ODE Philippe Rukimbira Department of Mathematics Florida International University PR (FIU) MAP 2302 1 / 92
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 2302 2 / 92
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 2302 3 / 92
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 2302 4 / 92
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 2302 5 / 92
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 2302 6 / 92
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 2302 7 / 92
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 2302 8 / 92
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 2302 9 / 92
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 2302 10 / 92
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 1 +... + u n y n is a particular solution of the nonhomogeneous equation. PR (FIU) MAP 2302 11 / 92
For that, one solves the following linear system: y 1 u 1 +... + y nu n = 0 y 1 u 1 +... + y nu n = 0. =. y (n 1) 1 u 1 +... + 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 2302 12 / 92
Example y + y = tan x y + y = sec x PR (FIU) MAP 2302 13 / 92
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 2302 14 / 92
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 2302 15 / 92
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 2302 16 / 92
5.1-5.4: 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 2302 17 / 92
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 2302 18 / 92
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 2302 19 / 92
Solving t 0 = π 2 Φ λ is the phase shift. The motion can be graphed now! λt + Φ = π 2 PR (FIU) MAP 2302 20 / 92
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 2302 21 / 92
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 2302 22 / 92
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 2302 23 / 92
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 2302 24 / 92
b 2 λ 2 = 0: Critically damped motion. x(t) = e bt (C 1 + C 2 t). PR (FIU) MAP 2302 25 / 92
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 2302 25 / 92
Example #2, page 208. PR (FIU) MAP 2302 26 / 92
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 2302 27 / 92
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 2302 28 / 92
We find that: ω 2 C + 2bωD + λ 2 C = f ω 2 D 2bωC + λ 2 D =0 PR (FIU) MAP 2302 29 / 92
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 2302 29 / 92
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 2302 30 / 92
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 2302 31 / 92
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 2302 32 / 92
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 2302 33 / 92
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 2302 34 / 92
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 2302 35 / 92
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 2302 36 / 92
Assignments: page 189, #10-12, Page 217 (Ross), # 4-7, Page 224, #1-3 PR (FIU) MAP 2302 37 / 92
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 2302 38 / 92
4 = 1 8K so K = 32 x(0) = 1 6 ẋ(0) = 0 4 = m.32 so, m = 1 8 1 + 32x = 2 cos 3t 8ẍ PR (FIU) MAP 2302 39 / 92
ẍ + 256x = 16 cos 3t x(0) = 0 ẋ(0) = 0 Solving the above initial value problem leads to x(t) = 151 16 cos 16t + cos 3t. 1482 247 PR (FIU) MAP 2302 40 / 92
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 2302 41 / 92
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 2302 42 / 92
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 2302 43 / 92
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 2302 44 / 92
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 2302 45 / 92
(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 2302 46 / 92
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 2302 47 / 92
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 2302 48 / 92
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 2302 49 / 92
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 2302 50 / 92
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 2302 50 / 92
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 2302 51 / 92
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 = 1 5.4.3 Claim: for n 2, c n = 2 n!. PR (FIU) MAP 2302 52 / 92
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 2302 53 / 92
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 2302 54 / 92
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 5 10 + x 6 20 + x 7 140 +... PR (FIU) MAP 2302 55 / 92
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 2302 56 / 92
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 2302 57 / 92
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 2302 58 / 92
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 2302 59 / 92
Example Find two linearly independent solutions for y xy + 2y = 0 PR (FIU) MAP 2302 60 / 92
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 2302 60 / 92
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 2302 61 / 92
We deduce that c n+2 = c 2 = c 0 (n 2)c n (n + 2)(n + 1) PR (FIU) MAP 2302 62 / 92
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 2302 62 / 92
c 1 is also arbitrary. c 3 = c 1 3.2 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 3 +.. = c 1 (x 1 6 x 3 +...) PR (FIU) MAP 2302 63 / 92
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 2302 64 / 92
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 2302 65 / 92
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 2302 66 / 92
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 2302 67 / 92
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 2302 68 / 92
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 2302 69 / 92
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 2302 70 / 92
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 2302 71 / 92
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 2302 72 / 92
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 2302 73 / 92
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 2302 74 / 92
(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 2302 75 / 92
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 2302 76 / 92
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 2302 77 / 92
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 2302 78 / 92
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 2302 79 / 92
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 2302 80 / 92
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 2 4 + x 4 64 x 6 2304 +... 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 2302 81 / 92
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 2302 82 / 92
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 2302 83 / 92
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 2302 84 / 92
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 2 0 2. x φ 2 = 1 + t(1 + t2 0 2 )dt = 1 + x 2 2 + ( x 2 2 φ 3 = 1 + u + u2 2 + u3 3! 2 )2 PR (FIU) MAP 2302 85 / 92
. 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 2302 86 / 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 2302 87 / 92
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 2302 88 / 92
The multi-step methods (using several previous approximations to find y k ) are also known as "continuing methods". PR (FIU) MAP 2302 89 / 92
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 2302 90 / 92
PR (FIU) MAP 2302 91 / 92
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 = 12 10 y 2 = 152 100 y 3 = 1984 1000 y 4 = 26168 10000 PR (FIU) MAP 2302 92 / 92