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)=∫Γh α0+α1x+…+αg-1xg-1/y dx,H(x,y)=y2/2+Ψ(x),NWhere Ψ(x)is a polynomial,αi are real constants,Γh is the closed orbit,Γh∈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)= α1Ji(h)+ α2J2(h)has at most one zero in △,where Ji(h)=∫Γh f1(x)/y dx,J2(h)=∫Γh 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)=∫Γh α0+α1x/y dx has at most one zero.In addition,this method can also deal with the integral I(h)= α1∫Γh f1(x)ydx+α2∫Γh 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)＝∫Γhα0f0(x)+α1f1(x)+α2f2(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-α2β2/2x2+α2β2+2α2β+2αβ2/3x3-2α+2β+1/6x6+1/7x7-2α2β+2αβ2+α2+4αβ+β2/4x4+α2+4αβ+β2+2α+2β/5x5.When(i)α=β=0,(ii)α＝β＝1,J(h)=∫Γh α0+α1x+α2x2/y dx has at most two zeros.