2 Series Solutions near a Regular Singular Point
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1 McGill University Math 325A: Differential Equations LECTURE 17: SERIES SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS II 1 Introduction Text: Chap. 8 In this lecture we investigate series solutions for the general linear DE a 0 xy n + a 1 xy n a n xy = bx, where the functions a 1, a 2,..., a n, b are analytic at x = x 0. If a 0 x 0 0 the point x = x 0 is called an ordinary point of the DE. In this case, the solutions are analytic at x = x 0 since the normalized DE y n + p 1 xy n p n xy = qx, where p i x = a i x/a 0 x, qx = bx/a 0 x, has coefficient functions which are analytic at x = x 0. If a 0 x 0 = 0, the point x = x 0 is said to be a singular point for the DE. If k is the multiplicity of the zero of a 0 x at x = x 0 and the multiplicities of the other coefficient functions at x = x 0 is as big then, on cancelling the common factor x x 0 k for x x 0, the DE obtained holds even for x = x 0 by continuity, has analytic coefficient functions at x = x 0 and x = x 0 is an ordinary point. In this case the singularity is said to be removable. For example, the DE xy + sinxy + xy = 0 has a removable singularity at x = 0. 2 Series Solutions near a Regular Singular Point In general, the solution of a linear DE in a neighborhood of a singularity is extremely difficult. However, there is an important special case where this can be done. For simplicity, we treat the case of the general second order homogeneous DE a 0 xy + a 1 xy + a 2 xy = 0, x > x 0, with a singular point at x = x 0. Without loss of generality we can, after possibly a change of variable x x 0 = t, assume that x 0 = 0. We say that x = 0 is a regular singular point if the normalized DE y + pxy + qxy = 0, x > 0, is such that xpx and x 2 qx are analytic at x = 0. A necessary and sufficient condition for this is that lim xpx = p 0, lim x 2 qx = q 0 x 0 x 0 exist and are finite. In this case xpx = p 0 + p 1 x + + p n x n +, x 2 qx = q 0 + q 1 x + + q n x n + 1
2 and the given DE has the same solutions as the DE x 2 y + xxpxy + x 2 qxy = 0. This DE is an Euler DE if xpx = p 0, x 2 qx = q 0. This suggests that we should look for solutions of the form y = x r a n x n = a n x n+r, with a 0 0. Substituting this in the DE gives n + rn + r 1a n x n+r + p n x n n + ra n x n+r + q n x n a n x n+r = 0 which, on expansion and simplification, becomes a 0 F rx r + n=1 {F n + ra n + [ n + r 1p 1 + q 1 an rp n + q n a 0 } x n+r = 0, where F r = rr 1 + p 0 r + q 0. Equating coefficients to zero, we get the indicial equation, and rr 1 + p 0 r + q 0 = 0, 1 F n + ra n = n + r 1p 1 + q 1 an 1 rp n + q n a 0 2 for n 1. The indicial equation 1 has two roots: r 1, r 2. Three cases should be discussed separately. 2.1 Case I: The roots r 1 r 2 N Two roots do nt differ by an integer. In this case, the above recursive equation 2 determines a n uniquely for r = r 1 and r = r 2. If a n r i is the solution for r = r i and a 0 = 1, we obtain the linearly independent solutions y 1 = x r 1 a n r 1 x n, y 2 = x r 2 a n r 2 x n. It can be shown that the radius of convergence of the infinite series is the distance to the singularity of the DE nearest to the singularity x = 0. If r 1 r 2 = N 0, the above recursion equations can be solved for r = r 1 as above to give a solution y 1 = x r 1 a n r 1 x n. A second linearly independent solution can then be found by reduction of order. However, the series calculations can be quite involved and a simpler method exists which is based on solving the recursion equation for a n r as a ratio of polynomials of r. This can always be done 2
3 since F n + r is not the zero polynomial for any n 0. If a n r is the solution with a 0 r = 1 and we let y = yx, r = x r a n rx n. 3 Thus, we have the following equality with two variables x, r: x 2 y + x 2 pxy + x 2 qxy = a 0 F rx r = r r 1 r r 2 x r Case II: The roots r 1 = r 2 In this case, from the equality 4 we get x 2 y + x 2 pxy + x 2 qxy = r r 1 2 x r. Differentiating this equation with respect to r, we get y y x 2 + x 2 px + x 2 qx y = 2r r 1 + r r 1 2 x r lnx. Setting r = r 1, we find that y 2 = y x, r 1 = x r1 a n r 1 x n lnx + x r1 a nr 1 x n = y 1 lnx + x r1 a nr 1 x n, where a nr is the derivative of a n r with respect to r. This is a second linearly independent solution. Since this solution is unbounded as x 0, any solution of the given DE which is bounded as x 0 must be a scalar multiple of y 1. Case III: The roots r 1 r 2 = N > 0 For this case, we let zx, r = r r 2 yx, r. Thus, from the equality 4 we get x 2 z + x 2 pxz + x 2 qxz = r r 1 r r 2 2 x r. Differentiating this equation with respect to r, we get z z x 2 + x 2 px + x 2 qx z = r r 2 [ r r 2 + 2r r 1 x r + r r 1 r r 2 2 x r lnx. Setting r = r 2, we see that y 2 = z x, r 2 is a solution of the given DE. Letting b n r = r r 2 a n r, we have F n + rb n r = [ n + r 1p 1 + q 1 bn 1 r rp n + q n b 0 r 5 and y 2 = lim r r 2 x r lnx b n rx n + x r b nrx n. 6 3
4 Note that a n r 2 0, for n = 1, 2,... N 1. Hence, we have b 0 r 2 = b 1 r 2 = b 2 r 2 = = b N 1 r 2 = 0. However, a N r 2 =, as F r 2 + N = F r 1 = 0. Hence, we have b N r 2 = lim r r 2 r r 2 a n r = a <, subsequently, Furthermore, lim x r lnxb N rx N = ax r 1 lnx. r r 2 F N r 2 b N+1 r 2 = F 1 + r 1 b N+1 r 2 = r 1 p 1 + q 1 bn r 2 r 2 p N+1 + q N+1 b 0 r 2 = r 1 p 1 + q 1 bn r 2 7 Thus, b N+1 r 2 = r 1p 1 + q 1 F 1 + r 1 b Nr 2 = a r 1p 1 + q 1 F 1 + r 1 = aa 1r 1. 8 Similarly, we have then we obtain F N r 2 b N+2 r 2 = F 2 + r 1 b N+2 r 2 b N+2 r 2 = a In general, we can write = [ 1 + r 1 p 1 + q 1 bn+1 r 2 r 1 p 2 + q 2 b N r 2 r 2 p N+2 + q N+2 b 0 r 2 = a [ 1 + r 1 p 1 + q 1 a1 r 1 a r 1 p 2 + q 2, 9 [ 1 + r1 p 1 + q 1 a1 r 1 + r 1 p 2 + q 2 F 2 + r 1 = aa 2 r b N+k r 2 = aa k r Substituting the above results to 6, we finally derive y 2 = ax r 1 a n r 1 x n lnx + x r 2 b nr 2 x n = ay 1 lnx + x r 2 b nr 2 x n. 12 This gives a second linearly independent solution. The above method is due to Frobenius and is called the Frobenius method. 4
5 Example 1. The DE 2xy + y + 2xy = 0 has a regular singular point at x = 0 since xpx = 1/2 and x 2 qx = x 2. The indicial equation is rr r = rr 1 2. The roots are r 1 = 1/2, r 2 = 0 which do not differ by an integer. We have r + 1r a 1 = 0, n + rn + r 1 2 a n = a n 2 for n 2, so that a n = 2a n 2 /r + n2r + 2n 1 for n 2. Hence 0 = a 1 = a 3 = a 2n+1 for n 0 and 2 a 2 = r + 22r + 3 a 2 0, a 4 = r + 42r + 7 a = r + 2r + 42r + 32r + 7 a 0. It follows by induction that a 2n = 1 2n r + 2r + 4 r + 2n2r + 32r + 4 2r + 2n 1 a 0. Setting, r = 1/2, 0, a 0 = 1, we get y 1 = x 2 n x 2n 5 9 4n + 1n!, y 2 = x 2n 3 7 4n 1n!. The infinite series have an infinite radius of convergence since x = 0 is the only singular point of the DE. Example 2. The DE xy + y + y = 0 has a regular singular point at x = 0 with xpx = 1, x 2 qx = x. The indicial equation is rr 1 + r = r 2 = 0. This equation has only one root x = 0. The recursion equation is The solution with a 0 = 1 is setting r = 0 gives the solution n + r 2 a n = a n 1, n 1. a n r = 1 n 1 r r r + n 2. y 1 = 1 n xn n! 2. Taking the derivative of a n r with respect to r we get, using a nr = a n r d dr ln [ a n r logarithmic differentiation,we get 2 a nr = r r a n r r + n 5
6 so that a n0 = 2 1 n n n! 2. Therefore a second linearly independent solution is y 2 = y 1 lnx n n n! 2 x n. n=1 The above series converge for all x. Any bounded solution of the given DE must be a scalar multiple of y 1. 6
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