The invariant ergodic measure for X shows that there is a finite probability to reach all states from any point when > 0, in contrast to the determinstic case = 0; the invariant measure also shows that these probabilites are exponentially small, proportional to e−V /. It is practical to relate reaction rates to exit times from domains: define forX
solving (9.3) and a given domainA∈Rd the exit time τ(X) = inf{t :Xt∈/ A}. We want to understand the exit probability
P(τ < T) =E[1τ <T] =:qτ as→0+.
The Kolmogorov-backward equation shows that
∂tqτ −V0·∂xqτ+∂xxqτ = 0 inA×(0, T)
Remark 9.9 (A useless solution). A naive try could be to remove the diffusion part ∂xxqτ in (9.4); that leads to the hyperbolic equation
∂tqτ−V0·∂xqτ = 0 inA×(0, T)
qτ = 1 on ∂A×(0, T)
qτ(·, T) = 0 on A× {T}
(9.11)
which can be solved by the characteristics ˙yt=−V0(yt):
d dtqτ(y t, t) =∂ tqτ+ dyt dt ·∂xqτ =∂tqτ −V 0·∂ xqτ = 0.
Since the equilibrium points are stable, it turns out that all characteristics leave the domain on the upper part t= T see Figure 9.3, whereqτ = 0, so that the solution of
(9.11) becomes qτ = 0, and that is a useless solution.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 !3 !2 !1 0 1 2 3 time X(t)
Figure 9.3: Four pathsXt starting with X0< x
0 in the domain of the global attractor
x−
The limit in Remark9.9needs to be refined to give something useful. The invariant measure with probabilities proportional toe−V / suggests a change of variablesq
τ(x, t) =
ew(x,t)/. The right way to studyq
τ as→0+ is to use the limit
lim
→0+logqτ = lim→0+w =:w
which we believe has a bounded non positive limit, using the invariant measure. Sinceqτ
is a probability we know that w ≤0 and (9.10) implies thatw solves the second order
Hamilton-Jacobi equation
∂tw−V0·∂xw+∂xw·∂xw+∂xxw = 0 inA×(0, T)
w(x,·) = 0 on∂A×(0, T)
A good way to understand this Hamilton-Jacobi equation is to view it as an optimal control problem. In the limit astends to zero, the optimal control problem becomes determinstic, see Theorem 8.10; assume that lim→0+w =:wto obtain the first order
Hamilton-Jacobi equation ∂tw−V0·∂xw+∂xw·∂xw | {z } =:H w(x),x = 0 inA×(0, T) w(x,·) = 0 on ∂A×(0, T) w(·, T) =−∞ on A× {T}.
Following Section 8.1.4, a useful optimal control formulation for this Hamilton-Jacobi equation is ˙ Yt=−V0(Yt) + 2αt max α:(0,T)→Rd − Z τ 0 |αt|2dt+g(Yτ, τ)
which has the right Hamiltonian sup
α∈Rd
λ· −V0(y) + 2α
− |α|2=H(λ, y) =−V0(y)·λ+|λ|2.
Here the final cost is zero, if the exit is on the boundary ∂A×(0, T), and minus infinity if the exit is on A× {T} (i.e. the path did not exit fromA):
g(x, t) =
0 on∂A×(0, T)
−∞ onA× {T}.
Theorem8.10shows that the limit lim→0+logqτ = lim→0+w =w satisfies
w(x, t) = sup α:(t,τ)→Rd − Z τ t |α|2dt+g(Yτ, τ) = sup α −1 4 Z τ t |Y˙t+V0(Yt)|2dt+g(Yτ, τ).
WhenT tends to infinity and X0 is an equilibrium point, this limit whas a simple
explicit solution showing that reaction rates are determined from local minima and saddle points of V, cf. Figure 9.4:
Theorem 9.10. Assume that y+ is a global attractive equilibrium in A. Let X0=y+,
then
lim
Proof. It is clear the optimal control paths starting in y+ need to exit through∂A, so
g(Yτ) = 0. The integral cost can be rewritten as
sup α −1 4 Z τ 0 |Y˙t+V0(Yt)|2dt = sup α −1 4 Z τ 0 |Y˙t−V0(Yt)|2 | {z } ≥0 dt− Z τ 0 ˙ Yt·V0(Yt)dt | {z } V(Yτ)−V(y +) . (9.13)
Here the last integral is minimal ifYτ exits through a point on∂A where V is minimal,
which is a saddle pointif we have chose A to be the largest domain wherey+ is a global
attractor. It remains to show that such an exit is compatible with having the first integral equal to zero; the first integral equals zero means that ˙Yt=V0(Yt), which implies that
Y moves orthogonal to the level lines of the V-potential. Such a path is possible by taking α=V0(Yt) and requiresT to be sufficiently large so that the time to reach the
boundary on the optimal path ˙Yt=V(Yt) is shorter, when X0 tends to y
+ this time
tends to infinity.
We see that the probability to exit from an equilibrium is exponentially tiny, pro- portional to e−(infy∈∂AV(y)−V(y+))/ astends to zero, and therefore such exits are rare
events. In the next section we show that the most probable path, the so called reaction paths, that gives such rare events are those where the stochastic pathsX closely follow the optimal control paths Y. Since is small and the control α is not, the Brownian motion must some time be large of order−1/2. Therefore the rare events of exits depend
on the rare events of such large deviation in the Brownian motion.
The Theorem relates to the basis of reaction theory in chemistry and statistical physics, where the probability to go from one state with energyV1 to another with energy
V2 > V1 is proportional to Boltzmanns rate e−(V2−V1)/(kBT); here kB is Boltzmanns
constant andT is the temperature. We see that, with =kBT and V the energy, the
simple model (9.3) can describe reactions and physical transition phenomena. A simple way to see that thereaction rate is qτ is to take N independent particles starting iny+.
After very long time N qτ of them have exited from the domain and the reaction rate
becomes the quotient N qτ/N =qτ.
Exercise 9.11. Show that the mean exit timeu(x, t) :=E[τ −t |Xt=x] satisfies lim
→0+logu(y+, t) = infy∈∂AV(y)−V(y+).
Exercise 9.12. Does
lim
→0+logqτ =V(X
0)− inf
y∈∂AV(y)
hold when X0 starts from a different point than the global attractor in A? Answer:
Exercise 9.11 shows that the product of the limits of the mean exit time and the probability to exit is equal to one, that is the mean exit time is exponentially large, roughly e(infy∈∂AV(y)−V(y+))/.