1. Series Solutions of Linear Differential Equations CHAPTER 5

2. Ch5_2 Contents  5.1 Solutions about Ordinary Points 5.1 Solutions about Ordinary Points  5.2 Solution about Singular Points 5.2 Solution about Singular Points  5.3 Special Functions 5.3 Special Functions

3. Ch5_3 5.1 Solutions about Ordinary Point  Review of Power Series Recall from that a power series in x – a has the form Such a series is said to be a power series centered at a.

4. Ch5_4  Convergence exists.  Interval of Convergence The set of all real numbers for which the series converges.  Radius of Convergence If R is the radius of convergence, the power series converges for |x – a| R.

5. Ch5_5  Absolute Convergence Within its interval of convergence, a power series converges absolutely. That is, the following converges.  Ratio Test Suppose c n  0 for all n, and If L 1, this series diverges, if L = 1, the test is inclusive.

6. Ch5_6  A Power Defines a Function Suppose then  Identity Property If all c n = 0, then the series = 0.

7. Ch5_7  Analytic at a Point A function f is analytic at a point a, if it can be represented by a power series in x – a with a positive radius of convergence. For example: (2)

8. Ch5_8  Arithmetic of Power Series Power series can be combined through the operations of addition, multiplication and division.

9. Ch5_9 Example 1 Write as one power series. Solution Since we let k = n – 2 for the first series and k = n + 1 for the second series,

10. Ch5_10 then we can get the right-hand side as (3) We now obtain (4) Example 1 (2)

11. Ch5_11  Suppose the linear DE (5) is put into (6) A Solution A point x 0 is said to be an ordinary point of (5) if both P and Q in (6) are analytic at x 0. A point that is not an ordinary point is said to be a singular point. DEFINITION 5.1

12. Ch5_12  Since P and Q in (6) is a rational function, P = a 1 (x)/a 2 (x), Q = a 0 (x)/a 2 (x) It follows that x = x 0 is an ordinary point of (5) if a 2 (x 0 )  0. Polynomial Coefficients

13. Ch5_13  A series solution converges at least of some interval defined by |x – x 0 | < R, where R is the distance from x 0 to the nearest singular point. If x = x 0 is an ordinary point of (5), we can always find two linearly independent solutions in the form of power series centered at x 0, that is, THEOREM 5.1 Criterion for an Extra Differential

14. Ch5_14 Example 2 Solve Solution We know there are no finite singular points. Now, and then the DE gives (7)

15. Ch5_15 Example 2 (2) From the result given in (4), (8) Since (8) is identically zero, it is necessary all the coefficients are zero, 2c 2 = 0, and (9) Now (9) is a recurrence relation, since (k + 1)(k + 2)  0, then from (9) (10)

16. Ch5_16 Example 2 (3) Thus we obtain

17. Ch5_17 and so on. Example 2 (4)

18. Ch5_18 Example 2 (5) Then the power series solutions are y = c 0 y 1 + c 1 y 2

19. Ch5_19 Example 2 (6)

20. Ch5_20 Example 3 Solve Solution Since x 2 + 1 = 0, then x = i, −i are singular points. A power series solution centered at 0 will converge at least for |x| < 1. Using the power series form of y, y’ and y”, then

21. Ch5_21 Example 3 (2)

22. Ch5_22 Example 3 (3) From the above, we get 2c 2 － c 0 = 0, 6c 3 = 0, and Thus c 2 = c 0 /2, c k+2 = (1 – k)c k /(k + 2) Then

23. Ch5_23 Example 3 (4) and so on.

24. Ch5_24 Example 3 (5) Therefore,

25. Ch5_25 Example 4 If we seek a power series solution y(x) for we obtain c 2 = c 0 /2 and the recurrence relation is Examination of the formula shows c 3, c 4, c 5, … are expresses in terms of both c 1 and c 2. However it is more complicated. To simplify it, we can first choose c 0  0, c 1 = 0. Then we have

26. Ch5_26 Example 4 (2) and so on. Next, we choose c 0 = 0, c 1  0, then

27. Ch5_27 Example 4 (3) and so on. Thus we have y = c 0 y 1 + c 1 y 2, where

28. Ch5_28 Example 5 Solve Solution We see x = 0 is an ordinary point of the equation. Using the Maclaurin series for cos x, and using, we find

29. Ch5_29 Example 5 (2) It follows that and so on. This gives c 2 = － 1/2c 0, c 3 = － 1/6c 1, c 4 = 1/12c 0, c 5 = 1/30c 1,…. By grouping terms we get the general solution y = c 0 y 1 + c 1 y 2, where the convergence is |x| < , and

30. Ch5_30 5.2 Solutions about Singular Points  A Definition A singular point x 0 of a linear DE (1) is further classified as either regular or irregular. This classification depends on (2)

31. Ch5_31 A singular point x 0 is said to be a regular singular point of (1), if p(x) = (x – x 0 )P(x), q(x) = (x – x 0 ) 2 Q(x) are both analytic at x 0. A singular point that is not regular is said to be irregular singular point. DEFINITION 5.2 Regular/Irregular Singular Points

32. Ch5_32 Polynomial Ciefficients  If x – x 0 appears at most to the first power in the denominator of P(x) and at most to the second power in the denominator of Q(x), then x – x 0 is a regular singular point.  If (2) is multiplied by (x – x 0 ) 2, (3) where p, q are analytic at x = x 0

33. Ch5_33 Example 1  It should be clear x = 2, x = – 2 are singular points of (x 2 – 4) 2 y” + 3(x – 2)y’ + 5y = 0 According to (2), we have

34. Ch5_34 Example 1 (2) For x = 2, the power of (x – 2) in the denominator of P is 1, and the power of (x – 2) in the denominator of Q is 2. Thus x = 2 is a regular singular point. For x = −2, the power of (x + 2) in the denominator of P and Q are both 2. Thus x = − 2 is a irregular singular point.

35. Ch5_35 If x = x 0 is a regular singular point of (1), then there exists one solution of the form (4) where the number r is a constant to be determined. The series will converge at least on some interval 0 < x – x 0 < R. THEOREM 5.2 Frobenius’ Theorem

36. Ch5_36 Example 2: Frobenius’ Method  Because x = 0 is a regular singular point of (5) we try to find a solution. Now,

37. Ch5_37 Example 2 (2)

38. Ch5_38 Example 2 (3) which implies r(3r – 2)c 0 = 0 (k + r + 1)(3k + 3r + 1)c k+1 – c k = 0, k = 0, 1, 2, … Since nothing is gained by taking c 0 = 0, then r(3r – 2) = 0(6) and (7) From (6), r = 0, 2/3, when substituted into (7),

39. Ch5_39 Example 2 (4) r 1 = 2/3, k = 0,1,2,… (8) r 2 = 0, k = 0,1,2,… (9)

40. Ch5_40 Example 2 (5) From (8) From(9)

41. Ch5_41 Example 2 (6) These two series both contain the same multiple c 0. Omitting this term, we have (10) (11)

42. Ch5_42 Example 2 (7) By the ratio test, both (10) and (11) converges for all finite value of x, that is, |x| < . Also, from the forms of (10) and (11), they are linearly independent. Thus the solution is y(x) = C 1 y 1 (x) + C 2 y 2 (x), 0 < x < 

43. Ch5_43 Indicial Equation  Equation (6) is called the indicial equation, where the values of r are called the indicial roots, or exponents.  If x = 0 is a regular singular point of (1), then p = xP and q = x 2 Q are analytic at x = 0.

44. Ch5_44 Thus the power series expansions p(x) = xP(x) = a 0 ＋ a 1 x ＋ a 2 x 2 ＋ … q(x) = x 2 Q(x) = b 0 ＋ b 1 x ＋ b 2 x 2 ＋ …(12) are valid on intervals that have a positive radius of convergence. By multiplying (2) by x 2, we have (13) After some substitutions, we find the indicial equation, r(r – 1) + a 0 r + b 0 = 0(14)

45. Ch5_45 Example 3 Solve Solution Let, then

46. Ch5_46 Example 3 (2) which implies r(2r – 1) = 0(15) (16)

47. Ch5_47 Example 3 (3) From (15), we have r 1 = ½, r 2 = 0. Foe r 1 = ½, we divide by k + 3/2 in (16) to obtain (17) Foe r 2 = 0, (16) becomes (18)

48. Ch5_48 Example 3 (4) From (17)From (18)

49. Ch5_49 Example 3 (5) Thus for r 1 = ½ for r 2 = 0 and on (0,  ), the solution is y(x) = C 1 y 1 + C 2 y 2.

50. Ch5_50 Example 4 Solve Solution From xP = 0, x 2 Q = x, and the fact 0 and x are their own power series centered at 0, we conclude a 0 = 0, b 0 = 0. Then form (14) we have r(r – 1) = 0, r 1 = 1, r 2 = 0. In other words, there is only a single series solution

51. Ch5_51 Three Cases (1) If r 1, r 2 are distinct and do not differ by an integer, there exists two linearly independent solutions of the form:

52. Ch5_52 (2) If r 1 – r 2 = N, where N is a positive integer, there exists two linearly independent solutions of the form:

53. Ch5_53 (3) If r 1 = r 2, there exists two linearly independent solutions of the form:

54. Ch5_54 Finding a Second Solution  If we already have a known solution y 1, then the second solution can be obtained by

55. Ch5_55 Example 5 Find the general solution of Solution From the known solution in Example 4, we can use (23) to find y 2 (x). Here please use a CAS for the complicated operations.

56. Ch5_56 Example 5 (2)

57. Ch5_57 5.3 Special Functions  Bessel’s Equation of order v (1) where v  0, and x = 0 is a regular singular point of (1). The solutions of (1) are called Bessel functions.  Lengender’s Equation of order n (2) where n is a nonnegative integer, and x = 0 is an ordinary point of (2). The solutions of (2) are called Legender functions.

58. Ch5_58 The Solution of Bessel’s Equation  Because x = 0 is a regular singular point, we know there exists at least one solution of the form. Then from (1), (3)

59. Ch5_59 From (3) we have the indicial equation r 2 – v 2 = 0, r 1 = v, r 2 = −v. When r 1 = v, we have (1 + 2v)c 1 = 0 (k + 2)(k + 2+ 2v)c k+2 + c k = 0 or (4) The choice of c 1 = 0 implies c 3 = c 5 = c 7 = … = 0, so for k = 0, 2, 4, …., letting k + 2 = 2n, n = 1, 2, 3, …, we have (5)

60. Ch5_60 Thus (6)

61. Ch5_61 We choose c 0 to be a specific value where  (1 + v) is the gamma function. See Appendix II. There is an important relation:  (1 +  ) =  (  ) so we can reduce the denominator of (6):

62. Ch5_62 Hence we can write (6) as

63. Ch5_63 Bessel’s Functions of the First Kind  We define J v (x) by (7) and (8) In other words, the general solution of (1) on (0,  ) is y = c 1 J v (x) + c 2 J -v (x), v  integer(9) See Fig 5.3

64. Ch5_64 Fig 5.3

65. Ch5_65 Example 1  Consider the DE We find v = ½, and the general solution on (0,  ) is

66. Ch5_66 Bessel’s Functions of the Second Kind  If v  integer, then (10) and the function J v (x) are linearly independent. Another solution of (1) is y = c 1 J v (x) + c 2 Y v (x).  As v  m, m an integer, (10) has the form 0/0. From L’Hopital’s rule, the function and J v (x) are linearly independent solutions of

67. Ch5_67 Hence for any value of v, the general solution of (1) is (11) Y v (x) is called the Bessel function of the second kind of order v. Fig 5.4 shows y 0 (x) and y 1 (x).

68. Ch5_68 Fig 5.4

69. Ch5_69 Example 2  Consider the DE We find v = 3, and from (11) the general solution on (0,  ) is

70. Ch5_70 DEs Solvable in Terms of Bessel Function  Let t =  x,  > 0, in (12) then by the Chain Rule,

71. Ch5_71  Thus, (12) becomes The solution of the above DE is y = c 1 J v (t) + c 2 Y v (t) Let t =  x, we have y = c 1 J v (  x) + c 2 Y v (  x)(13)

72. Ch5_72  Another equation is called the modified Bessel equation order v, (14) This time we let t = ix, then (14) becomes The solution will be J v (ix) and Y v (ix). A real-valued solution, called the modified Bessel function of the first kind of order v is defined by (15)

73. Ch5_73  Analogous to (10), the modified Bessel function of the second kind of order v  integer is defined by (16) and for any integer v = n, Because I v and K v are linearly independent on (0,  ), the general solution of (14) is (17)

74. Ch5_74  We consider another important DE: (18) The general solution of (18) is (19) We shall not supply the details here.

75. Ch5_75 Example 3 Find the general solution of on (0,  ) Solution Writing the DE as according to (18) 1 – 2a = 3, b 2 c 2 = 9, 2c – 2 = −1, a 2 – p 2 c 2 = 0 then a = −1, c = ½. In addition we take b= 6, p = 2. From (19) the solution is

76. Ch5_76 Example 4  Recall the model in Sec. 3.8 You should verify that by letting we have

77. Ch5_77 Example 4 (2) The solution of the new equation is x = c 1 J 0 (s) + c 2 Y 0 (s), If we resubstitute we get the solution.

78. Ch5_78 Properties  (1)  (2)  (3)  (4)

79. Ch5_79 Example 5 Derive the formula Solution It follows from (7)

80. Ch5_80 Example 5 (2)

81. Ch5_81  The result in example 5 can be written as which is a linear DE in J v (x). Multiplying both sides the integrating factor x -v, then (20) It can be shown (21) When y = 0, it follows from (14) that (22)

82. Ch5_82 Spherical Bessel Functions  When the order v is half an odd number, that is,  1/2,  3/2,  5/2, ….. The Bessel function of the first kind J v (x) can be expressed as spherical Bessel function: Since  (1 +  ) =  (  ) and  (1/2) =  ½, then

83. Ch5_83 Hence and

84. Ch5_84 The Solution of Legender Equation  Since x = 0 is an ordinary point of (2), we use After substitutions and simplifications, we obtain or in the following forms:

85. Ch5_85 Using (25), at least |x| < 1, we obtain

86. Ch5_86 Notices: If n is an even integer, the first series terminates, whereas y 2 is an infinite series. If n is an odd integer, the series y 2 terminates with x n.

87. Ch5_87 Legender Polynomials  The following are nth order Legender polynomials: (27)

88. Ch5_88 They are in turn the solutions of the DEs. See Fig 5.5 (28)

89. Ch5_89 Fig 5.5

90. Ch5_90 Properties  (1)  (2)  (3)  (4)  (5)

91. Ch5_91 Recurrence Relation  Without proof, we have (29) which is valid for k = 1, 2, 3, … Another formula by differentiation to generate Legender polynomials is called the Rodrigues’ formula: (30)