Environmental Engineering Reference
In-Depth Information
Remark 1. The two terms in the right-hand side of Eq. (2.93) clearly have the same
unit. The unit of the first term is, by viewing u as the displacement,
[
]= a 1
1
[
u
[ ϕ ]
d
ξ
d
η ]
t
2
2
2
(
at
)
( ξ
x
)
( η
y
)
T
L ·
1
T ·
1
L ·
L 2
=
L
·
=
L
,
which shows the correctness of the unit in Eq. (2.93).
Remark 2. In a polar coordinate system, Eq. (2.93) reads
at
r d r
0
1
ϕ (
x
+
r cos
θ ,
y
+
r sin
θ )
u
(
x
,
y
,
t
)=
d
θ
2
π
a
t
2
0
(
)
r 2
at
.
at
r d r
0
ψ (
x
+
r cos
θ ,
y
+
r sin
θ )
+
d
θ
(2.94)
0
2
r 2
(
at
)
Example 1 . Find the solution of
u tt =
a 2
Δ
u
, − <
x
,
y
< + ,
0
<
t
,
x 2
u
(
x
,
y
,
0
)=
(
x
+
y
) ,
u t (
x
,
y
,
0
)=
0
.
Solution. By Eq. (2.94), we obtain
at
r d r
0
2
1
a
(
x
+
r cos
θ )
(
x
+
y
+
r cos
θ +
r sin
θ )
u
(
x
,
y
,
t
)=
d
θ
2
π
t
0
2
r 2
(
at
)
at
1
a
r d r
x 2
r 2 cos 2
=
(
x
+
y
)
d
θ +
(
3 x
+
y
)
θ
d
θ
2
π
t
0
2
0
0
(
at
)
r 2
x 2
a 2 t 2
=
(
x
+
y
)+(
3 x
+
y
)
.
thus has unit L 3 . From the solution, we
If x and y have the unit of length, u
(
x
,
y
,
0
)
L 3
have
[
u
(
x
,
y
,
t
)] =
,
which agrees with the unit of u
(
x
,
y
,
0
)
.Otherwise,theremust
exist some errors in derivation.
Remark 3. By a similar approach we can also obtain, from Eq. (2.85), the solution
of
u tt =
a 2
Δ
u
+
f
(
x
,
y
,
t
) , − <
x
,
y
< + ,
0
<
t
,
u
(
x
,
y
,
0
)=
0
,
u t (
x
,
y
,
0
)=
0
.
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