Appendix: All Solutions to the Problems
833
From y(0) = 0, we get c 4 = 0; From y(x 1 ) = y 1 , we get c 3 = y 1 +
√
3x 1 .
Therefore
y =
√
3x
(0 ≤ x ≤ x c )
−
√
3x + y 1 +
√
3x 1 (x c ≤ x ≤ x 1 )
Because the two extremal curves intersect at the broken point, namely
√
3x =
−
√
3x + y 1 +
√
3x 1 , we get x = x c =
√
3y 1 +3x 1
6
, so the piecewise extremal curve is
y =
⎧
⎨
⎩
√
3x
0 ≤ x ≤
√
3y 1 +3x 1
6
−
√
3x + y 1 +
√
3x 1
√
3y 1 +3x 1
6
≤ x ≤ x 1
Similarly, B can be discussed. In a word, the extremal curve of the original functional has a broken point, it is the broken curve through two given boundary points
and composed of the straight line segments taking
√
3 and −
√
3 as the slopes.
4.21 Find the transversality condition of the functional J [y]
=
x 1
x 0
f (x, y)e
arctan y
1 + y 2 dx.
Solution Let F = f (x, y)e
arctan y
1 + y 2 , taking the partial derivative to y
F y = f (x, y)e
arctan y
1 + y 2
1 + y 2 +
y
1 + y 2
= f (x, y)e
arctan y
1 + y
1 + y 2
Let the curve of variable endpoint be y 1 = ϕ(x 1 ), from the transversality condition
there is
f (x, y)e
arctan y
1 + y 2 + (ϕ
− y
) f (x, y)e
arctan y
1 + y
1 + y 2
x=x 1
= 0
or
f (x, y)
1 + ϕ
− y
+ ϕ
y
1 + y 2
= 0
If f (x 1 , y 1 ) = 0, then
ϕ
−y
1+ϕ y = −1, namely when the variable boundary point
of the extremal curve y = y(x) slides along the curve y 1 = ϕ(x 1 ), the angle of
intersection of the tangential lines of two curves is
π
4
. The proof is as follows:
Let the angle of intersection tangent line of the extremal curve y = y(x) at the
intersection and the x axis be α, the angle of intersection of the tangent line of the
curve y 1 = ϕ(x 1 ) at the point and the x axis is β, then y
= tan α, ϕ
= tan β,
and it follows from the angle sum and difference identities of trigonometric function