Appendix: All Solutions to the Problems
763
y = x + c 1 ln x + c 2
From the boundary conditions y(x 0 ) = y 0 , y(x 1 ) = y 1 , we obtain
c 1 =
y 1 − y 0 − x 1 + x 0
ln x 1 − ln x 0
, c 2 =
(y 0 − x 0 ) ln x 1 − (y 1 − x 1 ) ln x 0
ln x 1 − ln x 0
Thus the extremal curve is
y = x +
(y 1 − y 0 ) − (x 1 − x 0 )
ln x 1 − ln x 0
ln x +
(y 0 − x 0 ) ln x 1 − (y 1 − x 1 ) ln x 0
ln x 1 − ln x 0
2.7 Find the extremal curve of the functional J [y] =
2
1 (x y
2
− y)dx, the boundary
conditions are y(1) = 0, y(2) = 1.
Solution The Euler equation of the functional is
−1 − 2y
− 2x y
= 0
or
dy
2y + 1
= −
dx
2x
Integrating, we get
(2y
+ 1)x = c 1
Integrating again, we get
2y + x = c 1 ln x + c 2
From the boundary conditions y(1) = 0, y(2) = 1, we get c 2 = 1, c 1 =
3
ln 2
.
Thus there is y =
3 ln x
2 ln 2
+
1−x
2
.
2.8 Find the extremal curve of the functional J [y] =
e
1 (x y
2
+ yy
)dx, the boundary
conditions are y(1) = 0, y(e) = 1.
Solution The Euler equation of the functional is
y
− y
− 2y
− 2x y
= 0
or
x y
+ y
= 0