266 



Mr. G. H. Darwin. 



[Mar. 18,. 



Hence the coefficient of a? 3 on the right-hand side must be zero, and 



therefore -i- — -i- — — - — P L —-^-=Q- 

 4a T 4b x 2(« 1 s + i g2) 



And 4-= 



8a! 8b x 4( ai 2 + /32) 

 Now when #=+00, arc tan — a =±7r, and when 33= — 00, it is 

 equal to — \ir. 



Hence when a =+00, j= A exp. [±w7i/S/8(a 1 2 + /3 2 )], i=—j; the 

 upper sign being taken for + 00 and the lower for — 00. 



Then since j \ tends to become constant when #=+00, and since 

 9 — - §-= y , therefore when x is very large e tends to vary as a?". 



If x be very small j has a finite value, and i varies as x, and e varies 

 as « 9 . 



J, i, and e all become infinite when x=h, and * also becomes infinite 

 when x=h. 



This analyti cal solution is so complex that it is not easy to under- 

 stand its physical meaning ; a geometrical illustration will, however, 

 make it intelligible. 



The method adopted for this end is to draw a series of curves, the 

 points on which have x as abscissa and i, j, e, n, B as ordinates. The 

 figure would hardly be intelligible if all the curves were drawn at once, 

 and therefore a separate figure is drawn for i, j, and e ; but in each 

 figure the straight line which represents n is drawn, and the energy 

 curve is also introduced in order to determine which way the figure is 

 to be read. The zero of energy is of course arbitrary, and therefore 

 the origin of the energy curve is in each case shifted along the vertical 

 axis, in such a way that the energy curve may clash as little as 

 possible with the others. 



It is not very easy to select a value of h which shall be suitable for 

 drawing these curves within a moderate compass, but after some con- 

 sideration I chose 7^ = 2*6, and figs. 2, 3, and 4 are drawn to illustrate 

 this value of h. If the cubic X 3 — 4\ — (2'6) 3 =0, be solved by Cardan's 

 method, it will be found that X= 2*5741, and using this value in the 

 formula for the roots of the biquadratic we have 



^4_ 2 .6a' 3 + l = (^-2-539)(^--826)[(^+ -382)3 + (-575) 2 ]. 



Hence a = 2'd39, b='826, «= — -382, /3='575, }fc=l'95, and 

 4a 1 =2'356, 4b 1 =4-496, ai =2-332, 

 Then we have 



