10.2 The Euler Equations of Functionals …
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The Euler equation of the functional is
v∇ × ∇ × a − J = 0
( 2 )
Example 10.2.10 The Ericksen’s simplified (or reduced) energy functional is
J =
V
ε|∇u|
2
+ |∇ × u|
2
+
1
2ε
(1 − |u|
2
)
2
dV , where, ε is a constant. Find its
Euler equations.
Solution The original functional can be written as
J =
V
ε∇u : ∇u + |∇ × u|
2
+
1
2ε
(1 − u · u)
2
dV
(1)
The integrand is F = ε∇u : ∇u + |∇ × u|
2
+
1
2ε
(1 − u · u)
2 , taking the variation
to u, the various partial derivatives are
∂ F
∂ u
= −
2
ε
(1 − u · u)u,
∂ F
∇u
= 2ε∇u,
∂ F
∇ × u
= 2∇ × u
(2)
The Euler equation of the functional is
−ε∇ · ∇u + ∇ × ∇ × u =
1
ε
(1 − u · u)u
(3)
or
−εεu + ∇ × ∇ × u =
1
ε
(1 − |u|
2
)u
(4)
where, is the Laplace operator, = ∇ · ∇.
Example 10.2.11 Find the Euler equations of the energy functional of a gradient
vector flow
J =
V
[μ(|∇u|
2
+ |∇v|
2
+ |∇w|
2
) + |∇ f |
2
|ui + v j + wk − ∇ f |
2
]dV
where, μ is a regularity parameter, and f is an edge map.
Solution The integrand is
F = μ(|∇u|
2
+ |∇v|
2
+ |∇w|
2
) + |∇ f |
2
|ui + v j + wk − ∇ f |
2
(1)
or
F = μ(|∇u| 2 + |∇v| 2 + |∇w| 2 ) + |∇ f | 2 (u 2 + v 2 + w 2 − 2u f x − 2v f y − 2w f z + |∇ f | 2 )
(2)