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DIFFERENTIAL EQUATIONS that is, unless ax and lx* are finite 'at #=00; which we thus Si by (aA + + (aB + 6D)r , and outside these circles will be -assume ; putting y = tr{l+ A1t+ ■ ■), we find for the index equation represented by [(aA + &C)P + 2(aB + &D)R]m;1 + [(«A + 6C)Q + («B + Equation at 33=00 the equation r(r-1)+ r (2 - aa:)0 +(6a:2)o = 0. bu)lw2. A single-valued branch of such integral can be obtained of the ^ there be finite singular points at • • £m, where we by making a barrier in the plane joining 00 to 0 and 1 to 00 • for assume instance, by excluding the consideration of real negative values of second «2->l, dealt with an(1the if cases m — Q, m=l being easily x and of real positive values greater than 1, and defining the phase order. > 4>{x) = 2 {x-^i) ■ ■ • {x - fm), we must have a • <p(x) and b ■ [^(a;)] finite for all finite values of of x and x — 1 for real values between 0 and 1 as respectivelyJ 0 ■x, equal say to the respective polynomials p(x) and 0(x), of and tt. which by the conditions at x—co, the highest respective orders We can form the Fuchsian equation of the second order with possible are m-1 and 2(m-l). The index equation at x = £1 is three arbitrary singular points ft, ft, and no singular point r(r-l) + rf(f1)/0'(f1) + (9(f1)/[^'(|1)]2 = O, and if a1( ft 2be its roots, at a: =00, and with respective indices aifta2fta3ft such we have aj + ^^l-and ajft = 0(f1)/[<?i'(^1)] - Thus by that “! + ft +1*2 + ft + a3 + ft = 1. This equation can Transforman elementary theorem of algebra, the sum 2(1 - af-ft)/(a:-&), then be transformed into the hypergeometric equation *flon of extended to the m finite singular points, is equal to f/(x)/<f>{x), in 24 ways ; for out of ft, ft, ft we can in six ways choose 1. e e<ju^m and the sum 2(1 — a-i — ft) is equal to the ratio of the coefficients of two, say ft, ft, which are to be transformed respectively -w 0 the highest powers of a; in j/{x) and <t>(x), and therefore equal to into 0 and 1, by {x - ft)/(a3 - ft)=<(« - 1); and then there 1 +a+ft where a, ft are the indices at x=oa . Further, if {x, l)m_2 are four possible transformations of the dependent variable which denote the integral part of the quotient 6(x)j(p(x), we have will reduce one of the indices at £ = 0 to zero and one of the indices 2aift0'(f<)/(a:-£i) equal to -(x, l)m_2 + d(x)/<p(x), and the co- at i! = l also to zero, namely, we may reduce either aj or ft at t = 0 efficient of xm~2 in (x, 1)TO_2 is aft Thus the differential equation and simultaneously either a2 or ft at t=. Thus the hypergeohas the form y"+y 2,(1 - ai - ft)/(a; - + y[(x,l)m_2 + 2aift^'(^)/ metric equation itself can be transformed into itself in 24 ways (x-£i)]l<p{x) = 0. If, however, we make a change in the dependent and from the expression F(X, ft 1 - Xj, x) which satisfies it follow 23 variable, putting y = (x- ^1)ai ■ ■ (x- !;m)XmV, it is easy to see that other forms of solution; they involve four series in each of the arguthe equation changes into one having the same singular points , ?—-, —Five of the 23 solutions agree about each of which it is regular, and that the indices at x=^ ments, XyX 1, -, become 0 and ft-a*, which we shall denote by Xf, for (x-fy)*! can with the fundamental solutions already described about a; =0, x = l be developed in positive integral powers of x-^, about
- by X=cc ; and from the principles
by which these were obtained it is this transformation the indices at x=co are changed to a + ax + clear that the 24 forms are, in value, equal in fours. + aTO,/3 + ft + •• -fftrewhich we shall denote by X,/*.. If we suppose immediately The quarter periods K, K' of Jacobi’s theory of elliptic functions this change to have been introduced, and still denote the independent variable by y, the equation has the form y" + y"2,(l -i)/ 2 (£-£i) + 2/(a;, l)TO_2/0(:r) = O, while X + yu + Xj-f • • Xm = m-1. Con- of which K= / (1 — A sin d)~ld9, and K' is the same function of versely, it is easy to verify that if X/t be the coefficient of ^m“2 in (x, l)m_2, this equation has the specified singular points and 1-h, can easily be proved to be the solutions of a hypergeometric equation of which h is the independent ^version. indices whatever be the other coefficients in {x, l)m_2Thus we see that (beside the cases m = 0, m=) the “Fuchsian variable. When K, K' are regarded as defined in terms Modular fuact,ons equation ” of the second order with two finite singular points is of h by the differential equation, the ratio K'/K is an infinitely many valued function of h. But it is remarkable that distin uis Hypergeo- w en g hed by the fact that it has a definite form Jacobi s own theory of theta functions leads to an expression for h in metric thecase singular points and the indices are assigned. terms of K'/K [see art. Functions, Analytic] in terms of singleInh that equation. gular . >arePuling (* - UK* = */(« -1),as the sin- valued functions. . We may then attempt to investigate, in general, 1 points transformed to 0,- 1, 00, and, is clear, without change of indices. Still denoting the independent variable in what cases the independent variable a? of a hypergeometric equaby x, the equation then has the form x(l - x)y" + y'[l --x tion is a single-valued function of the ratio s of two independent (1 + X +/i)] - X/xy = 0, which is the ordinary hypergeometric equa- integrals of the equation. The same inquiry is suggested by the tion. Provided none of XI, X2, y. be zero or integral, it has the problem of ascertaining in what cases the hypergeometric series solutions F(X, /*, l-Xj, x), a:xiF(X + X1( y + , l + X^ a:)abouta;=0x ; F(afiyx) is the expansion of an algebraic (irrational) function of x. about a:=l it has the solutions F(X,/q 1 - X2,1 - a), (l-a;) 2 In order to explain the meaning of the question, suppose that F(X + X2, y + X2, 1 + X2, 1 -x), where X + /i-(-XJ + X2=l ; about x= 00 the plane of x is divided along the real axis from - 00 to 0 it has the solutions arxF(X, X + Xj, X-yu + 1, a;-1), a;-^F(,a,/i + Xj, and from 1 to +00, and, supposing logarithms not to enter p.- + l,x where F(aj3yx) is the series l + a,8x/y + about a;=0, choose two quite definite integrals yx, y2 of the equation, say Vl = F(X/rl - x), y2 = ftiF(X + X1; ^ + Xj, 1 + Xj, x), with -—1 2 7(7+1) “ ^ whlcI1 convei'ges when x < 1, whatever the condition that the phase of x is zero when x is real and bea, ft 7 may be, converges for all values of x for which |x| = l tween 0 and 1. Then the value of s=^-2 is definite for all values provided the real part of 7 — a—^>0 algebraically, and converges . 2/1a single-valued monogenic for all these values except x= 1 provided the real part of y-a-(3 of x in the divided plane, s being > — 1 algebraically. In accordance with our general theory, branch of an analytical function existing and without singularities logarithms are to be expected in the solution when one of Xj,X2, all over this region. If, now, the values of s that so arise be X — /i is zero or integral. Indeed when Xj is a negative integer, plotted on to another plane, a value p + iq of s being reprenot zero, the second solution about x = Q would contain vanishing sented by a point (p, q) of this s-plane, and the value of x from factors in the denominators of its coefficients ; in case X or y be which it arose being mentally associated with this point of the one of the positive integers 1, 2, • ■( — Xj), vanishing factors occur s-plane, these points will fill a connected region therein, with a also in the numerators ; and then, in fact, the second solution continuous boundary formed of four portions corresponding to the about £c = 0 becomes x times an integral polynomial of degree two sides of the two barriers of the a;-plane. The question is (- ) 7 ^ or t°f degree (— Xj) - y. But when is a negative in- then, firstly, whether the same value of s can arise for two different teger including zero, and neither X nor y is one of the positive values of x, that is, whether the same point (p, q) of the s-plane integers 1, 2 ■ ■ ( —Xj), the second solution about x = 0 involves a can arise twice, or in other words, whether the region of the s-plane term having the factor log x. When Xj is a positive integer, not overlaps itself or not. Supposing this is not so, a second part of zero, the second solution about x=0 persists as a solution, in the question presents itself. If in the cc-plane the barrier joinaccordance with the order of arrangement of the roots of the ing -co to 0 be momentarily removed, and x describe a small circle index equation in our theory ; the first solution is then replaced with centre at x=0 starting from a point x= -h-ik, where h, k by an integral polynomial of degree — X or — y, when X or is one are small, real, and positive and coming back to this point, the of the negative integers 0, -1, - 2, • • -, 1 - X1; but otherwise con- original value s at this point will be changed to a value <r, which in tains a logarithm. Similarly for the solutions about x—1 or the original case did not arise for this value of x, and possibly not x=co •, it will be seen below how the results are deducible from at all. If, now, after restoring the barrier, the values arising by those for x~0. continuation from <r be similarly plotted on the s-plane, we shall Denote now the solutions about x = 0 by
- those about x= 1 again obtain a region which, while not overlapping itself, may
an by > d those about x=<x> by wq, w2; in the region (SoSj) quite possibly overlap the former region. In that case two values March common the circles S0, Sj of radius 1 whose centres of x would arise for the same value or values of the quotient ydy,, are le to of the P0exist ints xec—u0, x=l, all the first four are valid, arising from two different branches of this quotient. We shall and H tbere integral. where A, B, 0, DL are ations u1 = Av; 1 +inV,vthe D?;2 understand then, by the condition that x is to be a single-valued 2. Mregion 2 = Ct’i +(S^) constants function of x, that the region in the s-plane corresponding to any lying inside the circle Sj and outside the circle S0, those that are valid branch is not to overlap itself, and that no two of the regions are v^w-yw^, and there exist equations vx = Pwq + Qv;,2,v2 = Bwj + 1w2, corresponding to the different branches are to overlap. Now in where P, Q, B, T are constants; thus considering any integral describing the circle about x=0 from x= -h-ik to -h + ik, where whose expression within the circle S0 is aux + bu2, where a, b are h is small and k evanescent, s = 03xiF(X + Xj, y + x, 1+Xj, a:)/F(X, y, constants, the same integral will be represented within the circle 1 — X1( x) is changed to 0 = se2,r'xi. Thus the two portions of
