60 Mr Pocklington, On the Symbolic Integration of 



force near a Hertzian oscillator, SV<r = 0, Sao- = 0, where a is the 

 axis of the oscillator. These give <r -L V and a±a, therefore 



*=VclV .P=V.aVP. 



This is equivalent to the first step in the Cartesian solution of 

 the problem. 



2. On the other hand, the solution of V<r — as obtained by 

 the process described above is er = 0. This is certainly not correct, 

 unless we suppose (and we have no justification for this) the 

 equation to hold throughout the whole of space, and further 

 make stipulations as to the value of er at infinity. The correct 

 solution is a = VP, where P is subject to the condition V 2 P=0 

 throughout the space where the original equation holds. This 

 result can be obtained by applying in turn the second and first 

 results of § 1. 



Similarly in the case of the equation $aV . cr = 0, we expect to 

 find as the solution, a = 0, but the actual solution is 



=4-<) 



It is clear that before this method of integration can be safely 

 employed, the whole subject must be carefully investigated. 



3. All the equations considered above may be regarded 

 as being of the form <fxr = 0, where ^ is a linear and vector 

 function, the constituents of which contain V. The last equations 

 of § 1, for instance, may be written fiSVa + ySacr = 0, which is of 

 this form, <j> being /^V ( ) + ySa ( ). 



Let the cubic in $ be 



<jb 3 — m^ 2 + m 2 (f) — m 3 = 0, 



where m 1 ,m 2 ,m 3 are scalar functions of V, then if m 3 =}= 0, we have 

 a solution in the form 



$ 2 - mA> + w 2 



a = 0. 



ra 3 



In evaluating this, the inverse operation l/m 3 must be per- 

 formed first, as otherwise the result obtained is too general. We 

 preferably write this result 



o- = (<£ 2 — m^ + m 2 ) X, 

 where m,\ = 0. 



