Environmental Engineering Reference
In-Depth Information
2.4.2 Uniqueness
Suppose that
u
1
(
x
,
t
)
and
u
2
(
x
,
t
)
are two solutions of PDS (2.21). The difference
between them
w
(
x
,
t
)=
u
1
(
x
,
t
)
−
u
2
(
x
,
t
)
must be the solution of the mixed problem
⎧
⎨
a
2
w
xx
,
w
tt
=
(
0
,
l
)
×
(
0
,
+
∞
)
w
(
0
,
t
)=
w
(
l
,
t
)=
0
,
(2.22)
⎩
w
(
x
,
0
)=
w
t
(
x
,
0
)=
0
.
Its uniqueness will be established once we show
w
(
x
,
t
)
≡
0.
From a physical point of view,
w
(
x
,
t
)
in PDS (2.22) represents the string dis-
placement in the field of vibration. As
ϕ
=
ψ
=
f
=
0and
w
(
0
,
t
)=
w
(
l
,
t
)=
0,
there is no cause for vibration so
w
0. However, this kind of physical expla-
nation cannot serve as the formal proof that
w
(
x
,
t
)
≡
0.
The
u
in Eq. (2.20) is indeed the solution of PDS (2.21) if the conditions specified
in the existence theorem hold. As
(
x
,
t
)
≡
ϕ
=
0and
ψ
=
0 satisfy these conditions, the
solution
w
(
x
,
t
)
of PDS (2.22) can be obtained by applying
ϕ
=
0and
ψ
=
0into
Eq. (2.20) so that
w
0. Note that the well-posedness of solution (2.20) has
not been established yet. This can only serve as a demonstration of
w
(
x
,
t
)
≡
(
x
,
t
)
≡
0, not
a formal proof.
To formally prove that
w
(
x
,
t
)
≡
0, consider the energy of a vibrating string.
Without loss of generality, let
ρ
=
1. The kinetic energy of string segment d
s
is
1
2
w
t
Δ
d
K
=
x
.
Its potential energy is, under the assumption of negligible
w
x
,
T
1
1
1
2
Tw
x
Δ
w
x
−
d
U
=
T
(
d
s
−
Δ
x
)=
+
Δ
x
=
x
a
2
is the tension. The total energy of the whole string of length
l
at time
instant
t
is thus
where
T
=
l
w
t
+
a
2
w
x
d
x
1
2
E
(
t
)=
.
0
This is called the
energy integral
.
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