Appendix: All Solutions to the Problems
859
5.21 Let Γ be the fixed boundary of the domain D, find the Euler
equation and the natural boundary condition of the functional J [u] =
˜
D (u
2
x + u
2
y + u x ϕ + u y ψ)dxdy, where, ϕ, ψ ∈ C
1
(D).
Solution The integrand is F = u
2
x + u
2
y + u x ϕ + u y ψ = ∇u · ∇u + u x ϕ + u y ψ,
the various partial derivatives are
∂ F
∂∇u
= 2∇u, F u x = ϕ, F u y = ψ
The Euler equation of the functional is
2u − ϕ x − ψ y = 0 or u
2
xx + u
2
yy =
ϕ x + ψ y
2
For the two-dimensional problem, there are n x =
dy
dΓ
, n y = −
dx
dΓ
, thus the natural
boundary condition is
2∇u · n + ϕ
dy
dΓ
− ψ
dx
dΓ
= 0 or
∂u
∂n
=
1
2
ψ
dx
dΓ
− ϕ
dy
dΓ
Chapter 6: Solutions to the Problems in Variational
Problems in Parametric Forms
6.1 Verify whether the values of the following functionals have nothing to do with
the parameter form of the admissible curve.
(1) J [x, y, z] =
t 1
t 0
˙
x 2 + ˙
y 2 + ˙
z 2 dt; (2) J [x, y] =
t 1
t 0
x ˙
x ˙
y 2 dt;
(3) J [x, y] =
t 1
t 0
(x
2
˙
x + 3x y ˙
y
2
)dt; (4) J [x, y] =
t 1
t 0
(x ˙
y + y ˙
x
2
)dt;
(5) J [x, y] =
t 1
t 0
(x ˙
y + y ˙
x)
2 dt; (6) J [x, y] =
t 1
t 0
x ˙
y 2 + y ˙
x 2 dt.
Solution
(1) For an arbitrary positive number k, there is
F(x, y, k ˙
x, k ˙
y, k ˙
z) =
(k ˙
x) 2 + (k ˙
y) 2 + (k ˙
z) 2
= k
˙
x 2 + ˙
y 2 + ˙
z 2 = k F(x, y, ˙
x, ˙
y, ˙
z)
This function satisfies the condition of positively homogeneous function of order
one, so the value of the functional has nothing to do with the parameter form.
(2) For an arbitrary positive number k, there is
F(x, y, k ˙
x, k ˙
y) =
x(k ˙
x)(k ˙
y) 2 = k
kx ˙
x ˙
y 2 = k F(x, y, ˙
x, ˙
y)