4.1 Variational Problems of the Simplest Functional
251
Substituting the straight equation into the tangential equation of the circle, c 1 can
be obtained
c 1 =
b + y c
k
Substituting the above expression into the equation of the circle, we give c
2
2 =
x 0 +
b+y c
k
2 + (y 0 + y c )
2 . Thus, the extremal curve is
x +
b + y c
k
2
+ (y + y c )
2
=
x 0 +
b + y c
k
2
+ (y 0 + y c )
2
The tangential equation is rewritten as
(x 1 + c 1 )m = (y 1 + y c )
After squaring two ends of the above expression, substituting it into the equation
of the circle, the coordinates of the intersection can be obtained
x 1 = −c 1 ±
c 2
√
1 + m 2
= −
b + y c
k
±
(kx 0 + b + y c ) 2 + k 2 (y 0 + y c ) 2
k 2 (1 + k 2 )
y 1 = kx t + b = −y c ±
(kx 0 + b + y c ) 2 + k 2 (y 0 + y c ) 2
1 + k 2
For the extremal curve of the functional
J [y] =
x 1
x 0
1 + y 2
y
dx, and y(0) = 0, y 1 (x 1 ) = x 1 − 5
which is equivalent to the case of y c = 0, x 0 = 0, y 0 = 0, k = 1 and b = −5. c 1 and
c 2 can be obtained
c 1 =
b + y c
k
=
−5 + 0
1
= −5
c
2
2 =
x 0 +
b + y c
k
2
+ (y 0 + y c )
2
=
0 +
−5 + 0
1
2
+ (0 + 0)
2
= 25
Thus the equation of the circle can be obtained
(x − 5)
2
+ y
2
= 25