50 KAKL PEAESON 



or, by (iv), P n (r), the chance that an individual taking n flights from a centre 

 should be found within a distance r from that centre, is : 



Since *.M = 2^^^.«}. 



we have here the complete analytical solution in known functions — i.e. the 

 Bessel's functions with imaginary arguments — of my original problem of the 

 random walk. But this formal solution provides no better method for shortly 

 determining the dispersal curves than that already indicated in these pages. 



(16) Problem IX. To find the distribution after m migrations each of n 

 flights, there being originally a circular clearance ivhich is not kept sterile. 



The solution is found by writing mcr 2 for a\ putting the iV's for the v's 

 in Q t which becomes Q™, and multiplying by the factor (/u,A) m_1 assumed to 

 be constant. This ,can be done to any of the forms (lxxix) — (lxxxiii), or (lxxxvi). 

 If we write : 



e x = ejm and e 2 = e 2 /m 

 we find : 



m F n (c) = N<jiAr->Qre--->S^j"tfe-*dx } ) (xciii), 



or: m f n (c)^N(^) m - 1 Q t m e-'- I J (2iJIJc)e- x dx (xciv). 



Or again : 



m F n (c)=N(^) m - l Qre-^ + ^S^J^jj s (2iJi 1 e i ) (xcv), 



m f n (c) = N(^r-iQ t m ^m>cr's(u s (J^j F.(^L)(«+1)) (xcvi). 



Of these, I have found the first quite as convenient as any other to obtain 

 numerical results from. I shall now illustrate the circular patch formulae. 



Illustration I. A circular patch ^ mile radius is cleared of mosquitoes but 

 not kept sterile. To find the density at the centre, at ^ mile from the centre, 

 and at the margin after ten breeding cycles. 



We shall suppose as before £ = 200 yards, n = 6, and therefore 



1 a 2 

 0^=120,000 square yards. e 2 = - 2 = -3227. 



The second term in Q™ will be of the order ^ of the first and I shall 

 neglect it. Accordingly the solution may be taken 



