Zhan De-sheng

(Department of Basic,Maanshan Technical College,Maanshan,Anhui,243031)

Communicated by Wang Chun-peng

Abstract:This paper is concerned with the asymptotic behavior of the solution uε of ap-Ginzburg-Landau system with the radial initial-boundary data.The author proves that the zeros of uεin the parabolic domain B1(0)×(0,T]locate near the axial line{0}×(0,T].In particular,all the zeros converge to this axial line when the parameter ε goes to zero.

Key words:p-Ginzburg-Landau equation,initial-boundary value problem,location of zero

1 Introduction

Let n≥3 and B={x∈Rn;|x|<1}.Write BT=B×(0,T],where T∈(0,∞).We are concerned with the asymptotic behavior of the weak solution uεof the following problem:

when ε→0.Recall that a weak solution of(1.1)is a measurable function u:BT→Rn,such that

Moreover,if a function u is a weak solution of(1.1),andthen u solves(1.1)–(1.3)in weak sense.

In the case of p=n=2,the problem can be used to described the properties of vortices in the study of the phase transition,such as the theories of superconductor,super fluids and XY-magnetism(see[1]and the references therein).

We shall prove the following theorems:

Theorem 1.1Assume 1

0,there exists a constant h(independent of ε)such that

Remark 1.1Theorem 1.1 implies that all the zeros of uεare located near{0}×[0,T]when ε→ 0 and σ is sufficiently small.Namely,there does not exist any zero in the domain far away from{0}×[0,T]since

Theorem 1.2Under the same assumption of Theorem 1.1,for any σ>0,there exists a C>0 such that

By the same argument in[2],from(1.5)–(1.7)we can also derive the H¨older convergence of?uεwhen 2

Theorem 1.3Assume that maxand uεis the weak solution of(1.1)–(1.3).We have

2 Proof of Results

According to Proposition 2.1 in[2],we know that the weak solution of(1.1)–(1.3)is unique.It has the radial structure as follows:

In addition,fε,the module of the weak solution,can be viewed as the limit of a nonnegative function,which solves the problem

Hereafter,the solution of(1.1)–(1.3)with g(x)=x andis called the radial solution of(1.1)–(1.3).

Proposition 2.1(Energy inequality) Assume that uεis a radial solution with the form(2.1).Then there exists a constant C1>0 which is independent of ε and T,such that

Since(2.3)implies ft(0,t)=ft(1,t)=0,integrating by parts yields

Integrating it from σ to t for any 0< σ

where C1>0 is independent of ε andand

Letting M→∞and using(2.4)and(2.5),we can see(2.7)at last.The proof is completed.

Proposition 2.2Assume that p>1,and uεis a radial solution of(1.1)–(1.3).Then there is not its zero point near some partial parabolic boundary.

Proof.When p∈(2,n),by the same argument in[2],we can use the anisotropic Sobolev embedding theorem to deduce that,for any σ>0,

When p=n or p∈(1,2),by means of Proposition 2.4 in[2]we obtain easily that,for any ρ >0 and σ >0,

Proposition 2.3Let uεbe the weak radial solution of(1.1)–(1.3).Assume that εkis a positive sequence going to zero.For any T>0,there exists a C>0 such that as k sufficiently large,

Proof. The idea in[3]is used.We claim that for all T>0,there exists a C>0 such that as 00 sufficiently large,

Otherwise,we can find some t0>0 and k0>0,such that as ε∈ (0,εk0),

with?C>0 to be determined later.If denoting Eηε(u(x,t0),G)by ν(ε),then

Combining this with(2.10)we see that

Integrating it from ε to εk0yields

as long as ε is sufficiently small.Choosingwhen p=n,where C1and C2are the constants in(2.7),we deduce easily the result which is contradicted with(2.7).This completes the proof.

Hereafter,we always assume ε= εk∈ (0,εk0).

Proposition 2.4Assume that uεsolves(1.1)–(1.3).Then there exists a constant C0>0 which is independent of ε,such that for any x1,x2∈and t>0,

Proof. It is a direct corollary of Proposition 2.4 in[2].

By the same argument in[2],from Proposition 2.4 we also have the following Proposition.

Proposition 2.5Let uεbe the weak solution of(1.1)–(1.3),and write

where σ and ρ are the constants in Proposition 2.2.There exist positive constants λ, μ independent of ε and t,such that if

where B2lεis some ball of radius 2lε with l≥ λ,and σ is the constant in(2.8)and(2.9),then

We borrow the technique of[2]in the following argument,and the main idea comes from[4].To find the zero points based on Proposition 2.5,we may take(2.11)as the ruler to distinguish the cylinders B(x,λε)× I which contain these points.Hereafter,we always assume that λ and μ are the constants in Proposition 2.5.If

for all t∈ I,then B(xε,λε)×I is called a good cylinder.Otherwise B(xε,λε)×I is called a bad cylinder.

Now suppose that{B(xεi,λε),i∈ N}is a family of balls satisfying

Similar to the proof of Proposition 2.6 in[2],from(2.13),Proposition 2.3 and the definition of bad cylinders,we also deduce that the number of bad cylinders CardJε≤ N0,where N0>0 is an integer independent of ε.This means that,no matter how small the parameter ε,the number of the bad cylinders is always finite.Therefore,based on an analogous discussion in Ch III of[4],we also find a subset J?Jεand a constant h≥λ such that

Proof of Theorem 1.1According to the argument above,we may modify the family of bad cylinders,such that the new one,denoted by{B(,hε)× I;i∈ J},satisfies

The last condition implies that every two cylinders in the new family are not intersected.

Suppose that there exists a point

Hence,in virtue of Proposition 2.5,if x∈S0,then(x,t0)must be contained in bad cylinders.However,since|x0|>hε,{(x,t0);x ∈ S0}cannot be covered by a single bad cylinder.It has to be covered by at least two disjoint bad cylinders.This is impossible.The contradiction means

Combining this with(2.8)and(2.9)yields the conclusion of Theorem 1.1.

Proof of Theorem 1.2When p∈(1,n),we have already the uniform estimate(1.6),which is implied by(2.7).When p=n,(2.7)is not sufficient to derive the uniform estimate.We only need to deduce the local estimate(1.7)as p=n.The ideas in §3 of[5]or §3 of[6]are used here,since the consequences there were true not only for the n-Ginzburg-Landau minimizer,but also for all W1,n-function,as long as it satisfies(1.5)and Proposition 2.3.

Proposition 2.6Assume that p=n,and uεis a radial solution of(1.1)–(1.3).Then for any r∈ (0,1),we can fi nd C=C(r,σ,T)>0 such that(1.7)holds.

Proof. According to Theorems 3.9 and 3.10 in[6],from Proposition 2.3 and(1.5)we can deduce that for any given t?∈ [σ,T]and r ∈ (0,1),

with C4=C(r,σ,T)>0 independent of ε.Taking the in fi mum for t?∈ [σ,T]we have

Combining this with(2.7)we can see that

Thus,the proof is completed.

Proof of Theorem 1.3When p∈[2,n],the proof is same as in Sections 3 and 4 of[2]since(1.5)–(1.7)are still true.It is only need to handle the case ofRecalling the proof as p>2 in§3 of[2],we can deduce(3.5)by using the embedding inequality(3.3)with an appropriateHowever,the parameter q does not exist when p<2.Now,we shall replace the right hand side of(3.5)in[2]by a more suitable upper bound.Thus,we claim that for any σ>0,there exists a C>0 such that

In fact,if setting

Let(y,s)=(ε?1x,ε?pt)in(1.1),and de fi ne v(y,s)=u(x,t).Then(1.1)becomes vs=div(|?v|p?2?v)+v(1?|v|2),in weak sence on={(y,s);(x,t)∈BT}.According to Theorem 1 in[7],for any ρ>0,there exists a C>0 such that as|?v(y,s)|≤ C.Letting x=yε and t=sεpin the result above,we have|fr|≤ Cε?1.Combining this with(2.16),we obtain

On the other hand,by the definition of S and Proposition 2.3,it is easy to see that

which implies

Similar to the derivation of(2.17),we have

which,together with(2.17),implies(2.15).

Taking b=0,from(2.15)we can see that

The rest argument is same as in[2].