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A first order phase transition in the threshold-$thetage 2$ contact process on random $r$-regular graphs and $r$-trees

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 Publication date 2010
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and research's language is English




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We consider the discrete-time threshold-$theta ge 2$ contact process on a random r-regular graph on n vertices. In this process, a vertex with at least theta occupied neighbors at time t will be occupied at time t+1 with probability p, and vacant otherwise. We show that if $theta ge 2$ and $r ge theta+2$, $epsilon_1$ is small and p is at least $p_1(epsilon_1)$, then starting from all vertices occupied the fraction of occupied vertices stays above $1-2epsilon_1$ up to time $exp(gamma_1(r)n)$ with probability at least $1 - exp(-gamma_1(r)n)$. In the other direction, we show that for $p_2 < 1$ there is an $epsilon_2(p_2)>0$ so that if $p le p_2$ and the number of occupied vertices in the initial configuration is at most $epsilon_2(p_2)n$, then with high probability all vertices are vacant at time $C_2(p_2) log(n)$. These two conclusions imply that on the random r-regular graph there cannot be a quasi-stationary distribution with density of occupied vertices between 0 and $epsilon_2(p_1)$, and allow us to conclude that the process on the r-tree has a first order phase transition.



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A little over 25 years ago Pemantle pioneered the study of the contact process on trees, and showed that on homogeneous trees the critical values $lambda_1$ and $lambda_2$ for global and local survival were different. He also considered trees with periodic degree sequences, and Galton-Watson trees. Here, we will consider periodic trees in which the number of children in successive generation is $(n,a_1,ldots, a_k)$ with $max_i a_i le Cn^{1-delta}$ and $log(a_1 cdots a_k)/log n to b$ as $ntoinfty$. We show that the critical value for local survival is asymptotically $sqrt{c (log n)/n}$ where $c=(k-b)/2$. This supports Pemantles claim that the critical value is largely determined by the maximum degree, but it also shows that the smaller degrees can make a significant contribution to the answer.
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