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## Material Information- Title:
- Confluent cases of second order linear differential equations with four singular points
- Creator:
- Cundiff, Joyce Coleman, 1927-
- Publication Date:
- 1961
- Copyright Date:
- 1961
- Language:
- English
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- v, 117, 1 leaves : ; 28 cm.
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Differential equations ( jstor ) Equation roots ( jstor ) Equations ( jstor ) Exponents ( jstor ) Graduates ( jstor ) Integers ( jstor ) Mathematics ( jstor ) Recurrence relations ( jstor ) Taylor series ( jstor ) Differential equations, Linear ( lcsh ) Dissertations, Academic -- Mathematics -- UF Mathematics thesis Ph. D - Genre:
- bibliography ( marcgt )
non-fiction ( marcgt )
## Notes- Thesis:
- Thesis - University of Florida.
- Bibliography:
- Bibliography: leaf 116.
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- Manuscript copy.
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- Vita.
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CONFLUENT CASES OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH FOUR SINGULAR POINTS By JOYCE COLEMAN CUNDIFF A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA June. 1961 ACKNOWLEDGMENTS There are many people who have contributed to the preparation of this dissertation. First, the writer wishes to express her deep appreciation to the chairman of her supervisory committee, Professor Russell W. Cowan, for suggesting the topic of study and for his guidance and helpfulness during the research. She is also grateful to the members of her committee, Professors W.R. Hutcherson, J. T. Moore, W. P. Morse, C. B. Smith and A. Sobczyk. Warm thanks are due Professors F. W. Kokomoor and J. E. Maxfield for their encouragement and interest. The patience, cooperation and persistence of her typist, Sandra Fife, is gratefully recognized. She is humbly grateful to her husband for the less tangible but real assistance received by his encouragement, patience and enthusiasm. To Mother, ever uplifting and inspiring TABLE OF CONTENTS ACKNOWLEDGMENTS . . . . . . . . . . i DEDICATION . . . . . . . ... . .ill INTRODUCTION . . . . . . . ... .. .. 1 CHAPTER I Second Order Linear Differential Equations 1.1 Definitions . . . . . . . . . 2 1.2 Formal Solutions of Differential Equations . 3 1.3 Convergence of Formal Solutions . . . .. 10 1.4 Second Solution in Case where the Difference of the Exponents is an Integer. Example . .24 1.5 Solutions Valid for Large Values of Izi . . 35 1.6 Irregular Singular Points and Confluence . 38 CHAPTER II Confluent Differential Equations 2.1 Differential Equation with Four Singular Points . . . . . . . . . . 39 2.2 Differential Equation with Singular Points at z = 0,l, ,-o . . . . . . 44 2.3 Scheffe's Criteria applied to Confluent Case 50 2.4 Solutions of Confluent Case after Normalizing 59 2.5 Factored Solution W ,. Hypergeometric Solution W .. .. . . .. 67 2.6 Notation and Proofs involving 2F1(a,b;c;z) . 72 2.7 Recurrence Relations, nth Derivative, Sum and Product Formulas using W .k ......... 83 CL s CHAPTER III Related Differential Equations and Classification 3.1 Related Equations of Mathematical Physics Derived from an Equation having Four Singularities . . . . . .... ... . 90 (a) Lame's . . . . . . . . . 93 (b) Legendre's . . . . . . . . 94 (c) Jacobi's . . . . . . . ... .95 (d) Gegenbauer's. Derivation of Geganbauer's Equation . . . . . . . ... 96 (e) Laguerre's . . . . . . . 100 (f) Equation with Solution being Incomplete Gamma Function . . . . . . 101 (g) Gauss's . . . . . . . . .101 (h) Kummer's . . . . . . ... 102 (i) Whittaker's . . . . ...... .102 3.2 Classification of Differential Equations having Four Singular Points when the Exponent Difference is 2 ..........103 LIST OF REFERENCES .................. 116 BIOGRAPHICAL SKETCH . . . . .... . 117 INTRODUCTION This is a discussion of ordinary second order linear differential equations with four singular points and the con- fluent cases which occur by permitting their singular points to coalesce. Chapter I presents a second order differential equation, formal solutions with proof of convergence, solu- tions valid for large values of Izi, irregular singular points and confluence of singular points. Chapter II de- velops the general form for a differential equation having four regular singular points and then considers solutions of the confluent equation with singularities at z = O,1,o, o. Using the normalized form of the confluent equation we find a hypergeometric solution with recurrence relations, nth derivative,and sum and product formulas. In chapter III we derive some of the related equations of mathematical physics (see table of contents) from the general equation of chapter II and present a classification of this type equation when the exponent difference is i. Chapter I elaborates on pertinent material in [8, chapter 10] Except for the classification in chapter III, see [2, p. 499], the remaining work is believed to be entirely original. The numbers in brackets refer to the list of references. CHAPTER I Second Order Linear Differential Equations 1.1 Definitions. Let the standard form of an ordinary second order linear differential equation be represented as (1-1) d2u + p(z) du + q(z) u = 0, dz2 dz where p(z) and q(z) are assumed to be functions of z analytic [8, p. 85] in some domain D except at a finite number of poles. Any point in D at which p(z) and q(z) are both analytic will be called an ordinary point of the equation; any other points of D will be called singular points of the equation. If there exists a point z=c of D such that, when p(z) and q(z) or both have poles at z=c, the poles are of such order that (z-c)p(z) and (z-c)2q(z) are analytic, then z=c is called a regular singular point for the differential equation. Any poles of p(z) and q(z) that are not of this nature are called irregular singular points. If z=c is a regular singular point, the equation may be written (1-2) (z-c) du + (z-c)P(z-c) du + Q(z-c) u = 0 , dz2 dz where P(z-c) and Q(z-c) are analytic at z=c. Expanding these functions in a Taylor's Series about z=c we have 00 P(z-c) = c) c) + ... + pn(z-c)n + = Zpn(z-c)n n=o and Q(z-c) = q + q1(z-c) + ... + qn(-c)n + ... = q(z-c)n n=o where p ,p1,...,q9,q1,...are constants. These series con- verge in the domain Dc formed by a circle of radius r (center c) and its interior, where r is chosen sufficiently small so that c is the only singular point of the equation which is in DC. c Thus p(z) =P(-c) and q(z) a(z-c) z-c (z-c)2 in equation (1-1). 1.2 Formal Solutions of Differential Equations. Let us assume a formal solution of the differential equation to be u = (z-c)a 1 + Zan(z-c)n] n=1 where a,al,a2,... are constants to be determined. Assuming that the term-by-term differentiation and multi- plication of the series are valid, we have U' = c(Z-C)a-' 00oo [i+ an(z-c)n] + n=l CO (z-c) ann(z-c)n-i n=l = (z-c)-1 [a + c(a+n)an(z-c)n] n=1 and u" = (a-1)(z-c)-2, +I (a+n)a (z-c)n n=l + (z-c)-' Zn(a+n)a (z-c)n- n=i co = (z-c)a-2c (a-1) + (a+n)(a+n-l)a (z-c) n=l Now substituting these expressions for u,u' and u" into equation (1-2) we obtain cc (1-5) (z-c)C a(ca-1) + a (a+n)(a+n-1)(z-c)n n=l + (z-c) P(z-c)[a + za (a+n)(z-c)n n=l + (z-c)a Q(z-c)[ 1 + an(z-c)n1 n=l = 0. Next substitute the series for P(z-c) and Q(z-c) into equation (1-3): 5 (z-c) a(a-l) + YIa (u+n) (a+n-l) (z -c)nl + (z-c)a[ + p(z-c) + p1(z-cc) +...+ pn(z-c)" 00 +...]-.[ + an(a+n)(z-c)n n=l + (Z-c)aC[q + ql(z-c) + q2(z-c)' ...+ qn(Z-C)n 00 +7 an(Z-C)n = 0. n=1 Now equate to zero the coefficients of the successive powers of (z-c): (z-c)": a(a-1) + poa + qo = 0, (z-C)Q+l (1-4) a 2 + (po-1) a + qo = 0. : aa(a+l) + alpo(a+l) + apl + alqo + q = 0, a [2 + a + po(a+l) + q9o + ap, + q = 0, a (a+1)2-a-1 + p (a+l) + qo] + ap1 + q = O, 1) (-)( ) + + a + 0. + (p -1)(a+l) + qo] + op, + q 0. (z-c)a+2: a2(a+2)(a+l) + poa2(a+2) + plal(a+l) + P2a + qa2 + qlal + q2 = 0, a2[(a+2)(a+l) + po(a+2) + q] + a[Pl(a+l) + ql] + p2a + q2 = 0, a2[(a+2)2 a 2 + p (a+2) + + a lP(a+l) + q1 gq] + P2a + q = 0, a2 (a+2)' + (po-l)(a+2) + q + a [Pl(a+l) + q] And in general we can write: (z-c)+n: a (a+n)(a+n-l) + pa (a- + p a + q a n on n n-I + z q a- m=l + pa + q2 = 0. n-1 +n) + ma n(a+n-m) m=1 + q = 0, -n an (a+n)2 a n + po(a+n) + q 0 n-I + an-m Pm(a+n-m) + qm] + Pna + q = 0, m=l (1-5) a [(c+n)2 + (p-l1)(a+n) + q 0 n-1 + Z aMp (a+n-m) + q.] + ap + q = 0. m=l The first of these equations which is obtained by equating to zero the coefficients of the lowest power of (z-c) is called the indicial equation. This equation determines two values of a (which may or may not be dis- tinct). If z=c had been an irregular point, the indicial equation would have been at most of the first degree. To see this, suppose P(z-c) has a simple pole at z=c and expand it in a Taylor's series. P(z-c) = p_l(z-c)-l+ pO + P(z-c)+...+ Pn(z-c)n+. Pn(z-c)" p_ / 0, n=-l and as before 00 Q(z-c) qo + q1(z-c)+...+ qn(z-c)n +... = qn(z-c)n n=o Now multiply according to the following scheme [which is equivalent to substituting the series for P(z-c) and Q(z-c) into equation (1-3)] and obtain the indicial equation: q + ql(z-c) +... p 1+po(z-c) +... (z-c)2 u = (z-c)" [ + al(z-c) +...] S (-c) 1 a + (c+l)a,(z-c) +... u" = (z-c)-2[a(u-l)+(u+l)ual(z-c) +...] The indicial equation is (Z-c)a-1 p_la = 0. But a 5 0, and p_, 0 or z=c would not be a simple pole which contradicts our assumption. Or if we assume that Q(z-c) has a simple pole and expand it in a Taylor's series: at z=c Q(z-c) = q_l(z-c)-~ + qo + ql(z-c)+...+qn(z-c)n +... 03 = q(z-c)n, n=-l q_, O, and P(z-c) = po + p (z-c) + P2(z-c)2 +...+pn(z-c)n +.. p pn(z-c)n n=o Now multiplying similar to the previously mentioned scheme: q -(z-c)-'+ q +... u = (z-c)[ 1 + a (z-c) +... po(z-c)+ pi(z-c)2+... u' = (z-c)a-1 a+(a+l)al(z-c)+... (z-c)2 u" = (z-c)a-2 a(a) +...] The indicial equation is (z-c) -i: q-, = 0, which is a contradiction of our assumption. If both P(z-c) and Q(z-c) have simple poles at z=c, then q_i(z-c)- + qo+... u = (z-c)al + ai(z-c) +... pl(Z-C) + u ,a (z -1[ ] p_-+p.(z-c) +... u (z-c)a + (a+l)ai(z-c)+...] (z-c)2 u" = (z-c)a-2 a(a-l)+(a+l)aa (z-c)+... The indicial equation is (z-c)"-i: pla + q_ = 0, which is first degree in a. Hence, if z=c is an irregular singular point, the indicial equation is at most of first degree. Now the roots of the indicial equation are called the exponents of the differential equation at the point z=c. Returning to our indicial equation which we shall call F(a), (1-6) F(a) = a2 + (p -l)a + qo = 0, let a=?l and a=?2 be the roots. Each value for a deter- mines, in order, a set of coefficients a ,a2,..., thus pro- ducing two distinct series solutions, provided that 1 2 is not an integer (zero included). It will be shown later that when the difference is a non-zero integer then our proof for convergence of the series fails and if the difference is zero, the two solutions are obviously the same. 1.3 Convergence of Formal Solutions. We now pre- sent a proof by induction that these series solutions con- verge. Assuming that the exponents ?i and 2 are not equal, choose them so that Re(-l)aRe(X2), and let A1i-2 = s. Now since the exponents are a = 1 ,2 we can write (i-o), (1-7) F(a) = (a-1l)(a-z), then for a = 1 +n, F(A +n) = (A?+n-?A)(?l+n-h2) = n(n+s) . Let P(z-c) = p + p1(z-c) + p2(z-c)2 +..+ pn(z-c)n +. where p(n)(c) Pn n according to Taylor's expansion. By means of repeated differentiation [4, p. 70] (n) n_ n P(z) dz (z) 2= 1 "J Zn+l C If we let the curve C be a circle with center at the origin, z=rei1, then dz=rieie de. Let M1 be the maximum value of IP(z)l on C. Then n_ P(z)dz n j IP(z)l ldzl n+1 n+1 2Ti C zn+1 i C Izn+l n!2 M1 rie id8 f n+1.l 0O I 27T M1n! d M1n! 27T < -- I 0 -_ [9_ ] 21rr n 2rn [0 0 o M1n! r Thus and Similarly for qn: Iql s. 2r where M2 is the maximum value of IQ(z)I on C. We next show that -n I Pn + qnl M13r-n INPn + q 1 s Ix1l-IPnl+lqnl -n -n -n/ n SIN "M1r-n + Mr- = r (I.11 -M+M, = r M , where MS = I IM1 + M2. (z) n pn I Mr-n ^*l)5r i i * --n Choose M to be larger than M1,M2, and M3 and also M 1. Then (1-8) I|pn < Mr-n; nIq ( Mr-n; IlP + qn < Mr-n for n = 1,2,3,.... Now with the aid of equation (1-6), we can write equation (1-4) as a F(a+l) + ap + ql = 0 and -(apl + q ) a1 = SF(a+l) Now let a = X then iall l1 pl+ql < Mr M IF(?h+1)l ll+sl r since ll+sI > 1. That this is true can be seen by letting 1 = a + pi and X = 7 + 6i. Then s = (a+pi)-(y+6i) = (a-7)+i(3-6) and 11+s = V(l+a-y) + (p-6) . If P-6 = 0, or P = 6, then a / y for we have assumed from the beginning that A1 and Ag are not equal. By our choice of hi and X2 a > 7 and 1 + a y > 1 and |1+sl > 1. If a = y, then B 6 andVl + (p-6)2 > 1 or Il+sI > 1. We have now shown that an has a maximum for n = 1. Now we assume that lanl < Mnr- n = 1,2,...,m-l. From equations (1-5) and (1-6) we can write n-1 an F(a+n) + a n-pm(+nl-m) + qm] + pa + qn 0, m=l and letting a = \, n-1 ,a n-mPm +n-m) + qm] n qn m=1l a = n F( +n) Now replace n by m and replace m by t. Then m-1 am-t [Pt(h1n-t) + qt]- m q (1-9) lam = t= F(l +m) m-i m-1 Slam-t p Ip1t +qtl+I lPm+qml+ 7 (m-t)lam-t JiPt t=1 t=l m mn+s m-1 m-1 Z lamt IMr-t + Mr-O + (m-t)ia i '- Mr- t=1 t=1 \ ------------1,-- ----- m I1+smr For the first summation: m-1 m-1 (1-10) a -iM = M a -tIr t=1 t=1 SM [IaMi 1r-l+jam-2 Ir2+...+Ia r-m+2+ ar-m+l1 < M[Mm-1 -m m-2 -m M M r + M r +...+ M2r-m + Mr-m] < Mr -[M-l + Mm-2 +...+ M2 + M < Mr- (m-1)M m-1 < (m-l)r-M . For the second summation: m-1 (1-11) (M-t)lam-t Mr-t t=1 = Ml(m-1) F, r ni-1 -m m-2 -m < M (m-1)M r +(m-2)Mm-2 r +. < Mr-m[ (m-)Mm-l+(m-2)Mm-l+...+ < Mmr-m[(m-l)+(m-2)+...+ 2 + 1 < M'r-m M (m-1) ..+ Mr-] Mm-1I (m-l)r M + Mr- + Mmr -m (m-l) |aJ < ------- 2 -- m2 1+sm-1 M-r mm -mrm -m m -m -1 Mr M -r M + Mr + Mr (m-1) < m 2_2+SM - m'|+sm- mn'Il+smn Thus lam-lir- 1 +(M-2)la -21r- 2 +...+(l)lallr-nl+ll min -mm T ,nm- m m -n mr M r M + Mr + Mr (m-1) < 2 m' l+sm- | But ll+sm-l ) >1, for letting s = 1-A2 = a+3i (y+6i): si+ iT m+s m+s| 1+ !1 = m I = m' m m m and Im+s = (a-y+m)2 + (--6)' > m since 1i and ?2 are distinct. Thus Im+sl > 1 m and a I ( + rl) < r- M for m > 1. m 2m Therefore, by induction, anl < Mnr-n for all values of n. If the values of the coefficients corresponding to the exponent y2 be al,a ,... we want to show, by a proof similar to the previous induction proof, that la'l MnKr- , where K is the upper bound of Il-sl- Ii- sl-1 11- -1 ... Let us begin by showing that such a bound exists for s not a positive integer. If 0 < Is! < 1, then 1 1ls' ' H1-ls I , J = 1,2,3,..., J 1,2,3,... Proof: Let s = a+-i, (a,p real); then 0 < Isi = a2V+2 < 1. 1- = 1- = (1- )2 + () =-- 1 + a '+ 5'i -~i--i-i j 2e j J' = V l- +a N ), S1- Qa+f3 2 1 -JaV+02 = 1 IsI = 1-IsI J If Is) > 1, then 1 s Is 1-^ 1- Proof: Let Isl = t + 0, where t is the largest integer in Isl and 0 g 6 < 1. J2 j j2 S- 1- (+ ) + J- 1 + S- - Jj3 1 1l-Is I 11- s1 We desire I- s to have a minimum value larger than zero, where j=l,2,3,..., t-n,..., t-l,t,t+l,...t+ .... (n=l,2,3,..., t-l;=l,2,3, ...). Difficulty could arise, however, when j=t and if 0 = 0. Then the absolute value would be zero. First we show it+ad t+e 1-0 1- -- > and 1- -- t-n t t+tl -TT analytically, where n=0,1,2,..., t-l; t=1,2,3,..., 0 < e < 1. t+8 8 1- t-n n=0,1,2,..., t-l; 0 < 8 < 1. t-n-t-0 Q t-n ' n+O e nt + Ot O 8t n , nt -nG, true for n=0. If n/O, then t > -8. t+"t~ t+ .-e t+'t (t+l)(t- ) d( t+0) 1-8 t- t=1, 1-8 t+l ' 1-8 t+l (1-0)(t+d) , _ t+4-8t-8e , _ t+ , 1 t+0 , _ 1, which is true. 2,3,...; O < 8 < i. Therefore, +0 0 t+ 1-0 (1-12) 1- t+e and 1- -t+ t-n t+' ttT where n=0,1,2,..., t-1; ,=1,2,3,...' 0 < ( < 1. Inserting Isl for t+0 and j for t-n and t+L in (1-12) we see that, for Isl > 1 and 0 < 0 < 1, either i-- or 1- t- Jt j I t+1 depending on the values of 0 and t. Now consider the case where 0=0 (Isl=t) and j=t = a2+ep P30. We wish to find a 6 such that if s =a+pi, then 1s 1 I js +6 1- a+3i 6a +P2 + 5 -Va2 1- IVVa21'32 -a pi | 161 (Va'+P a) + p 62 , a'+B 2a Va2+132 + a'+p2 > 52 , 2t"-2at 62 , 2t(t-a) a 56 However, since Isl=t > 1 and t =Wa2+8 )> a, if we have given any value of s (Isl > 1, pBO), then a value of 6 can be determined such that 0 < 6 < 1. Referring to (1-12) with 0 replaced by 6, we have for Isl 1, 0 < 6 < 1, that t+6 6 a 1t+6 1-6 1- T and 1- I F- an t I t+t I ' where n=0,l,2..., t-1; k=1,2,3,... Summarizing our results we have: If 0 < Isl < 1, then 1- 1 1-is . For Isl > 1, Isl / an integer, (a) -if 0 < 0 < 1, then one of the following s I Ii = >- i _ 6 1 1 + 1 = 1 +6 1 f (b) if 9 = O, then one of the following t+1 where J=1,2,3,...; IsI=t+6 (where t is the largest integer in Isl and 0 0 ( 1); 6 chosen such that 0 < 6 < 1 and 5's 2t(t-2). We have thus proved the theorem that given any s (s not an integer) an upper bound x of the sequence -1 j=1,2,3,..., is the maximum of 1 t t+l t t+l 1;TT (6/0o); 19/ 0) ; 1- si (eo); 1- ( 6o), 6 1- 6 where 6 is chosen such that 62 2t(t-a) and 0 < 6 < 1. Now we present an inductive proof that la'l < Mn K r-n n Again using equation (1-6) with a now replaced by 22+ n, we have (1-13) F(h2+n) = (h2+n-h?1)(2 +n-2) = n(n-s) If -i is replaced by Nt, the inequalities of (1-8) will still be true. However, they will also be found valid if M 1 is chosen larger than as well as M1,M2,M and M 1. Let this be done. Thus, IpnI < Mr-n; Iq < Mr-n I2pn+q nI < Mr-n Also, since M > and M g 1, we have the additional inequalities which will be utilized later in the proof, MIc > 1, (Mc)"- )> 1, where m=1,2,3,... When a = h2 and n=l in equation (1-4), we can write a' = (h2+l)2+(po-l)(h,2+l)+q and with equation (1-6) this becomes a' = 2p 1 F(N2+1) Taking the absolute value of each side and using (1-13) with n=l, we obtain the inequality aI p +q < I Mr- IF(A +1)1 1l-sl Next, assume n=1,2,3,... m-l, and show that a' M r . Using (1-9) with A2 and a' replacing 'A and a respectively, m-i m- t[Pt (2- + q t] 2pm la'll t=l a'i =- m F(A2+m) Tm- m-i lam- lPt 2+qlt pm+q (m-t) a-tl Pt t=l t=l mim-si m-1 S .Mrm-t t + Mr + t=l m-1 i (m-t)'la tl' 'Mr- t=l m2 1-sm1 I lan'i n n -n a' < M K r Similar to (1-10) and (1-11) with Jam_t < (MK)lr-m M > 1, m-1 am a -tI Mr- ( (m-1)(Mc)M-1 Mr- t=1 < (m-l)r-mMKm- , and m-1 S(m-t)a I Mr- t=1 Thus, (m-l)r-WMmK'- + Mr-m la'l < m m m-1 -m < (m-1)(MK) mMr < r (m-l)r-mMmKm- 2 + M r -m (m-1) KIn- -1 m ll-sm I -im m-- -nm r-i -m i -mm iii mr- MmK -m-_r- M -1 + Mr-m + Mmr-m m (m-l) r- m I -sm- Now since (Mc)m- > 1, Mr-m Mm-m m-1 Mr Mr K = Mr-m(l, Mm-1 m-I) = Mr (1-M K ) = Mr 1-(MKc) i] is negative. Thus, Ia'l < m mr- M K -+ r- M (m-1) m- m2lsm-1 m |1-sm | -mm m-l1r+l Sr M (m+) 2mI 1-sm1 < r-mMmCm (ml) < r-mMmVm 2m for m > 1. Therefore, Ja'l < Mr ,n n for all values of n. The radius of convergence for the power series [4, p. 80], for the Mnr-(z-c)n is n=l Mn -n lim M n -o, Mn+ r-n-l n ->oo M r r r limn - M M . nj o That is, the series is absolutely convergent within the circle Iz-cl = .. Since laI Mr-", the series an(z-c)n also must converge within the circle Iz-cl n=i r = and is, therefore, uniformly convergent in the region Iz-cl ( [L, p. 951. Similarly an'(z-c)n converges uniformly within the n=l region Iz-cl < r .1- We have thus obtained two formal solutions u,(z) = (z-c) 1 + a (z-c)n n=l and u'(z) = (z-c) 1 + an(z-c)n] n=1 which are uniformly convergent series of analytic functions when Iz-c( < rM- and Iz-cl < rM1-l respectively, pro- vided that arg (z-c) is restricted in such a manner that the series are single-valued. Consequently, the formal substitution of these series into the differential equation is justified for -\ 2 not a positive integer. These solutions are valid in the vicinity of a regular singular point. 1.4 Second Solution in the Case where the Difference of the Exponents is an Integer. We now derive a second solution in the case where the difference of the exponents is an integer. When l-'2 = s is a positive integer or zero, the solution u2(z) may break down or coincide with u1(z). Try the change of variable u = ul(z)-. and substitute this into (z-c)2 u" + (z-c)P(z-c) u' + Q(z-c) u = 0 to determine the equation for ). U = U1 0, u' = u .)' + u 0, u" = u. *)" + 2u; 0' + u" 0. Substituting in we have (z-c)2 u1()" + 2uj' + u4] + (z-c)P(z-c)(u )' + u ) + Q(z-c) ul*) = 0, (z-c)2u 1n" + 2u (z-c)2 + (z-c)P(z-c) -uI 0 + (z-c)2 u1 + (z-c)P(z-c) u1 + Q(z-c) u1 0 = 0. Since u (z) (z-c)2 is a solution u"1 + (z-c)P(z-c) u! + Q(z-c) u1 = 0, and (z-c u 2(-c)2 +2 + (Z-C)P(z-c).uIj = 0. Dividing by u. we have (z-c)2 "+ [2 (C) + (z-c)P(Z-c) 0' = 0. To find a general solution let 0' = Y. (z-c) Y' + 2 u (z-c) + (z-c)P(z-c) Y = 0, (z-c)2 d 2 (z-c) + (z-c)P(z-c) Y = 0. Separating variables and integrating, -Y + I dYY f[ S+ u 1 In Y = In B 21n -2 Y =Bu- 1 P(z-c dz = In B, S(z-c) u (z- )- dz u/ P z-c) SP(z-c/ dz Tz-7 cy Now since Y = 0', O [B u-2 e - dz ] d FZ --d7 where P(z-c) = po+ pl(z-c) + pp(z-c)'+...+ p(z-c)+.. dz - P(z-c) dz P [p(z-c)1 + PI + P2(z-c) +.'. dz = poln(z-c) + pl(z-c) + p, 2(z +.. Hence, 0= A+B = A+B -2 [-p ln(z-c) pl(z-c)-...-pn(z-c)n-...] uI e u 2 (z-c)-Po e ( 2 +BC)- dz 00 ,P(z-c)n n n n=l = A + B dz . But since oo u,(z) = (z-c) 1 + an(z-c) n -1 ) = A + B (z-c) -2 g(z)dz, (z-c)n ri SPn n [ i-2 n=l where g(z) = 1 + a (z-c)n e n=1 A and B are arbitrary constants and g(z) is analytic throughout the interior of any circle C whose center is z=c, which does not contain any singularities of P(z-c) or -1 of (z-c) 1u(z), nor any zeros of the latter. g(c) = 1. Since g(z) is analytic in the interior of C, it can be expanded in a Taylor's series about its center z=c. Let 00 g(z) = 1 + g(z-c)nl n=1 We already have a2 + (p0-l) a + q = 0 (1-6) with roots a = Al,'2 and Al-A2 = s. The sum of the roots l+A 2 = -po, so we can write -p -2?, = 2 +~2-1-2 = -1 = -s-1. Then 00 (1-14) = A + B 1f + (zc)n (z-c)-1 dz n=1 = A + B [ s-1 (z-c)-s-ldz +J gn(-c)n--ldz n=l gs(z-c) 1dz CO + n= g ( c)n-+dz n=s+l (z-c ) sSjI- nS n=1 -n s-n (zC)n-s (z-c) + gs In(z-c) + in- (z-c)n-s n=s+l The general solution, analytic in C except at z=c, is u = ul(z).* = A ui(z) + B[ u(z).gs.ln(z-c) + (z)1 where (z-c)-r s-I n= c) s-n n=l 1 u(z) = u,(z)[- 1 Vs + n=s+l n- n-s and u1(Z) = (Z-C) 11 (ZC)n-] Co +Xa(zc)n] n=1 + = A + B[ Rewriting, we have u(z) = (z-c) 1 00 S-1 + a(z-c) (z-c)f- g-s(z-c)n- n=l n=l (z)n-s (Z-C) Cn + n--s n=s+l = (Z-c) 2 uo la(-c) n=1l s-1 n (z-c)n s s-n n=l (z-c)n] + n-- n=s+l = (z-c) - = (ZC) 2[ + h(Z-C)nJ n=l where h are constants: n Sg1 al s-a + and hn(z-c)n = - n=l an(z-c) + n=l 1 + an(z-c)n n=l s-1 g-n )n (z-c) n=- n=1 00 Zs gn Ln-s n=s+l (z c)nI (z-c) When s = 0, we can write (1-14) in the form CO S= A + B 1 + g (z-c)n] (z-c) dz n=l A + B [ (z-c)r dz + gn (z-c)ndz n=1 =A + B ln(z-c) + g(z-c) n=l Therefore, when 1 = -2, the second solution is u = ul(z).* = A ul(z) + B ul(z) ln(z-c) + n (z-c)n+1] n=l Thus it is seen that when ,1 -2 is an integer, the second solution involves a logarithm, except when gs = 0. A practical way of obtaining this second solution is to first obtain the solution u1(z), and then determine the coefficients in a function 00 Z ^.2+n U1(z) = bn(z-c) n=o by substituting u = ul(z)ln(z-c) + 5 (z) in the equation and equating to zero the coefficients of the various powers of z-c in the resulting expression. 31 Example: Find the solutions of the equation S 1 2 u" +- u' m u = 0 z regular near z = 0 [8, p. 201]. Equation (1-2) with c = 0 becomes (1-15) z2u" + zP(z) u' + Q(z) u = 0. 2 Thus, if we multiply our given equation by z we have 2 2 2 z u + zu' m z u = 0, 2 2 where P(z) = 1 and Q(z) = -m z Assuming 00oo a n u = z az , n=o then 2 2 a a+2n a+2n+2 -m z u = az +...+a z +a z +... a /0. 0 2n 2n+2, o z u' = a aza-l+...+(a+2n+2)a2n+2 a+2n+l+... z u2 = a a(a-1)z -2+...+(a+2n+2)(a+2n+l)a 2n2+2n+... 0 2n+2 . Equating the coefficients of powers of z-c to zero, z : a(a-l) + a = 0, 2 a = 0 and a = 0,0. a+2n+2 a 2n+(a+2n+2)a2n++((+2n+2)(a+2n+l)a2n+2 = 0. Z2 -m2+2n 2 m a2n+2 2 2n+2 (a+2n+2)2 2 m a2n 2n+2 22(n+ )2 If n = 0, 2 m ao a2 -2 2 212 If n = 1, 2 m a2 a4 2(22 2 (2)2 4 m ao 2 (2!)2 If n = 2, 2 m a4 a6 22() 6 m a 2 6(3!)2 2n an n 2"n !2 a n = 2 2 n !n 1) n=l,2,3, . Therefore, if we arbitrarily choose a = 1, o 2n 2n S2n(n U1 = n=o For the second solution, let u = ullnz + n=o and substitute this into (1-15). b z n u' = u'ln z + ul u" = u'ln z + 2u ( - 0) +- + nb z- n=l 00 + Zn(n-l)bnzn-2 n=2 00 00 n n ul'z21n z + 2u 2z u + n(n-l)bnn + u zln z + ul + nbnz n=2 n=l 00 2 2 2 n+2 - um z in z m b z2 n1 n n=o = 0. Since the coefficient of In z is 2 2 2 z u, + ulz m z u1 = O, we have left 00 00 2uz + n2bz mbn b zn+2 1 n= n= n=1 n=o = 0, 2uz + (n2 m2z2) b z = 0. n=o Now, taking the derivative of our first solution ul, we obtain 2n 2n-1 2n m z n=A 2n(n!)2 n=1 u z 2 2n 2n 4n m2nz z 22n(n!)2 n=1 n=o 2 =0. (n -m z )b z = 0. Equating to zero the coefficients of like powers of z, we have z: b 0 = 0, b = arbitrary constant. 1 b 0. z : b =0. 1 2n z : 4n m2 + n 2b 2n + 4n b2n 2 n(n!)2 b 2n 1 2 2n-2 2n n 4 - mb = 2n-2 2n n m 22 (n!)2 , n=1,2,jj, . Choose bo = 0. Then for n = 1, m2b b_ o 4 If n =2, 2 2 m -m 22(1!)2 22(1!)2 b -1 bm 2m4 b4 =22 [- b2 4 2 2 4 2 (2!) 2 24(1!)2 24 2 2 4 -m b4 4 -m 2 (2!) If n = 3, 1 b6 6 3 (1 + ). mb4 2 3m6 4 2 (3!)2 j so that -1 m6 (1 3m6 1 322 (T(2! 2 26(3!)2 b = -m i + 1+ 6 26(5)2 2 3 Finally, 2n 2n ( 1 1 m n -m 1 1\ mHn 2n 22(n)2 2 n2 1 1 1 H 1 + 1 + 1 +...+ . n 2 3 n For n odd, n 2 2 z : nb -mb = O, nbn n-2 2 b b n odd, n 2 n-2 n since bI = 0. Therefore, the second solution is Co 2n 2n Sm H z u(z) = u1ln z 2n 2 (n!2 n= 22n(n!)2 n=1 1.5 Solutions Valid for Large Values of Izl. Let us now consider solutions valid for large values of |z|. To do 1 this we let z = in equation (1-1). If the solution of the zi transformed equation is valid for sufficiently small values of I1zl, then the original equation is said to be valid for large values of izl. If the point z1 = 0 is an ordinary (or regular or irregular) point of the transformed equation, then the point at infinity is said to be an ordinary (or regular or irregular) point of the original equation. We begin with the change of variable z in the zi equation d2+ p(z) du + dz dz q(z) u = 0. 1 z z Z1 1 z = Z 1 z' dzl 2 dz. z du du dz du dz dzI dz z dzI du 2 du du dz 2 3 2 2 dz z dz1 z dzI dz 2 2 du 3 du 4 du 2z + z . dz dz+ dz2 1 1 Substituting in (1-1) we obtain 4 d u 3 2 / 11 du + z + 2z z p ~q- u = 0. 1 2 1 1 + dz 1 dz z 1 1 4 Dividing by z4 d u 2 1 du --1 1 dz 2u+ p +l zq u 0. dz 2 1 z 2dz 1 z 1 1 1 1 (1-1) 37 Now z = 0 (i.e. z = m) will be an ordinary point when du the coefficients of and u are both analytic. This re- dz. 1 quires p and q to have the following forms: 1 2 3 p = 2z + Az + A3a + . 1 4 5 6 q (1) = B4z + B5zl + B6zl +.', B / 0. Expanding about z = o: A A2 p(z) + + +...3 w + ~- " , = 0 wil z z B4 B B6 z) + +- +.. z z z .1 be a regular singular B4 / 0. point when 2 1p and q 2l Zi) 1 \ 1 1- are analytic at z = 0. 21 z2 1 1 This means that p and q must be: p1 z11) = q 1 = C11z + C2z2 + C321 +... 2 3 4 D z2 + D3a3 + D4z1 +... Expanding about z = c: C C C C + + +. P(z) = z' -2 + -3 z z q( D D3 D q(z) = z +- + + z z z C / 0. D2 / 0. C, /0. D / 0. .* Further, it might be helpful to know for what values of t, m and n z = o will be a regular point in the following equation having polynomial coefficients (po / 0 and 4, m, n non-negative integers). (poz + plzt' +...+ pd) u" + (q zm + qzm +...+ qm) u + (r zn + r z +...+ r ) u = 0. Dividing by the coefficient of u", q m m-- r)o n-- n-~ -1 u" + z-+ ELz +... u' + z + F z +... = O. PO O We must have m-t = -1 or 4 = m+l and n-i = -2, 4 = n+2. Therefore, if z = o is a regular singular point, then 4 = m+l = n+2. 1.6 Irregular Singular Points and Confluence. We have already shown in section 1.2 that near an irregular singular point a second order differential equation cannot have two solutions because the indicial equation is at most of the first degree; there may be one or no solutions of this form near this point. If a differential equation A is obtained from another equation B by making two or more singularities of equation B tend to coincidence, such a limiting process is called con- fluence. Equation A is called a confluent form of equation B. CHAPTER II Confluent Differential Equations 2.1 Differential Equation with Four Singular Points. We shall now develop a general form for a second order linear differential equation which has every point except a,, a2, a3 and a4 as ordinary points (z = being an ordi- nary point). These four points are to be regular singular points and let their exponents be ar, pr at ar (r=1,2,3,4). Then the form of our equation will have to be d2u + r + H(z) du+ + + J(z) u = , z -i ---a dz2 z-a)2 z-a) r dz r=l r=1 -ar where kr, r', mr (r=1,2,3,4) are constants to be determined, and H(z), J(z) are polynomials in z. Let u = ho(z-a)' + hl(z-al) +l+... h0 / 0, then 4 M Z+2 z-] + J(z) u = ho(z-al) + h (z-al) +... 4 Z r- + H(z) u'= ho(z-a )-+... z-ar0 r=l 1 u" = hoX(?-l)(z-a )-2+... The indicial equation for z = al is (z-a )-2: ho(A-1) + hok 1 + h o = 0, o o 1 o 1 ?2 + (k-1l)? + 1 = 0. Then so that a1 + pi = 1-k , k = 1 a1 oi' and ali = .1i Similarly for the other regular points, so that (2-1) d2u + dz2 r=l l-ar-Pr + H( 1 du z-ar I dz + +arr 2 + J(z) r~l (Zar)2 z-a L z-a r u = 0. 1 If z = o (or zi = 0, z = ) is to be an ordinary point, zi 2z-z p(z) and z4q(z) should be analytic at z = , 2 1 1 1 1 or - p( ) and -4- q( -) should be analytic at z = 0 1 z 1 z 1 1 1 (section 1.5). Now ( 21 1 Sz z 2 1 1 21 z l-oai -6(. r r 1 H 1 Z 1 z 2 z r=1 r 1 1 2 1 2 zp 1 Z1 41-a r-0r 1 z zl (-arzi) 3 H( r=l 1 z z 1 -Z1 1 = i -^ (- i H - 21 rr r z 1 2 J/ = r---i--I^ ^T 2 4_, 2 2 I- r 1+a Zl + aZ +. r=l 1 1 1 4 2 L (1- -r- ) r=1 n 1 ( r=l + h z H n 1 2 Z 1 n=o 1 where hn are constants. When zI = 0, we need H z( 0 and 4 2 (1-ar-r) = 0, r=l or 4 S(ar+fr) = 2. r=l Rewriting q(z) in the form (z-a1) (z-a2)(z-a3) (z-a2)2(z-a3)(z-a4) (z-a4 )2 z-al)(z-a ) (l-a z )2 (-a2 z)(l-a z ) (l-a2z)2(l1-a 3z)(1-a4zO) (l-a 4z )2(-a zl)(l-a 2z) (1-a z )2(1-a z )(l-a zl) 1 1 J J(1 z4 z 1 which is analytic at z = O, if J ( -) = 0. To evaluate nr (r=1,2,3,4), we have (z-a) 2(z-a2)(z-a3) (z-a 2 z-a3 )(z-a4) (z-a3)2(z-a4)(z-a1) (z-a )2(z-al)(z-a2) q(z) = (z-a ) (z-a4)(z-a ) then 1 -. q zi + J(z), 4 S[ar r r+ r= (-r)2 z-ar 1) Then clearing of fractions, n (z-az--a3 (z-a4 )2 + n2(z-a )2 (z-a3 )(z-a4) + n3(z-al)(z-a2)2(z-a4) + n4(z-al)(z-a )(z-a3)2 SY11z-a2z-a3 )2 )2 + a 22 (z-01)2(z-a3)2 a )2 +a33 (z-a)2 (z-a22(z-a4)2 + aq4 (z-al )2(z-a3 )(z-a )2 + m (z-a1)(z-a)2 (z-a3)2 (z-a)2 + m2(z-al)2(z-a2) (z-a3)2(z-a4) + m3(z-al)2 (z-a2)2 (z-a3)(-aa4)2 + m4(z-a )2(z-a2)2(z-a3)2(z-a4). To determine nl, let z = a : nl(al-a2)(a -a3)(al-a4)2 = alBl(al-a )(a-a2(a-a )2, nl "ll(ai-a )(a -a3). Similarly, n2 = 22 (a2 -a3)(a2-a4) n3 = a3 3(a3-a4)(a3-a), "4 = a44( a4-a)(a4-a2) Therefore, the general form for a second order linear differential equation with regular singular points at z = ar (r= 1,2,3,4) is 2 4 d + l1-ar-Pr du dzq z-ar dz a2 2(a2-3 )(a2-a4)_ (z-a )2(z-a )(z-a4) + 1(al-a )(al-a 3) 1 (z-a )2(z-a 2)z-a3) S33 (a3-a1) (a3-a,) (z-a )2 (z-a )(z-a ) a+ 4134(a4-a,)(a4-a ) (z-a4 )2(z-al)(z-a u = (z-a ) (z-a )(z-a ) where 4 r=l (a +p ) = 2, ar and (r being the exponents at z = ar' To express the fact that this type we will write al u = al S1 (2-3) u satisfies an equation of a2 a3 a4 a2 "3 "4 P2 33 34 2.2 Differential Equation with Singular Points at z = 0, 1, . Let us now consider the confluent case where a2 = 0, a3 = 1, a, = a4 become infinite. (2-2) z First, let a2 = 0 and a3 = 1, then (2-2) becomes 2 [ 1-a -p (2-4) a u + dz2 z-aL l-a ,-P 1-a3 -3 1-a 4-P4 du + ----- + +j z z-1 z-a4 + Lapiai(ai-l1) I(z-a) )z2z-l) a 22a4 +2(- + z2(z-i)(z-a4) S3P3(1-a,4)(-al) (z-1)2 (z-a4)(z-al) a 44 (a4-al)a4 ] . ----- ----- = 0 . (z-a4)2 z-al)z Before letting a4 become infinite, rewrite the equation in the form [ -al + [a 1al)(al-1) (z-a1 )2z(z-1) l-c2-P2 1-a3-3 1-a 4-P4 du z z-1 z-a4 dz (2 3 2 z (z-1) a4 (a4 44(1- ) ( --- 2(z-al)z- a(4 1/ Now let a4 become infinite: (2-5) d2 + dz2 1-a1-i 1-a -2 + z- + z z-a1 z 1-a3-031 du z-1 dz + iOlal(al-1) a 22 + a33(l-al) + a44 ] (z-a )2z(z) + -2(z- = (ZO (z-al)2z(z-1) z2(2-1) (z-1)2(z-al) (z-al)z 2 d2u d+ dz ( P -1 (1-a ) C33(3 a4 (z-1) -a4 1(z-al a4 which may be represented by a 0 1 oo (2-6) u= al a2 a3 z f1 P 2 z3 4 This equation is actually more general then one might assume, for by means of the linear transformation (a4-a3) (z-a2) (a2-a ) (z-a4) any three points z = a2, a3, a4 can be carried into U = 0, 1, 0. Finally, we want to let aI become infinite to obtain the confluent case with singular points O 1,, o. Rewrite the equation (2-5) in the form d u 1[ -al-P1 1-2- 2 1--3 3 du --+ -+ + ---- dz L z-a z z-1 dz [ a a,1) CXP 2 a3 f -l + + S-iz(z-l) z (z-l) (z-1)2 z -) "4P'4 u 0, + (z- = )z, (z-a1 )z 47 and when al becomes infinite we have the confluent equation (2-7) du 1-a 2-2 1-a3-P3B du CP a 22 (2-7) +-U + + l + 2(zl) dz2 z z-1 z(z-l) z(z-) + U = O, C 3P3 =0, (z-1) with singular points at z = 0, 1, oo. Let us proceed to find solutions of this equation. Multiplying the equation by z 2(z-l)2, we have 2 (2-8) Z2 (-l)2 d + (2---3-3)3 dz + (-3+2a +2 +a3+p3)z2 + (1-a 2- )z + [(ac + a A3)z2 + (-c,1 a2,B)z + ap2 u = 0. Assume the solution N ,+1 A+n-2 A+n-i u C z+ C +...+ Cn z + Cn z o 1 n-2 n-1 h+n + CnZ +...n C / 0, then u' = C Az 1 + Cl(A+l)z +...+ C (A+n-2)z 3 o 1 n-2 + C (+n-)z+n-2 + C (A+n)z +n +... nI-1 n u" = Co0A(-l)z- 2 +...+ Cn2 (+n-2) (+n-)z+n-4 + Cn_(+n-1)(-+n-2)zh+n-3 + C (+n)(+n-l)zh+n-2+... If these equations for u and its derivatives were now sub- stituted in equation (2-8) and the coefficients of the powers of z equated to zero,we would obtain the following results. The indicial equation is z: 7(N-1) + N(l-a2-P2) + CL22 = 0. (2-9) -2 (a +2)A + a2 = 0. Then = a 2)2' Finding Cn, z7+n: Cn-2(A+n-2)(A+n-3)-2 Cn-1 (+n-1)(A+n-2) + Cn(+n)(n-) Cn_2(+n-2)(2-a2-2 -3) + Cnl(?+n-l)(-3+2a2+2 2 +a3+3) + Cn(?v+n)(l-aC2-P2) + Cn 2(0a1 +03~3) + Cn-1(-a1P1- a22) +.CnC2 2 = 0. Cn = nC._ -(+n-2)(N+n-3)-(-+n-2)(2-a ,-P2-a,3-3 1)- (a L-3)1 + Cn-1 2(?+n-l)(N+n-2)-(A+n-l) (-3+2+2P +3+33) + J (2+n)(7+n-l)+(,+n)(l-a 2_ ) + 2 a Simplifying the denominator by using the indicial equation (2-9) we have C = Cn2[-(N+n-2)(+n-l-cx2-P2-a3-3) p I 033] + Cn-1 (?+n-1)(2\+2n-l-22-2a -2 -a3-3) + au 1 + a2P1 n(2-+n--1 -) n(27+n-a -p2) Let us compute a few terms of our series solution to see if a general term can be obtained. Co C (2+l-2a2-2P2-a3-p3) + a 11+ ~ P2] 1 (2A+1-a 2-2) Using the indicial equation (2-9) we can simplify and write Co[(1-03-p3) + al 1- a22] C1 = 22+1-a -- C2 = { [-(+l-a3-3) 3] + C[( )(2 +-2a2-2P2-a3-3) + a P + a P2 - S2(2)+2-a -6 ) Substituting C1 into this expression: C2 = C {(2+l-a-2)[-(l-a2-2-a3-~3)-all-a33 +[(1-a3-p3)7 + al"Bl-a2] [ (A+1)(2h+3-2a2-2P2-as3-3) 1 + a 1P + a P21. ++22 j 2(2X+l-a,-2_)(2X+2-a_-P2) It seems apparent from the appearance of C1 and C,2 and from the fact that Cn is a three term recurrence relation, that a real deal of work might be required to write these two solutions of (2-7) with a general term, thereby making available a form suitable for further development. Since equation (2-7) has five undetermined parameters (aucB a ,2V L3,33), let us consider the conditions necessary to insure a solution with a two term recurrence relation. This can be readily accomplished by applying Scheffe's Criteria [6, p. 240] which Crowson has presented and proved in his dis- sertation [1, p. 114]. 2.3 Scheff4's Criteria Applied to Confluent Case. The criteria to be applied is: Necessary and sufficient con- ditions for a solution of a second order linear ordinary differential equation, p2(z)u"(z) + )u) + ()( + po(z)u(z) = 0, to have a two term recurrence equation, relative to the point z = 0, is that in some neighborhood of the point z = 0, (2-10) Pj(z) = [Sj Tzh] zJ-m , where m is an integer, h is a positive integer, and Sj, Tj are constants such that Si Tj / 0 for some i = 0,1,2 and j = 0,1,2. In assigning values to the exponents, it can be seen from equation (2-4) that none of the exponents can be a function of a Since the sum of the exponents must be 2, if there are any exponents which are functions of a4 there must be at least two such exponents. ar, r (r = 1,2,3) could not be functions of a4, otherwise the coefficient of du will be undefined when a becomes infinite. If a and dz 4 4 04 are functions of a4, then the term CL4 (a4-al)a4 (z-a4)2(z-a )z would be undefined as a4 becomes infinite. Thus the exponents can not be functions of a4. Next, suppose that two or more exponents in equation (2-6) are functions of a Such an equation might be al 0 1 o0 u = aa^+f a i a -a1L- + t z a- 2 3 a a1 where the sum of the exponents is f + a2 + P2 + a3 + 3 + t + 4 = 2, f, t and t are arbitrary constants. Using (2-2) we can write this equation in the form 1 l -aF -f- a 1 i-a -P2 i-a - u" + + + z-al z z-1 OP2 3B3 (1-al) ,) z + + - z (z-1) (z-12 (z-a ) a1 U" + a a z -1 VaI / f (,+ L-)al(al-l) u' + al (z-az )z(z-l) (-a --+ t) (- a'- l a-"]u = 0. + -2-2 + u' z z-1 a2p L3 a1 -1 ( al 4 2 1 + u = 0. a- -1)z Letting al become infinite: u" + + + U-33 z z-1 + 2 2 + 33 4 z(z-) z 2(z-1) (z-1) Z U = . + 1- z 1)2 1 aW -1 z(z-1) Clearing of fractions we can write (2-11) z2(z-1)2u" + [Iz2(z-1)2+(,a 2-B2 )z(z)2 + (1-a3-P3)Z2(Z-l) U' +[ z(z-1)-a22(z-l) + a3B3z + Bz(z-l)2] u = O, where the singular points are at z = 0 1, o, . Applying the criteria (2-10) we have for J = 2 S[S ,,zh] 2-m 2 P2 = [ T z = z (z-l) , 2-m h-m+2 4 3 2 S2 T2z z -2z + z but this is impossible. However, this difficulty can be avoided by reducing the coefficient of u" by a factor of at least z-1. To do this a313 must equal zero, so let p3=0. Thus, with B3=0 and dividing equation (2-11) by z-1, we now have z (z-l)u" + Z2 )-( 2)(z-)+ -a3 + [zz-a 22 + P4'z(z-l) u = 0. (z 3 )u" + z3 +(2-a-- 2-a 3-)Z2 + (-l+2 +B2)z u' + [f4tz2 + (l-P4)tZ 22] U = 0. Now returning to our criteria (2-10), S2-m h-m+2 3 2 P2 = S2 Tz = z -Z , where h-m+2 = 3 and2-m = 2. Therefore, m = 0, h = 1, S2 = T2 = -1, S2T2 / 0. We need P, = [S-Tz ] = tz3 + (2-2- 2-a 3--)z2 + (-l+az +2)z so that t = O, T1 = a2+ 2 + a3 -2, S1 = a2+82-1 and p = (2-a2-32-a3)z2 + ( 2+0-l )z. Finally, we have PO [S-Toz = -2f2' where S = -a22 and T = 0. Thus, if (3 = 0 in equation (2-11) the resulting equation, which has a two term recurrence relation for a solution, is (2-12) z2(z-l)u" + [(2-a2-2-(3)z2 +(U2+2-1)z u'-a282 u = 0. The requirement that f + a +p0+a3+t+B4 =2 is actually not restrictive, since f, t and P. do not appear in our final result. Consider the alternative of having the exponents in (2-6) be functions of a and assume that neither al nor a4 appear in the exponents. Apply the criteria (2-10) to equation (-) +[i -a3-3 al [(1 (2-7) u" + I-a2-P2 + U' + z z-1 z(z-l) z 2 z (z-l) a3r3 1 + u_ = 0, (z-1) or clearing of fractions z (z-)2 u" + (1-La -2)z(z-1)2+(l-aa-a)z (z-1) u' + [alZz(z-l) a2P2(z-1) + c3P3Z2 u = 0, where I r=l (ar+pr) = 2. As before we must remove a factor of z-1, therefore let f3 = 0 so that a3g3 = 0. Then z(z-l)u" + (l-a--P2)Z(z-l)+(l-a3)z2]U '+ o1 lZ-a22 u (z3 zp2)u +[(2-a 2-P23)Z )z 2-+)Z u'+al 1P l- 2 21 Applying the criteria (2-10) we have when j = 2 p2 [S2 -T2zh] 2-m = 0, = 0. 3 2 = Z -Z where h-m+2 = 3 and 2-m = 2. Thence, m = O, h = 1, S2 = T2 = -1, S2T2 / 0. Further, P = [S-Tiz z = (2-a2-P2-a3)z2+(a2+f2-1)z and T = a 2+2 + a3 -2, S1 = 2a +2 1. Finally, Po = S Toz = a 11z a22 with T = -al~l, So = -ar2. The equation with a solution having a two term re- currence relation is, therefore, (2-13) z2 (z-)u" + (2-a 2-B2-a3)z2 + (a 2+0-l)z] u' + [ QPz a22] u = , where the only restrictions are that in equation (2-7) ar' Pr (r=1,2,3,4) contain no function of a1 or a4, and P3 = 0. Since equation (2-12) can be obtained from (2-13) by letting a181 = O, let us obtain solutions of the latter. aC 1z-a 2 u = Cz +...+C z +n+C z +n+... C / O, (2-a2-p2-a3)z2 u' = Coz-1 l+...+Cn_l(A+n-l)zh+n-2 (lzn+.+n- 1 . + (a2+B2-1)z + C .(+n)zhn- z -z u = CoNh(-l)z -2+... + Cn-1(+n-l)(h+n-2)z +n-3 + C (A+n)(A+n-l)zh+n-2 U . The indicial equation at z = 0 is z: -A(-1) + (ag+pg-l)N a 0 = 0, 2 (a2+p 2) + a2 2 = 0. The exponents are h = C2,P2 z +n: Cn-1(A+n-l)(-+n-2)-Cn (-+n) (h+n-1)+C_n- (X+n-1)(2-a2 -p2-3 ) + Cn (+n)(a2+P2-1) + Cn-la p Cna22 = 0. =-C n[(h+n-1)(?+n-2)+(\+n-1)(2-a 2- -a ) + alp C = -- h------------------ n -(2+n)(p+n-1)+(7+n)(2 +( -1l)-a 22 Cn-_l[nl; + (X+n-1)(X+n-a2-p2-a3)] n (A+n)2 (a2+P2)(?+n) + a2zB SCn-_1 [a + (?+n-1)(A+n-c2-P2-a3)] n (X+n-a2)(?+n-~,3) Let a = a : Cn=-l1 a + (a2+n-1)(n-P2-a3)] C = . n n(a 2-p+n) Co[ a + a2(1-P2-a3)] C = + 1 (a2-p2+1) C [a~l1 + (a2+1)(2-32-a3)] C 2 2 ((a2-Bg+2) Co[Cl + a2(l-P2-a3)] i1 [a + (a2+l)(2-P2- 3)] 2 2!(a2-2+1)(a2- 2+2) Therefore the solutions of (2-13) are: u = 2 [1 + z1(-2(-2-3) 1 o l!(a2-P2+1) [a p1+a2 (-132-3 ) ] [a2Pl+(a2+l) (2-p2-a3) ] 2 + Z +... 2!(a2-P2+1)(C2-P2+2) + [all+a2(l-2-a3) ..[a1+(a2+n-1)(n-p2-a3)] n + Z +... nl.(a2-_p2+l) ... (a2-32+n) (2-14) u1 = Coz2 1 + [.a 1+a 2(l-p-a 3)] .. .[a1p+(a2+n-1)(n-p2-a 3)] n n=1 n(a 2-P2+1) .. .(a2-P+n) and similarly (2-15) u2 = cz 2 1 + [ 1p1+2(l-a2-a3 )] ..[al+(P2+n-l)(n-a2-a3)] n n=1 n!(2p-a+1) ...(p2-a2+n) If a l = 0, then the solutions of (2-13) will be hypergeometric: (2-16) u, = Co z 2F1(a2,-p2-at3; 2-32; Z) and (2-17) U2 = C z 2F1(P2,-C2-2-a3; 2-a2; z) The hypergeometric form is (2-18) PF(a,b; c; z) = 1 + z a + a(ab(b+l z2 2 1''' !c 2!c c+l) + a(a+l)...(a+n-l)b(b+l)...(b+n-1) ... n! c(c+l)...(c+n-l) 00 Sa(a+l)...(a+n-l)b(b+l)...(b+n-1) n n=o n! c(c+l)...(c+n-l) 2.4 Solutions of Confluent Cases after Normalizing. Another approach to obtaining solutions of equation (2-7), which we will now use, is to remove the u' term and find solutions of this transformed equation. To normalize the general equation u" + p(z) u' + q(z) u = 0, let u = vw. Then we have u = vw. u' = v'w u" = v"w + vw'. + 2v'w' + vw". Hence, vw" + (2v' + vp) w' + (v" + v'p + vq) w = 0 . Let 2v' + vp = 0 and solve for v. 2 d -vp, dv 1 ---p dz. v 2 Integrating, In v = - p dz. v = e 2- p dz Therefore, to normalize the equation u" + p(z) u' + q(z) u = 0, let v = e (2-19) in (2-20) - p dz vw" + (v" + v'p + vq) w = 0. Now to normalize equation (2-7) we use (2-19) to find V = Z Then S(a -3) (a3+3-1) v' =. (2+p2-1)z 2 (z-1) 1 1 1 2+ (a(+ +-1) (a3+B-3) + (a3+B3-1)z (z-1) 2 1 1 (a +p -1) (a3+p3-1) (z-1) 2 (-1) 3 (a+2-1)(z-l) + (c+1--1)z] ( -1) f (03 +133-1 S1 -a -(a 2+a-5)z v = (a2+2-i) (a2+B2-3)z + 7(a2+p2- ) (3 3 -1 ) + -(a3 +3 -1) (a3+P -3)z 2(a2+02-3) !(a2+ 2-1) (z-1) (z-i) (z-i) (a +B,-5) (az+ -5) (Z-1) + [3 (a +3 -1)(a2+-3)(z-1)2 + (a +p2-1)(ca3+f-1)z(z-1) + (a 3+13-)(a3+z3-3)Z2 Substituting v into equation (2-20) with p and q from (2-7), 1 + Tz a2+0-5) -!(a(+3 -5) (Z-1)(2 (3+3 [ +(a -1)(a+ -3)(Z_1) 12 +f3 -1 2 -3 + (a + 2-1)(a3+43-1)z(z-l) + -(a3+33-1)(a3+t3-3)z2 1 (a 2+2-3) (a3+3-3) + z (z-1) ( z2+(a 2-1)(z-1) 1 V =- Z 2 V, 1 V = ~Z 2(a2+p2-3) 31( C +P 3 -) 1 3- 3 i(a3 +f3-5) 1 1 (:y 2+32-1)) ((c3+f3-1) (z-1) +(C +P-) / 1-c22 1-+ - (+ z +2- Z-1) (3 -)3- 3 12-2 +z (+-1zJ 1 1 (z-l)2 (z-) z(z-1) z (z-1) + )2 w = 0. S(a2 +02-5) (a3+P3-5) Dividing by z (z-1) and simplifying, z(z-l)2 w" + (a2+2-1)(a1+-3)(z-l)2 1 1( + 2 2+ (c2+2-1)(a3+p3-1)z(z-1) + T(Ca3+33-l)(a3+33-3)Z2 1 1-a 33 P - + -z(z-1) (a+P,-1l)(z-1)+(a3- l 1)z +2 3 3 z z-1 - 2 21 3 +23 3 + ( 2 )2 C2 + w 3 0. Z(z-l) 22(Z-1) (Z-)2 z (z-l) w" + ~-(a,+B-l)(a2,+1-3)(z-1)2 + (a 2+(2-)(a3+3-1)z(z-1) + 4(a3+0-1)(a3 .3)22 (a2+P2-1)(z-1) + (a3+f33-1)z + a l1z(z-1)-a P3(z-1) + a 3z 2 w = 0. z (z-)2 w" + r 1(a -1)(a--) +2- 33 T(a2++p2-1)(a2+p2-3) + (a 2+p2-1)(a3+P3-1) + (a3+P3-1)(a3+33-5) -(a2+p2-l) (a2+2-1)(C3--i) - (a 3+3-1)2 + a1B + a3 12Z2 + [- (a2-2-1)(a2+-2-3) - -(a2+2-1)(a3+,3-1) + (a2+,2-1)2 + (a +2-l)(a3+P3-) - a1 al2BZ + [-2+p2-1)(a2+p2-) +2-12 + aO22] Sw = 0. Therefore, the normalized form of equation (2-7) is (2-21) z2 (-1)2 + 1(az+B)2 T(3-33 ) 2(a2+ 2)(a3+33) +) +(a3-3) + i(a2+B 2)( 0 3- 3) + 2(a2+ B2) +p2()3 -F3) + 4l-1 +a3 ]z2 + 1( a+22) + -2(a2+2)(a3+P3) - 2(a. + ) -( 3) a aL 2Z + -(a 2+2 + + a$ 2] w = 0. To obtain the solutions of (2-21) let w = a + + +n-2 + a 1z+n-l + a Z+n+ w = az + az +...+ a z + a z + az +... 0 1 n-i n ao / 0, then 'A-2 +n-4 w" = ao(M-l1)z +...+ a n2(+n-2)(A+n-3)z +n-3 +n-2... + a _(+n-l)(A+n-2)z +n an(an)+n)(A+n-1)z . The indicial equation is z: X(A-1) (a+P2)2 + 1 + ca = 0, TIa+p +a0 (2-22) 2 + 1 (a- 2 = . N -N + = 0. The exponents are 1 +1 - 4 (a-1 2) 1 (2a 2) 2 Finding an: z?+n: an-2(A+n-2)(N+n-3) 2an_-l(+n-l)(?+n-2) + an(A+n)(h+n-l) + an2 [- -(a+p)2- (a3+3)2 (a2i+2)(a3 3) + -(a2+2) 3+) + 13 + a 3 + an-1 ()a2+2) + )(a1 3 1 +- 2- 1( S- )- C I1 a2 + an2 (22=02)2 1 p] = 0. + + 22 (2-23) a = a 2-(h+n-2)(+n-3) + ("282 + f(a3+ 2 1 1 1 + (2 .2 ) ( +,) (x2+.., ) -2(3t 3)-~~ a3c3 + a n 2(\A+n-1)(+n-2) (2+e2) (ac2+ )(al3+ 3) + (C2+,2) + -(a+3 3) + + 2 1+ cl1 + 2'2]} (h+n)(h+n-1) (a2-02) + T In order to simplify further work we can, without loss of generality, let (a282) + -(1 (+3)2 + .-(a2+2)(a3+) (a+B2) 2()3+3) a P2 a33 = k and - (a2+2) (a2-( )(a+3) + (a2z) + l(a4+33) + 1ap1 + a2P2 = m. Making these substitutions in (2-23) and simplifying the denominator by use of the indicial equation (2-22), we have an- [-(h+n-2)(7,+n-3)+k] + an- [2+-n-l) (+n-2) + m a =- n n(2?+n-l) a o 2h(-1) + m] 1(2X) ao(2?2 -2?+m) 1!(2A) a[-A(\-1) + k] + a [2(A+1)? + mI 2(2,+1) 2?(-_2 ++k) + (22 -2N+m)( 2? +27\+m) 2! 2?(2?+1) 4x4-2A3 + (4m-2)A2 + 2k? + m 2 21 (2?)(2?+1) a -( \+l)A + k + a2[2(?+2)( \+1) + m] a3 = a ( 2(-_2--+k)(2X2-2N+m)(2)+l)+(2X2+6+4 +m)[ 4X4-23 + (4m-2)?2 + 2kA + m2] , 31 (2\) (2A+1)(2\+2) a3 ao[8?6 + 12\5 + 4(3m-l)?4 + 6(3m+2k-2)?3 + 2(3m2+4m-2+4k) 2 + 2(3m2-m+3mk+2k)X + m(m +4m+2k)] a1 = S= a a2 = ao[ a2 = a 3! (2-)(2?+1)(27,+2) The solutions of equation (2-21) are (2-24) w = az + (22-2+) 1 1 (2?X) 44 -2X3 +(4m-2)N2 + 2kN + m 2 2 + Z z +... 2!(2) (2A+l) I+(a2-82) where A = +( 2 and k and m have the values: 2 )2 k = -(a +2+2 3+(a2+22+a32.3) a l 33' m = (a(+2-1)(a2+ 2 3) + i3 + a 2 2 2.5 Factored Solution Wk,m and Hypergeometric Solution W ,k. Next, we shall consider solutions of the equation 00 0 1 0O (2-25) u = I a 2k a4 z m 1 -a 0 P4 when the u' term has been removed. This can be obtained from equation (2-21) by letting alB = m, a2 = a, P2 = 1-a, a3 = 2k, P3=0. Although P3=0 in the normalized equation, this does not insure a two term recurrence relation as it would in the original equation (2-7). Our transformed equation, that is (2-25) in normalized form, is 68 (2-26) z2 (-l)2w" + (m+ k)z2+(a-a-m)z+(l-a) w = 0. Assume the solution w = Coz + + Cn-2z+n-2+ Cn- +n-1+ Cnz+n+... C 0, = 'n+...+C z C + CZ +... C / O, then w" = Co(A?-l)z -2+...+ Cn_(+n-2)( +n3)z+n-4 + Cn _l(+n-l)(A+n-2)zh4n-3+ Cn(N+n)(A+n-l)z+n-2 +... The indicial equation and exponents are z: ,(N-1) + a a2 =0, \2 7 + a(l-a) = 0, (A-a)(A+a--l) = 0, X = a, 1-a. Obtaining C : z7+n: Cn-2 +n-2)( +n-3)-2Cn-l+n-l)( +n-2) + Cn(A+n)(A+n-l) + Cn(m + k2) + Cn_(a2 a ) + C (a a2) = 0. -a-m0+ 69 Cn = -C (+-n-2)(h+n-3)+(m+ 1- -k2) +C- 2 (h+n-1)(A+n-2) (ac a m) 1 2 *------ S(A+n)(?+n-1) + a a Let ? = a: S= Cn -(a+n-2)(a+n-3)-m- + k2] + Cn_1a2 + (4n-5)a +2(n-1)(n-2) + m n- -- 1 S n(2a+n-l) C a -a2_ +m C1 1 C 1(2a) C -a(a-l)-m- + k2] + c 2a+ C2 01 ---------------------- 2 2(2a+l) [2a(-a2+a-m- +k2)+(a2-a+m)(a2+3a+m) 2= o 2!(2a)(2a+l) ' aa4+(2m-l)a2 + (2k2- ')a + m 2 2!(2a)(2a+l) C -a(a+l)-m- +k' + C[a2+7a+4+m C =3(2+2) I5(2a+2) C = CO 2(a2-a+m)(-a -a-m- 1 +k 2)(2a+l) + a 4+(2m-l)a2 + (2k2 )a+m2](a 2 +7a+4+m) 1 -- + (2 1) 3!(2a)(2a+l)(2a+2) C = Co a6+a +(35m+l4+m+l +(6+6k2- 9)a+(12k+3+3m25)a2 2 2 23 3 3 2 2 1 +(3m2+6k2+6mk2- m- )a+(m 2m2+2k- 1 ) 1 3!2a(2a+l)(2a+2) Thus one solution of (2-26) can be written with a few terms as: (2-27 -a+ a +(2m-l)a 2+(2k- Z)a+m2 2 (2-27) w = C z 1+ Z+ z + L 1! 2a 2. 2a(2a+l) J A second solution of (2-26) can be obtained by replacing a by 1-a: (2-28) w = Coz1-a+ 2-a+ S1(2-2a) a 4a 3+(5+2m)a 2+(-2k2-4m- 3)a+(m +2m- +2k2) 2 + +... 21 (2-2a)(3-2a) Although solution (2-27) has no obvious general term, it can be shown by a lengthy process of long division (which is omitted here), that it can be written in the factored form: 2+ (a 2+2ka+m)[ +(2k+2)a+2+2k+m+1 w=C z (l-z) 1+ 2k-z+ z W=0 (lz 11 2a 2! 2a(2a+l) (a2+2ka+m )[a2+(2k+2)a+2k+m+l] [a +(2k+4)a+4k+m+4 3 + z +... 35 2a(2a+l)(2a+2) + ((a 2+2ka+m) a +(2k+2)a+2k+m+l] ** a +2(k+n-l)a+m n + (n-l)(2k+n-l1) z+... n' 2a(2a+l)...(2a+n-1) Let this solution be indicated W k,m, since a,k,m are the only parameters; therefore (2-27) can be written (2-29) Wa,k,m = C0a(1-z)k+ 1 + (a2+2ka+m) n= [a2+2(k+l)a+2k+l+m] [a2+2(k+n-l)a+(n-l)(2k+n-l)+m] n n! 2a(2a+l)***(2a+n-1) If m = 0, then the solution (2-29) becomes hypergeo- metric. Let this solution be indicated W ,, a and k being the only parameters: Wa,k = Coza(1-z)k+ 1 S(a 2+2ka) a2+2(k+l)a+2k+l [a2+2(k+n-1)a+(n-1)(2k+n-l) n= nl 2a(2a+l) *.(2a+n-1) Factoring further we may write k+ -- Wk Co ZU(-z) I Sa(a+2k) (a+l)(a+2k+l) ]*[ (a+n-1)(a+2k+n-l) ] n=1 n! 2a(2a+l)...(2a+n-1) * Rearranging factors we arrive at the solution k+ W,k = Coza(l-z) 2 + a(a+l) '*(a+n-l(+2)( )(a+2k+l) *.(ac+2k+n-l) zn n= n! 2a(2a+1) **(2a+n-l) which is hypergeometric in form (2-18) and shall be denoted by Wa,k Therefore, when m = 0 in (2-26) we have the solution 1 k+- (2-30) W,k = (l-z) 2Fl(a,a+2k; 2a; z), where we have let Co = 1. A second solution can be obtained by replacing a by 1-a. 2.6 Notation and Proofs Involving 2Fl(a, b; c; z). Before presenting several interesting properties of Wa,k, we shall introduce notation and prove seven statements that will be useful in section 2.7. Notation: n,j are integers. (a)n = a(a+l)(a+2)...(a+n-1). n(n-1)...(n-J+l) = 1. DnW = dnW Dn, dzn W = W a (z). Wa,k a,kZ). If F = 2F(a, b; c; z) = F (a, b; c; z), then F(a+) = F(a+l, b; c; z), [5,p.50] , F(b+) = F(a, b+1; c; z), F(c+) = F(a, b; c+l; z), F(a-) = F(a-l b; c; z), etc., F(+n) = F(a+n, b+n; c+n; z), F(-1) = F(a-l,,b-1; c-1; z). We shall prove the following: n = 1,2,3,... F = 1 [aF(a+) + (1-c)F(c-)] a-c+1 L1P F = 1 F(a-) + (b-c)z F(c+)l 1-z L c J F(+l) = -- F(b+) F az I+ Ja DnF = (a)n(b)n (c)n F(+n); and for F = 2F1(a, a+2k; 2a; z) = F(a, a+2k; 2a; z), (2-35) F(+n) (2a) n = z (n,) -(l- z) I (1-Z) n+l n -k- 2 J=o (2-31) (2-32) (2-33) (2-34) (-1) (1-z) W ,k+ n- J) r~k (2-36) Dn z(1-z) ] k-n+ = z"(l-z) 2 (-1) (-a (-k-x) ,(z (-z j =o k(2-37) F(-l) zykzl1 (2-37) F(-1) = (1-z)z 2(2a-1)(1-z)-Wak+(-2k)W (2a-1)(2-z)L a a-j Finally, we state Leibniz's rule [9, p. 409] for finding the nth derivative but shall not prove it: (2-38) Dn(uv) = u D"v + (n)Du D n- + = V 1 ( D2u Dn- +... + >tDu D"n- v +...+ (D u)v n (n D= u Dn"-. t=o PROOFS (2-31) F = 1 [aF(a+) + (l-c)F(c-)] a-c+ L J Proof: aF(a+) + (l-c)F(c-) (a+l) (b) + (c) n! n co (1-c) Z n=o (a) (b) (c-1)n! (c-l) n! = [a(a+l) (b)n =o (c) n (a) (b) z (1-c) -L I (c-l)nn! n=0 c, S-- [a+n-(c+n-1) n=o (a)(b)n (c) n! S(a c + 1)F. Therefore, F -1 a?(a+) + (1-c)F(c-)] a-c+1 L 1 F = 1 1-z I c (b-c)z F(c+) C (a-1) (b) (C n n (c),nl (b-c)z c n=o (a) (b) n (c+1) nn (a-1)(b). n + (b-c) (c) n' c 1 (a-1) (b) (b- ) n=1 (c)nn! c =1 + [ (a-1) + n(b-c) S1 L a+n-1 (a+n-l)(b+r S1 + (a-l)(b+n-l)+n(b-c) n=l (a+n-1)(b+n-1) (a)n-1(b)n-1 n n=l(c+l)n-l(n-l)! (a)n(b)n z I-1)J (c) nn (a) (b)n (c) n! (2-32) Proof: F(a-) + 00 =L n=o 00 = 1 +L n=l (a)n(b)n n+1 ^ ~n ^+Z S(c+l) n! z n zo n =1 [1- n(c+n-1) (a)n(b)n n n=L (a+n-l)(b+n-l) (c) nn 0o FZ [ n(c+n-1) n=l (a+n.1)(b+n-1) (a) (b) (cn n (c) n! 0o F z n=l Cor F y n=o (a) n-1(b) n-1 (c)n-l(n-l)! (a)n(b)n n+1 (c) n! = F zF = (1-z)F. Thus, F(a-) + (b-c)z F(c+) = (l-z)F c F =1 F(a-) + (b-c)z F(c+ 1-z c I F(+l) = [ F(b+) F] az L J F(b+) F = n=o (a) (b+1)n (c) n! n n=o (2-33) Proof: (a)n(b)n (c) n! n Sb+n 7^ (a)n(b+ln-1 n (c)n(n-l))! 00 =1 n=l ,00 7-1 *L n=l az c n c n Hence, F(b+) F = az F(+1) c F(+1) = -c [F(b+) F] (a) (b) Dn = n F(+n). (c)n D [F(a, b; c; z)] =D [ (a)(b)n n=o (c) n! n=o ^n n(a) (b)n n- (c) n! (n+)(a) n+(b)n+ (c)n+l(n+l)! (a) (b+1)1 n Wn n-1 n (c)n(n-l)! S(a+)n-1 (b n-1 n-i S(c+l) (n-1)! z =1 n-1 1 (a+l(b+)n n L (c+1) n: =0 n (2-34) Proof: zn] 00 oz n=l n=o cn= (c+l)nn! ab F(+1). c Similarly, D2 [F(a, b; c; z)] 00co Sa(a+l)b(b+l) c(c+l) n=l n=1 (a+2)n-1(b+2)n-1 n-i Z (c+2) (n-l) (a) (b) w (c) nL 2 n=o (a+2) (b+2) n (c+2) n! n (a)2(b)2 F(+2). (c)2 In general, Dn [F(a, b; c; z)] (a(b)n F(+n), (c)n n=1,2,3,... If F = F(a, a+2k; 2a; z) = z-(l-z)--Wa,k then (2-35) F(+n) (2a) n+1 n 2n z-a-n(l-z)_k-- n-y Sc (1-z) 2 (-i ()n j=o j \ J2 1-z) Wk+ n-j 2 Note that when k is replaced by k + m in the premise, then F(a, a+2k+2m; 2a; z) = z-a(1-z)-k-m-2W ca,k+m Proof: We begin by applying (2-53) to F(+l) and express the result as a function of W a,k+: F(+l) = 2a [F(a, a+2k+l; 2a; z) PF az L I 2a -a( -k-1 -k-1 -- z (1--z) W (1-z)' 2W az a,k+ az(lz ,k . (a) ,k+ ,k Apply the principle of (2-33) to F(+2): (2-40) F(+2) =2+1) [F(a+l, a+2k+2; 2a+l; z) F(+l) (a+l)z J Using the expression (2-39) with k replaced by k + in (2-40), we have (2-41) F(+2) S(2a+l) a 1-a-1l -k-zW kW (a+l)z a ak+ ak+ 2a -a-1 1 w 1 a (1-lz)-- Wa ,k+ -(l-z)2W ,1 a L ak+ akJ/ )2z (1-2 ) a,+-2(1-z)W ak + (1-z)Wk] Let us find F(+3) before generalizing our results to F(+n). Upon applying (2-33) we have (2-42) F(+3) = (2a+22) FF(a+2, a+2k+3; 2a+2; z) F(+2) (a+2)z IJ . Replace k by k + in (2-41) to write (2-42) in the form (2+2) (2a)2 -a-2 k-2 F(+3) = z (-z)-k-2 W k+3 -2(-z) (a+2)z (a)2 2-k+l (2) 2 a-2(l k- S(1-z)Wa,k]- (a)--2 ( )- a,k+ 1 2(-z) W ,k2 + (l-z) Wak Simplifying further (2a)3 -a-3 k12 F(+3) = z (1-z) W,k+ -(1-z)2W, (a)3 a,k+1 + 3(1-z)W ,k 1 -(l-z)Wk] Finally, 2cx) n -a-n -k- n F(+n) z -a (k (1-) 1) k+ n-2 (a)nk+ , 2) a,k+ j ak+ n +...+ (1-z)W ,k . Therefore, (2a) an(l-z) n+l1 F(+n) n za-nz (-1) (j\1-z) W k+n-J (a)nJ=o >a,k+ n=1,2, 11, . (2-36) Dn [za(l-z)k+ n = a(1-z)k-n (-1) () -a(-k-)n z(l-z) . J=0 Using Leibniz's rule where u = za and v = (1-z)k+ find DIu and Dn-v. When u = za Du = aza-1; D2u = a(a-l)za-2; ...; DDu = (-l)t(-a)za-" If v = (1-z)k+, then 3 Dv = -(k+)(1-z)k-; D2v = (k+)(k-)(1-z) k- .. D"'v = (-k-1) (1-z)k+-n+ . Hence, the nth derivative is found to be Dn[za(l-z)k+] = ( (-)(-a) --k ) (l-zk+L-n+ n = za(1z)k-n+ (- (-)(-k-) z -z) 4t=o If F(-l) = F(a-l, a+2k-l; 2a-l; z) show that (2-37) ( -kz 1-a F(-l) = (-z) 2(2a-l)(l-z)2Wa- +(a-2k)Wa k- (2a- ( (2-z) L Apply (2-31) to F(-l): F(a-l, a+2k-1; 2a-l; z) -1 [(--l)P(a, a+2k-l; 2a-l; z) -a+1 1 + (2-2a)F(a-l, a+2k-l; 2a-2; z)] = -F(a, a+2k-l; 2a-l; z) + 2F(j-1, a+2k-l; 2a-2; z). Now apply (2-32) to the first term with the result that F(-l) -1 [(-1) + (2k-a)z F(a, a+2k-l; 2a; z)] 1-z 2a-1 + 2F(a-1, a+2k-l; 2a-2; z). Collecting the terms containing F(-l) we have (2-38) 1 + 1 F(-l) a-2k z F(a, a+2k-l; 2a; z) 1-z 2a-1 1-z + 2F(a-l, a+2k-l; 2a-2; z). Now, since (2-39) F(a, a+2k; 2a; z) = z-a(l-z)- k2W l,k if k is replaced by Ic -, then F(a, a+2k-l; 2a; z) = z-a(l-z)-kW If a is replaced by a-1 in (2-39), then F(a-l, a+2k-l; 2a-2; z) = z-a+1 (-z)-k- -1k Using these two results and solving equation (2-38) for F(-l), permits the result (1-z) a-2k _z z-a(lz)-kW ! (2-z) L2a-1 1-z aK2- + 2z (l-z)k 2 k] Sz- ( j- (a- W 1 + 2(l-z)\ (2-z) -2a-l ak-2 a-1k or F(-) (-z) z (a-2k)W' k-1 + 2(2a-l)(l-z)2W- (2a-1)(2-z) ) a 2 a-1,k . 2.7 Recurrence Relations, nth Derivative, Sum and Product Formulas for W a By utilizing the statements we have just verified, it is possible to obtain recurrence relations for W Several such relations will now be o,k ' presented. By replacing a by a-1 in equation (2-30), we may write (2-40) W = z-1(l-z)k+ F(a-1, (+2k-1; 2a-2; z). Since the parameters a and b in F(a, b; c; z) can be inter- changed without any lose in generality, we can also inter- change a and b in equation (2-32) to obtain the valid relationship F -1 F(b-)+ (a-c)z F(c+) l-z ce t Now, if this be applied to (2-40), we may conclude that (2-41) Wa-1,k = zal(1-z) 2 F(a-l, a+2k-2; 2a-2; z) + (1-a)z F(-l)] 2a-2 * Replacing a by a 1 and k by k - in (2-30), it is seen that F(a-1, a+2k-2; 2a-2; z) = z-a(l-z)-k a-,k-" This, together with (2-37) enables us to write (2-41) in the form a-1 k- 1-a -k Wa = z (1-z) z (-z)- Wa,k 1 (z)kz2-a 2(2a-1)(l1z)2W 2(2a-1) (2-z) a-l,k + (a-2k)Wak- = (1-z) W Z1 W =a-1,k (2-z) a-1,k (a-2k) z(l-z)- W ak 2(2a-1)(2-z) Then 1 + -z W= (I-z1_ Wa-,k- + (a-2k)z w,, i 2-z 2 a-lk2(2a-1)(z-2) ak Solving for W -1,k, we have the recurrence relation Wak (2-z) -z- (a-2k)z a-l,k 2 L2(2a-l)(z-2) (2-42) (-z 2(2a-1)(z-2) Wa- (2-2) Wa-,k 4(1-2a) -1,k-2 + (a-2k)z W kl] Another recurrence relation is found by first replacing a by a + 1 in (2-30), giving us W a+,k = Z+l(1l-z)k+2 F(a+l, a+2k+l; 2a+2; z). Then apply (2-33) to get a+1 k+ (2a+) Wa+ik z 1 (-z) [P(a, a+2k+l; 2a+l; z) F(a, a+2k; 2a+l; z)] When (2-31) is applied to each term, we may write Wa+ (2a+l) (Za(l-z) 1 [aF(a+l, a+2k+l; 2a+l; z) a+l, a -a 2aF(a, a+2k+l; 2a; z)] +1 [aF(a+l, a+2k; 2a+l; z) -2aF(a, a+2k; 2a; z)]. aL JJ (2-43) Wa+ k = (2a+l) za(-z)k+ -F(a+l, a+2k+l; 2a+l; z) + 2F(a, a+2k+l; 2a; z) + F(a+l, a+2k; 2a+l; z) 2F(a, a+2k; 2a; z)]. From (2-35), letting n = 1, we have F(+1) = 2z --(l-z)-- [W ,k+ -(i-z)YW ] k ,k , which can be used to rewrite the first and third terms of (2-43). Hence, (2a+l) z,(lz)k+-f 2z-a-1 z)-k-1 W S z -z-2-z) k- W+,k a k+ 1 2 (1-2) Wz- + + 2-a(1-)--1 W ,+ + 2z'- (1-z) l[wa, (1-z)l W a, j 2z-'(l-z) 2 W f- WI, - 2(2a+l) [- -1 -z)-- W1 + a a,2+y 1Z-1 ( lz) Wk .] Thus, we have a second recurrence relation (2-44 ) axzW C,1j = 2(2a+1) -(1-z)2 W Cx, kj + (2-z) W a~k To obtain the nth derivative of our solution (2-30) Wa,k = za(l-z)k+2 F(a, a+2k; 2a; z), we shall apply Leibniz's rule, where u = z(l-z)k+2 and v = P(a, a+2k; 2a; z). Thus, using (2-34) thru (2-36) we may write n D W ak t= D [z(1-z)k] D"-o [F(a, a+2k; 2a; z)] t,=o n 4 SZ [()"(l-,z)k-t+ (-, -) z- (1-z) t-=o J=o (a) (a+2k) (2n- ) n-t n-t = a(lz-z)k- t=o j=o f+_ _n-lCA 4 3 S*(a+2k) z--2 ( i ( -)2 / Wa,k+ n- .-4 J=o Therefore, (2-45) DnW ,k = z-n(-z)- i(l-z)- zt(a+2k)n- 'c=0 * (-) -a -k-J-j (1- J-=o n--4 S (-l)j n- (1z) Wk+ n--3i o a0k F(+n-LZ) (-k-.L j(-) 2 t,_j 88 Now that we have the nth derivative of V (z) we a, k can use Taylor's Theorem [8, p. 93] to write formulas for Wa,k (x+y) and hW (xy). Taylor's expansion can be ex- pressed in the form 2 n f(y+a) = f(a) + f'(a)y + f"(a) 1 +...+ f(n) Y + Now if a be replaced by x, we have 2 n f(y+x) = f(x) + f'(x)y + f"(x) 2 +...+ f(n)(x) y +.. 2! n! oo (2-46) f(y+x) = f () (x) n . n=o Using (2-46), replace y by (y-l)x to obtain 00 (2-47) f(xy) = f(n)(x) (Y- )nxn.n n=o Since Wa,k(z) is a convergent, analytic function for 0 have an addition theorem and a multiplication theorem: (2-48) Wk (x+y) = W( (x) n n = ji x (-x(1-x (a+2k) n=o t=o J=o n-, J n S (-1 J) n ) ( ) a,k+ n--j j=-o and (2-49) Wa,k(xy) = (x) ( nn n=O Sn(y )n (1-x) n x-(a+2k)_ n= n. t=o J=o 1 J=0 where lyl < 1. CHAPTER III Related Differential Equations and Classification 3.1 Related Equations of Mathematical Physics Derived from an Equation having Four Singularities. The most gener- al linear differential equation of the second order which has every point except a ,a2,a3 and o as an ordinary point with exponents ar'Pr at ar (r=1,2,3) and exponents 41,"2 at 0, is 3 3 2 3 r-ar-B3 du 3rr Az+B (3-2) 2 + (ar+)-2 + + A = 0. dz 2 z-a (z-ar) 3(z-a where A is such that tj and L2 are the roots of r=1 r=1 To verify this, we can begin with equation (2-1) and let a4 become infinite, thus obtaining 3 3 u [ 1-ar-r + H(z) u' + r + ---- + J(z) u=0, +-a + H I z-a ) z-a r=1 r r=1 +(Z-ar Z-ar where H(z) and J(z) are polynomials in z. We must now depart from the previous work of chapter II, for we desire z = o to be a regular singular point in this case. Referring to section 1.5, we see that z = ( (z = ) will be a regular 90 singular point when 2 z ) and 21 <1 are analytic at z1 = O. For the first one, 1l 11-1- a2 3I 3 1 22 p + + + -L1 z1 \Z1 Z1 1 a -- a a z1 1 z1 2 z1 3 1-a lp1 1-a 2-P2 1-a3 -i 1 l-[l-a11 -a2z l -a3z 11 which is analytic at zi = 0 if H = In the second, 1 c 1l a1P2 a 3 3 1 + -+ + + 2 -a1 z 11 1 1 a -a a P3 m m + + ++ z-- -a z- -a3 1 1 a1 22 3__ _3_ l-2alz +a 1z l-2a 2z+az1 1-2a3z +a1 1 i 2 3 1 - z 1-a z1 1-a2z 1 -a z z1 z , which is analytic at z = O, only if mi + m2 + m3 = 0 and <1 ) = 0. Now m m m, 1 + 2 3+ 1: + +[(m1- z-al z-a z-a = 2 1 S1 ]2 3 (z-a )(z-a )(z-a 3 where A and B are functions of mr and ar (r = 1,2,3). But since ml m +m3 = 0, we have equation (3-1). In order to show that A must satisfy the condition (3-2), we can make the substitution z =- in (3-1), then the indicial equation zi for z = 0 will be this condition. An alternative method of finding the Indicial equation at z = w (or zi = 0), and the procedure we shall use here, would be to assume the solution u = b0oz + blz--1 +. and equate the coefficient of the largest power of z (i.e. z 2) to zero to obtain the indicial equation. Noting the expansion of the following about z = o(or - = 0), we have Z z 2 a a 1 ( 1 = 1 1a/ 2ar -+ = 2+ +- + --+... where r = 1,2,5> and zar = a r Z + 1- - (z-a,) z-a,)( z 1 , z 1 a a a (z-a )(z-a2)(z-a3) z -- 1 1 Sz z 1 +a 2+a3 = -T- 1+ + .... z z" Then 3 - / 2aar Nz-+b 1+ +..) u = bz +b z j Z2 Z o 1 SAz+BA + al+a2+a3 z \I z +' 3 2 1-a -Br a a + -- + r +...) r= z z z r=l 1 -A-1 -b-2 u' =-b Az ? +b(--l)z +... u" = bo (h+l)z-N-2 + b, (+l)(,+2)z--3.. 1. . 3 3 z -: (h41) h (l-ar-r) + arr + A = 0. r=l r=l 3 3 N2 + 1[i- (1-ar-Pr)] X + Zarfr+ A = 0. r=l r=1 3 3 S2 + (ar+Pr)-2] N + Carr + A = O, r=1 r=1 where the exponents at z = c are = 1 and 2 We now proceed to show how many of the related differ- ential equations of mathematical physics can be obtained from a differential equation having four singularities, by assigning values to the exponents ar, Pr, the singularities ar (r = 1,2,3), and the constant B. (a) Lame's Equation [8, p. 2051. Beginning with (3-1) take ar = 1/2 and fr = 0 (r = 1,2,3), 2 = i + n + 2, and b / 0. i r--- - -l+ ... 1 B = - h, where h and n are constants. We can obtain A by using (3-2) and the fact that the sum of the roots equals 3 3 2 (ar+6r) = and the product of the roots is a arr r=l r=l + A = A, where the roots are 41 and ul + n + 1. Thus, 24 + n + 2 = 1 = n, and A = (1 +n--1) = A n(nin ) n(n+l). Substituting these values into (3-1), the result is Lame's Equation: du 2 du n(n+l)z + h dz2 + z-a dz i3 u - dz r 4 (z-a ) r=1 1 r r=l (b) Legendre's Equation. Let a = a = O, a3 = 1, a = 3 = = = 0 2 a +1+ i+2+2 = 3/2, B = n(n+l) 3 + al1 + a2ZB in (3-1). Since the product 1,2 = = ar r r=l + A from (3-2), we have A = -(alP1 + a2p2). Hence, d2u 1 du 1 al1I + a2 2 ++ L)+ + + 2 dz 2 dz z z-1z -(a lf + a P2 )Z 2 n(n+l) + a 1P + az2 2 u 0. + -- -- ^----L 1 2 2 u = 0. z (z-l) (3-3) d2 + ] +n(+l u = 0. dz2 + z z-1 dz 4z2(z-l) dz kz2(z-1) |

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PAGE 1 CONFLUENT CASES OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH FOUR SINGULAR POINTS By JOYCE COLEMAN CUNDIFF A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLNfENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA June, 1961 PAGE 3 acknov;ledgments There are many people who have contributed to the preparation of this dissertation. First, the vrrlter wishes to express her deep appreciation to the chairman of her supervisory committee, Professor Russell W. Cowan, for suggesting the topic of study and for his guidance and helpfulness during the research. She is also grateful to the members of her committee. Professors VJ.R. Hutcherson, J. T. Moore, W. P. Morse, C. B. Smith and A. Sobczyk . Warm thanks are due Professors P. W. Kokomoor and J. E. Maxfield for their encouragement and interest. The patience, cooperation and persistence of her typist, Sandra Fife, is gratefully recognized. She is humbly grateful to her husband for the less tangible but real assistance received by his encouragement, patience and enthusiasm. 11 PAGE 4 To Mother, ever uplifting and inspiring PAGE 5 TABLE OP CONTENTS ACKNOV/LEDGMENTS il DEDICATION ill INTRODUCTION 1 CHAPTER I Second Order Linear Differential Equations 1.1 Definitions 2 1.2 Formal Solutions of Differential Equations ... 3 1.^ Convergence of Formal Solutions 10 1.^ Second Solution in Case where the Difference of the Exponents is an Integer. Example . . . .2k 1.5 Solutions Valid for Large Values of | z | . . . .35 1.6 Irregular Singular Points and Confluence ... 38 CHAPTER II Confluent Differential Equations 2.1 Differential Equation with Pour Singular Points ^5 2.2 Differential Equation with Singular Points at z = 0,1,00,00 2j.i^ 2.3 Scheffe's Criteria applied to Confluent Case . 50 2.4 Solutions of Confluent Case after Normalizing . 59 2.5 Factored Solution W , . Hyp ergeome trie Solution W , . . .^'^'.^ 67 2.6 Notation and Proofs involving P (a,b;c;z) . . 72 2.7 Recurrence Relations, nth Derivative, Sum and Product Formulas using W 83 iv PAGE 6 CHAPTER III Related Differential Equations and Classification 3.1 Related Equations of Mathematical Physics Derived from an Equation having Four Singularities 90 (a) Larae's 93 (b) Legendre's 9^ (c) Jacobi's 95 ( d ) Gegenbauer ' s . Deriva tion of Geganbauer ' s Equation 96 (e) Laguerre's 100 (f) Equation with Solution being Incomplete Gamma Function 101 (g) Gauss's 101 (h) Kummer's 102 (i) Whittaker's 102 3.2 Classification of Differential Equations having Pour Singular Points when the Exponent Difference is i 103 LIST OP REFERENCES 116 BIOGRAPHICAL SKETCH 11? v PAGE 7 INTRODUCTION This is a discussion of ordinary second order linear differential equations with four singular points and the confluent cases which occur by permitting their singular points to coalesce. Chapter I presents a second order differential equation, formal solutions v/ith proof of convergence, solutions valid for large values of \z\, irregular singular points and confluence of singular points. Chapter II develops the general form for a differential equation having four regular singular points and then considers solutions of the confluent equation with singularities at z = 0,1, Â«, 00. Using the normalized form of the confluent equation we find a hypergeometrlc solution with recurrence relations, nth derivative, and sum and product fonnulas. In chapter III we derive some of the related equations of mathematical physics (see table of contents) from the general equation of chapter II and present a classification of this type equation when the exponent difference is i. Chapter I elaborates on pertinent material in [8, chapter 10]^. Except for the classification in chapter III, see [2, p. 499], the remaining work is believed to be entirely original. The numbers in brackets refer to the list of references PAGE 8 CHAPTER I Second Order Linear Differential Equations 1.1 Definitions . Let the standard form of an ordinary second order linear differential equation be represented as (1-1) ^ + P(z) ^ + q(z) u = 0, dz'^ dz where p(z) and q(z) are assumed to be functions of z analytic [8, p. 83] In some domain D except at a finite number of poles. Any point In D at which p(z) and q(z) are both analytic will be called an ordinary point of the equation; any other points of D will be called singular points of the equation. If there exists a point z=c of D such that, when p(z) and q(z) or both have poles at z=c, the poles are of such order that (z-c)p(z) and (z-c)*q(z) are analytic, then z=c Is called a regular singular point for the differential equation. Any poles of p(z) and q(z) that are not of this nature are called Irregular singular points . If z=c Is a regular singular point, the equation may be written (1-2) (z-c)Â» p^ + (z-c)P(z-c) ^ + Q(z-c) u = , dz* dz where P(z-c) and Q(z-c) are analytic at z=c. Expanding these functions In a Taylor's Series about z=c we have 2 PAGE 9 P(Z-C) = Po + Pi(z-C) + ... + Pj^(2-C)" + ... = 2^Pn(2" c) , n=o and Q(z-c) = q^ + qi(2-c) + .Â•Â• + qÂ„(?.-c)" + ... 2^ q^(z-c)". n^o where p ,p.,...,q ,qj^,...are constants. These series converge In the domain D^ formed by a circle of radius r (center c) and its Interior, where r is chosen sufficiently small so that c Is the only singular point of the equation which is in D^ . Thus (,) .LL^iLLl and z-c In equation (l-l) . f X Q (z-c) q(z) = -) ^ (z-c)^ 1 . 2 Formal Solutions of Differential Equations . Le t us assume a formal solution of the differential equation to be 00 u = (z-c)Â°' [l + ^a^(z-c)"j ^ n=i where a,a ,a2>... are constants to be determined. Assuming that the term-by-term differentiation and multiplication of the series are valid, we have PAGE 10 n-i n=l 09 = (2-c) a-1 a + \ (a+n)a^{z-c)" n=i and u" (a-l)(z-c) a-2 a + \ (a+n)a^(z-c)'^ n=i + (z-c)'^~ \ n(a+n)a^(z-c) n-l n=i 00 = (z-c)Â°'"^[a(a-l) + y (a+n){a+n-l)a^(z-c)"l n=i Now substituting these ej^resslons for u^u' and u" into equation (1-2) v:e obtain (1-5) (z-c)*^ a(a-l) + y a^(a+n)(a+n-l)(z-c)" n = l 00 (z-c)"* P(z-c) a -f-V a^(a+n)(z-c)"l n=l + (z-c)" Q(2-c)] 1 +ya^(z-c)'' = 0. n=i Next substitute the series for P(z-c) and Q(z-c) into equation (1-5): PAGE 11 a> (z-c)" [a(u-l) + ) a^^(a+n)(a+n-l)(z-c)' n-l + (z-c)''rp^ + Pi(z-c) + P2(z-c)' +...+ PÂ„(z-c)" eo + . . . Â•] a + ^ a^(a+n)(z-c)" n=i + (z-c)Â°'rq + qi(z-c) + q2(z-c)Â» +...+ qÂ„(z-c)" + . . . 1 + y a^(z-c)" = 0. n-l Now equate to zero the coefficients of the successive powers of (z-c) : (z-c)Â°': a(a-l) + p^a + q^ = 0, aÂ» + (Pq-I) a + q^ = 0. (z-c)Â°''^S a^a(a+l) + a^p^(ai+l) + ap^ + a^q^ + q^ = 0, a' + a + p^(a+l) + q^ + (xp^ + q^ = 0Â» a r(a+l)Â»-a-l + p^Ca+l) + q^ + ^^Pi + q^ = 0, (1-4) a^[(a+l)Â» + (p^-l)(a+l) + q^ + aPi + q^ 0. PAGE 12 .a+2 (z-c)" : a2(a+2)(a+l) + p^a.^{a+2) +p^a^(a+l) + T^a ^ %^2 -^ ^1^1 +^2=0' (a+2)(a+l) + p^(a+2) + q^ + a. p^(a+l) + q^ + P^o. + q2 = ^' a^ (0+2)=^ _ a 2 + p^(a+2) + q^ + a^|p^(a+l) + q^ + p^a + Qg = 0, a ,r{a+2)Â» + (p^-l)(a+2) + q^j + a. p^(a+l) + q^ + Ps^ + q2 0. And in general we can write: (z-c) : a (a+n)(a+n-l) + p a (a+n) + ) p a (a+n-m) n-1 m-i n-i + pa + qa +)qa +q =0, (a+n)Â» a n + p^(a+n) + q^ I n-l + } a n-m p (a+n-m) + q + p a + q = 0, m=l PAGE 13 (1-5) a n (a+n)Â« + (p^-l)(a+n) + q^ n-l + > ^ Â™' PÂ„,(Â°'+"-'^) *" Q., + cip + q =0. I_j n-rn, *^ra^ ^m *^n n The first of theoe equations which 1g obtained by equating to zero the coefficients of the lowest power of (z-c) Is called thf3 indlclal equation. This equation determines two values of a (which may or may not be distinct) . If z=c had been an Irregular point, the indlclal equation v/ould have been at most of the first degree. To see this, suppose P(z-c) has a simple pole at z=c and expand it in a Taylor's series. P(z-c) = p_^(z-c)* + p + Pj^(z-c) + . , .+ p (z-c)"+... 00 = 2^Pn(z-c)Â° , P.l ^ 0, n=-l End as before Q(z-c) * q^, + q^(2-c)+...+ q^(z-c)" +... = V q^(z-c)" n=o Now multiply according to the follov;lng scheme [which is equivalent to substituting the series for P(z-c) and Q(z-c) into equation (1-3)] and obtain the indlclal equation: PAGE 14 8 % "^ ci^{z-c) +. . . P_l+Po (z-c) +. . . (z-c) u == (z-c)*^ 1 + aj^(z-c) +... ,a-l u' ^ (z-c) " a + (a+l)a^(z-c) +... u" (z-c) a2 a(a-l) + (a+l)aaj^(z-c) +... The indiclal equation is / \a-i (z-c) : p_j^a = 0. But a ^ 0, and p_j^ ^ or z=c would not be a simple pole which contradicts our assumption. Or if we assume that Q(z-c) has a simple pole at z=c and expand it in a Taylor's series: Q(z-c) = q_j^(z-c)" + Qq + qi(z-c)+. . .+q^(z-c) +... = 2 ^n^^-^)"^ q.i 7^ 0, n=-i and P(z-c) = Pq + p^(z-c) + P2(z-c)' +. . .+p^(z-c)" +... n=o Now multiplying similar to the previously mentioned scheme: PAGE 15 -1 q_^(z-c)" + q^ p (z-c)+ p. (z-c)"+. . . (z-c) U = (2-C) 1 + a (z-c) +. . . u' = (z-c)Â°'"^ra+(a+l)a^(z-c) + . ..1 u" = (z-c)Â°'"^ra(a-l) +... The Indlcial equation is (z-c)"-': q_;, 0, which is a contradiction of our assumption. If both P(z-c) and Q(z-c) have simple poles at z=c. then ,-1 q_j^(z-c)+ q^+. . . P_l-4>o(z-c) +. Â• . (z-c) u (z-c)'^ 1 + aj^(z-c) +... I u' ^ (7-0)Â°^a + (a+l)aj^(z-c)+. . . ,a-2 u" = (z-c)'^-'' a(a-l) + (a+l)'aa^(z-c) + .. The indlcial equation is (z-c)Â°'"-^: p_^a + q_^ = 0, which is first degree in a. Hence, if z=c is an irregular singular point., the indlcial equation is at most of first degree. Now the roots of the indicial equation are called the exponents of the differential equation at the point z=c. Returning to our indicial equation which we shall call F(a), (1-6) F(a) = aÂ» + (pÂ„-l)a + qÂ„ = 0, PAGE 16 10 let a='h^ and 0=7^2 be the roots. Each value for a determines. In order, a set of coefficients aj^,a , ..., thus producing two distinct series solutions, provided that "K -"K A, Â£f is not an Integer (zero Included) . It will be shovm later that v;hen the difference is a non-zero Integer then our proof for convergence of the series fails and if the difference is zero, the two solutions are obviously the same . 1.3 Convergence of Formal Solutions . We now present a proof by induction that these series solutions converge . Assuming that the exponents Ik^ and \ are not equal, choose them so that Re(Xj^ ) kRe(X2) Â» ^rid let Xi->^2 ^ ^' ^Â°^ since the exponents are a = \ ,\ , we can write (i-o), (1-7) F(a) = {o.-\^){a--h^), then for a = A-+n, P(>v^+n) = (A^+n-A^)(X^+n-?v2) = n(n+s) . Let P(z-c) Pq + Pi(z-c) + P2(z-c)* +...+ p^(2-c)" +..., where p = ^^"^(c) "^ n: according to Taylor's expansion. By means of repeated differentiation [^, p. 70] PAGE 17 11 ,(n) (z) nl f P(z) di If we let the curve C be a circle with center at the origin, z=re'^^, then dz=rie^ dS . Let n^ oe the maximum value of |P(z) I on C. Then nl r P(2)dz , / n+l 2tt1 "-' ^ Z nl 2TTi . r \?u)\'\dz\ J7T 10 n'^ M,.|rie^Â°de| o ' 271 Mnl r M^n! 27r M nl _J__ / de = _J__ [9] -iÂ— n / Â„ n "^ -"o n 2Trr "^ 2TTr r o Thus and ^'% M^nl r |PÂ„I s M^r -n Similarly for q : -n qÂ„l s M^r where M^ is the maximum value of |Q(z) | on C. We next show that s UJ -M^r"" + M^r"" = r"**' (l\|-Mi+M^)= i-'X' where M3 = UilM^ + Mg. PAGE 18 12 Choose M to be larger than M^^^M , and M^ and also M s 1. Then (1-8) IpJ < Mr""; | q J < Mr""; |>.^p^ + qj < Mr""; for n = 1 ,2,3, ' Â• ' . Now with the aid of equation (1-6), we can write equation (l-^) as a^P(a+l) + ap^ + q^ = and -(ap^ + q^) a = F(a+1) Now let a = X , then la I = ^ ^ ^ / Mr X M ^ |F(Xi+l)l |l+s| r > since |l+s| > 1. That this is true can be seen by letting \^ = a + pi and 7^2 =7 + 61. Then s = (a+pl)-{7+6i) = (a-7)+i(f2-6) and |l+sl ='V(i+a-7)Â« + (p-6)' . If P-6 =0, or p =6, then a 7^ 7 for we have assumed from the beginning that \^ and Ag ^^^ ^^^ equal. By our choice of ?v^ and Tvg ' Â°' > '>' ^"^ 1 + a 7 > 1 and | l+s | > 1. If a = 7, then p ^ 6 andVl + (p-6)Â» > 1 or | l+s 1 > 1 . We have now shown that a has a maximum for n = 1. n Now we assume that |a^| < M^'r"" , n 1,2, ...,m-l. PAGE 19 13 Prom equations (1-5) &-'^"^ (1-^) '"'e can write n-l a F(a+n) + ) a m m = l and letting a = >\, > p (a+n-ra) + q m Iap + q = 0, a = n n-l n-l n-m p (X.+n-m) + q l^n ^n Now replace n by m and replace m by t. Then (1-9) Iaj jn-l mt p^(A^+m-t) + q^ "^iP q t = l P(A^+rn) ra-l m-l t = l t^l m|iii+s 1 m-l m-l a ^|Mr"* + Mr""^ + mÂ— t < t = l ^('n-t)|a^_J -Mr"* t-i m |l+sm" I For the first summation: m-l (1-10) yia^_jMr-* m-l L t=i t=l -t = M -.-ll^''-^l-.-2l^"'-^---^l^2l^""'''-^l^ll^"'"^' PAGE 20 14 < M M r + M r +...+ M'r + Mr < m ..:i-l ,.!3-2 Mr "I M" " + M"* "+...+ M* + M < Mr""(m-1)M'"^ -nL.m < (m-l)r~"M For the second summation: m-l (1-11) ^ {M-t)|a^^_J -Mr"^ t=i = M -1 -2 (m-l)|a^_Jr-V(m-2)la^_2|r-%... + (l)|aJr -m+l < Ml {m-l)M'"-^r""+(m-2)M"'-2r-"+...+ Mr -m < Mr -m m-l)M +(m-2)M +. . .+ M < M"'r"'"r(m-l) + (m-2) + ...+ 2 + 1 ThuÂ£ iÂ«J < < M r (m-l) . m-l)r M + Mr + M r (m-l) m^l+sm -1 Mr ""M"" -r "M"' + Mr "" + M**"r *" ^ (m-l) m*|l+sm" i PAGE 21 15 < mrM-rM+Mr +Mr ^ (Â•'n-l) m" I l+sm~ / r""rr(m+i) ^ I -1 I 2mll+srn | But il+sm~^i > 1, for letting s = ^^1-^2 " '^'^^ " (7+<5i): m ' m ' m and Im+s! VCa-T+m)' + 0-6) Â» > ni , since "K. and "K^ are distinct. m > 1 Thus Im+sl and |aj PAGE 22 16 If < |s! < 1, then 1s u-sr^i ' J ~ ^><-tj>'''f or -fi S |l-ls| I , J = 1,2,?,...^ Proof: Let s = a+pi, (a,p real); then < jsj =VaÂ»+p' < 1 ^-! = 1a pi 8 jo. a = V'^-l''^'7>'=V^-T^l^ Vl ^Vg'-t^' ^ g'-Hp* =A/(^1 Y^Li+fl! J J* V " J Â§ 1 vi!Â±ir , 1 .Va Sj-AS +PÂ« = 1 |s| = 1-|3 If |s| > 1, then Proof: Let Is I = t + 9, where t is the largest Integer in |s| and Â§ fc* < 1. 11 a ^i W (i-H)-.fii a , A Â« 1 1^ + a'+P " J J" 1 2"'/aÂ»+pÂ» _!_ g'+p' ^ _Vgf+pf. J 1- PAGE 23 17 We desire 1-'-r'to have a minimum value larger than zero, where J=l,2,3^.-.^ t-n,..., t-1, t, t+1, . . . t+-t, . . . (n=l,2,3, . . ., t-i;t=l,2,3, . . .) . Difficulty could arise. Then the absolute value however. PAGE 24 18 Tlierefore, (1-12) 1-r-rrl g T and 1t+9 t-nl " t I" t+^ where n=0,1^2,..., t-1; t=l,2,3, Â• . Â• ', < 9 < 1 1-0 t+T ' Inserting |s| for t+0 and J for t-n and t+-t in (1-12) we see that, for |s| > 1 and < < 1, either 1j e ^ t Â°^ 1J 1-0 tTl ' depending on the values of and t. Now consider the case where 0=0 (|B|=t) and J=t = Va'+3* , p>^0. We wish to find a 6 such that if s a+pl, then s 11g+pi 1 Ul+^ J VoM^' I VaÂ»+p" a pi| 1_ " Va'+P* + 5 2t"-2at 5 6Â» , 2t(t-a) a 6Â« . However, since 1 s | =t > 1 and t =Va'+pÂ« > a, if we have given any value of s (|s| > 1, Pt^O), then a value of 6 can be determined such that < 6 < 1. Referring to (1-12) with replaced by 6, we have for |s| >1, 0<6 <1, that PAGE 25 19 1t+6 6 Â„ , I , t+6 , _ and |1^^1-6 t+T where n=0,l,2..., t-i; 1=1,2 ,'j>, .. . Summarizing our results we have: If < Isl < 1, then -! ^ 1s For |s| > 1, |s| ^ an integer, (a) if < e < 1, then one of the following t ' 1i '-^ , t+e\ t+T ' (b) if e 0, then one of the following \ -! PAGE 26 20 Now we present an inductive proof that |a'| < MÂ° ^" r"" ' n ' ^ Again using equation (1-6) with a now replaced by 7^.^+ n, we have (1-15) P(^2+'^) = (X2+J^-^i)(^2+"-^2^ " n(n-s) . If A, is replaced by 7v , the inequalities of (1-8) will still be true. However, they will also be found valid if M is chosen larger than Â— as well as M, ,M ,MÂ„ and M ^ i. Let this be done. Thus, IpJ PAGE 27 21 Taking the absolute value of each fildc and using (1-135) v^ith n===l, v/e obtain the Inequallt.v i^2Pl+qi Mr -1 ' |P(^,+1)1 U-sl Â• Next, assume a^l < M K r n=l,2,3, . . ., ra-1, and show that m m Â— m laI PAGE 28 22 Similar to (l-lO) and (1-11) with la' .| < (M<)'V~"', M^ > 1, m-1 in-1 ., -m ^ |a;_tl'W^"* < (ni-l)(MK:)'""^ Mr t-i -mm jii-l (m-l)r M K , and m-l I (n-t)|a;_J.Mr-* Â».'n m-l -m,.m m-l ,. -m ,,m -m m / , \ n-1 mr M k: -r M < + Mr + M r ^ (m-l ) k m'|l-sm~ I ra-l Now since (M/c) ~ > 1, ,. Â— m ,,m Â— m m-l ,, Â— ra/., ,,mÂ— 1 mÂ— 1\ Mr M r K = Mr (l-M < ) = Mr -"'ri-(M<)'"-^l Is negative . Thus, |a'| < ' m ' ^ ^_-m m m-l , -ni.,m m / -, \ m-l mr M fc + r M ^ (m-l) ic m' I l-sm~ I PAGE 29 23 / f M K (m-fl) 23a 1 1 sm~ I 2m m.-jn m r M k: , for m > 1 Therefore, la'l PAGE 30 and 2h We have thus obtained two formal solutions u^(z) = (z-c) 1 + oo n=l u_(z) = (z-c) \^ 1 + Ja;(z-o)" n=l Which are uniformly convergent series of analytic functions when |z-cl < rM' and fz-c| < rM"'^^" , respectively, provided that arg (z-c) is restricted in such a manner that the series are single-valued. Consequently, the fonnal substitution of these series into the differential equation is justified for ^j^-^g not a positive integer. These solutions are valid in the vicinity of a regular singular point. 1 .^ Second Solution in the Case where the Difference of the Exponents is an Integer . We now derive a second solution in the case where the difference of the exponents is an Integer. When >^j^-X2 = a is a positive integer or zero, the solution ^2.^z) may break down or coincide with u^(z). Try the change of variable u = \i^{z) -^ and substitute this into (z-c)* u" + (z-c)P(z-c) u' + Q(z-c) u = to determine the equation for ({). PAGE 31 25 u = u^.(t), U' = Uj^ Â•^' + u^ <|), u" = Uj^ Â•(!)" + 2u^ (j)' + uJI ({). Substituting in we have r~ (z-c)' u^(l)" + 2u^(t>' + u^'0 + (z-c)P(z-c)(u^(t)' + u^(t)) + Q(z-c) u^-(l) = 0, (z-c)"u^0" + 2u^(z-c)Â» + (z-c)P(z-c)-u^ j (J)' + (z-c)* u^' + (z-c)P(z-c) u^ + Q(z-c) u^ I ({) = 0, Since Uj^(z) Is a solution (z-c)Â» u"^ + (z-c)P(z-c) u^ + Q(z-c) u^ = 0, and (z-c)2 u^(t)" + 2u^(z-c)2 + {z-c)P(z-c)-u^ 0' = 0. Dividing by u. we have (z-c)Â» (|)" + 2 Â— (z-c)Â» + (z-c)P(z.c) (J)' = 0. To find a general solution let ({) ' = Y. u: (z-c)Â« Y' + ^ ' dz 2 ~ (2-c)Â« + (z-c)P(z-c)j Y 2 ^ (2-c)Â» + (z-c)P(z-c) = 0, 0. PAGE 32 26 Separating variables and integrating, r _u m^i[^*%^\ dz = In B, In Y = In B 2ln u^ Y = B u^^ e z-c) ,Â„ Â— -f dz , Now since Y = 0' , d(J) B u^ e -/^M dz dz where P(z-c) = p^+ Pj^(z-c) + P2(z-c) '+Â• . .+ p^(z-c)%...^ = p^ln(2-c) + Pi(z-c) + p^ ^ 2^ "^"Â•. Hence, n [-P In(z-c) p (z-c). . .-p (z-c)"-. . .] = A + B / u-^ e Â° " dz r ^ r, [-Pi{z-c)-...-p (z-c)"-...] = A + B / u~^(z-c) Po e ^ "" dz n n = A + B / (z-c)"H> u^ -p -2 n=l dz . PAGE 33 27 But since >v r u,(z) = (z-c) '[l +^a^(2-c)" n^l '^ o 1 PAGE 34 28 s-l = A + B /(z-c)-'-'d2 +J 2^g^(E-c)"-'-'d: n-1 dz n = s+l A + B s-l (z-c)~ V ^n / nH-s . , V ^:r"2."^^^^"''^ + gg ln(2-c) n=l s n = s+l The general solution, analytic in C except at z=c, is u = Uj^(z)-(t) = A Uj^(z) + B u^(z) -gg 'InCz-c) + u{z) where u(z) = u^(z) s-l / \-s V^ ^n / \n-s n=l V ^ (z-c) L. n-s n=s+l n-s and u,(2) = (z-c) '[l +2^a^{z-c)Â° n=l PAGE 35 29 Rev/riting, we have 00 ^ ^"^ g n=l n=l 00 V ^n / \n-s + 2. ^ <'"' . n = s+l !-l , s^2ri +ya (2-c)" (z-c) i u " g. s l_^ s-n "" ' = (z-c) \n=l where h are constants: n K ' S-1 S and 00 00 oo n=l n=l n=l s-1 00 > Â— !i (z-c) + ) Â— -r(z-c) ^ n-s ^ ZL, ""^ n=l n=s+l PAGE 36 30 When s = 0, we can write (l-l4) in the form CO (J) = A + Bj^ 1 +yg^(z-c)Â° (z-c)"^ d n=l 00 = A + B r jT (z-c)"^d2 + J y g^(2-c)Â°dz I = A + B / \ n+1 gÂ„(2-c) -1 n=l Therefore, when 7\. ^ Ag* the second solution Is u = u^(z).(J) = A u^(z) + B u^(z)rin(z-c) ^^tIi (z-c)"-"' Thus It is seen that when Aj^-Xg is an integer, the second solution involves a logarithm, except when g = 0, A practical way of obtaining this second solution is to first obtain the solution Uj^(z), and then determine the coefficients in a function u^(z) = y b^(z-c) \^+n n=o by substituting u ^ u^(z)ln(z-c) + u,(z) in the equation and equating to zero the coefficients of the various powers of z-c in the resulting expression. PAGE 37 31 Example: Find the solutions of the equation 1 2 u"+Â— u' -mu = z regular near z = [8, p. 201]. Equation (1-2) with c = becomes (1-15) z^u" + zP(z) u' + Q(z) u = 0. Thus, If we multiply our given equation by z^ we have 2m, , 22 z u + zu Â• m z u = 0, where P(z) = 1 and Q(z) = -m^z^ . Assuming 00 u = z ) a z , n=0 then -m^z^ I u = a^z%. . .-3n^^""-2n.2^^-^'"'^Â• ., ^o ^ Â° z ; u' = a^az^-^+... + (a+2n+2)a2^^22'^"^^""*"^+-" z^ I u" = a^a(a-l)zÂ°'"^+... + (a4-2n+2)(a+2n+l)a2j^^22Â°'"^^"+.-. Equating the coefficients of powers of z-c to zero, 2 Â•Â• a(a-l) + a = 0, a = and a 0,0. ^a+2n+2 2 . / Â„ Â„ V , , , ^ Â• -^ 32n"*'^Â°''*"^""^^)^2n+2"*"(^"^2"+2)(Â°'"*'2n+l)a2^^2 = 0. PAGE 38 If n = 0, If n = 1, If n = 2, 32 2 m (a+2n+2; 2 ^ Â«2n a. 2^(n+l)' 2 ra a^ 2 4 ma ^ ^o 2 6 ma. ma 4 o a^ = 2^(3)^ 2^(31)' m a o 2n 2^n(^Â„^,j2 > Therefore, If we arbitrarily choose a =^ 1, ^y m^"z^" ""^ "A 2^"(nl)2 For the second solution, let 00 u = Uj^lnz + \ t)jjz" n=o and substitute this into (1-15). n=l,2,5, . PAGE 39 35 Znb z"~ u" = u^ln z + 2uj[ (^ I ^ u^ (^ -2 ^ + 2^ n(n-l)b n-2 Z 00 00 u"z^ln z + 2u^z u^ + ^ n(n-l)b^zÂ° + u^zln z + u^^ + \ nb^^z" n=2 n=l 00 2 2 ^ 2 V u n+2 ^ u.m zlnz-m >bz =0. n=o Since the coefficient of In z Is 2 2 2 z u'' + u'z m z u, =0, we have left en 00 2u'z +)nbz -m >bz =0, n=l n=o 00 2u'z + y (n^ m^z^) b z" = 0. n=o Now, taking the derivative of our first solution u^, v/e obtain 00 2n 2n-l Z PAGE 40 34 so that 00 / Â—9;; 5~ "*" ^ (n -m z ;b^2 ^ 0. n=o Equating to zero the coefficients of like powers of z, we have o z : b^'O = 0, b^ = arbitrary constant. 1 z : b^ = 0. 2n 2q Â— r7T + 4n bm bÂ„ Â„ = 0. 22n/ ,x2 2n 2n-2 2n 2 n ni b_ Â„ 2n 2n-2 n m 2^''{nlf n=:l,2,3, . Choose b^ = 0. Then for n = 1, 2 2, m b "2 = m -m 4 2^(1!)^ 2^(1!)' If n =2, b. = -, r 2 ,^4 rn_ , 2m 2' 1 1 1 2^11)2 ?^ b. = -m 4, ,2 (1 +|) 2 (21) '^ If n = 3, "^'7 2, ra b 4 TT^Tp" J PAGE 41 35 -1 m 2 (2!) J (1 +|) ^rs yn 2"(51)' 2Â«(3.)2 V'^2 3 Finally, -m 2n 2n 2^"(nl)2 2n,. For n odd, H =l+-5-+-=-+...+ -. n 2 3 n n z : ""V^'^n-a = 0' n n odd. since b = 0. Therefore, the second solution Is 00 2n-2n m H_z ~2H": Â— 72 n=l ^ ' 1.5 Solutions Valid for Large Values of \z\ . Let us now consider solutions valid for large values of | z | . To do this we let z = Â— in equation (l-l) . If the solution of the transformed equation is valid for sufficiently small values of Izj^l, then the original equation is said to be valid for PAGE 42 36 large values of | z | . If the point z^ == is an ordinary (or regular or irregular) point of the transformed equation, then the point at infinity is said to be an ordinary (or regular or irregular) point of the original equation. We begin with the change of variable z = Â— in the ^1 equation (1-1) 2 d u , / X du , / . Â„ Â— o + P(2) Â— + q(z) u 0. dz dz z Â— ; ^1 ^1 =z' ^^1 = Â— ^2 z du dz du dz. dz, dz 1 du 2 , z dz. d^u dz^ 2 du z dz. d^u . ^^1 2 o iJ z 6z^ dz 2 2 d u , 3 du , 4 d u Â— 2 = ^^1 Â— + ^1 TT dz dz, dz Substituting in (1-1 ) we obtain 4 d^u ^ dz. 2z, ^2 ^ ^1 P du dz, + q ) u = 0, Dividing by z^ d!u dz? 2_ LZ, -P ^ I, du dz. ^^i z. Â— u = 0. ^1 PAGE 43 37 Nov/ Zj^ = (l .e . z = ") will be an ordinary point when the coefficients of -jÂ—and u are both analytic. This requires p and q to have the following forms: E^qjanding about z = <Â»: P 2 3 P(z) z +-2 +-T+---, z z * B B B q(z) = -^ +_ + -g. +... B / 0. z z z ' Zj^ = will be a regular singular point when 2 i p i i ) and q ( i ;Â• i_ are analytic at z, =0. This means that p and q must be: Ci / 0, *l^lj = ^2^? + ^3^ + D^2^ *Â•Â•Â•, ^2 ^ 0. Expanding about z = Â»: P(2) =-^ +-J +-4 +... c, / 0, ^1 S . C, z z Â°2 ^3 ^4 q(2)=-^+Â— 3-+-^+... D/ PAGE 44 ^Â•8 Further, it might be helpful to know for what values of Â•t, m and n z = <Â» will be a regular point in the following equation having polynomial coefficients (p / and -t, m, n non-negative integers) . (X, -t" 1 \ II / Di inÂ— 1 \ I p^z + p^z +...+ p^) u + (q^z + q^z +...+ q^) u + (r z" + r,z"~^+...+ r ) u = 0. ^ o 1 n' Dividing by the coefficient of u", U+(-Â— Z +Ej^Z +...ju'+fÂ— Z +Fj^Z +...ju=0, We must have m--t = -1 or t = m+1 and n-t = -2, I = n+2. Therefore, if z = Â«> is a regular singular point, then I = ra+1 = n+2. 1.6 Irregular Singular Points and Confluence . We have already shown in section 1.2 that near an irregular singular point a second order differential equation cannot have two solutions because the indicial equation is at most of the first degree; there may be one or no solutions of this form near this point. If a differential equation A is obtained from another equation B by making two or more singularities of equation B tend to coincidence, such a limiting process is called confluence. Equation A is called a confluent form of equation B. PAGE 45 CHAPTER II Confluent Differential Equations 2-1 Differential Equation with Four Singular Points . We shall now develop a general form for a second order linear differential equation which has every point except ^1' ^2' ^3 ^"^ ^4 ^^ ordinary points (z = Â« being an ordinary point) . These four points are to be regular singular points and let tholr exponents be a , p at a (r=l,2,3,4). Then the form of our equation will have to be dz^ I^^H(^)]l^I m. (z-a^)' z-a + J(z) ) u = 0, where k^, l^, m^ (r-1,2,3,'^) are constants to be determined, and K{z), J(z) are polynomials In z. Let >v . ^ /_ xA+l u = h^(z-aj^ + h^(z-aj + . . . \ ^0> 4 PAGE 46 40 The Indlclal equation for z = a, is (z-a^)' Then so that h A(>v-1) + h k,>v + h t, = 0, o ^ ' o 1 o 1 ' A^ + (kj^-l)A + I =0. >^ = ^l>^l a^ + Pj^ = 1-k^, k^ = 1 ttj, p^. and Similarly for the other regular points, so that 1-a^-p, dz ^'Â—1 r jwtV 1 ^ a 3 + m, + J(z) > u = 0, r=l * r' If 2 = 00 (or z=0, z = Â— ) is to be an ordinary point, 2 4 2z-z p(z) and z q(z) should be analytic at z = oo, or 2 P( ? ) ^^^ ~T ^( -iT-) should be analytic at z, = (section 1.5) Now 2_ Z, 1_ z. 2_ Z, 1 r = l 1-a -p r r Z, -a PAGE 47 41 4 2_ V ""'"r-^r l_u( L. ^ _ y (l-".-Pr) / ^ 1 \ . 1 Â„ r i_^ r=l 1 4 2 (1-a^-p^) / 2 2 Z/_, Z, V r 1 r 1 r=l 1+a z, + a z, +. . . -t^H^ 4 2 y (l-a^.p^) ^ ^" ~r '' r' 00 *lK<-h^{k>' Z. 'Â—' Z, ^ 1 1 n=o 1 where hÂ„ are constants. When z, = 0, we need n 1 ' and or 2 ^(1-a^-pJ = 0, ^ (O^r-^r) = 2. r=i PAGE 48 k2 Rewriting q(z) in the fonn q(2) = n. n. (z-aj^) {z-a^){z-a^) {2.-a^y{z-tx^){z-a^) n. n. (2-33) (z-a^)(z-a^) (z-a^)^(2-a^)(z-a2) + J(2), then ~ ^ I z~ n. (l-a^zJ^Cl-a^zJCl-agzJ n. n. (l-a22j^(l-a3z^)(l-a^Zi) (l-a3Z J'^d-a^Zj ) (1-a^z^) n ' j' ' (l-a4Z^)^(l-a^z^)(l-a2zJ zj V ^i ' which is analytic at z^^ = 0, if J ( -^^ ) = 0. To evaluate n^ (r-1,2,3,'^) , we have n. n. (z-a^)^(z-a2)(z-a3) (z-a2)^(z-a^)(z-a^ ) '3/\"4 n. n. (z-ag) (z-a4)(z-a^) (z-a^ )"(z-a^ ) (z-a^) ^I r r r 2 z-a ri^-Cz-a^) '^ r PAGE 49 43 Then clearing of fractions. nj(z-a2)(z-a3)(z-a^) + HgCz-a^) (z-a3)(z-a4) + n3(z-aj^)(z-a2)^(z-a^) + n4(z-aj^) (z-a^) (z-ag)^ = a,p,(z-a^)2(z-a3)2(z-aj2 -^ a^^^ (z-ci^)^ {z-c^^f {z-a^f + a3P3(z-aj2(2.a2)^z-aJ^ + 04^^(2-3 J^(z-a2)^{z.a3 )2 + m^(z-a^)(z-a2)^(z-a3)^(z-a^)^ + m2(z-a^)'^(z-a2)(z-a3)^(z-a^)^ + m3(z-aj^{z-a2)^(z-a3){z-a4)^ + m4(z-aj2(2-a2)^z-a3)^(z-aj. To determine n., let z = a : ni(a,-a2)(a^-a3)(a^-aj2 ^ a,p, (3^-82)^(3^-33 )^a^-aj\ Similarly, "2 S^2(32-33)(32-34)> "^3 = Â«3P3(S-34)(33-Â«l)' ^4 Â°'4p4(^-^)(Â«4-32)- PAGE 50 44 Therefore, the general form for a second order linear differential equation with regular singular points at z a^ (r= l,2,p,4) is (2-2) H 4. V '^'^^'^ r du 2 l_^ z-a dz + r=l aj^p^(a^-a^)(a^-a3) _ (z-aj^)^(2-a2)(z-a3) + ^2^2^Q2-^3H^2-^4 ) _^ Â°'3^3(^3-^4)(S-^l) (z-a^) (z-a3)(z-aj (2-83) (2-a4)(z-a^) a^p^{a^-a^)(a4-a^) (z-a^) (z-a^)(z-a2) u = 0, where ^ (a^+Pj = 2, a and ^ being the exponents at z = a^* To express the fact that u satisfies an equation of this type we will write (2-3) ^1 ^2 ^3 ^4 ^ = \ O'l ^^2 ^'S Â°'4 . Pi ^2 ^3 ^4 Â• 2.2 Differential Equation with Singular Points at z =0^ 1, <Â»> gÂ°. Let us now consider the confluent case where a =0, ^3 = 1* a^ a. become infinite 4 PAGE 51 ^5 PlrsU, let a^ and a^ = 1, then (2-2) becomes (2-1,) 5-^ + r ^Â— ^-H ^-^ + Â—A^ -h -:^4Â— dz z-a. z-1 z-a. du _ a3P3(l-a,)(l-a^) (z-aj^z(z-l) z^Cz-DCz-a^) (z-l)^(z-a4)(z-a^) a^P^a^(a^-l) ^ a^Pga^ ^4^4(^4-^1^^^ u = 0. (z-a^) (z-a^)zJ Before letting a^ become infinite, rewrite the equation in the form dz 2-a, z-1 z-a. dz ^^2^2 Â«3P3(i--l>l-^) aiP,a^(a^-l) ^ ^ [(z-a,)^z(z-l) ,2(^_^)/ |_ _^\ ^ (^.,)2/ |_ .A,.3^) ^M^at ;> -1^ (z-a^)z-l u = 0. J Now let a. become infinite: 4 (2-5) ^4 dz L 1-a^-p, l-<^2-^2 l-^'a-Pa z-a, z-1 du dz a,P,a^(a^-l) L(z^^2^2 a3P3(l-a^) QI4P4 (z-aj^z(z-l) z^(z-l) (z-l)"(z-aj (z-aj: u = 0, PAGE 52 46 which may be represented oy (2-6) u / ^ a. 00 Â«2 ^3 ^4 2 V ^1 ^2 ^3 ^4 . This equation Is actually more general then one might assume, for by means of the linear transformation (84-33) (z-a2) u (82-33) (z-aj ' any three points z = Qg' ^3' ^4 ^^^ ^^ carried into u = 0, 1, 00. Finally, we want to let a^^ become Infinite to obtain the confluent case with singular points 0, 1, <Â», <Â», Rewrite the equation (2-5) in the form dz^ l-a,-p, ^ l-g^-p, ^ l-a3-P3 z-a. r^^k z-l ^'2^2 du dz L(|..l)z(z.l) ^(-^) (z-l)\|--l a^p 4^4 (z-aj^)z _ u = 0, PAGE 53 47 and when a. becomes Infinite we have the confluent equation (2-7) ^ + dz r l-^2"^2 ^-Â°^3-^3 Z-1 du dz l-z(z-l) z^(z-l) "3P3 (2-l)^J u = 0, with singular points at z =0, 1, <Â», Â«. Let us proceed to find solutions of this equation Multiplying the equation by z (z-1) , we have (2-8) z2(z-l)^i^ + dz (2-a,-p,-a^-pJz 2 ^2 3 "-^3 + (-3+2a2+2P2-><Â»3+P3)2^ + (l-<^2"^2^^ du dz [(^1^1 + ^3^)2^ -^ (-^1^1 ^2^2)^ *^^2^2] ^ = 0Assume the solution u = C z^ + r z^-^^ 4-...+ C z>^-^"-2 4. C ^z^-^"-^ o 1 n-2 n-1 + C^z^""" +... Co ^ 0, then u .A-l C "Kz"' + C, (X+l)z +...+ C ^(7v+n-2)z o 1 * ' n Â— ^ X+n-3 + C ,(?v+n-l)z^'*'""^ + C (X+n)z^'^"'^ +... n Â— i n PAGE 54 48 u" = C^-\{-k-l)z^'^ +...+ C^_2(X+n-2)(>.+n-3)z^"^""^ + C^_^(X+n-l)(X+n-2)z^+"-^ + CÂ„(A+n) (7v+n-l)z^+"-^+. n If these equations for u and Its derivatives were now substituted In equation (2-8) and the coefficients of the powers of z equated to zero, we would obtain the following results . The indlclal equation is z : (2-9) Then Finding C^^, 7v(^-l) + 7^(l-aÂ„-PÂ„) + a ft_ = 2 ^2 2'-^2 Â•K^ {a,^+^^)\ + 02^2 = ?v = oig.Pa2 Â•Â• CÂ„ Â„(?v+n-2)(?v+n-3)-2 C^ (X+n-l) (>.+n-2) + C^(X+n)(}v4n-l) + C^_2(X+n-2) (2-02-^2-03-^3) + C^.l(^+n-l)(-5+2a2+2P2+ot3+P3) + C^(X+n) (1-02-^2 ) + C^.^^^'iPi^af^a) + CÂ„_,(-a,p,-a2P2) *.C,a2P2 = 0'n-2 -(X+n-2)(A+n-3)-(X+n-2)(2-a2-P2-a3-p3)-(a^P^-H3i3^) + C n-l 2(X+n-l)(X+n-2)-(7v+n-l)(-5+2aÂ„+2p--Kx~+pÂ„) 2 ^2 "^3 "-^a + ^'if^l + =^2^2 (>v+n)(X+n-l) + (>\+n)(l-a..-pÂ„) + a,p 2 ^2 Â•2^2 PAGE 55 i+9 Simplifying the denominator by using the indlcial equation (2-9) v;c have n \ n-2 -(>v+n-2)(X+n-l-a^-P2-a3-P3) Ci^Pj^ 03^3 + C n-l (X+n-1) (2>+2n-l-2a^-2P2-a3-p3 ) + a^p^ + a^Pg n(2>v+n-a2-p2) Let us compute a few terms of our series solution to see If a general term can be obtained . Ci = X{2>^+l-2a2-2P2-a3-P3) + a^P^ + 33^2 (2^+l-aÂ„-pÂ„) 2 ^2 Using the indlcial equation (2-9) we can simplify and write Ci = (1-03-^3)7. + a,P, 02^2] 2AH-l-a2-^2 ^2 M ^o +3-2a2-2P2-cx3-p3) + a^p, + a232]| (A+1)(2X 2(2X+2-a2-62) PAGE 56 50 Substituting C. into this expression; C /(2X-hl-a,-pJ -A(?v+l-a2-p2-^3-P3 )-Â°'iPi-Â«3P3 (1-^3-^3)^ + a^^^-a^^, (7v+l)(2X+>2a,-2p_-a.-PÂ„) + '^iPl + Â«2^2 2(2X+l-a^-P^)(2X+2-a2-p2) It seems apparent from the appearance of C. and C., and from the fact that C^^ is a three term recurrence relation, that a greal deal of v/ork might be required to write these two solutions of (2-7) with a general term, thereby making available a form suitable for further development, oince equation (2-7) has five undetermined parameters (ciiPi>*^2'^2 013*^3), let us consider the conditions necessary to insure a solution with a two term recurrence relation. This can be readily accomplished by applying Schef f e ' s Criteria [6, p. 240] which Crowson has presented and proved in his dissertation [1, p. 114]. 2 . 3 Scheffe's Criteria Applied to Confluent Case . The criteria to be applied is: Necessary and sufficient conditions for a solution of a second order linear ordinary differential equation. P2(z)u"(z) + p^(z)u'(2) + p^(z)u(z) = 0; PAGE 57 51 to have a two term recurrence equation, relative to the point z = 0, is that in some neighborhood of the point z = 0, (2-10) Pj(z) = ^J h^ h z^, where m is an integer, h is a positive integer, and S,, T. are constants such that S^ T / for some i = 0,1,2 and J = 0,1,2. In assigning values to the exponents, it can be seen from equation (2-4) that none of the exponents can be a function of a . Since the sum of the exponents must be 2, if there are any e^qponents which are functions of a^ there must be at least two such exponents, a^t^j. (r = 1,2,3) could not be functions of a., otherwise the coefficient of -3Â— will be undefined when a. becomes infinite. If a, and dZ 4 4 p. are functions of a , then the terra (z-a^) (z-a^)z would be undefined as a^ becomes infinite. Thus the exponents can not be functions of a. . 4 Next, suppose that two or more exponents in equation (2-6) are functions of a^^ . Such an equation might be / n 1 00 U = a, a^-t+f 1 a. P. P, PAGE 58 52 v;here the sura of the exponents Is f + a2 + Pg + 03 + ^3 + t + ^4 =2, f, I and t are arbitrary constants. Using (2-2) we can write this equation in the form u" + 1 + 1 Â±. + z-a, z ^-^2-^^^ z-l u' + (^4-|-)a^(a^-l) _(z-a^)2z(z-l) ^'2^2 a3P3(l.a ) + + z"(z-l) (2-l)-{z-aJ ^4 (-^1^-17-^ t) (z-a^)z u = 0. u" + a. ^-f -^1 a 2 y ^1 + = = + l-a.,-p 3 *-3 2-1 U y 4+ YiJ a. 1~1 Z(2-1) ^'(z-1) -.Pa , ^^^< H -' (z-l)^(l--l 2 1 \, u = 0. Letting a^^ become infinite: u" + l-aÂ„-pÂ„ 1-a -p I + Â± Â— Â± + 1 Â— Â£ z z-l u ^2^2 Â°^3P3 ^4^ t + _ + Z(Z-I) Z^(Z-I) (z-1)^ 2 u = 0, PAGE 59 53 Clearing of fractions we can vn?ite (2-11) z^(2-l)^u" + tz^(z-l)%(l-a2-P2)z(z-l)^ + (l-a--pjz^(z-l) u { l,z(z-l)-a_pÂ„(z-l) + a^P-z + p.-tz(z-l) Â•2^2 u 0, where the singular points are at z = 0, 1, Â«, Â«. Applying the criteria (2-10) we have for J = 2 P2 = S TÂ„z' 2 2-in 2/ , s2 Z ^ Z (z-1) , Â„ 2-m _ h-m+2 4^3, 2 SgZ T^z = z -2z + z , but this Is Impossible. However, this difficulty can be avoided by reducing the coefficient of u" by a factor of at least z-1. To do this UgPg must equal zero, so let ^3=0, Thus, with ^3=0 and dividing equation (2-11) by z-1, we now have 2^(z-l)u" + tz^(z-l)+(l-a2-Pj2)z(z-l) + (l-a3)2^ u ^z-a p., + ^.lziz-1) u = 0. (z^-z^)u" + Lz^ +(2-a2-P2-a3-t)z^ + (-14^2+^2)2! u [ + P.^z + (l-p.)l,z a^p 2^2 u = PAGE 60 5^ Nov; returning to our criteria (2-10), P 2 ^2^'""" T^^^-"^-'' z'-E^ where h-m+2 = 5 and2-m = 2. Therefore, m = 0, h = 1, Sg = Tg = -1, S2T2 ^0. We need Pi = S,-T,z z = >tz + {Z-Qk^-^^-a^-CiT?" + (-l+a2+^2^2' so that -t = 0, Tj = Â°'2"^2 * Â°^3 '2' Sj^ = ^2^P2"-'^"^ Pi = (2-Qt2'^2'S)2^ + (a24^^-l)z. Finally, we have o o where S^ = *Â°'2^2 ^"^^ "^o "^ ^' = -^^2' Thus, if ^3 = in equation (2-11) the resulting equation, which has a two term recurrence relation for a solution, is (2-12) z"(z-l)u" + (2-aÂ„-p_-a3 )z^+(a,-^Â„-l) 2 ^2 u'-OgPg u = 0, Tlie requirement that f + aÂ„+pÂ„+ao+t+0. = 2 is actually not restrictive, since f, t and p^ do not appear in our final result. Consider the alternative of having the exponents in (2-6) be functions of a^^ and assume that neither a nor a^ appear in the exponents. Apply the criteria (2-10) to equation PAGE 61 55 (2-7) u" + 1-0^2-^2 ^ ^-^3-^3 Z-1 u' + Â«1P1 <^2^2 ;(z-l) 2^(z-l) a^P 3^3 (z-1)' u 0, or clearing of fractions 2 .,11 (z-1)^ u" + (1-a -pjz(z-l)2+(l-a3-fÂ„)z^(z-l) '2 '^2 U a^p^z(z-l) a2p2(2-l) + CaPgZ' u = 0, where Y (cir-^^r) = 2 r=l As before we must remove a factor of z-1, therefore let Pg = so that agPg = 0. Then z^{z-l)u" +r(l-a2-P2)2(z-l)+(l-a3)z^1u'+ra^P^z-a232j u = 0, (z^-z^)u" +r(2-a2-p2-Â°'3)2^+(a2"^^2-^^^V''*'[''l^l^''Â°'2P2V " Â°' Applying the criteria (2-10) we have when J = 2 Po = S2"'r22 2-m 3 2 Z = Z -Z > Where h-m+2 = 3 and 2-m = 2. Thence, m = 0, h = 1, Sg = T, = -1, S2T2 /^ 0. Further, Pi = S,-T,z z = (2-a_-pÂ„-a^)z'^+(a.+a_-l)z 2 ^2 3 2 '-^2 PAGE 62 55 and Tj^ = a^-^^ + a^ -2, S^ = a^+Pg 1. Finally, With T^ = -a,p,, S^ = -a^^^. The equation with a solution having a two term recurrence relation Is, therefore, (2-13) z'^(z-l)u" + (2-02-^2-^3)2 + {cL^+^^-l)z u ^'l^l^ ^2^2 u = 0, where the only restrictions are that In equation (2-7) a , P (r=l,2,3Â»^) contain no function of a^ or a^ , and P3 = 0. Since equation (2-12) can be obtained from (2-13) by letting oiiPjL "^ '^^ "^^^ ^^ obtain solutions of the latter. .A "1^12-^2^2 (2-a2-P2-Â«3)2' + (02+^2-^)2 3 2 z -z u c z"+...-k; ,z^"*'"'^+c z^"^Â°+... c / 0, o n-1 n ,0 .>^-l u' = Cq>vz'^" +. . .+C^_^(^+n-l); X+n-2 u" = C^A(>v-l)z^"^+... + C . (?v+n-l)(X+n-2)z X+n-3 + C^(X+n)(>.+n-l)z^"^Â°"^+... The Indlclal equation at z = is -?v(X-l) + {a^^^^'l)-K agPg = 0* )v^ (a-+p,)7v + a^p, = 0. zN 2^2 2*'^2 PAGE 63 The exponents are 57 ?v a^,jp.^. ^ ' C^_i(>^+n-l)(X+n-2)-C^{X+n)(^+n-l)+C^_^(>v+n-l)(2-a2 -^2-^3) -^ C^(>^+")K^2-1) + ^n-l^l^ ^n'^2p2 = 0, -c -1 c = (7s+n-l)(?v+n-2)+(>v+n-l)(2-a2-P2-Â°'3) * ^1^1 -(X+n)(?v+n-l) + (7v+n)(a_+pÂ„-l)-a^^ 2 ^2 '2^2 'n-l C = n a^P^ + {h+n-l){-h+n-a^-^^-a^) (A+n)^ (aÂ„+pj(7v+n) + a^p '2 ^2 2^2 'n-l C = n a^P^ + {},+n-l)i-K+n-a^-^^'a^) {Â•K+n'a^){'K+n-ii.^) Let X = a^: C = n Cn-i[^iPi + (a2+n-l)(n-P2-a3) 2 ^2 Cl = C2 = a^Pl + 02(1-32-^3) (a2-p2+l) 2(02-^2-*^) PAGE 64 58 CÂ„ = ^iPi + Â«2(l-P2-Â°'3) '^iPl + {a2+l)(2-(32-a3) Therefore the solutions of (2-13) are: a. u, = C z 1 o Ii(a2-P2+1) z +. . . [a^P,-Ki3(l-P2-a3)HaiPi-^(Â°^2-^^)(^-t^2-Â°^3)3 2l(a2-P2+l)(<^2"^2+2) [a^P^-Ki2(^"P2"Â°^3^^ Â•Â•Â•^Â°'l^l"^(Â°^2"*'""^^(""^2~Â°'3^^ n nl(aÂ„-pÂ„+l)...(a_-pÂ„4-n) z +. "2 ^2 2 ^2 (2-lif) U^ = C^ [a^P^-KigCl-Pg-aa ) ] Â• . .[a^p^+(a2+n-l) (n-Pg-^a ) ] ^^ 00 n=l and similarly n I (02-^2+1) "Â•iÂ°^2'^2'^^^ P. (2-15) U2 = C^z " < 1 [a^^^+^^il-QL^-a^)] . . .[a^p^+(P2+n-l)(n-a2-a3)] ^ nl(pÂ„-a-+l) ...(p.-a_+n) n-1 2 2 2 2 If Oj^Pj^ = 0, then the solutions of (2-13) will be hyp e rge ome trie: (2-16) u^ = C^ Â•^2 Z 2^l(Â°'2'-P2-Â°'3J Â°'2-P2^ 2) PAGE 65 59 and 3. (2-17) ^2 = C^Z 2^(P2'-'^2-Â°'3^ P2-Â°'2' 2)The hyp ergeome trie fomi Is (2-18) ,P,(a,b; c; z) = 1 ^ i^ z . Â°(gt^|Mb|l) ^ ^.,, . a(a+l) . . .(a+n-l)b(b+l) . . .(b-f-n-1) n nl c(c+l) . . .(c+n-1) eo -I n=o a(a+l) . . .(a+n-l)b(b+l) . . .(b+n-l) n nl c(c+l) . . . (c+n-1) 2.^ Solutions of Confluent Cases after Normalizing . Another approach to obtaining solutions of equation (2-7), which we will now use, is to remove the u' term and find solutions of this transformed equation. To normalize the general equation u" + p(2) u' + q(z) u = 0, let u ^ vw. Tlien we have q p 1 u = vw u ' = V ' w + vw ' . u" = v"w + 2v'w' + vw" Hence, Let vw" + (2v' + vp) w' + (v" + v'p + vq) w = 2v Â• + vp = PAGE 66 60 ancl solve for v . 2 H -vp. Â— = 2 P dz. Integrating^ In V = " -2 / P dz . ^ / p dz Therefore, to normalize the equation u" + p(s) u' + q(z) u 0, let l/p d2 (2-19) V = e In (2-20) vw" + (v" + v'p + vq) w = 0. Now to normalize equation (2-7) we use (2-19) to find V = z (z-1) Then . 1 , ^^ ,, i (Â°'2-^2-5) , . I (Â«3-^3-l) , 1 . ^ ,, -I K^P2-1) f ,A (^3-^3-^) + -2 (a3+p3-l)z (z-1) PAGE 67 61 1 |(Â«2-^P2-^) 2 ^ + (a3+a3-l)z (^-1) l^'^j-^s-^) (a^+P -l)(z-l) 2 '"^2 = T(Â°^2^^2-1)(Â°'2-^P2-3)Z t(Q^.-^,-5) . . ^(a,+P.-l) (z-1) 2 ^^3 ^3 + ^(a2-^P2-l)(Â°'3-^P3-^)' i(a,+p,-^) _ , ^(a^+^ -3) 2''^2 ^2 (2-1) 2 ^^3 "^3 1(^2+^2-1) , . 1(^1.^-^,-5) + r(o'.+P^-i)(Q'.+P,-3)z ^'^^'"^ (z-i) 2 ^^3 ^3 4 ^ 3 "^3 ^^ ^^3 ^^3 V = Â— Z 2 1 ^(Â°'2-^p2-5), ,i(<^3^p3-5) (z-1)' ^(a2+p2-l)(^2-^P2-^)(2-l)^ + (a^+P2-l)(a3+P3-l)z(z-l) + ^(a3+^3-l)(a3+P3-3)z^ Substituting v Into equation (2-20) with p and q from (2-7), iK-^2-1) , ,. 1(^3^3-1) ., z (z-1) w ^,1 |(Â°'2+V5). ^i(^3+^3-5) + < Â— z (z-1) i(a2+^2-l)(a2+P2-5)(z-l)2 + (a2+P2-l)(Â°'3-^3-l)2(z-l) +|(a3+P3-l)(a3+P3-3)z' 1 |(Â°'2+P2-^) , ,. |(Â°'3-^3-5) + 2^ (z-1) (a_+p -l)(z-l) PAGE 68 + (a_+p^-l)z 3 ^3 , l-aÂ„-p V 62 2 "2 ^ ^-^3~^3 ^ z-1 |(a,_+p -1) + z 2'' 2 ^2 (2-1) |(a3+P3-l) rcx,3 1^1 a 6 2^2 Lz(z-l) z'^Cz-l) S^3 (z-1)' Dividing by z w = 0, t(Â°'2-^2-5). . 4(<^.+P.-5) (2-1) 2^ 3 ^3 and simplifying. z^(z-l)2 w" +
PAGE 69 63 z^(z-l)^ w" + ^{a^+^^-l){a^+^.^-^) + i(a2-^^-l)(a3+P3-l) + i(a,+p,-l)(a,+p.,-5) Ua.-^^-lf (a,^-Ki,-l) {a34^3-l) 4^^3^3 ^'^^3^3 2^^2 '-^2 i(Â«3^3-l)^ -^ ^1^1 -^ ^3^3^ ^ [^(^'a-^a-^) (^2-^2-5) -^(aÂ„+p -l)(aÂ„+pÂ„-l) + (aÂ„+p^-l)^ + (a.+p^-l) (aÂ„+pÂ„-l) 2^ 2^2 "'^""3 ""^S 2 "^2 2^2 -^'^^3^3 ^'iPl ^2^2 2 + i(a24^2-l)(Â°'2^2-^) -1(^2-^^2-1)' + ^2^2 w = 0. Therefore, the normalized form of equation (2-7) is (2-21) z^(z-l)^ w" +{ -T(Â«2-^P2)' -t('-3-^3)' |(<^2-*-P2)(Â°'3^3) + ^('='2+^2) + â€¢(Â°^3-^3) + ^1^1 + ^'S^S ^(0^2+^2)^ +i(a2-^P2)(Â°'3-*-p3) |(Â°^2+^2) i(Â°^3-^3) ^1^1 V 2^2 t(Â°'2-^2)^ +T + Â°'2^2 w = 0, To obtain the solutions of (2-21) let w = a z^ + a,z^-*-V...4. a .z^-'"-" -f a , z^'^'^"' + a z^-*-%. , . o X nÂ—z nÂ— 1 n a / 0, PAGE 70 64 then w" = a^A(^-l)2^"^+...+ a^_2(X+n-2)(>+n-3)z^'*""'^ + a^_^(?v+n-l)(?v+n-2)z^"^Â°~^+ aÂ„(X+n) (7v+n-l)z'^'^''"^+. . . n The Indiclal equation Is z\ X(>v-1) jia^-^^r^f + J + 02^2 = 0* (2-22) The exponents are ^^ >^ +T ii^.-^of = 0. 4 4^ 2 ^2 + \ 1 4 T i(^2-^2y 1Â± (^2-^2) Finding a : .?^+n z""^": a _(7^+n-2)(7v+n-3) 2a , (X+n-l) (?v+n-2) + a ^(>.+n)(7^+n-l) + a^_2 [" T^V^2^'' ' T^^'a+Pa)' l(^2-*-^2^(^3-^^3^ + |(Â°'2-^P2) + ^(^3+^3) + Â°^lPl + a-,p ai^a + a n-l ^{a^+^^f +^{a^+^.,){cL^+^^) i(^2-^2) ^(Â°'3-^3)^1^1 ^2^2] + a i(^,+Pj^ 2 '"^2 + 4 + a^P^ 0. PAGE 71 65 (2-23) aÂ„ = a n-2 -(7v+n-2)(7v+n-3) + ii<^2^2^^ +t(Â«3+P3)' + i(a2+f32){a3-Hp3) -^{a^^.,) ' ^i^3^3)-<^i^i ^3^3 + a^_j2{>.+n-l)(X+n-2) ^-Cdg+Pg^ ICa^+li^) (013-^3) ^|(^2-^^2) +â€¢^^3-^3) 4V^2 "^2 In order to simplify further work we can, without loss of generality, let t(^2-^2)' -^tK^^s)" +|(Â«2-^P2)(Â°'3-^3) -â€¢^^2^^ and 1(^2+^2)" -|(Â°'2+^2)K-^'3) +^(^2^-2) +â€¢(^3-^3) -^ ^1^1 + 02^2 = Â»"Â• Making these substitutions in (2-23) and simplifying the denominator by use of the Indlcial equation (2-22), we have a Â„_2r-(X+n-2)(7v+n-3)+k + a n-l 2(X+n-l)(7^+n-2) + m a n(2A+n-l) PAGE 72 66 a 2>v(?^-l) + m ^1 = 1{2-K) l!(2>v) a ^2 = -X(X-l) + kj + aj^ 2(?v+l)A + m 2(2A+1) ^2 = a 2>(-?v^+X+k) + (2?v^-2^+m)(2?v^+2?v-Kn 2i 2'\(2X+1) ^]. ^2 = 4>v^-2-\^ + (4m-2)?v^ + 2k> + m^ 2i (2X)(2?v+l) a. -(X+l)>v + k + a. 2(A+2)(X+1) + m S = 5(2X+2) ag = a^ / 2(-X^->v+k)(2>v^-2X-Hn)(2X+l) + (2X^+6X44+m)Ux^-27v + (4m-2)X^ + 2kX + m^ 51 (2A)(2X+l)(2X+2) ^1 = ^Â« o o 8x^ + 12X'^ + 4(3m-l)x'* + 6(5m+2k-2)A^ + 2(3m^44m-2-+4k)X^ + 2(3ni^-nH-5ink+2k)X + in(m +4m+2k) 31 (2X)(2A+l)(2X+2) PAGE 73 67 The solutions of equation (2-21) are (2-24) w = a z' 1 + (2>^ -S^+Ti) ^ 11(2X) ^ 4 ?v^-2X^ + (4m-2)>v^ + 2k^ + m^ 2 ^ -^ I II I 1 2 "T" Â• Â• Â• 2l(2X)(2)v+l) lt{Â°'2-P2) where X = and k and m have the values: ^ = t(Â°'2-^P2-^3+P3)^ |(Â°'2-'^2-^3-^3) " ^^l^l " ^Z^Z' m = |(Â«2-^2-inÂ°'2-^P2-^3^3) + '^iPl + Â«2^22.5 Factored Solution W , and Hypergeometrlc Solution W , . Next, we shall consider solutions of the a,k ' equation (2-25) 00 a. m \ a a i^1-a 1 00 2k a^ P, v/hen the u' term has been removed. "Hiis can be obtained from equation (2-21) by letting a^Pj^ = m, Og = a, ^^ = l-Oj Qg = 2k, P3=0. Although ^3=0 in the normalized equation, this does not insure a two term recurrence relation as it would in the original equation (2-7). Our transformed equation, that is (2-25) in normalized form, is PAGE 74 68 (2-2G) z^(z-l)^w" + 1 9 ^ ^ (in+ ^ k )z +(a -a-m)z+a(l-a) w = 0, Assume the solution w = C.zV...+ C_ Â„z^-*-Â°-2+ C_ ,z^+Â°-V C z^-'V... C / 0, n , o ' ' 'n-2' 'n-1 then .^-2 w" = C^^(?y-l)z''"^ +...+ C^_2(^+n-2)(?v+n-5)z ?v+n-4 + CÂ„_^(X+n-l)(>.+n-2)z'^'^"-^+ C^(X+n) (>v+n-l)2'^"^""2 + . . . The Indlclal equation and exponents are z\ Â•K{-k-l) + a a^ 0, X ?v + a(l-a) = 0, (7v-a)(?v-Ki-l) = 0, X = a, 1-a. Obtaining C : .?v+n ^n-2^'^'*'"~2)(7v+n-3)-2C^_^(7v+n-l)(?v+n-2) + C^(>v+n)(>v+n-l) + C^_2("' + T k^) + CÂ„_^(a^ a ra) + Cj^(a a^) 0. PAGE 75 69 n 1 n-^ (?v+n-2)(A+n-3) + (m+ ^ -k" ) +C, n.l 2(X+n-l)(A+n-2) (a a m) (?v+n)(>v+n-l) + a a' Let ?v = a: C = < C n-z -(a+n-2)(a+n-5)-mt + ^' + C n-1 a + (4n-5)a +2(n-l)(n-2) + m n(2a+n-l) 2 a -a-Hn Ci = l(2a) ^2 = 1 2 -a(a-l)--mx + ^^^ + C, a +5a-Hn 2(2a+l) ^2 = 2a(-a^+a-mi +k^ ) + (a^-a+ni) (a^+3a+m) ' 2l(2a)(2a+l) r 4 ^2 = 2 1 a +{2m-l)a + (2k -^)a + m 2l(2a)(2a+l) ^3 = 1 2 -a(a+l)-m-j +k + C, a'^+7a+4+m 5(2a+2) PAGE 76 70 Cg = C^ ( 2(a^-a+m)(-a^-a-m^ +l<:^)(2a+l) + a'* + (2in-l)a^ (2k^ i)a+m^l(a^+7a-Â»4-Hn)) i ^ J J 3l(2a)(2 2a+l ) ( 2a+2 ) C3 = c^ a^+3a^ + (3nn-l)a'* + (6m+6k^Â•|)a^+(l2k^+3ni+3m^-5)ci + (3m^+6k^+6mk^| ra|-)a+(ra^-f2m^+2k^mi m) 3l2a(2a+l)(2a+2) Thus one solution of (2-25) can be written with a few terms as: (2-27) w = C z^ , o-o-Hti a +(2in-l)a +(2k 2-)a+m 2 II 2a 21 2a(2a+l) A second solution of (2-26) can be obtained by replacing a by 1-a: (2-28) w C^z i-a 1+ Â°Lz5LH!L_ z ll(2-2a) a'*-ifa^ + (5+2m)a^+(-2k^-4m|)a+(m^+2ini +2k^ ) 21 (2-2a)(3-2a) z +. . Although solution (2-27) has no obvious general term. It can be shown by a lengthy process of long division (which is omitted here), that it can be written in the factored form: PAGE 77 w=C z (1-z) \1+ 2+ Â° 1 11 2a 71 (a +2!ca+m) a +(2k+2)a+2k+raH (a +2ka-Hn) a + ( 2k+2 )a+2k+m+l 21 2a(2a+l) a +(2k-f4)a-+4k-i-m+4 31 2a(2a+l){2a+2) z + + ((a^+2ka-Hn) a^+(2k+2)a+2k+m+l + (n-l)(2k+n-l) ) . Â— a +2(k+n-l)a-Hn 2a(2a+l) . . .(2a+n-l) + . . . Let this solution be indicated W , , since a.k.m are the a,k,m ' ' only parameters; therefore (2-27) can be written a k+t (2-29) W k = CÂ„2"{l-z)*^-^2 / 1 + ) (a"+2ka-Hn) a +2(k+l)a+2k+l+m n=j a +2(k+n-l)a+(n-l)(2k+n-l)+m n! 2a(2a+l) Â• Â• Â•(2a+n-l) If m = 0, then the solution (2-29) becomes hypergeometric. Let this solution be indicated W . , a and k being the only parameters; n=l (a +2ka) a +2(k+l)a+2k+l a +2(k+n-l)a+(n-l)(2k+n-l) nl 2a(2a+l) Â• . .(2a+n-l) ') PAGE 78 72 Factoring further we may write W , = CÂ„zÂ°'(l-z) 00 a(a+2k) ^+1 'L (a+l)(a+2k+l) (a+n-l)(a+2k+n-l) n=l nl 2a(2a+l) Â• Â• -(aa+n-l) Â— z Rearranging factors we arrive at the solution W a,k CÂ„zÂ«(l-z) k+ + y a(a-H) Â• Â• Â•(a+n-l)(a+2k)(a+2k+l) Â• Â• -(g+gk+n-l) ^n J^^ n! 2a(2a+l) Â• Â• -(Sa+n-l) which Is hypergeometrlc In form (2-l8) and shall be denoted by W . . Therefore, when m = In (2-26) we have the solution (2-30) W a,k = z^'d-z) k.-i ,P^(a,a+2k; 2aj z). where we have let C^ = 1 . A second solution can be obtained by replacing a by 1-a. 2.6 Notation and Proofs Involving 2Pi^(a, b; c; z). Before presenting several Interesting properties of W , , we shall Introduce notation and prove seven statements that will be useful in section 2.7. Notation: n,J are integers. (a)j^ a(a+l)(a+2) Â• Â• -(a+n-l) . PAGE 79 75 ( "^ ) = n(n-l)'"(n-J-H) . f n \ ^ ^ dz W , = W , (z) . a,k: a,k^ ' If P ^P^Ca, b; cj z) = F (a, bj c; z), then P(a+) = F(a+1, b; c; z) , [5,P.5o] , P(b+) = P(a, b+lj c; z), P(c+) = P(a, b; c+1; z) , P(a-) = F(a-1 , b; c; z) , etc., P(+n) = F(a+n, b+n; c+n; z) , n = 1,2,3> P(-l) = P(a-1, b-1; c-1; z) . We shall prove the following: (2-31) (2-32) P = -AraF(a4-) + (l-c)F(c-)l a-c+1 L J' (c.)] ^ -^ rp(a-) 1-z L c (2-33) P(+l) = Â— az F(b+) F (2-34) d'Â»P = (a)n(b)n p(+n). (c)a and for P = 2Pi('^' a+2k; 2a; z) = P(a, a+2k; 2a; z)^ (2-35) F(+n) If^zÂ— (i-z)-^-^J (-i)^(5>-)i,,.ii^. (a), J=o PAGE 80 74 (2-36) DÂ° z^'d-z) 2 -, k-n+ ^(-l)j(ny.)^(_,4)^_^,-J(,.,)J, J=o (2-37) P(-l) =ii^^i_l (2a-l)(2-z)L 2(2a-l)(l-z)-^^_^^^+(a-2k)W^^^_,^ Finally, we state Leibniz's rule [9, p. 409] for finding the nth derivative but shall not prove it: "-'v +f^)D2u d"-2v +, (2-38) d"(uv) = u d"v + (^J)du d"-^ + Q: + r^Vu d"--^ V +...+ (DÂ°u)v laX" J Vu D^-^v, ^=0 PROOFS (2-31) P = a-c+1 L aP(a+) + (l-c)P(c-) Proof: aP(a+) + (l-c)P(c-) (a+1) (b) {a)Â„(b), El d-rj. jid; v-i \.o J {0 ) Â— ^^Â—^ z" + (1-c) y ial^ ^ (c)_nl ^_ (c-1) nl n=o ^""K (c-1) n =0 * ' n = I[ ra(a+l) (b) (a)Â„(b), Â« Â« (c) n! n =0 * ' n Â°' 'Â° + (1-c) _ial_iÂ£ I (c-1) nl J PAGE 81 75 CO > n=0 a+n-(c+n-l) (c)^nl (a c + 1)^. Tlierefore, F a-c+1 a?(a+) + (l-c)F(c-) (2-32) P = 1-z p(a-) + iJilÂ£k p(c+) Proof: ^^ (a-1) (b) ., V ^ (a) (b) = ; Z + -^ ' Â— > Z L Â« (c) nl n =0 * ' n ^^ (c+1) nl n =o ^ ' r, n 00 ^ V r-, (a-1) (b) /, V ^ Â— Â— Â— ^Â— z + -^ ', 00 (a) (b) ^ 'n^ ''n n+1 Z n=l (c)Â„nl c ^^ (c+1) n V^ (a-1) (b) ^K ^ \" (a) i (b) , = 1 + y _: " Â° z'' + (^-^) \ Â°-^ Â°-^ (a-1) + "(b-c) 1 + ; '-'. l-a+n-1 (a+n-1) (b+n-1) J 1 (a) (-) ^ 'n * ' n n Z (o)^nl = 1 + 00 (a-l)(b+n-l)+n(b-c) l (^^n(Â°^n n Â„ti L (a+n-l)(b+n-l) (c)^nl PAGE 82 76 1 + 1n(c+n-l) ^^^L (a+n-l)(b+n-l) (a) (b) (c) nl = F -I n(c+n-l) (a) (b) z = F nti^(a+n-l)(b+n-l) J (c)j^nl -I ntl (^)n-l("-l)V(a) (b) ^Â« (c) nl n =o ^ ' n = F zP = (l-z)P. Thus, F(a-) +i^Z^ F(c+) . (l-z)] or l-z L c J Â• (2-55) F(+l) =^ az P(b+) P Proof: P(b+) P = z n=o (^)n"(a) (b) ^ (c) nl n=o ^ 'n PAGE 83 77 Hence, Proof: w ' L n=l b+n b n n (a) (b+1) , n ^ ** n 1 n (c)Â„(n-l)l L (a) (b+1) , n^ ' n-1 ii (c)Â„(r.-l)! iÂ£V ^"Â•^'^n-l(^-^l)n-l n-1 L i (c+l)n-l("-l)' _ az f (^+^n(^+l)n n Â„^Â„ (c+1) n! n=o ^ ' n z . or (2-34) P(b+) F = -^ P(+l) c F(+l) = -^ az P(b+) P D"P = (a) (b) n* 'n (c), P(+n) D F(a, b; c; z) [^ ^o (<=)Â„"' (c) nn4 (c)Â„^.i(n+l)l PAGE 84 78 ab c (a+1) (b+1) z ^ (c+1) nl n=o n ab F(+l) Similarly, P(a, b; c; z) a(a+l)b(b+l) V ^^'^^^n-l^^'^^^n-l ^n-l c(c+l) ^4i (c+2)^_^(n-l)i (c)o F(+2) . In general. F(a, b; c; z) (a) (b) " Â°-F(+n), n=l,2,3. (c) If P = F(a, a+2k; 2a; z) = z~^(l-z)"^"^W^ , , then (2-35) P(+n) (2a) n 2-Â°'-Â°' n+1 {i-z)-''--r^(-ii(5)(i-.)X,,.ii^ Note that when k Is replaced by k + m In the premise, then P(a, a+2k+2mj 2aj z) = z"^(l-2)"^"'""2 W a,k-Hn PAGE 85 79 Proof: We begin by applying (2-33) to F(tl) and express the result as a function of W. F(+l) = 2ci az 2a as a,k-t-ni' P(a, a+2k+l; da; z) F z-Â°'(l-z)-^-^W a,k+5 z"'^(l-2)-^-2w a,k (2-39) F(+l) =iM z-Â°'-^(i-z)-^-^ (a) W , ^1 (l-z)2w , a,k+i ^ ' a,k Apply the principle of (2-33) to F(+2): ?(a+l, a+2k+2; 2a+l; z) F(+l) (2-itO) P(+2) = i2aÂ±ll (a+l)z Using the expression (2-39) with k replaced by k + | in (2-4c), we have (2-41) P(+2) = l2aÂ±ll/2a,-a-l(^.^)-k-| (a+l)z\Â°^ \,k+l-(l--)%,k+i 2a ^-a-i/^ x-k-i Â— z (l-z) (2a), , 3 -(^ z (l-z) 2 W ,^:-(l-z)2W , a,k+-| ^ ' a,k :'a,l:Â«-2(l-^)%,k.i + (l-z)W a,k Let us find P(+3) before generalizing our results to P(+n) . Upon applying (2-33) we have PAGE 86 80 (2-42) p(+3) = (^"+^) (ci+2)z F(a+2, a+21<+3; 2a+2; 2) F{+2) Replace k by k + i m (2-4l) to write (2-42) In the fom F(.3) = 120^21 /if^ ^.,.2 _k-: (a+2)zl (a). ^ ^ a,K+ Â— / a,k+l (20) (a). 2 Â„-a-2,, ,-k_ r (1-2)' (^-)^a,k^] 3 r w a,k+l Simplifying further '''^^'=^^"""''^-'"1v..i-'-^'*v k+i " ^(l-'^a.k^ -(l-)'''a,l Finally, (2a) p(+n) = iÂ± z-a-Â° (c^)Â„ (1-2) -kn+l ".,.. I -(;>!-) w,,,, ^ 2 \Jy < a,k+ n-J + . . .+ (l-2)\ , 1 Therefore, F(+n) = (2a) "7^ ^ ^-^'-"(l-z) -kn+l J=0 2 n=l,2,3, .. PAGE 87 81 (2-36) d"" :^(l.z)^^' = z^(l-z)^-'^^ Y (-l)''(^)(-a)j(-k-i)Â„_jZ-J(l-z)J J=o Using Leibniz's rule where u = z*^ and v = (1-z) ^ , find D'^'u and DÂ°"^v'. When u = z^ , Du = az ; D u = a(a-l)z ;...; D u = (-1) (-oi)^z If V = (1-z)'^'^^, then 3 Dv = -(k+i)(l-z)^-^; D^v = (k+i)(k-i)(l-z)^2;...; .n-t D"-S = (.k-i)^^(l-z) k+-t-n+^ Hence, the nth derivative is found to be .^(^.rr\^+i z"(l-2)' ^(;)(-l)^-a),z-^.k-|),.,(l-z)^-^-^ -t=o n = zÂ°'(l-z)^-Â°-^^V(-l)^(^)(-a)^(-k-^)Â„_^z-^{l-z)^ . t=^o If P(-l) = P{a-1, a+2k-l; 2a-l; z) show that (2-37) P /t \-k l-a (-1) = (1-^) z (2a-l)(2-z) Apply (2-31) to P(-l): 2(2a-l)(l-z)^^W^.,^j^+(a-2k)W^^^.. P(a-1, a+2k-l; 2a-l; z) PAGE 88 -a+1 t 82 (a-l)P(a, a+2k-l; 2a-l; z) + (2-2a)F{a-l, a+2k-lj 2a-2; z) = -P(a, a+2k-lj 2a-l; z) + 2F(j[-1, a+2k-l; 2a-2; z) Now apply (2-52) to the first term with the result that P(-l) = -1 1-z F(.l) + (2k-a)z p^^^ a+2k-l; 2aj z)l 2a-l J + 2F(a-l, a+2k-l; 2a-2; z) . Collecting the terms containing P(-l) we have (2-^8) (l + -I-') F(.i) = Â£21^ . _J_ ^ 1-z ^ 2a-l 1-z P(a, a+2k-l; 2a; z) + 2F(a-l, a+2k-l; 2a-2; z). Now, since (2-39) F(a, a+2k; 2a; z) z'^( l-z)'^'^W , , If k Is replaced by k 5, then P(a, a+2k-l; 2a; z) = z"Â°'(l-2)"^W , 1. a , KÂ— 2 If a Is replaced by a-1 In (2-39), then F(a-1, a+2k-l; 2a-2; z) z'^""^^ (l-z)"^'2y ^ Using these two results and solving equation (2-38) for F(-l), permits the result PAGE 89 85 F(-l) =iLzll a -2k z _-a/ ,-k. L2a-1 1-2 a,ic-2 Sz-'d-.j-X-V.,,] -a+1 LlzA rk (2-z) (at--^k) 2a "^ W ^i + 2(l-z)% _2 ^^'^ 2 a-1, k or F(-l) = ^^'^)" ^ [(a-2k)'W , ^ + 2(2a-l)(l-z)2W , , (2a-l)(2-z) L ^^'^-s a-l,k_ 2.7 Recurrence Relations, nth Derivative, Sum and Product Formulas for W , . By utilizing the statements v;e have Just verified. It Is possible to obtain recurrence relations for W^ ^. Several such relations will nov/ be presented . By replacing a by a-1 In equation (2-30), we may v/rlte (2-40) W^_^^,^ = z^-^l-z)^-^^F(a-l, a+2k-l; 2a-2; z) . Since the parameters a and b In F(a, bj c; z) can be Interchanged without any lose In generality, we can also Interchange a and b In equation (2-32) to obtain the valid relationship P = 1-z p(b-) + l^LZ^k p(c+) Now, if this be applied to (2-4o), we may conclude that PAGE 90 84 (2-41) W^_^^^ = z"~^(l-z)^"2rp(a-l, a+2k-2; 2a-2; z) 2a-2 J Â• Replacing a by a 1 and k by k ^ in (2-30), it is seen that P(a-1, a+2k-2j 2a-2j z) = z^"Â°'(l-z)"^W , , This, together with (2-37) enables us to write (2-4l) in the form Vi,. -^"-'(i-)"-*^-"" -k, ^''^(i-)"'^Vi,k2(2a-l)(2-z) /, \-k 2-a (1-z) z 2(2a-l)(l-z)^W^_^^ 1 = (1-z) W , , 1 Â— W , , ^ ' a-l,k-2 {2-7.) ^'^Â»^ k (a-2k) 2(2a-l)(2-z) ^(1-^)"' ^^a,k-i Then 1 + 2-z a-l,k L a-l,k-i 2(2a-l)(z-2) '^'^"^J ' Solving for W , ,.^ we have the recurrence relation a-i,k ^ ^ ' a-l,k-2 2(2a-l)(z-2) ^'^''^ or PAGE 91 85 (2-'+2) V/ , , = i (l-2r^r2(^a-l)(z-2) W , ^-^'^ 4(l-2a) L a-l,k-2 + (a-2Ic)z V; ,. 1 a,k-2 Another recurrence relation Is found by first replacing a by a + 1 In (2-^0), giving us ^W,k " z'^'^^Cl-z)^'*"^ F(a+1, a+2k+l; 2a+2; z) . Then apply (2-53) to get az F(a, a+2k+l; 2a+l; z) P{a, a+2k; 2a+l; z) When (2-31) Is applied to each term, we may write aF(a+l, a+2k+l; 2a+l; z) Vi,. = ^ ^"(1-)'^* {:t 2aF(a, a+2k+l; 2aj z) aF(a4-l, a+2kj 2a+l; z) -2aP(a, a+2k; 2a; z) f2-*J) "a+l,k = ^^^^ z^d-^j^^^ -F(a+1, a+2k+l; 2a+l; z) + 2P(a, a+2k+l; 2a; z) + P(a+1, a+2k; 2a+l; z) 2F(a, a+2k; 2a; z) From (2-35)* letting n =1, we have PAGE 92 86 '(-Hi) 2z-Â«-^(l.z)-^-^ [w^^ , .1 -(l-z)^W , which can be used to reT^rrite the first and third terms of (2-45). Hence, W a+1 _^ ,i2HÂ±ii ,a(i.,)X^/.2,-a-i(i.,,-k-i v; a,k+^ (l-z)2 w a>k . 2z-(l-z)-^W,^,, -a-1 + 2z """(1-z) -k-t .\,k (1-)" Wa,k.i. 2z"'^(l-z)"^"^ W a,k 2(2a-H) /r -1/, N-i /, = _i L{ _2 (1-z) 2 + (1z) W a,k+^ 2z-^ -1 -1 WÂ„ . z" (l-z)2 W , a,k ^ ' a,k2(2a+l) az -t^-)' Vk+i*(^-)'^a,l< -(1-)' Vk-i Thus, we have a second recurrence relatl on (2-44) azV/^^^^j^ . 2(2a-.l) [-(l-z)^ W^^^^, + (2-z) W^^ ,k (i-z)2 v; Â„ i a.k ,rt.-2 To obtain the nth derivative of our solution (2-30) W ^^j^ = zÂ°'(l-z)^-*-2 F(a, aH-2k; 2a; z). PAGE 93 87 we shall apply Leibniz's rule, where u = zÂ°'(l-z) ^ and V = P(a, a+2k; 2a; z). Thus, using (2-54) thru (2-36) we may write "X,. = i(t^* [zM-.)''^i^ ^=0 D n-t P(a, a+2k ; 2a; z)j t=o I = ^ [(;y (i.z)''-*^ ^ (.i)^(5)(-.),(-M),.,.-^i-z)^ (a) .(a+2k)n . ^--tJlzi^ F(+n-^) (2a) n-l -1=0 j =o Â•(a+2k)^.^z-<^-^^(l-z) Therefore , -kn--tfi n-l Â— y(-l)f-/)(l-.)\,,,n^ J=o '^ (2-45) D%^^j^ = z-"(l-z)2 y [(j)(l-z)" 2 z'^(a+2k)^_ ^(-i)Y"-^)(i-z)\,,,n::|^ J=o PAGE 94 88 Now that v;e have the nth derivative of V , (z) v^;o can use Taylor's Theorem [8, p. 93] to write formulas for W^ j^ (x+y) and W ^^ (xy) . Taylor's expansion can be expressed in the form f(y+a) = f(a) + f '(a)y + f"(a) ^ +...+ f(Â°^a) ^ +... Now if a be replaced by x, we have f(y+x) = f(x) + f'(x)y + f"(x) ^ +...+ f^Â°^x) ^ +... 00 (2-46) f(y+x) = Vf^") (x) C . n^O Using (2-46), replace y by (y-l)x to obtain (2-47) f(xy) = ^f(")(x) ^y-^r^" n=o Since W^ ^^(z) is a convergent, analytic function for <|z| <1, it can be used in place of f(x) above and we have an addition theorem and a multiplication theorem: 00 (2-48) WÂ„_^ (x^y) = ^ w(;) (X) ^ n=o "Â» r n=o * PAGE 95 89 n--t ^(-l)j(Y)(l-x)^W^,,,n^ J=o and (2-49) w^,,(xy) = l^l'^lMhzipl n=0 n=olt=o J=o n-t . I (-X)^("-*)(l-x)^ WÂ„^,^ n^; J=o where | y | < 1 . PAGE 96 CHAPTER III Related Differential Equations and Classification 3'1 Related Equations of Mathematical Physics Derived from an Equation having Four Singularities . The most general linear differential equation of the second order which has every point except a^^ja^ja^ and <Â» as an ordinary point with exponents o.^,^^ at a^ (r=l,2,5) and exponents [i--i>\i2 ^^ 00, Is (>1) ^ ^u , r ^-^'r-P. ^2 r'^i 2-a^ du dz ^a,P L r*^ r Az+B i(2"^r)^ if(z-ar) U = 0, r=l where A Is such that \i^ and n are the roots of 3 3 (>2) M. + MY^ (a^+P^)-2J + Y.^'^^r + A = 0. r=l r=l To verify this, we can begin with equation (2-1) and let a^ become Infinite, thus obtaining u 1-a -p r=l 2-ar H(z)Ju r=l a p m (z-a^)' z-a. + J(z)) u=0. where H(z) and J(z) are polynomials In z . We must now depart from the previous work of chapter II, for we desire z = Â» to be a regular singular point In this case . Referring to section 1.5, we see that z = Â« (z = Â— ) will be a regular 90 PAGE 97 91 alngular point when 2 pf Â— 1 and Â—5q( Â— ) are analytic at z =0. For the first one. ^-k PAGE 98 92 where A and B are functions of m^ and a^ (r = 1,2,3). But since m, +m^+mo = 0, we liave equation (5-1). In order to Show that A must satisfy the condition {'^-2), we can make the substitution z = Â— in (3-1), then the Indicial equation ^1 for z, = will be this condition. An alternative method of finding the indicial equation at z = Â» (or z^ = O) , and the procedure we shall use here, would be to assume the solution u = b^z"^ + bj^z"^"-^ +... and equate the coefficient of the largest pov^er of z (i.e. z" ) to zero to obtain the indicial equation. Noting the expansion of the following about z = oo(or =0), we have z l^-^^-i/l.li.!^.... z (z-aj.) z ^ z / z where r =1,2,3* and 1 1 (z-a^)(z-a2)(z-a3) '^ I^ ^ i_ li ^ i. fi A ^_ ^ y z z z Â— 2" I 1+ +, PAGE 99 Then 93 ^^'r^ L 2a. 1+ + . . . r=l ^AZJ^^.V^V^,.. Q 3 (1+ Â— + -l +...) z z u = b z"Vb,z"'^"V... bÂ„ /^ 0, f 1-^r-^r r=l ->v-2 u=-b^Xz-^-Vb^(-?.-l)z"^"^+ 3 : X(7v-H) X 2^ (1-a^-p^) + ^ ^^p^ + A = 0. >.2 -H r=l 3 1^ (l-a,-3j r=l r=l r "^ r A + y o,r^j.+ A = 0. r=l \^ + y (a^+P,)-2 7v ^^^^r^r + A = 0, 'r-1 r=l where the exponents at z ^ Â« are X = p-j^ and p.^ Â• We now proceed to show how many of the related differential equations of mathematical physics can be obtained from a differential equation having four singularities, by assigning values to the exponents a^., ^j., the singularities a (r = 1,2,5), and the constant B. (a) Lame 's Equation [8, p. 205]. Beginning with (3-1) take a^ = 1/2 and p^ =0 (r 1,2,3), M-g = M-^ + n + i, and PAGE 100 9^ B = Â— h, where h and n are constants. We can obtain A by using (3-2) and the fact that the sura of the roots equals 3 3 2 \ (a^+p^) = -g and the product of the roots Is ) a p r=l r=l + A = A, where the roots are p.and ^ + n + 2* Thus, and 2[Xj^ + n + i = i, ^Ll = in, A = Hj, (n^+n-Kl) = -I n(in-Hl) = ^ n(n+l) Substituting these values into (3-1), the result is Lame's Equation: d u V j_ du n(n+l)z + h ,, _ ^ , 2 "^ / z-a dz " .,3 , r. \ dz ^^, r 4jy (z-a^) r=l (b) Legendre's Equation . Let ^i = ^ = 0, ag = 1, ^3 = ^3 = 0> U-i = 0, \i^ = i, a^+Pj-hi2+^2 = V2, B = i n(n+l) 3 + cij^Pj^ + ^^2^2 '"" (^"1) Â• Since the product \i:^Vi2 = = ) a p r=l + A from (3-2), we have A -= -(ci^Pi + Â°'2^2^' Hence, d!u dz^ i^-L du dz r^i^i + ^2^2 i-z z-lJ -(a^P^ + 0^^2)2 + T n(n+l) + a^3^ + a^P, u 0. (3-3) dz^ Z Z-1 dH+_nInÂ±LL u ^ ^^ 4z2(z-l) PAGE 101 95 -2 Now let z = C dz ^ -2 r'cic, II = i c' . Then du _ du d^ dz df^ cTz dz 1 -3 du 2 t, ^ Â• d!u dz^ 2 -2 du 2 ^ dC 1 1.3 d u dc' (i r) = 3 -5 du ^ 1 -6 d u 4 dC ^ dC Substituting in (3-3): 4 ^ 2 ^ dC 4 ^ ~ 4 ^ ~ 2 ^ 1-r 5 -1 du ^ n(n+l)^ ^ ^ Q^ ^ ^6 dfu ._i4_du^n(n.l)f ^ ^ ^^ 4(1-;: ) Multiplying by Â— ^ Â— ^Â— ^ , \^e have the General Legendre equation: (1.^2) d^ _ 2^ du ^^(^^^j ^ ^ 0^ dC" dC (c) Jacob! 's Equation . If we let A oe an arbitrary finite constant, B = a3n(a+0+n+l) , a^ = -a, a = -p, pj^ pg 0, a^ = 1, a^ = -1 in equation (3-1), then ^'^ PAGE 102 96 Now let aÂ„ become infinite and we have dz 1-Kx 1+g.l du _ n(a->-p+n+l) ^ ^ q z-1 z+l-l dz (z-l)(z+l) This can be written in the form of Jacobi's equation: d^u ^ p-a-(a-Hp+2)z du , nfa+p+n+l) Â„ _ ^ + r 1 di "^ -^ 2 u 0. dz 1-z 1-z (d) Gegenbauer's Equation [8, p. 529]. We shall first derive Gegenbauer's equation which is (l-z^)u" (H-2v)zu' + n(n+2v)u 0, and then show how it can be obtained from equation (3-1). Let (() be defined: (>4) (}) = (l-2hz+h2)-'' = ^CÂ„(z) hÂ°. n=o Next, find M and ^ . ah bz ^ = -v(l-2hz+h^)"*'"^-2z+2h) bh 2v(z-h)(l-2hz+h^)*''""^ , hence (>5) ^ = 2v(z-h)0{l-2hz+h^)"^ . dh M = 2hv(l-2hz+h2)-^-' , bz (>6) ^ = 2hv 0(l-2hz+h^)'^ . dz Then from (3-5) and (3-6) we have PAGE 103 97 (>7) hM. (z-h)M . ^ dh dz Using the form 00 n=o we may also write (>8) f^= Vncy ^h n.i and (>9) f = 7c;h\ ^2 n=o Rewriting (3-7) with the use of (>8) and (3-9) gives us 00 CO Y^ncy = (z-h) ^C'h\ n=l n-o n 1 n Equating the coefficients of h ~ and h we obtain: h"-S (n-l)CÂ„., = zC;_, C'_2 , n=2,3,...^ (>10) C^_^ = sc;., (n-l)CÂ„_, . In (3-10), replace n by n+1, thus obtaining (3-11) C;_^ = zC; nC^ , n=l,2,.... 00 \ Â— ' B Using (3-8) In (3-3) and letting (j) = ; C h , we now have n=o PAGE 104 98 00 y nC h n=l n-1 2v(z-h)(l-2hz+h^)"^ V Cjjh" , n^O 00 {l-2hz+h^) VnG^h""^ 2v(z-h) V C^hÂ° . n-l n=o Equating coefficients of h n-l nCÂ„ 2(n-l)zCÂ„., + {n-2)C^,2 = 2vzC^-i "2 ^C^.2> and (3-12) nC 2(n-l+v)zC + (n-2+2v)C Â„ = 0. Differentiating (5-12) vjith respect to z gives us nC 2(n-l+v)C -2(n-l+v)zC' , + (n-2+2v)C' ,= 0. n ^ ' n-l * ' n-l ^ ' n-2 Using (3-lC) and simplifying, n C 2(n-l+v)C , -2(n-H-v)zC' , n ^ ' n-l ^ ' n-l + (n-2+2v) zC . (n-l)C , n-l ^ ' n-l = 0, nC n 2(n-l+v) + (n-l)(n-2+2v) 'n-l 2(n-l+v) + (n-2+2v) nC ' n(n+2v-l)C , nzC ' , = Oj n ^ ' n-l n-l ' C' (n+2v-l)C , zC' = 0. n ^ ' n-l n-l Now use (3-11) and simplify the result: 2CA.1 = 0, C' (n+2v-l)C , z C + nzC = 0, n ^ ' n-l n n ' PAGE 105 99 (>13) (l-z2)c; (n+2v-l)C^_^ + nzC^ = 0, Differentiating (3-13) with respect to z, we have -2zC^ + (l-z^)c; (n+2v-l)c;_j^ + nC^ + nzC n 0, (l-z2)c; + (n-2)zC^ (n+2v-l)C^_^ + nC^ = 0. With the use of (3-11) > we may write (l-z^)C; + (n-2)zC^ (n+2v-l)(zC^-nC^) + nC^ 0, (l-z^)C" (l+2v)zC' + n(n+2v)CÂ„ = 0. ^ ' n n n If C Is replaced by u, we have the desired form of Oegenbauer's equation, (l-z^)u" (H-2v)zu' + n(n+2v)u = 0. If we then let v = i, we have Legendre's equation, (l-z^)u"-2zu' + n(n+l)u = 0. Now to obtain Gegenbauer's equation, we use equation (3-1) > letting a^ = l,a2 = -l,a^ = ^^2 " ^ v, p^ = ^^ 0, A Is an arbitrary finite constant, B = a3n(n+2v) . Thus, d^U 1-l+v ^ 1-i+v z-1 z+1 z-a3 du dz '^3^3 Az+a,n(n+2v) 1(2-33)^ (z-l)(z+l)(z-a3)J u = 0. Now let 83 become infinite and we have the desired result: d^u . (H-2v)z du _ n(n+2v) ^ dz 2^-1 dz z^-l (l-z^) ^ (l+2v)z ^ + n(n+2v) u = dz dz PAGE 106 100 Before continuing further let us obtain the equation to be used In subsequent cases. Using the scheme / a, u = ^2 ^ a^^+f a^ a^ 00 1 a^ i_ a. ^2 P: P. we can write equation (2-2): 1 u" + 1-a^^-r3_ z-a. l-a,-3, l-a,-p 2 '"2 3 ^"3 Z-a, z-a. u r(^-|^){a,-a,)(a,-a3) ^^p^f^^.a,) a3P3(a3-a,) _ (z-a^)"{z-a2)(z-a3) (z-a^ )^(z-a3 ) (z-ag )^(z-a^ ) (z-a^)(z-a^) u = 0, where f + Cg + ^2 + ^^3 "Â•" ^3 "*" ^ + ^4 = 2j f,^ and t are arbitrary constants. Now let a^^ become Infinite, then (3-14 ) u" + 1-a -p l-a-,-3. I. + f Â— Â± + . '3 ^3 z-a. z-a. u' + L(z-a2)(z-a3) a,,p (a -a ) + _Ji_f Â— i Â— Â£ + ^'sPs ^^. u = 0, (2-a2)''(z-a3) (2-83)^ (z-a^)^ and there are no restrictions placed on the arbitrary constants . (e) Laguerre's Equation [5, p. 186] . The equation Is zu" + (l+a-z)u' + nu = 0. PAGE 107 101 If v;e assign the following values to the arbitrary constants In equation (3-1'+), v;e have the desired differential equation: a^ = 83 = 0, a^ = -a, p^ = 0, ^ = -1, 03 + P3 = 1, 03^3 = 1, ^^ = -n. u" + -1 . 1-KX u' + u = 0, thus zu" f (l+a-z)u' + nu = 0. (f ) The Equation having the Incomplete Gamma Functions as Solutions [7, p. 9C] . The equation is zu" + {l+a+z)u' + au = 0, with solutions 7(aiZ) = Â— iP^Ca; 1+a; -z) Using equation (3-l^)# let a^ = a^ = 0, a^ = 1-a, 3, = 0, a^ =^ -^3 =1, p^=a, l, = l, then u" + z z u' + 2 ~ 2 z z z u = 0, zu" + {l+a+z)u' + au = 0. (g) Gauss's Equation [7, p. 2]. Consider equation (3-I) with the substitutions: a = a^ ^ 0, 83 = 1^ a. = A+l-ab-c, 02 = B = ab-A, Og = -a-b+c, p^ = Pg = 0, p^ = 1 u" + ab+c-A A-ab 1-Ha+b-c z 2 z-1 u' + ab-A Az+ab-A 2 (z-1) u = 0, u" + c(z-l)+(l+a+b-c)z z(z-l) u ' + (ab-A) (z-l)+Az-i-ab-A ' z'(z-l) u = 0, PAGE 108 102 u" + c-(l+a+D)z 2(1-2) u' ab z(l-2) u = 0. Therefore, z(l-z)u" + c-(l+a+b)z u' abu = 0, which is Gauss's differential equation, often called the hypergeometric equation, and has the hypergeometrlc function 2^1 (Â®* ^' ^'' ^) ^^ ^ solution [8, p. 28^]. (h) Kummer's Equation [7, p. 2]. Use equation (3-1^+) with a^ a^ = 0, a^ = -b, p^ " ^^ <^3 = P3 = ^> ^4 = ^' t = -1, then u -1+ 1+b PAGE 109 103 Any general solution of Whlttaker's equation Is called a Whittaker function and can be expressed In terms of a confluent hyp ergoeme trie function P. (a;b;z). We begin with equation (3-1^), letting ^2 "^ ^3 " ^' T 9 , 2 T 3 2 ,, 1-r +m 1 -r -m +1 1+ Â— ^ Â— + Â— 1 u that u" + Thence J 2 (3-15) u" + u' +(^^ + 11^ ^ u k, -t = 1, so 4 -Hn 2^2 u 0. To normalize (3-15) use (2-19) to find V = z Then v' = -5 z" 2 . v" =1 z2 . Equation (2-20) now becomes 2 Â„ z w ^ fl z-^ i Z-* >kz^ + ^ z-U^-^^ y w = 0. 1 .2 Upon simplifying and multiplying by z^ , the equation can be written z w" + (-m^+kz) w = 0, which is Whittaker's differential equation. 3.2 Classification of Differential Equations having Four Singular Points when the Exponent Difference is ^. The PAGE 110 104 remaining part of this chpater will be devoted to obtaining a classification of differential equations having four singular points with exponent difference equal to i . We shall use equation (>1) with p^ = a^+ Hr=l,2,5) and ^i^ = M.i+i> thus u = 0, z-a ) J r=:l i-2a^ r f '^r^^'r^^) . Az+B Z ?-<=Â«r V^ ^-V-^-^s; A74-R 3 Â„ 3 3 2 r=l r=l r^l To verify this expression for A, we recall that the sum and product of the roots of (3-2) will be equal to the negative of the coefficient of mand the constant term, respectively, thus 3 (>18) 2n^ + i = ^ (2a^ +i) + 2 and r=l 3 (3-19) ^^(Hi+i) 2^a^(a^+i) + A, r=l where vie have used the fact that p = a^ + i (r = 1,2,3) and \X2 "^ V^i + iÂ» From (3-l8) we have 3 which v/hen substituted in (3-19) gives PAGE 111 105 ( 3 1 V^ Â— ; a = r r=l 3 3 r^l r-l Therefore , A = r-l 3 3 2 V \ Â• I Differential equations having four singular points can be classified by the number and nature of the singular points. The coalescence of tv;o singular points having exponent difference \ produces a regular singularity with an arbitrary exponent difference. For, let a^ (3-17), then a in (3-20) u" + l-2a^-2a2 i-2a. z-a. 2-83-. u a (a +4) + Â— r, + Az+B u = (z-ag) (z-a^)''{z-a3)J To find the exponent difference at z = ^-^t we let (3-21) u = h^(z-a^)^ + h^(z-aj^''' +... h^ j^ Then u = ,X-1 h^>^ ( z-a j^ )''Â•' + h^(?v+l)(z-a^)'' +, u h^X(7v-l)(z-a^)''~^ + h^(?.+l)?v(z-a^)^"^ +. Also note that, expanding about z =3]^* we may write Az+B 1 . Az+B (z-a^) (z-ag) (z-a^) z-ag 1 Aa^^+B agA+B (z-a^) 1-3^-33 (8^-83) 2 (2-^1) + PAGE 112 106 If (3-21) were then substituted into (3-20) the Indiclal equation would be found to be (z-a^)^"^: ?v(X-l) + (l-2ci^-2a2)X + a (a +i) + a (a^+i) + = 0, Aa +B -K^ 2{a^+cL^)-K + a^ia^+l) + a^{a^+i) + = 0. The exponents a and p at the confluent singularity z = a^^ are given by the equations a + ^ = 2{a^-Hi^), Aa^+B ap = a^(a^+i) + a2(a2+i) + . 2l"^3 From these equations it can be seen that the exponent difference at the confluent singularity is not i but may have any desired value depending on the choice of B. Further, the coalescence of 3 or more singular points with exponent difference ^ results in an irregular singularity. If ^1=22" ^3' ^^^" (-^-17) becomes 3 -5-2a -2a -2aÂ„ u" +2 1 2 3 ^, z-a 1 -aj^(a^+i)-Ki2(^2"*"^^"^3^Â°'3"*'^) Az+B ^ L (z-aj^ " (z-a^)'j Prom our definitions (section 1.1 ), since u = PAGE 113 (z-a,)^q(z) = (z-aj107 (z-a,)' , Az+B 3 Az+B -, z-a, r=l 1 is not analytic at z = a^^, vie have an irregular singular point at z = a. . The nature of an irregular singular point depends on the number of singularities with exponent difference i that coalesced to produce it. If three such singularities coalesce, the irregular singularity is said to be of the first species [2, p. ^96]; if four such singularities coalesce, it is of the second species. Now every linear differential equation of the second order with rational coefficients has a definite number of regular and irregular singular points associated with it [2, p. 495]. Since each singularity can be obtained by a confluence of an appropriate number of regular singularities with exponent difference 5, we can consider the equation as derived from one of four standard equations. Let any equation be characterized by a formula [2, p. 497] ( a, b, 0^^,02), where a = the number of its regular singularities having exponent difference ^, b = the number of its regular singularities with exponent difference not ^, PAGE 114 108 c, = the number of its irregular singularities of the first species, c = the number of Its irregular singularities of the second species. Any two equations having the same formula may differ from each other in (a) the location of the singular points, (b) the exponents relative to the regular singular points, and (c) certain arbitrary constants. We shall now take an example of each of the four types of equations that can be obtained from the equation v;ith singular points at z = Sj^* ^2' ^3' Â°Â°' ^^^^ ^^^^ normalized form, and after each present the linear transformations that will carry the general cases over into the particular example . The cases derived from (^+,0,0) arc: I. (2,1,0). II. (0,2,0). III. (1,0,1). rv. (0,0,0,1). I. (2,1,0). Tlic singular points will be at z -^ 0,1, CO, CO for our example, so let a ^0, a ^1, a_ oo in equation (3-17) to give us (3-22) u" + i-2a^ i-2cL^ u z-1 J B (z-1)^ z(z-l)J u = 0, where A is finite and B = -agB'. PAGE 115 109 Now to normalize (3-22) we have from (2-I9) that -i(i-2ajln z i(i-2a2)ln (z-l) 1 1 = Z (z-1) Further 5 a, -r 1 1 4 / , x~2 4 . /_ 1\_~1 4 / ,^~2 4 V = (a,-|)z ' ' (z-l) 2 ^ -. (a2-i)z V" = (ct,-i)(a^-|)zÂ°'^" ^ (z-l)''^" ^ + 2(a^-i)(a^. i)z''^' ' (^-l)""'" ' 1 a, -r ciÂ„(2-1) a 9 + (a,i)(c<,i)z"'" ^ (z-1)-^ " With p(z) and q(z) from equation (3-22) and the values of V, v', v" Just found substituted Into (2-20) we have zÂ°^' "(z-l)"^' ^ Â„' + /(a,i)(a,. |)zÂ°^" ' (z-1)-^ ' aiÂ„5 "14 + 2(a^-i)(a2|)z ^ ^ (z-l) 5 CtÂ„T 2 4 1 O.,-r * (".-7)(<^2-7)^"' ' 5 (a^4-)z (z-l) (z-l) 1 9 ^'a4 ciÂ„2 4 1 Â«1-T 4(a^^)z ^ ^z-1) Â°'24^ 2 -^Â° ^l 2 2a2 Â— Â— Â— Â— Â— -f" ^iM " ' ' z-l . ."' ^(z-i)"=^ r^ii^ . Â°^'"^^^' . B (z-l)" z(z.l)J w a 1 Dividing by z ^ '^(z-1) ^ ^ PAGE 116 110 Z^(2-l)\" + (a,i)(a^|)(z-l)^ + 2{a,i) (a^:^)2(z-l) + (^^2t)(^2|)^'-2(<^i^)^(z-l)^-2(a^^){a^^)z(z-l) 2(a,i)(a2i)z(2-l)-2(a2|)'z^-Kx^ (a,+ |)(2-l)^ + 0^(02+ j)z^ + B'z(z-l) w = 0. Z^(2-l)\" +{ 1 n2 (^1-t)(-1-|)^(-2-t)(-2-|)-2(-1-t) 2K^)(a,. i)-2(a2i)^ 4a,(a^+ ^)^,i^,^ |) + B 2(a,^)(a,|)+^(a,1)^ + 2(a,i) (a^i) 2a^(a^+ i)-B>lz+(a^i)(a^|)-2(a^i)24^^(a^+ i)\ Â„ = 0, z^(z-l)^w" + i(a^-Hi2) 2a^a2 +| + B 1(^1-^2) -^ 2a^Â°'2 T B z + TS" w = 0, The normalized form is z"(2-l)^w" + ^kz(z-l) -f -^ where k = -(oj^+a^) " 2aj^^2 *" T "^ ^ ' . w = 0, PAGE 117 Ill The equations with singular points at ^ = a,a,b,Â« and C = a,b,o>,<Â» can be transformed into the form of (;5-22) by taking C = a + b-a and C = (b-a)z + a, respectively . II. (0,2,0). Choose the singular points to be at z = 0,0, CO, 00, then with a^^ = ag = 0, a^ = Â« equation (3-17) becomes (3-25) u" + l-2a^-2a2 ra^{cL^+i)^^ia^+^) 3,1 u' + u = 0, for A a finite constant and B = -agB'. To normalize (3-25) use equation (2-I9) to obtain -i(l-2a -2a ) In z a +a -i V = e = z . V' = (a^-i w=0. PAGE 118 112 Dividing through by z ^1-^22 z\" + m2 (cii-Hi2-l)(cx,+a22) 2(a^+a2-i) + a^(a^+i) + 02(0^+1) + Bw = 0. z^w" + -(a^+a2-2)(ai+a2+i)-Hai + CI2 + i{a^-Hi^) + B' w = 0, 2 Â„ z w + -2a^a2 + ^(Â«i-*^2) +T *Â• B' w = 0, Therefore, z^w" + mw = 0, where m = -aoj^ttg + iCaj^+a ) + -5+ B' . The transfonnatlon C = z + a carries the equation with singular points ^ = a,a,Â«,Â« Into an equation of the form (j5-25). III. (1,0,1). Consider the equation with singular points at 2 = 0,00,00,00 which can be obtained from (3-17) by taking a^^ = 0, a^ a^ = 00, A a finite constant, B = a2a3B-: (3-26) u + u' + _2 "^ z u = 0. To normalize (3-26) use (2-19) to find V = z 1 PAGE 119 113 Then 5 V = (a^4)2 V" = (a,-^)(a,-|)z Substituting in (2-20), 9 5, "^l4 1 1 4 ti , Z W + a, -r / Ix / 5x ^1 4 ^/ 1x2 9_ 1x2 ^'l' 4 + a^(aj,+i)z 9 5 '^l" 4 Â°'l' 4 ^ ^ + B'z ^ w = 0, Dividing by z ^'l4 Z^-^" + (a,^)(a,|-)-2(a,i)2^^,(a,+i) + B'z] w = 0, (3-27) z^w" + f-|^ + B'z ) w = This can be further simplified by letting 1B' Then dz = ~r dC , dw ^ dw d_^ ^ g, dw d w ^ (31)2 d w dz^ dC^ dz dC dz dC Thus (3-27) can be written ^2 d w ^ + C) w = 0. PAGE 120 11^ The transformations C = z + a and C = a + Â— ^ ^ z take the equations with singular points at ^ = a, 00,00,00 and C = a, a, a, 00, respectively, into an equation of the form (3-26). IV. (0,0,0,1). In this case the only equation possible is one with all four singular points at z = Â«>. It will be, from (3-17), (3-28) d!u dz^ + B'u = 0, where A is finite and B = -a^a^a^B'. If we let z = 'B' then dz = .\ d^, Vb' du ^ d^A _d^ dz dC dz B du 2 2 d u _ Â„, d u ^2 Â° ,Â„2 dz dQ Thus, (3-28) can be written + u = 0. Summary of classification: Case I. (2,1,0). Normalized form; z^(z-l)^w" + kz(z-l) + yg w = 0, PAGE 121 115 k Is an arbitrary constant. Case II. (0,2,0). Normalized form: 2 z w" + mw = 0, m is an arbitrary constant. Case III. (1,0,1). Normalized form: z\" + ( Â±. + z^ w = 0. Case IV. (0,0,0,1). Normalized form; w" + w = 0. PAGE 122 LIST OF REFERilllCES 1. H. L. Crowson, A Study of a Linear Ordinary Second Order Differential Equation with Five Regular Singular Points , unpublished Ph , D. dissertation. University of Florida, 1959, PH^ Â• 2. E. L. Ince, Ordinary Differential Equations , Dover Publications, Inc., New York , 1 95 6 . 5. D. Jackson, Fourier Series and Orthogonal Polynomials , Carus Monograph No. G, Math. AssT of Am., Menasha, 19^1 . t. T. M. MacRobert, Functions of a Complex Variable , 4th ed . revised, Macmillan ?c Co. LTD., New York, 1954. 5. E. D. Ralnville, Special Functions , Macmillan Co., New York, i960. 6. H. Scheffe, Linear Differential Equations with a Two Term Recurrence Formula, J. Math. Physics M.I.T. vol. 21, 10'42, p. ^40. 7. L. J. Slater, Confluent H^'pe^^geometrlc Functions , Cambridge University Press, 19G0. 8. E. T. Whlttaker and G. N. Watson, A Course of Modern Analysis , 4th ed . revised, Cambridge University Press, 1955. 9. D. V. Wldder, Advanced Calculus , Prentice-Hall, Inc., New York, 1947 . 116 PAGE 123 BIOGRAPHICAL SKETCH Joyce Coleman Cundiff was born on December l4 , 192? in Zric, Pennsylvania. In May, 19^5^ she was c^^acluatecl from VJeslcyville High School. She received the degree of Bachelor of Science in rnathonatics from UeGtriinster College, Hew Wilmington, Pennsylvania, in June, 19^9After sixteen nonths with N.A.C.A. in Cleveland, Ohio, as a research computer, she joined the U.S. Air Force for two years. The lÂ£.st year of ser-vice v^as spent at Patrick /J'T. as a mathematician working with digital computers. Following her discharge she remained at this same Job and worked for Radio Corporation of America for two years, taking a six months leave of absence for graduate work at Radcliffe College. In February, 1956> she enrolled in the Graduate School of the University of Florida and received the degree of Master of Science in June, 1957. She held a graduate assiatantship from 1956 until I958. After her marriage to John L. Cundiff in June, 1958* she taught one year at Alabama Polytechnic Institute and one year at the University of Florida. In June, I960, she began studying full time. Besides being a member of the local mathematics and physics honor societies at Westminster College, she is a member of Phi Kappa Phi, Mathematical Association of America, National Council of Teachers of Mathematics and American Association of University Professors. 117 PAGE 124 This dissertation was prepared under the direction of the chairman of the candidate's supervisory committee and has been approved by all members of that committee. It was submitted to the Dean of the College of Arts and Sciences and to the Graduate Council, and was approved as partial fulfillment of the requirements for the degree of Doctor of Philosophy. June 5, 1961 Dean, College nces / Dean, Graduate School Supervisory Committee: Chairman / 1 J Ti (1 Â« f^> Xon^J ^^ ^\. 9r |