A Large Jump Asymptotic Framework for Solving Elliptic and

Parabolic Equations with Interfaces and Strong Coefficient

Discontinuities

I. Klapper and T. Shaw

Department of Mathematical Sciences,

Montana State University

Abstract. We address the solution of multiregion elliptic and parabolic problems with strongly discontinuouscoefficients across interfaces. The underlying idea is, roughly, to perturb around the infinitely discontinuouscoefficient solution, that is, to perturb the solution where one region is a perfect conductor. The result is partialdecoupling of the interface conditions. The algorithm requires a small number of well-conditioned solves in theindividual regions (in many cases only a single solve over each region is necessary) that are then assembledinto an accurate global solution. The error from the assembly step is asymptotically small in the ratio of thediscontinuous coefficients. Further, the framework can be extended to problems with moderately discontinuouscoefficients using a series expansion in the discontinuity ratio in a manner similar to a Schwarz alternatingmethod.

Key words: interface, discontinuous coefficients, elliptic equations, parabolic equations.

1 The interface problem

In this paper we present a large jump asymptotics (LJA) framework for reduction of an elliptic or parabolicproblem with strongly discontinuous coefficient jumps to a small number of subproblems, each with continuouscoefficients and each independent of the discontinuity amplitude. We focus on solving the Poisson equation

∇ · (β(x)∇u) = f(x) (1)

and the diffusion equation∂tu −∇ · (β(x, t)∇u) = f(x, t) (2)

together with appropriate boundary conditions on a given domain Ω = Ω+ ∪ Γ ∪ Ω− where Ω+ and Ω− aresubregions of Ω separated by a smooth interface Γ (see, e.g., Fig. 1). More general elliptic and parabolicproblems could be treated as well. In the remainder of this Section, we focus on equation (1) and thus suppresst-dependence of solutions, coefficients, and data, but the discussion extends to equation (2) straightforwardly.

The coefficient β(x) is assumed to take the piecewise form

β(x) =

β+α+(x), x ∈ Ω+

β−α−(x), x ∈ Ω− (3)

where α±(x) are dimensionless smooth positive functions bounded away from zero and β± are positive param-eters. The function f(x) likewise decomposes into f+(x) for x ∈ Ω+ and f−(x) for x ∈ Ω−. The solutions inregions Ω+, Ω−, are coupled across Γ through a pair of interface matching conditions. Allowing the possibilityof discontinuities in β and f across Γ as well as interface jumps, these conditions are

[u] = v, [βun] = w (4)

1

Ω+

Ω−

Γ

n

(a)

Ω+

Ω−

Γ

n

(b)

Figure 1: Multiregion domains Ω with interface Γ and normal vector n. (a) Layered geometry. (b) Bubblegeometry, with a single bubble.

for given functions v and w defined on Γ. Here and hereafter, subscripted n denotes the outward (from Ω−)normal derivative.

The aim is to approximate the solution

u(x) =

u+(x), x ∈ Ω+

u−(x), x ∈ Ω−

to (1) or (2) in the case of a large jump in β across the interface Γ. This type of problem is relevant in manyapplications including bubbly flow [24], oil reservoir simulation [20], groundwater flow [3], electromagnetics [5],plasmas [13], biomaterials [23], etc., where coefficients may have large to very large discontinuities across aninterface between regions with different material properties. In many of these problems the quantity β(x)∇u,the flux field of u, is physically important and in fact equations (1) and (2) may be regarded as conservationlaws for flux. If β = ∞ (a “perfectly conducting” material) then the flux field is necessarily zero, but often themore relevant concern is what happens to β(x)∇u as, say, β− goes to ∞ with β+ fixed. We wish to addressthis problem in particular with regards to approximation of solutions for large β− (or small β+) in this paper.

We begin by scaling (1) by β+ to obtain

∇ · (α+(x)∇u+) = (β+)−1f+(x), x ∈ Ω+

β ∇ · (α−(x)∇u−) = (β+)−1f−(x), x ∈ Ω−(5)

with β = β−/β+ assumed to be large. Define the flux field for the scaled problem by

F(x) =

α+(x)∇u+ x ∈ Ω+

βα−(x)∇u− x ∈ Ω−(6)

Scaling the interface conditions, we obtain the jump conditions (across Γ)

u+ − u− = v (7)

and

α+u+n − βα−u−

n =1

β+w. (8)

In (7) and (8) the quantities u±, α±, and u±n are limiting values at points of Γ with the limits taken from the

± sides of the interface.

2

A wide range of numerical approaches have been developed to handle geometries such as in Figure 1, e.g.[16], [4], [11]. However application of a standard interface solver to (5)-(8) typically slows or fails entirely due to

ill-conditioning resulting from the large ratio β (for example, [24] reports requiring 104 iterations for convergence

in a fluid problem with β = 103), though these difficulties can be eased with care [9]. The difficulty arises from

the fact that large variation in β manifests in the form of large coefficient variations in the matrix equation tobe solved. This matrix equation is generally ill-conditioned with condition number proportional to β. Earlierworks of Li [15] (together with [10],[25]) and Yang [26] proposed solutions to this problem in the context ofiterative application respectively of an immersed interface algorithm and a finite element domain decompositionalgorithm, in each case either explicitly or implicitly decoupling in some manner regions on either side of theinterface. The method presented here can be viewed as a modification and extension of those efforts. Both Li’sand Yang’s algorithms are rather complex and thus will not be described here. Notably, however, those authorsdiscovered in each case that their methods were relatively insensitive to grid resolution as β increased and that,interestingly, efficiency actually improved with increasing β. We suggest that a key step in both algorithms is,roughly, to require that the flux (6) through the high β region (the “good” conductor) be determined by theflux capacity of the low β region (the “poor” conductor) and not vice-versa. One can immediately observe an

advantage by using the flux interface condition [βun] = w to determine, stably for large β,

u−

n =1

β

α+

α−u+

n −1

β−α−w (9)

rather than, unstably for large β,

u+n = β

α−

α+u−

n +1

β+α+w.

In fact the significance of this observation goes deeper as will be argued in the following; the stable choice (9)

is a natural consequence of considering the large β problem to be a perturbation of the β = ∞ problem.In this paper we present a framework for approximating solutions to various problems of the sort mentioned

above. The underlying idea is to describe solutions on either side of the interface in terms of expansions in β.Within this framework, solutions of subproblems are formed separately on both sides of the interface and thenpieced together in a prescribed manner. The main advantages of our framework are first, any irregular regionsolver can be used for the individual region subproblems (e.g. [17], [18], [12], [7]), and second, these individual

region subproblems are independent of discontinuity parameter β and (9) is built in. The significance of this

latter property is that the large β conditioning issue is entirely absent.We focus here on problems (1) and (2) in the limit β → ∞. However a number of extensions of the LJA

framework are possible including solution at moderate β. Also, importantly, the framework to be proposed isindependent of dimension though we will only consider one and two dimensional problems. More general ellipticand parabolic problems than (1) and (2) can be treated.

2 The Asymptotic Poisson Problem

For definiteness, the following specifications are assumed (none are essential): we choose the domain Ω = [x, z] ∈ [0, 1]2 to be two-dimensional, rectangular, periodic in the horizontal direction, with entire top bound-ary of Ω contained in Ω+ and entire bottom boundary contained in Ω−. See Figure 1(a). A somewhat differentgeometry (Figure 1(b)) will also be used occaionally.

We consider equations (5) in the limit β → ∞. (More particularly we assume β+ fixed, β− → ∞; theopposite choice, β− fixed, β+ → 0, may differ in some details. See for example the test problem at the end ofthis Section.) Top and bottom boundary conditions in Ω take the mixed form aiu|z=i + biun|z=i = ci, i = 0, 1(we assume the coefficients a0, b0 and c0 for the “high conductivity” region boundary are constants for simplicity,a physically reasonable requirement). With motivation to be described below, let

u(x; β−1) =

u+0 (x), x ∈ Ω+

u−

0 (x) + β−1u−

1 (x), x ∈ Ω−(10)

3

be an approximation, the LJA, to the solution of (5)-(8) where u±

0 and u−

1 are independent of β. In fact, as a

remark (to be expanded upon later), (10) is a truncated Taylor series in β−1 of the exact solution. Substituting

u into (5)−(8) and equating powers of β−1 we obtain the subproblems

∇ · (α−∇u−

0 ) = 0 x ∈ Ω−

u−

0n = 0 x ∈ Γa0u

−

0 + b0u−

0n = c0 z = 0(11)

∇ · (α+∇u+0 ) =

1

β+f+ x ∈ Ω+

u+0 = u−

0 + v x ∈ Γa1u

+0 + b1u

+0n = c1 z = 1

(12)

∇ · (α−∇u−

1 ) =1

β+f− x ∈ Ω−

u−

1n =1

α−

(

α+u+0n −

1

β+w

)

x ∈ Γ

a0u−

1 + b0u−

1n = 0 z = 0

(13)

(11)-(13) are to be solved in succession, on region Ω−, then Ω+ and then back to Ω−. Problem (12) uses Dirichletdata extracted from u−

0 on Γ and problem (13) uses Neumann data extracted from u+0 on Γ. The resulting LJA

approximation (10) satisfies exactly (5)-(8), together with the imposed boundary conditions, with the exception

of the interface condition (7) which is satisfied up to an O(β−1) error (in particular, [u] − v = −β−1u−

1 on Γ),

arbitrarily small for β sufficiently large. We remark that inclusion of an additional term β−1u+1 in the LJA

approximation, while allowing (7) to be satisfied exactly, would instead however introduce an O(β−1) error inthe flux condition (8) while requiring an additional solve for u+

1 . Note that the apparent imbalance in (10)disappears when (10) is substituted into the expression (6)

Note that in (11)-(13) three standard elliptic solves on irregular domains appear to be required at first glance.In fact, if a0 6= 0, then (11) has solution u−

0 = c0/a0 and hence only two solves are needed, i.e., essentially asingle solve over the entire domain Ω is necessary. The case a0 = 0 must be treated somewhat differently.In this instance c0 = 0 necessarily and then u−

0 is again constant. However the value of this constant is notset directly from the boundary conditions. Rather it must be determined so that the solvability requirement∫

Γ α−u−

1n ds =∫

Ω−(β+)−1f−dx in problem (13) is satisfied. To accomplish this, two solutions of variants of

problem (12) are needed: first a solution u+00 with data u+

00 = v, x ∈ Γ, and a1u+00 + b1u

+00n = c1 at z = 1,

and second a solution u+01 with data u+

01 = 1, x ∈ Γ, and a1u+01 + b1u

+01n = 0 at z = 1. Given u+

00, u+01, a

linear combination u+0 = u+

00 +Cu+01 is constructed so that the interface condition in (13) satisfies the solvability

condition∫

Γ α−u−

1n ds =∫

Ω−(β+)−1f−dx, i.e., so that

∫

Γ

β+α+u+0n ds =

∫

Γ

w ds +

∫

Ω−

f−dx.

Observe then that u−

0 = C and also that in this case a total of three solves are, coincidentally, necessary. Similarissues arise in the bubble geometry, Fig. 1(b), see Section 6.

Two further remarks: first, note that the approximation (10) in fact improves as β increases. Also, the

individual problems for u±

0 and u−

1 are independent of β so that grid resolution is independent of β and thereis thus, importantly, no conditioning problem. These last properties are in agreement with [15], [26].

4

Observe that the required no-flux condition u−

0n = 0 in (11) entirely decouples u−

0 from the upper regionΩ+. With u−

0 thus determined, u+0 is consequently fixed on the interface Γ by the condition [u0] = v, in turn

effectively decoupling u+0 and thus allowing its determination. Then in turn u−

1 is fixed by flux balance withu+

0 as explained above. It is exactly this decoupling sequence which makes the scheme effective. The origin ofthe LJA decoupling can thus be explained as follows. Regarding (5) as a transport problem, we observe firstthat the highly conductive region Ω− carries flux much more efficiently than the Ω+ region. Hence any attemptto calculate the interface flux using data from Ω− is likely to overwhelm the O(1) flux-carrying capacity of theΩ+ region if errors are present. Conversely, the nearly perfect conductance of the Ω− region requires that uthere be constant plus a term of size O(β−1), i.e., very nearly uniform. Such uniformity is not the case in Ω+,indeed, a significant drop in u is necessary to force flux through this relatively poorly conducting region. Hence,any error in calculation of u at the interface using data from Ω+ may overwhelm calculations in Ω−. Together,these two observations suggests that Ω− data be used to calculate the interface solution and that Ω+ data beused to calculate interface flux.

We briefly consider the approximation error made in (10). Let u(x; β−1) be the exact solution to (1) with

conditions (4). (10) can be viewed as a Taylor approximation in β−1 to u, see Section 3 and Appendix A. Thus

the error u−u is O(β−1) even without the correction term β−1u−

1 in (10). However, in many and perhaps mostproblems of the sort considered here, the flux (6) is also important (e.g. if u is a potential function). Inclusion

of β−1u−

1 is then necessary so that the flux error β(x)∇u − β(x)∇u is O(β−1) as well. A quick examination

of the interface condition in problem (11) illustrates further why the O(β−1) term u−

1 is required: the β = ∞“perfect conductor” solution to (11) cannot allow any flux into the finitely conducting region, i.e., u−

0n = 0 on

Γ and hence, though β−1u−

1 (x) (in (10)) would at first glance appear to be negligible, it in fact provides thedominant contribution to the flux coupling across the interface. That is, without including u−

1 , the error in flux

in the lower region is O(1). An interesting consequence of this observation is that the flux from the β → ∞

limiting solution does not agree with the β = ∞ solution.For illustrative purposes we consider two examples, the first a simple 1D case. Let Ω = [0, 1] with interface

Γ = r where r ∈ (0, 1). Suppose f = 0, v = w = 0, and α = 1, and that u(0) = 0, u(1) = 1. Then the exactsolution to (5) is given by

u+(y) = 1 +1

1 − r

(

1 +1

β

r

1 − r

)−1

(y − 1), y ∈ (r, 1]

u−(y) =1

β

1

1 − r

(

1 +1

β

r

1 − r

)−1

y, y ∈ [0, r)

(14)

Alternatively, use of the LJA approximation (10) results in the reduction of problems (11)-(13) to

u−

0,yy = 0, u−

0,y(r) = 0, u−

0 (0) = 0,

for y ∈ [0, r) (with solution u−

0 (y) = 0) followed by

u+0,yy = 0, u+

0 (r) = u−

0 (r) = 0, u+0 (1) = 1, (15)

for y ∈ (r, 1] (with solution u+0 (y) = 1 + (y − 1)/(1 − r)) followed by

u−

1,yy = 0, u−

1,y(r) = u+0,y(r) =

1

1 − r, u−

1 (0) = 0, (16)

for y ∈ [0, r) (with solution u−

1 (y) = y/(1 − r)). We obtain the assembled global approximation

u(y; β) =

u+0 , y ∈ (r, 1]

u−

0 +1

βu−

1 , y ∈ [0, r) =

1 +1

1 − r(y − 1), y ∈ (r, 1]

1

β

1

1 − ry, y ∈ [0, r)

(17)

5

which satisfies the stated problem up to an error in the interface continuity condition [u] = −β−1(r/(1 − r)) =

O(β−1). As predicted, the term β−1u−

1 , while small, provides the O(1) flux contribution β∇u = β∇(β−1u−

1 ) =

(1 − r)−1 in [0, r). Note (17) is a low order Taylor approximation in β−1 to the exact solution (14).We make a few remarks here about numerical issues. First, (17) comprises the LJA for this example. To

compute the LJA numerically, one introduces a Poisson solver of some sort and then computes solutions to (15)and (16). Depending on choice of solver and discretization, numerical error would be of course introduced. Thefocus of this paper however is the global approximation (17); virtually no numerical error is introduce at this

step, only an O(β−1) truncation error (flux is also correct to an O(β−1) truncation error). That is, the LJAis really a framework for fitting together solutions computed using whatever solver the user favors. Second, akey point of stress is that problems (15) and (16) are independent of β and hence well-conditioned even for

large β. In contrast if we were to compute a solution for the example problem using, for example, the ghostfluid method [16], we would be forced to solve a matrix equation with O(β) condition number. This difficultybecomes troublesome especially in higher dimensions.

As a second example and to further illustrate these points, we consider a two dimensional test Poisson problemused in [1], [2]. The domain for this problem is Ω = [−1, 1]2 with Ω− = r < 1/2, Ω+ = r > 1/2

⋂

Ω (where

r =√

x2 + y2) and

β(x) =

β+ x ∈ Ω+

β−(r2 + 1) x ∈ Ω− .

The source term is f(x) = 8r2 +4 throughout Ω. We fix β− = 1 and vary β+. On the interface [u] = [βun] = 0.Dirichlet boundary condtions are applied on ∂Ω using the exact solution

u(x) =

1/4 + β(r4/2 + r2 − 9/32) x ∈ Ω+

r2 x ∈ Ω−, (18)

for this problem.The limit of interest here is β+ → 0, β → ∞. We start in this case from

β−1∇2u+ = (β−)−1f = 8r2 + 4, x ∈ Ω+

∇ · [(r2 + 1)∇u−] = (β−)−1f = 8r2 + 4, x ∈ Ω−

Note in Ω+u+ that we have ∇2u+ = β(8r2 + 4) and thus the source effectively is singular in the limit β → ∞.So we seek a solution of the form

u(x; β+) =

βu+−1 + u+

0 x ∈ Ω+

u−

0 x ∈ Ω−. (19)

This scaling differs from other problems in this paper in two important respects. First the singular source in Ω+

requires inclusion of a singular solution term of the form βu+−1. Second, the flux continuity condition for this

problem takes the form β+u+n = (5/4)u−

n and is not singular; hence a term of the form β−1u−

1 is not required.From expansion (19), we obtain a sequence of three problems, the first being

∇2u+−1 = 8r2 + 4 x ∈ Ω+,

u+−1 = 0 x ∈ Γ,

u+−1 = r4/2 + r2 − 9/32 x ∈ ∂Ω.

The solution of this problem is u+−1 = r4/2 + r2 − 9/32.

The second problem is then

∇ · [(r2 + 1)∇u−

0 ] = 8r2 + 4 x ∈ Ω−,u−

0n = (4/5)u+−1n x ∈ Γ,

with solution u−

0 = r2 + C. C = 0 is determined by the requirement that the flux into Ω− balances the sourceinside Ω−, see Section 6 for a discussion of related solvability issues.

6

The third and last problem is∇2u+

0 = 0 x ∈ Ω+,u+

0 = u−

0 x ∈ Γ,u+

0 = 1/4 x ∈ ∂Ω,

with solution u+0 = 1/4. We remark that although u+

0 is small relative to βu+−1, it is necessarily included to

avoid an O(1) error in the boundary conditions and in the interface condition [u] = 0.As it turns out for this particular problem, (19) is exact. Generally, for different data for example, such

would not be the case and we would expect an O(β−1) error in the interface condition [βun] = 0. We also notethat use of the scaling (19) effectively desingularizes the problem; that is the singular part of the problem isabsorbed into u+

−1.We make a few remarks on comparison to the multigrid solution methods in [1] and [2]. In fact, the method

in [1] fails at β+ = 5 ·10−4. [2] succeeds in finding a solution with slowly increaing work as β+ → 0. Of course adirect comparison to the methods used here is not really possible. We use a scaled expansion to turn the originaldiscontinuous problem into a series of smooth, well-behaved problems, each of which can be solved with whatevermethod is most convenient, a very problem-dependent criterion. In this example, explicit formulas are available.However, even if numerical solution was necessary, there are many easily implemented and effective methodsto solve smooth, β independent problems in the given geometry and produce very accurate approximations.Context of the problem would determine choice of solver.

3 Series Approximation

Results of the previous Section suggest generalization of the LJA framework to series of the form

uk(x; β) =

k∑

j=0

β−ju+j , x ∈ Ω+

k+1∑

j=0

β−ju−

j , x ∈ Ω−

(20)

in order to extend approximation of solutions of (5) to moderate values of β and also to connect LJA to Taylor

series expansion (see the Appendix). Substituting uk(x; β−1) into (5)-(8) we obtain, grouped by order in β−1,

[

∇ · (α+(x)∇u+0 ) −

1

β+f+(x)

]

+k

∑

j=1

β−j∇ · (α+(x)∇u+j ) = 0,

β∇ · (α−(x)∇u−

0 ) +

[

∇ · (α−(x)∇u−

1 ) −1

β+f−(x)

]

+

k+1∑

j=2

β−j∇ · (α−(x)∇u−

j ) = 0,

in regions Ω+, Ω− respectively together with interface conditions

(u+0 − u−

0 − v) +

k∑

j=1

β−j(u+j − u−

j ) = 0,

−βα−u−

0n + (α+u+0n − α−u−

1n − w) +

k+1∑

j=2

β−j(α+u+j−1,n − α−u−

jn) = 0.

The boundary conditions of the original problem (on ∂Ω) are applied to u±

0 . For j > 0, the u±

j all are subjectto homogeneous boundary conditions.

7

The partial decoupling seen previously extends to the series solution. We again obtain (11), (12), and (13)

for u±

0 and u−

1 . The remaining terms of (20) can be paired off in powers of β−1 resulting in complementaryDirichlet and Neumann (along Γ) problems for u+

j and u−

j+1, j ≥ 1:

∇ · (α+∇u+j ) = 0 x ∈ Ω+

u+j = u−

j x ∈ Γ

a1u+j + b1u

+jn = 0 z = 1

(21)

(note that u−

j is calculated prior to solution of (21)) followed by

∇ · (α−∇u−

j+1) = 0 x ∈ Ω−

α−u−

j+1,n = α+u+jn x ∈ Γ

a0u−

j+1 + b0u−

j+1,n = 0 z = 0

(22)

With each additional pair, the new approximation uk continues to satisfy exactly all conditions except continuityat the interface which is correct up to an O(β−(k+1)) error (in particular [u] − v = β−(k+1)u−

k+1 on Γ). Also,

as before, each problem is independent of β so that condition numbers and grid resolutions are independent ofβ. As a remark, we note that this iterative procedure is reminiscent of the Dirichlet-Neumann iterative domaindecomposition method presented in [8] for continuous β(x) problems.

(a)

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0.6

0.8

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U(X

,Z)

(b)

Figure 2: Solution to (1) with β piecewise constant in the limit β = ∞ using the LJA approximation (10). Highconducting region is the lower layer. f = v = w = 0 and u = 1 on the top boundary, u = 0 on the bottomboundary. Computations were carried out on a 256× 257 grid. (a) Limiting flux F(x, z), β → ∞. (b) Limiting

solution u(x, z), β → ∞.

For illustration we return to the 1-d example problem of the previous Section. In this example the higherorder paired problems (21)-(22) reduce to

u+j,yy = 0, u+

j (r) = u−

j (r), u+j (1) = 0,

for y ∈ (r, 1] andu−

j+1,yy = 0, u−

j+1,y(r) = u+j,y(r), u−

j+1(0) = 0,

for y ∈ [0, r) with solutions

u+j (y) =

1

1 − r

(

−r

1 − r

)j

(y − 1), y ∈ (r, 1]

u−

j+1(y) =1

1 − r

(

−r

1 − r

)j+1

y, y ∈ [0, r)

(23)

8

Assembling uk as defined in (20), we obtain a truncated Taylor series in β−1 of the exact solution (14). Note

convergence of uk, k → ∞, requires β to be sufficiently large depending on r. Such a restriction is a generalfeature – convergence of (20) occurs only for sufficiently large β depending on the geometry of the interface Γ.For example, geometries that produce locally strong flux, e.g. sharp bends in Γ, seem to require larger valuesof β for convergence. General convergence results are presented in the Appendix.

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0

0.2

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0.4

0.6

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,Z)

(c)

Figure 3: Solution to (1) with β piecewise constant, β = 15. High conducting region is the lower layer.f = v = w = 0 and u = 1 on the top boundary, u = 0 on the bottom boundary. Computations were carried outon a 256×257 grid. (a) LJA approximation (10), a poor approximation for β = 15. (b) Series approximation (20)

with k = 3, a much better approximation. (c) LJA approximation with β = 1000.

4 Two Dimensional Examples

In the computations to follow we employed finite differencing on a uniform rectangular grid using a level setfunction to represent the interface(s). Ghost nodes and Taylor series based interpolation are used along theinterface in a manner similar to [6]. All components of our solver were tested and seen to be approximately 2ndorder in the uniform norm for problems with piecewise constant coefficients.

4.1 Example

As an example we consider a two layer geometry (Fig. 2) with f = v = w = 0, u|z=0 = 0, u|z=1 = 1, andperiodic boundary conditions. The high conducting domain Ω− is the lower region. LJA approximation (10) to

problem (5) is computed. We then take the limit β → ∞. This limit, in the case of the quantity u, is

u =

u+0 (x) x ∈ Ω+

u−

0 (x) x ∈ Ω− (24)

where u−

0 (x) = 0 is the solution to (11) and u+0 (x) is the computed solution to (12), see Figure 2(b). However,

the flux field (6) has limit

F →

∇u+0 (x) x ∈ Ω+

∇u−

1 (x) x ∈ Ω− (25)

as β → ∞ (Fig. 2(a)). Note that the flux tends to a different limit from the β = ∞ perfect conductor flux field

F|β=∞

=

∇u+0 (x) x ∈ Ω+

0 x ∈ Ω−

9

1 2 3 4 5 6 7 8 9 1010

−25

10−20

10−15

10−10

10−5

100

Number of Terms

Max

imum

Am

plitu

de

151501500

(a)

101

102

103

104

105

106

107

10−7

10−6

10−5

10−4

10−3

10−2

10−1

100

log β

log

erro

r

(b)

Figure 4: (a) Infinity norm of the kth term of the solution to (1) for the example in Figure 3 for values β = 15,

β = 150, β = 1500. (b) Infinity norm of the error in the LJA approximation solution to (1) for the example in

Figure 3 as a function of β.

In fact, β → ∞ is a singular limit – conditions (7) and (8) are replaced by (7) at β = ∞ along with a no fluxinterface condition for the perfect conductor region.

We reiterate that two solves are performed to approximate u, one in the upper layer with Dirichlet conditionson the interface to find u+

0 and one the lower layer with Neumann conditions on the interface to find u−

1 .

Moreover, each solution is independent of β and hence does not suffer from large β conditioning problems.(24) and (25) are the zeroth order Taylor approximations to u and β∇u as β → ∞. For finite but sufficiently

large β, we can approximate u and β∇u by adding more terms in the Taylor series as described in Section 3.Figure 3 shows the computed solution to the same problem as in Figure 2 except at the finite value β = 15.Numerical computation of an increasing number of terms of (20) indicates convergence in this geometry for

β > β0 ∼ 5 approximately. Approximation (10) to (5) is shown in Figure 3(a). The O(β−1) error in thecontinuity matching condition is clearly visible. A four term approximation (k = 3 in (20)) is shown inFigure 3(b). In this case four solves were required in each of the two layers. The supremum of the next

terms in the series, max(sup(|β−4u+4 |), sup(|β−5u−

5 |)), is 4 · 10−3, approximately (β/β0)−5 ∼ 3−5 = .0041. We

plot maximum of the kth term for several different values of β in Figure 4(a). In comparison, we plot the

infinity norm error of the LJA approximation versus β in Figure 4(b). For the exact solution we use a five termapproximation.

Contrast the LJA approximation to the β = 15 solution to the LJA approximation to the β = 103 solutionwith k = 0 (Fig. 3(c)). In this case, again a single solve only is required in each region in order that the supremum

of the next terms in the series, max(sup(|β−1u+1 |), sup(|β−2u−

2 |)), is 0.0047, approximately (β/β0)−1 ∼ 200−1 =

.005.

5 The Asymptotic Diffusion Problem

We apply the LJA approximation (10) to the diffusion equation (2) under scaling as in (5). (Extension to higherorder as in Section 3 is straightforward and will not be repeated here.) The resulting equations, in order ofsolution, are

10

∇ · (α−∇u−

0 ) = 0 x ∈ Ω−

u−

0n = 0 x ∈ Γa0u

−

0 + b0u−

0n = c0 z = 0(26)

1β+ ∂tu

+0 −∇ · (α+∇u+

0 ) =1

β+f+ x ∈ Ω+

u+0 = u−

0 + v x ∈ Γa1u

+0 + b1u

+0n = c1 z = 1

u+0 (x, 0) = U0(x) x ∈ Ω+

(27)

∇ · (α−∇u−

1 ) =1

β+

(

−f− + ∂tu−

0

)

x ∈ Ω−

u−

1n =1

α−

(

α+u+0n −

1

β+w

)

x ∈ Γ

a0u−

1 + b0u−

1n = 0 z = 0

(28)

where U0(x) are the initial conditions in Ω+. Initial conditions in Ω− contribute either not at all or in a weak

form depending on the z = 0 boundary condition (see discussion below). We assume here that β = β−/β+ ismade large with β+ fixed, i.e., β− is large.

A few remarks: note u−

0 and u−

1 both satisfy Poisson equations (the same is still true for u−

j , j > 1) reflecting

the fact that in the large diffusivity Ω− region, the solution evolves quasistatically to first order in β−1. Hencetime scales faster than the Ω− diffusion time are not resolved – this is generally a desirable stiffness reducingfeature – time stepping occurs on the slow Ω+ diffusion time scale. Secondly, note that the Poisson problemfor u−

0 has constant solution with constant independent of time if a0 6= 0. (For the exceptional case a0 = 0, seediscussion below.) The Poisson problem for u−

1 must in principle be solved at each time step as the interfacecondition is time dependent. However, u−

1 decouples from the rest of the problem so in actuality (28) need onlybe solved at those times for which it is wanted. As a third remark, observe as before that each piece of theoverall solution is independent of β. Hence, again, grid resolution is independent of β and large β conditioningproblems are absent.

A consequence of the absence of a fast time scale is that the solution in Ω− is immediately smoothed. Forexample, given z = 0 boundary condition au + bun = c with a, b, c constants, the solution to (26) for t > 0 isu = c/a if a 6= 0. That is, because of perfect conductance in Ω− the solution is instantaneously smoothed to aconstant determined by the boundary conditions and independent of initial conditions. If a = 0, then the z = 0boundary condition must satisfy

∫

z=0 c/d dx = 0, i.e., c=0, so that un = 0. In this case, the solution to (26) isa constant whose value is determined by the initial conditions U0(x) in Ω−. In particular, for T > 0,

u−

0 |t=T =1

|Ω−|

[

∫

Ω−

U0 dx +

∫ T

0

∫

Γ

(

α+u+0n −

1

β+w

)

ds dt

]

(29)

where |Ω−| is the volume of Ω−, i.e., u−

0 evolves quasistatically as a constant equal to the average value ofthe initial conditions in Ω− plus accumulated flux into Ω− from Ω+. (29) is effectively a solvability conditionfor (28).

As an example we consider two versions of a 2D contamination problem, i.e., a problem with geometry asin Figure 1(a) in the limit β → ∞ with data u(1, t) = 1 and initial conditions u(x, 0) = 0, see Figure 5. Thebottom boundary condition is u(z = 0, t) = 0. In this case the solution to (26) is u−

0 = 0. Thus we solve (27)with interface condition u+

0 = 0. Computations were performed using the Crank-Nicolson method combined

11

with the spatial discretization scheme previously described. Solution of (28) is only performed at those timesfor which it is wanted – that is, u−

1 is not necessary for later (in t) computations so only need be computedfor those values of t at which the user wishes to report results. Flux of u into the bottom layer is immediatelycooled by the infinite “zero-temperature” bath below the z = 0 boundary.

(a)

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

10

0.2

0.4

0.6

0.8

1

XZ

U(X

,Z)

(b)

(c) 00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

10

0.2

0.4

0.6

0.8

1

XZ

U(X

,Z)

(d)

Figure 5: Solution to (2) with β piecewise constant in the limit β = ∞ using the LJA approximation (10). Highconducting region is the lower layer. f = v = w = 0 and u = 1 on the top boundary, u = 0 on the bottomboundary. Computations were carried out on a 256 × 257 grid. (a) Limiting flux F, β → ∞ at t = 0.01. (b)

Limiting solution u(x, z), β → ∞ at t = 0.01. (c) Limiting flux F, β → ∞ at t = 1.0. (d) Limiting solution

u(x, z), β → ∞ at t = 1.0.

In Figure 6 the bottom boundary condition is instead uz(z = 0, t) = 0, i.e., an insulating boundary. In thiscase the solution to (26) is given by (29) (with U0 = 0, w = 0). We solve (27) with the time dependent interfacecondition u+

0 = u−

0 . As for the previous example, computations were performed using the Crank-Nicolsonmethod combined with the spatial discretization scheme previously described. Flux of u into the bottom layeraccumulates until the “tank” is full. Again, solution of (28) is only performed at the times for which it is needed,which in this particular case are the four times for which the approximate solution is plotted.

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

10

0.2

0.4

0.6

0.8

1

XZ

U(X

,Z)

(a)

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

10

0.2

0.4

0.6

0.8

1

XZ

U(X

,Z)

(b)

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

10

0.2

0.4

0.6

0.8

1

XZ

U(X

,Z)

(c)

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

10

0.2

0.4

0.6

0.8

1

XZ

U(X

,Z)

(d)Figure 6: Solution to (2) with β piecewise constant in the limit β = ∞ using the LJA approximation (10). Highconducting region is the lower layer. f = v = w = 0 and u = 1 on the top boundary, u,z = 0 on the bottomboundary. Computations were carried out on a 256 × 257 grid. Figure shows the limiting solution u(x, z),

β → ∞ at (a) t = 0.01, (b) t = 0.1, (c) t = 0.2, (d) t = 1.0.

12

6 Bubble Geometries

Lastly we consider use of the LJA in the case of geometries with one or more included regions (e.g. Fig. 1(b)).There are two subcases: first, if the exterior region is the “minus” region with β− >> β+, then the LJAprocedure as already described can be applied essentially without alteration and will thus not be consideredfurther. If, however, the exterior region is the “plus” region with β+ << β− then a complication arises due tothe fact that in this case each interior problem is Neumann and hence the solutions on interior regions are onlydetermined up to a constant. That is, while u−

0 is constant in each bubble (generally a different constant frombubble to bubble), those constants are undetermined by the conditions of (11) and must in fact be calculatedas part of the solution of (12) in such a way as to satisfy the solvability conditions (for (13))

∫

Γj

β+α+u+0ndS =

∫

Ω−

j

f−dV +

∫

Γj

wdS (30)

where Ω−

j , j = 1, 2, . . . , n, is the jth bubble region of Ω and its boundary is denoted by Γj . In particular, we

wish to find u+0 such that u+

0 satisfies the given boundary conditions on ∂Ω and is constant on each componentof the interface Γ =

⋃

Γj while satisfying the n conditions (30). To accomplish this it is necessary to performn + 1 Poisson solves in the exterior region Ω+ in order to determine functions u+

00 and u+0j, j = 1, 2 . . . , n. The

problem to be solved for u+00 is

∇ · (α+∇u+00) =

1

β+f+ x ∈ Ω+

u+00 = v x ∈ Γ

a1u+00 + b1u

+00n = c1 x ∈ ∂Ω

The problems to be solved for u+0j, j = 1, 2 . . . , n are

∇ · (α+∇u+0j) = 0 x ∈ Ω+

u+0j = δij x ∈ Γi

a1u+0j + b1u

+0jn = 0 x ∈ ∂Ω

where δij is the Kronecker delta function. Note that these n + 1 solves are trivially parallelizable. Now, u+00

satisfies all solution requirements except that its interface fluxes

∫

Γj

β+α+u+0ndS,

j = 1, 2, . . . , n, through the interface components Γj are generally incompatible with the solvability condi-tions (30). We fix this difficulty by taking a linear combination u+

0 = u+00 +

∑

cju+0j with the n coefficients cj

chosen to satisfy the n conditions (30) and at the same time setting the constant solution components u−

0 |Ω−

j= cj .

Note that a similar process is required to satisfy solvability conditions for the O(β−1) solution components u−

1 ,u+

1 . If, however, only gradient information is required at this order then constants are unimportant and so thoseextra solves are unnecessary.

As an example, we solve for u in a two-bubble geometry (Fig. 7) with an applied gradient u|z=1 = 1,

u|z=0 = 0 and periodic boundaries on the sides. The β → ∞ solution is shown in Fig. 7(b) and the β → ∞ fluxfield is shown in Fig. 7(a). Three solutions of Dirichlet problems are required for the outer region as describedabove and one solution of a Neumann problem is required for the interior region. Note that despite the factthat u is nearly constant inside the bubbles for large β (and in fact exactly constant in the limit β → ∞) theflux through the bubbles actually appears dominant in the sense that “most” integral lines of the flux field passthrough one of the bubbles.

13

(a)

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

10

0.2

0.4

0.6

0.8

1

XZ

U(X

,Z)

(b)

Figure 7: Solution to (1) with β piecewise constant in the limit β = ∞ using the LJA approximation (10). Highconducting region is inside the two bubble. f = v = w = 0 and u = 1 on the top boundary, u = 0 on the bottomboundary. Computations were carried out on a 256× 257 grid. (a) Flux field F(x, y). (b) Solution u(x, y)

7 Summary and Conclusions

The perturbation framework described here presents three main virtues. First, it is efficient, requiring formoderate to large coefficient discontinuities a small number of equation solves. These solves are independentof discontinuity amplitude and hence well conditioned. Further, choice of solution method can be optimizedto the context of the particular problem. In the two layer problem, typically only one solve over all of Ω isnecessary for large discontinuities. (For diffusion equations, time integration is only necessary on one regionwith a single elliptic solve on the other region at the final time.) That is, one could almost say that the largediscontinuity problem is not inherently more difficult than the continuous one in this case. The most difficultinstance is that of a domain consisting of a number of high β inclusion regions. In this case extra Dirichlet solvesare required for each inclusion, though these solves are trivially parallelizable. The second virtue is that LJA issimple to implement – the user provides a favorite elliptic solver on irregular domains (in fact, different solversfor different regions can be used) along with a procedure to extract boundary data and then need only directthe code to solve the sequence of problems as presented here. Third, the method is independent of dimensionand extends to 3D without modification.

Underlying LJA is the presence of a singular limit. In the limit β → ∞, one of the interface matchingconditions (7) or (8) is lost, either (7) on the relatively “good” conducting side of the interface or (8) on therelatively “poor” conducting side. That is, the singular limit acts to partially decouple the two regions and (7)and (8) become boundary conditions rather than interface matching conditions. Hence, at least for sufficientlylarge discontinuity, iteration for the purpose of satisfying interface matching conditions is not necessary andthus we find ourselves in the position where a singular limit actually results in significant simplification.

8 Acknowledgements

This work was supported by the NIH award 5R01GM067245-02.

9 Appendix A: Real Analyticity of u With Respect to β−1

We provide a brief sketch of the argument that the solution to (5)-(8) has a convergent Taylor series in β−1 for

β−1 near 0. Consider first differentiability in β−1 of solution u(x; β−1) together with boundary conditions. We

14

assume all coefficients and data as well as interface Γ are smooth and independent of β−1. (For ease of notation

only we set α+ = α− = β+ = 1.) Define for β−1 > 0

U(x; β−1) =u(x; β−1) − u(x; 0)

β−1

(Here u(x; 0) is defined to be the solution of ∇2u = 0 in Ω− and Ω+ with [u] = v and u−n = 0 on the interface.)

Then U satisfies∇2U+ = 0, x ∈ Ω+

∇2U− = f−(x), x ∈ Ω− (31)

with interface conditionsU+ − U− = 0, x ∈ Γ

β−1U+n − U−

n = w − u+n (x; 0) x ∈ Γ

together with homogeneous boundary conditions. Since, by assumptions made above, u+n (x; 0) is smooth and

bounded, then the interface conditions for U are smooth and bounded and hence clearly U is continuous andbounded in β−1 near β−1 = 0 and U(x; β−1) → U(x; 0) as β−1 → 0. That is, u(x; β−1) is differentiable in β−1

at (and near) β−1 = 0 and u′(x; 0) = U(x; 0) where the prime refers to differentiation with respect to β−1.

The above argument can be repeated to find u′′ and in fact all order derivatives in β−1 of u at and nearβ−1 = 0. We must argue for uniformity to finish the real analyticity argument. The issue depends on theSteklov-Poincare operator, defined in this context as the operator which, given Dirichlet data on the interfaceΓ, solves the Lapace problem in the domain of interest and returns the resulting Neumann data on Γ. Theinverse operator in turn returns Dirichlet data given Neumann data. The Steklov-Poincare operator is linear,continuous, and coercive [21] and hence it follows that the derivatives of u with respect to β−1 grow at mostalgebraically. Real analyticity follows.

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