Going threedimensional (my first solution for the elliptic integral)

Creating a field of velocities with several arbitrary functions of z, we can write the circulation along an ellipse be proportional to the ellipse arc. Then we create a rotational tub, from the speed field having horizontal ellipses, as

M(z)x2+N(z) y2= 1 (1)

On their surface, circulation along a closed line will be zero (in the case of a triangle, the sum of 3 circulations) and we shall put the ellipse arc AB as function of this circulation ar BC amd CA. Choosing planes passing by z-ax, y=L(z) ·x with L defined by x1,y1 and x2,y2

extremes of arc.

We choose CD for the circulation be integrable, a surface made with straight lines y=L(z) ·x

( passing by z-ax). In the lines BC and CA the calculation is elemental because with

L=constant, we will have from (2)

dG=udx+vdy=Ap(ydx-xdy)=y2d(x/y)Ap GBC=0 circulation zero.

El velocities field can have the form

u= Apy v=-Apx w=Cxyp being p= (x2+Fy2)1/2 (2)

The rotor is:

Rx=dw/dy-dv/dz Ry=du/dz-dw/dx Rz=dv/dx-du/dy

Their components are

Rx=x/p(-Cx2-Cp2-AF´y2+A’p2)=x/p(x2(A´+C)+y2(2CF+ A´F-AF´/2)

Ry=y/p(x2A´-2Cp2)+Fy2(A´-C+AF´/2F)

Rz= -3Ap

We force than the rotor tub be like (1) then their tangent planes will be

xMdx+yNdy+1/2(M´x2+N´y2) dz= 0 The normal to this plane will be

n=[xM, yN, 1/2(M´x2+N´y2)]

that has to be perpendicular to rotor; id est , the scalar product Rxn=0

and (xMRx=) Mx2/p[x2(C+ A’)+AFy2(2C/a+ A’/A+F’/2F)] +…

(yNRy=)Ny2/p[x2(A’-2C)+AFy2(A’/A+F’/2F-C/A)]+...

-(3A/p)/2[M’x2+N’y2](x2+ Fy2)= 0 have to be identically null.

El term of x 4 is M(C+A´)=3AM´/2 i.e.

dln(M3/A2) =dl(Q)=2C/A Q=M3/A2 (3)

we have C(z) function of the rest.

El term of y4 is

NAF(A´/A+F´/2F-C/A)= 3/2(AFN´) i.e.

dln(A4F)=dln(M3N3) FA4=b2M3N3 (4)

(b us an arbitrary constant) A, can be write as function of the rest functions of z.

El term of x 2 y 2 will be

2FMdln(M3F1/2)+2Ndln(A3/M3)=3/2(FM´+N´) i.e.

dln(FM6/A2)+N/FMdln(A6/M6)=3dlnM+3N´/FM

that is N/FM[dln(N3)- dln(A6/M6)]= dln(FM3/A2) and then

writing N=fM N/FM [dln(M3/A2N3M3/A4)]= f/F d(QF) =d(QF)

we can see that FQ have to be constant, (because f=m when F is m2)

abd (4) now will be k=a2M3/Q C=(1/2)dlnQ=dQ/2Q

F=a/Q f=c/Q F=mf is F/Q2=b2f3= a/b2=(Qf)3 Q=1/f (4)

We have M(z), and A(z) Q=M3/A2 variables without conditions for the field and rotor.

For a integrad factor we shall have a freedom degree at L(z)

k= aM3/2Q-1/2 ½[3M’/M– Q’/Q]=(1/2)dl(M3/Q)=0 Q=M3

Now we are going to study at two separate summands, the circulation G

From the rotor tub x2(1+fL2)=M-1 results x3=(1+fL2)-3/2M-

3/2

dG1=udx+vdy and dG2=wdz=Cpxydz=-x3(L/2) (1+FL2)1/2[3M’/M– Q’/Q] dG1=ypdx-xpdy= py2d(x/y)= -aQ-1/2M-3/2(1+mfL2)1/2L2x3

dL/L2

dG1=-(a/2)M-3/2(1+mfL2)1/2(1+fL2)-3/2Q-1/2d(2L) We call R=(1+mfL2)1/2(1+fL2)-3/2

dG2=-(a/2)Q-1/2(dQ/Q)x3L(1+mfL2)1/2 =-(a/2)RQ-3/2M-3/2LdQ -(a/2)Q-3/2L(1+mfL2)1/2dQ(1+fL2)-3/2 and the sum

dG=-(a/2)(1+mfL2)1/2(1+fL2)-3/2Q-3/2[LdQ+2QdL]

dividing by QL the final bracket

dG=f3/2 R[f1/2L]dl(L2/f)=R 2dln(L/f1/2)=R·f d(L/f1/2)

There are solutions for f=ct and for R ct

******************if R is constant fL2 = za+2b if a+2b=0 z0 L/f1/2 = zb-a/2 =z-a b=-a/2 with b=1 a=-2

R=(1+mfL2)1/2(1+fL2)-3/2 there is solution for R ct dG=Kdz/z G=K·ln(z)

(f =constant) elliptic cylinder dG=R f d(L/f1/2) RdL=(1+mfL2)1/2/(1+fL2)3/2dL calling w a L

dG=(1/m+w2)1/2/(1+w2)3/2dw we do L=zb f=za

Now if f =ct dG=R f d(L/f1/2) RdL=(1+mfL2)1/2/(1+fL2)3/2dL elliptic cylinder dG=dL(A+L2)1/2/(B+L2)3/2 pseudo-elliptic 3dtype A=1/mf B=Am A(1-m)=N if N=1 A=1/(1-m) f=1-m A=B+N dG=dL/[(B+L2)-

2+N(B+L2)-3]1/2 L=B1/2shw

dG=d(sh(w/2)/[1/2+sh2(w/2)]1/2wch2(w/2)

dG=dz(1/2+z2)-1/2/(1+z2) with z=sha dG=da(1/2ch2a+sh2a/ch2a)1/2=da[(1/2+1/2)th2+th2a]1/2

dG=cha.da.(1+4sh2a)1/2/ch2a= d(v)(1/4+v2)1/2/(1+v2) v=(1/2)shb dG=db(3+ch2b-3)/(3+ch2b)

dG=db(1-3/(2+ch2b)=b-3db/(2+ch2b)= b-3ebd(eb)/(10e2b+e4b+1) e2b=w dw/(w2+10w+1)

-5+/-(24)1/2 (w+5)2-24 w+5=241/2chg dG=dgshg/sh2g G=∫dg/shg=ln[th(g/2)]