10.2 The Euler Equations of Functionals …
637
Example 10.2.13 Find the Euler equation of the functional J [u] =
V [|∇u| + f (x, y, z)u]dxdydz and natural boundary condition.
Solution Let the integrand F = |∇u| + f (x, y, z)u, then
∂ F
∂u
= f (x, y, z),
∂ F
∂|∇u|
=
1, the Euler equation of the functional is
f (x, y, z) − ∇ ·
∇u
|∇u|
= 0 or ∇ ·
∇u
|∇u|
= f (x, y, z)
(1)
The natural boundary condition is
∇u
|∇u|
· n
S
= 0
( 2 )
there S is the area that surrounds V.
Example 10.2.14 The mixed type functional is
J [u] =
V
[k(x, y, z)∇u · ∇u + 2u f (x, y, z)]dV +
S
u
2 h(x, y, z)dS
Find its Euler equation and natural boundary condition.
Solution Putting k = k(x, y, z), f = f (x, y, z), h = h(x, y, z), F = k∇u ·∇u +
2u f , the Euler equation of the functional is
∇ · k∇u =
∂ku x
∂ x
+
∂ku y
∂ y
+
∂ku z
∂z
= f (in V )
The natural boundary condition is
k∇u · n + hu = 0 or k(u x n x + u y n y + u z n z ) + hu = 0 or k
∂u
∂n
+ hu = 0 (on S)
This is exactly the same as the Euler equation and boundary condition of Example
5.4.3.
Example 10.2.15 Find the Euler equation and corresponding boundary condition
of the functional for the acoustic field J =
1
2
V (|∇ p|
2
− k
2 p
2
− 2jωρqp)dV +
S jωρv n pdS.
Solution Putting F =
1
2
(|∇ p|
2
− k
2 p
2
− 2jωρqp), G = jωρv n p, the various partial
derivatives are
F p = −k
2 p − jωρq,
∂ F
∂∇ p
= ∇ p, G p = jωρv n
The Euler equation of the functional is