Regularized solutions for some backward nonlinear partial differential ...

REGULARIZED SOLUTIONS FOR SOME BACKWARD NONLINEAR PARABOLIC EQUATIONS WITH STATISTICAL DATA

arXiv:1701.08459v2 [math.AP] 7 Feb 2017

MOKHTAR KIRANE, ERKAN NANE AND NGUYEN HUY TUAN

Abstract. In this paper, we study the backward problem of determining initial condition for some class of nonlinear parabolic equations in multidimensional domain where data are given under random noise. This problem is ill-posed, i.e., the solution does not depend continuously on the data. To regularize the instable solution, we develop some new methods to construct some new regularized solution. We also investigate the convergence rate between the regularized solution and the solution of our equations. In particular, we establish results for several equations with constant coefficients and time dependent coefficients. The equations with constant coefficients include heat equation, extended Fisher-Kolmogorov equation, Swift-Hohenberg equation and many others. The equations with time dependent coefficients include Fisher type Logistic equations, Huxley equation, Fitzhugh-Nagumo equation. The methods developed in this paper can also be applied to get approximate solutions to several other equations including 1-D Kuramoto-Sivashinsky equation, 1-D modified Swift-Hohenberg equation, strongly damped wave equation and 1-D Burger’s equation with randomly perturbed operator.

1. Introduction 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    ut + A(t, u)u = F (u(x, t)) + G(x, t), 0 < t < T, x ∈ Ω, u(x, t) = 0, x ∈ ∂Ω, (1.1)   u(x, T ) = H(x), x ∈ Ω

where the domain Ω = (0, π)d is a subset of Rd and x := (x1 , ...xd ). The functions F and G are called the source functions that satisfy the usual Lipschitz and growth conditions. The function H is given and is often called a final value data. The operator A is given by the Laplacian, or a function of the Laplacian defined by the spectral theorem. The problem (1.1) is called the backward problem for classical parabolic equation when A = −∆. It is applied in fields as the heat conduction theory [5], hydrology [4], groundwater contamination [33], to digitally remove blurred noiseless image [8] and also in many other practical applications of mathematical physics and engineering [37]. It is well-known that the backward parabolic problem is severely ill–posed in the sense of Hadamard [16]. Hence solutions do not always exist, and in the case of existence, the solutions do not depend continuously on the initial data. In fact, from small noise contaminated physical 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 uT = u(x, T ) = H(x). We suppose that the measurements are described as follows Gobs = G + ”noise”,

H obs = 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” (introduced in (1.2)) is considered as a deterministic quantity, it is natural to study what happens when knoisekL2 → 0. When the operator A(t) = A ( independent of t), regularization results were considered by many authors, we refer the reader to the survey paper of Tuan [38] and the references therein. 1

When A(t) depends on t and F = G = 0, Lions and Lattes [22] proposed the following quasireversibility method:  ′ uǫ (t) + A(t)uǫ + ǫA∗ (t)A(t)uǫ = 0 (1.3) uǫ (T ) = uǫT . However, Lions and Lattes did not study regularization results for this problem. The regularization result here is still open although some progress has been made. The first paper on this case seems to be that of Krein [12], where he used the log-convexity method to get stability estimates of H¨ older type. His method and results have been further developed by Hao and Duc [17].

When A(t) depends on t and u, to the best of our knowledge, there do not exist any results on the 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. Regularization results 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 e (x, 0) of u(x, 0) should be studied by modeled as a random quantity, the convergence of estimators u statistical methods. Methods applied to the deterministic cases cannot be applied directly for this case. e (x, 0) and to consider the The main idea in usinghthe random noise isi of finding suitable estimators u 2 expected square error E ke u(x, 0) − u(x, 0)k in a suitable space, also called the mean integrated square

error (MISE). There exist a considerable amount of literature on regularization methods for linear backward problem 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 uT = Ku0 + ”noise”. (1.4) where K is a bounded linear operator that does not have a continuous inverse. There are many well-known methods including spectral cut-off (or called truncation method) of Cavalier [6, 9], the Tiknonov method [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 backward heat 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 not transform the solution of problem (1.1) into (1.4), this makes the nonlinear problem more challenging. Furthermore, as introduced in the subsection 1.1, if A = A(t, u) then we can not transform the problem (1.1) into a nonlinear integral equation, then the previous methods can not be applied for regularizing the 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 xi = (xi1 , ...xid ) be grid points of Ω with index i = (i1 , i2 , ...id ) ∈ Nd , 1 ≤ ik ≤ nk for k = 1, d where  π(2i − 1) π(2i − 1) π(2id − 1)  2 1 , ik = 1, nk , k = 1, d. , , ... xi = (xi1 , ...xid ) = (1.5) 2n1 2n2 2nd

We consider the following nonparametric regression model of data as follows

ei = D e i1 ,i2 ,...i : = H(xi1 , ...xi ) + Λi1 ,i2 ,...i Υi1 ,i2 ,...i = H(xi ) + Λi Υi D d d d d

ei (t) = G ei1 ,i2 ,...i (t) : = G(xi1 , ...xi , t) + ϑΨi1 ,i2 ,...i (t) = G(xi , t) + ϑΨi (t), G d d d 2

(1.6) (1.7)

for ik = 1, nk , k = 1, d. Here Υi := Υi1 ,i2 ,...id ∼ N (0, 1) and Ψi (t) := Ψi1 ,i2 ,...id (t) are Brownian 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 sought solution) that belongs to an appropriate space. So our main purpose is to consider a regularized problem for finding an approximate solution in the random cases. 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 normp of difference between the regularized solution and the sought solution in L2 (Ω) tends to zero when |n| = n21 + ... + n2d → +∞. Our methods in this paper can be applied to solve the backward problem in the deterministic case that was introduced 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 A(t, u). Case 1: A(t, u) = A. In this case, we apply Fourier truncation method associated with knowledge on trigonometric theory in nonparametric regression for establishing regularized solutions. The process for finding the regularized solution is given by the following steps : First, we approximate the given data bβn defined by Theorem (2.1). Then, we express the b βn and G H and G by approximating functions H solution of Problem (1.1) into a nonlinear integral equation which is represented as Fourier series and then we give some regularized solutions which are defined by other nonlinear integral equations. In this case, we will derive rates of convergence under some a priori assumptions of the sought solution u. Main result in this case is given by Theorems 3.2. Case 2: A(t) depends only on t. As discussed before, regularized methods used in Case 1 cannot be 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]. We will not approximate directly the time dependents operator A(t) as introduced in [22]. Our 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 Pρn , in order to establish the well-posedness of bβn . Finding suitable regularized b βn and G the problem associated with the approximating functions H operators is important task in this section. Our main results in this case are Theorem 4.1 and 4.2. A special case of Theorem 4.1 is Corollary 4.1 which studies the extended Fisher-Kolmogorov equation. Case 3: A depends on t and u. Using the method in Case 2, we extend the results of Case 2 to the case when the coefficients of A depends on t and u. Special cases of the equation considered in 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 − θ1 ). See Remark 4.2 for more. Finally, we want to mention that the backward problem for some concrete nonlocal parabolic equations such as Ginzburg-Landau equation where the coefficients of these equations are perturbed by random noises can be studied with the methods in our paper. These equations include the 1-D Burger’s equation. 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. • Backward problem for 1-D Kuramoto-Sivashinsky equation: ut + d0 uxx + (d1 (x)uxx )xx = −uux + G(x, t),

(x, t) ∈ (0, π) × (0, T ),

with the conditions  u(0, t) = u(π, t) = uxx (0, t) = uxx (π, t) = 0, u(x, T ) = H(x), 3

t ∈ (0, T ), x ∈ (0, π),

(1.8)

(1.9)

where d0 , d1 are difussion coefficients. This nonlinear partial differential equation describes incipient instabilities in a variety of physical and chemical systems (see [3, 10]). • Backward problem for 1-D modified Swift-Hohenberg equation: ut + 2uxx + kuxxxx + au + u3 + |ux |2 + G(x, t) = 0,

(x, t) ∈ (0, π) × (0, T ),

(1.10)

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 toroidal devices [31], viscous film flow and bifurcating solutions of the Navier-Stokes [34]. • Backward problem for strongly damped wave equation:  utt (x, t) − α∆ut (x, t) − ∆u(x, t) = F (u(x, t)) + G(x, t), 0 < t < T, x ∈ Ω,     u(x, t) = 0, x ∈ ∂Ω, (1.11)  u(x, T ) = H(x), x ∈ Ω    ut (x, T ) = 0, x ∈ Ω,

where H ∈ L2 (Ω) is a given function and α is a positive constant. Strongly damped wave equation,occurs in a wide range of applications such as modeling motion of viscoelastic materials [18, 25, 29]. Some more physical applications of the equation (1.11) can be found in [30]. • 1-D Burger’s equation:  (x, t) ∈ Ω × (0, T ),  ut − (A(x, t)ux )x = uux + G(x, t), 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 dynamics, traffic flow [2]. The ill-posedness of the backward problem for Burgers equation has been 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 ek (t) = A(xk , t) + ϑξk (t), A

for

k = 1, n.

(1.13)

where xk = (2k−1)π , k = 1, n are the grid points in (0, π) and ξk (t), k = 1, n are independent 2n 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. Lemma 2.1. Let b : [0, T ] → R+ be a continuous function and C, D > 0 be constants that are independent of t, such that Z T b(τ )dτ, t > 0. b(t) ≤ C + D t

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 λp = |p|2 = p21 + p22 + · · · + p2d . The corresponding eigenfunctions are denoted respectively by r d 2 sin(p1 x1 ) sin(p2 x2 ) · · · sin(pd xd ). (2.1) ψp (x) = π 4

Thus the eigenpairs (λp , ψp ), p ∈ Nd , satisfy ( Aψp (x) = −λp ψp (x), ψp (x) = 0,

x∈Ω x ∈ ∂Ω.

The functions ψp are normalized so that {ψp }p∈Nd is an orthonormal basis of L2 (Ω). p We will use the following notation: |p| = |(p1 , · · · , pd )| = p21 + ... + p2d , |n| = |(n1 , · · · , nd )| = p 2 2 n1 + ... + nd . Definition 2.1. For γ > 0, we define ∞ ∞ n o X X Hγ (Ω) := h ∈ L2 (Ω) : ... |p|2γ < h, ψp >2 < ∞ . p1 =1

The norm on Hγ (Ω) is defined by

khk2Hγ (Ω) :=

∞ X

∞ X

...

p1 =1

(2.2)

pd =1

pd =1

|p|2γ < h, ψp >2 .

(2.3)

For any Banach space X, we denote by Lp (0, T ; X), the Banach space of measurable real functions v : (0, T ) → X such that !1/p Z T

kvkLp (0,T ;X) =

p

kv (·, t)kX dt

0

< ∞,

1 ≤ p < ∞,

kvkL∞ (0,T ;X) = esssup0 21 for any k = 1, d. Let us choose µ0 ≥ d max(µ1 , ...µd ). If H ∈ Hµ0 (Ω) and G ∈ L∞ (0, T ; Hµ0 (Ω)) then the following estimates hold

2

b − H E H

βn

L2 (Ω)

≤ C(µ1 , ...µd , H)βnd/2

2

b

E G βn (., t) − G(., t) ∞ L

(0,T ;L2 (Ω))

d Y

2

(nk )−4µk + 4βn−µ0 H µ H

k=1

≤ C(µ1 , ...µd , H)βnd/2

where

2 C(µ1 , ...µd , H) = 8π d Vmax

d Y

k=1

0 (Ω)

,

2

(nk )−4µk + 4βn−µ0 G ∞ L

2π d/2 16C 2 (µ1 , ...µd )π d/2

2 + .

H µ dΓ(d/2) dΓ(d/2) H 0 (Ω) 5

(0,T ;Hµ0 (Ω))

,

2



b

b Corollary 2.1. Let H, G be as in Theorem (2.1). Then the term E H + T E G βn − H βn − L2 (Ω)

2

G ∞ is of order 2 L

(0,T ;L (Ω))

d/2

max

βn

Qd

4µk k=1 (nk )

,

βn−µ0

!

.

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 = k = 1, d. Then for all q ∈ Nd , we have Rp,q = Qd

n1 X

1

k=1

nk

...

nd X

π(2ik −1) 2nk

for

ψp (xi1 , ...xid )ψq (xi1 , ...xid )

id =1

i1 =1

  1 , p ∓ q = 2l n , k = 1, d, k k k k πd =  0, otherwise.

(2.7)

If q ∈ Nd satisfies that qk := 1, nk − 1 then we have Rp,q

( 1 , pk = qk , k = 1, d, ψp (xi1 , ...xid )ψq (xi1 , ...xid ) = ... = Qd πd 0, ∃k = 1, d pk 6= qk , id =1 k=1 nk i1 =1 1

n1 X

nd X

Proof. The lemma is a direct consequence of Lemma 3.5 in [13].

(2.8)



Lemma 2.3. Let H ∈ L2 (Ω) and Hp =< H, ψp > be the Fourier coefficients of H for p = (p1 , ...pd ). k −1 , k = 1, d then the following equality holds Let’s recall that xik = 2i2n k πd Hp = Q d

k=1

nk

n1 X

i1 =1

...

nd X

id =1

H(xi1 , ...xid )ψp (xi1 , ...xid ) − Γn,p

(2.9)

where p satisfies that pk = 1, nk − 1, k = 1, d. Here Γn,p is defined by Γn,p =

l∈Nd ,l21 +..l2d 6=0

X

Hr .

r=2l·n±p

Proof. First, we have the expansion of the function H ∈ L2 (Ω) as the following Fourier series H(x) =

X

Hr ψr (x).

(2.10)

r∈Nd

where r = (r1 , ...rd ) ∈ Nd . Plug x = (xi1 , ...xid ) into the latter equation, we obtain H(xi1 , ...xid ) =

X

r∈Nd 6

Hr ψr (xi1 , ...xid ).

(2.11)

This implies that n1 X

1

Qd

k=1

nk

= Qd

k=1

=

r∈Nd

X

=

+

X

r∈Nd

|

where we denote

n1 X

H(xi1 , ...xid )ψp (xi1 , ...xid )

nd X

... 1

k=1

Hr Qd

nk

Hr Qd

Hr ψr (xi1 , ...xid ) ψp (xi1 , ...xid )

r∈Nd nd n 1 X X

...

ψr (xi1 , ...xid )ψp (xi1 , ...xid )

...

ψr (xi1 , ...xid )ψp (xi1 , ...xid )

i1 =1

id =1

n1 X

nd X

1 nk

k=1

!

X

id =1

i1 =1

Hr Q d

r∈N / d

|

nk

nd X

id =1

i1 =1

1

X

...

!

!

id =1

i1 =1

{z

:=H1

1

k=1

nk

n1 X

...

i1 =1

nd X

!

}

ψr (xi1 , ...xid )ψp (xi1 , ...xid ,

id =1

{z

}

:=H2

n o Nd := r = (r1 , r2 , ...rd ) ∈ Nd : rk = 1, nk − 1, k = 1, d .

(2.12)

Step 1. Consider H1 . If r ∈ / Nd then applying the first part of Lemma 2.2, we have  nd n1  1 , r ∓ p = 2l n , k = 1, d, X X 1 k k k k πd ψr (xi1 , ...xid )ψp (xi1 , ...xid ) = ... Qd  id =1 k=1 nk i1 =1 0, otherwise.

Hence, we deduce that

X

r ∈N / d

Hr Q d

1

k=1

nk

n1 X

i1 =1

...

nd X

1 πd

r=2l·n±p X

!

ψr (xi1 , ...xid )ψp (xi1 , ...xid )

id =1

  1 P / d Hr , rk ∓ pk = 2lk nk , k = 1, d, π d r∈N =  0, otherwise.

=

(2.13)

Hr .

(2.14)

r ∈N / d

where noting that the equation rk ∓ pk = 2lk nk is equavilent to r = 2l · n ± p. P For the sum rr=2l·n±p Hr on the right hand side of (2.14), since r ∈ / Nd , we can see that r is not different ∈N / d Pr=2l·n±p than p. This implies that l = (l1 , ...ld ) ∈ Nd in the sum r∈N Hr satisfies the following condition / d l12 + ... + ld2 6= 0.

Therefore, we can rewrite (2.14) as follows X

r∈N / d

Hr Qd

1

k=1

nk

n1 X

...

i1 =1

nd X

!

ψr (xi1 , ...xid )ψp (xi1 , ...xid )

id =1

1 = d π

l∈Nd ,l21 +..l2d 6=0

X

Hr .

(2.15)

r=2l·n±p

Step 2. Consider H2 . If r ∈ Nd then applying the second part of Lemma 2.2, we get

( 1 , rk = pk , k = 1, d, ψr (xi1 , ...xid )ψp (xi1 , ...xid ) = ... πd Qd 0, ∃k = 1, d rk 6= pk , id =1 k=1 nk i1 =1 1

n1 X

nd X

7

(2.16)

This leads to X

r∈Nd

Hr Q d

n1 X

1

k=1

nk

...

nd X

!

ψr (xi1 , ...xid )ψp (xi1 , ...xid )

id =1

i1 =1

=

1 Hp . πd

(2.17)

Combining (2.12), (2.15),(2.17), we obtain Qd

1

k=1

nk

n1 X

i1

nd X

1 H(xi1 , ...xid )ψp (xi1 , ...xid ) = d Hp + ... π i =1 =1

l∈Nd ,l21 +..l2d 6=0

d

X

!

Hr .

r=2l·n±p

This completes the proof of this Lemma.

(2.18) 

Using Lemma 2.3, we obtain the following Lemma Lemma 2.4. Assume that G ∈ C([0, T ]; C 1 (Ω)) then for t ∈ [0, T ], we have πd Gp (t) = Qd

k=1

nk

n1 X

...

nd X

id =1

i1 =1

G(xi1 , ...xid , t)ψp (xi1 , ...xid ) − Πn,p (t)

(2.19)

where p satisfies that pk = 1, nk − 1, k = 1, d and

l∈Nd ,l21 +..l2d =0

X

Πn,p (t) =

Gr (t).

r=2l·n±p

Next we consider the following Lemma Lemma 2.5. Assume that µ = (µ1 , ...µd ). Let us choose µ0 ≥ d max(µ1 , ...µd ). Then if H ∈ Hµ0 (Ω) then d

Y max(µ1 ,...µd )

k 2 |Hp | = |Hp1 ,...pd | ≤ d− p−µ , for p = (p1 , p2 , ...pd ). (2.20)

H µ k H

0 (Ω)

k=1

Proof. Using the Cauchy inequality

(a1 + ... + ad )d ≥ d for any ak ≥ 0, k = 1, d, we have d X

p2k

k=1

! d max(µ21 ,...µd )

d Y

d  max(µ12,...µd )  Y ≥ d p2k k=1

≥d

max(µ1 ,...µd ) 2

The left hand side of the latter inequality is bounded by d X

p2k

k=1

! d max(µ21 ,...µd )

ak

k=1

d Y

pµk k .

k=1

= |p|d max(µ1 ,...µd ) ≤ |p|µ0

The observations above imply that d Y

k=1

This leads to d Y

k=1

pµk k ≤ d−

max(µ1 ,...µd ) 2

|p|µ0 .

max(µ1 ,...µd ) max(µ1 ,...µd )

2 2 |p|µ0 < H, ψp > ≤ d− pµk k < H, ψp > ≤ d−

H

Hµ0 (Ω)

8

where we used the fact that

2



H µ H

∞ X

=

0 (Ω)

...

p1 =1

∞ X

pd =1

|p|2µ0 < H, ψp >2 ≥ |p|2µ0 < H, ψp >2 .

(2.21)

This completes the proof of Lemma.



Proof of Theorem 2.1. Using Lemma 2.3 and by a simple computation, we get

2

b

Hβn − H 2 L (Ω)

=4

X

p∈Wβn

"

+4

πd

Qd

k=1

nk

n1 X

p∈W / βn

8π 2d

≤ Q d |

+8 |

k=1 nk

2

...

id =1

i1 =1

X 2 Hp X

p∈Wβn

"

nd X

n1 X

...

Λi1 ,i2 ,...id Υi1 ,i2 ,...id ψp (xi1 , ...xid ) − Γn,p

nd X

Λi1 ,i2 ,...id Υi1 ,i2 ,...id

id =1

i1 =1

{z

{z

:=A1,2

}

|

p∈W / βn

{z

:=A1,3

#2 }

:=A1,1

2 X 2 X Hp Γn,p + 4

p∈Wβn

#2

(2.22)

}

where we used the inequality (a + b)2 ≤ 2a2 + 2b2 . The expectation of A1,1 is bounded by X Qd nk 8π 2d 2 k=1 Vmax EA1,1 ≤ Q 2 πd d p∈Wβn k=1 nk 8π d = Qd

k=1 nk

2 Vmax card (Wβn ) ,

(2.23)

above we used the fact that E2 Υi1 ,i2 ,...id = 1 and E (Υi1 ,i2 ,...id Υj1 ,j2 ,...jd ) = 0. Now we estimate the ! d n o X 2 d p = (p1 , ...pd ) ∈ N : pk ≤ βn = β(n1 , ...nd ) , (2.24) card (Wβn ) = card k=1

which is the number of p such that |p|2 ≤ βn . Let any p = (p1 , ...pd ) ∈ Nd such that us define rectangles Qp in Rd as follows o n Qp = z = (z1 , z2 , ...zd ) ∈ Rd : pk − 1 ≤ zk ≤ pk .

Let us define the set Q√βn as follows

[

Q√βn =

Pd

k=1

p2k ≤ βn . Let

Qp .

p2 ≤βn

It is easy to see that card (Wβn ) is equal to the volume of the set Q√βn and we can realize that d o n X Q√βn ⊂ Q′ √βn = z = (z1 , ...zd ) ∈ Rd : zk2 ≤ βn . k=1

  Hence card (Wβn ) is less than the volume of the set Q′ √βn which denoted by Vol Q′ √βn i.e,   card (Wβn ) ≤ Vol Q′ √βn . 9

  Now we find an upper bound of Vol Q′ √βn . First, we have  Z  dz1 dz2 ...dzd . Vol Q′ √βn = Q′ √β

(2.25)

n

We give the following coordinate system as follows z1 = r cos(θ1 ), z2 = r sin(θ1 ) cos(θ2 ), z3 = r sin(θ1 ) sin(θ2 ) cos(θ3 ), ... zd−1 = r sin(θ1 )... sin(θd−2 ) cos(θd−1 ) zd = r sin(θ1 )... sin(θd−2 ) sin(θd−1 ) where 1≤r≤

p βn , 0 ≤ θi ≤ π, i = 1.d − 2, 0 ≤ θd−1 < 2π.

From the Change of Variables formula that the rectangular volume element dz1 dz2 ...dzd can be written in spherical coordinates as  ∂x  i drdθ1 ...dθd−1 dz1 dz2 ...dzd = det ∂(r, θj ) =r

d−1

d−2

sin

1≤i≤d,1≤j≤d−1 (θ1 ) sind−3 (θ2 )... sin(θd−2 )drdθ1 ...dθd−1 .

Hence, applying Fubini’s theorem, we obtain that   Vol Q′ √βn Z = dz1 dz2 ...dzd Wρn (n)

=

Z

2π

0

=

Z 2π

=

Z

π

...

0

2π

Z

0

dθd−1 0



d/2 βn

d

π

Z

√

1

rd−1 sind−2 (θ1 ) sind−3 (θ2 )... sin(θd−2 )drdθ1 ...dθd−2 dθd−1 ! Z  Z  Z

√ βn

 Z 

βn

1

− 1 Z

π

rd−1 dr

0

π

sind−2 (θ1 )dθ1

0

 Z

0

π

 Z sind−3 (θ2 )dθ2 ...

π

0

0

π

sind−3 (θ2 )dθ2 ...

0

Thanks to page 245, [15], we know that Z π  Z π  Z sind−2 (θ1 )dθ1 sind−3 (θ2 )dθ2 ... 0

π

sind−2 (θ1 )dθ1

sin(θd−2 )dθd−2 0

 sin(θd−2 )dθd−2 .

π

sin(θd−2 )dθd−2

0



=



(2.26)

  d Vol Bd (1) . 2π

(2.27)

  Here Vol Bd (1) denotes the volume of of the unit d-ball. Using again [15] ( Proposition 4.2, page 246-247), we obtain that   π d/2 Vol Bd (1) = . (2.28) Γ( d2 + 1)

Combining (2.26), (2.27), (2.28) gives

 π d/2 β d/2  2π d/2 d/2 n Vol Q′ √βn ≤ = β dΓ(d/2) n Γ( d2 + 1)

which we have used the fact that Γ( d2 + 1) = d2 Γ( d2 ). This together with (2) leads to d o n X p2k ≤ βn = β(n1 , ...nd ) p = (p1 , ...pd ) ∈ Nd , pk ≥ 1, k = 1, d : card (Wβn ) =card k=1

2π d/2 d/2 β . ≤ dΓ(d/2) n

(2.29)

! (2.30)

10

This implies that d/2

2π d/2 βn . Q dΓ(d/2) dk=1 nk

2 EA1,1 ≤ 8π d Vmax

(2.31)

Next, in order to estimate A1,2 , we need to find an upper bound of Γn,p . Using Lemma 2.5, we estimate Γn,p as follows d

2

2

1 +..ld 6=0 l∈N ,lX Hr = Γn,p ≤

r=2l·n±p

−

≤d





H

max(µ1 ,...µd ) 2

H2l1 n1 ±p1 ,...2ld nd ±pd

X

l∈Nd ,l21 +..l2d 6=0

Hµ0 (Ω)

X

l∈Nd ,l21 +..l2d 6=0

d Y

k=1

(2lk nk ± pk )−µk .

(2.32)

Furthermore, for pk = 1, nk for k = 1, d, we have d Y

d Y

(2lk nk + pk )µk ≥

k=1

and d Y

k=1

(2lk nk − pk )µk ≥

k=1

Combining (2.33) and (2.34) gives d Y

d Y

(nk )µk

k=1

k=1

d Y

(2lk )µk

(2.33)

k=1

(2lk nk − nk )µk ≥

(2lk nk ± pk )−µk ≤

d Y

d Y

(nk )µk

k=1

d Y

(nk )−2µk

k=1

k=1

d Y

k=1

(2lk − 1)µk .

(2lk − 1)−2µk .

(2.34)

(2.35)

This together with (2.32) implies that

max(µ1 ,...µd )

2

H Γn,p ≤ d−

Hµ0 (Ω)

Since µk > 21 , we know that the series d− define this sum to be C(µ1 , ...µd ). Hence

d Y

X

(nk )−2µk

l∈Nd ,l21 +..l2d 6=0

k=1

P

max(µ1 ,...µd ) 2

l∈Nd ,l21 +..l2d 6=0





Γn,p ≤ C(µ1 , ...µd ) H

Hµ0 (Ω)

It follows from (2.30) that

A1,2 = 8

2 X Γn,p

d Y

Qd

d Y

k=1

(2lk − 1)−2µk .

k=1 (2lk

(2.36)

− 1)−2µk converges. So we

(nk )−2µk .

(2.37)

k=1

p∈Wβn

2

≤ 8C 2 (µ1 , ...µd ) H µ H

0 (Ω)

d Y

(nk )−4µk card (Wβn )

k=1

d

Y 16C 2 (µ1 , ...µd )π d/2

2 βnd/2 (nk )−4µk . ≤

H µ dΓ(d/2) H 0 (Ω)

(2.38)

k=1

For A1,3 on the right hand side of (2.22), noting that |p|2 ≥ βn if p ∈ / Wβn , we have the following estimate A1,3 = 4

X

p∈W / βn

2 2

|p|−2µ0 |p|2µ0 Hp ≤ 4βn−µ0 H µ H

11

0 (Ω)

.

(2.39)

Combining (2.22), (2.23), (2.38), (2.39), we obtain

2

b E H βn − H 2 L (Ω)

≤ EA1,1 + A1,2 + A1,3

d d/2

Y 2π d/2 16C 2 (µ1 , ...µd )π d/2 βn

2 d/2 + β (nk )−4µk

H

Q n d2d Γ(d/2) dk=1 nk dΓ(d/2) Hµ0 (2Ω) k=1

2

+ 4βn−µ0 H µ 2 ≤ 8π d Vmax

H

0 (Ω)

≤ C(µ1 , ...µd , H)βnd/2

d Y

2

(nk )−4µk + 4βn−µ0 H µ H

k=1

0 (Ω)

.

By a similar method used in the first part of the proof we can apply the previous results using Lemma 2.4, we immediately obtain the second error estimation.  3. Backward problem for parabolic equation with constant coefficients In this section, we consider the problem of recovering u(x, t), 0 ≤ t < T, such that    ut (x, t) + Au(x, t) = F (u(x, t)) + G(x, t), 0 < t < T, x ∈ Ω, u(x, t) = 0, x ∈ ∂Ω,   u(x, T ) = H(x),

(3.1)

where H ∈ L2 (Ω) is a given function. The operator A solves the following eigenvalue problem Aψp (x) = M(|p|)ψp (x),

for a non-decreasing function M and the eigenfunctions ψp (x) defined in (2.1). Now, we give some examples of operators A defined by the spectral theorem using the Laplacian in Ω and the corresponding eigenvalues. Example 3.1. (a) If A = −∆ in Ω with Dirichlet boundary conditions then Problem (3.1) is called nonlinear heat equation and its eigenvalues are M(|p|) = |p|2 = p21 + ... + p2d .

In this case the eigenfunctions are given by equation (2.1). (b) If A = ∆2 in Ω with Dirichlet boundary conditions then Problem (3.1) is called nonlinear biharmonic heat equation (see [27]) and using the spectral theorem its eigenvalues are  2 M(|p|) = |p|4 = p21 + ... + p2d .

In this case again, the eigenfunctions are given by equation (2.1). (c) If A = −∆ + ∆2 in Ω with Dirichlet boundary conditions then Problem (3.1) is called extended Fisher-Kolmogorov equation (see [19, 21]) and using the spectral theorem its eigenvalues are 2  M(|p|) = |p|2 + |p|4 = p21 + ... + p2d + p21 + ... + p2d .

In this case again, the eigenfunctions are given by equation (2.1). (d) If A = 2∆+∆2 in Ω with Dirichlet boundary conditions then Problem (4.1) is called Swift-Hohenberg equation (see [14]) and using the spectral theorem its eigenvalues are 2    M(|p|) = −2|p|2 + |p|4 = −2 p21 + ... + p2d + p21 + ... + p2d . In this case again, the eigenfunctions are given by equation (2.1).

Proposition 3.1. If Problem (3.1) has a unique solution u then it satisfies that Z T i Xh (T −t)M(|p|) e Hp − e(τ −t)M(|p|)Gp (τ )dτ ψp (x) u(x, t) = t

p∈Nd

12

−

X hZ

T

t

p∈Nd

i e(τ −t)M(|p|)Fp (u)(τ )dτ ψp (x).

(3.2)





 Where Gp (τ ) = G(·, τ ), ψp , Hp = H, ψp , Fp (u)(τ ) = F (u(·, τ )), ψp are the Fourier coefficients.

Proof. The proof is similar to Theorem 3.1, page 2975, [39] and we omit it here.



n Let’s define PM asothe operator of orthogonal projection onto the eigenspace spanned by the set ψp (x), 1 ≤ |p|2 ≤ M for a given M i.e, E X D PM v(x) := ψp (x), for all v ∈ L2 (Ω). (3.3) v, ψp L2 (Ω)

|p|2 ≤M

Our method in this subsection is described as follows: First, we approximate the two functions H and G by Hβn and Gβn which are defined in Theorem (2.1). Then we use the Fourier truncation method by adding the operator Pρn and introducing the following regularized problem  ∂U ρn ,βn (x, t)   + AU ρn ,βn (x, t) = Pρn F (U ρn ,βn (x, t)) + Pρn G(x, t), 0 < t < T,   ∂t (3.4) U ρn ,βn (x, t) = 0, x ∈ ∂Ω,     b βn (x), U ρn ,βn (x, T ) = Pρn H

where ρn is called the regularization parameter. The function ρ : Nd → R is a function that depend on βn and we use the notation ρn = ρ(n). Noting that limn→0 ρn = ∞. Define the set Wρn for any n = (n1 , ..nd ) ∈ Nd d n o X Wρn = p = (p1 , ...pd ) ∈ Nd : |p|2 = p2k ≤ ρn = ρ(n1 , ...nd ) .

(3.5)

k=1

Since the solution of Problem (3.4) depends on two terms βn and ρn , we denote it by U ρn ,βn . Theorem 3.2. Suppose that βn := β(n1 , n2 , ...nd ), ρn := ρ(n1 , n2 , ...nd ) are such that lim

|n|→+∞

βn =

lim

|n|→+∞

ρn = +∞,

lim

|n|→+∞

e2T M( Qd

√ ρn ) d/2 βn

4µk k=1 (nk )

=

lim

|n|→+∞

e2T M(

√ ρn ) −µ0 βn

= 0,

(3.6)

qP d 2 Suppose that where we recall |n| = k=1 nk . Assume that H, G, Hβn , Gβn are as in Theorem 2.1. ∞ F ∈ L (R) and F is a Lipschitz function, i.e. there exists a positive constant K such that |F (ξ1 ) − F (ξ2 )| ≤ K|ξ1 − ξ2 |,

∀ξ1 , ξ2 ∈ R.

(3.7)

2

The Problem (3.4) has a unique solution U ρn ,βn ∈ C([0, T ]; L (Ω)) which satisfies Z T i X h b βn ,p − bβn ,p (τ )dτ ψp (x) e(T −t)M(|p|) H e(τ −t)M(|p|)G U ρn ,βn (x, t) = p∈Wρn

−

X

p∈Wρn

t

"Z

T

e

(τ −t)M(|p|)

t

#

Fp (U ρn ,βn )(τ )dτ ψp (x).

(3.8)

E E D D b βn ,p = H b βn , ψp . bβn ,p (t) = G b βn (., t), ψp and H where G

(a) Assume that the problem (3.1) has a unique solution u such that X e1 < ∞, for all t ∈ [0, T ]. e2tM(|p|) u2p (t) := A

(3.9)

p∈Nd

2

Then as |n| → ∞, E U ρn ,βn (., t) − u(., t) e−2tM(

√ ρn )

max

is of order

L2 (Ω) ! √ d/2 √ e2T M( ρn ) βn 2T M( ρn ) −µ0 ,e βn , 1 . Qd 4µk k=1 (nk ) 13

(3.10)

(b) Assume that the problem (3.1) has unique solution u such that X M(|p|) α e2tM(|p|) u2p (t) := A e2 < ∞,

(3.11)

p∈Nd

2

for any α > 0 and t ∈ [0, T ]. Then as |n| → ∞, E U ρn ,βn (., t) − u(., t)

is of order

L2 (Ω)

e

√ −2tM( ρn )

e2T M( Qd

max

√ ρn ) d/2 βn

4µk k=1 (nk )

√ −2α √ , e2T M( ρn ) βn−µ0 , M( ρn )

!

.

(3.12)

for all t ∈ [0, T ]. (c) Assume that the problem (3.1) has a unique solution u such that X e3 < ∞, e2(t+δ)M(|p|)u2p = A

(3.13)

p∈Nd

2

for any real number δ ≥ 0 and t ∈ [0, T ]. Then as |n| → ∞, E U ρn ,βn (., t) − u(., t)

L2 (Ω)

e

√ −2tM( ρn )

e2T M( Qd

max

√ ρn ) d/2 βn

√ √ , e2T M( ρn ) βn−µ0 , e−2δM( ρn )

4µk k=1 (nk )

!

.

is of order (3.14)

Proof of Theorem 3.2. We divide the proof into some smaller parts. Part 1. The nonlinear integral equation (3.8) has unique solution U ρn ∈ C([0, T ]; L2 (Ω)). The proof is similar to [38]( See

Theorem 3.1, page 2975 [39]). Hence, we omit it here.

Part 2. The error estimate E U ρn ,βn − u 2 . L

First, using Parseval’s identity, equations (3.2), and (3.8) we get # "

2

 2  X

(T −t)M(|p|) b Hβn ,p − Hp e ≤4

U ρn ,βn − u L2 (Ω)

p∈Wρn

+4

X

"Z

X

T

  bβn ,p (τ ) − Gp (τ )dτ e(τ −t)M(|p|) G

e

(τ −t)M(|p|)

t

p∈Wρn

+4

T

t

p∈Wρn

+4

"Z

X 2 up .

#2


Fp (U ρn ,βn ) − Fp (u)(s) ds

#2 (3.15)

p∈W / ρn

Using the Cauchy-Schwartz inequality, the expectation of the right hand side of (3.15) is bounded by

2

E U ρn ,βn − u 2 L (Ω)    i2 X h √ b βn ,p − Hp  ≤ 4e2(T −t)M( ρn ) E  H p∈Wρn

+4

Z

T

e

√ 2(τ −t)M( ρn )

t

+ 4(T − t)

Z

T

dτ E 

e2(τ −t)M(

t

X 2 up . +4



X hZ

t

p∈Wρn

√ ρn )

X

p∈Wρn

T

E



h

 2 i b βn ,p (τ ) − Gp (τ ) dτ  G


i2 ds Fp (U ρn ,βn )(τ ) − Fp (u)(τ ) (3.16)

p∈W / ρn

14

It follows from the Lipschitz property of F that

2

2 √

b

E U ρn ,βn − u ≤ 4e2(T −t)M( ρn ) E H βn − H L2 (Ω) L2 (Ω)

2 √

b

+ 4T e2(T −t)M( ρn ) E G βn − G ∞ Z

L

(0,T ;L2 (Ω))

2

√

+ 4T K 2 e2(τ −t)M( ρn ) E U ρn ,βn (., τ ) − u(., τ ) 2 dτ L (Ω) t X 2 up . +4 T

(3.17)

p∈W / ρn

Now, we deal with the three cases. Case 1. Assume that the series

X

e2tM(|p|)u2p

(3.18)

p∈Nd √

e1 . Then multiplying both sides of the inequality (3.17) by e2tM( ρn ) , we obtain that converges to A

2 √

e2tM( ρn ) E U ρn ,βn (., t) − u(., t) 2 L (Ω)  

2

2

√

b

b e1 + 4A + T E G ≤ 4e2T M( ρn ) E H βn − H βn − G + 4T K 2

Z

L2 (Ω)

T

L∞ (0,T ;L2 (Ω))

2 √

e2τ M( ρn ) E U ρn ,βn (., τ ) − u(., τ )

L2 (Ω)

t

dτ,

where we used the fact that X X 2 2 √ e1 . up = e−2tM(|p|) e2tM(|p|) up ≤ e−2tM( ρn ) A p∈W / ρn

p∈W / ρn

Above we used the monotone increasing property of M. Using the above Gronwall’s inequality, we obtain that

2

√

e2tM( ρn ) E U ρn ,βn − u 2 L (Ω) 

2

2 √

b

b

≤ 4e2T M( ρn ) E H + T E G βn − H 2 βn − G ∞ L (Ω)

L

(0,T ;L2 (Ω))

e1 e4T K 2 (T −t) . + 4A

This implies that

2

E U ρn ,βn − u 2

L (Ω)

≤ 4e |

√ 4T K (T −t) 2(T −t)M( ρn ) 2

{z

C52

}e

e1 e4T K 2 (T −t) e−2tM( + 4A

√ ρn )



2

b

E H βn − H 2

L (Ω)

e

L (Ω)

max

Case 2. Suppose that the series

e2T M( Qd

√ ρn ) d/2 βn

4µk k=1 (nk )

15

2

(T −t)

(3.19)

L

(0,T ;L2 (Ω))

 (3.20)

is of order

√ , e2T M( ρn ) βn−µ0 , 1

2α X M(|p|) e2tM(|p|) u2p

p∈Nd

e4T K

2

b

+ T E G βn − G ∞

.

2

It follows from Corollary (2.1) that E U ρn ,βn (., t) − u(., t) 2 √ −2tM( ρn )



!

.

(3.21)

(3.22)

e2 . By a similar technique as in case 1 above and using the following estimate converges to A −2α 2α X X 2 2 up = e−2tM(|p|) M(|p|) e2tM(|p|) up M(|p|) p∈W / ρn

p∈W / ρn

we deduce that

2

E U ρn ,βn − u 2

√ −2α √ e2 , ≤ M( ρn ) e−2tM( ρn ) A

L (Ω)

2

4T K (T −t) 2(T −t)M( ≤ 4e {z }e |

√ ρn )



2

b − H E H

βn

L2 (Ω)

C52

2

b − G + T E G

βn

L∞ (0,T ;L2 (Ω))

√ −2α √ e−2tM( ρn ) . M( ρn )

2

It follows from Corollary (2.1) that E U ρn ,βn (., t) − u(., t) 2 e2 e4T K + 4A

2

(T −t)

L (Ω)

e

√ −2tM( ρn )

e2T M( Qd

max

√ ρn ) d/2 βn

4µk k=1 (nk )

Case 3. Suppose that the series

X

(3.23)

 (3.24)

is of order

√ −2α √ , e2T M( ρn ) βn−µ0 , M( ρn )

!

.

(3.25)

e2(t+δ)M(|p|) u2p

p∈Nd

e3 . By a similar technique as in case 1 above and using the following estimate converges to A X X 2 up = e−2(t+δ)M(|p|)e2(t+δ)M(|p|) u2p p∈W / βn

p∈W / ρn

≤ e−2(t+δ)M(

we deduce that

2

E U ρn ,βn − u 2

L (Ω)

2

4T K (T −t) 2(T −t)M( ≤ 4e {z }e |

√ ρn )

C52

√ βn )



2

b − H E H

βn

L2 (Ω)

√

e3 e4T K 2 (T −t) e−2(t+δ)M( ρn ) . + 4A

2

It follows from Corollary (2.1) that E U ρn ,βn − u

L2 (Ω)

e

√ −2tM( ρn )

max

e3 , A

e2T M( Qd

√ ρn ) d/2 βn

4µk k=1 (nk )

2

b − G + E G

βn

L∞ (0,T ;L2 (Ω))



is of order

√ √ , e2T M( ρn ) βn−µ0 , e−2δM( ρn )

!

.

Since

e2T M( Qd

k=1

ρn ) d/2 βn (nk )4µk

(3.27) 

Remark 3.1. We√ give one choice for βn and ρn which satifies (3.6). Let 0 < 2α0 < µ0 and e βn2α0 .

(3.26)

√ 2T M( ρn )

=

→ 0 when |n| → +∞, we can choose βn such that 2α0 +d/2

lim

|n|→+∞ 2α0 +d/2

Let us choose βn

=

Qd

k=1 nk then βn =

βn Qd

4µk k=1 (nk )

Q d

k=1 nk

 2α

= 0.

1 0 +d/2

α0 α0 √ log (βn ) = M( ρn ) = T T (2α0 + d/2) 16

(3.28)

and then we choose ρn such that ! d Y (3.29) log nk . k=1

2

In above theorem, for the case (b), E U ρn ,βn (., t) − u(., t)

L2 (Ω)

d Y

0t ! 4T−4α α +dT 0

nk

k=1

max

Qd

1

4µk −1 k=1 (nk )

,

d Y

2

In above theorem, for the case (c), E U ρn ,βn (., t) − u(., t) 2

2α0 −µ0 ! 2α +d/2

L (Ω)

d Y

k=1

0t ! 4T−4α α +dT 0

nk

max

Qd

1

4µk −1 k=1 (nk )

,

d Y

k=1

nk

d Y

0

nk

k=1

is of order

, log

−2α

nk

k=1

!!

.

(3.30)

is of order

2α0 −µ0 ! 2α +d/2 0

,

d Y

k=1

0δ ! ! 4T−4α α +dT 0

nk

.

(3.31)

4. The backward problem for parabolic equation with time dependent coefficients 4.1. The problem with coefficients that depend only on t. In this section, we consider the problem of constructing a solution u ∈ C([0, T ; V(Ω)]), u′ ∈ L2 (0, T ; L2(Ω)) such that u satisfies the following parabolic equation with time dependent coefficients    ut + A(t)u = F (u(x, t)) + G(x, t), x ∈ Ω, 0 < t < T, u(x, t) = 0, x ∈ ∂Ω, (4.1)   u(x, T ) = H(x), x ∈ Ω d

where H ∈ L2 (Ω). Here V(Ω) ֒→ L2 (Ω); i.e., V(Ω) ⊂ L2 (Ω) is continuously embedded into L2 (Ω). It means that there exists some constant m0 > 0 such that for all v ∈ V(Ω) kvkL2 (Ω) ≤ m0 kvkV(Ω) .

(4.2)

|F (u) − F (v)| ≤ KF (Q) |u − v| ,

(4.3)

 d Then L2 (Ω) ֒→ V ′ (Ω) via v 7→ v | · L2 (Ω) , where V ′ (Ω) denotes the dual space of V(Ω). In this section, we assume that the source function F : R → R is a locally Lipschitz function i.e, for each Q > 0 and for any u, v satisfying |u|, |v| ≤ Q, there holds where

  F (u) − F (v) : |u| , |v| ≤ Q, u 6= v < +∞. K(Q) := sup u−v

We note that the function Q → K(Q) is increasing and

(4.4)

lim K(Q) = +∞.

Q→+∞

First, we give the following definition

Definition 4.1. The pair of operators (A(t), P) satisfies Assumption (A) if the following conditions hold (a) For any v ∈ L2 (Ω), there exists an increasing function M : R → R+ such as X 

M (|p|) v, ψp L2 (Ω) ψp (x), v ∈ L2 (Ω). Pv(x, t) =

(4.5)

p∈Nd

(b) For u ∈ V(Ω), v ∈ V(Ω)

D

(P − A(t))u, v

E

L2 (Ω)

fa kukV(Ω) kvkV(Ω) , ≤M

(4.6)

ckvk2 . ≥M V(Ω)

(4.7)

fa > 0. for some constant M c > 0 such that (c) For any v ∈ V(Ω), there exists M D E (P − A(t))v, v 2

L (Ω)

Now we state the following lemma concerning an estimate of Pρn . 17

Lemma 4.1. Let Pρn be defined as follows X 

Pρn v(x) = M (|p|) v, ψp L2 (Ω) ψp (x),

for all

p∈Wρn

The operator Pρn is a linear, bounded operator, and satisfies that √ kPρn kL(L2 (Ω);L2 (Ω)) ≤ M ( ρn ), 2

2

v ∈ L2 (Ω).

(4.8)

(4.9) 2

2

where L(L (Ω); L (Ω)) is the space of all bounded linear operators from L (Ω) to L (Ω).

Proof. For v ∈ L2 (Ω), since M is a non-decreasing function, we have 2

X 2 kPρn vk2L2 (Ω) = M (|p|) v, ψp L2 (Ω) p∈Wρn

≤

√ 2 X

2 v, ψp L2 (Ω) ≤ M ( ρn ) p∈Wρn

√ 2 M ( ρn ) kvk2L2 (Ω) .

(4.10) 

2

Definition 4.2. Define a subspace of L (Ω) as follows ) ( 2γ X 2 2T M(|p|)

2 v, ψp L2 (Ω) < ∞ Gγ (Ω) := v ∈ L (Ω) : M (|p|) e

(4.11)

p∈Nd

for any γ ≥ 0. The norm of v ∈ Gγ (Ω) is given by v uX 2γ 2

u kvkGγ (Ω) = t M (|p|) e2T M(|p|) v, ψp L2 (Ω) .

(4.12)

p∈Nd

Lemma 4.2. For v ∈ G1+γ (Ω), γ ≥ 0 Proof. We have

−γ √

√ kPρn v − Pvk ≤ M (| ρn |) e−T M ( ρn ) v G1+γ (Ω) .

kPρn v − Pvk2L2 (Ω) = =

(4.13)

2

X 2 M (|p|) v, ψp L2 (Ω)

p∈W / ρn

−2γ 2+2γ X 2

e−2T M(|p|) M (|p|) e2T M(|p|) v, ψp L2 (Ω) M (|p|)

p∈W / ρn

−2γ √

2 √ e−2T M( ρn ) v G1+γ (Ω) . ≤ M (| ρn |)



4.1.1. The regularized solution and convergence rates. Since P is an unbounded operator on L2 (Ω), we approximate it by the following operator Pρn defined above in equation (4.8). Since A(t) is an unbounded operator, we approximate it by a new approximate operator A(t) − P + Pρn . Moreover since F is a locally Lipschitz source function, we approximate F by FQ defined by   w(x, t) > Q, F (Q), (4.14) FQ (w(x, t)) = F (w(x, t)), −Q ≤ w(x, t) ≤ Q,   F (−Q), w(x, t) < −Q.

for any Q > 0. In the sequel we use a parameter Qn := Q(n1 , n2 , ...nd ) → +∞ as |n| → +∞. So, when n large enough, we have that Qn ≥ kukL∞ (0,T ;L2 (Ω)) . Moreover, we also have FQn (u(x, t)) = F (u(x, t)), for |n| large enough.

(4.15)

kFQn (v1 ) − FQn (v2 )kL2 (Ω) ≤ 2K(Qn )kv1 − v2 kL2 (Ω) , v1 , v2 ∈ L2 (Ω).

(4.16)

Using observation on p. 1250 in [40], we also obtain that FQn is a globally Lipschitz source function in the following sense 18

We consider a regularized problem below  ∂U ρn ,βn   + A(t)U ρn ,βn − PU ρn ,βn + Pρn U ρn ,βn   ∂t    b βn (x, t), 0 < t < T, = FQn (U ρn ,βn (x, t)) + G    U ρn ,βn (x, t) = 0, x ∈ ∂Ω,     b βn (x). U ρn ,βn (x, T ) = H

(4.17)

In the following Theorem, we obtain the existence, uniqueness and continuous dependence of the solutions for the proposed problem. We state the error estimation between the regularized solution and the exact solution. Our main result in this section is as follows Theorem 4.1. Let H, G, Hβn , Gβn be as in Theorem 2.1. Let us choose βn , ρn such that lim

|n|→+∞

e2T M(

√

ρn ) d/2 βn

d Y

(nk )−4µk =

k=1

lim

|n|→+∞

e2T M (

√ ρn ) −µ0 βn

= 0.

(4.18)

Then Problem (4.17) has a unique solution U ρn ,βn ∈ C([0, T ]; L2(Ω)) ∩ L2 (0, T ; V(Ω)). Assume that Problem (4.1) has unique solution u ∈ C([0, T ]; L2 (Ω)) ∩ L∞ (0, T ; G1+γ (Ω)) for any γ ≥ 0. Choose Qn such that (4.19) lim e4K(Qn )T Π(n) = 0 |n|→+∞

where Π(n) = max

√ e2T M( ρn ) βnd/2

d Y

(nk )−4µk , e2T M(

k=1

Then as |n| → ∞ the error

2

E U ρn ,βn (., t) − u(., t) 2

is of order

L (Ω)

e



√

! −2γ √ ρn ) −µ0 . βn , M (| ρn |) 

4K(Qn )+2 (T −t) −2tM (√ρ ) n

e

Π(n).

Remark 4.1. Thanks to Remark (3.1), we give one choice for βn as follows 1 ! 2α +d/2 d 0 Y βn = nk

(4.20)

(4.21)

(4.22)

k=1

where 0 < α0
1, k = 1, d, using (4.24), we deduce that lim|n|→∞ Π(n) = 0. Hence, we need to choose n large enough such that Π(n) < 1. So, the equality (4.26) is suitable which leads to a chosen Qn . We state the two corollaries of the Theorem 4.1 next 19

Corollary 4.1. Let us take two functions Γ0 , Γ1 which are continuous functions on [0, T ]. Assume that   m0 = min min |Γ0 (t)|, min |Γ1 (t)| > 0 (4.27) 0≤t≤T

and

m1 = max



0≤t≤T

 max |Γ0 (t)|, max |Γ1 (t)| > 0.

0≤t≤T

0≤t≤T

(4.28)

In Problem (4.1), let A(t)u = −Γ0 (t)∆u + Γ1 (t)∆2 u and F (u) = u − u3 . Then we get the backward in time problem for extended Fisher-Kolmogorov equation with time dependent coefficients as follows  2 3   ut − Γ0 (t)∆u + Γ1 (t)∆ u = u − u + G(x, t), x ∈ Ω, 0 < t < T, u(x, t) = ∆u(x, t) = 0, x ∈ ∂Ω, 0 ≤ t ≤ T, (4.29)   u(x, T ) = H(x), x ∈ Ω.

From Definition (4.1), let the operator P be as follows X
Pv := m1 p2 + p4 < v, ψp > ψp .

(4.30)

p∈Nd

for any v ∈ L2 (Ω). It is easy to show that the pair of operators (A(t), P) as above satisfies
Assumption (A) in Defintion (4.1). It is easy to see that the eigenvalues of P are M (p) = m1 p2 + p4 . Next, we find the operator Pρn by truncating Fourier series in (4.63) and we have  X  (4.31) p2 + p4 < v, ψp > ψp (x). Pρn v := m1 |p|≤

q

ρn m1

Thanks to (4.17), a regularized problem for (4.29) is given below  ∂U ρn ,βn   − Γ0 (t)∆U ρn ,βn + Γ1 (t)∆2 U ρn ,βn − PU ρn ,βn + Pρn U ρn ,βn   ∂t    b βn (x, t), 0 < t < T, = FQn (U ρn ,βn (x, t)) + G    U ρn ,βn (x, t) = 0, x ∈ ∂Ω,     b βn (x), U ρn ,βn (x, T ) = H where

FQn

 Q − Q3n ,
 n
3 U ρn ,βn (x, t) = U ρn ,βn − U ρn ,βn ,   −Qn + Q3n ,

(4.32)

U ρn ,βn (x, t) > Qn ,

−Qn ≤ U ρn ,βn (x, t) ≤ Qn , U ρn ,βn (x, t) < −Qn .

Assume that H, G, u be as in Theorem (4.1). Let βn be as in (4.22). Let ρn and Qn be such that ! d Y 2α 0 ρn + ρ2n = log (4.33) nk m1 T (4α0 + d) k=1

and


δ0 − 1 (4.34) log Π(n) 4T respectively. The last two equations can be solved by a simple way that leads to the value of ρn and Qn . Then the error between the solution u of Problem (4.29) and the solution U ρn ,βn of Problem (4.32) is of order −4α0 δδ0 0 δ0 −µ0 δ0 0t ! 2α2α ! 2T α−2α " # d ! 4T α0 +dT d d 0 +d/2 0 +dT /2 Y Y Y 1 , max Qd nk nk , . (4.35) nk 4µk δ0 −δ0 k=1 (nk ) k=1 k=1 k=1 1 + 3Q2n =

It is easy to check that the term in (4.35) tends to zero as |n| → +∞. 20

4.1.2. Proof of the main results. Proof of Theorem 4.1. Step 1. The existence and uniqueness of the regularized problem. We refer to the proof of Theorem 4.2 where we prove existence and uniqueness for the more general operator A(t, u) which is more general than the operator A(t) in Theorem 4.1 and A(t, u) = A(t). Step 2. Regularity of the regularized solution U ρn ,βn . Let us define the function eρn ,βn (x, t) = eκn (t−T ) U ρn ,βn (x, t), (4.36) U eρn ,βn (x, t) with where κn is positive constant to be selected later. By taking the partial derivative of U respect to t, we obtain eρn ,βn ∂U ∂U ρn ,βn (x, t) = eκn (t−T ) (x, t) + κn eκn (t−T ) U ρn ,βn (x, t) ∂t ∂t  

= −eκn (t−T ) A(t) − P + Pρn U ρn ,βn (x, t) + eκn (t−T ) FQn (U ρn ,βn (x, t))

b ρn (x, t) + κn eκn (t−T ) U ρn ,βn (x, t) + eκn (t−T ) G  eρn ,βn (x, t) eρn (x, t) + κn U = − A(t) − P + Pρn U 

b βn (x, t) + eκn (t−T ) FQn (U ρn ,βn (x, t)). + eκn (t−T ) G

eρn ,βn (x, t) gives Taking the inner product of both sides with U 1 ∂ e kUρn ,βn (., t)k2L2 (Ω) 2 ∂tD E
eρn ,βn (., t) eρn ,βn (., t), U = P − A(t) U |

{z

=:J1

E D eρn ,βn (., t) eρn ,βn (., t), U + Pρn U L2 (Ω) L2 (Ω) } | {z } =:J2

eρn ,βn (., t)k2 2 + κn k U L (Ω) E D E D κn (t−T ) b κn (t−T ) e eρn ,βn (., t) + e G . + e FQn (U ρn ,βn (., t)), U β n (., t), Uρn ,βn (., t) L2 (Ω) L2 (Ω) {z } | {z } | =:J3

(4.37)

=:J4

It remains to estimate J1 , J2 and J3 . By the conditions for P in Assumption (A) in Definition 4.1, we obtain that ckU eρn ,βn (., t)k2 . (4.38) J1 ≥ M V(Ω) The term J2 can be estimated as follows

√ 2 eρn ,βn (., t)k2 2 e |J2 | ≤ kPρn kL(L2 (Ω);L2 (Ω)) kU L (Ω) ≤ M ( ρn )kUρn ,βn (., t)kL2 (Ω) ,

(4.39)

where we have used the inequality (4.10). Using Cauchy-Schwarz inequality and noting the globally Lipschitz property of the function FQn , we deduce that E D eρn ,βn (., t) |J3 | = eκn (t−T ) FQn (U ρn ,βn (., t)), U 2 L (Ω)

1 e 1 2 ≤ e2κn (t−T ) kFQn (U ρn ,βn (., t))k2L2 (Ω) + kU ρn ,βn (., t)kL2 (Ω) 2 2 2  1 eρn ,βn (., t)kL2 (Ω) + kF (0)kL2 (Ω) + 1 kU eρn ,βn (., t)k2 2 ≤ e2κn (t−T ) 2K(Qn )kU L (Ω) 2 2   eρn ,βn (., t)k2 2 + e2κn (t−T ) kF (0)k2 2 . ≤ 2K 2 Qn + 1 kU L (Ω)

(4.40)

L (Ω)

where we note that we have used above the fact that FQn (0) = F(0). Using Cauchy-Schwartz inequality, we have the bound of |J4 | as follows D E eρn ,βn (., t) b βn (., t), U |J4 | = eκn (t−T ) G L2 (Ω) (4.41) 1 2κn (t−T ) b 1 e 2 2 ≤ e kGβn (., t)kL2 (Ω) + kUρn ,βn (., t)kL2 (Ω) . 2 2 21

Combining (4.38), (4.39),(4.40),(4.41) we have that 1 ∂ e ckU eρn ,βn (., t)k2 kUρn ,βn (., t)k2L2 (Ω) ≥ M V(Ω) 2 ∂t √ 2 eρn ,βn (., t)k2 2 e + κn k U L (Ω) − M ( ρn )kUρn ,βn (., t)kL2 (Ω)   eρn ,βn (., t)k2 + e2κn (t−T ) kF (0)k2 2 . − 2K 2 (Qn ) + 1 kU H L (Ω)

Integrating the last inequality over [t, T ] yields

2 2κn (t−T ) b βn (., t)k2 2 eρn ,βn (., T )k2 2 EkG EkU L (Ω) + (T − t)kF (0)kL2 (Ω) L (Ω) + e # "Z T c eρn ,βn (., τ )k2 dτ eρn ,βn (., t)k2 2 kU ≥ EkU V(Ω) L (Ω) + 2M E t

+

Z

t

T



 √ eρn ,βn (., τ )k2 2 dτ. 2κn − 2M ( ρn ) − 4K 2 (Qn ) − 2 EkU L (Ω)

√ By choosing κn = M ( ρn ), we derive that  Z √ 2 2 2(t−T )M( ρn ) EkU ρn ,βn (., t)kL2 (Ω) ≤ 4K (Qn ) + 2 e b βn k2 2 + EkH L (Ω) + e

T

e2(τ −T )M(

t √ 2(t−T )M ( ρn )

+ (T − t)kF (0)k2L2 (Ω) . √ 2T M( ρn )

√

ρn )

eρn ,βn (., τ )k2 2 dτ EkU L (Ω)

b βn (., t)k2 2 EkG L (Ω)

(4.42)

, we get Multiplying both sides of the last inequality by e Z T   √ √ eρn ,βn (., τ )k2 2 dτ e2tM ( ρn ) EkU ρn ,βn (., t)k2L2 (Ω) ≤ 4K 2 (Qn ) + 2 e2τ M( ρn ) EkU L (Ω) +e

+e Noting that e2tM( we deduce that e2tM (

√ ρn )

√

ρn )

√ 2T M( ρn )

√ 2T M( ρn )

t

2tM ( b βn k2 2 EkH L (Ω) + e

(T − t)kF (0)k2L2 (Ω) .

2T M( b βn (., t)k2 2 EkG L (Ω) ≤ e

√

 Z EkU ρn ,βn (., t)k2L2 (Ω) ≤ 4K 2 (Qn ) + 2 + e2T M ( + e2T M (

√ ρn ) √ ρn )

√ ρn )

ρn )

T

b βn (., t)k2 2 EkG L (Ω)

b βn k2 ∞ EkG L (0,T ;L2 (Ω)) ,

e2τ M(

√ ρn )

t

eρn ,βn (., τ )k2 2 dτ EkU L (Ω)

2T M ( b βn k2 2 EkH L (Ω) + e

T kF (0)k2L2(Ω) .

√ ρn )

b βn k2 ∞ EkG L (0,T ;L2 (Ω))

(4.43)

Applying Gronwall’s inequality to the last inequality, we get √ ρn )

EkU ρn ,βn (., t)k2L2 (Ω)   h i √ 2 b βn k2 ∞ b βn k2 2 ≤ exp 4K 2 (Qn )(T − t) e2T M ( ρn ) EkH + T kF (0)k + Ek G 2 2 L (0,T ;L (Ω)) L (Ω) . L (Ω)

e2tM (

√

(4.44)

Multiplying both sides of the last inequality with e−2tM( ρn ) , we get the upper bound of EkU ρn (., t)k2L2 (Ω) which shows the stability of U ρn ,βn (., t) in the sense of the solution U ρn ,βn (., t) depend continuously on bβn and F . b βn , G the given data H Step 3. Error estimate between the regularized solution and the sought solution. It is easy to see that u satisfies

Putting

∂u(x, t) + Pρn u(x, t) = F (u(x, t)) + G(x, t) ∂t

+ Pρn − P u(x, t) + P − A(t) u(x, t). eρn ,βn (x, t) = U ρn ,βn (x, t) − u(x, t), Z 22

we have

Put


∂ e eρn ,βn (x, t) = FQn (U ρn ,βn (x, t)) − F (u(x, t)) − Pρn − P u(x, t) Zρn ,βn (x, t) + Pρn Z ∂t   bβn (x, t) − G(x, t). eρn ,βn (x, t) + G + P − A(t) Z Xρn ,βn (x, t) = eκn (t−T ) Zeρn ,βn (x, t).

Take the inner product of the both sides of the last equality by Xρn ,βn (x, t) and then by integrating with respect to the time variable, it follows that kXρn ,βn (., T )k2L2 (Ω) − kXρn ,βn (., t)k2L2 (Ω) Z TZ  Z T  = 2κn Pρn Xρn ,βn (x, τ ) Xρn ,βn (x, τ )dxdτ kXρn ,βn (., τ )k2L2 (Ω) dτ − 2 t {z } | t Ω + 2eκn (t−T ) |

+ 2eκn (t−T ) |

+ 2eκn (t−T ) | + 2eκn (t−T ) |

Z

T

t

Z

T

:=J5

Z

bβn (x, τ ) − G(x, τ )Xρn ,βn (x, τ )dxdτ G {z }

Ω

T

t

Z

FQn (U ρn ,βn (x, τ )) − F (u(x, τ ))Xρn ,βn (x, τ )dxdτ {z }

Ω

t

Z

:=J4

Z

:=J6

Z

eρn (x, τ )Xρn ,βn (x, τ )dxdτ (P − A(τ )) Z {z }

Ω

T

t

:=J7

Z

(Pρn − P) u(x, τ )Xρn ,βn (x, τ )dxdτ . {z }

Ω

(4.45)

:=J8

By the Cauchy-Schwartz inequality, the expectation of absolute of J4 is bounded by "Z s Z # 2  Z T   E J4 ≤ 2E |Xρn ,βn (x, τ )|2 dx dτ Pρn Xρn ,βn (x, τ ) dx t

Ω

√ ≤ 2M ( ρn )

Z

T

t

Ω

EkXρn ,βn (x, τ )k2L2 (Ω) dτ.

(4.46)

For |n| large enough, we recall that FQn (u) = F (u) and using the global Lipschitz property of FQn , we have the bound of J5 as follows by using Cauchy-Schwartz inequality "Z #



2 T



E J5 ≤ 2E FQn (U ρn ,βn (x, τ )) − F (u(x, τ )) 2 Xρn ,βn (x, τ ) dτ L (Ω)

t

= 2E

"Z

T

t

≤ 4K(Qn )

2



FQn (U ρn ,βn (x, τ )) − FQn (u(x, τ )) Xρn ,βn (x, τ )

Z

t

T

≤

dτ

EkXρn ,βn (., τ )k2L2 (Ω) dτ.

The term J6 is bounded by "Z # "Z 2  T Z b E J6 ≤ E Gβn (x, τ ) − G(x, τ ) dx dτ + E t

L2 (Ω)

#

Ω

t

2

b T E G βn (.) − G(.)

L∞ (0,T ;L2 (Ω))

+

Z

t

23

T

T

Z

Ω

(4.47)

 |Xρn ,βn (x, τ )| dx dτ

EkXρn ,βn (., τ )k2L2 (Ω) dτ.

2

# (4.48)

The term J7 is estimated using the Assumption (A) in Definition 4.1 as follows "Z # T D E E J7 = E (P − A(τ )) Xρn ,βn (., τ ), Xρn ,βn (., τ ) 2 dτ L (Ω)

t

cE ≥M

Z

T

kXρn ,βn (., τ )k2V(Ω) dτ

t

and using Lemma 4.2 "Z # Z T κn (τ −T ) E J8 = E 2e (Pρn − P) u(x, τ )Xρn (x, τ )dxdτ ≤

"Z

t

T

Ω

−2γ

2 √ √

e−2T M( ρn ) u(., τ ) M (| ρn |)

G1+γ (Ω)

t

#

dτ +

Z −2γ √

√ −2T M( ρn ) 2 e u L∞ (0,T ;G (Ω)) + ≤ T M (| ρn |) 1+γ

t

Z

T

t

T

EkXρn ,βn (., τ )k2L2 (Ω) dτ

EkXρn ,βn (., τ )k2L2 (Ω) dτ.

(4.49)

Combining (4.45), (4.46), (4.47),(4.48) (4.49), (4.49) gives

EkXρn ,βn (., T )k2L2 (Ω) − EkXρn ,βn (., t)k2L2 (Ω)  Z T √ ≥ 2κn − 2M ( ρn ) − 4K(Qn ) − 2 EkXρn ,βn (., τ )k2L2 (Ω) Z

t



2 −2γ √ √

cE +M kXρn ,βn (., τ )k2V(Ω) dτ − T M (| ρn |) e−2T M( ρn ) u ∞ L (0,T ;G1+γ (Ω)) t

2

b

− T E G . βn (.) − G(.) ∞ 2 T

L

(4.50)

(0,T ;L (Ω))

This leads to

2

e2κn (t−T ) E U ρn ,βn (., t) − u(., t) 2

L (Ω)

cE +M

Z  √ + 2κn − 2M ( ρn ) − 4K(Qn ) − 2

t

T

Z

t

T

kXρn ,βn (., τ )k2V(Ω) dτ

2

e2κn (τ −T ) E U ρn ,βn (., τ ) − u(., τ ) 2

L (Ω)

2

2

b

b

≤ E H − H + T E (.) − G(.) G

∞ βn βn L2 (Ω) L (0,T ;L2 (Ω)) −2γ √

√ 2 + T M (| ρn |) e−2T M( ρn ) u L∞ (0,T ;G1+γ (Ω)) .

dτ

√ √ Let us choose κn = M ( ρn ) and multiply both sides of the last inequality with e2T M( ρn ) , then we conclude that

2

√

e2tM( ρn ) E U ρn ,βn (., t) − u(., t) L2 (Ω)  

2

2 √

b

b

2T M( ρn ) ≤e E Hβn − H 2 + T E Gβn (.) − G(.) ∞ 2 L (Ω)

L

−2γ 2 √

+ T M (| ρn |)

u ∞ L (0,T ;G1+γ (Ω)) Z

2

  T √

+ 4K(Qn ) + 2 e2τ M( ρn ) E U ρn ,βn (., τ ) − u(., τ ) 2

(0,T ;L (Ω))

L (Ω)

t

The Gronwall’s inequality implies that

2

√

e2tM( ρn ) E U ρn ,βn (., t) − u(., t) L2 (Ω)   

2 4K(Qn )+2 (T −t) 2T M(√ρ )

b n E H e ≤e βn − H

L2 (Ω)

24

dτ.

2

b + T E G βn (.) − G(.)

L∞ (0,T ;L2 (Ω))

(4.51)



+e





4K(Qn )+2 (T −t)

−2γ 2 √

T M (| ρn |)

u

L∞ (0,T ;G1+γ (Ω))

Multiplying both sides of (4.52) with e−2tM( holds.

√ ρn )

.

(4.52)

and thanks to Corollary (2.1), we conclude that (4.21) 

4.2. General problem with coefficients that depend on t and u. Let H ∈ L2 (Ω). In this section, we consider the problem of constructing an u ∈ C([0, T ; V (Ω)]), u′ ∈ L2 (0, T ; L2 (Ω)) such that    ut + A(t, u)u = F (u(x, t)) + G(x, t), x ∈ Ω, 0 < t < T, u(x, t) = 0, x ∈ ∂Ω, (4.53)   u(x, T ) = H(x), x ∈ Ω. Let 0 < R < ∞. Define the following set n o BR (L2 (Ω)) := w ∈ L2 (Ω) : kwkL2 (Ω) ≤ R . (4.54) Definition 4.3. The pair of operators (A(t, w), P) satisfies Assumption (B) if the following conditions hold (a) For any v ∈ L2 (Ω), there exists a increasing function M such as X 

(4.55) M (|p|) v, ψp L2 (Ω) ψp (x). Pv(x) = p∈Nd

(b) For w ∈ BR (L2 (Ω)), u ∈ |mathcalV (Ω), v ∈ V(Ω) then D  E fa kukV(Ω)kvkV(Ω) . P − A(t, w) u, v ≤M L2 (Ω)

c > 0 such that (c) For w ∈ BR (L2 (Ω)) and any v ∈ V(Ω), there exists M D E ckvk2 . (P − A(t, w))v, v ≥M V(Ω) L2 (Ω)

f > 0 such that (d) There exists M D  E A(t, w1 ) − A(t, w2 ) u, v

L2 (Ω)

fkukV(Ω) kvkV(Ω) kw1 − w2 kL2 (Ω) , ≤M

(4.56)

(4.57)

(4.58)

for any u ∈ V(Ω), v ∈ V(Ω) and w1 , w2 ∈ BR (L2 (Ω)).

  Remark 4.2. We give some particular equations of the model (4.53). Let us choose A(t, u) = ∇ D(u)∇u

in the first equation of (4.53) then this equation is called a logistic reaction-diffusion equation. Here u represents the population density of species at location x and time t, D(u) is the density dependent diffusion coefficient, the notation ∇ is the usual gradient operator and F (u) is a logistic type source term. • When F (u) = au(1 − u) a > 0 , Problem (4.53) is called backward in time for Fisher-type logistic equations. (See [28]). • When F (u) = au2 (1−u) a > 0 , Problem (4.53) is called backward in time for Huxley equation. • When F (u) = au2 (1−u)(u−θ1) a > 0 , Problem (4.53) is called backward in time for FitzhughNagumo equation. Some more applications in biology of the above equations and generalized problem can be found in [7].

Using a similar method as in previous subsection, we present a regularized problem for Problem (4.53) as follows  ∂U ρn ,βn   + A(t, U ρn ,βn )U ρn ,βn − PU ρn ,βn + Pρn U ρn =   ∂t    b βn (x, t), 0 < t < T, FQn (U ρn ,βn (x, t)) + G (4.59)    U ρn ,βn (x, t) = 0, x ∈ ∂Ω,     b βn (x). U ρn ,βn (x, T ) = H 25

Theorem 4.2. Let H, G, u, βn , ρn be as Theorem 4.1. Then the system (4.59) has a unique solution U ρn ,βn ∈ C([0, T ]; L2 (Ω)) ∩ L2 (0, T ; V(Ω)). Choose Qn as in Theorem 4.1. Then for n large enough, the error   fM f

2 4M 0 (T −t) −2tM (√ρ ) 4K(Q )+2+ n

c M n e E U ρn ,βn (., t) − u(., t) 2 Π(n). (4.60) is of order e L (Ω)

Where the term Π(n) above is defined in equation (4.20).

Remark 4.3. One example for choices of βn , ρn , Qn are given in Remark (4.1). Corollary 4.2. Consider the following problem for Huxley equation    2    ut − ∇ D(u)∇u = u (1 − u) + G(x, t), x ∈ Ω, 0 < t < T, u(x, t) = 0, x ∈ ∂Ω,    u(x, T ) = H(x), x ∈ Ω

(4.61)

where the density dependent diffusion coefficient D satisfies that D0 ≤ D(w(x, t)) ≤ D1 for any w ∈ L2 (Ω) and D0 , D1 are positive numbers. We assume that D is a globally Lipschitz function, i.e, there exists f ≥ 0 such that M fkw1 − w2 kL2 (Ω) . kD(w1 ) − D(w2 )kL2 (Ω) ≤ M (4.62) From Definition (4.3), the operator P can be chosen as follows X p2 < v, ψp > ψp , (4.63) Pv := D1 ∆v = D1 p∈Nd

for any v ∈ L2 (Ω). It is easy to show that (A(t, w), P) as above, satisfies Assumption (B) in the sense of Definition (4.3). We can easily see that the eigenvalues of P are M (p) = D1 p2 . Next, we find the operator Pρn by truncating Fourier series in (4.63) and we have X Pρn v := m1 p2 < v, ψp > ψp . (4.64) |p|≤

q

ρn D1

Thanks to (4.59), a regularized problem for (4.61) is given below    ∂U ρn ,βn   − ∇ D(Uρn ,βn )∇Uρn ,βn − PU ρn ,βn + Pρn U ρn ,βn   ∂t    bβn (x, t), 0 < t < T, = FQn (U ρn ,βn (x, t)) + G    U ρn ,βn (x, t) = 0, x ∈ ∂Ω,     b βn (x), U ρn ,βn (x, T ) = H where

FQn

 Q2 − Q3n ,
 n
2
3 U ρn ,βn (x, t) = U ρn ,βn − U ρn ,βn ,   −Q2n + Q3n ,

(4.65)

U ρn ,βn (x, t) > Qn ,

−Qn ≤ U ρn ,βn (x, t) ≤ Qn , U ρn ,βn (x, t) < −Qn .

Assume that H, G, u are as in Theorem (4.1). Let βn be as in (4.22). Let ρn and Qn be such that v ! u d u Y 2α 0 ρn = t log (4.66) nk . m1 T (4α0 + d) k=1

and


δ0 − 1 (4.67) log Π(n) 4T respectively. Then the error between the solution u of Problem (4.61) and the solution U ρn ,βn of Problem (4.65) is of order −4α0 δδ0 0 δ0 −µ0 δ0 0t ! 2α2α " ! 4T # d ! 2T α−2α α0 +dT d d 0 +d/2 0 +dT /2 Y Y Y 1 , max Qd , . (4.68) nk nk nk 4µk δ0 −δ0 k=1 (nk ) k=1 k=1 k=1 2Qn + 3Q2n =

26

It is easy to check that the term in (4.68) tends to zero as |n| → +∞. Proof of Theorem 4.2. The proof is divided into some steps. Step 1. The existence and uniqueness of the solution to problem (4.59). Let V ρn ,βn (x, t) = U ρn ,βn (x, T − t) and define the following operator B(t, w) = P − A(t, w). Then it is obvious that V ρn ,βn (x, t) satisfies the following equation  ∂V ρn ,βn   + B(t, V ρn ,βn )V ρn ,βn − PU ρn ,βn   ∂t    b βn (x, t), 0 < t < T, = Pρn V ρn ,βn − FQn (V ρn ,βn (x, t)) − G (4.69)    V ρn ,βn (x, t) = 0, x ∈ ∂Ω,     b βn (x). V ρn ,βn (x, 0) = H By the assumptions on A above in (4.55), (4.56), (4.57), (4.58), it is easy to show that (i) For w ∈ BR (L2 (Ω)), u ∈ V(Ω), v ∈ V(Ω) then D E fa kukV(Ω) kvkV(Ω) . (B(t, w)u, v ≤M L2 (Ω)

(ii) For any v ∈ V(Ω) then

D

(B(t, w)v, v

E

L2 (Ω)

ckvk2 . ≥M V(Ω)

(iii) There exist a subspace V1 (Ω) ⊂ L2 (Ω) in which D  E fkukV(Ω) kvkV(Ω) kw1 − w2 kL2 (Ω) , ≤M B(t, w1 ) − B(t, w2 ) u, v L2 (Ω)

(4.70)

(4.71)

(4.72)

for any u ∈ V(Ω), v ∈ V(Ω) and w1 , w2 ∈ BR (L2 (Ω)). Hence, the conditions proved above for the operator B show that B satisfies the assumptions of Theorem 5.10 in [42] (page 252). With the help of Theorem 5.10 in of [42], we conclude that the Problem (4.69) 2 2 has unique solution V ρn ,βn ∈ C([0, T ];

L (Ω)) ∩ L (0, T ; V(Ω)).

Step 2. The error estimate for E U ρn ,βn (., t) − u(., t) 2 . L (Ω)

First, we have

and

∂u(x, t) + Pρn u(x, t) = F (u(x, t)) + G(x, t) ∂t

+ Pρn − P u(x, t) + P − A(t, u) u(x, t),

∂U ρn ,βn bβn (x, t) + Pρn U ρn ,βn (x, t) = FQn (U ρn ,βn (x, t)) + G ∂t

+ Pρn − P U ρn ,βn + P − A(t, U ρn ,βn ) U ρn ,βn .

(4.73)

Putting Z ρn ,βn (x, t) = U ρn ,βn (x, t) − u(x, t), we have

∂ Z ρ ,β (x, t) + Pρn Z ρn ,βn (x, t) ∂t n n
= FQn (U ρn ,βn (x, t)) − F (u(x, t)) − Pρn − P u(x, t)
b βn (x, t) − G(x, t) + P − A(t, U ρn ,βn ) Z ρn ,βn (x, t) + G   + A(t, u) − A(t, U ρn ,βn u(x, t).

Put Xρn ,βn (x, t) = eκn (t−T ) Z ρn ,βn (x, t), and taking the inner product of the last equality by Xρn ,βn (x, t), and integrating over (t, T ) we have kXρn ,βn (x, T )k2L2 (Ω) − kXρn ,βn (x, t)k2L2 (Ω) Z TZ Z T Pρn ,βn Xρn ,βn (x, τ )Xρn ,βn (x, τ )dxdτ = 2κn kXρn ,βn (x, τ )k2L2 (Ω) dτ − 2 t | t Ω {z } :=J4,4

27

+ 2eκn (t−T ) |

+ 2eκn (t−T ) |

+ 2eκn (t−T ) |

+ 2eκn (t−T ) | +2 |

Z

t

T

Z

T

t

Z

T

t

Z

T

t

Z

T

t

Z Z Z Z

Ω

FQn (U ρn ,βn (x, τ )) − F (u(x, τ ))Xρn ,βn (x, τ )dxdτ {z } :=J5,5

Ω

b βn (x, τ ) − G(x, τ )Xρn ,βn (x, τ )dxdτ G {z } :=J6,6


P − A(t, U ρn ,βn ) Z ρn ,βn (x, τ )Xρn (x, τ )dxdτ {z }

Ω

:=J7,7

Ω

(Pρn − P) u(x, τ )Xρn ,βn (x, τ )dxdτ {z } :=J8,8

Z



 eκn (τ −T ) A(τ, u) − A(τ, U ρn ,βn u(x, τ )Xρn ,βn (x, τ )dxdτ . Ω {z }

(4.74)

:=J9,9

By a similar techniques as in equation (4.46), we obtain Z T   √ E J4,4 ≥ −2M ( ρn ) EkXρn ,βn (., τ )k2L2 (Ω) dτ.

(4.75)

t

By a similar techniques as in equation (4.47), we obtain Z T   E J5,5 ≥ −4K(Qn ) EkXρn ,βn (., τ )k2L2 (Ω) dτ.

(4.76)

t

By a similar techniques as in equation (4.48), we obtain

2

 

b E J6,6 ≥ −T E G βn (.) − G(.)

L∞ (0,T ;L2 (Ω))

−

Z

T

t

EkXρn ,βn (., τ )k2L2 (Ω) dτ.

By a similar techniques as in equation (4.49), we obtain "Z T D  E
E J7,7 ) = E P − A(τ, U (ρn , βn )) Xρn (., τ ), Xρn ,βn (., τ ) t

cE ≥M

Z

t

T

L2 (Ω)

dτ

(4.77)

#

kXρn ,βn (., τ )k2V(Ω) dτ.

(4.78)

By a similar techniques as in equation (4.49), we obtain

2   √

E J8,8 ≥ −T e−2T M( ρn ) u ∞ L

(0,T ;G1+γ (Ω))

−

Z

t

T

EkXρn ,βn (., τ )k2L2 (Ω) dτ.

Now, we turn to estimate J9,9 . For τ ∈ (0, T ), using (4.58), we get Z   eκn (τ −T ) A(τ, u) − A(τ, U ρn ,βn u(x, τ )Xρn ,βn (x, τ )dx Ω

fku(., τ )kV(Ω) kXρn ,βn (x, τ )kV(Ω) kXρn ,βn (x, τ )kL2 (Ω) ≤M

fkukL∞ (0,T ;V(Ω)) kXρn ,βn (x, τ )kV(Ω) kXρn ,βn (x, τ )kL2 (Ω) ≤M

fM f0 kXρn ,βn (x, τ )kV(Ω) kXρn ,βn (x, τ )kL2 (Ω) ≤M

fM f0 ≤ M

ckXρn ,βn (x, τ )k2 M V(Ω) fM f0 2M

fM f0 kXρn ,βn (x, τ )kL2 (Ω) 2M + c M

28

!

(4.79)

f0 is a postive constant such that M f0 ≥ kukL∞ (0,T ;V(Ω)) . This implies that where M ! Z TZ   κn (τ −T ) A(τ, u) − A(τ, U ρn ,βn u(x, τ )Xρn ,βn (x, τ )dxdτ E J9,9 | = E 2 e t

cE ≤M

Ω

Z

t

T

kXρn ,βn (., τ )k2V(Ω) dτ

!

fM f0 4M E + c M

Z

t

T

kXρn ,βn (., τ )k2L2 (Ω) dτ

!

.

(4.80)

Combining (4.74), (4.75), (4.76),(4.77) (4.78), (4.79), (4.80) gives EkXρn ,βn (., T )k2L2 (Ω) − EkXρn ,βn (., t)k2L2 (Ω)  fM f0  Z T √ 4M ≥ 2κn − 2M ( ρn ) − 4K(Qn ) − 2 − EkXρn ,βn (., τ )k2L2 (Ω) c M t

2

2

√



b − T e−2T M( ρn ) u (.) − G(.) − T E G .

βn L∞ (0,T ;L2 (Ω)

L∞ (0,T ;G1+γ (Ω))

√ Letting κn = M ( ρn ) and by using Gronwall’s inequality as in the proof of Theorem 3.2, we conclude that

2

√

e2tM ( ρn ) E U ρn ,βn (., t) − u(., t) 2 L (Ω)   fM f

2 4M 4K(Qn )+2+ c 0 (T −t) 2T M(√ρ ) M n b βn − H e E H ≤e

2 L (Ω)   fM f

2

4M 4K(Qn )+2+ c 0 (T −t) 2T M(√ρ )

b M n e (.) − G(.) T E G +e

∞ βn L (0,T ;L2 (Ω))   fM f

4M 4K(Qn )+2+ c 0 (T −t) 2 M T u . (4.81) +e L∞ (0,T ;G1+γ (Ω))

Multiplying both sides of the last inequality with e−2tM ( (2.1), we get the desired result (4.60).

√ ρn )

and combining with the results in Corollary 

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