d
d kr
ð Þ
d Δp
ð Þ
d kr
ð Þ
þ
1
kr
d Δp
ð Þ
d kr
ð Þ
À 1 þ
1=2
ð
Þ
2
kr
ð Þ
2
"
#
Δp
ð Þ ¼ 0:
ð2:50Þ
Its solution is
Δp r
ð Þ ¼ C 1 I 1=2 kr
ð Þ þ C 2 K 1=2 kr
ð Þ ¼ C 2 K 1=2 kr
ð Þ,
ð2:51Þ
where I 1/2 and K 1/2 are the first- and second-kind modified Bessel functions of order
1/2:
I 1=2 x
ð Þ ¼
ffiffiffiffiffi
2
πx
r
sinh x, K 1=2 x
ð Þ ¼
ffiffiffiffiffi
2
πx
r
e
Àx
:
ð2:52Þ
C 1 and C 2 are undetermined constants. Since I 1/2 is divergent for large values of its
argument, we choose C 1 ¼ 0. To find u we need to solve Eq. (2.46) which takes the
following form after the use of Eqs. (2.51) and (2.45):
∂
2
∂r 2 þ
∂
r∂r
þ
1
r 2
∂
2
∂θ
2
u ¼
eq
ε c
C 2 K 1=2 kr
ð Þsin
θ
2
:
ð2:53Þ
Let
u r, θ
ð Þ ¼ u r
ð Þ sin
θ
2
:
ð2:54Þ
The substitution of Eq. (2.54) into Eq. (2.53) yields
∂
2
∂r 2 þ
∂
r∂r
À
1
4r 2
u ¼
eq
εc
C 2 K 1=2 k r
ð Þ:
ð2:55Þ
The homogeneous solution of Eq. (2.55) can be obtained easily. It can be verified
that its particular solution is proportional to K 1/2 . Hence the general solution to
Eq. (2.55) is
u r, θ
ð Þ ¼ C 3
ffiffi
r
p þ C 4
1
ffiffi
r
p þ
1
k
2
eq
εc
C 2 K 1=2 k r
ð Þ
!
sin
θ
2
¼ C 3
ffiffi
r
p þ C 4
1
ffiffi
r
p þ
1
k
2
eq
εc
C 2
ffiffiffiffiffiffiffi
2
πkr
r
e
Àkr
"
#
sin
θ
2
,
ð2:56Þ
where C 3 and C 4 are undetermined constants. If we want u to be bounded at the crack
tip, we must have
2.4 Antiplane Crack
25