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Ch 5.8: Bessel’s Equation
Bessel Equation of order ν:
Note that x = 0 is a regular singular point.
Friedrich Wilhelm Bessel (1784 – 1846) studied disturbances
in planetary motion, which led him in 1824 to make the first
systematic analysis of solutions of this equation. The
solutions became known as Bessel functions.
In this section, we study the following cases:
Bessel Equations of order zero: ν = 0
Bessel Equations of order one-half: ν = ½
Bessel Equations of order one: ν = 1
( ) 0222
=−+′+′′ yxyxyx ν
Bessel Equation of Order Zero (1 of 12)
The Bessel Equation of order zero is
We assume solutions have the form
Taking derivatives,
Substituting these into the differential equation, we obtain
( )
( )( )∑
∑ ∑
∞
=
−+
∞
=
∞
=
−++
−++=′′
+=′=
0
2
0 0
1
1)(
,)(,)(
n
nr
n
n n
nr
n
nr
n
xnrnraxy
xnraxyxaxy
( ) 0,0for,,)(
0
0∑
∞
=
+
>≠==
n
nr
n xaxaxrxy φ
022
=+′+′′ yxyxyx
( )( ) ( ) 01
0
2
00
=+++−++ ∑∑∑
∞
=
++
∞
=
+
∞
=
+
n
nr
n
n
nr
n
n
nr
n xaxnraxnrnra
Indicial Equation (2 of 12)
From the previous slide,
Rewriting,
or
The indicial equation is r2
= 0, and hence r1 = r2 = 0.
( )( ) ( ) 01
0
2
00
=+++−++ ∑∑∑
∞
=
++
∞
=
+
∞
=
+
n
nr
n
n
nr
n
n
nr
n xaxnraxnrnra
[ ] [ ]
( )( ) ( )[ ]{ } 01
)1()1()1(
2
2
1
10
=+++−+++
+++++−
+
∞
=
−
+
∑ nr
n
nn
rr
xanrnrnra
xrrraxrrra
( ){ } 0)1(
2
2
212
1
2
0 =+++++ +
∞
=
−
+
∑ nr
n
nn
rr
xanraxraxra
Recurrence Relation (3 of 12)
From the previous slide,
Note that a1 = 0; the recurrence relation is
We conclude a1 = a3 = a5 = … = 0, and since r = 0,
Note: Recall dependence of an on r, which is indicated by
an(r). Thus we may write a2m(0) here instead of a2m.
( )
,3,2,2
2
=
+
−= −
n
nr
a
a n
n
( ){ } 0)1(
2
2
212
1
2
0 =+++++ +
∞
=
−
+
∑ nr
n
nn
rr
xanraxraxra
,2,1,
)2( 2
22
2 =−= −
m
m
a
a m
m
First Solution (4 of 12)
From the previous slide,
Thus
and in general,
Thus
,2,1,
)2( 2
22
2 =−= −
m
m
a
a m
m
( ) ( )
,
1232
,
122244
,
2 26
0
624
0
22
0
2
2
42
0
2
⋅⋅
−=
⋅
==−=−=
a
a
aaa
a
a
a
( )
,2,1,
!2
)1(
22
0
2 =
−
= m
m
a
a m
m
m
( )
0,
!2
)1(
1)(
1
22
2
01 >




 −
+= ∑
∞
=
x
m
x
axy
m
m
mm
Bessel Function of First Kind,
Order Zero (5 of 12)
Our first solution of Bessel’s Equation of order zero is
The series converges for all x, and is called the Bessel
function of the first kind of order zero, denoted by
The graphs of J0 and several
partial sum approximations
are given here.
( )
0,
!2
)1(
1)(
1
22
2
01 >




 −
+= ∑
∞
=
x
m
x
axy
m
m
mm
( )
0,
!2
)1(
)(
0
22
2
0 >
−
= ∑
∞
=
x
m
x
xJ
m
m
mm
Second Solution: Odd Coefficients (6 of 12)
Since indicial equation has repeated roots, recall from Section
5.7 that the coefficients in second solution can be found using
Now
Thus
Also,
and hence
0)0(0)( 11 =′⇒= ara
( ){ } 0)()()1)(()(
2
2
212
1
2
0 =+++++ +
∞
=
−
+
∑ nr
n
nn
rr
xranrraxrraxrra
( )
,3,2,
)(
)( 2
2
=
+
−= −
n
nr
ra
ra n
n
0
)( =
′ rn ra
,2,1,0)0(12 ==′ + ma m
Second Solution: Even Coefficients (7 of 12)
Thus we need only compute derivatives of the even
coefficients, given by
It can be shown that
and hence
( ) ( ) ( )
1,
22
)1(
)(
2
)(
)( 22
0
22
22
2 ≥
++
−
=⇒
+
−= −
m
mrr
a
ra
mr
ra
ra
m
m
m
m





+
++
+
+
+
−=
′
mrrrra
ra
m
m
2
1
4
1
2
1
2
)(
)(
2
2

)0(
2
1
4
1
2
1
2)0( 22 mm a
m
a 



+++−=′ 
Second Solution: Series Representation (8 of 12)
Thus
where
Taking a0 = 1 and using results of Section 5.7,
( )
,2,1,
!2
)1(
)0( 22
0
2 =
−
−=′ m
m
a
Ha m
m
mm
m
Hm
2
1
4
1
2
1
+++= 
( )
0,
!2
)1(
ln)()( 2
1
22
1
02 >
−
+= ∑
∞
=
+
xx
m
H
xxJxy m
m
m
m
m
Bessel Function of Second Kind,
Order Zero (9 of 12)
Instead of using y2, the second solution is often taken to be a
linear combination Y0 of J0 and y2, known as the Bessel
function of second kind of order zero. Here, we take
The constant γ is the Euler-Mascheroni constant, defined by
Substituting the expression for y2 from previous slide into
equation for Y0above, we obtain
( )[ ])(2ln)(
2
)( 020 xJxyxY −+= γ
π
( ) 5772.0lnlim ≅−=
∞→
nHn
n
γ
( )
0,
!2
)1(
)(
2
ln
2
)( 2
1
22
1
00 >




 −
+





+= ∑
∞
=
+
xx
m
H
xJ
x
xY m
m
m
m
m
γ
π
General Solution of Bessel’s Equation,
Order Zero (10 of 12)
The general solution of Bessel’s equation of order zero, x > 0,
is given by
where
Note that J0→ 0 as x → 0 while Y0has a logarithmic
singularity at x = 0. If a solution which is bounded at the
origin is desired, then Y0 must be discarded.
( )
( ) 




 −
+





+=
−
=
∑
∑
∞
=
+
∞
=
m
m
m
m
m
m
m
mm
x
m
H
xJ
x
xY
m
x
xJ
2
1
22
1
00
0
22
2
0
!2
)1(
)(
2
ln
2
)(
,
!2
)1(
)(
γ
π
)()()( 0201 xYcxJcxy +=
Graphs of Bessel Functions,
Order Zero (11 of 12)
The graphs of J0and Y0 are given below.
Note that the behavior of J0and Y0 appear to be similar to sinx
and cosx for large x, except that oscillations of J0and Y0 decay
to zero.
Approximation of Bessel Functions,
Order Zero (12 of 12)
The fact that J0 and Y0 appear similar to sin x and cos x for
large x may not be surprising, since ODE can be rewritten as
Thus, for large x, our equation can be approximated by
whose solns are sin x and cos x. Indeed, it can be shown that
( ) 01
1
0 2
2
222
=





−+′+′′⇔=−+′+′′ y
x
v
y
x
yyvxyxyx
∞→





−





≅
∞→





−





≅
xx
x
xY
xx
x
xJ
as,
4
sin
2
)(
as,
4
cos
2
)(
2/1
0
2/1
0
π
π
π
π
,0=+′′ yy
Bessel Equation of Order One-Half (1 of 8)
The Bessel Equation of order one-half is
We assume solutions have the form
Substituting these into the differential equation, we obtain
( ) 0,0for,,)(
0
0∑
∞
=
+
>≠==
n
nr
n xaxaxrxy φ
0
4
122
=





−+′+′′ yxyxyx
( )( ) ( )
0
4
1
1
00
2
00
=−+
++−++
∑∑
∑∑
∞
=
+
∞
=
++
∞
=
+
∞
=
+
n
nr
n
n
nr
n
n
nr
n
n
nr
n
xaxa
xnraxnrnra
Recurrence Relation (2 of 8)
Using results of previous slide, we obtain
or
The roots of indicial equation are r1 = ½, r2 = - ½ , and note
that they differ by a positive integer.
The recurrence relation is
( ) 0
4
1
4
1
)1(
4
1
2
2
21
1
2
0
2
=






+





−++





−++





− +
∞
=
−
+
∑ nr
n
nn
rr
xaanrxarxar
( )( ) ( ) 0
4
1
1
0
2
0
=+





−++−++ ∑∑
∞
=
++
∞
=
+
n
nr
n
n
nr
n xaxanrnrnr
( )
,3,2,
4/1
)(
)( 2
2
=
−+
−= −
n
nr
ra
ra n
n
First Solution: Coefficients (3 of 8)
Consider first the case r1 = ½. From the previous slide,
Since r1 = ½, a1 = 0, and hence from the recurrence relation,
a1 = a3 = a5 = … = 0. For the even coefficients, we have
It follows that
and
( ) ( )
,2,1,
1224/122/1
22
2
22
2 =
+
−=
−+
−= −−
m
mm
a
m
a
a mm
m
,
!545
,
!3
02
4
0
2
aa
a
a
a =
⋅
−=−=
,2,1,
!)12(
)1( 0
2 =
+
−
= m
m
a
a
m
m
( ) ( ) 0
4
1
4
1
)1(4/1
2
2
21
1
2
0
2
=






+





−++





−++− +
∞
=
−
+
∑ nr
n
nn
rr
xaanrxarxar
Bessel Function of First Kind,
Order One-Half (4 of 8)
It follows that the first solution of our equation is, for a0 = 1,
The Bessel function of the first kind of order one-half, J½,
is defined as
0,sin
0,
!)12(
)1(
0,
!)12(
)1(
1)(
2/1
0
122/1
1
22/1
1
>=
>





+
−
=
>





+
−
+=
−
∞
=
+−
∞
=
∑
∑
xxx
xx
m
x
xx
m
xxy
m
m
m
m
m
m
0,sin
2
)(
2
)(
2/1
1
2/1
2/1 >





=





= xx
x
xyxJ
ππ
Second Solution: Even Coefficients (5 of 8)
Now consider the case r2 = - ½. We know that
Since r2 = - ½ , a1 = arbitrary. For the even coefficients,
It follows that
and
( ) ( ) 0
4
1
4
1
)1(4/1
2
2
21
1
2
0
2
=






+





−++





−++− +
∞
=
−
+
∑ nr
n
nn
rr
xaanrxarxar
( ) ( )
,2,1,
1224/122/1
22
2
22
2 =
−
−=
−+−
−= −−
m
mm
a
m
a
a mm
m
,
!434
,
!2
02
4
0
2
aa
a
a
a =
⋅
−=−=
,2,1,
!)2(
)1( 0
2 =
−
= m
m
a
a
m
m
Second Solution: Odd Coefficients (6 of 8)
For the odd coefficients,
It follows that
and
( ) ( )
,2,1,
1224/1122/1
12
2
12
12 =
+
−=
−++−
−= −−
+ m
mm
a
m
a
a mm
m
,
!545
,
!3
13
5
1
3
aa
a
a
a =
⋅
−=−=
,2,1,
!)12(
)1( 1
12 =
+
−
=+ m
m
a
a
m
m
Second Solution (7 of 8)
Therefore
The second solution is usually taken to be the function
where a0 = (2/π)½
and a1 = 0.
The general solution of Bessel’s equation of order one-half is
[ ] 0,sincos
0,
!)12(
)1(
!)2(
)1(
)(
10
2/1
0
12
1
0
2
0
2/1
2
>+=
>





+
−
+
−
=
−
∞
=
+∞
=
−
∑∑
xxaxax
x
m
x
a
m
x
axxy
m
mm
m
mm
0,cos
2
)(
2/1
2/1 >





=− xx
x
xJ
π
)()()( 2/122/11 xJcxJcxy −+=
Graphs of Bessel Functions,
Order One-Half (8 of 8)
Graphs of J½, J-½are given below. Note behavior of J½, J-½similar
to J0, Y0 for large x, with phase shift of π/4.
∞→





−





≅





−





≅






=





=−
xx
x
xYx
x
xJ
x
x
xJx
x
xJ
as,
4
sin
2
)(,
4
cos
2
)(
sin
2
)(,cos
2
)(
2/1
0
2/1
0
2/1
2/1
2/1
2/1
π
π
π
π
ππ
Bessel Equation of Order One (1 of 6)
The Bessel Equation of order one is
We assume solutions have the form
Substituting these into the differential equation, we obtain
( ) 0,0for,,)(
0
0∑
∞
=
+
>≠==
n
nr
n xaxaxrxy φ
( ) 0122
=−+′+′′ yxyxyx
( )( ) ( )
0
1
00
2
00
=−+
++−++
∑∑
∑∑
∞
=
+
∞
=
++
∞
=
+
∞
=
+
n
nr
n
n
nr
n
n
nr
n
n
nr
n
xaxa
xnraxnrnra
Recurrence Relation (2 of 6)
Using the results of the previous slide, we obtain
or
The roots of indicial equation are r1 = 1, r2 = - 1, and note that
they differ by a positive integer.
The recurrence relation is
( ) [ ] ( )[ ]{ } 011)1(1
2
2
21
1
2
0
2
=+−++−++− +
∞
=
−
+
∑ nr
n
nn
rr
xaanrxarxar
( )( ) ( )[ ] 011
0
2
0
=+−++−++ ∑∑
∞
=
++
∞
=
+
n
nr
n
n
nr
n xaxanrnrnr
( )
,3,2,
1
)(
)( 2
2
=
−+
−= −
n
nr
ra
ra n
n
First Solution: Coefficients (3 of 6)
Consider first the case r1 = 1. From previous slide,
Since r1 = 1, a1 = 0, and hence from the recurrence relation, a1
= a3 = a5 = … = 0. For the even coefficients, we have
It follows that
and
( ) ( )
,2,1,
12121
2
22
2
22
2 =
+
−=
−+
−= −−
m
mm
a
m
a
a mm
m
,
!2!32232
,
122 4
0
2
2
42
0
2
aa
a
a
a =
⋅⋅
−=
⋅⋅
−=
,2,1,
!!)1(2
)1(
2
0
2 =
+
−
= m
mm
a
a m
m
m
( ) [ ] ( )[ ]{ } 011)1(1
2
2
21
1
2
0
2
=+−++−++− +
∞
=
−
+
∑ nr
n
nn
rr
xaanrxarxar
Bessel Function of First Kind,
Order One (4 of 6)
It follows that the first solution of our differential equation is
Taking a0 = ½, the Bessel function of the first kind of order
one, J1, is defined as
The series converges for all x and hence J1 is analytic
everywhere.
0,
!!)1(2
)1(
1)(
1
2
201 >





+
−
+= ∑
∞
=
xx
mm
xaxy
m
m
m
m
0,
!!)1(2
)1(
2
)(
0
2
21 >





+
−
= ∑
∞
=
xx
mm
x
xJ
m
m
m
m
Second Solution (5 of 6)
For the case r1 = -1, a solution of the form
is guaranteed by Theorem 5.7.1.
The coefficients cn are determined by substituting y2 into the
ODE and obtaining a recurrence relation, etc. The result is:
where Hk is as defined previously. See text for more details.
Note that J1→ 0 as x → 0 and is analytic at x = 0, while y2 is
unbounded at x = 0 in the same manner as 1/x.
0,1ln)()( 2
1
1
12 >





++= ∑
∞
=
−
xxcxxxJaxy n
n
n
( ) 0,
!)1(!2
)1(
1ln)()( 2
1
2
11
12 >





−
+−
−+−= ∑
∞
=
−−
xx
mm
HH
xxxJxy n
m
m
mm
m
Bessel Function of Second Kind,
Order One (6 of 6)
The second solution, the Bessel function of the second kind
of order one, is usually taken to be the function
where γ is the Euler-Mascheroni constant.
The general solution of Bessel’s equation of order one is
Note that J1, Y1 have same
behavior at x = 0 as observed
on previous slide for J1and y2.
( )[ ] 0,)(2ln)(
2
)( 121 >−+−= xxJxyxY γ
π
0),()()( 1211 >+= xxYcxJcxy

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Ch05 8

  • 1. Ch 5.8: Bessel’s Equation Bessel Equation of order ν: Note that x = 0 is a regular singular point. Friedrich Wilhelm Bessel (1784 – 1846) studied disturbances in planetary motion, which led him in 1824 to make the first systematic analysis of solutions of this equation. The solutions became known as Bessel functions. In this section, we study the following cases: Bessel Equations of order zero: ν = 0 Bessel Equations of order one-half: ν = ½ Bessel Equations of order one: ν = 1 ( ) 0222 =−+′+′′ yxyxyx ν
  • 2. Bessel Equation of Order Zero (1 of 12) The Bessel Equation of order zero is We assume solutions have the form Taking derivatives, Substituting these into the differential equation, we obtain ( ) ( )( )∑ ∑ ∑ ∞ = −+ ∞ = ∞ = −++ −++=′′ +=′= 0 2 0 0 1 1)( ,)(,)( n nr n n n nr n nr n xnrnraxy xnraxyxaxy ( ) 0,0for,,)( 0 0∑ ∞ = + >≠== n nr n xaxaxrxy φ 022 =+′+′′ yxyxyx ( )( ) ( ) 01 0 2 00 =+++−++ ∑∑∑ ∞ = ++ ∞ = + ∞ = + n nr n n nr n n nr n xaxnraxnrnra
  • 3. Indicial Equation (2 of 12) From the previous slide, Rewriting, or The indicial equation is r2 = 0, and hence r1 = r2 = 0. ( )( ) ( ) 01 0 2 00 =+++−++ ∑∑∑ ∞ = ++ ∞ = + ∞ = + n nr n n nr n n nr n xaxnraxnrnra [ ] [ ] ( )( ) ( )[ ]{ } 01 )1()1()1( 2 2 1 10 =+++−+++ +++++− + ∞ = − + ∑ nr n nn rr xanrnrnra xrrraxrrra ( ){ } 0)1( 2 2 212 1 2 0 =+++++ + ∞ = − + ∑ nr n nn rr xanraxraxra
  • 4. Recurrence Relation (3 of 12) From the previous slide, Note that a1 = 0; the recurrence relation is We conclude a1 = a3 = a5 = … = 0, and since r = 0, Note: Recall dependence of an on r, which is indicated by an(r). Thus we may write a2m(0) here instead of a2m. ( ) ,3,2,2 2 = + −= − n nr a a n n ( ){ } 0)1( 2 2 212 1 2 0 =+++++ + ∞ = − + ∑ nr n nn rr xanraxraxra ,2,1, )2( 2 22 2 =−= − m m a a m m
  • 5. First Solution (4 of 12) From the previous slide, Thus and in general, Thus ,2,1, )2( 2 22 2 =−= − m m a a m m ( ) ( ) , 1232 , 122244 , 2 26 0 624 0 22 0 2 2 42 0 2 ⋅⋅ −= ⋅ ==−=−= a a aaa a a a ( ) ,2,1, !2 )1( 22 0 2 = − = m m a a m m m ( ) 0, !2 )1( 1)( 1 22 2 01 >      − += ∑ ∞ = x m x axy m m mm
  • 6. Bessel Function of First Kind, Order Zero (5 of 12) Our first solution of Bessel’s Equation of order zero is The series converges for all x, and is called the Bessel function of the first kind of order zero, denoted by The graphs of J0 and several partial sum approximations are given here. ( ) 0, !2 )1( 1)( 1 22 2 01 >      − += ∑ ∞ = x m x axy m m mm ( ) 0, !2 )1( )( 0 22 2 0 > − = ∑ ∞ = x m x xJ m m mm
  • 7. Second Solution: Odd Coefficients (6 of 12) Since indicial equation has repeated roots, recall from Section 5.7 that the coefficients in second solution can be found using Now Thus Also, and hence 0)0(0)( 11 =′⇒= ara ( ){ } 0)()()1)(()( 2 2 212 1 2 0 =+++++ + ∞ = − + ∑ nr n nn rr xranrraxrraxrra ( ) ,3,2, )( )( 2 2 = + −= − n nr ra ra n n 0 )( = ′ rn ra ,2,1,0)0(12 ==′ + ma m
  • 8. Second Solution: Even Coefficients (7 of 12) Thus we need only compute derivatives of the even coefficients, given by It can be shown that and hence ( ) ( ) ( ) 1, 22 )1( )( 2 )( )( 22 0 22 22 2 ≥ ++ − =⇒ + −= − m mrr a ra mr ra ra m m m m      + ++ + + + −= ′ mrrrra ra m m 2 1 4 1 2 1 2 )( )( 2 2  )0( 2 1 4 1 2 1 2)0( 22 mm a m a     +++−=′ 
  • 9. Second Solution: Series Representation (8 of 12) Thus where Taking a0 = 1 and using results of Section 5.7, ( ) ,2,1, !2 )1( )0( 22 0 2 = − −=′ m m a Ha m m mm m Hm 2 1 4 1 2 1 +++=  ( ) 0, !2 )1( ln)()( 2 1 22 1 02 > − += ∑ ∞ = + xx m H xxJxy m m m m m
  • 10. Bessel Function of Second Kind, Order Zero (9 of 12) Instead of using y2, the second solution is often taken to be a linear combination Y0 of J0 and y2, known as the Bessel function of second kind of order zero. Here, we take The constant γ is the Euler-Mascheroni constant, defined by Substituting the expression for y2 from previous slide into equation for Y0above, we obtain ( )[ ])(2ln)( 2 )( 020 xJxyxY −+= γ π ( ) 5772.0lnlim ≅−= ∞→ nHn n γ ( ) 0, !2 )1( )( 2 ln 2 )( 2 1 22 1 00 >      − +      += ∑ ∞ = + xx m H xJ x xY m m m m m γ π
  • 11. General Solution of Bessel’s Equation, Order Zero (10 of 12) The general solution of Bessel’s equation of order zero, x > 0, is given by where Note that J0→ 0 as x → 0 while Y0has a logarithmic singularity at x = 0. If a solution which is bounded at the origin is desired, then Y0 must be discarded. ( ) ( )       − +      += − = ∑ ∑ ∞ = + ∞ = m m m m m m m mm x m H xJ x xY m x xJ 2 1 22 1 00 0 22 2 0 !2 )1( )( 2 ln 2 )( , !2 )1( )( γ π )()()( 0201 xYcxJcxy +=
  • 12. Graphs of Bessel Functions, Order Zero (11 of 12) The graphs of J0and Y0 are given below. Note that the behavior of J0and Y0 appear to be similar to sinx and cosx for large x, except that oscillations of J0and Y0 decay to zero.
  • 13. Approximation of Bessel Functions, Order Zero (12 of 12) The fact that J0 and Y0 appear similar to sin x and cos x for large x may not be surprising, since ODE can be rewritten as Thus, for large x, our equation can be approximated by whose solns are sin x and cos x. Indeed, it can be shown that ( ) 01 1 0 2 2 222 =      −+′+′′⇔=−+′+′′ y x v y x yyvxyxyx ∞→      −      ≅ ∞→      −      ≅ xx x xY xx x xJ as, 4 sin 2 )( as, 4 cos 2 )( 2/1 0 2/1 0 π π π π ,0=+′′ yy
  • 14. Bessel Equation of Order One-Half (1 of 8) The Bessel Equation of order one-half is We assume solutions have the form Substituting these into the differential equation, we obtain ( ) 0,0for,,)( 0 0∑ ∞ = + >≠== n nr n xaxaxrxy φ 0 4 122 =      −+′+′′ yxyxyx ( )( ) ( ) 0 4 1 1 00 2 00 =−+ ++−++ ∑∑ ∑∑ ∞ = + ∞ = ++ ∞ = + ∞ = + n nr n n nr n n nr n n nr n xaxa xnraxnrnra
  • 15. Recurrence Relation (2 of 8) Using results of previous slide, we obtain or The roots of indicial equation are r1 = ½, r2 = - ½ , and note that they differ by a positive integer. The recurrence relation is ( ) 0 4 1 4 1 )1( 4 1 2 2 21 1 2 0 2 =       +      −++      −++      − + ∞ = − + ∑ nr n nn rr xaanrxarxar ( )( ) ( ) 0 4 1 1 0 2 0 =+      −++−++ ∑∑ ∞ = ++ ∞ = + n nr n n nr n xaxanrnrnr ( ) ,3,2, 4/1 )( )( 2 2 = −+ −= − n nr ra ra n n
  • 16. First Solution: Coefficients (3 of 8) Consider first the case r1 = ½. From the previous slide, Since r1 = ½, a1 = 0, and hence from the recurrence relation, a1 = a3 = a5 = … = 0. For the even coefficients, we have It follows that and ( ) ( ) ,2,1, 1224/122/1 22 2 22 2 = + −= −+ −= −− m mm a m a a mm m , !545 , !3 02 4 0 2 aa a a a = ⋅ −=−= ,2,1, !)12( )1( 0 2 = + − = m m a a m m ( ) ( ) 0 4 1 4 1 )1(4/1 2 2 21 1 2 0 2 =       +      −++      −++− + ∞ = − + ∑ nr n nn rr xaanrxarxar
  • 17. Bessel Function of First Kind, Order One-Half (4 of 8) It follows that the first solution of our equation is, for a0 = 1, The Bessel function of the first kind of order one-half, J½, is defined as 0,sin 0, !)12( )1( 0, !)12( )1( 1)( 2/1 0 122/1 1 22/1 1 >= >      + − = >      + − += − ∞ = +− ∞ = ∑ ∑ xxx xx m x xx m xxy m m m m m m 0,sin 2 )( 2 )( 2/1 1 2/1 2/1 >      =      = xx x xyxJ ππ
  • 18. Second Solution: Even Coefficients (5 of 8) Now consider the case r2 = - ½. We know that Since r2 = - ½ , a1 = arbitrary. For the even coefficients, It follows that and ( ) ( ) 0 4 1 4 1 )1(4/1 2 2 21 1 2 0 2 =       +      −++      −++− + ∞ = − + ∑ nr n nn rr xaanrxarxar ( ) ( ) ,2,1, 1224/122/1 22 2 22 2 = − −= −+− −= −− m mm a m a a mm m , !434 , !2 02 4 0 2 aa a a a = ⋅ −=−= ,2,1, !)2( )1( 0 2 = − = m m a a m m
  • 19. Second Solution: Odd Coefficients (6 of 8) For the odd coefficients, It follows that and ( ) ( ) ,2,1, 1224/1122/1 12 2 12 12 = + −= −++− −= −− + m mm a m a a mm m , !545 , !3 13 5 1 3 aa a a a = ⋅ −=−= ,2,1, !)12( )1( 1 12 = + − =+ m m a a m m
  • 20. Second Solution (7 of 8) Therefore The second solution is usually taken to be the function where a0 = (2/π)½ and a1 = 0. The general solution of Bessel’s equation of order one-half is [ ] 0,sincos 0, !)12( )1( !)2( )1( )( 10 2/1 0 12 1 0 2 0 2/1 2 >+= >      + − + − = − ∞ = +∞ = − ∑∑ xxaxax x m x a m x axxy m mm m mm 0,cos 2 )( 2/1 2/1 >      =− xx x xJ π )()()( 2/122/11 xJcxJcxy −+=
  • 21. Graphs of Bessel Functions, Order One-Half (8 of 8) Graphs of J½, J-½are given below. Note behavior of J½, J-½similar to J0, Y0 for large x, with phase shift of π/4. ∞→      −      ≅      −      ≅       =      =− xx x xYx x xJ x x xJx x xJ as, 4 sin 2 )(, 4 cos 2 )( sin 2 )(,cos 2 )( 2/1 0 2/1 0 2/1 2/1 2/1 2/1 π π π π ππ
  • 22. Bessel Equation of Order One (1 of 6) The Bessel Equation of order one is We assume solutions have the form Substituting these into the differential equation, we obtain ( ) 0,0for,,)( 0 0∑ ∞ = + >≠== n nr n xaxaxrxy φ ( ) 0122 =−+′+′′ yxyxyx ( )( ) ( ) 0 1 00 2 00 =−+ ++−++ ∑∑ ∑∑ ∞ = + ∞ = ++ ∞ = + ∞ = + n nr n n nr n n nr n n nr n xaxa xnraxnrnra
  • 23. Recurrence Relation (2 of 6) Using the results of the previous slide, we obtain or The roots of indicial equation are r1 = 1, r2 = - 1, and note that they differ by a positive integer. The recurrence relation is ( ) [ ] ( )[ ]{ } 011)1(1 2 2 21 1 2 0 2 =+−++−++− + ∞ = − + ∑ nr n nn rr xaanrxarxar ( )( ) ( )[ ] 011 0 2 0 =+−++−++ ∑∑ ∞ = ++ ∞ = + n nr n n nr n xaxanrnrnr ( ) ,3,2, 1 )( )( 2 2 = −+ −= − n nr ra ra n n
  • 24. First Solution: Coefficients (3 of 6) Consider first the case r1 = 1. From previous slide, Since r1 = 1, a1 = 0, and hence from the recurrence relation, a1 = a3 = a5 = … = 0. For the even coefficients, we have It follows that and ( ) ( ) ,2,1, 12121 2 22 2 22 2 = + −= −+ −= −− m mm a m a a mm m , !2!32232 , 122 4 0 2 2 42 0 2 aa a a a = ⋅⋅ −= ⋅⋅ −= ,2,1, !!)1(2 )1( 2 0 2 = + − = m mm a a m m m ( ) [ ] ( )[ ]{ } 011)1(1 2 2 21 1 2 0 2 =+−++−++− + ∞ = − + ∑ nr n nn rr xaanrxarxar
  • 25. Bessel Function of First Kind, Order One (4 of 6) It follows that the first solution of our differential equation is Taking a0 = ½, the Bessel function of the first kind of order one, J1, is defined as The series converges for all x and hence J1 is analytic everywhere. 0, !!)1(2 )1( 1)( 1 2 201 >      + − += ∑ ∞ = xx mm xaxy m m m m 0, !!)1(2 )1( 2 )( 0 2 21 >      + − = ∑ ∞ = xx mm x xJ m m m m
  • 26. Second Solution (5 of 6) For the case r1 = -1, a solution of the form is guaranteed by Theorem 5.7.1. The coefficients cn are determined by substituting y2 into the ODE and obtaining a recurrence relation, etc. The result is: where Hk is as defined previously. See text for more details. Note that J1→ 0 as x → 0 and is analytic at x = 0, while y2 is unbounded at x = 0 in the same manner as 1/x. 0,1ln)()( 2 1 1 12 >      ++= ∑ ∞ = − xxcxxxJaxy n n n ( ) 0, !)1(!2 )1( 1ln)()( 2 1 2 11 12 >      − +− −+−= ∑ ∞ = −− xx mm HH xxxJxy n m m mm m
  • 27. Bessel Function of Second Kind, Order One (6 of 6) The second solution, the Bessel function of the second kind of order one, is usually taken to be the function where γ is the Euler-Mascheroni constant. The general solution of Bessel’s equation of order one is Note that J1, Y1 have same behavior at x = 0 as observed on previous slide for J1and y2. ( )[ ] 0,)(2ln)( 2 )( 121 >−+−= xxJxyxY γ π 0),()()( 1211 >+= xxYcxJcxy