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Discrete Green’s functions ∗
Fan Chung†
University of California, San Diego
La Jolla, CA 92093-
S.-T. Yau
Harvard University
Cambridge, MA 02138
Dedicated to the memory of Gian-Carlo Rota Abstract We study discrete Green’s functions and their relationship with discrete Laplace equations. Several methods for deriving Green’s functions are discussed. Green’s functions can be used to deal with diffusion-type problems on graphs, such as chip-firing, load balancing and discrete Markov chains.
1 Introduction
Many combinatorial problems involve solving equations of the following general type. Let V denote a set of states (in the setting of Markov chains ) or a set of vertices ( as in a graph). Let g denote a given function g : V → R. The problem of interest is to find f satisfying the following discrete Laplace equation:
∆f (x) =
y
(f (x) − f (y))pxy = g(x) (1)
where pxy denote the transition probability from x to y. For a typical random walk in a graph, pxy is often taken to be 1/dx for y adjacent to x and 0 otherwise (where dx is the degree of x, defined to be dx =^ ∑ y dxy).
For some combinatorial games or diffusion processes, there are additional constraints for finding a solution f in (1). For a subset S of V , we define the boundary δS of S by
δS = {y 6 ∈ S : pxy 6 = 0 for some x ∈ S} ∗The original paper appeared in Journal of Combinatorial Theory (A), 91 , (2000), 191-214. Additional revisions were incorporated. † Research supported in part by NSF Grant No. DMS 98-
For a function σ : δS → R, we say f satisfies the boundary condition σ if f (x) = σ(x) for x in δS.
For example, the problem of evaluating the probability fx,y(z) of a Markov chain hitting x before hitting y can be formulated as the following problem of solving the Laplace equation with boundary conditions. We consider S = V − {x, y}, δS = {x, y} and σ(x) = 1, σ(y) = 0. Then fx,y(z) is the solution for the following equation: ∆f (z) = 0
for all z ∈ S and f satisfies the boundary condition σ.
Suppose δS 6 = ∅ and the subgraph induced by S is connected. It is not hard to see [6] that ∆ is nonsingular as an operator on the space of functions defined on S. The Green’s function is the left inverse operator of the Laplace operator ∆ (restricted to the subspace of functions defined on S):
G∆ = I
where I is the identity operator.
If we can determine the Green’s function G, then we can solve the Laplace equation in (1) by writing f = G∆f = Gg.
We will also consider Green’s functions for the case that there is no boundary. We will discuss a related example concerning the so-called “hitting time”, the expected number of steps for a Markov chain to reach a state y with an initial state x. It is worth mentioning that numerous diffusion-type problems can be treated in a similar way, including chip-firing games, load balancing algorithms and the mixing of random walks. Thus, Green’s functions provide a powerful tool in dealing with a wide range of combinatorial problems.
Green’s functions were introduced in a famous essay by George Green [16] in 1828 and have been extensively used in solving differential equations [2, 5, 15]. The concept of Green’s functions has had a pervasive influence in numerous areas. Many formulations of Green’s functions occur in a variety of topics. Articles on discrete Green’s functions or discrete analytic functions appear sporadically in the literature, most of which concern either discrete regions of a manifold or finite approximations of the (continuous) equations [3, 12, 17, 13, 19, 21]. In this paper, we consider Green’s functions for discrete Laplace equations defined on graphs.
It is easy to see that L is a symmetric matrix and we call L the normalized Laplacian. In this paper, we consider graphs without isolated vertices so that the dx are all nonzero.
For a subset S of vertices, the Dirichlet eigenvalues of L are exactly the eigenvalues of the submatrix LS with rows and columns restricted to those indexed by vertices in S. Let λ 1 ≤ λ 2 ≤
... ≤ λs denote the eigenvalues of LS , where s = |S|. It is not hard to check (also see [6]) that
λ 1 = inf g^ 〈g,〈g, g^ LS^ 〉g〉
= inf f
x,y∈S∪δS ∑ (f^ (x)^ −^ f^ (y))^2 wxy S f^2 (x)dx
where f and g range over all nontrivial functions satisfying the Dirichlet boundary condition:
f (x) = 0 = g(x) (3)
for all x in the boundary δS of S.
The celebrated matrix-tree theorem [18] states that the number of spanning trees in a graph Γ is equal to the determinant of LS , where S is any maximum proper subset of the vertex set. Therefore the number of spanning trees in a graph Γ is exactly ∏s i=1 λi
∑^ x∈S^ dx x∈S dx
We remark that in equation (2), the degrees dx are the degrees in the host graph Γ (not in the induced subgraph S). When the induced subgraph S is connected, we see from (2) that LS is nonsingular and λ 1 > 0 (see [6]). Thus the inverse of LS , denoted by G, is well-defined. We note that G is just a symmetric normalized version of the Green’s function G since
G = T 1 /^2 GT −^1 /^2
and we have T −^1 /^2 GT 1 /^2 ∆ = 0
For example, suppose we consider a path Pn which can be regarded as an induced subgraph of a cycle Cm , with m > n + 1. Suppose that the vertices of Pn are 1, 2 ,... , n where the boundary consists of two vertices 0 and n + 1. Then
∆f (x) =^12 (2f (x) − f (x − 1) − f (x + 1))
and ∆ = L = 12 L since dx = 2 for all x. The Dirichlet eigenvalues for Pn are 1 − cos (^) nkπ+1 and the corresponding eigenfunctions are
φk(j) =
n + 1 sin^
jkπ n + 1
for k = 1,... , n. The problem of determining the Green’s function G for a path will be discussed later.
For a given connected induced subgraph S of a graph Γ, and for a real parameter t ≥ 0, the Dirichlet heat kernel of S is defined by
Ht = e−tLS = I − tLS + t
2 2! L
Thus,
Ht(x, y) =
∑^ s i=
e−λitφi(x)φi(y) (4)
where λi’s are the eigenvalues of LS and φi’s are the corresponding orthonormal eigenfunctions. It follows from the definition (4) that Ht satisfies the following heat equation:
d dt Htf^ =^ −LS^ Htf^ (5)
for any f satisfying the Dirichlet boundary condition. Furthermore, we have H 0 = I and
tlim→∞ Ht(x, y) = 0^ (6)
Thus, Let A = I − LS satisfy A(x, y) = √wdxy xdy
We can express Ht in an alternative form:
Ht = e−tetA = e−t(I + tA + t
2 2! A^ +^.. .) = e−t^
k≥ 0
Pk(x, y) t
k k!
where Pk(x, y) is the sum of the weights of all paths of length k joining x and y. Here, the weight of a path is the product of all edge weights in the path. We use the convention that P 0 (x, x) = 1.
for given g defined on S ∪ δS. There are two main steps for deriving a solution f. In this section, we deal with the first part of finding a solution to ∆f = 0 satisfying the boundary condition σ, a function defined on the boundary δS.
Theorem 1 The solution f to the following equation
∆f (x) = 0
for x ∈ S, satisfying the boundary condition
f (y) = σ(y)
and for y ∈ δS, can be written as
f (z) =
i
λi
x∼^ xy∈∈SδS
dx−^1 /^2 φi(x)σ(y)
d− z 1 / (^2) φi.
for z in S where φi’s are the eigenfunctions of LS.
Proof: We consider f˜ (x) = T 1 /^2 f (x) and f˜ : S → R is the solution of the following equation:
LS f˜(x) = 0
for x ∈ S. We can write f˜ as a linear combination of the eigenfunctions φi of LS.
f˜ = ∑ i
aiφi.
which implies ai = 〈φi, f˜〉.
Now we consider the function
f 0 (x) =
{ (^0) if x ∈ S, σ(x) otherwise.
Let fS denote the function f restricted to S. Clearly, f −f 0 satisfies the Dirichlet boundary condition. We have
λiai = 〈LS φi, f˜〉 = 〈LS φi, ( f˜ − f˜ 0 )〉 = 〈φi, T −^1 /^2 LT 1 /^2 ( f˜ − f˜ 0 )〉 = 〈φi, T −^1 /^2 L(f − f 0 )S 〉 = 〈T 1 /^2 φi, ∆(f − f 0 )S 〉 = 〈T 1 /^2 φi, −(∆f 0 )S 〉 = −
x∈S
dxφi(x) d^1 x
y∼x
(f 0 (x) − f 0 (y))
=
x∈S
y^ y∈∼δSx
d− x 1 /^2 φi(x)σ(y).
Consequently,
ai = (^) λ^1 i
x∼^ xy∈∈SδS
dx−^1 /^2 φi(x)σ(y)
f˜ = ∑ i
λi
x∼^ xy∈∈SδS
dx−^1 /^2 φi(x)σ(y)
φi
and so,
fS (z) =
i
λi
x∼^ xy∈∈SδS
dx−^1 /^2 φi(x)σ(y)
d− z 1 / (^2) φi(z).
This completes the proof of Theorem 1.
Example 1 For a path Pn with vertex set { 1 , 2 ,... , n}, we assume the boundary condition σ(n +
- = 0 and σ(0) = 1. The solution f (x) to the equation ∆f = 0 satisfying the boundary condition σ is the probability of a walk starting from x hitting 0 before hitting n + 1. We can solve for f directly and get f (z) = 1 − (^) n + 1z.
4 Green’s function for a path
In the previous sections, there are several explicit formulas for the Green’s function. Instead, here we consider direct methods for evaluating the Green’s function for a path with Dirichlet boundary condition. The solutions we will obtain leads to intriguing equalities.
Let the vertex set of Pn be denoted by { 1 , 2 ,... , n} with boundary { 0 , n+1}. Since ∆ = L = L/2, we have LG = GL = I. Here we assume 1 ≤ x < y ≤ n. From LG = I, it follows that 1 2 (2G(x, y)^ −^ G(x^ −^1 , y)^ −^ G(x^ + 1, y)) = 0 From GL = I, we have 1 2 (2G(x, y)^ −^ G(x, y^ −^ 1)^ −^ G(x, y^ + 1)) = 0 with the convention that G(x, y) = 0 if either x or y is not in { 1 ,... , n}. Therefore we have
G(x, y) − G(x − 1 , y) = G(x − 1 , y) − G(x − 2 , y) = G(x − 2 , y) − G(x − 3 , y) =... = G(1, y).
This implies that
G(x, y) = xG(1, y).
In a similar way, we can get
G(1, y) = c(n + 1 − y)
for some constant c. Now, we use the fact that 1 2 (2G(x, x)^ −^ G(x^ −^1 , x)^ −^ G(x^ + 1, x)) = 1 to get c = (^) n+1^2 and G(x, x) = cx(n + 1 − x). Thus we have proved the following:
Theorem 3 For a path Pn with vertex set { 1 ,... , n} as an induced subgraph with boundary { 0 , n + 1}, its Green’s function satisfies
G(x, y) = (^) n + 1^2 x(n + 1 − y)
for 1 ≤ x ≤ y ≤ n.
As an immediate consequence of Theorem 3 and equation (10), we obtain the following (somewhat nontrivial) equality:
Corollary 1 The following equality holds for integers 1 ≤ x ≤ y ≤ n: ∑^ n k=
sin (^) nkxπ+1 sin (^) nkyπ+ 1 − cos (^) nkπ+1^ =^ x(n^ + 1^ −^ y).
5 Green’s functions for lattices
In this section, we describe a way to determine Green’s functions for cartesian product of graphs. In particular, this method can be used to evaluate Green’s functions for lattices.
We start with an induced subgraph S of a graph Γ. For α ∈ R, let Gα denote the symmetric matrix satisfying (LS + α)Gα = IS
where LS is the Dirichlet Laplacian for the induced subgraph S. Clearly,
Gα(x, y) =
i
λi + α φi(x)φi(y)
where φi’s are orthonormal eigenfunctions of LS associated with eigenvalues λi.
Now we consider two induced subgraphs S and S′^ of graphs Γ and Γ′, respectively. We let S × S′^ denote the induced subgraph of the cartesian product of Γ and Γ′^ by the subset of vertices (v, v′) where v ∈ S and v′^ ∈ S′. The cartesian product of two graphs (V, E) and (V ′, E′) has vertex set {(v, v′) : v ∈ V, v′^ ∈ V ′} and edges of the form {(v, v′), (v, u′)} or {(v, v′), (u, v′)} where {u, v} ∈ E, {u′, v′} ∈ E′.
Let C denote a contour in the plane, say, consisting of all α ∈ C satisfying | 2 − α| = 2. Let G and G′^ denote the Green’s functions of^ S^ and^ S′, respectively. Then we have the following:
Theorem 4 Suppose S and S′^ are induced subgraphs of two graphs Γ and Γ′, which are both regular of degrees d. The Green’s function G of the cartesian product S×S′^ with Dirichlet boundary condition is G((x, x′), (y, y′)) = (^) πi^1
C Gα(x, y)G−′α(x′, y′)dα
where C, G, G′^ are defined as above.
Its proof needs the following useful fact:
Theorem 7 For a path P with vertices 1 , 2 ,... , n and a real α, the Green’s function Gα satisfies
Gα(x, y) = 2(r
x (^) − r−x)(rn+1−y (^) − r−(n+1−y)) (r − r−^1 )(rn+1^ − r−(n+1))
where 2(1 + α) = r + r−^1.
Proof: For α = 0, we know from Theorem 3 that G 0 (x, y) = 2x(n + 1 − y)/(n + 1). For x < y, we have
0 = (L + α)Gα(x, y) = 12 (2(1 + α)Gα(x, y) − Gα(x + 1, y) − Gα(x − 1 , y)) = 12 ((r + r−^1 )Gα(x, y) − Gα(x + 1, y) − Gα(x − 1 , y)).
This implies
Gα(x + 1, y) − rGα(x, y) = (^1) r (Gα(x, y) − rGα(x − 1 , y)) =... = (^) rcyx.
For x ≤ y, we have
Gα(x, y) = (^) rcx−y 1 + rG(x − 1 , y) = (^) rcx−y 1 (1 + r^2 +... + r2(x−1)) = c
′ y (r 2 x (^) − 1) rx−^1
In a similar way, we get
Gα(x, y) = c(r^2 x^ − 1)(1 − r−2(n+1−y))r−x−y^.
To determine the value of c, we consider
1 = (L + α)G(x, x)
= 12 ((r + r−^1 )G(x, x) − G(x + 1, x) − G(x − 1 , x)) = c 2 ( (r^ +^ r
− (^1) )(r 2 x (^) − 1)(1 − r−2(n+1−x)) r^2 x^ −^
(r^2 x^ − 1)(1 − r−2(n−x)) r^2 x+1^ −^
(r2(x−1)^ − 1)(1 − r−2(n+1−x)) r^2 x−^1 ) = c 2 ( r
2 x(1 − r− (^2) )(1 − r−2(n+1−x)) r^2 x−^1 +
(r^2 x^ − 1)(−1 + r^2 )r−2(n+1−x) r^2 x+1^ ) = c 2 (r
(^2) − 1)(r 2 x(1 − r−2(n+1−x)) + r−2(n+1−x)(r 2 x (^) − 1)) r^2 x+ = c 2 (r
(^2) − 1)(r 2 x (^) − r−2(n+1−x)) r^2 x+ = c 2 (r
(^2) − 1)(1 − r−2(n+1)) r.
This implies c = (^) (r (^2) − 1)(1^2 −r r−2(n+1)).
Thus we have
Gα(x, y) = 2(r
2 x (^) − 1)(1 − r−2(n+1−y)) (r − r−^1 )(1 − r−2(n+1))rx+y = 2(r
x (^) − r−x)(rn+1−y (^) − r−(n+1−y)) (r − r−^1 )(rn+1^ − r−(n+1))
as claimed.
By using the above Theorem and the definitions of α, r, we have the following:
Corollary 2 For a real α 6 = 0 and 1 ≤ x ≤ y ≤ n, the Green’s function Gα for a path P with vertices 1 , 2 ,... , n satisfies
Gα(x, y) =^2 Ux−^1 (1 + Un^ (1 +α)Un α−)y^ (1 +^ α)
where U is the Chebyshev polynomial of the second kind.
Now we are ready to prove Theorem 6.
Proof of Theorem 6:
From Theorem 4, it is enough to determine the residues of Gα(x, y)G′−α(x′, y′) for α in the interior of the contour C. From Theorem 7, the poles of GαG′−α are exactly at r = eiπk/(n+1)
eigenvalue λ of Γ has multiplicity |V (Γ)|φ^2 j (x) where φ is the orthonormal eigenfunction of λ in P. The heat kernel H of Γ and the heat kernel h of P are related in a nice way [11]:
Ht(x, y) =
π−^1 (vr )ht(v 0 , vr ).
Thus Green’s functions for distance regular graphs can be deduced from Green’s functions for a weighted path.
We will treat a general weighted path which can then be used to deal with distance regular graphs. We consider a general weighted path with edge weights wk,k+1 = w(vk , vk+1), for k = 0,... , m. We will consider two situations with respect to the boundary. (The case with no boundary will be examined in Section 7.) In the first case the boundary consists of one single vertex v 0. In the second case the boundary is {v 0 , vm+1}. As a matter of fact, the first case can be viewed as a special case of the second in the sense that the edge weight of the last edge wm− 1 ,m is zero, although the Green functions for two cases are quite different.
6.1 The boundary has one single vertex
In this subsection, we consider the Green function G with Dirichlet boundary condition for the boundary v 0. We assume without loss of generality that all edge weights are nonzero. We will first consider the normalized Green function G(vi, vj ) = G(i, j) = G(j, i) which satisfies, for x 6 = y,
w √x− 1 ,x dx^ (^
G √(x, y) dx^ −
G( √x − 1 , y) dx− 1 ) =^
w √x,x+ dx^ (^
G( √x + 1, y) dx+1^ −
G √(x, y) dx^ )^ (14)
This implies that for x < y, G √(x, y) dy^ =^
G √(x, x) dx^. Since w √x− 1 ,x dx^ (^
G √(x, x) dx^ −
G( √x − 1 , x) dx− 1 ) = 1, we have wx− 1 ,x( G(x, x dx ) − G(x d^ −x−^11 , x) ) = 1
and G(x, x) dx^ =^
wx− 1 ,x^ +^
wx− 2 ,x− 1 +^...^ +^
w 0 , 1
We can then derive for G(y, x) =
√dx √dy G(y, x) and G(x, y) =
√dy √dx G(y, x).
6.2 The boundary has two endpoints
In this subsection, we consider the Green function G with Dirichlet boundary condition for the boundary {v 0 , vm+1}. From (14), we have, for x < y,
G(x, y) = cx
dy ( (^) wy,y^1 +1 +... + (^) wm,m^1 +1 ) = c
dxdy ( (^) wy,y^1 +1 +... + (^) wm,m^1 +1 )( (^) wx−^11 ,x + (^) wx−^12 ,x− 1 +... + (^) w^10 , 1 )
for some constant c. We can then compute c as follows:
1 = LG(x, x) = w^ √x−d^1 x,x ( G^ √(x, xdx) − G(^ √x^ −d^1 , x) x− 1 ) − w^ √x,xd+1x ( G(^ √x^ + 1d , x) x+ − G(^ √x, xdx) )
= c
y
wy− 1 ,y^.
Therefore we have c = (
y
wy− 1 ,y^ )
− 1
and
G(x, y) =
√d ∑ xdy y wy−^11 ,y
( (^) wy,y^1 +1 +... + (^) wm,m^1 +1 )( (^) wx−^11 ,x + (^) wx−^12 ,x− 1 +... + (^) w^10 , 1 ).
7 Green’s function with no boundary
In the remainder of the paper, we consider the case of Green’s functions with no boundary. This case is slightly more difficult than the case with non-empty boundary and Dirichlet boundary con- ditions. The Laplace operator ∆ or the normalized Laplacian L as defined in Section 2.1. has a zero eigenvalue. Again, we consider a connected finite graph Γ so there is exactly one zero eigenvalue (see [6]). The Green function G is a matrix with its entries, indexed by vertices x and y, defined by
G∆(x, y) = I(x, y) − (^) voldy
where vol is the sum of all degrees in Γ. Equivalently, the normalized Green function G = T −^1 /^2 GT 1 /^2 satisfies GL = LG = I − P 0 = I − φ∗ 0 φ 0 (15)
which implies, by equation (15),
(I − φ∗ 0 φ 0 )T 1 /^2 QT 1 /^2 = −vol G. (19)
By checking the (x, x)-entry of the above equation of matrices and using the fact that Q(x, x) = 0, we have
(φ∗ 0 φ 0 T 1 /^2 QT 1 /^2 )(x, x) = (^) vol^1
z
dxdz Q(z, x) = vol G(x, x). (20)
By checking the (x, y)-entry of (19), we get √ dxdy Q(x, y) − (^) vol^1
z
dxdy dz Q(z, y) = −vol G(x, y). (21)
By combining (20) and (21), we have
√ dxdy Q(x, y) −
dx dy^ vol^ G(y, y)^ =^ −vol^ G(x, y).^ (22)
Equivalently, we have
√d xdy Q(x, y)^ −
dx dy^ vol^ G(y, y)^ =^ −vol^ G(x, y)
√ (^) d y dx^ (23)
as desired. Next we evaluate the Green function for a weighted path with no boundary.
Theorem 9 Let the vertex set of a path Pn be { 1 , 2 ,... , n} and edge weights wx,x+1 for x = 1 ,... , n − 1. The normalized Green function G(x, y), x ≤ y, for Pn satisfies:
G(x, y) = G(y, x)
√d xdy vol 2 (^
z This implies that for x < y, G √(x, y) dx^ −
G( √x − 1 , y) dx− 1 =
√d y (dx− 1 +^...^ +^ d 1 ) vol wx− 1 ,x^ , G( √x, y + 1) dy+1^ −
G √(x, y) dy^ =^ −
√d x(dy+1 +^...^ +^ dn) vol wy,y+^. By telescoping and summing equations of the above types, it leads to the solution in (24). An alternative proof is to directly check that (24) satisfies (25).
Example 2 We consider a path P 3 with vertices 1, 2 , 3 and edge weights wj,j+1. Its Green function satisfies
G(1, 1) = (^) vold^1 2 ( (d^2 +^ d^3 )
2 w 1 , 2 +^
d^23 w 2 , 3 ) G(1, 2) =
√d 1 d 2 vol 2 (−^
d 1 (d 2 + d 3 ) w 1 , 2 +^
d^23 w 2 , 3 ) G(1, 3) =
√d 1 d 3 vol 2 (−^
d 1 (d 2 + d 3 ) w 1 , 2 −^
(d 1 + d 2 )d 3 w 2 , 3 ) G(2, 2) = (^) vold^2 2 ( d
(^21) w 1 , 2 +^
d^23 w 2 , 3 ).
Example 3 We consider a path Pn with edge weights wj,j+1 = 1 for j = 1,... , n − 1 and w 1 , 1 = wn,n = 1. Its Green function satisfies
G(x, y) = G(x, y) = (^) n^12
z 0.
G˜(x, y) = G(0^ (n, k) k
= 2 −^2 n(−
j
