ON A CHARACTERIZATION OF THE KERNEL OF THE

DIRICHLET-TO-NEUMANN MAP FOR A PLANAR REGION

DAVID INGERMAN AND JAMES A. MORROW

Abstract. We will show that the Dirichlet-to-Neumann map Λ for the electrical conductivity equation on a simply connected plane region has an alternating property, which may be considered as a generalized maximum principle. Using this property, we will prove that the kernel, K, of Λ satisfies a set of inequalities of the form (−1)^{\frac{n(n+1)}{2}} det K(x_{i}, y_{j}) > 0. We will show that these inequalities imply Hopf’s lemma for the conductivity equation. We will also show that these inequalities imply the alternating property of a kernel.

1. Introduction

In this paper we will derive some properties of the Dirichlet-to-Neumann map for the electrical conductivity equation in R^{2}. These properties are analogs of properties which characterize the Dirichlet-to-Neumann maps for electrical networks (see [1] and [2]). We recall some definitions. Let Ω be a relatively compact, simply connected open set in R^{2} with C^{2} boundary. Let γ(p) > 0 be a C^{2} function on \overline{Ω}. Let f be a function defined on ∂Ω. Then there is a unique function u, defined on \overline{Ω}, such that

∇(γ∇u) = 0 (1.1)

and u(p) = f(p), for p ∈ ∂Ω. (Equation (1.1) is the electrical conductivity equation and a function, u, that satisfies (1.1) is called a \underline{γ-harmonic function}.) Let \frac{∂u}{∂n}(p) be the directional derivative of u in the direction of the outward pointing unit normal n at the point p ∈ ∂Ω. Then the Dirichlet-to-Neumann map, Λ, is defined by the formula

Λf(p) = γ(p)\frac{∂u}{∂n}(p). (1.2)

The domain of Λ may be taken to be H^{\frac{1}{2}} (∂Ω) and the image is in H^{− \frac{1}{2}} (∂Ω). Λ is a pseudo-differential operator of order 1 and as such has a kernel, K(x, y), defined as a distribution on ∂Ω × ∂Ω. The kernel gives a representation of Λ by the formula

Λf(x) = ∫_{∂Ω} K(x, y)f(y)dy, (1.3) where x and y are arc length coordinates on ∂Ω. For the pseudo-differential operator Λ, the kernel K is a symmetric function, K(x, y) = K(y, x), and for a fixed x ∈ ∂Ω, lim_{y→x} |K(x, y)| = ∞. More precisely,

1991 Mathematics Subject Classification. Primary 94C15 .

Key words and phrases. conductivity, Dirichlet-to-Neuman, kernel.

K(x, y) = \frac{k(x, y)}{|x − y|^{2}} + D(x, y), (1.4)

where k is continuous on ∂Ω × ∂Ω, k(x, y) = k(y, x), k(x, x) ≠ 0, and D is a distribution supported on Δ = {(x, x) : x ∈ ∂Ω}. (In this formula, |x − y| is the separation in arc length of points with arc length coordinates x and y and the continuous term in this expansion has been incorportated into the term \frac{k(x,y)}{|x−y|^{2}} .) If x ∉ supp(f), then the integral is an ordinary integral and there are no convergence questions. Since we will be interested in the behaviour of K(x, y) for x ≠ y we will ignore D and will pretend that K(x, y) = \frac{k(x,y)}{|x−y|^{2}} . The expansion (1.4) follows from Lemma 3.7 of [6] or Theorem 0.1 in [7]. The boundary, ∂Ω, is a Jordan curve and hence is homeomorphic to a circle. Pick an orientation on ∂Ω. We say that (x_{1}, …, x_{n}; y_{1}, ⋯, y_{n}) is a \underline{circular pair} if there are points p, q ∈ ∂Ω which divide ∂Ω into two connected components, A, B such that {x_{1}, …, x_{n}} ⊂ A, {y_{1}, …, y_{n}} ⊂ B, and x_{1}, …, x_{n}, y_{1}, …, y_{n} are in circular order on ∂Ω. (Note that this definition is modified from the definition in [2].) The main theorem of this paper is the following theorem, which we prove to be equivalent to the alternating property stated in section 2.

Theorem 1.1. Let (x_{1}, …, x_{n}; y_{1}, ⋯, y_{n}) be a circular pair on ∂Ω. Let L = (l_{ij}) be the n × n matrix with entries defined by l_{ij} = K(x_{i}, y_{j}). Then

(−1)^{\frac{n(n+1)}{2}} det(L) > 0 (1.5)

We consider this to be a generalization of a result in [2]. We will see how it implies the classical Hopf lemma for the conductivity equation in dimension 2.

2. The Alternating Property

We first restate and prove a result of [2]. Suppose that ∂Ω = I ∪ J, where I and J are disjoint connected arcs. Then we have the following theorem.

Theorem 2.1. Let f be a smooth function on ∂Ω such that f = 0 on I. Suppose there is a sequence of points {p_{1}, …, p_{n}} ⊂ I in circular order such that

(−1)^{i+1}Λf(p_{i}) > 0. (2.1)

Then there is a sequence of points {q_{1}, …, q_{n}} ⊂ J in circular order such that

(−1)^{n}Λf(p_{i})f(q_{i}) > 0. (2.2)

Proof. Equation (2.2) is equivalent to

Λf(p_{i})f(q_{n+1−i}) < 0. (2.3)

We first describe how to pick the point q_{n}. Let u be the solution of (1.1) such that u = f on ∂Ω. By (2.1) \frac{∂u}{∂n}(p_{1}) > 0. Hence there is a small open line segment, α, such that α ⊂ Ω, p_{1} is one end of α and u < 0 on α. Let W be the connected component of {z ∈ Ω : u(z) < 0} that contains α. Suppose that \overline{W} ∩ J = ∅. Then u = 0 on ∂W. But this contradicts the maximum principle since u < 0 in W and W ≠ ∅. Thus \overline{W} ∩ J ≠ ∅. Now u = 0 at every point of ∂W that is in Ω. Using the maximum principle again we see that there is a q_{n} ∈ \overline{W} ∩ J such that f(q_{n}) < 0 and there is an open line segment β ⊂ W such that q_{n} is an end point of β. Now we can connect the ends of α and β that are inside W by a smooth curve in W. Hence there is a smooth curve C_{1} such that C_{1} is diffeomorphic to a line segment, has end points points p_{1} and q_{n}, and C_{1} − p_{1} − q_{n} ⊂ W. Then u(z) < 0 for all z ∈ C_{1} − p_{1}. We can repeat this argument to produce curves C_{j} such that C_{j} joins p_{j} to a point q_{n+1−j} ∈ J, C_{j} − p_{j} − q_{n+1−j} ⊂ Ω, and (−1)^{j}u(z) < 0 for all z ∈ C_{j} − p_{j}. These curves cannot intersect and by the Jordan curve theorem the points p_{1}, …, p_{n}, q_{1}, …, q_{n} must be in circular order on ∂Ω. It is easy to see that these points satisfy (2.3).

We have referred to this property as the alternating property. Elsewhere ([5]) a similar property has been called the variation diminishing property. See also section 6 of this paper.

3. The Weak Inequality

We first prove the weaker statement:

Theorem 3.1. Let (x_{1}, …, x_{n}; y_{1}, ⋯, y_{n}) be a circular pair on ∂Ω. Let L = (l_{ij}) be the n × n matrix with entries defined by l_{ij} = K(x_{i}, y_{j}). Then

(−1)^{\frac{n(n+1)}{2}} det(L) ≥ 0 (3.1)

Proof. The proof is by induction on n. We first consider n = 1. The proof goes by contradiction. Suppose that there are points p, q ∈ ∂Ω with p ≠ q and K(p, q) > 0. Then there is an ϵ > 0 such that p ∉ D_{ϵ} = {y : |y − q| < ϵ} and K(p, y) > 0 for y ∈ D_{ϵ}. Let f(y) be a continuous function on ∂Ω such that supp(f) ⊂ D_{ϵ} = {y : |y −q| < ϵ}, f(q) > 0, and f(s) ≥ 0 for all s ∈ ∂Ω. Then

γ(p)\frac{∂u}{∂n}(p) = Λf(p) = ∫_{D_{ϵ}} K(p, y)f(y)dy > 0,

where u satisfies (1.1) and u(s) = f(s), s ∈ ∂Ω But then there must be a point z near p in Ω such that u(z) < 0. This contradicts the maximum principle.

4 D. INGERMAN AND J. MORROW

Next we assume that the result is true for all (n − 1) × (n − 1) matrices and prove that it is true for n × n matrices. If the result is not true, then we have a circular pair (x_{1}, …, x_{n}; y_{1}, ⋯, y_{n}) such that (−1)^{\frac{n(n+1)}{2}} det(L) < 0. (3.2) Consider the matrix L^{−1} with entries (h_{ij}). Then

h_{ij} = (−1)^{i+j} \frac{det(L_{ij})}{det(L)}, (3.3)

where L_{ij} is the (i, j) minor of L. By induction, (3.2), and (3.3) (−1)^{i+j+ \frac{n(n−1)}{2} + \frac{n(n+1)}{2} +1}h_{ij} = (−1)^{i+j+n+1}h_{ij} ≥ 0. (3.4) Since L is nonsingular, for fixed i there must be some j for which (−1)^{i+j+n+1}h_{ij} > 0. (3.5) Now let w = [1, −1, 1, …, (−1)^{n+1}]^{T} be an n-vector with alternating signs. Let z = L^{−1}w. Then using (3.4) and (3.5) it is easy to verify that (−1)^{i+n}z_{i} > 0. (3.6) To summarize, we have a vector z such that

(−1)^{i+1} = w_{i} = ∑_{j=1}^{n} K(x_{i}, y_{j})z_{j} (3.7)

and (−1)^{n+1}z_{i}w_{i} > 0. (3.8) Now, choose small intervals D_{j} around the points y_{j} such that the D_{j} are disjoint and do not contain any of the points x_{i}. Choose the D_{j} so small that |K(x_{i}, y) − K(x_{i}, y_{j})| < ϵ, y ∈ D_{j}, i = 1, …, n. (3.9) Also choose functions f_{j} such that

supp(f_{j}) ⊂ D_{j}, z_{j}f_{j}(y) ≥ 0, and ∫_{D_{j}} f_{j} = z_{j}. (3.10)

Let f = ∑ f_{j}. Then

\begin{aligned}|Λf(x_{i}) − w_{i}| & = | ∫_{∂Ω} K(x_{i}, y)f(y)dy − ∑_{j=1}^{n} K(x_{i}, y_{j})z_{j}| \\ & = | ∫_{∂Ω}(K(x_{i}, y) − K(x_{i}, y_{j}))f(y)dy| \\ & ≤ ϵ ∑_{j=1}^{n} |z_{i}|. \end{aligned} (3.11)

Thus we conclude that for ϵ small enough Λf(x_{i}) has the same sign as w_{i}. By the alternating property, there would have to be a set of n points t_{i} in circular order such that

(−1)^{n}w_{i}f(t_{i}) > 0. (3.12)

For such a set of points we would have to have t_{i} ∈ D_{i} and hence f(t_{i}) would have the same signs as z_{i}. This contradicts (3.8).

4. The Strong Inequality

We now prove Theorem (1.1). We consider the cases n = 1 and n > 1 separately. Let us assume the arc length of ∂Ω is S and that points on ∂Ω are parametrized by the numbers in the interval [0, S). When n = 1, suppose there is a pair of points x_{1}, y_{1} with 0 ≤ x_{1} < y_{1} and K(x_{1}, y_{1}) = 0. By (1.4) there is no sequence of points z_{j} such that x_{1} < z_{j} < y_{1}, lim_{j→∞} z_{j} = x_{1}, and lim_{j→∞} K(x_{1}, z_{j}) = 0. Hence there is a point η_{2} with x_{1} < η_{2} < y_{1} such that

K(x_{1}, η_{2}) = 0, and K(x_{1}, η) < 0, for x_{1} < η < η_{2}.

Let x be any number such that x_{1} < x < η_{2} and choose η_{1} so that x < η_{1} < η_{2}. Then (x_{1}, x; η_{1}, η_{2}) is a circular pair and hence

|\begin{matrix}K(x_{1}, η_{1}) & K(x_{1}, η_{2}) \\ K(x, η_{1}) & K(x, η_{2}) \end{matrix}| ≤ 0. (4.1)

Since

K(x_{1}, η_{2}) = 0, K(x, η_{2}) ≤ 0, and K(x_{1}, η_{1}) < 0,

it follows that

K(x, η_{2}) = 0. (4.2)

This shows that for all x, with x_{1} < x < η_{2}, K(x, η_{2}) = 0. Hence we get the contradiction that lim_{x→η_{2}} K(x, η_{2}) = 0.

The proof for n > 1, makes use of the following result in [2]. It was later pointed out to us that Charles Dodgson (Lewis Carroll) used a version of this identity in [4]. Let (x_{1}, …, x_{n}; y_{1}, …, y_{n}) be a circular pair. We assume that the coordinates on ∂Ω are chosen so that 0 ≤ x_{1} < ⋯ < x_{n} < y_{1} < ⋯ < y_{n} < S. Let L be the matrix with i, j entry equal to K(x_{i}, y_{j}). We will use the notation

κ(x_{1}, …, x_{n}; y_{1}, …, y_{n}) = det(L). (4.3)

Lemma 4.1. Let (a_{1}, …, a_{n+1}; b_{1}, …, b_{n+1}) be a circular pair. Then

(4.4) κ(a_{1}, …, a_{n+1}; b_{1}, …, b_{n+1})κ(a_{1}, …, a_{n−1}; b_{3}, …, b_{n+1}) =

κ(a_{1}, …, a_{n}; b_{1}, b_{3}, …, b_{n+1})κ(a_{1}, …, a_{n−1}, a_{n+1}; b_{2}, …, b_{n+1}) − κ(a_{1}, …, a_{n}; b_{2}, …, b_{n+1})κ(a_{1}, …, a_{n−1}, a_{n+1}; b_{1}, b_{3}, …, b_{n+1})

Assume that

κ(x_{1}, …, x_{n}; y_{1}, …, y_{n}) = 0 (4.5)

for some circular pair. First we claim that there is no sequence of points z_{j} such that x_{n} < z_{j} < y_{1}, lim_{j→∞} z_{j} = x_{n}, and lim_{j→∞} κ(x_{1}, …, x_{n}; z_{j}, y_{2}, …, y_{n}) = 0. For this would imply that there are constants c_{k} (independent of j) so that

K(x_{n}, z_{j}) = ∑_{k<n} c_{k}K(x_{k}, z_{j}), (4.6)

and hence

lim_{j→∞} K(x_{n}, z_{j}) = ∑_{k<n} c_{k}K(x_{k}, x_{n}), (4.7)

contradicting (1.4). Thus there is a number η_{1} with x_{n} < η_{1} < y_{1} such that

κ(x_{1}, …, x_{n}; η_{1}, y_{2}, …, y_{n}) = 0 and (4.8)

κ(x_{1}, …, x_{n}; η, y_{2}, …, y_{n}) ≠ 0, for x_{n} < η < η_{1}. (4.9)

Let x be such that x_{n} < x < η_{1}. Then there is an η such that x < η < η_{1} and hence (x_{1}, …, x_{n}, x; η, η_{1}, y_{2}, …, y_{n}) is a circular pair. By (4.4), (4.5), and (3.1)

(4.10) 0 ≥ κ(x_{1}, …, x_{n}, x; η, η_{1}, y_{2}, …, y_{n})κ(x_{1}, …, x_{n−1}; y_{2}, …, y_{n}) = κ(x_{1}, …, x_{n}; η, y_{2}, …, y_{n})κ(x_{1}, …, x_{n−1}, x; η_{1}, y_{2}, …, y_{n}) − κ(x_{1}, …, x_{n}; η_{1}, y_{2}, …, y_{n})κ(x_{1}, …, x_{n−1}, x; η, y_{2}, …, y_{n}) = κ(x_{1}, …, x_{n}; η, y_{2}, …, y_{n})κ(x_{1}, …, x_{n−1}, x; η_{1}, y_{2}, …, y_{n}) ≥ 0.

Using this and (4.9) we see that

κ(x_{1}, …, x_{n−1}, x; η_{1}, y_{2}, …, y_{n}) = 0, for x_{n} < x < η_{1}. (4.11)

As above, this contradicts (1.4) and proves the theorem.

5. The Hopf Lemma

We now show how the fact that K(x, y) < 0 for x ≠ y implies the Hopf lemma (reference) for the conductivity equation.

Theorem 5.1. Let u be a non constant solution of ∇(γ∇u) = 0, and let p ∈ ∂Ω be a point where u assumes a minimum. Then

\frac{∂u}{∂n}(p) < 0 (5.1)

Proof. We may assume that u(p) = 0. Let f = u|_{∂Ω}. Since u is not constant, supp(f) is not empty. Thus there is an interval D around p in ∂Ω such that supp(f) − D is not empty. Let ψ be a smooth function on ∂Ω such that ψ = 1 on supp(f) − D, ψ = 0 on an interval around p, and 0 ≤ ψ ≤ 1. Let g = ψf and let v be the solution of ∇(γ∇v) = 0 with v|_{∂Ω} = g. Since f ≥ g it follows that u ≥ v. It is also true that g ≥ 0. Since p ∉ supp(g) and K(p, y) < 0,

0 > ∫_{∂Ω} K(p, y)g(y)dy = γ(p)\frac{∂v}{∂n}(p) ≥ γ(p)\frac{∂u}{∂n}(p), (5.2)

which proves the theorem.

6. The Variation Diminishing Property

We will use the following notation. Let M(x, y) be a continuous function on [c, d]×[a, b]. Let c ≤ x_{1} < x_{2} < ⋯ < x_{n} ≤ d, a ≤ y_{1} < y_{2} < ⋯ < y_{n} ≤ b. Let T be the n × n matrix with i, j entry equal to M(x_{i}, y_{j}). Let

μ(x_{1}, x_{2}, …, x_{n}; y_{1}, y_{2}, …, y_{n}) = det(T).

The following lemma from [5] is sometimes paraphrased by saying that the kernel M has the variation diminishing property. It will be used to show that the inequalities (1.5) imply the alternating property.

Lemma 6.1. Let f be a continuous, not identically 0, function defined on the interval [a, b], such that f changes its sign on this interval no more than n − 1 times. Let M(x, y), x, y ∈ [c, d] × [a, b], be a continuous kernel with the property that

μ(x_{1}, x_{2}, …, x_{n}; y_{1}, y_{2}, …, y_{n}) > 0, (6.1)

whenever c ≤ x_{1} < x_{2} < ⋯ < x_{n} ≤ d, a ≤ y_{1} < y_{2} < ⋯ < y_{n} ≤ b. Then the function

g(x) = ∫_{a}^{b} M(x, y)f(y)dy

vanishes in [c, d] no more than n − 1 times.

By saying that function f changes its sign k times on the interval [a, b] we mean that there are k + 1 points x_{1} < x_{2} < ⋯ < x_{k+1} in [a, b] such that for i = 1, 2, …, k

f(x_{i})f(x_{i+1}) < 0. (6.2)

Proof. By hypothesis there are points a = s_{0} < s_{1} < s_{2} < ⋯ < s_{n−1} < s_{n} = b such that in each interval (s_{i−1}, s_{i}), i = 1, 2, …, n function f does not change its sign and is not identically 0. For i = 1, 2, …, n let

g_{i}(x) = ∫_{s_{i−1}}^{s_{i}} M(x, y)f(y)dy. (6.3)

Then

g(x) = ∑_{i=1}^{n} g_{i}(x). (6.4)

For any c ≤ x_{1} < x_{2} < ⋯ < x_{n} ≤ d the determinant

det({g_{i}(x_{j}}) = ∫_{s_{n−1}}^{s_{n}}⋯ ∫_{s_{0}}^{s_{1}} μ(x_{1}, x_{2}, …, x_{n}; y_{1}, y_{2}, …, y_{n})f(y_{1}) … f(y_{n})dy_{1} … dy_{n} (6.5)

is not 0 since the integrand is not identically zero and has constant sign. This shows that there is no non-trivial linear combination of g_{i}’s vanishing at n points and hence that g(x) = ∑_{i=1}^{n} g_{i}(x) cannot vanish at n points.

We note that this proof only used the fact that μ(x_{1}, x_{2}, …, x_{n}; y_{1}, y_{2}, …, y_{n}) has constant sign. We need one more lemma before coming to the proof of the alternating principal.

Let K(x, y) be a kernel on ∂Ω×∂Ω. We assume that K(x, y) is continuous when x ≠ y, but we don’t assume anything about K on the diagonal of ∂Ω × ∂Ω. Let κ(x_{1}, …, x_{n}; y_{1}, …, y_{n}) be defined as in section 4.

Lemma 6.2. Suppose that κ(x_{1}, …, x_{n}; y_{1}, …, y_{n}) is never zero and has constant sign for all circular n-pairs (x_{1}, …, x_{n}; y_{1}, …, y_{n}). Let ∂Ω = I ∪ J where I and J are disjoint connected arcs. Let f be a continuous function on ∂Ω with supp(f) ⊂ J. Let

g(x) = ∫_{∂Ω} K(x, y)f(y)dy. (6.6)

Then if there is a sequence of n + 1 points in I in circular order at which g alternates in sign, then there is a sequence of at least n + 1 points in J in circular order at which f alternates in sign.

Proof. If there is no sequence of n + 1 points of J at which f alternates in sign, then f can change its sign no more than n − 1 times in J. By Lemma 6.1, g can vanish no more than n − 1 times in I. But we are assuming that g has n + 1 alternations of sign in I and hence at least n zeros in I. This contradiction proves the lemma.

We now state and prove the theorem.

Theorem 6.1. Using the notation of Lemma 6.2, suppose that

(−1)^{\frac{n(n+1)}{2}} κ(x_{1}, …, x_{n}; y_{1}, …, y_{n}) > 0 (6.7)

for all n > 0 and all circular n-pairs (x_{1}, …, x_{n}; y_{1}, …, y_{n}). Let f be a continuous function on ∂Ω with supp(f) ⊂ J. Let

g(x) = ∫_{∂Ω} K(x, y)f(y)dy. (6.8)

Suppose there is a sequence of points {p_{1}, …, p_{n}} ⊂ I in circular order such that

(−1)^{i+1}g(p_{i}) > 0 (6.9)

Then there is a sequence of points {q_{1}, …, q_{n}} ⊂ J in circular order such that

(−1)^{n}g(p_{i})f(q_{i}) > 0. (6.10)

Proof. By Lemma 6.2 there is a sequence of points in J at which f alternates in sign. If there is no sequence with the desired alteration property then J is a disjoint union of subintervals J_{i}, in circular order, such that

1. f is not identically 0 on J_{i}, i = 1, …, n,

2. f does not change its sign on J_{i}, i = 1, …, n

3. for some z_{i} ∈ J_{i},

(−1)^{n+i}f(z_{i}) > 0. (6.11)

We use the idea of Lemma 6.1. For i = 1, 2, …, n let

g_{i}(x) = ∫_{J_{i}} K(x, y)f(y)dy. (6.12)

Then

g(x) = ∑_{i=1}^{n} g_{i}(x). (6.13)

Let

G = [\begin{matrix}g_{1}(x_{1}) & g_{2}(x_{1}) & … & g_{n}(x_{1}) \\ g_{1}(x_{2}) & … & & g_{n}(x_{2}) \\ ⋮ & & & ⋮ \\ g_{1}(x_{n}) & … & & g_{n}(x_{n}) \end{matrix}] . (6.14)

Let u be the n-vector with u_{i} = 1, i = 1, …, n. Then

Gu = [g(x_{1}) ; g(x_{2}) ; ⋮ ; g(x_{n})] . (6.15)

Using (6.11) we will show that the signs of u are all negative. This contradiction will prove the theorem. We need to compute the signs of the entries of G^{−1}. Rather than get lost in a cloud of indices, we will give the proof in the case that n = 3 and leave the general proof to the reader. In this case the assumption (6.11) implies that f(y) ≥ 0 in J_{1}, f(y) ≤ 0 in J_{2}, and f(y) ≥ 0 in J_{3}. As in section 3 we compute the signs of the cofactors of G. First we have

det(G) = ∫_{J_{1}} ∫_{J_{2}} ∫_{J_{3}} κ(x_{1}, x_{2}, x_{3}; y_{1}, y_{2}, y_{3})f(y_{1})f(y_{2})f(y_{3})dy_{1}dy_{2}dy_{3} < 0. (6.16)

We find that

|\begin{matrix}g_{2}(x_{2}) & g_{3}(x_{2}) \\ g_{2}(x_{3}) & g_{3}(x_{3}) \end{matrix}| = ∫_{J_{2}} ∫_{J_{3}} κ(x_{2}, x_{3}; y_{2}, y_{3})f(y_{2})f(y_{3})dy_{2}dy_{3} > 0. (6.17)

Hence (G^{−1})_{11} < 0. Next we compute that

(−1)^{1+2} |\begin{matrix}g_{1}(x_{2}) & g_{3}(x_{2}) \\ g_{1}(x_{3}) & g_{3}(x_{3}) \end{matrix}| = ∫_{J_{1}} ∫_{J_{3}} κ(x_{2}, x_{3}; y_{1}, y_{3})f(y_{1})f(y_{3})dy_{1}dy_{3} > 0, (6.18)

and thus (G^{−1})_{21} < 0. Continuing the calculation we find that the signs of G^{−1} are as follows

G^{−1} = [\begin{matrix}− & + & − \\ − & + & − \\ − & + & − \end{matrix}] . (6.19)

This yields the contradiction

[\begin{matrix}− & + & − \\ − & + & − \\ − & + & − \end{matrix}] [\begin{matrix}+ \\ − \\ + \end{matrix}] = [\begin{matrix}1 \\ 1 \\ 1 \end{matrix}] . (6.20)

7. Remarks and Conjectures

We have not tried to state the most general hypotheses under which our results are valid, but have stated them in such a way that the essential ideas of the proofs are clear. We can also prove determinant inequalities for for certain “blocks” in Dirichlet-to-Neumann kernels for multiply connected plane domains. We can differentiate our inequalities to get a set of inequalities involving determinants of derivatives of the Dirichlet-to-Neumann kernel. These inequalities are equivalent to the set of inequalities (1.5). In our arguments we seem to need to assume that γ is in C^{2}(Ω), however it is possible that weaker assumptions would suffice.

We would like to single out the following conjecture on characterizing the kernel of a Dirichlet-to-Neumann map.

Conjecture 7.1. Let Ω be a relatively compact, simply connected region in the plane with C^{2} boundary. Let K(x, y) = \frac{k(x,y)}{|x−y|^{2}} , where (x, y) ∈ ∂Ω × ∂Ω − Δ, k is continuous on ∂Ω × ∂Ω, k(x, x) ≠ 0, and K satisfies (1.5). Then there is a distribution D(x, y) on ∂Ω × ∂Ω, supported on the diagonal, Δ, and a regularization of K as a distribution on ∂Ω × ∂Ω, so that L = K + D is the kernel of the Dirichlet-to-Neumann map for some conductivity, γ, on Ω. The distribution D is determined by the property that

∫_{∂Ω} L(x, y)dy = 0. (7.1)

Equation (7.1) is analogous to the fact the the Dirichlet-to-Neumann matrix for an electrical network has row sums equal to zero. This implies that the diagonal is determined by the off-diagonal terms. This is true as well in the continuous case.

We have discussed these results with many people and their suggestions and advice have been of great benefit. Among these people are Ed Curtis, John Sylvester, and Gunther Uhlmann. The importance of the alternating property was recognized some time ago by Ed Curtis.

References

[1] Y. Colin de Verdiere, I. Gitler, and D. Vertigan, Réseaux électriques planaire. II, Comment. Math. Helv. 71 (1996), 144–167.

[2] E. B. Curtis, D. Ingerman and J. Morrow, Circular planar graphs and resistor networks, submitted.

[3] E. B. Curtis, E. Mooers and J. Morrow, Finding the conductors in circular networks from boundary measurements, Math. Modelling and Numerical Mathematical Modelling and Numerical Analysis 28 (1994), 781-813.

[4] C. L. Dodgson, Condensation of Determinants, Royal Society of London, Proceedings 15 (1866), 150-155.

[5] F. R. Gantmakher and M. G. Krein, Oscillation Matrices and Kernels and Small Vibrations of Mechanical Systems, Moscow, 1950.

[6] J. Sylvester, An Anisotropic Inverse Boundary Value Problem, Communications on Pure and Applied Mathematics XLIII (1990), 201-232.

[7] J.Sylvester and G. Uhlmann, Inverse Boundary Value Problems at the Boundary – Continuous Dependence, Communications on Pure and Applied Mathematics XLI (1988), 197-219.

[8] M. Tsuji Potential Theory in Modern Function Theory, Chelsea, 1975.

Department of Mathematics, University of Washington

E-mail address: ingerman@math.washington.edu

Department of Mathematics, University of Washington

E-mail address: morrow@math.washington.edu