ON LAGRANGE’S BALLISTIC PROBLEM. 
205 
The first line of this expression, with the terms contributed by A,, A-, , makes up 
3 V [ F x (<r + ^) \ 
da) | <x j 
and the remaining lines are unaltered when — s is written for r. 
The terms contributed by Kj and Lj are the same as 
K1 + L1W—|Ki 
+ A A 
S^CBa + sY 
(T 
? 
and thus the Z of the second middle wave is expressed entirely as the sum of the Z of 
the first reflected wave from the left and a function of the form 
A JLY j 0a ( <T—U) 
\a da) \ a 
Further, noting that with equal pistons h = L,, we see that (f>., (a—u) contains no 
terms of degree higher than the ninth in s or \ (a—u). Also we'see that it can be 
expressed as a rational integral function of (s— s 1 )/(R 1 +s 1 ) of the ninth degree, and that 
it contains no terms of degree lower than the fourth. Since Z and dZ/da are continuous 
at s = s 1 with the Z and dZ/da belonging to the first reflected wave from the left, the 
function <y 2 can contain no terms of the fourth or fifth degree in (s—s^/fR, +S,). The 
vanishing of the coefficients of these terms does not introduce any new condition. On 
replacing a 0 , ... by — A 0 , ..., we have the result that in the second middle wave 
Z = 
Ki + \i\U + 
where £ is written for (s—$i)/(Ri + .§i), and = R x == S 2 , while >; 6 , >/ 7 , >/ 8 , >/ 9 are given by 
the equations 
>76 = — 3 
(Kj — ki) ^ ))— —— (^r') + 280£ 0 + 140^i+ 62(, 2 +23^3+ 6v,t—(6 
2*2 'Ll Cl \2v2/ 
>n 
= 4(K 1 -^ 1 )^i- 9 - 6 ^-^J 0 -240^-120f 1 -52^-18f 3 -4f 4 -^ 
= — I (Ki — ki) - Y ——— ( a) +70f 0 + 35{i+ 15(, 2 +5^3 +^4 — is, 
A 2 2 / Cl \ Ag/ 
>79 
