2.8 Variational Problems Depending on Functions of Several Variables
169
∂
∂r
F w r = 2
−
1
r 2
∂w
∂r
+
1
r
∂
2 w
∂r 2 − 2
1
r 3
∂
2 w
∂θ 2 +
1
r 2
∂
3 w
∂r ∂θ 2
+ 2μ
∂
3 w
∂r 3
∂
∂θ
F w θ = 4(1 − μ)
−
1
r 2
∂
3 w
∂r ∂θ 2 +
1
r 3
∂
2 w
∂θ 2
∂
2
∂r 2 F w rr = 2
2
∂
3 w
∂r 3 + r
∂
4 w
∂r 4
+ 2μ
∂
3 w
∂r 3 + 2
1
r 3
∂
2 w
∂θ 2 − 2
1
r 2
∂
3 w
∂r ∂θ 2 +
1
r
∂
4 w
∂r 2 ∂θ 2
∂
2
∂r ∂θ
F w r θ = 4(1 − μ)
−2
1
r 2
∂
3 w
∂r ∂θ 2 +
1
r
∂
4 w
∂r 2 ∂θ 2 + 2
1
r 3
∂
2 w
∂θ 2
∂
2
∂θ 2 F w θθ = 2
1
r 2
∂
3 w
∂r ∂θ 2 +
1
r 3
∂
4 w
∂θ 4
+ 2μ
1
r
∂
3 w
∂r 2 ∂θ 2
Substituting the above related equations into the Ostrogradsky equation, after
management, we obtain
∂ 4 w
∂r 4 +
2
r
∂ 3 w
∂r 3 −
1
r 2
∂ 2 w
∂r 2 +
1
r 3
∂w
∂r
+
1
r 4
∂ 4 w
∂θ 4 + 2
2
1
r 4
∂ 2 w
∂θ 2 −
1
r 3
∂ 3 w
∂r ∂θ 2 +
1
r 2
∂ 4 w
∂r 2 ∂θ 2
=
q
D
Making use of Eq. (1.4.15), the above equation can be changed into
D
2 w = q
This is the control differential equation for thin plate (or slab) bending in the
form of operator.
Example 2.8.9 Write the Euler-Ostrogradsky equation of the functional
J [u] =
¨
D
[u
4
x + u
4
y + 12u f (x, y)]dxdy.
Solution The Euler-Ostrogradsky equation of the functional is
12 f (x, y) −
∂
∂ x
4u
3
x −
∂
∂ y
4u
3
y = 0
or
u
2
x u xx + u
2
y u yy = f (x, y)
Example 2.8.10 Using the relationships between rectangular coordinates and polar
coordinates x = r cos θ , y = r sin θ , the differential of arc length ds =
˙
x 2 + ˙
y 2 dθ ,
there is the functional
J [y] =
B
A
(x
2
+ y
2
)ds