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Series Solutions of Linear Differential Equations

CHAPTER 5. Series Solutions of Linear Differential Equations. Contents. 5.1 Solutions about Ordinary Points 5.2 Solution about Singular Points 5.3 Special Functions. 5.1 Solutions about Ordinary Point.

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Series Solutions of Linear Differential Equations

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  1. CHAPTER 5 Series Solutions of Linear Differential Equations

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

  3. 5.1 Solutions about Ordinary Point • Review of Power SeriesRecall from that a power series in x – a has the formSuch a series is said to be a power series centered at a.

  4. Convergence exists. • Interval of ConvergenceThe set of all real numbers for which the series converges. • Radius of ConvergenceIf R is the radius of convergence, the power series converges for |x – a| < R and diverges for |x – a| > R.

  5. Absolute ConvergenceWithin its interval of convergence, a power series converges absolutely. That is, the following converges. • Ratio Test Suppose cn 0for all n, and If L < 1,this series converges absolutely, if L > 1,this series diverges, if L = 1,the test is inclusive.

  6. A Power Defines a FunctionSuppose then • Identity PropertyIf all cn= 0,then the series = 0.

  7. Analytic at a PointA 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. Arithmetic of Power SeriesPower series can be combined through the operations of addition, multiplication and division.

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

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

  11. DEFINITION 5.1 A point x0 is said to be an ordinary point of (5) if both P and Q in (6) are analytic at x0. A point that is not an ordinary point is said to be a singular point. A Solution • Suppose the linear DE(5)is put into(6)

  12. Polynomial Coefficients • Since P and Q in (6) is a rational function, P = a1(x)/a2(x), Q = a0(x)/a2(x)It follows that x = x0is an ordinary point of (5) if a2(x0)  0.

  13. THEOREM 5.1 If x = x0is an ordinary point of (5), we can always find two linearly independent solutions in the form of power series centered at x0, that is, Criterion for an Extra Differential • A series solution converges at least of some interval defined by |x – x0| < R, where R is the distance from x0to the nearest singular point.

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

  15. Example 2 (2) From the result given in (4), (8)Since (8) is identically zero, it is necessary all the coefficients are zero, 2c2 = 0,and (9)Now (9) is a recurrence relation, since (k + 1)(k + 2)  0,then from (9) (10)

  16. Example 2 (3) Thus we obtain

  17. Example 2 (4) and so on.

  18. Example 2 (5) Then the power series solutions are y = c0y1 + c1y2

  19. Example 2 (6)

  20. Example 3 Solve SolutionSince x2+ 1 = 0,then x = i, −i are singular points. A power series solution centered at 0 will converge at least for |x| < 1.Usingthe power series form of y, y’ and y”, then

  21. Example 3 (2)

  22. Example 3 (3) From the above, we get 2c2－c0 = 0, 6c3 = 0 , andThus c2 = c0/2, ck+2 = (1 – k)ck/(k + 2)Then

  23. Example 3 (4) and so on.

  24. Example 3 (5) Therefore,

  25. Example 4 If we seek a power series solution y(x)forwe obtain c2 = c0/2 and the recurrence relation isExamination of the formula shows c3, c4, c5, … are expresses in terms of both c1 and c2. However it is more complicated. To simplify it, we can first choose c0  0, c1 = 0.Then we have

  26. Example 4 (2) and so on. Next, we choose c0 = 0, c1 0, then

  27. Example 4 (3) and so on. Thus we have y = c0y1 + c1y2, where

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

  29. Example 5 (2) It follows thatand so on. This gives c2 =－1/2c0, c3 =－1/6c1, c4 = 1/12c0, c5 = 1/30c1,…. By grouping terms we get the general solution y = c0y1 + c1y2, where the convergence is |x| < , and

  30. 5.2 Solutions about Singular Points • A DefinitionA singular point x0 of a linear DE(1)is further classified as either regular or irregular. This classification depends on(2)

  31. DEFINITION 5.2 A singular point x0 is said to be a regular singular point of (1), if p(x) = (x – x0)P(x), q(x) = (x – x0)2Q(x) are both analytic at x0 . A singular point that is not regular is said to be irregular singular point. Regular/Irregular Singular Points

  32. Polynomial Ciefficients • If x – x0appears 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 – x0 is a regular singular point. • If (2) is multiplied by (x – x0)2,(3)where p, q are analytic at x = x0

  33. Example 1 • It should be clear x = 2, x = – 2are singular points of(x2 –4)2y” +3(x –2)y’ + 5y = 0According to (2), we have

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

  35. THEOREM 5.2 If x= x0 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 – x0< R. Frobenius’ Theorem

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

  37. Example 2 (2)

  38. Example 2 (3) which implies r(3r –2)c0 = 0 (k + r + 1)(3k + 3r + 1)ck+1 – ck= 0, k = 0, 1, 2, …Since nothing is gained by taking c0 = 0, then r(3r – 2) = 0 (6)and (7)From (6), r = 0, 2/3,when substituted into (7),

  39. Example 2 (4) r1 = 2/3, k = 0,1,2,… (8) r2 = 0, k = 0,1,2,… (9)

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

  41. Example 2 (6) These two series both contain the same multiple c0. Omitting this term, we have (10) (11)

  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 isy(x) = C1y1(x) + C2y2(x), 0 < x < 

  43. Indicial Equation • Equation (6) is called the indicial equation, where the values of r are called the indicial roots, or exponents. • If x = 0is a regular singular point of (1), then p = xP and q = x2Q are analytic at x = 0.

  44. Thus the power series expansionsp(x) = xP(x) = a0＋a1x＋a2x2＋…q(x) = x2Q(x) = b0＋b1x＋b2x2＋… (12)are valid on intervals that have a positive radius of convergence. By multiplying (2) by x2, we have (13)After some substitutions, we find the indicial equation, r(r – 1) + a0r + b0 = 0(14)

  45. Example 3 Solve SolutionLet , then

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

  47. Example 3 (3) From (15), we have r1 = ½ , r2 = 0.Foe r1 = ½ ,we divide by k + 3/2in (16) to obtain (17) Foe r2 = 0 ,(16) becomes (18)

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

  49. Example 3 (5) Thus for r1 = ½ for r2 = 0 and on (0, ), the solution is y(x) = C1y1+ C2y2.

  50. Example 4 Solve SolutionFrom xP = 0, x2Q = x, and the fact 0 and x are their own power series centered at 0, we conclude a0 = 0, b0 = 0.Then form (14) we have r(r – 1) = 0, r1= 1,r2 = 0.In other words, there is only a single series solution

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