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versão impressa ISSN 0001-3765versão On-line ISSN 1678-2690

An. Acad. Bras. Ciênc. vol.88 no.4 Rio de Janeiro out./dez. 2016  Epub 01-Dez-2016

http://dx.doi.org/10.1590/0001-3765201620150326

Mathematical Sciences

Translation Hypersurfaces with Constant Sr Curvature in the Euclidean Space

1Universidade Federal do Piauí, Campus Ininga, Centro de Ciências da Natureza, Departamento de Matemática, Bairro Ininga, 64049-550 Teresina, PI, Brazil

2Universidade Federal do Cariri, Campus Juazeiro do Norte, Centro de Ciências e Tecnologia, Bairro Cidade Universitária, 63048-080 Juazeiro do Norte, CE, Brazil

Abstract

The main goal of this paper is to present a complete description of all translation hypersurfaces with constant r -curvature Sr , in the Euclidean space n+1 , where 3rn-1 .

Key words Euclidean space; Scherk's surface; Translation hypersurfaces; r-Curvature

INTRODUCTION

It is well known that translation hypersurfaces are very important in Differential Geometry, providing an interesting class of constant mean curvature hypersurfaces and minimal hypersurfaces in a number of spaces endowed with good symmetries and even in certain applications in Microeconomics. There are many results about them, for instance, Chen et al. (2003), Dillen et al. (1991), Inoguchi et al. (2012), Lima et al. (2014), Liu (1999), López (2011), López and Moruz (2015), López and Munteanu (2012), Seo (2013) and Chen (2011), for an interesting application in Microeconomics.

Scherk (1835) obtained the following classical theorem: Let M:={(x,y,z):z=f(x)+g(y)} be a translation surface in 3 , if is minimal then it must be a plane or the Scherk surface defined by

z(x,y)=1aln|cos(ay)cos(ax)|,

where a is a nonzero constant. In a different aspect, Liu (1999) considered the translation surfaces with constant mean curvature in 3 -dimensional Euclidean space and Lorentz-Minkowski space and Inoguchi et al. (2012) characterized the minimal translation surfaces in the Heisenberg group Nil3 , and López and Munteanu, the minimal translation surfaces in Sol3 .

The concept of translation surfaces was also generalized to hypersurfaces of n+1 by Dillen et al. (1991), who obtained a classification of minimal translation hypersurfaces of the (n+1) -dimensional Euclidean space. A classification of the translation hypersurfaces with constant mean curvature in (n+1) -dimensional Euclidean space was made by Chen et al. (2003).

The absence of an affine structure in hyperbolic space does not permit to give an intrinsic concept of translation surface as in the Euclidean setting. Considering the half-space model of hyperbolic space, López (2011), introduced the concept of translation surface and presented a classification of the minimal translation surfaces. Seo (2013) has generalized the results obtained by Lopez to the case of translation hypersurfaces of the (n+1) -dimensional hyperbolic space.

Definition 1. We say that a hypersurface Mn of the Euclidean space n+1 is a translation hypersurface if it is the graph of a function given by

F(x1,,xn)=f1(x1)++fn(xn)

where (x1,,xn) are cartesian coordinates and each fi is a smooth function of one real variable for i=1,,n.

Now, let Mnn+1 be an oriented hypersurface and λ1,,λn denote the principal curvatures of Mn . For each r=1,,n , we can consider similar problems to the above ones, related with the r -th elementary symmetric polynomials, Sr , given by

Sr=1i1<<irnλi1λir

In particular, S1 is the mean curvature, S2 the scalar curvature and Sn the Gauss-Kronecker curvature, up to normalization factors. A very useful relationship involving the various Sr is given in the [Proposition 1, Caminha (2006)]. This result will play a central role along this paper.

Recently, some authors have studied the geometry of translational hypersurfaces under a condition in the Sr curvature, where r>1 . Namely, Leite (1991) gave a new example of a translation hypersurface of 4 with zero scalar curvature. Lima et al. 2014 presented a complete description of all translation hypersurfaces with zero scalar curvature in the Euclidean space n+1 and Seo 2013 proved that if M is a translation hypersurface with constant Gauss-Kronecker curvature GK in n+1 , then M is congruent to a cylinder, and hence GK=0 .

In this paper, we obtain a complete classification of translation hypersurfaces of n+1 with Sr=0 . We prove the following

Theorem 1. Let Mn(n3) be a translation hypersurface in n+1 . Then, for 2<r<n , Mn has zero Sr curvature if, and only if, it is congruent to the graph of the following functions

• F(x1,,xn)=i=1n-r+1aixi+j=n-r+2nfj(xj)+b,

on n-r+1×Jn-r+2××Jn , for certain intervals Jn-r+2,,Jn , and arbitrary smooth functions fi:Ji . Which defines, after a suitable linear change of variables, a vertical cylinder, or

• A generalized periodic Enneper hypersurface given by

F(x1,,xn)
=
i=1n-r-1aixi
+
k=n-rn-1βakln|cos(-an-ran-1σr-1(an-r,,an-1)βxn+bn)cos(akβxk+bk)|+c

on n-r-1×In-r××In , with a1,,an-r,,an-1,bn-r,,bn and c are real constants where an-r,,an-1 and σr-1(an-r,,an-1) nonzero, β=1+i=1n-r-1ai2 , Ik(n-rkn-1) are open intervals defined by the conditions |akβxk+bk|<π/2 while In is defined by |-an-ran-1σr-1(an-r,,an-1)βxn+bn|<π/2 .

Theorem 2. Any translation hypersurface in n+1(n3) with Sr constant, for 2<r<n , must have Sr=0 .

Finally, we observe that, when one considers the upper half-space model of the (n+1) -dimensional hyperbolic space n+1 , that is,

+n+1={(x1,,xn,xn+1)n+1:xn+1>0}

endowed with the hyperbolic metric ds2=1xn+12(dx12++dxn+12) then, unlike in the Euclidean setting, the coordinates x1,,xn are interchangeable, but the same does not happen with the coordinate xn+1 and, due to this observation, López 2011 and Seo 2013 considered two classes of translation hypersurfaces in n+1 :

A hypersurface Mn+1 is called a translation hypersurface of type I (respectively, type II) if it is given by an immersion X:Unn+1 satisfying

X(x1,,xn)=(x1,,xn,f1(x1)++fn(xn))

where each fi is a smooth function of a single variable. Respectively, in case of type II,

X(x1,,xn)=(x1,,xn-1,f1(x1)++fn(xn),xn)

Seo proved

Theorem 3(Theorem 3.2, Seo 2013). There is no minimal translation hypersurface of type I in n+1 .

and with respect to type II surfaces he proved

Theorem 4(Theorem 3.3, Seo 2013). Let M3 be a minimal translation surface of type II given by the parametrization X(x,z)=(x,f(x)+g(z),z) . Then the functions f and g are as follows:

f(x)
=
ax+b,
g(z)
=
1+a2cz21-c2z4𝑑z,

where a , b , and c are constants.

We emphasize that the result proved by Seo, Theorem 3.2 of Seo 2013, implies that our result (Theorem 2) is not valid in the hyperbolic space context.

PRELIMINARIES AND BASIC RESULTS

Let M¯n+1 be a connected Riemannian manifold. In the remainder of this paper, we will be concerned with isometric immersions, Ψ:MnM¯n+1 , from a connected, n -dimensional orientable Riemannian manifold, Mn , into M¯n+1 . We fix an orientation of Mn , by choosing a globally defined unit normal vector field, ξ , on M . Denote by A , the corresponding shape operator. At each pM , A restricts to a self-adjoint linear map Ap:TpMTpM . For each 1rn , let Sr:Mn be the smooth function such that Sr(p) denotes the r -th elementary symmetric function on the eigenvalues of Ap , which can be defined by the identity

det(Ap-λI)=k=0n(-1)n-kSk(p)λn-k. (1)

where S0=1 by definition. If pMn and {el} is a basis of TpM , given by eigenvectors of Ap , with corresponding eigenvalues {λl} , one immediately sees that

Sr=σr(λ1,,λn),

where σr[X1,,Xn] is the r -th elementary symmetric polynomial on X1,,Xn . Consequently,

Sr=1i1<<irnλi1λir,wherer=1,,n.

In the next result we present an expression for the curvature Sr of a translation hypersurface in the Euclidean space. This expression will play an essential role in this paper.

Proposition 1. Let F:Ωn be a smooth function, defined as F(x1,,xn)=i=1nfi(xi), where each fi is a smooth function of one real variable. Let Mn be the graphic of F, given in coordinates by

φ(x1,,xn)=i=1nxiei+F(x1,,xn)en+1. (2)

The Sr curvature of Mn is given by

Sr=1Wr+21i1<<irnnf¨i1f¨ir(1+1mnmi1irf˙m2), (3)

where the dot represents derivative with respect to the corresponding variable, that is, for each j=1,,n, one has f˙j=dfjdxj(xj)=Fxj(x1,,xn) and W2=1+|F|2

Proof. Let F be as stated in the Proposition, denote by F=i=1nFxiei the Euclidean gradient of F and , the standard Euclidean inner product. Then, we have

F=i=1nf˙iei

and the coordinate vector fields associated to the parametrization given in (2) have the following form

φxm=em+f˙men+1,m=1,,n.

Hence, the elements Gij of the metric of Mn are given by

Gij=φxi,φxj=δij+f˙if˙j,

implying that the matrix of the metric G has the following form

G=In+(F)tF,

where In is the identity matrix of order n . Note that the i -th column of G , which will be denoted by Gi , has the expression given by the column vector

Gi=ei+f˙iF. (4)

An easy calculation shows that the unitary normal vector field ξ of Mn satisfies

Wξ=en+1-F,

where W2=1+|F|2 . Thus, the second fundamental form Bij of Mn satisfies

WBij=Wξ,2φxixj=en+1-F,δijf¨ien+1=δijf¨i,

implying that the matrix of B is diagonal

B=1Wdiag(f¨1,,f¨n),

with i -th column given by the column vector

Bi=f¨iWei. (5)

If A denotes the matrix of the Weingarten mapping, then A=G-1B . In (1), changing λ by λ-1 gives

det(λA-I)=i=1n(-1)n-iSiλi.

Thus, we conclude that the expression for curvature Sr can be found by the following calculation

(-1)n-rSr=1r!drdλr|λ=0det(λA-I).

Note that

(-1)n-rdetGSr=1r!detGdrdλr|λ=0det(λA-I)=1r!drdλr|λ=0det(λB-G).

Due to the multilinearity of function det , on its n column vectors, it follows immediately that

ddλ|λ=0det[λB1-G1,,λBn-Gn]=i=1n(-1)n-1det[G1,,Bii-th term,,Gn],

drdλr|λ=0det(λB-G)=r!1i1<<irnn(-1)n-rdet[G1,,Bi1,,Bir,,Gn]

and thus

Sr=1detG1i1<<irnndet[G1,,Bi1,,Bir,,Gn]. (6)

Now, applying the expressions (4) and (5) in (6) we reach to the expression

Sr=1detGWr1i1<<irnnf¨i1f¨irdet[e1+f˙1F,,ei1,,eir,,en+f˙nF]. (7)

Calculating the determinant on the right in the equality above, we get

det[e1+f˙1F,
,ei1,,eir,,en+f˙nF]=
=
1+ii1,,irf˙idet[e1,,ei1,,Fi-th term,,eir,,en]
=
1+1inii1,,irf˙i2.

Consequently, the expression for Sr in (7) assumes the following form

Sr=1detGWr1i1<<irnnf¨i1f¨ir(1+1inii1,,irf˙i2).

Finally, using that detG=W2 we obtain the desired expression

Sr=1Wr+21i1<<irnnf¨i1f¨ir(1+1inii1,,irf˙i2).

RESULTS

In order to prove Theorem 1 we need the following lemma.

Lemma 1. Let f1,,fr be smooth functions of one real variable satisfying the differential equation

k=1rf1¨(x1)fk¨(xk)^fr¨(xr)(β+fk˙2(xk))=0, (8)

where β is a positive real constant and the big hat means an omitted term. If fi¨0, for each i=1,r then

k=1rfk(xk)=k=1r-1βakln|cos(-a1ar-1σr-2(a1,,ar-1)βxr+br)cos(akβxk+bk)|+c

where ai,bi,c, i=1,r are real constants with ai,σr-2(a1,,ar-1)0.

Since the derivatives f¨i0 it follows that f1¨(x1)fr¨(xn)0 . Thus dividing (8) by this product we get the equivalent equation:

k=1rβ+fk˙2(xk)fk¨(xk)=0,

which implies, after taking derivative with respect to xl for each l=1,r , that (β+fl˙2(xl)fl¨(xl))=0 , thus β+fl˙2(xl)fl¨(xl)=a~l for some non null constant a~l . Thus, setting al=1a~l

fl¨(xl)β+fl˙2(xl)=al for eachl=1,,r

which can be easily solved to give:

arctan(fl˙(xl)β)=alβx+bl for some constantbl

and consequently

fl(xl)=-1alβln|cos(alβxl+bl)|+cl,l=1,,r. (9)

Now, since k=1r1ak=0 it implies that 1ar=-σr-2(a1,,ar-1)a1ar-1 , from (9) it follows that

fr(xr)=σr-2(a1,,ar-1)a1ar-1βln|cos(arβxr+br)|+cr.

Consequently

k=1rfk(xk)=k=1r-11akβln|cos(-a1ar-1σr-2(a1,,ar-1)βxr+br)cos(akβxk+bk)|+c,

where c=c1++cr . ∎

With this lemma at hand we can go to the proof of Theorem 1.

Proof of the Theorem 1 From Proposition 1, we have that Mn has zero Sr curvature if, and only if,

1i1<<irnf¨i1f¨ir(1+1knk{i1,ir}f˙k2)=0. (10)

In order to ease the analysis, we divide the proof in four cases.

Case 1: Suppose f¨i(xi)=0 , i=1,,n-r+1 . In this case, we have no restrictions on the functions fn-r+2,,fn . Thus

Ψ(x1,,xn)=(x1,,xn,i=1n-r+1aixi+j=n-r+2nfj(xj)+b)

where ai,b and for l=n-r+2,,n , the functions fl:Il are arbitrary smooth functions of one real variable. Note that the parametrization obtained comprise hyperplanes.

Case 2: Suppose f¨i(xi)=0 , i=1,,n-r , then, there are constants αi such that f˙i=αi , for i=1,,n-r . From (10) we have

f¨n-r+1f¨n(1+α12++αn-r2)=0,

from which we conclude that f¨k=0 for some k{n-r+1,n} and thus, this case is contained in the Case 1 .

Case 3: Now suppose f¨i(xi)=0 , i=1,,n-r-1 and f¨k(xk)0 , for every k=n-r,,n . Observe that if we had f¨k(xk)=0 for some k=n-r,,n the analysis would reduce to the Cases 1 and 2. In this case, there are constants αi such that f˙i=αi for any 1in-r-1 . From (10) we have

k=n-rnf¨n-rf¨k^f¨n(β+f˙k2)=0

where β=1+k=1n-r-1αk2 and the hat means an omitted term. Then, from Lemma 1 we have that

k=n-rnfk(xk)=k=n-rn-1βakln|cos(-an-ran-1σr-1(an-r,,an-1)βxn+bn)cos(akβxk+bk)|+c

where an-r,,an-1,bn-r,,bn and c are real constants, and an-r,,an-1 , and σr-1(an-r,,an-1) are nonzero.

Case 4: Finally, suppose f¨i(xi)=0 , where 1ik and n-kr+2 , and f¨i(xi)0 for any i>k . We will show that this case cannot occur. In fact, note that for any fixed lk+1

k+1i1<<irnf¨i1f¨ir(1
+
1mnmi1,irf˙m2)
=
f¨lk+1i1<<ir-1ni1,,ir-1lf¨i1f¨ir-1(1+1mnml,i1,,ir-1f˙m2)
+
k+1i1<irni1,,irlf¨i1f¨ir(1+1mnmi1,irf˙m2)

Derivative with respect to the variable xl(lk+1) , in the above equality, gives

f˙˙˙lk+1i1<<ir-1ni1,,ir-1lf¨i1f¨ir-1(1
+
1mnml,i1,,ir-1f˙m2)
+ (11)
2f˙lf¨lk+1i1<<irni1,,irlf¨i1f¨ir=0.

That is, if we set

Al
=
k+1i1<<ir-1ni1,,ir-1lf¨i1f¨ir-1(1+1mnml,i1,,ir-1f˙m2) and
Bl
=
k+1i1<<irni1,,irlf¨i1f¨ir

then, it follows that Al,Bl do not depend on the variable xl and we can write

Alf˙˙˙l+2Blf˙lf¨l=0. (12)

We have two possible situations to take into account: Case I. Al0,lk+1 , and Case II. there is an lk+1 such that Al=0 .

Case I. Al0 : Under this assumption, there are constants αl(l=k+1,,n) such that equation (12) becomes f˙˙˙l+2αlf˙lf¨l=0 . Furthermore, it can be shown that for {l1,,lr+1}{k+1,,n}

r+1Gr(f1,,fn)xl1xlr+1=2k=1r+1(f˙lkf¨lkm=1mkr+1f˙˙˙lm) (13)

where

Gr(fk+1,,fn):=Wr+2Sr=k+1i1<<irnnf¨i1f¨ir(1+1inii1,,irf˙i2).

Since Sr=0 it follows that Gr=0 , and using that k=1r+1f˙lkf¨lk0 we obtain

k=1r+1(m=1mkr+1f˙˙˙lmf˙lmf¨lm) (14)
=
s=1r+1(f˙lsf¨lsm=1msr+1f˙˙˙lm)k=1r+1f˙lkf¨lk=0.

Now, for l=l1,lr+1 , substitute f˙˙˙l+2αlf˙lf¨l=0 in (14) to obtain the identity

σr(αl1,,αlr,αlr+1)=0 (15)

for any l1,,lr,lr+1{k+1,,n} . Hence we conclude that,

σr(αk+1,,αn)
=
0
σr+1(αk+1,,αn)
=
0.

These equalities, from [Proposition 1, Caminha (2006)], imply that at most r-1 of the constants αl(lk+1) are nonzero. If αl10,,αlm0 with mr-1 , in the expression obtained for Bl , making ll1,,lm and taking derivatives with respect to the variables xl1,,xlm we get

j=l1lmf˙˙˙jσr-m(f¨k+1,,f¨l^,,f¨^l1,,f¨^lm,,f¨n)=0

for all l{k+1,,n}{l1,,lm} . As f˙˙˙j0 for all j{l1,lm} , we obtain that

σr-m(f¨k+1,,f¨^l,,f¨^l1,,f¨^lm,,f¨n)=0

for all l{k+1,,n}{l1,,lm} . Consequently,

σr-m(f¨k+1,,,f¨^l1,,f¨^lm,,,f¨n)
=
0
σr-m+1(f¨k+1,,,f¨^l1,,f¨^lm,,,f¨n)
=
0.

Since (n-k-m)-(r-m)=n-k-r2 , at most r-m-1 of the functions f¨l are nonzero, for k+1ln and ll1,,lm , leading to a contradiction. So, αj=0 for all j{l1,lr-1} , which implies that f¨l is constant for all l{k+1,,n} . Now, again from equation (11) we get

k+1i1<<irni1,,irlf¨i1f¨ir=0,for anyl{k+1,,n}.

From which, we conclude that

σr(f¨k+1,,f¨n)
=
0
σr+1(f¨k+1,,f¨n)
=
0.

Therefore, at most r-1 of the functions f¨l(k+1ln) are nonzero, leading to a contradiction. Thus, it follows that Case 4 cannot occur, if Al0 for every l .

Case Al=0 :In this case, we have Blfl˙fl¨=0 implying

Al
=
k+1i1<<ir-1ni1,,ir-1lf¨i1f¨ir-1(1+1mnml,i1,,ir-1f˙m2)=0 and
Bl
=
k+1i1<<irni1,,irlf¨i1f¨ir=0.

Derivative of Al with respect to variable xs , for s=k+1,,n and sl , gives

f˙˙˙sk+1i1<<ir-2ni1,,ir-2l,sf¨i1f¨ir-2(1+k+1mnml,s,i1,,ir-2f˙m2)
+2f˙sf¨sk+1i1<<ir-1ni1,,ir-1l,sf¨i1f¨ir-1=0. (16)

Now, for i1,,ir{k+1,,n} with i1,,ir,l distinct indices, taking the derivatives of Bl with respect to xi1,,xir gives

f˙˙˙i1f˙˙˙ir=0.

Consequently, for at most r-1 indices, say i1,,ir-1 , we can have f˙˙˙im0 , (m=1,,r-1) , and f˙˙˙j=0 for every j=k+1,,n , with jl,i1,,ir-1 . Thus f˙˙˙im0 , with iml , together with equation Blxim=0 implies that the sum

k+1i1<<ir-1ni1,,ir-1l,imf¨i1f¨ir-1=0.

Now, if f˙˙˙j=0 we have by equation (16) that

k+1i1<<ir-1ni1,,ir-1l,jf¨i1f¨ir-1=0.

Therefore,

k+1i1<<ir-1ni1,,ir-1l,jf¨i1f¨ir-1=0,j=k+1,,nandjl

From which, we conclude that

σr-1(f¨k+1,,f¨^l,,f¨n)
=
0
σr(f¨k+1,,f¨^l,,f¨n)
=
0.

Thus, for at most r-2(r3) indices we must have f¨j0 , for every j=k+1,,n , and jl . This contradicts the hypothesis assumed in Case 4. Hence, Al=0 cannot occur. Since the case Al0 , cannot occur as well, it follows that Case 4 is not possible. This completes the proof of the theorem. ∎

[Proof of the Theorem 2] Let Mnn+1 be a translation hypersurface with constant Sr curvature. First, note that

mWr+2xi1xim (17)
=
j=1m(r+4-2j)k=1mf˙ikf¨ikWr+2-2m.

We have as a consequence of the proof of Theorem 1, see (13), the identity

r+1Gr(f1,,fn)xl1xlr+1=2k=1r+1(f˙lkf¨lkm=1mkr+1f˙˙˙lm)

where Gr(f1,,fn)=1i1<<irnnf¨i1f¨ir(1+1inii1,,irf˙i2) . With this we conclude, by Proposition 1, that

j=1r+1(r+4-2j)k=1r+1f˙lkf¨lkW-rSr
=
r+1(Wr+2Sr)xl1xlr+1
= (18)
2k=1r+1(f˙lkf¨lkm=1mkr+1f˙˙˙lm).

Now, we have two cases to consider: r odd and r even.

Case r odd: Suppose that there are l1,,lr+1 such that k=1r+1f˙lkf¨lk0 . Then,

Qr:=j=1r+1(r+4-2j)W-rSr
=
2s=1r+1(f˙lsf¨lsm=1msr+1f˙˙˙lm)k=1r+1f˙lkf¨lk
=
2k=1r+1(m=1mkr+1f˙˙˙lmf˙lmf¨lm).

Therefore,

r+1Qrxl1xlr+1=0.

On the other hand, using (17) we obtain

r+1Qrxl1xlr+1=j=1r+1(r+4-2j)i=1r+1(-r+2-2i)k=1r+1f˙lkf¨lkW-3r-2Sr.

Since r is odd, we conclude that r+4-2j0 and -r+2-2j0 , for any j and, therefore, Sr=0 .

Now, if for at most r indices we have f¨j0 for example j=l1,,lr then

Wr+2Sr=f¨l1f¨lrα,

for some constant α0 . Thus,

(r+2)Wrf˙l1f¨l1Sr=f˙˙˙l1f¨l2f¨lrα.

If f˙˙˙l1=0 , then Sr=0 . Otherwise,

(r+2)Wr+2f˙l1f¨l1Sr=f˙˙˙l1f¨l2f¨lrW2α (r+2)f˙l1(f¨l1)2=f˙˙˙l1W2.

As r>1 implies that W does not depend on the variables xl2,,xln , it follows that f¨l2==f¨ln=0 leading to a contradiction.

Case r even: In this case, there is a natural q2 such that r=2q . Then r+1q+2 and consequently

k=1r+1(r+4-2k)=0.

Therefore, by (18) we get

k=1r+1(f˙lkf¨lkm=1mkr+1f˙˙˙lm)=0.

Suppose that there are l1,,lr+1 such that k=1r+1f¨lk0 . In this case,

k=1r+1(m=1mkr+1f˙˙˙lmf˙lmf¨lm)=0.

We conclude that for each li there is a constant αli such that f˙˙˙li=αlif˙lif¨li . Now, it is easy to verify (see (11)) that

(r+2)f˙lr+1f¨lr+1WrSr
=
Gr(f1,,fn)xlr+1
=
f˙˙˙lr+1Gr-1(f1,,f^l,,fn)
+
2f˙lr+1f¨lr+11i1<<irni1,,irlr+1f¨i1f¨ir.

Therefore,

(r+2)WrSr=αlr+1Gr-1(f1,,f^lr+1,,fn)+21i1<<irni1,,irlr+1f¨i1f¨ir.

Differentiating this identity with respect to the variable xlr+1 , gives

(r+2)rf˙lr+1f¨lr+1Wr-2Sr=0 implying that Sr=0.

Finally, suppose that for any (r+1) -tuple of indices, say l1,,lr+1 it holds that k=1r+1f¨lk=0 . Then,

σr+1(f¨1,,f¨n)
=
0
σr+2(f¨1,,f¨n)
=
0.

Implying that at least n-r derivatives f¨l vanish, i.e., there are at most r functions such that f¨j0 for example j=l1,,lr . Thus, by Proposition 1

Wr+2Sr=f¨l1f¨lrα

for some constant α0 . We conclude that Sr=0 analogously to the way it was presented for the case r odd. ∎

Acknowledgments

The authors would like to express their gratitude to referee for valuable suggestions on the improvement of the paper. The first, second and third author are partially supported by Coordenação de Aperfeiçoamento de Pessoal de Nível Superior (CAPES). The fourth author is partially supported by CAPES and Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq).

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Received: May 12, 2015; Accepted: October 20, 2015

Correspondence to: Paulo Alexandre Araújo Sousa. E-mail: paulosousa@ufpi.edu.br