Bessels Differential equation continued

1
Bessels Differential equation (continued)
2
In this lecture we study properties of Bessels
functions, which are solutions of Bessels
equation.
3
Zeros of Bessel Functions
Fact
(i) If 0 ? p lt 1/2, then every solution of the
Bessels equation has an infinite number of
positive zeros and the distance between
successive zeros is lt ? and ?? as x ??.
(ii) If p 1/2, then every solution of the
Bessels equation has an infinite number of
positive zeros and the distance between
successive zeros is ?.
4
(iii) If p gt 1/2, then every solution of the
Bessels equation has an infinite number of
positive zeros and the distance between
successive zeros is gt ? and ?? as x ??.
5
Problem 5 Page 357
Prove (i)
(ii)
6
Proof of (i)
We know
Now
7
(No Transcript)
8
Hence
9
Hence
The other part is similarly proved.
10
Properties of Bessel functions of the First kind
We note the definition of Jp(x)
(1)
(2)
11
Differentiating term by term, we get
12
RHS
13
(3)
14
Differentiating term by term, we get
Replace n by n1
15
In particular, we note that
16
Recurrence relation for Bessel functions
We have
Differentiating the LHS, we get
Dividing throughout by xp, we get
.(4)
17
We have
Differentiating the LHS, we get
Multiplying throughout by xp, we get
. (5)
18
(6)
(4)(5) gives
(4)-(5) gives
(7)
From Eq. (7), we get
the Recurrence relation for Jp(x), namely
19
For example, p 1/2 gives
For example, p 2 gives
20
Fact The positive zeros of Jp(x) and Jp1(x)
alternate on the positive x-axis.
Proof Let x1 and x2 be two successive zeros of
Jp(x). Hence Jp(x1) Jp(x2) 0 and Jp(x) is
not zero for any x between x1 and x2.
Hence
are of opposite
gives
signs. Hence
are of opposite signs.
21
Or Jp1(x) is zero at some point between x1 and
x2.
Now let x1 and x2 be two successive zeros of
Jp1(x). Hence Jp1(x1) Jp1(x2) 0 and
Jp1(x) is not zero for any x between x1 and x2.
are of opposite
Hence
sign. Now (4) (with p replaced by p1) gives
22
are of opposite signs.
Hence
Or Jp(x) is zero at some point between x1 and x2.
That is between any two successive zeros of
Jp(x), there is a zero of Jp1(x) and between
any two successive zeros of Jp1(x), there is a
zero of Jp(x).
Thus the positive zeros of Jp(x) and Jp1(x)
alternate on the positive x-axis.
23
Look at the differential equation
Changing the independent variable x to t by the
substitution a x t, the above equation becomes
the Bessels equation
whose solution is
24
i.e.
c1, c2 arbitrary constants.
In particular, Jp(ax) is a solution of the
differential equation
25
Orthogonality of Bessel functions
Let
be the positive zeros of Jp(x) arranged in
increasing order. Then
26
The first part of the above result states that
Jp(?mx) and Jp(?nx) are orthogonal over the
interval 0, 1 with respect to the weight
function x.
Proof of the Orthogonality properties.
By the remarks made earlier, u Jp(ax) satisfies
the differential equation
27
i.e.
Similarly if v Jp(bx), then
Multiplying the first equation by v, the second
by u and subtracting, we get
i.e.
28
i.e.
or
Integrating with respect to x from 0 to 1,we get
At the lower limit x 0, the RHS is clearly 0.
29
Hence if a ?n and b ?m are distinct positive
zeros of Jp(x), then u(1) v(1) 0. Hence
As a ? b,
30
Proof of the second part
u Jp(ax) satisfies the differential equation
i.e.
Multiplying by 2u?, we get
i.e.
31
Integrating with respect to x from 0 to 1,we get
If p gt 0,
If p 0,
Thus the RHS is zero at the lower limit x 0.
Also
32
if a is a positive zero of Jp(x).
Hence the RHS a2J?p(a)2 at the upper limit x
1.
Thus
or
33
We know
Hence when x a , a positive zero of Jp(x),
So we get
34
Fourier-Bessel Series Expansion
Let
be the positive zeros of Jp(x) arranged in
increasing order. Then given a piecewise
continuous function f(x), we can expand it as a
series
called the Fourier-Bessel series of f(x).
35
Multiplying both sides by x Jp(?nx) and
integrating with respect to x, from 0 to 1, we
get (noting that Jp(?nx) and Jp(?mx) are
orthogonal if n ? m)
n 1, 2,
Thus the coefficients ans are all found.
36
Problem 5 Page 364
Find the Fourier-Bessel series expansion of the
function
in terms of the functions
( ?ns are the positive zeros of
where
37
Now
Put t ?nx. We thus get
gives
(n 1,2,)
38
Thus the Fourier-Bessel series expansion of xp
is
Next slide
39
Problem 4 Page 363
Find the Fourier-Bessel series expansion of the
function
in terms of the functions
( ?ns are the positive zeros of
40
where
Now
Put t ?nx. We thus get
41
gives
(n 1,2,)
Thus the Fourier-Bessel series expansion of f(x)
is
End of Lecture