160
2 Variational Problems with Fixed Boundaries
u = u(x, y) + εη(x, y)
(2.8.5)
where, ε is a sufficiently small parameter in absolute value; η(x, y) is an arbitrarily
differentiable function, and satisfies on the boundary
η(x, y)| L = 0
(2.8.6)
Let
δu = εη(x, y)
(2.8.7)
then there is
δu| L = 0
(2.8.8)
For Eq. (2.8.7), taking the partial derivatives with respect to x and y
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
∂
∂ x
(δu) = δ
∂u
∂ x
= εη x (x, y)
∂
∂ y
(δu) = δ
∂u
∂ y
= εη y (x, y)
(2.8.9)
Because the functional J [u(x, y)] attains extremum on u = u(x, y), then the
corresponding first variation δ J is equal to zero, that is
δ J =
¨
D
(F u δu + F u x δu x + F u y δu y )dxdy = 0
(2.8.10)
The second term and third term on the right side of Eq. (2.8.10) are recast as
F u x δu x + F u y δu y =
∂
∂ x
(F u x δu) +
∂
∂ y
(F u y δu) −
∂
∂ x
F u x +
∂
∂ y
F u y
δu (2.8.11)
Substituting Eq. (2.8.11) into Eq. (2.8.10), we obtain
δ J =
¨
D
F u −
∂
∂ x
F ux −
∂
∂ y
F u y
δudxdy +
¨
D
∂
∂ x
(F ux δu) +
∂
∂ y
(F u y δu)
dxdy
(2.8.12)
Using the Green formula, the second integral on the right side of Eq. (2.8.12) is
changed into
¨
D
∂
∂ x
(F u x δu) +
∂
∂ y
(F u y δu)
dxdy =
L
(F u x δudy − F u y δudx)
From Eq. (2.8.8), δu| L = 0, so the integral of the above expression vanishes.
Thus, by Eq. (2.8.12), we obtain