Bessel s Equation. MATH 365 Ordinary Differential Equations. J. Robert Buchanan. Fall Department of Mathematics
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1 Bessel s Equation MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018
2 Background Bessel s equation of order ν has the form where ν is a constant. x 2 y + xy + (x 2 ν 2 )y = 0 Bessel s equation arises frequently in the mathematical modeling of vibration in polar, cylindrical, or spherical coordinates.
3 Vibrating Circular Membrane (i.e. a drum) Wave equation: u tt = c 2 (u xx + u yy ) for x 2 + y 2 a 2. Boundary condition: u(x, y, t) = 0 for x 2 + y 2 = a 2.
4 Polar Coordinates Since x = r cos θ and y = r sin θ we can re-write the boundary value problem as ( 1 Wave equation: u tt = c 2 r (r u r ) r + 1 ) r 2 u θθ for r a. Boundary condition: u(r, θ, t) = 0 for r = a.
5 Isotropy Assume the solution is independent of θ, i.e. the motion is the same in every direction. Wave equation: u tt = c2 r (r u r ) r for r a. Boundary condition: u(r, t) = 0 for r = a.
6 Isotropy Assume the solution is independent of θ, i.e. the motion is the same in every direction. Wave equation: u tt = c2 r (r u r ) r for r a. Boundary condition: u(r, t) = 0 for r = a. We will assume the solution u(r, t) = R(r)T (t). For PDEs this is called the method of separation of variables.
7 Separation of Variables u tt = c2 r (r u r ) r R(r)T (t) = c2 r (r R (r)) T (t) R(r)T (t) = c2 r (R (r) + r R (r))t (t) 1 T (t) c 2 T (t) = R (r) + r R (r) r R(r)
8 Separation of Variables u tt = c2 r (r u r ) r R(r)T (t) = c2 r (r R (r)) T (t) R(r)T (t) = c2 r (R (r) + r R (r))t (t) 1 T (t) c 2 T (t) = R (r) + r R (r) r R(r) Since the left-hand side is a function of t only and the right-hand side is a function of r only, they must be constant.
9 Constant R (r) + r R (r) r R(r) = λ 2 R (r) + r R (r) = λ 2 r R(r) r R (r) + R (r) + λ 2 r R(r) = 0 r 2 R (r) + r R (r) + λ 2 r 2 R(r) = 0
10 Change of Variable Define x = λr and y(x) = R ( x λ) then R (r) = dy dx λ R (r) = d 2 y dx 2 λ2 and substituting into the ODE yields r 2 R (r) + r R (r) + λ 2 r 2 R(r) = 0 r 2 d 2 y dx 2 λ2 + r dy dx λ + λ2 r 2 y = x 2 y + xy + x 2 y = 0 which is Bessel s equation of order 0.
11 Bessel s Equation of Order Zero x 2 y + xy + x 2 y = 0 has a regular singular point at x = 0. The exponents of singularity are r 1 = r 2 = 0. Two linearly independent solutions are: where H n = y 1 (x) = 1 + n=1 y 2 (x) = y 1 (x) ln x + n k=1 1 k. ( 1) n ( x ) 2n (n!) 2 2 n=1 ( 1) n+1 H ( n x ) 2n (n!) 2 2
12 Remarks Solution y 1 (x) is called the Bessel function of the first kind of order zero and denoted J 0 (x). The Bessel function of the second kind of order zero is denoted Y 0 (x) and defined as Y 0 (x) = 2 π [y 2(x) + (γ ln 2)J 0 (x)] where γ is the Euler Máscheroni constant and γ = lim n (H n ln n) General solution to Bessel s equation of order zero is written y(x) = c 1 J 0 (x) + c 2 Y 0 (x).
13 Graphs 1.0 y x J 0 (x) Y 0 (x)
14 Asymptotic Properties (1 of 2) when x. 0 = x 2 y + xy + (x 2 ν 2 )y = y + 1 ) x y + (1 ν2 y y + y x 2
15 Asymptotic Properties (1 of 2) when x. 0 = x 2 y + xy + (x 2 ν 2 )y = y + 1 ) x y + (1 ν2 y y + y Since cos(x c) and sin(x c) solve y + y = 0 we should expect J 0 (x) and Y 0 (x) to be oscillatory. In fact x 2 as x. J 0 (x) Y 0 (x) ( 2 πx ( 2 πx ) 1/2 ( cos x π ) 4 ) 1/2 ( sin x π ) 4
16 Asymptotic Properties (2 of 2) y J 0 (x) 2 π x cos x - π x
17 Bessel s Equation of Order 1/2 ( x 2 y + x y + x 2 1 ) y = 0 4 has a regular singular point at x = 0. Indicial equation: ( x ) p 0 = lim x x 0 x 2 q 0 = lim x 0 x 2 = 1 ) ( x x 2 = = r(r 1) + r 1 4 = r Exponents of singularity: r 1 = 1 2 and r 2 = 1 2.
18 Solution (1 of 8) Assuming y(x) = a n x r+n, differentiating, and substituting into the ODE we obtain 0 = x 2 (r + n)(r + n 1)a n x r+n 2 + x (r + n)a n x r+n 1 = = ( + x 2 1 ) a n x r+n 4 (r + n)(r + n 1)a n x r+n + (r + n)a n x r+n 1 4 a n x r+n + a n x r+n+2 [ (r + n)(r + n 1) + (r + n) 1 ] a n x r+n + 4 a n x r+n+2
19 Solution (2 of 8) 0 = = = = [ (r + n)(r + n 1) + (r + n) 1 ] a n x r+n + 4 [ (r + n) 2 1 ] a n x r+n + 4 ( r a n x r+n+2 [ (r + n) 2 1 ] a n x r+n + 4 n=2 ) ( a 0 x r + (r + 1) n=2 [ (r + n) 2 1 ] a n x r+n + 4 a n 2 x r+n ) a 1 x r+1 a n 2 x r+n n=2 a n x r+n+2
20 Solution (3 of 8) 0 = = ( r 2 1 ) ( a 0 x r + (r + 1) 2 1 ) a 1 x r n=2 ( r n=2 [ (r + n) 2 1 ] a n x r+n + 4 ) ( a 0 x r + (r + 1) n=2 a n 2 x r+n ) a 1 x r+1 ([ (r + n) 2 1 ] ) a n + a n 2 x r+n 4 Recurrence relation: a n (r) = a n 2(r) (r + n) 2 1 4
21 Solution (4 of 8) When r = r 1 = 1 2 then the expression ( (r + 1) 2 1 ) a 1 x r+1 = 0 4 when a 1 = 0.
22 Solution (4 of 8) When r = r 1 = 1 2 then the expression ( (r + 1) 2 1 ) a 1 x r+1 = 0 4 when a 1 = 0. Thus we will choose a 0 = 1 and a 1 = 0. a 1 = 0 implies a 2n+1 = 0 for all n N.
23 Solution (5 of 8) a 2 (1/2) = a 0 (2)(3) = 1 3! a 4 (1/2) = a 2 (4)(5) = 1 5! a 6 (1/2) = a 4 (6)(7) = 1 7!. a 2n (1/2) = ( 1) n (2n + 1)! y 1 (x) = x 1/2 [1 + = x 1/2 sin x. n=1 ] ( 1) n x 2n = x 1/2 ( 1) n x 2n+1 (2n + 1)! (2n + 1)!
24 Solution (6 of 8) Note: r 1 r 2 = 1 N but when r = r 2 = 1 2 then the expression ( (r + 1) 2 1 ) a 1 x r+1 = 0 4 for any value of a 1.
25 Solution (6 of 8) Note: r 1 r 2 = 1 N but when r = r 2 = 1 2 then the expression ( (r + 1) 2 1 ) a 1 x r+1 = 0 4 for any value of a 1. Thus we will again choose a 0 = 1 and a 1 = 0. a 1 = 0 implies a 2n+1 = 0 for all n N.
26 Solution (7 of 8) y 2 (x) = x 1/2 [1 + a 2 ( 1/2) = a 0 (2)(1) = 1 2! a 4 ( 1/2) = a 2 (4)(3) = 1 4! a 6 ( 1/2) = a 4 (6)(5) = 1 6! a 2n ( 1/2) = ( 1)n (2n)! = x 1/2 cos x. n=1. ] ( 1) n x 2n = x 1/2 ( 1) n x 2n (2n)! (2n)!
27 Solution (8 of 8) Bessel function of the first kind of order one-half J 1/2 (x) = ( ) 2 1/2 ( ) 2 1/2 y 1 (x) = sin x. π πx Bessel function of the second kind of order one-half J 1/2 (x) = ( ) 2 1/2 ( ) 2 1/2 y 2 (x) = cos x. π πx
28 Graphs y x J 1 (x) 2 Y 1 (x)
29 Bessel s Equation of Order One x 2 y + xy + (x 2 1)y = 0 The value x = 0 is a regular singular point since ( x ) lim x x 0 x 2 = 1 = p 0 ( x lim x 2 2 ) 1 = 1 = q 0. x 0 The exponents of singularity are the solutions of so r 1 = 1 and r 2 = 1. x 2 0 = F(r) = r(r 1) + r 1 = r 2 1
30 Solution (1 of 12) Assuming y(x) = a n x r+n, differentiating, and substituting into the ODE we obtain 0 = x 2 (r + n)(r + n 1)a n x r+n 2 + x (r + n)a n x r+n 1 = = + (x 2 1) a n x r+n (r + n)(r + n 1)a n x r+n + (r + n)a n x r+n a n x r+n + a n x r+n+2 [(r + n)(r + n 1) + (r + n) 1] a n x r+n + a n x r+n+2
31 Solution (2 of 12) 0 = = = [(r + n)(r + n 1) + (r + n) 1] a n x r+n + a n x r+n+2 [ ] (r + n) 2 1 a n x r+n + [ ] (r + n) 2 1 a n x r+n + = (r 2 1)a 0 x r + + n=2 ( (r + 1) 2 1 a n x r+n+2 a n 2 x r+n n=2 [ ] (r + n) 2 1 a n x r+n + ) a 1 x r+1 a n 2 x r+n n=2
32 Solution (3 of 12) 0 = = ( ) ( ) r 2 1 a 0 x r + (r + 1) 2 1 a 1 x r+1 + n=2 [ ] (r + n) 2 1 a n x r+n + a n 2 x r+n n=2 ( ) ( ) r 2 1 a 0 x r + (r + 1) 2 1 a 1 x r+1 ([ ] ) + (r + n) 2 1 a n + a n 2 x r+n n=2 Recurrence relation: a n (r) = a n 2(r) (r + n) 2 1 for n 2.
33 Solution (4 of 12) When r = r 1 = 1 then the expression ( ) (r + 1) 2 1 a 1 x r+1 = 0 when a 1 = 0.
34 Solution (4 of 12) When r = r 1 = 1 then the expression ( ) (r + 1) 2 1 a 1 x r+1 = 0 when a 1 = 0. Thus we will choose a 0 = 1 and a 1 = 0. a 1 = 0 implies a 2n+1 = 0 for all n N.
35 Solution (5 of 12) a 2 (1) = a 0 (2)(4) = 1 4(2!)(1!) a 4 (1) = a 2 (4)(6) = (3!)(2!) a 6 (1) = a 4 (6)(8) = (4!)(3!). a 2n (1) = y 1 (x) = x J 1 (x) = ( 1) n 4 n (n + 1)!(n!) [ 1 + n=1 ] ( 1) n x 2n 4 n (n + 1)!(n!) ( 1) n ( x ) 2n+1. n!(n + 1)! 2
36 Solution (6 of 12) Since r 1 r 2 = 2 the second solution will be of the form where y 2 (x) = ay 1 (x) ln x + x 1 [1 + ] c n ( 1)x n n=1 a = lim r 1 [(r ( 1))a 2(r)] = lim r 1 = lim r 1 = lim r 1 (r + 1)a 0 (r + 2) 2 1 (r + 1) r 2 + 4r + 3 (r + 1) (r + 3)(r + 1) 1 = lim r 1 r + 3 = 1 2
37 Solution (7 of 12) Recall: the coefficients of the power series c n (r 2 ) = d dr [(r r 2)a n (r)] r=r2, a 1 (±1) = 0 which implies a 1 ( 1) = 0, this in turn implies c 1 ( 1) = 0, from the recurrence relation then c 2n+1 ( 1) = 0 for n N.
38 Solution (8 of 12) From the recurrence relation a n (r) = a n 2(r) (r + n) 2 1 for n 2, and the fact that a 1 ( 1) = 0 then a 2n+1 ( 1) = 0 and a 2n+1 ( 1) = 0 for n N. a 2 (r) = 1 (r + 1)(r + 3) a 4 (r) = 1 (r + 1)(r + 3)(r + 3)(r + 5). a 2n (r) = ( 1) n (r + 1) (r + 2n 1)(r + 3) (r + 2n + 1)
39 Solution (9 of 12) Thus (r + 1)a 2 (r) = 1 r (r + 1)a 4 (r) = (r + 3)(r + 3)(r + 5) (r + 1)a 2n (r) =. ( 1) n (r + 3) (r + 2n 1)(r + 3) (r + 2n + 1)
40 Solution (10 of 12) Lemma If f (x) = (x α 1 ) β 1(x α 2 ) β 2 (x α n ) βn and x / {α 1, α 2,..., α n } then This implies f (x) f (x) = β 1 + β β n. x α 1 x α 2 x α n f (x) = f (x) [ β1 x α 1 + β 2 x α β n x α n ].
41 Solution (11 of 12) Applying the previous lemma we get [ d 2 dr [(r + 1)a 2n(r)] r= 1 = ( 1) n (r + 1)a 2n (r) r r + 5 ] r + 2n r + 2n + 1 ( 1) n+1 [ 2 = 2 2n 1 (n 1)!(n!) n ] 2n = c 2n ( 1) = ( 1) n+1 4 n (n 1)!(n!) r= 1 [ n n ( 1) n+1 4 n (n 1)!(n!) (H n 1 + H n ) ]
42 Solution (12 of 12) Finally, the second solution takes the form [ ] y 2 (x) = J 1 (x) ln x + 1 ( 1) n (H n 1 + H n ) ( x ) 2n 1. x n!(n 1)! 2 n=1 The Bessel function of the second kind of order one is defined to be Y 1 (x) = 2 π ( y 2(x) + (γ ln 2)J 1 (x)).
43 Graphs y x J 1 (x) Y 1 (x)
44 Useful Bessel Function Identities d dx [x n J n (x)] = x n J n 1 (x) d [ x n J n (x) ] dx = x n J n+1 (x) J n+1 (x) = 2n x J n(x) J n 1 (x)
45 Homework Read Section 5.7 Exercises: 5 8
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