VECTOR ANALYSIS. 39 



which V.Vu = 0, u will be constant in this surface, and the surface 

 will be contiguous to a region in which V.Vu = Q and u has a greater 

 value than in the surface, or else a less value than in the surface. 

 Let us imagine a sphere lying principally on the other side of the 

 surface, but projecting slightly into this region, and let us particularly 

 consider the surface-integral of Vu for the small segment cut off by 

 the surface Vu = 0. The integral for that part of the surface of the 

 segment which consists of part of the surface Vu = will have the 

 value zero, the integral for the spherical part will have a value either 

 greater than zero or else less than zero. Therefore the integral for 

 the whole surface of the segment cannot have the value zero, which 

 is demanded by the general condition, V.Vu = 0. 



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

 and which may extend to infinity) 



and in all the bounding surfaces 



u = constant = a, 



and (in case the space extends to infinity) if at infinite distances 

 within the space u = a, then throughout the space 



Vu = 0, and u = a. 



For, if anywhere in the interior of the space Vu has a value 

 different from zero, we may find a point P where such is the case, 

 and where u has a value b different from a, to fix our ideas we will 

 say less. Imagine a surface enclosing all of the space in which u < b. 

 (This must be possible, since that part of the space does not reach to 

 infinity.) The surface-integral of Vu for this surface has the value 

 zero in virtue of the general condition V.Vu = 0. But, from the 

 manner in which the surface is defined, no part of the integral can be 

 negative. Therefore no part of the integral can be positive, and the 

 supposition made with respect to the point P is untenable. That the 

 supposition that b > a is untenable may be shown in a similar 

 manner. Therefore the value of u is constant. 



This proposition may be generalized by substituting the condition 

 V.[tVu] = Q for V.Vu = Q, t denoting any positive (or any negative) 

 scalar function of position in space. The conclusion would be the 

 same, and the demonstration similar. 



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

 and which may extend to infinity) 



and in all the bounding surfaces the normal component of Vu vanishes, 



fllJL 



and at infinite distances within the space (if such there are) r 2 ^- =0, 



