On the global maximum of the solution to a stochastic heat equation with compact-support initial data

0.18 MB·English

by  Mohammud Foondun

On the global maximum of the solution
to a stochastic heat equation with
9
0
0 compact-support initial data
∗
2

n
Mohammud Foondun Davar Khoshnevisan
a
J
4 January 19, 2009
2

]
R
Abstract
P
h. Considerastochastic heatequation∂tu=κ∂x2xu+σ(u)w˙ foraspace-
timewhitenoisew˙ andaconstantκ>0. Undersomesuitableconditions
t
a on theinitial function u0 and σ,we show that thequantity
m
[ limsupt−1lnE sup|ut(x)|2
t→∞ „x∈R «
1
v is bounded away from zero and infinity byexplicit multiples of 1/κ. Our
4 proofworksbydemonstratingquantitativelythatthepeaksofthestochas-
1 tic process x 7→ ut(x) are highly concentrated for infinitely-many large
8 values of t. In the special case of the parabolic Anderson model—where
3
σ(u) = λu for some λ > 0—this “peaking” is a way to make precise the
.
1 notion of physicalintermittency.
0
Keywords: Stochastic heat equation, intermittency.
9
0
: AMS 2000 subject classification: Primary: 35R60, 37H10, 60H15; Sec-
v
ondary: 82B44.
i
X
Running Title: Thestochastic heat equation.
r
a
1 Introduction
We consider the stochastic heat equation,
∂u (x) ∂2u (x)
t =κ t +σ(u (x))w˙(t,x) for t>0 and x R, (1.1)
∂t ∂x2 t ∈
where κ > 0 is fixed, σ : R R is Lipschitz continuous with σ(0) = 0, w˙
→
denotes space-time white noise, and the initial data u :R R is nonrandom.
0
→
∗ResearchsupportedinpartbyNSFgrantDMS-0704024.
1
Thereareseveralareastowhich(1.1)hasdeepandnaturalconnections;perhaps
chief among them are the stochastic Burgers’ equation [9] and the celebrated
KPZ equation of statistical mechanics [10,11]; see also [12, Chapter 9].
Itiswellknownthat(1.1)hasanalmost-surelyunique,adaptedandcontin-
uoussolution ut(x) t≥0,x∈R [5,Theorem6.4,p.26]. Inaddition,thecondition
{ }
that σ(0)=0 implies that if u L2(R), then u L2(R) a.s. for all t 0; see
0 t
∈ ∈ ≥
Dalang and Mueller [6]. Note that our conditions on σ ensure that
σ(u) Lip u for all u R, (1.2)
| |≤ σ| | ∈
where
σ(x) σ(x′)
Lip := sup − . (1.3)
σ −∞<x<x′<∞(cid:12) x−x′ (cid:12)
(cid:12) (cid:12)
Our goal is to establish the following gene(cid:12)ral growth est(cid:12)imate.
(cid:12) (cid:12)
Theorem 1.1. Suppose there exists L (0, ) such that σ(u) L u for
σ σ
∈ ∞ | | ≥ | |
all u R. Suppose also that u 0 is Ho¨lder-continuous of order 1/2,
0
∈ 6≡ ≥
nonnegative, and supported in [ K,K] for some finite K >0. Then, (1.1) has
−
an almost-surely unique, continuous and adapted solution ut(x) t≥0,x∈R such
{ }
that u L2(R) a.s. for all t 0, and
t
∈ ≥
L4 Lip4
σ limsupt−1lnE sup u (x)2 σ. (1.4)
t
8κ ≤ t→∞ (cid:18)x∈R| | (cid:19)≤ 8κ
Because of Mueller’s comparison principle [13], the nonnegativity of u im-
0
plies that sup E(u (x)) =sup E(u (x)), and this quantity has to be finite
t,x | t | t,x t
because u is bounded; confer with (1.6). Consequently,
0
sup u (x) supu (x) as t . (1.5)
x∈Rk t kL1(P) ≪(cid:13)x∈R t (cid:13)L2(P) →∞
(cid:13) (cid:13)
(cid:13) (cid:13)
When Lip =L , (1.1) becomes(cid:13)the well-s(cid:13)tudied parabolic Anderson model
σ σ
[1,3]. And (1.5) makes precise the physical notion that the solution to (1.1)
concentrates near “very high peaks” [1,3,10,11].
In order to explain the idea behind our proof, we introduce the following.
Definition 1.2. We say that a continuous random field f := f(t,x) t≥0,x∈R
{ }
haseffectively-compact support ifthereexistsanonrandommeasurablefunction
p:R R of at-most polynomial growth such that:
+ +
→
(a) limsup t−1ln E(f(t,x)2)dx>0; and
t→∞ |x|≤p(t) | |
(b) limsup t−1lnR E(f(t,x)2)dx<0.
t→∞ |x|>p(t) | |
We might refer to the fuRnction p as the radius of effective support of f.
OneoftheideashereistouseMueller’scomparisonprinciple[13]tocompare
sup u (x) with the L2(R)-norm of x u (x), which is easier to analyze.
x∈R| t | 7→ t
2
We carry these steps out in Lemma 3.3. We also appeal to the fact that the
compact-support property of u implies that u (x) has an effectively-compact
0 t
support[Proposition3.7]. Thiscanbeinterpretedasakindofoptimalregularity
theorem. However, these matters need to be handled delicately, as “effectively
compact” cannot be replaced by “compact”; see Mueller [13].
Our method for establishing aneffectively-compactsupportproperty is mo-
tivated strongly by ideas of Mueller and Perkins [14]. In the cases that u (x)
t
denotes the density of some particles at x at time t, our effectively-compact
support property implies that most of the particles accumulate on a very small
set. This method might appealto the readerwho is interestedin mathematical
descriptions of physical intermittency.
Throughout this paper we use the mild formulation of the solution, in ac-
cordance with Walsh [15]. That is, u is the a.s.-unique adapted solution to
t ∞
u (x)=(p u )(x)+ p (y x)σ(u (y))w(dsdy), (1.6)
t t 0 t−s s
∗ −
Z0 Z−∞
wherep (z):=(4κτπ)−1/2exp( z2/(4κτ))denotes theheatkernelcorrespond-
τ
−
ing to the operator κ∂2/∂x2, and the stochastic integral is understood in the
sense of Walsh [15]. Some times we write X in place of E(X p) 1/p.
p
k k { | | }
2 A preliminary result
Asmentionedintheintroduction,thestrategybehindourproofofTheorem1.1
istorelatetheglobalmaximumofthesolutiontoa“closed-formquantity”that
resemblessup u (x) forlargevalues oft. Thatclosed-formquantityturnsout
x| t |
to be the L2(R)-norm of x u (x). Our next result analyses the growth of
t
7→
the mentioned closed-form quantity. We related it to sup u (x) in the next
x| t |
section. ThemethodsofthissectionfollowcloselytheclassicalideasofChoquet
and Deny [4] that were developed in a determinstic setting.
Theorem 2.1. Suppose σ : R R is Lipschitz continuous, σ(0) = 0, and
→
there exists L (0, ) such that L u σ(u) for all u R. If u L2(R)
σ σ 0
∈ ∞ | |≤ | | ∈ ∈
and u 0, then (1.1) has an almost-surely unique, continuous and adapted
0
6≡
solution ut(x) t≥0,x∈R such that ut L2(R) a.s. for all t 0, and
{ } ∈ ≥
L4 Lip4
σ limsupt−1lnE u 2 σ. (2.1)
8κ ≤ k tkL2(R) ≤ 8κ
t→∞
(cid:16) (cid:17)
Proof. It suffices to establish (2.1). Note that
E u (x)2
t
| |
(cid:16) (cid:17) t ∞ (2.2)
= (p u )(x)2+ ds dy E σ(u (y))2 p (y x)2.
t 0 s t−s
| ∗ | | | ·| − |
Z0 Z−∞ (cid:16) (cid:17)
3
Because σ(u) L u,
σ
| |≥ | |
E u 2
k tkL2(R)
(cid:16) (cid:17) t ∞
= p u 2 + ds dy E σ(u (y))2 p 2
k t∗ 0kL2(R) | s | ·k t−skL2(R) (2.3)
Z0 Z−∞ (cid:16) (cid:17)
t
p u 2 +L2 E u 2 p 2 ds.
≥k t∗ 0kL2(R) σ· k skL2(R) ·k t−skL2(R)
Z0 (cid:16) (cid:17)
We can multiply the preceding by exp( λt) throughout and integrate [dt] to
−
find that if
∞
U(λ):= e−λtE u 2 dt, (2.4)
k tkL2(R)
Z0 (cid:16) (cid:17)
then
∞ ∞
U(λ) e−λt p u 2 dt+L2 U(λ) e−λt p 2 dt. (2.5)
≥ k t∗ 0kL2(R) σ· · k tkL2(R)
Z0 Z0
According to Plancherel’s theorem, the following holds for all finite Borel mea-
sures µ on R:
1 ∞
p µ 2 = µˆ(ξ)2e−2κtξ2dξ. (2.6)
k t∗ kL2(R) 2π | |
Z−∞
Therefore, Tonelli’s theorem ensures that
∞ 1 ∞ µˆ(ξ)2
e−λtkpt∗µk2L2(R)dt= 2π λ|+2κ|ξ2 dξ. (2.7)
Z0 Z0
Weapplythisidentitytwicein(2.3): Oncewithµ:=δ ;andoncewithdµ/dx:=
0
u . This leads us to the following.
0
1 ∞ uˆ (ξ)2 1 ∞ 1
U(λ) | 0 | dξ+L2 U(λ) dξ
≥ 2π λ+2κξ2 σ· · 2π λ+2κξ2
Z0 Z0 (2.8)
1 ∞ uˆ (ξ)2 1
= | 0 | dξ+L2 U(λ) .
2π λ+2κξ2 σ· · 2√2κλ
Z0
Since u 0, the first [Fourier] integral is strictly positive. Consequently, the
0
6≡
above recursive relation shows that U(λ) = if λ L4/(8κ). This and a
∞ ≤ σ
real-variable argument together imply the first inequality in (2.1); see [8] for
more details.
For the other bound we use a Picard-iteration argument in order to obtain
an a priori estimate. Let u(0)(x):=u (x) and iteratively define
t 0
t ∞
u(n+1)(x):=(p u )(x)+ p (y x)σ(u(n)(y))w(dsdy). (2.9)
t t∗ 0 t−s − s
Z0 Z−∞
Since kpt∗u0kL2(R) ≤ku0kL2(R) and |σ(u)|≤Lipσ|u|,
2
(n+1)
E u
t L2(R)
(cid:18)(cid:13) (cid:13) (cid:19) (2.10)
(cid:13) (cid:13) t 2
(cid:13) (cid:13)u 2 +Lip2 E u(n) p 2 ds.
≤k 0kL2(R) σ·Z0 (cid:18)(cid:13) s (cid:13)L2(R)(cid:19)·k t−skL2(R)
(cid:13) (cid:13)
(cid:13) (cid:13)
4
Therefore, if we set
2
M(k)(λ):=sup e−λtE u(k) , (2.11)
t≥0(cid:20) (cid:18)(cid:13) t (cid:13)L2(R)(cid:19)(cid:21)
(cid:13) (cid:13)
then it follows that (cid:13) (cid:13)
∞
M(n+1)(λ) u 2 +Lip2 M(n)(λ) e−λ(t−s) p 2 ds
≤k 0kL2(R) σ· · k t−skL2(R)
Z0 (2.12)
Lip2
= u 2 + σ M(n)(λ).
k 0kL2(R) 2√2κλ
Thus, in particular, sup M(n)(λ) < if λ > Lip4/(8κ). We can argue
n≥0 ∞ σ
similarly to show also that if λ>Lip4/(8κ), then
σ
2 1/2
sup e−λtE u(n+1) u(n) < . (2.13)
Xn t≥0(cid:20) (cid:18)(cid:13) t − t (cid:13)L2(R)(cid:19)(cid:21) ∞
(cid:13) (cid:13)
In particular, uniqueness shows(cid:13)that if λ>Lip(cid:13)4/(8κ), then
σ
2
lim sup e−λtE u(n) u =0. (2.14)
n→∞t≥0(cid:20) (cid:18)(cid:13) t − t(cid:13)L2(R)(cid:19)(cid:21)
(cid:13) (cid:13)
Consequently, if λ>Lip4/(8κ), then(cid:13) (cid:13)
σ
sup e−λtE u 2 = lim M(n)(λ)
t≥0 k tkL2(R) n→∞
h (cid:16) (cid:17)i (2.15)
supM(k)(λ)< .
≤ ∞
k≥0
The second inequality of (2.1) follows readily from this bound.
3 Proof of Theorem 1.1
Our proof of Theorem 1.1 hinges on a number of steps, which we develop sepa-
rately. First we recall the following.
Proposition 3.1 (Theorem 2.1 and Example 2.9 of [8]). If u is bounded and
0
measurable, then u (x) Lp(P) for all p [1, ). Moreover, γ(p)< for all
t
p [1, ) and γ(2) L∈ip4/(8κ), where ∈ ∞ ∞
∈ ∞ ≤ σ
γ(p):=limsupt−1suplnE(u (x)p). (3.1)
t
t→∞ x∈R | |
Next, we record a simple though crucial property of the function γ.
Remark3.2. SupposeX isanonnegativerandomvariablewithfinitemoments
ofallorder. ByHo¨lder’sinequality,p lnE(Xp)isconvexon[1, ). Itfollows
7→ ∞
that γ is convex—in particular continuous—on [1, ).
∞
5
Now we begin our analysis, in earnest, by deriving an upper bound on the
Lk(P)-norm of the solution u (x) that includes simultaneously a sharp decay
t
rate in x and a sharp explosion rate in t.
Lemma 3.3. Suppose that u 0, and u is supported in [ K,K] for some
0 0
6≡ −
finite constant K >0. Then, for all real numbers k [1, ) and p (1, ),
∈ ∞ ∈ ∞
x2 k+1 (1/p) γ(kp)
limsupt−1sup + − lnE u (x)k . (3.2)
t→∞ x∈R(cid:18)4t2 k | t | (cid:19)≤ p
(cid:0) (cid:1)
Proof. According to Mueller’s comparison principle ([13]; more specifically, see
[5, Theorem 5.1, p. 130]), the solution to (1.1) has the following nonnegativity
property: Because u 0 then outside a single null set, u 0 for all t 0.
0 t
≥ ≥ ≥
And therefore,
1 K
u (x) =(p u )(x)= e−(x−y)2/(4κt)u (y)dy. (3.3)
k t k1 t∗ 0 √4κπt 0
Z−K
Because (x y)2 (x2/2) K2,
− ≥ −
u (x) const e−x2/(4t) for all x R and t 1. (3.4)
t 1
k k ≤ · ∈ ≥
The constantappearing in the above display depends on K. Next we note that
for every θ (0, ),
∈ ∞
E u (x)k θk+E u (x)k;u (x) θ
t t t
| | ≤ | | ≥ (3.5)
(cid:0) (cid:1) θk+ E(cid:0) u (x)kp 1/p (P(cid:1)u (x)>θ )1−(1/p).
t t
≤ | | · { }
Proposition 3.1 implies tha(cid:0)t (cid:0) (cid:1)(cid:1)
sup E u (x)kp 1/p exp t γ(kp)+o(1) , (3.6)
t
x∈R | | ≤ (cid:18) p (cid:19)
(cid:0) (cid:0) (cid:1)(cid:1)
whereo(1) 0ast . Also,wecanapply(3.4)togetherwiththeChebyshev
→ →∞
inequality to find that
x2 1
(P u (x)>θ )1−(1/p) const θ−1+(1/p)exp 1 . (3.7)
t
{ } ≤ · −4t · − p
(cid:18) (cid:20) (cid:21)(cid:19)
Taking into consideration (3.6) and (3.7), inequality (3.5) reduces to
E ut(x)k inf θk+αθ−1+(1/p) , (3.8)
| | ≤θ>0
(cid:0) (cid:1) (cid:16) (cid:17)
where
x2 1 γ(kp)+o(1)
α:=exp 1 +t . (3.9)
−4t · − p p
(cid:18) (cid:20) (cid:21) (cid:19)
6
Somecalculus showsthat the functiong(θ):=(θk+αθ−1+1/p)1 attains its
(θ>0)
minimum at θ :=((p 1)/kp)p/(kp+p−1). This yields
−
kp/(kp+p−1)
p 1 1 p kp
E u (x)k αkp/(kp+p−1) − − − .
t
| | ≤ kp · 1 p
(cid:18) (cid:19) (cid:18) − (cid:19)
(cid:0) (cid:1)
We now divide both sides of the above display by αkp/(kp+p−1) and take the
appropriate limit to obtain the result.
Our next lemma is a basic estimate of continuity in the variable x. It is not
entirely standard as it holds uniformly for all times t 0. We emphasize that
≥
the constant p is assumed to be an integer. We will deal with this shortcoming
subsequently.
Lemma 3.4. Suppose that the initial function u is Ho¨lder continuous of order
0
1/2. Then, for all integers p 1 and β > γ(2p) there exists a constant
≥ ≥
A (0, ) such that the following holds: Simultaneously for all t 0,
p,β
∈ ∞ ≥
u (x) u (x′)
sup sup t − t A eβt/(2p). (3.10)
j∈Zj≤x<x′≤j+1(cid:13) |x−x′|1/2 (cid:13)2p ≤ p,β
(cid:13) (cid:13)
(cid:13) (cid:13)
Proof. Burkholder’s inequality(cid:13) [2] and Minko(cid:13)wski’s inequality together imply
that
u (x) u (x′)
k t − t k2p
(p u )(x) (p u )(x′)
t 0 t 0
≤| ∗ − ∗ |
t ∞ 1/2
+z ds dy σ(u (y))2 p (y x) p (y x′)2
2p s t−s t−s
| | ·| − − − | (3.11)
(cid:13)Z0 Z−∞ (cid:13)p
(cid:13) (cid:13)
(p (cid:13)u )(x) (p u )(x′) (cid:13)
≤| t∗(cid:13)0 − t∗ 0 | (cid:13)
t ∞ 1/2
+z′ ds dy u (y)2 p (y x) p (y x′)2 ,
2p | s | ·| t−s − − t−s − |
(cid:13)Z0 Z−∞ (cid:13)p
(cid:13) (cid:13)
(cid:13) (cid:13)
where zp i(cid:13)s a positive and finite constant that depend only on(cid:13)p, and zp′ :=
z Lip .
p σ
On one hand, we have the following consequence of Young’s inequality:
sup sup (p u )(x) (p u )(x′) sup u (a) u (b)
t 0 t 0 0 0
t≥0|x−x′|≤δ| ∗ − ∗ |≤|a−b|≤δ| − | (3.12)
const δ1/2.
≤ ·
On the other hand, the generalized Ho¨lder inequality suggests that if p 1
≥
is an integer, then for all s ,...,s 0 and y ,...,y R,
1 p 1 p
≥ ∈
p p
p 2
E u (y ) u (y ) . (3.13)
 sj j ≤ sj j 2p
j=1 j=1
Y(cid:12) (cid:12) Y(cid:13) (cid:13)
 (cid:12) (cid:12)  (cid:13) (cid:13)
7
Therefore,
t ∞
ds dy u (y)2 p (y x) p (y x′)2
s t−s t−s
| | ·| − − − |
(cid:13)Z0 Z−∞ (cid:13)p (3.14)
(cid:13) (cid:13)
(cid:13) t ∞ (cid:13)
(cid:13) ds dy u (y) 2 p (y x) (cid:13)p (y x′)2.
≤ k s k2p·| t−s − − t−s − |
Z0 Z−∞
[Write the p-th power of the left-hand side as the expectation of a product and
apply (3.13).]
A proof by contradictionshows that Proposition3.1 gives the following [see
[8] for more details]:
c :=supsup e−βsE u (y)2p < for all β >γ(2p). (3.15)
β s
s≥0y∈R | | ∞
(cid:2) (cid:0) (cid:1)(cid:3)
Consequently,
t ∞
ds dy u (y)2 p (y x) p (y x′)2
s t−s t−s
| | ·| − − − |
(cid:13)Z0 Z−∞ (cid:13)p
(cid:13) (cid:13)
(cid:13) t ∞ (cid:13)
(cid:13) c1/p ds dy eβs/p p (y x) p(cid:13) (y x′)2 (3.16)
≤ β · ·| t−s − − t−s − |
Z0 Z−∞
∞ ∞
c1/peβt/p ds e−βs/p dy p (y x) p (y x′)2.
≤ β · | s − − s − |
Z0 Z−∞
Sincepˆ (ξ)=exp( κsξ2),Plancherel’stheoremtellsusthattheright-handside
s
−
of the preceding inequality is equal to
c1/peβt/p ∞ ∞
β ds e−βs/p dξ e−2κsξ2[1 cos(ξ(x x′))]
π · − −
Z0 Z−∞ (3.17)
2c1/peβt/p ∞ [1 cos(ξ(x x′))]
β
= − − dξ.
π · (β/p)+2κξ2
Z0
Because 1 cosθ min(1,θ2), a direct estimation of the integral leads to the
− ≤
following bound:
t ∞
ds dy u (y)2 p (y x) p (y x′)2
s t−s t−s
(cid:13)Z0 Z−∞ | | ·| − − − | (cid:13)p (3.18)
(cid:13) (cid:13)
(cid:13) const eβ(cid:13)t/p x x′ ,
(cid:13) ≤ · (cid:13) ·| − |
wheretheimpliedconstantdependsonlyonp,κ,andβ. This,(3.12),and(3.11)
together imply the lemma.
The preceding lemma holds for all integers p 1. In the following, we
≥
improve it [at a slight cost] to the case that p (1,2) is a real number.
∈
8
Lemma 3.5. Suppose the conditions of Lemma 3.4 are met. Then for all
p (1,2) and δ (0,1) there exists a constant B (0, ) such that the
p,δ
∈ ∈ ∈ ∞
following holds: Simultaneously for all t 0 and x,x′ R with x x′ 1,
≥ ∈ | − |≤
E u (x) u (x′)2p B x x′ p e(1+δ)λpt, (3.19)
t t p,δ
| − | ≤ ·| − | ·
where (cid:0) (cid:1)
λ :=(2 p)γ(2)+(p 1)γ(4). (3.20)
p
− −
Proof. We start by writing
E u (x) u (x′)2p =E u (x) u (x′)2(2−p) u (x) u (x′)4(p−1) .
t t t t t t
| − | | − | | − |
(cid:0) (cid:1) (cid:16) (cid:17)
We can apply Ho¨lder’s inequality to conclude that for all p (1,2), t 0,
∈ ≥
and x,x′ R,
∈
E u (x) u (x′)2p
t t
| − | (3.21)
(cid:0) E (cid:1)ut(x) ut(x′)2 2−p E ut(x) ut(x′)4 p−1.
≤ | − | | − |
We now use Lemm(cid:2)a 3(cid:0).4 to obtain the f(cid:1)o(cid:3)llowi(cid:2)ng(cid:0): (cid:1)(cid:3)
E u (x) u (x′)2 2−p x x′ (2−p)A2(2−p)eβ1(2−p)t
| t − t | ≤| − | 1,β1
(cid:2) (cid:0) (cid:1)(cid:3)
and
E u (x) u (x′)4 p−1 x x′ 2(p−1)A4(p−1)eβ2(p−1)t,
| t − t | ≤| − | 2,β2
where A1,β(cid:2)1,(cid:0)A2,β2 ∈ (0,∞) an(cid:1)d(cid:3) β1 > γ¯(2) and β2 > γ¯(4) are fixed and finite
constants. The proof now follows by combining the above and choosing β and
1
β such that (1+δ)γ¯(2)>β >γ¯(2) and (1+δ)γ¯(4)>β >γ¯(4).
2 1 2
The preceding lemma allows for a uniform modulus of continuity estimate,
which we record next.
Lemma 3.6. Suppose the conditions of Lemma 3.4 are met. Then for all
p (1,2) and ǫ,δ (0,1) there exists C (0, ) such that simultaneously
p,ǫ,δ
∈ ∈ ∈ ∞
for all t 0,
≥
u (x) u (x′)2
sup sup | t − t | C e(1+δ)λpt, (3.22)
j∈Z(cid:13)j≤x<x′≤j+1 |x−x′|1−ǫ (cid:13)p ≤ p,ǫ,δ·
(cid:13) (cid:13)
(cid:13) (cid:13)
where λ was de(cid:13)fined in (3.20). (cid:13)
p
Proof. The proof consists of an application of the Kolmogorovcontinuity theo-
rem. Recall that the spatial dimension is 1. Since p>1 in Lemma 3.5, we can
use a suitable versionof Kolmogorovcontinuity theorem, for example Theorem
4.3 of reference [5, p. 10], to obtain the result. The stated dependence of the
constant, C is consequence of the explicit form of inequality (3.19) and the
p,ǫ,δ
proof of Theorem 4.3 in [5].
9
Before we begin our proofof Theorem1.1,we provethat under some condi-
tion the L2(P)-norm of the solution has an effectively-compact support.
Proposition 3.7. If the conditions of Theorem 1.1 are met, then there exists a
finiteandpositive constantm suchthat u (x) has an effectively-compact support
t
with radius of effective support p(t)=mt.
Proof. We begin by noting that for all m,t>0,
u (x)2dx u (x)dx+ u (x)2dx. (3.23)
t t t
| | ≤ |x|>mt | |
Z|x|>mt Z|x|>mt Zut(x)≥1
Therefore,
E u (x)2dx
t
Z|x|>mt| | ! (3.24)
(p u )(x)dx+ E u (x)2; u (x) 1 dx.
t 0 t t
≤ ∗ | | ≥
Z|x|>mt Z|x|>mt
(cid:0) (cid:1)
Since u has compact support, (3.4) implies that
0
(p u )(x)dx=O e−m2t/2 as t . (3.25)
t 0
∗ →∞
Z|x|>mt (cid:16) (cid:17)
Next we estimate the final integral in (3.24).
Thanks to (3.4) and Chebyshev’s inequality,
P u (x) 1 const e−x2/(4t), (3.26)
t
{ ≥ }≤ ·
uniformly for all x R and t 1. Also, from Proposition 3.1, there exists a
∈ ≥
constant b (0, ) such that
∈ ∞
supE u (x)4 bebt/4 for all t 1. (3.27)
t
x∈R | | ≤ ≥
(cid:0) (cid:1)
Using the preceding two inequalities, the right-hand side of inequality (3.24)
reduces to
E u (x)2dx
t
Z|x|>mt| | !
O e−m2t/2 +const E(u (x)4)e−x2/(8t)dx (3.28)
t
≤ · | |
(cid:16) (cid:17) Z|x|>mtp
O e−m2t/2 +const b1/2ebt/8 e−x2/(8t)dx.
≤ · ·
(cid:16) (cid:17) Z|x|>mt
We now choose and fix m>√b to obtain from the preceding that
limsupt−1lnE u (x)2dx <0. (3.29)
t
t→∞ Z|x|>mt| | !
10