170
2 Variational Problems with Fixed Boundaries
the form of its polar coordinates is
J [r (θ )] =
θ 1
θ 0
r
2
r 2 + r 2 dθ
where, θ 0 < θ 1 , r
=
dr
dθ
; the coordinates of the fixed points A and B are (r 0 , θ 0 ) and
(r 1 , θ 1 ) respectively. Prove: The first integral of the Euler equation for J [r (θ )] is
−3dθ =
c 1 dρ
1 − c
2
1 ρ 2
The family of the extremal curves is
r
3 sin(3θ + c 2 ) = c 1
Proof Since
x
2
+ y
2
= (r cos θ)
2
+ (r sin θ)
2
= r
2
dx = cos θ dr − r sin θ dθ, dy = sin θ dr + r cos θ dθ
(dx)
2
= (dr )
2 cos
2
θ − 2r sin θ cos θ dr dθ + r
2 sin
2
θ(dθ)
2
(dy)
2
= (dr )
2 sin
2
θ + 2r sin θ cos θ dr dθ + r
2 cos
2
θ(dθ)
2
ds =
(dx) 2 + (dy) 2 =
(dr ) 2 + r 2 (dθ) 2 =
r 2 + r 2 dθ
Moreover the coordinates of the fixed points A and B are (r 0 , θ 0 ) and (r 1 , θ 1 )
respectively, so it can be deduced that
J [y] =
B
A
(x
2
+ y
2
)ds = J [r (θ )] =
θ 1
θ 0
r
2
r 2 + r 2 dθ
Due to the integrand does not contain θ , the Euler equation has the first integral
F − r
F r = r
2
r 2 + r 2 −
r
2 r
2
√
r 2 + r 2
= c 1
or
r
4
√
r 2 + r 2
= c 1
Squaring the two ends, we obtain
r
8
= c
2
1 (r
2
+ r
2
)