1889.] Equations with Boundary Conditions. 355 
Solving these equations we obtain 
_ diy +a? + 8%) + ay, fy2— (@' — BY} 
eT Cnet tC 
b(y2+a? +b%) + ay, fy2-(e—b)P 
b {(a* — b*)? — y,"(a* + b°)} 
Thus we have, if we adopt the upper sign, the relations 
dy, \(a* — bY)’ — y,*(a* + b*)} 
b(y? +a" +b’) + ay, {y2—(@ — BP 
dy, {(a” — b*)* — y," (a+ B°)} 
bys +a +0) + ay, fy2—-(@— By 
dx, {(a’ — b’)? — «,7(a® +b")} 
b (e+ a? + b*) — ax, {an — (a —b*)}? 
da, {(a* — b*)’ — #,'(a" + b*)} 
(ae +a +0") — aa,{a; —(@ — Ye’ 
w= 
the latter two expressions being deduced from the former two 
with the aid of the equations a, +4 y, = 0 and a, + y, = 0. 
In a similar manner, starting with the second and third of 
our original equations, we should obtain 
da, {(a? — B’ — a2(a +b) 
b (a? +a" +b) + az, ja? (a — 0)? 
da, {(a” — b*)’ — x,"(a* + b*)} 
~ b(ae +a’ +b) + aa, fa,’ — (a — 6)? 
dy, {(a* — b*) — y,*(a@ + B)} 
(bY! +a +B) — ay, (92 —@ BF 
dy, {(a" i b*)” a Yn (a ar b*) 
b(y,' + a? + b*) — ay, {yy — (a? — b°)}? 
If we multiply these two sets of relations together, and take 
the square root of the result, we have 
dy, ((a* — BY — y2 (a? + BY} 
(By, + at + BY — aby, fy = (a — BY 
dy, (a? — BY — y. (a + B)} 
[DG + a +0 a'y? ly — (@ OP 
262 
