4.1 Variational Problems of the Simplest Functional
253
Fig. 4.4 Example 4.1.5
graph
O
x
y
y =
1
x
y =x −1
y =
1
x
This equation shows that the extremal curve y = y(x) and the the curve y 1 =
ψ(x 1 ) are orthogonal at the intersection, the transversality condition is converted to
the orthogonal condition.
If an endpoint of the curve is fixed, then the transversality condition holds only
to the undetermined endpoint.
Example 4.1.5 Find the shortest curve connecting the two curves y = −x − 1 and
y =
1
x
on a plane, see Fig. 4.4.
Solution Let the plane curve be y = y(x), according to the meaning of the problem,
the shortest curve is the minimum finding the functional
J [y] =
x 1
x 0
1 + y dx
where x 0 and x 1 are respectively on the curve C 0 and curve C 1 , namely
y = ϕ(x) = −x − 1, A(x 0 , ϕ(x 0 )) ∈ C 0
y = ψ(x) =
1
x
, B(x 1 , ϕ(x 1 )) ∈ C 1
Since F =
1 + y is only the function of y
, therefore the integral of the Euler
equation for the functional is a straight line y = c 1 x + c 2 . At the moment the system
of equations about four unknown numbers x 0 , x 1 , c 1 and c 2 can be written as