On Chen's biharmonic conjecture for hypersurfaces in R 5
aa r X i v : . [ m a t h . DG ] J u l ON CHEN’S BIHARMONIC CONJECTURE FORHYPERSURFACES IN R YU FU, MIN-CHUN HONG, AND XIN ZHAN
Abstract.
A longstanding conjecture on biharmonic submanifolds, proposedby Chen in 1991, is that any biharmonic submanifold in a Euclidean space isminimal . In the case of a hypersurface M n in R n +1 , Chen’s conjecture wassettled in the case of n = 2 by Chen and Jiang around 1987 independently.Hasanis and Vlachos in 1995 settled Chen’s conjecture for a hypersurface with n = 3. However, the general Chen’s conjecture on a hypersurface M n remainsopen for n >
3. In this paper, we settle Chen’s conjecture for hypersurfaces in R for n = 4. Introduction
In 1983, Eells and Lemaire [15] introduced the concept of biharmonic mapsbetween Riemannian manifolds. Since then, biharmonic maps have been extensivelystudied by mathematicians. In particular, geometers investigated a special class ofbiharmonic maps named biharmonic immersions . An immersion φ : ( M n , g ) −→ ( N m , h ) is biharmonic if and only if its mean curvature vector field −→ H satisfies thefourth-order semi-linear elliptic equation (e.g. [7], [33], [34])∆ −→ H + trace R N ( dφ, −→ H ) dφ = 0 , (1.1)where ∆ is the Laplacian of M , −→ H is the mean curvature vector of the immersionand R N is the curvature tensor of N .Biharmonic submanifolds have attracted a lot of attentions and many results onbiharmonic submanifolds were obtained in past two decades (e.g. [3-5, 8, 16-30, 32,35, 38]).In the Euclidean ambient space, biharmonic submanifolds are defined via thegeometric condition ∆ −→ H = 0, or equivalently ∆ φ = 0, which was originally pro-posed by B. Y. Chen in his pioneering work of finite type theory in the middle of1980s (c.f. [9], [10]). It is easy to see from the definition that minimal submanifoldsare automatically biharmonic.B. Y. Chen [9] in 1991 proposed a well-known conjecture in the following: Chen’s conjecture : Any biharmonic submanifold in the Euclidean space R m is minimal .Although partial results on Chen’s conjecture were obtained for low dimensionsand with additional geometric conditions (e.g. [1], [28], [39], [37]), Chen’s conjectureis widely open.Since the most important biharmonic submanifolds are hypersurfaces in Eu-clidean spaces, Chen’s conjecture on hypersurfaces in Euclidean spaces is a basic Mathematics Subject Classification.
Primary 53C40, 58E20; Secondary 53C42.
Key words and phrases.
Biharmonic maps, Biharmonic submanifolds, Chen’s conjecture, one and has been investigated by many mathematicians. Let M n be a biharmonichypersurface in the Euclidean space R n +1 . Chen in 1986 and Jiang [24] in 1987independently made a pioneering contribution to prove that a biharmonic surface M in R is minimal. Dimitri´c [14] extended this result to hypersurfaces withat most two distinct principal curvatures in R n +1 . In 1995, Hasanis and Vla-chos [22] made an important progress and settled the conjecture for the case n = 3.Defever [13] in 1998 reproved the conjecture for n = 3 through a new tensorialanalysis approach. Later, Chen and Munteanu [11] confirmed Chen’s conjecturefor δ (2)-ideal and δ (3)-ideal hypersurfaces in R n +1 . The first author [17] in 2015showed that Chen’s conjecture is true for hypersurfaces with three distinct prin-cipal curvatures in R n +1 . Recently, Montaldo, Oniciuc and Ratto [31] confirmedthe conjecture for the G − invariant hypersurfaces of cohomogeneity one in R n +1 .Koiso and Urakawa [25] also verified the conjecture for generic hypersufaces withirreducible principal curvature vector fields in R n +1 . However, Chen’s conjectureremains open until now for hypersurfaces M n for n ≥ R n states that for n ≥
2, any entire solution f : R n → R of the minimal graph equation n X i,j =1 (cid:16) δ ij − f i f j |∇ f | (cid:17) f ij = 0is an affine function.The Bernstein problem was first investigated by Bernstein in 1915 for n = 2,solved by De Giorgi [12] in 1965 for n = 3 and Almgren [2] in 1966 for n = 4.Simons [40] in 1968 solved it for n ≤
7. Finally, Bombieri-De Giorgi-Giusti [6] in1969 affirmatively resolved the Bernstein problem by constructing a counterexamplethat the Bernstein problem is not true for n ≥ R n . Chen’s conjecture for biharmonic graphsstates that for n ≥
2, any entire solution f : R n → R of the biharmonic graphequations (c.f. [36]) ( ∆(∆ f ) = 0 , (∆ f k )∆ f + 2 h∇ f k , ∇ ∆ f i = 0 , k = 1 , . . . n is minimal; i.e. ∆ f = − div( ∇ f ) = 0.In contrast to the Bernstein problem, Chen’s conjecture holds true for n = 2(Chen [9], Jiang 1987 [24]) and for n = 3 (Hasanis-Vlachos, 1995 [22]). However,Chen’s conjecture for the case n ≥ n ≥ n = 4. More precisely, we prove Theorem 1.1.
Every biharmonic hypersurface in the Euclidean space R is mini-mal. The main approach for the proof of Theorem 1.1 is a continuation of the programdeveloped in [20]. However, in [20], an extra condition of constant scalar curvature is assumed. In this paper, we overcome the difficulty and remove the condition for n Chen’s biharmonic conjecture for hypersurfaces 3 n = 4 by dealing with the differential equations related to biharmonicity. By trans-ferring the biharmonic equations into a system of algebraic differential equations,we developed a new method to determine the behavior of the principal curvaturefunctions via investigating the solution of the system of algebraic differential equa-tions. This new approach in the current paper provides an interesting insight forus to understand the geometric structure of biharmonic hypersurfaces.We would like to outline our proofs here. Assume that the mean curvature H is non-constant. Through Proposition 2.1, the immersion φ : M n → R n +1 of abiharmonic hypersurface M n in R n +1 satisfies two geometric equations (2.2). Thesecond one of (2.2) tells us that grad H is an eigenvector of the Weingarten operator A with the corresponding principal curvature λ = − H . For the non-constantmean curvature H , the multiplicity of the principal curvature λ is one (see Lemma2.2). By using the Gauss and Codazzi equations, biharmonic equations reduce tothree differential equations (2.3)-(2.5) on principal curvature functions λ i . Noticingthat the equations in Lemma 2.3 are an over-determined system of differentialequations related to the principal curvatures and the coefficients of connection, wetransfer the equations (2.3)-(2.5) into five equations concerning P i =2 ( ω ii ) k in termsof λ and T (see details in Section 3). In the case of n = 4, we derive two equationsconcerning T and λ without ω ii . Utilizing the elimination method via differentialequations, we show that T is also a smooth function depending only on one variable t . Through relations between λ i and ω kij in Lemmas 4.1 and 4.2, we distinguish ω kij in two cases. For each case, we deal with different algebraic equations by acomplicated deduction and get a contradiction finally. Remark . It should be pointed out that Chen’s conjecture for hypersurfaces inthe case of n >
Acknowledgement:
The authors would like to thank Professors Bang-Yen Chenand Cezar Oniciuc for their interest and useful comments. The first author issupported by Liaoning Provincial Science and Technology Department Project(No.2020-MS-340), and Liaoning BaiQianWan Talents Program. The partial re-search of the second author was supported by the Australian Research Councilgrant (DP150101275). 2.
Preliminaries
A biharmonic map φ between an n -dimensional Riemannian manifold ( M n , g )and an m -dimensional Riemannian manifold ( N m , h ) is a critical point of the bi-energy functional E ( φ ) = 12 Z M | τ ( φ ) | dv g , where τ ( φ ) = trace ∇ d φ is the tension field of φ that vanishes for a harmonic map.The concept of biharmonic maps was introduced in 1983 by Eells and Lemaire [15] Fu, Hong and Zhan with the aim to study k − harmonic maps. The Euler-Lagrange equation associatedto the bi-energy is stated as τ ( φ ) = − ∆ τ ( φ ) − trace R N ( dφ, τ ( φ )) dφ = 0 , where τ ( φ ) is the bitension field of φ , and R N is the curvature tensor of N m (e.g. [23], [24]). Hence, φ is called a biharmonic map if its bitension field τ ( φ )vanishes identically.In a special case, an immersion φ : ( M n , g ) −→ ( N m , h ) is biharmonic if andonly if its mean curvature vector field −→ H fulfills the fourth-order semi-linear ellipticequation (1.1). It is well-known that any minimal immersion is harmonic. Thebiharmonic immersions are called proper biharmonic if they are not harmonic.Let φ : M n → R n +1 be an isometric immersion of a hypersurface M n in theEuclidean space R n +1 and let X , Y , Z be the tangent vector fields of M n . Denotethe Levi-Civita connections of M n and R n +1 by ∇ and ˜ ∇ respectively. Then theGauss and Codazzi equations are written as R ( X, Y ) Z = h AY, Z i AX − h AX, Z i AY, ( ∇ X A ) Y = ( ∇ Y A ) X, where A is the Weingarten operator, and R is the curvature tensor of M n .We recall that the mean curvature vector field −→ H can be defined by −→ H = 1 n trace h, (2.1)where h is the second fundamental form. For the mean curvature H , choose ξ tobe the unit normal vector field of M n satisfying −→ H = Hξ .The sufficient and necessary conditions for a hypersurface M n to be biharmonicare given (see [8], [10]), which are the basic characterizations of a biharmonic hy-persurface in R n +1 in the following: Proposition 2.1.
The immersion φ : M n → R n +1 of a hypersurface M n in theEuclidean space R n +1 is biharmonic if and only if H and A satisfy (2.2) ( ∆ H + H trace A = 0 , A ∇ H + nH ∇ H = 0 , where the Laplacian operator ∆ applied on a function f is given by∆ f = − div( ∇ f ) = − n X i =1 h∇ e i ( ∇ f ) , e i i = − n X i =1 ( e i e i − ∇ e i e i ) f. We collect two results from [20] concerning principal curvatures for biharmonichypersurfaces, whose proofs are standard (see also [25]).
Lemma 2.2. ([20])
Let M n be an orientable biharmonic hypersurface with non-constant mean curvature in R n +1 and assume the mean curvature H is non-constant.Then the multiplicity of the principal curvature λ (= − nH/ is one, i.e. λ j = λ for ≤ j ≤ n . Using the Gauss and Codazzi equations, biharmonic equations (2.2) can be sum-marized into a system of 2 n − n Chen’s biharmonic conjecture for hypersurfaces 5 Lemma 2.3. ([20],[25])
Assume that H is non-constant. Then the smooth real-valued principal curvature functions λ i and the coefficients of connection ω ii ( i =2 , . . . , n ) satisfy the following differential equations e e ( λ ) = e ( λ ) (cid:16) n X i =2 ω ii (cid:17) + λ S, (2.3) e ( λ i ) = λ i ω ii − λ ω ii , (2.4) e ( ω ii ) = ( ω ii ) + λ λ i , (2.5) where λ = − nH/ , e = ∇ H/ |∇ H | and S is the squared length of the secondfundamental form h of M . Some key lemmas
From now on, we study the biharmonicity of a hypersurface M n in a Euclideanspace R n +1 with n = 4. Since M with at most three distinct principal curvatureseverywhere in a Euclidean space R is minimal (see [17]), we work only on the casethat the connected component of M A has different principal curvatures. In general,the set M A of all points of M , at which the number of distinct eigenvalues ofthe Weingarten operator A (i.e. the principal curvatures) is locally a constant, isopen and dense in M . Meanwhile, on each connected component, the principalcurvature functions of A are always smooth. Assume that, on a component, themean curvature H is non-constant. Then there exists a neighborhood U p of p suchthat ∇ H = 0.It follows from (2.2) that ∇ H is an eigenvector of the Weingarten operator A with the corresponding principal curvature − H . Choosing locally e such that e isparallel to ∇ H , and with respect to some suitable orthonormal frame { e , e , e , e } ,the Weingarten operator A of M is given by A = diag( λ , λ , λ , λ ) , where λ i are the principal curvatures and λ = − H . Therefore, it follows from(2.1) that P i =1 λ i = 4 H , and hence λ + λ + λ = − λ . (3.1)Denote by S the squared length of the second fundamental form h of M . It followsfrom (3.1) that S is given by S = trace A = X i =1 λ i = X i =2 λ i + λ . (3.2)As ∇ H = P i =1 e i ( H ) e i and e is parallel to the direction of ∇ H , we have that e ( H ) = 0 , e i ( H ) = 0 , ≤ i ≤ e ( λ ) = 0 , e i ( λ ) = 0 , ≤ i ≤ . (3.3)Setting ∇ e i e j = P k =1 ω kij e k (1 ≤ i, j ≤ ∇ e k h e i , e i i = 0 and ∇ e k h e i , e j i = 0 ( i = j ) yields respec-tively that ω iki = 0 , ω jki + ω ikj = 0 , i = j. Fu, Hong and Zhan
It follows from using the Codazzi equation that e i ( λ j ) = ( λ i − λ j ) ω jji , ( λ i − λ j ) ω jki = ( λ k − λ j ) ω jik for distinct i, j, k .Due to (3.3), we consider an integral curve of e passing through p = γ ( t ) as γ ( t ), t ∈ I . It is easy to show that there exists a local chart ( U ; t = x , x , x , x )around p , such that λ ( t, x , x , x ) = λ ( t ) on the whole neighborhood of p .Set f k = ( ω ) k +( ω ) k +( ω ) k , for k = 1 , . . . ,
5. In the following, an interestingsystem of algebraic equations will be derived.
Lemma 3.1.
With the notions f k , the following two relations hold f − f f + 3 f + 8 f f − f = 0 , (3.4) f − f f + 5 f f + 5 f f − f = 0 . (3.5) Proof.
It is easy to check that(3.6) f − f = (cid:0) X i =2 ω ii (cid:1) − X i =2 ( ω ii ) = 2( ω ω + ω ω + ω ω )and(3.7) f − f = (cid:0) X i =2 ω ii (cid:1) − X i =2 ( ω ii ) = 2 (cid:8) ( ω ω ) + ( ω ω ) + ( ω ω ) (cid:9) . Combining (3.6) with (3.7) gives( f − f ) − f − f )(3.8) = 4( ω ω + ω ω + ω ω ) − (cid:8) ( ω ω ) + ( ω ω ) + ( ω ω ) (cid:9) = 8 (cid:8) ( ω ) ω ω + ω ( ω ) ω + ω ω ( ω ) (cid:9) = 8 f ω ω ω . Similarly, we have f − f = ( ω + ω + ω ) − (cid:8) ( ω ) + ( ω ) + ( ω ) (cid:9) (3.9) = 3 (cid:8) ( ω ) ω + ( ω ) ω + ( ω ) ω + ( ω ) ω + ( ω ) ω + ( ω ) ω (cid:9) + 6 ω ω ω = 3 X i =2 ( ω ii ) ( f − ω ii ) + 6 ω ω ω = 3 f f − f + 6 ω ω ω . Eliminating ω ω ω from (3.8) and (3.9), we get (3.4). n Chen’s biharmonic conjecture for hypersurfaces 7 A direct computation shows that f f = (cid:16) X i =2 ω ii (cid:17)(cid:16) X i =2 ( ω ii ) (cid:17) = X i =2 ( ω ii ) + ω n ( ω ) + ( ω ) o + ω n ( ω ) + ( ω ) o + ω n ( ω ) + ( ω ) o = f + ω n(cid:2) ( ω ) + ( ω ) (cid:3) − ω ) ( ω ) o + ω n(cid:2) ( ω ) + ( ω ) (cid:3) − ω ) ( ω ) o + ω n(cid:2) ( ω ) + ( ω ) (cid:3) − ω ) ( ω ) o = f + X i =2 ω ii (cid:16) f − ( ω ii ) (cid:17) − ω ω ω (cid:0) ω ω + ω ω + ω ω (cid:1) , which together with (3.6) yields(3.10) f f = 2 f + f f − f f − ( f − f ) ω ω ω . Eliminating ω ω ω from (3.9) and (3.10) again, one gets6 f f = 12 f + 6 f f − f f − ( f − f )( f − f f + 2 f ) . (3.11)Moreover, eliminating the terms of f from (3.4) and (3.11) gives (3.5). (cid:3) For simplicity, we write λ = λ ( t ), f = T , T ′ = e ( T ), T ′′ = e e ( T ), T ′′′ = e e e ( T ) and T ′′′′ = e e e e ( T ). Then the functions f , · · · , f are expressed inthe terms of λ and T in the following: Lemma 3.2. f , f , f , f , and f can be written as f = T,f = T ′ + 3 λ ,f = T ′′ − λ T + 6 λλ ′ ,f = T ′′′ − λ T ′ − λλ ′ T + 2 λ ′ + 4 λλ ′′ − λ ,f = T ′′′′ − λ T ′′ − λλ ′ T ′ − (13 λλ ′′ + λ ′ − λ ) T +2 λλ ′′′ + λ ′ λ ′′ − λ λ ′ . (3.12) Proof.
Since e ( λ ) = 0, λ is not constant. From (2.3), one has f = e e ( λ ) − λSe ( λ ) = λ ′′ λ ′ − λλ ′ S =: T. (3.13)Taking the sum of i from 2 to 4 in (2.5)-(2.4) respectively ang using (3.1), we have f =3 λ + e ( f ) = T ′ + 3 λ , (3.14) g := X i =2 λ i ω ii (3.15) = λT − e ( λ ) = λT − λ ′ . Fu, Hong and Zhan
Multiplying by ω ii both sides of equation (2.5), we have12 e (cid:0) ( ω ii ) (cid:1) = ( ω ii ) + λλ i ω ii . Taking the sum of i in the above equation gives f = 12 e ( f ) − λg = 12 T ′′ − λ T + 6 λλ ′ . (3.16)Differentiating (3.15) along e , using (2.4) and (2.5) we have e ( g ) = 2 X i =2 λ i (cid:0) ω ii (cid:1) + λ X i =2 λ i − λ X i =2 (cid:0) ω ii (cid:1) . (3.17)Hence, it follows from (3.1), (3.2) and (3.13) that g := X i =2 λ i (cid:0) ω ii (cid:1) = 12 (cid:8) e ( g ) − λ (cid:0) S − λ (cid:1) + λf (cid:9) = 12 (cid:8) e ( g ) − λ ′′ + λ ′ T + λ + λf (cid:9) . Using (3.14) and (3.15), the above expression reduces to g = λT ′ + λ ′ T − λ ′′ + 2 λ . (3.18)Multiplying ( ω ii ) on both sides of equation (2.5), we have13 e (cid:0) ( ω ii ) (cid:1) = ( ω ii ) + λλ i ( ω ii ) . Taking the sum of i from 2 to 4 in the above equation, we obtain f = 13 e ( f ) − λg (3.19) = 16 T ′′′ − λ T ′ − λλ ′ T + 2 λ ′ + 4 λλ ′′ − λ . Multiplying equation (2.4) by λ i gives λ i ω ii = 12 e ( λ i ) + λλ i ω ii , which together with (3.2) yields g : = X i =2 λ i ω ii = 12 e ( S − λ ) + λg (3.20) = 12 (cid:16) λ ′′ − λ ′ Tλ − λ (cid:17) ′ + λg = − λ ′ λ T ′ + (cid:16) λ − λ ′′ λ − λ ′ λ (cid:17) T − λλ ′ + λ ′′′ λ − λ ′′ λ ′ λ . Differentiating (3.18) with respect to e and using (2.4) and (2.5), we have e ( g ) = 3 X i =2 λ i (cid:0) ω ii (cid:1) − λ X i =2 (cid:0) ω ii (cid:1) + 2 λ X i =2 λ i ω ii , n Chen’s biharmonic conjecture for hypersurfaces 9 which leads to g := X i =2 λ i (cid:0) ω ii (cid:1) (3.21) = 13 (cid:0) e ( g ) + λf − λg (cid:1) = 12 λT ′′ + λ ′ T ′ + 13 (2 λ ′′ − λ − λ ′ λ ) T − λ ′′′ + 203 λ λ ′ + λ ′′ λ ′ λ . Multiplying ( ω ii ) on both sides of equation (2.5), we have14 e (cid:0) ( ω ii ) (cid:1) = ( ω ii ) + λλ i ( ω ii ) . After taking the sum of i in the above equation, we have f = 14 e ( f ) − λg (3.22) = 124 T ′′′′ − λ T ′′ − λλ ′ T ′ −
112 (13 λλ ′′ + λ ′ − λ ) T + 2 λλ ′′′ + 53 λ ′ λ ′′ − λ λ ′ . (cid:3) Lemma 3.3.
Let M be an orientable biharmonic hypersurface with simple distinctprincipal curvatures in R . Then the function T depends only on the variable t .Proof. In the case T = 0 in a region, it follows from (3.13) that S depends only on t . In the following, we assume that T = 0 .Substituting (3.12) into (3.4) and (3.5) yields − T ′′′ + 4 T T ′′ + 3 T ′ + ( − T + 26 λ ) T ′ + ( T − λ T (3.23) + 58 λλ ′ T ) + 39 λ − λλ ′′ − λ ′ = 0 , − T ′′′′ + 10 T ′ T ′′ + (10 T + 50 λ ) T ′′ − (20 T + 20 λ T (3.24) − λλ ′ ) T ′ + (4 T − λ T + 120 λλ ′ T − λ T + 26 λλ ′′ T + 2 λ ′ T ) + 568 λ λ ′ − λλ ′′′ − λ ′ λ ′′ = 0 . Differentiating (3.23) with respect to e , we have − T ′′′′ + 4 T T ′′′ + 10 T ′ T ′′ + ( − T + 26 λ ) T ′′ − T T ′ (3.25) + (4 T − λ T + 110 λλ ′ ) T ′ + ( − λλ ′ T + 58 λλ ′′ T + 58 λ ′ T ) + 156 λ λ ′ − λ ′ λ ′′ − λλ ′′′ = 0 . Eliminating the terms on T ′′′′ in (3.24)-(3.25), we get4 T T ′′′ − (16 T + 24 λ ) T ′′ − T T ′ + (24 T − λ T − λλ ′ ) T ′ (3.26) + ( − T + 80 λ T − λλ ′ T + 84 λ T + 32 λλ ′′ T + 56 λ ′ T ) − λ λ ′ − λ ′ λ ′′ + 24 λλ ′′′ = 0 . Moreover, we can eliminate the terms on T ′′′ of (3.23) and (3.26). Then we obtain − λ T ′′ + (18 λ T − λλ ′ ) T ′ (3.27) + ( − λ T + 15 λλ ′ T + 60 λ T − λλ ′′ T + 2 λ ′ T ) − λ λ ′ − λ ′ λ ′′ + 6 λλ ′′′ = 0 . Differentiating the above equation along e , one sees − λ T ′′′ + (18 λ T − λλ ′ ) T ′′ + 18 λ T ′ (3.28) + ( − λ T + 66 λλ ′ T + 60 λ − λ ′ − λλ ′′ ) T ′ + ( − λλ ′ T + 15 λ ′ T + 15 λλ ′′ T + 240 λ λ ′ T − λ ′ λ ′′ T − λλ ′′′ T ) − λ λ ′ − λ λ ′′ − λ ′′ + 4 λ ′ λ ′′′ + 6 λλ ′′′′ = 0 . Note that both equations (3.23) and (3.26) have a non-zero term of T , but (3.28)does not involve any term of T . Therefore, we conclude that (3.28) are entirely dif-ferent from equations (3.23) and (3.26), which are third-order differential equationswith respect to T .Next, we eliminate the terms of T ′′′ , T ′′ , T ′ and derive a non-trivial equation of T .To eliminate the terms of T ′′′ from (3.23) and (3.28), we have(6 λ T + 27 λλ ′ ) T ′′ + ( − λ T − λλ ′ T (3.29) + 96 λ + 13 λ ′ + 31 λλ ′′ ) T ′ + (6 λ T + 12 λλ ′ T − λ T − λ ′ T − λλ ′′ T + 108 λ λ ′ T + 12 λ ′ λ ′′ T + 16 λλ ′′′ T ) + 234 λ + 237 λ λ ′ − λ λ ′′ + 2 λ ′′ − λ ′ λ ′′′ − λλ ′′′′ = 0 . Note that equations (3.29) and (3.27) are entirely different. Then, multiplying2 λT + 9 λ ′ to (3.27) and 2 λ to (3.29), we eliminate the terms of T ′′ to obtain a T ′ − a T + a T + a = 0 , (3.30)where a =62 λ λ ′′ − λλ ′ + 192 λ ,a =44 λ λ ′′′ − λλ ′ λ ′′ + 550 λ λ ′ + 18 λ ′ ,a = − λ λ ′′′′ + 46 λλ ′ λ ′′′ − λ λ ′′ + 4 λλ ′′ − λ ′ λ ′′ − λ λ ′ + 468 λ . If a = 0 in a region, then (3.30) becomes an equation of Ta T + a = 0 . If a = 0, we have a = 0 as well since T = 0. Hence the conclusion followsobviously. If a = 0, then the above equation yields that a = 0. We will derive acontradiction. Let us consider the equations a = 0 and a = 0:62 λλ ′′ − λ ′ + 192 λ = 0 , (3.31) 44 λ λ ′′′ − λλ ′ λ ′′ + 550 λ λ ′ + 18 λ ′ = 0 . (3.32) n Chen’s biharmonic conjecture for hypersurfaces 11 It is easy to see the two equations (3.31) and (3.32) are entirely different. Differen-tiating (3.31) gives 31 λλ ′′′ − λ ′ λ ′′ + 384 λ λ ′ = 0 , which together with (3.32) reduces to − λλ ′′ + 279 λ ′ + 77 λ = 0 . (3.33)After eliminating the terms of λ ′′ between (3.31) and (3.33), one has − λ ′ + 22163 λ = 0 . (3.34)Differentiating the above equation leads to − λ ′′ + 22163 λ = 0 . (3.35)Substituting (3.34) and (3.35) into (3.31), we find1066545669 λ = 0 , which is a contradiction since λ is non-constant. Hence we have a = 0.Differentiating (3.30) along e again, we deduce (cid:0) λ − λλ ′ + 62 λ λ ′′ (cid:1) T ′′ − (cid:0) λ T + 124 λ λ ′′ T (3.36) − λλ ′ T − λ λ ′ − λ λ ′′′ + 218 λλ ′ λ ′′ + 91 λ ′ (cid:1) T ′ − (cid:0) λ λ ′ + 62 λ λ ′′′ − λλ ′ λ ′′ − λ ′ (cid:1) T + (cid:0) λ λ ′′ + 2200 λ λ ′ + 44 λ λ ′′′′ − λλ ′ λ ′′′ − λλ ′′ − λ ′ λ ′′ (cid:1) T + 3276 λ λ ′ − λ λ ′ − λ λ ′′′ − λ λ ′ λ ′′ − λ λ ′′′′′ + 22 λλ ′ λ ′′′′ + 54 λλ ′′ λ ′′′ + 28 λ ′ λ ′′′ − λ ′ λ ′′ = 0 . Multiplying 192 λ − λ ′ + 62 λλ ′′ and 6 λ on the both sides of equations (3.27)and (3.36) respectively, we thus get (cid:0) T λ + 372 T λ λ ′′ − T λ λ ′ + 6180 λ λ ′ (3.37) + 636 λ λ ′′′ − λ λ ′ λ ′′ + 1089 λλ ′ (cid:1) T ′ − (cid:0) λ + 372 λ λ ′′ − λ λ ′ (cid:1) T − (cid:0) λ λ ′ + 372 λ λ ′′′ − λ λ ′ λ ′′ + 981 λλ ′ (cid:1) T + (cid:0) λ + 3948 λ λ ′′ + 7044 λ λ ′ + 264 λ λ ′′′′ − λ λ ′ λ ′′′ − λ λ ′′ + 1448 λλ ′ λ ′′ − λ ′ (cid:1) T − λ λ ′ + 660 λ λ ′′′ − λ λ ′ λ ′′ + 3073 λ λ ′ − λ λ ′′′′′ + 132 λ λ ′ λ ′′′′ + 696 λ λ ′′ λ ′′′ − λλ ′ λ ′′′ − λλ ′ λ ′′ + 218 λ ′ λ ′′ = 0 . Since (3.37) has a non-zero term of T , (3.37) is different from (3.30). Combining(3.30) with (3.37), we obtain a non-trivial polynomial equation concerning T withthe coefficients depending only on the variable t (3.38) b T + b = 0 , where b =209088 λ + 174078 λ λ ′′ − λ λ ′ + 8064 λ λ ′′′′ − λ λ ′ λ ′′′ − λ λ ′′ + 302157 λ λ ′ λ ′′ − λ λ ′ + 2604 λ λ ′′ λ ′′′′ − λ λ ′′′ − λ λ ′ λ ′′′′ + 18354 λ λ ′ λ ′′ λ ′′′ − λ λ ′′ − λ λ ′ λ ′′′ + 1350 λ λ ′ λ ′′ − λλ ′ λ ′′ + 520 λ ′ ,b = − λ λ ′ − λ λ ′′′ − λ λ ′ λ ′′ + 361623 λ λ ′ − λ λ ′′′′′ + 12438 λ λ ′ λ ′′′′ + 28338 λ λ ′′ λ ′′′ − λ λ ′ λ ′′′ − λ λ ′ λ ′′ + 120509 λ λ ′ λ ′′ − λ λ ′′ λ ′′′′′ + 954 λ λ ′′′ λ ′′′′ + 19795 λ λ ′ + 981 λ λ ′ λ ′′′′′ − λ λ ′ λ ′′ λ ′′′′ − λ λ ′ λ ′′′ + 5076 λ λ ′′ λ ′′′ − λ λ ′ λ ′′′′ + 1050 λ λ ′ λ ′′ λ ′′′ − λ λ ′ λ ′′ + 360 λλ ′ λ ′′′ + 415 λλ ′ λ ′′ − λ ′ λ ′′ . If b = 0 and b = 0 in a region, similarly, after eliminating the terms of λ ′′′′′ , λ ′′′′ , λ ′′′ , λ ′′ , λ ′ item by item, one gets a non-trivial polynomial equation of λ withconstant coefficients, which implies that λ is a constant. This is a contradiction.Therefore, we conclude that T depends only on the variable t . (cid:3) Lemma 3.4.
Let M be an orientable biharmonic hypersurface with non-constantmean curvature in R . Then e i ( λ j ) = 0 for ≤ i, j ≤ , that is, all principalcurvatures λ i depend only on the variable t .Proof. If the number m of distinct principal curvatures is 3 or 2, it has been provedin [17], so we only need to consider the case of four distinct principal curvatures.Since λ i = λ j at any point in U p , it follows from (2.4) and (2.5) that there exists aneighborhood V ⊂ U p such that ω ii = ω jj for i = j .According to Lemmas 3.3, (3.12) implies that f k for k = 1 , . . . , t , that is, e i ( f k ) = 0 for 2 ≤ i ≤
4. Hence, differentiating both sides ofequations f k = P i =2 ( ω ii ) k for k = 1 , , e i (2 ≤ i ≤ e i ( ω ) + e i ( ω ) + e i ( ω ) = 0 ,ω e i ( ω ) + ω e i ( ω ) + ω e i ( ω ) = 0 , ( ω ) e i ( ω ) + ( ω ) e i ( ω ) + ( ω ) e i ( ω ) = 0 . (3.39)Since ω , ω , ω are mutually different at V and the determinant of the coefficientmatrix of (3.39) is the Vandermonde determinant with order 3, it follows that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ω ω ω ( ω ) ( ω ) ( ω ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( ω − ω )( ω − ω )( ω − ω ) = 0 . According to Cramer’s rule in linear algebra, one gets e i ( ω ) = e i ( ω ) = e i ( ω ) = 0 . Furthermore, for j = 2 , , e i e ( ω jj ) − e e i ( ω jj ) = [ e i , e ]( ω jj ) = X l =2 ( ω li − ω l i ) e l ( ω jj ) , n Chen’s biharmonic conjecture for hypersurfaces 13 we get e i e ( ω jj ) = 0 . Differentiating (2.5) with respect to e i , taking into account the above equation and e i ( ω jj ) = 0, we derive e i ( λ j ) = 0for any 1 ≤ j ≤ ≤ i ≤
4. Therefore, we complete a proof of Lemma 3.4. (cid:3)
Remark . Note that our method developed in Lemmas 3.2-3.4 works also for n = 2 and n = 3, which were proved with different approaches in [24], [22], or [13].4. Proof of the main theorem
We first recall some relations concerning the coefficients of connection and princi-pal curvature functions verified in [20] or [19]. Actually, they are the four dimensionversion of Lemmas 3.5 and 3.6 in [20] while Lemma 3.4 holds.
Lemma 4.1. ([20])
For three distinct principal curvatures λ i , λ j and λ k (2 ≤ i, j, k ≤ , we have the following relations: ω ( λ − λ ) = ω ( λ − λ ) = ω ( λ − λ ) , (4.1) ω ω + ω ω + ω ω = 0 , (4.2) ω ( ω − ω ) = ω ( ω − ω ) = ω ( ω − ω ) . (4.3) Lemma 4.2. ([20])
Under the assumptions as above, we have ω ω − ω ω = − λ λ , (4.4) ω ω − ω ω = − λ λ , (4.5) ω ω − ω ω = − λ λ . (4.6) The proof of Theorem 1.1.
If the mean curvature H is constant, the first equa-tion of (2.2) reduces H = 0 immediately. Assume that the mean curvature H isnon-constant on U p . According to Lemma 4.1, we distinguish it into the followingcases A and B. Case A . ω = 0, ω = 0 and ω = 0.In this case, equations (4.1) and (4.3) reduce to ω − ω λ − λ = ω − ω λ − λ = ω − ω λ − λ , and hence there exist two smooth functions α and β depending on t such that ω ii = αλ i + β (4.7)for i = 2 , ,
4. Differentiating with respect to e on both sides of equation (4.7),using (2.4) and (2.5) we get e ( α ) = λ ( α + 1) + αβ, (4.8) e ( β ) = β ( αλ + β ) . (4.9)Taking a sum on i in (4.7) and using (3.1), one has X i =2 ω ii = − αλ + 3 β. (4.10) Taking into account (4.2), equations (4.4), (4.5) and (4.6) lead to ω ω + ω ω + ω ω = − λ λ − λ λ − λ λ , which together with (4.7) further reduces to(1 + α )( λ λ + λ λ + λ λ ) + 2 αβ ( λ + λ + λ ) + 3 β = 0 . (4.11)Since S − λ = P i =2 λ i and − λ = P i =2 λ i , it follows from (4.11) that(1 + α ) ×
12 (10 λ − S ) − αβλ + 3 β = 0and hence (1 + α ) S = 10(1 + α ) λ − αβλ + 6 β . (4.12)Differentiating (4.12) with respect to e and using (4.8)-(4.9), one has(1 + α ) e ( S ) =4 (cid:8) α ) λ − αβ (cid:9) e ( λ )(4.13) + 2(10 αλ − αS − βλ ) (cid:8) λ ( α + 1) + αβ (cid:9) + 12 β ( β − αλ )( αλ + β ) . Moreover, differentiating (3.1) with respect to e and using (2.4), we get − e ( λ ) = X i =2 ( λ i − λ ) ω ii . Using (4.7) and (3.1), the above equation reduces to − e ( λ ) = X i =2 ( λ i − λ )( αλ i + β )(4.14) = X i =2 (cid:8) αλ i + ( β − λ α ) λ i − βλ (cid:9) = α ( S − λ ) − β − λ α ) λ − βλ = α (2 λ + S ) − βλ . By using (4.12) and (4.14), we eliminate S to get e ( λ ) = −
11 + α (4 λ α − λ α β + 2 αβ + 4 λ α − λ β ) . (4.15)On the other hand, differentiating (4.14) with respect to e , it follows from (4.8)and (4.9) that − e e ( λ ) =(4 αλ − β ) e ( λ ) + αe ( S )(4.16) + (2 λ + S ) (cid:8) λ ( α + 1) + αβ (cid:9) − βλ ( αλ + β ) . Substituting (4.10) into (2.3) gives e e ( λ ) = 3( − αλ + β ) e ( λ ) + λ S. (4.17)Eliminating the terms of e e ( λ ) between (4.16) and (4.17) yields( − αλ + 3 β ) e ( λ ) + αe ( S ) + (2 λ + S ) (cid:8) λ ( α + 1) + αβ (cid:9) (4.18) − βλ ( αλ + β ) + 3 λ S = 0 . n Chen’s biharmonic conjecture for hypersurfaces 15 Combining (4.18) with (4.14) we may eliminate e ( λ ) and hence3 αe ( S ) =( − αλ + 3 β ) (cid:8) (2 λ + S ) α − βλ (cid:9) (4.19) − λ + S ) (cid:8) λ ( α + 1) + αβ (cid:9) + 18 βλ ( αλ + β ) − λ S. Also, combining (4.13) with (4.14), we eliminate e ( λ ) to obtain3(1 + α ) e ( S ) =4 (cid:8) α ) λ − αβ (cid:9)(cid:8) βλ − (2 λ + S ) α (cid:9) (4.20) + 6(10 αλ − αS − βλ ) (cid:8) λ ( α + 1) + αβ (cid:9) + 36 β ( β − αλ )( αλ + β ) . Eliminating the terms of e ( S ) in (4.19)-(4.20) yields (cid:8) (1 + α )(5 αλ + β ) − α β (cid:9)(cid:8) βλ − (2 λ + S ) α (cid:9) (4.21) + (22 α λ − αβλ + 2 λ + S − α S ) (cid:8) λ ( α + 1) + αβ (cid:9) + 6 β (2 αβ − α λ − λ )( αλ + β ) + 3(1 + α ) λ S = 0 . Applying (4.12) to eliminate the terms of S in (4.21), we derive (cid:8) (1 + α )(5 αλ + β ) − α β (cid:9)(cid:8) βλ − αλ (cid:9) + (22 α λ − αβλ + 2 λ ) (cid:8) λ ( α + 1) + αβ (cid:9) + 6 β (2 αβ − α λ − λ )( αλ + β )+ (cid:8) − α (5 αλ + β ) + (1 − α ) λ + 3 λ (cid:9)(cid:8) α ) λ − αβλ + 6 β (cid:9) + αβ (1 + 3 α )1 + α (cid:8) α ) λ − αβλ + 6 β (cid:9) = 0 , which reduces to 8 α λ − α βλ + 16 α β λ − α β (4.22) + 9 α λ − α βλ + 8 α β λ − αβ − α λ + 6 αβλ − β λ − λ = 0 . Differentiating (4.22) along e and using (4.8), (4.9) and (4.15), we have n λ (8 α + 9 α − α −
7) + 2 λ (6 αβ − α β − α β )+ (16 α β + 8 α β − β ) on λ α − λ α β + 2 αβ + 4 λ α − λ β o − (1 + α ) n λ ( α + 1) + αβ on λ α − λ α β + 4 α (16 λ β + 9 λ ) − α (4 β + 14 λ β ) + 2 α (8 λ β − λ ) + (6 λ β − β ) o − (1 + α )( λ αβ + β ) n − β (4 α + 2 α ) + 2 β (16 λ α + 8 λ α − λ )+ (6 λ α − λ α − λ α ) o = 0 , which leads to 24 α λ − α βλ + 188 α β λ − α β λ (4.23) + 36 α λ − α βλ + 191 α β λ − α β λ − α λ + 76 α βλ − α β λ + 15 α β λ − α λ + 123 α βλ − αβ λ + 9 β λ − αλ + 18 βλ + 28 α β + 24 α β = 0 . To simplify the notations, (4.22) and (4.23) can be respectively rewritten as P β + P β + P β + P = 0 , (4.24) P β + P β + P β + P β + P = 0 , (4.25)where P = − α − α, P = 16 α λ + 8 α λ − λ ,P = − α λ − α λ + 6 αλ , P = 8 α λ + 9 α λ − α λ − λ ,P = 28 α + 24 α , P = − α λ − α λ + 15 α λ + 9 λ ,P = 188 α λ + 191 α λ − α λ − αλ ,P = − α λ − α λ + 76 α λ + 123 α λ + 18 λ ,P = 24 α λ + 36 α λ − α λ − α λ − αλ . Multiplying (4.24) by P β and (4.25) by P , we eliminate the terms of β toobtain( P P − P P ) β + ( P P − P P ) β + ( P P − P P ) β − P P = 0 , which can be rewritten as(4.26) P β + P β + P β + P = 0 , where P = P P − P P = ( − α − α − α − α + 18 α ) λ ,P = P P − P P = (192 α + 268 α + 22 α − α − α ) λ ,P = P P − P P = ( − α − α + 62 α + 304 α + 150 α + 36 α ) λ ,P = − P P = (96 α + 192 α − α − α − α − α ) λ . Multiplying (4.24) by P and (4.26) by P , we eliminate the terms on β to get(4.27) P β + P β + P = 0 , where P = P P − P P = ( − α − α − α − α + 60 α − α ) λ ,P = P P − P P = (320 α + 1944 α + 2580 α + 788 α + 12 α + 180 α ) λ ,P = P P − P P = ( − α − α − α − α − α − α − α ) λ . n Chen’s biharmonic conjecture for hypersurfaces 17 Multiplying (4.27) by P β and using (4.24) again, one has P β + P β + P = 0 , (4.28)where P =( − α − α − α − α + 5288 α + 1480 α − α + 288 α ) λ ,P =(3712 α + 20864 α + 32176 α + 11296 α − α − α + 780 α − α ) λ ,P =( − α − α − α − α + 13320 α + 11736 α + 1456 α + 12 α + 504 α ) λ . Eliminating the terms of β in (4.27)-(4.28) gives( P P − P P ) β = − ( P P − P P ) , (4.29)where P P − P P = n α + 964608 α + 3273216 α + 6086784 α + 7013568 α + 5538720 α + 3434064 α + 1872528 α + 852864 α + 289728 α + 77904 α + 18576 α + 2592 α o λ , − ( P P − P P ) = n λ α + 842496 λ α + 1857024 λ α − λ α − λ α − λ α − λ α − λ α − λ α − λ α − λ α − λ α − λ α o λ . By substituting (4.29) into (4.27), one derives P ( P P − P P ) − P ( P P − P P )( P P − P P )(4.30) + P ( P P − P P ) = 0 , namely, n α + 12599707041792 α + 168848312500224 α + 1539102781734912 α + 10358767180578816 α + 52901614335688704 α + 208041234970705920 α + 638282602571268096 α + 1546611187674415104 α + 2991538078614835200 α + 4659369092304187392 α + 5883550040047592448 α + 6057015443197185024 α + 5112599987173082112 α + 3566610599450135040 α + 2084666842179740160 α + 1044195650378735616 α + 461769373555605504 α + 184830143553160704 α + 67312785426974208 α + 22101468989783040 α + 6548262801067008 α + 1779181049894400 α + 441651900169728 α + 96378291542016 α + 18243537610752 α + 3112242227712 α + 445229250048 α + 38268370944 α + 846526464 α o λ = 0 . Since λ = 0, the above equation is a non-trivial polynomial equation concerning α with constant coefficients. This implies that α must be a constant. It follows from(4.8) that(4.31) β = − α + 1 α λ . Substituting (4.31) into (4.9) and (4.15), we have e ( λ ) = λ (cid:16) αλ − α α λ (cid:17) = − α λ , (4.32) e ( λ ) = − α + 6 α ( α + 1) + α ) α + 4 α + α ) α α λ (4.33) = − α + 1) α λ . Combining (4.32) with (4.33) yields that 4 α + 1 = 0, which is impossible. Case B. ω = ω = ω = 0 at any point p in V .In this case, it follows from (4.4), (4.5), (4.6) that ω ω = − λ λ , (4.34) ω ω = − λ λ , (4.35) ω ω = − λ λ . (4.36)Then we divide Case B into the two subcases B.1 and B.2. Case B.1.
All principal curvatures λ , λ , λ are nonzero.It follows from (4.34), (4.35), (4.36) that all ω , ω , ω are nonzero as well.Combining (4.34) with (4.35) gives ω ω = λ λ , which together with (4.36) gives ( ω ) + λ = 0 . It follows that λ = ω = 0, which is a contradiction. n Chen’s biharmonic conjecture for hypersurfaces 19 Case B.2.
At least one of principal curvatures λ , λ , λ is zero.Without loss of generality, we assume λ = 0 at some point p on M . It followsfrom (2.4) that ω = 0 at p . Hence (3.1) becomes λ + λ = − λ . (4.37)Differentiating (4.37) with respect to e and using (2.4), we have − e ( λ ) = ( λ − λ ) ω + ( λ − λ ) ω . (4.38)Differentiating (4.38) with respect to e , we apply (2.4)-(2.5) to get − e e ( λ ) =2( λ − λ )( ω ) + 2( λ − λ )( ω ) (4.39) − e ( λ )( ω + ω ) + ( λ − λ ) λ λ + ( λ − λ ) λ λ . Furthermore, (2.3) becomes e e ( λ ) = e ( λ )( ω + ω ) + λ ( λ + λ + λ ) . (4.40)Eliminating the terms of e e ( λ ) between (4.39) and (4.40), it gives2 e ( λ )( ω + ω ) + 2( λ − λ )( ω ) + 2( λ − λ )( ω ) (4.41) + ( λ − λ ) λ λ + ( λ − λ ) λ λ + 3 λ ( λ + λ + λ ) = 0 . Multiplying ω or ω to the both sides of equation (4.38) respectively, it followsfrom using (4.34) that( λ − λ )( ω ) = − e ( λ ) ω + ( λ − λ ) λ λ , ( λ − λ )( ω ) = − e ( λ ) ω + ( λ − λ ) λ λ . Substituting the above two equation into (4.41), we obtain − e ( λ )( ω + ω ) + 2( λ − λ ) λ λ + 2( λ − λ ) λ λ (4.42) + ( λ − λ ) λ λ + ( λ − λ ) λ λ + 3 λ ( λ + λ + λ ) = 0 . Taking into account (4.37) and eliminating λ in (4.42), one has2 e ( λ )( ω + ω ) = 9 λ λ + 27 λ λ + 21 λ . (4.43)It follows from (4.38), (4.34) and (4.37) that (4.43) is simplified to2( λ − λ )( ω ) + 2( λ − λ )( ω ) = − λ λ − λ λ − λ . (4.44)Differentiating (4.43) with respect to e , it follows from using (4.2) and (4.3) that2 e e ( λ )( ω + ω ) + 2 e ( λ ) n ( ω ) + ( ω ) + λ λ + λ λ o = (9 λ + 54 λ λ + 63 λ ) e ( λ ) + (18 λ λ + 27 λ )( λ − λ ) ω , which, together with (4.40), (4.37) and (4.34), implies e ( λ ) n ω ) + 4( ω ) − λ − λ λ − λ o (4.45) + (47 λ + 3 λ λ − λ λ ) ω + (20 λ + 12 λ λ + 4 λ λ ) ω = 0 . Eliminating the terms of e ( λ ) between (4.43) and (4.45) yields(9 λ λ + 27 λ λ + 21 λ ) n ω ) + 4( ω ) − λ − λ λ − λ o + 2( ω + ω )(47 λ + 3 λ λ − λ λ ) ω + 2( ω + ω )(20 λ + 12 λ λ + 4 λ λ ) ω = 0 , which, together with (4.34), leads to2(4 λ λ + 57 λ λ + 89 λ )( ω ) + 2(22 λ λ + 66 λ λ + 62 λ )( ω ) (4.46) =(9 λ λ + 27 λ λ + 21 λ )(69 λ + 42 λ λ + 5 λ ) − λ (3 λ + λ )( − λ λ + 15 λ λ + 67 λ ) . Equations (4.34), (4.44) and (4.46) can be rewritten in the following forms: ω ω = L, (4.47) M ( ω ) + N ( ω ) = K , (4.48) M ( ω ) + N ( ω ) = K , (4.49)where L = 3 λ λ + λ ,M = 2 λ − λ , N = − λ − λ ,K = − λ λ − λ λ − λ ,M = 8 λ λ + 114 λ λ + 178 λ ,N = 44 λ λ + 132 λ λ + 124 λ ,K = 1449 λ + 2343 λ λ + 1636 λ λ + 543 λ λ + 65 λ λ . Similarly to the above cases, we eliminate the terms of ω and ω from (4.47),(4.48) and (4.49). At last, we obtain M ( M N − M N ) L + N ( M K − M K ) (4.50) − K ( M N − M N )( M K − M K ) = 0 , which reduces to − λ λ − λ λ − λ λ − λ λ − λ λ − λ λ − λ λ − λ λ − λ λ + 135723600 λ λ + 162996624 λ λ + 62868960 λ = 0 . Therefore, we get a polynomial equation concerning k = λ λ with constant coeffi-cients as follows:62868960 k + 162996624 k + 135723600 k − k − k − k − k − k − k − k − k − k = 0 . The above equation shows that k = λ λ must be a constant. Applying λ = λ k and(4.37) to (2.4), it gives ω = e ( λ ) λ − λ = e ( λ )(1 − k ) λ , (4.51) ω = e ( λ ) λ − λ = (3 k + 1) e ( λ )(4 k + 1) λ . (4.52) n Chen’s biharmonic conjecture for hypersurfaces 21 It follows from (4.51) and (4.52) that (2.3) and (4.34) become e e ( λ ) = (3 k − k − e ( λ )(4 k + 1)( k − λ + (10 k + 6 k + 2) λ k , (4.53) e ( λ ) = (1 − k )(4 k + 1) k λ . (4.54)Taking into account e ( λ ) = 0 and differentiating (4.54) yield(4.55) e e ( λ ) = 2(1 − k )(4 k + 1) k λ . Substituting (4.54)-(4.55) into (4.53) gives15 k + 6 k + 2 = 0 . Obviously, the above quadratic equation has no real root, which contradicts to ourassumption. Therefore, we complete a proof of Theorem 1.1. (cid:3)
Remark . In [21], the authors claimed that they proved the minimality of bihar-monic hypersurfaces in R . Unfortunately, there is a crucial error in their proofs.In Page 11 of [21]: Since a = − b = − c , the equation (3.43) in [21] should becorrected as − H ( λ − λ )(2 λ − λ − λ ) a + b + b − c − c = k − k . (4.56)Combining (3.41), (3.42) with (4.56) in [21], one has4 H n ( λ − λ ) − ( λ − λ ) − ( λ − λ )(2 λ − λ − λ ) o a = 0 . (4.57)However, (4.57) gives no further information due to the fact that( λ − λ ) − ( λ − λ ) − ( λ − λ )(2 λ − λ − λ ) ≡ . In fact, the expression k − k + k − k + k − k ≡ References [1] K. Akutagawa, S. Maeta, Biharmonic properly immersed submanifolds in Euclidean spaces,Geom. Dedicata 164 (2013), 351–355.[2] F. J. Almgren, Some interior regularity theorems for minimal surfaces and an extension ofBernstein’s theorem, Ann. Math., Second Series, 84 (1066), 277–292.[3] A. Balmus, S. Montaldo and C. 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Math., Second Series, 88 (1968),62–105. School of Mathematics, Dongbei University of Finance and Economics, Dalian 116025,P. R. China
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