Appendix: All Solutions to the Problems
783
Solution The Euler equations of the functional are
y − ¨
x = 0
x − ¨
y = 0
Deriving the second equation two times and then substituting it into the first
equation, we obtain
d
4 y
dt 4 − y = 0
Solving we get
y = c 1 e
T
+ c 2 e
−t
+ c 3 cos t + c 4 sin t
From the boundary conditions x(0) = 0, x
π
2
= 1, y(0) = 0, y
π
2
= −1, we
get c 1 = c 2 = c 3 = 0, c 4 = −1, thus the extremal curve is y = − sin t, x = sin t.
2.54 Find the extremal curve of the functional J [y, z] =
1
0 (y
2
+ z
2
+ 2y)dx, the
boundary conditions are y(0) = 1, y(1) =
3
2
, z(0) = 0, z(1) = 1.
Solution The Euler equations of the functional are
2 − 2y
= 0, 2z
= 0
Integrating we get
y =
x
2
2
+ c 1 x + c 2 , z = c 3 x + c 4
The integral constants that conform to the boundary conditions are c 1 = 0, c 2 = 1,
c 3 = 1, c 4 = 0. Therefore the extremal curves are
y =
x
2
2
+ 1, z = x
2.55 Find the extremal curve of the functional J [y, z]
=
x 1
x 0
(2yz − 2y
2
+ y
2
− z
2
)dx.
Solution This is the functional with two functions, F = 2yz − 2y
2
+ y
2
− z
2 , the
extremal curve should satisfy the Euler equations, there are
⎧
⎪ ⎨
⎪ ⎩
F y −
d
dx
F y = 2z − 4y − 2y
= 0
F z −
d
dx
F z = 2y + 2z
= 0