POISSON EQUATION IN NONSYMMETRIC NETWORKS

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IN NONSYMMETRIC NETWORKS

VICTOR ANANDAM

Kirchoff problem in a finite electrical network leads to the problem of solving the discrete Poisson equation ∆u = f in a finite network X with symmetric conductance. In this note, without assuming thatX has symmetric conductance, we give a necessary and sufficient condition for the existence of a solution of the Poisson equation inX. Later is given a sufficient condition for the existence of such a solution whenX is an infinite network.

AMS 2010 Subject Classification: 31C20, 31A30.

Key words: Kirchhoff problem, discrete Poisson equation, finite or infinite networks.

1. INTRODUCTION

The Kirchhoff problem in a finite electrical network X with symmetric conductance is equivalent to the problem of solving the discrete Poisson equation ∆u=−f when f is a real-valued function onX (see for example Soardi [8, Theorem 1.24]). A necessary and sufficient condition for the existence of a solution u is P

x∈X

f(x) = 0; since the function f in the Kirchhoff problem satisfies this condition, the problem has a solution. To prove this necessary and sufficient condition, one depends on the fact that the conductancec(x, y) between the nodes x and y is symmetric, that is c(x, y) =c(y, x) for all pairs of x and y inX; that is, the Laplace matrix ∆ is symmetric.

In this note we investigate the problem of solving ∆u = −f where the Laplace matrix ∆ is not symmetric, but singular. This leads to the consider- ation of solving ∆u = −f in an infinite network X without symmetric conductance. We prove that if a restriction is placed on the infinite network X (X is then called a network nonsingular at infinity), then for any real-valued function f on X,the equation ∆u=−f has a solution u inX.

REV. ROUMAINE MATH. PURES APPL.,56(2011),4, 253–259

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2. PRELIMINARIES

Brelot proved in [4]: Ifµ≥0 is a Radon measure in an open setωinRⁿ, n≥2,then there exists a superharmonic function sin ω such that ∆s=−µ in the sense of distributions. For this, he uses a Runge-type harmonic approx- imation theorem in Rⁿ.Our intention here is to obtain a discrete analogue of this result in networks.

A network X consists of a countable (finite or infinite) set of points, called vertices, some of them pairwise joined by edges. The symbol x ∼ y means that there is an edge joining x and y; in that case, x and y are called neighbours. A vertexx is said to beterminal if it has only one neighbour. We assume that given any two vertices xandy,there exists an associated number t(x, y)≥0 calledconductance such thatt(x, y)>0 if and only ifx∼y;t(x, y) and t(y, x) may be different. We assume that there is no self-loop in X;X is connected; and eachxinXhas only a finite number of neighbours. A network X is called a tree if there is no cycle in X.

For any subsetE ofX, we write

◦

E ={x: x and all the neighbours ofx are inE}

and ∂E=E\E.^◦ For a real-valued functionu on E,write

∆u(x) =X

y∼x

t(x, y)[u(y)−u(x)] = X

y∈X

t(x, y)[u(y)−u(x)]

for any x ∈ E.^◦ We say that u is superharmonic on E if ∆u(x) ≤0 for every x∈E.^◦

LetT be an infinite tree without positive potentials (that is, every positive superharmonic function inT is constant). Assume thatT has no terminal vertices. Then Bajunaid et al. [2, Proposition 4.2] and [3] give a construc- tion which can be used to prove that if h is a harmonic function on the set Bn = {x :|x| ≤ n} in T, then there exists a harmonic function v on T such thatv =honBn.Using this method of harmonic extension and following the method of Brelot’s indicated above, it is easy to show: Let T be an infinite tree without terminal vertices; there may or may not be any positive potential inT. Then given any real-valued functionf onT, there exists a functionuon T such that ∆u(x) =−f(x) for any x inT.

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3. POISSON EQUATION IN FINITE NETWORKS

If X is a finite network consisting of vertices x₁, x₂, . . . , x_n, then write t_ij = t(x_i, x_j) and t_ii = −P

j6=i

t_ij. The Laplace operator ∆ in this case is represented by the singular matrix ∆ = (t_ij)_1≤i,j≤n which need not be a symmetric matrix. In this section we assume that X is a finite network.

Lemma 3.1. Let E be a proper subset of X. Suppose u is a real-valued function on E such that ∆u(x) = 0 if x ∈ E and u(x) = 0 if x 6∈ E. Then u= 0.

Proof. If the lemma is false, without loss of generality, we can assume that u(x) ≤ u(y₀) = M > 0 for each x ∈ X and some y₀ ∈ E. Let z ∈ X\E. Then there exists a path{y₀, y1, . . . , yk=z} connecting y0 toz. Since

∆u(y₀) =P

t(y₀, y)[u(y)−u(y₀)] = 0 and u(y₀) =M is the maximum value, u(y) =M for all y∼y0.In particular, u(y1) =M.

Letibe the smallest index such thaty_i ∈Efor allj≤iandy_i+1∈X\E.

Then by the above argument repeated, we find u(y₀) =u(y₁) =· · ·=u(y_i) = M; henceu(yi+1) =M sinceyi+1 ∼yi. Howeveru(yi+1) = 0 sinceyi+1∈X\E.

This contradiction shows that u ≤ 0 on X. Similarly, we prove that u ≥ 0

on X.

Lemma 3.2. The rank of ∆is (n−1).

Proof. LetE =X\{x₁}and ∆₁ be the matrix of order (n−1) obtained from ∆ by suppressing the first row and the first column. Letu= (u2, . . . , un)^t be a column vector such that ∆₁u = 0.Let v = (v₁, v₂, . . . , v_n)^t be a column vector such thatv1 = 0 andvi=uiifi≥2.Consider the column vector ∆v(x).

It can be calculated that ∆v(x) = ∆1u(x) for each x ∈ E. Thus v is a function on X such that ∆v = 0 on E and v = 0 on X\E. Hence by Lemma 3.1,v= 0 on X and consequentlyu= 0.Thus we have proved that if u is a solution of the homogeneous system ∆₁u= 0, thenu = 0.This means that ∆1 is a nonsingular matrix. However ∆ is singular. Hence the rank of ∆ is (n−1).

Remark3.3. Suppose ∆ represents the transition matrix of a finite Markov chain X.Generally ∆ is not a symmetric matrix (Markov chains that are not necessarily reversible). Let A and B be disjoint sets in X. Then by adopting the method used in the proof of Lemma 3.2, we can construct a function u on X such that u(x) = 1 if x ∈ A, u(x) = 0 if x ∈ B and ∆u(x) = 0 if x ∈ X\(A∪B).We show (as in the proof of Lemma 3.1) that 0 ≤u(x) ≤1 for x ∈X. Thus u(x) represents the probability of a random walker starting at a vertex xreaches A before reaching any vertex in B,see Lawler [5].

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Lemma3.4. A necessary and sufficient condition for the system of equations

n

P

j=1

aijxj = bi (i = 1, . . . , m) to have a solution is that every solution of the system of equations

m

P

i=1

a_ijy_i = 0 (j = 1, . . . , n) should also satisfy

m

P

i=1

b_iy_i= 0 (see Mirsky [6, Problem 39, p. 166]).

Proof. SupposeX and Y are two Banach spaces. Let T :X → Y be a bounded linear operator and T^∗ :Y^∗ →X^∗ be its adjoint. If N(T^∗) denotes the null space of T^∗ and R(T) denotes the range of T, then R(T)^⊥ =N(T^∗) (Rudin [7, Theorem 4.12]). In the particular case when X and Y are finite dimensional. T can be represented by a matrix [T]; in that case [T^∗] can be taken as the transpose of [T]. Consequently, the above lemma is a finite dimensional reformulation of the result R(T)^⊥=N(T^∗).

Theorem 3.5. Let X be a finite network and∆ be the Laplace matrix.

Let ∆^t denote the transpose of∆.Then

i. The subspace {x: ∆^tx= 0} is of dimension1 consisting of vectors of the form λk where k= (k1, k2, . . . , kn)^t.

ii. For a real-valued function f on X, ∆u = −f has a solution if and only if

n

P

i=1

kif(xi) = 0.

Proof. i. Since the rank of ∆ is (n−1) by Lemma 3.2, rank of ∆^t = rank of ∆ = n−1. Hence the subspace A = {x : ∆^tx = 0} is of dimension n−(n−1) = 1.

ii. Let k = (k1, k2, . . . , kn)^t be an element in A. Let f be an arbitrary real-valued function. Then by Lemma 3.4, ∆u=−f has a solution if and only ifk^tf =

b

P

i=1

k_if(x_i) = 0.

Remark 3.6. Suppose ∆ is a symmetric matrix as in the case of a finite electrical network. In this case, the vector k in the theorem above can be taken as (1,1, . . . ,1)^tso that a solution to the equation ∆u=−f exists if and only if

n

P

i=1

f(xi) = 0 (Kirchhoff).

4. POISSON EQUATION IN INFINITE NETWORKS LetXbe an infinite network in this section. Fix a vertex eand measure distances frome.Thus forx∈X,|x|= dist(e, x) =nwhere{e, x₁, x₂, . . . , x_n= x}is a shortest path joining etox.

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Definition 4.1. A vertexx inX, |x|=n≥1,is said to be nonsingular if the following two conditions are satisfied:

i. there is only one vertexy such thaty∼x and|y|=n−1; and ii. the set of verticesz such thatz∼x and |z|=n+ 1 contains at least one vertex that is nonterminal.

Remark 4.2. 1) The condition i is always verified in an infinite tree T since there is no closed path in T.

2) Suppose T is an infinite tree in which each nonterminal vertex has at least 2 nonterminal vertices as neighbours (in particular, any tree without terminal vertices like homogeneous trees, binary trees etc.). Then any vertex in T other than the fixed vertexeis nonsingular.

Definition 4.3. A networkXis said to be nonsingular at infinity, if there exists an integer m ≥ 1 such that any nonterminal vertex x, |x| = n ≥ m, is nonsingular in X. The smallest such m, if it exists, is referred to as the nonsingular index of X.

Theorem4.4. Let X be a network that is nonsingular at infinity. Then given any real-valued function f on X, there exists a function u on X such that ∆u(x) =−f(x) for all x in X.

Proof. We know that for any vertexy inX, there exists a functionq_y(x) on X such that ∆qy(x) = −δ_y(x) for all x in X (see for example [1, Theo- rem 3.2.6 and Theorem 3.4.5]). Let m be the nonsingular index of X. Then v(x) = P

|y|≤m

f(y)q_y(x) is a well-defined function on X such that ∆v(x) =

−f(x) for allx,|x| ≤m.Letube the restriction ofvto the set|x| ≤m.Then u is a function on B_m = {x : |x| ≤ m} such that ∆u(x) = −f(x) for every x ∈ B^◦_m. We shall now extend u to the set B_m+1 such that ∆u(x) = −f(x) for every x∈B^◦_m+1.

Let x0 ∈ ∂Bm. Then |x₀|= m and x0 is nonterminal, by definition. If z_i ∼x₀,|z_i|=m+ 1 and z_i is terminal, then defineu(z_i) =µ_i where

t(z_i, x₀)[µ_i−v(x₀)] =f(z_i),

so that ∆u(zi) =−f(zi); ifyj ∼x0,|y_j|=m+ 1 andyj is not terminal, then take u(y_j) =λwhere λis a constant so determined to satisfy the equation

−f(x₀) = ∆u(x₀) = X

|a|≤m

t(x₀, a)[u(a)−u(x₀)] + X

|b|=m+1

t(x₀, b)[u(b)−u(x₀)].

Here λis the only unknown which can be easily calculated. Thusu is defined now at all the neighbouring vertices of x0.

Supposex₁ is another vertex in∂B_m.Remark thatx₀andx₁ are nonsingular. Hence whatever be the position ofx1 on∂Bm,the neighbours ofx1 and

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the neighbours of x0 outside Bm will not have any common vertex because of the condition i in Definition 4.1. Thus the above mentioned extension of u to all the neighbouring vertices of x1 can be carried out independent of the indicated extensionuto all the neighbouring vertices ofx0.Repeating this process with respect to each vertex on∂B_m+1,we get an extension ofu toB_m+1 such that ∆u(x) =−f(x) for everyx∈B_m+1^◦ .(Recall that all terminal verticesz,

|z|=m+ 1, are in

◦

B_m+1.) Consequently, we can construct an extension ofu toX such that ∆u(x) =−f(x) for everyx inX.

Corollary 4.5. Let X be a network nonsingular at infinity. Then any real-valued function f on X can be written as a difference of two superharmonic functions s₁ and s₂ on X, that is f = s₁−s₂. This representation is unique up to an additive superharmonic function in X.

Proof. Let ∆f =g.Then there exists⁰₁ands⁰₂onXsuch that ∆s⁰₁=−g⁺ and ∆s⁰₂ = −g⁻; clearly s⁰₁ and s⁰₂ are superharmonic on X. Consequently,

∆(s⁰₁−s⁰₂) =−∆f; that is, for a harmonic functionhonX, −f =s⁰₁−s⁰₂+h.

Write s₁=s⁰₂ and s₂ =s⁰₁+h.Then f =s₁−s₂.

For the uniqueness, letf =t1−t2 be another such representation. Then g = ∆f = ∆t₁−∆t₂. Since ∆t₁ ≤0 and ∆t₂ ≤0,then −∆t₂ =g⁺+v and

−∆t₁ =g⁻+v where v≥0. Let ∆q =−v on X, so that q is superharmonic on X. Consequently, −∆t₂ =−∆s⁰₁−∆q so that t2=s⁰₁+q+H whereH is harmonic on X. Hence t₂ = (s₂−h) +q+H.Writeu=q+H−h so that u is superharmonic on X andt2 =s2+u; then t1=s1+u.

Corollary4.6. LetX be a network with nonsingular indexm.Letϕ(x) be a real-valued function defined on∂Bn,n≥m.Then there exists a harmonic function h on X such that h(x) =ϕ(x) on∂B_n.

Proof. Let u be the Dirichlet solution [1, Theorem 3.1.7] in Bn with boundary values ϕ on ∂B_n. Then as in the proof of Theorem 4.4, u can be extended to B_n+1 as functionv such that ∆v= 0 at every vertex in∂B_n and v =u on B_n, so that ∆v = 0 at every vertex in

◦

B_n+1 ⊃B_n. Continuing this procedure, we constructhonXsuch thath=uonBnand ∆h= 0 onX.

REFERENCES

[1] V. Anandam, Harmonic functions and potentials on finite or infinite networks. Lect.

Notes Unione Mat. Ital.12(2011), Springer.

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Math.30(2003), 706–745.

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[3] I. Bajunaid, J.M. Cohen, F. Colonna and D. Singman, Corrigendum to Trees as Brelot spaces.Adv. Appl. Math. (2010), doi.10.1016/j.aam. 2010. 09. 004.

[4] M. Brelot,Fonctions sousharmoniques associ´ees `a une mesure.Stud. Cercet. S¸tiint¸. Mat.

Ia¸si2(1951), 114–118.

[5] G.F. Lawler, Introduction to Stochastic Processes.Chapman & Hall Probability Series, 1995.

[6] L. Mirsky,An Introduction to Linear Algebra.Oxford University Press, 1955.

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Received 6 November 2011 The Institute of Mathematical Sciences C.I.T Campus, Taramani

Chennai 600 113, India vanandam@hotmail.com