70 Solution of a Functional Equation. 



We propose to determine the nature of <f>x by rigorous analysis. 



From the nature of the function (p[x-\-z)-\-'f{x~z), which 

 constitutes the right hand member of (1), we know that its se- 

 cond differential, with reference to x, is the same as its second 

 differential, with reference to z ; therefore the second differential 

 of qixcpz, which constitutes the left hand member of (1), with re- 

 ference to X, is the same as its second differential, with reference 

 to z ; hence we have the following condition : 

 d^.q>x d^,(pz 



, . , . , , . rf^.qca: d^.ffz 



or, which IS the same thmg, -7—^-^^^— fi -i -^9Z (3). 



Now, since the left hand member of (3) is a function of x alone, 



and the right hand member is a function of z alone, it follows 



that each member is equal to a constant quantity, which we will 



d^ .(fx 

 denote by a^ ; then we shall have ~r-^-:~(px=a^ (4). 



d.(fx d.<px d^.fpx 



Multiply (4) by 2q^x.-^, and we get 2. -^ • -^^ = 



d. cpx 

 2a=<r^.-^ (5). 



[d.<px 

 Integratmg (5) and addmg the constant c, we get 1 ~7~ 



a^{q>xy-\-c (6), or, by a shght reduction, we obtain dx— 



d.cpx 



— (7) 



Va-{(pxy-\-c ^ '' 



Integrating (7) we get a;=-log. c' ys/ a'^{(fxY-\'C-\ra.(fx\ (8), 



where d is another constant. 



Multiplying (8) by a, and passing from logarithms to exponen- 



ax I \ 



tials, we get e —d W a^i^fxy -\-c-\-a.(fx\ (9), where e is 



such, that hyp. log.e=l. 



Dividing (9) by c' and transposing, we get \^a^{fpxY-\-c=- 



X ax 



-7-6 —a.(fx (10). 



Squaring ( 10) and reducing, we get 



2a ax \ 2ax 



--j.cpx ,e =7J.e -c (11). 



