Lecture 22 Power series solutions for 2nd order linear ODE s (not necessarily with constant coefficients) Recall a few facts about power series: a n z n This series in z is centered at z 0. Here z can be real or complex. Denote b z its absolute value, which is a non-negative real number. The following questions arise: (Q) When does this converge absolutely, i.e., when does n a n z n converge? Moreover, is there a real number R > 0 such that a n z n converge absolutely z with z < R, but diverges for z > R? (Q2) When can we differentiate a n z n term by term, i.e., when do we have the equality ( an z n ) na n z n? (Q3) Given a function f(z) when can we express it as a power series a n bz n, for z in some region, say for z < R? Given a nz n, we look at ( ) an+ z n+ L lim n a n z ( ) n a n+ lim z n a n If L makes sense, it will be a non-negative real number. A basic result (cf. Chapter 2 of the Notes for Maa) is the following: Theorem a n z n converges absolutely iff L <. Definition The radius of convergence R 0 is the largest non-negative real number such that
a n z n converges absolutely for z < R, and a n z n diverges for z > R. We will allow R to be, in which case a n z n converges z. By this definition, z < R lim a So R lim n n a n+. This answers Q. a n n a n+ z. For Q2, here is another basic result (cf. Chap. 2 of the Notes for Maa): Theorem 2 For z < R, we have If z R, anything can happen. ( a n z n ) a n nz n. To understand Q3 better, recall (from Maa) that if f is infinitely differentiable at z 0, then we have the associated Taylor series n 0 a n z n, with a n f (n) (0). n! Examples of infinitely differentiable functions at z 0: e z, sin(z), cos(z), polynomials, and rational functions P (z)/q(z), with P, Q polynomials and Q(0) 0. A Subtle Fact: The Taylor series of a function f (around z 0) may not equal f(z), for f infinitely differentiable. The simplest example of such a function is φ(x) which is e /x if x > 0 and 0 if x 0. You may check that φ is differentiable to any order at x 0, but with φ (n) (0) 0 for all n 0, making the Taylor expansion around 0 to be identically 0. But φ(x) is > 0 for x > 0, however small, and so cannot be represented by the Taylor expansion. 2
However, pathology like this does not happen if there is an R > 0 such that for all z with z < R, f(z) has a(n absolutely convergent) power series expansion f(z) a n z n. n In such a case, f is said to be an analytic function at z 0, in fact in z < R. Since we can differentiate under the sum sign for z < R (by Theorem 2), it follows (by evaluating each series expression for f (n) (z) at z 0) that we must have Examples of analytic functions () (2) Applying the Ratio test, ( L lim ( f : analytic a n f (n) (0). n! lim n e z a n+ n a n (n + )! n 0 z n n!. ) z <, z in C ) n! lim n n + 0 R, i.e., e z is represented by this power series for all z; so f(z) e z is analytic z, and f (n) (0). n! n! sin(z) z z3 3! + z5 5! + ( )n z 2n+ (2n + )! +... cos(z) z2 2! + z4 4! + ( )n z 2n (2n)! +... (3) f(z) z n, a rational function regular at z 0, i.e., the z denominator of f(z) is non-zero at z 0. a n+ L lim z z, n a n }{{}, as a n, n 3
implying that L < iff z <. So R, and our function is analytic in z <. (4) f(z) ln(z), z real and positive. As z 0, ln(z). This is nevertheless analytic near z, since we can write f(z) a n (z ) n, n which is a convergent power series around z. In fact we claim that that R in this case. In particular, the series expansion is valid for any small z, as long as it is non-zero. To check the Claim, write log(z) f(z) as ln( + (z )). Then, with u z, we have f (z) z + (z ) + u, and + u u + u2 u 3 + u 4..., which is valid, by (3), for u < (as R ). The series expression for can, by Theorem 2, be integrated term by term for u <. This +u way we get a power series for ln( + u), namely ln( + u) u u2 2 + u3 3 u4 4 +..., valid in u <. Since f(z) ln( + u), we get ln z n (z )n ( ), n n which holds for all z with z <. Hence the Claim. 4
( ) Ordinary points and Singular points Suppose we have a linear ODE of 2nd order: c (x) d2 u dx 2 + c 2(x) du dx + c 3(x)u 0, where the coefficients are not (necessarily) constant! The fact that ( ) has second order means c (x) is not identically zero. However, it could be zero for special values of x, where we need to exercise some care. The values of x where c (x) 0 are called the singular points of the ODE, while those x where c (x) 0 are called the ordinary points. Often there are only a finite number of singular points, but this is not always the case, as seen by looking at the example c (x) sin x, which vanishes at nπ for any integer n. with At the ordinary points, one can divide ( ) by c (u) to get a monic ODE: d 2 u dx + p(x)du + q(x)u 0, 2 dx p(x) c 2(x) c (x), and q(x) c 3(x) c (x). We will look for power series solutions! 5
Lecture 23 Last time we ended by looking at the general 2nd order linear homogeneous ODE c (x) d2 y dx 2 + c 2(x) dy dx + c 3(x)y 0, (*) with c (x) not identically zero. We are interested in the case when c (x), c 2 (x), c 3 (x) are not constants. Definition. A point x 0 is an ordinary point for ( ) iff c (x) 0. Otherwise x 0 is called a singular point. More precisely, when c (x) 0 we can divide by it, so ( ) becomes d 2 y dx 2 + p(x)dy dx + q(x)y 0, p(x) c 2(x) c (x), q(x) c 3(x) c (x). To say x 0 is an ordinary point means p, q are analytic at x 0, i.e., p(x) α n (x x 0 ) n, q(x) β n (x x 0 ) n, valid in an interval around x 0. Recall the following key examples of analytic functions at x 0 : ) Polynomials in x 2) Rational functions f(x) g(x), f, g polynomials, with g(x 0) 0 3) Exponential functions e xt, e x2 +2,... 4) sin x, cos x (5) tan x, if x 0 (2n + ) π 2 6
The Basic Principle: Suppose x 0 is an ordinary point for d 2 y dx + p(x)dy + q(x)y 0. (*) 2 dx Then it has a basis of (analytic) solutions of the form y (x) + b 2 (x x 0 ) 2 + + b n (x x 0 ) n, n2 y 2 (x) (x x 0 ) + c 2 (x x 0 ) 2 + (x x 0 ) + Hence the general solution is of the form y a 0 y + a y 2 a n (x x 0 ) n, c n (x x 0 ) n. where a 0, a are arbitrary constants, and for n 2, a n a 0 b n + a c n. Examples () d 2 y dx 2 + y 0 This ODE even has constant coefficients, every point is an ordinary point for this ODE. We already know that a basis of solutions is given by {sin x, cos x}, but we want to derive this using power series. Try a power series solution for y at x 0 0: y a n x n If the radius of convergence R > 0, then we can differentiate term by term, i.e., dy dx d dx (a nx n ) a n nx n n 2 a m+ (m + )x m, m n, 7
d 2 y dx d 2 dx ( ) dy dx m(m + )a m+ x m m (k + )(k + 2)a k+2 x k. k0 So d 2 y dx + y (n + )(n + 2)a 2 n+2 x n + a n x n ((n + )(n + 2)a n+2 + a n ) x n For the above sum to be 0 at all points in an open interval around 0, we must have (n + )(n + 2)a n+2 + a n 0, n 0. In other words, to have such a power series solution for the ODE, we must satisfy the recursive formula for all n 0: a n+2 (n + )(n + 2) a n. Hence a 2 2 a 0, a 4 2 a 2 a 6 5 6 a 4 2 3 4 a 0, 5 6 4 3 a 2 6! a 0,... a 2n ( )n (2n)! a 0, n 0. Similarly, a 3 2 3 a, a 5 2 3 4 5 a,..., implying a 2n+ ( )n (2n + )! a, n 0. 8
Thus y a n x n y a 0 n even 0 a n x n + a 2m x 2m + n odd 0 a n x n a 2m+ x 2m+ ( ) m x 2m +a (2m)! } {{ } cos x ( ) m x 2m+ (2m + )! } {{ } sin x (2) So the general solution is (as expected) of the form y a 0 cos x + a sin x, where a 0, a are arbitrary constants. In the notation of the Basic Principle, y cos x and y 2 sin x. Finally, since we know that the power series expressions are valid for all x (with R ), the power series solutions we found are also valid everywhere. y + xy + y 0 This is a linear ODE with non-constant coefficients. Since p(x) x, q(x), they are analytic everywhere, and so any x 0 is an ordinary point. Let x 0 0. Try y a nx n and assume R > 0. Then x dy dx x d 2 y dx 2 na n x n na n x n (m + )(m + 2)a m+2 x m So y + xy + y ((n + )(n + 2)a n+2 + na n + a n )x n, 9
which will be 0 iff all the coefficients are 0, giving rise to the recursive formula: (n + )(n + 2)a n+2 (n + )a n a n+2 a n for all n 0 n + 2 (We are able to cancel n + from both sides because it is non-zero for n 0.) Explicitly, yielding a 2 a 0 2, a 4 a 2 4 a 0 4 2, a 6 a 0 6 4 2,..., a 2m ( ) m a 0 (2m)(2m 2)(2m 4)... 4(2) ( )m a 0. 2 m m! The odd coefficients all depend on a. If we put a 0, a 0, we get one fundamental solution, and if we put a 0 0, a, we get the other fundamental solution Let a 0, a 0. Then y a 2m x 2m ( ) m x 2m 2 m m! ( ) x 2 m 2 m! So y e x2 /2. Note that when a linear ODE has constant coefficients, the fundamental solutions involve terms like e λx, but here we have a quadratic exponent of the exponential, caused by the ODE having non-constant coefficients. It could even be more complicated in that a fundamental solution may not be simply expressible - at all (as with the case of y 2 below) in terms of an exponential function. Let a, a 0 0. We get y 2 a 2m+ x 2m+, with a 3 a 3, a 5 a 3 5 a 5 3,..., a 2m+ 0 ( ) m (2m + )(2m )(2m 3)... (3),...
Thus and a 2m+ ( )m (2m)(2m 2)... 4(2) (2m + )! y 2 ( 2) m m!x 2m+. (2m + )! ( 2)m m! (2m + )!, It will be left as an exercise to check, as we did in Example, that the power series representing y and y 2 have positive radii of convergence. (Is R in each case?) Regular Singular Points Definition: A singular point x 0 for an ODE y + p(x)y + q(x) 0 is called a regular singular point iff (x x 0 )p(x) and (x x 0 ) 2 q(x) are both analytic at x x 0. Example: x 2 y + sin xy + y 0, which obviously has x 0 as the unique singular point. Now p(x) sin x x 2, q(x) x 2 xp(x) sin x x x2 3! + x4 x x3 + x5... 3! 5! x 5!... x 2n ( ) n 2n + )! Since x 2 q(x), it is obviously analytic (everywhere). We need to check that the power series for xp(x) converges around x 0. For this we apply the Ratio test: lim n ( ) n+ x 2(n+) /(2(n + ) + )! ( ) n x 2n /(2n + )! lim n x 2 (2n + 3)(2n + 2) 0, for any x. So R, and x 2 p(x) is analytic everywhere, not just around x 0. So x 0 is a regular singular point of this ODE, and every other point is ordinary!