8 - Series Solutions of Differential Equations

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1 8 - Series Solutions of Differential Equations 8.2 Power Series and Analytic Functions Homework: p # ü Introduction Our earlier technques allowed us to write our solutions in terms of elementary functions (such as polynomial functions, exponential functions, trigonometric functions, etc). In most cases we are not able to do so. We can approximate the solution using numerical methods (Euler s method, etc.), or we can use power series. ü Power Series A power series about the point x 0 is an expression of the form (1) an Hx - x 0 L n = a 0 + a 1 Hx - x 0 L + a 2 Hx - x 0 L 2 +, where x is a variable and the a n s are constants. We say that (1) converges at the point x = c if the infinite series (of real numbers) an Hc - x 0 L n N converges; that is, the limit of the partial sums lim NØ an Hc - x 0 L n exists (as a finite number). If this dimit does not exist, the power series is said to diverge at x = c. Note that (1) converges at x = x 0. Theorem 1 - Radius of Convegence For each power series of the form (1), there is a number r (0 r ), called the radius of convergence of the power series, such that (1) converges absolutely for x - x 0 <r and diverges for x - x 0 >r. If the series (1) converges for all values of x, then r=. When the series (1) converges only at x 0, then r=0. Note that at the interior points of the interval, the power series converges absolutely, that is, the series an Hx - x 0 L n converges for every x in the interior. Recall that we had to test the endpoints of the interval individually. At the endpoints, the power series could diverge, converge conditionally, or converge absolutely. For the geometric series x n, the radius of convergence is r=1, since 1 = 1 + x x x2 + = x n for -1 < x < 1. So the radius of convergence is r=1 and the interval of convergence is -1 < x < 1. Recall that if the series was not a geometric series, we used the ratio test to find the interval of convergence.

2 2 CH_08_notes.nb Theorem 2 - Ratio Test for Power Series If, for n large, the coefficients a n are nonzero and satisfy lim nø a n = L (0 L ) a n+1 then the radius of convergence of the power series an Hx - x 0 L n is r=l ü Example Determine the convergence set of the given power series. 3 n n! xn Theorem 3 - Power Series Vanishing on an Interval If an Hx - x 0 L n = 0 for all x in some open interval, then each coefficient a n equals zero. Given two power series f HxL = an Hx - x 0 L n and ghxl = bn Hx - x 0 L n with nonzero radii of convergence. The sum is given by f HxL + ghxl = Han + b n LHx- x 0 L n for all x in the common interval of convergence for both power series. The product, given by f HxL ghxl = cn Hx - x 0 L n n where c n = k=0 a k b n-k, for all x in the common open interval of convergence for both power series. Theorem 4 - Differentiation and Integration of Power Series If the series f HxL = an Hx - x 0 L n has a positive radius of convergence r, then f is differentiable in the interval x - x 0 <r and termwise differentiation gives the power series for the derivative: f ' HxL = nan Hx - x 0 L n-1 for x - x 0 <r n=1 Furthermore, termwise integration gives the power series for the integral of f: Ÿ f HxL x = an n + 1 Hx - x 0L n+1 + C for x - x 0 < r

3 CH_08_notes.nb 3 ü Shifting the Summation Index Note that an Hx - x 0 L n = ak Hx - x 0 L k = a j Hx - x 0 L j Occasionally we will need to shift the summation index to determine the terms of a power series. ü Example Express the given power series as a series with generic term x k. k=0 j=0 (a) nhn - 1L an x n+2 n=2 (b) nan Hx - al n an Hx - al n n=1 ü Analytic Functions Definition 1 - Analytic Function A function f is said to be analytic at x 0 if, in an open interval about x 0, this funciton is the sum of a power series an Hx - x 0 L n that has a positive radius of convergene.

4 4 CH_08_notes.nb Elementary functions such as x, sin x, and cos x are analytice for all x, while ln x is analytic for x > 0. A rational function P HxL is an analytic funtion except at those x QHxL 0 for which QHx 0 L = 0. Recall the representations of some of these functions: (10) x = 1 + x + x2 + x3 + = x n 2! 3! n! (11) sin x = x - x3 + x5 + = H-1L n n+1 x2 3! 5! H2 n+1l! (12) cos x = 1 - x2 + x4 + = H-1L n 2! 4! H2 nl! x2 n (13) ln x = Hx - 1L - 1 Hx - 2 1L2 + 1 Hx - H-1L n-1 3 1L3 + = Hx - 1L n n n=1 The inportance of f being analytice is the fact that if f HxL i analytic at x 0, then it is the sum of some power series that converges in a neighborhood of x 0 : f HxL = an Hx - x 0 L n Any power series regardless of how it is derived that converges in some neighborhood of x 0 to a function has to be the Taylor series of that function.

5 CH_08_notes.nb Power Series Solutions to Linear Differential Equations Homework: p # ü Introduction We start by considering the linear differential equation (1) a 2 HxL y'' + a 1 HxL y' + a 0 HxL y = 0 that has been written in the standard form (2) y'' + phxl y' + qhxl y = 0 Definition 2 - Ordinary and Singular Points A point x 0 is called an ordinary point of equation (1) if both p = a 1 and q = a 0 a 2 a 2 ordinary point, it is called a singular point of the equation. ü Example Determine all singularpoints of the given differential equation. Ix 2 + xm y'' + 3 y' - 6 xy= 0 are analytic at x 0. If x 0 is not an At an ordinary point x 0 of equation (1) or (2), the coefficient functions phxl and qhxl are analytic. It runs out, in a eighborhood of an ordinary point x 0, the solutions to (1) or (2) can be expressed as a power series about x 0. Thus, we want to find a power series solution of the form (4) yhxl = a 0 + a 1 Hx - x 0 L + a 2 Hx - x 0 L 2 + = an Hx - x 0 L n Note that Note that y' HxL = a a 2 Hx - x 0 L + 3 a 3 Hx - x 0 L 2 + = nan Hx - x 0 L n-1 n=1 y'' HxL = 2 a a 3 Hx - x 0 L + 12 a 4 Hx - x 0 L 2 + = nhn - 1L an Hx - x 0 L n-2 n=2

6 6 CH_08_notes.nb ü Example Find at least the first four nonzero terms in a power series expansion about x = 0 for a general solution to the given differential equation: Ix 2 + 1M y'' - xy' + y = 0

7 CH_08_notes.nb 7 ü Example Find at least the first four nonzero terms in a power series expansion about x = 0 for the solution to the given initial value problem. y'' + Hx - 2L y' - y = 0; yh0l =-1, y' H0L = 0

8 8 CH_08_notes.nb 8.4 Equations with Analytic Coefficients Homework: p # ü Introduction We have intrduced a procedure for finding the power series solution of (1) y'' + phxl y' + qhxl y = 0 Theorem 5 - Existence of Analytic Solutions Suppose x 0 is an ordinary point for equation (1). Then (1) has two linearly independent analytic solutions of the form (2) yhxl = an Hx - x 0 L n Moreover, the radius of convergence of any power series solutionfhe form given by (2) is at least as large as the distance from x 0 to the nearest singular pont (real or complex-valued) of equation (1) ü Example Find a minimum value for the radius of convergence of a power series solution about x 0 = 0. Ix 2-5 x + 6M y'' - 3 xy' - y = 0

9 CH_08_notes.nb 9 ü Example Find at least the first four nonzero terms in a power series expansion about x = 2 for a general solution to the given differential equation: x 2 y'' - xy' + 2 y = 0; yh0l =-1, y' H0L = 0

10 10 CH_08_notes.nb 8.6 Method of Frobenius Homework: p # ü Introduction Definition 3 - Regular Singular Points A singular point x 0 of (7) y'' HxL + phxl y' HxL + qhxl yhxl = 0 is said to be a regular singular point if both Hx - x 0 L phxl and Hx - x 0 L 2 qhxl are analytic at x 0. Otherwise x 0 is called an irregular singular point. ü Example Classify each singular point (real or complex) of the given equation as regular or irregular. Ix 2-4M 2 y'' + 3 Hx - 2L y' + 5 y = 0 Let us assume that x = 0 is a regular singular point of y'' HxL + phxl y' HxL + qhxl yhxl = 0. Then xphxl and x 2 qhxl are analytic functions at x = 0. Which means that in some open interval about x = 0 (4) xphxl = p 0 + p 1 x + p 2 x 2 + = pn x n (5) x 2 qhxl = q 0 + q 1 x + q 2 x 2 + = qn x n One result of (4) and (5) is (6) lim xø0 xphxl = p 0 and lim xø0 x 2 qhxl = q 0 Note that (4) and (5) can be rewritten as follows: (9) phxl = pn x n-1 and qhxl = qn x n-2

11 CH_08_notes.nb 11 Frobenius (inspired by the solutions to Cauchy-Euler equations) considered solutions of y'' HxL + phxl y' HxL + qhxl yhxl = 0 at the regular singular point x = 0 to have the form: (10) whr, xl = x r an x n = an x n+r, x > 0 Differentiating w with respect to x gives (11) w' Hr, xl = Hn + rl an x n+r-1 (12) w'' Hr, xl = Hn + rlhn + r - 1L an x n+r-2 Substituting these (9), (10), (11), and (12) into y'' HxL + phxl y' HxL + qhxl yhxl = 0 are regrouping terms by powers of x we obtain - 1L + p 0 r + q 0 D a 0 x r-2 + 1L ra 1 + p 0 Hr + 1L a 1 + p 1 ra 0 + q 0 a 1 + q 1 a 0 D x r-1 + = 0 Since each coefficient is equal to zero, we consider the equation - 1L + p 0 r + q 0 D a 0 = 0 Definition 4 - Indicial Equation If x 0 is a regular sinular point of y'' + py' + qy= 0, then the indicial equation for this point is (16) rhr - 1L + p 0 r + q 0 where p 0 := lim xøx0 Hx - x 0 L phxl and q 0 := lim xøx0 Hx - x 0 L 2 qhxl The roots of the indicial equation are called the exponents (indices) of the singularity x 0. ü Example Find the indicial equation and the exponents for the singularity x = 0 of 3 xy'' + y' - y = 0. To find the coefficients we will be looking for a recursive relationship as before. To do this, we will use the larger root of the indicial equation.

12 12 CH_08_notes.nb ü Example Find a series expansion about the regular singular point x = 0 for a solution to 3 xy'' + y' - y = 0.

13 CH_08_notes.nb 13 Method of Frobenius To derive a series solution about the singular point x 0 of (29) a 2 HxL y'' HxL + a 1 HxL y' HxL + a 0 HxL yhxl = 0, x > x 0 (a) Set phxl := a 1HxL 0HxL a 2 HxL a 2 HxL 0L phxl and Hx - x 0 L 2 qhxl are analytic at x 0, then x 0 is a regular singular point and the remaining steps apply. (b) Let (30) whr, xl = Hx - x 0 L r an Hx - x 0 L n = an Hx - x 0 L n+r and, using termwise differentiation, substitute whr, xl into equation (29) to obtain an equation of the form A 0 Hx - x 0 L r+j + A 1 Hx - x 0 L r+j+1 + = 0 (c) Set the coefficients A 0, A 1, A 2,... equal to zero. [Notice that the equation A 0 = 0 is just a constant multiple of the indicial equation rhr - 1L + p 0 r + q 0 = 0.] (d) Use the system of equations A 0 = 0, A 1 = 0, A 2 = 0,... to find the recurrence relation involving a k and a 0, a 1,..., a k-1. (e) Take r = r 1, the larger root of the indicial equation and use the relation obtained in step (d) to determine a 1, a 2,..., recursively in terms of a 0 and r 1. (f) A series expansion of a solution to (29) is (31) whr 1, xl = Hx - x 0 L r 1 an Hx - x 0 L n, x > x 0 where a 0 is arbitrary and the a n s are defined in terms of a 0 and r 1. What is the radius of convergence of the power series in (31)? Theorem 6 - Frobenious s Theorem If x 0 is a regular sinular point of equation (29), then there exists at least one series solution of the form (30), where r = r 1 is the larger root of the associated indicial equation. Moreover, this series conrges for all x such that 0 < x - x 0 < R, where R is the distance from x 0 to the nearest other singular point (real or complex) of (29). For simplicity, will consider only expansions about the regular singular point x = 0 and only those equations for which the associated indicial equation has real roots.

14 14 CH_08_notes.nb ü Example Find a series expansion about the regular singular point x = 0 for a solution to x 2 y'' - xy' + H1 - xl y = 0, x > 0.

15 CH_08_notes.nb Finding a Second Linearly Independent Solution Homework: p # ü Introduction In the last section, we used the method of Frobenius to find one series solution, how do we find the second one, if it exists? Theorem 7 - Form a Second Linearly Independent Solution Let x 0 be a regular singular point for y'' + py' + qy= 0 and let r 1 and r 2 be the roots of the associated indicial equation, where Re r 1 Re r 2. (a) If r 1 - r 2 is not an integer, then there exist two linearly independent solutions of the form (11) y 1 HxL = an Hx - x 0 L n+r 1, a0 0 (b) (c) (12) y 2 HxL = bn Hx - x 0 L n+r 2, b0 0 If r 1 = r 2, then there exist two linearly independent solutions of the form (13) y 1 HxL = an Hx - x 0 L n+r 1, a0 0 (14) y 2 HxL = y 1 HxL lnhx - x 0 L + bn Hx - x 0 L n+r 1 n=1 If r 1 - r 2 is a positive integer, then there exist two linearly independent solutions of the form (15) y 1 HxL = an Hx - x 0 L n+r 1, a0 0 (16) y 2 HxL = Cy 1 HxL lnhx - x 0 L + bn Hx - x 0 L n+r 2, b0 0 where C is a constant that could be zero. ü Example Find the first few terms in the series expansion about the regular singular point x = 0 for two linearly independent solutions to x 2 y'' - xy' + H1 - xl y = 0, x > 0.