lol 



so that 



v=(— k^ -d( (a(kd-4-k,)) (t + 1 2(a)' nog((a)'^t— 1) ((a) '/^t+l) + (5 



(28) 



Eliminating ^H" between (25) and (28) we have the the geodesic lines for 



r =: d given by 



— (u=k + k,)' - (kd-' + k,)' ■' d(u2k+kj)' '--uikd-' + k , ' -' + <5 



V = log 



u.d 2d2 d(u2k+kJV 24-u(kd-+k,)V, 2 



(The above equation is equation 29.) 



When r — d (26) gives rise to an ellii)tic integral for the reduction of 



which we recall from the general tlieory of elliptic integrals. (See Note 1. ) 



R(x) = Ax^ + 4Bx3 + 6Cx= + 4B'x + A' 



g, = A A' — 4BB' + 30 = 



g3 = ACA' + 2B0B' — A^B2 AB'-' — 0=* 



In this case we have, 



R(t) =: abt^ — (a+b)t= + I 



g, = ab + (a+b)^ 12 



g3 = (-ab(a+b)) 6 + (a + b)3 216 

 We also have 



E,'(t) =4abt3- 2 a + b t 



R'^(t) = 12abt= — 2(a + b) 

 Substituting in (26) 



t = . + (1 4R'(0> (pu — 1 24R^^(e)) (See Note 2) (30) 



Where f is one of the roots of R(t) = 0. In this ease take f = l/(a) V - 

 then, R'(l (a)' -) = (2(b— a) )/ (a) V/ = 



R"' (1 (a)' 2) = 2( b — a) 



So that (30) may be written, 



t = l {^)\' + ((b-a),(2(a)'/-)) (pu-pv) 



when pv := (1 12) (5b — a) and therefore 



abt2 = b+ (b(b— a))/(pu— pv) + (1 4)((b(b -a) =)/(pu — pv) - 



Recalling now that, 



(p'v) ' = 4p3v - gopv — gj (31) 



p^rv =6pv — 1 2g, (32) 



and also, 



(p'v)2/(pt!. — pv)2 4- (pu— p"v)'(pv) = 



p(u + V) + p(u — V) — 2pv (33) 



Note 1.— Kleiu, Modular, Fuiictioueu, Vol. I, p. 15. 

 Note 2.— Enneper, Elliptische Funetioneii, p. 30. 



