6 Proceedings of the Royal Irish Academy. 



§ 3. On Fkedholm Solutions. 



If a conductor is placed in a field of force a, a distribution e will be on 

 the surface which will satisfy the condition that the field produced inside the 

 conductor is annulled by <j, which gives the quaternion integral equation 



<7 = 27rf/i/e + / Tdv'a, [e'] T{p - /)-'. 

 Our object is to obtain equations of the Fredholm type from this. If we take 

 the real part, we find 



£ = 2wUvc + V J TcW [_e"\ T(p - p'Y' 



+ cr'dldt j Teh' [,'] Tip - p')-\ 



If in place of e we had a force £ exp {U), where h may be complex, and where £ 

 is independent of t, then e is replaced by c exp Qd), and [c'] l3y 



eGn-^lkt-kC'Tip-py]. 



Writing / for exp [kt - kc' T(p - p')] . T{p - p')-\ the equation thus becomes 



i = 2wUvc + W I Tdv'e'f + kc-' J TcWi'f. 



We thus get a scalar integral equation 



-SsUv^- 2ttc + J e'Tdv'SUvV'f - kc-' J Tdv'SiUvf. 



The second integral requires to be transformed into a function of e. This is 

 effected by the aid of Stokes's theorem in the form \ F{Vdv'^') = 0, where 

 F(Vdv'A') denotes a hnear function of Vdv'A', and the siu'face-integral 

 extends over any closed sm-face, with certain conditions as to the singularities 

 of the function. We have the equation of continuity on the surface 



S{Vdv'V.i'Uv')^-cTdV. 



If we now take a function g, which has no infinities except at the point p, and 

 which satisfies the equation f = (Uv VUv'Vy-g, and if we take the function X 

 such that X = Vv'VUv'V.g, the integral 



! Tdv'i'f = ji'Uv'Vdv'V'.X, 



so that / Tdv'Si' TJv .f = J Si Uv' Frf./V. A Uv, 



which, by Stokes's theorem, 



= lSUv\VdvV'.Uv'.L 



= - J e'Tdv'SUvX + J SUv\ V( VdvV. Uv'i'). 



If (' is such that V{Vd.v'V.Uv'i) = 0, then the transformation is effected. 

 This latter condition is 



I'Sdv'V'W - Uv'Sdv'Vi' = 0, 



or Sdv'Vt' = 0, which can be expressed in various ways such as that Si dp' 



