280 Prof. Challis on a Problem in 



Up to this point the reasoning is the same as that in the July 

 Number. The subsequent reasoning there proceeds on the as- 

 sumption that the centre of the circle might be placed so as to 

 make the curve pass through the given points. But Mr. Tod- 

 hunter has pointed out that in that case the equation A = is 

 not satisfied. I find, in fact, on making the trial, that the equa- 

 tion is satisfied only when the centre of the circle is on the straight 

 line joining the two points. Hence it follows that the line is a 

 semicircle the extremities of which are coincident with the points. 

 But this cannot be the case unless the given surface is equal to 

 that of a sphere of which the interval between the points is a 

 diameter; so that we have thus solved only a very restricted 

 case of the general problem. It is to be observed that this result 

 has been obtained by using exclusively the equation A = 0, and 

 therefore without assuming, as is done when the equation Ap = Q 

 is used, that the line is symmetrical with respect to the axis of 

 revolution. Thus, considering the premises of the argument, we 

 may conclude from it that, so far as is indicated by the equation 

 A = 0, the problem admits of no exact algebraic solution except 

 the restricted one just mentioned. 



With respect to the deductions I made from the equation 

 Ap = 0, I have ascertained that a logical fault was committed 

 where I used the equation y = c instead of the equation y = } 

 the latter having been obtained by the integration of the com- 

 plete differential equation Ap = } and therefore not admitting of 

 being superseded by integrating p — 0. Also the before-men- 

 tioned condition of symmetry would seem to require that the 

 straight line should be coincident with the axis of x. Having 

 abandoned the solution depending on the equation y = c, I have 

 no occasion to meet the objection that it did not satisfy A = as 

 well as Ap — 0; but the present argument necessitates the re- 

 mark that this objection remains valid against the solution 

 which Mr. Todhunter advocates in the Magazine for June, viz. 

 that in which the required line consists partly of a straight line 

 coincident with the axis, and partly of a semicircle. He himself 

 shows that, so far as regards the straight line, A = 2X, and then 

 proceeds to meet the difficulty by arguing a posteriori that at any 

 rate a maximum is obtained. Without* denying the truth of this 

 conclusion, I assert that the difficulty is in this way evaded 

 rather than met and overcome. The real explanation may, I 

 believe, be derived from the result of the foregoing discussien 

 of the equation A = 0. It is there shown that that equation is satis- 

 fied by no definite algebraic integral fulfilling the condition of a 

 maximum. This being the case, suppose that we have obtained 

 from the equation Ap — a definite algebraic solution; this eo- 

 lutipn cannot satisfy A = throughout', for if so, the latter equa- 



