792
Appendix: All Solutions to the Problems
Solution The Euler equation of the functional is
y
− y
− y
(6)
= 0
or
y
(6)
= 0
Integrating six times, we get
y = c 1 x
5
+ c 2 x
4
+ c 3 x
3
+ c 4 x
2
+ c 5 x + c 6
Using the boundary conditions y(0) = 0 and y
(0) = 1, we obtain c 6 = 0 and
c 5 = 1, then using other four boundary conditions, we obtain c 1 =
3
2
, c 2 = c 4 = 0,
c 3 = −
5
2
. Thus the extremal curve is y =
3x
5 −5x
3
2
+ x.
2.72 Find the variations and the Euler equations of the following functionals
(1) J [u] =
˜
D (u
2
x − u
2
y )dxdy; (2) J [u] =
˜
D (u
2
x − u
2
yy )dxdy.
Solution
(1) Let the integrand F = u
2
x − u
2
y , the first variation of the functional is
δ J =
¨
D
δ(u
2
x − u
2
y )dxdy =
¨
D
(2u x δu x − 2u y δu y )dxdy
The Euler equation of the functional is
F u −
d
dt
F u x −
d
ds
F u y = −2u xx + 2u yy = 0
namely u xx − u yy = 0.
(2) Let the integrand F = u
2
x − u
2
yy , the first variation of the functional is
δ J =
¨
D
δ(u
2
x − u
2
yy )dxdy =
¨
D
(2u x δu x − 2u yy δu yy )dxdy
The Euler equation of the functional is
−
d
dx
F u x +
d
2
dy 2 F u yy = −2u xx − 2u yyyy = 0
namely u xx + u yyyy = 0 or
∂
2 u
∂ x 2 +
∂
4 u
∂ y 4 = 0.