0387986995Basic TheoryC

Using this result, it can be shown that (FS) also satisfies the system

Step 3. Upon applying Exercise XIII-9 to (E), the convergence of t%i(x, y) is proved.

XIII-11. Complete the proof of Theorem XIII-7-6 by verifying rigorously statements (a), (b), and (c) in the proof.

XIII-12. Show that the series

is a formal solution of the differential equaton d2-xZ -xy = 0, where r is the Gamma-

function.

XIII-13. Show that the differential equation d2-xZ - xy= 0 has a unique solution

O(x) such that (1) b(x) is entire in x and (2) ¢(x)exp [3x3/21 admits the formal

series p(x) exp [x3I2J as its asymptotic exansion as x -oc in the sector I arg xj <

r, where p(x) is givenby (FS).

Remark. Ai(x) = 2(--X) is called the Airy function (cf. [AS, p. 446), [Wasl, pp.

v/Fr

124-1261, and [01, pp. 392-394]).

XIII-14. Using the same notations in Exercises XIII-12 and XIII-13, show that if

xp (2ril

w = el3 J then ¢(w'ix) and Q(wx) are two solutions of equation (S). Also,

(i) derive asymptotic expansions of ¢(w-lx) and m(wx),

(ii) show that {Q(x),y5(w-lx)}, {0(w-1x),O(wx)}, and {d(wx),Q(x)} are three fundamental sets of solutions of (S),

is equal to the 2 x 2 identity matrix,

(iv) using (iii), show that cl = -w-1.

XIII-15. Show that if O(x, A) is an eigenfunction of the eigenvalue problem

(ii)all negative odd integers are eigenvalues of (EP) and there is no other eigen- value,

(iii) for every non-negative integer n, Hn(x) = (-1)ne=2 Un (e-y2) is a polyno-

Remark. The polynomials Hn(x) are called the Hermit polynomials (cf. [AS, p.

775] and [01, p. 49]).

XIII-16. Construct Green's function of the boundary-value problem

00

Hint. See §VI-4.

n-1 h

XIII-17. Consider a differential operator C[y] _ dxn + 1: an-h(x) h , where

m=-mj

equal to zero identically, while mj = -oo if aj(x) is zero identically. In the (X, Y)- plane, consider the points Pj = (j, mj) (j = 1,... , n). Construct a convex polygon

II whose vertices are (0, 0), pi, p2, ... , p, such that each pk is one of those points

Pj, and that all other points Pj are situated below the polygon. Set po = (0, 0) and Pk = (ak, Pk) (k = 1) ... , s), where an = n. Denote by pk the slope of the

segment k k+1 (k = 0, ... , s - 1). Then, Po > Pl > P2 > ... > p,-,. Assume that pk>-1 (k<vo)and pk<-1 (k>vo). Show that

(i) the differential equation £[y] = 0 has n - at,,, linearly independent formal so-

Me

lutions of the form ye(x) = xe" E Req(x)(logx)q (e = a,,. + 1,... , n), where

q=o

Rrq(x) E C[[x-1]], b E C, and the Mt are non-negative integers,

(ii) if k < vo, then the differential equation £[y] = 0 has ak+1 - ak linearly

me

independent formal solutions of the form yr = e^<(x)x6, E Rrq(x)(log x)q (e = q-0

1 + ok,... , nk+1), where At(x) = Arxl+1 b + terms of lower degree, At E C,

A, 96 0, bt E C, M, are non-negative integers, and Req E C[[x-lIP]], p being a positive integer.

Hint. Compare the polygon in this problem with the Newton polygon at x=0 defined in §XIII-7.

XIII-18. Consider a differential operator C[yl = x9+1 dx + 1(x)y', where q is a

00

positive integer and f2(x) _ > xlf2, with Hi being n x n constant matrices. As in

t=o

§V-4, the operator C can be represented by the matrix

Here, 0, is the nr x oo zero matrix and I,,,, is the oc x oo identity matrix. Let

Qo=So+No and A=S+N

be the S-N decomposition of 0o and that of A in the sense of §V-3. Also, let

Po, P1, P2, ... , Pm, ... be n x n constants matrices and let AO be an n x n diagonal matrix such that det Po 0 0 and that

where

P1

P2

P0

P=

P.

Note that At are eigenvalues of So. Hence, A, are eigenvalues of 00. Show that

(a)the matrix S represents a multiplication operation: y -+ o(x)y for an n x n matrix o(x) = So + O(x) whose entries are formal power series in x,

(b)the matrix N represents a differential operator Lo[yM = x4+1 fy + v(x)y for an n x n matrix v(x) = No +O(x) whose entries are formal power series in x,

(c)C[f, = £o[yl + o(x)y and Go[o(x)yj = a(x)Go[yl,

+00

(d) if we set P(x) _ F xmP,,,, then P(x)-1o(x)P(x) = A0,

m=0

(e)if we set /C[uZ = P(x)-'C[P(x)uZ and Ko[u] = P(x)-1Go[P(x)ul, then K[u"j =

Ko[ul + Aou,

(f)if we set

(cf. [HKS]).

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