Environmental Engineering Reference
In-Depth Information
u
x
(
0
,
t
)=
0. PDS (2.71) is equivalent to PDS (2.70) in 0
<
x
<
+
∞
,0
<
t
. Following
the same approach as for the odd continuation, we obtain
⎧
⎨
x
+
at
ϕ
(
x
+
at
)+
ϕ
(
x
−
at
)
1
2
a
x
a
,
+
ψ
(
ξ
)
d
ξ
,
t
≤
2
x
−
at
x
+
at
ϕ
(
x
+
at
)+
ϕ
(
at
−
x
)
1
2
a
u
(
x
,
t
)=
(2.72)
+
ψ
(
ξ
)
d
ξ
⎩
2
at
−
x
0
1
2
a
x
a
,
+
ψ
(
ξ
)
d
ξ
,
t
>
0
which shows that the incoming and the reflected waves have the same phase at the
end
x
=
0; thus there is no semi-wave loss.
Method of Continuation for Mixed Problems in Finite Domains
When equations are not separable, we cannot use the Fourier method to solve mixed
problems. While we may sometimes solve such problems by using integral transfor-
mations, it is often quite involved to perform the inverse transformations. Here we
discuss the method of continuation for such problems by considering a PDS arising
in mechanical machining
⎧
⎨
v
2
u
tt
+
2
vu
tx
+(
−
1
)
u
xx
=
f
(
x
,
t
)
,
0
<
x
<
1
,
0
<
t
,
u
(
0
,
t
)=
u
(
1
,
t
)=
0
,
(2.73)
⎩
u
(
x
,
0
)=
u
t
(
x
,
0
)=
0
.
Based on the given boundary conditions, consider an odd continuation of the non-
homogeneous term
f
as
the nonhomogeneous term after the continuation. This leads to the Cauchy problem
u
tt
(
x
,
t
)
. For convenience and conciseness, we still use
f
(
x
,
t
)
v
2
R
1
+
2
vu
tx
+(
−
1
)
u
xx
=
f
(
x
,
t
)
,
×
(
0
,
+
∞
)
,
v
>
1
,
(2.74)
u
(
x
,
0
)=
u
t
(
x
,
0
)=
0
.
=(
±
)
+
The equation in (2.74) is hyperbolic with characteristic curves
x
v
1
t
C
(
C
is a constant). Consider a variable transformation
ξ
=
x
−
(
v
+
1
)
t
,
η
=
x
−
(
v
−
1
)
t
,
PDS (2.74) is thus reduced into
⎧
⎨
4
f
1
−
(
v
−
1
)
ξ
+(
v
+
1
)
η
,−
ξ
−
η
2
u
ξη
=
F
(
ξ
,
η
)
,
F
(
ξ
,
η
)=
,
2
(2.75)
u
η
ξ
=
η
=
⎩
u
|
ξ
=
η
=
0
.
Its solution can be obtained by integrating the equation twice with respect to
ξ
and
η
respectively and applying the initial conditions,
η
η
η
ξ
(
ξ
,
η
)
ξ
.
u
(
ξ
,
η
)=
−
d
F
d
(2.76)
ξ
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