192 Mr. R. Hargreaves on a Diffraction Problem 



edge. We therefore try a function of form e tk( - Yt+!/) ^(r-i-y) y 

 and with p for r + y the differential equation yields 



^- + 27 + t * = '° r P^=Ce-^, 



which corresponds to a function of the type in (1). 



The function <j>(r, y) vanishes for r+y = 0, i. e. on the 

 axis OY'. The use of r + y involves a certain disability in 

 respect to change of sign, which must be directly imposed 



if demanded by the continuity of ^y, ^- or —-. The first 

 two vanish on OY', but 



■d$ = /I *cos{l+KVt-r)} ^ 



3* \ nr'~ 2r \lr~+~y 



while x= + ^(r + y^fr—y*) according as a; is positive or 

 negative. Thus on the two sides of OY' we have 



a finite quantity except at r = 0. The continuity of =~ 



therefore requires the change from +<fr(r, y) to — <£(r, y) 

 in crossing OY'. 



Corresponding to the reflected wave we have a similar 

 function with —y for y, and here the change of sign occurs 

 on the axis OY. We have now the material for construct- 

 ing the solution, which, for the condition ^ = on the 

 barrier, is 



^r=i cos k(Vt +y)—i cos k(Vt— y) + <£(r, y)—<l>(r, —y) 



in 1st quadrant (region A) 



= „ » » +*(»•* y) + 0(r,-y) 



in 2nd & 3rd quadrants (region C) 



= „ v » # ~<#>Ky) + </>(n -?/) 



in 4th quadrant (region B) 



For the problem with zero velocity on the barrier, the 

 signs of terms ^ cos k(Vt—y) and <£(r, -—y) must be changed 

 throughout. In the solution (2) it will be noted that for 



y = in C, ^- has the value due to the incident wave only. 



In the other solution tfr has the value for incident wave only 

 for y = in C. 



