2.10 Invariance of the Euler Equation
181
Substituting the expressions of x, y of Eqs. (2.10.13), (2.10.14) and (2.10.16) into
Eq. (2.10.5), and take note that ϕ u , ϕ v , ψ u and ψ v are still the function of ϕ and ψ,
we obtain
J [v(u)] =
u1
u0
F
ϕ(u, v), ψ(u, v),
ψ u + ψ v v u
ϕ u + ϕ v v u
(ϕ u + ϕ v v u )du =
u1
u0
F 1 (u, v, v
)du
(2.10.17)
where
F 1 (u, v, v
) = F
ϕ(u, v), ψ(u, v),
ψ u + ψ v v u
ϕ u + ϕ v v u
(ϕ u + ϕ v v u )
(2.10.18)
This can be concluded that if y = y(x) is the extremal curve the functional
(2.10.5) on Oxy plane, then v = v(u) is the extremal curve the functional (2.10.17)
on Ouv plane.
For the variational problem depending on the function of several variables, there
is still the invariance of Euler equation.
Example 2.10.1 Try to make the functional
J [u] =
¨
D
F(x, y, u, u x , u y )dxdy
be written in the form expressed by polar coordinate.
Solution Let U (r, θ) = u(r cos θ, r sin θ), then there is
x = r cos θ, y = r sin θ
The both sides of the above two expressions are taken the partial derivatives with
respect to x, y, we give
1 =
∂r
∂ x
cos θ − r sin θ
∂θ
∂ x
, 0 =
∂r
∂ x
sin θ + r cos θ
∂θ
∂ x
0 =
∂r
∂ y
cos θ − r sin θ
∂θ
∂ y
, 1 =
∂r
∂ y
sin θ + r cos θ
∂θ
∂ y
Solving for the various partial derivatives, we obtain
∂r
∂ x
= cos θ,
∂θ
∂ x
= −
sin θ
r
,
∂r
∂ y
= sin θ,
∂θ
∂ y
=
cos θ
r
Thus
∂u
∂ x
=
∂U
∂r
∂r
∂ x
+
∂U
∂θ
∂θ
∂ x
=
∂U
∂r
cos θ −
∂U
∂θ
sin θ
r