Environmental Engineering Reference
In-Depth Information
yields
c
1
−
h
1
c
2
=
0
,
c
1
(
1
+
h
2
l
)+
h
2
c
2
=
0
.
=
1
−
h
1
Its solution is
c
1
=
c
2
=
0, because
Δ
=
h
2
+
h
1
(
1
+
h
2
l
)
=
0.
1
+
h
2
l
2
We thus obtain the trivial solution
X
(
x
)
≡
0; therefore
λ
cannot be zero.
If
λ
<
0, the general solution of the equation in (6.43) reads
c
1
e
−
ax
c
2
e
ax
X
(
x
)=
+
,
a
2
where
−
=
λ
and
a
>
0. Applying the boundary conditions leads to
−
ac
1
+
ac
2
=
0
,
a
e
−
la
a
e
la
−
+
=
0
.
a
2
e
−
la
e
la
=
Its solution is
c
1
=
c
2
=
0, because
Δ
=
−
0. We thus again have the
trivial solution
X
(
x
)
≡
0; therefore the eigenvalues of (6.43) must be positive.
2
For positive
λ
=
β
>
0, the general solution of the equation in (6.43) reads
X
(
x
)=
c
1
cos
β
x
+
c
2
sin
β
x
.
Applying the boundary conditions yields
β
c
2
−
h
1
c
1
=
0
,
or
c
2
=
h
1
c
1
/
β
−
c
1
β
sin
β
l
+
c
2
β
cos
β
l
+
h
2
c
1
cos
β
l
+
h
2
c
2
sin
β
l
=
0
,
so that
c
1
=
0 to have a nontrivial solution. Its solution is, by noting that
h
1
and
h
2
are physically positive values,
0,
c
2
=
l
2
h
1
h
2
β
1
h
1
+
h
1
h
2
β
1
=
β
−
=
−
.
cos
β
l
β
l
h
2
(
h
1
+
h
2
)
l
l
Let
x
l
2
h
1
h
2
x
1
f
(
x
)=
cot
x
−
−
,
(6.44)
l
(
h
1
+
h
2
)
2
,
β
l
thus represent the zero points of
f
(
x
)
.Since
f
(
x
)
is an odd function and
λ
=
β
we wish to find the positive zero points of
f
(
x
)
only. Let
μ
m
be the
m
-th positive
zero point of
f
(
x
)
. Hence, we have eigenvalues
μ
m
l
2
2
m
λ
m
=
β
=
,
m
=
1
,
2
,....
Without taking account of the arbitrary constant
c
2
, the corresponding eigenfunc-
Search WWH ::
Custom Search