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Appendix: All Solutions to the Problems
=
x 1
x 0
(x + 2y − 4yy
+ 4yy
)δydx − 2y
2
δy
x 1
x 0
2.4 Find the extremal curve of the functional J [y] =
1
0 (y
2
+ 12x y)dx, the
boundary conditions are y(0) = 0, y(1) = 1.
Solution Let the integrand F = y
2
+ 12x y, the Euler equation of the functional is
12x − 2y
= 0
Integrating two times, we get
y = x
3
+ c 1 x + c 2
From the boundary conditions y(0) = 0, y(1) = 1, we get c 1 = 0, c 2 = 0, thus
the extremal curve is y = x
3 .
2.5 Find the extremal curve of the functional J [y] =
x 1
x 0
(x y + y
2
)dx.
Solution Let the integrand F = x y + y
2 , the Euler equation of the functional is
x − 2y
= 0
Integrating two times, we get
y =
x
3
12
+ c 1 x + c 2
2.6 Find the extremal curve of the functional J [y] =
x 1
x 0
(2y + x y
2
)dx, the boundary
conditions are y(x 0 ) = y 0 , y(x 1 ) = y 1 .
Solution Let the integrand F = 2y + x y
2 , the Euler equation of the functional is
2 − 2y
− 2x y
= 0
Separating the variables, we obtain
dy
y − 1
= −
dx
x
Integrating, we get
(y
− 1)x = c 1
Integrating once more, we get