50 PROCEEDINGS OF THE AMERICAN ACADEMY. 



The same determinant occurs a second time, and a second time only, in 

 Z, -rr. It comes, namely, from i ik also, if we replace therein the 



elements of the yth column by their derivatives with regard to ik, — 

 the same determinant, that is, except perhaps as to sign; and it is 

 easily seen that the signs in the two cases are opposite, so that the two 

 determinants cancel each other. Thus the m (m — 1) determinants, 



-TT- may be written, cancel each other in pairs ; 



i . 

 and the latter expression is, as asserted, zero. 



V. Conditions for <f>-L(u) being (— 1)" times its Adjoint. 



§ 16. The Conditions. 



The remainder of this paper will be devoted 1o a study of the prob- 

 lem : What are the conditions that a differential expression should pos- 

 sess the property of its being possible, by multiplying it by a suitable 

 function, (j>, of the independent variable or variables, to make (f> ' L(u) 

 equal to (— l) n times its adjoint ? 8 After a discussion of ordinary differ- 

 ential expressions I shall give a complete solution of the problem for 

 partial differential expressions of the second order, obtaining also cer- 

 tain results for those of higher order. 



Before attacking the problem, let us notice that the property in 

 question is an invariant property. It is, of course, invariant for a multi- 

 plication of L(u) by a function of the independent variables. It is in- 

 variant for a change of independent variables. For let L(u) go over, 

 under such a change of variables into L(u). Now <f)-L(u) and 

 (— 1)" <f> ■ L(v) are adjoint. Therefore, by the proposition last proved, 



j times the transformed of §-L{u) and -j times the transformed of 



(— l) n <f> ■ L{v) are adjoint. That is, j (p ■ L(u) and (— l) ra y <fr ■ L(y) are 

 ad oint. That is, L(u) can be made equal to (— l) n times its adjoint 

 by multiplying it by -= (f). In the same way, by making use of proposi- 

 tion 6, page 19, we may show that the property n question is inva- 



8 This problem is solved, in the case of partial differential expressions of 

 the second order, in two independent variables, by du Bois-Reymond in the 

 article referred to in the last note. The fact that the expression, whose vanish- 

 ing forms the condition for the possibility of a solution, is an invariant, is not, 

 however, noticed. 



