Gradient Estimates of Solutions to the Conductivity Problem with Flatter Insulators

YanYan Li and Zhuolun Yang

Department of Mathematics,Rutgers University,110 Frelinghuysen Rd,Piscataway,NJ 08854,USA

Abstract.We study the insulated conductivity problem with inclusions embedded in a bounded domain in R n.When the distance of inclusions,denoted byε,goes to 0,the gradient of solutions may blow up.When two inclusions are strictly convex,it was known that an upper bound of the blow-up rate is of orderε?1/2 for n=2,and is of orderε?1/2+βfor someβ>0 when dimension n≥3.In this paper,we generalize the above results for insulators with flatter boundaries near touching points.

Key Words:Conductivity problem,harmonic functions,maximum principle,gradient estimates.

1 Introduction and main results

whereφ∈C2(?Ω)is given,and

refers to conductivities.The solution ukand its gradient?ukrepresent the voltage potential and the electric fields respectively.From an engineering point of view,It is an interesting problem to capture the behavior of?uk.Babuˇska,et al.[3]numerically analyzed that the gradient of solutions to an analogous elliptic system stays bounded regardless of ε,the distance between the inclusions.Bonnetier and Vogelius[5]proved that for a fixed k,|?uk|is bounded for touching disks D1and D2in dimension n=2.A general result was obtained by Li and Vogelius[11]for general second order elliptic equations of divergence form with piecewise H¨older coefficients and general shape of inclusions D1and D2in any dimension.When k is bounded away from 0 and∞,they established a W1,∞bound of ukinΩ,and a C1,αbound in each region that do not depend onε.This result was further extended by Li and Nirenberg[10]to general second order elliptic systems of divergence form.Some higher order estimates with explicit dependence on r1,r2,k andε were obtained by Dong and Li[7]for two circular inclusions of radius r1and r2respectively in dimension n=2.There are still some related open problems on general elliptic equations and systems.We refer to p.94 of[11]and p.894 of[10].

When the inclusions are insulators(k=0),it was shown in[6,9,13]that the gradient of solutions generally becomes unbounded,asε→0.It was known that(see e.g.,Appendix of[4])when k→0,ukconverges to the solution of the following insulated conductivity problem:

Hereνdenotes the inward unit normal vectors on?Di,i=1,2.

The behavior of the gradient in terms ofεhas been studied by Ammari et al.in[1]and[2],where they considered the insulated problem on the whole Euclidean space:

for some positive constant C independent ofε.They also showed that the upper bounds are optimal in the sense that for appropriate H,

In fact,the equation

was studied there,and the estimates derived have explicit dependence on r1,r2,k andε.

The above upper bound of?u was localized and extended to higher dimensions by Bao,Li and Yin in[4],where they considered problem(1.2)and proved

The upper bound is optimal for n=2 as mentioned earlier.For dimensions n≥3,the upper bound was recently improved by Li and Yang[12]to

for someβ>0.

Yun[16]considered the problem(1.3)in R3,with unit disks

D1=B1(0,0,1+ε/2), D2=B1(0,0,?1?ε/2),

and a harmonic function H.He proved that for some positive constant C independent of ε,

He also showed that this upper bound of|?u|on theε-segment connecting D1and D2is optimal for H(x)≡x1.

λ≤a(x)≤Λ for x∈～Ω,

for some positive constantsλ,Λ.Letν=(ν1,···,νn)denote the unit normal vector on?D1and?D2,pointing towards the interior of D1and D2.We consider the following insulated conductivity problem:

for someλ1,λ2,λ3>0.Let a(x)∈

whereφ∈C2(?Ω)is given.For 0

Since the blow-up of gradient can only occur in the narrow region between D1and D2,we will focus on the following problem near the origin:

whereν=(ν1,···,νn)denotes the unit normal vector onΓ+andΓ?,pointing upward and downward respectively.

Remark 1.1.For m=2,(1.10)was proved in[4]and[12]for n=2 and n≥3,respectively.

Therefore,a corollary of Theorem 1.1 is as follows.

2 Proof of Theorem 1.1

Our proof of Theorem 1.1 is an adaption of the arguments in our earlier paper[12]for m=2,and follows closely the arguments there.

We fix aγ∈(0,1),and let r0>0 denote a constant depending only on n,m,γ,R0,λ1,λ2,‖f‖C2and‖g‖C2,whose value will be fixed in the proof.For any x0∈Ω0,r0,we define

and perform a change of variables by setting

for s,t>0.We will show that the Jacobian matrix of the change of variables(2.3),denoted by?xy,and its inverse matrix?yx satisfy

Let v(y)=u(x),then v satisfies

where the matrix(bij(y))is given by

(?xy)tis the transpose of?xy.

It is easy to see that(2.5)implies,usingλ≤(aij)≤Λ,

From(2.3),one can compute that

(?xy)ij=0 for 1≤i≤n?1, j/=i.

By(1.6b),one can see that

Since|yn|≤hr,by using(1.6a)and(1.6b),we have that,for 1≤i≤n?1,

Next,we will show that

Indeed,by(1.6b),we have

Since|y′|>r/4,it is easy to see

On the other hand,since|y′|<2r and|x′0|

Therefore,

and(2.9)is verified.

We have shown(?xy)ii～1 for all i=1,···,n,and|(?xy)ij|≤Cδ(1?γ)for i/=j.We further require r0to be small enough so that off-diagonal entries of?xy are small.Therefore(2.5)follows.As mentioned earlier,(2.8)follows from(2.5).

Now we define,for any integer l,

We also define the corresponding coefficients,for k=1,2,···,n?1,

and for other indices,

In particular,we have

which is(2.2)after reversing the change of variables.

Remark 2.1.Lemma 2.1 does not hold for dimension n=2,since Q2,1Q1/4,1?R2is the union of two disjoint rectangular domains,and the Harnack inequality cannot be applied on it.Therefore,we will separate the cases n=2 and n≥3 in our proof of Theorem 1.1.

where C1>1 is a constant independent of r.Since both u1and u2satisfy Eq.(1.9),by the maximum principle,

for i=1,2.Therefore,

Adding up the above two inequalities,we have

We start with r=r0=δ1?γ/2,and set ri+1=ri/2.Keep iterating(2.11)k+1 times,where k satisfies 5δ≤rk<10δ,we will have

Since

we have

2?(k+1)<10δγ

and hence(2.10)follows immediately.

We make a change of variables again by

where the matrix b(z)=(bij(z))is given by

Similar to the proof of Lemma 2.1,we will show that the Jacobian matrix of the change of variables(2.14),denoted by?yz,and its inverse matrix?zy satisfy

From(2.14),one can compute that

First we will show that

where in the last line,we have used the same arguments in showing(?yz)nn≤C earlier.

We have shown(?yz)ii～1 for all i=1,···,n,and|(?yz)ij|≤Cδfor i/=j.We further require r0to be small enough so that off-diagonal entries are small.Therefore(2.17)follows.As mentioned earlier,(2.18)follows from(2.17).

Next,we will show

By a straightforward computation,we have,for any i=1,···,n?1,

where in the last line,(1.6b)and(1.6c)have been used.For any i=1,···,n?1,by(1.6b)and(1.6c),

Finally,we compute,for i=1,···,n?1,

Therefore,(2.21)is verified,and hence(2.20)follows as mentioned above.

Now we define

for any integer l,and

We also define the corresponding coefficients,for k=1,2,···,n?1,

and for other indices,

and,by(2.20),

Apply Lemma 2.1 in[12]on S with N=1,we have

It follows that

In particular,this implies

and it concludes the proof of Theorem 1.1 for the case n≥3 after takingβ=γσ/2.

For the case n=2,we work with u instead of v,and repeat the argument in deriving the first inequality in(2.22),we have

In particular,

This concludes the proof of Theorem 1.1 for the case n=2.

Acknowledgements

The first author is partially supported by NSF Grants DMS-1501004,DMS-2000261,and Simons Fellows Award 677077.The second author is partially supported by NSF Grants DMS-1501004 and DMS-2000261.