576 Mr. C. E. Weatherburn : ProUe 



ms in 



This is a Fredholm equation from which (f>(s) may be deter- 

 mined by the usual formula. To show that it admits a 

 unique solution it is sufficient to show that the corresponding- 

 homogeneous equation with second member zero does not 

 admit a non-zero solution. If it did we should have a 

 solution to the problem when both yjr(s) and ot(s) are zero; 

 that is, when there are no fixed external charges and no 

 volume distribution, but only the two strata fju(t) and </>($)» 

 The condition (15) would then read 



_ dV , _. ,dV , + v A 

 Multiplying by V and integrating over 2 we have 



Now since rj is supposed zero the potential is zero through- 

 out the .conductors. Hence in the last equation the integra- 

 tion may be made over 2 and ® without affecting the 

 equation. But each of the integrals is then an essentially 

 positive quantity*, and their sum equal to zero. This is 

 impossible, so that <£(V), V(p), and fi(t) are all zero. Since 

 then the homogeneous integral equation does not admit a 

 solution, (18) gives a unique value for 4>(s). This value 

 substituted in (16) gives fi(t). The required potential V(p) 

 at any external point p is then 



VO) = I g(pr)p(r)dr +J g(pq)v(q)dq 



+ 1 g(pt) ?(*)&+ \ g(ps)(l>(s)ds, 



cr-p(r) being the density of the fixed external charge at the 



point r. 



§ 5. Third Problem. — Suppose next that there is only a 

 single conductor and that we are given not its potential but 



its total charge t- The volume density -^-.^(p) is unknown 



as well as the surface density — fi(t). The conductor is- 



under the influence of the same fixed external charges as 

 before, and there is the same surface 2 of discontinuity. 



* Weatherburn, loc. cit. First paper, § 3 (14). 



t Picard has treated the simple case in which there is no surface 2 of 

 discontinuity in the field. Of. Ami. Ecole Normale, t. 25, p. 585 (1908). 



