DSpace at VNU: A nonlinear parabolic equation backward in time: Regularization with new error estimates

Based on the fundamental solution to the parabolic equation, we propose to solve this problem by the Fourier truncated method, which generates a well-posed integral equation.. Our regula

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Nonlinear Analysis

journal homepage:www.elsevier.com/locate/na

A nonlinear parabolic equation backward in time: Regularization with new error estimates

aDepartment of Mathematics, SaiGon University, 273 AnDuongVuong, HoChiMinh city, Viet Nam

bDepartment of Mathematics, University of Natural Science, Vietnam National University, 227 Nguyen Van Cu, Q.5, HoChiMinh City, Viet Nam

a r t i c l e i n f o

Article history:

Received 21 April 2009

Accepted 6 May 2010

MSC:

35K05

35K99

47J06

47H10

Keywords:

Backward parabolic problem

Ill-posed problem

Nonlinear parabolic equation

Truncation method

Error estimate

a b s t r a c t

Consider a nonlinear backward parabolic problem in the form

u t+Au(t) =f(t,u(t)), u(T) =g,

where A is a positive self-adjoint unbounded operator Based on the fundamental solution

to the parabolic equation, we propose to solve this problem by the Fourier truncated method, which generates a well-posed integral equation Then the well-posedness of the proposed regularizing problem and convergence property of the regularizing solution to the exact one are proven Our regularizing scheme can be considered a new regularization, with the advantage of a relatively small amount of computation compared with the quasi-reversibility or quasi-boundary value regularizations Error estimates for this method are provided together with a selection rule for the regularization parameter These errors show that our method works effectively

© 2010 Elsevier Ltd All rights reserved

1 Introduction

We consider the backward time problem for the following nonlinear PDE:

where A is any non-negative, self-adjoint operator which does not depend on f and g is a given vector in a Hilbert space H.

An example for problem(1)–(2)is the backward heat problem

(u t−1u=f(x,t,u), (x,t) ∈Ω× (0,T)

u(x,t) =0, (x,t) ∈ ∂Ω× (0,T)

u(x,T) = ϕ(x), x∈Ω

whereΩ = (0, π)N ⊂R N This is an example for the problem(1)–(2)corresponding to H=L2(Ω)and A= −∆(associated with the homogeneous Dirichlet boundary condition) which has the eigenbasisφp(x) = (2

π)N/ 2sin(p1x1) sin(p N x N)and the eigenvalueλp =p2 Here we denote p= (p1, p N) ∈N N , x= (x1,x2, ,x p) ∈ R N and p2=p2+ · · ·p2

N The backward parabolic problems can be applied to several practical areas such as image processing, mathematical finance, and physics (See [1–3]) However, the ill-posed nature of the problems requires certain types of regularization

∗Corresponding author.

E-mail address:tuanhuy_bs@yahoo.com (N.H Tuan).

0362-546X/$ – see front matter © 2010 Elsevier Ltd All rights reserved.

techniques One approach to regularize the ill-posed problems is based on the use of eigenfunction expansion, where the eigenpairs of the corresponding elliptic operator are available Another approach is to use the method of quasi-reversibility [4–15] Other approaches include the least squares method with Tikhonov-type regularization [16] and the use of heat kernels Carasso et al [17] introduced a method of transforming the problem(1)–(2)into a second-order in time problem Later Carasso [18,19] introduced the concept of a supplementary constraint such as that of slow evolution from the continuation boundary (SECB) The methods of quasi-reversibility and Tikhonov regularization introduce artificial contamination, which is not from noises, to the numerical solutions For a recent survey on backward parabolic equations (BPE), we refer the reader to [20,21] Although there are many papers on the linear homogeneous case of the backward problem, we only find a few papers on nonhomogeneous and nonlinear cases of BPE

In 2007, in [22], the authors considered the problem(1)–(2)in the case A is the Laplace operator and H = L2(0, π)by using the quasi-boundary value method They gave the approximated problem as follows

u

t −u

xx=

∞ X

n= 1

e−tn2

t

T +e−tn2f n( u)(t)sin nx, (x,t) ∈ (0, π) × (0,T), (3)

u(x,0) +u(x,T) = ϕ(x) −

∞ X

n= 1

Z T

0



s

T +e−sn2f n( u)(s)ds



and obtain the error estimation

Very recently, in [23], we considered the problem(1)–(2)by using the method of stabilized quasi-reversibility By replacing

the operator A by f(A), chosen later under some better conditions, we approximated the problem(1)–(2)as follows:

u

with 0<  <1 We gave the following error estimate

Here u and u are solutions to the ill-posed and approximate problems, respectively, and C is a computable constant

independent ofβ()whereβ() →0 if →0 It is clear that the errors(6)and(9)in the time t =0, are not given

To improve these above disadvantages, in this paper, we develop a new regularization method which is called the Fourier method for solving the problem(1)–(2) As far as we know, there have not been any results of the Fourier series method for treating nonlinear BPE until now Meanwhile, we will establish some faster convergence error estimates In particular,

the convergence of the approximate solution at t = 0, is also proved This is an improvement of many recent results in [7,24,25,22,26,23,27]

Namely, assume that A admits an orthonormal eigenbasis{ φp }p≥1in H, associated with the eigenvalues such that

0< λ1≤ λ2≤ λ3≤ · · · lim

p→∞λp = ∞

Informally, problem(1)–(2)can be transformed to an integral equation having the form

u(t) =

∞

X

p= 1



e(T−t)λp g p−

Z T

t

e−(t−s)λp f p( u)(s)ds



where

g=

∞

X

p= 1

g pφp,

f(t,u(t)) =

∞ X

p= 1

f p( u)(t)φp.

Since t<T , we know from(10)that, when p become large, exp{ (T−t)λp }increase rather quickly Thus, the term e− (t−T)λp

is the instability cause In [26], this term is replaced by the better term

e−tλp

 +e−Tλp.

In [23], we give a different better term having the form

(λp +e−Tλp)t−T

.

So, to regularize problem(1)–(2), we hope to recover the stability of problem(10)by filtering the high frequencies with a suitable method The essence of our regularization method is just to eliminate all high frequencies from the solution, and instead consider(10)only forλp ≤m, where mis an appropriate positive constant satisfying lim→ 0m = ∞ We note

that m is a constant which will be selected appropriately as the regularization parameter This regularization method is rather simple and convenient for dealing with some ill-posed problems The present paper is devoted to establishing such

a method for problem(1)–(2)

2 Fourier regularization and the main results

Let g and gdenote the exact and measured data at t=T , respectively, which satisfy

kg−gk ≤ .

Let

P = {p≥1,p∈N, λp ≤m} ,

Q= {p≥1,p∈N, λp >m}

where m is a constant which will be selected appropriately as the regularization parameter For everyvin H having the

expansionv = P∞

p= 1vpφp, vp ∈R,p=1,2, we define the operators S(t),S(t)as follows

S(t)v =

∞

X

p= 1

e−tλpvpφp,

S(t)v = X

p∈P

We have the following approximation problem

u(t) =S(T−t)g−

Z T

t

S(s−t)f(s,u(s))ds

p∈P



e(T−t)λp g p−

Z T

t

e(s−t)λp f p( u)(s)ds



Theorem 1 Let g ∈H and let f :R×H→H be a Lipschitz continuous function in both variables, i.e.

kf(t1, w) −f(t2, v)k ≤k(|t1−t2| + k w − vk),

for k>0 independent of w, v ∈H,t∈R Then problem(12)has uniquely a solution u ∈C([0,T];H)for any positive.

Proof First, it is easy to prove that

Thus, for everyvin H having the expansionv = P∞

p= 1vpφp, we have

kS(t)vk2 = X

p∈P

e2tλpv2

p≤e2tmX

p∈P

v2

p

≤e2tm

∞ X

p= 1

v2

p =e2tmk vk2.

Hence,(13)is proved 

Forw ∈C([0,T];H), consider the function G(w)(t)defined by

G(w)(t) =S(T −t)g−

Z T

t

S(s−t)f(s, w(s))ds.

It is clear that t→G(w)(t)is continuous from[0;T] →H and G(w)(T) =g.

We claim that, for everyw, v ∈C([0,T];H)we have

kG n(w)(t) −G n(v)(t)k ≤k n(T−t)neTnm C

n

where C=max{T,1}and||| |||is sup norm in C([0,T];H)

We shall prove the latter inequality by induction

For n=1, using(13)and the Lipschitz property of f , we have

kG(w)(t) −G(v)(t)k =

Z T

t

S(s−t)(f(s, w(s))) −f(s, v(s))ds

≤ ke(s−t)m

Z T

t

k w(s) − v(s)kds

≤ Cke Tm(T−t)|||w − v|||.

Suppose that(14)holds for n=j We prove that(14)holds for n=j+1 In fact, we have

kG j+1(w)(t) −G j+1(v)(t)k =

Z T

t

S(s−t)(f(s,G jw)(s) −f(s,G jv)(s))ds

≤k(T−t)eTm

Z T

t

kG j(w)(s) −G j(v)(s)k2ds

≤k j+1eTm(j+ 1 )(T−t)j+ 1

(j+1)! C j

+ 1||| w − v|||.

Therefore, by the induction principle, we have

kG n(w)(t) −G n(v)(t)k ≤ (CkT e Tm)n

for allw, v ∈C([0,T];H)

We consider G:C([0,T];H) →C([0,T];H) Since

lim

n→∞

(CkT e Tm)n

n! =0,

there exists a positive integer number n0such that G n0is a contraction It follows that the equation G n0(w) = whas a unique

solution u∈C([0,T];H)

We claim that G(u) = u In fact, one has G(G n0(u)) = G(u) Hence G n0(G(u)) = G(u) By the uniqueness of the

fixed point of G n0, one has G(u) =u, i.e., the equation G(w) = whas a unique solution u ∈C([0,T];H)

Remark 2.1 In the Hilbert space, let H = L2(0, π)and let A = −∆ is the Laplace operator We takeλn = n2,

φn =

q

2

πsin(nx)are eigenvalues and orthonormal eigenfunctions, which form a basis for H Letk kHbe the norm of L2(0, π)

The function f satisfies the condition ofTheorem 1which is given by an example

f(u) =















− e10

e−1u+

e21

e−1 u∈ (e10,e11]

e10

e−1u+

e21

e−1 u∈ (−e11, −e10]

0 |u| >e11.

It is not difficult to check that the above function satisfying the Lipschitz condition ofTheorem 1 However, the present paper

can not treat for example a nonlinear term, such as f(t,u) =u q for any q This is the serious disadvantage of this paper We hope the big class of function f will be studied in future reports.

Theorem 2 The solution of the problem(12)depends continuously on g in H.

Proof of Theorem 2 Let u andvbe two solutions of(12)corresponding to the values g and h We have

ku(t) − v(t)k = S(T −t)(g−h) − Z T

t

S(s−t)(f(s,u(s)) −f(s, v(s))) (16) Using(13), we obtain

ku(t) − v(t)k ≤ kS(T−t)kkg−hk +

Z T

t

kS(s−t)kkf(s,u(s)) −f(s, v(s))kds

≤e(T−t)mkg−hk +k

Z T

Then, we obtain

etmku(t) − v(t)k ≤eTmkg−hk +k

Z T

t

esmku(s) − v(s)kds.

Using Gronwall’s inequality we have

etmku(t) − v(t)k ≤ek(T−t)eTmkg−hk

Hence, we obtain

This completes the proof of the theorem 

Remark 2.2 (1) If we select m = 1, since(18), the stability of the method becomes E1() =eC1, which is the same as the quasi-reversibility method in [6,8,14,15]

(2) If m = 1

Tln(1

), the stability magnitude is E2(,t) =C2t

T− 1, is the same as the quasi-boundary value method of some related results in [7,11,22]

(3) If m= 1

Tln( T

( 1 + ln (T

 ))), the stability magnitude is better than E3() =C3 T

( 1 + ln (T

 )), which is given in [24,26].

(4) If m= 1

Tln



T

 ln (Te

 )



, the stability magnitude is

E4(,t) =C4e(T−t)m =C4T t− 1 T−1ln(Te− 1)
T t− 1

which is similar to [27]

Note that

lim

→ 0

E4(,t)

E2(,t) =0

and

lim

→ 0

E4(,t)

E3(,t) =→lim0



 + lnT



t T

=



0, 0<t<T,

1, t =0.

So, the error E4(,t)is better than E1(),E2(,t),E3() To our knowledge, this is the best known result

Theorem 3 Let f,g be as in Theorem 1 Assume the exact solution u(t) of the problem(1)–(2)such that

Z T

0

∞

X

p= 1

Then one has

ku(t) −u(t)k ≤2

√

for every t∈ [0,T], where

M= ku(0)k2+T

Z T

0

∞ X

p= 1

e2sλp f p2(u)(s)ds, (21)

and uis the unique solution of problem(12)corresponding to.

Proof of Theorem 3 The functions u(t)and u(t)can be written as the forms

u(t) =

∞

X

p= 1



e(T−t)λp g p−

Z T

t

e(s−t)λp f p( u)(s)ds

 φp,

u(t) = S(T −t)g−

Z T

t

S(s−t)f(s,u(s))ds

p∈P



e(T−t)λp g p−

Z T

t

e(s−t)λp f p( u)(s)ds

 φp.

u(t) −u(t) = X

p∈Q



e(T−t)λp g p−

Z T

t

e(s−t)λp f p( u)(s)ds



φp + X

p∈P

Z T

t

e(s−t)λp(f p( u)(s) −f p( u)(s))ds

 φp

=I+J

where

p∈Q



e(T−t)λp g p−

Z T

t

e(s−t)λp f p( u)(s)ds

 φp

p∈P

Z T

t

e(s−t)λp(f p( u)(s) −f p( u)(s))ds

 φp

=

Z T

t

S(s−t)(f(s,u(s)) −f(s,u(s)))ds.

Since p∈Q, we have

e−tλp ≤e−tm,

this implies that the term I can be estimated as follows

kIk2 = X

p∈Q



e(T−t)λp g p−

Z T

t

e(s−t)λp f p( u)(s)ds

2

≤2X

p∈Q

e−2tλp



eTλp g p−

Z T

0

esλp f p( u)(s)ds

2

+2X

p∈Q

e−2tλp

Z t

0

esλp f p( u)(s)ds

2

≤2e−2tm

∞ X

p= 1



eTλp g p−

Z T

0

esλp f p( u)(s)ds

2

+2T e−2tm

∞ X

p= 1

Z T

0

e2sλp f p2(u)(s)ds

=2e−2tmku(0)k2+2T e−tm

∞ X

p= 1

Z T

0

e2sλp f p2(u)(s)ds

=e−2tm 2ku(0)k2+2T

∞ X

p= 1

Z T

0

e2sλp f p2(u)(s)ds

!

=2e−2tm M.

Using the Lipschitzian hypothesis on f and(13), we obtain

kJk2 ≤ (T−t) Z T

t

kS(s−t)k2k (f(s,u(s)) −f(s,u(s)))k2ds

≤k2T

Z T

t

e2(s−t)mku(s) −u(s)k2ds.

Therefore we obtain

ku(t) −u(t)k2 ≤2(kIk2+ kJk2)

≤4e−2tm M+2k2T

Z T

t

e2(s−t)mku(s) −u(s)k2ds.

It follows that

e2tmku(t) −u(t)k2≤4M+2k2T

Z T

t

e2smku(s) −u(s)k2ds.

By using Gronwall’s inequality, we obtain

e2tmku(t) −u(t)k2≤e2k2T(T−t)4M,

which implies that

ku(t) −u(t)k ≤2

√

Me k2T(T−t)e−tm. 

Remark 2.3 (1) Choosing m= 1

Tln(1

), we obtain the error estimate

ku(t) −u(t)k ≤2

√

Me k2T(T−t)t

This error is given in [28,22,23] And if the function f(t,u(t)) =0, the error is given in [7

(2) If we select m =ln(1

)1

ln1

s

 (s≥0), we have the another error

ku(t) −u(t)k ≤2

√

Me k2T(T−t)t

T



ln1



−st

Note that if t >0,t

T and ln 1

−st

tend to 0 when →0 Hence, this error is better than the error in(22) And in the case

s=0, this error is also(22).(20)is generalization of many related errors in some previous papers such as [28,15,22,26,23]

(3) In the case f(t,u(t)) =f(t), to find the convergence of approximate solution, we don’t ask the condition(19).Thus, we have

u(t) −u(t) = X

p∈Q



e(T−t)λp g p−

Z T

t

e(s−t)λp f p( s)ds

 φp.

Then

ku(t) −u(t)k2 = X

p∈Q



e(T−t)λp g p−

Z T

t

e(s−t)λp f p( s)ds

2

p∈Q

u2p= kuk2− X

p∈P

u2p.

It is easy to prove that

lim

→ 0

X

p∈P

u2p= kuk2.

Then

lim

→ 0

ku(t) −u(t)k2=0.

If there exist s>0 such that

∞

X

p= 1

λs

p u2p< ∞

then

ku(t) −u(t)k2= X

p∈Q

1

λs p

λs

p u2p≤ 1

m s

 X

p∈Q

λs

p u2p≤ 1

m s



∞ X

p= 1

λs

p u2p.

(4) In most known results, such as [28,22,23], the errors between the exact solution and approximate solution are calculated

in the form Ct

T Notice that the convergence estimate in this theorem is useless at t=0 Let us take t =0 in(11), the error estimate is

ku(0) −u(0)k ≤ √Me k2T2,

which does not tend to zero when →0 In the next theorem, we will give another estimation which the error in the case

t =0 is considered

(5) As we know, mis a constant which will be easily selected appropriately as regularization parameter In [23],β()is also the regularization parameter However, in that paper, it is not easy to selectβ()because it depends on the operator A and approximated operator f(A) For details, the authors selectedβ()satisfying two conditions

(a)|f(α)| ≤ −1

T ln(β()), α ∈ [0, ∞).

(b)k (−A+f(A))uk ≤ β()keTA uk

This proves that our method is more effective than the method in [23]

(6) In this theorem, we require a condition on the expansion coefficient f pin(19) We note that the solution u depends on the nonlinear term f and therefore f p, f p( u)is very difficult to assign a value Such obscurity makes this theorem hard to use

for numerical computations Hence, we require another condition as follows

∞

X

p= 1

In this case, we only require the assumption of u, there is no need to compute the function f p( u) In fact, in the homogeneous case of problem(1)–(2), i.e f =0, then the left hand side of(24)is equal toku(0)k2 Hence, the condition(24)

is natural and acceptable And we note(24)is weaker than the condition in theorem 3.1 (page 7, [23])

Theorem 4 Suppose problem(1)–(2)has a unique solution u(t)satisfying(24)and f,g are as in Theorem 3 Then we obtain another estimation

for every t∈ [0,T], where

P(,t) = X

p∈Q



eTλp g p−

Z T

t

esλp f p( u)(s)ds

2

p∈Q

R(,t) =

s



ke 2k2T(T−t)Z T

0

P(,t)dt+P(,t)



and uis the unique solution of problem(12)corresponding to.

Proof of Theorem 4 It is easy to see that P(,t) →0 when →0 Since the proof ofTheorem 3, we get

kIk2 = X

p∈Q



e−(t−T)λp g p−

Z T

t

e−(t−s)λp f p( u)(s)ds

2

p∈Q

e−2tλp



eTλp g p−

Z T

t

esλp f p( u)(s)ds

2

≤e−2tm X

p∈Q



eTλp g p−

Z T

t

esλp f p( u)(s)ds

2

=e−2tm P(,t).

This follows that

ku(t) −u(t)k2 ≤2(kIk2+ kJk2)

≤2e−2tm P(,t) +2k2T

Z T

t

e2(s−t)mku(s) −u(s)k2ds.

Applying Gronwall’s inequality, we get

e2tmku(t) −u(t)k2≤ke−2k2Tt

Z T

t

e2k2Ts P(,s)ds+P(,t).

Finally

ku(t) −u(t)k2 ≤



ke 2k2T(T−t)Z T

0

P(,t)dt+P(,t)



e−2tm

=R2(,t)e−2tm. 

Remark 2.4 (1) In the right hand side of(25), if t>0 then both term e−tm and R(,t)tend to zero So, the convergence in

(25)is better than in(20)

(2) When t=0,(25)becomes

ku(0) −u(0)k2≤ke 2k2T2

Z T

0

Noting that the right hand side of(28)tends to zero when →0

(3) In the simple linear case f(t,u(t)) =0, this means that k=0, then the term P(,t)can be estimated

P(,t) = X

p∈Q

e2Tλp g2≤

∞ X

p= 1

e2Tλp g2= ku(0)k2. (29) So

R(,t) =

s



ke 2k2T(T−t)Z T

0

P(,t)dt+P(,t)



(30)

Theorem 5 Let f,g be as in Theorem 1 Assume that the exact solution u of(1)–(2)corresponding to g be defined as in Theorem 1 Let g ∈H be measured data such that

kg−gk ≤ .

Let us select m =ln(1

)1

(ln1)−s 2T



and let ube the solution of problem(12)corresponding to g.

(a) If u(t)satisfies(19), then

ku(t) −u(t)k ≤exp(k2(T−t)2)t

T



ln1



−s(T−t)

2T

+Me k2T(T−t)t

T



ln1



st 2T

(b) If u(t)satisfies(24), then

ku(t) −u(t)k ≤exp(k2(T−t)2)T t



ln1



−s(T−t)

2T

+R(,t)ek2T(T−t)T t



ln1



st 2T

for every t∈ [0,T]and M,R(,t)is defined in Theorems 3and4respectively.

Proof of Theorem 5 We shall prove (a), and that (b) is similar to (a) Thus, letv be the solution of problem(12)

corre-sponding to g.

UsingTheorems 2and4, we get

ku(t) −u(t)k ≤ ku(t) − v(t)k + kv(t) −u(t)k

≤e(T−t)mexp(k2(T−t)2)kg−gk +2

√

Me k2T(T−t)e−tm

≤exp(k2(T−t)2)T t



ln1



−s(T−t)

2T

+2

√

Me k2T(T−t)T t



ln1



st 2T

,

≤



2

√

M+1



ek2T(T−t)t

T



ln1



−s(T−t)

2T

1+



ln1



s

2

! ,

for every t∈ (0,T)

This completed the proof of the theorem 

Remark 2.5 (1) If the function f(t,u(t)) =0, using(32)we obtain the following error

ku(t) −u(t)k ≤exp(k2(T−t)2)t

T



ln1



−s(T−t)

2T

+2ku(0)kek2T(T−t)t

T



ln1



st 2T

Noting that the convergence in t =0 is not given here

(2) When s=0, the estimate(33)becomes

ku(t) −u(t)k ≤ (R(,t) +1)ek2T(T−t)t

This error is similar to the results in [29,28,15]

(3) In the above error if we let t=0 then

ku(0) −u(0)k ≤exp(k2T2)



ln1



−s

2

Since s>0 we get

lim

→ 0 exp(k2T2)



ln1



−s

2

+R(,0)ek2T2

!

=0.

(4) From(36)we know lim→ 0R(,0) =0 But the term R(,0)is not often computed in practice, so we can’t estimate it

by the term depending on This is disadvantage of the error(36) To improve this, let us select

m = 1

T ln

T ln T

s

 + ln T

! ,

and using the proof ofTheorem 5(a), we obtain

ku(t) −u(t)k ≤ ku(t) − v(t)k + kv(t) −u(t)k

≤e(T−t)mexp(k2(T−t)2)kg−gk +2

√

Me k2T(T−t)e−tm

≤exp(k2T2)e(T−t)m



1+e−Tm

 2

√

M

 ,

≤exp(k2T2)t

T



1+ln(T− 1)

T



t

T− 1

lnT



s (t

T− 1)

where

I =1+e−Tm

 2

√

M=1+2

√

M T

1+lnT

lnT

s

If s>1 then

lim

→ 0

1+lnT

lnT

s =0

and thus lim→ 0I =1 Recall that in [22], the convergence of the approximated solution is Ct

T



1 + ln (T − 1 )

T

t

T− 1

So, the convergence in(37)is quicker than that given in [22] We also note that the approximation in the case t = 0 is proved Moreover, comparing(37)with the result obtained in [7,24,25,15,22,26,23] we know estimate(37)is sharp and the best known estimate

Acknowledgements

The authors would like to thank Professor Ravi P Agarwal for his valuable help in the presentation of this paper The authors are also grateful to the anonymous referees for their valuable comments leading to the improvement of our paper

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