Chapter 5.8: Bessel s equation

Transcription

1 Chapter 5.8: Bessel s equation Bessel s equation of order ν is: x 2 y + xy + (x 2 ν 2 )y = 0. It has a regular singular point at x = 0. When ν = 0,, 2,..., this equation comes up when separating variables in the Laplacian and is one of the most important equations in physics and engineering. We will study the solutions when ν = 0, /2,.

2 Bessel equation of order 0 This is x 2 y + xy + x 2 y = 0. It already has the form of an Euler equation, except that the coefficient of y is x 2, rather than a constant. The indicial equation at x = 0 is r(r ) + r = 0 r 2 = 0 = r = 0. Thus, the indicial equation has a double root. 2

3 J 0 (x) The first solution, denoted J 0 (x) is a power series: J 0 (x) = We have: L[J 0 ](x) = a n x n. [n(n )a n +na n ]x n + We shift down in the third term to get a n x n+2. We get: a x + n=2 n=2 a n 2 x n. [n(n )a n + na n + a n 2 ]x n = 0 Note that n(n )a n + na n = n 2 a n. 3

4 J 0 (x) The term a x is stranded by itself, and has to be zero. We can see this from the equation. If we divide by x it says: Thus, xy + y + xy = 0 = y (0) = 0 = a. a = a 3 = a 5 = = 0. Hence J 0 is an even function. Its recursion relation is a 2n a 2(n+) = (2n + 2) 2. It is easy to solve this recursion relation: a 2n = ( ) n 2 2n (n!) 2a 0. 4

5 J 0 (x) = ( ) n 2 2n (n!) 2x2n Indeed, each step up 2n 2(n + ), i.e. n n+, puts in a new ( ), a new 2 2 and a new n 2. As we proceed down from 2n, 2(n ),..., 0 we get ( ) n 2 2n (n!) 2. For a 0 =, this gives the stated formula. The other solution has the form Y 0 (x) = J 0 (x) ln x + b n x 2n. The coefficients b n can be found by reduction of order or by using the special methods for repeated roots. It is also important, but is left to the exercises. 5

6 Asymptotics as x x 0 An important notion is how the solution behaves as x x 0. We use the symbol f(x) (x x 0 ) r to mean that f(x) = (x x 0 ) r a n (x x 0 ) n, and say f is asymptotic to (x x 0 ) r as x x 0. If there are repeated roots, the asymptotics involve the log factor. Thus at a regular point x 0, a solution y of an ODE satisfies y C, x x 0. At a singular point, y (x x 0 ) r if y is the solution corresponding to the root r of the indicial equation at x 0. Thus, J 0 (x), x 0; Y 0 (x) ln x, x 0. 6

7 Asymptotics as x This slide is only for cultural interest: Bessel functions have asymptotics as x as well as x 0. Although you need to know asymptotics as x 0, we only give the asymptotics as x for cultural reasons it is one of the most important fact about Bessel functions, but it is beyond the scope of this course to prove what we claim. Namely, as x, the Bessel functions behave as follows: J 0 (x) Cx /2 cos(x π/4), Y 0 (x) Cx /2 sin(x π/4). Here, means is asymptotic to, i.e behaves like as x. In many ways, asymptotics are more useful than exact formulae, and I hope you will learn about them in a future course. 7

8 Bessel equation of order /2 This is the equation x 2 y + xy + (x 2 )y = 0. 4 The indicial equation is now r(r )+r 4 = 0 r2 4 = 0 r = ± 2. This is the problematic case where the roots differ by an integer: /2 ( /2) =. It is easier to find the solution for the larger root. 8

9 The root r = /2 We seek a solution of the form: y = a n x n+r, r = /2. Plugging into the formula gives: [a n (n + r)(n + r ) + a n (n + r)]x n+r + a n x n+2+r 4 a nx n+r. We need to down shift by 2 in the second to last term. The rest adds up easily and we get a n [(n + r) 2 4 ]xn+r + n=2 a n 2 x n+r. 9

10 Hanging chad There are two terms hanging over in the first sum that don t match any terms in the second one (n = 0, ). They equal a 0 [r 2 4 ]xr + a [( + r) 2 4 ]xr+. The first term cancels because r = /2. The second does not and forces a = 0. The recursion relation is then We see then that a n = [(n + r) 2 4 ]a n 2. a = a 3 = a 5 = 0 and hence that the power series is even. 0

11 Enter J /2 (x) The solution with r = /2 is then the even power series with recursion relation a 2n = [(2n+r) 2 4 ]a 2(n ) = [(2n+r /2)(2n+r+/2)] a 2(n ). Recalling that r = /2 it simplifies to: Thus, we have: a 2n = (2n)(2n + ) a 2(n ). a 2n = ( ) n (2n + )! a 0. The solution is therefore J /2 (x) = x /2 ( ) n (2n + )! x2n.

12 Enter J /2 (x) = x /2 sin x We can simplify the formula J /2 (x) = x /2 by re-writing it as ( ) n (2n + )! x2n J /2 (x) = x /2 ( ) n (2n + )! x2n+. We then observe that the infinite series is the Taylor series of sin x. This gives the displayed formula for J /2. Note that J /2 x /2 as x 0. 2

13 The second solution We now find the second solution, which behaves like, x /2 as x 0. This can be difficult when the roots differ by an integer, but in this case it fortunately is not. We seek a solution of the form: y = a n x n+r, r = /2. As before, [a n (n + r)(n + r ) + a n (n + r)]x n+r + a n x n+2+r 4 a nx n+r. We down shift by 2 in the second to last term as before and get a n [(n + r) 2 4 ]xn+r + n=2 a n 2 x n+r. 3

14 Hanging chad There are two terms hanging over in the first sum that don t match any terms in the second one (n = 0, ). They equal a 0 [r 2 4 ]xr + a [( + r) 2 4 ]xr+. The first term cancels because r = /2. The second also does! The recursion relation is a n = [(n + r) 2 4 ]a n 2. 4

15 Enter Y /2 (x) x /2 as x 0 The solution with r = /2 is then power series with recursion relation a 2n = = [(2n+r) 2 4 ]a 2(n ) [(2n+r /2)(2n+r+/2)] a 2n 2. Substituting r = /2 gives Thus, we have: a 2n = [(2n)(2n )] a 2n 2. a 2n = ( ) n (2n)!. The solution is therefore x /2 {a 0 ( ) n (2n)! x2n +a ( ) n (2n + )! x2n+ }. 5

16 Y /2 (x) = x /2 cos x The first term is x /2 cos x and the second is the solution previously obtained. general solution is: Thus the a 0 x /2 cos x + a x /2 sin x. Interestingly, the Bessel functions are familiar functions and we can immediately see how they behave as x 0 and x. 6

17 Bessel equation of order This is the equation x 2 y + xy + (x 2 )y = 0. The indicial equation is now r(r )+r = 0 r 2 = 0 r = ±. Again this is a problematic case where the roots differ by an integer: ( ) = 2. We first look at the larger root r =. 7

18 The root r = We seek a solution of the form: y = a n x n+r, r =. We have done the general case, so again we get: [a n (n + r)(n + r ) + a n (n + r)]x n+r + a n x n+2+r a n x n+r. We down shift by 2 in the second to last term to get a n [(n + r) 2 x n+r + n=2 a n 2 x n+r. 8

19 Hanging chad There two hanging terms are a 0 [r 2 ]x r + a [( + r) 2 ]x r+. The first term cancels because r =. The second does not and forces a = 0. The recursion relation is then We see then that a n = [(n + r) 2 ] a n 2. a = a 3 = a 5 = 0 and hence that the solution is even. 9

20 Enter J (x) x as x 0 The solution with r = is then the even power series with recursion relation a 2n = = [(2n+r) 2 ] a 2n 2 [(2n+r )(2n+r+)] a 2n 2. Recalling that r = it simplifies to: a 2n = [(2n)(2n + 2)] a 2n 2 = 2 2 (n + )n a 2n 2. Thus, we have: a 2n = ( ) n 2 2n n!. The solution is therefore J (x) = x ( ) n 2 2n n!(n + )! x2n. 20

21 Y (x) x, x 0 The second solution exhibits the full complexity of solutions differing by integer roots and has the form: Y (x) = J (x) ln x + x for certain coefficients b n. b n x n Further details left to the exercises. 2

22 Bessel functions and vibrating drums Bessel functions are closely related to modes of vibration of a drum. By a drum we mean a circular membrane which is fixed at its rim. The modes of vibration are the periodic (in time) motions of the membrane. The height of the membrane at time t above the disc in the x y plane at a point (x, y) is denoted u(x, y, t). If it is a periodic vibration it has the form u(x, y, t) = e iλt φ(x, y). Here, λ is the frequency of vibration. The profile mode φ(x, y) solves the equation where = 2 x y 2. φ = λ 2 φ 22

23 Radial part of vibration The modes can be written in polar coordinates as φ(r, θ) = f(r)e inθ for some f(r) and integer n. The equation for f(r) turns out to be r d dr (rdf dr ) + (λ2 n2 )f(r) = 0. r2 Multiplying by r 2 and simplifying gives: r 2 f + rf + (λ 2 r 2 n 2 )f = 0. If we put x = rλ and rewrite the equation in terms of ρ and f(r) = y(x) = y(λr), we get Bessel s equation of order n : x 2 y + xy + (x 2 n 2 )y = 0. 23

24 Frequencies of vibrations of a drum Thus, the modes of vibration of a drum have the form e inθ J n (λr). The frequencies λ are determined by the rule that the drum is still at r = (the rim), so J n (λ) = 0. Thus, the zeros of Bessel functions give the frequencies of vibration of a circular drum. Exercise 4 is about the Bessel function J 0 (λr) as radial modes of vibrations of a drum. The n = 0 means that only the diameter lines of the drum are vibrating, no waves are travelling around the rim. After 20 years of acoustics, the zeros or frequencies are still not understood. 24

25 Why does a drum sound like a drum? The frequencies with which a drum vibrates are the zeros {λ j } of the Bessel functions J 0, J,.... If the vibration is purely radial, they are the zeros of J 0. A vibrating string has frequencies nω where ω 2 is the bass note. I.e. the frequencies are integer multiples of one bass frequency. The higher harmonics are in harmony with each other. We hear music to our ears. For a drum, the frequencies {λ j } are far from being an arithmetic progression. In fact, they behave like random numbers, i.e. numbers put out by a random number generator. We hear the bass note of a drum; the drum s muffled sound is due to the randomness of the higher frequencies. 25

26 Problem 4 We recall that J 0 (0) =. Therefore J(λ j x) =, x = 0 0, x =. The function J 0 (λx) for any λ satisfies the equation: y + x y + λ 2 y = 0. We claim that if λ j λ k are two zeros of J 0 then 0 xj 0(λ j x)j 0 (λ k x)dx = 0. 26

27 Orthogonality Write L[y] = y + x y. We note that L = x [xy ]. The key formula to show is that L[y ]y 2 xdx = y L[y 2 ]xdx 0 0 if y () = y 2 () = 0. We show this by integrating by parts twice. Then let y = J 0 (λ k x), y 2 = J 0 (λ j x). Since L[J 0 (λ x)] = λ 2, L[J 0(λ 2 x)] = λ 2 2, we get (λ 2 j λ2 k ) 0 xj 0 (λ j x)j 0 (λ k x)dx = 0. This is the desired orthogonality This is related to the fact that J 0 (λ j r) represent modes of vibration of a drum where all of the vibrations are radial, i.e. outward toward the rim. 27