Power Series Solutions to the Bessel Equation

Transcription

1 Power Series Solutions to the Bessel Equation Department of Mathematics IIT Guwahati

2 The Bessel equation The equation x 2 y + xy + (x 2 α 2 )y = 0, (1) where α is a non-negative constant, i.e α 0, is called the Bessel equation of order α.

3 The Bessel equation The equation x 2 y + xy + (x 2 α 2 )y = 0, (1) where α is a non-negative constant, i.e α 0, is called the Bessel equation of order α. The point x 0 = 0 is a regular singular point. We shall use the method of Frobenius to solve this equation.

4 The Bessel equation The equation x 2 y + xy + (x 2 α 2 )y = 0, (1) where α is a non-negative constant, i.e α 0, is called the Bessel equation of order α. The point x 0 = 0 is a regular singular point. We shall use the method of Frobenius to solve this equation. Thus, we seek solutions of the form y(x) = a n x n+r, x > 0, (2) with a 0 0.

5 Differentiation of (2) term by term yields y = (n + r)a n x n+r 1.

6 Differentiation of (2) term by term yields y = Similarly, we obtain (n + r)a n x n+r 1. y = (n + r)(n + r 1)a n x n+r 2.

7 Differentiation of (2) term by term yields y = Similarly, we obtain (n + r)a n x n+r 1. y = (n + r)(n + r 1)a n x n+r 2. Substituting these into (1), we obtain (n + r)(n + r 1)a n x n+r + (n + r)a n x n+r + a n x n+r+2 α 2 a n x n+r = 0.

8 This implies x r [(n + r) 2 α 2 ]a n x n + x r a n x n+2 = 0.

9 This implies x r [(n + r) 2 α 2 ]a n x n + x r a n x n+2 = 0. Now, cancel x r, and try to determine a n s so that the coefficient of each power of x will vanish. For the constant term, we require (r 2 α 2 )a 0 = 0. Since a 0 0, it follows that r 2 α 2 = 0,

10 This implies x r [(n + r) 2 α 2 ]a n x n + x r a n x n+2 = 0. Now, cancel x r, and try to determine a n s so that the coefficient of each power of x will vanish. For the constant term, we require (r 2 α 2 )a 0 = 0. Since a 0 0, it follows that r 2 α 2 = 0, which is the indicial equation.

11 This implies x r [(n + r) 2 α 2 ]a n x n + x r a n x n+2 = 0. Now, cancel x r, and try to determine a n s so that the coefficient of each power of x will vanish. For the constant term, we require (r 2 α 2 )a 0 = 0. Since a 0 0, it follows that r 2 α 2 = 0, which is the indicial equation. The only possible values of r are α and α.

12 Case I. For r = α, the equations for determining the coefficients are: [(1 + α) 2 α 2 ]a 1 = 0 and, [(n + α) 2 α 2 ]a n + a n 2 = 0, n 2.

13 Case I. For r = α, the equations for determining the coefficients are: [(1 + α) 2 α 2 ]a 1 = 0 and, [(n + α) 2 α 2 ]a n + a n 2 = 0, n 2. Since α 0, we have a 1 = 0.

14 Case I. For r = α, the equations for determining the coefficients are: [(1 + α) 2 α 2 ]a 1 = 0 and, [(n + α) 2 α 2 ]a n + a n 2 = 0, n 2. Since α 0, we have a 1 = 0. The second equation yields a n 2 a n = (n + α) 2 α = a n 2 2 n(n + 2α). (3)

15 Case I. For r = α, the equations for determining the coefficients are: [(1 + α) 2 α 2 ]a 1 = 0 and, [(n + α) 2 α 2 ]a n + a n 2 = 0, n 2. Since α 0, we have a 1 = 0. The second equation yields a n 2 a n = (n + α) 2 α = a n 2 2 n(n + 2α). (3) Since a 1 = 0, we immediately obtain a 3 = a 5 = a 7 = = 0.

16 For the coefficients with even subscripts, we have a 0 a 2 = 2(2 + 2α) = a (1 + α), a 4 = a 6 = a 2 4(4 + 2α) = ( 1) 2 a !(1 + α)(2 + α), a 4 6(6 + 2α) = ( 1) 3 a !(1 + α)(2 + α)(3 + α),

17 For the coefficients with even subscripts, we have a 0 a 2 = 2(2 + 2α) = a (1 + α), a 4 = a 6 = and, in general a 2n = a 2 4(4 + 2α) = ( 1) 2 a !(1 + α)(2 + α), a 4 6(6 + 2α) = ( 1) 3 a !(1 + α)(2 + α)(3 + α), ( 1) n a 0 2 2n n!(1 + α)(2 + α) (n + α). Therefore, the choice r = α yields the solution ) y α (x) = a 0 x (1 α ( 1) n x 2n n n!(1 + α)(2 + α) (n + α) n=1 Note: The ratio test shows that the power series formula converges for all x R.

18 x < 0: Put x = t, where t > 0, and set z(t) = y(x),

19 x < 0: Put x = t, where t > 0, and set z(t) = y(x), x 2 y + xy + (x 2 α 2 )y = 0, x < 0 (4) = t 2 z + tz + (t 2 α 2 )z = 0, t > 0

20 x < 0: Put x = t, where t > 0, and set z(t) = y(x), x 2 y + xy + (x 2 α 2 )y = 0, x < 0 (4) = t 2 z + tz + (t 2 α 2 )z = 0, t > 0 = z(t) = t r a n t n, r 2 α 2 = 0.

21 x < 0: Put x = t, where t > 0, and set z(t) = y(x), For r = α, z α (t) = a 0 t α (1 + x 2 y + xy + (x 2 α 2 )y = 0, x < 0 (4) = t 2 z + tz + (t 2 α 2 )z = 0, t > 0 = z(t) = t r a n t n, r 2 α 2 = 0. n=1 ) ( 1) n t 2n, t > 0 2 2n n!(1 + α)(2 + α) (n + α)

22 x < 0: Put x = t, where t > 0, and set z(t) = y(x), For r = α, z α (t) = a 0 t α (1 + Therefore x 2 y + xy + (x 2 α 2 )y = 0, x < 0 (4) = t 2 z + tz + (t 2 α 2 )z = 0, t > 0 = z(t) = t r a n t n, r 2 α 2 = 0. n=1 y α(x) = a 0( x) α (1 + is a solution of (4). ) ( 1) n t 2n, t > 0 2 2n n!(1 + α)(2 + α) (n + α) n=1 ) ( 1) n x 2n, x < n n!(1 + α)(2 + α) (n + α)

23 Therefore, the function y α (x) is given by ) y α (x) = a 0 x (1 α ( 1) n x 2n + 2 2n n!(1 + α)(2 + α) (n + α) n=1 is a solution of the Bessel equation valid for all real x 0.

24 Therefore, the function y α (x) is given by ) y α (x) = a 0 x (1 α ( 1) n x 2n + 2 2n n!(1 + α)(2 + α) (n + α) n=1 is a solution of the Bessel equation valid for all real x 0. Qn: What about the second linearly independent solution?

25 Therefore, the function y α (x) is given by ) y α (x) = a 0 x (1 α ( 1) n x 2n + 2 2n n!(1 + α)(2 + α) (n + α) n=1 is a solution of the Bessel equation valid for all real x 0. Qn: What about the second linearly independent solution? When you can find the second linearly independent solution y α?

26 Therefore, the function y α (x) is given by ) y α (x) = a 0 x (1 α ( 1) n x 2n + 2 2n n!(1 + α)(2 + α) (n + α) n=1 is a solution of the Bessel equation valid for all real x 0. Qn: What about the second linearly independent solution? When you can find the second linearly independent solution y α?

27 Case II. For r = α, determine the coefficients from [(1 α) 2 α 2 ]a 1 = 0 and [(n α) 2 α 2 ]a n + a n 2 = 0. These equations become (1 2α)a 1 = 0 and n(n 2α)a n + a n 2 = 0.

28 Case II. For r = α, determine the coefficients from [(1 α) 2 α 2 ]a 1 = 0 and [(n α) 2 α 2 ]a n + a n 2 = 0. These equations become (1 2α)a 1 = 0 and n(n 2α)a n + a n 2 = 0. If α ( α) = 2α is not an integer, these equations give us a n 2 a 1 = 0 and a n = n(n 2α), n 2. Again a 3 = a 5 = a 7 = = 0.

29 Case II. For r = α, determine the coefficients from [(1 α) 2 α 2 ]a 1 = 0 and [(n α) 2 α 2 ]a n + a n 2 = 0. These equations become (1 2α)a 1 = 0 and n(n 2α)a n + a n 2 = 0. If α ( α) = 2α is not an integer, these equations give us a n 2 a 1 = 0 and a n = n(n 2α), n 2. Again a 3 = a 5 = a 7 = = 0. Note that this formula is same as (3), with α replaced by α. Thus, the solution is given by y α (x) = a 0 x α (1 + n=1 which is valid for all real x 0. ( 1) n x 2n 2 2n n!(1 α)(2 α) (n α) ),

30 Therefore when 2α is not an integer the Bessel equation x 2 y + xy + (x 2 α 2 )y = 0 has two linearly independent solutions ) y α (x) = a 0 x (1 α ( 1) n x 2n +, 2 2n n!(1 + α)(2 + α) (n + α) y α (x) = a 0 x α (1 + n=1 n=1 both are valid for all real x 0 ) ( 1) n x 2n, 2 2n n!(1 α)(2 α) (n α)

31 Therefore when 2α is not an integer the Bessel equation x 2 y + xy + (x 2 α 2 )y = 0 has two linearly independent solutions ) y α (x) = a 0 x (1 α ( 1) n x 2n +, 2 2n n!(1 + α)(2 + α) (n + α) y α (x) = a 0 x α (1 + n=1 n=1 both are valid for all real x 0 ) ( 1) n x 2n, 2 2n n!(1 α)(2 α) (n α) Note: 2α is a not an integer means α 1, α 3, etc. and 2 2 α integer.

32 Therefore when 2α is not an integer the Bessel equation x 2 y + xy + (x 2 α 2 )y = 0 has two linearly independent solutions ) y α (x) = a 0 x (1 α ( 1) n x 2n +, 2 2n n!(1 + α)(2 + α) (n + α) y α (x) = a 0 x α (1 + n=1 n=1 both are valid for all real x 0 ) ( 1) n x 2n, 2 2n n!(1 α)(2 α) (n α) Note: 2α is a not an integer means α 1, α 3, etc. and 2 2 α integer. Note: In fact when α / Z + the function y α (x) defined above forms a solution of Bessel equation. Why?

33 Now we concentrate only in the case when x > 0.

34 Euler s gamma function and its properties For s R with s > 0, we define Γ(s) by Γ(s) = 0 t s 1 e t dt. The integral converges if s > 0 and diverges if s 0.

35 Euler s gamma function and its properties For s R with s > 0, we define Γ(s) by Γ(s) = 0 t s 1 e t dt. The integral converges if s > 0 and diverges if s 0. Integration by parts yields the functional equation Γ(s + 1) = sγ(s).

36 Euler s gamma function and its properties For s R with s > 0, we define Γ(s) by Γ(s) = 0 t s 1 e t dt. The integral converges if s > 0 and diverges if s 0. Integration by parts yields the functional equation In general, Γ(s + 1) = sγ(s). Γ(s + n) = (s + n 1) (s + 1)sΓ(s), for every n Z +.

37 Euler s gamma function and its properties For s R with s > 0, we define Γ(s) by Γ(s) = 0 t s 1 e t dt. The integral converges if s > 0 and diverges if s 0. Integration by parts yields the functional equation In general, Γ(s + 1) = sγ(s). Γ(s + n) = (s + n 1) (s + 1)sΓ(s), for every n Z +. Since Γ(1) = 1, we find that Γ(n + 1) = n!. Thus, the gamma function is an extension of the factorial function from integers to positive real numbers. Therefore, we write Γ(s + 1) = sγ(s), s R +.

38 When s < 0 and s is not a negative integer we define Γ(s+1) if 1 < s < 0, s Γ(s+1) Γ(s) = if 2 < s < 1, s

39 When s < 0 and s is not a negative integer we define Γ(s+1) if 1 < s < 0, s Γ(s+1) Γ(s) = if 2 < s < 1, s In fact Γ(s) is defined for all s R \ {0} Z

40 Using this gamma function, we shall simplify the form of the solutions of the Bessel equation.

41 Using this gamma function, we shall simplify the form of the solutions of the Bessel equation.with s = 1 + α, we note that (1 + α)(2 + α) (n + α) = Γ(n α). Γ(1 + α)

42 Using this gamma function, we shall simplify the form of the solutions of the Bessel equation.with s = 1 + α, we note that (1 + α)(2 + α) (n + α) = Γ(n α). Γ(1 + α) Choosing a 0 = 2 α, the expression for y Γ(1+α) α, can be written as ( x ) α ( 1) n ( x ) 2n J α (x) =, x > 0. 2 n!γ(n α) 2

43 Using this gamma function, we shall simplify the form of the solutions of the Bessel equation.with s = 1 + α, we note that (1 + α)(2 + α) (n + α) = Γ(n α). Γ(1 + α) Choosing a 0 = 2 α, the expression for y Γ(1+α) α, can be written as ( x ) α ( 1) n ( x ) 2n J α (x) =, x > 0. 2 n!γ(n α) 2 The function J α defined above for x > 0 and α 0 is called the Bessel function of the first kind of order α.

44 What about J α?

45 What about J α? Of course when α ( α) = 2α is not an integer then J α is defined as below and both J α and J α are linearly independent. ( x ) α ( 1) n ( x ) 2n J α (x) =, x > 0. 2 n!γ(n + 1 α) 2 i.e.,j α is nothing but y α with a 0 = 2α. Γ(1 α)

46 What about J α? Of course when α ( α) = 2α is not an integer then J α is defined as below and both J α and J α are linearly independent. ( x ) α ( 1) n ( x ) 2n J α (x) =, x > 0. 2 n!γ(n + 1 α) 2 i.e.,j α is nothing but y α with a 0 = 2α. Γ(1 α) In fact, J α is defined as above for α 0, α / Z + is a solution of Bessel equation for x > 0. Why?...

47 What about J α? Of course when α ( α) = 2α is not an integer then J α is defined as below and both J α and J α are linearly independent. ( x ) α ( 1) n ( x ) 2n J α (x) =, x > 0. 2 n!γ(n + 1 α) 2 i.e.,j α is nothing but y α with a 0 = 2α. Γ(1 α) In fact, J α is defined as above for α 0, α / Z + is a solution of Bessel equation for x > 0. Why?... Conclusion: If α / Z + {0}, J α (x) and J α (x) are linearly independent on x > 0. The general solution of the Bessel equation for x > 0 is y(x) = c 1 J α (x) + c 2 J α (x).

48 When α is a non-negative integer, say α = p, the Bessel function J p (x) is given by ( 1) n ( x ) 2n+p J p (x) =, (p = 0, 1, 2,...). n!(n + p)! 2 Figure: The Bessel functions J 0 and J 1.

49 Useful recurrence relations for J α d dx (xα J α (x)) = x α J α 1 (x). { d dx (xα J α (x)) = d dx } x α ( 1) n ( x ) 2n+α n! Γ(1 + α + n) 2 { = d } ( 1) n x 2n+2α dx n! Γ(1 + α + n)2 2n+α ( 1) n (2n + 2α)x 2n+2α 1 = n! Γ(1 + α + n)2 2n+α. Since Γ(1 + α + n) = (α + n)γ(α + n), we have d ( 1) n 2x 2n+2α 1 dx (xα J α (x)) = n! Γ(α + n)2 2n+α = x α = x α J α 1 (x). ( 1) n ( x ) 2n+α 1 n! Γ(1 + (α 1) + n) 2

50 The other relations involving J α are: d dx (x α J α (x)) = x α J α+1 (x). α x J α(x) + J α(x) = J α 1 (x). α x J α(x) J α(x) = J α+1 (x). J α 1 (x) + J α+1 (x) = 2α x J α(x). J α 1 (x) J α+1 (x) = 2J α(x). Note: Workout these relations. *** End ***