Going through a passport control with wife, or sequential adsorption at extremes.

1. Going through a passport control with wife, or sequential adsorption at extremes. Stanislav Volkov University of Bristol Bath, 4 March 2009

2. CSA model description • Particles arrive sequentially, at times t=1,2,3,… and are adsorbed by one of N+1 sites equidistantly distributed on a circumference • The number of particles already adsorbed at site i at time t is denoted as i(t) where i=1,2,…,N+1 • New particle gets adsorbed at site i with probability proportional to where 0 is a fixed constant • Here and Ujdenotes a certain neighbourhood of site j. • Examples: • [trivial] Uj={ j } • [asymmetric] Uj={ j-1, j } • [symmetric] Uj={ j-1, j , j+1 }

3. Questions • What happens as t? • Do all of i(t) go to infinity? • If yes, do they go at the same speed? • But first introduce auxiliary Markov chain where i=1,2,…,N

4. Results • Theorem 1 [trivial case]: suppose Uj={j}. Then MC is ergodic for <1 and transient for >1. • Theorem 2 [asymmetric]: suppose Uj={j,j+1}. Then MC  is transient for >1. Now suppose N=2. Then for 0<<1 MC  is ergodic, while for >1 it is transient and a.s. the trajectory of (1(t),2(t)) is a spiral. • Theorem 3 [symmetric]: suppose Uj={j-1,j,j+1}. Then MC  is transient for any 0<1 and N3. Now suppose >1. Then a.s. there is a k{1,2,…,N+1} such that limt i(t)= if and only if i=k or k+1; limt k(t) / k-1(t) = m for some finite integer m.

5. Case = (adsorption at max) • In symmetric case, at most two peaks can grow • In asymmetric case, only one peak grows • Proofs are easy here

6. Case =0 (adsorption at min) • Theorem 4 [asymmetric]: suppose Uj={j,j+1} and . Then MC  is ergodic for even N and null recurrent for odd N. Moreover, a.s. there is a T such that for all t  T |k(t)-k+2(t)|2 for k=1,2,…,N+1. Moreover, where |k(t)|  2 and for some >0 as n   where B(t) is a standard Brownian motion

7. Case =0 (min, asymmetric) • Example 1: case N odd Example 2: case N even

8. Case =0 (min, asymmetric) • Example 3

9. Case =0 (min, symmetric) • Theorem 5 [symmetric]: suppose Uj={j-1j,j+1} and . Then MC  is transient. Moreover, for the empty initial configuration a.s. there is a limit limt (t) / t =(x1,…,xN+1) This limit takes finitely many possible values, each satisfying: • there is an >0 such that xi{0, ½ , } • if xi = 0 then xi-1 > 0 and xi+1 > 0 • if xi =  then xi-1 =0 and xi+1 = 0 • if xi = ½  then it  (…,  , 0 , ½  , ½  , 0 ,  ,…). •  xi = 1.

10. Case =0 (min, symmetric) • Example 1

11. Case =0 (min, symmetric) • Example 2

12. Proof for min, asymmetric • Let us work with the potential uk(t)=k(t)+ k+1(t) • Set Ek(t)={uk(t)++, uk-1(t)++} and min(t)=mink{uk(t)} Then • Now set vk(t)=uk(t) - min(t)  0, k=1,2,…,N+1 • Proposition 1: if N+1 is even then v1(t)+v3(t)+v5(t)+…= v2(t)+v4(t)+v6(t)+…

13. Proof for min, asymmetric • Define tj=min{t>tj-1: min(tj)>min(tj-1)} and set • Proposition 2: S(j+1)S(j). Moreover, there is an integer K and an >0 such that on the event S(j)>0 • Proposition 3: S(j)=0 for all large j. Additionally, state 0=(0,0,…,0) is recurrent for Markov chain v(t).

14. Proof for min, asymmetric • Now observe that vi+1-vi=ui+1-ui= i+2- i hence |j- i|  2 (N+1) for all i,j (where i-j even if N+1 even). Next use the renewal times for v(t) to show convergence to Brownian motion.

15. And finally, going through control… 