42 VECTOR ANALYSIS. 



and u = constant for its boundary, which will be a single surface for a 

 continuous aperiphractic space. Hence throughout the space 



88. If throughout an acyclic space contained within finite boundaries 

 but not necessarily continuous 



Vxr = Vxfc> and V.T = V.a>, 



and in all the bounding surfaces the normal components of T and w 

 are equal, then throughout the whole space 



T = ft). 



Setting VU = T ft), we have V.Vw = throughout the space, and 

 the normal component of Vu at the boundary equal to zero. Hence 

 throughout the whole space Vu = T CD = 0. 



89. If throughout a certain space (which need not be continuous, 

 and which may extend to infinity) 



V.V 

 and in all the bounding surfaces 



= ft) 



and at infinite distances within the space (if such there are) 



T = ft>, 



then throughout the whole space 



T = ft). 



This will be apparent if we consider separately each of the scalar 

 components of T and w. 



Minimum Values of the Volume-integral fffuw. to dv. 

 (Thomson's Theorems.) 



90. Let it be required to determine for a certain space a vector 

 function of position &> subject to certain conditions (to be specified 

 hereafter), so that the volume-integral 



fffu w.wdv 



for that space shall have a minimum value, u denoting a given 

 positive scalar function of position. 



a. In the first place, let the vector o> be subject to the conditions 

 that V.o) is given within the space, and that the normal component 

 of eo is given for the bounding surface. (This component must of 

 course be such that the surface-integral of ft) shall be equal to the 

 volume-integral fV. <o dv. If the space is not continuous, this must 

 be true of each continuous portion of it. See No. 57.) The solution 

 is that Vx(itft)) = 0, or more generally, that the line-integral of uw for 

 any closed curve in the space shall vanish. 



