4.4 Variational Problems of Functionals with Functions of Several Variables
289
Proof Converting the functional (4.4.9) into parameter form and quoting the
Jacobian determinant
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
J 1 = J (x, y) =
∂(x,y)
∂(ξ,η)
=
x ξ x η
y ξ y η
= x ξ y η − x η y ξ
J 2 = J (y, u) =
∂(y,u)
∂(ξ,η)
=
y ξ y η
u ξ u η
= y ξ u η − y η u ξ
J 3 = J (u, x) =
∂(u,x)
∂(ξ,η)
=
u ξ u η
x ξ x η
= u ξ x η − u η x ξ
(1)
where, ξ and η are parameters.
Let J (x, y) = 0, according to the implicit function existence theorem, then the
integrand of the functional (4.4.9) can be written as
F(x, y, u, u x , u y ) = F(x, y, u, p, q) = F
x, y, u, −
J 2
J 1
, −
J 3
J 1
(2)
where, p = u x = −
J (y,u)
J (x,y)
= −
J 2
J 1
, q = u y = −
J (u,x)
J (x,y)
= −
J 3
J 1
.
The integrand in the parameters form is
G = F J (x, y) = F J 1
(3)
The various partial derivatives of the integral G are
G x ξ = F J 1x ξ + F x ξ J 1 = y η F + J 1 F p ×
J 2 y η
J
2
1
− J 1 F q ×
J 1 u η − J 3 y η
J
2
1
= y η (F − pF p ) − F q ×
x η J 2
J 1
= y η (F − pF p ) + x η pF q
(4)
G x η = F J 1x η + F x η J 1 = −y ξ F − J 1 F p ×
J 2 y ξ
J
2
1
− J 1 F q ×
J 1 u ξ + J 3 y ξ
J
2
1
= −y ξ (F − pF p ) + F q ×
x ξ J 2
J 1
= −y ξ (F − pF p ) − x ξ pF q
(5)
G u ξ = F J 1u ξ + F u ξ J 1 = 0 + J 1 F p ×
y η
J 1
− J 1 F q ×
x η
J 1
= y η F p − x η F q (6)
G y ξ = F J 1y ξ + F y ξ J 1 = −x η F + J 1 F p ×
u η J 1 + J 2 x η
J
2
1
− J 1 F q ×
J 3 x η
J
2
1
= −x ξ (F − q F q ) + F p ×
y η J 3
J 1
= −x η (F − q F q ) − y η q F p
(7)
G y η = F J 1y η + F y η J 1 = x ξ F + J 1 F p ×
u ξ J 1 + J 2 x ξ
J
2
1
+ J 1 F q ×
J 3 x ξ
J
2
1
= x ξ (F − q F q ) − F p ×
y ξ J 3
J 1
= x ξ (F − q F q ) + y ξ q F p
(8)