# A Variational Problem and Classification Theorem for Extremal Hypersurfaces in the Space Form

Authors: Yu-Chung Chang
Journal: Journal of Mathematics
 (2022)

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## Abstract

The classification of the isoparametric extremal hypersurfaces in the space form is obtained in this paper. We also derived the Euler-Lagrange equation for extremal hypersurface and obtained Simons’ type integral inequality. When the integral equality holds, we can obtain the characteristics of isoparametric extremal hypersurfaces by using this classification theorem. The classification of isoparametric extremal hypersurfaces in the space form is the first study that has not been seen in the previous literature.


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## Body

## 1. Introduction

LetNn+1c be a space form with constant sectional curvature c, we choose c=−1, c=0, and c=1 in this paper to represent the hyperbolic space ℍn+1, Euclidean space ℝn+1, and unit sphere Sn+1, respectively. Let x:Mn⟶Nn+1c be an n-dimensional hypersurface, choose an orthonormal frame field e1,e2,…,en,en+1 in Nn+1c such that e1,e2,…,en is tangent to M, en+1 is normal to M, and θ1,θ2,…,θn,θn+1 is represented as the dual frame. We use the following convention on the range of indices:(1)1≤A,B,C,…≤n+1,1≤i,j,k,…≤n.The first fundamental form is the inner product of tangent vectorsIv,w=v,w and often written by the metric tensor gij. The second fundamental form is the symmetric bilinear form on the tangent space hv,w=Spv,w, where Sp is the shape operator (see [1]). Let hij=Spei,ej; then, the second fundamental form can be denoted by(2)h=∑i,jhijθiθj.The mean curvature is defined byH=1/n∑ihii. We define a tensor as(3)h0=∑i,jh˜ijθiθj,where h˜ij=hij−Hδij. Let S and ρ2 be the square length of h and h0, respectively. Then,(4)S=∑i,jhij2,ρ2=∑i,jh˜ij2,ρ2=S−nH2.IfM is an n-dimensional Riemannian manifold in n+p-dimensional Riemannian manifold Nn+P, the Willmore functional is well-known (see [2, 3]; also refer to [4–6] in Sn+P)(5)Wn/2x=∫MS−nH2n/2dμ=∫Mρndμ,where S=∑α,i,jhijα2, α=n+1,…,n+p. A Willmore submanifold is a critical point of Wn/2x. Consider p=1, the Willmore functional of hypersurface is the same as before with S=∑i,jhij2, and a critical point of Wn/2x is called a Willmore hypersurface.In the research on holographic entanglement entropy of physics, Willmore functional is a newer tool [7]. Recently, Deng and Li consider the Willmore functional of spacelike hypersurface in Lorentzian space ℝ1n+1; its critical point is called a Willmore spacelike hypersurface [8]. In [9], the following nonnegative functional Fx was studied.(6)Fx=∫MS−nH2dμ=∫Mρ2dμ.Whenn = 2, the functional Fx is consistent with the Willmore functional.Definition 1.
(see [9]). If M is a hypersurface in the space form Nn+1c, then M is called extremal if it is a critical point of(7)Fx=∫Mρ2dμ.
In this paper, we shall study the classification and variation problems for extremal hypersurface in the space formNn+1c. Firstly, we shall derive the Euler-Lagrange equation for extremal hypersurface in Nn+1c and then classify the isoparametric extremal hypersurfaces in the space form. Finally, we shall consider a compact extremal hypersurface in Nn+1c and obtain an integral inequality for this compact extremal hypersurfaces. More precisely, we have the following three theorems (see Sections 3–5, respectively).Theorem 1.
Ifx:Mn⟶Nn+1c is an n-dimensional compact hypersurface without boundary in the space form Nn+1c, then M is an extremal hypersurface of Fx given by (6) if and only if(8)n−1ΔH+∑i,j,kh˜ijh˜ikh˜kj−n−42ρ2H=0,where ΔH=∑iHii.Theorem 2.
Suppose thatx:Mn⟶Nn+1c is an isoparametric extremal hypersurface with distinct principal curvatures κ1>κ2>⋯>κs. The multiplicity of κi is denoted by mi, i=1,2,…,s. Then the following properties hold.(1)
Ifs=1, then ρ2=0. In other words, M is total umbilical.(2)
Ifs=2, assume that the two distinct principal curvatures κ1=λ and κ2=μ with multiplicity k and n−k, respectively, 1≤k≤n−1, and then either ρ2=0, or n=2k, or(9)ρ2=nkn−kn−424n−2k2H2.(a)
Ifn=2k=4, then M is the hypersurface S2coth2θ−1×ℍ2tanh2θ−1 for c=−1, or S2r×ℝ2 for c=0, or S2r×S21−r2 for c=1, where λ˜=λ−H, μ˜=μ−H, and θ>0.(b)
Ifn=2k and n≠4, then Nn+1c=Sn+1, M=Sk1/2×Sk1/2, and ρ2=n.(c)
Ifn≠4, then H=0 if and only if n=2k.(3)
Ifs>2, then the space form Nn+1c is a sphere Sn+1c and s=3,4 or 6.(a)
Ifs=3, then either n=6 and M has distinct principal curvatures(10)κ1=λ,κ2=λ−31+3λ,κ3=λ+31−3λ,with multiplicitiesm1=m2=m3=2, or M is minimal and has distinct principal curvatures(11)κ1=3,κ2=0,κ3=−3,with multiplicitiesm1=m2=m3=1,4 or 8 (n=3,12, or 24).(b)
Ifs=4, then m1=m3, m2=m4,(12)κ1=λ,κ2=λ−1λ+1,κ3=−1λ,κ4=−λ+1λ−1,andA=λ−1/λ is the positive solution of the equation(13)m1−5m13−m127m2−10−2m1m2m2−3+4m22A6+4m1−m13+m22m2+4+2m1m25m2−9+2m124m2+1A4−16m2m13−m2−2m22+2m1m24m2−9+2m125m2+2A2+64m22m12m2−2+5m2−2m22+m1m27m2−6=0,with multiplicitiesm1,m2=2,2, or m1,m2=4,5, or 2φm1−1k=m1+m2+1 for some integer k. Here φx is a number of integers y with 1≤y≤x and y≡0,1,2,4 mod 8.(c)
Ifs=6, then either n=12 and M has distinct principal curvatures(14)κ1=λ,κ2=3λ−1λ+3,κ3=λ−31+3λ,κ4=−1λ,κ5=λ+31−3λ,κ6=1+3λ3−λ,with multiplicitiesm1=⋯=m6=2, or n=6 and M is minimal and has distinct principal curvatures(15)κ1=2+3,κ2=1,κ3=2−3,κ4=−2+3,κ5=−1,κ6=−2+3,with multiplicitiesm1=⋯=m6=1.Theorem 3.
Ifx:Mn⟶Nn+1c is a compact extremal hypersurface without boundary in the space form Nn+1c, then(16)∫Mρ2nc+n−22H2−ρ2dμ≤0.
In particular, if(17)0≤ρ2≤nc+n−22H2,then either ρ2≡0 or ρ2≡nc+n−2/2H2. If ρ2≡nc+n−2/2H2, then either n=2k or(18)k=n2±12n2H2n−42n2H2+16c.
Ifn=2k=4 and H≠0, then M is the hypersurface S2coth2θ−1×ℍ2tanh2θ−1 for c=−1, or S2r×ℝ2 for c=0, or S2r×S21−r2 for c=1. If n=2k and n≠4, or n=2k=4 and H=0, we have that Nn+1c=Sn+1 and M is the hypersurface Sk1/2×Sk1/2.


## 2. Preliminaries

Ifx:Mn⟶Nn+1c is a hypersurface in Nn+1c, choose a unit basis e1,…,en,en+1 such that e1,…,en forms a local orthonormal frame tangent to M with dual frame ω1,…,ωn and en+1 is normal to M. Let d be the differential operator in M and the structure equations as follows:(19)dx=∑iθiei,dei=∑jθijej+∑jhijθjen+1−θix,den+1=−∑i,jhijθjei,where we use the ranges of indices(20)1≤A,B,C,…≤n+1,1≤i,j,k,…≤n.The covariant derivatives ofhij are defined by(21)∑khijkθk=dMhij+∑khkjθik+∑khikθjk,∑lhijklθl=dMhijk+∑lhljkθil+∑lhilkθjl+∑lhijlθkl.We need the following lemmas. Related lemmas and calculations can also refer to [5, 6, 9–11].Lemma 1.
(see [11]). If x:Mn⟶Nn+1c, we have the following Codazzi equation and Gauss equation:(22)hijk=hikj,Rijkl=cδikδjl−δilδjk+hikhjl−hilhjk.Proof.
Refer to Lemma 2.1 in [11] with p=1.Lemma 2.
(see [11]). If x:Mn⟶Nn+1c, the following Ricci identities hold:(23)hijkl−hijlk=∑mhmjRmikl+∑mhimRmjkl.
More precisely,(24)hijkl−hijlk=cδilhkj−δikhlj+cδjlhik−δjkhli+∑mhmjhmkhil−hmjhmlhik+∑mhimhmkhjl−himhmlhjk.Proof.
Refer to Lemma 2.2 in [11] with p=1.Lemma 3.
(see [12]). Suppose that x:Mn⟶Nn+1c is a hypersurface in the space form Nn+1c. Consider a one-parameter family xt=X⋅,t:Mn×−ε,ε⟶Nn+1c with x0=x. Denoted by ξ and en+1 the deformation vector field dxt/dt⌋t=0 and the unit normal field in Nn+1c, respectively. Set η=ξ,en+1, where , is the metric on Nn+1c. For any smooth function f of n variables,(25)ddtt=0∫MfS1,…,Sndμ=∫Mη−S1fS1,…,Sn+∑r=1nSrS1−r+1Sr+1DrfS1,…,Sn+∑i,j,r=1nDrfS1,…,Sn,ijTr−1ij+∑r=1nDrfS1,…,Snn−r+1Sr−1dμ,where Sr is the r-th elementary symmetric function of the eigenvalues κ1,…,κn of hij=κiδij, DrfS1,…,Sn=∂f/∂Sr, DrfS1,…,Sn,ij is the covariant derivative, and Trij, r=0,1,…,n−1, are defined by(26)T0ij=δij,Tr+1ij=Sr+1δij−∑kTrikhkj.
More precisely,(27)S0=1,S1=κ1+⋯+κn,…,Sn=κ1⋯κn.Proof.
Refer to Theorem A in [12].


## 3. The Euler-Lagrange Equation for Extremal Hypersurface

In this section, we will use Lemmas 1–4 to derive the Euler-Lagrange equation for extremal hypersurface in the space formNn+1c. More precisely, we have the following theorem.Theorem 4.
Ifx:Mn⟶Nn+1c is an n-dimensional compact hypersurface without boundary in the space form Nn+1c, then M is an extremal hypersurface of Fx given by (6) if and only if(28)n−1ΔH+∑i,j,kh˜ijh˜ikh˜kj−n−42ρ2H=0,where ΔH=∑iHii.Proof.
ChoosefS1,…,Sn=n−1S12−2nS2 in Lemma 3. By a direct calculation, we can easily obtain that(29)S1=nH,S2=12S12−S,S3=13∑iκi3−S1S+S1S2.
SinceS12=κ1+⋯+κn2=n2H2 and S12−2S2=κ12+⋯+κn2=S, we have(30)fS1,…,Sn=nS12−2S2−S12=nS−nH2=nρ2,D1f=2n−1S1,D2f=−2n,Drf=0,r≥3.
From (25), we can obtain(31)nddtt=0∫Mρ2dμ=ddtt=0∫Mn−1S12−2nS2dμ=∫Mηn−1S13−4n−1S1S2+6nS3+2n−1ΔS1dμ.
Then, we have the Euler-Lagrange equation ofFx as follows:(32)n−1S13−4n−1S1S2+6nS3+2n−1ΔS1=0.
By using (29) and S=ρ2+nH2, we can conclude that(33)−nn+2ρ2H−2n2H3+2n∑iκi3+2nn−1ΔH=0.
This implies that(34)n−1ΔH−n+22ρ2H−nH3+∑iκi3=0.
Sinceh˜ij=κ˜iδij and h˜ij=hij−Hδij, we have ∑iκ˜i3=∑iκi3−3∑iκi2H+3∑iκiH2−nH3. By using ∑iκi2=S=ρ2+nH2 and ∑iκi=nH, we can obtain ∑iκi3=∑iκ˜i3+3ρ2H+nH3. Then the Euler-Lagrange equation of Fx can be written by(35)n−1ΔH−n−42ρ2H+∑iκ˜i3=0.
Now, by using∑i,j,kh˜ijh˜ikh˜kj=∑iκ˜i3, we have that(36)n−1ΔH−n−42ρ2H+∑i,j,kh˜ijh˜ikh˜kj=0.
In other words,M is an extremal hypersurface of Fx if and only if(37)n−1ΔH−n−42ρ2H+∑i,j,kh˜ijh˜ikh˜kj=0.
Hence, Theorem4 holds.


## 4. The Classification of Isoparametric Extremal Hypersurfaces in the Space Form

In this section we shall classify the isoparametric extrema hypersurfaces ofFx in the space form Nn+1c. We need Lemma 4; Lemma 4 is a comprehensive arrangement of many theorems.Lemma 4.
(see [5, 13–19]). Suppose that Nn+1c is a space form and x:Mn⟶Nn+1c is an isoparametric hypersurface. Assume that M has distinct principal curvatures κ1>κ2>⋯>κs with multiplicities m1,m2,…,ms.(1)
IfNn+1c is the Euclidean space ℝn+1, then s≤2 (see [13, 14]).(2)
IfNn+1c is a hyperbolic space ℍn+1c, then s≤2 (see [15]).(3)
IfNn+1c is a sphere Sn+1c, then s=1,2,3,4, or 6 and the following holds (see [5, 16–19]):(a)
Ifs=1, then ρ2=0.(b)
Ifs=2, then M=Skr×Sn−k1−r2.(c)
Ifs=3, then the three multiplicities mi are the same and mi=2k, k=0,1,2,3.(d)
Ifs=4, then m1=m3 and m2=m4. Moreover, either m1,m2=2,2 or 4,5, or 2φm1−1k=m1+m2+1 for some integer k. Here φx is a number of integers y with 1≤y≤x and y≡0,1,2,4 mod 8.(e)
Ifs=6, the six multiplicities mi are the same and mi=1 or 2.(f)
There exists an angleθ, 0<θ<π/s, such that κt=cotθ+t−1/sπ, t=1,…,s.Proof.
Refer to [5, 13–19].Lemma 5.
Ifx:Mn⟶Nn+1c is isoparametric and extrema, then(38)∑iκ˜i3−n−42ρ2H=0,where κ˜i=h˜ijδij.Proof.
IfM is isoparametric, then H is constant. Equation (38) is an immediate consequence of Theorem 4 and the fact H is constant.Lemma 6.
(see [20]). If x:Mn⟶Nn+1c is an n-dimensional complete connected hypersurface in Nn+1c and M has two distinct principal curvatures, assume that n≥3, ρ2 is constant, and the principal curvatures λ and μ of M have multiplicities k and n−k, respectively.(1)
If1<k<n−1, then M is isometric to one of the following: Skcoth2θ−1×ℍn−ktanh2θ−1 for c=−1, or Skr×ℝn−k for c=0, or Skr×Sn−k1−r2 for c=1, where θ>0 and(39)λ1=⋯=λk=cothθ,λk+1=⋯=λn=tanhθ.(2)
If one of the two distinct principal curvatures is simple and the sectional curvature ofM is nonnegative, then M is isometric to one of the following: S1coth2θ−1×ℍn−1tanh2θ−1 for c=−1, or S1r×ℝn−1 for c=0, or S1r×Sn−11−r2 for c=1.Proof.
Refer to Theorem 1.1 in [20].
Ifx:Mn⟶Nn+1c is an isoparametric extremal hypersurface with distinct principal curvatures κ1>κ2>⋯>κs. The multiplicity of κi is denoted by mi, i=1,2,…,s. Then we have the following propositions.Proposition 1.
Ifs=1, Mn has only one distinct principal curvature, then ρ2=0, and M is total umbilical.Proof.
Ifs=1, the principal curvatures κ1=⋯=κn=λ. Choose λ˜=λ−H; since(40)nλ˜=∑iλ˜=∑iλ−H=∑iλ−nH,by using ∑iλ=nH, we have λ˜=0. This implies that ρ2=nλ˜2=0, and M is total umbilical.Proposition 2.
Ifs=2, assume that the two distinct principal curvatures κ1=λ and κ2=μ with multiplicity k and n−k, respectively, 1≤k≤n−1, and then either ρ2=0, or n=2k, or(41)ρ2=nkn−kn−424n−2k2H2.(a)
Ifn=2k=4, then M is the hypersurface S2coth2θ−1×ℍ2tanh2θ−1 for c=−1, or S2r×ℝ2 for c=0, or S2r×S21−r2 for c=1, where λ˜=λ−H, μ˜=μ−H, and θ>0.(b)
Ifn=2k and n≠4, then Nn+1c=Sn+1, M=Sk1/2×Sk1/2, and ρ2=n.Proof.
Ifs=2, the principal curvatures λ and μ have multiplicities k and n−k, respectively, let λ˜=λ−H, and μ˜=μ−H. Then, we have(42)H=kλ+n−kμn,λ˜=n−knλ−μ,μ˜=−knλ−μ,(43)ρ2=kλ˜2+n−kμ˜2=kn−knλ−μ2.
From equation (38), we have(44)kλ˜3+n−kμ˜3=kn−kn−2kn2λ−μ3=n−42ρ2H.
Ifn=4 or H=0 then n=2k. If n≠4 and n≠2k, from (42) and (43), we can get(45)kλ˜3+n−kμ˜3=kn−k3n3λ−μ3−n−k−k3n3λ−μ3=kn−kn2λ−μ3n−2k,n−42ρ2H=n−42kn−knλ−μ2kλ+n−kμn=kn−kn2λ−μ2n−42kλ+n−kμ.
From (44), we have(46)kn−kn2λ−μ3n−2k=kn−kn2λ−μ2n−42kλ+n−kμ.
This implies that(47)λ−μn−2k=n−42kλ+n−kμ.
Hence, we can obtain(48)k−2λ=−n−k−2μ.
Sincen≠4 (k≠2), we have(49)λ=−n−k−2k−2μ.
Substitute into (42); we have(50)H=kλ+n−kμn=−2n−2kμnk−2=2n−2kλnn−k−2.
Substitute (43); we have(51)ρ2=kn−knn−42k−22μ2=kn−knn−42n−k−22λ2=nkn−kn−424n−2k2H2.(a)
Ifn=4, then n=2k. By using Lemma 6, M=S2coth2θ−1×ℍ2tanh2θ−1 for c=−1, or S2r×ℝ2 for c=0, or S2r×S21−r2 for c=1, where θ>0.(b)
Ifn=2k and k≠2, from (44), H=0. By using Lemma 6, M=Skcoth2θ−1×ℍktanh2θ−1 for c=−1, or Skr×ℝk for c=0, or Skr×Sk1−r2 for c=1, where θ>0. Since H=0, if c=−1, we have λ=cothθ, μ=tanhθ, H=cothθ+tanhθ/2=0. Then cothθ=−tanhθ. This equation has no solution, the contradiction shows that M≠Skcoth2θ−1×ℍn−ktanh2θ−1. If c=0, from λ=−μ and H=0, then these two principal curvatures are the same. This contradiction shows that M≠Skr×ℝk. If c=1, we have(52)λ1=⋯=λk=1−r2r,λk+1=⋯=λn=−r1−r2.
SinceH=0, we can obtain r=1/2. Hence, M=Sk1/2×Sk1/2 and ρ2=n.Proposition 3.
Ifs>2, then the space form Nn+1c is a sphere Sn+1c and s=3,4 or 6.(a)
Ifs=3, then either n=6 and M has distinct principal curvatures(53)κ1=λ,κ2=λ−31+3λ,κ3=λ+31−3λ,with multiplicitiesm1=m2=m3=2, or M is minimal and has distinct principal curvatures(54)κ1=3,κ2=0,κ3=−3,with multiplicitiesm1=m2=m3=1,4 or 8 (n=3,12, or 24).(b)
Ifs=4, then m1=m3, m2=m4,(55)κ1=λ,κ2=λ−1λ+1,κ3=−1λ,κ4=−λ+1λ−1,andA=λ−1/λ is the positive solution of the equation(56)m1−5m13−m127m2−10−2m1m2m2−3+4m22A6+4m1−m13+m22m2+4+2m1m25m2−9+2m124m2+1A4−16m2m13−m2−2m22+2m1m24m2−9+2m125m2+2A2+64m22m12m2−2+5m2−2m22+m1m27m2−6=0,with multiplicitiesm1,m2=2,2, or m1,m2=4,5, or 2φm1−1k=m1+m2+1 for some integer k. Here φx is a number of integers y with 1≤y≤x and y≡0,1,2,4 mod 8.(c)
Ifs=6, then either n=12 and M has distinct principal curvatures(57)κ1=λ,κ2=3λ−1λ+3,κ3=λ−31+3λ,κ4=−1λ,κ5=λ+31−3λ,κ6=1+3λ3−λ,with multiplicitiesm1=⋯=m6=2, or n=6, and M is minimal and has distinct principal curvatures(58)κ1=2+3,κ2=1,κ3=2−3,κ4=−2+3,κ5=−1,κ6=−2+3,with multiplicitiesm1=⋯=m6=1.Proof.
Ifs>2, from Lemma 4, we have that the space form Nn+1c is a sphere Sn+1c and s=3,4 or 6.(a)
Ifs=3, from Lemma 4, m1=m2=m3=m,n=3m, m=1,2,4,8, and(59)κ1=cotθ=λ,κ2=cotθ+π3=λ−31+3λ,κ3=cotθ+2π3=λ+31−3λ.Then, we have(60)H=κ1+κ2+κ33=λλ2−33λ2−1,(61)κ˜1=2λ1+λ23λ2−1,κ˜2=3−λ1+λ23λ2−1,κ˜3=−3+λ1+λ23λ2−1,whereκ˜i=κi−H. By a direct calculation, we can obtain(62)ρ2=6m1+λ233λ2−12,(63)∑iκ˜i3=6mλλ2−31+λ233λ2−13.From equation (38) in Lemma 5,(64)0=∑iκ˜i3−n−42ρ2H=6−3m26mλλ2−31+λ233λ2−13.Then, eitherm=2 or λ=3. If m=2, then n=6 and M has distinct principal curvatures:(65)κ1=λ,κ2=λ−31+3λ,κ3=λ+31−3λ,with multiplicitiesm1=m2=m3=2. If λ=3, then n=3,12, or 24 and M has distinct principal curvatures κ1=3, κ2=0, and κ3=−3.(b)
Ifs=4, from Lemma 4, m1=m3, m2=m4, and(66)κ1=cotθ=λ,κ2=cotθ+π4=λ−1λ+1,κ3=cotθ+2π4=−1λ,κ4=cotθ+3π4=−λ+1λ−1.Let(67)A=λ−1λ,B=λ−1λ+1−λ+1λ−1.Sincen=2m1+m2 and AB=−4, we have(68)H=m1A+m2B2m1+m2=m1A2−4m22m1+m2A,(69)ρ2=m15m1+2m2A4+4m12−4m1m2+m22A2+16m22m1+5m22m1+m2A2,(70)∑iκ˜i3=A2+42m1m2m1+2m2A2−42m1+m22m1+m22A3.From Lemma5, ∑iκ˜i3−n−4/2ρ2H=0; we can obtain(71)m1−5m13−m127m2−10−2m1m2m2−3+4m22A6+4m1−m13+m22m2+4+2m1m25m2−9+2m124m2+1A4−16m2m13−m2−2m22+2m1m24m2−9+2m125m2+2A2+64m22m12m2−2+5m2−2m22+m1m27m2−6=0.Ifm1,m2=2,2, then A=λ−1/λ is positive solution of A6−12A4+48A2−64=0. This implies that A=2,(72)κ1=1+2,κ2=2−1,κ3=−2−1,κ4=−1+2,H=0, and ρ2=24=3n. If m1,m2=4,5, the A=λ−1/λ is positive solution of 175A6−1473A4+6465A2−21020=0.(c)
Ifs=6, from Lemma 4, m1=m2=⋯=m6=m=1 or 2, and(73)κ1=cotθ=λ,κ2=cotθ+π6=3λ−1λ+3,κ3=cotθ+2π6=λ−31+3λ,κ4=cotθ+3π6=−1λ,κ5=cotθ+4π6=λ+31−3λ,κ6=cotθ+5π6=1+3λ3−λ.Then, we can obtain(74)H=λ−1λ+1λ2−4λ+1λ2+4λ+12λλ2−33λ2−1,(75)ρ2=15m1+λ262λ2λ2−323λ2−12,wherem=1 or 2. From Lemma 5, ∑iκ˜i3−n−4/2ρ2H=0; we can obtain(76)1512−nm1+λ26λ−1λ+1λ2−4λ+1λ2+4λ+18λ3λ2−333λ2−13=0.Then eitherm=2, n=12, and M has distinct principal curvatures(77)κ1=λ,κ2=3λ−1λ+3,κ3=λ−31+3λ,κ4=−1λ,κ5=λ+31−3λ,κ6=1+3λ3−λ,orm=1, n=6, and M is minimal with distinct principal curvatures(78)κ1=2+3,κ2=1,κ3=2−3,κ4=−2+3,κ5=−1,κ6=−2+3.This completes the proof of Proposition3.
Now, combining Propositions 1–3, the following classification theorem of isoparametric extrema hypersurfaces in a space form can be obtained.Theorem 5.
Suppose thatx:Mn⟶Nn+1c is an isoparametric extremal hypersurface with distinct principal curvatures κ1>κ2>⋯>κs. The multiplicity of κi is denoted by mi, i=1,2,…,s. Then the following properties hold.(1)
Ifs=1, then ρ2=0. In other words, M is total umbilical.(2)
Ifs=2, assume that the two distinct principal curvatures κ1=λ and κ2=μ with multiplicity k and n−k respectively, 1≤k≤n−1, and then either ρ2=0, or n=2k, or(79)ρ2=nkn−kn−424n−2k2H2.(a)
Ifn=2k=4, then M is the hypersurface S2coth2θ−1×ℍ2tanh2θ−1 for c=−1, or S2r×ℝ2 for c=0, or S2r×S21−r2 for c=1, where λ˜=λ−H, μ˜=μ−H, and θ>0.(b)
Ifn=2k and n≠4, then Nn+1c=Sn+1, M=Sk1/2×Sk1/2, and ρ2=n.(c)
Ifn≠4, then H=0 if and only if n=2k.(3)
Ifs>2, then the space form Nn+1c is a sphere Sn+1c and s=3,4, or 6.(a)
Ifs=3, then either n=6 and M has distinct principal curvatures(80)κ1=λ,κ2=λ−31+3λ,κ3=λ+31−3λ,with multiplicitiesm1=m2=m3=2, or M is minimal and has distinct principal curvatures(81)κ1=3,κ2=0,κ3=−3,with multiplicitiesm1=m2=m3=1,4, or 8 (n=3,12, or 24).(b)
Ifs=4, then m1=m3, m2=m4,(82)κ1=λ,κ2=λ−1λ+1,κ3=−1λ,κ4=−λ+1λ−1,andA=λ−1/λ is the positive solution of the equation(83)m1−5m13−m127m2−10−2m1m2m2−3+4m22A6+4m1−m13+m22m2+4+2m1m25m2−9+2m124m2+1A4−16m2m13−m2−2m22+2m1m24m2−9+2m125m2+2A2+64m22m12m2−2+5m2−2m22+m1m27m2−6=0,with multiplicitiesm1,m2=2,2, or m1,m2=4,5, or 2φm1−1k=m1+m2+1 for some integer k. Here φx is a number of integers y with 1≤y≤x and y≡0,1,2,4, mod 8.(c)
Ifs=6, then either n=12 and M has distinct principal curvatures(84)κ1=λ,κ2=3λ−1λ+3,κ3=λ−31+3λ,κ4=−1λ,κ5=λ+31−3λ,κ6=1+3λ3−λ,with multiplicitiesm1=⋯=m6=2, or n=6, and M is minimal and has distinct principal curvatures(85)κ1=2+3,κ2=1,κ3=2−3,κ4=−2+3,κ5=−1,κ6=−2+3,with multiplicitiesm1=⋯=m6=1.Proof.
It is a combining consequence of Propositions1–3.
The classification theorem of isoparametric extremal hypersurfaces is too complicated. For the subsequent discussion, we have the following corollary.Corollary 1.
Suppose thatx:Mn⟶Nn+1c is an isoparametric extremal hypersurface with distinct principal curvatures κ1>κ2>⋯>κs. The multiplicity of κi is denoted by mi, i=1,2,…,s. Assume that(86)ρ2=nc+n−22H2.
Then,M has at most two distinct principal curvatures, i.e., s=1 or 2.(1)
Ifs=1, then ρ2=0.(2)
Ifs=2, assume that the two distinct principal curvatures κ1=λ and κ2=μ with multiplicity k and n−k respectively, 1≤k≤n−1, and then either ρ2=0 or n=2k, or(87)k=n2±12n2H2n−42n2H2+16c.(a)
Ifn=2k=4 and H≠0, then M is the hypersurface S2coth2θ−1×ℍ2tanh2θ−1 for c=−1, or S2r×ℝ2 for c=0, or S2r×S21−r2 for c=1.(b)
Eithern=2k and n≠4, or n=2k=4 and H=0, we have that Nn+1c is not a hyperbolic space, ρ2=0 for Nn+1c=ℝn+1, and ρ2=n for Nn+1c=Sn+1. If Nn+1c=Sn+1, M is the hypersurface Sk1/2×Sk1/2.Proof.
By using Theorem5, we can easily check this corollary.(1)
Ifs=1, then ρ2=0.(2)
Ifs=2, the principal curvatures λ and μ have multiplicities k and n−k, respectively; let λ˜=λ−H and μ˜=μ−H. Then either ρ2=0, or n=2k, or(88)ρ2=nkn−kn−424n−2k2H2.(a)
Ifn=2k=4 and H≠0, from Theorem 5, M is the hypersurface S2coth2θ−1×ℍ2tanh2θ−1 for c=−1, or S2r×ℝ2 for c=0, or S2r×S21−r2 for c=1. If c=−1, then we have λ=cothθ, μ=tanhhθ, H=cothθ+tanhθ/2, and ρ2=cothθ−tanhθ2. If c=0, then λ=r, H=r/2, and ρ2=r. If c=1, then(89)λ=1−r2r,μ=−r1−r2,H=1−2r22r1−r2,ρ2=1r21−r2.(b)
Eithern=2k and n≠4, or n=2k=4 and H=0; we have that ρ2=2kλ˜2=2kμ˜2=2kλ2=2kμ2. From (86), if Nn+1c=ℍn+1, then c=−1 and ρ2=−n<0. This contradiction shows that Nn+1c is not a hyperbolic space. If Nn+1c=ℝn+1, then c=0 and ρ2=0. If Nn+1c=Sn+1, then c=1 and ρ2=n. Then we have ρ2=n=2kλ2=nλ2. This implies that λ=1, μ=−1, and M is the hypersurface Sk1/2×Sk1/2.Ifρ2≠0,n≠4, n≠2k, and H≠0, by using Theorem 5 and equation (60), we have(90)ρ2=nkn−kn−424n−2k2H2=nc+n−22H2.Then,k is the positive integer solution of the equation(91)n2H2+16ck2−nn2H2+16ck+2n2n−2H2+2c=0.SinceH≠0, if n2H2+16c=0, we have c≠0 and n−2H2+2c=0. Then(92)H2=−16cn2=−2cn−2.This implies thatn=4. The contradiction shows that n2H2+16c≠0. Hence, k is a positive integer and(93)k=n2±12n2H2n−42n2H2+16c.(3)
Ifs>2, from Theorem 5, the space form Nn+1c is a sphere Sn+1c and s=3,4, or 6. Then we can conclude that c=1 and(94)ρ2=n1+n−22H2.Now, we shall show thats≠3, s≠4, and s≠6.(a)
Assume thats=3; from Theorem 5, either m=2, n=6, and M has distinct principal curvatures(95)κ1=λ,κ2=λ−31+3λ,κ3=λ+31−3λ,with multiplicitiesm1=m2=m3=2, or M is a Cartan minimal hypersurface with distinct principal curvatures(96)κ1=3,κ2=0,κ3=−3,and multiplicitiesm1=m2=m3=1,4, or 8 (n=3,12, or 24). If M is a Cartan minimal hypersurface, then ρ2=6m1=2n. It contradicts equation (94); ρ2=n. This contradiction shows that M is not a Cartan minimal hypersurface.Ifn=6, then(97)H=λλ2−33λ2−1,ρ2=121+λ233λ2−12.By using equation (62),(98)ρ2=62λ6−3λ4+12λ2+13λ2−12=121+λ233λ2−12.We haveλ=3 and the principal curvatures are(99)κ1=λ=3,κ2=0,κ3=−3.Hence,M is a Cartan minimal hypersurface with ρ2=2n. This contradiction shows that s≠3.(b)
Ifs=4, from Theorem 5, m1=m3, m2=m4, and(100)κ1=λ,κ2=λ−1λ+1,κ3=−1λ,κ4=−λ+1λ−1.Let(101)A=λ−1λ,B=λ−1λ+1−λ+1λ−1.Sincen=2m1+m2 and AB=−4, we have(102)H=m1A+m2B2m1+m2=m1A2−4m22m1+m2A,ρ2=m15m1+2m2A4+4m12−4m1m2+m22A2+16m22m1+5m22m1+m2A2.From equation (94),(103)ρ2=2m1+m21+m1A2−4m22m1+m2−14m1+m22A2.Combine these two equations; we haveA=λ−1/λ is positive solution of(104)m1m12+m1m2−6−2m2A4−8m1m2m1+m2−4A2+16m2m1m2−2+m2m2−6=0.By a direct calculation, we have(105)A2=4m1m2m1+m2−4T1−42m1m2T1+T2+2m1m2m1+m2−4T1,whereT1=−6m12+m13−2m1m2+m12m2 and T2=−6m22+m23−2m1m2+m1m22. From Theorem 5 or Proposition 3, A=λ−1/λ is the positive solution of equation (56). But substituting (105) into equation (56), the equation does not hold. This contradiction shows that s≠4.(c)
Assume thats=6; then m1=m2=⋯=m6=m=1 or 2. Either n=12 and M has distinct principal curvatures(106)κ1=λ,κ2=3λ−1λ+3,κ3=λ−31+3λ,κ4=−1λ,κ5=λ+31−3λ,κ6=1+3λ3−λ,with multiplicitiesm1=⋯=m6=2, or n=6, and M is minimal and has distinct principal curvatures(107)κ1=2+3,κ2=1,κ3=2−3,κ4=−2+3,κ5=−1,κ6=−2+3,with multiplicitiesm1=⋯=m6=1. If n=12, we have(108)H=λ−1λ+1λ2−4λ+1λ2+4λ+12λλ2−33λ2−1,ρ2=151+λ26λ2λ2−323λ2−12,whereλλ2−33λ2−1≠0. By using equation (94),(109)ρ2=12c+5λ−12λ+121−4λ+λ221+4λ+λ224λ2λ2−323λ2−12.Combine these two equations; we haveλ43λ2−14λ2−34c−5=0. This contradiction shows that n≠12. If n=6, M is a minimal hypersurface and ρ2=30m=5n. From equation (94), we have ρ2=n. This contradiction shows that s≠6.
Ifx:Mn⟶Nn+1c is a minimal isoparametric extrema hypersurface in the space form Nn+1c, we have Corollary 2.Corollary 2.
Suppose thatx:Mn⟶Nn+1c is a minimal isoparametric extrema hypersurface with distinct principal curvatures κ1>κ2>⋯>κs. The multiplicity of κi is denoted by mi, i=1,2,…,s.(1)
Ifs=1, then ρ2=0.(2)
Ifs=2, assume that the two distinct principal curvatures κ1=λ and κ2=μ with multiplicity k and n−k, respectively, 1≤k≤n−1, and then either ρ2=0 or n=2k. In the latter case, Nn+1c=Sn+1, ρ2=n, and M=Sk1/2×Sk1/2.(3)
Ifs>2, then the space form Nn+1c is a sphere Sn+1c and s=3,4, or 6.(a)
Ifs=3, then M is a Cartan minimal hypersurface with ρ2=2n.(b)
Ifs=4, then m1=m3, m2=m4. Moreover, ρ2=3n.(c)
Ifs=6, then m1=m2=⋯=m6=1 or 2. Moreover, ρ2=5n.Proof.
SinceM is minimal, H=0; we prove this corollary by using Theorem 5.(1)
By using Theorem5, if s=1, then ρ2=0 and M is totally umbilical.(2)
Ifs=2, since H=0, from Theorem 5, either ρ2=0 or n=2k. If n=2k, then ρ2=n and M=Sk1/2×Sk1/2.(3)
Ifs>2, from Theorem 5, the space form Nn+1c is a sphere Sn+1c and s=3,4, or 6.(a)
Ifs=3, from (60)(110)H=λλ2−33λ2−1=0.We haveλ=3 and the principal curvatures(111)κ1=λ=3,κ2=0,κ3=−3.From (62), ρ2=2n.(b)
Ifs=4, from Theorem 5, m1=m3, m2=m4. Since M is minimal, by using equation (68)(112)H=m1A2−4m22m1+m2A=0,whereA=λ−1/λ. Then we have A2=−4m2/m1 and m1λ4−2m1+2m2λ2+m1=0. This implies that(113)λ2=m1+2m2+2m2m1+m2m1=m2m1+m1+m2m12.By using Lemma5 and equations (69) and (70), we can conclude that(114)m1−m2m12+m1m2+m1m2m1m1+m24m12+2m1m2+2m1m2m1m1+m2=0.Then we have that eitherm1=m2 or m12+2m1m2+2m1m2m1m1+m2=0. Substituting these solutions into equation (69),(115)ρ2=6m2m12+m1m2+m1m2m1m1+m22m12+2m1m2+2m1m2m1m1+m2m12m1m1m2+2m2m1m2+m1m1+m22,then eitherρ2=0 or m1=m2. If m1=m2, by a direct calculation of the above equation, we have that ρ2=12m1=3n.(d)
Ifs=6, then m1=m2=⋯=m6=m=1 or 2. Since M is minimal, from (74),(116)H=λ−1λ+1λ2−4λ+1λ2+4λ+12λλ2−33λ2−1=0.Then we can conclude thatλ=2+3 and the principal curvatures(117)κ1=2+3,κ2=1,κ3=2−3,κ4=−2+3,κ5=−1,κ6=−2+3.From (75), we have ρ2=30m=5n.


## 5. Integral Inequality for Compact Extremal Hypersurfaces in the Space Form

LetNn+1c be a space form and x:M⟶Nn+1c is a compact hypersurface; from the structure equations and Lemmas 1 and 2, we have the following lemmas.Lemma 7.
Ifx:Mn⟶Nn+1c is a compact hypersurface in Nn+1c, then(118)12Δρ2=∑i,j,kh˜ijk2+n∑i,jh˜ijHij+ρ2nc+H2−ρ2+n∑i,j,kHh˜ijh˜ikh˜kj.Proof.
From Lemmas1 and 2, by a direct calculation, we have(119)∑i,jhijΔhij=∑i,j,khijhijkk=n∑i,jhijHij+∑i,j,k,lhijhilRlkjk+∑i,j,k,lhijhlkRlijk.
Sinceh˜ij=hij−Hδij, we can obtain(120)∑i,jh˜ijΔh˜ij=n∑i,jh˜ijHij+∑i,j,k,lh˜ijh˜ilRlkjk+∑i,j,k,lh˜ijh˜lkRlijk.
By using equation (23),(121)∑i,j,k,lh˜ijh˜ilRlkjk+∑i,j,k,lh˜ijh˜lkRlijk=nc+H2ρ2+n∑i,j,kHh˜ijh˜ikh˜kj−ρ4.
Then,(122)12Δρ2=∑i,j,kh˜ijk2+∑i,jh˜ijΔh˜ij=∑i,j,kh˜ijk2+n∑i,jh˜ijHij+ρ2nc+H2−ρ2+n∑i,j,kHh˜ijh˜ikh˜kj.
We can conclude that Lemma7 holds.Lemma 8.
Ifx:Mn⟶Nn+1c is a compact extrema hypersurface in Nn+1c, then(123)12Δρ2=∑i,j,kh˜ijk2+n∑i,jh˜ijHij+ρ2nc+n−22H2−ρ2−nn−1HΔH.Proof.
By using Theorem4, if M is an extremal hypersurface,(124)nH∑i,j,kh˜ijh˜ikh˜kj=−nn−1HΔH+nn−42ρ2H2.
From Lemma7,(125)12Δρ2=∑i,j,kh˜ijk2+n∑i,jh˜ijHij+ρ2nc+n−22H2−ρ2−nn−1HΔH.
Hence, Lemma8 holds.
For the following estimate, also see [4, 11].Lemma 9.
LetM be a compact hypersurface in the space form Nn+1c; then we have the following estimates:(126)∇h2≥3n2n+2∇H2,∇h˜2≥2nn−1n+2∇H2,where ∇h2=∑i,j,khijk2, ∇h˜2=∑i,j,kh˜ijk2, ∇H2=∑kHk2, and Hk=∇kH. Equality holds if and only if all the functions hij are constant for all i,j.Proof.
Following the method of Lemma 4.1 in [4], construct the tensor as(127)Fijk=hijk−nn+2Hiδjk+Hjδik+Hkδij.
Then, we can obtain(128)F2=∑i,j,kFijk2=∇h2−3n2n+2∇H2=∇h˜2−2nn−1n+2∇H2≥0.
Equality holds if and only ifhij and H are constant for all i,j.
By using Lemmas 7–9, we have the following integral inequality for compact extremal hypersurfaces inNn+1c.Theorem 6.
Ifx:Mn⟶Nn+1c is a compact extremal hypersurface without boundary in the space form Nn+1c, then(129)∫Mρ2nc+n−22H2−ρ2dμ≤0.
In particular, if(130)0≤ρ2≤nc+n−22H2,then either ρ2≡0 or ρ2≡nc+n−2/2H2. If ρ2≡nc+n−2/2H2, then either n=2k or(131)k=n2±12n2H2n−42n2H2+16c.
Ifn=2k=4 and H≠0, then M is the hypersurface S2coth2θ−1×ℍ2tanh2θ−1 for c=−1, or S2r×ℝ2 for c=0, or S2r×S21−r2 for c=1. If n=2k and n≠4, or n=2k=4 and H=0, we have that Nn+1c=Sn+1 and M is the hypersurface Sk1/2×Sk1/2.Proof.
From Lemma8, we have(132)∫M12Δρ2dμ=∫M∑i,j,kh˜ijk2+n∑i,jh˜ijHij+ρ2nc+n−22H2−ρ2−nn−1HΔHdμ.
SinceM is compact without boundary, by Stokes’ theorem and ∑ih˜iiα=0,(133)0=∫M∑i,j,kh˜ijk2+ρ2nc+n−22H2−ρ2dμ≥∫M2nn−1n+2∇H2+ρ2nc+n−22H2−ρ2dμ≥∫Mρ2nc+n−22H2−ρ2dμ,where we used Lemma 8, ∇H2≥0, and(134)∫Mn∑i,jh˜ijHij−nn−1HΔHdμ=∫M−n∑i,jh˜ijjHi+nn−1∇⊥H2dμ=∫M−n∑α,i,jh˜ijj+Hiδjj−HjδijHi+nn−1∇⊥H2dμ=∫M−nn−1∇⊥H2+nn−1∇⊥H2dμ=0.
Then, we can obtain the integral inequality(135)∫Mρ2nc+n−22H2−ρ2dμ≤0.
In particular, if(136)0≤ρ2≤nc+n−22H2,then either ρ2=0 (i.e., M is totally umbilical) or ρ2=c+n−2/2H2. If ρ2≠0, by using Corollary 1, when n=4 and H≠0, then M is the hypersurface S2coth2θ−1×ℍ2tanh2θ−1 for c=−1, or S2r×ℝ2 for c=0, or S2r×S21−r2 for c=1. In other cases, Nn+1c=Sn+1 and M is the hypersurface Sk1/2×Sk1/2.


## 6. Comparison of Integral Inequalities for Compact Extremal Hypersurfaces

Guo and Li studied a compact extremal submanifold without boundary in the unit sphereSn+p, and they obtained an integral inequality. This integral inequality is described in the compact extremal hypersurface as the following theorem.Theorem 7.
(see [9]). If x:Mn⟶Sn+1 is a compact extremal hypersurface without boundary in a sphere, then(137)∫Mρ2n−ρ2dμ≤0.
In particular, if(138)0≤ρ2≤n,then either ρ2≡0 or ρ2≡n. If ρ2≡n, then n=2k and M=Sk1/2×Sk1/2.Proof.
Refer to Theorem 1.3 in [9].
The integral inequality,(139)∫Mρ2nc+n−22H2−ρ2dμ≤0,is better and more useful than the integral inequality(140)∫Mρ2n−ρ2dμ≤0.
When isoparametric extremal hypersurfaces are with(141)n≤ρ2≤n1+n−22H2,
the integral inequality,(142)∫Mρ2n−ρ2dμ≤0,tells us nothing. But the integral inequality,(143)∫Mρ2nc+n−22H2−ρ2dμ≤0,tells us either ρ2=0 (i.e., M is totally umbilical) or ρ2=c+n−2/2H2. In the latter case, M is the hypersurface S2coth2θ−1×ℍ2tanh2θ−1, or S2r×ℝ2, or S2r×S21−r2, or Sk1/2×Sk1/2. So, the sharp estimate in (130) is meaningful, if M is not minimal and(144)n≤ρ2≤n1+n−22H2.
There are many special compact extremal hypersurfaces that Guo and Li have not considered. The following examples state that there exist many isoparametric extremal hypersurfaces which is different formSk1/2×Sk1/2 and has the properties ρ2=nc+n−2/2H2 with H≠0. Example 1 states that if n=4, even if n=2k, there exist isoparametric extremal hypersurfaces that satisfy the sharp estimate which is different form Sk1/2×Sk1/2. Example 2 states that there exist isoparametric extremal hypersurfaces that satisfy the sharp estimate with n≠2k.Example 1.
Suppose thatr is a constant and r≠1/2. The tori(145)M=S2r×S21−r2,has two district principal curvatures and constant mean curvature(146)κ1=κ2=1−r2r,κ3=κ4=−r1−r2,H=1−2r22r1−r2≠0.
Let(147)κ˜1=κ˜2=κ1−H=12r1−r2,κ˜3=κ˜4=κ3−H=−12r1−r2.
Then,(148)ρ2=1r21−r2,since(149)∑iκ˜i3=0=n−42ρ2H.
We can obtain thatM=S2r×S21−r2 is an extrema hypersurface in S5. Now,(150)nc+n−22H2=41+H2=41+1−4r2+4r44r21−r2=1r21−r2=ρ2.
Hence,M=S2r×S21−r2 is an extrema hypersurface in S5 with ρ2=nc+n−2/2H2 and H≠0.
SinceM=S2r×S21−r2 is an extrema hypersurface, we have the integral inequality by using Theorem 7,(151)∫Mρ2n−ρ2dμ≤0.
From Theorem6, we have the integral inequality(152)∫Mρ2nc+n−22H2−ρ2dμ≤0.
In fact,(153)∫Mρ2n−ρ2dμ<∫Mρ2nc+n−22H2−ρ2dμ≤0.
This means that the integral inequality of Theorem6 is better and more useful than the integral inequality of Theorem 7.Example 2.
The tori(154)M=S4r×S31−r2=S423×S333,has two district principal curvatures and constant mean curvature(155)κ1=⋯=κ4=22,κ5=⋯=κ7=−2,H=−27.
Let(156)κ˜1=⋯=κ˜4=κ1−H=9214,κ˜5=⋯=κ˜7=κ5−H=−627.
Then,(157)ρ2=547,since(158)∑iκ˜i3=−81249=n−42ρ2H.
We can obtain thatM=S42/3×S33/3 is an extrema hypersurface in S8. Now,(159)nc+n−22H2=71+52H2=547=ρ2.
In particular,(160)k=n2±12n2H2n−42n2H2+16c=4.
Hence,M=S42/3×S33/3 is an extrema hypersurface in S8 with ρ2=nc+n−2/2H2 and H≠0.
SinceM=S42/3×S33/3 is an extrema hypersurface, we have the integral inequality by using Theorem 7,(161)∫Mρ2n−ρ2dμ≤0.
From Theorem6, we have the integral inequality(162)∫Mρ2nc+n−22H2−ρ2dμ≤0.
The fact(163)∫Mρ2n−ρ2dμ<∫Mρ2nc+n−22H2−ρ2dμ≤0,shows that the integral inequality of Theorem 6 is better and more useful than the integral inequality of Theorem 7.


## 7. Conclusion

In this paper, we derive the Euler-Lagrange equation for extremal hypersurfaceMn in the space form Nn+1c as Theorem 1 and then classify the isoparametric extremal hypersurfaces in the space form as Theorem 2. Since the classification theorem of isoparametric extremal hypersurfaces is too complicated, we decompose Theorem 5 into Propositions 1–3. We also reduce Theorem 5 to two corollaries when ρ2=nc+n−2/2H2 and when Mn is a minimal. Finally, we obtain an integral inequality for the compact extremal hypersurfaces as Theorem 6. Comparing with the integral inequality in Theorem 7 obtained by Guo and Li, we can find that the integral inequality in Theorem 6 is better and more useful than Guo and Li’s integral inequality.



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*Source: 5881125-2022-09-16.xml*