Example Ordinary Point Regular Singular Point - Study Guide

Example: Ordinary Point 1. Solve y 00 − xy 0 + 2y = 0 about the ordinary point x = 0 (we also did this in class) Substituting y = ∞ X cn xn into the differential equation, n=0 ∞ X n(n − 1)cn xn−2 − n=2 ∞ X ncn xn + 2 n=1 ∞ X cn xn = 0 n=0 Rewrite so that every sum begins with the same power of x. In this case, we start with x1 because of the middle sum. Rewriting the first two sums gives us: ! ! ∞ ∞ ∞ X X X 2c2 + (k + 2)(k + 1)ck+2 xk − kck xk + 2c0 + 2 ck xk = 0 k=1 k=1 And simplify: 2(c2 + c0 ) + ∞ X k=1 ((k + 2)(k + 1)ck+2 − (k − 2)ck ) xk = 0 k=1 This gives us the recurrence relation for the coefficients: c2 = −c0 and ck+2 = k−2 ck , (k + 2)(k + 1) k = 1, 2, 3, . . . With c0 = 1, and c1 = 0, we should find that c2 = −1 and the rest of the coefficients are zero. Therefore, the first solution is: y1 (x) = 1 − x2 With c0 = 0 and c1 = 1, we should find that 1 1 c2 = 0, c3 = − , c4 = 0, c5 = − ,... 6 120 Now, 1 5 1 x + ... y2 (x) = x − x3 − 6 120 and the general solution is y = c0 y1 + c1 y2 . 2. (Problem 18, Section 6.1 for Extra Practice) Solve y 00 − xy = 0 about x = 0. Going through our substitution, we get: ∞ X n(n − 1)cn xn−2 − n=2 ∞ X cn xn+1 = 0 n=1 Rewrite so that the sums both start with x1 (because of the second sum): 2c2 + ∞ X (k + 2)(k + 1)ck+2 xk − k=1 And simplify: 2c2 + ∞ X k=1 ∞ X ck−1 xk = 0 k=1 ((k + 2)(k + 1)ck+2 − ck−1 ) xk = 0 Therefore, c2 = 0 and ck+2 = 1 ck−1 , k = 1, 2, 3, . . . (k + 2)(k + 1) To get our linearly independent solutions, first let c0 = 1, c1 = 0 to get: c3 = 1 1 , c4 = 0, c5 = 0, c6 = 6 180 And now let c0 = 0, c1 = 1 to get the second solution: c3 = 0, c4 = 1 1 , c5 = c6 = 0, c7 = 12 504 The two solutions are: 1 6 1 y1 = 1 + x3 + x + h.o.t, 6 180 y2 = x + 1 4 1 7 x + x + h.o.t 12 504 Examples: Regular Singular Point 3. Solve y 00 − 1 0 2x y + 1+x 2x2 y = 0 using the power series method about x = 0. First, we see that: 1+x 1 , Q(x) = 2x 2x2 and we verify that x = 0 is a regular singular point. We do this by checking that xP (x) and x2 Q(x) are analytic at x = 0: 1 1 xP (x) = − , x2 Q(x) = (1 + x) 2 2 Both of these are analytic at x = 0, so x = 0 is a regular singular point. P (x) = − Our technique will be to first multiply the equation by x2 , and we’ll multiply by 2 to simplify the algebra: 2x2 y 00 − xy 0 + (1 + x)y = 0 Now assume the form of the solution to be the following. We’ve also include the first few terms for later: r y=x ∞ X n cn x = n=0 ∞ X cn xr+n = c0 xr + c1 xr+1 + c2 xr+2 + . . . n=0 Previously, when we differentiated, we re-started the index (because the constant terms drop out), but now the constant terms will not drop out: y0 = ∞ X cn (r + n)xr+n−1 , y 00 = n=0 ∞ X cn (r + n)(r + n − 1)xr+n−2 n=0 Substitute into the differential equation: 2x2 ∞ X cn (r + n)(r + n − 1)xr+n−2 − x n=0 ∞ X cn (r + n)xr+n−1 + (1 + x) n=0 ∞ X cn xr+n = 0 n=0 Simplifying: ∞ X n=0 cn (r + n)(r + n − 1)xr+n − ∞ X n=0 cn (r + n)xr+n + ∞ X n=0 cn xr+n + ∞ X n=0 cn xr+n+1 = 0 The last sum could be rewritten (let k = n + 1, then let n = k) as ∞ X cn−1 xr+n n=1 The first three sums begin with xr , the last one starts with xr+1 . We split off the first entry from the first three sums, and combine the remaining sums to get: (2c0 r(r − 1) − c0 r + c0 ) + ∞ X (2cn (r + n)(r + n − 1) − (r + n)cn + cn + cn−1 ) xr+n = 0 n=1 The constant term tells us what r should be (THE INDICIAL EQUATION): c0 (2r(r − 1) − r + 1) = 0 ⇒ 2r2 − 2r − r + 1 = 0 ⇒ 2r2 − 3r + 1 = 0 ⇒ r = 1, 1 2 We will get two linearly independent solutions by using the two values of r. First, for r = 1, we are back to our usual situation: ∞ X (2cn (n + 1)n − ncn + cn−1 )xn+1 = 0 n=1 which simplifies: ∞ X (ncn (2n + 1) + cn−1 )xn+1 = 0 n=1 which implies that each coefficient must be zero: cn = − cn−1 n(2n + 1) Notice that, rather than depending on both c0 and c1 , we will now only depend on one of them. Let c0 = 1, and compute the remaining coefficients, written out so you can see the pattern: c1 = − 1 , 3·1 c2 = − c1 1 = , 2·5 (3 · 5) · (1 · 2) c3 = − c2 1 =− 3·7 (3 · 5 · 7) · (1 · 2 · 3) For the actual series, simplify to: 1 1 1 c0 = 1, c1 = − , c2 = , c3 = − ,... 3 30 630 Similarly, solve when r = 12 . In this case, we have:       ∞  √ X 1 1 1 x 2cn +n n− − n+ cn + cn + cn−1 xn = 0 2 2 2 n=1 which simplifies to: ∞ √ X x (cn (2n − 1) n + cn−1 ) xn = 0 n=1 so that: cn = − cn−1 n(2n − 1) Again by taking c0 = 1, we get: c1 = − 1 = −1, 1 c2 = − c1 1 = , 2·3 6 c3 = − c2 1 = − ,... 3·5 90 We now have y1 and y2 . Putting it all together, the general solution is: y(x) = C1 y1 + C2 y2 or    √ 1 1 2 1 3 1 2 1 3 y(x) = C1 x 1 − x + x − x + h.o.t + C2 x 1 − x + x − x + h.o.t 3 30 630 6 90  In Maple, verify this: Order:=4; de:=2*x^2*diff(y(x),x$2)-x*diff(y(x),x)+(1+x)*y(x)=0; dsolve(de,y(x),type=series); 4. (Problem 16, Sect 6.2) Solve: 2xy 00 + 5y 0 + xy = 0 about the point x = 0. We multiply through by x, although that is not necessary, to get: 2x2 y 00 + 5xy 0 + x2 y = 0 Now substitute the power series for y, y 0 , y 00 to get: 2x2 ∞ X cn (r + n)(r + n − 1)xr+n−2 + 5x n=0 ∞ X cn (r + n)xr+n−1 + x2 n=0 ∞ X cn xr+n = 0 n=0 Simplifying, ∞ X 2cn (r + n)(r + n − 1)xr+n + n=0 ∞ X 5cn (r + n)xr+n + n=0 ∞ X cn xr+n+2 = 0 n=0 The last sum starts with xr+2 , so pull the first two terms off the first two sums: (2c0 r(r − 1) + 5c0 r) xr + (2c1 (r + 1)r + 5c1 (r + 1)) xr+1 + ∞ X (2cn (r + n)(r + n − 1) + 5cn (r + n) + cn−2 )xr+n = 0 n=2 The first term gives the indicial equation, r(2r + 3) = 0 ⇒ r = 0, r = −3 2 The second term implies that c1 = 0 for either r, and for the rest of the coefficients, cn = − cn−2 , n = 2, 3, 4, . . . (n + r)(2n + 2r + 3) Choosing r = −3/2 first, the formula simplifies to: cn = − cn−2 (2n − 3)n cn = − cn−2 n(2n + 3) Choosing r = 0, we get: In either case, set c0 = 1, and we get:     1 1 1 1 4 y = C1 x−3/2 1 − x2 + x4 + . . . + C2 1 − x2 + x + ... 2 40 14 616