4.2 Variational Problems of Functionals with Several Functions
259
+
x 1
x 0
[F(x, y + δy, z + δz, y
+ δy
, z
+ δz
) − F(x, y, z, y
, z
)]dx
(4.2.2)
Applying the mean value theorem to the first integral on the right side of the above
expression and considering the continuity of F, meanwhile separating out its main
linear part in the second integral, there is
δ J = F| x=x 1 δx 1 + F y δy
x=x 1
+ F z δz| x=x 1
+
x 1
x 0
F y −
d
dx
F y
δy +
F z −
d
dx
F z
δz
dx
(4.2.3)
Because the extremum of J [y, z] can only be obtained on the extremal curve,
therefore it must satisfy the Euler equations
F y −
d
dx
F y = 0, F z −
d
dx
F z = 0,
(4.2.4)
The general solution of the equations contains four arbitrary constants, since point
B can change, it has one more unknown x 1 , in order to make the functional (4.2.1)
have unique a set of solutions, will need to determine the five constants will need
to be determined. Because point A is fixed, from y(x 0 ) = y 0 and z(x 0 ) = z 0 , the
two constants can be determined. The remaining three constants can be determined
according to the extremal condition of the functional δ J = 0.
According to Eq. (4.2.4) and the extremum condition δ J = 0, the expression
(4.2.3) can be written as
δ J = F| x=x 1 δx 1 + F y δy
x=x 1
+ F z δz| x=x 1 = 0
(4.2.5)
According to the discussion in Sect. 4.1, there is
δy| x=x 1 = δy 1 − y
(x 1 )δx 1 , δz| x=x 1 = δz 1 − z
(x 1 )δx 1
(4.2.6)
Substituting Eq. (4.2.6) into Eq. (4.2.5), we get
δ J = (F − y
F y − z
F z )
x=x 1
δx 1 + F y
x=x 1
δy 1 + F z | x=x 1 δz 1 = 0
(4.2.7)
In the expression (4.2.7), δx 1 , δy 1 and δz 1 are arbitrary, namely point B can change
in any way. According to the relations among y 1 , z 1 and x 1 , which can be divided
into four kinds of situations to discuss.