Additional material: Linear Differential Equations

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1 Chapter 5 Additional material: Linear Differential Equations 5.1 Introduction The material in this chapter is not formally part of the LTCC course. It is included for completeness as it contains proofs and motivations of some of the claims made at the beginning of the next chapter Although the general solution of a first-order linear equation can be given as a quadrature, higher-order equations are, in general, much more complicated. This chapter is concerned with producing series expansions of solutions to linear differential equations. We will show that the only place where a solution to a linear equation can have a singularity is where one or more of the coefficient functions has a singularity. Definition 1 The point = 0 is called an ordinary point of the linear ODE d n y d n + p 1() dn 1 y d n p n()y = 0 (5.1) if the coefficient functions p k, k =1,...,n, are analytic in a neighbourhood of = 0. Otherwise = 0 is called a singular point of the equation. We know that we can find a Taylor series expansion of any solution to equation (5.1) about any ordinary point so in this chapter we will be concerned with the behaviour of solutions in the neighbourhood of a singular point. In general, solutions can behave very badly around singular points and it can be very difficult to obtain convergent series expansions of solutions about such points. We will only consider solutions in neigbourhoods of points where the coefficient functions have poles. Even in this case, solutions can in general have essential singularities, which are difficult to analyse. This leads us to the consideration of so-called regular singularities, where solutions may be branched but there are no essential singularities. We will describe a series expansion method due to Frobenius. Many of the classical special functions are either solutions of linear differential equations with regular singular points or certain limits of such solutions. 5.2 Analytic continuation of solutions We begin by proving the following theorem. Theorem 2 Let p j, j = 1,...,n, be analytic in 0 ρ. For any constants c k, k =0,...,n 1, equation (5.1) together with the initial conditions y (k) = c k, k =0,...,n 1 has a unique solution which is analytic on 0 <ρ. 22

2 CHAPTER 5. ADDITIONAL MATERIAL: LINEAR DIFFERENTIAL EQUATIONS 23 Proof: Without loss of generality we will take 0 = 0 and ρ = 1. Let y() = a k k, k=0 where a k = c k /k!, k =0,...,n 1. We need to show that the radius of convergence of this series is at least one. This series is majorised by where Y solves the initial value problem Y () = A k k, (5.2) k=0 d n Y d n = P 1() dn 1 Y d n P n()y, (5.3) where the series for P j majorises the series for p j at = 0 and Y (k) (0) = c k +, k =0,...,n 1, for any >0. Using Cauchy s inequality p (m) j (0) M j ρ m = M j, M j max 1 p j(), we choose the majorising series P j () = M j m = m=0 M j 1. Then equation (5.3) becomes (1 ) Y (n) = M 1 Y (n 1) + M 2 Y (n 2) + + M n Y. The coefficients A m in the expansion of Y satisfy the recurrence relation (j + 1)(j + 2) (j + n)a j+n =(M 1 + j)(j + 1)(j + 2) (j + n 1)A j+n 1 +M 2 (j + 1)(j + 2) (j + n 2)A j+n M n 1 (j + 1)A j+1 + M n A j. (5.4) The initial conditions show guarantee that A 0,...,A n 1 are positive. It follows from induction using equation (5.4) that A j > 0 for all j. Furthermore, dividing equation (5.4) by (j + 1) (j + n)a j+n 1 gives A j+n = M 1 + j A j+n 1 n + j + n k=2 (j + n k)! A j+n k M k, j =0, 1,... (5.5) (j + n 1)! A j+n 1 Choose M 1 >n. Then we see that A j+n >A j+n 1 for all j =0, 1,...and so A j+n k /A j+n 1 < 1 for k 2. It follows from equation (5.5) that lim (A j+n /A j+n 1 ) = 1. Hence the radius of j convergence of (5.2) is 1. Theorem 2 allows us to analytically continue any solution of equation (5.1) along any path that avoids the singularities of the p j s. Note that in general such solutions will be branched at these singularities.

3 CHAPTER 5. ADDITIONAL MATERIAL: LINEAR DIFFERENTIAL EQUATIONS First order equations Consider the first-order linear homogeneous ODE dy + Q()y =0, (5.6) d where Q is analytic in some neighbourhood of the point = a. The general solution of equation (5.6) is y() =κ exp Q(ζ)dζ, (5.7) γ where κ = y(a) and in general the value of y() depends on the path of integration γ from a to. If Q is everywhere analytic then y will have no singularities. Suppose that Q has a pole of order q at = 0. On substituting the expansion Q() = Q n ( 0 ) n, n= q Q 0 =0, into equation (5.7) we see that, up to an overall constant factor, where f() =exp m=0 h() =exp y() =f()g()h(), Q m m +1 ( 0) m+1 2 m= q Q m m +1 ( 0) m+1, g() =( 0 ) r, and r = Q 1. Clearly f is analytic at = 0, g is either meromorphic or branched at = 0, and h has an essential singularity at = 0 (assuming q 2). In the above analysis we were able to solve equation (5.6) explicitly. In general, this will not be the case when we consider higher-order equations. For this reason we would like to construct some kind of useful series representation of the solution y near a pole of Q without using the explicit solution. The above argument shows that if q = 1 then y() has a series expansion about = 0 of the form y() = a n ( 0 ) n+r, a 0 =0. (5.8) If q>1then( 0 ) r y() has an essential singularity at = 0 and the generation of local series expansions directly from the equation itself becomes much more complicated. If q = 1 then = 0 is said to be a regular singular point of equation (5.6). 5.4 Regular singular points In this section we will consider second-order equations of the form d 2 y d 2 + P ()dy + Q()y =0, (5.9) d

4 CHAPTER 5. ADDITIONAL MATERIAL: LINEAR DIFFERENTIAL EQUATIONS 25 where at least one of the coefficient functions has a pole at = 0. expansions about = 0 of the form Hence P and Q have P () = P n ( 0 ) n p, Q() = Q n ( 0 ) n q, (5.10) where P 0 and Q 0 are not ero and where p and q are integers satisfying max(p, q) > 0. Let us find necessary conditions on P and Q such that any solution of equation (5.9) has a series expansion of the form (5.8). Substituting the y() =a 0 ( 0 ) r + in equation (5.9) and keeping only the possible dominant terms (i.e., the terms of the form ( 0 ) s where s has the most negative real part) gives This gives r(r 1)a 0 ( 0 ) r P 0 ra 0 ( 0 ) r p Q 0 a 0 ( 0 ) r q + =0. r(r 1) + + P 0 r( 0 ) 1 p + + Q 0 ( 0 ) 2 q + =0. (5.11) The lowest power of ( 0 ) in equation (5.11) is min(0, 1 p, 2 q). Since equation (5.9) is second-order, it has two independent solutions. Generically, we would expect to find two values of r such that there is a solution with an expansion of the form (5.8). In order that equating the coefficients of the lowest power of ( 0 ) in equation (5.11) will give a quadratic equation in r (as opposed to a linear equation), we must have 0 min(1 p, 2 q). The point = 0 is then called a regular singular point and the quadratic equation for r is called the indicial equation at = 0. Note that even though the indicial equation is a quadratic equation for r, it could have a single repeated root. We will see in the next chapter that we can still find two series expansions in this case, although the second expansion is more general than (5.8). Definition 3 An equation of the form (5.9) is said to have a regular singularity at = 0 if and only if the equation is singular at = 0 and ( 0 )P () and ( 0 ) 2 Q() are analytic in a neighbourhood of = 0. In other words, a linear homogeneous second-order ODE for y() has a regular singularity at = 0 if and only if it has the form d 2 y d 2 + n 1 dy p n ( 0 ) d + q n ( 0 ) n 2 y =0, (5.12) where at least one of the constants p 0, q 0, q 1 is not ero. The indicial equation is More generally, we have the following definition. r(r 1) + p 0 r + q 0 =0. (5.13) Definition 4 Let = 0 be a singular point of the linear homogeneous ODE d N N 1 y d N + p N n () dn y =0. (5.14) dn The singularity at = 0 is said to be regular if ( 0 ) n p n () is regular in a neighbourhood of = 0 for n =1,...,N 1.

5 CHAPTER 5. ADDITIONAL MATERIAL: LINEAR DIFFERENTIAL EQUATIONS 26 Exercise 5 Identify all singular points of the ODE d 2 y d dy ( + 1) 2 d ( 2) 2 y =0, in the finite complex plane and classify each singular point as regular or irregular. For an equation of the form (5.14), the point at infinity is called an ordinary point, a regular singular point, or an irregular singular point if the resulting linear equation for w(x) = y(), x =1/ has an ordinary point, a regular singular point, or an irregular singular at x =0 respectively. In particular, under this change of variables, equation (5.9) becomes d 2 w 2 dx 2 + x P (1/x) dw x 2 dx + Q(1/x) x 4 w =0. It follows that equation (5.9) has an ordinary point at infinity if and only if P () = O 2, Q() =O 4,. (5.15) If equation (5.9) has a singularity at infinity then this singularity is regular if and only if 1 1 P () =O, Q() =O 2,. Exercise 6 Determine whether the point = is an ordinary point, a regular singular point, or an irregular singular point of the equation 5.5 Series expansions d2 y d 2 +5dy ( + 1)y =0. d In this section we will demonstrate how to find series expansions of solutions about regular singularities. In particular, we will consider the example of Bessel s equation. A general theorem will be given later. Bessel s equation is 2 d2 y d 2 + dy d +(2 ν 2 )y =0, (5.16) where ν is an arbitrary constant. Without loss of generality, we take (ν) 0. Exercise 7 Note that =0is a regular singular point for Bessel s equation. Show that the roots of the indicial equation for Bessel s equation (5.16) about =0are r = ν and r = ν. Assume that equation (5.16) has a solution that can be represented in a neighbourhood of = 0 by a convergent series of the form (5.8), where a 0 = 0. Then equation (5.16) becomes (n + r)(n + r 1)an n+r +(n + r)a n n+r + a n n+r+2 ν 2 a n n+r =0. That is, (n + r)(n + r 1) + (n + r) ν 2 a n n+r + a n n+r+2 =0. (5.17)

6 CHAPTER 5. ADDITIONAL MATERIAL: LINEAR DIFFERENTIAL EQUATIONS 27 We would like to combine all terms into a single sum of terms proportional to n+r.tothisend we rewrite the second sum in (5.17) as a n n+r+2 = a n 2 n+r. Note that the re-indexed sum begins at n = 2 so equation (5.17) becomes n=2 + [r 2 ν 2 ]a 0 r +[(r + 1) 2 ν 2 ]a 1 r+1 [(n + r) 2 ν 2 ]a n + a n 2 n+r =0. (5.18) n=2 Now we equate the coefficients of different powers of to ero, r : [r 2 ν 2 ]a 0 =0, (5.19) r+1 : [(r + 1) 2 ν 2 ]a 1 =0, (5.20) n+r : [(n + r) 2 ν 2 ]a n + a n 2 =0, n =2, 3,... (5.21) Since a 0 = 0, equation (5.19) is just the indicial equation at = 0, which gives r = ±ν. Let us begin by considering the case r = r 1 := ν. Equation (5.20) becomes (2ν + 1)a 1 = 0, which gives a 1 =0since(ν) 0. Similarly, setting r = ν in equation (5.21) gives a n 2 a n =, n =2, 3,..., (5.22) n(n +2ν) since the denominator never vanishes. Equation (5.22) is a recurrence relation for the a n s. Recalling that a 1 = 0, equation (5.22) shows that a 2k 1 =0, k =1, 2,... For n = 2k even, the recurrence relation (5.22) becomes and so a 2k = a 2k = a 2(k 1), k =1, 2,... 4k(k + ν) ( 1) k a 0 4 k, k =1, 2,... k!(ν + 1)(ν + 2) (ν + k) Hence, for any ν with (ν) 0, a one-parameter family of solutions of Bessel s equation is given by y 1 () =a 0 1+ ν ( 1) k 2k 4 k. (5.23) k!(1 + ν)(2 + ν) (k + ν) k=1 Next we consider the second solution r = r 2 := ν of the indicial equation. First we note that if ν =0thenr 2 = r 1 = 0 and we have just seen that there is only a one parameter family of solutions to equation (5.16) of the form (5.8). We will return to the problem of finding a second independent solution in this case. On substituting r = ν in equations (5.20) and (5.21), we obtain (2ν 1)a 1 = 0, (5.24) n(2ν n)a n = a n 2, n =2, 3,... (5.25)

7 CHAPTER 5. ADDITIONAL MATERIAL: LINEAR DIFFERENTIAL EQUATIONS 28 Note that if ν = N/2 for some positive integer N, then equations (5.24) and (5.25) do not allow us to uniquely determine a N in terms of a N 2. For now we will assume that 2ν is not an integer. Then Equation (5.24) shows that a 1 =0 and so it follows from equation (5.25) that a 2k 1 = 0, k =1, 2,... For n =2k even, equation (5.25) gives the recurrence relation a 2k = a 2(k 1) n(n 2ν), and so we have found a second independent solution to equation (5.16) given by y 2 () =a 0 1+ ν ( 1) k 2k 4 k. (5.26) k!(1 ν)(2 ν) (k ν) k=1 Note that y 2 can be obtained from y 1 by replacing ν with ν. Bessel s equation remains unchanged under ν ν. The solutions y 1 and y 2 are clearly linearly independent since y 1 a 0 ν and y 2 a 0 ν as Resonance We have seen that we can always find a series representation of a solution about a regular singular point 0 of a second order ODE of the form ( 0 ) r multiplied by an analytic function, and in fact we can generically find a second independent solution of the same form. We encountered a problem when the roots of the indicial equation differed by an integer. This phenomenon is referred to as resonance. We will learn how to find a series expansion for a second independent solution. This series may, however, involve logs. Resonance is extremely important both from the point of view of forming a complete theory of regular singular points, and also from the point of view of applications. It is extremely important when we come to consider nonlinear equations. In this section we return to Bessel s equation (5.16) in the case ν = N/2 wheren =0, 1, 2,... Recall that in this case the solution y 1 given by (5.23) exists but the recurrence relation for y 2 breaks down. We begin by establishing some more general theory. Using the method of reduction of order, it can be shown that given any solution y 1 of equation (5.9), a second solution is given by y 2 () =y 1 () y1 2 ()E() d, E() :=exp P () d. (5.27) We could use equation (5.27) to calculate the series expansion of y 2 directly from the series expansion of y 1, however, for our purposes it is more useful to use equation (5.27) to understand the general form of the series expansion of y 2, which we can then calculate directly. Exercise 8 If equation (5.9) has a regular singular point at = 0,showthatE has a series expansion of the form E() =( 0 ) p 0 F (), where p 0 = lim ( 0 )P () and F is is analytic 0 in a neighbourhood of = 0. Using the expansions y 1 () = a n ( 0 ) n+r 1 and y1 2 ()E() = b n ( 0 ) n p 0 2r 1,

8 CHAPTER 5. ADDITIONAL MATERIAL: LINEAR DIFFERENTIAL EQUATIONS 29 a 0,b 0 = 0, equation (5.27) becomes y 2 = a m ( 0 ) m+r 1 m=0 b n ( 0 ) n p 0 2r 1 d. (5.28) If p 0 +2r 1 n = 1 for all n =0, 1, 2,..., then equation (5.28) shows that y 2 has an expansion of the form c n ( 0 ) n p 0 r On the other hand, if (p 0 1) + 2r 1 = N, (5.29) is a non-negative integer, then equation (5.28) shows that y 2 is the sum of a term of the form b N y 1 ()ln plus a series of the form c n ( 0 ) n+s,wherec 0 = 0 and s =1 p 0 r 1 if N = 0 and s =2 p 0 r 1 if N = 0. Note that if N =0theny 2 necessarily contains a log term in its expansion. For N =1, 2,..., y 2 will only contain a log term if b N = 0, which is a condition that depends on the precise form of the equation. Now we wish to interpret equation (5.29). Recall that r 1 and r 2 are roots of the indicial equation r 2 +(p 0 1)r + q 0 =0. It follows that r 1 + r 2 =1 p 0, and so equation (5.29) becomes r 1 r 2 = N, wheren is an integer. We have proved the following. Theorem 9 (Frobenius) Consider the differential equation d 2 y d 2 + p() dy 0 d + q() y =0, (5.30) ( 0 ) 2 where p and q are analytic on 0 <ρand r 1,r 2 are the roots of the indicial equation at = 0 satisfying (r 1 r 2 ) 0. Then equation (5.30) has a solution with series expansion y 1 () = a n ( 0 ) n+r 1, a 0 =0, for near 0. A series expansion can also be developped for an independent solution y 2 according to whether r 1 r 2 is an integer. All of these series converge on a covering surface of 0 < 0 <ρ. Case 1: r 1 r 2 is not an integer. In this case any independent solution has a series expansion of the form y 2 () = b n ( 0 ) n+r 2, b 0 =0, about = 0. Case 1: r 1 r 2 is an integer. In this case any independent solution has a series expansion of the form y 2 () =κy 1 ()ln + b n ( 0 ) n+r 2, about = 0, where κ is a constant (possibly 0). If r 1 = r 2 then κ = 0and without loss of generality, we can choose κ =1.

9 CHAPTER 5. ADDITIONAL MATERIAL: LINEAR DIFFERENTIAL EQUATIONS 30 Example 10 The indicial equation about the regular singular point =0for the equation 2 2 d2 y ( + 1)dy d2 d + y =0, has roots r 1 =1, r 2 =1/2. Since r 1 r 2 =1/2 is not an integer, it follows that the general solution does not contain a logarithm. Example 11 The indicial equation about the regular singular point =1for the equation ( 1) 2 d2 y + 5( 1)dy +( + 3)y =0, d2 d has a repeated root r 1 = r 2 = 2. Therefore, the general solution necessarily contains a log term. Example 12 Recall that the roots of the indicial equation for Bessel s equation are r = ±ν. Hence, if ν =0then the second independent solution necessarily contains a logarithm. From the recurrence relation (5.24), we see that if ν =1/2, then there is no logarithm in the second series solution of bessel s equation. Exercise 13 For each of the following equations, determine whether the series expansion about =0of the general solution contains a logarithm. 1. d2 y 3 d dy 2 1 d y =0, 2. d2 y 2 dy d d y =0. Further Exercises 1. Find all singular points of the following equations in the extended complex plane. Classify each of these singular points as either regular or irregular. (a) y + y 1 y =0, (b) y 3 2 y =0, (c) y =0, (d) y + y =0, (e) y +3y + y =0. 2. Show that the series (5.23) converges for all = Any second-order homogeneous linear ordinary differential equation, can be mapped to the standard form y + p()y + q()y =0, (5.31) w + I()w = 0 (5.32) by a transformation of the form w() =h()y(), for some h. Show that I() =q 1 dp 2 d 1 4 p2. Equation (5.32) is called the normal form of equation (5.31) and the function I is called its invariant.

10 CHAPTER 5. ADDITIONAL MATERIAL: LINEAR DIFFERENTIAL EQUATIONS Show that = is an irregular singularity for any constant coefficient linear ODE of any order n unless the equation is y (n) = 0. Explain why this is so with reference to the form of solutions to such equations. 5. Find series expansions around = 0 of two independent solutions to the ODE Show that these series converge for all. d 2 y d dy 2 d y =0. 6. For each of the following equations, state whether the series expansion about = 0 of the general solution contains a logarithm. (a) d2 y +6 dy 2 +9 d 2 + d + 2 y = 0, (b) d2 y d dy d + +3 (c) d2 y +(1 2)dy +( 1)y = 0, d2 d (d) d2 y d dy d +42 y = Verify that y() =1/ is a solution of 2 2 y = 0, (1 ) d2 y + 2(1 2)dy 2y =0. d2 d Use reduction of order to find the general solution. 8. For which choices of the parameter α is the general solution of equation d 2 y d 2 + α 3 dy 3 d y =0 analytic at = 0?

11 Chapter 6 The Hypergeometric Equation 6.1 Introduction We will study some particular linear second-order ordinary differential equations d 2 y d 2 + p()dy + q()y =0, (6.1) d where p and q are analytic in some open set Ω. For any 0 Ω and any c 0 and c 1 C, there is a unique solution of equation (6.1) satisfying y( 0 )=c 0 and y ( 0 )=c 1, which is analytic in a neighbourhood of = 0. Furthermore this solution can be analytically continued along any curve on which p and q are analytic. In general, however, these solutions are branched around the singularities of p and q. Any point at which p or q is singular such that p has at most a simple pole and q has at most a double pole, is called a regular or Fuchsian singularity. The hypergeometric equation is (1 ) d2 y +[c (a + b + 1)]dy ab y =0, (6.2) d2 d where a, b, and c are constants. It is one of the most important equations in mathematics and arises in many applications. The hypergeometric equation is an example of a Fuchsian equation. An ODE is said to be Fuchsian if all its singular points in the extended complex plane are Fuchsian. Many of the great early mathematical analysts including Gauss, Riemann, Fuchs, Kummer, and Klein produced detailed studies of Fuchsian equations. The hypergeometric equation (6.2) has singular points at = 0, = 1, and =. Fuchsian equations with fewer than three singular points can be solved explicitly (see the exercises at the end of this chapter). We will see that any Fuchsian equation with exactly three singular points can be mapped to equation (6.2) by a simple change of variables. Fuchsian equations with four or more singular points are much more difficult to analyse than those with only three. We will consider the global consequences of the branching of solutions of the hypergeometric equation. In particular, we wish to see how a solution transforms as it is analytically continued around any closed curve in the plane which avoids the singularities at = 0, = 1, and =. 6.2 Riemann s equation In this section we will derive the general form of a Fuchsian equation with three (distinct) regular singular points at x 1, x 2, and x 3 (initially we will assume that x 1, x 2, and x 3 are finite.) Such 32

12 CHAPTER 6. THE HYPERGEOMETRIC EQUATION 33 an equation necessarily has the form d 2 u dx 2 + p(x) du (x x 1 )(x x 2 )(x x 3 ) dx + q(x) [(x x 1 )(x x 2 )(x x 3 )] 2 u =0, where p and q are polynomials. Since x = is an ordinary point it follows from condition (5.15) that for x, the coefficient of du/dx must be 2x 1 + O(x 2 ) and the coefficient of u must be O(x 4 ). Hence d 2 u dx 2 + A1 + A 2 + A 3 x x 1 x x 2 x x 3 B1 + + B 2 + B 3 x x 1 x x 2 x x 3 du dx u =0, (6.3) (x x 1 )(x x 2 )(x x 3 ) where A 1 + A 2 + A 3 = 2. The indicial equation for (6.3) at the point x = x i is r 2 +(A i 1)r + B i j=i (x i x j ) =0. Hence if r i and s i are the roots of the indicial equation at x i, then r i + s i r i s i = B i / j=i (x i x j ). Hence equation (6.3) becomes the Riemann equation = 1 A i and + d 2 u 1 dx 2 + r1 s r 2 s r 3 s 3 du x x 1 x x 2 x x 3 dx r1 s 1 (x 1 x 2 )(x 1 x 3 ) + r 2s 2 (x 2 x 3 )(x 2 x 1 ) + r 3s 3 (x 3 x 1 )(x 3 x 2 ) x x 1 x x 2 x x 3 u =0, (6.4) (x x 1 )(x x 2 )(x x 3 ) where r 1 + r 2 + r 3 + s 1 + s 2 + s 3 = 1. We derived equation (6.4) under the assumption that x 1, x 2, and x 3 are finite. However, if we take the limit x 3 inequation (6.4) then we obtain the general form of a Fuchsian equation with regular singular points at x 1, x 2, and. In particular, if we choose x 1 = 0, x 2 = 1, x 3 =, r 1 = 0 and r 2 = 0, then equation (6.4) becomes the hypergeometric equation (6.2), where we have set s 1 =1 c, s 2 = c a b, r 3 = a, s 3 = b, and x =. The general solution of Riemann s equation (6.4) is denoted by x 1 x 2 x 3 P r 1 r 2 r 3 x, s 1 s 2 s 3 where r i and s i are the roots of the indicial equation at the regular singular point x i. So the general solution of the hypergeometric equation (6.2) is denoted by 0 1 P 0 0 a. (6.5) 1 c c a b b We will now show that any Fuchsian equation with three singular points (i.e., Riemann s equation) can be mapped to the hypergeometric equation (6.2). The Möbius transformation x = (x 2 x 3 )(x x 1 ) (x 2 x 1 )(x x 3 ) (6.6)

13 CHAPTER 6. THE HYPERGEOMETRIC EQUATION 34 maps the regular singular points x 1, x 2 and x 3 to 0, 1 and respectively while maintaining the Fuchsian nature of the equation. Let y() = u(x) be the general solution of the transformed equation. Note that x 2 x 3 (x 3 x 1 )(x 1 x 2 ) (x x 1)+O (x x 1 ) 2 ; x x 1, := x 1+ 3 x 1 (x 1 x 2 )(x 2 x 3 ) (x x 2)+O (x x 1 ) 2 ; x x 2, (x 2 x 3 )(x 3 x 1 ) x 1 x 2 (x x 3 ) 1 + O(1) x. It follows that the roots of the indicial equations for equation (6.4) at the singular points x = x 1, x = x 2 and x = x 3 are the same as those for the transformed equation at = 0, = 1, and = respectively. Furthermore, one of the roots of the indicial equation at = 0 can be normalised to 0 by dividing y() by r 1. Similarly one of the roots of the indicial equation at = 1 can be normalised to 0 by dividing y() by( 1) r 2. This shows that the general solution of Riemann s equation (6.4) is x 1 x 2 x P r 1 r 2 r 3 x = r 1 ( 1) r 2 P 0 0 r 1 + r 2 + r 3, s 1 s 2 s 3 s 1 r 1 s 2 r 2 r 1 + r 2 + s 3 which can be written in terms of the solution of the hypergeometric equation as x r1 0 1 x1 x r2 x2 (x 2 x 3 )(x x 1 ) P 0 0 a, (6.7) x x 3 x x 3 (x 2 x 1 )(x x 3 ) 1 c c a b b where a = r 1 + r 2 + r 3, b = r 1 + r 2 + s 3, and c =1+r 1 s Series solutions In this section we will find series expansions of solutions of the hypergeometric equation (6.2)) about its singular points. We begin by finding series expansions for solutions about = 0. On substituting y() = α n n+r into equation (6.2) we obtain 0 = = α n [(n + r)(n + r 1) n+r 1 (n + r)(n + r 1) n+r +c(n + r) n+r 1 (a + b + 1)(n + r) n+r ab n+r ] αn (n + r)(n + r + c 1) n+r 1 α n {(n + r)(n + r + a + b)+ab} n+r = α 0 r(r + c 1) r 1 + αn+1 (n + r + 1)(n + r + c) α n (n + r + a)(n + r + b) n+r Equating the coefficient of r+n to ero gives the recurrence relation α n+1 = (n + r + a)(n + r + b) (n + r + 1)(n + r + c) α n n =0, 1,... (6.8)

14 CHAPTER 6. THE HYPERGEOMETRIC EQUATION 35 For the case r = r 01 = 0 this becomes yielding α n+1 = α 1 = ab 1 c α 0, α 2 = α n = (n + a)(n + b) (n + 1)(n + c) α n n =0, 1,..., (a + 1)(b + 1) a(a + 1)b(b + 1) α 1 = 2 (c + 1) 1 2 c(c + 1), a(a + 1) (a + n 1) b(b + 1) (b + n 1) α 0. (6.9) n!c(c + 1) (c + n 1) So two independent solutions are F (a, b, c; ) and 1 c F (a c +1,b c +1, 2 c; ), where F (a, b, c; ) :=1+ n=1 (a) n (b) n n!(c) n n, and (c) n = Γ(c + n)/γ(c) (i.e.,(c) 0 = 1 and (c) n = c(c + 1) (c + n 1), n =1, 2,...). Hence the hypergeometric series can also be written as F (a, b, c; ) = Γ(c) Γ(a)Γ(b) Γ(a + n)γ(b + n) n. n!γ(c + n) From equation (6.9) we have α n+1 n+1 α n n = (n + a)(n + b) (n + 1)(n + c) as n. So the series converges absolutely for < 1 by the Ratio Test. Example 1 F (1, 1, 2; ) = 1 log(1 + ). Substituting r = 1 c in the recurrence relation (6.8) gives ˆα n+1 = Repeated use of this recurrence relation gives the series y 02 () = 1 c (a c + 1)(b c + 1) !( c + 2) (n + a c + 1)(n + b c + 1) ˆα n. (6.10) (n + 1)(n +2 c) (a c + 1)(a c + 2)(b c + 1)(b c + 2) 2 + 2!( c + 2)( c + 3) where c = 1, 2, 3, 4,... Note that we can get the recurrence relation (6.10) by replacing a with a c + 1, b with b c + 1, and c with 2 c, in the recurrence relation (6.8). This gives y 02 () = 1 c F (a c +1,b c +1, 2 c; )., So y 01 () = α n n = F (a, b, c; ), and y 02 () = 1 c ˆα n n = 1 c F (a c +1,b c +1, 2 c; ).

15 CHAPTER 6. THE HYPERGEOMETRIC EQUATION 36 Exercise 2 Express the general solution of in terms of F. 12(1 ) d2 y +(6 11)dy d2 d + y =0, Series expansions about the other singularities show that two independent solutions near = 1 are y 11 = F (a, b, a + b c + 1; 1 ), (6.11) y 12 = (1 ) c a b F (c a, c b, c a b + 1; 1 ) (6.12) and two independent solutions near = are 6.4 Kummer s solutions y 1 = a F (a, a c +1,a b + 1; 1 ), (6.13) y 2 = b F (b, b c +1,b a + 1; 1 ). (6.14) In section 6.2 we saw that any second order Fuchsian equation with three regular singular points (6.4) can be transformed to the hypergeometric equation (6.2) by first using a Möbius transformation to map the three singular points to 0, 1 and and then replacing u by r 1 ( 1) r 2 y() where r 1 and r 2 are roots of the indicial equations at = 0 and =1respectively. There are exactly six Möbius transformations that map the set {0, 1, } onto itself. These transformations form a group called the anharmonic group and are given in table 6.1. So starting from the hypergeometric equation (6.2) with fixed values of a, b, and c, we can write down six equations of the form (6.4) that arise from the six permutations of (0, 1, ) given by the anharmonic transformations. For each of these six equations there are (generically) two roots r 01, r 02 of the indicial equation at = 0 and two roots r 11, r 12 of the indicial equation at = 1, giving us four factors of the form y() = r 0m ( 1) r 1n u(), m, n =1, 2, such that y again satisfies a hypergeometric equation (6.2), however, the parameters a, b, and c will have changed in general. This gives us a total of 24 solutions to the hypergeometric equation that can be expressed in terms of any given solution, say y 01 () =F (a, b, c; ). Table 6.1: Anharmonic transformations Transformation ζ() ζ(0) ζ(1) ζ() For the hypergeometric equation there are six ways of labelling the points 0, 1, as 1, 2, 3, corresponding to the elements of the anharmonic group. There are two ways of labelling the roots of the indicial equation at =0(i.e. eitherr j = 0 and s j =1 c or s j = 0 and

16 CHAPTER 6. THE HYPERGEOMETRIC EQUATION 37 r j =1 c.) Similarly there are two ways of labelling the roots of the indicial equation at =1 (0 and c a b.) In this way we generate 24 solutions of the hypergeometric equation. We have r1 0 1 P r 1 r 2 r 3 = 1 r2 2 P 0 0 â s 1 s 2 s ĉ ĉ â ˆb ˆb ( 2 3 )( 1 ), ( 2 1 )( 3 ) where â = r 1 + r 2 + r 3, ˆb = r 1 + r 2 + s 3, and ĉ =1+r 1 s 1. Note that this is a relationship between general solutions so we ignore overall multiplicative constants. When we take a limit, e.g., 3, we can rescale the right side with a factor of r 1+r 2 3. Let us follow the above procedure to find three other expressions for y 01 = F (a, b, c; ). This solution is characterised by the fact that it is analytic at = 0, where it has the value 1. If we choose the Möbius transformation to be the identity (i.e., 1 = 0, 2 = 1, 3 =, then in order to get a solution analytic at = 0 we should choose r 1 = 0. The choice r 2 = 0 does not give any new solution. The choice r 2 = c a b gives the solution ( 1) c a b F (c b, c a, c; ). This solution is analytic at = 0. Multiplication by ( 1) c a b gives a solution which is 1 at = 0 (for an appropriate choice of branch.) Apart from the identity, the only anharmonic transformation that maps = 0toitselfis. On making this transformation in the hypergeometric equation and multiplying by 1 one of two powers of (1 ), we again obtain a solution of the hypergeometric equation (with different parameters) which is 1 at = 0. So we have four different expressions for y 01. Using similar reasoning for the other solutions, we find a total of 24 solutions which are naturally grouped into six sets of identical functions. y 01 = F (a, b, c; ) (6.15) = (1 ) c a b F (c a, c b, c; ) (6.16) = (1 ) a F a, c b, c; (6.17) 1 = (1 ) b F b, c a, c; (6.18) 1 y 02 = 1 c F (a c +1,b c +1, 2 c; ) (6.19) = 1 c (1 ) c a b F (1 a, 1 b, 2 c; ) (6.20) = 1 c (1 ) c a 1 F a c +1, 1 b, 2 c; (6.21) 1 = 1 c (1 ) c b 1 F b c +1, 1 a, 2 c; (6.22) 1

17 CHAPTER 6. THE HYPERGEOMETRIC EQUATION 38 y 11 = F (a, b, a + b c + 1; 1 ) (6.23) = 1 c F (a c +1,b c +1,a+ b c + 1; 1 ) (6.24) = a F a, a c +1,a+ b c + 1; 1 (6.25) = b F b, b c +1,a+ b c + 1; 1 (6.26) y 12 = (1 ) c a b F (c a, c b, c a b + 1; 1 ) (6.27) = (1 ) c a b 1 c F (1 a, 1 b, c a b + 1; 1 ) (6.28) = (1 ) c a b a c F 1 a, c a, c a b + 1; 1 (6.29) = (1 ) c a b b c F 1 b, c b, c a b + 1; 1 (6.30) y 1 = a F (a, a c +1,a b +1, 1 ) (6.31) 1 c a b = a F (1 b, c b, a b +1, 1 ) (6.32) = a 1 = a 1 a F a, c b, a b + 1; 1 1 c a 1 F a c +1, 1 b, a b + 1; 1 1 (6.33) (6.34) y 2 = b F (b, b c +1,b a +1, 1 ) (6.35) 1 c a b = b F (1 a, c a, b a +1, 1 ) (6.36) = b 1 = b 1 a F b, c a, b a + 1; 1 1 c a 1 F b c +1, 1 a, b a + 1; 1 1 (6.37) (6.38) 6.5 Integral representations The hypergeometric equation (6.2) can be written as Ly = 0, where L = (1 ) 2 2 +[c (a + b + 1)] ab. From the identity L t b 1 (1 t) c b 1 (1 t) a = a t b (1 t) c b (1 t) a 1 t it follows that γ t b 1 (1 t) c b 1 (1 t) a dt (6.39)

18 CHAPTER 6. THE HYPERGEOMETRIC EQUATION 39 solves equation (6.2) provided the integral converges and the path of integration γ in equation (6.39) is chosen such that either γ is closed on the Riemann surface of the integrand or terminates at points where t b (1 t) c b (1 t) a 1 vanishes. In particular, if (c) > (b) > 0, then y() := 1 0 t b 1 (1 t) c b 1 (1 t) a dt (6.40) is a solution of the hypergeometric equation. Note that y is analytic in a neighbourhood of = 0, so y() =κf (a, b, c; ), for some constant κ. Setting = 0 and noting that F (a, b, c; 0) = 1 gives Hence Theorem 3 Exercise 4 Show that κ = y(0) = F (a, b, c; ) = 1 0 t b 1 (1 t) c b 1 dt = Γ(c) Γ(b)Γ(c b) F (a, b, c; 1) = 1 0 Γ(b)Γ(c b). Γ(c) t b 1 (1 t) c b 1 (1 t) a dt. (6.41) Γ(c)Γ(c a b) Γ(c a)γ(c b). (6.42) 6.6 Monodromy of the hypergeometric equation In this section we will consider the result of analytically continuing a solution y of the hypergeometric equation around any closed curve in the complex plane that does not pass through 0, 1, or. In general, if we start at some point 0 on γ and analytically continue the solution along γ back to 0, we will not return to the same value due to branching at = 0, = 1, and =. Note that we are free to deform γ so long as it does not cross these singular points. Hence, to analyse the analytic continuation of y along any closed curve, we need only consider its analytic continuation around = 0 and = 1. Analytic continuation around = is equivalent to analytic continuation around = 0 and = 1. We will denote by γ 0 and γ 1 the circles of radius (where 0 <<1) traversed in the positive (anti-clockwise) direction, with centres = 0 and = 1 respectively. We begin by considering analytic continuation around = 0. For this purpose, it is convenient to use the basis of solutions y 01 () = F (a, b, c; ), y 02 () = 1 c F (a c +1,b c +1, 2 c; ). Recall that F is analytic in a neighbourhood of = 0, so y 01 is unchanged on analytic continuation around γ 0. However, on analytic continuation around γ 0, the original value of the factor 1 c is multiplied e 2πic. We summarise this by writing γ 0 : y01 y 02 y01 e 2πic y 02. Similarly, near = 1 it is convenient to use the basis of solutions y 11 () = F (a, b, a + b c + 1; 1 ), y 12 () = (1 ) c a b F (c a, c b, c a b + 1; 1 ).

19 CHAPTER 6. THE HYPERGEOMETRIC EQUATION 40 The circle γ 1 can be parameterised as =1+e iθ,whereθ ranges from 0 to 2π. The only branching in the continuation of y 11 and y 12 around γ 1 comes from the factor (1 ) c a b.this leads to γ 1 : y11 y 12 y11 e 2πi(c a b) y 12. In order to track the behaviour of a particular solution around a curve looping around both = 0 and = 1, we need to choose a single pair of basis solutions and relate all others to this pair. A suitable pair is y 01 and y 11, which are chosen because they are both branched around only one of the two singular points (it remains to be seen that these two solutions are independent). Since y 11 and y 12 form a basis, there exist constants α and β such that This is equivalent to y 01 = αy 11 ()+βy 12 (). (6.43) F (a, b, c; ) = αf (a, b, a + b c + 1; 1 )+ β(1 ) c a b F (c a, c b, c a b + 1; 1 ). (6.44) In the case (c a b) > 0, we substitue = 1 in equation (6.44) to give F (a, b, c; 1) = αf (a, b, a + b c + 1; 0). Using F (a, b, c; 0) = 1 and equation (6.42), we find α = Γ(c)Γ(c a b) Γ(c a)γ(c b). From equation (6.43), we see that after y 01 is continued along γ 1,wehave γ 1 : y 01 αy 11 + βe 2πi(c a b) y 12 =e 2πi(c a b) y 01 + α 1 e 2πi(c a b) y 11. where Similarly, it can be shown that γ = Hence, after analytic continuation along γ 0,wehave Now using the identity we have αγ = So finally, if we choose y 11 = γy 01 + δy 02, (6.45) Γ(a + b c + 1)Γ(1 c) Γ(b c + 1)Γ(a c + 1). (6.46) γ 0 : y 11 γ(1 e 2πic )y 01 +e 2πic y 11. (6.47) Γ(1 )Γ() = π sin π, sin π(c a) sinπ(c b) sin πc sin π(c a b) = (1 e2iπ(c a) )(1 e 2iπ(b c) ) e 2iπb (1 e 2iπ(c a b) )(1 e 2iπc ). y A () :=y 01 () and y B = α 1 e2iπ(c a b) 1 e 2iπ(c a) y 11 ()

20 CHAPTER 6. THE HYPERGEOMETRIC EQUATION 41 and let Y := (y A,y B ), then where and γ 0 : Y YM 0, γ 1 : Y YM 1, M 0 := M 1 := 1 e 2πib (1 e 2iπ(b c) ) 0 e 2πic e 2πi(c a b) 0 1 e 2πi(c a) 1 are the monodromy matrices around = 0 and =1respectively. Note that every closed loop that avoids = 0, = 1, and = can be divided into successive loops γ 0 and γ 1, which may be traversed in the positive or negative directions. After analytic continuation along such a curve, Y returns to the value for some integers p 1,...,p n and q 1,...,q n. Further Exercises Y YM, M = M p 1 0 Mq 1 1 Mp 2 0 Mq 2 1 Mpn 0 Mqn 1, 1. Show that and d n d F (a, b, c; ) =abf (a +1,b+1,c+ 1; ), d c d n F (a, b, c; ) =(a) n(b) n F (a + n, b + n, c + n; ). (c) n 2. Show that F (1, 1/2, 3/2; 2 )= 1 tan Derive some of Kummer s solutions. 4. Express the general solution of the equation in terms of F. x(x 2) d2 u dx x 1 du dx u =0 5. Consider the equation a() d2 y d 2 + b()dy d where a, b, and c are polynomials. + c()y =0, (6.48) (a) Show that if the only singular points of equation (6.48) are regular singular points at = 0 and =, then the equation is Euler s equation where k 1 and k 2 are constants. 2 d2 y d 2 + k 1 dy d + k 2y =0, (6.49) (b) Find the indicial equation for Euler s equation (6.49) for the singularity at = 0. Call the roots r 1 and r 2.

21 CHAPTER 6. THE HYPERGEOMETRIC EQUATION 42 (c) Find the general solution of Euler s equation in the case where r 1 = r 2. (d) In the case when the roots of the indicial equation are both equal (r 1 = r 2 = r, say), let y 1 be any non-ero solution of Euler s equation. Substitute y() =u()y 1 () into Euler s equation and find an ODE for u. (e) Hence find the general solution to Euler s equation in the case r 1 = r (a) Show that the only second order linear homogeneous ODE with rational coefficients that has regular singularities at = a and = b (a = b) and no singularity at = (i.e., = is an ordinary point) is d 2 y d µ ( a)( b) where µ and ν are constants. (b) Solve the equation d 2 y d2 ( 1) dy d + ν ( a) 2 ( b) 2 y =0, dy d ( 1) 2 y =0 by using a fractional linear (i.e. Möbius) transformation to map the regular singular points at = 0 and = 1 to 0 and respectively and solving the resulting Euler equation. 7. Show that the hypergeometric equation can be written as where ϑ = d d. (ϑ + a)(ϑ + b)y = ϑ(ϑ + c 1)y, 8. Derive equations (6.45) and (6.46) and hence show that after analyic continuation around γ 0, the solution y 11 transforms as in equation (6.47). 9. Find the monodromy matrix with respect to the basis of solutions Y := (y A,y B ) corresponding to analytic continuation in the positive direction around the square of side length 4 centred at the origin with sides parallel to the real and imaginary axes.