Environmental Engineering Reference
In-Depth Information
One-Dimensional Wave Equations
To solve, using the method of descent based on Eq. (2.84),
u
tt
=
a
2
u
xx
+
f
(
x
,
t
)
, −
∞
<
x
<
+
∞
,
0
<
t
,
(2.96)
u
(
x
,
0
)=
ϕ
(
x
)
,
u
t
(
x
,
0
)=
ψ
(
x
)
.
Consider the spherical surface
S
at
of center
M
(
x
)
and radius
at
. Its intersecting
2
2
,
x
circumference with plane-
Oxy
is denoted by
C
at
:
2
(
ξ
−
x
)
+
η
=(
at
)
−
at
≤
at
. For a spherical zone on
S
at
,
ξ
≤
x
+
2
1
2
2
ξ
d
S
=
2
π
|
η
|
d
s
=
(
at
)
+(
ξ
−
x
)
+
η
d
ξ
=
2
π
at
d
ξ
,
x
−
at
≤
ξ
≤
x
+
at
.
Therefore,
⎡
⎣
∂
∂
⎤
⎦
M
)
at
M
)
at
1
ϕ
(
ψ
(
u
(
x
,
t
)=
d
S
+
d
S
4
π
a
t
S
at
S
at
∂
∂
x
+
at
x
+
at
1
=
ϕ
(
ξ
)
2
π
d
ξ
+
ψ
(
ξ
)
2
π
d
ξ
4
π
a
t
x
−
at
x
−
at
x
+
at
=
ϕ
(
x
+
at
)+
ϕ
(
x
−
at
)
1
2
a
+
ψ
(
ξ
)
d
ξ
,
(2.97)
2
x
−
at
which is the D'Alembert formula of one-dimensional wave equations. Similarly, for
the one-dimensional case, Eq. (2.85) reduces to
⎡
⎤
t
⎣
M
,
τ
)
⎦
1
f
(
u
(
x
,
t
)=
d
S
d
τ
4
π
a
a
(
t
−
τ
)
0
S
a
(
t
−
τ
)
x
+
a
(
t
−
τ
)
d
t
1
f
(
ξ
,
τ
)
=
−
τ
)
·
2
π
a
(
t
−
τ
)
d
ξ
τ
4
π
a
a
(
t
0
x
−
a
(
t
−
τ
)
t
x
+
a
(
t
−
τ
)
1
2
a
=
d
τ
f
(
ξ
,
τ
)
d
ξ
.
(2.98)
0
x
−
a
(
t
−
τ
)
The superposition of Eqs. (2.97) and (2.98) is, by the principle of superposition, the
solution of PDS (2.96).
Example 2.
Prove that the solution of
u
tt
a
2
u
xx
=
+
f
(
t
)
, −
∞
<
x
<
+
∞
,
0
<
t
,
u
(
x
,
0
)=
u
t
(
x
,
0
)=
0
.
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