948
Appendix: All Solutions to the Problems
∇ · (e
ψ
∇ϕ) = 0
e
ψ
2(∇ϕ)
2
+
(∇ψ)
2
2
− 2k
2
− ∇ · (e
ψ
∇ψ) = 0
(1)
Taking note that ∇e
ψ
=
∂e
ψ
∂ x i
= e
ψ ∂ψ
∂ x i
= e
ψ
∇ψ, and using the property of the
divergence ∇ · ua = ∇u · a + u∇ · a, Eq. (1) can be transformed into
∇ϕ · ∇ψ + ϕ = 0
2(∇ϕ)
2
−
(∇ψ)
2
2
− ψ − 2k
2
= 0
(2)
The natural boundary conditions are ∇ϕ · n| S = 0, ∇ψ · n| S = 0.
10.24 In the process of adiabatic compression heating for the plasma of magnetic
confinement, the Lagrange function is
L =
(∇ψ) 2
2R 2
1 −
F 2
ρ
−
1
2
ρ
ρ − F 2
I
R
+ R FΩ
2
−
ρ R 2 Ω 2
2
+
ρ γ S
γ − 1
− ρ H
R
where, F, I , Ω and S are all arbitrary functions of ψ, find the Euler equation and
natural boundary condition of the functional.
Solution Taking the various partial derivatives, we have
∂ L
∂ψ
= −R
I
R
+ R FΩ
ρ
ρ − F 2
I ψ
R
+ R(F ψ Ω + FΩ ψ )
+
I
R
+ R FΩ
ρ F F ψ
(ρ − F 2 )
− ρ R 3 ΩΩ ψ +
Rρ γ
γ − 1
S ψ − Rρ H ψ −
(∇ψ) 2
R
F
ρ
F ψ
∂ L
∂∇ψ
=
1
R
1 −
F 2
ρ
∇ψ
The Euler equation corresponding to the Lagrange function is
− R
I
R
+ R FΩ
ρ
ρ − F 2
I ψ
R
+ R(F ψ Ω + FΩ ψ )
+
I
R
+ R FΩ
ρ F F ψ
(ρ − F 2 )
− ρ R 3 ΩΩ ψ +
Rρ γ
γ − 1
S ψ − Rρ H ψ −
(∇ψ) 2
R
F
ρ
F ψ − ∇ ·
1
R
1 −
F 2
ρ
∇ψ = 0
The natural boundary conditions are
1
R
1 −
F
2
ρ
∇ψ · n
S
= 0.
10.25 The non-Darcy flow problem. The functional is J
=
˜
A K (|∇φ|)(φ
2
x + φ
2
y )dxdy, where, φ is the piezometric head; K (|∇φ|) is the
water conductance. Find its Euler equation.
Solution The functional is rewritten as