Robin Boundary Value Problem for One-Dimensional Landau...

Acta Mathematica Sinica, English Series

Oct., 2012, Vol. 28, No. 10, pp. 2033–2066

Published online: July 5, 2012

DOI: 10.1007/s10114-012-0016-4

Http://www.ActaMath.com

Acta Mathematica Sinica, English Series© Springer-Verlag Berlin Heidelberg & The Editorial Office of AMS 2012

Robin Boundary Value Problem for One-Dimensional

Landau–Lifshitz Equations

Shi Jin DING Jin Rui HUANG Xiao E LIUSchool of Mathematical Sciences, South China Normal University,

Guangzhou 510631, P. R. China

E-mail : [email protected] huangjinrui [email protected] lxe [email protected]

Abstract In this paper, we are concerned with the existence and uniqueness of global smooth solu-

tion for the Robin boundary value problem of Landau–Lifshitz equations in one dimension when the

boundary value depends on time t. Furthermore, by viscosity vanishing approach, we get the existence

and uniqueness of the problem without Gilbert damping term when the boundary value is independent

of t.

Keywords Landau–Lifshitz equation, Robin boundary value problem, existence, uniqueness, smooth

solution.

MR(2000) Subject Classification 35D10, 35K50

1 Introduction

In this paper, we are concerned with the existence and uniqueness of smooth solution to the1-dimensional Robin boundary value problem for the Landau–Lifshitz equation as follows:

Zt = −εZ × (Z × Zxx) + Z × Zxx, (1.1)

Z(x, 0) = φ(x), (1.2)

− α1Z(0, t) + α2Zx(0, t) = α(t), (1.3)

β1Z(l, t) + β2Zx(l, t) = β(t), (1.4)

where |φ(x)| ≡ 1, ∀x ∈ [0, 1] and Z(x, t) = (Z1(x, t), Z2(x, t), Z3(x, t)) ∈ S2, αi, βi > 0 (i = 1, 2)and the compatibility conditions hold:

− α1φ(0) + α2φ′(0) = α(0), (1.5)

β1φ(l) + β2φ′(l) = β(0). (1.6)

Landau–Lifshitz equation which describes the evolution of spin fields in continuum ferro-magnets bears a fundamental role in the understanding of non-equilibrium magnetism. Forone-dimensional equation of (1.1) with periodic initial-boundary conditions, Zhou et al. [1]have shown the existence and uniqueness of global smooth solution for ε ≥ 0; Dong and Guo [2]

Received January 12, 2010, accepted May 5, 2011

Supported by National Basic Research Program of China (973 Program) (Grant No. 2011CB808002), National

Natural Science Foundation of China (Grant Nos. 11071086 and 11128102), the University Special Re- search

Foundation for Ph.D. Program (Grant No. 20104407110002)

2034 Ding S. J., et al.

have shown the existence and uniqueness of smooth solution for the inhomogeneous Dirichletboundary value problem of Landau–Lifshitz equation with Gilbert damping term in one di-mension if the initial function is smooth and the boundary value depends on t; Zhou et al. [3]studied another system of ferromagnetic spin chain as follows: Zt = εZxx +Z×Zxx +f(x, t, Z)with the initial condition and the following nonlinear boundary conditions:

Zx(0, t) = ∇ψ0(t, Z(0, t)),

− Zx(l, t) = ∇ψ1(t, Z(l, t)),

where ψ0(t, Z) and ψ1(t, Z) are vector-valued functions of t and Z. They got the existence ofweak solution for ε ≥ 0; Guo et al. [4] obtained the existence and uniqueness of smooth solutionfor (1.1) with the homogeneous Neumann boundary value problem for ε = 0. Chen [5] provedthat there exists a partially regular solution of the two-dimentional problem with Dirichletboundary value for ε > 0, where the boundary value depends only on x, independent of t. Formore results about Landau–Lifshitz equation, we refer to, for instance, [6–9] and referencestherein.

However, there are few results about Robin boundary value problem of Landau–Lifshitzequations. The paper consists of three parts.

Firstly, we establish the existence and uniqueness of local smooth solution to problem (1.1)–(1.4) for ε > 0 by the differential-difference method. Since we have |Zj | = |Z(xj , t)| = 1 onlywhen j = 1, 2, . . . , J − 1, we should deal with the term of j = 0 and j = J independently.

Secondly, from (1.1), we have

Zt = εZxx + ε|Zx|2Z + Z × Zxx, (1.7)

since A × (B × C) = (A · C)B − (A · B)C and |φ(x)| ≡ 1. It is convenient for us to makethe global estimates for the equation in this form. Applying an energy method, we obtain theexistence of global smooth solution when ε > 0.

Thirdly, we make use of the viscosity vanishing method to establish the existence anduniqueness of global smooth solution when ε = 0, provided that α(t) and β(t) are both constantvectors in time. Throughout this paper, our difficulties mainly lie in how to deal with theboundary terms.

More precisely, the first result of this paper is

Theorem 1.1 Let φ(x) ∈ Hk((0, l)), and α(t), β(t) ∈ H[ k2 ]+1

loc ((0,+∞)). Then for any givenT > 0, problem (1.1)–(1.4) admits a unique global smooth solution Z(x, t) :

Z(x, t) ∈ G(T ) =( [ k

2 ]⋂s=0

W s∞(0, T ;Hk−2s((0, l)))

)∩

( [ k+12 ]⋂

s=0

Hs(0, T ;Hk+1−2s((0, l)))).

On the other hand, we may establish some priori estimates for the solution of (1.1)–(1.4)uniform in ε by constructing some new energy laws with special boundary condition in whichα(t) and β(t) are constant vectors α, β. Then we establish the existence and uniqueness ofsmooth solution to the following problem:

Zt = Z × Zxx, (1.8)

Z(x, 0) = φ(x), (1.9)

Robin Boundary Value Problem for One-Dimensional Landau–Lifshitz Equations 2035

− α1Z(0, t) + α2Zx(0, t) = α, (1.10)

β1Z(l, t) + β2Zx(l, t) = β. (1.11)

In estimates, the orthogonal bases Z,Zx, Z×Zx and Z,Zt, Z×Zt are used to deal with thehighest order terms. Again, since |Z| = 1, many other useful identities can be derived.

The second result is

Theorem 1.2 Let φ(x) ∈ Hk((0, l)). Then for any given T > 0, problem (1.8)–(1.11) admitsa unique smooth solution Z(x, t) :

Z(x, t) ∈[ k2 ]⋂

s=0

W s∞(0, T ;Hk−2s((0, l))).

At present, we cannot get the uniform estimates in ε if α, β depend on t. The reasons areas follows: for H2 estimate, we meet a term like

∫ l

0Zt · (Z × Zxxt). To dealing with this term,

we integrate it by parts, then we have to estimate the term Zt · (Z ×Zxt)|x=lx=0. However, if α, β

depend on t, these terms lack of control.

2 Local Smooth Solution

We need the following well-known lemmas.

Lemma 2.1 ([10]) Let p, q, r be real numbers and j, m be integers such that 1 ≤ p, q, r ≤ ∞,0 ≤ j < m. If u ∈Wm,p(Ω) ∩ Lq(Ω), Ω ⊂ R

N , then

‖Dju‖Lr(Ω) ≤ C(‖u‖Lp(Ω) + ‖Dmu‖Lp(Ω))θ‖u‖1−θLq(Ω),

where ‖ · ‖p = ‖ · ‖Lp(Ω),jm ≤ θ ≤ 1 and

1r− j

N= θ

(1p− m

N

)+ (1 − θ)

1q.

Lemma 2.2 ([11]) Let Ω ⊂ RN be arbitrary bounded domain with piecewise smooth bound-

aries. Then the following inequality is valid for every function v ∈W 1,m(Ω):

‖v‖q ≤ C(‖v‖1 + ‖Dv‖αm‖v‖1−α

r ),

where m ≥ 1, r ≥ 1, and α = ( 1r − 1

q )( 1r − 1

m + 1N )−1; moreover, if m < N and N > 1, then

q ∈ [r, mNN−m ] for r ≤ mN

N−m , and q ∈ [ mNN−m , r] for r ≥ mN

N−m . If m ≥ N > 1, then q ∈ [r,+∞)is arbitrary ; moreover, if m > N , then q ∈ [r,+∞]. If m ≥ N = 1, then q ∈ [r,+∞]. Thepositive constant C depends on N , m, r, α and the domain Ω but independent of the functionv.

Lemma 2.3 ([12]) Let η(·) be a nonnegative, absolutely continuous function on [0, T ], whichsatisfies for a.e. t the differential inequality

η′(t) ≤ φ(t)η(t) + ψ(t),

where φ(t) and ψ(t) are nonnegative, summable functions on [0, T ]. Then

η(t) ≤ e∫ t0 φ(s)ds

[η(0) +

∫ t

0

ψ(s)ds].


Lemma 2.4 ([13]) Assume f, h ∈ C[a, b], g ≥ 0, and g ∈ L1(a, b). Furthermore,

f(t) ≤ h(t) +∫ t

a

g(s)f(s)ds, t ∈ [a, b],

then

f(t) ≤ h(t) +∫ t

a

g(s)h(s)[exp

∫ t

s

g(u)du]ds, t ∈ [a, b].

Lemma 2.5 ([2]) Let p be real number and k, m be integers such that 2 ≤ p ≤ ∞, 0 ≤ k ≤ m.Then

‖δkuh‖p ≤ C‖uh‖1−r2 (‖δmuh‖2 + ‖uh‖2)r,

where uh = uj = u(xj) (j = 0, 1, 2, . . . , J) is any discrete function on the interval [0, l], xj = jh,h = l

J , r = 1m (k + 1

2 − 1p ), C is a constant which is independent of uh and h, and

‖δkuh‖p =( J−k∑

i=0

∣∣∣∣Δk+ui

hk

∣∣∣∣p

h

) 1p

, ‖δkuh‖∞ = max0≤j≤J−k

∣∣∣∣Δk+ui

hk

∣∣∣∣.Lemma 2.6 ([4]) For any discrete functions uj and vj , we have

(1)∑J

j=1 ujΔ−vj = −∑J−1j=0 vjΔ+uj − u0v0 + uJvJ ,

(2)∑J−1

j=1 ujΔ+Δ−vj = −∑J−1j=0 Δ+ujΔ+vj − u0Δ+v0 + uJΔ−vJ ,

(3) Δ+(ujvj) = uj+1Δ+vj + vjΔ+uj ,

where Δ+, Δ− denote the forward and backward differences respectively.

In this section, we will prove the local existence of smooth solution to (1.1)–(1.4) for smoothdata by the differential-difference method. For simplicity, we let ε = 1 without loss of generalityin this section. We establish the following difference-differential equation:

d

dtZj = −Zj ×

(Zj × Δ+Δ−Zj

h2

)+ Zj × Δ+Δ−Zj

h2, j = 1, 2, . . . , J − 1, (2.1)

Zj |t=0 = φ(jh) = φj , j = 1, . . . , J − 1, (2.2)

− α1Z0 + α2Δ+Z0

h= α(t), (2.3)

β1ZJ + β2Δ−ZJ

h= β(t), (2.4)

where h = lJ , J > 0, Zh = {Zj = Zj(t) = Z(xj , t)|j = 0, 1, 2, . . . , J}.

It is clear that the initial value problem for ordinary differential equation (2.1)–(2.4) admitsa local smooth solution. For such solution, we shall give some estimates uniformly in h andthen get a local smooth solution to problem (1.1)–(1.4). In this section, we always denote asmooth solution of (2.1)–(2.4) by Zj .

Lemma 2.7 If φ(x) ∈ S2, φ(x) ∈ H1([0, l]) and α(t), β(t) ∈ H1loc((0,+∞)), then there are

constants T0 > 0, C > 0 independent of h such that

sup0≤t≤T0

(‖δZh(t)‖2 + ‖Zh‖∞) +∫ T0

0

‖δ2Zh(t)‖22dt ≤ C. (2.5)

Proof We will decompose this proof into two steps.

Step 1 Taking the inner product of (2.1) with Zj , we obtain Zj ·Zjt = 0 (j = 1, 2, . . . , J−1).Since Zj |t=0 = φj ∈ S2, we have Zj(t) ∈ S2 (j = 1, 2, . . . , J − 1).


According to (2.3) and (2.4), there exists a constant C1 such that

|Z0| ≤∣∣∣∣ α2

α1h+ α2

∣∣∣∣ · |Z1| +∣∣∣∣ α(t)hα1h+ α2

∣∣∣∣ ≤ |Z1| + C1,

|ZJ | ≤∣∣∣∣ β2

β1h+ β2

∣∣∣∣ · |ZJ−1| +∣∣∣∣ β(t)hβ1h+ β2

∣∣∣∣ ≤ |ZJ−1| + C1.

Since |Zj | = 1 for j = 1, 2, . . . , J − 1, we have ‖Zh‖∞ ≤ C.

Step 2 Taking the inner product of (2.1) with (Δ+Δ−Zj

h2 )h and summing over j from 1 toJ − 1, we have

J−1∑j=1

(Δ+Δ−Zj

h2

)h · Zjt =

J−1∑j=1

∣∣∣∣Zj × Δ+Δ−Zj

h2

∣∣∣∣2

h. (2.6)

By Lemma 2.6 (2), the left-hand side of (2.6) is equal to

J−1∑j=1

(Δ+Δ−Zj

h2

)h · Zjt = −

J−1∑j=0

(Δ+Zj

h

)·(

Δ+Zjt

h

)h− Z0t · Δ+Z0

h+ ZJt · Δ−ZJ

h

= −12d

dt

J−1∑j=0

∣∣∣∣Δ+Zj

h

∣∣∣∣2

h− α′(t) − α2Δ+Z0t

h

−α1· Δ+Z0

h

+β′(t) − β2

Δ−ZJt

h

β1· Δ−ZJ

h

= −12d

dt

J−1∑j=0

∣∣∣∣Δ+Zj

h

∣∣∣∣2

h+1α1α′(t) · Δ+Z0

h− α2

2α1

d

dt

∣∣∣∣Δ+Z0

h

∣∣∣∣2

+1β1β′(t) · Δ−ZJ

h− β2

2β1

d

dt

∣∣∣∣Δ−ZJ

h

∣∣∣∣2

.

Therefore, we have

12d

dt‖δZh‖2

2 +J−1∑j=1


h2

∣∣∣∣2

h =1α1α′(t) · Δ+Z0

h− α2

2α1

d

dt

∣∣∣∣Δ+Z0

h

∣∣∣∣2

+1β1β′(t) · Δ−ZJ

h− β2

2β1

d

dt

∣∣∣∣Δ−ZJ

h

∣∣∣∣2

. (2.7)

Integrating (2.7) with respect to t, we have

12‖δZh‖2

2 +∫ t

0

J−1∑j=1


h2

∣∣∣∣2

hdt

=12‖δZh(0)‖2

2 +1α1

∫ t

0

α′(t) · Δ+Z0

hdt− α2

2α1

(∣∣∣∣Δ+Z0

h

∣∣∣∣2

−∣∣∣∣Δ+φ0

h

∣∣∣∣2)

+1β1

∫ t

0

β′(t) · Δ−ZJ

hdt− β2

2β1

(∣∣∣∣Δ−ZJ

h

∣∣∣∣2

−∣∣∣∣Δ−φJ

h

∣∣∣∣2)

≤ 12‖δZh(0)‖2

2 +1α1

∫ t

0

α′(t) · Δ+Z0

hdt+

α2

2α1

∣∣∣∣Δ+φ0

h

∣∣∣∣2

+1β1

∫ t

0

β′(t) · Δ−ZJ

hdt+

β2

2β1

∣∣∣∣Δ−φJ

h

∣∣∣∣2

, (2.8)


where φ0 = φ(0), φJ = φ(l).From the facts that

‖δZh(0)‖22 =

J−2∑j=1

∣∣∣∣Δ+Zj(0)h

∣∣∣∣2

h+∣∣∣∣Δ+Z0(0)

h

∣∣∣∣2

h+∣∣∣∣Δ+ZJ−1(0)

h

∣∣∣∣2

h

=J−2∑j=1

∣∣∣∣Δ+φj

h

∣∣∣∣2

h+∣∣∣∣Δ+Z0(0)

h

∣∣∣∣2

h+∣∣∣∣Δ−ZJ (0)

h

∣∣∣∣2

h

=J−2∑j=1

∣∣∣∣Δ+φj

h

∣∣∣∣2

h+∣∣∣∣α(0) + α1Z0(0)

α2

∣∣∣∣2

h+∣∣∣∣β(0) − β1ZJ (0)

β2

∣∣∣∣2

h

≤ ‖δφh‖22 + C ≤ C

and ∣∣∣∣Δ+Z0(t)h

∣∣∣∣ =∣∣∣∣α(t) + α1Z0(t)

α2

∣∣∣∣ ≤ C,

∣∣∣∣Δ−ZJ (t)h

∣∣∣∣ =∣∣∣∣β(t) − β1ZJ (t)

α2

∣∣∣∣ ≤ C,

we get12‖δZh‖2

2 +∫ t

0

J−1∑j=1


h2

∣∣∣∣2

hdt ≤ 12‖δφh‖2

2 + C ≤ C.

By Lemma 2.4, we obtain

sup0≤t≤T

‖δZh‖ ≤ C,

∫ t

0

J−1∑j=1


h2

∣∣∣∣2

hdt ≤ C.

At the same time, we know from the fact (A×B) · (C ×D) = (A · C)(B ·D) − (A ·D)(B · C)that

J−1∑j=1

(Δ+Δ−Zj

h2

)2

h =J−1∑j=1


h2

∣∣∣∣2

h+J−1∑j=1

(Zj · Δ+Δ−Zj

h2

)2

h

andJ−1∑j=1

(Zj · Δ+Δ−Zj

h2

)2

h =J−2∑j=2

(Zj · Δ+Δ−Zj

h2

)2

h+(Z1 · Δ+Δ−Z1

h2

)2

h

+(ZJ−1 · Δ+Δ−ZJ−1

h2

)2

h.

Since Zj ∈ S2 (j = 1, 2, . . . , J − 1), we get

Zj · Δ+Δ−Zj

h2= −1

2

(Δ+Zj

h· Δ+Zj

h+

Δ−Zj

h· Δ−Zj

h

), j = 2, . . . , J − 2,

and henceJ−2∑j=2

(Zj · Δ+Δ−Zj

h2

)2

h =J−2∑j=2

[− 1

2

(Δ+Zj

h· Δ+Zj

h+

Δ−Zj

h· Δ−Zj

h

)]2

h

≤J−2∑j=1

(Δ+Zj

h

)4

h

= ‖δZh‖44 ≤ C‖δZh‖3

2(‖δ2Zh‖2 + ‖δZh‖2)

≤ 12‖δ2Zh‖2

2 + C. (2.9)


Furthermore, we have(Z1 · Δ+Δ−Z1

h2

)2

h =(Z1 · 1

hΔ+

(Δ+Z0

h

))2

h =(Z1 · 1

hΔ+

(α(t) + α1Z0

α2

))2

h

=(α1

α2Z1 · Δ+Z0

h

)2

h =(α1

α2Z1 · α(t) + α1Z0

α2

)2

h

=[α1

α22

Z1 · α(t) +(α1

α2

)2

Z1 · Z0

]2

h

≤ C‖Zh‖2∞ + C ≤ C, (2.10)(

ZJ−1 · Δ+Δ−ZJ−1

h2

)2

h =(ZJ−1 · 1

hΔ−

(Δ−ZJ

h

))2

h =(ZJ−1 · 1

hΔ−

(β(t) − β1ZJ

β2

))2

h

=(β1

β2ZJ−1 · Δ−ZJ

h

)2

h =(β1

β2ZJ−1 · β(t) − β1ZJ

β2

)2

h

=[β1

β22

ZJ−1 · β(t) −(β1

β2

)2

ZJ−1 · ZJ

]2

h

≤ C‖Zh‖2∞ + C ≤ C. (2.11)

Therefore, we have∫ T0

0‖δ2Zh(t)‖2

2 ≤ C by observing thatJ−1∑j=1

∣∣∣∣Δ+Δ−Zj

h2

∣∣∣∣2

h =J−1∑j=1

∣∣∣∣Δ2+Zj−1

h2

∣∣∣∣2

h =J−2∑j=0

∣∣∣∣Δ2+Zj

h2

∣∣∣∣2

h = ‖δ2Zh‖22.

Then Lemma 2.7 is proved. �

Lemma 2.8 If φ(x) ∈ S2, φ(x) ∈ H2([0, l]) and α(t), β(t) ∈ H2loc((0,+∞)), then there are


sup0≤t≤T0

(‖Zht(t)‖2 + ‖δ2Zh(t)‖2 + ‖δZh‖∞) +∫ T

0

‖δZht‖22 ≤ C. (2.12)

Proof It follows from (2.1) that

Zjtt = −Zjt ×(Zj × Δ+Δ−Zj

h2

)− Zj ×

(Zjt × Δ+Δ−Zj

h2

)− Zj ×

(Zj × Δ+Δ−Zjt

h2

)

+ Zjt × Δ+Δ−Zj

h2+ Zj × Δ+Δ−Zjt

h2. (2.13)

Multiplying (2.13) by Zjth and summing over j from 1 to J − 1, we haveJ−1∑j=1

Zjtt · Zjth = −J−1∑j=1

[Zj ×

(Zjt × Δ+Δ−Zj

h2

)]· Zjth

−J−1∑j=1

[Zj ×

(Zj × Δ+Δ−Zjt

h2

)]· Zjth

+J−1∑j=1

[Zj × Δ+Δ−Zjt

h2

]· Zjth. (2.14)

The left-hand side of (2.14) is equal toJ−1∑j=1

Zjtt · Zjth =12d

dt

J−1∑j=1

|Zjt|2h =12d

dt

( J∑j=0

|Zjt|2h− |Z0t|2h− |ZJt|2h).


For the first term on the right-hand side of (2.14), we have

−J−1∑j=1

[Zj ×

(Zjt × Δ+Δ−Zj

h2

)]· Zjth

= −J−1∑j=1

[(Zj · Δ+Δ−Zj

h2

)Zjt − (Zj · Zjt)

Δ+Δ−Zj

h2

]· Zjth

= −J−1∑j=1

(Zj · Δ+Δ−Zj

h2

)|Zjt|2h.

For the second term on the right-hand side of (2.14), we have

−J−1∑j=1

[Zj ×

(Zj × Δ+Δ−Zjt

h2

)]· Zjth

= −J−1∑j=1

[(Zj · Δ+Δ−Zjt

h2

)Zj − (Zj · Zj)

Δ+Δ−Zjt

h2

]· Zjth

=J−1∑j=1

(Δ+Δ−Zjt

h2· Zjt

)h

= −J−1∑j=0

Δ+Zjt

h· Δ+Zjt

hh− Z0t

Δ+Z0t

h+ ZJt

Δ−ZJt

h

= −‖δZht‖22 − Z0t · α

′(t) + α1Z0t

α2+ ZJt · β

′(t) − β1ZJt

β2

= −‖δZht‖22 −

1α2Z0t · α′(t) − α1

α2|Z0t|2 +

1β2ZJt · β′(t) − β1

β2|ZJt|2.

For the third term on the right-hand side of (2.14), by Lemma 2.6, we have

J−1∑j=1

[Zj × Δ+Δ−Zjt

h2

]· Zjth

=J−1∑j=1

(Zjt × Zj) · Δ+Δ−Zjt

h2h

= −J−1∑j=0

Δ+(Zjt × Zj) · Δ+Zjt

h− (Z0t × Z0) · Δ+Z0t

h+ (ZJt × ZJ ) · Δ−ZJt

h

= −J−1∑j=0

(Zj+1,t × Δ+Zj + Δ+Zjt × Zj) · Δ+Zjt

h− (Z0t × Z0) · Δ+Z0t

h

+ (ZJt × ZJ ) · Δ−ZJt

h

= −J−1∑j=0

(Δ+Zj

h× Δ+Zjt

h

)· Zj+1,th− (Z0t × Z0) · α

′(t) + α1Z0t

α2

+ (ZJt × ZJ ) · β′(t) − β1ZJt

β2


= −J−1∑j=0

(Δ+Zj

h× Δ+Zjt

h

)· Zj+1,th− 1

α2(Z0t × Z0) · α′(t) +

1β2

(ZJt × ZJ ) · β′(t).

Thus (2.14) becomes

12d

dt‖Zht‖2

2 + ‖δZht‖22 =

12d

dt(|Z0t|2h+ |ZJt|2h) −

J−1∑j=1

(Zj · Δ+Δ−Zj

h2

)|Zjt|2h

− 1α2Z0t · α′(t) − α1

α2|Z0t|2 +

1β2ZJt · β′(t) − β1

β2|ZJt|2

−J−1∑j=0

(Δ+Zj

h× Δ+Zjt

h

)· Zj+1,th

− 1α2

(Z0t × Z0) · α′(t) +1β2

(ZJt × ZJ ) · β′(t). (2.15)

Integrating (2.15) with respect to t, we have

12‖Zht‖2

2 +∫ t

0

‖δZht‖22dt

=12‖Zht(0)‖2

2 +12(|Z0t|2h+ |ZJt|2h) − 1

2(|Z0t(0)|2h+ |ZJt(0)|2h)

−∫ t

0

J−1∑j=1

(Zj · Δ+Δ−Zj

h2

)|Zjt|2hdt− 1

α2

∫ t

0

Z0t · α′(t)dt

− α1

α2

∫ t

0

|Z0t|2dt+1β2

∫ t

0

ZJt · β′(t)dt− β1

β2

∫ t

0

|ZJt|2dt

−∫ t

0

J−1∑j=0

(Δ+Zj

h× Δ+Zjt

h

)· Zj+1,thdt− 1

α2

∫ t

0

(Z0t × Z0) · α′(t)dt

+1β2

∫ t

0

(ZJt × ZJ ) · β′(t)dt

≤ 12‖Zht(0)‖2

2 +12(|Z0t|2h+ |ZJt|2h) −

∫ t

0

J−1∑j=1

(Zj · Δ+Δ−Zj

h2

)|Zjt|2hdt

− 1α2

∫ t

0

Z0t · α′(t)dt+1β2

∫ t

0

ZJt · β′(t)dt

−∫ t

0

J−1∑j=0

(Δ+Zj

h× Δ+Zjt

h

)· Zj+1,thdt− 1

α2

∫ t

0

(Z0t × Z0) · α′(t)dt

+1β2

∫ t

0

(ZJt × ZJ ) · β′(t)dt. (2.16)

For the first term on the right-hand side of (2.16), we have

12‖Zht(0)‖2

2 =12

J∑j=0

|Zjt(0)|2h =12

( J−1∑j=1

|Zjt(0)|2h+ |Z0t(0)|2h+ |ZJt(0)|2h)

=12

( J−1∑j=1

∣∣∣∣ − φj ×(φj × Δ+Δ−φj

h2

)+ φj × Δ+Δ−φj

h2

∣∣∣∣2

h


+ |Z0t(0)|2h+ |ZJt(0)|2h)

≤ C‖δ2φh‖22 + C,

where we have used the following facts

|Z0t|2h =∣∣∣∣ α2

α1h+ α2Z1t − hα′(t)

α1h+ α2

∣∣∣∣2

h ≤ C|Z1t|2h+ C

≤ C

∣∣∣∣ − Z1 ×(Z1 × Δ+Δ−Z1

h2

)+ Z1 × Δ+Δ−Z1

h2

∣∣∣∣2

h+ C

≤ C

∣∣∣∣Δ+Δ−Z1

h2

∣∣∣∣2

h+ C = C

∣∣∣∣Δ2+Z0

h2

∣∣∣∣2

h+ C

= C

∣∣∣∣Δ+(α(t) + α1Z0)h

∣∣∣∣2

h+ C = C

∣∣∣∣Δ+Z0

h

∣∣∣∣2

h+ C

= C

∣∣∣∣α(t) + α1Z0

α2

∣∣∣∣2

h+ C ≤ C

and

|ZJt|2h =∣∣∣∣ β2

β1h+ β2ZJ−1,t +

hβ′(t)β1h+ β2

∣∣∣∣2

h ≤ C|ZJ−1,t|2h+ C

≤ C

∣∣∣∣ − Z1 ×(ZJ−1 × Δ+Δ−ZJ−1

h2

)+ ZJ−1 × Δ+Δ−ZJ−1

h2

∣∣∣∣2

h+ C

≤ C

∣∣∣∣Δ+Δ−ZJ−1

h2

∣∣∣∣2

h+ C = C

∣∣∣∣Δ2−ZJ

h2

∣∣∣∣2

h+ C

= C

∣∣∣∣Δ−(β(t) − β1ZJ )h

∣∣∣∣2

h+ C = C

∣∣∣∣Δ−ZJ

h

∣∣∣∣2

h+ C

= C

∣∣∣∣β(t) + β1ZJ

α2

∣∣∣∣2

h+ C ≤ C.

For the second term on the right-hand side of (2.16), using the same method as the first term,we have

12(|Z0t|2h+ |ZJt|2h) ≤ C.

For the third term on the right-hand side of (2.16), we have

−∫ t

0

J−1∑j=1

(Zj · Δ+Δ−Zj

h2

)|Zjt|2hdt

= −∫ t

0

[ J−2∑j=2

(Zj · Δ+Δ−Zj

h2

)|Zjt|2h+

(Z1 · Δ+Δ−Z1

h2

)|Z1t|2h

+(ZJ−1 · Δ+Δ−ZJ−1

h2

)|ZJ−1,t|2h

]dt

= −∫ t

0

J−1∑j=1

[− 1

2

(Δ+Zj

h· Δ+Zj

h+

Δ−Zj

h

Δ−Zj

h

)]|Zjt|2hdt

+∫ t

0

(Z1 · Δ+Δ−Z1

h2

)|Z1t|2hdt


+∫ t

0

(ZJ−1 · Δ+Δ−ZJ−1

h2

)|ZJ−1,t|2hdt

≤ C

∫ t

0

‖δZh‖2∞‖Zht‖2

2dt+ C

∫ t

0

|Z1t|2hdt+ C

∫ t

0

|ZJ−1,t|2hdt

≤ C

∫ t

0

‖δZh‖2∞‖Zht‖2

2dt+ C

∫ t

0

‖Zht‖22dt

≤ C

∫ t

0

‖δZh‖2(‖δZh‖2 + ‖δ2Zh‖2)‖Zht‖22dt+ C

∫ t

0

‖Zht‖22dt

≤ C

∫ t

0

(1 + ‖δ2Zh‖22)‖Zht‖2

2dt.

For the fourth and fifth terms on the right-hand side of (2.16), we have

− 1α2

∫ t

0

Z0t · α′(t)dt+1β2

∫ t

0

ZJt · β′(t)dt ≤ C

∫ t

0

‖Zht‖2∞dt+ C

≤ C

∫ t

0

‖Zht‖2(‖Zht‖2 + ‖δZht‖2)dt+ C

≤ C

∫ t

0

‖Zht‖22dt+ θ

∫ t

0

‖δZht‖22dt+ C.

For the sixth term on the right-hand side of (2.16), we have∫ t

0

J−1∑j=0

(Δ+Zj

h× Δ+Zjt

h

)· Zj+1,thdt ≤

∫ t

0

‖δZh‖∞‖δZth‖2‖Zth‖2dt

≤ θ

∫ t

0

‖δZht‖22dt+ C

∫ t

0

‖δZh‖2∞‖Zth‖2

2dt

≤ θ

∫ t

0

‖δZht‖22dt+ C

∫ t

0

(1 + ‖δ2Zh‖22)‖Zht‖2

2dt.

For the seventh and the eighth terms on the right-hand side of (2.16), we have

− 1α2

∫ t

0

(Z0t × Z0) · α′(t)dt+1β2

∫ t

0

(ZJt × ZJ ) · β′(t)dt

≤ C

∫ t

0

‖Zht‖2∞dt+ C ≤ C

∫ t

0

‖Zht‖2(‖Zht‖2 + ‖δZht‖2)dt+ C

≤ C

∫ t

0

‖Zht‖22dt+ θ

∫ t

0

‖δZht‖22dt+ C.

Putting these estimates together, and choosing θ small enough, we have

‖Zht‖22 +

∫ t

0

‖δZht‖22dt ≤ C

∫ t

0

(1 + ‖δ2Zh‖22)‖Zht‖2

2dt+ C. (2.17)

By Lemma 2.4, we get

sup0≤t≤T0

‖Zht‖2 ≤ C,

∫ T0

0

‖δ3Zht‖22 ≤ C.

Taking the inner product of (2.1) with Δ+Δ−Zj

h2 h, and then summing over j from 1 to J−1,we have

J−1∑j=1

Zjt · Δ+Δ−Zj

h2h = −

J−1∑j=1

(Zj · Δ+Δ−Zj

h2

)2

h+J−1∑j=1

(Δ+Δ−Zj

h2

)2

h. (2.18)


By (2.17)–(2.18) and the Cauchy inequality, we get

‖δ2Zh‖22 ≤ C‖Zht‖2

2 + C ≤ C.

So we have

sup0≤t≤T0

‖δ2Zh‖22 ≤ C. �

Lemma 2.9 If φ(x) ∈ S2, φ(x) ∈ H3([0, l]), and α(t), β(t) ∈ H2loc((0,+∞)), then there are


sup0≤t≤T0

(‖δZht(t)‖2 + ‖δ3Zh(t)‖2) +∫ T0

0

‖δ2Zht(t)‖22 ≤ C, (2.19)

sup0≤t≤T0

(‖Zht(t)‖∞ + ‖δ2Zh(t)‖∞) ≤ C. (2.20)

Proof Multiplying (2.13) by −Δ+Δ−Zjt

h2 h, and summing over j from 1 to J − 1, we have

−J−1∑j=1

Zjtt · Δ+Δ−Zjt

h2h =

J−1∑j=1

[Zjt ×

(Zj × Δ+Δ−Zj

h2

)]· Δ+Δ−Zjt

h2h

+J−1∑j=1

[Zj ×

(Zjt × Δ+Δ−Zj

h2

)]· Δ+Δ−Zjt

h2h

+J−1∑j=1

[Zj ×

(Zj × Δ+Δ−Zjt

h2

)]· Δ+Δ−Zjt

h2h

−J−1∑j=1

(Zjt × Δ+Δ−Zj

h2

)· Δ+Δ−Zjt

h2h. (2.21)

The left-hand side of (2.21) is equal to

−J−1∑j=1

Zjtt · Δ+Δ−Zjt

h2h

=J−1∑j=0

Δ+Zjtt

h· Δ+Zjt

hh+ Z0tt · Δ+Z0t

h− ZJtt · Δ−ZJt

h

=12d

dt

J−1∑j=0

∣∣∣∣Δ+Zjt

h

∣∣∣∣2

h+α′′(t) − α2

Δ+Z0tt

h

−α1· Δ+Z0t

h− β′′(t) − β2

Δ−ZJtt

h

β1· Δ−ZJt

h

=12d

dt‖δZht‖2

2 −1α1α′′(t) · Δ+Z0t

h+

α2

2α1

d

dt

∣∣∣∣Δ+Z0t

h

∣∣∣∣2

− 1β1β′′(t) · Δ−ZJt

h

+β2

2β1

d

dt

∣∣∣∣Δ−ZJt

h

∣∣∣∣2

=12d

dt‖δZht‖2

2 −1α1α′′(t) · α

′(t) + α1Z0t

α2+

α2

2α1

d

dt

∣∣∣∣Δ+Z0t

h

∣∣∣∣2

− 1β1β′′(t) · β

′(t) − β1ZJt

β2+

β2

2β1

d

dt


h

∣∣∣∣2

=12d

dt‖δZht‖2

2 −1

α1α2α′′(t) · α′(t) − 1

α2α′′(t) · Z0t +

α2

2α1

d

dt

∣∣∣∣Δ+Z0t

h

∣∣∣∣2


− 1β1β2

β′′(t) · β′(t) +1β2β′′(t) · ZJt +

β2

2β1

d

dt


h

∣∣∣∣2

.

For the first term on the right-hand side of (2.21), we haveJ−1∑j=1

[Zjt ×

(Zj × Δ+Δ−Zj

h2

)]· Δ+Δ−Zjt

h2h

≤ C‖Zh‖∞‖Zht‖∞‖δ2Zh‖2‖δ2Zht‖2

≤ C‖Zht‖∞‖δ2Zh‖2‖δ2Zht‖2

≤ θ‖δ2Zht‖22 + C‖Zht‖2

∞‖δ2Zh‖22

≤ θ‖δ2Zht‖22 + C‖Zht‖2(‖Zht‖2 + ‖δZht‖2)‖δ2Zh‖2

2

≤ θ‖δ2Zht‖22 + C‖δZht‖2

2 + C.

For the second term on the right-hand side of (2.21), we haveJ−1∑j=1

[Zj ×

(Zjt × Δ+Δ−Zj

h2

)]· Δ+Δ−Zjt

h2h

≤ ‖Zh‖∞‖Zht‖∞‖δ2Zh‖2‖δ2Zht‖2

≤ ‖Zht‖∞‖δ2Zh‖2‖δ2Zht‖2

≤ θ‖δ2Zht‖22 + C‖Zht‖2

∞‖δ2Zh‖22


2

≤ θ‖δ2Zht‖22 + C + C‖δZht‖2

2.

SinceJ−1∑j=1

(Zj · Δ+Δ−Zjt

h2

)2

h =J−2∑j=2

(Zj · Δ+Δ−Zjt

h2

)2

h+(Z1 · Δ+Δ−Z1t

h2

)2

h

+(ZJ−1 · Δ+Δ−ZJ−1t

h2

)2

h (2.22)

and |Zj | = 1 (j = 1, 2, . . . , J − 1), we have

Zj · Δ+Δ−Zjt

h2= −Zjt · Δ+Δ−Zj

h2−

(Δ+Zj

h· Δ+Zjt

h+

Δ−Zj

h· Δ−Zjt

h

), j = 2, . . . , J − 2.

Then we haveJ−2∑j=2

(Zj · Δ+Δ−Zjt

h2

)2

h =J−2∑j=2

[− Zjt · Δ+Δ−Zj

h2−

(Δ+Zj

h· Δ+Zjt

h+

Δ−Zj

h· Δ−Zjt

h

)]2

h

≤ C‖Zht‖2∞‖δ2Zh‖2

2 + C‖δZh‖2∞‖δZht‖2

2

≤ C‖Zht‖2(‖Zht‖2 + ‖δZht‖2)

+ C‖δZh‖2(‖δZh‖2 + ‖δ2Zh‖2)‖δZht‖22

≤ C + C‖δZht‖22,

we also have(Z1 · Δ+Δ−Z1t

h2

)2

h =[Z1 · 1

hΔ+

(Δ+Z0t

h

)]2

h =[Z1 · 1

hΔ+

(α′(t) + α1Z0t

α2

)]2

h


=[α1

α2Z1 · Δ+Z0t

h

]2

h =[α1

α2Z1 · α

′(t) + α1Z0t

α2

]2

h

≤ C + C‖Zht‖2∞h ≤ C + C‖Zht‖2(‖Zht‖2 + ‖δZht‖2)h

≤ C + C‖δZht‖22,(

ZJ−1 · Δ+Δ−ZJ−1,t

h2

)2

h =[ZJ−1 · 1

hΔ−

(Δ−ZJt

h

)]2

h

=[ZJ−1 · 1

hΔ−

(β′(t) − β1ZJt

β2

)]2

h

=[β1

β2ZJ−1 · Δ−ZJt

h

]2

h =[β1

β2ZJ−1 · β

′(t) − β1ZJt

β2

]2

h

≤ C + C‖Zht‖2∞h ≤ C + C‖Zht‖2(‖Zht‖2 + ‖δZht‖2)h

≤ C + C‖δZht‖22.

So we getJ−1∑j=1

(Zj · Δ+Δ−Zjt

h2

)2

h ≤ C + C‖δZht‖22.

Furthermore, we can have the estimate about the third term of (2.21) as follows:J−1∑j=1

[Zj × Zj × Δ+Δ−Zjt

h2

]· Δ+Δ−Zjt

h2h =

J−1∑j=1

(Zj · Δ+Δ−Zjt

h2

)2

h−J−1∑j=1

(Δ+Δ−Zjt

h2

)2

h

≤ C + C‖δZht‖22 − ‖δ2Zht‖2

2,

where we have used the fact thatJ−1∑j=1

(Δ+Δ−Zjt

h2

)2

h =J−1∑j=1

∣∣∣∣Δ2+Zj−1,t

h2

∣∣∣∣2

h =J−2∑j=0

∣∣∣∣Δ2+Zjt

h2

∣∣∣∣2

h = ‖δ2Zht‖22.

For the fourth term on the right-hand side of (2.21), we have

−J−1∑j=1

(Zjt × Δ+Δ−Zj

h2

)· Δ+Δ−Zjt

h2h

≤ ‖Zht‖∞‖δ2Zh‖2‖δ2Zht‖2 ≤ θ‖δ2Zht‖22 + C‖Zht‖2

∞‖δ2Zh‖22


2

≤ θ‖δ2Zht‖22 + C + C‖δZht‖2

2.

Therefore, we have12d

dt‖δZht‖2

2 + ‖δ2Zht‖22

≤ 1α1α2

α′′(t) · α′(t) +1α2α′′(t) · Z0t − α2

2α1

d

dt|Δ+Z0t

h|2 +

1β1β2

β′′(t) · β′(t)

− 1β2β′′(t) · ZJt − β2

2β1

d

dt


h

∣∣∣∣2

+12‖δ2Zht‖2

2 + C + C‖δZht‖22,

provided that θ is small enough. Integrating the result with respect to t, one gets

‖δZht‖22 +

∫ t

0

‖δ2Zht‖22 ≤ ‖δZht(0)‖2

2 +1

α1α2

∫ t

0

α′′(t) · α′(t) +1α2

∫ t

0

α′′(t) · Z0t


+α2

2α1

∣∣∣∣Δ+Z0t(0)h

∣∣∣∣2

+1

β1β2

∫ t

0

β′′(t) · β′(t)

− 1β2

∫ t

0

β′′(t) · ZJt +β2

2β1

∣∣∣∣Δ−ZJt(0)h

∣∣∣∣2

+ C + C

∫ t

0

‖δZht‖22

≤ ‖δZht(0)‖22 + C + C

∫ t

0

‖Zht‖2∞ +

α2

2α1

∣∣∣∣Δ+Z0t(0)h

∣∣∣∣2

+β2

2β1

∣∣∣∣Δ−ZJt(0)h

∣∣∣∣2

+ C

∫ t

0

‖δZht‖22. (2.23)

So we have ∫ t

0

‖Zht‖2∞ ≤ C

∫ t

0

‖Zht‖22 + C

∫ t

0

‖δZht‖22. (2.24)

From (2.1), we have

Δ+Zjt

h= −Zj+1 ×

(Zj+1 ×

Δ2+Δ−Zj

h3

)− Zj+1 ×

(Δ+Zj

h× Δ+Δ−Zj

h2

)

− Δ+Zj

h×

(Zj × Δ+Δ−Zj

h2

)+ Zj+1 ×

Δ2+Δ−Zj

h3+

Δ+Zj

h× Δ+Δ−Zj

h2. (2.25)

So we haveJ−2∑j=1

(Δ+Zjt(0)

h

)2

h ≤ C.

Since we have the facts that∣∣∣∣Δ+Z0t(0)h

∣∣∣∣2

h =∣∣∣∣α

′(0) + α1Z0t(0)α2

∣∣∣∣2

h ≤ C + C|Z0t(0)|2h ≤ C,

∣∣∣∣Δ+ZJ−1t(0)h

∣∣∣∣2

h =∣∣∣∣Δ−ZJt(0)

h

∣∣∣∣2

h =∣∣∣∣β

′(0) − β1ZJt(0)β2

∣∣∣∣2

h ≤ C + C|ZJt(0)|2h ≤ C,

where we have used the same way as (2.16), so we have

‖δZht(0)‖22 =

J−1∑j=0

(Δ+Zjt(0)

h

)2

h =J−2∑j=1

(Δ+Zjt(0)

h

)2

h+∣∣∣∣Δ+Z0t(0)

h

∣∣∣∣2

h+∣∣∣∣Δ+ZJ−1,t(0)

h

∣∣∣∣2

h

≤ C. (2.26)

From (2.3) and (2.4), we have

α2

2α1

∣∣∣∣Δ+Z0t(0)h

∣∣∣∣2

=α2

2α1

∣∣∣∣α′(0) + α1Z0t(0)

α2

∣∣∣∣2

≤ C + C|Z0t(0)|2

≤ C + C

∣∣∣∣Z1t(0) − α′(0)α2

h

∣∣∣∣2

≤ C + C

∣∣∣∣ − φ1 ×(φ1 × Δ+Δ−φ1

h2

)+ φ1 × Δ+Δ−φ1

h2+α′(0)α2

h

∣∣∣∣2

≤ C + C‖δ2φh‖2∞ ≤ C + C‖δ2φh‖2(‖δ2φh‖2 + ‖δ3φh‖2)

≤ C + C|δ2φh‖22 + C‖δ3φh‖2

2 ≤ C, (2.27)

β2

2β1

∣∣∣∣Δ−ZJt(0)h

∣∣∣∣2

=β2

2β1

∣∣∣∣β′(0) − β1ZJt(0)

β2

∣∣∣∣2

≤ C + C|ZJt(0)|2


≤ C + C

∣∣∣∣ZJ−1,t(0) +β′(0)β2

h

∣∣∣∣2

≤ C + C

∣∣∣∣ − φJ−1 ×(φJ−1 × Δ+Δ−φJ−1

h2

)

+ φJ−1 × Δ+Δ−φJ−1

h2+β′(0)β2

h

∣∣∣∣2

≤ C + C‖δ2φh‖2∞ ≤ C + C‖δ2φh‖2(‖δ2φh‖2 + ‖δ3φh‖2)

≤ C + C‖δ2φh‖22 + C‖δ3φh‖2

2 ≤ C. (2.28)

Putting (2.26)–(2.28) together, we have

‖δZht‖22 +

∫ t

0

‖δ2Zht‖22 ≤ C

∫ t

0

‖δZht‖22 + C. (2.29)


sup0≤t≤T0

‖δZht‖2 ≤ C,

∫ T0

0

‖δ2Zht‖22 ≤ C.

Multiplying (2.25) by Δ2+Δ−Zj

h3 h, and then summing over j from j = 1 to J − 2, we have

J−2∑j=1

Δ+Zjt

h· Δ2

+Δ−Zj

h3h = −

J−2∑j=1

(Zj+1 ·

Δ2+Δ−Zj

h3

)2

h−J−2∑j=1

(Δ2

+Δ−Zj

h3

)2

h

−J−2∑j=1

[Zj+1 ×

(Δ+Zj

h× Δ+Δ−Zj

h2

)]· Δ2

+Δ−Zj

h3h

−J−2∑j=1

[Δ+Zj

h×

(Zj × Δ+Δ−Zj

h2

)]· Δ2

+Δ−Zj

h3h

+J−2∑j=1

(Δ+Zj

h× Δ+Δ−Zj

h2

)· Δ2

+Δ−Zj

h3h. (2.30)

Using the fact that Zj ∈ S2 (j = 1, 2, . . . , J − 1), we get

Zj+1 ·Δ2

+Δ−Zj

h3= −Δ+Zj+1 + Δ−Zj+1

2h· Δ+Δ−Zj+1

h2− 3Δ+Zj + Δ−Zj

2h· Δ+Δ−Zj

h2(2.31)

for j = 2, . . . , J − 3. So we have

J−3∑j=2

(Zj+1 ·

Δ2+Δ−Zj

h3

)2

h ≤ C‖δZh‖2∞‖δ2Zh‖2

2 ≤ C. (2.32)

By (2.3) and (2.4), we have

Z2 ·Δ2

+Δ−Z1

h3= Z2 ·

Δ2+Δ+Z0

h3= Z2 · 1

h2Δ2

+

(α(t) + α1Z0

α2

)

=α1

α2Z2 · 1

hΔ+

(α(t) + α1Z0

α2

)=

(α1

α2

)2

Z2 · α(t) + α1Z0

α2, (2.33)

ZJ−1 ·Δ2

+Δ−ZJ−2

h3= ZJ−1 ·

Δ2−Δ−ZJ−1

h3= ZJ−1 · 1

h2Δ2

−

(β(t) − β1ZJ

β2

)


=β1

β2ZJ−1 · 1

hΔ−

(β(t) − β1ZJ

β2

)=

(β1

β2

)2

ZJ−1 · β(t) − β1ZJ

β2. (2.34)

From (2.30)–(2.34) and the Cauchy inequality, we have

‖δ3Zh‖22 ≤ C‖δZth‖2

2 + C ≤ C. �

Lemma 2.10 Let φ(x) ∈ Hk(Ω) (k ≥ 1), α(t), β(t) ∈ H[ k2 ]+1

loc ((0,+∞)). Then there exist aconstant T0 > 0 and a solution such that

sup0≤t≤T0

‖δk−2sZhts‖2 +∫ T0

0

‖δk+1−2sZhts‖22dt ≤ C,

where the constant C is independent of h.

From the above uniform estimates in h of Zj(t) and the standard method, we can provethat the solutions of (2.1)–(2.4) approximate the solutions of (1.1)–(1.4), and then we get theexistence of local smooth solutions to (1.1)–(1.4).

3 Global Smooth Solution

In Section 2, we have obtained a local smooth solution of (1.1)–(1.4) when ε > 0 is fixed. In thissection, we intend to prove the global existence of smooth solution to problem (1.1)–(1.4) forfixed ε > 0 by deriving the global (in time) estimates. In the following, we always suppose thatZ(x, t) is a global smooth solution of problem (1.1)–(1.4) for ε > 0. For simplicity, we let ε = 1without loss of generality in this section. We intend to derive the following global estimates.

Lemma 3.1 If φ(x) ∈ H1((0, l)) and α(t), β(t) ∈ H1loc(0,+∞), suppose that Z(x, t) is global

smooth solution of problem (1.1)–(1.4), then for any given T > 0, we have

|Z| = 1, (3.1)

sup0≤t≤T

‖Zx(·, t)‖2 +∫ T

0

‖Zxx(·, t)‖22 ≤ C. (3.2)

Proof Multiplying (1.1) by Z(x, t), we have Z(x, t) · Zt(x, t) = 0. This implies

|Z| = 1, ∀ (x, t) ∈ [0, l] × [0, T ].

In order to get (3.2), we divide the proof into two steps.

Step 1 Taking the inner product of (1.1) with Zxx and by using integration by parts, wehave

‖Zx(x, t)‖22 + 2

∫ t

0

∫ l

0

|Z × Zxx|2 = ‖φ′(x)‖22 + 2

∫ t

0

Zt · Zx|x=lx=0. (3.3)

By (1.3) and (1.4), we have

2∫ t

0

Zt · Zx|x=lx=0 = 2

∫ t

0

[Zt(l, t) · Zx(l, t) − Zt(0, t) · Zx(0, t)]

= 2∫ t

0

[β′(t) − β2Zxt(l, t)

β1· Zx(l, t) − α(t) − α2Zxt(0, t)

−α1· Zx(0, t)

]

=2β1

∫ t

0

β′(t) · Zx(l, t) − β2

β1(|Zx(l, t)|2 − |Zx(l, 0)|2)

+2α1

∫ t

0

α′(t) · Zx(0, t) − α2

α1(|Zx(0, t)|2 − |Zx(0, 0)|2)


≤ 2β1

∫ t

0

β′(t) · Zx(l, t) +β2

β1|Zx(l, 0)|2 +

2α1

∫ t

0

α′(t) · Zx(0, t) +α2

α1|Zx(0, 0)|2

=2β1

∫ t

0

β′(t) · β(t) − β1Z(l, t)β2

+β2

β1

∣∣∣∣β(t) − β1Z(l, 0)

β2

∣∣∣∣2

+2α1

∫ t

0

α′(t) · α(t) + α1Z(0, t)α2

+α2

α1

∣∣∣∣α(t) + α1Z(0, 0)α2

∣∣∣∣2

≤ C + C

∫ t

0

‖Z‖2∞ ≤ C, (3.4)

where we have used the fact that |Z| = 1. So we get

‖Zx(x, t)‖22 + 2

∫ t

0

∫ l

0

|Z × Zxx|2 = ‖φ′(x)‖22 + 2

∫ t

0

Zt · Zx|x=lx=0 ≤ C.


sup0≤t≤T

‖Zx‖2 ≤ C,

∫ T

0

∫ l

0

|Z × Zxx|2 ≤ C.

Step 2 Taking the scalar product of (1.7) with Zxx and integrating the result over [0, l]×[0, t],

we have

‖Zx(x, t)‖22 + 2

∫ t

0

∫ l

0

|Zxx|2 = ‖φ′(x)‖22 + 2

∫ t

0

Zt · Zx|x=lx=0 + 2

∫ t

0

∫ l

0

|Zx|4

≤ C + C

∫ t

0

‖Zx‖44

≤ C + C

∫ t

0

(‖Zx‖42 + ‖Zxx‖2‖Zx‖3

2)

≤ C + C

∫ t

0

(1 + ‖Zxx‖2)

≤ C +∫ t

0

‖Zxx‖22,

where we have used (3.4). This implies

‖Zx(x, t)‖22 +

∫ t

0

∫ l

0

|Zxx|2 ≤ C.

Then (3.2) follows. �

Lemma 3.2 Let φ(x) ∈ H2((0, l)) and α(t), β(t) ∈ H2loc((0,+∞)), suppose that Z(x, t) is a

global smooth solution of problem (1.1)–(1.4). Then for any given T > 0, we have

sup0≤t≤T

(‖Zt(·, t)‖2 + ‖Zxx(·, t)‖2 + ‖Zx(·, t)‖∞) +∫ T

0

‖Zxx(·, t)‖22dt ≤ C. (3.5)


Ztt = Zxxt + 2(Zx · Zxt)Z + |Zx|2Zt + Zt × Zxx + Z × Zxxt. (3.6)

Multiplying (3.6) by Zt, and integrating the result over (0, l), we have∫ l

0

Zt · Zttdx =∫ l

0

Zt · [Zxxt + 2(Zx · Zxt)Z + |Zx|2Zt + Zt × Zxx + Z × Zxxt]dx


=∫ l

0

Zt · Zxxtdx+∫ l

0

|Zx|2|Zt|2dx+∫ l

0

Zt · (Z × Zxxt)dx. (3.7)

The first term of the right-hand side of (3.7) is equal to∫ l

0

Zt · Zxxtdx = Zt · Zxt|x=lx=0 −

∫ l

0

|Zxt|2dx

= Zt(l, t) · Zxt(l, t) − Zt(0, t) · Zxt(0, t) −∫ l

0

|Zxt|2dx

= Zt(l, t) · β′(t) − β1Zt(l, t)

β2− Zt(0, t) · α

′(t) + α1Zt(0, t)α2

−∫ l

0

|Zxt|2dx

=1β2Zt(l, t) · β′(t) − β1

β2|Zt(l, t)|2 − 1

α2Zt(0, t) · α′(t)

− α1

α2|Zt(0, t)|2 −

∫ l

0

|Zxt|2dx

≤ C‖Zt‖∞ −∫ l

0

|Zxt|2dx

≤ C‖Zt‖122 (‖Zt‖2 + ‖Zxt‖2)

12 −

∫ l

0

|Zxt|2dx

= C‖Zt‖2 + C‖Zt‖122 ‖Zxt‖

122 − ‖Zxt‖2

2

≤ C‖Zt‖22 +

110

‖Zxt‖22 − ‖Zxt‖2

2 + C.

The second term of the right-hand side of (3.7) is equal to∫ l

0

|Zx|2|Zt|2 ≤ ‖Zt‖2∞‖Zx‖2

2 ≤ C‖Zt‖2∞ ≤ C‖Zt‖2(‖Zt‖2 + ‖Zxt‖2)

= C‖Zt‖22 + C‖Zt‖2‖Zxt‖2 ≤ 1

10‖Zxt‖2

2 + C‖Zt‖22.

The third term of the right-hand side of (3.7) is equal to∫ l

0

Zt · (Z × Zxxt)dx = Zt · (Z × Zxt)|x=lx=0 −

∫ l

0

Zt · (Zx × Zxt),

where

Zt · (Z × Zxt)|x=lx=0 = Zt(l, t) · (Z(l, t) × Zxt(l, t)) − Zt(0, t) · (Z(0, t) × Zxt(0, t))

= Zt(l, t) ·(Z(l, t) × β′(t) − β1Zt(l, t)

β2

)

− Zt(0, t) ·(Z(0, t) × α′(t) + α1Zt(l, t)

α2

)

=1β2Zt(l, t) · (Z(l, t) × β′(t)) − 1

α2Zt(0, t) · (Z(0, t) × α′(t))

≤ C‖Zt‖∞‖Z‖∞≤ C‖Zt‖

122 (‖Zt‖2 + ‖Zxt‖2)

12

≤ C‖Zt‖22 +

110

‖Zxt‖22 + C


and

−∫ l

0

Zt · (Zx × Zxt) ≤ C‖Zt‖∞‖Zx‖2‖Zxt‖2

≤ C‖Zt‖∞‖Zxt‖2

≤ C‖Zt‖2∞ +

110

‖Zxt‖22

≤ C‖Zt‖2(‖Zt‖2 + ‖Zxt‖2) +110

‖Zxt‖22

≤ C‖Zt‖22 +

15‖Zxt‖2

2.

Putting the estimates together, we have

d

dt

∫ l

0

|Zt|2dx+∫ l

0

|Zxt|2dx ≤ C‖Zt‖22 + C.

So we have sup0≤t≤T ‖Zt‖ ≤ C and∫ t

0

∫ l

0|Zxt|2dxds ≤ C.

Multiplying (1.7) by Zxx and integrating over [0, l], we get∫ l

0

ZtZxx =∫ l

0

|Zxx|2 −∫ l

0

|Zx|4.

Following by ‖Zx‖4 ≤ C(‖Zx‖2 + ‖Zxx‖142 ‖Zx‖

344 ) and the Cauchy inequality, we have

‖Zxx‖2 ≤ C(‖Zt‖2 + 1) ≤ C, ∀ t ∈ [0, T ]. (3.8)

Then we see sup[0,T ] ‖Zxx‖22 ≤ C.

Finally, (3.5) follows by Lemma 2.2. �

Lemma 3.3 Let φ(x) ∈ H3((0, l)) and α(t), β(t) ∈ H2loc((0,+∞)), suppose that Z(x, t) is a

global smooth solution of problem (1.1)–(1.4). Then for any given T > 0, we have

sup0≤t≤T

(‖Zxt(·, t)‖2 +‖Zxxx(·, t)‖2 +‖Zxx(·, t)‖∞ +‖Zt(·, t)‖∞)+∫ T

0

‖Zxxt(·, t)‖22dt ≤ C. (3.9)

Proof Since we have the fact that 12

ddt

∫ l

0|Zxt|2 = Zxt · Ztt|x=l

x=0 −∫ l

0Zxxt · Ztt, we get

12‖Zxt‖2

2 −12‖Zxt(x, 0)‖2

2 =∫ t

0

Zxt · Ztt|x=lx=0 −

∫ t

0

∫ l

0

Zxxt · Ztt. (3.10)

The first term of the right-hand side of (3.10) is equal to∫ t

0

Zxt · Ztt|x=lx=0

=∫ t

0

[Zxt(l, t) · β

′′(t) − β2Zxtt(l, t)β1

− Zxt(0, t) · α′′(t) − α2Zxtt(0, t)

−α1

]

=1β1

∫ t

0

Zxt(l, t) · β′′(t) − β2

2β1(|Zxt(l, t)|2 − |Zxt(l, 0)|2)

+1α1

∫ t

0

Zxt(0, t) · α′′(t) − α2

2α1(|Zxt(0, t)|2 − |Zxt(0, 0)|2)

=1β1

∫ t

0

β′(t) − β1Zt(l, t)β2

· β′′(t) − β2

2β1(|Zxt(l, t)|2 − |Zxt(l, 0)|2)


+1α1

∫ t

0

α′(t) + α1Zt(0, t)α2

α′′(t) − α2

2α1(|Zxt(0, t)|2 − |Zxt(0, 0)|2)

=1

β1β2

∫ t

0

β′(t) · β′′(t) − 1β2

∫ t

0

Zt(l, t) · β′′(t) − β2

2β1(|Zxt(l, t)|2 − |Zxt(l, 0)|2)

+1

α1α2

∫ t

0

α′(t) · α′′(t) − 1α2

∫ t

0

Zt(0, t) · α′′(t) − α2

2α1(|Zxt(0, t)|2 − |Zxt(0, 0)|2)

≤ C + C

∫ t

0

‖Zt‖2∞ +

β2

2β1|Zxt(l, 0)|2 +

α2

2α1|Zxt(0, 0)|2

≤ C + C

∫ t

0

[‖Zt‖2(‖Zt‖2 + ‖Zxt‖2)]

≤ C + C

∫ t

0

‖Zxt‖22,

where we have used

Zxt(0, 0) =α′(0) + α1Zt(0, 0)

α2=α′(0)α2

+α1

α2(−φ(0) × (φ(0) × φ′′(0)) + φ(0) × φ′′(0)),

Zxt(l, 0) =β′(0) − β1Zt(l, 0)

β2=β′(0)β2

+β1

β2(−φ(l) × (φ(l) × φ′′(l)) + φ(l) × φ′′(l)).

The second term of the right-hand side of (3.10) is equal to

−∫ t

0

∫ l

0

Zxxt · Ztt

= −∫ t

0

∫ l

0

Zxxt · [Zxxt + 2(Zx · Zxt)Z + |Zx|2Zt + Zt × Zxx + Z × Zxxt]

= −∫ t

0

∫ l

0

|Zxxt|2 − 2∫ t

0

∫ l

0

(Zx · Zxt)(Z · Zxxt) −∫ t

0

∫ l

0

|Zx|2(Zt · Zxxt)

−∫ t

0

∫ l

0

(Zt × Zxx) · Zxxt. (3.11)

Since we have the facts that

− 2∫ t

0

∫ l

0

(Zx · Zxt)(Z · Zxxt) ≤ C

∫ t

0

‖Zx‖∞‖Z‖∞‖Zxt‖2‖Zxxt‖2

≤ C

∫ t

0

‖Zxt‖2‖Zxxt‖2 ≤ C

∫ t

0

‖Zxt‖22 +

16

∫ t

0

‖Zxxt‖22,

−∫ t

0

∫ l

0

|Zx|2(Zt · Zxxt) ≤ C

∫ t

0

‖Zx‖2∞‖Zt‖2‖Zxxt‖2

≤ C

∫ t

0

‖Zxxt‖2 ≤ 16

∫ t

0

‖Zxxt‖22 + C,

−∫ t

0

∫ l

0

(Zt × Zxx) · Zxxt ≤ C

∫ t

0

‖Zt‖∞‖Zxx‖2‖Zxxt‖2 ≤∫ t

0

‖Zt‖∞‖Zxxt‖2

≤ 16

∫ t

0

‖Zxxt‖22 + C

∫ t

0

‖Zt‖2∞

≤ 16

∫ t

0

‖Zxxt‖22 + C

∫ t

0

‖Zt‖2(‖Zt‖2 + ‖Zxt‖2)

≤ 16

∫ t

0

‖Zxxt‖22 + C

∫ t

0

‖Zxt‖22 + C,


we have

‖Zxt‖22 +

∫ t

0

∫ l

0

|Zxxt|2 ≤ C

∫ t

0

‖Zxt‖22 + C.

By Lemma 2.4, we get sup0≤t≤T ‖Zxt‖2 ≤ C,∫ t

0

∫ l

0|Zxxt|2 ≤ C.

From (1.7), we have

Zxt = Zxxx + 2(Zx · Zxx)Z + |Zx|2Zx + Zx × Zxx + Z × Zxxx. (3.12)

Multiplying (3.12) by Zxxx, integrating over (0, l), and noticing that

‖Zx‖6 ≤ C(‖Zx‖2 + ‖Zxx‖132 ‖Zx‖

232 ),

we get‖Zxxx‖2 ≤ C‖Zxt‖2 + C. (3.13)

So we have sup0≤t≤T ‖Zxxx‖2 ≤ C.Finally, (3.9) follows by Lemma 2.2. �Using the mathematical induction method, we have the following lemma.

Lemma 3.4 Let φ(x) ∈ Hk((0, l)) and α(t), β(t) ∈ H[ k2 ]+1

loc ((0,+∞)), suppose Z(x, t) is aglobal smooth solution of problem (1.1)–(1.4). Then for any given T > 0, we have

sup0≤t≤T

‖∂st ∂

k−2sx Z(·, t)‖2 +

∫ T

0

‖∂st ∂

k+1−2sx Z(·, t)‖2 ≤ C, 0 ≤ s ≤

[k

2

]. (3.14)

The process of induction is divided into two parts. One proves (3.14) when k is even. Theother proves the case when k is odd.

Combining the local existence obtained in Section 2 and the global in time estimates inLemma 3.4, we can get the global existence of smooth solution to problem (1.1)–(1.4) for fixedε > 0 in the following sense.

Theorem 3.5 Let φ(x) ∈ Hk((0, l)) and α(t), β(t) ∈ H[ k2 ]+1

loc ((0,+∞)). Then for any givenT > 0, (1.1)–(1.4) admits at least one global smooth solution Z(x, t) in (0, l) × [0, T ] :

Z(x, t) ∈ G(T ) =( [ k

2 ]⋂s=0

W s∞(0, T ;H(k−2s)((0, l)))

)∩

( [ k+12 ]⋂

s=0

Hs(0, T ;Hk+1−2s((0, l)))).

4 Uniform Estimates with ε and Global Solution with Constant Boundary Con-ditions for ε = 0

In Section 3, we have obtained a global smooth solution for (1.1)–(1.4) when ε > 0 is fixed. Inthis section, we intend to prove the global existence of smooth solution to problem (1.8)–(1.11)which implies constant boundary. To meet the need in sending ε to zero, we want to establishsome estimates which are both global in time and uniform in ε. The difficulty in the uniformestimate is overcome by constructing some new energy laws and by the fact that Z,Zx, Z ×Zx

is a base of R3, so is Z,Zt, Z × Zt. We assume ε < 1 without loss of generality in this section.

Lemma 4.1 If φ(x) ∈ H1(0, l), suppose that Z(x, t) is a global smooth solution of problem(1.1), (1.2), (1.10) and (1.11). There are constant C > 0 independent of ε. We have

|Z(x, t)| = 1, ∀ (x, t) ∈ [0, l] × [0, T ], (4.1)


sup0≤t≤T

‖Zx(·, t)‖2 + ε

∫ T

0

‖Zxx(·, t)‖22dt ≤ C. (4.2)

Proof Multiplying (1.1) by Z(x, t), we have Z(x, t) · Zt(x, t) = 0. This implies

|Z(x, t)| = 1, ∀ (x, t) ∈ [0, l] × [0, T ].

Multiplying (1.1) by Zxx, and integrating the result over (0, l) × (0, t), we have

‖Zx(x, t)‖2 + 2ε∫ t

0

∫ l

0

|Z × Zxx|2dx = ‖φ′(x)‖2 + 2∫ t

0

Zt · Zx|x=lx=0. (4.3)

By (1.10) and (1.11), we get

2∫ t

0

Zt · Zx|x=lx=0 = 2

∫ t

0

[Zt(l, t) · Zx(l, t) − Zt(0, t) · Zx(0, t)]

= 2∫ t

0

[−β2Zxt(l, t)β1

· Zx(l, t) − −α2Zxt(0, t)−α1

· Zx(0, t)]

= −β2

β1(|Zx(l, t)|2 − |Zx(l, 0)|2) − α2

α1(|Zx(0, t)|2 − |Zx(0, 0)|2)

≤ β2

β1|φ′(l)|2 +

α2

α1|φ′(0)|2 ≤ C.

So we have

‖Zx(x, t)‖22 + 2ε

∫ t

0

∫ l

0

|Z × Zxx|2dx ≤ C.

Multiplying (1.7) by Zxx and integrating the result over (0, l), we have

‖Zx(x, t)‖22 + 2ε

∫ t

0

‖Zxx‖22

= ‖φ′(x)‖22 + 2

∫ t

0

Zt · Zx|x=lx=0 + 2ε

∫ t

0

‖Zx‖44

= ‖φ′(x)‖22 + 2

∫ t

0

[Zt(l, t) · Zx(l, t) − Zt(0, t) · Zx(0, t)] + 2ε∫ t

0

‖Zx‖44

≤ C + Cε

∫ t

0

‖Zx‖44

≤ C + Cε

∫ t

0

(‖Zx‖42 + ‖Zxx‖2‖Zx‖3

2)

≤ C + Cε

∫ t

0

(1 + ‖Zxx‖2)

≤ C + ε

∫ t

0

‖Zxx‖22.

Finally, we have

‖Zx‖22 + ε

∫ t

0

‖Zxx‖2 ≤ C.

Then (4.2) follows. �

Lemma 4.2 If φ(x) ∈ H2(0, l), suppose that Z(x, t) is a global smooth solution of problem(1.1), (1.2), (1.10) and (1.11), then for any given T > 0, there are constant C > 0 independent


of ε such that

sup0≤t≤T

(‖Zt(·, t)‖2 + ‖Zxx(·, t)‖2 + ‖Zx(·, t)‖∞) + ε

∫ T

0

‖Zxt(·, t)‖22dt ≤ C. (4.4)

Proof We shall use the following identities which follow from the fact |Z(x, t)| = 1:

Z · Zt = 0, Z · Zx = 0, Z · Zxx = −|Zx|2.It follows from (1.7) that

Ztt = εZxxt + 2ε(Zx · Zxt)Z + ε|Zx|2Zt + Zt × Zxx + Z × Zxxt. (4.5)

Multiplying (4.5) by Zt and integrating the result over (0, l) × (0, t), we have

‖Zt‖22 = ‖Zt(x, 0)‖2

2 + 2ε∫ t

0

∫ l

0

Zt · Zxxt + 2ε∫ t

0

∫ l

0

|Zx|2|Zt|2

+∫ t

0

(Z × Zxt) · Zt|x=lx=0 −

∫ t

0

∫ l

0

Zt · (Zx × Zxt), (4.6)

where ‖Zt(x, 0)‖22 = ‖εφ′′(x) + ε|φ′(x)|2φ(x) + φ(x) × φ′′(x)‖2

2 and∫ t

0

(Z × Zxt) · Zt|x=lx=0

=∫ t

0

[(Z(l, t) × Zxt(l, t)) · Zt(l, t) − (Z(0, t) × Zxt(0, t)) · Zt(0, t)]

=∫ t

0

[(Z(l, t) × −β1Zt(l, t)

β2

)· Zt(l, t) −

(Z(0, t) × α1Zt(0, t)

α2

)· Zt(0, t)

]= 0.

For the second term of the right-hand side of (4.6), we have

2ε∫ t

0

∫ l

0

Zt · Zxxt = 2ε∫ t

0

Zxt · Zt|x=lx=0 − 2ε

∫ t

0

∫ l

0

|Zxt|2

= 2ε∫ t

0

(− β1

β2|Zt(l, t)|2 − α1

α2|Zt(0, t)|2

)− 2ε

∫ t

0

∫ l

0

|Zxt|2

= −2ε∫ t

0

(β1

β2|Zt(l, t)|2 +

α1

α2|Zt(0, t)|2

)− 2ε

∫ t

0

∫ l

0

|Zxt|2

≤ −2ε∫ t

0

∫ l

0

|Zxt|2.

Since W 1,1(0, l) ↪→ L∞(0, l), we get

2ε∫ t

0

∫ l

0

|Zx|2|Zt|2 ≤ 2ε∫ t

0

‖Zx‖2∞‖Zt‖2

2 ≤ Cε

∫ t

0

(1 + ‖Zxx‖22)‖Zt‖2

2.

Meanwhile, if |Zt| = 0, then the vectors Z,Zt, Z × Zt form an orthogonal basis of R3. Let

Zxt = a1Z + b1Zt + c1Z × Zt,

and it is easy to see that

a1 = −Zx · Zt, b1 =Zxt · Zt

|Zt|2 , c1 =(Z × Zt) · Zxt

|Zt|2 .

Therefore, we have

−∫ t

0

∫ l

0

Zt · (Zx × Zxt)


= −∫ t

0

∫ l

0

Zt · (Zx × (a1Z + c1Z × Zt))

=∫ t

0

∫ l

0

(Zx · Zt)(Zt · (Zx × Z))

=∫ t

0

∫ l

0

(Zx · Zt)((εZxx + ε|Zx|2Z + Z × Zxx) · (Zx × Z))

= ε

∫ t

0

∫ l

0

(Zx · Zt)(Zxx · (Zx × Z)) +∫ t

0

∫ l

0

(Zx · Zt)((Z × Zxx) · (Zx × Z)). (4.7)

For the first term of (4.7), we have

ε

∫ t

0

∫ l

0

(Zx · Zt)(Zxx · (Zx × Z)) ≤ ε

∫ t

0

‖Zx‖2∞‖Zt‖2‖Zxx‖2 ≤ Cε

∫ t

0

(1 + ‖Zxx‖22)(‖Zt‖2

2 + 1).

For the second term on the right-hand side of (4.7), we have∫ t

0

∫ l

0

(Zx · Zt)((Z × Zxx) · (Zx × Z))

= −12

∫ t

0

(Zx · Zt)|Zx|2|x=lx=0 +

12

∫ t

0

∫ l

0

(Zx · Zxt)|Zx|2 +12

∫ t

0

∫ l

0

(Zxx · Zt)|Zx|2, (4.8)

where we have used the fact that

(A×B) · (C ×D) = (A · C)(B ·D) − (A ·D)(B · C).

Since the first term of (4.8) is equal to

− 12

∫ t

0

(Zx · Zt)|Zx|2|x=lx=0

= −12

∫ t

0

[(Zx(l, t) · −β2Zxt(l, t)

β1

)|Zx(l, t)|2 −

(Zx(0, t) · −α2Zxt(0, t)

−α1

)|Zx(0, t)|2

]

= − β2

4β1(|Zx(l, t)|4 − |Zx(l, 0)|4) − α2

4α1(|Zx(0, t)|4 − |Zx(0, 0)|4)

≤ β2

4β1|φ′(l)|4 +

α2

4α1|φ′(0)|4 ≤ C,

the second term of (4.8) is equal to

14

∫ t

0

∫ l

0

(|Zx|2)t|Zx|2 =18‖Zx‖4

4 −18‖Zx(x, 0)‖4

4 ≤ 18‖Zx‖4

4.

The third term of (4.8) is equal to

12

∫ l

0

|Zx|2[Zxx · (εZxx + ε|Zx|2Z + Z × Zxx)] =ε

2

∫ t

0

∫ l

0

|Zx|2|Zxx|2 − ε

2

∫ t

0

∫ l

0

|Zx|6

≤ ε

2

∫ t

0

‖Zx‖2∞‖Zxx‖2

2 +ε

2

∫ t

0

‖Zx‖66.

Taking the cross product of (1.7) with Zxx, and then by using the Cauchy inequality andLemma 4.1, we have

‖Zxx‖22 ≤ C‖Zt‖2

2 + C‖Zx‖44 + C.


Then by noticing that (4.7) and the following inequalities:

‖Zx‖∞ ≤ C(‖Zx‖2 + ‖Zxx‖122 ‖Zx‖

122 ),

‖Zx‖6 ≤ C(‖Zx‖2 + ‖Zxx‖132 ‖Zx‖

232 ),

‖Zx‖4 ≤ C(‖Zx‖2 + ‖Zxx‖142 ‖Zx‖

342 ),

we finally have

‖Zt‖22 + 4ε

∫ t

0

‖Zxt‖22 ≤ C + Cε

∫ t

0

(1 + ‖Zxx‖22)(1 + ‖Zt‖2

2)ds.

Since ε∫ T

0‖Zxx‖2

2 ≤ C from Lemma 4.1 and by Lemma 2.4, we get

sup0≤t≤T

‖Zt‖2 ≤ C, ε

∫ T

0

‖Zxt‖22 ≤ C.

Lemma 4.2 follows by (3.8) and Lemma 2.2. �

Lemma 4.3 If φ(x) ∈ H3(0, l), suppose that Z(x, t) is a global smooth solution of problem(1.1), (1.2), (1.10) and (1.11), then for any given T > 0, there are constant C > 0 independentof ε such that

sup0≤t≤T

(‖Zxt(·, t)‖2+‖Zxxx(·, t)‖2+‖Zt(·, t)‖∞+‖Zxx(·, t)‖∞)+ε∫ T

0

‖Zxxt(·, t)‖22dt ≤ C. (4.9)

Proof Since we have the fact that 12

ddt

∫ l

0|Zxt|2 = Zxt · Ztt|x=l

x=0 −∫ l

0Zxxt · Ztt, we get

12

∫ l

0

|Zxt|2 − 12

∫ l

0

|Zxt(x, 0)|2 =∫ t

0

Zxt · Ztt|x=lx=0 −

∫ t

0

∫ l

0

Zxxt · Ztt. (4.10)

For the first term of the right-hand side of (4.10), we have∫ t

0

Zxt · Ztt|x=lx=0

= − β2

2β1(|Zxt(l, t)|2 − |Zxt(l, 0)|2) − α2

α1(|Zxt(0, t)|2 − |Zxt(0, 0)|2)

≤ β2

2β1|Zxt(l, 0)|2 +

α2

α1|Zxt(0, 0)|2

=β2

2β1

∣∣∣∣ − β1Zt(l, 0)β2

∣∣∣∣2

+α2

α1

∣∣∣∣α1Zt(0, 0)α2

∣∣∣∣2

=12|Zt(l, 0)|2 +

12|Zt(0, 0)|2

=12|εφ′′(l) + ε|φ′(l)|2φ(l) + φ(l) × φ′′(l)|2 +

12|εφ′′(0) + ε|φ′(0)|2φ(0) + φ(0) × φ′′(0)|2

≤ C.

For the second term of the right-hand side of (4.10), we have

−∫ t

0

∫ l

0

Zxxt · Ztt

= −∫ t

0

∫ l

0

Zxxt · (εZxxt + 2ε(Zx · Zxt)Z + ε|Zx|2Zt + Zt × Zxx + Z × Zxxt)


= −ε∫ t

0

∫ l

0

|Zxxt|2 − 2ε∫ t

0

∫ l

0

(Zx · Zxt)(Z · Zxxt) − ε

∫ t

0

∫ l

0

|Zx|2(Zt · Zxxt)

−∫ t

0

∫ l

0

(Zt × Zxx) · Zxxt. (4.11)

For the second term and third term of the right-hand side of (4.11), we know that

− 2ε∫ t

0

∫ l

0

(Zx · Zxt)(Z · Zxxt) ≤ ε

∫ t

0

‖Zx‖∞‖Zxt‖2‖Zxxt‖2 ≤ ε

4

∫ t

0

‖Zxxt‖22 + C

∫ t

0

‖Zxt‖22,

− ε

∫ t

0

∫ l

0

|Zx|2(Zt · Zxxt) ≤ Cε

∫ t

0

‖Zx‖2∞‖Zt‖2‖Zxxt‖2 ≤ ε

4

∫ t

0

‖Zxxt‖22 + C.

Then if |Zt| = 0, let Zxxt = a2Z + b2Zt + c2Z × Zt. It is easy to see that

a2 = Z · Zxxt = −2Zx · Zxt − Zt · Zxx, b2 =Zt · Zxxt

|Zt|2 , c2 =(Z × Zt) · Zxxt

|Zt|2 .

Then we have

−∫ t

0

∫ l

0

(Zt × Zxx) · Zxxt

= 2∫ t

0

∫ l

0

(Zx · Zxt)((Zt × Zxx) · Z) +∫ t

0

∫ l

0

(Zt · Zxx)((Zt × Zxx) · Z)

−∫ t

0

∫ l

0

((Zt × Zxx) · (Z × Zt))(Z × Zt) · Zxxt

|Zt|2 . (4.12)

Noticing that

2∫ t

0

∫ l

0

(Zx · Zxt)((Zt × Zxx) · Z) +∫ t

0

∫ l

0

(Zt · Zxx)((Zt × Zxx) · Z)

≤ C

∫ t

0

‖Zx‖∞‖Zt‖∞‖Zxx‖2‖Zxt‖2 +∫ t

0

‖Zt‖2∞‖Zxx‖2

2

≤ C

∫ t

0

(‖Zxt‖322 + 1),

and the third term of the right-hand side of (4.12) is equal to∫ t

0

∫ l

0

(Zxx · Z)((Z × Zt) · Zxxt)

=∫ t

0

∫ l

0

|Zx|2((Z × Zxxt) · Zt)

= |Zx|2(Z × Zxt)|x=lx=0 − 2

∫ t

0

∫ l

0

(Zx · Zxx)((Z × Zxt) · Zt) −∫ t

0

∫ l

0

|Zx|2((Zx × Zxt) · Zt)

=∣∣∣∣β − β1Z(l, t)

β2

∣∣∣∣2(Z(l, t) × −β1Zt

β2

)−

∣∣∣∣α+ α1Z(0, t)α2

∣∣∣∣2(Z(0, t) × α1Zt

α2

)

− 2∫ t

0

∫ l

0

(Zx · Zxx)((Z × Zxt) · Zt) −∫ t

0

∫ l

0

|Zx|2((Zx × Zxt) · Zt)

≤ C

∫ t

0

‖Zt‖∞ + C

∫ t

0

‖Zx‖∞‖Zt‖∞‖Zxx‖2‖Zxt‖2 + C

∫ t

0

‖Zx‖3∞‖Zt‖2‖Zxt‖2

≤ C

∫ t

0

(‖Zxt‖322 + 1),


where we have used the fact ‖Zt‖∞ ≤ C(‖Zt‖2 + ‖Ztx‖122 ‖Zt‖

122 ), we finally get

‖Ztx‖22 + ε

∫ t

0

‖Zxxt‖22 ≤ C

∫ t

0

(‖Ztx‖22 + 1).

By Gronwall’s equality, we have

sup0≤t≤T

‖Ztx‖2 ≤ C, ε

∫ t

0

‖Zxxt‖22 ≤ C.

Then (4.9) follows by (3.13) and Lemma 2.2. �

Lemma 4.4 If φ(x) ∈ H4(0, l), suppose that Z(x, t) is a global smooth solution of problem(1.1), (1.2), (1.10) and (1.11), then for any given T > 0, there is constant C > 0 independentof ε such that

sup0≤t≤T

(‖Ztt(·, t)‖2 + ‖Zxxt(·, t)‖2 + ‖Zxxxx(·, t)‖2 + ‖Zxt(·, t)‖∞ + ‖Zxxx(·, t)‖∞)

+ ε

∫ T

0

‖Zxtt(·, t)‖22dt ≤ C. (4.13)


Zttt = εZxxtt + 2ε|Zxt|2Z + 2ε(Zx · Zxtt)Z + 4ε(Zx · Zxt)Zt + ε|Zx|2Ztt

+ Ztt × Zxx + 2Zt × Zxxt + Z × Zxxtt. (4.14)

Then we have∫ t

0

∫ l

0

Zttt · Ztt = ε

∫ t

0

∫ l

0

Zxxtt · Ztt + 2ε∫ t

0

∫ l

0

|Zxt|2(Z · Ztt) + 2ε∫ t

0

∫ l

0

(Zx · Zxtt)(Z · Ztt)

+ 4ε∫ t

0

∫ l

0

(Zx · Zxt)(Zt · Ztt) + ε

∫ t

0

∫ l

0

|Zx|2|Ztt|2

+ 2∫ t

0

∫ l

0

(Zt × Zxxt) · Ztt +∫ t

0

∫ l

0

(Z × Zxxtt) · Ztt. (4.15)

The first term of the right-hand side of (4.15) is equal to

ε

∫ t

0

∫ l

0

Zxxtt · Ztt = ε

∫ t

0

Zxtt · Ztt|x=lx=0 − ε

∫ t

0

∫ l

0

|Zxtt|2dx

= ε

∫ t

0

−β1Ztt(l, t)β2

· Ztt(l, t) − ε

∫ t

0

α1Ztt(l, t)α2

· Ztt(0, t) − ε

∫ t

0

∫ l

0

|Zxtt|2dx

= −εβ1

β2

∫ t

0

|Ztt(l, t)|2 − εα1

α2

∫ t

0

|Ztt(0, t)|2 − ε

∫ t

0

∫ l

0

|Zxtt|2dx

≤ −ε∫ t

0

∫ l

0

|Zxtt|2dx.

Besides, we have the facts that

2ε∫ t

0

∫ l

0

|Zxt|2(Z · Ztt) = −2ε∫ t

0

∫ l

0

|Zxt|2|Zt|2 ≤ 2ε∫ t

0

‖Zt‖2∞‖Zxt‖2

2 ≤ C

and

2ε∫ t

0

∫ l

0

(Zx · Zxtt)(Z · Ztt) + 4ε∫ t

0

∫ l

0

(Zx · Zxt)(Zt · Ztt) + ε

∫ t

0

∫ l

0

|Zx|2|Ztt|2


≤ 2ε∫ t

0

‖Zx‖∞‖Ztt‖2‖Zxtt‖2 + 4ε∫ t

0

‖Zx‖∞‖Zt‖∞‖Zxt‖2‖Ztt‖2 +∫ t

0

‖Zx‖2∞‖Ztt‖2

≤ ε

2

∫ t

0

‖Zxtt‖22 + C

∫ t

0

(‖Ztt‖22 + 1).

Taking the cross product of (4.5) with Zxxt, and noticing that Z ·Zxxt = −2(Zt ·Zxt)−Zx ·Ztt,we get

‖Zxxt‖2 ≤ C(‖Ztt‖2 + 1), (4.16)

then we have

2∫ t

0

∫ l

0

(Zt × Zxxt) · Ztt ≤ 2∫ t

0

‖Zt‖∞‖Ztt‖2‖Zxxt‖2 ≤ C

∫ t

0

(‖Ztt‖22 + 1).

The last term of the right-hand side of (4.15) is equal to∫ t

0

∫ l

0

(Z × Zxxtt) · Ztt

=∫ t

0

(Z × Zxtt) · Ztt|x=lx=0 −

∫ t

0

∫ l

0

(Zx × Zxtt) · Ztt

=∫ t

0

Z(l, t) × −β1Ztt(l, t)β2

· Ztt(l, t) − Z(0, t) × α1Ztt(0, t)α2

· Ztt(0, t)

−∫ t

0

∫ l

0

(Zx × Zxtt) · Ztt

= −∫ t

0

∫ l

0

(Zx × Zxtt) · Ztt.

If |Zx| = 0, let Zxtt = a3Z + b3Zx + c3Z × Zx. We can find out

a3 = −2Zt · Zxt − Zx · Ztt, b3 =Zxtt · Zx

|Zx|2 , c3 =(Z × Zx) · Zxtt

|Zx|2 .

Then

−∫ t

0

∫ l

0

(Zx × Zxtt) · Ztt

= −∫ t

0

∫ l

0

(Zx × (a3Z + b3Z × Zt)) · Ztt

= 2∫ t

0

∫ l

0

(Zt · Zxt)((Zx × Z) · Ztt) +∫ t

0

(Zx · Ztt)((Zx × Z) · Ztt)

−∫ t

0

∫ l

0

((Z × Zx) · Zxtt)(Ztt · Z)

= 2∫ t

0

∫ l

0

(Zt · Zxt)((Zx × Z) · Ztt) +∫ t

0


+∫ t

0

∫ l

0

[(Zxtt × Z) · |Zt|2Zx]. (4.17)

For the first and the second terms of (4.17), we have

2∫ t

0

∫ l

0

(Zt · Zxt)((Zx × Z) · Ztt) +∫ t

0



≤ C

∫ t

0

(‖Zt‖∞‖Zx‖∞‖Zxt‖2‖Ztt‖2 + ‖Zx‖2∞‖Ztt‖2

2

≤ C

∫ t

0

(‖Ztt‖22 + 1).

The third term of (4.17) is equal to∫ t

0

∫ l

0

[(Zxtt × Z) · |Zt|2Zx] = −∫ l

0

(Zxt × Z) · |Zt|2Zx|t=tt=0 +

∫ t

0

(Zxt × Zt) · |Zt|2Zx

+ 2∫ t

0

∫ l

0

(Zt · Ztt)[(Zxt × Z) · Zx].

Since we have the facts that

Zxt(x, 0) = εφ′′′(x) + 2ε(φ′(x) · φ′′(x))φ(x) + ε|φ′(x)|2φ′(x) + φ′(x) × φ′′(x) + φ(x) × φ′′′(x),

Zt(x, 0) = εφ′′(x) + ε|φ′(x)|2φ(x) + φ(x) × φ′′(x),

we get

−∫ l

0

(Zxt × Z) · |Zt|2Zx|t=tt=0

= −∫ l

0

(Zxt × Z) · |Zt|2Zx +∫ l

0

(Zxt(x, 0) × φ(x)) · |Zt(x, 0)|2φ′(x)

≤ C‖Z‖∞‖Zt‖2∞‖Zx‖2‖Zxt‖2 + C ≤ C.

From Lemmas 4.1–4.3, we have∫ t

0

(Zxt × Zt) · |Zt|2Zx ≤ C

∫ t

0

‖Zxt‖2‖Zt‖3∞‖Zx‖2 ≤ C,

2∫ t

0

∫ l

0

(Zt · Ztt)[(Zxt × Z) · Zx] ≤ C

∫ t

0

‖Zx‖∞‖Zt‖∞‖Zxt‖2‖Ztt‖2 ≤ C

∫ t

0

‖Ztt‖22 + C.

In conclusion, we have

‖Ztt‖22 + ε

∫ t

0

‖Zxtt‖22 ≤ C

∫ t

0

‖Ztt‖22 + C.

Here we have used the following facts

Ztt(x, 0) = εZxxt(x, 0) + 2ε(φ′(x) · Zxt(x, 0))Z

+ ε|φ′(x)|2Zt(x, 0) + Zt(x, 0) × φ′′(x) + φ′(x) × Zxxt,

Ztxx(x, 0) = εφ′′′′(x) + 2ε|φ′′(x)|2φ(x) + 2ε(φ′(x) · φ′′′(x))φ+ 4ε(φ′(x) · φ′′(x))φ′(x)+ ε|φ′(x)|2φ′′(x) + 2φ′(x) × φ′′′(x) + φ× φ′′′′(x),

Zxt(x, 0) = εφ′′′(x) + 2ε(φ′(x) · φ′′(x))φ(x) + ε|φ′(x)|2φ′(x) + φ′(x) × φ′′(x) + φ(x) × φ′′′(x),

Zt(x, 0) = εφ′′(x) + ε|φ′(x)|2φ(x) + φ(x) × φ′′(x).

Then following by Gronwall’s inequality, one has

sup0≤t≤T

‖Ztt‖2 ≤ C, ε

∫ T

0

‖Zxtt‖22 ≤ C.

From (4.16), we havesup

0≤t≤T‖Zxxt‖2 ≤ C.


Similarly from (1.7), we get

Zxxt = εZxxxx + 2ε|Zxx|2Z + 2ε(Zx · Zxxx)Z

+ 4ε(Zx · Zxx)Zx + ε|Zx|2Zxx + 2Zx × Zxxx + Z × Zxxxx. (4.18)

Taking the cross product of (4.18) with Zxxxx and noting that Z ·Zxxxx = −4Zx ·Zxxx−3|Zxx|2,we get

sup0≤t≤T

‖Zxxxx‖2 ≤ C.

(4.13) follows by Lemma 2.2. �So by induction, we have the following lemma.

Lemma 4.5 If φ(x) ∈ Hk(0, l), suppose that Z(x, t) is a global smooth solution of problem(1.1), (1.2), (1.10) and (1.11), then for any given T > 0, there is a constant C > 0 independentof ε such that

sup0≤t≤T

‖∂st ∂

k−2sx Z(·, t)‖2 ≤ C, 0 ≤ s ≤

[k

2

]. (4.21)

The proof is similar to the induction given in [4]. It is enough for us to make some modifi-cations.

Combining with the local existence obtained in Section 2 and the global in time and uniformin ε estimates in Lemma 4.5, we can pass to the limit ε → 0 in equation (1.1) and then to theglobal smooth solution of problem (1.8)–(1.11) in the following sense.

Theorem 4.6 Let φ(x) ∈ Hk((0, l)). Then for any given T > 0, problem (1.8)–(1.11) admitsat least one smooth solution Z(x, t) :

Z(x, t) ∈[ k2 ]⋂

s=0

W s∞(0, T ;Hk−2s((0, l))).

5 Uniqueness

Theorem 5.1 The global smooth solution of (1.1)–(1.4) obtained in Theorem 3.5 and theglobal smooth solution of (1.8)–(1.11) obtained in Theorem 4.6 are unique.

Proof The proof is similar to the procedure showed in [8]. Let Z1(x, t) and Z2(x, t) be smoothsolutions of (1.1)–(1.4). Let W (x, t) = Z1(x, t)−Z2(x, t), and we assume ε = 1. From (1.7), wehave

Wt = Wxx + Z2((Z1x + Z2x) ·Wx) + |Z1x|2W + Z2 ×Wxx +W × Z1xx, (5.1)

W (x, 0) = 0, (5.2)

− α1W (0, t) + α2Wx(0, t) = 0, (5.3)

β1W (l, t) + β2Wx(l, t) = 0. (5.4)

From (5.3) and (5.4), we get

Wx(0, t) =α1

α2W (0, t), (5.5)

Wx(l, t) = −β1

β2W (l, t). (5.6)


Multiplying (5.1) by W , we have

12d

dt

∫ l

0

|W |2 −∫ l

0

Wxx ·W =∫ l

0

|Z1x|2|W |2 +∫ l

0

(Z2 ×Wxx) ·W

+∫ l

0

((Z1x + Z2x) ·Wx) · (Z2 ·W ). (5.7)

For the second term of the left-hand side of (5.7), we have

−∫ l

0

Wxx ·W = −Wx ·W |x=lx=0 +

∫ l

0

|Wx|2

= −Wx(l, t)W (l, t) +Wx(0, t)W (0, t) +∫ l

0

|Wx|2

=β1

β2|W (l, t)|2 +

α1

α2|W (0, t)|2 +

∫ l

0

|Wx|2.

Since Z2 is smooth, we have the fact that∫ l

0

(Z2 ×Wxx) ·W = Z2 ×Wx ·W |x=lx=0 −

∫ l

0

(Z2x ×Wx) ·W

= Z2(l, t) ×(− β1

β2W (l, t)

)·W (l, t) − Z2(0, t) ×

(− α1

α2W (0, t)

)·W (0, t)

−∫ l

0

(Z2x ×Wx) ·W

= −∫ l

0

(Z2x ×Wx) ·W

≤ 14

∫ l

0

|Wx|2 + C

∫ l

0

|W |2.

Using Holder’s inequality again and noticing that Z1 is smooth too, we have∫ l

0

((Z1x + Z2x) ·Wx) · (Z2 ·W ) ≤ 14

∫ l

0

|Wx|2 + C

∫ l

0

|W |2.

Therefore, we getd

dt

∫ l

0

|W |2 +∫ l

0

|Wx|2 ≤ C

∫ l

0

|W |2. (5.8)

Using Lemma 2.3 and noting that W (x, 0) = 0, we can get the uniqueness when ε > 0.In the second step, we consider the problem (1.8)–(1.11). Let Z1(x, t) and Z2(x, t) be smooth

solutions of (1.8)–(1.11). Let W (x, t) = Z1(x, t) − Z2(x, t). Similarly as above, we have

Wt = Z2 ×Wxx +W × Z1xx. (5.9)

Therefore, we have

12d

dt

∫ l

0

|W |2

=∫ l

0

(Z2 ×Wxx) ·W = (Z2 ×Wx) ·W |x=lx=0 −

∫ l

0

(Z2x ×Wx) ·W

= (Z2(l, t) ×Wx(l, t)) ·W (l, t) − (Z2(0, t) ×Wx(0, t)) ·W (0, t)


−∫ l

0

(Z2x ×Wx) ·W

=(Z2(l, t) × −β1

β2W (l, t)

)·W (l, t)

−(Z2(0, t) × α1

α2W (0, t)

)·W (0, t) −

∫ l

0

(Z2x ×Wx) ·W

= −∫ l

0

(Z2x ×Wx) ·W ≤ C

∫ l

0

(|W |2 + |Wx|2). (5.10)

On the other hand, multiplying (5.9) by Wxx, we have

12d

dt

∫ l

0

|Wx|2 = −∫ l

0

(W × Z1xx) ·Wxx

= −(W × Z1xx) ·Wx|x=lx=0 +

∫ l

0

(W × Z1xxx) ·Wx

=∫ l

0

(W × Z1xxx) ·Wx

≤ C

∫ l

0

(|W |2 + |Wx|2), (5.11)

where we have applied (5.5)–(5.6). Putting (5.10) and (5.11) together, we obtain

12d

dt

∫ l

0

(|W |2 + |Wx|2) ≤ C

∫ l

0

(|W |2 + |Wx|2).

Using Lemma 2.3 and W (x, 0) = 0,Wx(x, 0) = 0, we get the uniqueness when ε = 0. �

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