ODE. Philippe Rukimbira. Department of Mathematics Florida International University PR (FIU) MAP / 92

Transcription

1 ODE Philippe Rukimbira Department of Mathematics Florida International University PR (FIU) MAP / 92

2 4.4 The method of Variation of parameters 1. Second order differential equations (Normalized, standard form!). y + P(x)y + Q(x)y = f (x) Suppose y 1 and y 2 form a fundamental set of solutions on an interval I for y + P(x)y + Q(x)y = 0 We seek functions u 1 (x) and u 2 (x) such that: y p = u 1 y 1 + u 2 y 2 is a particular solution of the nonhomogeneous differential equation. PR (FIU) MAP / 92

3 In that case, y p = u 1 y 1 + y 1u 1 + u 2y 2 + y 2u 2 We can impose the additional condition on u 1 and u 2 : y 1 u 1 + y 2u 2 = 0 That is equivalent to y p = u 1 y 1 + u 2y 2 PR (FIU) MAP / 92

4 From there, y p = u 1 y 1 + u 1y 1 + u 2 y 2 + u 2y 2 Now, substitute for y p, y p and y p into the nonhomogeneous differential equation" which becomes: y p + Py p + Qy p = f (x) u 1 y 1 + u 1y 1 + u 2 y 2 + u 2y 2 + Pu 1y 1 + Pu 2y 2 + Qu 1y 1 + Qu 2 y 2 = f (x) PR (FIU) MAP / 92

5 Reorganizing leads to u 1 y 1 + u 1(y 1 + Py 1 + Qy 1) + u 2 y 2 + u 2(y 1 + Py 2 + Qy 2) = f (x). and finally, one obtains a second condition on u 1 and u 2 y 1 u 1 + y 2 u 2 = f (x) PR (FIU) MAP / 92

6 What we have now is a system of two equations involving (the derivatives of ) u 1 and u 2 y 1 u 1 + y 2u 2 = 0 y 1 u 1 + y 2 u 2 = f (x) PR (FIU) MAP / 92

7 Observe that the determinant of the linear system in no other than the Wronskian W (y 1, y 2 ) 0 by assumption. Hence, the system has a unique solution (u 1, u 2 ). ( ) 0 y det 2 u 1 = f (x) y 2 W (y 1, y 2 ) = y 2f (x) y 1 y 2 y 1 y 2 u 2 = f (x)y 1 y 1 y 2 y 1 y 2 From u 1 and u 2, we obtain u 1 and u 2 by integration. PR (FIU) MAP / 92

8 Example y 3y + 2y = e3x 1 + e x Notice that undetermined coefficient methods does not work in this example! Using the characteristic equation technique for instance, we find that the general solution to the associated homogeneous equation is y c = C 1 e x + C 2 e 2x. PR (FIU) MAP / 92

9 The method of variation of parameters tells us that we can find a particular solution y p = u 1 e x + u 2 e 2x by solving e x u 1 + e2x u 2 = 0 e x u 1 + 2e2x u 2 = e 3x 1 + e x PR (FIU) MAP / 92

10 e2x u 1 = 1 + e x u 2 = ex 1 + e x u 1 = (e x + 1) + ln (e x + 1) u 2 = ln (e x + 1) A particular solution is y p = ( (e x + 1) + ln (e x + 1))e x + ln (e x + 1)e 2x and, finally, the general solution is y = C 1 e x + C 2 e 2x + e 2x (ln (e x + 1)) + e x (ln (e x + 1) (e x + 1)). PR (FIU) MAP / 92

11 Variation of parameters for higher order equations y (n) + P 1 y (n 1) P n y = f (x). Let y 1,..., y n be n linearly independent solutions of y (n) + P 1 y (n 1) P n y = 0 Look for u 1,..., u n such that u 1 y u n y n is a particular solution of the nonhomogeneous equation. PR (FIU) MAP / 92

12 For that, one solves the following linear system: y 1 u y nu n = 0 y 1 u y nu n = 0. =. y (n 1) 1 u y (n 1) u n = f (x) u i = W i(y 1,..., y n ) W (y 1,..., y n ) where W i (y 1,..., y n ) is the Wronskian in which the column i has been replaced by the column (0,...0, f (x)). PR (FIU) MAP / 92

13 Example y + y = tan x y + y = sec x PR (FIU) MAP / 92

14 4.5 The Cauchy-Euler equation Standard form, n th order: Second order example: a n x n d n y dx n a 1x dy dx + a 0y = g(x). x 2 y 2xy + 2y = x 3 ln x Observation: The substitution t = ln x reduces Cauchy-Euler equation to an equation with constant coefficients! PR (FIU) MAP / 92

15 From dy dx = dy dt dt dx = 1 dy x dt and d 2 y dx 2 = 1 dy x 2 dt + 1 d 2 y dt x dt 2 dx = 1 x 2 ( dy dt + d 2 y dt 2 ) Substitution leads to d 2 y dt 2 3dy dt + 2y = te3t. This can be solved using the method of undetermined coefficients or variation of parameters! PR (FIU) MAP / 92

16 y(t) = C 1 e 2t + C 2 e t + ( 1 2 t 3 4 )e3t y(x) = C 1 x 2 + C 2 x + ( 1 2 ln x 3 4 )x 3 PR (FIU) MAP / 92

17 : Applications: Spring vibrations, electric circuits problems Second order differential equations with constant coefficients Consider a spring with natural length L. Suspend a mass with weight F g = mg. The spring is stretched by l. Choose an orientation vector i downward. Hooke s Law mg = Kl where K is the constant of the spring. PR (FIU) MAP / 92

18 Disturb the mass m with initial position x 0 and velocity v 0. If x denotes the displacement from equilibrium position, then mẍ i = Kx i mẍ = Kx ẍ + λ 2 x = 0 where λ 2 = K m. The mass executes a free, undamped motion. x(t) = C 1 cos λt + C 2 sin λt. PR (FIU) MAP / 92

19 The simple, harmonic motion. K λ = m is the angular velocity, x(t) = A cos (λt + Φ) T = 2π λ is the natural period of the motion f = λ 2π = 1 T is the natural frequency. A = C1 2 + C2 2 is the amplitude and Φ from tan Φ = C 2 C 1 is the phase angle. PR (FIU) MAP / 92

20 Solving t 0 = π 2 Φ λ is the phase shift. The motion can be graphed now! λt + Φ = π 2 PR (FIU) MAP / 92

21 Example #1, page 197. Example: A 16 lb weight is placed upon the lower end of a coil spring suspended vertically from a fixed support. The weight comes to rest in its equilibrium position, thereby stretching the spring 6 in. Determine the resulting displacement as a function of time in each of the following cases: a) If the weight is then pulled down 4 in. below its equilibrium position and released at t=0 with initial velocity of 2 ft/sec directed downward. b) If the weight is then pulled down 4 in. below its equilibrium position and released at t=0 with an initial velocity of 2 ft/sec directed upward. c) If the weight is then pushed up 4 in. above its equilibrium position and released at t=0 with an initial velocity of 2 ft/sec. directed downward. PR (FIU) MAP / 92

22 16 = K 6 12 = K 2, so K = 32. ẍ + K 16 x = 0, M = M 32 = 1 2 slugs Finish the example! ẍ + 64x = 0 PR (FIU) MAP / 92

23 Damped motion Mẍ = Kx βẋ; F R = βẋ ẍ + λ 2 x + 2bẋ = 0 where 2b = β M. ẍ + 2bẋ + λ 2 x = 0 Free, damped motion The motion is not necessarily periodic anymore. PR (FIU) MAP / 92

24 m 2 + 2bm + λ 2 = 0 b 2 λ 2 > 0: Over-damped motion. m = b ± b 2 λ 2. b x(t) = e bt [C 1 e 2 λ 2t + C 2 e b 2 λ 2t ]. Observe the damping factor e bt! PR (FIU) MAP / 92

25 b 2 λ 2 = 0: Critically damped motion. x(t) = e bt (C 1 + C 2 t). PR (FIU) MAP / 92

26 b 2 λ 2 = 0: Critically damped motion. x(t) = e bt (C 1 + C 2 t). b 2 λ 2 < 0: Under-damped motion. x(t) = e bt [C 1 cos λ 2 b 2 t + C 2 sin λ 2 b 2 t]. Observe a periodic factor and a damping factor! PR (FIU) MAP / 92

27 Example #2, page 208. PR (FIU) MAP / 92

28 Forced motion Mẍ + βẋ + Kx = F cos ωt ẍ + 2bẋ + λ 2 x = f cos ωt where f = F M, 2b = β M, λ2 = K M. PR (FIU) MAP / 92

29 Assume we are in the under-damped situation, so that x c (t) = Ae bt cos ( λ 2 b 2 t + φ). We look for a particular solution using the undetermined coefficients method: x p (t) = C cos ωt + D sin ωt PR (FIU) MAP / 92

30 We find that: ω 2 C + 2bωD + λ 2 C = f ω 2 D 2bωC + λ 2 D =0 PR (FIU) MAP / 92

31 We find that: ω 2 C + 2bωD + λ 2 C = f ω 2 D 2bωC + λ 2 D =0 or equivalently (λ 2 ω 2 )C + 2λωD = f 2bωC + (λ 2 ω)d =0 PR (FIU) MAP / 92

32 By Cramer s method for instance, f (λ 2 ω 2 ) C = (λ 2 ω 2 ) 2 + 4b 2 ω 2 2bωf D = (λ 2 ω 2 ) 2 + 4b 2 ω 2 PR (FIU) MAP / 92

33 x p = f (λ 2 ω 2 ) (λ 2 ω 2 ) 2 + 4b 2 ω cos ωt + 2bωf 2 (λ 2 ω 2 ) 2 sin ωt + 4b 2 ω2 We will express x p as x p = A cos (ωt + φ) PR (FIU) MAP / 92

34 f (λ 2 ω 2 ) A cos φ = (λ 2 ω 2 ) 2 + 4b 2 ω 2 2bωf A sin φ = (λ 2 ω 2 ) 2 + 4b 2 ω 2 So and finally A = f [(λ 2 ω 2 ) 2 + 4b 2 ω 2 ] 1/2 x p (t) = f [(λ 2 ω 2 ) 2 + 4b 2 ω 2 cos (ωt + φ) ] 1/2 PR (FIU) MAP / 92

35 da dω = 2(λ2 ω 2 )ω 4b 2 ω [(λ 2 ω 2 ) 2 + 4b 2 ω 2 ] 3/2 f The critical ω s are ω = 0 and ω 2 = λ 2 2b 2. x p (t) achieves the maximum amplitude when ω = ω R = K λ 2 2b 2 = M β2 2M 2 PR (FIU) MAP / 92

36 ω R is called the resonance frequency. Observe that the resonance frequency ω R λ 2 b 2 = smaller than the frequency of the free motion! The graph of y = A(ω) is called the resonance curve! K M β2 4M 2 is PR (FIU) MAP / 92

37 x(t) = x c + x p The function x c (t) eventually becomes negligible as time goes on, this is the transient state. x p (t) which remains forever, is called the steady state solution. PR (FIU) MAP / 92

38 In the undamped situation, b = 0, x p (t) = f λ 2 cos (ωt + φ) ω2 We can see that as ω approaches the natural frequency λ, the amplitude of the steady state blows up! This phenomenon is known as Pure resonance. It always has destructive effects on the systems. PR (FIU) MAP / 92

39 Assignments: page 189, #10-12, Page 217 (Ross), # 4-7, Page 224, #1-3 PR (FIU) MAP / 92

40 Example: Forced motion A mass weighting 4 lb stretches a spring 1.5 in. The mass is displaced 2 in. in the positive direction and released with zero initial velocity. assuming that there is no damping, and that the mass is acted upon by an external force of 2 cos 3t lb, formulate the initial value problem describing the motion of the mass and solve it. PR (FIU) MAP / 92

41 4 = 1 8K so K = 32 x(0) = 1 6 ẋ(0) = 0 4 = m.32 so, m = x = 2 cos 3t 8ẍ PR (FIU) MAP / 92

42 ẍ + 256x = 16 cos 3t x(0) = 0 ẋ(0) = 0 Solving the above initial value problem leads to x(t) = cos 16t + cos 3t PR (FIU) MAP / 92

43 Analogous systems Mass on a spring: mẍ + βẋ + Kx = f (t) Inductor-Resistor-Capacitor (L-R-C) series circuit: L q + R q + 1 C q = E(t) (E(t) is the electric potential) PR (FIU) MAP / 92

44 Put E(t) = E 0 sin ωt Write the circuit equation as LI + RI + 1 C I = ωe 0 cos ωt Solving: I(t) = I tr + I sp where lim I tr = 0 t + I sp = Steady periodic solution PR (FIU) MAP / 92

45 I sp = E 0 cos (ωt α) R 2 + (ωl 1 ωc )2 α = tan 1 Z = ωrc 1 LCω 2, 0 α π R 2 + (ωl 1 ωc )2 (in Ohms) is called the impedance of the circuit. The amplitude of the signal is I 0 = E 0 Z PR (FIU) MAP / 92

46 Electrical resonance I 0 = E 0 Z = E 0 R 2 +(ωl 1 ωc )2 attains its maximum when ω = 1 LC This is the resonance frequency!. Tuning a radio receiver consists essentially in modifying the value of C so as to match the frequency of the incoming radio signal! PR (FIU) MAP / 92

47 6. Series solutions Examples Solve dy dx 2xy = 0 We look for a solution of the form y = c n x n n=0 Then and dy dx = nc n x n 1 n=0 n=0 dy dx 2xy = nc n x n 1 2 c n x n+1 = 0 n=0 PR (FIU) MAP / 92

48 (m + 1)c m+1 x m 2 c m 1 x m = 0 m=0 m=1 c 1 + [(m + 1)c m+1 2c m 1 ]x m = 0 m=1 PR (FIU) MAP / 92

49 c 1 = 0 and (m + 1)c m+1 2c m 1 = 0 which implies that for m 2. c 1 = 0 and c m+1 = 2 c m 1 m + 1 We will use the recurrence formula for c m to generate all coefficients in the power series. PR (FIU) MAP / 92

50 c 0 c 1 = 0 arbitrary c 2 = 2 c 0 2 = c o c 3 = 2 c 1 3 = 0 c 4 = 2 c 2 4 = c 0 2 c 5 = 0 PR (FIU) MAP / 92

51 In general, c 2n+1 = 0 for n = 0, 1, 2... and c 2n = 2 c 2(n 1) 2n = 2. =. = c 0 n! c 2(n 2) (n 2)(n 1) = c 2(n 2) n(n 1) PR (FIU) MAP / 92

52 y = c 0 (1 + m=1 ) ( x 2m = c 0 m! m=0 y = c 0 e x 2. (x 2 ) m ) m! PR (FIU) MAP / 92

53 y = c 0 (1 + m=1 ) ( x 2m = c 0 m! m=0 y = c 0 e x 2. (x 2 ) m ) Check this solution using the separable equation technique! m! PR (FIU) MAP / 92

54 Example 2 (x 1)y xy + y = 0, y(0) = 2, y (0) = 6 y = n 0 c n x n y = n 0 nc n x n 1 y = n 0 n(n 1)c n x n 2 0 = (x 1)y xy + y = xy y xy + y = [(n + 1)nc n+1 (n + 2)(n + 1)c n+2 + (1 n)c n ]x n n 0 PR (FIU) MAP / 92

55 So c n+2 = (1 n)c n + (n + 1)nc n+1 (n + 2)(n + 1) c 0 = y(0) = 2 c 1 = y (0) = 6 Using the above recurrence formula, we see that c 2 = 1, c 3 = 1 3, c 4 = 1 4.3, c 5 = Claim: for n 2, c n = 2 n!. PR (FIU) MAP / 92

56 proof True for c 2 and c 3. Assume it is true for n 1 and n 2, then c n = (1 n)c n 2 + (n 1)(n 2)c n 1 n(n 1) = 2 2 (1 n) (n 2)! + (n 1)(n 2) (n 1)! n(n 1) = 2 n! PR (FIU) MAP / 92

57 y = 2 + 6x 2 n 2 x n n! = 8x 2 2x 2 n 2 = 8x 2( n 0 y = 8x 2e x x n n! ) x n n! Check by substitution that this is indeed the solution to the initial value problem. PR (FIU) MAP / 92

58 Example 2 (x 1)y (2 x)y + y = 0, y(0) = 2, y (0) = 1 The recurrence relation is: c 0 = y(0) = 2, c 1 = y (0) = 1 c n+2 = c n + (n 2)c n+1 n + 2 y = 2 x + 2x 2 x 3 + x 4 2 x x x PR (FIU) MAP / 92

59 Series solutions around ordinary points y + P(x)y + Q(x)y = 0 The standard form for a second order linear differential equation. Definition A point x 0 is said to be an ordinary point for the above differential equation if P(x) and Q(x) are analytic at x 0 ; that is, both have power series in (x x 0 ) with positive radius of convergence. PR (FIU) MAP / 92

60 A point that is not an ordinary point is said to be a singular point. Note: For power series solutions at an ordinary point, the radius of convergence is at least equal to the distance to the nearest singular point. PR (FIU) MAP / 92

61 Example takes the standard form (x 2 1)y + 4xy + 2 x 2 1 y = 0 y + 4x x 2 1 y 2 + (x 2 1) 2 y = 0 1 and 1 are singular points for this equation. Any other point is a regular point, according to the above definition. PR (FIU) MAP / 92

62 Theorem If x = x 0 is an ordinary point of the differential equation y + P(x)y + Q(x)y = 0, we can always find two linearly independent power series solutions of the form y = c n (x x 0 ) n n=0 PR (FIU) MAP / 92

63 Example Find two linearly independent solutions for y xy + 2y = 0 PR (FIU) MAP / 92

64 Example Find two linearly independent solutions for y xy + 2y = 0 Look for y = c n x n n=0 Then: y = y = nc n x n 1 n=1 n(n 1)c n x n 2 n=2 PR (FIU) MAP / 92

65 Therefore, y xy + 2y = 0 implies that 0 = n=2 n(n 1)c nx n 2 n=1 nc nx n + n=0 2c nx n = n=0 (n + 2)(n + 1)c n+2x n + n=0 2c nx n n=1 nc nx n = 2c 2 + 2c 0 + n=1 [(n + 2)(n + 1)c n+2 + (2 n)c n ]x n PR (FIU) MAP / 92

66 We deduce that c n+2 = c 2 = c 0 (n 2)c n (n + 2)(n + 1) PR (FIU) MAP / 92

67 We deduce that c n+2 = c 2 = c 0 (n 2)c n (n + 2)(n + 1) c 0 : Arbitrary c 2 = c 0 c 4 = 0 c 2n = 0, n > 2 PR (FIU) MAP / 92

68 c 1 is also arbitrary. c 3 = c c 2n+1 = (2n 3)c 2n 1 (2n + 1)(2n), 2n + 1 > 3 y 1 = c 0 c 0 x 2 = c 0 (1 x 2 ) y 2 = c 1 x c 1 6 x = c 1 (x 1 6 x ) PR (FIU) MAP / 92

69 Another example y xy y = 0, x 0 = 1 Find two linearly independent power series solutions. Look for for y = c n (x 1) n n=0 y (x 1)y y y = 0 PR (FIU) MAP / 92

70 6.2. Solutions about singular points: Frobenious Method Definition For a differential equation Py + Qy + Ry = 0 A singular point x 0 is said to be a regular singular point if (x x 0 ) Q P and (x x 0 ) 2 R P are analytic at x 0. Otherwise, the regular point x 0 is said to be irregular singular. PR (FIU) MAP / 92

71 Example: The Euler Equation x 2 y + αxy + βy = 0 x = 0 is a regular singular point. As an alternate method of solution, we look for y = x r Then x r (x r ) αx(x r ) + βx r = 0 = x r (r(r 1) + αr + β) Any solution of F(r) = r(r 1) + αr + β = 0 leads to a solution y = x r of the Euler equation. PR (FIU) MAP / 92

72 Real, Distinct Roots If r 1 and r 1 are the 2, real distinct roots, then y = Ax r 1 + Bx r 2 Example: 2x 2 y + 3xy y = 0, x > 0 PR (FIU) MAP / 92

73 Double Roots F(r) = (r r 1 ) 2 In this case y 1 = x r 1 is a solution and the method of reduction of order shows that y 2 = x r 1 ln x is also a solution. (Check this directly!) Example: x 2 y + 5xy + 4y = 0, x > 0 PR (FIU) MAP / 92

74 Complex-Conjugate Roots r 1 = λ + iµ, r 2 = λ iµ z 1 = x λ+iµ = e (λ+iµ) ln x = x λ (cos (µ ln x) + i sin (µ ln x)) z 2 = x λ (cos (µ ln x) i sin (µ ln x)) PR (FIU) MAP / 92

75 y 1 = 1 2 (z 1 + z 2 ) = x λ cos (µ ln x) and y 2 = 1 2i (z 1 z 2 ) = x λ sin (µ ln x) are two linearly independent real solutions. Example: x 2 y + xy + y = 0 PR (FIU) MAP / 92

76 More Generally Example: 2x 2 y xy + (1 + x)y = 0 0 is a regular singular point. We look for y = x r n=0 a n x n = a n x n+r Substituting into the differential equation and grouping like terms from lowest power of x to higher powers: n=0 F(r)x r+k + n 1[G(r, n)]x n+r+k = 0 PR (FIU) MAP / 92

77 Solve the indicial equation F(r) = 0 and obtain recurrence relations from G(r, n) = 0 If r 1 r 2 is not an integer, we always obtain two linearly independent solutions. PR (FIU) MAP / 92

78 Example 2 x 2 y + (x 2 + 4x)y + (2x + 2)y = 0 y = c n x n+r n=0 y = (n + r)c n x n+r 1, y = (n + r)(n + r 1)c n x n+r 2 PR (FIU) MAP / 92

79 x 2 y = (n + r)c n x n+r+1 = (m 1 + r)c m 1 x m+r n=0 m=1 2xy = 2c n x n+r+1 = 2c m 1 x m+r n=0 m=1 PR (FIU) MAP / 92

80 (n + r)(n + r 1)cn x n+r + (n + r)c n x n+r+1 + 4(n + r)c n x n+r + 2c n x n+r+1 + 2c n x n+r = 0 c 0 (r(r 1) + 4r + 2)x r + m=1 [[(m + r)(m + r + 3) + 2]c m + (m + r + 1)c m 1 ] x m+r = 0 F(r) = r(r + 3) + 2 = 0 = r 2 + 3r + 2 (m + r + 1)c m 1 c m = (m + r)(m + r + 3) + 2 PR (FIU) MAP / 92

81 Work with the smaller root 2 first: r 1 = 2, r 2 = 1 c n = c n 1 n Inductively, we see that c n = ( 1)n c 0 n! This leads to ( 1) n y 1 = c 0 x n 2 = c 0 x 2 ( 1) n x n n! n! n=0 n=0 = c 0 x 2 e x PR (FIU) MAP / 92

82 Form r = 1, we obtain Inductively: c n = c n 1 n + 1 c n = ( 1) n c 0 (n + 1)! Y 2 = c 0 ( 1) n (n + 1)! x n 1 = c 0 x 2 n=0 Check directly that x 2 and x 2 e x are solutions! The general solution is y = Ax 2 e x + Bx 2 ( 1) n+1 (n + 1)! x n+1 = c 0 x 2 (e x 1) PR (FIU) MAP / 92

83 6.3 Bessel s Equation and Bessel s functions Bessel s Equation of order p. x 2 y + xy + (x 2 p 2 )y = 0 Any solution is called a Bessel function of order p. Theses functions occur in connection with problems of Physics and Engineering. PR (FIU) MAP / 92

84 The case p = 0 0 is a regular, singular point. xy + y + xy = 0 Look for y = c n x n+r n=0 r 2 c 0 x r 1 + (1 + r)c 1 x r + [(n + r) 2 c n + c n 2 ]x n+r 1 = 0 n=2 The indicial equation is r 2 = 0 with double root r 1 = 0 = r 2. PR (FIU) MAP / 92

85 Then and (1 + r) 2 c 1 = 0 (n + r) 2 c n + c n 2 = 0, n 2 r = 0 c 1 = 0 r = 0 n 2 c n + c n 2 = 0 c n = c n 2 n 2 PR (FIU) MAP / 92

86 c 2n+1 = 0 c 2n = ( 1)n c 0 (n!) 2 2 2n, n 1 y = c 0 n=0 J 0 (x) = ( 1) n (n!) 2 (x 2 )2n = y 1 (x). n=0 ( 1) n (n!) 2 (x 2 )2n Bessel Function of the first kind of order zero. J 0 (x) = 1 x x 4 64 x A second solution must be of the form (See Theorem 6.3) y = x cnx n + J 0 (x) ln x PR (FIU) n=0 MAP / 92

87 8.3. Successive approximations Picard s Method y = f (x, y), y(x 0 ) = y 0. y = φ(x), the d.e. is just specifying the slope of the tangent line to the graph of the solution! Zeroth approximation: φ 0 = y 0. First approximation: φ 1 (satisfying a different d.e.) φ 1 (x) = f (x, φ 0(x)), φ 1 (x 0 ) = y 0. x φ 1 (x) = y 0 + f (t, φ 0 )dt x 0 PR (FIU) MAP / 92

88 Second approximation: φ 2. (Satisfying a different d.e.) φ 2 (x) = f (x, φ 1), φ 2 (x 0 ) = y 0.. n th approximation: x φ 2 (x) = y 0 + f (t, φ 1 (t))dt x 0 φ n (x) = x x 0 f (t, φ n 1 (t))dt PR (FIU) MAP / 92

89 One has a sequence of functions φ 0, φ 1,..., φ n,... The exact solution is given by: φ = lim n φ n Picard used this method to prove existence of solutions! Observe that φ (x) = lim n φ n(x) = lim n f (x, φ n 1 (x)) = f (x, φ). PR (FIU) MAP / 92

90 Example y = xy, y(0) = 1 Let y = φ(x) where u = x 2 2. φ 0 = 1 x φ 1 = 1 + tdt = 1 + x x φ 2 = 1 + t(1 + t2 0 2 )dt = 1 + x ( x 2 2 φ 3 = 1 + u + u2 2 + u3 3! 2 )2 PR (FIU) MAP / 92

91 . Where u = x 2 2. φ n (x) = n i=1 u i i! φ(x) = lim φ n (x) = e x 2 n 2. PR (FIU) MAP / 92

92 8.4. Numerical Methods: The Euler Method Approximating the solution of y = f (x, y), y(x 0 ) = y 0. Let x 1 = x 0, x 2 = x 0 + h,... x N = x N 1 + h If φ(x) is the exact solution, let φ(x 1 ),..., φ(x N ) be the evaluation of φ at points x k. PR (FIU) MAP / 92

93 A numerical method will use the IVP to estimate φ(x k ), k = 1, 2,..., N. Let y 1, y 2,..., y N be approximations to φ(x 1 ),, φ(x 2 ),..., φ(x N ). A one-step method uses y k 1 to find y k using the differential equation. The method has also an alternate name of starting method". PR (FIU) MAP / 92

94 The multi-step methods (using several previous approximations to find y k ) are also known as "continuing methods". PR (FIU) MAP / 92

95 The Euler Method φ the exact solution of y = f (x, y), y(x 0 ) = y 0. y n+1 = y n + hf (x n, y n ). Geometrically, the segment of then graph of y = φ(x) between (x n, φ(x n )) and (x n+1, φ(x n+1 )) is replaced by the line segment joining (x n, y n ) and (x n+1, y n + hf (x n, y n )). y 0 = φ(x 0 ) PR (FIU) MAP / 92

96 PR (FIU) MAP / 92

97 Example y = 2x + y; y(0) = 1 Use h = 0.2. x 0 = 0 x 1 = 0.2 x 3 = 0.4 x 3 = 0.6 x 4 = 0.8 y 0 = 1 y 1 = y 2 = y 3 = y 4 = PR (FIU) MAP / 92