274
4 Problems with Variable Boundaries
Proof Because endpoint (x 1 , y 1 ) moves freely only on the straight line x = x 1 ,
δx 1 = 0, δy 1 and δy
1 can be arbitrary, from the formula (4.3.9), if making δ J = 0
hold, the coefficients of δy 1 and δy
1 must be zero, therefore Eq. (4.3.14) holds. Quod
erat demonstrandum.
Example 4.3.1 Discuss when y(0) = 0, y
(0) = 1, y(1) = 1, y
(1) is arbitrary, the
extremal situation of the functional J [y] =
1
0 (a + y
2
)dx.
Solution F = a + y
2 , F y = 0, F y = 0, F y = 2y
, Euler-Poisson equation is
d
2
dx 2 (2y
) = 0
( 1 )
or
y
(4)
= 0
( 2 )
The general solution is
y = c 1 + c 2 x + c 3 x
2
+ c 4 x
3
(3)
From the boundary condition y(0) = 0, we get c 1 = 0, from the boundary
condition y
(0) = 1, we get c 2 = 1. From the boundary condition y(1) = 1, we get
c 3 + c 4 = 0
( 4 )
According to the arbitrariness of y
(1), it can be seen from the third formula of
Eq. (4.3.10) that F y
x=1
= 0, consequently there is y
(1) = 0. Thus from Eq. (3),
we give
y
(1) = 2c 3 + 6c 4 = 0
( 5 )
Solving simultaneously Eqs (4) and (5), we give c 3 = 0, c 4 = 0. Thus the
extremum of the functional can only be achieved on the straight line y = x.
Example 4.3.2 Let the functional J [y] =
1
0 (y
2
− 2x y)dx, the boundary conditions are y(0) = 0, y
(0) = 0, y(1) = 1, according to δ J = 0, determine the
extremal curve of the functional.
Solution The Euler equation of the functional is
−2x + 2y
(4)
= 0
or
y
(4)
= x