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Appendix: All Solutions to the Problems
yy
= c
Integrating we get
y
2
= c 1 x + c 2
where, c 1 = 2c. From the boundary conditions y(0) = 0, y(1) = 1, we solve for
c 1 = 1, c 2 = 0. Thus y =
√
x.
2.44 Find the extremal curve of the functional J [y] =
1
0 (y
2
+ y
2
)dx, the boundary
conditions are y(0) = 0, y(1) = 1.
Solution The Euler equation of the functional can be expressed as
y
− y = 0
The general solution of the Euler equation is
y = c 1 e
x
+ c 2 e
−x
From the boundary conditions y(0) = 0, y(1) = 1, solve for c 1 =
1
e 1 −e −1 = −c 2 ,
thus the extremal curve is
y =
e
x
− e
−x
e 1 − e −1 =
sinh x
sinh 1
2.45 Find the extremal curve of the functional J [y] =
1
0 (y
2
+ 4y
2
)dx, the
boundary conditions are y(0) = e
2 , y(1) = 1.
Solution The Euler equation of the functional is
8y − 2y
= 0
The general solution of the Euler equation is
y = c 1 e
2x
+ c 2 e
−2x
From the boundary conditions y(0) = e
2 , y(1) = 1, we get c 1 = 0, c 2 = e
2 ,
therefore the extremal curve is y = e
2(1−x) .
2.46 Find the extremal curve of the functional J [y] =
x 1
x 0
(ay + by
2
)dx, the
boundary conditions are y(x 0 ) = y 0 , y(x 1 ) = y 1 .
Solution The Euler equation of the functional is