2.6 Variational Problems Depending on Several Functions of One Variable
143
⎧
⎨
⎩
y = c 1 cos 2x + c 2 sin 2x
z = −
x
2
2
+ c 3 x + c 4
Using the boundary conditions, we obtain c 1 = 0, c 2 = 1, c 3 =
32+π
2
8π
, c 4 = 0,
therefore the extremal curves are
⎧
⎨
⎩
y = sin 2x
z = −
x
2
2
+
32 + π
2
8π
x
Example 2.6.3 Find the extremal curve of the functional J [y, z] =
x 1
x 0
(y
2
+ z
2
+ z
2
)dx, the boundary conditions are y(x 0 ) = y 0 , y(x 1 ) = y 1 ,
z(x 0 ) = z 0 , z(x 1 ) = z 1 .
Solution The Euler equations of the functional are
y
= 0
z − z
= 0
Integration gives
y = c 1 x + c 2 , z = c 3 e
x
+ c 4 e
−x
Using the boundary conditions, we obtain c 1 =
y 1 −y 0
x 1 −x 0
, c 2 = y 0 −
y 1 −y 0
x 1 −x 0
x 0 , c 3 =
z 1 e
x 1 −z 0 e
x 0
e 2x 1 −e 2x 0 , c 4 =
z 1 e
−x 1 −z 0 e
−x 0
e −2x 1 −e −2x 0 . Thus the extremal curves are
⎧
⎪ ⎨
⎪ ⎩
y = y 0 +
y 1 − y 0
x 1 − x 0
(x − x 0 )
z =
z 1 e
x 1 − z 0 e
x 0
e 2x 1 − e 2x 0
e
x
+
z 1 e
−x 1 − z 0 e
−x 0
e −2x 1 − e −2x 0
e
−x
Example 2.6.4 Find the extremal curve of the functional J [y, z] =
x 1
x 0
f (y
, z
)dx.
Solution The Euler equations of the functional present the following form
F y y y
+ F y z z
= 0, F y z y
+ F z z z
= 0
When F y y F z z − (F y z )
2
= 0, from the two equations, we obtain y
= 0 and
z
= 0, integration gives y = c 1 x + c 2 , z = c 3 x + c 4 . This is a family of straight
lines in space.
Example 2.6.5 Let the functional be
J [ϕ(x), ϕ(y)] =
¨
D
K (x, y)ϕ(x)ϕ(y)dxdy +
b
a
ϕ(x)[ϕ(x) − 2 f (x)]dx