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Electronic Journal of Qualitative Theory of Differential Equations 2013, No. 7, 1-15;http://www.math.u-szeged.hu/ejqtde/

Time-periodic solution for a

fourth-order parabolic equation describing crystal surface growth

Xiaopeng Zhao, Bo Liu, Ning Duan

College of Mathematics, Jilin University, Changchun, 130012, P. R. China

Abstract. In this paper, by using the Galerkin method, the existence and uniqueness of time-periodic generalized solutions to a fourth-order parabolic equation describing crystal surface growth are proved.

Keywords. Time-periodic solution, fourth-order parabolic equation, Galerkin method.

1 Introduction

In the study of crystal surface growth, there arises the following diffusion equa- tion

ut=−jx+f(x, t),

where u(x, t) denotes the variation of height from the average, j is the atom current parallel to the surface, andf(x, t) is a noise term caused by shot noise in the incoming flux. Takingj=uxxx+1+|uuxx|2, we obtain the well-known BCF model (see [4, 5, 7, 8, 11, 13])

ut+uxxxx+

ux

1 +|ux|2

x

=f(x, t), in (0,1)×R. (1.1) During the past years, many authors have paid much attention to the equation (1.1). It was Rost and Krug [13] who studied the unstable epitaxy on singular surfaces using equation (1.1) with a prescribed slopedependent surface current.

In their paper, they derived scaling relations for the late stage of growth, where power law coarsening of the mound morphology is observed. In [11], in the limit of weak desorption, O. Pierre-Louis et al. derived the equation (1.1) for a vicinal surface growing in the step flow mode. This limit turned out to be singular, and nonlinearities of arbitrary order need to be taken into account.

Recently, H. Fujimura and A. Yagi [4] considered the equation of (1.1). In their paper, the uniqueness local solutions and the global solutions are obtained, a dynamical system determined from the initial-boundary value problem of the model equation was also constructed. In [5], H. Fujimura and A. Yagi continued a study on the model equation (1.1). They considered the asymptotic behavior of trajectories of the dynamical system by constructing exponential attractors and a Lyapunov function. There is much literature concerned with the Eq.(1.1), for more results we refer the reader to [6, 14] and the references therein.

This work was supported by Graduate Innovation Fund of Jilin University (Project 20121059).

Corresponding author, E-mail address: zxp032@gmail.com.

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Furthermore, several authors have paid attention to the time-periodic prob- lems [1, 19, 20]. But, to the best of our knowledge, only a few papers deal with time periodic solutions of fourth-order diffusion equations. In [10, 17], the exis- tence of time periodic solutions for the Cahn-Hilliard type equation and viscous Cahn-Hilliard equation with periodic concentration dependent potentials and sources has been investigated. In [15, 16], Wang et. al. considered the existence and uniqueness of time-periodic generalized solutions and time-periodic classi- cal solutions to the generalized Ginzburg-Landau model equation in 1D and 2D case. In [3], by using the Galerkin method and the Leray-Schauder fixed point theorem, Fu and Guo studied the existence and uniqueness of a time periodic solution for the viscous Camassa-Holm equation. There are also many papers were denoted to the periodic problems, for example [9, 12, 18] and so on.

Here, we investigate the existence and uniqueness of time-periodic general- ized solutions to the equation (1.1) in one spatial dimension together with the condition

u(x+ 1, t) =u(x, t), t∈R, (1.2) and the time-periodic condition

u(x, t+ω) =u(x, t), x∈(0,1), t∈R, (1.3) whereω >0 is a constant andf(x, t) isω-periodic functions with respect to the timet, which also satisfies

Z

f(x, t)dx= 0.

Throughout this paper, we use the following notations.

LetX be a Banach space,Cωk(R;X) denotes the set ofX-valuedω-periodic functions onRwith continuous derivatives up to orderk. The norm inCωk(R;X) is defined as

kukCk

ω(R;X)= sup

0≤t≤ω

{

k

X

i=0

kDtiukX},

whereDt=∂t,k · kX is the norm inX. We also defineLpω(R;X) (1≤p≤ ∞) as the set ofω-periodicX-valued measurable functions onRsuch that

kukLpω(R;X)= Z ω

0

kukpXdt 1p

<∞, where 1≤p <∞, kukLpω(R;X)= ess sup

0≤t≤ω

kukX<∞, wherep=∞.

LetWωk,p(R;X) denote the set of functions which belong toLpω(R;X) together with their partial derivatives with respect totup to the orderk.

In the following, we frequently use the Poincar´e inequality (see [2]):

kuk2≤1

2kuxk2, where Z 1

0

u(x, t)dx= 0.

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Denotek ·kL2(0,1)byk ·k,k ·kL(0,1)byk ·k,k ·kLp(0,1)byk ·kpandk ·kHm(0,1)

byk · kHm, respectively.

2 Integration estimations and existence of the approximate solutions for problem (1.1)-(1.3)

Let{yj(x)} (j= 1,2,· · ·) be the orthonormal base in L2(0,1) being composed of the eigenfunctions of the eigenvalue problem

y′′+λy= 0, y(0) =y(1) = 0, corresponding to eigenvaluesλj (j= 1,2,· · ·).

Suppose that uN(x, t) = PN

j=1uN j(t)yj(x) is the Galerkin approximate solution to the problem (1.1)-(1.3), where a group of function uN j(t) (j = 1,2,· · ·, N)∈C1(ω,R),N is a natural number.

Performing the Galerkin procedure for the equation (1.1), we obtain uN t+uN xxxx+

uN x 1 +|uN x|2

x

=f, in (0,1)×R. (2.1) with

uN(x+ 1, t) =uN(x, t), t∈R, (2.2) and the time-periodic condition

uN(x, t+ω) =uN(x, t), x∈(0,1), t∈R, (2.3) Lemma 2.1 Suppose thatf ∈Cω(R;L2(0,1)),M1= sup0≤t≤ωkf(·, t)k. Then, there exists a approximate solutionuN for problem (2.1)-(2.3), which satisfies

sup

0≤t≤ω

kuN(·, t)k2≤c0M12, wherec0 is a positive constant independent of N,M1.

Proof. Using Poincar´e’s inequality and H¨older’s inequality, we have kuNk2≤1

2kuN xk2=−1 2

Z 1 0

uNuN xxdx≤ 1

4kuNk2+1

4kuN xxk2, that is

kuNk2≤ 1

3kuN xxk2. (2.4)

Multiplying both sides of (2.1) byuN, and integrating it over (0,1), making use of H¨older’s inequality and (2.4), we obtain

1 2

d

dtkuNk2+kuN xxk2

= Z 1

0

u2N x

1 +|uN x|2dx+ Z 1

0

f uNdx≤ kuN xk2+1

8kuNk2+ 2kfk2

≤ 1

2kuN xxk2+5

8kuNk2+ 2kfk2≤17

24kuN xxk2+ 2M12.

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Then,

d

dtkuNk2+ 7

12kuN xxk2≤4M12. (2.5) Integrating (2.5) over [0, ω], we get

Z ω 0

kuN xxk2≤48

7 M12ω. (2.6)

It then follows from (2.6) that there is at1∈(0, ω) such that kuN xx(·, t1)k2≤48

7 M12. (2.7)

Adding (2.4) and (2.7) together gives kuN(·, t1)k2≤ 16

7 M12. (2.8)

Integrating (2.5) again over [t1, t+ω] (∀t∈[0, ω]), using (2.8), we get sup

0≤t≤ω

kuN(·, t)k2≤ kuN(·, t1)k2+ 8ωM12= (16

7 + 8ω)M12. (2.9) Settingc0= 167 + 8ω, we complete the proof.

Employing the Leray-Schauder fixed-point argument, we can prove that there exists at least one solutionuN(t) =PN

j=1uN jyj(x) for the problem (2.1)- (2.3).

Lemma 2.2 Suppose that the assumptions of Lemma 2.1 hold and f ∈Cω(R;H2(0,1)), ft∈Cω(R;L2(0,1)).

Then

sup

0≤t≤ω

(kuN t(·, t)k2H2+kuN(·, t)k2H4)≤c1(M2),

where M2 = sup0≤t≤ω(kfkH2 +kftk). Here and in the sequel, ci(M2)(i = 1,2,· · ·) is nondecreasing with respect to M2 andlimM2→0ci(M2) =ci(0) = 0, ci(M2)is independent of N.

Proof. Multiplying both sides of (2.1) by−uN xx, and integrating it over (0,1), we obtain

1 2

d

dtkuN xk2+kuN xxxk2+ Z 1

0

uN xuN xxx 1 +|uN x|2dx+

Z 1 0

f uN xxdx= 0, Then, using H¨older’s inequality, we get

1 2

d

dtkuN xk2+kuN xxxk2

≤ 1

4kuN xxxk2+ Z 1

0

u2N x

(1 +|uN x|2)2dx+1

4kfk2+kuN xxk2

≤ 1

4kuN xxxk2+kuN xk2+1

4kfk2+kuN xxk2.

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By Nirenberg’s inequality, we have kuN xk2

c1kuN xxxk31kuNk23 +c2kuNk2

≤1

8kuN xxxk2+c2(M2).

and

kuN xxk2

c1kuN xxxk23kuNk13 +c2kuNk2

≤1

8kuN xxxk2+c3(M2).

Summing up, we get d

dtkuN xk2+kuN xxxk2≤c4(M2), (2.10) wherec4(M2) = 2c2(M2) + 2c3(M2) +12M22. Integrating (2.10) over [0, ω], we have

Z ω 0

kuN xxx(·, t)k2dt≤c4(M2)ω. (2.11) It then follows from (2.11) that there exists a timet2∈(0, ω) such that

kuN xxx(·, t2)k2≤c4(M2). (2.12) Then

kuN x(·, t2)k2≤1

8kuN xxx(·, t2)k2+c2(M2)≤ 1

8c4(M2) +c2(M2). (2.13) Integrating (2.10) again over [t2, t+ω] (∀t∈[0, ω]), using (2.13), we obtain

sup

0≤t≤ω

kuN x(·, t)k2≤ kuN x(·, t2)k2+ 2c4(M2)ω=c5(M2). (2.14) Based on Sobolev’s embedding theorem, we have

kuN(·, t)kC[0,1] ≤ckuNkH1 ≤c6(M2), t∈[0, ω]. (2.15) Multiplying both sides of (2.1) by uN xxxx, and integrating it over (0,1), we obtain

1 2

d

dtkuN xxk2+kuN xxxxk2= Z 1

0

uN x

1 +|uN x|2

x

uN xxxxdx+ Z 1

0

f uN xxxxdx.

Then, using H¨older’s inequality, noticing thats≤12(1 +s2), we get 1

2 d

dtkuN xxk2+kuN xxxxk2

≤ 1

4kuN xxxxk2+ Z 1

0

uN x

1 +|uN x|2

2

dx+1

4kuN xxxxk2+ Z 1

0

f2dx

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≤ 1

2kuN xxxxk2+ Z 1

0

uN xx

1 +|uN x|2− 2|uN x|2uN xx

(1 +|uN x|2)2 2

dx+ Z 1

0

f2dx

≤ 1

2kuN xxxxk2+ 2 Z 1

0

uN xx

1 +|uN x|2 2

dx +2

Z 1 0

2|uN x|2uN xx (1 +|uN x|2)2

2

dx+ Z 1

0

f2dx

= 1

2kuN xxxxk2+ 2 Z 1

0

u2N xx 1

(1 +|uN x|2)2dx +8

Z 1 0

u2N xx |uN x|4

(1 +|uN x|2)4dx+ Z 1

0

f2dx

≤ 1

2kuN xxxxk2+ 2kuN xxk2+1

2kuN xxk2+M2. Using Nirenberg’s inequality, we derive that

5

2kuN xxk2≤5

2(c1kuN xxxxk13kuN xk23 +c2kuN xk)2≤1

4kuN xxxxk2+c7(M2), Summing up, we deduce that

d

dtkuN xxk2+1

2kuN xxxxk2≤c8(M2), (2.16) wherec8(M2) = 2c7(M2) + 2M22. Integrating (2.16) over [0, ω], we have

Z ω 0

kuN xxxx(·, t)k2dt≤2c8(M2)ω. (2.17) It then follows from (2.17) that there exists a timet3∈(0, ω) such that

kuN xxxx(·, t3)k2≤2c8(M2). (2.18) Then

kuN xx(·, t3)k2≤ 1

10kuN xxxx(·, t3)k2+2

5c7(M2)≤1

5c8(M2) +2

5c7(M2).(2.19) Integrating (2.16) again over [t3, t+ω] (∀t∈[0, ω]), using (2.19), we obtain

sup

0≤t≤ω

kuN xx(·, t)k2≤ kuN xx(·, t3)k2+ 2c8(M2)ω=c9(M2). (2.20) Based on Sobolev’s embedding theorem, we have

kuN x(·, t)kC[0,1]≤ckuNkH2≤c10(M2), t∈[0, ω]. (2.21) Multiplying both sides of (2.1) by uN xxxxxx, and integrating it over (0,1), we obtain

1 2

d

dtkuN xxxk2+kuN xxxxxk2+ Z 1

0

uN x

1 +|uN x|2

xx

uN xxxxxdx= Z 1

0

fxuN xxxxxdx.

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Using H¨older’s inequality, we get 1

2 d

dtkuN xxxk2+kuN xxxxxk2

≤ Z 1

0

uN x

1 +|uN x|2 2

xx

dx+ Z 1

0

fx2dx+1

2kuN xxxxxk2

≤ 4 Z 1

0

u2N xxx

(1 +|uN x|2)2dx+ 144 Z 1

0

u2N xu4N xx

(1 +|uN x|2)4dx+ 16 Z 1

0

u4N xu2N xxx (1 +|uN x|2)4dx +256

Z 1 0

u6N xu4N xx (1 +|uN x|2)6dx+

Z 1 0

fx2dx+1

2kuN xxxxxk2

≤ 5kuN xxxk2+ 40kuN xxk44+M22+1

2kuN xxxxxk2. By Nirenberg’s inequality, we have

5kuN xxxk2≤5(c1kuN xxxxxk13kuN xxk23+c2kuN xxk)2≤ 1

8kuN xxxxxk2+c11(M2), and

40kuN xxk44≤40(c1kuN xxxxxk121kuN xxk1112+c2kuN xxk)4≤1

8kuN xxxxxk2+c12(M2).

Summing up, we deduce that d

dtkuN xxxk2+1

2kuN xxxxxk2≤c13(M2), (2.22) wherec13(M2) = 2c11(M2) + 2c12(M2) + 2M22. Integrating (2.22) over [0, ω], we have

Z ω 0

kuN xxxxx(·, t)k2dt≤2c13(M2)ω. (2.23) It then follows from (2.23) that there exists a timet4∈(0, ω) such that

kuN xxxxx(·, t4)k2≤2c13(M2). (2.24) Then

kuN xxx(·, t4)k2 ≤ 1

40kuN xxxxx(·, t4)k2+1

5c11(M2)

≤ 1

20c13(M2) +1

5c11(M2). (2.25) Integrating (2.22) again over [t4, t+ω] (∀t∈[0, ω]), using (2.25), we obtain

sup

0≤t≤ω

kuN xxx(·, t)k2≤ kuN xxx(·, t4)k2+ 2c13(M2)ω=c14(M2). (2.26) Based on Sobolev’s embedding theorem, we have

kuN xx(·, t)kC[0,1]≤ckuNkH3 ≤c15(M2), t∈[0, ω]. (2.27)

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Multiplying both sides of (2.1) byuN xxxxxxxx, and integrating it over (0,1), we obtain

1 2

d

dtkuN xxxxk2+kuN xxxxxxk2+ Z 1

0

uN x

1 +|uN x 2

xxx

uN xxxxxxdx

= Z 1

0

fxxuN xxxxxxdx.

Using H¨older’s inequality, we get 1

2 d

dtkuN xxxxk2+kuN xxxxxxk2

≤ 1

2kuN xxxxxxk2+ Z 1

0

uN x

1 +|uN x 22

xxx

dx+ Z 1

0

fxx2 dx

≤ 1

2kuN xxxxxxk2+ Z 1

0

fxx2 dx+c Z 1

0

uN xxxx 1 +|uN x|2

2 dx +c

Z 1 0

uN xuN xxuN xxx (1 +|uN x|2)2

2

dx+c Z 1

0

|uN xx|3 (1 +|uN x|2)2

dx +c

Z 1 0

|uN x|2uN xx

(1 +|uN x|2)2

dx+c Z 1

0

|uN x|2|uN xx|3 (1 +|uN x|2)3

dx

+c Z 1

0

|uN x|3uN xxuN xxx

(1 +|uN x|2)3

dx+c Z 1

0

|uN x|4|uN xx|3 (1 +|uN x|2)4

dx

≤ 1

2kuN xxxxxxk2+ Z 1

0

fxx2 dx+c(kuN xxxxk2+kuN xxk44 +kuN xxk66+kuN xxxk44+kuN xxk2).

By Nirenberg’s inequality, we get

ckuN xxxxk2 ≤ c(c1kuN xxxxxxk31kuN xxxk23 +c2kuN xxxk)2

≤ 1

16kuN xxxxxxk2+c16(M2),

ckuN xxk44 ≤ c(c1kuN xxxxxxk161kuN xxk1516 +c2kuN xxk)4

≤ 1

16kuN xxxxxxk2+c17(M2),

ckuN xxk66 ≤ c(c1kuN xxxxxxk121kuN xxk1112 +c2kuN xxk)6

≤ 1

16kuN xxxxxxk2+c18(M2), and

ckuN xxxk44 ≤ c(c1kuN xxxxxxk121 kuN xxxk1112 +c2kuN xxxk)2

≤ 1

16kuN xxxxxxk2+c19(M2).

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Summing up, we deduce that d

dtkuN xxxxk2+1

2kuN xxxxxxk2≤c21(M2), (2.28) wherec21(M2) = 2(c16(M2) +c17(M2) +c18(M2) +c19(M2) +M22+cc9(M2)).

Integrating (2.28) over [0, ω], we have Z ω

0

kuN xxxxxx(·, t)k2dt≤2c21(M2)ω. (2.29) It then follows from (2.23) that there exists a timet5∈(0, ω) such that

kuN xxxxxx(·, t5)k2≤2c21(M2). (2.30) Then

kuN xxxx(·, t5)k2 ≤ 1

16ckuN xxxxxx(·, t5)k2+1

cc16(M2)

≤ 1

8cc21(M2) +c16(M2)

c . (2.31)

Integrating (2.28) again over [t5, t+ω] (∀t∈[0, ω]), using (2.31), we obtain sup

0≤t≤ω

kuN xxxx(·, t)k2≤ kuN xxxx(·, t5)k2+ 2c21(M2)ω=c22(M2). (2.32) Based on Sobolev’s embedding theorem, we have

kuN xxx(·, t)kC[0,1]≤ckuNkH4 ≤c22(M2), t∈[0, ω]. (2.33) Multiplying both sides of (2.1) byuN t, and integrating it over (0,1), we obtain

kuN tk2

= (−uN xxxx−( uN x

1 +|uN x|2)x+f, uN t)

≤ 1

2kuN tk2+3

2kuN xxxxk2+3

2kfk2+3 2

Z 1 0

uN x

1 +|uN x|2 2

x

dx

≤ 1

2kuN tk2+3

2kuN xxxxk2+3 2kfk2 +3

Z 1 0

uN xx

1 +|uN x|2 2

dx+ 3 Z 1

0

2u2N xuN xx

(1 +|uN x|2)2 2

dx

≤ 1

2kuN tk2+3

2kuN xxxxk2+3

2kfk2+ckuN xxk2

≤ 1

2kuN tk2+3

2c22(M2) +3

2M22+cc9(M2).

Hence

kuN tk2≤c23(M2)≡3c22(M2) + 3M22+ 2cc9(M2). (2.34)

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Differentiating (2.1) with respect tot, we get uN tt+uN xxxxt+ ( uN x

1 +|uN x|2)xt=ft, (x, t)∈(0,1)×R. (2.35) Multiplying both sides of (2.35) by uN xxt, and integrating it over (0,1), we obtain

1 2

d

dtkuN xtk2+kuN xxxtk2

= Z 1

0

uN x 1 +|uN x|2

t

uN xxxtdx− Z 1

0

ftuN xxtdx.

≤ 1

4kuN xxxtk2+ 2 Z 1

0

uN t

1 +|uN x|2 2

dx +2

Z 1 0

2u2N xuN xt

(1 +|uN x|2)2 2

dx+1

4kuN xxtk2+M22

≤ 1

4kuN xxxtk2+5

2kuN xtk2+1

4kuN xxtk2+M22. Using Nirenberg’s inequality, we obtain

5

2kuN xtk2≤ 5

2(c1kuN xxxtk13kuN tk23 +c2kuN tk)2≤ 1

8kuN xxxtk2+c24(M2), and

1

4kuN xxtk2≤1

4(c1kuN xxxtk23kuN tk13+c2kuN tk)2≤1

8kuN xxxtk2+c25(M2).

Summing up, we have d

dtkuN xtk2+kuN xxxtk2≤c26(M2), (2.36) wherec26(M2) = 2c24(M2) + 2c25(M2) + 2M22. Integrating (2.36) over [0, ω], we have

Z ω 0

kuN xxxt(·, t)k2dt≤c26(M2)ω. (2.37) It then follows from (2.37) that there exists a timet6∈(0, ω) such that

kuN xxxt(·, t6)k2≤c26(M2). (2.38) Then

kuN xt(·, t6)k2≤ 1

20kuN xxxt(·, t6)k2+2

5c24(M2)≤ 1

20c26(M2) +2

5c24(M2).(2.39) Integrating (2.36) again over [t6, t+ω] (∀t∈[0, ω]), using (2.39), we obtain

sup

0≤t≤ω

kuN xt(·, t)k2≤ kuN xt(·, t6)k2+ 2c26(M2)ω=c27(M2). (2.40)

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Based on Sobolev’s embedding theorem, we have

kuN t(·, t)kC[0,1]≤ckuN tkH1≤cc27(M2), t∈[0, ω]. (2.41) Multiplying both sides of (2.35) byuN xxxxt, and integrating it over (0,1), we obtain

1 2

d

dtkuN xxtk2+kuN xxxxtk2=− Z 1

0

uN x 1 +|uN x|2

xt

uN xxxxtdx+ Z 1

0

ftuN xxxxtdx.

Therefore 1 2

d

dtkuN xxtk2+kuN xxxxtk2

≤ 1

4kuN xxxxtk2+ 2 Z 1

0

uN x 1 +|uN x|2

2 xt

dx+ 2 Z 1

0

ft2dx

≤ 1

4kuN xxxxtk2+ 8 Z 1

0

uN xxt

1 +|uN x|2 2

dx+ 8 Z 1

0

6uN xuN xxuN xt

(1 +|uN x|2)2 2

dx +8

Z 1 0

2u2N xuN xxt (1 +|u2N x)2

dx+ 8

Z 1 0

4u3N xuN xxuN xt (1 +|uN x|2)3

2

dx+ 2 Z 1

0

ft2dx

≤ 1

4kuN xxxxtk2+ckuN xxtk2+ckuN xtk2+ 2M22. Nirenberg’s inequality gives

ckuN xxtk2≤c(c1kuN xxxxtk31kuN xtk23 +c2kuN xtk)2≤1

4kuN xxxxtk2+c28(M2).

Summing up, we get d

dtkuN xxtk2+kuN xxxxtk2≤c29(M2), (2.42) wherec29(M2) = 2cc27(M2) + 2cc28(M2) + 4M22. Integrating (2.42) over [0, ω], we have

Z ω 0

kuN xxxxt(·, t)k2dt≤c30(M2)ω. (2.43) It then follows from (2.43) that there exists a timet7∈(0, ω) such that

kuN xxxxt(·, t7)k2≤c30(M2). (2.44) Then

kuN xxt(·, t7)k2≤ 1

4ckuN xxxxt(·, t7)k2+c28(M2)

c ≤ 1

4cc30(M2) +c28(M2) c (2.45). Integrating (2.42) again over [t7, t+ω] (∀t∈[0, ω]), using (2.45), we obtain

sup

0≤t≤ω

kuN xxt(·, t)k2≤ kuN xxxxt(·, t7)k2+ 2c30(M2)ω=c31(M2). (2.46)

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Based on Sobolev’s embedding theorem, we have

kuN xt(·, t)kC[0,1]≤ckuN tkH2 ≤cc21(M2), t∈[0, ω]. (2.47) Combining (2.9), (2.16), (2.21), (2.27), (2.33), (2.35), (2.40) and (2.46) to- gether, we complete the proof of Lemma 2.2.

3 Existence and uniqueness of solutions for the problem (1.1)-(1.3)

Theorem 3.1 Suppose that the assumptions in Lemma 2.2 is satisfied, then there exists a generalized time-periodic solution

u(x, t)∈L2ω(R;H4(0,1)), ut(x, t)∈L2ω(R;H2(0,1)), (3.1) for problem (1.1)-(1.3), which satisfies

Z ω 0

Z 1 0

ut+uxxxx+ ( ux

1 +|ux|2)x−f

ϕdxdt= 0, ∀ϕ∈L2ω(R;L2(0,1)).(3.2) Especially, ifM2 is sufficiently small, the solution is unique.

Proof. Based on Lemma 2.2 and Sobolev’s embedding theorem, we obtain the following estimate

sup

0≤t≤ω

(kuN t(·, t)kC1[0,1]+kuNkC3[0,1])≤c32(M2). (3.3) It then follows from (3.3) and Ascoli-Arzel´a’s theorem that there exists a func- tionu(x, t) and a subsequence of {uN(x, t)}, still denoted by{uN(x, t)}, such that, whenN →+∞,{uN(x, t)}, uN x(x, t)} uniformly converge tou(x, t) and ux(x, t) on [0, ω]×(0,1). Based on the result of Lemma 2.2, whenN →+∞, the subsequences{uN xx},{uN xxx},{uN xxxx}, {uN t},{uN xt}and {uN xxt} weakly converge touxx,uxxx,uxxxx,ut,uxt anduxxtin L2ω(R;L2(0,1)). Set

W ={u|u∈L2ω(R;H4(0,1)), ut∈L2ω(R;H2(0,1))}.

Aubin’s compact lemma implies that the embeddingW ֒→L2ω(R;H2(0,1)) is compact. Owing to the assumptions, we know that there exists a subsequence of{uN(x, t)} still denoted by{uN(x, t)} such that, whenN→+∞,{uN(x, t)}

is convergent inL2ω(R;H3(0,1)).

Setting F(s) = 1+|s|s 2, according to the previous subsequences {uN(x, t)}, we conclude that{[F(uN x)]x}={

uN x

1+|uN x|2

x}weakly converges to [F(ux)]x= ux

1+|ux|2

x in L2ω(R;L2(0,1)). In fact, for any ω ∈ L2ω(R;L2(0,1)), by (3.3),

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we have

| Z ω

0

([F(uN x)]x−[F(ux)]x, w)dt|

≤ Z ω

0

Z 1 0

(|F(uN x)−F(ux)||uN xx|+|F(ux)||uN xx−uxx|)|ω|dxdt

≤ Z ω

0

Z 1 0

(|F′′(θuN x+ (1−θ)ux)||uN x−ux||uN xx||ω|)dxdt +

Z ω 0

Z 1 0

(|F(ux)||uN xx−uxx||ω|)dxdt

≤ c32(M2) Z ω

0

Z 1 0

(|uN x−ux|+|uN xx−uxx|)|ω|dxdt

≤ c32(M2)[kuN x−uxkL2((0,ω)×(0,1))+kuN xx−uxxkL2((0,ω)×(0,1))]

·kωkL2((0,ω)×(0,1)), (3.4)

whereθ∈(0,1). By (3.4), we know that there exists a subsequence{uN(x, t)}

such that{[F(uN x)]x} weakly converges to [F(ux)]x in L2(R;L2(0,1)). Then, problem (1.1)-(1.3) admits a generalized time-periodic solution u(x, t), which satisfies (3.1) and (3.2).

Now, we are going to prove the uniqueness of the solution. Suppose that u(x, t) andv(x, t) are two solutions of (1.1)-(1.3). Letξ(x, t) =u(x, t)−v(x, t), thenξ(x, t) satisfies the following problem

ξtxxxx+ [F(ux)]x−[F(vx)]x= 0, x∈(0,1), t∈R, ξx(0, t) =ξx(1, t) =ξxxx(0, t) =ξxxx(1, t), t∈R, ξ(t+ω) =ξ(x, t), t∈R.

(3.5) Multiplying both sides of the equation of (3.5) byξ, integrating the products over (0,1) and using the mean value theorem, we obtain that

1 2

d

dtkξk2+kξxxk2= Z 1

0

(F(ux)−F(vx))ξxdx

= Z 1

0

(F(θux+ (1−θ)vxxxxdx≤c33(M2)kξxxkkξxk

≤ 1

4kξxxk2+ [c33(M2)]2xk2= 1

4kξxxk2−[c33(M2)]2(ξ, ξxx)

≤ 1

2kξxxk2+ [c33(M2)]4kξk2. (3.6) Using Poincar´e’s inequality, we get

kξk2≤ 1

2kξxk2=−1

2(ξ, ξxx)≤ 1

4kξxxk2+1 4kξk2, which means

kξk2≤ 1 3kξxxk2.

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It then follows form (3.6) and the above inequality that d

dtkξk2+ (3−2[c32(M2)]4)kξk2≤0. (3.7) Taking M2 sufficiently small such that 3−2[c32(M2)]4 > 0, using Gronwall’s inequality, we have

kξ(·, t)k2≤ kξ(·,0)k2e−(3−2[c32(M2)]

4)t

, ∀t >0.

Sinceξ(x, t) is time-periodic, for anyt∈R, there exists a natural number N0

such thatt+N0ξ >0 and

kξ(·, t)k2=kξ(·, t+N0ω)k2≤ kξ(·,0)k2e−(3−2[c32(M2)]

4)N ω

, ∀N > N0, that is

kξ(·, t)k2= 0, ∀t∈R. Then, Theorem 3.1 is proved.

Acknowledgment

The authors thank the referees for their valuable comments and suggestions about this article.

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(Received July 8, 2012)

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