BOUNDARY PROBLEMS AND DEVELOPMENTS. 69 



b 



But -Z(t) = J 'D(x)H(t,z)dx 



a 



b 



and Y(x)-= JG(x,t)B(t) 



■dt 



from formulas (13) and (15). 

 Hence equation (16) is equivalent to 



b b 



(17) f f -D(x) { H (t, x) + G(x, t) } B(t)- dx dt = 0. 



a a 



This result, having been obtained without reference to the nature of 

 B' and -D must, moreover, hold for all possible choices of these 

 vectors. We shall proceed to choose a particular set apposite to the 

 proof in hand. 



Let any point (x , t ) of the region a ^ x,t ^ b,x^ t, be arbitrarily 

 chosen and draw the surrounding small rectangle As whose sides are 

 t = t ± A t and x — x =*= A x. Then choose B • and • D so that 

 b t (t) = 0, when I =(= j , b io (t) 3= in As, b J0 (t) = outside of As, 

 d k (x) = 0, when k =f= H, d io (x) =£ in As, d{ (x) = outside of As. 



For this choice equation (17) is equivalent to 



(18) J J d io (x) {h ioJO (t, x) + g iok (x, t)} b J0 (t) dxdt = 0, 



to- At aro-Az 



and inasmuch as d io (x) b io (t) =j= in As, clearly {h ioJO (t, x) + Qi j a {x, t)) 

 changes sign, i.e. vanishes somewhere in this region. Now let Ax and 

 At approach zero. Then in the limit we have 



hi j (to, .T ) + #i ; (.T , *o) = 0. 



From this it is seen that G(x , U) -\- H(t , x ) = 0, and since (x , t ) 

 was any point not on the diagonal it follows that 



G(x, t) + //(/, x) = 0, x =j= t. Q. E. D. 



By direct reference to formula (12) it is readily seen that the Green's 

 function possesses the following characteristics: 



