4.2 Variational Problems of Functionals with Several Functions
263
1
ϕ x
=
y
ϕ y
= −
z
1
This condition shows that the tangent vector t(1, y
, z
) found the extremal curve
at point B(x 1 , y 1 , z 1 ) runs parallel to each other with the normal vector N(ϕ x , ϕ y , −1)
of the surface at the same point. Therefore the transversality condition becomes that
the extremal curve is at the orthogonal condition with the surface z = ϕ(x, y).
Example 4.2.2 Find the extremal curve of the functional J [y(x), z(x)] =
x 1
0 (y
2
+ z
2
+ 2yz)dx, one endpoint is given, y(0) = 0, z(0) = 0, another point
(x 1 , y 1 , z 1 ) is undetermined on the plane x = x 1 .
Solution The Euler equations of the functional are
⎧
⎪ ⎨
⎪ ⎩
F y −
d
dx
F y = 0
F z −
d
dx
F z = 0
or
z
− y = 0
y
− z = 0
Solve simultaneously the equations, and get
y = c 1 e
x
+ c 2 e
−x
+ c 3 cos x + c 4 sin x
z = c 1 e
x
+ c 2 e
−x
− c 3 cos x − c 4 sin x
From the boundary conditions y(0) = 0, z(0) = 0, we get
c 1 = −c 2 , c 3 = 0
Therefore
y = c 1 e
x
− c 1 e
−x
+ c 4 sin x
z = c 1 e
x
− c 1 e
−x
− c 4 sin x
The transversality conditions of the variable boundary point are
F y
x=x 1
= 0, F z | x=x 1 = 0
By the above mentioned conditions, we give
F y
x=x 1
= 2y
x=x 1
= 0, F z | x=x 1 = 2z
x=x 1
= 0