Chapter 5.2: Series solution near an ordinary point We now look at ODE s with polynomial coefficients of the form: P (x)y + Q(x)y + R(x)y = 0. Thus, we assume P (x), Q(x), R(x) are polynomials in x. Why? Because many, many important equations are of this type: Bessel s equation: x 2 y +xy +(x 2 ν 2 )y = 0. Here, ν is a real number. Legendre s equation: (1 x 2 )y 2xy + α(α + 1)y = 0. Here, α is a real number. 1
Ordinary point versus singular point We assume P, Q, R have no common factors (otherwise we simplify by dividing out). A point x 0 such that P (x 0 ) 0 is called an ordinary point. A point x 0 such that P (x 0 ) = 0 is called a singular point. Singular points are much more complicated than ordinary points. So we first consider ordinary points. 2
Solutions near an ordinary point We could divide by P and rewrite our equation in the form y +p(x)y +q(x)y, p(x) = Q(x) R(x), q(x) = P (x) P (x). But it is usually simpler not to, when the coefficients are polynomials. Our goal: solve equations P (x)y + Q(x)y + R(x)y = 0 near ordinary points x 0 by power series solutions y = centered at x 0. n=0 a n (x x 0 ) n 3
Example Take the old favorite y + y = 0. Try to solve with y = n=0 a n x n. Take two derivatives under the summation sign: y = n=0 a n (x n ) = n=0 a n n(n 1)x n 2. The n = 0, 1 terms look odd at first, but actually they are zero. So we can write this as y = n=2 a n n(n 1)x n 2. 4
Example (cont.) Now for the infamous change of index of summation. We would like the series to be in powers of x n, not x n 2. So we put m = n 2 and rewrite the series as y = m=0 a m+2 (m + 2)(m + 1)x m. This follows from writing n = m + 2 everywhere. We can also replace the letter m by n. Our equation becomes: y +y = m=0 Equivalently, [a m+2 (m+2)(m+1)+a m ]x m = 0. a m+2 (m + 2)(m + 1) + a m = 0, for all m. 5
Example (concluded) This equation is called a recurrence relation. We can write it as: 1 a m+2 =, for all m. (m + 2)(m + 1) The recurrence relation goes up in steps of two: if we know a m, we can solve for a m+2. If we know a 0 = y(0), then we know a 2, a 4,..., a 2k,... a 1 = y (0), then we know a 3, a 5, a 7,..., a 2k+1,... 6
Example (concluded) So let s write the recurrence relation separately for even and odd indices: a 2k+2 = 1 (2k+2)(2k+1) a 2k. a 2k+3) = 1 (2k+3)(2k+2) a 2k+1. So far we haven t solved anything, we ve just rewritten the problem. At some point, we have to solve the recurrence relation. This is the hard part of the problem. 7
Even terms Let s solve for the even coefficients first. We need to find the pattern in: a 2 = 1 2 a 0, a 4 = 1 4 3 a 2 = 1 4! a 0 a 6 = 1 6 5 a 4 = 1 6 5 1 4! a 0 = 1 6! a 0. It is fairly clear that the sign of a 2k alternates. Also, that a 2k has the factor 1. So a good (2k)! guess is that a 2k = ( 1) k 1 (2k)!. Pluggin into the power series gives a 0 ( 1) k x2k k=0 (2k)! = a 0 cos x. 8
Odd terms The pattern is similar for the odd coefficients. We have a 3 = 1 3 2 a 1, a 5 = 1 5 4 a 3 = 1 5! a 1 a 7 = 1 7 6 a 5 = 1 7 6 1 5! a 0 = 1 7! a 0. The sign of a 2k+1 alternates and a 2k has the factor 1. So a good guess is that (2k+1)! a 2k+1 = ( 1) k 1 (2k + 1)!. Plugging into the power series gives a 1 k=0 ( 1) k x 2k+1 (2k + 1)! = a 1 sin x. 9
Observations As the E-U theorem predicts, and as we know from other solution methods, there are two free parameters a 0, a 1, which are y(0), y (0). So we see the general solution y = a 0 cos x + a 1 sin x emerge from the power series method too. 10
Airy s equation Now let s do a new equation which is very important in physics: Airy s equation y xy = 0. It models rainbows, for reasons we will see. Since x = 0 is an ordinary point, we again try to solve with As before, y = y = n=2 n=0 a n x n. a n n(n 1)x n 2. We put m = n 2 and rewrite the series as y = m=0 a m+2 (m + 2)(m + 1)x m. 11
Airy s equation: The linear term We also have: xy = n=0 a n x n+1 = Our equation becomes m=0 a m 1 x m. y +xy = m=0 [a m+2 (m+2)(m+1)+a m 1 ]x m = 0. Equivalently, a m+2 (m + 2)(m + 1) + a m 1 = 0, for all m. 12
3-term recurrence relation The recurrence relation is between a m+2 and a m 1, which are three apart, rather than two as in our first example. This suggests that we have three undetermined coefficients, a 0, a 1, a 2. How can this be? The EU theorem tells us there is a unique solution if we know y(0), y (0). The key is that y (x) = xy(x). The equation forces y (0) = 0 = a 2. So the only free coefficients are a 0, a 1, as the EU theorem says. 13
Airy s equation: The linear term Since the recurration relation goes up in steps of three, a 0 determines a 3, a 6, a 9,..., a 3k,.... a 1 determines a 4, a 7, a 10,..., a 3k+1, All coefficients a 2 = a 5 = a 8 = = a 3k+2 = 0. 14
The coefficients The recurrence relation says: 1 a 3k+3 = (3k + 3)(3k + 2) a 3k. It almost looks like we re getting (3k + 3)! in the denominator as we unravel the expression. However, we always miss the (3k + 1) factors. There is not much better we can say than a 3k = ( 1) k a 0 (3k)(3k 1)(3k 3)(3k 4) 3 2. Similarly, a 3k+1 = ( 1) k a 1 (3k + 1)(3k)(3k 1)(3k 2) 4 3. 15
The solution Thus, the general solution is y = a 0 k=0 ( 1) k x 3k (3k)(3k 1)(3k 3)(3k 4) 3 2 +a 1 k=0 ( 1) k x 3k+1 (3k+1)(3k)(3k 1)(3k 3) 4 3. The famous Airy function is the solution given by: 1 3 2/3 π Here, n=0 Γ( 1 3 (n + 1)) (3 1/3 x) n 2(n + 1)π sin( ). n! 3 a 0 = 1 3 2/3 π Γ(1 3 ) sin(2π 3 ), a 1 = 1 3 2/3 π Γ(2 3 ) sin(4π 3 ). 16