Appendix: All Solutions to the Problems
949
J =
¨
A
K (|∇φ|)(φ
2
x + φ
2
y )dxdy =
¨
A
K (|∇φ|)∇φ · ∇φdxdy
=
¨
A
K (|∇φ|)|∇φ|
2 dxdy
Putting F = K (|∇φ|)|∇φ|
2 , Finding the various partial derivatives, we have
∂ F
∂ K (|∇φ|)
∂ K (|∇φ|)
∂|∇φ|
∇φ
|∇φ|
=
∂ K (|∇φ|)
∂|∇φ|
|∇φ|∇φ,
∂ F
∂|∇φ|
= 2K (|∇φ|)|∇φ|,
The Euler equation of the functional is
∇ ·
∂ K (|∇φ|)
∂|∇φ|
|∇φ|∇φ + 2∇ · K (|∇φ|)∇φ = 0
If omitting the first term, then
∂(K φ x )
∂ x
+
∂(K φ y )
∂ y
= 0
The given result is
∂(K φ x )
∂ x
+
∂(K φ y )
∂ y
= 0
10.26 Find the Euler equation of the functional
J [u]
=
V
|∇u| ln(1 + |∇u|) + λ 1
f
u
+ λ 2
f
u
+ a
2
dV .
Solution Putting F = |∇u| ln(1 + |∇u|), the various partial derivatives are
∂ F
∂u
= −λ 1
f
u 2 − 2λ 2
f
u
+ a
f
u 2 = −(λ 1 + 2λ 2 a)
f
u 2 − 2λ 2
f
u 3
∂ F
∂|∇u|
= ln(1 + |∇u|) + |∇u|
1
1+|∇u|
=
|∇u|+(1+|∇u|) ln(1+|∇u|)
1+|∇u|
The Euler equation of the functional is
∇ ·
|∇u| + (1 + |∇u|) ln(1 + |∇u|)
(1 + |∇u|)|∇u|
∇u + (λ 1 + 2λ 2 a)
f
u 2 + 2λ 2
f
u 3 = 0
10.27 Find the Euler equation of the functional J [u] =
V
|∇u − a|
2 dV .
Solution Putting F = |∇u − a|
2
= (∇u − a) · (∇u − a), the partial derivative of
F with respect to ∇u is
∂ F
∂∇u
= 2(∇u − a), the Euler equation of the functional is
u − ∇ · a = 0.