636
10 Variational Problems of Functionals with Vector, Tensor …
Determine the various partial derivatives
∂ F
∂u
= 2(u − f x )|∇ f |
2
,
∂ F
∂v
= 2(v − f y )|∇ f |
2
,
∂ F
∂w
= 2(w − f z )|∇ f |
2
,
∂ F
∂∇u
= 2μ∇u,
∂ F
∂∇v
= 2μ∇v,
∂ F
∂∇w
= 2μ∇w
Substituting the aforementioned various expressions into the Euler equation, we
obtain
⎧
⎨
⎩
μμu − (u − f x )|∇ f |
2
= 0
μμv − (v − f y )|∇ f |
2
= 0
μμw − (w − f z )|∇ f |
2
= 0
(3)
Example 10.2.12 The Plateau(’s) problem. The functional of the Plateau’s problem
is
A =
D
1 + u 2
x (x, y) + u 2
y (x, y)dx dy =
D
√
1 + ∇u · ∇udx dy
Find its Euler equation and natural boundary condition.
Solution 1 Let the integrand F =
√
1 + ∇u · ∇u, then
∂ F
∂∇u
=
∇u
√
1+∇u·∇u
, the Euler
equation of the functional is
∇ ·
∇u
√
1 + ∇u · ∇u
= 0
( 1 )
The natural boundary condition is
∇u
√
1 + ∇u · ∇u
· n
Γ
= 0
( 2 )
where, Γ is the boundary of D.
Solution 2 Let the integrand F =
√
1 + ∇u · ∇u =
1 + |∇u|
2 , then
∂ F
∂|∇u|
=
|∇u|
√
1+|∇u|
2
, the Euler equation of the functional is
∇ ·
|∇u|
1 + |∇u|
2
∇u
|∇u|
= ∇ ·
∇u
1 + |∇u|
2
= 0
( 3 )
The natural boundary condition is
∇u
1 + |∇u|
2
· n
Γ
= 0
( 4 )
10 Variational Problems of Functionals with Vector, Tensor …
Determine the various partial derivatives
∂ F
∂u
= 2(u − f x )|∇ f |
2
,
∂ F
∂v
= 2(v − f y )|∇ f |
2
,
∂ F
∂w
= 2(w − f z )|∇ f |
2
,
∂ F
∂∇u
= 2μ∇u,
∂ F
∂∇v
= 2μ∇v,
∂ F
∂∇w
= 2μ∇w
Substituting the aforementioned various expressions into the Euler equation, we
obtain
⎧
⎨
⎩
μμu − (u − f x )|∇ f |
2
= 0
μμv − (v − f y )|∇ f |
2
= 0
μμw − (w − f z )|∇ f |
2
= 0
(3)
Example 10.2.12 The Plateau(’s) problem. The functional of the Plateau’s problem
is
A =
D
1 + u 2
x (x, y) + u 2
y (x, y)dx dy =
D
√
1 + ∇u · ∇udx dy
Find its Euler equation and natural boundary condition.
Solution 1 Let the integrand F =
√
1 + ∇u · ∇u, then
∂ F
∂∇u
=
∇u
√
1+∇u·∇u
, the Euler
equation of the functional is
∇ ·
∇u
√
1 + ∇u · ∇u
= 0
( 1 )
The natural boundary condition is
∇u
√
1 + ∇u · ∇u
· n
Γ
= 0
( 2 )
where, Γ is the boundary of D.
Solution 2 Let the integrand F =
√
1 + ∇u · ∇u =
1 + |∇u|
2 , then
∂ F
∂|∇u|
=
|∇u|
√
1+|∇u|
2
, the Euler equation of the functional is
∇ ·
|∇u|
1 + |∇u|
2
∇u
|∇u|
= ∇ ·
∇u
1 + |∇u|
2
= 0
( 3 )
The natural boundary condition is
∇u
1 + |∇u|
2
· n
Γ
= 0
( 4 )
