Mem. Fac. Sci. Eng. Shimane Univ. Series B: Mathematical Science 32 (1999), pp. 75–84.
Y UKIHIRO SHOGENJI AND M ARETSUGU Y A M A S A K I (Received: December 25, 1998) Abstract. The smallest eigenvalue of a weighted discrete Laplacian is closely related to a generalized Hardy's inequality on networks. We shall estimate the smallest eigenvalue by using a discrete Kuramochi potential with some numerical experiments.
1. PROBLEM SETTING
Let X be a finite set of nodes, Y be a finite set of arcs and K be the node-arc incidence matrix. Assume that the graph G := {X,Y,K} is connected and has no self-loop.  For every two nodes a , b ∈ X , denote by ρ ( a , b ) the geodesic distance between a and b , i.e., the minimum number of arcs in the paths between a and b. For a strictly positive real-valued function r, N := { G, r } is called a network. Denote by L(X) the set of all real valued functions on X, by L+(X) the set of all nonnegative u ∈ L(X). For u ∈ L(X) , the discrete derivative du , the discrete Laplacian ∆ u(x) and the Dirichlet sum D ( u ) of u on N are defined by
Let A 0 be a nonempty subset of X such that X \ A 0 is connected and let m ∈ L ( X ) satisfy m(x) = 0 on A 0 and m(x) > 0 on X \ A 0 . A generalized Hardy's inequality is to find the best possible constant C m > 0 such that x∈ m(x)u(x)² ≤ CmD(u) for all u ∈ L(X) such that u(x) = 0 on A 0 . By special choices of N , A 0 and m , we obtain Wirtinger's inequality and Hardy's inequality in [2] and [3].  We shall show that 1/C m is equal to the smallest eigenvalue of an eigenvalue problem. We shall also give an esimation of this value by using a discrete Kuramochi potential studied in [4] and [5].
Let us put
For simplicity, let us put
We shall consider the extremum problem ( H m ): Find :=   inf{Xm(u); u ∈ L(X; Ao)} λm inf{D(u); u ∈ L(X; A0), ‖ u‖ m = 1}. Proposition 2.1. There exists an optimal solution ũ of problem (H m ), i.e., λm = D(ñ),ñ ∈ L(X; A 0 ) and ‖ñ‖m = 1. Proof. Let { vk } be a sequence in L(X; A 0 ) such that Xm( vk ) → λ m as k → ∞ . Put uk = vk/||vk||m. Then ||uk||m = 1 and Xm(uk) = D(uk) = D(vk)/||vk||² = Xm(vk). Since { u k ( x )} is bounded for each x ∈ X , we may assume that { u k } converges pointwise to a function й ∈ L(X; A 0 ). We have ‖ й‖ m = 1 and so that χm(¥) = λm
Denote by S ( λ m ) be the set of all optimal solutions of problem ( H m ), i.e., S(λm) := {u ∈ L(X; A 0 ); Xm(u) = λm}. Consider the following eigenvalue problem of finding a number μ and a nonzero function u ∈ L(X; A 0 ) which satisfy ( Eig ) Δu(x) = −μm(x)u(x) on X \ A 0 . Denote by E m (Δ) the set of all μ satisfying ( Eig ) and by EV m (μ) the set of nonzero functions u satisfying ( Eig ) with μ ∈ Em (Δ). For every u ∈ EVm(μ), we have
Since D ( u ) is positive definite on the set L ( X ; A 0 ), we see that E m ( ∆ ) consists of positive real numbers. By the above observation, we have Proposition 2.2.  λm = min{μ;μ ∈ Em(Δ)}. Lemma 2.1. S(λm) = EVm(λm). Proof. By the above observation, it suffices to show S(λm) ⊂ EVm(λm). Let u ∈ S ( λ m ). Denote by ε x ∈ L ( X ) the characteristic function of the set { x }. For any real number t and x ∈ X \ A 0 , we have λm = Xm(u) ≤ Xm(u + tε x ) or λm||u + tε x || ² ≤ D(u + tε x ). Noting the relation
we obtain D(u, ε x ) = λm((u, ε x ))m· Since D(u,ε x ) = −∆u(x) and ((u,ε x ))m = m(x)u(x), we conclude that u ∈ ■ EVm(λm) Lemma 2.2. Assume that u ∈ S(λm). Then |u| ∈ S(λm) and u(x1)u(x2) ≥ 0 for every x 1 , x 2 ∈ X \ A 0 with ρ(x 1 , x 2 ) = 1. Proof. Let v = |u|. Then v ∈ L(X;A 0 ) and D(v) ≤ D(v) holds (cf.[9]). Since ||v||m = ||u||m, we have λm ≤ Xm(v) ≤ Xm(u) = λm, and hence ν ∈ S(λm).  Suppose that there exist x 1 , x 2 ∈ X \ A 0 such that ρ( x 1 , x 2 ) = 1 and u( x 1 ) u( x 2 ) < 0. Let y ∈ Y be an arc whose endpoints are x 1 and x².  Then
so that D(v) < D(v). Thus λm = χm(v) < χm(v) = λm. This is a contradiction. ■ Corollary 2.1. If u ∈ S(λm), then either u = | u | or u = − | u | . Lemma 2.3. If u ∈ S(λ m ) is non-negative, then u(x) > 0 on X \ A 0 . Proof. Let u ∈ S ( λm ) be nonnegative. By Lemma 2.1, Δu(x) = −λmm(x)u(x) ≤ 0   on X \ A 0 . Namely u is superharmonic on X \ A 0 . By the minimum principle (cf. [9]), we have u(x) > 0 on X \ Ao. ● Corollary 2.2. If u ∈ S ( λ m ), then either ∆ u ( x ) < 0 on X \ A 0 or ∆ u ( x ) > 0 on X \ Ao. Lemma 2.4. The dimension of EV m ( λ m ) is one. Namely, if u 1 , u 2 ∈ EV m ( λ m ), then u1 and u2 are proportional. Proof. Assume that there exist u 1 , u 2 ∈ EV m ( λm ) such that they are not propor- tional.  Choose numbers α and β such that |α|+|β| > 0 and αυ 1 (x 1 )+βu 2 (x 1 ) = 0 for some x 1 ∈ X \ A 0 . Let u = α u 1 + β u 2 . Then u̸= 0, since u 1 and u 2 are not proportional. We have Δu(x) = αΔu1(x)+βΔu2(x) — λmm(x)u1(x) — λmm(x)u2(x) = -\mm(x)u(x) Namely u ∈ EVm(λm) = S(λm). We have △u(x 1 ) = λmm(x)u(x 1 ) = 0. This contadicts Corollary 2.2. Summing up these results, we obtain Theorem 2.1.  There exists a unique ũ ∈ L(X;A 0 ) such that λm = D(ñ) and ||ñ||m = 1; (1) й(x) > 0 on X \ A0; (2) Δ˜ı˜(x) = −λmm(x)˜(x) on X \ Ao. (3) 3. ESTIMATION OF Am Let us put D(N; A o ) := {u ∈ L(X; A o ); D(u) < ∞}. Since N is a finite network, we see that D( N ; A 0 ) = L( X ; A 0 ).  Notice that D( N ; A 0 ) is a Hilbert space with the inner product D( u, v ) (cf. [9]). The Kuramochi function ğ x of N with pole at x ∈ X \ A 0 is defined by the reproducing property. u(x) = D(u, ğ x)   for all u ∈ D(N; A 0 ) (cf. [4]). For each nonempty subset B of X \ A 0 , let us put d( A 0 , B ) := inf { D ( u ); u ∈ D ( N ; A 0 ), u = 1 on B }. We have
Lemma 3.1.  ğx has the following properties:
Now we shall estimate the value of λ m . Our idea is to use the discrete Ku- ramochi function studied in [4] and [5].  A similar idea can be founded in [8] to estimate Lyapunov's inequality. The Kuramochi potential Gm(x) of m is defined by
Lemma 3.2. Let ũ be as in Theorem 2.1. Then
Proof. By the reproducing property of the Kuramochi function and Lemma 3.1, we have
Theorem 3.1. The following estimation holds:
Proof. Let ŭ be as in Theorem 2.1. There exists b ∈ X \ A 0 such that ŭ( b ) = max{й(х);х ∈ Х}. Then we have by Lemma 3.2
We can prove the right hand side inequality similarly.
x∈x m(x).  Then the following estimation holds: min{d(A 0 , {x}); x ∈ X \ A 0 } ≤ m(X)λ m ≤ max{d(A 0 , {x}); x ∈ X \ A 0 }. Proof. Let ŭ be as in Theorem 2.1. There exists b ∈ X \ A 0 such that ŭ( b ) = max{й(х);х ∈ Х}. Then we have by Lemma 3.2
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4. CLASSICAL HARDY'S INEQUALITY
In this section, we consider the following special finite network N = {X,Y,K,r} defined by:
and K(x,y) = 0 for any other pair. Notice that the graph {X,Y,K} is a subgraph of the one-dimesional lattice domain Z. For simplicity, we set
Furthermore
We shall prove
for every u ∈ L(X; A 0 ).
Therefore we have
so that
□
Corollary 4.2. Assume that A0 = {x0} and rk = 1 for all k. Then
for all uk (k = 0, 1, · · · , n) with uo = 0. Notice that ρ ( x 0 , x k ) = R k and ρ ( x 0 , x k ) = k in Corollary 4.2, this inequality can be found in [2], page 239. We may expect that Corollary 4.2 would also holds in the general case. However it is not true as shown by Table 4 in the next section. Hereafter in this section we always assume that A 0 = {x 0} and m(xk) := r k / R 2 . In order to obtain the value λ m , we calculate the minimum eigenvalue of ( Eig ) numerically:
In order to study λ m as a function of the size n of N , we denote it by λ( n ) := λm(n). Some numerical experiments are given in the next section. In the present case, the Kuramochi function is given by
We estimate λ m by using the Kuramochi potential Ãm :
It is easily seen that
By Theorem 3.1, we have
Some numerical experiments for these quantities are also given in the next section.
5. NUMERICAL EXPERIMENTS
Let G = {X,Y,K} be the same graph as in Section 4. The graph can be drawn as follows: We take A 0 = {x 0} and m(x k ) := r k / R 2 except in Table 4. Table 1: The case where r k = 1 for all k
Table 2 : The case where rk = 1/k for all k .
Calculus of the minimum eigenvalue:
82 Table 3: The case where rk = 2¹−k for all k.
Calculus by Mathematica shows that
Calculus of the minimum eigenvalue:
Calculus by Mathematica shows that λ( n ) becomes strange if n ≥ 18.
Calculus by Mathematica shows that λ m (n) becomes strange if n ≥ 51. We remark that as n → ∞, so that λm(n) –