Appendix: All Solutions to the Problems
785
y = c 1 e
x
+ c 2 x
−x
+ c 3 cos x + c 4 sin x + 1, z
= c 1 e
x
+ c 2 x
−x
− c 3 cos x − c 4 sin x + 1
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
.
Proof Let the integrand F = y
2
+ 2yz
+ 2zy
+ z
2 , the Euler equations of the
functional are
⎧
⎪ ⎨
⎪ ⎩
F y −
d
dx
F y = −2y
+ 2z
− 2z
= 0
F z −
d
dx
F z = −2y
+ 2y
− 2z
= 0
Integrating, we get
y = c 1 x + c 2 , z = c 3 x + c 4
From the boundary conditions y(x 0 ) = y 0 , y(x 1 ) = y 1 , 得c 1 =
y 1 −y 0
x 1 −x 0
, c 2 =
y 0 x 1 −y 1 x 0
x 1 −x 0
. Substituting c 1 and c 2 into the first solution of the Euler equations, we
obtain
y =
y 1 − y 0
x 1 − x 0
x +
y 0 x 1 − y 1 x 0
x 1 − x 0
Subtracting y 0 on the both sides, we get y − y 0 =
y 1 −y 0
x 1 −x 0
(x − x 0 ). Similarly we get
z − z 0 =
z 1 −z 0
x 1 −x 0
(x − x 0 ). Thus there is
x−x 0
x 1 −x 0
=
y−y 0
y 1 −y 0
=
z−z 0
z 1 −z 0
, this is a space straight
line equation. Quod erat demonstrandum.
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.
Solution Because the integrand F =
˙
x 2 + ˙
y 2
x−k
does not contain t, the Euler equation
has the first integral
F − ˙
x F ˙
x =
˙
y
2
√
x − k
˙
x 2 + ˙
y 2
= c,
d
dt
F ˙
y =
d
dt
˙
y
√
x − k
˙
x 2 + ˙
y 2
= 0
˙
y
2
= c
√
x − k
˙
x 2 + ˙
y 2 , ˙
y
2
= c
2
1 (x − k)( ˙
x
2
+ ˙
y
2
)
From the latter equation, we solve for
˙
x
˙
y
, and we get
dx
dy
=
˙
x
˙
y
=
1
c
2
1 (x−k)
− 1.
Separating the variables and putting x =
1
c
2
1
cos
2 t + k, integrating we get