Series Solutions to Differential Equations-Basics

Transcription

1 Series Solutions to Differential Equations-Basics April 10, 2015 A power series can be used to study differential equations This is only an introduction to a much richer and beautiful area of mathematics With our current tools we begin to see how one can show the existence and uniqueness of functions with specified properties, even if the functions themselves are not easily expressed in terms of familiar functions The Case of: f = kf In modern applications of calculus, one often begins by knowing how a function behaves and then wishes to determine what functions satisfy this behavior A differential equation is one way that this occurs For example, if we know that the instantaneous rate of change of a function is always proportional to the value of the function, we can express this as saying that f (x = kf(x for some value k 0 Suppose that we don t already know that the solution of this is f(x = Ae kx for some A We could start by asking if there is some power series P (x centered at 0 that satisfies this equation That is, P (x = a n x n P (x = kp (x na n x n 1 = k a n x n 1

2 For two power series to be equal, the coefficients must be equal for every power of x n So we must have a 1 = ka 0 2a 2 = ka 1 3a 3 = ka 2 Using this, we can express every value of a n in terms of a 0 a 1 = ka 0 a 2 = 1 2 ka 1 = 1 2 k(ka 0 = k2 2 a 0 a 3 = 1 3 ka 2 = 1 ( k 2 3 k 2 a 0 = k3 3! a 0 a 4 = 1 4 ka 3 = 1 ( k 3 4 k 3! a 0 = k4 4! a 0 = a n = 1 n ka n 1 = 1 ( k n 1 n k (n 1! a 0 = kn n! a 0 Thus, the power series P (x must have the form k n P (x = n! a 0x n (kx n = a 0 n! It is just our luck that the power series (kx n has the nice name e kx and so n! P (x = a 0 e kx Notice that our power series solution was only determined once we picked some value a 0 and this is P (0 = a 0 Of course we might expect this, since understanding the speed of a function is not enough, we also need to know its value at some point (like where it began to understand its value at other points The Case of: f = kf Suppose we now wish to determine which functions satisfy f (x = kf(x For example, if k = 1, then f(x = sin x and f(x = cos x both would satisfy this condition Are there other solutions for k = 1? Again, suppose we seek a power series solution to this problem Then if P (x = a n x n P (x = n(n 1a n x n 2 2 n=2

3 and so we want to know if there is a P (x so that n(n 1a n x n 2 = k a n x n n=2 By equating like powers of x this means we want 2a 2 = ka 0 6a 3 = ka 1 4(3a 4 = ka 2 We cannot express all of these values in terms of a 0 alone, but we can do it using only a 0 and a 1 as follows a 2 = k 2 a 0 a 3 = 1 (32 ka 1 a 4 = 1 (43 ka 2 = k2 4! a 0 a 5 = k (54 a 3 = k2 5! a 1 a 6 = 1 (65 ka 4 = k3 6! a 0 a 7 = k (76 a 5 = k3 7! a 1 = a 2n = = kn (2n! a 0 a 2n+1 = k n (2n + 1! a 1 Thus, ( ( k n x 2n k n x 2n+1 P (x = a 0 + a 1 (2n! (2n + 1! Recalling the power series expansions of sin x and cos x cos x = ( 1 n x2n (2n! sin x = ( 1 n x 2n+1 (2n + 1! we can rewrite our solution when k < 0 as follows Write k = α 2, then ( ( ( α 2 n x 2n ( α 2 n x 2n+1 P (x = a 0 + a 1 (2n! (2n + 1! ( ( = a 0 ( 1 n (αx2n + a 1 ( 1 n (αx2n+1 (2n! α (2n + 1! = a 0 cos(αx + a 1 α sin(αx 3

4 Theorem 01 Every analytic solution of the differential equation for α 0 f (x = α 2 f(x is a linear combination of sin(αx and cos(αx, that is with a 0 = f(0 and a 1 = f (0 f(x = a 0 cos(αx + a 1 α sin(αx What about when k > 0? Then k = β 2 and we are trying to solve f (x = β 2 f(x We can rewrite our solution ( ( (β 2 n x 2n (β 2 n x 2n+1 P (x = a 0 + a 1 (2n! (2n + 1! ( ( (βx 2n = a 0 + a 1 (βx 2n+1 (2n! β (2n + 1! Recalling that the power series expansion of e x is x n and using that (1 + n! ( 1 n is 2 when n is even and 0 when n is odd we see that e x + e x = x n n! + ( xn n! = (1 + ( 1 n xn n! = 2 x 2n (2n! The hyperbolic cosine is defined as cosh x = ex +e x 2 and we see that cosh x = x 2n (2n! Similarly, the hyperbolic sine is defined as sinh x = ex e x 2 and its power series is sinh x = 4 x 2n+1 (2n + 1!

5 and so when k = β 2 0, P (x = a 0 cosh(βx + a 1 β sinh(βx Theorem 02 Every analytic solution of the differential equation for β 0 f (x = β 2 f(x is a linear combination of sinh(βx and cosh(βx, that is with a 0 = f(0 and a 1 = f (0 f(x = a 0 cosh(βx + a 1 β sinh(βx Aside Notice that we have shown that the only analytic solution to f (x = f(x with f(0 = 1 and f (0 = 0 is cos x and the only solution to f (x = f(x with f(0 = 0 and f (0 = 1 is sin x At the same time, the only solution to f (x = f(x with f(0 = 1 and f (0 = 0 is cosh x and the only solution to f (x = f(x with f(0 = 0 and f (0 = 1 is sinh x More General Equations We do not need to insist on just constant terms to solve differential equations with power series, for example Example 01 Determine all solutions which are analytic at 0 such that f (x = xf(x Solution If we look for a power series solution P (x = a nx n then we must have na n x n 1 = x a n x n = a n x n+1 5

6 By equating like powers of x this tells us that a 1 = 0 2a 2 = a 0 3a 3 = a 1 4a 4 = a 2 and so in particular a 2n+1 = 0 and a 2n = 1 2n a 0 and Recall the series of P (x = a 0 + a 0 2 x2 + a 0 4 x4 + a 0 6 x6 + = a 0 + a 0 ln(1 x = for x < 1 which we can use to re-express our solution ( x 2n a 0 +a 0 2n = a ( (x 2 n = a n 2 ln(1 x2 = a 0 (1 + ln 1 x 2 for x < 1 Observation Earlier, when we considered differential equations with constant terms, we obtained solutions which were convergent for all R However, in Example 01, when we considered a more general differential equation our solution had a radius of convergence R = 1 Example 02 Determine all analytic solutions to xf (x = f(x Solution If we let P (x = a nx n be a solution, then x n n x ( a 1 + 2a 2 x + 3a 3 x 2 + = a 0 + a 1 x + a 2 x 2 + and so by equating like power of x this tells us that we must have a 0 = 0, a 1 = a 1 2a 2 = a 2 3a 3 = a 3 x 2n 2n and so a 1 is arbitrary but all other a n = 0 and so f(x = a 1 x 6

7 Example 03 Determine all series solution centered at 0 such that f (x = xf(x Solution If we look for a power series solution P (x = a nx n then we must have n(n 1a n x n 2 = x a n x n = a n x n+1 n=2 By equating like powers of x this tells us that 2(1a 2 = 0 3(2a 3 = 0 4(3a 4 = a 0 5(4a 5 = a 1 6(5a 6 = a 2 7(6a 7 = a 3 8(7a 8 = a 4 9(8a 9 = a 5 = and so in particular a 2+4n = 0 and a 3+4n = 0 and P (x = a 0 x 4n (4n(4n 1(4(n 1(4(n 1 1 (4(3 + a 1 x 4n+1 (4n + 1(4n(4(n 1 + 1(4(n 1 (5(4 which has radius of convergence R = 1 by the Ratio Test Notice that even though this is an explicit solution, it is not particularly enlightening to our understanding of the function in this form The following discussion is not expected on exams, but might help ones overall understanding Sample exercises begin on page 11 The Case of: f (k+1 + c k f (k + + c 0 f = 0 What if we wish to consider an example like f (x + bf (x + cf(x = 0? When b = 0, this reduces to the case f (x = cf(x that we solved before Starting once again with P (x = a nx n, in order for P (x to be a solution we must have n(n 1a n x n 2 + b na n x n 1 + c a n x n = 0 n=2 7

8 and equating like powers of x this gives us the relation n(n 1a n + b(n 1a n 1 + ca n 2 = 0 1 a n = n(n 1 (b(n 1a n 1 + ca n 2 = b n a c n 1 n(n 1 a n 2 This is a generating relation Once you know a n 1 and a n 2, the next term a n will be completely determined Thus, given a 0 and a 1, all the other a n can be found by using the generating formula Though complicated, it should be clear that we can generate a power series solution to this equation Perhaps what may be a little surprising is that this power series will converge for all x In fact, if we started with a relation of the form f (k+1 (x + c k f (k (x + + c 1 f (x + c 0 f(x = 0 Then a power series solution P (x would have the generating formula for the a n, n k + 1 in terms of the previous values of a k The fact that the resulting power series for any choices of c 0,, c k and initial values a 0, a 1,, a k has an infinite radius of convergence can be shown by making a limit comparison test with an exponential series In fact, if we let c > 1 be larger than any of the c i and let M be larger than any of the t! cat for 0 t k then (k+1 t a n x n (k + 1x n M n! = Me (k+1 x and so the series P (x generated will absolutely converge for all x Theorem 03 For any numbers c 0, c k and a 0,, a k there exists a unique analytic function f on all of R such that f (k+1 (x + c k f (k (x + + c 1 f (x + c 0 = 0 subject to the initial conditions f (n (0 = a n for 0 n k Note We do have an explicit formula for computing the power series of a solution which a machine can quickly calculate for us In practice, however, such solutions are not generally useful in themselves except in unusually nice situations Theorem 03 is 8

9 primarily about the existence and uniqueness of a solution and is the first of a large family of similar results for differential equations studied in a more advanced course A common manner in which Theorem 03 is used, is to guess solutions to special cases and then to show that the known solutions can be used to generate all possible solutions For example, for the case f (x = kf(x, for k = ±1 we have solutions sin x, cos x, sinh x and cosh x and using power series we showed that all solutions are obtained by linear combinations of these Determining when a collection of known solutions can generate all solutions without constantly resorting to power series is another powerful tool developed in an advanced course in differential equations Example 04 Show that if β > 0, then every solution to f (x = β 2 f(x can be written as a linear combination of e βx and e βx Solution We can use the power series analysis as done for Theorem 02, but instead, we can simply use Theorem 03 directly First observe that for any c 0 and c 1, if we let f(x = c 0 e βx + c 1 e βx then f (x = c 0 βe βx + c 1 ( βe βx f (x = c 0 β 2 e βx + c 1 ( β 2 x βx = β 2 ( c 0 e βx c 1 e βx = β 2 f(x Given the initial conditions that f(0 = a 0 and f (0 = a 1, using we see that if we let f(0 = c 0 + c 1 = a 0 f (0 = βc 0 βc 1 = a 1 c 0 = 1 2 ( a 0 + a 1 β c 1 = 1 2 ( a 0 a 1 β then f must be the unique analytic solution to the differential equations with the given initial conditions Observation Theorem 03 says that one has a unique solution but it does not imply that every solution must appear the same Using Theorem 02 and Example 04 we see that every solution to f (x = β 2 f(x subject to f(0 = a 0 and f (0 = a 1 can be written as either f(x = a 0 cosh(βx + a 1 β sinh(βx = c 0e βx + c 1 e βx 9

10 f(x = cosh x, f(x = sinh x f(x = e x, f(x = e x and hence as linear combinations of two different sets of functions Graphing these functions, it may not be immediately clear that this is the case and part of the study of the general theory of differential equations is about finding good representations for solutions that help understand the solutions for a particular application y 20 y x x Example 05 Let E(x be the unique solution to f (3 (x = f(x with f(0 = 0, f (0 = 0 and f (0 = 1 Show that every solution to f (3 (x = f(x is a linear combination of E, E and E Solution By Theorem 03 we know that there is a unique power series solution to f (3 (x = f(x subject to the given the initial conditions a 0, a 1 and a 2 We know the solution must begin P (x = a 0 + a 1 x + a 2 2! x2 + by the relation P (3 (x = P (x we can read off the rest of the terms of this solution to be P (x = a 0 +a 1 x+ a 2 2! x2 + (a 0 x 3 3! + a 1 x 4 4! + a 2 x 5 x + (a 6 0 5! 6! + a x 7 1 7! + a 2 x 8 +( + 8! Since we know that this solution is absolutely convergence for all R, we can rearrange the terms without changing the sum and rewrite this as P (x = a 0 x 3n 3n! + a 1 x 3n+1 (3n + 1! + a 2 = a 0 E (x + a 1 E (x + a 2 E(x x 3n+2 (3n + 2! 10

11 Sample Exercises Determine the the unique analytic solution for each of the following differential equations and initial conditions using a power series expansion and recursive relation on the coefficients 1 y = 3y; y(0 = 8 y = 8e 3x = 8 3 n x n n! 2 y = 4y; y(0 = 0; y (0 = 1 y = 1 sin(2x = ( 4n 2 ( 1 (2n+1! x2n+1 3 y = 4y; y(0 = 2; y (0 = 0 y = 2 cos(4x = 2 4n (2n! x2n 4 y = xy; y(0 = 1 y = 5 y = x 2 y; y(0 = 1 y = x 2n 2 n n! x 3n 3 n n! 6 y = x 2 y; y(0 = 1; y (0 = 0 y = y = x 2 y; y(0 = 0; y (0 = 1 y = x + 8 xy + y = 0; y(0 = 0; y (0 = 1 y = x 4n 4 n n! ( (4n 1 x 4n+1 4 n n! ( (4n + 1 ( 1 n+1 (n 1! n! xn 9 (Bessel Functions-too hard for exam, but for your possible interest Bessel functions, first defined by Daniel Bernoulli and generalized by Friedrich Bessel are the canonical solution to (Bessel x 2 d2 y dx + xdy dx + (x2 α 2 y = 0 We will consider the case when α is an integer These show up in applications such as heat conduction, vibrations in thins membranes, diffusion problems and the dynamics of floating objects (a Let y(x = a n x n be an analytic solution to (Bessel Show that the coefficients must satisfy the recursive relation (n 2 α 2 a n = a n 2 11

12 (b Use part (a to show that a n = 0 for n < α and that a α is arbitrary (c Show the relation (α + 2n 2 α 2 = 2 2 n(α + n and use it to show that f(x = a α ( 1 n α! n!(n + α! ( x 2 2n+α is a solution to (Bessel By dividing out a α α! we obtain the Bessel functions of the first kind J α (x = ( 1 n ( x 2n+α n!(n + α! 2 (d What is the radius of convergence for J α (x? (e For x < 2, by the Alternating Series Estimation, what polynomial of minimal degree will ensure an estimate of J 1 (x to an error no greater than 10 5? 12