Appendix: All Solutions to the Problems
835
y
(x 2 ) = ϕ(x 2 )
or
c 1 = ±
−(x 2 − 5)
9 − (x 2 − 5) 2
At the tangential point, there is y(x 2 ) = ϕ(2), we get
c 1 x 2 = ±
9 − (x 2 − 5) 2
Solving simultaneously the above two equations, we obtain
x =
16
5
, c 1 = ±
3
4
Therefore the extremal curve is y = ±
3
4
x
0 ≤ x ≤
16
5
.
Similarly, It can be obtained y = ∓
3
4
(x − 10)
34
5
≤ x ≤ 10
. It follows that
the extremal curve is composed of the straight line segment that is contingence
with the circumference, the circular arc and the another straight line segment that
is contingence with the circumference. This problem can also be solved by directly
applying the formula in Example 4.6.1.
4.24 Find the function that can make the functional J [y] =
π
4
0 (y
2
− y
2
)dx reach
extremum, one boundary point is fixed, y(0) = 0, another boundary point can slide
on the straight line x =
π
4
.
Solution Let F = y
2
− y
2 , the Euler equation of the functional is
2y + 2y
= 0
The general solution of the Euler equation
y = c 1 cos x + c 2 sin x
From y(0) = 0, we get c 1 = 0, so that y = c 2 sin x, y
= c 2 cos x, F y = −2y
=
−2c 2 cos x. When another boundary point slides on the straight line x =
π
4
, there is
F y
x=
π
4
= −2c 2 cos x| x=
π
4
= −
√
2c 2 = 0
Getting c 2 = 0, thus y ≡ 0.
4.25 Only using the necessary condition δ J = 0, find the curve that can make
the functional J [y] =
x 1
0
√
1+y 2
y
dx attain extremum, one boundary point is fixed,