4.2 Variational Problems of Functionals with Several Functions
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Integrating the above equations, we get
y = c 1 x + c 2
z = c 3 x + c 4
namely the found extremal curve of the functional is a family of straight lines in
space.
Since the integrand in the example is the case of f (x, y, z) = 1 in Example 4.2.1,
Only the ends of the boundary conditions are variable, the transversality conditions
at point (x 0 , y 0 , z 0 ) and point (x 1 , y 1 , z 1 ) are converted to the orthogonal conditions.
Therefore, It can only get extremum in straight line perpendicular to each other with
the surface z = ϕ(x, y) at point (x 0 , y 0 , z 0 ) and the surface z = ψ(x, y) at point
(x 1 , y 1 , z 1 ).
Example 4.2.4 Find the shortest distance from point M(x 0 , y 0 , z 0 ) to the straight
line
y = mx + p
z = nx + q
Solution This problem boilds down to finding the extremum of the functional
J [y, z] =
x 1
x 0
1 + y 2 + z 2 dx
(1)
The boundary point M(x 0 , y 0 , z 0 ) of the extremal curve of the functional is given,
while another boundary point can move along the given straight line
y = mx + p
z = nx + q
(2)
At the moment, the functions ϕ and ψ are respectively
ϕ = mx + p
ψ = nx + q
(3)
Since the functional is only the function of y
and z
, therefore the general solution
of the Euler equations is the straight line, that is
y = c 1 x + c 2
z = c 3 x + c 4
(4)
The transversality condition is