MATH 2280 - LECTURE 22DYLAN ZWICK1. Regular Singular PointsLast time we looked at how to solve linear OD Es of the form:A(x)y′′+ B(x)y′+ C(x)y = 0.The first thing we do is rewrite the ODE as:y′′+ P (x)y′+ Q(x)y = 0,where, of course,P (x) =B(x)A(x), and Q(x) =C(x)A(x).Now, if P (x) and Q(x) a re a nalytic around the point a then we havetwo linearly independent solutions of the form:y(x) =∞Xn=0cn(x − a)nwhere the raddi of convergence are at least as great as the distancein the complex plane from a to the nearest singular point of either P (x)or Q(x).1.1. Ordinary, Regular, and Irregular Points. We first state with-out proof the fact tha t either P (x) and Q(x) are analytic at x = a ofapproach ±∞ as x → a.Now, o f course, we must ask the question of what we do if either P (x)or Q(x) is not analytic at a? Well, it turns out we have methods fordealing with this if they’re not analytic in the “right way”. We’ll alsorestrict ourselves to dealing with the case a = 0, but we note that byjust shifting our coordinates this restriction incurs no loss of generality.Date: Spring 2009.12 DYLAN ZWICKWe divide singular p oints into two types: regular singular points,and irregular singular points. A regular singular point is a singularpoint where, if we rewrite:y′′+ P (x)y′+ Q(x)y = 0asy′′+p(x)xy′+q(x)x2y = 0the functions p (x) and q(x) are analytic. We will go over how to solvesecond order linear ODEs around regular singular points using seriesmethods. We will not discuss how to solve ODEs around irregularsingular points, as that is a much more difficult and advanced topic.Example - Determine whether x = 0 is an ordinary point, a regularsingular point, or an irregular singular point of the ODE:x2y′′+ (6 sin x)y′+ 6y = 0Solution - If we divide through by x2we get:y′′+6 sin xx2y′+6x2y = 0which has limx→0= ∞ for both coefficient functions, so it’s not anordinary point. However, if we rewrite this in the format above, wesee:p(x) =6 sin xx, q(x) = 6,both of which are analytic at x = 0. So, x = 0 is a regular singularpoint of the ODE.We again state a fact without proof, this time tha t if the limits:limx→0p(x) and limx→0q(x)exist, are finite, and are not 0 then x = 0 is a regular singular point.If both limits are 0 then x = 0 may be a regular singular point or a nordinary point. If either limit fails to exists or is ±∞ then x = 0 is anirregular singular point. So, this gives us a useful way of testing if asingular point is regular.MATH 2280 - LECTURE 22 31.2. The Method of Frobenius. Now we’ll figure out how to ac-tually solve these ODEs around regular singular points. We start byexamining the simplest such ODE:x2y′′+ p0xy′+ q0y = 0where p0, q0are both constants. This ODE is solved by y = xr, wherer satisfies the quadratic:r(r − 1) + p0r + q0= 0.Using this as our starting point, in general we assume our solutionhas the form:y(x) = xr∞Xn=0cnxn.Note - This is NOT a power series if r /∈ Z+.This is called a Frobenius series. Now, we want to take a look atwhat this constant r needs to be and at what kind of series we get. So,assume that we have a solution in this form. In this case we have:y(x) = xr∞Xn=0cnxn=∞Xn=0cnxn+r,y′(x) =∞Xn=0cn(n + r)xn+r−1,y′′(x) =∞Xn=0cn(n + r)(n + r − 1)xn+r−2.If we substitute this into:x2y′′+ xp(x)y′+ q(x)y = 0,where p(x) and q(x) are analytic around x = 0 and so have a powerseries representa t io n of the form:p(x) = p0+ p1x + p2x2+ · · ·q(x) = q0+ q1x + q2x2+ · · ·then plugging everything in we get:4 DYLAN ZWICK[r(r − 1)c0xr+ (r + 1)rc1xr+1+ · · · ] + [p0x + p1x2+ · · · ] · [rc0xr−1+(r + 1)c1xr+ · · · ] + [q0+ q1x + · · · ] · [c0xr+ c1xr+1+ · · · ] = 0 .Now, if we look at the xrterm we get, assuming (as we of coursealways can and should) that c06= 0, we get the relation:r(r − 1) + p0r + q0= 0.This is called the indicial equation of the ODE, and it must, ac-cording to the identity principle, be satisfies for our solution to work.This is, of course, only a necessary condition, and we certainly haven’tproven it’s sufficient. That’s where the next theorem comes in:Theorem - Suppose that x = 0 is a regular singular point of theODE:x2y′′+ xp(x)y′+ q(x)y = 0.Let ρ > 0 denote the minimum of the radii of convergence of thepower series:p(x) =∞Xn=0pnxnand q(x) =∞Xn=0qnxn.Let r1and r2be the real roots (we’ll always be assuming our rootsare real), of the indicial equation with r1≥ r2. Then(1) For x > 0, there exists a solution of our ODE of the form:y1(x) = xr1∞Xn=0anxn, a06= 0,corresponding to the larger root r1.(2) If r1− r2is neither zero nor a positive integer, then there existsa second linearly independent solution for x > 0 of the form:y2(x) = xr2∞Xn=0bnxn, b06= 0,corresponding to the smaller root r2.The radii o f convergence of y1and y2are at least ρ. We determinethe coefficients by plugging our series into:x2y′′+ xp(x)y′+ q(x)y = 0.MATH 2280 - LECTURE 22 5Example - Use the metho d of Frobenius to solve the ODE:2x2y′′+ 3xy′− (x2+ 1)y = 0around the regular singular point x = 0.Proof - Rewriting this we get:y′′+32xy′+−12−12x2x2y = 0and so p0=32and q0= −12.This gives us the indicial equation:r(r − 1) +32r −12= (r −12)(r + 1) = 0,and so our two roots are r1=12and r2= −1. So, our theoremguarantees two linearly independent Fro benius type solutions.Frequently it’s easier to work out our solutions without plugging inspecific values of r until the end. That’s what we’ll do here. Now, ifwe have a solution of the form:y(x) =∞Xn=0xn+r,then if we plug this form into our ODE we get the relation:2∞Xn=0(n + r)(n + r − 1)cnxn+r+ 3∞Xn=0(n + r)xn+r−∞Xn=0cnxn+r+2−∞Xn=0cnxn+r= 0.If we shift the third series over by 2 we get:2∞Xn=0(n + r)(n + r − 1)cnxn+r+ 3∞Xn=0(n + r)xn+r−∞Xn=2cn−2xn+r−∞Xn=0cnxn+r= 0.Now, if we lo ok at t he xrand xr+1coefficients we get the relations:6 DYLAN ZWICK[2r(r − 1) + 3r − 1]c0= 2(r2+12r −12)c0= 0[2(r + 1)r + 3(r + 1) − 1]c1= 0.If we plug in our values for r we see that the first of these is auto-matically satisfies for any c0, as the multiplier of c0is just a constantmultiplied by the indicial equation. On the other hand, if we plug inour values for r we see that the second equation is only satisfies forc1= 0.As for the other coefficients we get the relations:2(n + r)(n + r − 1)cn+ 3(n + r)cn− cn−2− cn= 0,which simplify