958
Appendix: All Solutions to the Problems
∇ × p∇ × v − ω
2 qv = −g
The boundary condition is
( p∇ × v) × n| S = 0
10.45 Find the Euler equation and boundary condition of the functional
J [ A] =
V
1
2
1
μ
(∇ × A)
2
+
1
μ
(∇ · A)
2
+ (σ + jεω) jω A
2
− 2 A · J f
dV
−
S 2
(H × n) S 2 · AdS −
S 2
1
μ
∇ · A
S 2
A · ndS
where, μH = ∇ × A.
Solution Putting F =
1
2
1
μ
(∇ × A)
2
+
1
μ
(∇ · A) + (σ + jεω) jω A
2
− 2 A · J f
,
the various partial derivatives are
∂ F
∂ A
= (σ + jεω) jω A − J f ,
∂ F
∂∇ × A
=
1
μ
∇ × A,
∂ F
∂∇ · A
=
1
μ
∇ · A
The Euler equation of the functional is
∇ ×
1
μ
(∇ × A) − ∇
1
μ
(∇ · A) + (σ + jεω) jω A − J f = 0
(1)
The boundary conditions are
H × n = (H × n) S 2 ,
1
μ
∇ · A =
1
μ
∇ · A
S 2
(2)
10.46 The functional of the nonlinear generalized variational principle (potential
energy form) is
Π
J 1
G [U, S, u] =
V
[Σ(U) + S : (R
T
· F − U) − u · F]dV −
S σ
u · PdS
−
S u
(u − u) · R · S · NdS
Write the Euler equations and boundary conditions of the functional.
Solution The integrand is F = Σ(U)+ S : (R
T
· F−U)−u· F, where, F = I +u∇,
S : R
T
· F = R · S : F = R · S : (I + u∇), finding the various partial derivatives