1. 1 Chapter 3 Linear Differential Equations of Second Order and Higher 3.1 Introduction (a) Linear equations Second order and higher 1.Constant coefficients and homogeneous (f=o)(Sec.3.4) 2.Constant coefficients and non-homogeneous(Sec.3.7) 3.Nonconstant coefficients (Sec. 3.6, Chapter 4) First order: (b) Nonlinear equations First order: : Second order and higher:

2. 2 3.2 Linear dependence and linear independence Definition: A set of functions {u 1,u 2,…,u n } is said to linear dependence on an interval I if at least one of them can be expressed as a linear combination of the others on I. If none can be so expressed, then the set is linear independence. Example 1 Theorem 3.2.1 Test for linear dependence/independence A fine set of functions {u 1,u 2,…,u n } is LD on an interval I if and only if there exist scalars  j, not all zero, such that identically on I, If (2) is true only if all the  ’s are zero, then the set is LI on I. (2)

3. 3 How to determine a set of functions is LD or LI ? For a set of function Define Wronskian determinant If there is any value of x in I, say x 0, such that Then all the  ’s must be zero, so the set {u 1,u 2,…,u n } is LI. Example 2 Show whether the given set is LD or LI

4. 4 Theorem 3.2.3 A necessary and sufficient condition for linear dependence If u 1,u 2,…,u n are solutions of an nth-order linear homogeneous differential equation Where the coefficients p j (x) are continuous on an interval I, then W[ u 1,…,u n ](x)=0 on I is both necessary and sufficient for the linear dependence of the set { u 1,…,u n } on I. Example 3Use Theorem 3.2.3 to determine if the set is LD or LI. (10)

5. 5 Theorem 3.3.1 3.3 Homogeneous equation: General solution General solution: A family of solutions that contains every solution of Eq. (4) on that interval. Particular solution: One member of that family of solutions. (4) Considering Existence and uniqueness for initial-value problem If p 1 (x), p 2 (x),…,p n (x) are continuous on a closed interval I, then the initial-value problem consisting of the differential equation (4) together with the initial conditions y(a)=b 1, y’(a)=b 2, …., y (n-1) (a) =b n where the initial point a is in I, has a solution on I, and that solution is unique.

6. 6 Considering the nth-order linear homogeneous equation Which can be re-writing in the compact form Where is called an nth-order differential operator The differential operator is valid for an equation combined any finite number of functions, say u 1,…,u n, that is (4)

7. 7 Theorem 3.3.2 Superposition of solutions of Eq.(4) If y 1,…,y k, are solutions of Eq. (4), then C 1 y 1 +…+C k y k is too, for any constants C 1,…,C k. Prove Eq. (9) at p.84 Example 2 Assume L is a second order operator, that is L=

8. 8 Theorem 3.3.3 General solutions of Eq.(4) If p 1 (x),.…,P n (x) be continuous on an open interval I. Then the nth-order linear homogeneous differential Eq. (4) admits exactly n LI solutions; that is, at least n and no more than n. If y 1 (x),..., y n (x) is such a set of LI solutions on I, then the general solution of Eq. (4) on I is Example 4 Find the solution in series form using Taylor series method Taylor series: Any such set of n LI solutions is called a basis, or fundamental set, of solution of the differential equation.

9. 9 3.4 Solution of Homogeneous equation: Constant coefficient Euler’s formula

10. 10 Define

11. 11 3.4.2 Exponential solutions Question: If higher-order equations admit exponential solutions too? Could we find a solution in the form ? If the equation has a solution of (19), then it must be true that (18) (19) and to be such that (22) Eq. (22) is called characteristic equation. (find , find the solution)

12. 12 Case 1 Distinct real root 1, 2 if (a 1 2 -4a 2 )>0 Example 1

13. 13 Case 2 Repeated root, if (a 1 2 -4a 2 )=0 Supplement

14. 14 Example 2 Integrating, we have g(x) =a 0 +a 1 x+a 2 x 2 +….+a n-1 x n-1 According to previous inference, we have

15. 15 Case 3 if (a 1 2 -4a 2 )<0 Example 3

16. 16 Example 4

17. 17 3.5 Application to Harmonic Oscillator: Free Oscillation For the case where f(t)=0, which is called free oscillation. The corresponding characteristic equation is with roots of (3)

18. 18 Case 1 : No damping, c=0 Eq. (3) becomes and Eq.(4) gives and the solution of Eq.(5) is (5) where  =(k/m) 1/2 is the so-called natural frequency, in rad/sec of the system (6) Eq. (6) can be re-write as (7) (8) C and D have no special physical significance.

19. 19 Solution: For the Eq. (7) subjected to the initial conditions increases with k, decreases with m, and is independent of the initial condition x 0 and x o ’, and hence the amplitude is (7)

20. 20 Case 2 Underdamped vibration (c < c cr ) causes the oscillation to damp out as t  ∞

21. 21 Case 3 Critically damped vibration (c = c cr ) As c increased further, the system becomes so sluggish that when c attains the critical value c cr the oscillation ceases altogether. In this case Eq. (4) gives the repeated root and http://www.youtube.com/watch?v=_q7ojtFWDBU&feature=player_detailpage 俄伏爾加河大橋

22. 22 Case 4 Overdamped vibration (c > c cr ) As c increase beyond c cr, Eq. (4) gives two distinct roots, but they are both real and negative, so If one or both roots were positive then we could have exponential growth, which would make no sense, physically.

23. 23 3.6 Solution of Homogeneous Equation: Nonconstant Coefficients 3.6.1 Cauchy-Euler equation where c j ’s are constant, it is called Cauchy-Euler equation, and is called an equidimensional equation. when n=2, the Eq. is reduced to (3) Try to solve (3) by seeking y in the form where is a yet-to-be determined constant, and Eq. (3) becomes (4) Let us consider the x interval to be 0<x<∞

24. 24 Try to solve (3) by seeking y in the form and Eq. (3) becomes Define the discriminate Case 1 △ > 0 Two distinct real roots, 1 and 2 Example 1

25. 25 Case 2 △ = 0 Repeated real roots We have the solution Ax 1, the missing solution is needed to be found from the known solution. Let Example 2

26. 26 Example 3 Case 3 △ < 0 Complex roots (16)

27. 27 Which is the same as Eq. (3). The function lnx and its derivatives are not defined at x=0, nor for x<0. To deal with the case where x < o, it is more convenient to make the change of variable x=-ξ in Eq. (3), so that ξwill be positive. Letting y(x)=y(- ξ )=Y( ξ ) (18) and Eq. (3) becomes (19)

28. 28 Theorem 3.6.1 For the case of distinct real roots, for instance Both of these forms are accounted for by the single expression Second-Order Cauchy-Euler Equation Example 4

29. 29 3.6.2 Reduction of order Consider a second-order Eq. If one solution, Y(x), is known, the idea behind Lagrange’s method of reduction of order is to seek the missing solution in the form where A(x) is to be determined. (29) (28) Substituting Eq.(29) into Eq.(28), we have (30) (31)

30. 30 Letting A’=p, Eq. (30) can be reduced to (32) Integrating Eq. (30), we have (33)

31. 31 Example 5 Legendre’s Equation

32. 32 3.6.3 Factoring the operator For the nth-order linear homogeneous equation where (36) Considering n=2, Eq. (36) can be reduced to

33. 33 Example 6

34. 34 Example 7

35. 35 Example 8

36. 36 3.7 Solution of Nonhomogeneous Equation Theorem 3.7.1 General Solution of If y h (x) and y p (x) are homogeneous and particular solutions of the L[y]=f, respectively, on an interval I, then a general solution of the differential Eq. on I is Theorem 3.7.2 General Solution of If y h (x) is a general solution of L[y]=0, on an interval I, and y p1 (x), …, y pk (x) are particular solutions of L[y]=f 1, …., L[y]=f k on I, respectively, then a general solution of L[y]=f 1 +…+f k on I is (8) (9) This result is a superposition principle. 3.7.1 General Solution

37. 37 3.7.2 Undetermined coefficients f(x) y p (x)

38. 38 Example 1

39. 39 Example 2

40. 40 Example 3

41. 41 Example 4 Example 5

42. 42 3.7.3 Variation of parameters For the general linear first-order equation For an nth-order linear differential equation L[y]=f has a homogeneous solution The homogeneous solution can be easily obtained, and the particular solution can be derived from the homogeneous solution by multiplying an undetermined function A(x) According to the method of variation of parameters we seek a particular solution in the form (38) (39)

43. 43 Let us carry out the procedure for the linear second-order equation with the known general homogeneous Assume the particular solution is in the form of and the differentiation of Eq.(42) is (40) (41) (42) Let The Eq. (43) can be reduced and its differentiation is (43) (44)

44. 44 and Eq. (40) becomes The two parenthetic groups vanish by virtue of y 1 and y 2 being solutions of the homogeneous equation L[y]=0, so Eq.(46) simplifies to The result, together with Eq. (44), gives the Eqs. of (47) The Wronskian determinant of y 1 and y 2 is (48) (45) (46)

45. 45 Solving Eq. (47) by Cramer’s rule gives Therefore, the particular solution of Eq. (40) can be expressed as (49)

46. 46 3.7.4 variation of parameters for higher-order equations Taking the same procedure as the previous section, we have (68) (58) (57) (56)

47. 47 Let and

48. 48 Example 5

49. 49 3.8 Application to harmonic Oscillator: Forced Oscillation 3.8.1 Undamped case (c=0) (1) (2) (3) (4) (5)

50. 50 Nonresonant oscillation Putting Eq. (5) into the left side of Eq. (3) gives and Equating the coefficients of cos Ω t and sin Ω t, we have (7) Substituting the results into Eq. (3), we have

51. 51 (9) (8) It is natural to regard m and k (and hence w) as fixed, and F o and  as controllable parameters. From Eq. (7), it is observed that the response is at the same frequency as the forcing function, . More interesting is the amplitude varied with , which is sketched in Fig. 2. Eq. (7) can be re- expressed in the equivalent form. (10)

52. 52 where the phase angle Φ is 0 for  and  for  > . The resulting amplitude- and phase-response curves are shown in Fig. 3. From Fig. 3a, observe that as the driving frequency approaches the natural frequency the amplitude tends to infinity. Further, as Ω→∞ the amplitude tends to zero. Fig. 3b shows that the response is in-phase (Φ=0) with the forcing function for Ω w it is 180 0 out-of-phase. This discontinuous jump is striking since only an infinitesimal change in produces a discontinuous change in the response. Fig. 3 Amplitude- and phase-response curves

53. 53 Resonant oscillation For the special case where Ω= , the x p(t) in Eq. (4) can be revised as Putting that form into Eq. (2), we find that C=0 and D=F 0 /(2m  ), x p can be expressed by Eq. (11) and plotted in Fig. 4. (11) (12) In the special case the response is not a harmonic oscillation but a harmonic function times t, which factor causes the magnitude to tend to infinity as t →∞.

54. 54 Beats What is the solution x(t) as Ω approaches  ? To investigate the phenomenon, let’s use the simple initial conditions x(0)=0 and x’(0)=0. Eq. (8) can be expressed as (13) Eq. (13) can also be expressed as (14) Suppose that Ω is close to (but not equal to) the natural frequency ω. Then the frequency of the second sinusoid in Eq. (14) is very small compared to that of the first, so the sin[(ω-Ω)t/2] factor amounts, essentially, to a slow “amplitude Modulation” of the relatively high frequency sin[(ω+Ω)t/2] Factor The phenomenon is known as beats.

55. 55 Fig. 5 Beats, and approach to resonance

56. 56 3.8.2 Damped case For the harmonically driven oscillator, this time with a cx’ damping term included (c>0) The homogeneous solution is

57. 57 Figure 6 Amplitude response curves or

58. 58 Problems for Chapter 3 Exercise 3.2 3.(e) 、 (j) 4.(c) 、 (g) 、 (h) 5.(a) Exercise 3.3 2.(a) 、 (b) 、 (d) 8.(a) 、 (d) 、 (f) 9.(a) 、 (b) 、 (d) 11.(a) 、 (b) 、 (c) Exercise 3.4 1. (d) 4. (a) 、 (d) 、 (h) 、 (j) 6. (b) 、 (d) 、 (h) 、 (k) 9. (a) 、 (b) 、 (c) 11. (a) 12. (b) 、 (d) 、 (h) 、 (k) Exercise 3.6 1.(b) 、 (d) 、 (h) 、 (m) 7. (a) 、 (e) 、 (f) 8. (a) 10. (b) 11. (a) 、 (b) 15. 17. (a) 、 (b) Exercise 3.7 2.(b) 、 (c) 、 (g) 、 (h) 、 (m) 4. (d) 、 (h) 、 (m) Exercise 3.8 12.(a) 、 (b) 、 (d) 、 (g) 13.(e)