4.6 One-Sided Variational Problems
307
Using the expression (4.1.15), we have
δy(x M ) = δy M − y
(x M )δx M = [ϕ
(x M ) − y
(x M )]δx M
(4.6.18)
δy(x N ) = δy N − y
(x N )δx N = [ϕ
(x N ) − y
(x N )]δx N
(4.6.19)
Substituting the formula (4.6.18) and formula (4.6.19) into the expression (4.6.16),
taking note that at x = x M and x = x N , we have y(x k ) = ϕ(x k ), where, k = M, N ,
and from the arbitrariness of δx M and δx N , we obtain
[F(x, ϕ, y
) − F(x, ϕ, ϕ
) − (y
− ϕ
)F y (x, ϕ, y
)]
x=x M
= 0
(4.6.20)
[F(x, ϕ, y
) − F(x, ϕ, ϕ
) − (y
− ϕ
)F y (x, ϕ, y
)]
x=x N
= 0
(4.6.21)
By the differential mean value theorem, if y
(x M ) = ϕ
(x M ), then there exists p
between y
(x M ) and ϕ
(x M ), such that expression (4.6.20) is changed into
[(ϕ
− y
)(F y (x, ϕ, p) − F y (x, ϕ, y
)]
x=x M
= 0
(4.6.22)
By the differential mean value theorem again, there exists q between p and y
(x M ),
such that
[(ϕ
− y
)( p − y
)F y y (x, ϕ, q)]
x=x M
= 0
(4.6.23)
Due to y
(x M ) = ϕ
(x M ), p = y
(x M ), getting F y y (x M , ϕ(x M ), q) = 0, this is
in contradiction with assumption F y y = 0, therefore there must be at x = x M
y
(x M ) = ϕ
(x M )
(4.6.24)
The fomula (4.6.24) shows that the curve AM and the curve y = ϕ(x) have the
common tangent at point M. Similarly it can be concluded that the curve NB and the
curve y = ϕ(x) also have the common tangent at point N. Thus the extremal curve
and the curve y = ϕ(x) are tangent at two points M and N. Quod erat demonstrandum.
Example 4.6.1 Under the condition that the admissible curve can not pass through
the interior of the circle domain surrounded by the circumference (x −a)
2
+ y
2
= R
2 ,
Find the curve that makes the functional J [y] =
x 1
x 0
y
3 dx attain extremum, the
boundary conditions are y(x 0 ) = 0, y(x 1 ) = 0, where x 0 < a − R, a + R < x 1 .
In addition find the extremal curve when a = 13, R = 5, x 0 = 0, x 1 = 26, see
Fig. 4.13.
Solution Because the integrand F = y
3 is only the function of y
, the integral of
the Euler equation of the functional is