294
4 Problems with Variable Boundaries
∂
2 w
∂ x 2 +
∂
2 w
∂ y 2 =
1
r
∂w
∂r
+
∂
2 w
∂r 2 +
1
r 2
∂
2 w
∂θ 2
(14)
Using the above equations, the Ostrogradsky Eq. (12) can be written as
1
r
∂w
∂r
+
∂
2 w
∂r 2 +
1
r 2
∂
2 w
∂θ 2 +
q
N
= 0
(15)
The boundary conditions are:
(1) On Γ 1 , namely when r = R, there is
w(R) = 0
(16)
(2) On Γ 2 , namely when r = R 0 , R 0 is the radius of the circle for the contact
domain, there is
w(R 0 ) = d
(17)
(3) On the variable boundary Γ 2 , supposing that w is axisymmetric, there is w θ = 0,
by Eq. (13), Eq. (9) can be written as
∂w
∂r
r =R 0
= w
(R 0 ) = 0
(18)
Because w is axisymmetric, Eq. (15) can be simplified to
1
r
d
dr
r
dw
dr
+
q
N
= 0
(19)
The general solution of Eq. (19) is
w = −
1
4
q
N
r
2
+ c 1 ln r + c 2
(20)
where, c 1 , c 2 are undetermined integral constants.
Using the boundary conditions (16), (17) and (18), the three equations determining
c 1 , c 2 and R 0 are obtained, namely
⎧
⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎩
w(R) = −
1
4
q
N
R
2
+ c 1 ln R + c 2 = 0
w(R 0 ) = −
1
4
q
N
R
2
0 + c 1 ln R 0 + c 2 = d
w
(R 0 ) = −
1
2
q
N
R 0 +
c 1
R 0
= 0
(21)
Solving the above equations, we get