certain Differential Equations in Quaternions. 61 



As an example of this process, consider the equation 

 <pa = FVo- - <r = 0, 



which expresses the condition that a vector is equal to its curl. 

 The cubic is easily found to be 



<f> 3 + 3<£ 2 + (3 - V 2 ) </> + (1 - V 2 ) = 0, 



giving o- = (</> 2 + 30 + 3 - V 2 ) \ 



= \+VVX-VSV\ 



where \ is any vector satisfying 



(1-V 2 )\=0. 



As a particular case, take \ = a m , then 



Sin To /COS To sin Tp\ ._ _, rr tt o T r \ 



o- = a -^ - (-^^ - -^J (2pflfop + F. pFap - ^Fap). 



4. Another problem that can be successfully attacked by this 

 process is that of finding the electromagnetic field in the neigh- 

 bourhood of a Hertzian vibrator when the surrounding dielectric 

 is anisotropic as regards specific inductive capacity, but has two 

 of its principal specific inductive capacities equal (e.g. a uniaxial 

 crystal). 



Let the electric displacement be a-e u>pt , the electric force 

 (a<r + biSitr) &*& and the magnetic force re^ 1 , where 27r/p is the 

 period of the oscillation, i is a unit vector parallel to the direction 

 of greatest or least specific inductive capacity, and a> = */(—!). 

 The equations to be solved are 



V Vt = wpar \ 



VV(aa- + biSia) = -o)pT\ (1). 



SV<r = 0, SVt = J 



Eliminating a- from the first two, and using $Vt = 0, we have 

 XT = (p 2 - aV 2 ) r + b ViVS . iVr = 0. 



(If we leave the term aVSVr in this equation the resulting 

 solution identically satisfies SVr = 0, but the working is some- 

 what more complicated. The final result is the same.) Now 

 bViVS ,iV( ) satisfies a quadratic equation which is easily found, 

 and from this equation we can immediately deduce that of ^, 

 namely, 



% 2 - (2p 2 - 2aV 2 + bVHV) x + (p 2 - aV 2 ) (p 2 - aV 2 + bV 2 iV) = 0. 



