Computation of maximum eigenvalue and spatial entropy 3

A_nis a 2ⁿ× 2ⁿmatrix, so computing ρ(A_n) is usually quite difficult when n is large. However, for a class of A₂, the recursive formulae for ρ(A_n) can be computed explicitly, along with a limiting equation to ρ^∗ =exp(h(A₂)), as in [3]. This class of A₂ has the form of  A B

The results in [3] are recalled as follows.

Lemma 1.1. Let A and B be non-negative and non-zero m × m matrices, respectively, and α and β are positive numbers. The maximum eigenvalue of

 A αB

βB A



is then the maximum eigenvalue of A +p

αβB. (1.13)

Theorem 1.2. Assume that A₂ =

 A B

Then

λ_n= α_n−1+ β_n−1,

where α_k and β_k satisfy the following recursive relations:

α_k+1 = aα_k+ bβ_k, β_k+1 = p

(a₂α_k+ b₂β_k)(a₃α_k+ b₃β_k), for k ≥ 0, and α₀ = β₀ = 1.

Furthermore, the spatial entropy h(A₂) is equal to

logξ_∗, where ξ_∗ is the maximum root of the following polynomials Q(ξ):

(I) if a₂ = a₃ = 1,

Q(ξ) ≡ 4ξ²(ξ − a)²+ (γ²− 4δ)(ξ − a)²− γ²ξ²− 2γ(2b − aγ)ξ − (2b − aγ)², where

γ = b₂+ b₃ and δ = b₂b₃. (II) if a₂a₃ = 0 and a₂b₃+ a₃b₂ = 1,

Q(ξ) ≡ ξ³− aξ²− δξ + aδ − b.

Moreover, if a2a3 = 0 and a2b3+ a3b2 = 0, then h(A2) = 0.

The proofs of above two theorems are shown in [3].

2. Three-Symbols Problems

In this section,we focus our study on three-symbols problems. We try to generalize the result of Theorem 1.2 to the three-symbols cases.

2.1. Transition Matrices and Spatial Entropy

By the same reason as two symbols on lattice Z2×2, we take a transition

4

matrix A₂ of three symbols on lattice Z2×2 as

The recursive formulae for n+1-th order transition matrices A_n+1 defined on Z2×(n+1) are

for any n ≥ 2.

The definition of spatial entropy h(A₂) of three symbols on lattice Z2×2

is the same as two symbols which can also be proved as h(A₂) = lim

n→∞

1

n log ρ(A_n), (2.4)

see [3].

2.2. Computation of Maximum Eigenvalues and En-tropy

For three symbols, A_n is a 3ⁿ× 3ⁿ matrix, so computing the maximum eigenvalue of A_n (ρ(A_n)) is harder than it is for two symbols. We begin with the study of A₂ of the form

Let λn be the eigenvalue of An, and Un be the corresponding eigenvector of λ_n,

Assume

u_n = v_n= w_n, (2.10)

then (2.9) implies

(A_n+ B_n+ C_n)u_n = λ_nu_n. (2.11) Conversely, under the assumption (2.10), (2.11) implies (2.9). Therefore, (2.8) and (2.11) are equivalent.

We first prove the following lemma which is a generalization of Lemma 1.1. By the assumption (2.12), (2.13) follows. The proof is complete.

Now, we can prove our first theorem.

7

Theorem 2.2. Assume (2.12) holds, and a + b + c ≥ 1, then

h(A₂) = log(a + b + c). (2.14) Proof . Under the assumption (2.12) and by Lemma 2.1,

An+ Bn+ Cn= (a + b + c)

Combining (2.15) with (2.16),

λ_n = 3(a + b + c)ⁿ⁻¹. (2.17) (2.14) follows. The proof is complete.

Remark 2.3. (i) It is of interest to study the case when A₂ is of the form (2.5) but (2.12) fails. A lemma like Lemma 1.1 need to be established, some progress has been made.

(ii) Result of Theorem 2.2 also holds for any number of symbols provides that the assumptions like (2.12) hold.

Lemma 2.4. Let A and B be non-negative and non-zero n × n matrices, re-spectively, and a1, a2, a3, and a4 are positive numbers. The maximum

 is then the maximum eigenvalue of

a₁A +a₄+pa²₄+ 8a₂a₃

2 B. (2.18)

8

Proof . Consider

a₁A − λ a₂B a₂B a₃B a₄B − λ a₁A a3B a1A a4B − λ

= 0.

There are two cases:

Case I. If |a1A − λ| = 0, it is clear that λ = a1A. and we could simplify it to

|I − {[a1A − a2a3B(a1A − λ)⁻¹B][(a4B − λ) − a2a3B(a1A − λ)⁻¹B]⁻¹}²| = 0.

Then, we have

|I + {[a1A − a2a3B(a1A − λ)⁻¹B][(a4B − λ) − a2a3B(a1A − λ)⁻¹B]⁻¹}| = 0 or |I − {[a₁A − a₂a₃B(a₁A − λ)⁻¹B][(a₄B − λ) − a₂a₃B(a₁A − λ)⁻¹B]⁻¹}| = 0.

Since A and B are non-negative and a₁, a₂, a₃, and a₄ are positive, veri-fying that the maximum eigenvalue λ of



2 B are equal is relatively easy. The proof is complete.

Theorem 2.5. Assume that A2 =

 let λ_n be the largest eigenvalue of

|A_n− λ| = 0.

9

Then

λ_n = α_n−1+ β_n−1, (2.19)

where α_k and β_k satisfy the following recursive relations:

α_k=a₁α_k−1+ b₁β_k−1, (2.20)

β_k=1

2{(a₄α_k−1+ b₄β_k−1) + [(a₄α_k−1+ b₄β_k−1)²+

8(a₂α_k−1+ b₂β_k−1)(a₃α_k−1+ b₃β_k−1)]¹²}, (2.21) for k = 1, 2, . . . , n − 1, and

α₀ = 1, β₀ = 2. (2.22)

Furthermore, the spatial entropy h(A₂) is equal to logξ∗, where ξ∗ is the maximum root of the following polynomials Q(ξ):

(I) if a1 = b1 = 1,

Q₁(ξ) ≡ξ⁴− (2 + b₄)ξ³+ (1 − a₄+ 2b₄− 2b₂b₃)ξ²+

[(a4− b4) − 2b2(a3− b3) − 2b3(a2− b2)]ξ − 2(a2− b2)(a3− b3).

(2.23) (II) if a₁ = 0, b₁ = 1,

Q2(ξ) ≡ ξ⁴− b4ξ³− (a4+ 2b2b3)ξ²− 2(a2b3+ a3b2)ξ − 2a2a3. (2.24) (III) if a₁ = 1, b₁ = 0,

Q₃(ξ) ≡ ξ²− b₄ξ − 2b₂b₃. (2.25) Proof . Since the structure of A₂ is special, it is easy to show that for any k ≥ 2, we get

H_k =





A_k B_k B_k Bk Bk Ak

B_k A_k B_k



, and

Hk+1 =





A_k+1 B_k+1 B_k+1 B_k+1 B_k+1 A_k+1 B_k+1 A_k+1 B_k+1



, here

A_k+1 = H_k A =





a1Ak a2Bk a2Bk

a₃B_k a₄B_k a₁A_k a₃B_k a₁A_k a₄B_k



, (2.26)

10

and

B_k+1 = H_k B =





b₁A_k b₂B_k b₂B_k b₃B_k b₄B_k b₁A_k b₃B_k b₁A_k b₄B_k



, (2.27)

A₂ = A and B₂ = B. We know that |A_n+1− λ_n+1| = 0, so

|A_n+1+ 2B_n+1− λ_n+1| = 0. (2.28) Let

α₀ = 1 and β₀ = 2. (2.29)

By induction on k, 1 ≤ k ≤ n, and using (2.26),(2.27),(2.28) and Lemma 2.4, it is straightforward to derive

|α_kA_n−k+1+ β_kB_n−k+1− λ_n+1| = 0, (2.30) with α_k and β_k satisfy (2.20) and (2.21). In particular,

αn=a1αn−1+ b1βn−1, (2.31)

β_n=1

2{(a₄α_n−1+ b₄β_n−1) + [(a₄α_n−1+ b₄β_n−1)²+

8(a₂α_n−1+ b₂β_n−1)(a₃α_n−1+ b₃β_n−1)]¹²}, (2.32) and

λ_n+1 = α_n+ β_n. This proves the first part of the theorem.

The remainder of the proof, demonstrates that h(A₂) = logλ_∗ where λ_∗ is the maximum root of Q(λ). There are three cases:

Case I. From (2.31), if a₁ = b₁ = 1, we have

β_n−1 = α_n− α_n−1. (2.33)

Substituting (2.33) into (2.21), yields αn+1− αn=1

2{[(a4− b4)αn−1+ b4αn] + [((a4 − b4)αn−1+ b4αn)²+ 8((a₂− b₂)α_n−1+ b₂α_n)((a₃− b₃)α_n−1+ b₃α_n)]¹²}. (2.34) Now, let

ξ_n= α_n αn−1

(2.35)

11

and after dividing (2.34) by α_n, we have ξ_n+1− 1 =1

2{[(a₄ − b₄)1 ξn

+ b₄] + [((a₄− b₄)1 ξn

+ b₄)

2

+ 8((a₂− b₂) 1

ξ_n + b₂)((a₃− b₃) 1

ξ_n + b₃)]¹²}. (2.36) (2.36) can be written as the following iteration map:

ξ_n+1= G₁(ξ_n), (2.37)

where

G1(ξ) =1 + 1

2{[(a4− b4)1

ξ + b4] + [((a4− b4)1 ξ + b4)

2

+ 8((a₂− b₂)1

ξ + b₂)((a₃− b₃)1

ξ + b₃)]¹²}. (2.38) We first observe the fixed point ξ∗ of G₁(ξ), i.e., ξ∗ = G₁(ξ∗), is a root of Q(ξ).

Indeed, by letting ξ_n= ξ_n+1= ξ∗ in (2.36), we have ξ∗− 1 =1

2{[(a₄− b₄)1 ξ∗

+ b₄] + [((a₄− b₄)1 ξ∗

+ b₄)

2

+ 8((a₂ − b₂)1

ξ_∗ + b₂)((a₃− b₃)1

ξ_∗ + b₃)]¹²},

which gives us Q(ξ∗) = 0. It can be proven that the maximum fixed point of G₁(ξ) or the maximum root ξ_∗ of Q(ξ) = 0 satisfies 1 ≤ ξ_∗ ≤ 2 and

ξ_n→ ξ∗ as n → ∞. (2.39)

Details are omitted here for brevity. By (2.19),(2.33) and (2.35), we can also prove

λ_n+1

λ_n → ξ∗ as n → ∞. (2.40)

Hence, h(T2) = log ξ∗.

Table 3.1 is to evaluate the value of λ∗, where 4a₂+2a₃+a₄+1 = i, 1 ≤ i ≤ 8, and 4b₂+ 2b₃+ b₄+ 1 = j, 1 ≤ j ≤ 8.

12

ij 1 2 3 4 5 6 7 8 8 2 2.2938 2.2056 2.5214 2.2056 2.5214 2.6180 3

7 1.7900 2 2 2.2599 2 2.2599 2.4142 2.7693

6 1.6180 2 2 2.3593 1.6180 2 2.3028 2.7321

5 1 1 1.6956 2 1 1 2 2.4142

4 1.6180 2 1.6180 2 2 2.3593 2.3028 2.7321

3 1 1 1 1 1.6956 2 2 2.4142

2 1.6180 2 1.6180 2 1.6180 2 2 1

1 1 1 1 1 1 1 1.4142 2

Table 3.1

Case II. If a₁ = 0 and b₁ = 1, then, from (2.31), we have

α_n= β_n−1. (2.41)

Again, substituting (2.41) into (2.21) and letting ξ_n= _β^βn

n−1 lead to ξ_n=1

2{(a₄ 1

ξ_n−1 + b₄+ [(a₄ 1

ξ_n−1 + b₄)²+ 8(a₂ 1

ξ_n−1 + b₂)(a₃ 1

ξ_n−1 + b₃)]¹²}, (2.42) i.e., ξ_n = G₂(ξ_n−1), where

G2(ξ) =1 2{(a4

1

ξ + b4+ [(a4

1

ξ + b4)²+ 8(a₂1

ξ + b₂)(a₃1

ξ + b₃)]¹²}. (2.43) The maximum fixed point ξ∗ of (2.43) is the maximum root of Q(ξ) = 0 in (2.24). It can be also be proven that (2.39) and (2.40) holds in this case.

Table 3.2 is to show the value of λ_∗, where 4a₂+ 2a₃+ a₄+ 1 = i, 1 ≤ i ≤ 8, and 4b₂+ 2b₃+ b₄+ 1 = j, 1 ≤ j ≤ 8.

ij 1 2 3 4 5 6 7 8

8 1.4142 1.8536 1.6956 2.1234 1.6956 2.1234 2.2695 2.7321 7 1.1892 1.5437 1.4945 1.8737 1.4945 1.8737 2.0907 2.5346

6 1 1.6180 1.5214 2 1 1.6180 2 2.5115

5 0 1 1.2599 1.6956 0 1 1.7693 2.2695

4 1 1.6180 1 1.6180 1.5214 2 2 2.5115

3 0 1 0 1 1.2599 1.6956 1.7693 2.2695

2 1 1.6180 1 1.6180 1 1.6180 1.7321 2.3028

1 0 1 0 1 0 1 1.4142 2

13

Table 3.2

Case III. If a₁ = 1 and b₁ = 0, then, from (2.31), we have

α_n= α_n−1. (2.44)

Repeating the above steps, hence we get

ξ_n = b₄+pb²₄+ 8a₂a₃

2 . (2.45)

The maximum fixed point ξ∗ of (2.45) is the maximum root of Q(ξ) = 0 in (2.25). The proof is complete.

Table 3.3 is to show the value of λ∗.

3. Spatial Entropy of Cyclic Cases

In this section, we study the spatial entropy of A₂ when A₂ has certain cyclic structure. Consider Now, we have the following theorem for cyclic A₂.

14

Theorem 3.1. Assume A₂ is of the form (3.1). Then λ∗ ≡ lim

n→∞λ

1

nn, (3.3)

satisfies the limiting equation Q(λ) as follows:

(I)

A B C limiting equation Q(λ) λ∗

(i) E I J Q2(λ) λ^∗₂

(ii) E I J⁰ Q₂(λ) λ^∗₂

(iii) E J I Q₂(λ) λ^∗₂

(iv) E J J⁰ Q1(λ) λ^∗₁

(v) E J⁰ I Q₃(λ) λ^∗₃

(vi) E J⁰ J Q₁(λ) λ^∗₁

(II)

A B C limiting equation Q(λ) λ∗

(i) I E J Q₂(λ) λ^∗₂

(ii) I E J⁰ Q₁(λ) λ^∗₁

(iii) I J E Q1(λ) λ^∗₁

(iv) I J⁰ E Q₂(λ) λ^∗₂

(III)

A B C limiting equation Q(λ) λ∗

(i) J E I Q3(λ) λ^∗₃

(ii) J E J⁰ Q₂(λ) λ^∗₂

(iii) J I E Q₁(λ) λ^∗₁

(iv) J J⁰ E Q3(λ) λ^∗₃

(IV)

A B C limiting equation Q(λ) λ∗

(i) J⁰ E I Q₁(λ) λ^∗₁

(ii) J⁰ E J Q3(λ) λ^∗₃

(iii) J⁰ I E Q₃(λ) λ^∗₃

(iv) J⁰ J E Q₂(λ) λ^∗₂

where,

Q₁(λ) = λ − 1, λ^∗₁ = 1 Q2(λ) = λ²− λ − 1, λ^∗₂ = ¹⁺

√ 5

2 ∼= 1.618033988 Q₃(λ) = λ³− λ²− λ − 1, λ^∗₃ ∼= 1.839286755.

Proof of Case(I)(i). Since the structure of A₂ is similar to (2.5), it is easy to verify that (2.11) is also right to (3.1) and for any k ≥ 2, we have

A_k =





Ek Ik Jk

J_k E_k I_k I_k J_k E_k



.

15

By (2.11), |A_n− λ_n| = 0, so

|E_n+ I_n+ J_n− λ_n| = 0. (3.4) Let

α0 = 1 and β0 = 1. (3.5)

By induction on k, 1 ≤ k ≤ n, and using (2.11), it is straightforward to derive

|αkEn−k+ βkIn−k+ Jn−k − λn| = 0, (3.6) where α_k and β_k satisfy the following recursive relations:

α_k =α_k−1+ β_k−1, (3.7)

β_k =α_k−1+ 1, (3.8)

and

λ_n= 3α_n−2+ β_n−2+ 1. (3.9) By recursive formulae of (3.7) and (3.8), we get

α_n = 5 +√

Substituting (3.10) and (3.11) into (3.9), we have λ_n = 11 + 5√

The proofs of the following cases is similar to Case(I)(i). The proof is com-plete.

16

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在文檔中 二維網格模型中多符號的花樣生成問題 (頁 9-0)