Appendix: All Solutions to the Problems
951
Solution Putting F =
√
1 + ∇u · ∇u + λu, the partial derivatives of F with respect
to ∇u and u are
∂ F
∇u
=
∇u
√
1+∇u·∇u
,
∂ F
∂u
= λ, the Euler equation of the functional is
λ − ∇ ·
∇u
√
1 + ∇u · ∇u
= 0
The boundary condition is
∇u
√
1 + ∇u · ∇u
· n + 1
C
= 0
10.32 Find the Euler equation and boundary condition of the functional J =
V [ p(∇u)
2
+ qu
2
− 2 f u]dV +
S (σ u
2
− 2ku)dS.
Solution Putting F = p(∇u)
2
+ qu
2
− 2 f u, the various partial derivatives are
∂ F
∂u
=
2qu − 2 f ,
∂ F
∂∇u
= 2 p∇u, the Euler equation of the functional is
−∇ · ( p∇u) + qu = f
(1)
The boundary condition is
( p∇u · n + σ u)| S = k or
p
∂u
∂n
+ σ u
S
= k
(2)
10.33 Find the Euler equations and boundary conditions of the functional
J [u, v] =
V
αu
2 (v str − c)
2
1
|∇v| 2 + ε
+ β|∇u × ∇v|
2 + γ [δ + (1 − δ)|∇v|](u − u ind )
2
dV
Solution Putting F = αu
2
(v str −c)
2
1
|∇v|
2 +ε
+β|∇u × ∇v|
2
+γ [δ+(1−δ)|∇v|](u−
u ind )
2 , finding the various partial derivatives
∂ F
∂u = 2αu(v str − c) 2 1
|∇v| 2 +ε
+ 2γ [δ + (1 − δ)|∇v|](u − u ind ), ∂ F
∂∇u×∇v = β∇u × ∇v
∂ F
∂|∇v| = −αu 2 (v str − c) 2 2|∇v|
(|∇v| 2 +ε) 2 + γ (1 − δ)(u − u ind ) 2
Using the result of problem 10.1, the Euler equations of the functional are
⎧
⎨
⎩
2αu(v str − c) 2 1
|∇v| 2 +ε
+ 2γ [δ + (1 − δ)|∇v|](u − u ind ) − ∇ · [∇ × v × (β∇u × ∇v)] = 0
∇ ·
αu 2 (v str − c) 2 2∇v
(|∇v| 2 +ε) 2 − γ (1 − δ)(u − u ind ) 2 ∇v
|∇v| + ∇ × u × (β∇u × ∇v)
= 0
Using the result of problem 10.1, the natural boundary conditions are