monotonic function from Lemma 2.1.

Lemma 2.2 For any radial function F(r(x)) that satisfies the partially monotonicity con-dition on BΩ ,

∫

Ω

F dx ≤∫

BΩ

F dx (2.6)

with equality if and only if Ω = BΩ . For any strictly increasing radial function F(r(x)),

∫

Ω

F dx ≥∫

BΩ

F dx (2.7)

with equality if and only if Ω = BΩ .


Lemma 2.3 For all s > 0, we have

Λi(τ ,Ω) = s4Λi(s–2τ , sΩ

), i = 1, . . . , n, (2.8)

where sΩ = {x ∈Rn : x/s ∈ Ω} for s > 0.

Proof For any u ∈ H2(Ω) with

u = 0 and∫

Ω

u dx =∫

Ω

uu1 dx = · · · =∫

Ω

uui–1 dx = 0, i = 1, . . . , n,

let u(x) = u(x/s), then u is a valid trial function on sΩ and so

Qs–2τ ,sΩ [u] =∫

sΩ (|∇2u|2 + s–2τ |∇u|2) dx∫

sΩ u2 dx

=∫

sΩ (|s–2(∇2u)(x/s)|2 + s–2τ |s–1(∇u)(x/s)|2) dx∫

sΩ u(x/s)2 dx

=s–4+n ∫

Ω(|(∇2u)|2 + τ |(∇u)|2) dy

sn∫

Ωu2 dy

(substituting y = x/s)

= s–4Qτ ,Ω [u]. (2.9)

The lemma follows from (1.6). �

3 Proofs of the main resultsIn this section, we give the proofs of the main results of this paper.

Theorem 3.1 Let Ω be a bounded domain in an n-dimensional Euclidean space Rn and

let BΩ be the ball of same volume as Ω , then the first (n – 1) eigenvalues of (1.5) in Ω satisfy

n–1∑

i=1

1Λi(Ω)

≥ n – 1Λ1(BΩ )

, (3.1)

with equality if and only if Ω is a ball.

Proof Assume that the volume of Ω is equal to that of the unit ball B. Letting ϕi = ρ(r)xir ,

we know that∫

Ω

ϕi(r) dx = 0, for i = 1, . . . , n,

which means ϕi is perpendicular to u0 = 1/√|Ω|, which is the first eigenfunction of (1.5).

Letting {uj}∞j=0 be an orthonormal set of eigenfunctions of (1.5) on Ω , next we will showthat there exists new coordinate functions {x′

i}ni=1 such that

∫

Ω

ρ(r)x′i

ruj dx = 0, (3.2)

for j = 1, . . . , i – 1 and i = 2, . . . , n. To see this, we define an n × n matrix A = (aij), whereaji =

∫

Ωϕiuj dx =

∫

Ω

ρ(r)r xiuj dx, for i, j = 1, 2, . . . , n. Using the orthogonalization of Gram


and Schmidt (QR-factorization theorem), we know that there exist an upper-triangularmatrix T = (Tji) and an orthogonal matrix B = (bji) such that T = UQ, i.e.,

Tij =n∑

k=1

bikakj =∫

Ω

n∑

k=1

ρ(r)r

bikxkuj dx = 0, 1 ≤ j < i ≤ n.

Letting x′i =

∑nk=1 bikxk , i = 1, . . . , n, we get (3.2). Since B = (bji) is an orthogonal matrix,

{x′i}n

i=1 is also a set of coordinate functions. Therefore, denoting x′i, i = 1 . . . , n still by xi,

i = 1 . . . , n, and ϕi = ρ(r)r xi, we have

ϕi = 0 and∫

Ω

ϕi dx =∫

Ω

ϕiu1 dx = · · · =∫

Ω

ϕiui–1 dx = 0, i = 1, . . . , n.

It follows from the Rayleigh–Ritz inequality that

Λi(Ω)∫

Ω

ϕ2i dx ≤

∫

Ω

(∣∣∇2ϕi

∣∣2 + τ |∇ϕi|2

)dx, i = 1, . . . , n, (3.3)

which implies that

∫

Ω

ϕ2i dx ≤ 1

Λi(Ω)

∫

Ω

(∣∣∇2ϕi

∣∣2 + τ |∇ϕi|2

)dx, i = 1, . . . , n. (3.4)

Summing over i from 1 to n, we have

n∑

i=1

∫

Ω

ϕ2i dx ≤

n∑

i=1

1Λi(Ω)

∫

Ω

(∣∣∇2ϕi

∣∣2 + τ |∇ϕi|2

)dx. (3.5)

Since∑n

i=1 |∇2ϕi|2 = (ρ ′′)2 + 3(n–1)r4 (ρ – rρ ′)2, for any point p ∈ Ω , by a transformation of

coordinates if necessary, we have |∇2ϕi|2 ≤ (ρ′′)2

n–1 + 3r4 (ρ – rρ ′)2, i = 1, . . . , n. Then

n∑

i=1

1Λi(Ω)

∣∣∇2ϕi

∣∣2

=n–1∑

i=1

1Λi(Ω)

∣∣∇2ϕi

∣∣2 +

1Λn(Ω)

∣∣∇2ϕi

∣∣2

=n–1∑

i=1

1Λi(Ω)

∣∣∇2ϕi

∣∣2 +

1Λn(Ω)

((ρ ′′)2 +

3(n – 1)r4

(ρ – rρ ′)2 –

n–1∑

j=1

∣∣∇2ϕj

∣∣2

)

≤n–1∑

i=1

1Λi(Ω)

∣∣∇2ϕi

∣∣2 +

n–1∑

j=1

1Λj(Ω)

( (ρ ′′)2 + 3(n–1)r4 (ρ – rρ ′)2

n – 1–

∣∣∇2ϕj

∣∣2

)

=1

n – 1

((ρ ′′)2 +

3(n – 1)r4

(ρ – rρ ′)2

) n–1∑

i=1

1Λi(Ω)

. (3.6)


Similarly, we have

n∑

i=1

1Λi(Ω)

|∇ϕi|2 =n–1∑

i=1

1Λi(Ω)

|∇ϕi|2 +1

Λn(Ω)|∇ϕi|2

=n–1∑

i=1

1Λi(Ω)

|∇ϕi|2 +1

Λn(Ω)

(n – 1

r2 ρ2 +(ρ ′)2 –

n–1∑

j=1

|∇ϕj|2)

≤n–1∑

i=1

1Λi(Ω)

|∇ϕi|2 +n–1∑

j=1

1Λj(Ω)

( n–1r2 ρ2 + (ρ ′)2

n – 1– |∇ϕj|2

)

=1

n – 1

(n – 1

r2 ρ2 +(ρ ′)2

) n–1∑

i=1

1Λi(Ω)

. (3.7)

On the other hand,

n∑

i=1

|ϕi|2 = ρ2. (3.8)

Substituting (3.6)–(3.8) into (3.5), we have

1n – 1

n–1∑

i=1

1Λi(Ω)

≥∫

Ωρ2 dx

∫

Ω((ρ ′′)2 + 3(n–1)

r4 (ρ – rρ ′)2 + τ ( n–1r2 ρ2 + (ρ ′)2)) dx

=∫

Ωρ2 dx

∫

ΩN[ρ] dx

≥∫

BΩρ2 dx

∫

BΩN[ρ] dx

=1

Λ1(BΩ ), (3.9)

the last step is deduced by Lemma 2.2. If the equality holds, then equality holds in (3.9),which implies Ω must be a unit ball. By Lemma 2.3, for any domain Ω in R

n, we get

1n – 1

n–1∑

i=1

1Λi(Ω)

≥ 1Λ1(BΩ )

. (3.10)

This completes the proof of Theorem 3.1. �

Theorem 3.2 Let Ω be a bounded domain in an n-dimensional Euclidean space Rn and

let BΩ be the ball of same volume as Ω , then the first (n – 1) eigenvalues of (1.5) in Ω satisfy

1n – 1

n–1∑

i=1

Λi(Ω) ≤ (1 – ηn,τ ,|Ω|A2(Ω)

)Λ1(BΩ ). (3.11)

Proof Case 1. Ω is a bounded domain in Rn of class C1 with the same measure as the unit

ball B. By a similar argument as in the proof of Theorem 3.1, we have

Λi(Ω)∫

Ω

ϕ2i dx ≤

∫

Ω

(∣∣∇2ϕi

∣∣2 + τ |∇ϕi|2

)dx, i = 1, . . . , n. (3.12)


Summing over i from 1 to n, we have

n∑

i=1

Λi(Ω)∫

Ω

ϕ2i dx ≤

n∑

i=1

∫

Ω

(∣∣∇2ϕi

∣∣2 + τ |∇ϕi|2

)dx =

∫

Ω

N[ρ] dx. (3.13)

Since∑n

i=1 ϕ2i = ρ2, for any point p ∈ Ω , by a transformation of coordinates if necessary,

we have ϕ2i ≤ ρ2

n–1 , i = 1, . . . , n. Then

n∑

i=1

Λi(Ω)ϕ2i =

n–1∑

i=1

Λi(Ω)ϕ2i + Λn(Ω)ϕ2

n

=n–1∑

i=1

Λi(Ω)ϕ2i + Λn(Ω)

(

ρ2 –n–1∑

j=1

ϕ2j

)

≥n–1∑

i=1

Λi(Ω)ϕ2i +

n–1∑

j=1

Λj

(ρ2

n – 1– ϕ2

j

)

=n–1∑

i=1

Λiρ2

n – 1. (3.14)

Substituting (3.13) into (3.14), we have

1n – 1

n–1∑

i=1

Λi(Ω) ≤∫

ΩN[ρ] dx

∫

Ωρ2 dx

. (3.15)

On the other hand, we have

Λ1(B) =∫

B N[ρ] dx∫

B ρ2 dx. (3.16)

Combining (3.15) and (3.16), we have

Λ1(B)∫

Bρ2 dx –

1n – 1

n–1∑

i=1

Λi(Ω)∫

Ω

ρ2 dx ≥∫

BN[ρ] dx –

∫

Ω

N[ρ] dx. (3.17)

From equation (16) in [4], we know that

Λ1(B)∫

Bρ2 dx – Λ1(Ω)

∫

Ω

ρ2 dx ≤ C(1)n,τ

(Λ1(B) – Λ1(Ω)

),

where C(1)n,τ = nωn

∫ 10 ρ2(r)rn–1 dr. Then we have

Λ1(B)∫

Bρ2 dx –

1n – 1

n–1∑

i=1

Λi(Ω)∫

Ω

ρ2 dx ≤ C(1)n,τ

(

Λ1(B) –1

n – 1

n–1∑

i=1

Λi(Ω)

)

. (3.18)

From (15) and (20) in [4], we know that

Λ1(B)∫

Bρ2 dx – Λ1(Ω)

∫

Ω

ρ2 dx ≥∫

B/B1

N(ρ) dx –∫

B2/BN(ρ) dx,


and∫

B/B1

N(ρ) dx –∫

B2/BN(ρ) dx = C(2)

n,τ α2,

where B1 and B2 are two balls centered at the origin with radii r1, r2 such that |Ω ∩ B| =|B1| = ωnrn

1 and |Ω/B| = |B2/B| = ωn(rn2 – 1). Then we have

∫

BN[ρ] dx –

∫

Ω

N[ρ] dx ≥ C(2)n,τ

|Ω�B||Ω| , (3.19)

where C(2)n,τ = nωn((3 + τ )(R(1) – R′(1))2 + 2τR′(1)(R(1) – R′(1)))cn.

Combining (3.18) and (3.19), we have

Λ1(B) –1

n – 1

n–1∑

i=1

Λi(Ω) ≥ C(2)n,τ

C(1)n,τ

A2(Ω),

which implies that

1n – 1

n–1∑

i=1

Λi(Ω) ≤ Λ1(B)(

1 –C(2)

n,τ

Λ1(B)C(1)n,τ

A2(Ω))

. (3.20)

Case 2. Ω is the generic domain in Rn of class C1. Since

Λi(τ ,Ω) = s4Λi(s–2τ , sΩ

), i = 1, . . . , n, (3.21)

for all s > 0. Taking s = (ωn/|Ω|) 1n , for any domain Ω in R

n of class C1, we infer from (3.21)that

1n – 1

n–1∑

i=1

Λi(τ ,Ω) = s4 1n – 1

n–1∑

i=1

Λi(s–2τ , sΩ

)

≤ s4Λ1(s–2τ , B

)(

1 –C(2)

n,s–2τ

Λ1(s–2τ , B)C(1)n,s–2τ

A2(sΩ))

= Λ1(s–2τ , B

)(

1 –C(2)

n,s–2τ

Λ1(s–2τ , B)C(1)n,s–2τ

A2(Ω))

.

Setting ηn,τ ,|Ω| =C(2)

n,s–2τ

Λ1(s–2τ ,B)C(1)n,s–2τ

, we have (1.10). This completes the proof of Theo-

rem 3.2. �

AcknowledgementsThe authors would like to thank the referee for his or her careful reading and valuable comments such that the articleappears in its present version.

FundingThis work is supported by NSF of Hubei Provincial Department of Education (Grant Nos. B2019211, D20184301,B2016261), Research Team Project of Jingchu University of Technology (Grant No. TD202006), Research Project of JingchuUniversity of Technology (Grant No. QDB201608) and Hubei Key Laboratory of Applied Mathematics (Hubei University).


Availability of data and materialsNot applicable.

Competing interestsAll authors declare that they have no competing interests.

Authors’ contributionsAll authors have equal contribution. All authors read and approved the final manuscript.

Publisher’s NoteSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 28 May 2019 Accepted: 5 August 2020

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Isoperimetric inequalities of the fourth order Neumann