be the law L U on D
c
. To show that L ˜u
t
→ ν it suffices to show that the limit does not depend on the choice of sequence t
n
. Suppose s
n
is another sequence increasing to infinity. If r
n
is a third increasing sequence containing all the elements of s
n
and t
n
then the above argument shows that
L ˜u
r
n
is convergent and hence the limits of L ˜u
s
n
and L ˜u
t
n
must coincide.
Remark We do not yet know that the limit ν
c
is supported on D. We must rule out the possibility
that the wavefronts get wider and wider and the limit ν
c
is concentrated on flat profiles. We do this by a moment estimate in the next section. Once this is known, standard Markovian arguments in
section 3.3 will imply that ν = ν
c
|
D
, the restriction to D, is the law of a stochastic travelling wave.
3.2 A moment bound
We will require the following simple first moment bounds. Under hypothesis 2 we may choose K
1
, K
2
so that −K
1
1 − x ≤ f x ≤ K
2
x for all x
∈ [0, 1].
Lemma 11. Let u be a solution to 1 with initial condition u0 = φ ∈ D.
i For any T 0 there exist CT ∞ so that
Z
R
E
ut, x − φx
d x
≤ CT
1 + Z
R
φ1 − φx d x
for t ≤ T .
ii When the initial condition is Hx = I x 0 we have for x 0
E[ut, x] ≤ min{1, e
K
2
t
x
−1
G
t
x}, E[1
− ut, −x] ≤ min{1, e
K
1
t
x
−1
G
t
x}, and there exists CK
1
, K
2
, a ∞ so that
E[ |Γut|] ≤ CK
1
, K
2
, a + 2K
1 2
1
+ K
1 2
2
t for all t ≥ 0.
Proof For part i we may, by translating the solution if necessary, assume that φ crosses 12 at the
origin, that is φx ≤ 12 for x ≥ 0 and φx 12 for x 0. Taking expectations in 1 and
applying f x ≤ K
2
x we find that E[ut, x] solves ∂ ∂ tE[ut, x] ≤ ∆E[ut, x] + K
2
E[ut, x]. This leads to
E[ut, x] ≤ e
K
2
t
Z
R
G
t
x − yφ y d y
448
where G
t
x = 4πt
−12
exp −x
2
4t. Hence for t ≤ T Z
∞
E[ut, x] d x ≤ e
K
2
t
Z
∞
Z
R
G
t
x − yφ y d y d x =
e
K
2
t
Z
−∞
Z
∞
+ Z
∞
Z
∞
G
t
x − yφ y d x d y ≤ e
K
2
t
Z
−∞
Z
∞
G
t
x − y d x d y + e
K
2
t
Z
∞
φ y Z
R
G
t
x − yd x d y ≤ CT
1 +
Z
∞
φ y d y
. The process vt, x = 1
− ut, −x solves 1 with f u and gu replaced by − f
1 − v and −g1 − v. So the same argument estimates E[1 − ut, −x] and yields the bound with a possibly
different constant depending on K
1
Z
−∞
E[1 − ut, x] d x ≤ CT
1 + Z
−∞
1 − φ y d y .
Note that Z
R
ut, x − φx
d x ≤
Z
∞
ut, x + φx d x +
Z
−∞
1 − ut, x + 1 − φx d x. Combining this with the bounds above and also
Z
∞
φx d x + Z
−∞
1 − φx d x ≤ 2 Z
R
φ1 − φx d x which uses that
φ crosses 12 at the origin completes the proof of part i. For part ii we have more explicit bounds. Use a Gaussian tail estimate to the bound for x
E[ut, x] ≤ e
K
2
t
Z
∞ x
G
t
y d y ≤ 2t x
−1
e
K
2
t
G
t
x. In some regions the estimate E[ut, x]
≤ 1 is better. Since {Γut ≥ x} = {ut, x ≥ a} almost
449
surely for t 0, we have via Chebychev’s inequality
E[Γut
+
] =
Z
∞
P[Γut ≥ x] d x
= Z
∞
P[ut, x ≥ a] d x
≤ Z
2K
1 2
2
t
d x + Z
∞ 2K
1 2
2
t
a
−1
E[ut, x] d x ≤ 2K
1 2
2
t + Z
∞ 2K
1 2
2
t
2ta x
−1
e
K
1
t
x
−1
G
t
x d x ≤ 2K
1 2
2
t + a
−1
e
K
2
t
4πt
−12
Z
∞ 2K
1 2
2
t
x2K
2
te
−x
2
4t
d x =
2K
1 2
2
t + aK
2 −1
4πt
−12
. Similarly
{Γut ≤ −x} = {1 − ut, −x ≥ 1 − a} and the same argument yields the bound on E[1
− ut, −x] and also E[Γut
−
] ≤ 2K
1 2
1
t + aK
1 −1
4πt
−12
. These estimates combine to control E[
|Γut|] as desired for t ≥ 1. A slight adjustment bounds the region t ≤ 1. We briefly sketch a simple idea from [5] for the deterministic equation u
t
= u
x x
+ u1 − u started at H, which we will adapt for our stochastic equation. The associated centered wave satisfies
˜ u
t
= ˜ u
x x
+ ˜ u
x
˙ γ + ˜
u1 − ˜u
where γ
t
is the associated wavefront marker. Integrating over −∞, 0] × [t
, t], for some 0 t
t yields the estimate
≥ Z
−∞
[˜ ut, x
− ˜ut , x] d x
= Z
t t
˜ u
x
s, 0 ds − 1 − aγ
t
− γ +
Z
t t
Z
−∞
˜ u1
− ˜us, x d x ds. This allows one, for example, to control the size of the back tail
R
−∞
1 − ˜ut, x d x. Integrating over [0,
∞ gives information on the front tail. The following lemma gives the analogous tricks for the stochastic equation.
Lemma 12. Let u be the solution to 1 started from Hx = Ix 0. Let ˜
u be the solution centered
450
at height a ∈ 0, 1. Then for 0 t
t, almost surely, Z
∞
˜ ut, x
− ˜ut , x
d x =
− Z
t t
˜ u
x
s, 0 ds − aΓut − Γut +
Z
t t
Z
∞
f ˜
us, x d x ds + Z
t t
Z
∞
g˜ us, x d x dW
s
, 12
Z
−∞
˜ ut, x
− ˜ut , x
d x =
Z
t t
˜ u
x
s, 0 ds − 1 − aΓut − Γut +
Z
t t
Z
−∞
f ˜
us, x d x ds + Z
t t
Z
−∞
g˜ us, x d x dW
s
. 13
Proof Integrating 6 first over [t , t] and then over [0, U] we find
Z
U
˜ ut, x
− ˜ut , x
d x =
Z
U
Z
t t
˜ u
x x
s, x ds d x + Z
U
Z
t t
˜ u
x
s, x ◦ dΓus d x +
Z
U
Z
t t
f ˜
us, x ds d x + Z
U
Z
t t
g˜ us, x dW
s
d x =
Z
t t
˜ u
x
s, U − ˜u
x
s, 0 ds +
Z
t t
˜ us, U
− a ◦ dΓus +
Z
t t
Z
U
f ˜
us, x , d x ds + Z
t t
Z
U
g˜ us, x d x dW
s
. 14
The interchange of integrals uses Fubini’s theorem path by path for the first and third terms on the right hands side and a stochastic Fubini theorem for the second and fourth term for example the
result on p176 of [9] applies directly for the fourth term and also the second term after localizing at the stopping times
σ
n
= inf{t ≥ t : sup
y ∈[0,U]
|˜u
x
s, y| ≥ n}. To prove the lemma we shall let U
→ ∞ in each of the terms. Bound | f z| ≤ Cz1 − z for some C. Using the first moment bounds from Lemma 11 ii we see that
Z
t
Z
R
E[ | f
us, x|] d x ds ≤ C
Z
t
Z
R
E[u1 − us, x] d x ds
≤ C Z
t
Z
∞
E[us, x] d x ds + C Z
t
Z
−∞
E[1 − us, x] d x ds ∞.
451
This gives the domination that justifies R
t t
R
U
f ˜
us, x d x ds →
R
t t
R
∞
f ˜
us, x d x ds almost surely. The first moments also imply
Z
t
E
Z
∞
us, xd x
2
ds
= Z
t
Z
∞
Z
∞
E[us, xus, y] d x d y ds ≤
Z
t
Z
∞
Z
∞
E[us, x] ∧ E[us, y] d x d y ds ∞.
A similar argument bounds the integral over −∞, 0]. Bounding |gz| ≤ Cz1 − z we find that
E[ R
t
R
R
|gus, x|d x
2
ds] is finite. This shows that the integral R
t
R
∞
g˜ us, x d x dW
s
makes sense and the Ito isometry allows one to check that the L
2
convergence Z
t
Z
U
g˜ us, x d x dW
s
→ Z
t
Z
∞
g˜ us, x d x dW
s
. Theorem 3 iv shows that E[
R
t t
sup
x
|u
x
s, x| ds] ∞ and this gives the required domination to turn ˜
u
x
s, U → 0 into R
t t
˜ u
x
s, U ds → 0. This leaves the second term in 14 and the lemma will follow once we have shown that
R
t t
˜ us, U
◦ dΓus
→ 0 almost surely as U → ∞. To see this expand the integral using the decomposition for dΓut = d mt, a in 5 to see that
Z
t t
˜ us, U
◦ dΓus = Z
t t
˜ us, U
m
x x
s, a m
2 x
s, a − f am
x
s, a
ds −ga
Z
t t
˜ us, Um
x
s, a dW
s
− 1
2 ga
Z
t t
˜ us, U d
˜ u
·, Um
x
·, a, W
s
15 where we have converted from a Stratonovich to an Ito integral and we are writing [
· , ·]
t
for the cross quadratic variation. We claim that each of these terms converge to zero almost surely.
Note that the strict negativity of the derivative u
x
t, x and the relations 7 imply that the path s
→ m
x x
s, am
2 x
s, a − f am
x
s, a is almost surely continuous on [t , t]. So the first term
on the right hand side of 15 converges almost surely to zero by dominated convergence using ˜
us, U → 0 as U → ∞. The second term in 15 also converges to zero by applying the same
argument to the quadratic variation g
2
a R
t t
˜ u
2
s, Um
2 x
s, a ds. A short calculation leads to the explicit formula for the cross variation
d ˜
u ·, Um
x
·, a, W
t
= g˜
ut, Um
x
t, a d t − gam
x x
t, a˜ ut, U d t
−ga˜ut, Um
2 x
t, a d t − g
′
am
x
t, a˜ ut, U d t.
Again, since also g˜ ut, U
→ 0 as U → ∞, a dominated convergence argument shows that the final term in 15 converges to zero as U
→ ∞. This completes the proof of the first equation in the lemma. The second is similar by integrating
over [ −L, 0] and letting L → ∞.
452
Proposition 13. Suppose that f is of KPP, Nagumo or unstable type. In the latter two cases suppose
that f a = 0 and ga 6= 0. Let u be the solution to 1 started from Hx = Ix 0 and let ν
c
be the limit law constructed from u in Proposition 10. Then
ν
c
D = 1. In the KPP and Nagumo cases we have the increasing limits as t
↑ ∞ E
Z u1
− ut, x d x
↑ Z
D
Z
R
φ1 − φx d x ν
c
dφ ∞. 16
In the unstable case Z
D
Z
∞
φa − φx d x ν
c
dφ ≤ sup
t
E Z
∞
˜ ua
− ˜ut, x d x
∞, Z
D
Z
−∞
1 − φφ − ax d x ν
c
dφ ≤ sup
t
E
Z
−∞
1 − ˜u˜u − at, x d x
∞. 17
Proof We start with the case where f is of KPP type. In this case there is a constant C so that
C f x ≥ x1 − x on [0, 1]. In a similar but easier way to Lemma 12, one may integrate 1 over
s ∈ [t
, t] and then x ∈ R to find
Z
R
ut, x − ut
, x d x =
Z
t t
Z
R
f us, x d x ds +
Z
t t
Z
R
gus, x d x dW
s
. Taking expectations and rearranging one finds
t
−1
Z
t t
Z
R
E[u1 − us, x] d x ds ≤ C t
−1
Z
t t
Z
R
E[ f us, x] d x ds
≤ C t
−1
Z
R
E[ut, x − ut
, x] d x ≤ C t
−1
Z
∞
E[ut, x + ut , x] d x
+C t
−1
Z
−∞
E[1 − ut, x + 1 − ut
, x] d x. Using the first moments from Lemma 11 ii on each of the four terms of the right hand side we
find that lim sup t
−1
Z
t t
Z
R
E[u1 − us, x] d x ds ∞.
For example Z
∞
E[ut, x] d x ≤
Z
∞
min {1, e
K
2
t
x
−1
G
t
x} d x ≤
Z p
2K
2
t
1 d x + Z
∞
p
2K
2
t
e
K
2
t
x 2K
2
t
2
G
t
x d x =
p 2K
2
t + 1
2K
2
t .
453
The other terms are similar. Writing z
→ mt, z for the inverse function to x → ut, x we have Z
R
u1 − ut, x d x = −
Z
1
z1 − zm
z
t, z dz. The stochastic ordering of
L ut and Lemma 5 iv show that t → E[m
z
t, z] is decreasing for t
0. Thus t → R
R
E[u1 − us, x]d x is increasing and we conclude that
sup
t ≥0
Z
R
E[u1 − ut, x] d x ∞.
For fixed N 0 the functionals φ →
R
N −N
φ1 − φx d x are bounded and continuous on D
c
. So by the convergence of
L ˜ut to ν
c
in M D
c
we see that Z
N −N
E[˜ u1
− ˜ut, x] d x → Z
D
c
Z
N −N
φ1 − φx d x ν
c
dφ. The last two displayed equations imply that
Z
D
c
Z
R
φ1 − φx d x ν
c
dφ ≤ lim
t ↑∞
Z E[u1
− ut, x] d x ∞. 18
This in turn implies that ν
c
only charges D.
For 0 ≤ N ≤ M the function
I N , t = Z
M N
E[˜ ut, x] d x
is increasing in M and also in t since L ˜ut are stochastically increasing. We may therefore
interchange the t and M limits to see that lim
t ↑∞
Z
∞ N
E[˜ ut, x] d x
= lim
t ↑∞
lim
M ↑∞
Z
M N
E[˜ ut, x] d x
= lim
M ↑∞
lim
t ↑∞
Z
M N
E[˜ ut, x] d x
= lim
M ↑∞
Z
D
Z
M N
φx d x ν
c
dφ =
Z
D
Z
∞ N
φx d x ν
c
dφ. Similarly, as t
↑ ∞ Z
−N −∞
E[1 − ˜ut, x] d x ↑
Z
D
Z
−N −∞
1 − φ d x ν
c
dφ.
454
Since R
D
R
R
φ1 − φx d x ν
c
dφ ∞ we may, for any ε 0, choose N ε so that
sup
t
Z
∞ N
ε
E[˜ ut, x] d x
≤ Z
D
Z
∞ N
ε
φ d x ν
c
dφ ≤ ε, sup
t
Z
−N ε
−∞
E[1 − ˜ut, x] d x ≤
Z
D
Z
−N ε
−∞
1 − φ d x ν
c
dφ ≤ ε. 19
This control on the tails allows us to improve on 18 to the desired result 16. Now we consider the case where f
is of Nagumo type, and this is the only place we exploit the bi-stability of f
that is f
′
0, f
′
1 0. We may fix a smooth strictly concave h : [0, 1] → R satisfying h0 = h1 = 0 and h
′
a = 0, h
′
0 0, h
′
1 0. A running example to keep in mind is the case f
= x1 − xx −
1 2
and h = x1 − x. Then dhu = h
′
uu
x x
d t + h
′
f +
1 2
h
′′
g
2
u d t + h
′
gu dW
t
. 20
The properties of h and the fact that f is of Nagumo type together imply that h
′
f ≤ 0 on [0, 1] and
h
′
f only vanishes at 0, a, 1. Since ga
0 we have h
′
f +
1 2
h
′′
g
2
0 on 0, 1. The derivatives at x = 0, 1 are non-zero and this implies that here is an
ε 0 so that h
′
f + h
′′
g
2
≤ −εh. The aim is to obtain a differential inequality of the form d m
≤ C − εm for mt = R
R
E[hut, x] d x. Integrate 20 over [t
, t] and then [ −N, N] to obtain
Z
N −N
E[hut, x] d x −
Z
N −N
E[hut , x] d x
= Z
t t
Z
N −N
E[h
′
us, x u
x x
s, x] d x ds + Z
t t
Z
N −N
E[h
′
f +
1 2
h
′′
g
2
us, x] d x ds =
Z
t t
E[h
′
us, N u
x
s, N ] − E[h
′
us, −N u
x
s, −N] ds
− Z
t t
Z
N −N
E[h
′′
us, x u
2 x
s, x] d x ds + Z
t t
Z
N −N
E[h
′
f +
1 2
h
′′
g
2
us, x] d x ds where we have integrated by parts in the last equality. Letting N
→ ∞ is justified and is similar but simpler than Lemma 12 and we find
Z
R
E[hut, x] d x −
Z
R
E[hut , x] d x
≤ kh
′′
k
∞
Z
t t
Z
R
E[u
2 x
s, x] d x ds − ε Z
t t
Z
R
E[hus, x] d x ds ≤ kh
′′
k
∞
Z
t t
E[sup
z
|u
x
s, z|] ds − ε Z
t t
Z
R
E[hus, x] d x ds where in the last step we use that
Z
R
u
2 x
s, xd x ≤ sup
z
|u
x
s, z| Z
R
|u
x
s, x| d x = sup
z
|u
x
s, z|. 455
The stochastic monotonicity of s → L ˜us and Lemma 5 iv imply that the supremum
sup
z
|u
x
s, x| is stochastically decreasing. Since E[sup
z
|u
x
t , z
|] is finite by Theorem 3 iv we have the desired differential inequality for mt =
R
R
E[hut, x] d x. This implies that m stays bounded and since Chz
≥ z1 − z for some C we find sup
t ≥t
Z
R
E[u1 − ut, x] d x ≤ C sup
t ≥t
Z
R
E[hus, x] d x ∞.
As in the previous KPP case this implies 16 and that ν
c
only charges D.
Now we consider the case where f is of unstable type. Rearranging the conclusion of Lemma 12
we see, after taking expectations, that t
−1
Z
t t
Z
∞
E[ f ˜
us, x] d x ds ≤ t
−1
Z
∞
E[˜ ut, x + ˜
ut , x] d x
+t
−1
Z
t t
E[ |˜u
x
s, 0|] ds +at
−1
E |Γut| + |Γut
| .
21 We claim that the limsup as t
→ ∞ is finite for all three terms on the right hand side. The first term can bounded using
Z
∞
˜ us, x d x
≤ 1 − a
−1
Z
R
˜ u1
− ˜us, x d x =
1 − a
−1
Z
R
u1 − us, x d x
≤ 1 − a
−1
Z
∞
us, x d x + 1 − a
−1
Z
−∞
1 − us, x d x and then controlled by first moments as in the KPP case. For the second term the claim follows from
the fact that s → E[|˜u
x
s, 0|] is decreasing and finite from Theorem 3 iv. For the third term the claim follows from Lemma 11. We conclude that the limsup of the left hand side of 21 is finite.
Applying a similar argument to the second equation of Lemma 12 we have
lim sup
t →∞
t
−1
Z
t t
Z
∞
E[ f ˜
us, x] d x ds ∞,
lim sup
t →∞
t
−1
Z
t t
Z
−∞
E[ f ˜
us, x] d x ds ∞.
Note that f is of a single sign on each of the intervals [0, a] and [a, 1]. Indeed there exists C so
that C
| f x| ≥ xa − x for x ∈ [0, a] and C| f
x| ≥ 1 − xx − a for x ∈ [a, 1].
456
Therefore we have lim sup
t →∞
t
−1
Z
t t
Z
∞
E[˜ ua
− ˜us, x] d x ds ∞, lim sup
t →∞
t
−1
Z
t t
Z
−∞
E[1 − ˜u˜u − as, x] d x ds ∞.
We have Z
∞
˜ ua
− ˜ut, x d x = − Z
a
za − zm
z
t, z dz Z
−∞
1 − ˜u˜u − at, x d x = − Z
1 a
1 − zz − am
z
t, z dz. For fixed N
0 the functionals φ → R
N
a − φφ d x and φ → R
−N
1 − φφ − a d x are bounded and continuous on
D
c
. The same reasoning as in the previous cases yields 17. The construction of ν
c
in Lemma 10 as the law of a variable U shows that φx ≥ a for x 0 and φx ≤ a for x 0, for ν
c
almost all φ.
This and 17 imply that ν
c
charges only D or the single point φ ≡ a.
The argument that there is no mass on the point φ ≡ a is a little fiddly, and we start with a brief
sketch. We argue that if φ ≡ a with ν
c
positive mass then there are arbitrarily wide patches in ˜
ut, for large t, that are flattish, that is lie close to the value a. But the height of this large flatish patch will evolve roughly like the SDE d Y = f
Y d t + gY dW with Y = a. Since ga 6= 0 the
sde will move away from the value a with non-zero probability and this would lead to an arbitrary large value of E[
R
R
|1 − ˜u| |˜u − a| ˜u d x] for all large times which contradicts 17. To implement this argument we will use the following estimate.
Lemma 14. Let u be a solution to 1 driven by a Brownian motion W . Let Y be the solution to the SDE d Y = f
Y d t + gY dW with Y = a. Then there exists a constant c
T so that for all η ∈ 0, 1 Z
R
e
−η|x|
E[ |ut, x − Y
t
|
2
] d x ≤ c T
Z
R
e
−η|x|
E[ |u0, x − a|
2
] d x for t
∈ [0, T ]. Note that the constant c
does not depend on η ∈ 0, 1. Considered as a constant function in x, the
process Y
t
is a solution to 1. This lemma therefore follows by a standard Gronwall argument in order to estimate the L
2
difference between two solutions for an equation with Lipshitz coefficients. The use of weighted norms for equations on the whole space, that is the norm
R
R
e
−η|x|
E[ |ut, x −
vt, x |
2
] d x, is also standard - see, for example, the analogous estimate in the proof of Shiga [11] Theorem 2.2 for the harder case of an equation driven by space-time white noise.
Suppose aiming for a contradiction that ν
c
φ ≡ a = δ
1
0. By the convergence L ˜ut → ν
c
we have, for any η 0,
P Z
R
|˜ut, x − a|
2
e
−η|x|
d x ≤ 1
≥ E
1
− Z
R
|˜ut, x − a|
2
e
−η|x|
d x
+
→ Z
D
c
1
− Z
R
|φx − a|
2
e
−η|x|
d x
+
ν
c
dφ ≥ δ
1
. 457
Suppose the solution u is defined on a filtered space Ω, F , F
t
, P and with respect to an F
t
Brownian motion W . Then for t ≥ T η, we may choose sets Ω
t
∈ F
t
satisfying P[Ω
t
] = δ
1
2 and Z
R
|ut, x − a|
2
e
−η|x|
d x ≤ 1
on Ω
t
. Fix t
≥ T η. Let Y
s
: s ∈ [t, t + 1] solve the SDE dY = f
Y d t + gY dW with initial condition Y
t
= a. Since ga 6= 0 there must exist t
∗
∈ 0, 1, δ
2
, δ
3
0 satisfying 4δ
2
a ∧ 1 − a such that P[
|Y
t
∗
− a| ∈ δ
2
, 2 δ
2
] δ
3
. The requirement that 4 δ
2
a ∧ 1 − a ensures that when |Y
t
∗
− a| ∈ δ
2
, 2 δ
2
then Y
t
∗
≥
a 2
and 1 − Y
t
∗
≥
1 −a
2
. Then Z
R
E[u |1 − u| |u − a|t + t
∗
, x] d x ≥
Z
R
e
−η|x|
E[u |1 − u| |u − a|t + t
∗
, xIΩ
t
] d x ≥
Z
R
e
−η|x|
E[Y
t+t
∗
|1 − Y
t+t
∗
| |Y
t+t
∗
− a|IΩ
t
] d x −L
Z
R
e
−η|x|
E[ |ut + t∗, x − Y
t+t
∗
|IΩ
t
] d x =:
I − I I
22 where L
is the Lipschitz constant of z |1 − z||a − z| on [0, 1]. We now estimate the terms I and I I.
Firstly, I
= 2
η E
Y
t+t
∗
|1 − Y
t+t
∗
| |Y
t+t
∗
− a|IΩ
t
= 2
η P[Ω
t
]E
Y
t
∗
|1 − Y
t
∗
| |Y
t
∗
− a| Y
= a
by the Markov property ≥
a1 − aδ
1
4 η
E
|Y
t
∗
− a| Y
= a
≥ a1
− aδ
1
δ
2
δ
3
4 η
.
458
Secondly, using Cauchy-Schwarz, I I
= L
E
I Ω
t
E
Z
R
e
−η|x|
|ut + t
∗
, x − Y
t+t
∗
| d x F
t
≤ L r
2 η
E
I Ω
t
E
Z
R
e
−η|x|
|ut + t
∗
, x − Y
t+t
∗
|
2
d x F
t
1 2
≤ L
c 1
r 2
η E
I Ω
t
Z
R
e
−η|x|
|ut, x − a|
2
d x
1 2
by Lemma 14
≤ L c
1 r
2 η
E[I Ω
t
] by the choice of Ω
t
= L
c 1
r 2
η δ
1
2 .
Thus, substituting these estimates into 22, we find for t ≥ T η
Z
R
E ˜
u |1 − ˜u| |˜u − a|t + t
∗
, x d x =
Z
R
E u|1 − u| |u − a|t + t
∗
, x d x
≥ a1
− aδ
1
δ
2
δ
3
4 η
− L c 1
r 2
η δ
1
2 .
By taking η small this bound can be made arbitrarily large, which contradicts 17.
3.3 Proof of Theorem 1
