156
2 Variational Problems with Fixed Boundaries
If F does not explicitly contain x, then there is F x = 0, thus, there is
d
dx
F − y
F y −
d
dx
F y
− y
F y
= F y y
+ F y y
+ F y y
− y
F y −
d
dx
F y
− y
d
dx
F y −
d
dx
F y
− y
F y − y
d
dx
F y
= F y y
− y
d
dx
F y −
d
dx
F y
= y
F y −
d
dx
F y +
d 2
dx 2 F y
= 0
Therefore
F − y
F y −
d
dx
F y
− y
F y = C
Quod erat demonstrandum.
Example 2.7.8 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) = sinh 1, y
(0) = 1, y
(1) = cosh 1.
Solution The Euler equation of the functional is
y
(4)
− y
= 0
The characteristic equation of the Euler equation is
r
4
− r
2
= r
2
(r − 1)(r + 1) = 0
The roots of the characteristic equation are 0, 0, 1, −1. The general solution of
the Euler equation is
y = c 1 e
x
+ c 2 e
−x
+ c 3 x + c 4
According to the boundary conditions y(0) = 0, y(1) = sinh 1, y
(0) = 1,
y
(1) = cosh 1, the following equations can be listed
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
c 1 + c 2 + c 4 = 0
c 1 e + c 2 e
−1
+ c 3 + c 4 = sinh 1
c 1 − c 2 + c 3 = 1
c 1 e − c 2 e
−1
+ c 3 = cosh 1
Solve for c 1 =
1
2
, c 2 = −
1
2
, c 3 = c 4 = 0. Thus the extremal vurve is
y =
e
x
− e
−x
2
= sinh x