Environmental Engineering Reference
In-Depth Information
we arrive at
c
1
(
c
2
(
x
)
y
1
+
x
)
y
2
=
f
(
x
)
.
(6.38)
The solution of Eq. (6.37) and Eq. (6.38) is
−
y
1
(
)
(
)
(
)
(
)
y
2
x
f
x
x
f
x
c
1
(
x
)=
d
x
,
c
2
(
x
)=
d
x
,
Δ
Δ
y
1
(
x
)
y
2
(
x
)
where
Δ
=
is the
Wronskian
of
y
1
(
x
)
and
y
2
(
x
)
. A substitution of this
y
1
(
y
2
(
)
)
x
x
into Eq. (6.35) will yield the general solution of Eq. (6.34).
6.3 Separation of Variables
for One-Dimensional Mixed Problems
A general form of the one-dimensional mixed problem reads
⎧
⎨
A
2
u
xx
+
B
2
u
txx
+
u
t
/
τ
0
+
u
tt
=
f
(
x
,
t
)
,
(
0
,
l
)
×
(
0
,
+
∞
)
,
−
b
1
u
x
(
0
,
t
)+
k
1
u
(
0
,
t
)=
0
,
b
2
u
x
(
l
,
t
)+
k
2
u
(
l
,
t
)=
0
,
(6.39)
⎩
u
(
x
,
0
)=
ϕ
(
x
)
,
u
t
(
x
,
0
)=
ψ
(
x
)
,
where
b
i
and
k
i
(
i
=
1
,
2
)
are nonnegative real constants satisfying
b
i
+
k
i
=
0
,
(
i
=
1
,
2
)
. We can first find the solution with
f
=
ϕ
=
0, i.e.
⎧
⎨
A
2
u
xx
+
B
2
u
txx
,
u
t
/
τ
0
+
u
tt
=
(
0
,
l
)
×
(
0
,
+
∞
)
,
−
b
1
u
x
(
0
,
t
)+
k
1
u
(
0
,
t
)=
0
,
b
2
u
x
(
l
,
t
)+
k
2
u
(
l
,
t
)=
0
,
(6.40)
⎩
u
(
x
,
0
)=
0
,
u
t
(
x
,
0
)=
ψ
(
x
)
.
Once the solution
W
ψ
(
of PDS (6.40) is available, we can easily write out the
solution of PDS (6.39) by using the solution structure theorem.
x
,
t
)
6.3.1 Eigenvalue Problems
Consider the nontrivial solution of separation of variables of PDS (6.40)
u
=
X
(
x
)
T
(
t
)
,
(
)
(
)
where
X
are functions of the only variables present to be determined.
Substituting this into the equation of PDS (6.40) yields
x
and
T
t
1
τ
0
X
T
(
T
(
A
2
X
T
B
2
X
T
(
(
x
)
t
)+
X
(
x
)
t
)=
(
t
)+
t
)
,
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