ON LAGRANGE’S BALLISTIC PROBLEM. 
19 
22. Integration of the Equation for X .— We write tlie equation 
a 2 x 5 /ax , ax\ _ n 
"1—^— ) — U, 
dr 8 s r+s \ 3 r 3s 
and consider also a function Y which satisfies the adjoint equation 
Y 
a ’ Y ■ 5 ( A+ 3) 
+ 
= 0. 
Then the integral 
dr ds \dr . 3 s/ \r + sj 
/3X 5XY1 3 [ X /3Y | 5Y 
dr \ds r + s/} 06* l \dr ' r + s. 
dr ds 
taken over any area in the plane of (r, s) 
vanishes, and therefore the integral 
[Y( S #- — ) ds + x( ~ + —) dr 
J \ 3s r + s/ \ dr r + sJ 
taken round the boundary of the area also 
vanishes. 
We take the area of integration to be 
bounded by an arc of the locus along which 
X = 0, and two lines parallel to the axes of s and r and meeting at the point P. where 
r = r’ and s = s f . Let these be the lines PA and PC. Then we have 
A + ^U+f Y("P_iX U+x /3Y^Y 
= 0, 
Fig. 3 . 
[Y x] a _ [yx] p - j pi x^i + Jac Y^ - 
+ f x(A + ^)*- 
or, since X = 0 on the arc AC, 
[YX] P = j Y jfds— J x(^ + iL)*+f X(^Y 
Jac 3s J pa \ds r + s1 Jcp \dr r 
cp \dr r + s 
3Y 5Y 
+ s 
dr 
We choose Y so that, at P, Y = 1, along PA, where r = /, 3Y/3 s — — 5Y/(r + s), 
and along CP, where s = s f , cYjdr — — 5Y j(r + s). Then the value of X at P is 
The requisite form of Y is 
Y = 
Jac 
r' + s'\ 5 
Y A ds. 
j ds 
r + s 
F(0, -5, 1 , D, 
2 E 
VOL. CCXXII.- A . 
