Environmental Engineering Reference
In-Depth Information
2.5.2 Circular Domain
Consider the mixed problem
⎧
⎨
a
2
u
tt
=
Δ
u
+
f
(
x
,
y
,
t
)
,
D
×
(
0
,
+
∞
)
L
(
u
,
u
n
)
|
∂
D
=
0
,
(2.37)
⎩
u
(
x
,
y
,
0
)=
ϕ
(
x
,
y
)
,
u
t
(
x
,
y
,
0
)=
ψ
(
x
,
y
)
,
where
D
stands for the domain:
x
2
y
2
a
0
,
+
<
∂
D
is its boundary,
t
∈
(
0
,
+
∞
)
,and
L
(
u
0 can be the first, the second or the third kind of boundary condition.
We first develop the solution for the case of
f
,
u
n
)
|
∂
D
=
0 by using the method
of separation of variables. Considering a solution of the form
u
=
ϕ
=
=
v
(
x
,
y
)
T
(
t
)
,we
substitute it into the wave equation in PDS (2.37) to obtain
Δ
v
+
λ
v
=
0
,
L
(
v
,
v
n
)
|
∂
D
=
0
,
(2.38)
T
(
a
2
T
t
)+
λ
(
t
)=
0
,
where
is the separation constant.
As the boundary condition in Eq. (2.38) is not separable for
x
and
y
, we cannot
apply separation of variables to solve Eq. (2.38) in the Cartesian coordinate system.
In order to solve Eq. (2.38), consider polar coordinate transformation
x
−
λ
=
r
cos
θ
,
y
=
r
sin
θ
. Equation (2.38) is thus transformed into
⎧
⎨
⎩
2
v
2
v
∂θ
∂
1
r
∂
v
r
2
∂
1
k
2
v
r
2
+
r
+
2
+
=
00
<
r
<
a
0
∂
∂
(2.39)
L
(
v
,
v
r
)
|
r
=
a
0
=
0
,
v
(
r
,
θ
+
2
π
)=
v
(
r
,
θ
)
where
k
2
=
λ
. Now the boundary condition in Eq. (2.39) is separable for
r
and
θ
.If
we assume a solution of the form
v
=
R
(
r
)
Θ
(
θ
)
, Eq. (2.39) yields
Θ
+
μΘ
=
0
,
Θ
(
θ
+
2
π
)=
Θ
(
θ
)
,
(periodic condition)
(2.40)
k
2
−
r
2
R
1
r
R
+
R
+
=
0
,
L
(
R
,
R
r
)
|
r
=
a
0
=
0
(2.41)
where
μ
is the separation constant. To satisfy the periodic condition in Eq. (2.40),
μ
=
n
2
,
n
=
0
,
1
,
2
, ···
so
Θ
n
(
θ
)=
A
n
cos
n
θ
+
B
n
sin
n
θ
,
where
A
n
and
B
n
are arbitrary constants.
Substituting
n
2
into Eq. (2.41) yields a boundary-value problemwith param-
μ
=
k
2
eter
λ
=
⎧
⎨
k
2
r
2
R
n
n
2
1
r
R
n
(
R
n
(
r
)+
r
)+
−
(
r
)=
0
,
0
<
r
<
a
0
(2.42)
⎩
R
n
)
|
r
=
a
0
=
R
n
(
L
(
R
n
,
0
, |
R
n
(
0
)
| <
∞
, |
0
)
| <
∞
.
Search WWH ::
Custom Search