Environmental Engineering Reference
In-Depth Information
Its general solution is
y
=
C
1
J
2
(
t
)+
C
2
Y
2
(
t
)=
C
1
J
2
(
3
x
)+
C
2
Y
2
(
3
x
)
.
Example 2.
Find the solution of
⎧
⎨
4
x
2
y
1
9
x
2
y
+
xy
+
−
=
0
,
⎩
y
(
0
.
3
)=
2
.
where
y
(
x
)
is continuous at
x
=
0.
Solution.
Consider a variable transformation
t
=
2
x
. The equation is transformed
1
3
-order. Therefore the general solution is
y
into a Bessel equation of
=
C
1
J
3
(
2
x
)+
C
2
J
3
(
2
x
)
.
1
−
The continuity of
y
(
x
)
at
x
=
0 thus yields
C
2
=
0. Applying
y
(
0
.
3
)=
2 leads to
2
J
3
(
=
)
=
.
.
C
1
2
857
0
.
6
Thus
y
=
2
.
857
J
3
(
2
x
)
.
Remark 3.
The linear combination of Bessel functions of the first and second kinds
H
(
1
)
H
(
2
)
(
x
)=
J
γ
(
x
)+
i
Y
γ
(
x
)
,
(
x
)=
J
γ
(
x
)
−
i
Y
γ
(
x
)
γ
γ
=
√
−
are called the
Bessel function of the third kind
or the
Hankel function
.Herei
1.
In solving some PDS we arrive at
1
x
2
y
2
1
x
y
−
+
γ
y
+
=
.
0
(A.16)
By a variable transformation
t
=
i
x
, it is transformed into
1
y
2
1
t
y
+
−
γ
y
+
=
0
.
t
2
Its general solution reads
y
=
C
1
J
γ
(
i
x
)+
C
2
Y
γ
(
i
x
)
.
)
γ
+
2
m
m
i
γ
. Thus
Note that
(
i
=(
−
1
)
x
2
γ
+
2
m
i
γ
+
∞
1
m
=
0
J
γ
(
i
x
)=
m
!
Γ
(
γ
+
m
+
1
)
and
x
2
γ
+
2
m
+
∞
m
=
0
1
i
−
γ
J
γ
(
I
γ
(
)=
)=
x
i
x
m
!
Γ
(
γ
+
m
+
1
)
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