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2 Variational Problems with Fixed Boundaries
Example 2.4.5 Find the extremal curve of the functional J [y]
=
x 1
x 0
(x
2 y
2
+ 2y
2
)dx.
Solution The Euler equation of the functional is
4y − 4x y
− 2x
2 y
= 0
or
x
2 y
+ 2x y
− 2y = 0
The linear ordinary differential equation is called the Euler equation. Let x = e
t
or t = ln x, and use d to represent the derivative with respect to t, then the original
equation can be converted into
d(d − 1)y + 2dy − 2y = 0
The characteristic equation of the equation is
r (r − 1) + 2r − 2 = 0
namely
r
2
+ r − 2 = 0
The two roots of the characteristic equation are r 1 = 1, r 2 = −2. Thus the
extremal curve of the functional is
y = c 1 e
t
+ c 2 e
−2t
Substituting e
t back to the original variable, there is
y = c 1 x + c 2 x
−2
The two integral constants of the equation are determined by the boundary
conditions.
Example 2.4.6 Find the extremal curve of the functional J [y]
=
1
0 (y
2
− y
2
− y)e
2x dx, he boundary conditions are y(0) = 0, y(1) = e
−1 .
Solution The Euler equation of the functional is
−e
2x
− 2e
2x y − 4e
2x y
− 2e
2x y
= 0
or