the backward problem for ginzurg-landau-type equation · [email protected] nguyen dang minh...

Acta Math VietnamDOI 10.1007/s40306-015-0135-y

The Backward Problemfor Ginzurg-Landau-Type Equation

Dang Duc Trong1 ·Bui Thanh Duy2 ·Nguyen Dang Minh3

Received: 3 December 2013 / Revised: 28 July 2014 / Accepted: 4 August 2014© Institute of Mathematics, Vietnam Academy of Science and Technology (VAST) and Springer

Science+Business Media Singapore 2015

Abstract Let � be a bounded domain of Rn. In this paper, we consider a final value

problem for the nonlinear parabolic equation

ut = �u + au − bu3 + h(x, t), x ∈ �, t ∈ (0, T ) ,

u = 0, x ∈ ∂�, t ∈ (0, T ) ,

u(T ) = g, x ∈ �,

where g, h are given functions and the numbers a, b (b > 0) are modeling parameters. Theproblem does not fulfill Hadamard’s postulates of well posedness: it might not have a solu-tion in the strict sense; its solutions might not be unique or might not depend continuouslyon the data. Hence, its mathematical analysis is subtle. However, it has many applicationsin physics and other fields. For this reason, a regularization for the problem is proposed.In our problem, the function f (u) = au − bu3 is not globally Lipschitz. So, we cannotapply directly recent methods that have been used in Trong et al. (Zeitschrift Analysis undihre Anwendungen 26(2), 231–245, 2007), Trong et al. (Electron. J. Differ. Equ. 2009(109),

� Bui Thanh [email protected]

Dang Duc [email protected]

Nguyen Dang [email protected]

1 Faculty of Mathematics and Computer Science, University of Science, Ho Chi Minh City NationalUniversity, No. 227 Nguyen Van Cu Street, Ward 4, District 5, Ho Chi Minh City, Vietnam

2 Faculty of Fundamental Science, Ho Chi Minh City Architecture University, No. 196 PasteurStreet, Ward 6, District 3, Ho Chi Minh City, Vietnam

3 Center for Mathematical Science, University of Science, Ho Chi Minh City National University,No. 227 Nguyen Van Cu Street, Ward 4, District 5, Ho Chi Minh City, Vietnam

mailto:[email protected]



D. D. Trong et al.

1–16, 2009), and Trong and Tuan (Electron. J. Differ. Equ. 2009(77), 1–13, 2009). Wehave to approximate the function f by a globally Lipschitz function and use an approxi-mate equation to find the regularization solution of the problem. Error estimations betweenthe exact solution and the approximate solution, established from noise data gε , aδ , bδ , aregiven.

Keywords Backward heat problem · Nonlinear ill-posed problems · Quasi-boundarymethods · Ginzburg-Landau equation · Lipschitz function · Fourier series and coefficients

Mathematics Subject Classification (2010) 35K05 · 35K99 · 47J06 · 35Q56 · 26A16

1 Introduction

The equation ut = �u + au − bu3 is the well-known Ginzburg-Landau equation. It hasbeen used to model phenomena in many areas in physics, including phase transitions innon-equilibrium systems, the instability in hydrodynamic systems, chemical turbulence, andthermodynamics (see [6, 8]).

In 2000, Ames [1] carried out a research on the continuous dependence of solutionson the modeling parameter b, non-existence results for the real Ginzburg-Landau equationut = c�u + au − bu3 in � × [0, T ] and its backwardness in time. Both in the forwardproblem and backward problem (with some additional assumptions), the author showed thatthe solution depends continuously on the parameter b.

Following [1], we give a method to regularize our problem and consider the continuousdependence on two modeling parameters a, b in the backward in time problem. This is anext step from the research of Ames. In fact, we consider the problem

ut = �u + au − bu3 + h(x, t), x ∈ �, t ∈ (0, T ) , (1)

u = 0, x ∈ ∂�, t ∈ (0, T ) , (2)

u(x, T ) = g(x), x ∈ �, (3)

where � is a bounded domain of Rn, a, b ∈ R and b > 0, g ∈ L2(�), h ∈ L∞(� × [0, T ])are given functions. We have a backward problem for Ginzburg-Landau-type equation, andwe will give a regularization method for this problem.

In [12], the authors considered a problem of finding the function of temperaturedistribution u such that

ut = cuxx + f (x, t, u(x, t)) , (x, t) ∈ (0, π) × (0, T ) ,

u (0, t) = u (π, t) = 0, t ∈ (0, T ) ,

u(x, T ) = g(x), x ∈ (0, π),

where f (x, t, u(x, t)) and g are given functions. The quasi-boundary value method and thequasi-reversibility method have been used to regularize this problem. Note that the functionf ∈ L∞([0, π ] × [0, T ] × R) in [12] has to satisfy a globally Lipschitz condition withrespect to the quantity u.

The Lipschitz condition was assumed in [13] when the authors also studied this problemby another method. This method was called a modified quasi-boundary method and better

The Backward Problem for Ginzurg-Landau-Type Equation

estimation was given. The Lipschitz condition appeared once again in [14]. In that paper,the authors used a truncated Fourier series method to regularize the following problem

ut − uxx − uyy = f (x, y, u(x, y, t)), (x, y, t) ∈ � × (0, T ),

u(x, y, T ) = g(x, y), (x, y) ∈ �,

where � = (0, π) × (0, π) and the data g was given approximately.Recently, some results on backward problem for operator equations have been published.

For example, Trong and Tuan in [15] have considered the problem

ut + Bu = h(t, u(t)), t ∈ (0, T ),

u(T ) = ϕ,

whereB is a non-negative, self-adjoint operator and the function h is also globally Lipschitz.In the abovementioned papers, we can learn different methods to regularize a backward

problem under globally Lipschitz conditions. The case of locally Lipschitz source has stillnot been considered. In our problem, the function f (u) = au − bu3 is one of the source. Itis not easy to solve directly by the methods given in [12–14]. Hence, an improved regular-ization method for the case is needed. In this paper, we perturb not only the final datum g

but also two modeling parameters a and b to emphasize a continuing step for the researchof Ames in [1]. Finally, we give error estimations and numerical experiments to prove theefficiency of our methods.

In the next section, we will introduce main results of this paper.

2 Main Results

We consider the following bilinear continuous mapping A(., .) : H 10 (�) × H 1

0 (�) → R

where

A(u, v) =n∑

i=1

∫

�

∂u

∂xi

∂v

∂xi

dx.

We can verify that the map A(., .) is coercive and symmetric. Let {ϕn} be an orthonor-mal eigenbasis in L2(�) corresponding to eigenvalues {λn} of A. We have A(ϕn, v) =λn 〈ϕn, v〉L2(�) ∀v ∈ H 1

0 (�). Without loss of generality, we will assume that

0 < λ1 < λ2 < λ3 < ... < limn→+∞ λn = +∞.

Using the separation of variables method, we can transform the system (1)–(3) into theequation

u(x, t) =+∞∑

n=1

(gne

λn(T −t) −∫ T

t

[f a,b

n (u)(s) + hn(s)]eλn(s−t)ds

)ϕn(x),

D. D. Trong et al.

where f a,b(u) = au − bu3, gn = 〈g, ϕn〉L2(�), hn = 〈h, ϕn〉L2(�) and fa,bn (u) =

〈f a,b(u), ϕn〉L2(�). From now on, we will denote the solution u of the latter equation byu(g, a, b). We can rewrite the equation as

u(g, a, b)(x, t) =+∞∑

n=1

gneλn(T −t)ϕn(x)

−+∞∑

n=1

(∫ T

t

[f a,b

n (u(g, a, b))(s) + hn(s)]eλn(s−t)ds

)ϕn(x).

Before regularizing the backward problem, we note that the function f a,b(u) is locallyLipschitz, i.e., for M > 0, t � 0 and w, v ∈ C0(�) = {ψ ∈ C(�),ψ |∂� = 0} satisfying‖w(t)‖C0(�) � M, ‖v(t)‖C0(�) � M , we have

∥∥∥f a,b(w(., t)) − f a,b(v(., t))

∥∥∥C0(�)

� KM ‖w(., t) − v(., t)‖C0(�) ,

where KM = (|a|+3bM2). We also denote KM,δ = (|aδ|+3bδM2). Thus, for establishing

an approximation for f a,b, we consider the following globally Lipschitz function

fa,bM (z) =

⎧⎨

⎩

aM − bM3, z > M,

az − bz3, −M � z � M,

−aM + bM3, z < −M.

This is necessary for applying methods considered in [12–14].Let δ, ε ∈ R be positive, sufficiently small and let aδ ∈ R, bδ > 0, gε ∈ L2(�) be such

that |aδ − a| � δ, |bδ − b| � δ, ‖gε − g‖L2(�) � ε. Using the idea of quasi-boundarymethod in [12], we consider the following integral equation

uα(gε, aδ, bδ)(x, t) (4)

=+∞∑

n=1

eλn(T −t)

1 + αeT λngε,nϕn(x)

−+∞∑

n=1

(∫ T

t

eλn(s − t)

1 + αsT esλn

[f

aδ

M,n, bδ (uα(gε, aδ, bδ)) (s) + hn(s)]ds

)ϕn(x),

where α,M > 0 and

faδ,bδ

M,n (uα(gε, aδ, bδ))(s) =∫

�

faδ,bδ

M (uα(gε, aδ, bδ)(x, s))ϕn(x)dx,

hn(s) =∫

�

h(x, s)ϕn(x)dx, gε,n =∫

�

gε(x)ϕn(x)dx.

The main result of our regularization is the following theorem.

Theorem 1 For ε, δ, q > 0, put

α = max{δT |�| 12 , ε

}, M := M(α, δ) = 4

√q

8T 2H 2δ

ln

(ln

1

α

),

where Hδ = max{|a|, |aδ|, 3b, 3bδ}. If u(g, a, b)(., 0) ∈ C0(�) and+∞∑

n=1

(∫ T

0

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds

)2< +∞


then, for 0 < q < min{

tT

, 14

}(t ∈ (0, T )) and δ, ε small enough, we have

‖uα(gε, aδ, bδ)(., t) − u(g, a, b)(., t)‖L2(�) � E1(α) lnq

(1

α

)α

tT ∀t ∈ (0, T ), (5)

with

E1(α) =q ln

(ln(1α

))

4T 2H 2δ ln

12

(1α

) + √2 + √

P ,

P = 4 ‖u(g, a, b)(., 0)‖2L2(�)

+ 16+∞∑

n=1

(∫ T

0

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds

)2.

In addition, if u(g, a, b)(., 0) ∈ H 10 (�) then

‖uα(gε, aδ, bδ)(., tα) − u(g, a, b)(., 0)‖L2(�) � E2(α)4√8T

(ln

(1

α

))q− 14

+CM4√8T

(ln

(1

α

))− 14

. (6)

Here, tα is the (unique) positive solution of the equation√

t = αtT , 0 < q < min

{tαT

, 14

}

and

CM = max{√

Pe2T2K2

M , N}

, E2(α) =q ln

(ln(1α

))

4T 2H 2δ ln

12

(1α

) + 1,

N = max {1, |a|} ‖u(g, a, b)(., 0)‖2H 10 (�)

+ a2

2b|�| + b |�| ‖u(g, a, b)(., 0)‖4C0(�)

+T |�| ‖h‖2L∞(�×[0,T ]) .

Note that the right-hand side in (5), (6) tend to zero as α tends to zero. Therefore, theregularization for our problem is finished with the choice for α and the parameter M .

Remark 1 The set {u: u is a solution of problem (1)–(3),∑+∞

n=1(∫ 10 |fn(u)(s) + hn(s)|

esn2ds)2 < +∞} is not empty. In fact, let us show the following simple example

ut = uxx + f (u) + h(x, t), x ∈ (0, π), t ∈ (0, 1) ,

u(0, t) = u(π, t) = 0, t ∈ (0, 1) ,

u(x, 1) = sin x, x ∈ (0, π),

where h(x, t) = (1 + 3t34 ) sin x − t3 sin 3x

4 , f (u) = u − u3.The solution of the problem is u(x, t) = t sin x. Then, f (u)+h(x, t) = u−u3+h(x, t) =

sin x + t sin x. We deduce that

fn(u)(s) + hn(s) = 2

π

∫ π

0(sin x + s sin x) sin nxdx.

D. D. Trong et al.

We obtain that fn(u)(s) + hn(s) = 0 ∀n > 1 and f1(u)(s) + h1(s) = 1 + s. Hence,

+∞∑

n=1

(∫ 1

0|fn(u)(s) + hn(s)| esn2ds

)2

=(∫ 1

0(1 + s)esds

)2

= e2 < +∞.

So the source condition in Theorem 1 is reasonable.

From the proof of Theorem 1, we get some following results with respect to f a,b.

Corollary 1 If α, M are chosen as in Theorem 1 then

‖u(g, aδ, bδ)(., t) − u(g, a, b)(., t)‖L2(�) � E3(α)

(ln

(1

α

))q

αtT ,

where 0 < q < min{

tT

, 14

}(t ∈ (0, T )) and

‖u(g, aδ, bδ)(., tα) − u(g, a, b)(., 0)‖L2(�) � E4(α)4√8T

(ln

(1

α

))q− 14

+Qδ4√8T

(ln

(1

α

))− 14

,

where 0 < q < min{

tαT

, 14

}. Here,

E3(α) =q ln

(ln(1α

))

4T 2H 2δ ln

12

(1α

) +√Pδ + √P , E4(α) = √Pδ +

q ln(ln(1α

))

4T 2H 2δ ln

12

(1α

) ,

P = 4 ‖u(g, a, b)(., 0)‖2L2(�)

+16+∞∑

n=1

(∫ T

0

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds

)2,

Pδ = 4 ‖u(g, aδ, bδ)(., 0)‖2L2(�)

+16+∞∑

n=1

(∫ T

0

∣∣∣f aδ,bδn (u(g, aδ, bδ))(s) + hn(s)

∣∣∣ esλnds

)2,

Qδ = max

{√Pδ

(ln

(1

α

))q

, Nδ

},

Nδ = max {1, |aδ|} ‖u(g, aδ, bδ)(., 0)‖2H 10 (�)

+ a2δ

2bδ

|�|+b |�| ‖u(g, aδ, bδ)(., 0)‖4C0(�) + T |�| ‖h‖2L∞(�×[0,T ]) .

Corollary 2 Let L1, L2 > 0 and assume that |u(g, a, b)(x, t)| � L1, |u(g, aδ, bδ)(x, t)|� L2 for all x ∈ �, 0 � t � T . Putting L = max {L1, L2}, we have the estimation

‖u(g, aδ, bδ)(., t) − u(g, a, b)(., t)‖L2(�) � E5(δ)δtT , t ∈ (0, T ),

with E5(δ) = √Pδe

2T 2K2L,δ + T (L+L3)|�| 12

ln12 1

δ

e2T 2K2

L,δ + √Pe2T

2K2L and P , Pδ defined as in

Corollary 1.


Furthermore, if we assume u(g, a, b)(., 0), u(g, aδ, bδ)(., 0) ∈ H 10 (�) ∩ C0(�), then

‖u(g, aδ, bδ)(., 0) − u(g, a, b)(., 0)‖L2(�)

�(

CL,δ + T (L + L3)|�| 12ln

12 1

δ

eT 2K2

L,δ + CL

)4√8T

(ln

(1

δ

))− 14

,

where

CL,δ = max{√

Pδe2T 2K2

L,δ , Nδ

}, CL = max

{√Pe2T

2K2L,N

},

Nδ = max {1, |aδ|} ‖u(g, aδ, bδ)(., 0)‖2H 10 (�)

+ a2δ

2bδ

|�|+b |�| ‖u(g, aδ, bδ)(., 0)‖4C0(�) + T |�| ‖h‖2L∞(�×[0,T ]) .

In the case that a, b are not perturbed by aδ and bδ , we have the corollary.

Corollary 3 Assume that |u(g, a, b)(x, t)| � R1, |uε(gε, a, b)(x, t)| � R2 for all x ∈ �,

0 � t � T . From Theorem 1, if we put R = max {R1, R2} and choose α = ε, M = R, then

‖uε(gε, a, b)(., t) − u (g, a, b) (., t)‖L2(�) �(√

2 + √P)

e2K2RT 2

εtT ∀t ∈ (0, T )

and‖uε(gε, a, b)(., tε) − u (g, a, b) (., 0)‖L2(�) �(eK2

RT 2 + CR

)4√8T

(ln

(1

ε

))− 14

,

where ε is small enough and P , CR are defined as in Theorem 1.

Remark 21. We do not consider the case b = 0 in this paper. In fact, if b = 0, our problem is a

nonhomogeneous linear backward heat problem. We only note that, using our method,we get again the results of [12].

2. We do not know whether our estimates are genuinely optimal or not. However, we cancompare our convergence rate with the one of linear cases. Tautenhahn and Schroter[10] proved that the best possible worst-case error for identifying u(t) is given by ε

tT .

This is also the rate of our estimates in Corollary 3. We can see the same rate in [2, 5].We note that at t = 0, the quantity ε

tT does not tend to zero. Hence, sharper estimates

are needed. As is well-known, in most of related results, the error at t = 0 is of logarith-mic rate, which has been studied in many recent papers. In 2007, Fu et al. [4] gave an

estimate with rate O

((ln 1

ε

)− 85). Moreover, in that paper, the authors proved that it is

impossible to get the error asymptotically better than logarithmic rate. So our estimatesis reasonable. Finally, to get a stability estimate of Holder type for whole [0, T ], i.e.,the rate εk (0 < k < 1), we need stronger assumptions on the exact solution. Readerscan find an error of Holder type in [9] and [16].

3 Proofs

Lemma 1 For M > 0, we have∥∥∥f a,b

M (w(., t)) − fa,bM (v(., t))

∥∥∥C0(�)

� KM ‖w(., t) − v(., t)‖C0(�) ∀w, v ∈ C0(�),

D. D. Trong et al.

where we recall that KM = (|a| + 3bM2).

Proof For t > 0, if w(x, t) < −M, v(x, t) < −M ,

|f a,bM (w(x, t)) − f

a,bM (v(x, t))| = 0.

If w(x, t) < −M � v(x, t) � M ,

fa,bM (w(x, t)) − f

a,bM (v(x, t)) = −aM + bM3 − av(x, t) + b[v(x, t)]3.

Hence, we get

|f a,bM (w(x, t)) − f

a,bM (v(x, t))| �

(|a| + 3bM2

)|M + v(x, t)| � KM |w(x, t) − v(x, t)|.

If w(x, t) < −M < M < v(x, t), f a,bM (w(x, t))−f

a,bM (v(x, t)) = −2aM +2bM3 then

|f a,bM (w(x, t)) − f

a,bM (v(x, t))| � 2M

(|a| + bM2

)� KM |w(x, t) − v(x, t)|.

If w(x, t), v(x, t) ∈ [−M, M] then|f a,b

M (w(x, t))−fa,bM (v(x, t))| = |f a,b(w(x, t))−f a,b(v(x, t))| � KM |w(x, t)−v(x, t)|.

Combining these three cases gives∥∥∥f a,b

M (w(., t)) − fa,bM (v(., t))

∥∥∥C0(�)

� KM ‖w(., t) − v(., t)‖C0(�) ∀w, v ∈ C0(�).

The proof is complete.

Because of Lemma 1, the globally Lipschitz property of the function fM is verified.Therefore, we have

Proposition 1 Problem (4) has a unique solution in C([0, T ]; L2(�)). Moreover, we have

‖uα(g, aδ, bδ)(., t) − uα(w, aδ, bδ)(., t)‖L2(�) �√2α

tT

−1eK2

M,δT2 ‖g − w‖L2(�) .

Proof Readers can see [12] for the proof.

The following results are helpful for us to give an estimation of uα(gε, aδ, bδ) −u(g, a, b).

Proposition 2 We have

‖uα(gε, a, b)(., t) − uα(gε, aδ, bδ)(., t)‖L2(�) � δ√

R(α)eT 2K2

M,δαtT , (7)

where R(α) = T 2(M+M3

)2|�|α2 ln 1

α

. Hence,

‖uα(g, a, b)(., t) − uα(gε, aδ, bδ)(., t)‖L2(�) � δ√

R(α)eT 2K2

M,δαtT + ε

√2α

tT

−1eK2

M,δT2.

(8)

Proof We have

uα(gε, a, b)(x, t) =+∞∑

n=1

uα,n(gε, a, b)(t)ϕn(x),


where

uα,n(gε, a, b)(t) = eλn(T −t)

1 + αeT λngε,n −

∫ T

t

eλn(s−t)

1 + αsT esλn

[f

a,bM,n(uα(gε, a, b))(s) + hn(s)

]ds.

It follows thatuα,n(gε, a, b)(t) − uα,n(gε, aδ, bδ)(t)

=∫ T

t

eλn(s−t)

1 + αsT esλn

[f

aδ,bδ

M,n (uα(gε, aδ, bδ))(s) − fa,bM,n(uα(gε, a, b))(s)

]ds.

Therefore, ∣∣uα,n(gε, a, b)(t) − uα,n(gε, aδ, bδ)(t)∣∣

�∫ T

t

eλn(s−t)

1 + αsT esλn

∣∣∣f aδ,bδ


∣∣∣ ds.

To get the desired estimation, we need

Lemma 2 For s > t � 0 and β > 0, we have eλn(s−t)

1+βesλn� β

ts−1.

Proof For s > t and β > 0, we have

eλn(s−t)

1 + βesλn= e−λnt

β + e−λns= 1(βesλn + 1

) ts(β + e−λns

)1− ts

� βts−1.

The proof of this lemma is complete.

Now, we turn to the proof of Proposition 2. Applying Lemma 2 with β = αsT , we obtain

∣∣uα,n(gε, a, b)(t) − uα,n(gε, aδ, bδ)(t)∣∣

�∫ T

t

αtT

− sT

∣∣∣f aδ,bδ


∣∣∣ ds.

This inequality implies

α− 2tT ‖uα(gε, a, b)(., t) − uα(gε, aδ, bδ)(., t)‖2L2(�)

�+∞∑

n=1

(∫ T

t

α− sT

∣∣∣f aδ,bδ


∣∣∣ ds

)2

� (T − t)

+∞∑

n=1

∫ T

t

α− 2sT

∣∣∣f aδ,bδ


∣∣∣2ds

= T

∫ T

t

α− 2sT

∥∥∥f aδ,bδ

M (uα(gε, aδ, bδ))(., s) − fa,bM (uα(gε, a, b))(., s)

∥∥∥2

L2(�)ds.

Moreover, we have the inequality


� 2T∫ T

t

α− 2sT

∥∥∥f aδ,bδ

M (uα(gε, aδ, bδ))(., s) − faδ,bδ

M (uα(gε, a, b))(., s)

∥∥∥2

L2(�)ds

+2T∫ T

t

α− 2sT

∥∥∥f aδ,bδ

M (uα(gε, a, b))(., s) − fa,bM (uα(gε, a, b))(., s)

∥∥∥2

L2(�)ds.

D. D. Trong et al.

Because∣∣∣f aδ,bδ

M (uα(gε, a, b))(x, s) − fa,bM (uα(gε, a, b))(x, s)

∣∣∣ �(M + M3

)δ

for all (x, s) in � × [0, T ], we obtain


� 2T

T∫

t

α− 2sT

∥∥∥f aδ,bδ

M (uα(gε, aδ, bδ))(., s) − fa,bM (uα(gε, a, b))(., s)

∥∥∥2

L2(�)ds

+2T(M + M3

)2δ2 |�|

T∫

t

α− 2sT ds

� δ2T 2(M + M3

)2 |�|α2 ln 1

α

+2T

T∫

t

α− 2sT

∥∥∥f aδ,bδ

M (uα(gε, aδ, bδ))(., s) − faδ,bδ

M (uα(gε, a, b))(., s)

∥∥∥2

L2(�)ds.

Putting R(α) = T 2(M+M3)2|�|α2 ln 1

α

, we have


� δ2R(α) + 2T K2M,δ

∫ T

t

α− 2sT ‖uα(gεa, b)(., s) − uα(gε, aδ, bδ)(., s)‖2L2(�)

ds.

We get, in view of the Gronwall inequality


� δ2R(α)e2T (T −t)K2

M,δ .

Therefore, (7) holds. The triangle inequality gives

‖uα(g, a, b)(., t) − uα(gε, aδ, bδ)(., t)‖L2(�) � δ√

R(α)eT 2K2

M,δαtT + ε

√2α

tT

−1eK2

M,δT2.

(9)This completes the proof of the proposition.

Proposition 3 If∑+∞

n=1(∫ T

0 |f a,bn (u(g, a, b))(s) + hn(s)|esλnds)2 < ∞, u(g, a, b)(., 0)

∈ C0(�), then there is an α0 such that

‖uα(g, a, b)(., t) − u(g, a, b)(., t)‖L2(�) �√

Pe2T2K2

M αtT ∀t ∈ (0, T ) (10)

for all 0 < α < α0.In addition, if u(g, a, b)(., 0) ∈ H 1

0 (�), then there exists a tα > 0 such that

‖uα(g, a, b)(., tα) − u(g, a, b)(., 0)‖L2(�) � CM4√8T

(ln

(1

α

))− 14

. (11)


Here,

P = 4 ‖u(g, a, b)(., 0)‖2L2(�)

+ 16+∞∑

n=1

(∫ T

0

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds

)2,

CM = max{√

Pe2T2K2

M ,N}

,

N = max {1, |a|} ‖u(g, a, b)(., 0)‖2H 10 (�)

+ a2

2b|�| + b |�| ‖u(g, a, b)(., 0)‖4C0(�)

+T |�| ‖h‖2L∞(�×[0,T ]) .

In order to prove Proposition 3, we need the following lemma.

Lemma 3 If u(g, a, b)(., 0) ∈ C0(�), then

u(g, a, b) ∈ C ([0, T ]; C0(�)) ∩ C((0, T ], H 1

0 (�))

∩ C1((0, T ]; H 1

0 (�))

,

�u(g, a, b) ∈ C((0, T ]; L2(�)),

sup�×[0,T ]

|u(g, a, b)(x, t)| � e(k+1)T(

‖u(g, a, b)(., 0)‖C0(�) + b121

),

where k = |a| + 12 and b1 = 1

2 ‖h‖2L∞(�×[0,T ]).

Proof Put φ(x) = u(g, a, b)(x, 0), F(x, t, z) = az − bz3 + h(x, t). From Theorem 5.2.1in [11, p. 64], we can find a T (φ) > 0 such that the system

ut − �u = F(x, t, u), x ∈ �, t > 0,

u(x, 0) = φ(x)

has a unique solution u ∈ C([0, θ ]; C0(�)) ∩ C((0, θ ], H 10 (�)) ∩ C1((0, θ ]; H 1

0 (�)) forevery θ ∈ (0, T (φ)). Moreover, we have either T (φ) = ∞ or limt→T (φ) ‖u(., t)‖C0(�) =∞. Now, for 0 < θ < T (φ), we use Theorem 9.5 in [7, p. 214], a quasilinear version of theestimate in Theorem 2.11 in [7, p. 14] to claim

sup�×[0,θ]

|u(g, a, b)(x, t)| � e(k+1)θ(

‖u(g, a, b)(., 0)‖C0(�) + b121

). (12)

In fact, we have to verify the condition zF (x, t, z) ≤ kz2 + b1 of Theorem 2.11 in [7, p.140]. We have

z(az2 − bz3 + h

)� az2 − bz4 + zh � |a| z2 + |z| |h| ≤ |a| z2 + z2 + h2

2.

It implies

z(az2 − bz3 + h

)�(

|a| + 1

2

)z2 + 1

2‖h‖2L∞(�×[0,θ])

and (12) follows. By (12), we have T (φ) = ∞. Hence, we can choose θ = T and get ourlemma.

Proof of Proposition 3 We have

uα(g, a, b)(x, t) =+∞∑

n=1

uα,n(g, a, b)(t)ϕn(x),

D. D. Trong et al.

where

uα,n(g, a, b)(t) = eλn(T −t)

1 + αeT λngn −

∫ T

t

eλn(s−t)

1 + αsT esλn

[f

a,bM,n(uα(g, a, b))(s) + hn(s)

]ds.

On the other hand,

u(g, a, b)(x, t) =+∞∑

n=1

un(g, a, b)(t)ϕn(x),

where

un(g, a, b)(t) = gneλn(T −t) −

∫ T

t

[f a,b

n (u(g, a, b))(s) + hn(s)]eλn(s−t)ds.

It implies

un(g, a, b)(0) = gneλnT −

∫ T

0

[f a,b

n (u(g, a, b))(s) + hn(s)]eλnsds.

So,

gn = un(g, a, b)(0)e−λnT −∫ T

0

[f a,b

n (u(g, a, b))(s) + hn(s)]eλn(s−T )ds. (13)

Now, we consider

un(g, a, b)(t) − uα,n(g, a, b)(t) =(1 − 1

1 + αeT λn

)gne

λn(T −t)

+∫ T

t

eλn(s−t)

1 + αsT esλn

[f


]ds

−∫ T

t

[f a,b

n (u(g, a, b))(s) + hn(s)]eλn(s−t)ds.

From (13), we deduce

un(g, a, b)(t) − uα,n(g, a, b)(t)

= αun(0)eλn(T −t)

1 + αeT λn+∫ T

0

αeλn(T −t)

1 + αeT λn

[f a,b

n (u(g, a, b))(s) + hn(s)]eλnsds

+∫ T

t

eλn(s−t)

1 + αsT esλn

[f


]ds

−∫ T

t

[f a,b

n (u(g, a, b))(s) + hn(s)]eλn(s−t)ds

= αun(g, a, b)(0)eλn(T −t)

1 + αeT λn+∫ T

0

αeλn(T −t)

1 + αeT λn

[f a,b

n (u(g, a, b))(s) + hn(s)]eλnsds

+∫ T

t

eλn(s−t)

1 + αsT esλn

[f

a,bM,n(uα(g, a, b))(s) − f

a,bM,n(u(g, a, b))(s)

]ds

+∫ T

t

eλn(s−t)

1 + αsT esλn

[f

a,bM,n(u(g, a, b)(s) − f a,b

n (u(g, a, b))(s)]ds

−∫ T

t

esλnαsT eλn(s−t)

1 + αsT esλn

[f a,b

n (u(g, a, b)(s) + hn(s)]ds.


From Lemma 2, we get

eλn(T −t)

1 + αeT λn� α

tT

−1,eλn(s−t)

1 + αsT esλn

� αtT

− sT . (14)

Using (14), we have∣∣un(t) − uα,n(t)

∣∣

� αtT |un(g, a, b)(0)| + α

tT

∫ T

0

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds

+αtT

∫ T

t

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds

+αtT

∫ T

t

α− sT

∣∣∣fM,n(uα(g, a, b))(s) − fa,bM,n(u(g, a, b))(s)

∣∣∣ ds

+αtT

∫ T

t

α− sT

∣∣∣f a,bM,n(u(g, a, b))(s) − f a,b

n (u(g, a, b))(s)

∣∣∣ ds.

Thus,∣∣un(t) − uα,n(t)

∣∣ � An + Bn + Cn + Dn,

where

An = αtT |un(g, a, b)(0)| ,

Bn = 2αtT

∫ T

0

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds,

Cn = αtT

∫ T

t

α− sT

∣∣∣f a,bM,n(uα(g, a, b))(s) − f


∣∣∣ ds,

Dn = αtT

∫ T

t

α− sT

∣∣∣f a,bM,n(u(g, a, b))(s) − f a,b

n (u(g, a, b))(s)

∣∣∣ ds.

So, we get

‖u(g, a, b)(., t) − uα(g, a, b)(., t)‖2L2(�)

� 4

(+∞∑

n=1

A2n +

+∞∑

n=1

B2n +

+∞∑

n=1

C2n +

+∞∑

n=1

D2n

),

(15)and we deduce

α− 2tT ‖u(g, a, b)(., t) − uα(g, a, b)(., t)‖2

L2(�)

� 4 ‖u(g, a, b)(., 0)‖2L2(�)

+ 16+∞∑

n=1

(∫ T

0

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds

)2

+4+∞∑

n=1

(∫ T

t

α− sT

∣∣∣f a,bM,n(uα(g, a, b))(s) − f


∣∣∣ ds

)2

+4+∞∑

n=1

(∫ T

t

α− sT

∣∣∣f a,bM,n(u(g, a, b))(s) − f a,b

n (u(g, a, b))(s)

∣∣∣ ds

)2.

D. D. Trong et al.

Using the Holder inequality

α− 2tT ‖u(g, a, b)(., t) − uα(g, a, b)(., t)‖2

L2(�)

� 4 ‖u(g, a, b)(., 0)‖2L2(�)

+ 16+∞∑

n=1

(∫ T

0

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds

)2

+4 (T − t)

∫ T

t

α− 2sT

∥∥∥f a,bM (uα(g, a, b))(., s) − f

a,bM (u(g, a, b))(., s)

∥∥∥2

L2(�)ds

+4 (T − t)

∫ T

t

α− 2sT

∥∥∥f a,bM (u(g, a, b))(., s) − f a,b(u(g, a, b))(., s)

∥∥∥2

L2(�)ds.

Put

P = 4 ‖u(g, a, b)(., 0)‖2L2(�)

+ 16+∞∑

n=1

(∫ T

0

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds

)2.

Using Lemma 1, we deduce

α− 2tT ‖uα(g, a, b)(., t) − u(g, a, b)(., t)‖2

L2(�)

� P + 4 (T − t) K2M

T∫

t

α− 2sT ‖uα(g, a, b)(., s) − u(g, a, b)(., s)‖2

L2(�)ds

+4 (T − t)

T∫

t

α− 2sT

∥∥∥f a,bM (u(g, a, b))(., s) − f a,b(u(g, a, b))(., s)

∥∥∥2

L2(�)ds.

Put

M0 = e(|a|+ 12+1)T

(‖u(g, a, b)(., 0)‖C0(�) + 1√

2‖h‖L∞(�×[0,T ])

).

Lemma 3 implies that sup�×[0,T ] |u(g, a, b)(x, t)| � M0. Because M → +∞ as α → 0+,there exists an α0 > 0 such that M > M0 for all α ∈ (0, α0). Then, f

a,bM (u(g, a, b))

= f a,b(u(g, a, b)) and we get

α− 2tT ‖uα(g, a, b)(., t) − u(g, a, b)(., t)‖2

L2(�)

� P + 4T K2M

∫ T

t

α− 2sT ‖uα(g, a, b)(., s) − u(g, a, b)(., s)‖2

L2(�)ds.

The Gronwall inequality gives

‖uα(g, a, b)(., t) − u(g, a, b)(., t)‖L2(�) ≤ √Pe2T

2K2M α

tT . (16)

Now, we multiply the equation ut = �u + f (u) + h by ut and integrate the result thusobtained on � to get

∫

�

u2t dx =∫

�

�uutdx +∫

�

(auut − bu3ut

)dx +

∫

�

hutdx,

where u = u(g, a, b) for short. It follows that∫

�

u2t dx = −1

2

d

dt‖∇u(., t)‖2

L2(�)+∫

�

(auut − bu3ut

)dx +

∫

�

hutdx.


Integrating both sides of the equation on [0, t], we obtain∫ t

0

∫

�

u2s dxds + 1

2‖∇u(., t)‖2

L2(�)+ b

4

∫

�

[u(x, t)]4 dx

= 1

2‖∇u(., 0)‖2

L2(�)+ a

2

∫

�

[u(x, t)]2 dx − a

2‖u(., 0)‖2

L2(�)+ b

4

∫

�

[u(x, 0)]4 dx

+∫ t

0

∫

�

husdxds.

Hence,∫ t

0

∫

�

u2s dxds + 1

2‖∇u(., t)‖2

L2(�)+ b

4

∫

�

[u(x, t)]4 dx

� 1

2‖∇u(., 0)‖2

L2(�)+∫

�

a√2b

√b√2[u(x, t)]2 dx + |a|

2‖u(., 0)‖2

L2(�)

+b

4

∫

�

[u(x, 0)]4 dx +∫ t

0

∫

�

h2 + u2s

2dxds

� max {1, |a|}2

‖u(., 0)‖2H 10 (�)

+∫

�

(a2

4b+ b

4[u(x, t)]4

)dx + b

4

∫

�

[u(x, 0)]4 dx

+∫ t

0

∫

�

h2 + u2s

2dxds.

It follows that

1

2

∫ t

0

∫

�

u2s dxds + 1

2‖∇u(., t)‖2

L2(�)

� max {1, |a|}2

‖u(., 0)‖2H 10 (�)

+ a2

4b|�| + b

2|�| ‖u(., 0)‖4C0(�)+

T

2|�| ‖h‖2L∞(�×[0,T ]) .

Thus, we can get∫ t

0

∥∥∥∥∂u

∂s(., s)

∥∥∥∥2

L2(�)

ds

� max {1, |a|} ‖u(., 0)‖2H 10 (�)

+ a2

2b|�| + b |�| ‖u(., 0)‖4C0(�) + T |�| ‖h‖2L∞(�×[0,T ]) .

On the other hand,

u(., t) − u(., 0) =∫ t

0

∂u

∂s(., s)ds.

The Holder inequality gives

|u(., t) − u(., 0)|2 � t

∫ t

0

∣∣∣∣∂u

∂s(., s)

∣∣∣∣2

ds.

It follows that

‖u(., t) − u(., 0)‖2L2(�)

� t

∫ t

0

∥∥∥∥∂u

∂s(., s)

∥∥∥∥2

L2(�)

ds � N2t, (17)

where we recall u(., t) = u(g, a, b)(., t) and

N = max {1, |a|} ‖u(., 0)‖2H 10 (�)

+ a2

2b|�| + b |�| ‖u(., 0)‖4C0(�) + T |�| ‖h‖2L∞(�×[0,T ]) .

D. D. Trong et al.

For all t ∈ (0, T ), we have the inequality

‖uα(g, a, b)(., t) − u(g, a, b)(., 0)‖L2(�)

� ‖uα(g, a, b)(., t) − u(g, a, b)(., t)‖L2(�) + ‖u(g, a, b)(., t) − u(g, a, b)(., 0)‖L2(�) .

Putting CM = max{√Pe2T2K2

M ,N}, we obtain, in view of (16), (17)

‖uα(g, a, b)(., t) − u(g, a, b)(., 0)‖L2(�) � CM

(√t + α

tT

). (18)

For 0 < α < 1, the equation√

t = αtT has a unique solution tα and ln tα

tα= 2 lnα

T. Using the

inequality ln t > − 1tfor all t > 0, we have

2 lnα

T= ln tα

tα> − 1

t2α

and√

tα <4

√T

2

(ln

(1

α

))− 14

. (19)

From (18), (19), we obtain

‖uα(g, a, b)(., tα) − u(g, a, b)(., 0)‖L2(�) � CM

(√tα + α

tαT

)

� CM4√8T

(ln

(1

α

))− 14

.

We complete this proof.

Proof of Theorem 1 For all t ∈ (0, T ), we have

‖uα(gε, aδ, bδ)(., t) − u(g, a, b)(., t)‖L2(�)

� ‖uα(gε, aδ, bδ)(., t) − uα(g, a, b)(., t)‖L2(�) + ‖uα(g, a, b)(., t) − u(g, a, b)(., t)‖L2 (�).

Combining (8) in Proposition 2 and (10) in Proposition 3 gives


� δ√

R(α)eT 2K2

M,δαtT + ε

√2α

tT

−1eT 2K2

M,δ + √Pe2T

2K2M α

tT .

Furthermore, when 0 < α < 1, there exists uniquely a tα such that√

tα = αtαT . At this time,

we have

‖uα(gε, aδ, bδ)(., tα) − u(g, a, b)(., 0)‖L2(�)

� ‖uα(gε, aδ, bδ)(., tα) − uα(g, a, b)(., tα)‖L2(�)

+‖uα(g, a, b)(., tα) − u(g, a, b)(., 0)‖L2(�).

From (8), (11), and (19), we deduce


�(δ√

R(α)eT 2K2

M,δ + εα−1eT 2K2

M,δ + CM

)4√8T

(ln

(1

α

))− 14

.

Choosing α = max{δT |�| 12 , ε} gives

δ√

R(α) = δT (M + M3)|�| 12αln

12

(1α

) � M + M3

ln12

(1α

) .


Therefore, we obtain the results


�(

M + M3

ln12 1

α

e2T 2K2

M,δ + √2e2T

2K2M,δ + √

Pe2T2K2

M

)α

tT ,


�(

M + M3

ln12 1

α

+ 1

)e2T 2K2

M,δ4√8T

(ln

(1

α

))− 14 + CM

4√8T

(ln

(1

α

))− 14

.

On the other hand, we also have

K2M =

(|a| + 3bM2

)2� 4H 2

δ M4, K2M,δ =

(|aδ| + 3bδM

2)2

� 4H 2δ M4.

Moreover, because of the choice

M := Mα = 4

√q

8T 2H 2δ

ln

(ln

1

α

), (20)

where 0 < q < min{

tT

, 14

}, t ∈ (0, T ), we obtain


�

⎛

⎝q ln

(ln(1α

))

4T 2H 2δ ln

12

(1α

) + √2 + √

P

⎞

⎠ lnq

(1

α

)α

tT ,


�

⎛

⎝q ln

(ln(1α

))

4T 2H 2δ ln

12

(1α

) + 1

⎞

⎠ 4√8T

(ln

(1

α

))q− 14 + CM

4√8T

(ln

(1

α

))− 14

,

where CM = max{√

Pe2T2K2

M ,N}and the proof of Theorem 1 is complete.

Proof of Corollary 1 For 0 < t < T , we have

‖u(g, aδ, bδ)(., t) − u(g, a, b)(., t)‖L2(�)

� ‖u(g, aδ, bδ)(., t) − uα(g, aδ, bδ)(., t)‖L2(�)

+‖uα(g, aδ, bδ)(., t) − uα(g, a, b)(., t)‖L2(�)

+‖uα(g, a, b)(., t) − u(g, a, b)(., t)‖L2(�).

From (10), (7), we have

‖u(g, aδ, bδ)(., t) − uα(g, aδ, bδ)(., t)‖L2(�) �√

Pδe2T 2K2

M,δαtT ,

‖uα(g, aδ, bδ)(., t) − uα(g, a, b)(., t)‖L2(�) � δ√

R(α)eT 2K2

M,δαtT ,

‖uα(g, a, b)(., t) − u(g, a, b)(., t)‖L2(�) �√

Pe2T2K2

M αtT ,

D. D. Trong et al.

where

KM = |a| + 3bM2, KM,δ = |aδ| + 3bδM2, R(α) = T 2

(M + M3

)2 |�|α2 ln 1

α

,

P = 4 ‖u(g, a, b)(., 0)‖2L2(�)

+16+∞∑

n=1

(∫ T

0

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds

)2,

Pδ = 4 ‖u(g, aδ, bδ)(., 0)‖2L2(�)

+16+∞∑

n=1

(∫ T

0

∣∣∣f aδ,bδn (u(g, aδ, bδ))(s) + hn(s)

∣∣∣ esλnds

)2.

Hence,

‖u(g, aδ, bδ)(., t) − u(g, a, b)(., t)‖L2(�)

�√

Pδe2T 2K2

M,δαtT + δ

√R(α)e

T 2K2M,δα

tT + √

Pe2T2K2

M αtT ,

and the result is

‖u(g, aδ, bδ)(., t) − u(g, a, b)(., t)‖L2(�)

�

⎛

⎝q ln

(ln(1α

))

4T 2H 2δ ln

12

(1α

) +√Pδ + √P

⎞

⎠(ln

(1

α

))q

αtT .

On the other hand, for α > 0 small enough, we have

‖u(g, aδ, bδ(., tα) − u(g, a, b)(., 0)‖L2(�)

� ‖u(g, aδ, bδ)(., tα) − uα(g, aδ, bδ)(., tα)‖L2(�)

+‖uα(g, aδ, bδ)(., tα) − uα(g, a, b)(., tα)‖L2(�)

+‖uα(g, aδ, bδ)(., tα) − u(g, a, b)(., 0)‖L2(�).

From (7), (10), (11), we get

‖u(g, aδ, bδ)(., tα) − uα(g, aδ, bδ)(., tα)‖L2(�) �√

Pδe2T 2K2

M,δαtαT ,

‖uα(g, aδ, bδ)(., tα) − uα(g, a, b)(., tα)‖L2(�) � δ√

R(α)eT 2K2

M,δαtαT ,

‖uα(g, a, b)(., tα) − u(g, a, b)(., 0)‖L2(�) � CM,δ

(√tα + α

tαT

),

where

CM,δ = max{√

Pδe2T 2K2

M,δ , Nδ

},

Pδ = 4 ‖uδ(., 0)‖2L2(�)

+16+∞∑

n=1

(∫ T

0

∣∣∣f aδ,bδn (u)(s) + hn(s)

∣∣∣ esλnds

)2,

Nδ = max {1, |aδ|} ‖u(g, aδ, bδ)(., 0)‖2H 10 (�)

+ a2δ

2bδ

|�|+b |�| ‖u(g, aδ, bδ)(., 0)‖4C0(�) + T |�| ‖h‖2L∞(�×[0,T ]) .


Therefore, we deduce

‖u(g, aδ, bδ)(., tα) − u(g, a, b)(., 0)‖L2(�)

�√

Pδe2T 2K2

M,δαtαT + δ

√R(α)e

T 2K2M,δα

tαT + CM,δ

(√tα + α

tαT

).

So,

‖u(g, aδ, bδ)(., tα) − u(g, a, b)(., 0)‖L2(�)

�(√

Pδe2T 2K2

M,δ + M + M3

ln12 1

α

e2T 2K2

M,δ + CM,δ

)4√8T

(ln

(1

α

))− 14

.

The choice for M implies

‖u(g, aδ, bδ)(., tα) − u(g, a, b)(., 0)‖L2(�)

�

⎛

⎝√

Pδ +q ln

(ln(1α

))

4T 2H 2δ ln

12

(1α

)

⎞

⎠ 4√8T

(ln

(1

α

))q− 14 + Qδ

4√8T

(ln

(1

α

))− 14

,

where

Qδ = max

{√Pδ

(ln

(1

α

))q

, Nδ

}.

The proof of Corollary 1 is complete.

Proof of Corollary 2 By the same method as in Corollary 1, for 0 < t < T , we have

‖u(g, aδ, bδ)(., t) − u(g, a, b)(., t)‖L2(�) � ‖u(g, aδ, bδ)(., t) − uα(g, aδ, bδ)(., t)‖L2(�)

+‖uα(g, aδ, bδ)(., t)−uα(g, a, b)(., t)‖L2(�)

+‖uα(g, a, b)(., t) − u(g, a, b)(., t)‖L2(�).

Then, we deduce that

‖u(g, aδ, bδ)(., t) − u(g, a, b)(., t)‖L2(�) �√

Pδe2T 2K2

M,δαtT + δ

√R(α)e

T 2K2M,δα

tT

+√Pe2T

2K2M α

tT . (21)

We choose α = δ and M = L. Then, from (21), we infer

‖u(g, aδ, bδ)(., t) − u(g, a, b)(., t)‖L2(�)

�(√

Pδe2T 2K2

L,δ + T (L + L3)|�| 12ln

12 1

δ

e2T 2K2

L,δ + √Pe2T

2K2L

)δ

tT , t ∈ (0, T ).

Moreover, if α is small enough then

‖u(g, aδ, bδ)(., 0) − u(g, a, b)(., 0)‖L2(�)

� ‖u(g, aδ, bδ)(., 0) − uα(g, aδ, bδ)(., tα)‖L2(�)

+‖uα(g, aδ, bδ)(., tα) − uα(g, a, b)(., tα)‖L2(�)

+‖uα(g, a, b)(., tα) − u(g, a, b)(., 0)‖L2(�).

D. D. Trong et al.

Because

‖uα(g, aδ, bδ)(., tα) − u(g, aδ, bδ)(., 0)‖L2(�) � CM,δ4√8T

(ln

(1

α

))− 14

,

‖uα(g, a, b)(., tα) − u(g, a, b)(., 0)‖L2(�) � CM4√8T

(ln

(1

α

))− 14

,

‖uα(g, a, b)(., t) − uα(g, aδ, bδ)(., t)‖L2(�) � δ√

R(α)eT 2K2

M,δαtT ,

we have the following result

‖u(g, aδ, bδ)(., 0) − u(g, a, b)(., 0)‖L2(�)

� CM,δ4√8T

(ln

(1

α

))− 14 + δ

√R(α)e

T 2K2M,δα

tαT + CM

4√8T

(ln

(1

α

))− 14

.

We also have

αtαT = √

tα <4

√T

2

(ln

(1

α

))− 14

.

So, we deduce

‖u(g, aδ, bδ)(., 0) − u(g, a, b)(., 0)‖L2(�)

�(

CM,δ + δT (M + M3)|�| 12αln

12 1

α

eT 2K2

M,δ + CM

)4√8T

(ln

(1

α

))− 14

.

Choosing α = δ and M = L, we have the following estimation at t = 0

‖u(g, aδ, bδ)(., 0) − u(g, a, b)(., 0)‖L2(�)

�(

CL,δ + T (L + L3)|�| 12ln

12 1

δ

eT 2K2

L,δ + CL

)4√8T

(ln

(1

δ

))− 14

,

where

CL,δ = max{√

Pδe2T 2K2

L,δ , Nδ

}, CL = max

{√Pe2T

2K2L,N

},

Nδ = max {1, |aδ|} ‖u(g, aδ, bδ)(., 0)‖2H 10 (�)

+ a2δ

2bδ

|�|+b |�| ‖u(g, aδ, bδ)(., 0)‖4C0(�) + T |�| ‖h‖2L∞(�×[0,T ]) .

The proof of Corollary 2 is complete.

Proof of Corollary 3 The triangle inequality gives

‖uα(gε, a, b)(., t) − u (g, a, b) (., t)‖L2(�) � ‖uα(gε, a, b)(., t) − uα(g, a, b)(., t)‖L2(�)

+‖uα(g, a, b)(., t)−u (g, a, b) (., t)‖L2(�).

For t ∈ (0, T ) and from the result of Proposition 1, the proof of Proposition 3, we obtainthat

‖uα(gε, a, b)(., t) − u (g, a, b) (., t)‖L2(�)

�√2α

tT

−1eK2MT 2‖g − gε‖L2(�) + √

Pe2T2K2

M αtT .


With δ = 0, α = ε and M = R, we have

‖uε(gε, a, b)(., t) − u (g, a, b) (., t)‖L2(�) �√2ε

tT

−1eK2RT 2

ε + √Pe2T

2K2Rε

tT .

So, we deduce

‖uε(gε, a, b)(., t) − u (g, a, b) (., t)‖L2(�) �(√

2 + √P)

e2K2RT 2

εtT ,

where ε is small enough. Moreover, we also have

‖uα(gε, a, b)(., tα) − u (g, a, b) (., 0)‖L2(�) � ‖uα(gε, a, b)(, tα) − uα(g, a, b)(., tα)‖L2(�)

+‖uα(g, a, b)(., tα) − u (g, a, b) (., 0)‖L2(�).

The result of Proposition 1 and the proof of Proposition 3 imply

‖uα(gε, a, b)(., tα) − u (g, a, b) (., 0)‖L2(�)

�√2α

tαT

−1eK2RT 2‖g − gε‖L2(�) + CR

4√8T

(ln

(1

α

))− 14

.

So, we get

‖uε(gε, a, b)(., tε) − u (g, a, b) (., 0)‖L2(�) �(eK2

RT 2 + CR

)4√8T

(ln

(1

ε

))− 14

and the proof is complete.

4 Numerical Experiment

In this section, we examine a specific problem to illustrate the effect of the regularizationmethod

ut = �u + au − bu3 + h(x, t), x ∈ (0, π), t ∈ (0, 1) ,

u(0, t) = u(π, t) = 0, t ∈ (0, 1) ,

u(x, T ) = g(x), x ∈ [0, π ],

Fig. 1 Measured data: error 10−1 (left) and 10−2 (right)

D. D. Trong et al.

Table 1 Approximated tα

ε 10−1 10−2 10−3 10−4 10−5 10−6

tα 0.28 0.18 0.14 0.12 0.10 0.09

where a = b = T = 1 and

g(x) = 0.6

(sin(x) + 1

2sin(2x)

),

h(x, t) = 0.6(1 − t)

(sin(x) + 1

2sin(2x)

)− 0.63t

(sin(x) + 1

2sin(2x)

)3

+0.6t (sin(x) + 2 sin(2x)) .

The exact solution of the system is

uex(x, t) = 0.6t

(sin(x) + 1

2sin(2x)

).

Therefore, we can check the source condition in Theorem 1

+∞∑

n=1

(∫ T

0

∣∣∣f a,bn (u(g, a, b))(s) + hn(s)

∣∣∣ esλnds

)2

=(∫ T

00.6(1 + s)esds

)2+(∫ T

00.6(1 + 2s)e4sds

)2< +∞. (22)

We focus on investigating the case that condition data are measured with noise model

gε(x) = (1 + ε)g(x).

We also denote the measured data by dot and the exact data by line in the figures.We divide the time interval [0, T ] by the points tj = jT

nt, j = 1, . . . , nt and the spatial

interval [0, π ] by points xi = iπnx

, i = 1, . . . , nx . At the same time, to simulate a noisegenerated during the measurement, we use a random number generator with the uniformdistribution in [0, 1]. Here, we denote the approximate solution by dot line and the exact

Fig. 2 Regularization solution at t = 0.50 with ε = 10−1, . . . , 10−6


Fig. 3 Regularization solution at t = 0.25 with ε = 10−1, . . . , 10−6

solution by line in figures

uε(xj , T ) = 0.6

(sin(x) + 1

2sin(2x)

)(1 + ε.uni())

where ε is the level noise. Figure 1 below illustrates the data corresponding to noise levelsε = 10−1 and ε = 10−2.

We choose α = ε. From (20), we have

Mα = 4

√q

8T 2H 2δ

ln

(ln

1

α

), 0 < q < min

{t

T,1

4

}∀t ∈ (0, T ),

where Hδ = max{a, 3b} = max{1, 3} = 3. Choosing q = min{

tT

, 14

}− 0.001, we have

Mε = Mα = 4

√q

72ln

(ln

1

ε

).

Fig. 4 Regularization solution at t = 0 with ε = 10−1, . . . , 10−6

D. D. Trong et al.

Table 2 Comparing of errors in case Mε as in (20)

ε Mε Iterate Error at t = 0.50 Error at t = 0.25 Error at t = 0

10−1 0.23198 5 0.28157 0.48435 0.56492

10−2 0.26985 7 0.12630 0.39052 0.64163

10−3 0.28621 8 0.10781 0.31495 0.60378

10−4 0.29631 8 0.07049 0.19129 0.45833

10−5 0.30349 9 0.05703 0.14167 0.41867

10−6 0.30900 9 0.05201 0.12182 0.40423

The eigenvalues and eigenfunctions corresponding to the problem are λn = n2 and φn(x) =√2πsin(nx). Because of

en2(T −t)

1 + εT n2= e−n2t

e−n2T + ε≤ 1

εe−n2t ,

we have

en2(T −t)

1 + εT n2≤ 1

εe−n2tα , t ≥ tα,

where tα is the solution of equation√

tα = εtα (Table 1).

If ε = 10−6 and tα ≈ 0.09, we get 1εe−tαn2 < 10−6 when n >

√log(1012)

0.09 = 17.52174.Hence, we truncate the Fourier series for n ≥ 18. The approximation solution can be seenas the solution of

uε(x, t) =18∑

n=1

(en2(T −t)

1 + εeT n2gε,n −

∫ T

t

en2(s−t)

1 + εsT

esn2[f1,1Mε,n

(uε)(s) + hn(s)]ds

)sin(nx).

The iteration formula is

uk+1ε (x, t) =

18∑

n=1

(en2(T −t)

1 + εeT n2gn −

∫ T

t

en2(s−t)

1 + εsT esn2

[f1,1Mε,n

(ukε)(s) + hn(s)

]ds

)sin(nx)

Fig. 5 Regularization solution at t = 0.50,Mε = 1 with ε = 10−1, . . . , 10−5


Fig. 6 Regularization solution at t = 0.25,Mε = 1 with ε = 10−1, . . . , 10−6

where u0ε = sin(x). It will stop when

‖uk+1ε − uk

ε‖L2(�) < ε−10.

Here, instead of L2(0, π), we change L2 - error by root mean squared error

‖u − v‖L2(0,π) ≈ RMSE(u, v) =√√√√ 1

nx

nx∑

i=1

(u(xi) − v(xi))2.

To compute the Fourier coefficients in the formula (4), we use the numerical integration ofthe highly oscillatory function proposed by Filon in [3]. For 0 ≤ x0 < x2n ≤ π and f

defined on [x0, x2n], we have∫ x2n

x0

f (x) sin(px)dx = h[α(f0 cos(px0) − f2n cos(px2n)) + βS2n + γ S2n−1

]+ Es

Fig. 7 Regularization solution at t = 0,Mε = 1 with ε = 10−1, . . . , 10−6

D. D. Trong et al.

Table 3 Comparing of errors in case Mε = 1

ε Mε Iterate Error at t = 0.50 Error at t = 0.25 Error at t = 0

10−1 1.0 6 0.19516 0.31044 0.48310

10−2 1.0 6 0.06503 0.28417 0.44955

10−3 1.0 8 0.04697 0.16393 0.30481

10−4 1.0 9 0.01263 0.05634 0.14636

10−5 1.0 9 0.00245 0.01939 0.08698

10−6 1.0 9 0.00146 0.00737 0.06391

where

fi = f (xi), h = xi+1 − xi, θ = ph,

α = 1

θ+ sin(2θ)

2θ2− 2 sin2(θ)

θ3,

β = 2

[1 + cos2(θ)

θ2− sin(2θ)

θ3

],

γ = 4

[sin(θ)

θ3− cos(θ)

θ2

],

S2n−1 =n∑

i=1

f2i−1 sin(px2i−1),

S2n =n∑

i=1

f2i sin(px2i−1) − 1

2

[f0 sin(px0) + f2p sin(px2n)

],

Es = 1

90nh5f (4)(ξ) + O(ph7).

To calculate integrals with respect to the time variable t , we use the extended Simpsonformula for 0 ≤ t1 < t2 < · · · < tn ≤ T

∫ tn

t1

w(t)dt

≈ h

(3

8w(t1) + 7

6w(t2) + 23

24w(t3) +

n−3∑

i=3

w(ti) + 23

24w(tn−2) + 7

6w(tn−1) + 3

8w(tn)

).

Figures 2, 3, and 4 show the graphs of the approximation solutions and the exact solutioncorresponding to the noise: ε = 10−1, . . . , ε = 10−6 at t = 0.5, t = 0.25 and t =0, respectively. In each time slice, the line represents the exact solution and the dot linecorresponds to the approximation solution. Comparing of the errors in Table 2 at t = 0,t = 0.25, t = 0.50, we can see that the convergence slows down when time is close to zero.This reflects the behavior of rate of convergence in the theory. However, in experiments,physicists can know a priori information about the initial datum ‖u(g, a, b)(., 0)‖C0(�).

Therefore, we can choose Mε = e(k+1)T (‖u(g, a, b)(., 0)‖C0(�) + b1/21 ) as in Lemma 6.

From this idea, we can significantly improve the speed of convergence of the method ofadjustment. We can see this fact on Figs. 5, 6, and 7 and Table 3.


5 Conclusion

In the paper, we have considered the problem of finding a function u satisfying (1)–(3). Thisis a backward problem for Ginzburg-Landau-type equation, and the problem is ill-posed. Toregularize the problem, we give a method to approximate the function f (u) = au − bu3

by a globally Lipschitz function fM . Using the idea of quasi-boundary method, we findthe regularization solution from the approximate equation (4). Error estimations betweenthe exact solution and the approximate solution, established from noise data gε , aδ , bδ , aregiven. The paper extends the work by the authors in [1, 12–14]. Here, we obtain the error at0 < t < T is of Holder type (the rate ε

tT ). However, the error at t = 0 is of logarithmic rate,

so the convergence rate is very slow, we also mentioned that in Remark 2. To get a betterestimate, we have to add stronger assumptions on the exact solution and we will improve itin the future papers.

Acknowledgments We would like to thank the anonymous referee for constructive criticisms leading tothe improvement of our paper. The paper is supported by Vietnam National University - HoChiMinh City,Project B2014-18-01.

References

1. Ames, K.A.: Continuous dependence on modelling and non-existence results for a Ginzburg-Landauequation. Math. Meth. Appl. Sci. 23, 1537–1550 (2000)

2. Ames, K.A., Clark, G.W., Epperson, J.F., Oppenheimer, S.F.: A comparison of regularizations for anill-posed problem. Math. Comp. 67, 1451–1471 (1998)

3. Filon, L.N.G.: On a quadrature formula for trigonometric integrals. Proc. Roy Soc. Edinburgh (1928)4. Fu, C.L., Qian, Z., Shi, R.: A modified method for a backward heat conduction problem. Appl. Math.

Comput. 185, 564–573 (2007)5. Hetrick, B.C., Hughes, R.J.: Continuous dependence on modeling for nonlinear ill-posed problems. J.

Math. Anal. Appl. 349, 420–435 (2009)6. Katou, K.: Asymptotic spatial patterns on the complex time-dependent Ginzburg-Landau equation. J.

Phys. A 19, 1063–1066 (1986)7. Lieberman, G.M.: Second order parabolic differential equations. World Scientific Publishing Co. Pte.

Ltd (1996)8. Maugin, G.A., Muschik, W.: Thermodynamics with internal variables Part I. General Concepts. J. Non-

Equib. Thermodyn. 19, 217–249 (1994)9. Nam, P.T.: An approximate solution for nonlinear backward parabolic equations. J. Math. Anal. Appl.

367(2), 337–349 (2010)10. Tautenhahn, U., Schroter, T.: On optimal regularization methods for the backward heat equation.

Zeitschrift Analysis und ihre Anwendungen 15, 475–493 (1996)11. Thierry, C., Alain, H., Yvan, M.: An introduction to semilinear evolution equations. Clarendon Press,

Oxford (1998)12. Trong, D.D., Quan, P.H., Tuan, N.H., Khanh, T.V.: A nonlinear case of the 1D backward heat problem:

regularization and error estimate. Zeitschrift Analysis und ihre Anwendungen 26(2), 231–245 (2007)13. Trong, D.D., Quan, P.H., Tuan, N.H.: A quasi-boundary value method for regularizing nonlinear ill-

posed problems. Electron. J. Differ. Equ. 2009(109), 1–16 (2009)14. Trong, D.D., Tuan, N.H.: Remarks on a 2D nonlinear backward heat problem using a truncated Fourier

series method. Electron. J. Differ. Equ. 2009(77), 1–13 (2009)15. Trong, D.D., Tuan, N.H.: Stabilized quasi-reversibility method for a class of nonlinear ill-posed

problems. Electron. J. Differ. Equ. 2008(84), 1–12 (2008)16. Tuan, N.H.: Stability estimates for a class of semi-linear ill-posed problems. Nonlinear Anal. Real World

Appl. 14, 1203–1215 (2013)

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