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Extinction barrier Q is determined by Ns ( x , t ) for x in the extinction hyperplane, x$_{E}$ = 0. In the spirit of the eikonal approximation, we have to find such ( x , t ) in this hyperplane that s ( x , t ) be minimal. The minimum de- termines the boundary conditions for the optimal Hamil- tonian trajectory of extinction, ( x$_{opt}$ ( t ) , p$_{opt}$ ( t )). The condition that s ( x , t ) is minimal with respect to x$_{i}$$_{̸}$$_{=}$$_{E}$ on the extinction hyperplane means that p$_{i}$ = ∂$_{x}$$_{i}$s → 0 for i ̸ = E as the trajectory ( x$_{opt}$ ( t ) , p$_{opt}$ ( t )) approaches the hyperplane. The minimum of s ( x , t ) with respect to t within the period of modulation is reached if H ( x , p , t ) → 0 as the trajectory approaches the hyperplane. 

A consequence of conditions H ( x , p , t ) → 0 and p$_{i}$$_{̸}$$_{=}$$_{E}$ → 0 is that the momentum component p$_{E}$ remains bounded on trajectory ( x$_{opt}$ ( t ) , p$_{opt}$ ( t )). Indeed, near the extinction hyperplane, x$_{E}$ ≪ 1, we have 

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Here, we assumed that w ( x , r , t ) is nonsingular at x$_{E}$ → 0 and, since w = 0 for x$_{E}$ = 0 and r$_{E}$ ̸ = 0 [cf. Eq. (2)], we expanded w in x$_{E}$ to the lowest order. Let us assume now that | p$_{E}$ | → ∞ for x$_{E}$ → 0. Then only the term with maximal − r$_{E}$ ≡ − r$_{Em}$ should be kept in the sum over r$_{E}$ in Eq. (7); it is also clear that p$_{E}$ should be negative, otherwise the trajectory would not approach x$_{E}$ = 0. From the Hamilton equation for p$_{E}$ and Eq. (7) it follows that dp$_{E}$/dx$_{E}$ ≈ − 1 /x$_{E}$r$_{Em}$ . This relation, along with the explicit form of the Hamiltonian H , show that, if p$_{E}$ were diverging for x$_{E}$ → 0, the Hamiltonian would not become equal to zero but would remain ≈ ∂w/∂x$_{E}$ with the derivative calculated for x$_{E}$ = 0 and r$_{E}$ = r$_{Em}$ . This contradiction shows that the assumption | p$_{E}$ | → ∞ is wrong, p$_{E}$ remains limited for x$_{E}$ → 0. 

Equation(7) shows that x$_{E}$ → 0 exponentially as t → ∞ . As x$_{E}$ approaches zero, variables x$_{i}$$_{̸}$$_{=}$$_{E}$ are approaching the equilibrium position in the hyperplane x$_{E}$ = 0. This happens because p$_{i}$$_{̸}$$_{=}$$_{E}$ → 0 and the dy- namics of x$_{i}$$_{̸}$$_{=}$$_{E}$ in the hyperplane is described by the mean-field equations, Eq. (3). Therefore, 

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Function p$_{S}$ ( t ) is periodic in time, with p$_{i}$$_{̸}$$_{=}$$_{E}$ = 0 and with hitherto unknown component p$_{E}$ ( t ), which is dis- cussed below. 

The optimal trajectory ( x$_{opt}$ ( t ) , p$_{opt}$ ( t )) is a hetero- clinic Hamiltonian trajectory that goes from periodic or- bit ( x$_{A}$ ( t ) , p = 0 ) to periodic orbit ( x$_{S}$ ( t ) , p$_{S}$ ), and the action for extinction s$_{ext}$ is calculated along this trajec- tory. The trajectory x$_{opt}$ ( t ) is the optimal path to disease extinction: it describes the most probable sequence of el- ementary transitions leading to extinction. We note that, in periodically modulated systems, there is one optimal path per period, whereas in stationary systems trajecto- ries ( x$_{opt}$ ( t ) , p$_{opt}$ ( t )) are time-translation invariant. 

## 3. The t → ∞ value of the momentum on the Hamilton trajectory 

The momentum component p$_{E}$ is generically nonzero, as found for stationary systems [9, 13, 14, 32]. For peri- odically modulated systems, one can show that p$_{E}$ ̸ = 0 by extending the arguments presented in Ref. [13, 32]. This amounts to showing that the stable manifold of the periodic orbit ( x$_{S}$ ( t ) , p = 0 ) lies entirely in the invariant hyperplane x$_{E}$ = 0, p$_{i}$$_{̸}$$_{=}$$_{E}$ = 0, and, as a consequence, does not intersect the unstable manifold of the periodic orbit ( x$_{A}$ ( t ) , p = 0 ) . Such intersection is necessary in order to have a heteroclinic trajectory that would go from ( x$_{A}$ ( t ) , p = 0 ) to ( x$_{S}$ ( t ) , p = 0 ). 

The hyperplane x$_{E}$ = 0, p$_{i}$$_{̸}$$_{=}$$_{E}$ = 0 is formed by trajec- tories 

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The invariance of this hyperplane is a consequence of Eq. (2), which leads to ˙ p$_{i}$$_{̸}$$_{=}$$_{E}$ = 0 and ˙ x$_{E}$ = 0 for p$_{i}$$_{̸}$$_{=}$$_{E}$ = 0 and x$_{E}$ = 0. 

To prove that the stable manifold of ( x$_{S}$ ( t ) , p = 0 ) lies entirely in the invariant hyperplane x$_{E}$ = 0, p$_{i}$$_{̸}$$_{=}$$_{E}$ = 0, we first show that the trajectories, which are described by Eq. (9) and which start close to the state ( x$_{S}$ ( t ) , p = 0 ), approach this state for t → ∞ . Then, since the dimen- sion of the hyperplane x$_{E}$ = 0, p$_{i}$$_{̸}$$_{=}$$_{E}$ = 0 is equal to the dimension of the stable manifold of ( x$_{S}$ ( t ) , p = 0 ), we conclude that the stable manifold indeed lies in the hyperplane. 

Equations (9) for x$_{i}$$_{̸}$$_{=}$$_{E}$ are the mean-field equations in the extinction hyperplane x$_{E}$ = 0, cf. Eqs. (3), and therefore x$_{i}$ → ( x$_{S}$ ( t ))$_{i}$ for t → ∞ . Linearization of Eq. (9) for p$_{E}$ about ( x$_{S}$ ( t ) , p = 0 ) gives 

We compare this equation with the mean-field equa- tion for x$_{E}$ near x$_{S}$ ( t ). The latter has the form ˙ x$_{E}$ = x$_{E}$ ∑ r r$_{E}$ [ ∂$_{x}$$_{E}$w ( x , r , t )]$_{x}$$_{S}$$_{(}$$_{t}$$_{)}$. Since the state x$_{S}$ ( t ) is unstable in x$_{E}$ direction in the mean-field approxima- tion, from Eq. (10) ˙ p$_{E}$/p$_{E}$ < 0. Therefore, all trajecto- ries on the hyperplane x$_{E}$ = 0, p$_{i}$$_{̸}$$_{=}$$_{E}$ = 0 close to the state ( x$_{S}$ ( t ) , p = 0 ) approach this state asymptotically as t → ∞ , and thus the stable manifold of ( x$_{S}$ ( t ) , p = 0 ) lies in this hyperplane. 

From the above analysis one concludes that there are no Hamiltonian trajectories that would go from ( x$_{A}$ ( t ) , p = 0 ) to ( x$_{S}$ ( t ) , p = 0 ) . Therefore the opti- mal trajectory leading to extinction should go to a state ( x$_{S}$ ( t ) , [ p$_{S}$ ( t )]$_{E}$ ) with [ p$_{S}$ ( t )]$_{E}$ ̸ = 0. 

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