BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR
HEAT EQUATION
S.R.S.VARADHAN AND N. ZYGOURAS
Abstract. We consider a semilinear heat equation in one space dimension,
with a random source at the origin. We study the solution, which describes
the equilibrium of this system, and prove that, as the space variable tends to
infinity, the solution becomes a.s. asymptotic to a steady state. We also study
the fluctuations of the solution around the steady state.
1. Introduction.
One of the natural questions to ask in the study of random partial differential
equations is the effect of randomness on the long term or asymptotic behavior. For
linear equations, diffusions in a random environment is a typical case and the large
time behavior has a long history. The case when the motion in the environment is
either reversible or balanced it has been investigated in [5] and [6] among others.
On the other hand even for the simplest of nonlinear equations the effect of random
initial or boundary data, at locations far away from the random input is hard to
analyze. Turbulence, where ”far away” is in the frequency domain, has been a
particularly challenging problem, with some recent progress. In [2], for the NavierStokes equation with random forcing in two space dimensions, the existence of a
unique equilibrium is proved, though very little information is provided about its
properties.
In this article we provide a detailed description of the equilibrium state of a
randomly perturbed, nonlinear system. In particular, we consider a nonlinear heat
equation in one space dimension, with a source at the origin that is random in
time and with quadratic dissipation. There is a unique stochastic stationary state
where there is equilibrium between the source and dissipation. The solution, which
describes this state, decays far away from 0 in space and we are concerned here
with the effect of randomness in the source at the origin on the decay at infinity of
the solution.
We will consider the heat equation on R × R, with a quadratic dissipative term
and a stationary random source at the origin.
(1.1)
ut + uxx − u2 + λ(t)δ0 (x) = 0, −∞ < x < ∞, −∞ < t < ∞
where λ(t) = λ(t, ω) = λ(θt ω) is a stationary process represented through an ergodic
measure-preserving action of the translation group ω → θt ω acting on a probability
Date: October 11, 2007.
2000 Mathematics Subject Classification. 60H15, 60Fxx.
Key words and phrases. Random semilinear heat equation, law of large numbers, central limit
theorem.
1
2
S.R.S.VARADHAN AND N. ZYGOURAS
space (Ω, F, P ). We assume, in order to simplify the analysis, that λ(t) satisfies
0 < λ1 ≤ λ(t) ≤ λ2 < ∞.
We can think of the solution of this equation as describing, under suitable rescaling, the density of a system of annihilating Brownian particles. The particles appear
at the origin with an intensity determined by the source term λ(t), performing subsequently independent Brownian motions and they are killed as a result of binary
collisions resulting in the term −u2 . This fact has been made rigorous in the case
of dimension greater or equal to two and in the absence of the source term in [7].
It is worth pointing out that similar methods apply to nonlinear equations of
the form
ut + uxx − up + λ(t)δ0 (x) = 0, −∞ < x < ∞, −∞ < t < ∞.
or even more generally to equations of the form
ut + uxx − f (u) + λ(t)δ0 (x) = 0, −∞ < x < ∞, −∞ < t < ∞.
with f (u) ' up as u → 0.
It is not hard to show that there is a unique positive and bounded solution
u(t, x, ω) of the above equation on R × R × Ω, and this solution is covariant in the
sense that u(t + τ, x, ω) = u(t, x, θτ ω). The sketch of this proof goes as follows (
see [9] for details). Since the unique solution will be symmetric around x = 0, it is
more convenient to consider it as a solution of
(1.2)
ut + uxx − u2 = 0, 0 < x < ∞, −∞ < t < ∞
with Neumann boundary data i.e. ux (t, 0) = − 12 λ(t). The solution is smooth for
x > 0. To prove its existence we only need to notice that
a) If we prescribe terminal data, say u(T, x) = 0 at t = T , then the solutions
uT (t, x) ↑ as T ↑ ∞ by the maximum principle.
b) By the maximum principle, the solutions uT (t, x) are all dominated by any
positive solution of
(1.3)
ut + uxx − u2 = 0;
1
ux (t, 0) = − λ2
2
c) There are positive solutions of equation (1.3), that do not depend on t, given by
ua (x) =
6
;
(x + a)2
a>0
satisfying
(3)
uxx − u2 = 0,
ux (t, 0) = −
12
a3
1
Therefore ua (x), with a = ( λ242 ) 3 is a positive solution of equation (1.3) which
provides the necessary bound to prove existence. Uniqueness within the class of
bounded solutions is easily established. Again by the maximum principle it is not
hard to see that if λ(t) satisfies the bounds 0 < λ1 ≤ λ(t) ≤ λ2 < ∞, then
6
6
≤ u(t, x) ≤
(x + a1 )2
(x + a2 )2
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
3
1
3
where ai = ( 24
λi ) . We write
u(t, x) =
6
(x + a(t, x))2
The goal of this article is to establish the following two theorems.
Theorem 1.1. If λ(·) is ergodic then there exists a constant ā such that
lim a(0, x) = ā
x→∞
exists P almost surely.
For the second theorem we will assume that there exists 0 < A < ∞, such that
λ(·) has the property {λ(s) : s ≤ a} and {λ(s) : s ≥ b} are independent if b−a ≥ A.
We will use the standard notation E P [ ] to denote expectation with repect to P .
This condition is stronger than what we need, but we assume it in order to simplify
the proofs.
Theorem 1.2. If in addition λ(·), satisfies the mixing condition stated above, then
as x → ∞, x(a(0, x) − ā) has a limiting normal distribution with mean 0 and
variance
σ 2 = lim E P [x2 (a(0, x) − ā)2 ]
x→∞
We can show that σ 2 > 0 for a large class of examples.
Notice, that Theorems 1.1 and 1.2 imply that the solution u(t, x) has, for every
t, the following asymptotic expansion in x, for x tending to infinity :
6
12 a 12 G(t)
− 3 +
,
x2
x
x4
with G(t) a gaussian random variable.
u(t, x) ∼
Idea of proof:
We first write the equation satisfied by a. After some calculation
ut + uxx − u2 = 0
becomes
(1.4)
at + axx −
6ax
3a2x
−
=0
(x + a) (x + a)
The operator
3a2x
6ax
−
(x + a) (x + a)
is viewed as a perturbation of the Bessel operator
axx −
6
ax
x
by a Feynman-Kac term c(t, x) and we write equation (1.4) as
axx −
(1.5)
at + axx −
6
ax − c(t, x)a = 0
x
4
S.R.S.VARADHAN AND N. ZYGOURAS
where c is given by
(1.6)
c(t, x) =
3a2x
6ax
−
a(x + a) x(x + a)
We will consider the equation (1.5), in a domain (t, x) ∈ (−∞, ∞)×[`, ∞). If ` > 0,
there are no singularities. We will have to choose ` to be sufficiently large, and this
choice will be made later on.
Let us consider the space time Bessel process Qt,x with generator
∂
∂2
6 ∂
+
−
2
∂t ∂x
x ∂x
starting from (t, x) and the exit time τ` from (`, ∞),
τ` = inf{s : x(s) ≤ `}.
Then
·
a(t, x, ω) = E
Qt,x
¸
a(τ` , `, ω)R(t, τ` , ω)
where
R(t, σ, ω) = exp[−r(t, σ, ω)]
is the Feynman-Kac term, with
Z σ
r(t, σ, ω) =
c(s, x(s), ω)ds.
t
We will in Theorem 6.2, prove a uniform bound of the form supx,t,ω |xax | ≤ C.
This in turn will imply a uniform bound of the form supx,t,ω x3 |c(t, x)| ≤ C. We can
Rτ
then control E Qt,x [R(t, τ` )] in terms of E Qt,x [exp[ t ` V (x(s))ds] with V (x) = xC3 ,
which in turn can be controlled if ` = `(C) is large enough. The expectation is with
respect to the Bessel process Qt,x and R(t, τ` ) = R(t, τ` , ω) depends on the random
boundary conditions λ(t, ω) at x = 0. If we define, for a fixed suitably large `, the
conditional expectation
¯
£
¤
¤
τ,` £
(1.7)
g(t, x, τ, `, ω) = E Qt,x R(t, τ` , ω)¯τ` = τ = E Qt,x R(t, τ, ω)
where Qτ,`
t,x is the Bessel Bridge conditioned to exit from [`, ∞) at time τ , then we
can represent
£
¤
a(t, x, ω) = E Qt,x a(τ` , `, ω)g(t, x, `, τ` , ω)
We will show that, for fixed `, as x → ∞ and t → −∞ so long as |t| ∼ x2 ,
g(t, x, τ, `, ω) is nearly independent of t, x and can be approximated by a function
h(τ, `, ω) of τ which is covariant, i.e. h(τ, `, ω) = h(0, `, θτ ω). It will then follow
that
Z
£
¤
(1.8) a(0, x, ω) ' E Q0,x a(τ` , `, ω)h(τ` , `, ω) = a(τ, `, ω)h(τ, `, ω)p(x, `, τ )dτ
where p(x, `, τ ) is the density of the hitting time τ` of the Bessel process Q0,x . This
will give us a law of large numbers with
Z
lim a(0, x, ω) = ā = a(0, `, ω)h(0, `, ω) dP
x→∞
and if the approximation in (1.8) is good enough then
Z ∞
x(a(0, x) − ā) ' x
[(a(τ, `, ω)h(τ, `, ω) − ā)]p(x, `, τ )dτ
0
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
5
Since p(x, `, τ ) the density of the distribution of the exit time τ` of the Bessel process
Q0,x , admits a diffusive scaling limit
τ
1
p(x, `, τ ) ' 2 p(1, 0, 2 )
x
x
if a(t, `, ω)h(t, ω) is sufficiently mixing in t, then a CLT can be proved with the
required scaling. The fixed constant ` > 0 will be chosen later. We consider
the space-time Bessel process starting from x > ` at time t < τ , and condition
it so that the exit time from the interval (`, ∞) is τ , i.e. τ` = τ and denote
the conditioned process (Bessel bridge) by Qτ,`
t,x . Since the Bessel process is not
explicitly time dependent, Qτ,`
is
covariant
with
respect to time translations i.e.
t,x
τ,`
τ −t,`
Qt,x (A) = Q0,x (θt A). As t and x go to −∞ and ∞ appropriately i.e. t ∼ x2
this process will have a limit on the space C[(−∞, 0], [`, ∞)] that we will call Qτ,`
∗ .
Then with
Z σ
£
¤
R(−∞, σ, ω, x(·)) = exp −
c(s, x(s), ω)ds
−∞
and
(1.9)
τ,`
h(τ, `, ω) = E Q∗ [R(−∞, τ, ω, x(·))] = h(0, `, θτ ω)
g(t, x, τ, `, ω) will be approximated by h(τ, `, ω) = h(0, `, θτ ω). Therefore the solution a(0, x, ω) will be approximated by b
a(0, x), defined by
Z
b
a(0, x, ω) = a(τ, `, ω)h(τ, `, ω) p(x, `, τ )dτ
For fixed `, a(τ, `, ω) and h(τ, `, ω) are stationary processes in τ . While the density p(x, `, τ ) of the distribution of τ` under Q0,x is not explicit, its Fourier transform is in closed form and in addition, it has the scaling property x2 p(x, `, x2 t) =
p(1, x` , t). Its behavior, as x → ∞, is therefore easy to analyse. This asymptotic
behavior is combined with the standard ergodic theorem to prove that
(1.10)
lim b
a(0, x, ω) = ā = E P [a(0, `, ω)h(0, `, ω)]
x→∞
a.s. The estimate that plays the central role in our analysis is
lim x2 E P [|b
a(0, x, ω) − a(0, x, ω)|2 ] = 0
x→∞
This estimate reduces the CLT to proving a CLT for x(b
a(0, x, ω) − ā). This will
further require us to prove some mixing properties for h(τ, `, ω) and a(τ, `, ω) as
processes in τ . We will show that they inherit these from similar properties of the
source process λ(·, ω)
2. Bessel process.
We will now define the entrance process Qτ,`
∗ . Because of its covariant nature,
we need only define Q0,`
∗ .
Let us recall that Q0,`
t,x is the distribution of the Bessel process starting at time
t < 0 from x > ` conditioned to exit from (`, ∞) at time 0, i.e. τ` = 0. This defines
a Markov process on C[[t, 0]; [`, ∞)] with transition probability densities, s < σ < 0
and y, z > `
q D (σ − s, y, z)p(z, `, −σ)
q 0,` (s, y, σ, z) =
p(y, `, −s)
6
S.R.S.VARADHAN AND N. ZYGOURAS
where q D is the transition density of the Bessel process with Dirichlet boundary
condition at x = `.
The entrance process Q0,`
∗ is characterized by two properties. For every x > `,
the distribution of the exit time τx from (x, ∞) under Q0,`
∗ is given by the density
f 0,` (x, t) = p(x, `, −t) and the conditional distribution of Q0,`
∗ after time τx , given
the past is given by Q0,`
.
Uniqueness
within
the
class
of
measures
supported on
τx ,x
paths {x(·) : x(t) → ∞ as t → −∞} is an easy consequence of τx → −∞ as x → ∞.
Existence will follow from Kolmogorov’s consistency theorem provided {f 0,` (x, t)}
is consistent as a family of distributions of exit times. Under any Q0,`
t,x the successive
hitting times {τy } of levels x > y > ` are Markovian with the transition density
given by
p0,` (y1 , τ1 , y2 , τ2 ) =
(2.1)
and
p(y1 , y2 , τ2 − τ1 )p(y2 , `, −τ2 )
p(y1 , `, −τ1 )
Z
Z
p(x, y, t − s)p(y, `, −t)
p(x, `, −s)
ds
p(x, `, −s)
Z
= p(x, y, t − s)p(y, `, −t)ds = p(y, `, −t)
f 0,` (x, s) p0,` (s, x, t, y)ds
= f 0,` (y, t)
which proves the necessary consistency.
If we condition the hitting time of level ` to be any arbitrary time τ then the
0,`
corresponding Qτ,`
∗ is just the time shift by τ of Q∗ .
One of the quantities we will need an estimate on is
Q0,`
∗ [x(−s) ≤ y]
Theorem 2.1. For y ≥ `, s > 0
Z
Q0,`
∗ [x(−s) ≤ y] ≤ c
where
Z
∞
c−1 =
∞
9
1
e− 4t t− 2 dt
s
y2
1
9
e− 4t t− 2 dt
0
In particular for any r > 0 there is a constant Cr such that
0,`
r
E Q∗ [x(−s)−r ] ≤ Cr s− 2
For small s we can use the bound x(−s) ≥ `. In addition we have the moment
estimates
Z
(2.2)
τ k p(x, `, τ ) dτ ≤ Ck x2k
for k = 1, 2, 3.
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
Proof. Clearly
Q0,`
∗ [x(−s)
≤ y] ≤
Z
Q0,`
∗ [τy
7
∞
≤ −s] =
p(y, `, t)dt = Q0,y [τ` ≥ s] ≤ Q0,y [τ0 ≥ s]
s
and from the scaling properties of the Bessel process,
Z ∞
1
9
s
Q0,y [τ0 ≥ s] = Q0,1 [τ0 ≥ 2 ] = c
e− 4t t− 2 dt
s
y
2
y
Since τ` ≤ τ0 , the estimates (2.2) on the moments of τ` follow from the scaling
1
9
property of the distribution of τ0 and the tail behavior of p(1, 0, t) = ce− 4t t− 2 . We
can also bound
Z ∞
0,`
E Q∗ [x(−s)−r ] =
y −r dQ0,`
∗ [x(−s) ≤ y]
`
Z ∞
=r
y −(r+1) Q0,`
∗ [x(−s) ≤ y]dy
`
Z ∞
Z ∞
1
9
−(r+1)
≤Cr
y
dy
e− 4t t− 2 dt
s
y2
`
Z
∞
≤Cr
`
= Crs
− r2
7
s
y −(r+1) (1 + 2 )− 2 dy
y
Z ∞
y 6−r
7 dy
(y 2 + 1) 2
`
r
= C r s− 2
We will need the following estimates on p(x, `, τ ) the probability density of the
exit time τ` , from (`, ∞) of the Bessel process starting from level x > ` at time 0.
Theorem 2.2. There is a constant C such that for all x ≥ 2` and σ ≥ 0,
Z ∞
Cσ
(2.3)
|p(x, `, τ + σ) − p(x, `, τ )|dτ ≤ 2
x
0
and for x ≥ 2y, y ≥ `,
Z
(2.4)
∞
|p(x, `, τ ) − p(x, y, τ )|dτ ≤
0
Cy 2
x2
Proof. We note that the Laplace transform
Z
√
√
x3 3
6x2
λ + λ2 ]
ψ(x, 0, λ) = e−λτ p(x, 0, τ )dτ = e−x λ [1 + x λ +
15
15
can be explicitly calculated as the solution of
6
ψx = λψ
x
with ψ(0) = 1, ψ(∞) = 0. It is easy to see that for 0 ≤ a < b,
Z
ψ(b, 0, λ)
ψ(b, a, λ) = e−λτ p(b, a, τ )dτ =
ψ(a, 0, λ)
ψxx −
8
S.R.S.VARADHAN AND N. ZYGOURAS
By analytic continuation, setting λ = −iξ, the Fourier transform
Z ∞
ψ(1, 0, −iξ)
pb(1, a, ξ) =
ei ξ τ p(1, a, τ )dτ =
ψ(a,
0, −iξ)
0
is seen to satisfy
Z
∞
sup
0≤a≤ 12
−∞
[1 + ξ 2 ]|Dξr pb(1, a, ξ)|2 dξ ≤ C
for r = 1, 2, 3. The only subtle point here is the differentiability of ψ(1, 0, λ) at
λ = 0, which follows from an explicit
√ expansion of the form ψ(1, 0, λ) = 1 + c1 λ +
7
c2 λ2 + c3 λ3 + O(λ 2 ) in powers of λ. The Fourier transform pb therefore has three
continuous derivatives at ξ = 0. This implies by Plancherel identity
Z ∞
(2.5)
sup
(1 + τ 2 )3 |pτ (1, a, τ )|2 dτ ≤ C
0≤a≤ 12
0
Therefore
(2.6)
Z
∞
sup
0≤a≤ 12
(1 + |τ |) |pτ (1, a, τ )|dτ ≤ C
0
which in turn implies that for 0 ≤ a ≤ 21 ,
Z ∞
|p(1, a, τ ) − p(1, a, τ + σ)|dτ ≤ Cσ
0
The scaling relation p(x, `, τ ) = x12 p(1, x` , xτ2 ) immediately establishes the first part
of the theorem. As for the second part
Z ∞
p(x, `, τ ) − p(x, y, τ ) =
[p(x, y, τ − σ) − p(x, y, τ )]p(y, `, σ)dσ
0
and using (2.2)
Z ∞
Z
|p(x, `, τ ) − p(x, y, τ )|dτ ≤
0
0
∞
Z
∞
|p(x, y, τ − σ) − p(x, y, τ )|p(y, `, σ)dσdτ
Z
0
C
σp(y, `, σ)dσ
x2
Cy 2
≤ 2
x
=
(2.7)
3. The Main Estimate.
We will now estimate the difference between g(t, x, τ, `, ω) and h(τ, `, ω). But
before that we will need some preliminary estimates.
We start with Khasminski’s lemma which we state and prove for completeness.
Lemma 3.1. Let {Px } be a Markov family on a state space X with x(t) as its
trajectory. τ is the exit time from a set G and V (x) ≥ 0 is a non-negative function
on X. If
Z
£ τ
¤
Px
V (x(s))ds ≤ θ
sup E
x∈G
0
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
then for n ≥ 2
(3.1)
E
Px
· Z
£
τ
¸
· Z
¤n
£
Px
V (x(s))ds
≤ nθE
0
¤n−1
V (x(s))ds
¸
0
Moreover if α > 0 satisfies αθ < 1,
·
Z
£
Px
exp α
(3.2)
E
τ
V (x(s))ds
¸
¤
0
Proof.
τ
9
≤
1
1 − αθ
·Z τ
¸
Ex [
V (x(s))ds]n
0
·Z
¸
Z
= n!Ex
···
V (x(s1 )) · · · V (x(sn ))ds1 · · · dsn
0≤s1 ···<sn <τ
·Z τ
¸
≤ nθEx [
V (x(s))ds]n−1
0
By induction
(3.3)
·Z
Ex [
τ
¸
·Z
V (x(s))ds]n ≤ n!θn−1 Ex [
0
τ
¸
V (x(s))ds] ≤ n!θn
0
If αθ < 1, then summing the exponential series we get
·
¸ X
Z τ
∞
£
¤
E Px exp α
V (x(s))ds ≤
(αθ)n =
0
n=0
1
1 − αθ
We now look at the space-time Bessel process with generator
∂
∂
6 ∂
+
−
∂t ∂x2
x ∂x
in the domain (−∞, ∞) × [`, ∞) stopped when it exits at x = `. For the function
C
C
− 4x
we see that
u(x) = 4`
A=
(Au)(x) = −
and therefore with V (x) =
C
x3
C
4`
τ`
and θ =
Z
£
Qt,x
sup E
(3.4)
t,x≥`
C
; u(`) = 0
x3
¤
V (x(s))ds ≤ θ
t
Lemma 3.2. There exists ` such that for all x > y ≥ `, and n ≥ 1
· Z τy
¸
£
¤n
n Qt,x
(3.5)
sup y E
|c(s, x(s), ω)|ds
ω,x,t
t
· Z τy
¸
Z
¤n
τ,` £
|c(s, x(s), ω)|ds p(x, `, τ − t)dτ
= sup y n E Qt,x
ω,x,t
t
≤ Cn
If we replace the supremum over ω by the mean value then
10
S.R.S.VARADHAN AND N. ZYGOURAS
(3.6)
·
Z
£
lim sup sup y n E P E Qt,x
y→∞
¸
¤n
|c(s, x(s), ω)|ds
x,t
t
· Z τy
·Z
¸
¸
¤n
τ,` £
E Qt,x
|c(s, x(s), ω)|ds p(x, `, τ − t)dτ
= lim sup sup y n E P
y→∞
τy
x,t
t
=0
The quantities really do not depend on t, due to stationarity.
Moreover given α > 0, there exist `(α) and a constant C = C(α) such that if
x > ` ≥ `(α), then
Z τ`
£
¤
(3.7)
sup E Qt,x exp[α
|c(s, x(s), ω)|ds]
ω,x,t
t
Z τ
Z
£
¤
Qτ,`
|c(s, x(s), ω)|ds] p(x, `, τ − t)dτ
= sup E t,x exp[α
ω,x,t
t
≤ C(α)
Proof. Since from (6.7) we have the uniform bound |c(s, x, ω)| ≤ xC3 , (3.3), (3.2)
and (3.4) can be applied and this proves (3.5) and (3.7). Once we have (3.5) to
prove (3.6) it is enough to prove it for n = 1. For n = 1, if we are allowed to average
with respect to ω, then by stationarity, E P [|c(s, x)|] is independent of s. We use
the uniform bound supx G(x, z) ≤ Cz on the Green’s function
¡
¢
1
G(x, z) = z −6 (min{x, z})7 − y 7
7
of the Bessel operator Dxx − x6 Dx in the domain (t, ∞) × (y, ∞), to conclude that
Z τy
Z
£
¤
£ τy P
¤
E P E Qt,x [
|c(s, x(s), ω)|ds] = E Qt,x
E [|c(s, x(s), ω))|]ds
t
Z ∞ t
1
≤C
z E P |c(s, z, ω))|dz = o( )
y
y
in view of (6.9), and so (3.6) follows. We will need the estimates (3.7) for α ≤ 4
and this will dictate the choice of `.
Lemma 3.3. Estimates similar to Lemma 3.2. are valid for the entrance processes
Qτ,`
∗ . There exists ` such that for y > `,
· Z τy
¸
¤n
τ,` £
(3.8)
sup y n E Q∗
|c(s, x(s), ω)|ds
≤ Cn
ω,τ
−∞
If we replace the supremum over ω by the mean value then
¸
·
Z τy
¤n
τ,` £
=0
|c(s, x(s), ω)|ds
(3.9)
lim sup sup y n E P E Q∗
y→∞
τ
−∞
The quantities are again independent of τ .
Moreover given α > 0, there exist ` = `(α) and a constant C = C(α) such that
if x > ` > `(α), then
Z τ
¤
τ,` £
(3.10)
sup E Q∗ exp[α
|c(s, x(s), ω)|ds] ≤ C(α)
ω,τ
−∞
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
11
Proof. Since the entrance process is covariant with respect to translations we can
assume that τ = 0. Let us estimate
· Z τy
¸
¤n
0,` £
y n E Q∗
|c(s, x(s), ω)|ds
τx
uniformly in x > y. If we condition with respect to τx = t, then using the bound
of V (x) = xC3 for |c(s, x, ω)|
¸ Z
· Z τy
¸
· Z τy
¤n
£
¤n 0,`
£
Q0,`
Q`,0
∗
x,t
E
|c(s, x(s), ω)|ds
≤ E
V (x(s))ds f (x, t)dt
τx
t
· Z τy
¸
Z
¤n
`,−t £
= E Qx,0
V (x(s))ds f 0,` (x, t)dt
· Z 0τy
¸
Z
£
¤n
`,t
= E Qx,0
V (x(s))ds p(x, `, t)dt
· Z τy0
¸
£
¤n
= E Qx,0
V (x(s))ds
0
This proves (3.8) and (3.9). The proof of (3.10) is similar.
We now turn to the proof of the main estimate of this section. For functions g
of ω (and possibly other variables) we define the norm kgk2,P by
·
¸ 12
P
2
kgk2,P = E |g(ω)|
Theorem 3.4. Let g(0, x, τ, `, ω) and h(τ, `, ω) be as defined earlier in (1.7) and
(1.9). Then
¸1
Z ∞·
£
¤ 2
P
2
lim x
p(x, `, τ )dτ = 0
E |g(0, x, τ, `, ω) − h(τ, `, ω)|
x→∞
0
Therefore
lim xkb
a(0, x, ω) − a(0, x, ω)k2,P = 0
x→∞
Proof. The proof consists of several steps. First we truncate r(−∞, τ` , ω) and
replace it with r(τx , τ` , ω). For j ≥ 1, we define
σj = τ2j ` = inf{t : x(t) ≤ 2j `}
be the hitting time of the level yj = 2j `. R(τa , τ` , ω) = e−r(τa ,τ` ,ω) . Let n be the
largest integer such that 2n ` ≤ x. R(τx , τ` , ω) = e−r(τx ,τ` ,ω) will be successively
replaced by R(σj , τ` , ω) = e−r(σj ,τ` ,ω) with j = 1, . . . , n.
STEP A. In order to estimate the successive differences we will need to estimate the size of the difference between R(τa , τ` , ω) = e−r(τa ,τ` ,ω) and R(τb , τ` , ω) =
e−r(τb ,τ` ,ω) where a > b > `. More precisely if Qτ,`
0,x is the Bessel Bridge, then we
will need an estimate of
Z
° τ,` £
°2
H(x, a, b) = °E Q0,x R(τa , τ` ) − R(τb , τ` )]°2,P p(x, `, τ ) dτ
First we use the bound |ex − ey | ≤ emax{x,y} |x − y|, to estimate
|R(τa , τ` ) − R(τb , τ` )| ≤ eξ |r(τa , τb )|
12
S.R.S.VARADHAN AND N. ZYGOURAS
where
Z
Z
τ`
ξ≤
0
and V (x) =
τ`
|c(s, x(s), ω)|ds ≤
V (x(s))ds
0
C
x3
¤
¤
τ,` £
τ,` £
E Q0,x |R(τa , τ` ) − R(τb , τ` )|2 ≤ E Q0,x e2ξ |r(τa , τb )|2
¸1 ·
·
¸1
£ 4ξ ¤ 2
¤ 2
τ,` £
Qτ,`
Q
4
≤ E 0,x e
E 0,x |r(τa , τb )|
Therefore
¸¸ 12
· ·Z
£ Qτ,` 4ξ ¤
P
0,x
[e ] p(x, `, τ ) dτ
H(x, a, b) ≤ E
E
· ·Z
¸¸ 21
£ Qτ,` £
¤
× EP
E 0,x |r(τa , τb )|4 p(x, `, τ ) dτ
· ·
¸¸ 1 · ·
¸¸ 1
£
£
¤ 2
¤ 2
£
= E P E Q0,x [e4ξ ]
× E P E Q0,x |r(τa , τb )|4
δ(b)
b2
where δ(b) → 0 as b → ∞ in view of the estimate (3.6). In what follows we will use
δ(b) to denote quantities that are o(1) as b → ∞. Since the estimate is uniform in
x and a, we have the same estimate on
° τ,` £
¤°2
δ(b)
(3.12)
H(b) = sup °E Q∗ R(−∞, τ` ) − R(τb , τ` ) °2,P ≤ 2
b
τ
(3.11)
≤
We now replace R(−∞, τ` ) in the definition of
τ,`
h(τ, `, ω) = E Q∗ [R(−∞, τ` )]
by R(τx , τ` ) and define
τ,`
hx (τ, `, ω) = E Q∗ [R(τx , τ )] = hx (0, `, θτ ω)
Then, from (3.12)
sup kh(τ, `, ω) − hx (τ, `, ω)k2,P ≤
τ
δ(x)
x
STEP B. Now we take up the difference between hx (τ, `, ω) and g(t, x, τ, `, ω).
¤
¤
τ,` £
τ,` £
g(t, x, τ, `, ω) − hx (τ, `, ω) = E Q0,x R(τx , τ` ) − E Q∗ R(τx , τ` )
We represent
R(τx , τ` ) = 1 +
n
X
[R(σj , τ` ) − R(σj−1 , τ` )] + [R(τx , τ` ) − R(σn , τ` )]
j=1
to obtain
g(0, x, τ, `, ω) − hx (τ, `, ω) =
n
X
[gj (0, x, τ, `, ω) − hj (τ, `, ω)]
j=1
+ [gn+1 (0, x, τ, `, ω) − hx,n+1 (τ, `, ω)]
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
where for 1 ≤ j ≤ n
¤
τ,` £
gj (0, x, τ, `, ω) = E Q0,x R(σj , τ` ) − R(σj−1 , τ` )
¤
τ,` £
hj (τ, `, ω)E Q∗ R(σj , τ` ) − R(σj−1 , τ` )
and
¤
τ,` £
gn+1 (0, x, τ, `, ω) = E Q0,x R(0, τ` ) − R(σn , τ` )
¤
τ,` £
hx,n+1 (τ, `, ω) = E Q∗ R(τx , τ` ) − R(σn , τ` )
STEP B1. First we estimate ∆(x, n + 1), which is defined as
Z
°
°
∆(x, n + 1) = °gn+1 (0, x, τ, `, ω) − hx,n+1 (x, τ, `, ω)°2,P p(x.`, τ )dτ
Z
° ¤
°
£°
°gn+1 (0, x, τ, `, ω)°
+ khx,n+1 (τ, `, ω)°2,P p(x, `, τ )dτ
≤
2,P
Recalling that yj = 2j ` and that n is defined by yn ≤ x < 2yn .
Z
p
°
°
°gn+1 (0, x, τ, `, ω)° p(x, `, τ )dτ = H(x, x, yn )
2,P
and
Z
p
p
°
khx,n+1 (τ, `, ω)°2,P p(x, `, τ )dτ ≤ H(x) + H(yn )
Therefore
∆(x, n + 1) ≤
p
p
p
δ( x )
δ(yn ) δ(x)
H(x, x, yn ) + H(x) + H(yn ) ≤ 2
+
≤5 2
yn
x
x
STEP B2. We will now estimate for 1 ≤ j ≤ n, the quantity
Z
°
°
∆(x, j) = °gj (0, x, τ, `, ω) − hj (τ, `, ω)°2,P p(x, `, τ )dτ
By conditioning we can write
gj (0, x, τ, `, ω) − hj (τ, `, ω)
Z
£ Qτ,`
£
¤¤
E σj ,yj R(σj , τ` ) − R(σj−1 , τ` )
× [pτ,` (0, x, yj , σj ) − p(yj , `, τ − σj )] dσj
and
Z
kgj (0, x, τ, `, ω) − hj (τ, `, ω)k2,P ≤
¯¯ Qτ,`
£
¤¯¯
¯¯E σj ,yj R(σj , τ` ) − R(σj−1 , τ` ) ¯¯
2,P
× |pτ,` (0, x, yj , σj ) − p(yj , `, τ − σj )| dσj
Z Z
∆(x, j) ≤
° Qτ,`
£
¤°
°E σj ,yj R(σj , τ` ) − R(σj−1 , τ` ) °
τ,`
× |p
2,P
(0, x, yj , σj ) − p(yj , `, τ − σj )|dσj p(x, `, τ )dτ
Recall from (2.1) that
pτ,` (0, x, yj , σj ) =
p(x, yj , σj )p(yj , `, τ − σj )
p(x, `, τ )
13
14
S.R.S.VARADHAN AND N. ZYGOURAS
If we define for a > b
° τ,` £
¤°
F (a, b, τ ) = °E Q0,a R(0, τ` ) − R(τb , τ` ) °2,P
then by stationarity
∆(x, j)
Z Z
≤
F (yj , yj−1 , τ − σj )|pτ,` (0, x, yj , σj ) − p(yj , `, τ − σj )|p(x, `, τ )dσj dτ
Z
= F (yj , yj−1 , τ − σj )|p(x, yj , σj ) − p(x, `, τ )| p(yj , `, τ − σj ) dσj dτ
Z
= F (yj , yj−1 , τ )|p(x, yj , σj ) − p(x, `, τ + σj )| p(yj , `, τ ) dσj dτ
We can use the estimates (2.3) and (2.4) to obtain
Z
(yj2 + τ )
|p(x, yj , σj ) − p(x, `, σj + τ )|dσj ≤ C
x2
and arrive at
Z
C
∆(x, j) ≤ 2
F (yj , yj−1 , τ )(yj2 + τ )p(yj , `, τ )dτ
x
Moreover in view of (3.11)
Z
|F (yj , yj−1 , τ )|2 p(yj , `, τ )dτ
Z
¯ Qτ,` £
¤¯2 ¤
= E P [¯E 0,yj R(0, τ` ) − R(τyj−1 , τ` ) ¯ p(yj , `, τ )dτ
Z
h
¤2 i
Qτ,` £
P
≤E
E 0,yj R(0, τ` ) − R(τyj−1 , τ` ) p(yj , `, τ )dτ
h£
¤2 i
= E P E Q0,yj R(0, τ` ) − R(τyj−1 , τ` )
= H(yj , yj , yj−1 )
≤
δ(yj−1 )
2
yj−1
and from (2.2)
Z
(yj2 + τ )2 p(yj , `, τ )dτ ≤ Cyj4
Applying now Schwartz’s inequality,
∆(x, j) ≤ C
we finally have have
Z
yj2 δ(yj−1 )
yj δ(yj−1 )
≤ 2C
x2 yj−1
x2
°
°
°g(0, x, τ, `, ω) − h(τ, `, ω)°
2,P
≤
p(x, `, τ )dτ
n
δ( x )
C X
δ(yj−1 ) yj + 2
2
x j=1
x
1
= o( )
x
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
15
4. Law of Large Numbers.
We will now provide a proof of the law of large numbers stated in Theorem 1.1.
Our proof will rely on the main estimate of Theorem 3.4. Though, let us mention
that the law of large numbers can be derived by much easier estimates, bypassing
the much stronger one of Theorem 3.4. We prefer to give the proof by means of
Theorem 3.4, because, on the one hand it provides an expression for the limit ā,
(1.10), and on the other hand the representation that Theorem 3.4 provides will be
crucial for the proof of the central limit theorem later on.
In view of the bound |xax (t, x, ω)| ≤ C, it is sufficient to prove
Z
lim a(0, n, ω) = a(τ, `, ω)g(0, n, τ, `, ω)p(`, n, τ )dτ = ā
n→∞
with probability 1. Since
Z
b
a(0, n, ω) =
a(τ, `, ω)h(τ, `, ω)p(`, n, τ )dτ
from the estimate of Theorem 3.4 it is clear that
E P [|a(0, n, ω) − b
a(0, n, ω)|2 ] = o(n−2 )
and hence by an application of the Borel-Cantelli lemma, |a(0, n, ω)−b
a(0, n, ω)| → 0
with probability 1. It is therefore sufficient to prove
Z
b
a(0, n, ω) = a(τ, `, ω)p(n, `, τ )dτ → ā
a.e. P . If x(t) is an arbitrary stationary stochastic process with finite first moment,
and we wish to prove almost sure convergence to E[x(t)] of averages of the form
Z
Xn = x(t)f (n, t)dt
the following conditions on {f (n, t)} are sufficient.
R
a) f (n, t) ≥ 0 and f (n, t)dt = 1.
R
b) For every h > 0, |f (n, t) − f (n, t + h)|dt → 0 as n → ∞
c)
Z
sup
n
(1 + |t|)|
∂f (n, t)
|dt ≤ C
∂t
The standard proof of the ergodic theorem proceeds by proving it for x(t+h)−x(t),
and then approximating x(t), when E[x(t)] = 0 by linear combinations of these in
L1 . Finally it all comes down to the maximal lemma, [8], Section 6.1. Since we know
that the maximal lemma is valid
R for the standard averages, the problem reduces to
estimating, for x(t) ≥ 0, supn x(t) f (n, t)dt in terms of supτ z(τ ) where
Z
1 τ
x(t) dt
z(τ ) =
τ 0
16
S.R.S.VARADHAN AND N. ZYGOURAS
Z
Z
d
[tz(t)]f (n, t)dt
dt
Z
df (n, t)
= − tz(t)
dt
dt
Z
∂f (n, t)
≤ sup z(τ ) (1 + |t|)|
|dt
∂t
τ
x(t) f (n, t)dt =
We need a uniform bound
Z
¯ ∂p(n, `, τ ) ¯
¯ dτ ≤ C
(1 + |τ |)¯
∂τ
which follows from (2.6) by rescaling.
5. Central Limit Theorem.
Our goal is to prove a central limit theorem for the random variable
Z
Z
x [a(s, `, ω)h(s, `, ω) − ā]p(x, `, s)ds = x f (θs ω)p(x, `, s)ds
as x → ∞ where f (ω) = [a(0, `, ω)h(0, `, ω) − ā]. First we replace p(x, `, s) by
p(x, 0, s) which has the scaling property x2 p(x, 0, x2 s) = p(1, 0, s) and the difference
Z
∆x = x [p(x, `, s) − p(x, 0, s)]f (θs ω)ds
can be estimated easily with the help of (2.7).
Z
k∆x k2,P ≤ xkf (ω)k2,P |p(x, `, s) − p(x, 0, s)|ds
C `2
kf (ω)k2,P
x
The next step is to reduce it to the standard type of central limit theorem. If
Z t
ξ(t) =
f (θs ω)ds
≤
0
we have the identity
Z ∞
Z
x
f (θs ω)p(x, 0, s)ds =x p(x, 0, s)dξ(s)
0
Z ∞
= −x
ξ(s)ps (x, 0, s)ds
Z0 ∞
ξ(x2 s)ps (x, 0, x2 s)ds
= −x3
0
Z ∞
=−
x−1 ξ(x2 s)ps (1, 0, s)ds
0
−1
2
If x ξ(x t) converges to a Brownian motion with variance σ 2 and we have an
estimate
E P [|ξ(t)|2 ] ≤ Ct
then the tails
Z
∞
eT (ω) =
T
x−1 ξ(x2 s)ps (1, 0, s)ds
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
17
can be controlled in L2 (P ) uniformly in x. It is now easy
R to derive from a cen−1
2
tral
limit
theorem
for
x
ξ(x
t),
similar
theorems
for
x
f (θs ω)p(x, 0, s)ds and
R
x f (θs ω)p(x, `, s)ds. The limiting variance will now be equal to
Z ∞
2
σ
|p(1, 0, s)|2 ds
0
We summarize all of this as
Theorem 5.1. Let f (ω) be a function in L2 (P ) with mean 0 such that for some
constant C,
Z
¯ t
¯2
(5.1)
E P [¯
f (θs ω)ds¯ ] ≤ Ct
0
and
Z
x2 t
x−1
f (θs ω)ds
0
satisfies a central limit theorem with variance σ 2 . Then a central limit theorem
holds for
Z ∞
x
f (θs ω)p(x, `, s)ds
0
with limiting variance
Z
σ2
∞
|p(1, 0, s)|2 ds
0
We assume now that that our source process has the property that {λ(s, ω) :
s ≤ a} and {λ(s, ω) : s ≥ b} are independent if b − a ≥ A for some A < ∞. We will
prove the CLT under this condition. First we establish a general theorem.
Let (Ω, F, P ) be a probability space and θt a one parameter family of measure
preserving of transformations. Assume that {Fts ; s ≤ t} is a family of sub σ0
fields (corresponding to events observable during time [s, t]) satisfying Fts ⊂ Fts
s+τ
0
s
s
0
s
if t ≥ s ≥ s, Ft0 ⊂ Ft if s ≤ t ≤ t and θτ Ft = Ft+τ . Let Ft = ∨s≤t Fts
and F s = ∨t≥s Fts . Finally F = ∨s F s = ∨t Ft . Assume that for some finite A,
F t and Fs are independent under P provided t > s + A. It isR not very hard to
that if f (ω) is measurable with respect to some Fba with f (ω)dP = 0 and
Rshow
f 2 (ω)dP < ∞ then
Z tx2
−1
2
−1
x ξ(x t) = x
f (θs ω)ds
0
satisfies a central limit theorem as x → ∞ and converges to a Brownian motion
with a limiting covariance variance σ 2 s ∧ t where σ 2 is given by
Z ∞ Z
2
σ =
[ f (ω)f (θt ω)dP ] dt
−∞
A+(b−a)
Z
=
Z
[
Z
f (ω)f (θt ω)dP ] dt ≤ 2(A + b − a)
f 2 (ω)dP
−A−(b−a)
If f is only F measurable, then it needs to be well approximated by functions
measurable with respect to Fba .
Rt
First we obtain an estimate on the variance of ξ(t) = 0 f (θs ω)ds.
18
S.R.S.VARADHAN AND N. ZYGOURAS
Theorem 5.2. Assume that fk = E P [f (ω)|Fk−k ] satisfies E P [|f − fk |2 ] ≤ Ck −α
for k ≥ 1 with α > 0. Then

1− α

if 0 < α < 1
Ct√ 2
kξ(t)k2,P ≤ C t log t if α = 1

 √
if α > 1
C t
Proof. Let us write
f = f1 +
∞
X
[f2j − f2j−1 ]
j=1
with the corresponding integrals
ξ(t) =
∞
X
ηj (t)
j=0
with
Z
(5.2)
ηj (t) =
0
t
[f2j (θs ω) − f2j−1 (θs ω)]ds
for j ≥ 1, and
Z
t
η0 (t) =
f1 (θs ω)ds
0
Then
kξ(t)k2,P ≤
∞
X
kηj (t)k2,P
j=0
Since the correlations E P [f2j (s)f2j (s + t)] vanish if |t| ≥ A + 2j , it is easy to obtain
the bound
kηj (t)k22,P ≤ Ct min{t, A + 2j }kf2j − f2j−1 k22,P
≤ Ct min{t, A + 2j }2−jα
≤ Ct min{t, 2j }2−jα
where C is now a constant that depends on A. An easy estimation of the sum
X
1
[min{t, 2j }2−jα ] 2
j
proves the the thorem.
We note that if α > 1, we have the estimate
min{t, 2j }2−jα ≤ C 2−j(α−1)
and can use it to prove the central limit theorem for ξ(t).
Theorem 5.3. Let f ∈ L2 (P ). Assume that fk (ω) = E P [f (ω)|Fk−k ] satisfies
E P [|f − fk |2 ] ≤ Ck −α
for some α > 1. Then the central limit theorem is valid for
Z tx2
−1
f (θs ω)ds
x
0
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
19
with the limiting Brownian Motion having covariance σ 2 s ∧ t, where
Z ∞
Z ∞
2
P
σ = lim
E [f2k (ω)f2k (θs ω)]ds =
E P [f (ω)f (θs ω)]ds
k→∞
−∞
−∞
Proof. If we write
f = f1 +
X
[f2j − f2j−1 ]
j≥1
since each finite sum statisfies the central limit theorem ([3]), it is clearly sufficient
to show that ηj (t) defined in equation (5.2) satisfies
√
kηj (t)k2,P ≤ δj t
P
1−α
for some δj satisfying j δj < ∞. Clearly δj = 2−j 2 works.
Remark 5.4. Because of Theorem 5.1 the CLT holds for
Z
x [a(s, `, ω)h(s, `, ω) − ā]p(x, `, s)ds
provided we get an estimate of the type used in Lemma 5.1 with α > 1, for f (ω) =
a(0, `, ω)h(0, `, ω).
Since a and h are bounded functions of ω, it is sufficient to prove the estimate
seperately on a and h.
Theorem 5.5. Let f = a(0, `, ω) and fk = E P [f |Fk−k ]. Then
E P [|f − fk |2 ] ≤ Ck −4
6
Proof. Since u(t, `) = (`+a(t,`))
2 , and we have uniform bounds on u, a(t, `) depends
in a Lipschitz manner on u(t, `) with a Lipschitz constant that depends on `. For
a fixed `, we can therefore take f = u(0, `, ω) instead of a(0, `, ω). According to
Theorem 6.4, the total variation of u(0, `, ω) if we vary λ(·) arbitrarily outside
[−k, k] is dominated by C ` k −2 . The theorem is now an easy consequence.
A corresponding estimate for f (ω) = h(0, `, ω) is more complicated. We do it in
several steps, with each step formulated as a lemma.
Starting from the measure P on the space Ω of functions λ(t) on (−∞, ∞), we
define a measure P k on the subset of Ω × Ω consisting of functions (ω1 , ω2 ) =
(λ1 (t), λ2 (t)) such that λ1 (t) = λ2 (t) if |t| ≤ k. The measure P k is uniquely defined
by the property that the distribution of the process λ(t) = λ1 (t) = λ2 (t) in the
interval [−k, k] is the same as under P and for |t| ≥ k, λ1 (t), λ2 (t) are conditionally
independent given {λ(t) : |t| ≤ k}, each component having the same conditional
distribution as that of {λ(t) : |t| ≥ k} given {λ(t) : |t| ≤ k} under P . This gurantees
that under P k the marginal distributions of both λ1 (·) and λ2 (·) are P , the two
components are almost surely identical for |t| ≤ k, and are conditionally indpendent
given their common values on [k, k].
Then it is easy to see that
E P [|f (ω) − E[f |Fk−k ]|2 ] =
1 Pk
E [|f (ω 1 ) − f (ω 2 )|2 ]
2
20
S.R.S.VARADHAN AND N. ZYGOURAS
Lemma 5.6. Let c(t, x, ω) be as in equation (1.6) and f (ω) = h(0, `, ω). Then for
any β > 0,
Z 0
¯2
k
k
0,` ¯
E P [|f (ω 1 ) − f (ω 2 )|2 ] ≤ CE P E Q∗ [¯
[c(t, x(t), ω 1 ) − c(t, x(t), ω 2 )]dt¯
−∞
Z
k
0,`
≤ Cβ E P E Q∗ [
0
(1 + |t|)1+β |c(t, x(t), ω 1 ) − c(t, x(t), ω 2 )|2 dt]
−∞
Proof. We start with the formula
0,`
f (ω i ) = E Q∗
£
Z
0
exp[−
¤
c(t, x(t), ω i )dt]
−∞
and use the inequality
|ex − ey | ≤ emax{x,y} |x − y|.
From (3.10) we have the bounds
Z 0
¤
0,` £
sup E Q∗ exp[2
|c(t, x(t), ω i )|dt] ≤ C
ω
−∞
Hence we obtain,
1
2
2
|f (ω ) − f (ω )| ≤ CE
Q0,`
∗
≤ Cβ E
¯
[¯
Q0,`
∗
Z
£
0
¯2
[c(t, x(t), ω 1 ) − c(t, x(t), ω 2 )]dt¯ ]
−∞
0
Z
¯
¯2 ¤
(1 + |t|)1+β ¯c(t, x(t), ω 1 ) − c(t, x(t), ω 2 )¯ dt
−∞
We use the estimates on the difference of two solutions provided in Theorem 6.5
to continue with our estimation. We get a preliminary estimate which is not good
enough, but will bootstrap from it to a better one.
Lemma 5.7. For any β > 0, there exists a constant Cβ such that
k
E P [|f (ω 1 ) − f (ω 2 )|2 ] ≤ Cβ k −1+β
Proof. On the difference |c(t, x, ω 1 ) − c(t, x, ω 2 )| we have two different estimates.
For |t| ≤ k2 , from Theorem 6.5
|c(t, x, ω 1 ) − c(t, x, ω 2 )| ≤
C
kx
and, from the bounds (6.7) on c(t, x, ω)
C
x3
We now use the estimates from Theorem 2.1, and the fact that x ≥ `, to conclude
that, for |t| ≤ k2
|c(t, x, ω 1 ) − c(t, x, ω 2 )| ≤
(5.3)
0,`
E Q∗
£
¤
|c(t, x(t), ω 1 ) − c(t, x(t), ω 2 )|2 ≤
C
k 2 (1 + |t|)
£
¤
|c(t, x(t), ω 1 ) − c(t, x(t), ω 2 )|2 ≤
C
(1 + |t|)3
and for all t,
(5.4)
0,`
E Q∗
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
21
If we now use the estimates (5.3) for t ≤ k and (5.4) for t ≥ k, it is easy to see that
Z 0
¯
¯2 ¤
0,` £
E Q∗
(1 + |t|)1+β ¯c(t, x(t), ω 1 ) − c(t, x(t), ω 2 )¯ dt ≤ Cβ k −1+β
−∞
Lemma 5.8. Given any γ < 2, there is a constant Cγ such that
E P [|a(0, x, ω) − ā|2 ] ≤ Cγ x−γ
Proof. Applying the second part of Theorem 5.2 to f (ω) = a(0, `, ω)h(0, `, ω), we
see that the lemma is valid with γ = 2(1 − β).
Lemma 5.9. There exists α > 1 and a constant C such that
k
E P [|f (ω 1 ) − f (ω 2 )|2 ] ≤ Ck −α
Proof. Now we go back and improve on the basic estimate
C
sup |c(t, x, ω)| ≤ 3
x
t,ω
in the mean by getting a better estimate in the mean on ax (t, x). From
E P [|a(t, x, ω) − ā|2 ] ≤ Cβ x−2(1−β)
one sees, using Theorem 6.1, that
E P [|ax (t, x, ω)|2 ] ≤ Cβ x−4+2β
which in turn improves the estimate on c(t, x, ω) to
E P [|c(t, x, ω)|2 ] ≤ x−8+2β
Now we return to the proof of Lemma 5.6 and use (5.3) for |t| ≤ k δ where δ < 1.
The new estimate is
k
E P [|f (ω 1 ) − f (ω 2 )|2 ] ≤ Cβ k δ(1+β)−2 + Cβ k −δ(4−2β)
One can pick δ < 1 and β > 0 such that this works out to
k
E P [|f (ω 1 ) − f (ω 2 )|2 ] ≤ Ck −α
for some α > 1.
This completes the proof of the central limit theorem. We finally establish the
positivity of the variance in a special class of examples.
Let y(t) be a stationary Gaussian process with mean 0 and covariance ρ(t −
s) = E P [y(s)y(t)]. We assume that ρ(·) ≥ 0, has compact support and therefore
satisfies our mixing condition. Let φ(y) be a smooth function on R satisfying
λ1 ≤ φ(y) ≤ λ2 . Assume further that φ(y) is monotone and φ0 (y) > 0. Use the φ
to construct the random source as λ(t) = φ(y(t)). Then the limiting variance
σ 2 = lim x2 E P [|a(0, x, ω) − ā|2 ] > 0
x→∞
The proof proceeds by a perturbation argument. Suppose for each x > 0,
f (x, t) ≥ 0 is a function with compact support in t. We denote by P x the Gaussian
process with mean
Z
m(x, t) =
ρ(t − s)f (x, s)ds
22
S.R.S.VARADHAN AND N. ZYGOURAS
and covariance ρ(t − s). Then the Radon-Nikodym derivative R(x, ω) of P x with
respect to P is given by
Z
Z Z
£
¤
dP x
1
= exp
ρ(t − s)f (x, s)f (x, t)dsdt
f (x, t)y(t)dt −
dP
2
and
Z
[R(x, ω)]2 dP = eH(x)
where
Z Z
H(x) =
ρ(t − s)f (x, s)f (x, t)dsdt
The resulting source process can also be represented as as
λx (t) = φ(y(t) + m(x, t))
where y(·) has P for its distribution. Because
£
¤1
x
1
E P [x|a(0, x, ω) − ā|] ≤ e 2 H(x)] E P [x2 |a(0, x, ω) − ā|2 ] 2
if σ 2 = 0 and f (x, t) is chosen so that H(x) remains bounded as x → ∞, then
x
E P [x|a(x, 0, ω) − ā|] → 0
On the other hand we can solve
ut + uxx − u2 = 0
with different Neumann boundary data corresponding to λ(t) = φ(y(t)) and λx (t) =
φ(y(t) + m(x, t)) and get two solutions u1 and u2 . Since m(x, t) ≥ 0 and φ(·) is
monotone it follows that u2 ≥ u1 and the difference v will satisfy
1
vt + vxx − (u1 + u2 )v = 0; vx (t, 0) = − [φ(y(t) + m(x, t)) − φ(y(t)]
2
6
From the relation u(t, x) = (x+a(t,x))
2 , and the lower bounds u(t, x) ≥
not hard to obtain an estimate of the form (see (6.13))
6
(x+a1 )2
it is
a1 (t, x) − a2 (t, x) ≥ c2 x3 (u2 (x) − u1 (x)) = c2 x3 v(t, x)
x
If σ 2 = 0, both E P [x|a(x, 0, ω) − ā|] and E P [x|a(x, 0, ω) − ā|] → 0 and this implies
E P [x4 v(0, x, ω)] ≤ C E P [x[a1 (0, x, ω) − a2 (0, x, ω)]] → 0
as x → ∞ and we will show that does not happen for a suitable choice of f (t, x).
12
We have an upper bound of g = (x+a
2 on u1 + u2 . This will provide a lower
2)
bound on v,
Z
v(0, x, ω) ≥ [φ(y(t) + m(x, t)) − φ(y(t))]qg (0, x, t)dt
where qg is as in the proof of Theorem 6.4. We pick f (x, t) = x1 f ( xt2 ) where f > 0
is compactly supported and equal to 1 on [−T, T ]. It is easy to verify that
Z Z
t
s
ρ(t − s)f ( 2 )f ( 2 )dtds < ∞
sup H(x) = sup x−2
x
x
x≥1
x≥1
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
23
On the other hand we have
Z
Z
£ P
¤
1
s
E P [x4 v(0, x, ω)] ≥ x4
E [φ(y(t) +
ρ(t − s)f ( 2 )ds) − φ(y(t))] qg (0, x, t)dt
x
x
Z
Z
s
' x3 E P [φ0 (y(t))]qg (0, x, t)dt ρ(t − s)f ( 2 )ds
x
From (6.11) it follows that for large x,
Z
C1
qg (0, x, t)dt ≥ 3
x
Z
t2 qg (0, x, t)dt ≤ C2 x
This implies that if T is chosen large enough, then for all x ≥ 1,
Z x2 T
C1
C2 x
C3
qg (0, x, t) dt ≥ 3 − 4 2 ≥ 3
x
x T
x
0
By stationarity, E P [φ0 (y(t))] is a positive constant and
Z
Z x2 T
s
inf
inf 2
ρ(t − s)f ( 2 ) ds ≥
ρ(s)ds > 0
x≥1 0≤t≤x T
x
0
establishing the required contradiction.
6. Appendix.
Since our goal is to control the Feynman-Kac term we will need some estimates
on c(t, x, ω). Some of them will be uniform in ω and some will be only in the mean.
In addition we will need some estimates on the transition probabilities of the Bessel
process. We will collect all of these estimates in this section.
We will need the following estimate from the theory of parabolic partial differential
equation.
Theorem 6.1. For any t0 and x0 > 0, let Dtr0 ,x0 = {(t, x) : |x − x0 | ≤ r, t0 ≤ t ≤
t0 + r2 }. If v(t, x) is a solution of an equation of the form
(6.1)
vt + vxx + b(t, x)vx + q(t, x)v = 0
in G ⊂ (−∞, ∞) × [0, ∞), and Dt10 ,x0 ⊂ Dt20 ,x0 ⊂ G. Assume that b and q are
bounded by a constant B. Then
sup
(s,y)∈Dt1
|vx (s, y)| ≤ Ckv(t, x)kL2 [Dt2
0 ,x0
]
≤ 4C
0 ,x0
sup
|v(s, y)|
(s,y)∈Dt2
0 ,x0
where C depends only on B. In particular if v(t, x) is solution of an equation of
the form (6.1), with b(t, x) and q(t, x) satisfying
sup x|b(t, x)| ≤ C
t,x
sup x2 |q(t, x)| ≤ C
t,x
sup xα |v(t, x)| ≤ C
t,x
24
S.R.S.VARADHAN AND N. ZYGOURAS
then
sup xα+1 |vx (t, x)| ≤ C
t,x
Proof. A proof of the first half can be found in [4]. To see the second half, let v
be solution of (6.1) in (−∞, ∞) × [0, ∞). If we define v k (t, x) = k1 v(k 2 t, kx), then
v k (t, x) satisfies
k
+ bk (t, x)vxk (t, x) + q k (t, x)v k (t, x) = 0
vtk (t, x) + vxx
where bk (t, x) = kb(k 2 t, kx) and q k (t, x) = k 2 q(k 2 t, kx). In particular if we assume
that x|b(t, x)| ≤ C and x2 |q(t, x)| ≤ C for x ≥ 4, then bk (t, x) and q k (t, x) are
r
⊂ (−∞, ∞) × (4, ∞)
uniformly bounded on [1, ∞) × (−∞, ∞) for k ≥ 1 and Dt,x
for r = 1, 2. We can now get a bound on vx
|vx (k 2 t, kx)| ≤
C
C
2
kv(k 2 s, ky)k2,Dt,x
≤
k
k
sup
|v(k 2 s, ky)|
2
(s,y)∈Dt,x
Theorem 6.2. Consider a smooth solution u of
(6.2)
ut + uxx − u2 = 0,
x > 0, −∞ < t < ∞
with ux (t, 0) = − 12 λ(t). Let a(t, x) be the corresponding solution
s
6
(6.3)
a(t, x) =
−x
u(t, x)
of
(6.4)
at + axx −
3a2x
6ax
−
=0
(x + a) (x + a)
on G = [−∞, ∞) × (0, ∞). The following bounds are valid.
(6.5)
0 < a2 ≤ a(t, x) ≤ a1 < ∞
for all t, x and ω.
(6.6)
sup |xax (t, x, ω)| ≤ C < ∞
t,x,ω
In particular the function c(t, x, ω) defined in (1.6) satisfies the uniform bound
(6.7)
sup |c(t, x, ω)| ≤
t,ω
C
x3
Proof. We saw already that (6.5) is valid by maximum principle. We will establish
(6.6). We start with equation (6.2) and differentiate it with respect to x. Then for
the derivative v = ux , satisfies
vt + vxx − 2uv = 0 ; |v(t, 0)| =
We also have an estimate on u of the form
u(t, x) ≥
6
(x + a2 )2
λ2
1
λ(t) ≤
2
2
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
and w =
1
(x+a2 )3
25
solves the equation
12
1
w = 0 ; w(t, 0) = 3
(x + a2 )2
a2
wt + wxx −
Standard comparison using the maximum principle gives the bound.
To derive (6.6) we rewrite the equation for a in the form
(6.8)
at + axx − b(t, x)ax = 0
where
b(t, x) =
6 + 3ax (t, x)
(x + a)
satisfies |xb(t, x)| ≤ C. It follows from Theorem 6.1 that
sup |xax (t, x, ω)| ≤ C
t,x,ω
which proves (6.6). (6.7) is now an easy consequence of the formula (1.6) for
c(t, x, ω).
We now improve slightly on this estimate but in the mean with respect to ω.
Theorem 6.3. There is a finite constant C such that
·Z ∞
¸
(6.9)
EP
x|ax (0, x, ω)|2 dx ≤ C < ∞
0
Proof. Multiply the equation (6.4) for a by xa and take expectations.
E P [xaaxx ] = E P [
3xaa2x
6xaax
] + EP [
]
(x + a)
(x + a)
Integrate with respect to x from y1 to y2 and integrate by parts the trem on the
left.
Z y2
Z y2
Z y2
Z y2
3xaa2x
6xaax
E P [xaax ]|yy21 −
]dx+
]dx
E P [xa2x ]dx−
EP [
E P [aax ]dx =
EP [
(x + a)
(x + a)
y1
y1
y1
y1
Let us split the term
6xaax
6a2 ax
= 6aax −
(x + a)
(x + a)
and note that
Z
y2
aax dx =
All the terms except
finally get
R y2
y1
y1
1 2
[a (y2 ) − a2 (y1 )]
2
E P [xa2x ]dx are seen to remain bounded as y2 → ∞. We
Z
0
∞
E P [xa2x ]dx ≤ C
We now begin estimating the difference between two solutions of (6.2).
26
S.R.S.VARADHAN AND N. ZYGOURAS
Theorem 6.4. Let v(t, x) = u1 (t, x) − u2 (t, x) be the difference of two solutions
corresponding to two different initial values λ1 (·) and λ2 (·), that agree on {t : |t| ≤
k}. Then, for x > 0 and |t| ≤ k2 , v satisfies
Cx
k2
C
|v(t, x)| ≤
kx
C
|vx (t, x)| ≤
kx2
|v(t, x)| ≤
Proof. Let v = u1 − u2 be the difference of two solutions with boundary data
− 21 λ1 (·) and − 21 λ2 (·) that agree on {t : |t| ≤ k}. Then with σ(t) = λ1 (t) − λ2 (t)
1
vt + vxx − (u1 + u2 )v = 0 : vx (t, 0) = − σ(t)
2
Any bounded solution w(t, x) of an equation of the form
1
wt + wxx − g w = 0 : wx (t, 0) = − σ(t)
2
with a positive g has, by the maximum principle, a representation
Z ∞
(6.10)
w(t, x) =
σ(τ )qg (t, x, τ )dτ
t
with a nonnegative qg . A formula for qg can be written down in terms of a FeynmanKac formula
Z τ
x2
1
qg (t, x, τ ) = p
e− 4(τ −t) Ex,t,τ [exp[−
g(s, x(s))ds]
2 π(τ − t)
t
where Ex,t,τ is expectation with respect to Brownian motion (with variance 2)
starting from x > 0 at time t and conditioned to exit (0, ∞) at time τ . In particular
our solution v is given by
Z ∞
v(t, x) =
σ(τ )qu1 +u2 (t, x, τ )dτ
t
6
Since we have a lower bound of the form ui ≥ (x+a
2 , we can bound qu1 +u2 by qg ,
1)
12
corresponding to g = (x+a1 )2 . We estimate qg by exhibiting special solutions of the
form
a(t)
b(t)
(6.11)
w(t, x) =
+
+ c(t)(x + a1 )
3
(x + a1 )
(x + a1 )
w will satisfy
wt + wxx −
12
w=0
(x + a1 )2
provided a0 (t) = 10b(t), b0 (t) = 12c(t) and c0 (t) = 0. This yields the estimates
Z ∞
(τ − t)2 qu1 +u2 (t, x, τ )dτ ≤ Cx
t
Z ∞
qu1 +u2 (t, x, τ )dτ ≤ C x−3
t
BEHAVIOR OF THE SOLUTION OF A RANDOM SEMILINEAR HEAT EQUATION
27
for x ≥ `. By interpolation we also have
Z ∞
|τ − t|qu1 +u2 (t, x, τ )dτ ≤ C x−1
t
This takes care of the estimates.
We now translate these into estimates on a1 (t, x) − a2 (t, x)
Let λi (·) be two different initial conditions at x = 0 that agree in |t| ≤ k. Let
ui (t, x) be the corresponding solutions
uit (t, x) + uixx (t, x) − ui (t, x)2 = 0; ui (t, 0) = λi (t)
with
s
ai (t, x) =
6
−x
ui (t, x)
and
2
ci (t, x) =
(6.12)
3aix (t, x)
6aix (t, x)
−
ai (t, x)(x + ai (t, x)) x(x + ai (t, x))
We have the following estimates.
Theorem 6.5. For x ≥ ` and |t| ≤
k
2
C x2
k
Cx
1
2
|ax (t, x) − ax (t, x)| ≤
k
C
|c1 (t, x) − c2 (t, x)| ≤
kx
|a1 (t, x) − a2 (t, x)| ≤
Proof. Since
(6.13)
√ 2
6(u (t, x) − u1 (t, x))
p
p
p
u1 (t, x) u2 (t, x)( u1 (t, x) + u2 (t, x))
a1 (t, x) − a2 (t, x) = p
and from Theorem 6.4, for v = u1 − u2 we have the estimates
|v(t, x)| ≤
C
xk
and from equation (6.13)
|a1 (t, x) − a2 (t, x)| ≤ Cx3 |u1 (t, x) − u2 (t, x)|
which proves the first estimate. From Theorem 6.1 it follows that
C
x2 k
and this in turn implies the second estimate. The last estimate follows from equation (6.12) and the first two estimates of the theorem.
|u1x (t, x) − u2x (t, x)| ≤
28
S.R.S.VARADHAN AND N. ZYGOURAS
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Thesis, NYU, New York, 2004
Courant Institute of Mathematical Sciences, New York University, New York, NY,
10012, USA. e-mail: varadhan@cims.nyu.edu
Department of Mathematics, University of Southern California, Los Angeles, CA,
90089, USA. e-mail: zygouras@usc.edu