184 Major MacMahon and Mr Darlwcf 



if X {x, t) =f{x) t'\ e (x, t) = { f{x)Y'-'t''', 



and in particular if 



X {x, t) =f{x) t^, e {x, t) = [\ {x, t)Y-''. 

 A solution may also be obtained when k (x) = [f, in which case 



K[e[y,e{x,t)\'\.K[d{z, 6'(^, 0}] = e^'•^'^^''■'*^^'^^'''^^''•*^^• 

 Putting d{x,t)=f{a^ + lt, 

 we have 



e [y, d (x, t)] + e{z,e (x, t)] =/(y/) +f(z) + f(x) + ^t, 

 which is of the symmetrical form required. 



6. In the cases investigated above the kernels of the several 

 integral equations have been functions of the same form. It is, 

 however, easy to extend the method to the case where the kernels 

 are functions of different form. Thus if 



/i (x) /ci {0 (x, t)\ dx = -v/tj (t), 



bo 

 /a (x) K. {6 (x, t)} dx = ylr.2 (t), 



we are led to the condition 



K, [6 [y, \ (x, t)]] = K, [6 [x, \ (y, t)]]. 



Case 1. Let Ki(z)—z, k2{1/z) = z; then the condition becomes 

 0{y,X(x,t)].d{x,\(y,t)}^l; 

 a solution of which is 



d(x, t)^xWi^\ (A (OK % {0(0, F{^)i 



where \ {x, t) = (jr^ F {x), 



and ;j^ is any function. 



Case 2. Let k^ (z) = z, k.,(-z)^z; then the condition is 

 0{y,X(x,t)} + e{x,\{y, 01=0; 

 a solution of which is 



d (x, t) = x {F{x)., c/, (01 -x\<^ (0. F{x)\. 

 Case 3. Let k^ {z) = z\ k,_ (s) = (1 - ^0'" ; then the condition is 



\0[y,\{x,t)\J^ld{xMy^m=^\ 



a solution of which is 



e {x, t) = x[F{x), </>(0} ^ lixWia^), 4>{i)]y + (%{</> (0, F{x)]yr- 



