376 VIII. NEWTONIAN POTENTIAL FUNCTION. 



let us draw on each side of the surface S surfaces at distances 

 equal to e from S, and exclude that portion of space lying between 

 these, which we will call S and $ 2 . 



If the normals are drawn into r we have 



> '- 



The surface integrals are to be taken over both surfaces S 1 and 

 $ 2 and the volume integrals over all space except the thin layer 

 between S 1 and 8 2 . This is the only region where there is discon- 

 tinuity, hence in x the theorem applies, and 



is, 4,r-j (i) , +j r (1) 



Now let us make e infinitesimal, then the surfaces 8 lf $ 2 approach 

 each other and 8. V is continuous at 8, that is, is the same on 



both sides, hence, since = () = - ( ), in the limit the first 



3*V\r/ dn^\rj 



two terms destroy each other. This is not so for the next two. 



3 V 7}V 



for ~ is not equal to -= because of the discontinuity. 

 In the limit, then 



The volume integral, as before, denotes the potential / / / - : dr 

 due to the volume distribution, while the surface integral denotes 

 the potential of a surface distribution / / - > where 



dV 



- 



. Hence we get a new proof of Poisson's surface condition, 

 127, 94). 



132. Kelvin and Dirichlet's Principle. We shall now 

 consider a question known on the continent of Europe as Dirichlet's 

 Problem. 



Given the values of a function at all points of a closed surface 

 8 - - is it possible to find a function which, assuming these values 

 on the surface, is, with its parameters, uniform, finite, continuous, 

 and is itself harmonic at all points within 8? 



