Asymptotics at regular singular points

Transcription

1 Ma 7, 2011 Asmptotics at regular singular points Paul Garrett garrett/ Introduction 1. Examples 2. Regular singular points 3. Regular singular points at infinit 4. Examples reprise 5. Appendix: dinar points 6. Appendix: Euler-Cauch equations 7. Appendix: Abel s theem on differentiation of convergent power series Bibliograph Differential equations [1] x 2 u + bxu + cu = 0 with constants b, c have eas-to-understand solutions on 0, + : linear combinations of x α, x β f α, β solutions of the indicial equation XX 1 + bx + c = 0 Therefe, we imagine that a differential equation of the fm x 2 u + xbxu + cxu = 0 with b, c analtic near 0 has solutions asmptotic, as x 0 +, to solutions of the differential equation x 2 u + b0xu + c0u = 0 obtained b freezing the coefficients bx, cx of the iginal at x = 0 +. That is, solutions of the variable-coefficient equation should be asmptotic to x α f solutions α to the indicial equation XX 1 + b0x + c0 = 0. An equation of that fm, with b, c analtic near 0, is said to have a regular singular point at 0. Discussion below explains the behavi of solutions to such equations. 1. Examples We give two useful examples from the non-euclidean geometr on the upper half-plane. Recall that the SL 2 R-invariant [2] Laplacian on the upper half-plane H is [3] H = 2 2 x [1] These differential equations are examples of Euler-tpe Cauch-tpe equations, which are well understood. See the appendix. [2] a b As usual, SL 2 R acts b linear fractional transfmations z = az + b on H. In particular, there are c d cz + d 1 t t 0 real translations z = z + t f t R, and positive real dilations / z = tz f t > 0. t [3] It is not trivial to verif that this differential operat is SL 2 R invariant. Better, this operat is obtained b computing in codinates the image of the Casimir operat f SL 2 R. We accept the outcome f the present discussion. 1

2 Paul Garrett: Asmptotics at regular singular points Ma 7, 2011 [1.1] Translation-equivariant eigenfunctions We ask f H -eigenfunctions fz of the special fm fx + i = e 2πix u That is, such an eigenfunction is equivariant under translations: fz + t = e 2πix+t u = e 2πit e 2πix u = e 2πit fz with t R and z H The eigenfunction condition is the partial differential equation H λ e 2πix u = 0 Since the dependence on x is completel specified, this partial differential equation simplifies to the dinar differential equation [4] 2 u 4π λ u = 0 The point = 0 is not an dinar point f this equation, because in the fm u 4π 2 + λ 2 u = 0 the coefficient of u has a pole at 0. But = 0 is a regular singular point, because that pole is of der at most 2. Thus, following the idea to freeze 2 u + bu + c to 2 u + b0u + c0u, the indicial equation of the frozen equation is XX 1 λ = 0 Expressing λ as λ = ss 1, the roots of the indicial equation are s, 1 s. The frozen equation has distinct solutions s and 1 s f s 1 2. Thus, we could hope that solutions would have asmptotics as 0+ beginning u = A s 1 + O + B 1 s 1 + O as 0 + Indeed, this is the case, as we see below. It seems me difficult to obtain the asmptotics at 0 + from integral representations of solutions of the differential equation. [1.1.1] Remark: As we discuss below, 2 u 4π λu = 0 has an irregular singular point at +, so other methods are needed to obtain asmptotics f solutions as +. [1.1.2] Remark: Up to choices of nmalizations, the function u above, depending on the spectral parameter λ s, is called a Whittaker function Bessel function, and the enjo an enmous literature. One point here is to have direct access to their properties, as examples of simple general phenomena. [1.2] Dilation-equivariant eigenfunctions F complex β, we can consider a dilation-equivariance condition of H -eigenfunctions f. Thus, ft z = t β fz f t > 0 and z H x x fx + i = f + i = β f + i [4] This equation is a tpe of Bessel equation, with solutions which are K-tpe and I-tpe Bessel functions. 2

3 Paul Garrett: Asmptotics at regular singular points Ma 7, 2011 With ux = fx + i, the eigenfunction condition is H λ β u x = 0 To expand this, first dispatch one labious computation: 2 β 2 u x = β β 1 u x β x 2 u x = ββ 1 β 2 u x 2β β 1 x 2 u x + β 2x 3 u x + β x2 4 u x = β 4 x 2 u + 2 β 3 x1 βu + ββ 1 β 2 u Thus, keeping in mind that the arguments of u, u, u are x/, H λ β u x = β u + β 2 x 2 u + 2 β 1 x1 βu + ββ 1 β u λu = β 1 + x 2u + 2 x 1 βu + ββ 1 λ u Thus, dividing through b β and setting = 1, the eigenfunction condition H λ β ux/ = 0 becomes an dinar differential equation in x: letting λ β = ββ 1, 1 + x 2 u + 2x1 βu + λ β λu = 0 F this equation, x = 0 is an dinar point, so solutions admit convergent power series expansions there, and their behavi is clear. Behavi as x + can be explained b verifing that + is a regular singular point, b converting to codinates z = 1/x at infinit. Let ux = v1/x. Then Putting z = 1/x, this is u x = 1 x 2 v 1/x and u x = 1 x 4 v 1/x + 2 x 3 v 1/x u = z 2 v and u = z 4 v + 2z 3 v with u = ux, v = vz, z = 1/x The equation becomes z 2 z 4 v + 2z 3 v + 2 z 1 β z 2 v + λ β λv = 0 which is z 2 + 1z 2 v + 2zz 2 + βv + λ β λv = 0 z 2 v + z 2z2 + β z v + λ β λ z v = 0 The coefficients bz = 2z 2 + β/z and cz = λ β λ/z are analtic near z = 0, so this equation has a regular singular point at z = 0. The indicial equation is 0 = XX 1 + b0x + c0 = XX 1 + 2βX + λ β λ 3

4 Paul Garrett: Asmptotics at regular singular points Ma 7, 2011 Writing λ = λ s = ss 1, the roots of the indicial equation are 1 2 β ± 1 2 β2 + ss 1 ββ 1 = 1 2 β ± w 1 2 = β + w β + 1 w Thus, solution to the differential equation should be asmptotic to z β+s and z β+1 s as z 0, that is, to x β s and x β 1+s as x +. We will see that this is crect. [1.3] An irregular singular point Returning to the translation-equivariant eigenfunctions on H, we check that = + is not an dinar point n a regular singular point. Given u 4π 2 + λ 2 u = 0 again let ux = v1/x and put z = 1/x, obtaining z 4 v + 2z 3 v 4π 2 + λz 2 v = 0 4π z 2 v + 2zv 2 z 2 + λ v = 0 Since the coefficient of v has a pole at z = 0, this equation falls outside the present discussion. Instead, a different freezing idea succeeds: letting + freezes the iginal equation at +, giving a constantcoefficient equation u 4π 2 u = 0 with easil-understood solutions e ±2π. Happil the solutions to the iginal equation do have asmptotics with main terms e ±2π. Further details and proofs will be given later, in a discussion of irregular singular points. 2. Regular singular points A homogeneous dinar differential equation of the fm x 2 u + xbxu + cxu = 0 with b, c analtic near 0 is said to have a regular singular point [5] at 0. Similarl, x x o 2 u + x x o bxu + cxu = 0 with b, c analtic near x o has a regular singular point at x o. Obviousl it suffices to treat x o = 0, and is notationall convenient. The coefficients in an expansion of the fm ux = x α a n x n with a 0 0, α C [5] I must have learned about regular singular points first from [Ahlfs 1966]. While the latter mentions several attributions b name, it has no bibliograph whatsoever. Few current complex analsis textbooks in English discuss regular singular points. [Whittaker-Watson 1926] has extensive bibliographic notes, and treats man useful examples. 4

5 Paul Garrett: Asmptotics at regular singular points Ma 7, 2011 are determined recursivel, but we see below that this recursion succeeds onl when α satisfies the indicial equation αα 1 + b0α + c0 = 0 Further, when the two roots α, α of the indicial equation have a relation n + α α = 0 f 0 < n Z, the recursion f α ma fail, although the recursion f α will succeed. These conditions are easil discovered, as in the following discussion. The convergence of the recursivel defined series is imptant both because it produces a genuine function, and because it can be differentiated termwise, b Abel s theem. [2.1] The recursion The equation is x α+2 n + αn + α 1a n x n 2 + bx x α+1 n + αa n x n 1 + cx x α a n x n = 0 Dividing through b x α and grouping, n + αn + α 1a n x n + bx n + αa n x n + cx a n x n = 0 The vanishing of the sum of coefficients of x 0, and a 0 0, give the indicial equation. The coefficients a n with n > 0 are obtained recursivel, from the expected [ n + αn + α 1 + b0n + α + c0 ] an = in terms of a 0, a 1,..., a n 1 The coefficient of a n simplifies b invoking the indicial equation and the fact that the sum of the two roots α, α is 1 b0: n + αn + α 1 + b0n + α + c0 = nn + 2α 1 + b0 = nn + α α That is, nn + α α a n = in terms of a 0, a 1,..., a n 1 f n > 0 Since n > 0, the recursion can fail onl when n + α α = 0 f some 0 < n Z [2.2] Convergence To complete the proof of existence, we prove convergence. Let A, M 1 be large enough so that bx = n 0 b n x n with b n A M n cx = n 0 c n x n with c n A M n Inductivel, suppose that a l CM l, with a constant C 1 to be determined in the following. Then n nn+α α a n A n i+α M i CM n n i +A M i CM n i AM n C n 1 nn + 1 +n α +n 2 i=1 5 i=1

6 Paul Garrett: Asmptotics at regular singular points Ma 7, 2011 Dividing through b n n + α α, this is This motivates the choice a n AM n n 1 n α + 2 C 2 n + α α C n α + 2 sup 1 n Z 2 n + α α which gives a n ACM n, and a positive radius of convergence. [2.2.1] Remark: In fact, in light of the Monodrom theem, this estimate is far from best possible, but it is infeasible and pointless to tr f sharper estimates. Let ux = v1/x. Then Putting z = 1/x, this is 3. Regular singular points at infinit u x = 1 x 2 v 1/x and u x = 1 x 4 v 1/x + 2 x 3 v 1/x u = z 2 v and u = z 4 v + 2z 3 v with u = ux, v = vz, z = 1/x A differential equation u + pxu + qxu = 0 becomes z 4 v + 2z 3 v + px z 2 v + qxv = 0 z 2 v + z 2 p1/z z The point z = 0 is a regular singular point when the coefficients v + q1/z z 2 v = 0 2 p1/z z q1/z z 2 are analtic at 0. That is, z = 0 is a regular singular point when p, q have expansions of the fms p 1 z q 1 z = p 1 z + p 2 z = q 2 z 2 + q 3 z , equivalentl px = p 1 x + p 2 x qx = q 2 x 2 + q 3 x

7 Paul Garrett: Asmptotics at regular singular points Ma 7, Examples reprise We return to the two earlier examples from non-euclidean geometr on the upper half-plane. [4.1] Translation-equivariant eigenfunctions We ask f eigenfunctions fz of the special fm fx + i = e 2πix u which simplifies to the dinar differential equation 2 u 4π λ u = 0 with regular singular point at = 0. The indicial equation is XX 1 λ = 0 With λ = ss 1, the roots of the indicial equation are s, 1 s. B now we know that, unless s 1 s is an integer, the equation has solutions of the fm u s = s a l l l 0 u 1 s = 1 s b l l with coefficients a l and b l determined b the natural recursions. We emphasize that these power series have positive radius of convergence, so certainl give asmptotics as 0 +. Further, convergent series can be differentiated termwise, b Abel s theem. We execute a few steps of the recursion f the coefficients f s. The equation simplifies to l 0l + sl + s 1a l l 4π λ l 0 l 0 a l l = 0 ll + 2s 1 a l = 4π 2 a l 2 f l 1 with a 1 = 0 b convention, and a 0 = 1. Thus, the odd-degree terms are all 0, and Similarl, replacing s b 1 s, u s = s 1 + 4π s + 4π s s +... u 1 s = 1 s 1 + 4π s + 4π s 45 2s +... F Res 1 2, one of these solutions is obviousl asmptoticall larger than the other. F Res = 1 2, the are the same size, so some cancellation can occur. Write s = iν, so 1 s = 1 2 iν, and rewrite the expansions in those codinates: u 1 2 +iν = 1 2 +iν 1 + π2 2 u 1 2 iν = 1 2 iν 1 + iν + π iν 22 + iν π2 2 1 iν + π iν 22 iν

8 Paul Garrett: Asmptotics at regular singular points Ma 7, 2011 Then, f example, visibl u 1 2 +iν + u 1 2 iν = coslog + O 3 2 u 1 2 +iν u 1 2 iν = sinlog + O 3 2 Further, behavi of the higher terms as functions of ν is clear. [4.2] Dilation-equivariant eigenfunctions Returning to dilation-equivariant eigenfunctions fx + i = β ux/, we have the differential equation 1 + x 2 u + 2x1 βu + λ β λu = 0 with λ β = ββ 1 It was verified earlier that + is a regular singular point, b converting to codinates z = 1/x at infinit: with ux = v1/x and z = 1/x, the equation becomes The indicial equation is z 2 v + z 2z2 + β z v + λ β λ z v = 0 0 = XX 1 + b0x + c0 = XX 1 + 2βX + λ β λ With λ = ss 1 the roots of the indicial equation are β + s, β + 1 s. F s 1 s not an integer, there are solutions asmptotic to z β+s and z β+1 s as z 0 +. That is, solutions to 1 + x 2 u + 2x1 βu λ λ β u = 0 are asmptotic to x β s and x β 1+s as x +. We execute a few steps of the recursion f β = 0. Let vz = z s a l z l l 0 Using the equation becomes 1 z = 1 z2 + z 4 z f z near 0 l + sl + s 1a l z l 2z2 z 2 l + sa l z l λ + 1 z 2 a l z l = l 0 l 0 ll + 2s 1a l = 2 λ l 2 + sa l 2 + l 4 + sa l l 0 Thus, with a l = 0 f l < 0, and with a 0 = 1, all the odd-degree coefficients are 0, and the expansions are something like v = z s 1 + u = x s λs s z2 + 2 λ s 2 λs s + s z λs 1 2 λ 2 λs s x s s + s x still with λ = ss 1. These series have positive radius of convergence. as z 0 + as x + 8

9 Paul Garrett: Asmptotics at regular singular points Ma 7, Appendix: dinar points The following discussion is well-known, although the convergence discussion is often omitted. This is the simpler case extended b the discussion of the regular singular points. [5.1] Ordinar points A homogeneous dinar differential equation of the fm u + bxu + cxu = 0 with b, c analtic near 0 is said to have an dinar point at 0. The coefficients in a proposed expansion of the fm ux = a n x n with a 0 0 are determined recursivel from a 0 and a 1, as follows. The equation is nn 1a n x n 2 + bx na n x n 1 + cx a n x n = 0 nn 1a n x n 2 + bx n 1a n 1 x n 2 + cx a n 2 x n 2 = 0 The coefficients a n with n 2 are obtained recursivel, from the expected nn 1 a n = in terms of a 0, a 1,..., a n 1 To complete the proof of existence, we prove convergence. Take A, M 1 large enough so that bx = n 0 b n x n with b n A M n cx = n 0 c n x n with c n A M n Inductivel, suppose that a l CM l, with a constant C 1 to be determined in the following. Then n n nn 1 a n A n im i 1 CM n i + A M i 2 CM n i AM n 1 C n 1 nn i=1 Dividing through b nn 1, this is This motivates taking i=2 a n AM n 1 C n 1 n2 + 3n 2 nn 1 n 2 + 3n 2 C A sup 2 n Z nn 1 + n 1 which gives a n CM n. In particular, f arbitrar a 0 and a 1 the resulting power series has a positive radius of convergence. In particular, these series can be differentiated termwise, b Abel s theem. 9

10 [5.2] Ordinar points at infinit Let ux = v1/x and z = 1/x. Then Paul Garrett: Asmptotics at regular singular points Ma 7, 2011 u x = 1 x 2 v 1/x and u x = 1 x 4 v 1/x + 2 x 3 v 1/x u = z 2 v and u = z 4 v + 2z 3 v with u = ux, v = vz, z = 1/x A differential equation u + bxu + cxu = 0 becomes z 4 v + 2z 3 v + px z 2 v + qxv = 0 2z p 1 q 1 v + z z 2 v + z z 4 v = 0 The point z = 0 is an dinar point when the coefficient of v is analtic and vanishes to first der at 0, and the coefficient of v is analtic. That is, z = 0 is an dinar point when p, q have expansions at infinit of the fm p 1 z = 2z + p 2 z 2 + p 3 z 3... q 1 z = q 4 z 4 + q 5 z [5.3] Not-quite-dinar points Consider a differential equation with coefficients having poles of at most first der at 0: u + bx x u + cx x u = 0 with b, c analtic at 0. The coefficients in a proposed expansion of the fm ux = a n x n with a 0 0 are determined recursivel as follows. The equation is nn 1a n x n 2 + bx na n x n 2 + cx a n x n 1 = 0 nn 1a n x n 2 + bx na n x n 2 + cx a n 1 x n 2 = 0 We expect to determine the coefficients a n with n 2 recursivel, from nn 1 + b0n an = in terms of a 0, a 1,..., a n 1 f n 1 F b0 not a non-positive integer, the recursion succeeds, and a 0 determines all the other coefficients a n. F b0 = 0, so that the coefficient of v has no pole, the relation from the coefficient of x 1, b0a 1 + c0 a 0 = 0 10

11 Paul Garrett: Asmptotics at regular singular points Ma 7, 2011 implies that either c0 = 0 and the coefficient of v has no pole, returning us to the dinar-point case, a 0 = 0, and there is no non-zero solution of this fm. F b0 a negative integer l, the recursion f a l gives a l the coefficient 0, and imposes a non-trivial relation on the pri coefficients a n. To complete the proof of existence, we prove convergence, assuming b0 is not a non-positive integer. Dividing through b a constant if necessar, we can take M 1 large enough so that bx = n 0 b n x n with b n M n cx = n 0 c n x n with c n M n Inductivel, suppose that a l CM l, with a constant C 1 to be determined in the following. Then nn 1 + b0n an = n n im i 1 CM n i + i=1 Dividing through b nn 1 + b0n, this is This motivates taking n i=1 a n M n 1 C n 1 n 2 + 3n nn 1 + b0n C sup 2 n Z n 2 + 3n nn 1 + b0n M i 1 CM n i M n 1 C n 1 nn which gives a n CM n. In particular, f arbitrar a 0 the resulting power series has a positive radius of convergence. F example, the series can be differentiated termwise, b Abel s theem. 6. Appendix: Euler-Cauch equations The differential operat x d dx has readil-understood eigenfunctions on 0, + : from xu = λu we have u /u = λ/x, then log u = λ log x + C, and Differential operats u = const x λ f x > 0 x 2 d2 dx 2 + bx d + c with constants, b, c dx x k dk dx k + c k 1 dk 1 k 1x dx k c 1x d dx + c 0 where the power of x matches the der of differentiation can be understood as composites of operats of the fm x d α. These differential operats are of Euler tpe, Cauch tpe, Euler-Cauch tpe. In dx the der-two case, x d dx α x d dx β = x 2 d2 d + 1 α βx dx2 dx + αβ That is, given coefficients b, c C, the parameters α, β are solutions of the indicial equation XX 1 + bx + c = 0 + n Then the differential equation x 2 u + bxu + cu = 0 11

12 Paul Garrett: Asmptotics at regular singular points Ma 7, 2011 has solutions x α and x β. When the roots α, β coincide, a second solution f x > 0 is x α log x. This can be verified b computation, we can use a me general principle, as follows. F brevit, let D = x d dx. Suppose D αu = 0. Viewing u as a function of the spectral parameter α as well as the phsical variable x, differentiating with respect to α gives That is, Then 0 = D αu α = u + D α u α D α u α = u 0 D α 2 u α = D αu = 0 That is, u/ α is a solution of D α 2 v = 0 and not a solution of D αv = 0. In particular, u α xα = log x x α The same discussion shows that x x α k+1 log xk x α = 0 while x x α klog xk x α 0 7. Appendix: Abel s theem on power series [7.0.1] Theem: Abel Let fz = n 0 c n z z o n be a power series in one real complex variable z. Suppose that the series is absolutel convergent f z z o < r. Then the function given b fz is differentiable f z z < r, and the derivative is given b the absolutel convergent series nc n z n 1 n 0 [7.0.2] Collar: B repeated differentiation, f k z = n 0 nn 1... n k + 1 c n z n k In particular, f k z o = kk 1... k k + 1 c k = k! c k, so the power series coefficients of fz are uniquel determined. /// Proof: Without loss of generalit, z o = 0. Fix 0 < ρ < r, and ζ < ρ, z < r. The obvious candidate f the derivative is gz = nc n z n 1 n 0 Then fz fζ z ζ gζ = n 1 c n z n ζ n z ζ nζ n 1 12

13 Paul Garrett: Asmptotics at regular singular points Ma 7, 2011 F n = 1, the expression in the parentheses is 1. F n > 1, it is z n 1 + z n 2 ζ + z n 3 ζ zζ n 2 + ζ n 1 nζ n 1 = z n 1 ζ n 1 +z n 2 ζ ζ n 1 +z n 3 ζ 2 ζ n z 2 ζ n 3 ζ n 1 +zζ n 2 ζ n 1 +ζ n 1 ζ n 1 = z ζ [ z n ζ n 2 + ζz n ζ n ζ n 3 z + ζ + ζ n ] n 2 = z ζ k + 1 z n 2 k ζ k F z and ζ both smaller than ρ, the latter sum is dominated b k=0 n 2 nn 1 z ζ ρ < n 2 z ζ ρ n 2 2 Thus, fz fζ z ζ gζ z ζ c n n 2 ρ n 2 n 2 Since ρ < r the latter series converges absolutel, so the left-hand side goes to 0 as z ζ. /// Bibliograph: [Ahlfs 1966], L. Ahlfs, Complex Analsis, McGraw-Hill, [Whittaker-Watson 1927] E.T. Whittaker, G.N. Watson, A Course of Modern Analsis, Cambridge Universit Press, 1927 man subsequent editions and printings. 13