Let r0 be an initial condition on the surface curve S. Since θ˙ = 1 the first return to S occurs after a time of flight t = 2π.

Then, according to the definition of the Poincare map

r1 = P(r0)

Where ∫r0r11r(1-r2)dr = ∫02πdt

∫r0r11r(1-r2)dr = 2π

Using partial fraction method, we can write the integral as

1r(1-r2) = Ar+B1- r+C1 + r1 = A(1-r)(1+r)+B(1+r)r+C(1-r)r1 = A(1-r2)+B(r2+r)+C(r-r2)1 = (-A+B-C)r2+(B+C)r+A

Comparing the coefficients gives:

-A + B - C = 0B + C = 0A = 1

Put A = 1 in -A + B - C = 0, then B - C = 1

Add B - C = 1 and B + C = 0, This gives

B = 12

Put this value in the equation B - C = 1. Then it gives

C = -12

Hence,

1r(1-r2) = 1r+12(1- r)−12(1 + r)

Insert it in the integral

2π = ∫r0r1[1r+12(1- r)−12(1 + r)]dr

2π = [ln|r| - 12ln|r-1| - 12ln|r+1|]r0r1

2π = (ln|r1| - 12ln|r1-1| - 12ln|r1+1|) - (ln|r0| - 12ln|r0-1| - 12ln|r0+1|)

2π = ln|r1r12-1|- ln|r0r02-1|2π = ln|r1r12-1|ln|r0r02-1|          

Using logarithmic identity, the above equation can be written as

12ln|r1(r02-1)r0(r12-1)| = 2πln|r1(r02-1)r0(r12-1)| = 4π

Since r = 1 is a limit cycle, the absolute value can be dropped. Then the above equation becomes

r1(r02-1)r0(r12-1) = e4π

Rearrange it as

r12(r02 - 1) = e4πr02(r12 - 1)r12(r02 - 1 - e4πr02) = - e4πr02r12 = e4πr02(-r02 + 1 + e4πr02)

Divide numerator and denominator of the right-hand side expression by e4πr02

r12 = 1(-e-4π+ e-4πr0-2 + 1)    = 11 + -e-4π(r0-2-1)

Take the square root of both sides

r1 = 11 + -e-4π(r0-2 - 1)   = (1 + -e-4π(r0-2 - 1))-12

Hence, r1 = (1 + -e-4π(r0-2 - 1))-12 is proved.

According to the definition of the Poincare map,

rn+1 = P(rn) 

r1 = P(r0) 

Hence,

P(r0) = (1+-e-4π(r0-2-1))-12

P(r) = (1+-e-4π(r−2-1))-12

Differentiate it with respect to r

P'(r) = e-4πr-3(1+e-4π(r-2-1))32

Put r = 1. Then,

P'(1) = e-4π

Hence, P'(1) = e-4π is proved.

A Poincare function can be written using a Poincare map definition and partial fractions method.