196
2 Variational Problems with Fixed Boundaries
2.41 Find the extremal curve of the functional J [y] =
2
1 (y
2
+ 2yy
+ y
2
)dx, the
boundary conditions are y(1) = 1, y(2) = 0.
2.42 Find the extremal curve of the functional J [y] =
1
0 yy
2 dx, the boundary
conditions are y(0) = 1, y(1) =
3
√
4.
2.43 Find the extremal curve of the functional J [y] =
1
0 y
2 y
2 dx, the boundary
conditions are y(0) = 0, y(1) = 1.
2.44 Find the extremal curve of the functional J [y] =
1
0 (y
2
+ y
2
)dx, the
boundary conditions are y(0) = 0, y(1) = 1.
2.45 Find the extremal curve of the functional J [y] =
1
0 (y
2
+ 4y
2
)dx, the
boundary conditions are y(0) = e
2 , y(1) = 1.
2.46 Find the extremal curve of the functional J [y] =
x 1
x 0
(ay + by
2
)dx, the
boundary conditions are y(x 0 ) = y 0 , y(x 1 ) = y 1 .
2.47 Find the extremal curve of the functional J [y] =
x 1
x 0
(y
2
+ 2yy
− 16y
2
)dx.
2.48 Find the extremal curve of the functional J [y] =
x 1
x 0
1+y
2
y 2 dx.
2.49 Find the extremal curve of the functional J [y] =
x 1
x 0
1+y
2
y dx through points
(0,0) and (1,1).
2.50 Find the extremal curve of the functional J [y] =
x 1
x 0
y(1 + y 2 )dx.
2.51 Find the extremal curve of the functional J [y] =
x 1
x 0
√
1+y 2
y+k
dx, where, k is a
constant.
2.52 Let a car of quality m move on a horizontal orbit, at the beginning (t 0 = 0)
the velocity is zero. Neglecting the frictional resistance, at time t = t 1 , the
velocity of the car is v 1 , find the control law namely the external force F(t),
such that the velocity error and control energy are minimum, that is to say that
the functional J [F(t)] =
t 1
t 0
F
2
(t) + a[v 1 − v(t)]
2
dt obtains minimum,
where, a is a positive constant.
2.53 Find the extremal curve of the functional J [x, y] =
π
2
0 ( ˙
x
2
+ 2x y + ˙
y
2
)dt,
the boundary conditions are x(0) = 0, x
π
2
= 1, y(0) = 0, y
π
2
= −1.
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.
2.55 Find the extremal curve of the functional
J [y, z]
=
x 1
x 0
(2yz − 2y
2
+ y
2
− z
2
)dx.
2.56 Find the extremal curve of the functional J [y, z] =
x 1
x 0
(y
2
+ z
2
+ y
z
)dx.
2.57 Find the extremal curve of the functional
J [y, z]
=
x 1
x 0
(y
2
+ z
2
− 2yz + 2y + 2z)dx.
2.58 Prove that the extremal curve of the functional J [y, z]
=
x 1
x 0
(y
2
+ 2yz
+ 2zy
+ z
2
)dx satisfying the boundary conditions
y(x 0 ) = y 0 , y(x 1 ) = y 1 , z(x 0 ) = z 0 , z(x 1 ) = z 1 is the spatial straight
line
x−x 0
x 1 −x 0
=
y−y 0
y 1 −y 0
=
z−z 0
z 1 −z 0
.
2.59 Find the extremal curve of the functional J [x, y] =
t 1
t 0
˙
x 2 + ˙
y 2
x−k
dt, where, k is
a constant.
2.60 Find the extremal curve of the functional J [y] =
x 1
x 0
(16y
2
− y
2
+ x
2
)dx.