186
6 Two-Dimensional Problems in Elasticity …
We find the following ordinary differential equation to determine f (r ).
i.e.,
d
2
dr 2 +
1
r
d
dr
−
4
r 2
d
2 f
dr 2 +
1
r
d f
dr
−
4 f
r 2
= 0
On simplification, we get
d
4 f
dr 4 +
2
r
d
3 f
dr 3 −
9
r 2
d
2 f
dr 2 +
9
r 3
d f
dr
= 0
This is an ordinary differential equation, which can be reduced to a linear differential equation with constant coefficients by introducing a new variable “t” such that
r = e
t .
Also,
d f
dr
=
d f
dt
dt
dr
=
1
r
d f
dt
d
2 f
dr 2 =
1
r 2
d
2 f
dt 2 −
d f
dt
d
3 f
dr 3 =
1
r 3
d
3 f
dt 3 − 3
d
2 f
dt 2 + 2
d f
dt
d
4 f
dr 4 =
1
r 4
d
4 f
dt 4 − 6
d
3 f
dt 3 + 11
d
2 f
dt 2 − 6
d f
dt
on substitution, we get.
1
r 4
d
4 f
dt 4 − 6
d
3 f
dt 3 + 11
d
2 f
dt 2 − 6
d f
dt
+
2
r 4
d
3 f
dt 3 − 3
d
2 f
dt 2 + 2
d f
dt
−
9
r 4
d
2 f
dt 2 −
d f
dt
+
9
r 4
d f
dt
= 0
or
d
4 f
dt 4 − 4
d
3 f
dt 3 − 4
d
2 f
dt 2 + 16
d f
dt
= 0.
Let
d f
dt
= m
∴ m
4
− 4m
3
− 4m
2
+ 16m = 0
m
3
(m − 4) − 4m(m − 4) = 0
∴ (m − 4)
m
3
− 4m
= 0
or m
m
2
− 4
(m − 4) = 0
∴ m = 0, m = ±2, m = 4