Appendix: All Solutions to the Problems
773
2.34 Find the extremal curve of the functional J [y] =
x 1
x 0
√
1+y 2
x+k
dx, where, k is a
constant.
Solution Since the integrand F =
√
1+y 2
x+k
does not depend on y, at this time the
Euler equation is
d
dx
y
(x + k)
1 + y 2
= 0
The first integral is
y
(x + k)
1 + y 2
= c 1
Putting y
= tan t, substituting it into the above equation, we get y
cos t =
(x + k)c 1 , namely x + k =
sin t
c 1
, dx =
cos tdt
c 1
, dy =
sin tdt
c 1
, integrating we get
y = −
cos t
c 1
+ c 2 . Therefore the solutions with t the parameter can be obtained
x + k =
sin t
c 1
, y = −
cos t
c 1
+ c 2
Eliminating the parameter t, we obtain
(x + k)
2
+ (y − c 2 )
2
=
1
c
2
1
namely the extremal curves are a family circles that the centers are on the ordinate
axis.
2.35 Find the extremal curve of the functional J [y] =
x 1
x 0
(ax + b)
1 + y 2 dx.
Solution Because the integrand F = x
1 + y 2 does not depend on y, at this time
the Euler equation is
d
dx
(ax + b)y
1 + y 2
= 0
The first integral is
(ax + b)y
1 + y 2
= c 1
Putting y
= tan t, substituting it into the above equation, we get (ax + b) sin t =
c 1 , thus