Regularized solutions for some backward nonlinear parabolic equations with statistical data

0.38 MB·

by  Mokhtar Kirane

REGULARIZED SOLUTIONS FOR SOME BACKWARD NONLINEAR PARABOLIC
EQUATIONS WITH STATISTICAL DATA
MOKHTARKIRANE,ERKANNANEANDNGUYENHUYTUAN
7
1
0 Abstract. Inthispaper,westudythebackwardproblemofdetermininginitialconditionforsomeclass
2 ofnonlinearparabolicequationsinmultidimensionaldomainwheredataaregivenunderrandomnoise.
Thisproblemisill-posed,i.e.,thesolutiondoesnotdependcontinuouslyonthedata. Toregularizethe
b
instable solution, we develop some new methods to construct some new regularized solution. We also
e
investigate theconvergence ratebetween the regularizedsolutionandthesolutionofour equations. In
F
particular, we establish results for several equations with constant coefficients and time dependent co-

7 efficients. Theequations withconstant coefficients includeheat equation, extended Fisher-Kolmogorov
equation, Swift-Hohenberg equation and manyothers. The equations withtime dependent coefficients

] include Fisher type Logistic equations, Huxley equation, Fitzhugh-Nagumo equation. The methods
P developed inthispaper can alsobeappliedtoget approximate solutions toseveralother equations in-
A cluding1-D Kuramoto-Sivashinskyequation,1-D modifiedSwift-Hohenbergequation,stronglydamped
waveequation and1-DBurger’sequationwithrandomlyperturbedoperator.
.
h
t
a
1. Introduction
m
[ In this paper, we focus on the problem of finding the initial functions u(x,0) = u0(x) such that u
satisfies the following nonlinear parabolic equation
2
v u + (t,u)u=F(u(x,t))+G(x,t), 0<t<T,x Ω,
t
A ∈
9
u(x,t)=0, x ∂Ω, (1.1)
5  ∈
4 u(x,T)=H(x),x Ω
8 ∈
0 where the domain Ω =(0,π)d is a subset of Rd and x := (x1,...xd). The functions F and G are called
1. the sourcefunctions that satisfy the usualLipschitz andgrowthconditions. The function H is givenand
0 isoftencalledafinalvaluedata. Theoperator isgivenbytheLaplacian,orafunctionoftheLaplacian
A
7 defined by the spectral theorem.
1
The problem (1.1) is called the backward problem for classical parabolic equation when = ∆. It
: A −
v is applied in fields as the heat conduction theory [5], hydrology [4], groundwater contamination [33], to
i digitallyremoveblurrednoiselessimage[8]andalsoinmanyotherpracticalapplicationsofmathematical
X
physics and engineering [37]. It is well-known that the backward parabolic problem is severely ill–posed
r
a in the sense of Hadamard [16]. Hence solutions do not always exist, and in the case of existence, the
solutionsdonotdependcontinuouslyonthe initialdata. Infact,fromsmallnoisecontaminatedphysical
measurements, the corresponding solutions might have large errors. More details on ill-posedness of the
problem with random noise can be found in [26].
The analysis of regularization methods for the stable solution (in the sense of Hadamard) of problem
(1.1) depends on the mathematical model with the noise term on the source function G and the final
value data u =u(x,T)=H(x). We suppose that the measurements are described as follows
T
Gobs =G+”noise”, Hobs =H +”noise”. (1.2)
1.1. Background in the deterministic case. The Problem (1.1) is a generalized form of a class of
backward parabolic equations. We give a short history of this problem in the deterministic case. If the
”noise”(introducedin(1.2))isconsideredasadeterministicquantity,itisnaturaltostudywhathappens
when noise 0.
L2
k k →
When the operator (t) = ( independent of t), regularization results were considered by many
A A
authors, we refer the reader to the survey paper of Tuan [38] and the references therein.
1
When (t) depends on t and F = G = 0, Lions and Lattes [22] proposed the following quasi-
A
reversibility method:
u (t)+ (t)u +ǫ (t) (t)u = 0
′ǫ A ǫ A∗ A ǫ (1.3)
u (T) = uǫ .
(cid:26) ǫ T
However,LionsandLattesdidnotstudyregularizationresultsforthisproblem. Theregularizationresult
here is stillopenalthoughsomeprogresshas beenmade. The firstpaper onthis caseseems to be thatof
Krein[12],wherehe usedthelog-convexitymethodto getstabilityestimatesofHo¨ldertype. His method
and results have been further developed by Hao and Duc [17].
When (t) depends on t and u, to the best of our knowledge, there do not exist any results on the
A
backward problem. Regularization results in here are very difficult because one can not represent the
solution with a nonlinear integral as previously done by others. Hence, the regularized solution can not
be obtained with the previous techniques based on nonlinear integral method. Regularizationresults for
problem (1.1) in the deterministic case are still open.
1.2. Background on problem with random noise. If the errors are generated from uncontrollable
sources as wind, rain, humidity, etc, then the model is random. If the ”noise” (introduced in (1.2)) are
modeled as a random quantity, the convergence of estimators u(x,0) of u(x,0) should be studied by
statistical methods. Methods applied to the deterministic cases cannot be applied directly for this case.
The main idea in using the random noise is of finding suitableeestimators u(x,0) and to consider the
expected square errorE u(x,0) u(x,0) 2 in a suitable space, also called the mean integratedsquare
k − k
e
error (MISE). h i
Thereexistaconsiderabeleamountofliteratureonregularizationmethodsforlinearbackwardproblem
with random noise. When F(u) = 0, the problem (1.1) is linear and its solution can be defined by a
linear operator with random noise
u =Ku +”noise”. (1.4)
T 0
whereK isaboundedlinearoperatorthatdoesnothaveacontinuousinverse. Therearemanywell-known
methods including spectralcut-off(or calledtruncationmethod) ofCavalier[6,9], the Tiknonovmethod
[11], iterative regularization methods [12]. Mair and Ruymgaart [24] considered theoretical formulae
for statistical inverse estimation in Hilbert spaces and applied the method to solve the backward heat
problem. Recently, Hohage et al. [20] applied spectral cut-off (truncation method) and Tikhonov-type
method to solve linear statistical inverse problems including backwardheat equation (See p. 2625, [20]).
Recently, Problem (1.1) in the case of F =0 has been studied in [26] in the plane domain.
Until now, to the best of the authors’ knowledge, there are only a few results in the case of random
source, or random final value observations for nonlinear backward parabolic equation. And there are
no results on Problem (1.1). This is our motivation in the present paper. In a few sentences, we give
explanation why the nonlinear problem is difficult to investigate. Indeed, when F depends on u, we can
nottransformthesolutionofproblem(1.1)into(1.4),thismakesthenonlinearproblemmorechallenging.
Furthermore, as introduced in the subsection 1.1, if = (t,u) then we can not transform the problem
A A
(1.1)intoanonlinearintegralequation,thenthepreviousmethodscannotbeappliedforregularizingthe
problem. So, our task in this paper is developing and establishing new methods for solving
this problem.
1.3. Outline of the article. In this paper, inspired by the random model in [26], we introduce the
following random model in Rd.
Let Ω = (0,π)d Rd for d 1. Let us recall the functions H and G from equation (1.1). Let
x =(x ,...x ) be g⊂rid points of≥Ω with index i=(i ,i ,...i ) Nd,1 i n for k =1,d where
i i1 id 1 2 d ∈ ≤ k ≤ k
π(2i 1) π(2i 1) π(2i 1)
1 2 d
xi =(xi1,...xid)= 2n− , 2n− ,... 2n− , ik =1,nk, k =1,d. (1.5)
1 2 d
(cid:16) (cid:17)
We consider the following nonparametric regressionmodel of data as follows
D =D :=H(x ,...x )+Λ Υ =H(x)+Λ Υ (1.6)
i i1,i2,...id i1 id i1,i2,...id i1,i2,...id i i i
G (t)=G (t):=G(x ,...x ,t)+ϑΨ (t)=G(x ,t)+ϑΨ (t), (1.7)
i e i1e,i2,...id i1 id i1,i2,...id i i
2
e e
for i = 1,n , k = 1,d. Here Υ := Υ (0,1) and Ψ (t) := Ψ (t) are Brownian
k k i i1,i2,...id ∼ N i i1,i2,...id
motions. We assume furthermore that they are mutually independent.
Our main goal in this paper is to provide some regularized solutions that are called estimators for
approximating u(x,t), 0 t < T. In this paper, we do not investigate the existence and uniqueness
≤
of the solution of backward problem (1.1). The uniqueness of backward parabolic has attracted the
attention of many authors; see, for example, [23, 32, 41]. It is also a challenging and open problem, and
should be the topic for another paper. In this paper, we assume that the backward problem (1.1) has a
unique solution u (called soughtsolution) that belongs to anappropriate space. So our main purpose is
to considera regularizedproblemforfinding anapproximatesolutioninthe randomcases. Furthermore,
error estimates with the speed of convergence between the regularized solution and the sought solution
under some a priori assumptions on the sought solution are also our primary purpose. In particular,
the main purpose in our error estimates is to show that the norm of difference between the regularized
solutionandthesoughtsolutioninL2(Ω)tendstozerowhen n = n2+...+n2 + . Ourmethods
| | 1 d → ∞
inthis papercanbe appliedto solvethe backwardprobleminthedeterministic casethatwasintroduced
p
in subsection 1.1.
For the purpose of capturing the main points of the paper, we consider Problem (1.1) and describe
our main results for the three cases of (t,u).
A
Case 1: (t,u) = . In this case, we apply Fourier truncation method associated with knowledge
A A
on trigonometric theory in nonparametric regression for establishing regularized solutions. The process
for finding the regularizedsolutionis givenby the followingsteps : First, we approximatethe givendata
H and G by approximating functions H and G defined by Theorem (2.1). Then, we express the
βn βn
solution of Problem (1.1) into a nonlinear integral equation which is represented as Fourier series and
then we give some regularized solutionsbwhich arebdefined by other nonlinear integral equations. In this
case,we will derive rates of convergenceunder some a prioriassumptions of the soughtsolution u. Main
result in this case is given by Theorems 3.2.
Case 2: (t) depends only on t. As discussed before, regularized methods used in Case 1 cannot be
A
applied in this case. Hence, we need to figure out a new regularized method to establish a regularized
solution. Our main idea in this case is that of applying a modified Quasi-reversibility method given by
Lions[22]. Wewillnotapproximatedirectlythetimedependentsoperator (t)asintroducedin[22]. Our
A
method is of finding the unbounded time independent operator P that satisfies conditions in Definition
4.1. Then, we approximate P by a bounded operator P , in order to establish the well-posedness of
ρn
the problem associated with the approximating functions H and G . Finding suitable regularized
βn βn
operators is important task in this section. Our main results in this case are Theorem 4.1 and 4.2. A
specialcaseofTheorem4.1isCorollary4.1whichstudiestheebxtendedbFisher-Kolmogorovequation.
Case 3: depends on t and u. Using the method in Case 2, we extend the results of Case 2 to
A
the case when the coefficients of depends on t and u. Special cases of the equation considered in
A
Theorem 4.2 are Fisher type Logistic equations with F(u)=au(1 u), or Huxley equation with
−
F(u)=au2(1 u), or Fitzhugh-Nagumo equation with F(u)=au2(1 u)(u θ ). See Remark 4.2
1
− − −
for more.
Finally,wewanttomentionthatthebackwardproblemforsomeconcretenonlocalparabolicequations
suchasGinzburg-Landau equationwherethecoefficientsoftheseequationsareperturbedbyrandom
noises can be studied with the methods in our paper. These equations include the 1-D Burger’s equa-
tion. Furthermore, our analysis and methods in this paper can be applied to get approximate solutions
for many well-known equations. We state some examples below. We will work on these problems in a
forthcoming paper.
Backwardproblem for 1-D Kuramoto-Sivashinsky equation:
•
u +d u +(d (x)u ) = uu +G(x,t), (x,t) (0,π) (0,T), (1.8)
t 0 xx 1 xx xx − x ∈ ×
with the conditions
u(0,t) =u(π,t)=u (0,t)=u (π,t)=0, t (0,T),
xx xx ∈ (1.9)
u(x,T) =H(x), x (0,π),
(cid:26) ∈
3
where d ,d are difussion coefficients. This nonlinear partial differential equation describes in-
0 1
cipient instabilities in a variety of physical and chemical systems (see [3, 10]).
Backwardproblem for 1-D modified Swift-Hohenberg equation:
•
u +2u +ku +au+u3+ u 2+G(x,t)=0, (x,t) (0,π) (0,T), (1.10)
t xx xxxx x
| | ∈ ×
with condition (1.9). The Swift-Hohenberg equation is one of the universal equations used in
the description of pattern formation in spatially extended dissipative systems, (see [35]), which
arise in the study of convective hydrodynamics [36], plasma confinement in toroidaldevices [31],
viscous film flow and bifurcating solutions of the Navier-Stokes [34].
Backwardproblem for strongly damped wave equation:
•
u (x,t) α∆u (x,t) ∆u(x,t)=F(u(x,t))+G(x,t), 0<t<T, x Ω,
tt t
− − ∈
u(x,t)=0, x ∂Ω,
 ∈ (1.11)
u(x,T)=H(x), x∈Ω
u (x,T)=0, x Ω,
t
∈
whereH L2(Ω) is a given function and α is a positive constant. Strongly damped wave
∈
equation,occursin awide rangeofapplicationssuchasmodeling motionofviscoelasticmaterials
[18, 25, 29]. Some more physical applications of the equation (1.11) can be found in [30].
1-D Burger’s equation:
•
u (A(x,t)u ) =uu +G(x,t), (x,t) Ω (0,T),
t x x x
− ∈ ×
u(x,t) =0, x ∂Ω, (1.12)
 ∈
u(x,T) =H(x), x Ω,

∈
where Ω=(0,π). The Burgers equation is a fundamental partial differential equation occurring

in various areas of applied mathematics, such as fluid mechanics, nonlinear acoustics, gas dy-
namics,trafficflow[2]. The ill-posednessofthe backwardproblemforBurgersequationhasbeen
introduced by E. Zuazua et al [18] . The model here is as follows:
Assume the time dependent coefficient A(x,t) is noisy by random data
A (t)=A(x ,t)+ϑξ (t), for k =1,n. (1.13)
k k k
where xk = (2k2−n1)π,k =e 1,n are the grid points in (0,π) and ξk(t),k = 1,n are independent
Brownian motions.
2. Constructing a function from discrete random data
In this section, we develop a new theory for constructing a function in L2(Ω) from the given discrete
random data.
We first introduce notation, and then we state the main results of this paper.
We will occasionally use the following Gronwall’s inequaly in this paper.
Lemma2.1. Letb:[0,T] R+ beacontinuousfunctionandC,D >0beconstantsthatareindependent
→
of t, such that
T
b(t) C+D b(τ)dτ, t>0.
≤
Zt
Then we have
b(t) CeD(T t).
−
≤
Next we define fractional powers of the Dirichlet Laplacian
Af := ∆f.
−
Since A is a linear, densely defined self-adjoint and positive definite elliptic operator on the connected
bounded domain Ω with Dirichlet boundary condition, using spectral theory, it is easy to show that
the eigenvalues of A are given by λ = p2 = p2+p2+ +p2. The corresponding eigenfunctions are
p | | 1 2 ··· d
denoted respectively by
d
2
ψ (x)= sin(p x )sin(p x ) sin(p x ). (2.1)
p 1 1 2 2 d d
π ···
(cid:18)r (cid:19)
4
Thus the eigenpairs (λ ,ψ ), p Nd, satisfy
p p
∈
Aψ (x)= λ ψ (x), x Ω
p p p
− ∈
(ψp(x)=0, x ∂Ω.
∈
The functions ψp are normalized so that ψp p Nd is an orthonormal basis of L2(Ω).
{ } ∈
We will use the following notation: p = (p , ,p ) = p2+...+p2, n = (n , ,n ) =
| | | 1 ··· d | 1 d | | | 1 ··· d |
n2+...+n2.
1 d p
pDefinition 2.1. For γ >0, we define
∞ ∞
γ(Ω):= h L2(Ω): ... p2γ <h,ψ >2 < . (2.2)
p
H ∈ | | ∞
n pX1=1 pXd=1 o
The norm on γ(Ω) is defined by
H
∞ ∞
h 2 := ... p2γ <h,ψ >2 . (2.3)
k kHγ(Ω) | | p
pX1=1 pXd=1
For any Banach space X, we denote by L (0,T;X), the Banach space of measurable real functions
p
v :(0,T) X such that
→
1/p
T
v = v(,t) p dt < , 1 p< ,
k kLp(0,T;X) Z0 k · kX ! ∞ ≤ ∞
v =esssup v(,t) < , p= .
k kL∞(0,T;X) 0<t<T k · kX ∞ ∞
Let β :Nd R be a function. Now we state our first main result which gives error estimate between
→
H and H , and error estimate between G and G.
βn βn
Theorem 2.1. Define the set for any n=(n ,..n ) Nd
b Wβn b 1 d ∈
d
= p=(p ,...p ) Nd : p2 = p2 β =β(n ,...n ) (2.4)
Wβn 1 d ∈ | | k ≤ n 1 d
n kX=1 o
where β satisfies
n
lim β =+ .
n
n + ∞
| |→ ∞
For a given n and β we define functions that are approximating H,G as follows
n
πd n1 nd
H (x)= ... D ψ (x ,...x ) ψ (x) (2.5)
βn p∈XWβn" dk=1nk iX1=1 iXd=1 i1,i2,...id p i1 id # p
b e
and Q
πd n1 nd
G (x,t)= ... G (t)ψ (x ,...x ) ψ (x). (2.6)
βn p∈XWβn" dk=1nk iX1=1 iXd=1 i1,i2,...id p i1 id # p
Letµ=(µ1,...µdb)∈Rd withµk > 21Qforanyk=1,d. Leteuschooseµ0 ≥dmax(µ1,...µd). IfH ∈Hµ0(Ω)
and G L∞(0,T; µ0(Ω)) then the following estimates hold
∈ H
d
2 2
E Hβn −H L2(Ω) ≤C(µ1,...µd,H)βnd/2 (nk)−4µk +4βn−µ0 H µ0(Ω),
(cid:13) (cid:13) kY=1 (cid:13) (cid:13)H
(cid:13)(cid:13)b (cid:13)(cid:13) 2 d (cid:13)(cid:13) (cid:13)(cid:13) 2
E(cid:13)Gβn(.,t)−G(.,t)(cid:13)L∞(0,T;L2(Ω)) ≤C(µ1,...µd,H)βnd/2kY=1(nk)−4µk +4βn−µ0(cid:13)G(cid:13)L∞(0,T;Hµ0(Ω)),
whe(cid:13)(cid:13)reb (cid:13)(cid:13) (cid:13)(cid:13) (cid:13)(cid:13)
2πd/2 16 2(µ ,...µ )πd/2 2
C(µ ,...µ ,H)=8πdV2 + C 1 d H .
1 d maxdΓ(d/2) dΓ(d/2) µ0(Ω)
(cid:13) (cid:13)H
5 (cid:13) (cid:13)
(cid:13) (cid:13)
2
Corollary 2.1. Let H,G be as in Theorem (2.1). Then the term E H H + TE G
βn − L2(Ω) βn −
2 (cid:13) (cid:13) (cid:13)
G is of order (cid:13)b (cid:13) (cid:13)b
(cid:13) (cid:13) (cid:13)
L∞(0,T;L2(Ω))
(cid:13)
(cid:13)
(cid:13) βd/2
max dk=1(nnk)4µk, βn−µ0!.
Q
To prove this Theorem, we need some preliminary results.
Lemma 2.2. Let p,q∈Nd and p=(p1,...pd), q=(q1,...qd) with pk =1,nk−1 and xik = π(22inkk−1) for
k =1,d. Then for all q Nd, we have
∈
1 n1 nd
R = ... ψ (x ,...x )ψ (x ,...x )
p,q d n p i1 id q i1 id
k=1 k iX1=1 iXd=1
1
Q , p q =2l n , k =1,d,
= πd k∓ k k k (2.7)

0, otherwise.


If q Nd satisfies that q :=1,n 1 then we have
k k
∈ −
1 n1 nd 1 , p =q , k =1,d,
Rp,q = dk=1nk iX1=1...iXd=1ψp(xi1,...xid)ψq(xi1,...xid)=( πd0, ∃kk =1k,d pk 6=qk, (2.8)
Q
Proof. The lemma is a direct consequence of Lemma 3.5 in [13]. (cid:3)
Lemma 2.3. Let H L2(Ω) and H =< H,ψ > be the Fourier coefficients of H for p = (p ,...p ).
p p 1 d
∈
Let’s recall that xik = 2i2kn−k1, k =1,d then the following equality holds
πd n1 nd
H = ... H(x ,...x )ψ (x ,...x ) Γ (2.9)
p d n i1 id p i1 id − n,p
k=1 k iX1=1 iXd=1
Q
where p satisfies that p =1,n 1, k =1,d. Here Γ is defined by
k k n,p
−
l Nd,l2+..l2=0
∈ 1 d6
Γ = H .
n,p r
r=2ln p
X· ±
Proof. First, we have the expansion of the function H L2(Ω) as the following Fourier series
∈
H(x)= H ψ (x). (2.10)
r r
r Nd
X∈
where r=(r ,...r ) Nd. Plug x=(x ,...x ) into the latter equation, we obtain
1 d ∈ i1 id
H(x ,...x )= H ψ (x ,...x ). (2.11)
i1 id r r i1 id
r Nd
X∈
6
This implies that
1 n1 nd
... H(x ,...x )ψ (x ,...x )
d n i1 id p i1 id
k=1 k iX1=1 iXd=1
Q 1 n1 nd
= ... H ψ (x ,...x ) ψ (x ,...x )
dk=1nk iX1=1 iXd=1 rX∈Nd r r i1 id ! p i1 id
Q 1 n1 nd
= H ... ψ (x ,...x )ψ (x ,...x )
rX∈Nd r dk=1nk iX1=1 iXd=1 r i1 id p i1 id !
Q 1 n1 nd
= H ... ψ (x ,...x )ψ (x ,...x )
rX∈/Nd r dk=1nk iX1=1 iXd=1 r i1 id p i1 id !
Q :=H1
| 1 n1 nd{z }
+ H ... ψ (x ,...x )ψ (x ,...x ,
rX∈Nd r dk=1nk iX1=1 iXd=1 r i1 id p i1 id!
Q :=H2
where we denote | {z }
:= r=(r ,r ,...r ) Nd :r =1,n 1, k =1,d . (2.12)
d 1 2 d k k
N ∈ −
Step 1. Consider H . n o
1
If r / then applying the first part of Lemma 2.2, we have
d
∈N
1
1 n1 ... nd ψ (x ,...x )ψ (x ,...x )= πd, rk∓pk =2lknk, k =1,d, (2.13)
d n r i1 id p i1 id 
k=1 k iX1=1 iXd=1  0, otherwise.
Hence,Qwe deduce that

1 n1 nd
H ... ψ (x ,...x )ψ (x ,...x )
rX∈/Nd r dk=1nk iX1=1 iXd=1 r i1 id p i1 id !
1
Q
H , r p =2l n , k =1,d,
= πd r∈/Nd r k∓ k k k

P 0, otherwise.

r=2ln p
1 · ±
= H . (2.14)
πd r
rX∈/Nd
where noting that the equation r p =2l n is equavilent to r=2l n p.
k k k k
tFhoarnthpe.sTumhisPimrr=∈/pN2llid·ens±tphHatrlo=nt(hl1e,r..∓i.gldh)t∈haNnddisnidtehoefs(u2m.14),rrs=∈/inN2lcd·en±rp∈/HNrds·,awtie±sficeasntsheeeftohlalotwrinisgncootnddiifftieornent
l2+...+l2 =P0.
1 d 6
Therefore, we can rewrite (2.14) as follows
1 n1 nd 1 l∈Nd,l21+..l2d6=0
H ... ψ (x ,...x )ψ (x ,...x ) = H . (2.15)
rX∈/Nd r dk=1nk iX1=1 iXd=1 r i1 id p i1 id ! πd r=2Xl·n±p r
Step 2. Consider QH .
2
If r then applying the second part of Lemma 2.2, we get
d
∈N
1 n1 nd 1 , r =p , k =1,d,
dk=1nk iX1=1...iXd=1ψr(xi1,...xid)ψp(xi1,...xid)=( πd0, ∃kk =1k,d rk 6=pk, (2.16)
Q 7
This leads to
1 n1 nd 1
H ... ψ (x ,...x )ψ (x ,...x ) = H . (2.17)
rX∈Nd r dk=1nk iX1=1 iXd=1 r i1 id p i1 id ! πd p
Combining (2.12), (2.15),(2.Q17), we obtain
1 n1 nd 1 l∈Nd,l21+..l2d6=0
... H(x ,...x )ψ (x ,...x )= H + H . (2.18)
dk=1nk iX1=1 iXd=1 i1 id p i1 id πd p r=2Xl·n±p r!
This completeQs the proof of this Lemma. (cid:3)
Using Lemma 2.3, we obtain the following Lemma
Lemma 2.4. Assume that G C([0,T];C1(Ω)) then for t [0,T], we have
∈ ∈
πd n1 nd
G (t)= ... G(x ,...x ,t)ψ (x ,...x ) Π (t) (2.19)
p d n i1 id p i1 id − n,p
k=1 k iX1=1 iXd=1
where p satisfies that p =1Q,n 1, k =1,d and
k k
−
l Nd,l2+..l2=0
∈ 1 d
Π (t)= G (t).
n,p r
r=2ln p
X· ±
Next we consider the following Lemma
Lemma 2.5. Assume that µ = (µ1,...µd). Let us choose µ0 dmax(µ1,...µd). Then if H µ0(Ω)
≥ ∈ H
then
d
|Hp|=|Hp1,...pd|≤d−max(µ12,...µd) H µ0(Ω) p−kµk, for p=(p1,p2,...pd). (2.20)
(cid:13) (cid:13)H kY=1
(cid:13) (cid:13)
Proof. Using the Cauchy inequality (cid:13) (cid:13)
d
(a +...+a )d d a
1 d k
≥
k=1
Y
for any a 0,k =1,d, we have
k
≥
dmax(µ1,...µd)
d 2 d max(µ1,...µd)
p2 d p2 2
k! ≥ k
kX=1 (cid:16) kY=1 (cid:17)
d
dmax(µ12,...µd) pµk.
≥ k
k=1
Y
The left hand side of the latter inequality is bounded by
dmax(µ1,...µd)
d 2
p2 = pdmax(µ1,...µd) pµ0
k! | | ≤| |
k=1
X
The observations above imply that
d
pµkk ≤d−max(µ12,...µd)|p|µ0.
k=1
Y
This leads to
d
pµkk <H,ψp > ≤d−max(µ12,...µd)|p|µ0 <H,ψp > ≤d−max(µ12,...µd) H µ0(Ω)
kY=1 (cid:12)(cid:12) (cid:12)(cid:12) 8(cid:12)(cid:12) (cid:12)(cid:12) (cid:13)(cid:13)(cid:13) (cid:13)(cid:13)(cid:13)H
where we used the fact that
2 ∞ ∞
H = ... p2µ0 <H,ψp >2 p2µ0 <H,ψp >2 . (2.21)
µ0(Ω) | | ≥ | |
(cid:13) (cid:13)H pX1=1 pXd=1
(cid:13) (cid:13)
This completes the(cid:13)pro(cid:13)of of Lemma. (cid:3)
Proof of Theorem 2.1. Using Lemma 2.3 and by a simple computation, we get
2
H H
βn − L2(Ω)
(cid:13) (cid:13)
(cid:13)(cid:13)b (cid:13)(cid:13) πd n1 nd 2
=4 ... Λ Υ ψ (x ,...x ) Γ
p∈XWβn" dk=1nk iX1=1 iXd=1 i1,i2,...id i1,i2,...id p i1 id − n,p#
Q 2
+4 H
p
p∈/XWβn(cid:12)(cid:12) (cid:12)(cid:12)
8π2d (cid:12) (cid:12) n1 nd 2
... Λ Υ
≤ dk=1nk 2 p∈XWβn"iX1=1 iXd=1 i1,i2,...id i1,i2,...id#
(cid:16) (cid:17)
Q :=A1,1
2 2
+8| Γ +4 {Hz } (2.22)
n,p p
p∈XWβn(cid:12)(cid:12) (cid:12)(cid:12) p∈/XWβn(cid:12)(cid:12) (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
:=A1,2 :=A1,3
where we used the|inequ{azlity (a}+b|)2 2{az2+2b}2. The expectation of A is bounded by
1,1
≤
8π2d d n
EA k=1 kV2
1,1 ≤ dk=1nk 2 p∈XWβn Q πd max
(cid:16)Q8πd (cid:17)
= V2 card( ), (2.23)
d n max Wβn
k=1 k
above we used the fact that E2Υi1,i2,...idQ=1 and E(Υi1,i2,...idΥj1,j2,...jd)=0. Now we estimate the
d
card( )=card p=(p ,...p ) Nd : p2 β =β(n ,...n ) , (2.24)
Wβn 1 d ∈ k ≤ n 1 d !
n Xk=1 o
which is the number of p such that p2 β . Let any p=(p ,...p ) Nd such that d p2 β . Let
| | ≤ n 1 d ∈ k=1 k ≤ n
us define rectangles Q in Rd as follows
p
P
Q = z=(z ,z ,...z ) Rd :p 1 z p .
p 1 2 d k k k
∈ − ≤ ≤
n o
Let us define the set Q as follows
√βn
Q = Q .
√βn p
p2[≤βn
It is easy to see that card( ) is equal to the volume of the set Q and we can realize that
Wβn √βn
d
Q Q = z=(z ,...z ) Rd : z2 β .
√βn ⊂ ′√βn 1 d ∈ k ≤ n
n kX=1 o
Hence card( ) is less than the volume of the set Q which denoted by Vol Q i.e,
Wβn ′√βn ′√βn
(cid:16) (cid:17)
card( ) Vol Q .
Wβn ≤ ′√βn
9 (cid:16) (cid:17)
Now we find an upper bound of Vol Q . First, we have
′√βn
(cid:16) (cid:17)
Vol Q′√βn = dz1dz2...dzd. (2.25)
(cid:16) (cid:17) ZQ′√βn
We give the following coordinate system as follows
z =rcos(θ ), z =rsin(θ )cos(θ ),z =rsin(θ )sin(θ )cos(θ ),...
1 1 2 1 2 3 1 2 3
z =rsin(θ )...sin(θ )cos(θ )
d 1 1 d 2 d 1
− − −
z =rsin(θ )...sin(θ )sin(θ )
d 1 d 2 d 1
− −
where
1 r β , 0 θ π,i=1.d 2,0 θ <2π.
n i d 1
≤ ≤ ≤ ≤ − ≤ −
From the Change of Variables formpula that the rectangular volume element dz1dz2...dzd can be written
in spherical coordinates as
∂x
i
dz dz ...dz = det drdθ ...dθ
1 2 d 1 d 1
(cid:12) ∂(r,θj) (cid:12) −
(cid:12) (cid:16) (cid:17)(cid:12)1 i d,1 j d 1
(cid:12) (cid:12) ≤≤ ≤ ≤ −
=r(cid:12)(cid:12)d−1sind−2(θ1)(cid:12)(cid:12)sind−3(θ2)...sin(θd−2)drdθ1...dθd−1.
Hence, applying Fubini’s theorem, we obtain that
Vol Q′√βn
(cid:16) (cid:17)
= dz dz ...dz
1 2 d
ZWρn(n)
2π π π √βn
= ... rd−1sind−2(θ1)sind−3(θ2)...sin(θd 2)drdθ1...dθd 2dθd 1
Z0 Z0 Z0 Z1 − − −
2π √βn π π π
= dθ rd 1dr sind 2(θ )dθ sind 3(θ )dθ ... sin(θ )dθ
d 1 − − 1 1 − 2 2 d 2 d 2
(cid:18)Z0 − (cid:19) Z1 !(cid:18)Z0 (cid:19)(cid:18)Z0 (cid:19) (cid:18)Z0 − − (cid:19)
2π βnd/2 1 π π π
= (cid:16) d − (cid:17)(cid:18)Z0 sind−2(θ1)dθ1(cid:19)(cid:18)Z0 sind−3(θ2)dθ2(cid:19)...(cid:18)Z0 sin(θd−2)dθd−2(cid:19). (2.26)
Thanks to page 245, [15], we know that
π π π d
sind−2(θ1)dθ1 sind−3(θ2)dθ2 ... sin(θd 2)dθd 2 = Vol Bd(1) . (2.27)
(cid:18)Z0 (cid:19)(cid:18)Z0 (cid:19) (cid:18)Z0 − − (cid:19) 2π (cid:16) (cid:17)
Here Vol Bd(1) denotes the volume of of the unit d-ball. Using again [15] ( Proposition 4.2, page
246-247),(cid:16)we obt(cid:17)ain that
πd/2
Vol Bd(1) = . (2.28)
Γ(d +1)
(cid:16) (cid:17) 2
Combining (2.26), (2.27), (2.28) gives
πd/2βd/2 2πd/2
Vol Q n = βd/2 (2.29)
′√βn ≤ Γ(d +1) dΓ(d/2) n
(cid:16) (cid:17) 2
which we have used the fact that Γ(d +1)= dΓ(d). This together with (2) leads to
2 2 2
d
card( )=card p=(p ,...p ) Nd, p 1,k=1,d: p2 β =β(n ,...n )
Wβn 1 d ∈ k ≥ k ≤ n 1 d !
n Xk=1 o
2πd/2
βd/2. (2.30)
≤ dΓ(d/2) n
10