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
2π
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
2π
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
2π
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
)
−
⎧
⎨
⎫
⎬
2π
at
2π
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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