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) 路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) 路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 2
2 1/2
(x +Fy )
u= Apy v=-Apx w=Cxyp being (2)
p=
The rotor is: Rx=dw/dy-dv/dz du/dy
Ry=du/dz-dw/dx
Rz=dv/dx-
Their components are Rx=x/p(-Cx2-Cp2-AF´y2+A’p2)=x/p(x2(A´+C)+y2(2CF+ A´FAF´/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 this plane will be
The normal to
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’/2FC/A)]+... -(3A/p)/2[M’x2+N’y2](x2+ Fy2)= 0 identically null. El term of x4 is
M(C+A´)=3AM´/2
have to be i.e.
dln(M3/A2) =dl(Q)=2C/A (3)
Q=M3/A2
we have C(z) function of the rest. El term of y4 is NAF(A´/A+F´/2F-C/A)= 3/2(AFN´) dln(A4F)=dln(M3N3)
i.e.
FA4=b2M3N3
(4)
(b us an arbitrary constant) A, can be write as function of the rest functions of z. El term of x2y2
will be
2FMdln(M3F1/2)+2Ndln(A3/M3)=3/2(FM´+N´)
i.e.
dln(FM6/A2)+N/FMdln(A6/M6)=3dlnM+3N´/FM N/FM[dln(N3)- dln(A6/M6)]= dln(FM3/A2)
that is and then
writing N=fM =d(QF)
N/FM [dln(M3/A2N3M3/A4)]= f/F d(QF)
we can see that FQ have to be constant, (because f=m when F is m2) abd (4) now will be F=a/Q Q=1/f (4)
k=a2M3/Q
f=c/Q F=mf
C=(1/2)dlnQ=dQ/2Q
is F/Q2=b2f3=
We have M(z), and A(z) Q=M3/A2 conditions for the field and rotor.
a/b2=(Qf)3
variables without
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
Now we are going to study at two separate summands, the circulation G
Q=M3
x2(1+fL2)=M-1 results x3=(1+fL2)-3/2M-
From the rotor tub 3/2
dG1=udx+vdy Q’/Q]
dG2=wdz=Cpxydz=-x3(L/2) (1+FL2)1/2[3M’/M–
and
dG1=ypdx-xpdy= py2d(x/y)= dL/L2
-aQ-1/2M-3/2(1+mfL2)1/2L2x3
dG1=-(a/2)M-3/2(1+mfL2)1/2(1+fL2)-3/2Q-1/2d(2L) R=(1+mfL2)1/2(1+fL2)-3/2
We call
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 a=-2
2
a+2b
fL = z
****************** if a+2b=0 z0 L/f1/2 = zb-a/2 =z-a b=-a/2 with b=1
R=(1+mfL2)1/2(1+fL2)-3/2 there is solution for R ct G=K·ln(z)
dG=Kdz/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)]