Appendix: All Solutions to the Problems
799
dx = cos θ dr − r sin θ dθ, dy = sin θ dr + r cos θ dθ
(ds)
2
= (dx)
2
+ (dy)
2
= (dr )
2
+ r
2
(dθ)
2
If endpoints x 0 , x 1 corresponds to the value α and β of θ , α < β, then the
functional can be written as
J [r ] =
β
α
r
r 2 + r 2 dθ
Because F = r
√
r 2 + r 2 does not contain θ , the Euler equation has the first
integral
F − r
F r = r
r 2 + r 2 − r
rr
√
r 2 + r 2
=
r
3
√
r 2 + r 2
= c 1
Putting r
= r tan t, then
√
r 2 + r 2 = r sec t, substituting it into the above
equation, we obtain
r
2 cos t = c 1 , dr =
r tan tdt
2
=
r
dt
2
, dθ =
dr
r =
dt
2
Integrating the last equation, we get 2θ = t + c 2 . Therefore, the extremal curve
is r
2 cos(2θ − c 2 ) = c 1 . Furthermore
cos(2θ − c 2 ) = cos 2θ cos c 2 + sin 2θ sin c 2 = (cos
2
θ − sin
2
θ) cos c 2
+ 2 sin θ cos θ sin c 2 =
(x
2
− y
2
) cos c 2
r 2
+
2x y sin c 2
r 2
Thus, the extremal curve can also be written as
(x
2
− y
2
) cos c 2 + 2x y sin c 2 = c 1
2.84 Write the polar coordinate form of the Laplace equation u xx + u yy = 0.
Solution The corresponding functional of the Laplace equation u xx + u yy = 0
is J [u(x, y)] =
˜
D (u
2
x + u
2
y )dxdy. Through the relationship between the rectangular coordinates and polar coordinates x = r cos θ , y = r sin θ , the derivative
relationships between r, θ and x, y can be found
∂r
∂ x
=
x
r
= cos θ,
∂r
∂ y
=
y
r
= sin θ,
∂θ
∂ x
= −
sin θ
r
,
∂θ
∂ y
=
cos θ
r
The functional expressed in polar coordinates can be written as