Problems in the Theory of Elasticity. 35 



it is sufficient to show that the reduced problem is incom- 

 patible ; i. e, that the homogeneous equation 



Y(t)+$Y(s).V(st)d8=0 .... (54') 



does not admit any solution but zero. Suppose that it does 

 admit one, y { (s). Then the potential W^ji) of the double 

 stratum of moment Yi(s) vanishes at the boundary of the 

 inner region, and therefore identically throughout that 

 region, because zero is the only regular solution of (1) 

 vanishing over X. The surface traction for the displace- 

 ment W 1 (p) is therefore zero for the inner region, and 

 being continuous at the boundary is zero for S' also. The 

 value of Wi(p) must then be identically zero throughout S'. 

 It cannot be the displacement due to a simple translation or 

 rotation, for a double stratum displacement vanishes at 

 infinity. It follows that 



2v 1 (()=w 1 (i + )-W 1 (r)=0. 



Thus the homogeneous equation (54/) does not admit any 

 solution but zero, and X= — 1 is not a singular value. The 

 equation (54) then gives a unique finite and continuous 

 moment y(s), which", defines a double stratum potential 

 representing the displacement at any pointy of the body S, 

 assuming the value f(t) at the boundary. As a particular 

 case of (31) the solution may be written 



W( j p)=jf(s)'.H_ 1 (^)^ .... (55) 



where the suffix denotes the value of \ involved. In virtue 

 of (50) this may be given the alternative form 



w( / »)=-ijf(0-Tr_ 1 (i»ft, 



the: index -f having the usual significance. 



. & — ~ & - 



§ 14. If the body occupies the infinite outer region S' the 

 parameter value for the problem is A=+l, and the dis- 

 placement W(p) of the body, assuming the surface value 

 — f(f), is expressible as the potential of a double stratum of 

 moment y(s) given by the integral equation 



Y(t)-$Y{s).'V[si)d$=f(t). . . . (56) 



Hi Now the parameter value X=-rl is singular for the 

 kernel ^(st), the homogeneous equation 



v(t) = $v{s).y(st)ds ..... (56') 



admitting certain non-zero solutions. To see this we revert 



D 2 



