The Upper Bounds On The Number Of Zeros Of Some Abelian Integrals

In this paper,we consider hyperelliptic integral of the first kind as following,J(h)=??h ?0+?1x+…+?g-1xg-1/y dx,H(x,y)=y2/2+?(x),NWhere ?(x)is a polynomial,?i are real constants,?h is the closed orbit,?h?C {(x,y)|H(x,y)? h,h1<h<h2}.Let ?(hi,h2)= {h?H(x,y)= h contains an oval}.In the first part of this paper,we present a new method to prove that the integral J(h)= ?1Ji(h)+ ?2J2(h)has at most one zero in ?,where Ji(h)=??h f1(x)/y dx,J2(h)=??h f2(x)/y dx.In other words,we prove the monotonicity of the J2(h)/J1(h)(J1(h)?0).As application,when the first integral has the following form H(x,y)=y2/2+xb+1/b+1-xb/b+1/b(b+1)or y2/2+x2m/2m-x2m+1/2m+1,J(h)=??h ?0+?1x/y dx has at most one zero.In addition,this method can also deal with the integral I(h)= ?1??h f1(x)ydx+?2??h f2(x)y dx,and the associated result generalize the previous result.In the second part of this paper,we consider the number of the isolated zeros of the integral J(h)???h?0f0(x)+?1f1(x)+?2f2(x)/y dx.If the first integral has the following form H(x,y)= y2/2+xa(bx + c),we introduce a new transformation which efficiently simplifies the calculation.Using this transformation,we get some new conclusions,For example:Consider the first integral as following,H(x,y)=y2/2-?2?2/2x2+?2?2+2?2?+2??2/3x3-2?+2?+1/6x6+1/7x7-2?2?+2??2+?2+4??+?2/4x4+?2+4??+?2+2?+2?/5x5.When(i)?=?=0,(ii)????1,J(h)=??h ?0+?1x+?2x2/y dx has at most two zeros.