486 



BELL SYSTEM TECHNICAL JOURNAL 



From the symmetry of the equation, it is evident that u must be an odd 

 function of y and hence that the solution must pass through the origin. 

 The boundary condition in this case will be that u — > Uo for y — > ± co so that 

 there will be no space charge far from the junction. We can conveniently use 

 the origin as the point at which the solution from y = + ^ joins that from 

 •y = — 00 ; from symmetry, this requires merely that u = when y = 0. 



Fig. Al — Behavior of the solution of Equation (2.16) or (A7.8). 



For large negative y,u ^ smh~^ y and duldy = 1/cosh «o so that duldy 

 is small. It is at once evident that, for large values of K, u must lie above «o 

 so that the integral 



(sinh u - sinh Wo) dy = j- (A7.9) 



will be large enough to make the solution u{y) pass through the origin. If 

 w — Wo > 2 over the region of largest diflference, the space charge will be 

 largely uncompensated and the solution will correspond to that used in 

 equation (2.18). On the other hand, as iT -^ 0, the requirement that u{y) 

 pass through the origin leads to the conclusion that w — wo must be small for 

 all values of y. The possibility that u oscillates about wo need not be con- 

 sidered since it may readily be seen that, if for any negative value of y, 

 say ji , both n{y\) and u'{yi) are less than no{y\) and u'{yi), then u{y) and 

 u'iy) are progressively less than wo(y) and wo(y) as y increases from Vi to 0. 

 Hence, if for negative y the u curve goes below the Wo curve, it cannot pass 

 through the origin. 



