Characterization of the gradient blow-up of the solution to the conductivity equation in the presence of adjacent circular inclusions
aa r X i v : . [ m a t h - ph ] D ec Characterization of the gradient blow-up of the solution tothe conductivity equation in the presence of adjacent circularinclusions ∗ Mikyoung Lim † Sanghyeon Yu † August 11, 2018
Abstract
We consider the conductivity problem in the presence of adjacent circular inclusions hav-ing arbitrary constant conductivity. When two inclusions get closer and their conductivitiesdegenerate to zero or infinity, the gradient of the solution can be arbitrary large. We character-ize the gradient blow-up by deriving an explicit formula for the singular term of the solutionin terms of the Lerch transcendent function. This derivation is valid for inclusions havingarbitrary constant conductivity. We illustrate our results with numerical calculations.
AMS subject classifications.
Key words.
Conductivity equation; Anti-plane elasticity; Gradient blow-up; Bipolar coordinates; Lerch transcen-dent function
We consider the blow-up phenomena of the gradient of the solution to the conductivity problemwhen two conductors are closely located to each other in R . Let B and B be two disks withconductivity k and k , respectively, embedded in the background with conductivity 1. The con-ductivities k and k are assumed to be 0 < k , k = 1 < ∞ . Let σ denote the conductivitydistribution, i.e. , σ = k χ ( B ) + k χ ( B ) + χ ( R \ ( B ∪ B ) , (1.1)where χ is the characteristic function. Consider the following conductivity equation: ( ∇ · σ ∇ u = 0 in R ,u ( x ) − H ( x ) = O ( | x | − ) as | x | → ∞ , (1.2)where H is a given entire harmonic function in R . On the boundary of inclusions, the solution u to (1.2) satisfies u (cid:12)(cid:12) − ∂B j = u (cid:12)(cid:12) + ∂B j and k j ∂u∂ν (cid:12)(cid:12)(cid:12) − ∂B j = ∂u∂ν (cid:12)(cid:12)(cid:12) + ∂B j , j = 1 , . (1.3)The superscripts + and − denote the limit from inside and outside ∂B j , respectively. Letting ǫ := dist( B , B ) , ∗ This research was supported by the Korean Ministry of Science, ICT and Future Planning through NRF grantsNo. 2013012931 and No. 2013003192. † Department of Mathematical Sciences, Korea Advanced Institute of Science and Technology, Daejeon 305-701,Korea ([email protected], [email protected]).
1e estimate ∇ u in terms of ǫ as ǫ tends to 0, where the shape of B and B are fixed. This problemarises in relation with the computation of electromagnetic fields in the presence of fibers and thestress in composite materials.For the conductivities finite and strictly positive, it was proved by Li and Vogelius [16] that |∇ u | is bounded independently of ǫ , and Li and Nirenberg [15] have extended this result to ellipticsystems. Bonnetier and Vogelius showed in [9] that |∇ u | is bounded for circular touching inclusions.However, if k , k degenerate to ∞ , then the gradient may blow-up as ǫ tends to 0. The generic rateof gradient blow-up is ǫ − / in two dimensions [2, 3, 4, 5, 6, 8, 10, 14, 20, 21], while it is | ǫ log ǫ | − inthree dimensions [6, 7, 13, 17]. Ammari, Kang and Lim [4] and Ammari, Kang, Lee, Lee and Lim[2] obtained an optimal bound for |∇ u | when B and B are disks in R for 0 < k , k = 1 < ∞ .Yun [20, 21] obtained the blow-up rate for the perfectly conducting general shaped inclusions. Inthree dimensions, the blow-up rate was obtained by Bao, Li and Yin [6, 7] for perfect conductorsof general shape. For perfect conductors of spherical shape, Lim and Yun [17] obtained an optimalbound for |∇ u | , and Kang, Lim and Yun [13] derived an asymptotic of ∇ u . The insulating casehas the same blow-up rate as the perfectly conducting case in two dimensions. However, it is stillopen problem to clarify whether the gradient of u may blow-up or not for the insulating cases inthree dimensions.In this paper, we are interested in the dependence of the singular part of u on the conductivity k and k as well as ǫ . To state the related results in detail, let us introduce notations. For j = 1 , B j be the disk with radius r j centered at c j and R j be the reflection with respect to B j , i.e. , R j ( x ) = r j ( x − c j ) | x − c j | + c j , j = 1 , . Then the combined reflections R ◦ R and R ◦ R have unique fixed points, say p and p ,respectively. It is easy to show that, for j = 1 , p j = (cid:16) ( − j r ∗ √ ǫ + O ( ǫ ) (cid:17) n + p , n = p − p | p − p | , where p is the middle point of the shortest line segment connecting ∂B and ∂B . We also set r ∗ = p (2 r r ) / ( r + r ) , τ = τ τ , and τ j = ( k j − / ( k j + 1) , j = 1 , . (1.4)In [3], it was shown that the gradient of u may blow-up only when the linear term of H isnonzero. Hence u can be decomposed as u = u s + u r , (1.5)where u s is the solution to (1.2) with H replaced by ˜ H ( x ) = ∇ H ( p ) · ( x − p ) and the gradient of u r stays bounded regardless of ǫ in a bounded domain. Using the representation of u in terms ofthe single-layer potential, one can actually show that the gradient of ( u r − H + ˜ H ) is bounded in R . The following estimates of |∇ u s | were derived in [2, 3, 4]: if k , k >
1, then for any boundedset containing the inclusions, there are positive constants C and C independent of k , k , r , r , ǫ such that (cid:12)(cid:12) ∇ u s (cid:12)(cid:12) + ( x j ) ≥ C |∇ H ( p ) · n | − τ + r ∗ min( r ,r ) √ ǫ , j = 1 , , (1.6) (cid:13)(cid:13) ∇ u s (cid:13)(cid:13) L ∞ (Ω) ≤ C |∇ H ( p ) · n | − τ + r ∗ max( r ,r ) √ ǫ , (1.7)where x j , j = 1 ,
2, is the point on ∂B j closest to the other disk. For 0 < k , k <
1, we have (1.6)and (1.7) with n replaced by t and k j by 1 /k j , j = 1 ,
2, where t is the rotation of n by π/ k = k = ∞ , the singular term u s can be explicitlycharacterized using the singular function h defined as the solution to ∆ h = 0 in R \ B ∪ B ,h = constant on ∂B j , j = 1 , , Z ∂B j ∂h∂ν ( j ) ds = ( − j +1 , j = 1 , ,h ( x ) = O ( | x | − ) as | x | → ∞ , (1.8)where ν ( j ) is outward unit normal to ∂B j . Remind the circle of Apollonius: for a disk B r ( c ) withradius r centered at c , the circle ∂B r ( c ) is the locus of x satisfying the ratio condition | x − p || x − R ( p ) | = | p − c | r , for a fixed p / ∈ B r ( c ) , (1.9)where R is the reflection with respect to B r ( c ). Since R ( p ) = p and R ( p ) = p , we have2 πh ( x ) = ln | x − p | − ln | x − p | . (1.10)In [12], it was shown that u ( x ) = 2 πc n h ( x ) + u r ( x ) with c n = r ∗ (cid:0) ∇ H ( p ) · n (cid:1) , (1.11)where the gradient of u r is bounded regardless of ǫ in any bounded set. The solution h to (1.8) playsan essential role in the characterization of the blow-up feature not only for the circular inclusions.By investigating h , it was derived the optimal bounds of the gradient of u for convex domains intwo dimensions [20, 21] and for balls in three dimensions [17, 18]. In [1], for convex inclusions intwo dimensions, the blow-up of ∇ u was characterized using h corresponding to the disk osculatingto inclusions. It is worth to mention that, as shown in [1], h is an eigenfunction corresponding toeigenvalue 1/2 of a Neumann-Poincar´e operator.The purpose of this paper is to generalize (1.11) to the case of circular inclusions with arbitraryconstant conductivity. More precisely speaking, we characterize the singular term of u when twocircular inclusions with the conductivity 0 < k , k = 1 < ∞ gets closer. To do this, we expandthe solution u in bipolar coordinate system ( ξ, θ ) with poles located at p and p . In other words,for x = ( x, y ), the coordinates are defined as e ξ + iθ = z − ˜ p ˜ p − z , z = x + iy, (1.12)where ( ℜ{ ˜ p j } , ℑ{ ˜ p j } ) = p j , j = 1 .
2. It is worth to note that ξ ( x ) = 2 πh ( x ) . Applying (1.9), the first coordinate ξ takes the constant value ( − j ξ j on ∂B j with ξ := ln | p − c | − ln( r ) and ξ := ln | p − c | − ln( r ) . (1.13)To state the main theorems we also define a function q ( x ; β, τ , τ ):= 12 ( τ + τ ) L (cid:0) e − ( ξ + iθ ) − ξ ; β (cid:1) − ( τ + τ ) L (cid:0) e ( ξ + iθ ) − ξ ; β (cid:1) in R \ ( B ∪ B ) , ( τ + τ ) L (cid:0) e ξ − iθ ; β (cid:1) − ( τ + τ ) L (cid:0) e ( ξ + iθ ) − ξ ; β (cid:1) in B , ( τ + τ ) L (cid:0) e − ( ξ + iθ ) − ξ ; β (cid:1) − ( τ + τ ) L (cid:0) e − ξ + iθ ; β (cid:1) in B , (1.14)3here β = r ∗ ( − ln τ )4 √ ǫ , (1.15) L ( z ; β ) := − Z zt β zt dt, z ∈ C , | z | < . The followings are the main results in this paper. Theorem 1.2 can be proved similarly asTheorem 1.1.
Theorem 1.1
For k , k > , the solution u to (1.2) can be expressed as u ( x ) = c n ℜ{ q ( x ; β, τ , τ ) } + H ( x ) + u b ( x ) , (1.16) where c n is in (1.11) and there is a constant C independent of ǫ , k and k such that k∇ u b k L ∞ ( R ) ≤ C. Here |∇ u b | on ∂B ∪ ∂B could be its limit from the interior or the exterior. Proof . If τ < .
5, the gradient of ( u − H ) is bounded independently of ǫ from (1.7) and so does ∇ q from Lemma 4.1. Hence we can suppose that τ ≥ .
5. We prove the theorem using (2.10),(2.11), Lemma 3.1 and Lemma 3.2. (cid:3)
Theorem 1.2
For < k , k < , the solution u to (1.2) can be expressed as u ( x ) = c t ℑ{ q ( x ; β, − τ , − τ ) } + H ( x ) + u b ( x ) , c t = r ∗ ( ∇ H ( p ) · t ) , (1.17) and there is a constant C independent of ǫ , k and k such that k∇ u b k L ∞ ( R ) ≤ C. Here |∇ u b | on ∂B ∪ ∂B could be its limit from the interior or the exterior. The function q is continuous in R and harmonic except on ∂B j , and it decays to a constant atinfinity, see Lemma 3.3. If the inclusions have the extreme property ( k = k = ∞ or k = k = 0),then β = 0, L ( z ; β ) = − log(1 + z ) and q ( x ; β, | τ | , | τ | ) = ξ + iθ + b, x ∈ R \ B ∪ B , where ∇ b is uniformly bounded independently of ǫ , for the proof see (4.16). Hence we can consider L as a generalized complex logarithm and (the real part of) q as a generalization of h in (1.10). Asit will be shown in section 2.2, q and ∇ q can be represented in terms of the so called Lerch tran-scendent function, of which numerical calculation has been intensively studied and implementedin commercial softwares.The decomposition of u in the main results is valid for the interior as well as the exterior ofinclusions. Using these, we can derive not only the optimal bounds but also the singular term ofthe gradient of u explicitly. Let us consider gradient of q on the boundary of inclusions. In section4, we show that (cid:12)(cid:12) ∇ q ( x ; β, | τ | , | τ | ) (cid:12)(cid:12) + ∂B j = ( | τ | + | τ | + 2 τ ) cosh ξ j + cos θ r ∗ √ ǫ (cid:12)(cid:12)(cid:12) ℜ{ P (cid:0) e − ( ξ j + iθ ) ; β (cid:1) } (cid:12)(cid:12)(cid:12) + O (1) , where P ( z ; β ) := ( − z ) ∂∂z L ( z ; β ) . (1.18)4specially at x j , i.e. , θ = 0, we have (cid:12)(cid:12) ∇ q ( x j ; β, | τ | , | τ | ) (cid:12)(cid:12) + ≈ const. ǫ − / ℜ{ P (cid:0) e − ξ j ; β (cid:1) } . From thedefinition of L which is an integral whose integrand contains t β term, ℜ{ P (cid:0) e − ξ j ; β (cid:1) } is of order of1 / ( β + 1). This is in accordance with (1.6) and (1.7), see Remark 4.5.While β is zero for the extreme cases, it could be arbitrary large as well as small for highlyconducting or almost insulating cases. For example, β ≈ ǫ / if k , k ≈ ǫ − / or k , k ≈ ǫ / .Similarly, β ≈ ǫ − / if k , k ≈ ǫ − / or k , k ≈ ǫ / . As it will be discussed in more detail later inthe paper, one can derive the asymptotic of P when β is either large or small in comparison with ǫ . Especially, for β = O ( ǫ / ), (cid:12)(cid:12) ∇ ( u − u ∞ ) (cid:12)(cid:12) + ∂B j = const. 1 ǫ / (cosh ξ j + cos θ ) ln[2(cosh ξ j + cos θ )] + O (1) , where u ∞ is the solution of the perfectly conducting equation (4.15). In a practical computationof the electric field, (4.15) is often used instead of (1.3) when the inclusions are highly conducting.It is worth to emphasize that the error can be arbitrary large if so. More discussion is in section4.3.The paper is organized as follows. In section 2, we review the bipolar coordinate system andderive a summation lemma which is essential to prove the main results. Section 3 is to derivethe series expansion of u in terms of bipolar coordinate system, and in section 4 we obtain theasymptotic of ∇ u . We consider the conductivity problem defined in a bounded region in section 5and illustrate the main results with numerical calculations in section 6. Let us put α := | p − p | r ∗ √ ǫ + O ( ǫ ) . (2.1)In other words, α = p ǫ (2 r + ǫ )(2 r + ǫ )(2 r + 2 r + ǫ )2 r + 2 r + 2 ǫ . After rotation and shifting if necessary, it can be assumed that the centers of p and p are onthe x -axis and p = ( − α,
0) and p = ( α, . (2.2)We assume so in what follows.Each point x = ( x, y ) in the Cartesian coordinate system corresponds to ( ξ, θ ) ∈ R × ( − π, π ]in the bipolar coordinate system through the equations x = α sinh ξ cosh ξ + cos θ and y = α sin θ cosh ξ + cos θ (2.3)with a positive number α , see [19]. Setting α as (2.1), (2.3) means ξ ( x ) = ln(Γ / Γ ) and θ ( x ) ≡ ϕ − ϕ (mod 2 π ) (2.4)for x + iy + α = Γ e iϕ and α − ( x + iy ) = Γ e iϕ . The magnitude r of x satisfies r ( ξ, θ ) = α p (cosh ξ − cos θ ) / (cosh ξ + cos θ ) . | x | → + ∞ if and only if ( ξ, θ ) → (0 , π ), and if this is the case (cid:12)(cid:12)(cid:12) r ( ξ, θ ) p cosh ξ + cos θ (cid:12)(cid:12)(cid:12) ≤ α. (2.5)From the definition, we can derive that the coordinate curve { ξ = c } and { θ = c } are, respec-tively, the zero-level set of f ξ ( x, y ) = (cid:18) x − α cosh c sinh c (cid:19) + y − (cid:16) α sinh c (cid:17) , (2.6) f θ ( x, y ) = x + (cid:16) y + α cos c sin c (cid:17) − (cid:16) α sin c (cid:17) . Note that {| θ | ≤ π/ } is the disk with the diameter [ − α, α ], and { π/ ≤ | θ | ≤ π − ǫ / } iscontained {| x | ≤ (2 α ) / sin( ǫ ) } , whose radius is of magnitude ǫ / . Reversely, {| x | ≤ ǫ / } \ ( B ∪ B ) is contained in {| θ | ≤ π − θ ǫ } with θ ǫ ≈ ǫ / . That means the narrow region in betweentwo inclusions cover almost all angle in the bipolar coordinate system. The graph in Figure 2.1illustrates coordinate curves of a bipolar coordinate system. Θ = Π (cid:144) Θ = Θ = -Π (cid:144) Θ = -Π (cid:144) Θ = Π (cid:144) Ξ = Θ = Π (cid:144) Ξ = Ξ Ξ = -Ξ Ξ = - Ξ Θ = Π Θ = - Π (cid:144) Ξ = Ξ - - B B - - - - Figure 2.1: Bipolar coordinate system for r = 2 . , r = 3. The distance ǫ between disks are 2 inthe left and 0.1 in the right figure; ( ξ , ξ ) is (0 . , . . , . ǫ decreases, so do ξ and ξ in the order of √ ǫ , see (2.7). In theleft figure, solid lines are the coordinate curves of ξ -variable and the dotted ones are that of θ . Theshadow region in the right figure is {− π + ǫ / ≤ θ ≤ π − ǫ / } . All points outside the shadowregion have the bipolar coordinate θ close to ± π .Since ∂B j is the coordinate curve { ξ = ( − j ξ j } for ξ j defined as (1.13), we have ξ j = sinh − (cid:18) αr j (cid:19) and c j = α (cid:0) ( − j coth ξ j , (cid:1) , for j = 1 , . (2.7)Hence ξ j = α/r j + O ( ǫ √ ǫ ), j = 1 ,
2, and αξ + ξ = r ∗ O ( ǫ ) . (2.8)From (2.6), the outward unit normal ν to the circle ξ = c for nonzero c is ν ξ = c = ∇ f ξ |∇ f ξ | = sgn( c ) (cid:18) − − cos θ cosh c cosh c + cos θ , − sin θ sinh c cosh c + cos θ (cid:19) , where sgn( c ) takes the value of 1 or -1 as c is positive or negative, respectively. We define thetangential vector T as the rotation of ν by π/ ν ξ = c = − sgn( c )ˆ e ξ , T ξ = c = − sgn( c )ˆ e θ , { ˆ e ξ , ˆ e θ } are defined asˆ e ξ := ∂ x /∂ξ | ∂ x /∂ξ | and ˆ e θ := ∂ x /∂θ | ∂ x /∂θ | . It can be easily shown that the gradient of any scalar valued function g is written in thefollowing form: ∇ g = cosh ξ + cos θα (cid:18) ∂g∂ξ ˆ e ξ + ∂g∂θ ˆ e θ (cid:19) . (2.9)Hence the normal- and tangential derivatives of a function u in bipolar coordinates are ∂u∂ν (cid:12)(cid:12)(cid:12) ξ = c = ∇ u · v ξ = c = − sgn( c ) (cid:18) cosh c + cos θα (cid:19) ∂u∂ξ (cid:12)(cid:12)(cid:12) ξ = c (2.10) ∂u∂T (cid:12)(cid:12)(cid:12) ξ = c = − sgn( c ) (cid:18) cosh c + cos θα (cid:19) ∂u∂θ (cid:12)(cid:12)(cid:12) ξ = c . (2.11)The bipolar coordinate system is an orthogonal coordinate system and admits a general sepa-ration of variables solution to the harmonic function f as follows: f ( ξ, θ ) = a + b ξ + c θ + ∞ X n =1 (cid:2) ( a n e nξ + b n e − nξ ) cos nθ + (cid:0) c n e nξ + d n e − nξ ) sin nθ (cid:3) , where a n , b n , c n and d n are constants. Especially, the two linear functions x and y can be expandedas the following. For ξ >
0, we havesinh ξ − i sin θ cosh ξ − cos θ = e ζ + e − ζ e ζ − e − ζ = 1 + 2 ∞ X n =1 e − nξ (cos nθ − i sin nθ ) , (2.12)with ζ = ξ + iθ . Plugging ( θ + π ) instead of θ and using (2.3), x = sgn( ξ ) α (cid:2) ∞ X n =1 ( − n e − n | ξ | cos nθ (cid:3) ,y = − α ∞ X n =1 ( − n e − n | ξ | sin nθ. The Lerch transcendent function Φ is defined asΦ( z, s, β ) := ∞ X n =0 z n ( n + β ) s for s ∈ C , | z | < β = 0 , − , − , .... (2.13)The Lerch transcendent function has an integral representationΦ( z, , β ) = Z t β − − zt dt since we have Φ( z, , β ) = ∞ X n =0 z n Z t n + β − dt = Z t β − ∞ X n =0 (cid:0) zt (cid:1) n dt. | z | <
1. Applying the change of variables,Φ( z, , β ) = Z ∞ e − βt − ze − t dt. (2.14)One can easily show that L ( z ; β ) = − z Φ( − z, , β + 1)and, from (1.18), P ( z ; β ) = z z − βz Φ( − z, , β + 1) . (2.15)The function P is crucial to understand the blow-up feature of ∇ u . To understand better, let usreplace the summation in P by an integral in terms of the function p θ ( t ) := 11 + e − t + iθ defined for t > , θ ∈ ( − π, π ] . Using (2.14), we can express P as P ( e − s + iθ ; β ) = − p θ ( s ) + β Z ∞ e − βt p θ ( t + s ) dt, s > . (2.16)Applying the integration by parts, it becomes P ( e − s + iθ ; β ) = Z ∞ e − βt p ′ θ ( t + s ) dt. (2.17) P Recall that p θ ( t ) = e t + cos θ − i sin θ t + cos θ ) = 12 + sinh t − i sin θ t + cos θ ) , (2.18)and the derivative of p θ satisfies p ′ θ ( t ) = e − t + iθ (1 + e − t + iθ ) = 1 + cosh t cos θ + i sinh t sin θ t + cos θ ) . (2.19)Since (cid:12)(cid:12) e − t + iθ (cid:12)(cid:12) = p e − t (cosh t + cos θ ), we have | p θ ( t ) | ≤ √ cosh s + cos θ , for 0 < s ≤ t < , for t ≥ , (2.20) | p ′ θ ( t ) | = 12(cosh t + cos θ ) , t > . (2.21) Lemma 2.1
Set β, s, s , s > and s < s . For all θ ∈ ( − π, π ] , (cid:12)(cid:12) P ( e − s + iθ ; β ) (cid:12)(cid:12) ≤ β (cosh s + cos θ ) , (2.22) (cid:12)(cid:12) P ( e − s + iθ ; β ) (cid:12)(cid:12) ≤ √ cosh s + cos θ , (2.23) (cid:12)(cid:12) P ( e − s + iθ ; β ) − P ( e − s + iθ ; β ) (cid:12)(cid:12) ≤ s − s cosh s + cos θ . (2.24)8 roof . If s >
1, then we have | P ( θ ; s, β ) | ≤ s <
1, then | p θ ( t ) | ≤ √ cosh s + cos θ for s ≤ t < . Applying (2.16), it becomes (cid:12)(cid:12) P ( e − s + iθ ; β ) (cid:12)(cid:12) ≤ √ cosh s + cos θ (cid:18) β Z − s e − βt dt (cid:19) + β Z ∞ − s e − βt dt ≤ √ cosh s + cos θ + 2 . It proves (2.23). Moreover, from (2.17) and (2.21), we prove (2.22).Let t ≥
0. From the mean value property, p θ ( t + s ) − p θ ( t + s ) = ( s − s ) p ′ θ ( c ) for some c ∈ ( t + s , t + s ) . Applying (2.21), | p θ ( t + s ) − p θ ( t + s ) | ≤ s − s s + cos θ ) . Therefore we have | p θ ( s ) − p θ ( s ) | ≤ s − s s + cos θ ) , (cid:12)(cid:12)(cid:12)(cid:12) β Z ∞ e − βt p θ ( t + s ) dt − β Z ∞ e − βt p θ ( t + s ) dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ β Z ∞ e − βt s − s s + cos θ ) dt. This proves (2.24). (cid:3)
Lemma 2.2
Let β, ξ > . For all θ ∈ ( − π, π ] , we have (cid:12)(cid:12)(cid:12) (cosh ξ + cos θ ) P ( e − ξ + iθ ; β ) (cid:12)(cid:12)(cid:12) ≤ e ξ β + 1) , (2.25)14( β + 1) ≤ ℜ{ P ( e − ξ ; β ) } ≤ β + 1 , at θ = 0 . (2.26) Proof . When θ = 0, from (2.19) and (2.17),14 Z ∞ e − ( β +1) t dt ≤ ℜ{ P ( e − ξ ; β ) } ≤ Z ∞ e − ( β +1) t dt = 1 β + 1 . From (2.16) and (2.18), ℜ{ P ( e − ξ + iθ ; β ) } = − sinh ξ ξ + cos θ ) + β Z ∞ e − βt sinh( t + ξ )2(cosh( t + ξ ) + cos θ ) dt. Hence0 < (cosh ξ + cos θ ) ℜ{ P ( e − ξ + iθ ; β ) } + sinh ξ β Z ∞ e − βt (cosh ξ + cos θ ) sinh( t + ξ )2(cosh( t + ξ ) + cos θ ) dt ≤ β Z ∞ e − βt (cosh ξ + 1) sinh( t + ξ )2(cosh( t + ξ ) + 1) dt = (cosh ξ + 1) ℜ{ P ( e − ξ ; β ) } + sinh ξ . (cid:3) .4 Summation formula In this section we prove a lemma which is essential to prove the main results in this paper.
Lemma 2.3
Fix a a > . For a < a , < τ < and θ ∈ ( − π, π ] , we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a ∞ X m =1 τ m − e − ma + a + iθ (1 + e − ma + a + iθ ) − P ( e − ( a − a )+ iθ ; − (ln τ ) /a ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ a cosh( a − a ) + cos θ . Proof . Fix a θ . Thanks to (2.19), a ∞ X m =1 τ m − e − ma + a + iθ (1 + e − ma + a + iθ ) = a ∞ X m =0 τ m p ′ θ ( ma + s ) , s = a − a > . (2.27)To rewrite the right-hand side of (2.27), we consider f ( t ) := τ t p ′ θ ( a t + s ). From (2.21) andthe fact that τ ∈ (0 , (cid:12)(cid:12) τ t p ′ θ ( a t + s ) (cid:12)(cid:12) ≤ s + cos θ ) , t ≥ , (2.28)and the derivative of f is as follows: ddt (cid:16) τ t p ′ θ ( a t + s ) (cid:17) = τ t (cid:16) (ln τ ) p ′ θ ( a t + s ) + sp ′′ θ ( a t + s ) (cid:17) . (2.29)Note that, from (2.19), the real and imaginary part of e a t + s ) (cid:16) (ln τ ) p ′ θ ( a t + s )+ a p ′′ θ ( a t + s ) (cid:17) are quadratic polynomials of e a t + s . Hence the real and imaginary part of ddt (cid:16) τ t p ′ θ ( a t + s ) (cid:17) changes the sign at most 4 times on (0 , ∞ ). Therefore, from (2.21), (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a ∞ X m =0 τ m p ′ θ ( ma + s ) − a Z ∞ τ t p ′ θ ( a t + s ) dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ t ≥ | p ′ θ ( t ) | ≤ a cosh s + cos θ . By a change of variables, we get a Z ∞ τ t p ′ θ ( a t + s ) dt = Z ∞ e − bt p ′ θ ( t + s ) dt = − p θ ( s ) + b Z ∞ e − bt p θ ( t + s ) dt with b = − log τa . This proves the lemma. (cid:3) u in terms of the bipolar system From (1.5), we only need to consider linear functions for H . We can also assume B and B satisfy(2.2) without loss of generality. Define the bipolar system ( ξ, θ ) and ξ j , j = 1 ,
2, as in section 2.1.So ξ = − ξ and ξ = ξ represents ∂B and ∂B , respectively. Set ξ M := max( ξ , ξ ) and ξ s := min( ξ , ξ ) , then − ξ − ξ , ξ − ξ ∈ ( − ξ M − ξ s , − ξ s ) , x ∈ R \ B ∪ B ξ ∈ ( −∞ , − ξ ) , x ∈ B ξ ∈ ( ξ , ∞ ) , x ∈ B . (3.1)10 emma 3.1 Define a complex valued function U ( x, y ) := C + ∞ X n =1 (cid:16) A n e n ( ξ + iθ ) + B n e − n ( ξ + iθ ) (cid:17) , x ∈ R \ ( B ∪ B ) ∞ X n =1 (cid:16) A n e n ( ξ + iθ ) + B n e n (2 ξ + ξ − iθ ) (cid:17) , x ∈ B ∞ X n =1 (cid:16) A n e n (2 ξ − ξ + iθ ) + B n e − n ( ξ + iθ ) (cid:17) , x ∈ B (3.2) where C = − P ∞ n =1 ( A n + B n ) cos nπ and A n = 2 α ( − n τ − e n ( ξ + ξ ) − (cid:18) − τ e nξ − (cid:19) , B n = 2 α ( − n τ − e n ( ξ + ξ ) − (cid:18) τ e nξ (cid:19) . Then ( x + ℜ{ U } ) is the solution to (1.2) for H ( x, y ) = x , and ( y + ℑ{ U } ) is the solution for H ( x, y ) = y with /k j in the place of k j , j = 1 , . We prove the lemma after the following one.
Lemma 3.2
Suppose τ ≥ . . There is a constant C depending only on r , r such that (cid:12)(cid:12)(cid:12)(cid:12) ∂U∂ξ ( x ) − r ∗ ∂∂ξ q ( x ; β, τ , τ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C cosh ξ s + cos θ , x ∈ R , where q is defined as (1.14). We have the same equation for the partial derivative in θ variable. Proof . Note that A n and B n can be represented as A n = 2 α ( − n ∞ X m =1 τ m (cid:18) − τ e nξ − (cid:19) e − nm ( ξ + ξ ) ,B n = 2 α ( − n ∞ X m =1 τ m (cid:18) τ e nξ (cid:19) e − nm ( ξ + ξ ) . Firstly, set − ξ ≤ ξ ≤ ξ . By interchanging the order of summation, which is possible due tothe absolute convergence, we have ∞ X n =1 n (cid:16) A n e n ( ξ + iθ ) − B n e − n ( ξ + iθ ) (cid:17) = − α ∞ X m =1 τ m " τ ∞ X n =1 n (cid:16) − e − m ( ξ + ξ )+2 ξ + ξ + iθ (cid:17) n + ∞ X n =1 n (cid:16) − e − m ( ξ + ξ )+ ξ + iθ (cid:17) n + 1 τ ∞ X n =1 n (cid:16) − e − m ( ξ + ξ )+2 ξ − ξ − iθ (cid:17) n + ∞ X n =1 n (cid:16) − e − m ( ξ + ξ ) − ξ − iθ (cid:17) n . (3.3)Using the following formula P ∞ n =1 nz n = z (1 − z ) , for | z | < , z ∈ C , we have ∞ X n =1 n (cid:16) A n e n ( ξ + iθ ) − B n e − n ( ξ + iθ ) (cid:17) = 2 ατ ∞ X m =1 τ m − " τ e − m ( ξ + ξ )+2 ξ + ξ + iθ (1 + e − m ( ξ + ξ )+2 ξ + ξ + iθ ) + e − m ( ξ + ξ )+ ξ + iθ (1 + e − m ( ξ + ξ )+ ξ + iθ ) + 1 τ e − m ( ξ + ξ )+2 ξ − ξ − iθ (1 + e − m ( ξ + ξ )+2 ξ − ξ − iθ ) + e − m ( ξ + ξ ) − ξ − iθ (1 + e − m ( ξ + ξ ) − ξ − iθ ) . (3.4)11e apply Lemma 2.3 to this series with a = 2( ξ + ξ ). From (3.3), (3.4), Lemma 2.3, and Lemma2.1, ∞ X n =1 n (cid:16) A n e n ( ξ + iθ ) − B n e − n ( ξ + iθ ) (cid:17) = r ∗ " τ P ( e − ξ + ξ + iθ ; b ) + τ P ( e − ξ + ξ )+ ξ + iθ ; b ) + τ P ( e − ξ − ξ − iθ ; b )+ τ P ( e − ξ + ξ ) − ξ − iθ ; b ) + r ∗ ( τ + τ + 2 τ τ )2 O ( √ ǫ )cosh ξ s + cos θ , where b = − ln τ ξ + ξ ) . Note that b = β + O ( √ ǫ ) where β is defined as (1.15). Since b, β >
0, applying the mean valueproperty for a fixed t >
0, we have be − bt = βe − βt + ( b − β )(1 − st ) e − st , for a s ∈ (min( b, β ) , max( b, β )) . (3.5)Hence be − bt − βe − βt = ( ( b − β ) O (1) , for t > , ( b − β ) O ( t ) e − min( b,β ) t , for t ≥ . (3.6)Therefore, using (2.20) as well, (cid:12)(cid:12)(cid:12)(cid:12) b Z ∞ e − bt p θ ( t + ξ s ) dt − β Z ∞ e − βt p θ ( t + ξ s ) dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ | b − β | Z O (1) 1 √ cosh ξ s + cos θ dt + | b − β | Z ∞ e − min ( b,β ) t O ( t ) dt ≤ C √ ǫ ξ s + cos θ + O ( √ ǫ ) . Here the remainder term O ( √ ǫ ) is uniform with respect to k , k , ξ and θ . Therefore we can replace b by β . Using (2.8), we prove the lemma for − ξ ≤ ξ ≤ ξ .Similarly, for ξ ≤ − ξ , we can derive that ∞ X n =1 n (cid:16) A n e n ( ξ + iθ ) ± B n e nξ e n ( ξ − iθ ) (cid:17) = r ∗ " ( τ + τ ) P ( e ξ + iθ − ξ ; β ) ∓ ( τ + τ ) P ( e ξ − iθ ; β ) + O ( √ ǫ )cosh ξ s + cos θ , and, for ξ ≥ ξ , ∞ X n =1 n (cid:16) ∓ A n e ξ e − n ( ξ − iθ ) − B n e − n ( ξ + iθ ) (cid:17) = r ∗ " ∓ ( τ + τ ) P ( e − ξ + iθ ; β ) + ( τ + τ ) P ( e − ξ − iθ − ξ ; β ) + O ( √ ǫ )cosh ξ s + cos θ . This proves the lemma for ξ ≤ − ξ and for ξ ≥ ξ . (cid:3) roof of Lemma 3.1 It can be easily shown that the function u defined as (3.2) satisfies thetransmission condition of (1.3). Hence it is only need to show that there is a positive constant M such that lim sup | x |→∞ | x || ( u − H )( x ) | ≤ M. (3.7)From (2.5), it is enough to show thatlim sup ( ξ,θ ) → (0 ,π ) | ( u − H )( x ) |√ cosh ξ + cos θ ≤ M when the distance ǫ is fixed.Plugging ξ = 0 and θ = π into (3.3), it becomes ∞ X n =1 (cid:2) ( A n + B n ) cos nπ (cid:3) = 2 ατ ∞ X m =1 τ m − " τ − e − m ( ξ + ξ )+2 ξ − e − m ( ξ + ξ )+2 ξ + − e − m ( ξ + ξ ) − e − m ( ξ + ξ ) − τ − e − m ( ξ + ξ )+2 ξ − e − m ( ξ + ξ )+2 ξ − − e − m ( ξ + ξ ) − e − m ( ξ + ξ ) . Note that for ˜ ξ = 0 , ξ , ξ and | ξ | ≤ ξ s . Hence we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e − m ( ξ + ξ )+˜ ξ + ξ + iθ e − m ( ξ + ξ )+˜ ξ + ξ + iθ − − e − m ( ξ + ξ )+˜ ξ − e − m ( ξ + ξ )+˜ ξ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( e ξ + iθ + 1) e − m ( ξ + ξ )+˜ ξ (1 + e − m ( ξ + ξ )+˜ ξ + ξ + iθ )(1 − e − m ( ξ + ξ )+˜ ξ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ e − m ( ξ + ξ )+˜ ξ + ξ s (1 − e − m ( ξ + ξ )+˜ ξ + ξ s ) | e ξ + iθ + 1 | . Let us define a constant S := 2 ατ ∞ X m =1 τ m − " (cid:18) τ (cid:19) e − m ( ξ + ξ )+2 ξ + ξ s (1 − e − m ( ξ + ξ )+2 ξ + ξ s ) + e − m ( ξ + ξ )+ ξ s (1 − e − m ( ξ + ξ )+ ξ s ) + (cid:18) τ (cid:19) e − m ( ξ + ξ )+2 ξ + ξ s (1 − e − m ( ξ + ξ )+2 ξ + ξ s ) + e − m ( ξ + ξ )+ ξ s (1 − e m ( ξ + ξ )+ ξ s ) . Applying the same procedure as in the proof of Lemma 3.2 , we have | ( u − H )( x ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X n =1 (cid:2) ( A n e nξ + B n e − nξ ) cos nθ − ( A n + B n ) cos nπ (cid:3)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ S | e ξ + iθ + 1 | . Hence | ( u − H )( x ) | ≤ C | ξ + i ( θ − π ) | , as ( ξ, θ ) → (0 , π ) , with a constant C independent of ξ and θ . (cid:3) Lemma 3.3
The singular function q in (1.14) satisfies q ( x ; β, τ , τ ) − C q = O ( | x | − ) , as | x | → ∞ , Where C q is a constant defined as C q = n ( τ + τ ) L ( − e − ξ ; β ) − ( τ + τ ) L ( − e − ξ ; β ) o . roof . From the definition of L , we have (cid:12)(cid:12)(cid:12) L ( e − ( ξ + iθ ) − ξ ; β ) − L ( e − ξ ; β ) (cid:12)(cid:12)(cid:12) ≤ Z t β (cid:12)(cid:12)(cid:12)(cid:12) e − ( ξ + iθ ) − ξ te − ( ξ + iθ ) − ξ − − e − ξ − te − ξ (cid:12)(cid:12)(cid:12)(cid:12) dt = Z t β (cid:12)(cid:12)(cid:12)(cid:12) ( e − ( ξ + iθ ) + 1) te − ξ (1 + te − ( ξ + iθ ) − ξ )(1 − te − ξ ) (cid:12)(cid:12)(cid:12)(cid:12) dt = (cid:12)(cid:12)(cid:12) e − ( ξ + iθ ) + 1 (cid:12)(cid:12)(cid:12) Z t β (cid:12)(cid:12)(cid:12)(cid:12) te − ξ (1 − te − ξ ) (cid:12)(cid:12)(cid:12)(cid:12) dt ≤ C | ξ + i ( θ − π ) | , as ( ξ, θ ) → (0 , π ) , where a constant C is independent of ξ and θ . Therefore, we have L ( e − ( ξ + iθ ) − ξ ; β ) − L ( e − ξ ; β ) = O ( | x | − ) , as | x | → ∞ . Similarly, we prove the deacay property of L ( e ( ξ + iθ ) − ξ ; β ) − L ( e − ξ ; β ). (cid:3) ∇ u q The main term of u in both Theorem 1.1 and Theorem 1.2 is expressed in terms of q ( x ; β, τ , τ )with positive τ and τ . We assume that τ and τ are positive in this section.For notational sake, the normal- and the tangential derivatives at ( c, θ ) mean (2.10) and (2.11),respectively, on the coordinate level curve { ξ = c } . From (1.14) and (1.18), the derivative of q ( x ; β, τ , τ ) satisfies ∂q∂ξ = 12 ( τ + τ ) P (cid:0) e − ( ξ + iθ ) − ξ ; β (cid:1) + ( τ + τ ) P (cid:0) e ( ξ + iθ ) − ξ ; β (cid:1) in R \ ( B ∪ B ) , − ( τ + τ ) P (cid:0) e ξ − iθ ; β (cid:1) + ( τ + τ ) P (cid:0) e ξ + iθ − ξ ; β (cid:1) in B , ( τ + τ ) P (cid:0) e − ξ − iθ − ξ ; β (cid:1) − ( τ + τ ) P (cid:0) e − ξ + iθ ; β (cid:1) in B , (4.1) ∂q∂θ = i ( τ + τ ) P (cid:0) e − ( ξ + iθ ) − ξ ; β (cid:1) + ( τ + τ ) P (cid:0) e ( ξ + iθ ) − ξ ; β (cid:1) in R \ ( B ∪ B ) , ( τ + τ ) P (cid:0) e ξ − iθ ; β (cid:1) + ( τ + τ ) P (cid:0) e ξ + iθ − ξ ; β (cid:1) in B , ( τ + τ ) P (cid:0) e − ξ − iθ − ξ ; β (cid:1) + ( τ + τ ) P (cid:0) e − ξ + iθ ; β (cid:1) in B , where P is defined as (1.18). From (2.10), (2.11), (2.23) and (2.8)( ∇ q · ˆ e ξ ) (cid:12)(cid:12) ξ = c = cosh c + cos θr ∗ √ ǫ ∂q∂ξ (cid:12)(cid:12)(cid:12) ξ = c + O (1) , (4.2)( ∇ q · ˆ e θ ) (cid:12)(cid:12) ξ = c = cosh c + cos θr ∗ √ ǫ ∂q∂θ (cid:12)(cid:12)(cid:12) ξ = c + O (1) . (4.3) Lemma 4.1
There is a constant C independent of ǫ , k , and k such that k∇ q k L ∞ ( R ) ≤ C if τ < . Proof . Applying (3.1) to (2.23) for x ∈ B ∪ B and (2.25) for x ∈ R \ B ∪ B , we can provethan there is a C independent of ǫ , k and k such that | ( ∇ q · ˆ e ξ ) | , | ( ∇ q · ˆ e θ ) | ≤ C − ln τ , (4.4)for c = − ξ , ξ . At c = ( − j ξ j , j = 1 ,
2, (4.4) is satisfied for either outward or inward derivatives. (cid:3) roposition 4.2 Suppose τ , τ > . The tangential derivative of ℜ{ q ( x ; β, τ , τ ) } is remainbounded regardless of ǫ for all x ∈ R . The normal derivative of ℜ{ q ( x ; β, τ , τ ) } is bounded in B ∪ B . For x / ∈ B ∪ B , ∂ ℜ{ q } ∂ν (cid:12)(cid:12)(cid:12)(cid:12) ξ = c = − sgn ( c ) ( τ + τ + 2 τ ) cosh ξ s + cos θ r ∗ √ ǫ ℜ n P (cid:0) e − ( ξ s + iθ ) ; β (cid:1)o + O (1) . (4.5) Proof . From the definition of P , P ( z ; β ) = P ( z ; β ) , for all z ∈ C , | z | < . (4.6)Hence ℜ{ P ( z ) } = ℜ{ P ( z ) } and ℑ{ P ( z ) } = −ℑ{ P ( z ) } . (4.7)From (4.1), (4.7), (2.24) and (4.2), we have ∂ ℜ{ q } ∂ν (cid:12)(cid:12)(cid:12)(cid:12) ξ = c = − sgn( c ) cosh ξ s + cos θ r ∗ √ ǫ ( τ + τ + 2 τ ) ℜ (cid:8) P (cid:0) e − ξ s + iθ ; β (cid:1)(cid:9) + O (1) , x ∈ R \ B ∪ B cosh ξ + cos θ r ∗ √ ǫ ( τ − τ ) ℜ (cid:8) P (cid:0) e ξ + iθ ; β (cid:1)(cid:9) + O (1) , x ∈ B cosh ξ + cos θ r ∗ √ ǫ ( τ − τ ) ℜ (cid:8) P (cid:0) e − ξ + iθ ; β (cid:1)(cid:9) + O (1) , x ∈ B ,∂ ℜ{ q } ∂T (cid:12)(cid:12)(cid:12)(cid:12) ξ = c = − sgn( c ) cosh c + cos θ r ∗ √ ǫ ( τ − τ ) ℜ n P (cid:0) e − max( ξ s , | c | ) − iθ ; β (cid:1)o + O, x ∈ R . Let us prove the boundedness part in the corollary. From (2.22), we have (cid:12)(cid:12)(cid:12)(cid:12) ∂∂ν ℜ{ q ( x ; β, τ , τ ) } (cid:12)(cid:12)(cid:12) − ∂B j (cid:12)(cid:12)(cid:12)(cid:12) ≤ C | τ − τ | √ ǫ β O (1) + O (1) . Since | t − | ≤ − log t for all 0 < t < | τ − τ |− log τ ≤ | τ − | + (cid:12)(cid:12) − τ | (cid:12)(cid:12) − log τ − log τ ≤ . Hence the normal derivative inside B and B is bounded. Similarly, we prove the tangentialderivative is bounded. (cid:3) Following the proof of Proposition 4.2, it can be easily shown the following.
Proposition 4.3
Suppose τ , τ > . The normal derivative of ℑ{ q ( x ; β, τ , τ ) } is remain boundedregardless of ǫ for x ∈ R \ B ∪ B . For x ∈ B ∪ B , ∂ ℑ{ q } ∂ν (cid:12)(cid:12)(cid:12)(cid:12) ξ = c = cosh c + cos θ r ∗ √ ǫ ( τ + τ + 2 τ ) ℜ n P (cid:0) e −| c | + iθ ; β (cid:1)o + O (1) . The tangential derivative satisfies ∂ ℑ{ q } ∂T (cid:12)(cid:12)(cid:12)(cid:12) ξ = c = − sgn ( c ) ( τ + τ + 2 τ ) cosh c + cos θ r ∗ √ ǫ ℜ n P (cid:0) e − max( ξ s , | c | ) − iθ ) ; β (cid:1)o + O (1) , x ∈ R . (4.8)15 orollary 4.4 Suppose k , k > From Theorem 1.1, Theorem 1.2, Proposition 4.2 and Propo-sition 4.3, we have (cid:12)(cid:12) ∇ ( u − H ) (cid:12)(cid:12) + ( x ) = | c n | ( | τ | + | τ | + 2 τ ) cosh ξ j + cos θ r ∗ √ ǫ (cid:12)(cid:12)(cid:12) ℜ{ P (cid:0) e − ( ξ j + iθ ) ; β (cid:1) } (cid:12)(cid:12)(cid:12) + O (1) on ∂B j , j = 1 , . Especially at x j , which is the point on ∂B j closest to the other disk, we have θ = 0 and (cid:12)(cid:12) ∇ ( u − H ) (cid:12)(cid:12) + ( x j ) = | c n | ( | τ | + | τ | + 2 τ ) cosh ξ j + 12 r ∗ √ ǫ (cid:12)(cid:12) ℜ{ P (cid:0) e − ξ j ; β (cid:1) } (cid:12)(cid:12) + O (1) . The above asymptotic equations are still valid for < k , k < if n is replaced by t . Remark 4.5
Applying Lemma 2.2, it follows that there is a constant C and C such that |∇ u | + ( x j ) ≥ C r ∗ ( ∇ H ( p ) · n ) √ ǫ ( β + 1) , j = 1 , , (4.9) k∇ u k L ∞ ( ∂B j ) ≤ C r ∗ ( ∇ H ( p ) · n ) √ ǫ ( β + 1) , (4.10) where p is the middle point of the shortest line segment connecting ∂B and ∂B . Since β = r ∗ √ ǫ (1 − τ + O ((1 − τ ) ) , r ∗ √ ǫ ( β + 1) ≈ − τ + r ∗ √ ǫ . Recall that r ∗ / max ( r , r ) ≤ /r ∗ ≤ r ∗ / min ( r , r ) . Hence (4.9) and (4.10) are in accordance withthe upper and lower bounds (1.6) and (1.7). β In this section we let us consider the case when β is extremely small or large. Recall that β in(1.15) can be an arbitrary positive number even when ln τ is very small because the denominatoris small parameter. For example, β = r ∗ √ ǫ (1 − τ + O ((1 − τ ) ) ≈ ǫ / if k j ≈ ǫ − / , and β ≈ ǫ − / if k j ≈ ǫ − / .From the definition P in (2.15), it can be written as P ( z ; β ) = − z + 1 + β Φ( − z, , β ) . Set | z | <
1, namely, z = − e − ξ + iθ with ξ >
0. For β ≪ z, , β ) = 1 β + ∞ X k =0 z k k − β ∞ X k =1 z k k (1 + βk ) = 1 β − log(1 − z ) + O ( β ) (4.11)and ℜ{ Φ( z, , β ) } = 1 β − ln q e − ξ (cosh ξ + cos θ ) + O ( β ) . If β ≫
1, applying the integration by parts twice on (2.14), we haveΦ( z, , β ) = 1 β − z + 1 β − z (1 − z ) + 1 β r ( z, β )(cosh ξ + cos θ ) , (4.12)with | r ( z, β ) | ≤ . It is worth to mention that there are more complete results on the asymptoticexpansions of Φ( z, s, β ) for large and small β done by Ferreira and Lop´ez [11].16pplying the asymptotic of the Lerch transcendent in (4.11) and (4.12), we have ℜ{ P ( e − ξ j − iθ ; β ) } = 12 − sinh ξ j ξ j + cos θ ) − β ln | e − ξ j − iθ | + O ( √ ǫ )= 12 (cid:16) − β ln[2(cosh ξ j + cos θ )] (cid:17) + O ( √ ǫ )cosh ξ j + cos θ , for β = O ( ǫ / ) , (4.13)and ℜ{ P ( e − ξ j − iθ ; β ) } = 1 β e − ξ j − iθ ( e − ξ j − iθ + 1) + O ( √ ǫ )(cosh ξ j + cos θ ) = 1 β ξ j cos θ ξ j + cos θ ) + O ( √ ǫ )(cosh ξ j + cos θ ) , for β = O ( ǫ − / ) . One can easily show that there is a constant C independent of ǫ such that |∇ ( u − H )( x ) | ≤ C, for x ∈ R \ ( B ∪ B ) satisfying {| θ | > π − √ ǫ } . (4.14)Note that {| θ | > π − √ ǫ } is a region whose distance from the touching point is bigger than apositive constant independent of ǫ . Here √ ǫ is not the optimal rate for the region where thegradient ∇ ( u − H ) is bounded independently of ǫ . For the case of β = O ( ǫ / ) we can prove thefollowing lemma. Lemma 4.6
For ǫ, k , k satisfying β = O ( ǫ / ) , there is a constant C independent of C such that |∇ ( u − H )( x ) | ≤ C, for x ∈ R \ ( B ∪ B ) satisfying {| θ | > π − ǫ / } . Proof . If | θ | > π − ǫ / , then 1 + cos θ = O ( √ ǫ ). Hence (cosh ξ s + cos θ ) /α = O (1) . From (4.13),we prove the lemma. (cid:3) u k to u ∞ For the perfectly conducting case, k = k = ∞ , the conductivity problem becomes ∆ u = 0 in R \ B ∪ B ,u = constant on ∂B j , j = 1 , , Z ∂B j ∂u∂ν ( j ) ds = 0 , j = 1 , ,u ( x ) − H ( x ) = O ( | x | − ) as | x | → ∞ , (4.15)where ν ( j ) is outward normal to ∂B j . For this case, we can represent q ( β ; 0 , ,
1) in terms of linearcombination of ξ and θ . Using (2.7) and (2.2), we have q ( x ; 0 , ,
1) = − log (cid:16) e − ξ − ( ξ + iθ ) (cid:17) + log (cid:16) e − ξ +( ξ + iθ ) (cid:17) = − log (cid:18) e − ξ α − zα + z (cid:19) + log (cid:18) e − ξ α + zα − z (cid:19) . Hence q ( x ; 0 , ,
1) = − log (cid:18) (1 − e − ξ )( z + α coth ξ ) α + z (cid:19) + log (cid:18) (1 − e − ξ )( z + α coth ξ ) α − z (cid:19) = log (cid:18) α + zα − z (cid:19) + log (cid:18) z + ˜ c z − ˜ c (cid:19) + log 1 − e ξ − e − ξ = ξ + iθ + b, (4.16)17here ( ℜ{ ˜ c j } , ℑ{ ˜ c j } ) = c j , j = 1 .
2, and ∇ b is uniformly bounded independently of ǫ .Let us denote u k the solution to (1.2) with k = k = k and u ∞ to (4.15). In Appendix of [6],it was shown the weak H -convergence of u k to u ∞ for fixed ǫ when the background domain isbounded and inclusions are of strictly convex shape. For circular inclusions we have the follows. Lemma 4.7
Let H ( x ) = x and u ∞ be the solution to (4.15). Then, for fixed ǫ > , we have u k → u ∞ in W , ∞ ( R ) , as k → ∞ . Proof . For H ( x ) = x , as in the same way in the proof of Lemma 3.1, one can see that u ∞ = x + ℜ{ U ′ } where U ′ is defined as the following: U ′ ( x, y ) = C ′ + ∞ X n =1 (cid:16) A ′ n e n ( ξ + iθ ) + B ′ n e − n ( ξ + iθ ) (cid:17) , x ∈ R \ ( B ∪ B ) ∞ X n =1 (cid:16) A ′ n e n ( ξ + iθ ) + B ′ n e n (2 ξ + ξ − iθ ) (cid:17) , x ∈ B ∞ X n =1 (cid:16) A ′ n e n (2 ξ − ξ + iθ ) + B ′ n e − n ( ξ + iθ ) (cid:17) , x ∈ B where C ′ = − P ∞ n =1 ( A ′ n + B ′ n ) cos nπ , A ′ n = 2 α ( − n e n ( ξ + ξ ) − (cid:0) − e nξ − (cid:1) , and B ′ n = 2 α ( − n e n ( ξ + ξ ) − (cid:0) e nξ (cid:1) . By the maximum principle, without loss of generality, it is enough to consider x ∈ ∂B ∪ ∂B .Moreover, we may consider only x ∈ ∂B . To estimate k u k − u ∞ k L ∞ ( ∂B ) for k goes to infinity,we need to do the following quantities: ∞ X n =1 ( A n − A ′ n ) e − nξ cos nθ, ∞ X n =1 ( B n − B ′ n ) e nξ cos nθ, and ( C − C ′ ) . We compute | A n − A ′ n | = (cid:12)(cid:12)(cid:12)(cid:12) τ − e nξ + 1 τ − e n ( ξ + ξ ) − − e nξ + 1 e n ( ξ + ξ ) − (cid:12)(cid:12)(cid:12)(cid:12) ≤ M (cid:0) | τ − − τ − | + | − τ − | + | − τ − | (cid:1) e n (2 ξ + ξ ) ( e n ( ξ + ξ ) − ≤ M k e n (2 ξ + ξ ) ( e n ( ξ + ξ ) − . (4.17)Therefore we have | P ∞ n =1 ( A n − A ′ n ) e − nξ cos nθ | ≤ M/k.
Similarly, we can prove the convergencefor P ∞ n =1 ( B n − B ′ n ) e nξ cos nθ and ( C − C ′ ).Let us now consider k∇ ( u − u ∞ ) k L ∞ ( ∂B ) . For x ∈ ∂B , we have |∇ ( u k − u ∞ )( x ) | ≤ C ∞ X n =1 n (cid:0) | A n − A ′ n | e − nξ + | B n − B ′ n | e nξ (cid:1) ≤ Mk .
This completes the proof. (cid:3)
Lemma 4.8
Suppose /k = O ( ǫ ) and /k ≫ ǫ . There is a positive constant C independent of ǫ such that k∇ ( u k − u ∞ )( x j ) k ≥ Ckǫ , j = 1 , , where H ( x ) = x and x j is the point on ∂B j closest to the other disk. roof . From the definition, β = O ( ǫ / ). Applying Lemma 4.7, (4.13) and (1.11), ∂∂ν ( u k − u ∞ ) (cid:12)(cid:12)(cid:12) + ∂B j = c n cosh ξ j + cos θr ∗ √ ǫ β ln[2(cosh ξ j + cos θ )] + O (1) . (4.18)At x j , where θ = 0, ∂∂ν ( u k − u ∞ ) is order of magnitude 1 / ( kǫ ). (cid:3) The above lemma says that the convergence in Lemma 4.7 is not uniform in terms of ǫ . Thereis a practical implication of it in computing the electric field. The perfectly conducting boundarycondition gives a good approximation if ǫ is bigger than 1 /k , where |∇ u k | in the proximity of objectsis as big as ǫ − / and |∇ ( u k − u ∞ ) | is bounded. However, as ǫ ≪ /k , |∇ ( u k − u ∞ ) | becomes asbig as 1 / ( kǫ ) while |∇ u k | is still of magnitude ǫ − / . Hence the L ∞ -error in the computation ofelectric field can be arbitrary large.Let us visualize the non-uniform convergence with an example. We set H ( x ) = x , r = 2 , r = 3and k = k = ǫ − / . The centers of inclusions are located such that (2.2) is satisfied. In thefirst of Figure 4.1, we plot the singular term of ∂ ( u k − u ∞ ) /∂ν (cid:12)(cid:12) + ∂B in (4.18) for ǫ = 10 − and k = k = ǫ − / ≈ ǫ decreases. - - - Θ ¶ H u k - u ¥ L (cid:144) ¶ Ν ´ - ´ - ´ - ´ - ´ - Ε ¶ H u k - u ¥ L (cid:144) ¶ Ν H x L Figure 4.1: The left figure is the graph of ∂ ( u k − u ∞ ) /∂ν (cid:12)(cid:12) + ∂B when k = k = ǫ − / and ǫ = 10 − .The normal derivative of u ∞ is approximately 800. The right is the graph of ∂ ( u k − u ∞ ) /∂ν (cid:12)(cid:12) + ∂B at x in terms of ǫ .Since the harmonic conjugate of the solution u of (1.2) is the solution to (1.2) with k j replacedby 1 /k j , j = 1 ,
2, we have the similar non-uniform convergence of the gradient of u for the almostinsulating case. Let Ω be a bounded domain in R with C -boundary containing two circular conducting inclusions B and B . We assume that the inclusions are located away from ∂ Ω and the distance betweenthem is ǫ . We consider the following boundary value problem: ( ∇ · σ ∇ u = 0 in Ω ,∂u∂ν (cid:12)(cid:12)(cid:12) ∂ Ω = g ( g ∈ L (Ω)) , (5.1)where ν is the outward unit normal to ∂ Ω.For a bounded domain B with C boundary, define the single- and double layer potentials as S B [ ϕ ]( x ) = 12 π Z ∂B ln | x − y | ϕ ( y ) dσ ( x ) , x ∈ R , D B [ ϕ ]( x ) = − π Z ∂B h x − y , ν ( y ) | x − y | ϕ ( y ) dσ ( y ) , x ∈ R \ ∂B. u by series of solutions tothe free space solution, we get the following theorem. Theorem 5.1
Suppose k , k > . The solution u to (5.1) satisfies u ( x ) = c n ℜ{ q ( x ; β, τ , τ ) } + u r ( x ) , where c n is defined with H ( x ) = −S Ω [ g ]( x ) + D Ω [ u | ∂ Ω ]( x ) , x ∈ Ω (5.2) and |∇ u r ( x ) | is uniformly bounded for ǫ . Theorem 5.2
Suppose < k , k < . The solution u to (5.1) satisfies u ( x ) = c t ℑ{ q ( x ; β, − τ , − τ ) } + u r ( x ) , where c t is defined with H in (5.2) and |∇ u r ( x ) | is uniformly bounded for ǫ . The characterization of the singular term of u finds a very good application in the computationof electric fields. Computation of the gradient ∇ u (or u ) in the presence of adjacent inclusionswith large or small conductivity value is a challenging problem due to the blow-up feature of theelectric field. Modifying the algorithm in [12], where the electric field in the presence of inclusionswith extreme conductivities (perfectly conducting or insulating) was computed using h , to use q inthe place of h , one can accurately compute u for the inclusions of arbitrary constant conductivity. In this section we demonstrate the main results for some examples. We compare the gradient of thesolution u to (1.2) with that of the singular term which is derived in Theorem 1.1 and Theorem 1.2.For notational convenience, let us denote the singular terms in Proposition 4.5 and Proposition 4.8as follows: Q ∂B ( θ ) = r ∗ ( | τ | + | τ | + 2 τ ) cosh ξ + cos θ √ ǫ ℜ n P (cid:0) e − ( ξ + iθ ) ; β (cid:1)o . Outside inclusions, only the normal derivative of u blows-up in case of highly conducting inclusionswhile the tangential derivative does in case of almost insulating case.For all examples, r = 3, r = 2, and the centers of them are located satisfying (2.2). Since thegraph shows the similar behavior either on ∂B or on ∂B , we plot graphs only on ∂B . Data Acquisition
To compute u , we use the series expansion in Lemma 3.1. Using (2.10) and(2.11), both the normal and the tangential derivative of u can be represented as series as well as u .It is worth to remark the difficulty in these computation for small ǫ . Note that e − n ( ξ + ξ ) term isinvolved in Lemma 3.1 and, hence, the cost in numerical computation becomes very high for small ǫ >
0. For instance, in Example 1, we evaluate the summation for n ≤ ∗ to compute withina tolerance 10 − . If we change ǫ to 10 − , then 10-times more summation is required for the sametolerance.The singular function Q ∂B can be represented in terms of Lerch transcendent function Φ. Us-ing its analytic properties (such as integral representation), very efficient numerical algorithms arealready implemented in many commercial softwares. In this paper, it is used ’LerchPhi’ built-infunction in Mathematica. Moreover, for the case of extremely small and large β , the Lerch tran-scendent Φ can be approximated in terms of elementary functions. Therefore the computationalcost of computing the singular terms in ∇ u (or in u ) becomes much smaller than the one based onits series representation. 20 xample 1 . In Figure 6.1, we show the numerical illustration for normal derivative for highlyconducting case ( ∇ u of almost insulating case with H ( x ) = y haves the same feature). Fixing theconductivities of inclusions ( k = 1500 and k = 1200), we change ǫ ( ǫ = 0 . , . , . H ( x ) = x . We plot the graph of the normal derivative ∂ ( u − H ) /∂ν (cid:12)(cid:12) + ∂B and Q ∂B in gray and black, respectively. It shows that the difference between two terms areremained almost unchanged while the magnitude of them increases as ǫ decreases. - - - - Θ the normalderivativeon ¶ B - - - Θ the normalderivativeon ¶ B - - - Θ the normalderivativeon ¶ B Figure 6.1: Highly conducting case. k = 1500 , k = 1200, and ǫ is 0 . , . , . ∂ ( u − H ) /∂ν (cid:12)(cid:12) + ∂B while the black line does Q ∂B . Example 2 . In this example, we set ǫ = 0 .
01 and H ( x ) = x . The conductivities are ( k , k ) =(7 , , (70 , , (7000 , ∂ ( u − H ) /∂ν (cid:12)(cid:12) + ∂B and Q ∂B in gray and black,respectively. The difference between two terms are remained almost unchanged. - - - - Θ the normalderivativeon ¶ B - - - Θ the normalderivativeon ¶ B - - - Θ the normalderivativeon ¶ B Figure 6.2: The distance ǫ is fixed to be 0.01, and conductivities are ( k , k ) is(7 , , (70 , , (7000 , Example 3 . In this example, we draw the equipotential lines for the singular term in u and |∇ u | outside inclusions when the circular inclusions are highly conducting or almost insulating. Moreprecisely, we set ( k , k ) = (100 , , (0 . , .
02) and fix ǫ = 0 .
1. The background potential H ( x )is x for highly conducting inclusions and y for almost insulating ones.In Figure 6.3, we show contour plot of ℜ{ q } and (cid:12)(cid:12) ℜ{∇ q } (cid:12)(cid:12) which are singular terms of u and |∇ u | , respectively. For both cases, in the narrow region between the inclusions, |∇ u | changesfast in y -direction but slowly in x -direction which is in accordance with Proposition 4.2. Awayfrom the touching region, |∇ u | changes slowly, see Lemma 4.6. The contour of |∇ u | is continuousacross the boundary of inclusions for almost insulating case while it is discontinuous for highlyconducting, and |∇ u | blows-up inside inclusions for almost insulating case while it is bounded forhighly conducting case as ǫ goes to zero. References [1] H. Ammari, G. Ciraolo, H. Kang, H. Lee, and K. Yun, Spectral analysis of the Neumann-Poincar´e operator and characterization of the stress blow-up in anti-plane elasticity. Archiveon Rational Mechanics and Analysis, 208 (2013), 275-304.21 - - - - - - - - - - - - - - - - - - - - - - - Figure 6.3: Contour plot of the singular term of u , the first column, and of |∇ u ||