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PHYSICAL REVIEW E 91, 052122 (2015)

Recursion-transform method for computing resistance of the complex resistor network with three arbitrary boundaries Zhi-Zhong Tan ()* Department of Physics, Nantong University, Nantong 226019, China (Received 23 December 2014; published 14 May 2015) We develop a general recursion-transform (R-T) method for a two-dimensional resistor network with a zero resistor boundary. As applications of the R-T method, we consider a significant example to illuminate the usefulness for calculating resistance of a rectangular m × n resistor network with a null resistor and three arbitrary boundaries, a problem never solved before, since Green’s function techniques and Laplacian matrix approaches are invalid in this case. Looking for the exact calculation of the resistance of a binary resistor network is important but difficult in the case of an arbitrary boundary since the boundary is like a wall or trap which affects the behavior of finite network. In this paper we obtain several general formulas of resistance between any two nodes in a nonregular m × n resistor network in both finite and infinite cases. In particular, 12 special cases are given by reducing one of the general formulas to understand its applications and meanings, and an integral identity is found when we compare the equivalent resistance of two different structures of the same problem in a resistor network. DOI: 10.1103/PhysRevE.91.052122

PACS number(s): 05.50.+q, 84.30.Bv, 89.20.Ff, 02.10.Yn

I. INTRODUCTION

The computation of two-point resistance in networks has been a classical problem in circuit theory and graph theory since the German scientist Kirchhoff found the node current law and the circuit voltage law in 1845 [1]. Modeling the resistor network has become a basic method to solve a series of complicated problems in the fields of engineering, science, and technology. The computation of resistances is relevant to a wide range of problems ranging from classical transport in disordered media [2], electromigration phenomena in metallic lines [3], first-passage processes [4], lattice Greens functions [5], random walks [6], and resistance distance [7] to hybrid graph theory [8]. For an up-to-date list of references, see Ref. [9]. Modeling the resistor network can also help to carry on the research of exact finite-size corrections in critical systems [10–15]. In physics and probability theory, mean field theory (also known as self-consistent field theory) studies the behavior of large and complex stochastic models by studying a simpler model such as a finite disordered network [2,3,10–15]. Such models consider a large number of small interacting individual components which interact with each other. The effect of all the other individuals on any given individual is approximated by a single averaged effect, thus reducing a many-body problem to a one-body problem. The mean field theory is widely used in statistical mechanics, condensed system research of complex systems, magnetism, structure transition, and other fields [2,3,10–15]. As is known, many physical systems or real-world systems can be modeled by disordered networks. The disordered structure is characterized by elemental resistors of different resistances ri distributed in certain ranges of values. However, in the case of weak heterogeneity of the network (relatively small range of values for ri ), the mean field approach can be used, namely, a mean-field like approximation

*

[email protected]; [email protected]

1539-3755/2015/91(5)/052122(11)

is valid. In this case it can be a very useful theoretical estimate of the overall resistance between an arbitrary pair of nodes, even obtained by neglecting the spread of the ri and assuming only two values, r and r0 , for horizontal and vertical resistors. Therefore, within this approximation, a sufficiently large class of physical systems can be approximately described by the simple model of an m × n resistor network with arbitrary boundaries. In this paper we obtain the exact analytical solutions of the nonregular m × n resistor network. The exact analytical solution without numerical errors will provide the significant applications for the research of mean field theory and other interdisciplinary fields. The construction of and research on the models of resistor networks therefore make sense for theories and applications. By investigating the development history of the resistor network model, we find it is usually very difficult to obtain the exact resistance in a complex m × n resistor networks with arbitrary boundary because the boundary is like a wall or trap, which affects the behavior of the finite network, and requires not only the circuit theory but also the innovation of mathematical theories and methods [16–30]. For the explicit computation of two-point resistances in a resistor network, Cserti [16] evaluated the two-point resistance using the lattice Green’s function. His study is mainly confined to regular lattices of infinite size. In 2004 Wu [17] formulated a different approach and derived an expression for the two-point resistance in arbitrary finite and infinite lattices in terms of the eigenvalues and eigenvectors of the Laplacian matrix. The Laplacian analysis has also been extended to impedance networks after a slight modification of the formulation of [18]. The Laplacian method has solved a variety of types of resistance network with the normal boundary. Besides the previous cases [17,18], there is also the recent Tan-Zhou-Yang conjecture [19], and a globe [20]. But the applications of the Laplacian approach require a complete knowledge of the eigenvalues and eigenvectors of the Laplacian straightforward to obtain for regular lattices. However, the actual computation depends crucially on the geometry of the network and,

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©2015 American Physical Society

ZHI-ZHONG TAN

PHYSICAL REVIEW E 91, 052122 (2015)

for nonregular lattices such as a network with an arbitrary boundary; it is difficult to solve the eigenvalue problem with arbitrary elements. In recent years, a new method is created by Tan [21], in which it is easy to resolve the problem of nonregular lattices of the cobweb and globe [22–25]. This method splits the derivation into three parts. The first part creates a matrix relation between the current distributions on three successive transverse lines. The second part derives a recursion relation between the current distributions on the boundary. The third part diagonalizes the matrix relation to produce a recurrence relation involving only variables on the same transverse line. Basically, the method reduces the problem from two dimensions to one dimension. Here we refer to this method as the recursion-transform (R-T) method which was named for the first time is in Ref. [26]. Using the R-T method to compute the equivalent resistance relies on just one matrix along one direction, and the resistance is expressed by single summation [21–30]. Especially, the RT method is always valid to calculate the resistance when two opposing boundaries are arbitrary resistors. For example, the R-T method is used solving a cobweb network in 2013 [21,22,28,30], next a globe network [23], and a fen and a cobweb network [24]. Very recently a cobweb network with 2r boundary is resolved [25], a resistor network with an arbitrary boundary is resolved [26], and a hammock network with free boundary is resolved [27]. Actually many real-life problems of resistor networks with complex geometry may have complicated boundaries, such as a nonregular m × n resistor network with a zero resistor and three arbitrary resistors r1 ,r2 , and r3 distribute on boundaries as shown in Fig. 1, where r and r0 are, respectively, resistors in the horizontal and vertical directions except for four boundaries. The difference in the resistance values, r and r0 , of the resistors along the two orthogonal directions represents the anisotropy of crystal lattice, and three arbitrary resistors r1 ,r2 , and r3 expressed the realism and the complexity of boundary conditions. This difficult problem has not been solved before since it is a complex network with five different resistor elements. This paper focus on developing the R-T

FIG. 1. An m × n resistor network with a null resistor and three arbitrary boundaries. Bonds in the horizontal and vertical directions are resistors r and r0 except for four boundaries.

method to derive the general resistance between any two nodes in the two-dimensional m × n resistor network with complex boundaries. Our study shows the R-T method is very interesting and useful for resolving the complex binary resistor network. The organization of this paper is as follows: In Sec. II, we illustrate the R-T theory and approach by making use of the network of Fig. 1. In Sec. III, we give three examples to illustrate the applications of the R-T method by calculating the exact resistance between any two nodes in a nonregular m × n resistor network. In Sec. IV, we give several special cases of one of the main formulas to understand its application and meaning. Section V gives a summary and discussion of the method and results. II. A GENERAL THEORY AND APPROACH

Figure 1 shows a representation of a rectangular m × n resistor network with a null resistor boundary and three other boundaries made by resistors with arbitrary resistances r1 ,r2 , and r3 . Here r and r0 , respectively, are the resistances of the resistors in the horizontal and vertical directions (except for the boundaries), while m and n, respectively, are the numbers of meshes along the vertical and horizontal directions. According to discussions in Secs. I and V, this model can describe several real-world systems and thus can be of interest for several applications. In the following we detail the problem to illustrate the general R-T method. A. Overall train of thought and design

We consider A0 is the origin of the rectangular coordinate system, and we denote nodes of the network by coordinate (x,y). We assume the electric current J is constant and goes from the input d1 (x1 ,y1 ) to the output d2 (x2 ,y2 ). We denote the currents in all segments of the network as shown in Fig. 2. The resulting currents passing through all m + 1 row horizontal resistance of r, respectively, (1) (2) (3) (n) ,Ix,k ,Ix,k , . . . Ix,k ,(0  k  m); the resulting currents are Ix,k passing through all n + 1 column vertical resistance of r0 , respectively, are Ik(1) ,Ik(2) ,Ik(3) , . . . Ik(m) ,(0  k  n).

FIG. 2. Segment of resistor with current directions and parameters.

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RECURSION-TRANSFORM METHOD FOR COMPUTING . . .

Since we are considering a linear resistance of a twodimensional network we can use Ohm’s law to find the resistance Rm×n (d1 ,d2 ). All nodes on the bottom boundary can be uniformly called O as all resistors of the bottom boundary have zero resistance. The voltages between two nodes O and d1 , and two nodes O and d2 are, respectively, Um×n (O,d1 ) = r0

y1 

Ix(j1 )

Um×n (O,d2 ) = r0

and

j =1

y2 

Ix(j2 ) .

j =1

Using Ohm’s law, from Fig. 1 we have ⎞ ⎛ y2 y1 r0 ⎝ (j )  (j ) ⎠ Rm×n (d1 ,d2 ) = I − Ix1 . J j =1 x2 j =1 (j )

(1)

B. Modeling the matrix equation

A segment of the rectangular network is shown in Fig. 2. Using Kirchhoff’s law [1] to analyze the resistor network, the nodes’ current equations and the meshes’ voltage equations can be achieved from Fig. 2. We focus on the four rectangular meshes and nine nodes; this gives the relations,

(i) Ik+1 (m) Ik+1

= (2 +

h)Ik(1)

= (2 +

2h)Ik(i)

= (2 + h +

−

hIk(2)

−

−

hIk(i−1)

h3 )Ik(m)

−

(1) Ik−1 ,

−

i = 1,

hIk(i+1)

hIk(m−1)

−

−

(i) Ik−1 ,

(m) Ik−1 ,

1 < i < m,

i = m,

(2)

where h3 = r3 /r0 and h = r/r0 . Equation (2) can be written in a matrix form after considering the input and output of currents, Ik+1 = Am Ik − Ik−1 − J Hx δk,x ,

(3)

where Ik and Hx are, respectively, m × 1 column matrix, and reads T  (4) Ik = Ik(1) ,Ik(2) ,Ik(3) , · · · ,Ik(m) , (H1 )i = h(−δi,y1 + δi,y1 +1 ) and

obtain two matrix equations to model the boundary currents, I1 = [Am − (2 − h1 )E]I0

and

h1 I0 + I2 = Am I1 , (7)

In−1 = [Am − (2 − h2 )E]In

and

h2 In + In−2 = Am In−1 , (8)

where E is the m × m identity matrix, and matrix Am is given by (6). Above Eqs. (3)–(8) are all the equations we need to calculate the equivalent resistance of the m × n resistor network with three arbitrary boundaries. However, it is impossible for us to obtain the solution of the matrix current Ik directly from Eqs. (3)–(8). For this, we established the following approach of the matrix transformation to solve this problem indirectly. C. Approach of the matrix transformation

(j )

How to calculate the local currents Ix1 and Ix2 is the key to the problem. We are going to resolve the problem by modeling the matrix equation in terms of Kirchhoff’s law.

(1) Ik+1

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(H2 )i = h(δi,y2 − δi,y2 +1 ), (5)

where [ ]T denotes matrix transpose, and (Hk )i is the elements of Hx with the injection of current J at d1 (x1 ,y1 ) and the exit of current J at d2 (x2 ,y2 ). And ⎛ ⎞ 2+h −h 0 ··· 0 2 + 2h −h ··· ··· ⎜ −h ⎟ ⎜ . ⎟ .. . . . ⎟. . . . . Am = ⎜ . . . . ⎜ . ⎟ ⎝ 0 ⎠ ··· −h 2 + 2h −h 0 ··· 0 −h 2 + h + h3

In this section we rebuild a new difference equation and solve Eq. (3) indirectly by means of matrix transformations. We consider the solution of (3) in the absence of an injected current, namely, J = 0. The eigenvalues ti (i = 1,2, . . . ,m) of Am are the m solutions of the equation, det |Am − tE| = 0.

Since Am is a special tridiagonal matrix which is Hermitian it can be diagonalized by a similarity transformation to yield Pm Am (Pm )−1 = Tm ,

(10)

where Tm = diag{t1 ,t2 , . . . ,tm } is a diagonal matrix with eigenvalues ti of Am in the diagonal, and column vectors of (Pk )−1 are eigenvectors of Am . To find the exact eigenvalues ti and the exact eigenvectors of the transformation matrix (Pm )−1 is the key to the problem. Normally, it is difficult to find them. However, we can obtain the exact eigenvalues and the exact eigenvectors in the case of h3 = {0,h,2h} which will be examples in the following section. Next we apply Pm on the left-hand side of (3) and write Pm Ik+1 = Pm Am Ik − Pm Ik−1 − J Pm Hx δk,x .

(11)

By (10) and (11) we define Pm Ik = Xk

or

Ik = P−1 m Xk ,

(12)

where Xk is an m × 1 column matrix, and reads T  Xk = Xk(1) ,Xk(2) , . . . ,Xk(i) , . . . ,Xk(m) ,

(13)

where [ ]T denote matrix transpose. Assuming the row vectors of matrix Pm is Pi = [pi,1 ,pi,2 ,pi,3 , · · · ,pi,m ],

(14) −1

with the column vectors of the inverse matrix (Pm ) , 1 [pi,1 ,pi,2 , . . . ,pi,m ]T , (15) (Pi )−1 = (Pi ,Pi )m  2 where (Pi ,Pi )m = m j =1 pi,j . Thus we convert (3) into a new matrix equation after making use of (11) and (12),

(6) Next let us consider the boundary conditions of the left and right edges in the network. Applying Kirchhoff’s laws to a single mesh adjacent to each of the left and right boundaries we

(9)

(i) (i) Xk+1 = ti Xk(i) − Xk−1 − J hδk,x ζx,i ,

(16)

where (read ζx1 ,i as ζ1,i , ζx2 ,i as ζ2,i )

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ζ1,i = −pi,y1 + pi,y1 +1 and ζ2,i = pi,y2 − pi,y2 +1 .

(17)

ZHI-ZHONG TAN

PHYSICAL REVIEW E 91, 052122 (2015)

Supposing λi and λ¯ i are the roots of the characteristic equation for Xk(i) , from (16) we obtain

  1 1 λi = ti + ti2 − 4 and λ¯ i = ti − ti2 − 4 . (18) 2 2 Next, we consider the solution of (16) when the injection of current J is at d1 (x1 ,y1 ) and the exit of current J is at d2 (x2 ,y2 ). We obtain the piecewise solution, (i) (i) − X1(i) Fk−2 , Xk(i) = X2(i) Fk−1

1  k  x1 ,

Xx(i)1 +1 = ti Xx(i)1 − Xx(i)1 −1 − hJ ζ1,i ,

III. APPLICATIONS OF THE R-T METHOD A. Several definitions

In order to facilitate the expression of the equivalent resistance of resistor network, we define variables Sk,i , ti , and λi ,λ¯ i for later uses expressed as S1,i = sin(y1 θi ), S2,i = sin(y2 θi ), ti = λi + λ¯ i = 2(1 + h) − 2h cos θi ,

λi = 1 + h − h cos θi + (1 + h − h cos θi )2 − 1,

λ¯ i = 1 + h − h cos θi − (1 + h − h cos θi )2 − 1,

(19) (20)

(i) (i) Xk(i) = Xx(i)1 +1 Fk−x − Xx(i)1 Fk−x , x1  k  x2 , 1 1 −1

Xx(i)2 +1 = ti Xx(i)2 − Xx(i)2 −1 − hJ ζ2,i ,

(21) (22)

(i) (i) Xk(i) = Xx(i)2 +1 Fk−x − Xx(i)2 Fk−x , x2  k < n, 2 2 −1

(23)

and

and

(i) (i) (i) (i) αu,x = Fx(i) + (hu − 1)Fx−1 , βk,s = α1,x α (i) , k 2,n−xs (i) (i) (i) G(i) n = Fn+1 + (h2 − 1)(h1 − 1)Fn−1 + (h1 + h2 − 2)Fn .

h1 X0(i) + X2(i) = ti X1(i) , (i) (i) h2 Xn(i) + Xn−2 = ti Xn−1 .

(25) The above Eqs. (3)–(25) are suitable for all cases when h3 is an arbitrary constant. In particular, from the above equations we can obtain after some algebra and reduction the two exact solutions of Xx(i)1 and Xx(i)2 needed in our resistance calculation (1) when we take special values of h3 = {0,h,2h}. D. Derivation of the resistance formula (j )

(31) We consider A0 is the origin of the rectangular coordinate system, and we denote nodes of the network by coordinate {x,y}; we have the following main results in three cases of h3 = {0,h,2h}. B. The main results in the case of h3 = h

When h3 = h(r3 = r), the equivalent resistance between any two nodes d1 (x1 ,y1 ) and d2 (x2 ,y2 ) in a nearly m × n resistor network with two arbitrary boundaries can be written as

(j )

According to Eq. (1), the key currents Ix1 and Ix2 must be calculated for deducing the equivalent resistance Rm×n (d1 ,d2 ). (j ) (j ) From (12) we can easily obtain Ix1 and Ix2 by making use of (j ) (j ) Xx1 and Xx2 , m  1 (j ) Ik = X(i) pi,j , (26) (Pi ,Pi )m i=1 k where pi,j is defined in (14), and from (1),

Rm×n (d1 ,d2 ) (i) 2 (i) (i) 2 m − 2β1,2 S1,i S2,i + β2,2 S2,i 2r0  β1,1 S1,i = , (32) (i) 2m + 1 i=1 (1 − cos θi )Gn (i) where θi = (2i − 1)π /(2m + 1), and Sk,i ,βk,s ,G(i) are, n respectively, defined in (28) and (31). In particular, when n → ∞,x1 ,x2 → ∞ with x1 − x2 finite, we have results in the semi-infinite case,

Rm×n (d1 ,d2 )

⎞ ⎛ y2 y1 m m     r0 ⎝ X(i) pi,j − Xx(i)1 pi,j ⎠ . = (Pi ,Pi )m J i=1 x2 j =1 i=1 j =1

(30)

 (i) Fk(i) = λki − λ¯ ki /(λi − λ¯ i ),Fk(i) = Fk+1 − Fk(i) ,

(24) (i) = (ti + h2 − 2)Xn(i) Xn−1

(29)

where θi is determined by h3 in different cases, which will be explicitly given in the following sections. Define variables (i) (i) ,βk,s , and G(i) Fk(i) ,αu,x n for later uses by

where we define Fk(i) = (λki − λ¯ ki )/(λi − λ¯ i ). We conduct the same matrix transformation as used in (11), applying Pm to (7) and (8) on the left-hand sides; we are led to X1(i) = (ti + h1 − 2)X0(i)

(28)

Rm×∞ (d1 ,d2 ) =

|x −x | m 2 2 + S2,i − 2λ¯ i 2 1 S1,i S2,i 2r  S1,i  . 2m + 1 i=1 (1 + h − h cos θi )2 − 1

(33)

(27) Finally, we obtain the resistance Rm×n (d1 ,d2 ). The above approach is called the recursion-transform method. Obviously, the above method of computation is simple and clear, and is easy to understand, which has good theoretical significance and application value for science and technology. As applications, we give several new exact formulas by the following three examples in terms of the R-T method.

The derivation process of the two formulas are as follows. 1. The exact eigenvalues and eigenvectors

When h3 = h, solving (9) and (10) yields the exact eigenvalues ti ,

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ti = 2(1 + h) − 2h cos θi ,

(i = 1,2, . . . m),

(34)

RECURSION-TRANSFORM METHOD FOR COMPUTING . . .

where θi = (2i − 1)π/(2m + 1), and ⎛

 cos 1 − 12 θ1  ⎜ ⎜ cos 1 − 12 θ2 Pm = ⎜ .. ⎜ ⎝ . 1 cos 1 − 2 θm

PHYSICAL REVIEW E 91, 052122 (2015)

 cos 2 − 12 θ1  cos 2 − 12 θ2 .. . 1 cos 2 − 2 θm

 ⎞ cos m − 12 θ1  ⎟ cos m − 12 θ2 ⎟ ⎟, .. ⎟ ⎠ .  cos m − 12 θm

··· ··· .. . ···

(35)

with the following inverse matrix, (Pm )−1 =

4 [Pm ]T , 2m + 1

(36)

where [ ]T denote matrix transpose. Substituting (35) into (17) gives       1 1 1 θi − cos y1 − θi = −2 sin ζ1,i = cos y1 + θi sin(y1 θi ), 2 2 2       1 1 1 θi − cos y2 + θi = 2 sin θi sin(y2 θi ). ζ2,i = cos y2 − 2 2 2

(37)

Substituting (34) into (18), we obtain (30).

2. General solutions of the matrix equations

In this section we are going to derive the exact solution of Xx(i)k by (19)–(25) together with (34). Substituting (24) into (19) to eliminate X1(i) and X2(i) , we obtain (i) (i) Xk(i) = α1,k X0 , 1  k  x1 ,

(38)

(i) (i) = Fx(i) + (hu − 1)Fx−1 is defined in (31). Using (38) with k = x1 − 1,x1 , together with (20) and (21), we obtain where αu,x (i) (i) (i) Xk(i) = α1,k X0 − hJ Fk−x ζ , x1  k  x2 , 1 1,i

(39)

(i) (i) (i) where Fk−x α (i) − Fk−x α (i) = α1,k is used. Using (25) to eliminate Xn(i) yields 1 1,x1 −1 1 +1 1,x1 (i) (i) (ti qi − h2 )Xn−1 = qi Xn−2 ,

(40)

where qi = ti + (h2 − 2) is defined. To obtain the key parameters Xx(i)1 and Xx(i)2 needed in our resistance calculation (27), we also need several independent (i) (i) ,Xn−1 satisfy (23), we therefore obtain four independent equations. Together equations. Since Xx(i)2 −1 ,Xx(i)2 satisfy (39), and Xn−2 (i) (i) with (22) and (40) we have six independent equations relating the six unknowns X0(i) ,Xx(i)2 −1 ,Xx(i)2 ,Xx(i)2 +1 ,Xn−2 ,Xn−1 , namely, ⎞ ⎛ (i) ⎞ ⎛ (i) ⎛ (i) ⎞ X0 α1,x2 −1 −1 0 0 0 0 Fx2 −x1 −1 ζ1,i ⎟ ⎜ (i) ⎟ ⎜ (i) ⎜ (i) ⎟ ⎜X ⎟ ⎜ α1,x 0 −1 0 0 0⎟ ⎜ Fx2 −x1 ζ1,i ⎟ 2 ⎟ ⎜ x2 −1 ⎟ ⎜ ⎜ ⎟ (i) ⎟ ⎜ ⎟ ⎜ 0 ⎜ ⎟ −1 ti −1 0 0 ⎟ ⎜ Xx2 ⎟ ⎜ ζ2,i ⎜ ⎟, (41) = hJ ⎜ ⎟ ⎟ ⎜ (i) ⎜ ⎟ (i) (i) ⎜ ⎟ ⎜ 0 0 Fn−x2 −3 −Fn−x2 −2 1 0 ⎟ ⎜ ⎟ 0 ⎟ ⎜Xx2 +1 ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ (i) ⎟ ⎜ (i) (i) ⎝ ⎠ 0 0 Fn−x −Fn−x 0 1 ⎠ ⎝ Xn−2 ⎠ ⎝ 0 2 −2 2 −1 (i) 0 0 0 0 0 qi ci Xn−1 where ci = h2 − ti qi . Solving (41), we obtain after some algebra and reduction the solutions of X0(i) and Xx(i)2 needed in the process of resistance calculation, X0(i)

=

Xx(i)2

(i) (i) ζ + α2,n−x ζ α2,n−x 1 1,i 2 2,i

(ti − 2)G(i) n =

(i) (i) β1,2 ζ1,i + β2,2 ζ2,i

(ti − 2)G(i) n 052122-5

hJ,

hJ,

(42)

(43)

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PHYSICAL REVIEW E 91, 052122 (2015)

(i) (i) where αu,x , βk,x , and G(i) n are defined in (31). Substituting (42) into (38) with k = x1 yields

Xx(i)1 =

(i) (i) β1,1 ζ1,i + β1,2 ζ2,i

(ti − 2)G(i) n

hJ.

(44)

3. Derivation of the resistance formula

y From (27) it is clear that the key summation j =1 pi,j must be calculated for deducing the equivalent resistance Rm×n (d1 ,d2 ). By (35) and (36) we achieve the following equation:   y y   1 sin(yθi ) θi = pi,j = cos j − . (45) 2 2 sin(θi /2) j =1 j =1 Substituting (45) with y = y1 and y2 into (27), we obtain Rm×n (d1 ,d2 ) =

 m   sin(y2 θi ) sin(y1 θi ) 2r0 Xx(i)2 − Xx(i)1 . J (2m + 1) i=1 sin(θi /2) sin(θi /2)

(46)

After substituting Xx(i)1 and Xx(i)2 from (43) and (44) into (46), finally, we obtain the main result (32).

4. Derivation of formula (33)

In the case of n → ∞ with m finite, from (30) and (31) we have (i) β1,2

(ti − 2)G(i) n

=

1 −x2 −1 1 −x2 −1 1 −x2 1 −x2 + b1 b2 λ¯ n+x + a2 b1 λn−x + a1 b2 λ¯ n−x a1 a2 λn+x i i i i ,  (λi − λ¯ i ) a1 a2 λn−1 − b1 b2 λ¯ n−1 i i

(47)

where ak = (λi + hk − 1) and bk = (λ¯ i + hk − 1). By (30) we know λi > 1 > λ¯ i > 0, thus    ¯ (i) (i) |x | β1,2 β1,2 λxi 1 −x2 λ¯ i 1 −x2 1 λi + h1 − 1 ¯ x1 +x2 −1 x1 −x2 λ , lim lim = + = = . (x1 < x2 ). λ i i (i) n→∞ n→∞ (t − 2)G(i) λ i + h1 − 1 λi − λ¯ i λi − λ¯ i λi − λ¯ i x→∞ (ti − 2)Gn n i (48)  Substituting (48) into (32), we obtain (33) after λi − λ¯ i = 2 (1 + h − h cos θi )2 − 1 is used.

C. The main results in the case of h3 = 2h

When h3 = 2h(r3 = 2r), the equivalent resistance between any two nodes d1 (x1 ,y1 ) and d2 (x2 ,y2 ) in the m × n resistor network with two arbitrary boundaries is Rm×n (d1 ,d2 ) =

(i) 2 (i) (i) 2 m − 2β1,2 S1,i S2,i + β2,2 S2,i r0  β1,1 S1,i , (i) m i=1 (1 − cos θi )Gn

(49)

where θi = (2i − 1)π /(2m). In particular, when n → ∞,x1 ,x2 → ∞ with x1 − x2 finite, we have the result in the semi-infinite case, Rm×∞ (d1 ,d2 ) =

|x −x | m 2 2 + S2,i − 2λ¯ i 2 1 S1,i S2,i r  S1,i  . m i=1 (1 + h − h cos θi )2 − 1

The derivation process of the two formulas are as follows. When h3 = 2h, from (9) and (10) we obtain (29) and θi = (2i − 1)π/(2m), and    ⎞ ⎛ cos 1 − 12 θ1 cos 2 − 12 θ1 · · · cos m − 12 θ1    ⎟ ⎜ ⎜ cos 1 − 12 θ2 cos 2 − 12 θ2 · · · cos m − 12 θ2 ⎟ ⎜ ⎟, Pm = ⎜ ⎟ .. .. .. .. ⎝ ⎠ . . 1 . 1 . 1 cos 1 − 2 θm cos 2 − 2 θm · · · cos m − 2 θm 052122-6

(50)

(51)

RECURSION-TRANSFORM METHOD FOR COMPUTING . . .

PHYSICAL REVIEW E 91, 052122 (2015)

with the inverse matrix, 2 (52) [Pm ]T . m Although (35) and (51) have the same structure, but in which the elements θi are different from each other. When we implement the same calculation process as used in Sec. III B, formulas (49) and (50) are proved. (Pm )−1 =

D. The main results in the case of h3 = 0

When h3 = 0(r3 = 0), the equivalent resistance between any two nodes d1 (x1 ,y1 ) and d2 (x2 ,y2 ) in the m × n resistor network with two arbitrary boundaries can be written as Rm×n (d1 ,d2 ) =

m 2 2 − 2β1,2 S1,i S2,i + β2,2 S2,i h1 h2 (y2 − y1 )2 r0  β1,1 S1,i , r0 + m[(n − 1)h1 h2 + h1 + h2 ] m i=2 (1 − cos θi )G(i) n

(53)

where θi = (i − 1)π /m. In particular, when n → ∞,x1 ,x2 → ∞ with x1 − x2 finite, we have the result in the semi-infinite case, Rm×∞ (d1 ,d2 ) =

|x −x | m 2 2 + S2,i − 2λ¯ i 2 1 S1,i S2,i r  S1,i  . m i=2 (1 + h − h cos θi )2 − 1

(54)

Note that Eq. (54) is different from Eq. (50) because their θi are different from each other. The derivation process of the two formulas are as follows. When h3 = 0, from (9) and (10) we obtain (29) and θi = (i − 1)π/m, and √ √ √ ⎞ ⎛ 1/ 2 1/ 2 ··· 1/ 2    ⎟ ⎜ ⎜ cos 1 − 12 θ2 cos 2 − 12 θ2 · · · cos m − 12 θ2 ⎟ ⎟ ⎜ Pm = ⎜ ⎟, .. .. .. .. ⎟ ⎜ . . . . ⎠ ⎝    1 1 1 cos 1 − 2 θm cos 2 − 2 θm · · · cos m − 2 θm

with the inverse matrix, 2 [Pm ]T . (56) m Although (52) and (56) have the same structure, the elements θi are different from each other. Here the calculation process is partly different from Sec. III B, where we must deal with a singular boundary condition in the case of i = 1. When i = 1, we can’t work out Xk(1) from the above equations since the null resistor on the up-down boundaries is a singular condition. When i = 1, from θi = (i − 1)π/m and (29) we have the eigenvalue t1 = 2; substituting in (19)–(23), we obtain (Pm )−1 =

(1) (1) X1(1) = X2(1) = · · · = Xn−2 = Xn−1 ,

y2 , by conservation of current the summation over segments at a given horizontal axis k must yield J for y1 < k  y2 and zero otherwise,  n  J, y1 < i  y2 , (60) Ik(i) = 0, otherwise. k=0

Substituting (60) into (59), we obtain   1 1 J (1) Xk(1) = √ |y2 − y1 |,(k = 0,n). (n − 1)Xk + + h1 h2 2 As a result,

(57)

Xk(1) =

and using (24) and (25) yields X0(1) =

1 (1) X and h1 1

Xn(1) =

1 (1) X . h2 n−1

(58)

k=0

1   (i) =√ Ik . 2 i=1 k=0 m

Xk(1)

√ ,(k = 0,n). [(n − 1)h1 h2 + h1 + h2 ] 2

X0(1) =

n

Xn(1)

(59)

The summations in (59) are taken over all longitudinal current segments on the network. Since the current J flows from a node at horizontal axis y1 to a node at horizontal axis

J h1 h2 |y2 − y1 |

(61)

When k = 0,n, from (58) and (61) we have

Using (12) and (55) we have n 

(55)

J h2 |y2 − y1 |

√ ,(k = 0), [(n − 1)h1 h2 + h1 + h2 ] 2 J h1 |y2 − y1 | = √ ,(k = n). [(n − 1)h1 h2 + h1 + h2 ] 2

(62)

When i  2, we implement the same calculation process as used in Sec. III B; we obtain (43) and (44) with i  2. Applying (55) and (56) to (26) with k = x1

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yields y1  i=1

Ix(i)1 =

2 m



   y1 m y1 (1)  (i)  1 Xx1 cos k− θi . (63) √ Xx1 + 2 2 i=2 k=1

When x1 = {0,n}, substituting (61) into (63) yields  y1  J h1 h2 y1 |y2 − y1 | 1 (i) Ix1 = m (n − 1)h1 h2 + h1 + h2 i=1  m  (i) sin y1 θi + Xx1 . sin(θi /2) i=2

(64)

Similarly, we have y2  i=1

Ix(i)2



J h1 h2 y2 |y2 − y1 | (n − 1)h1 h2 + h1 + h2  m  (i) sin y2 θi + Xx2 . sin(θi /2) i=2

1 = m

According to the identity transform of a trigonometric function, we have 2 2 + S2,i = 1 − cos(y1 + y2 )θi cos(y1 − y2 )θi , S1,i 1 S1,i S2,i = [cos(y1 − y2 )θi − cos(y2 + y2 )θi ]. 2 When m → ∞ and y1 ,y2 → ∞, we must transform   m m (y1 ,y2 ) = + p, + q , 2 2

where p,q  m are integers, and p − q = y1 − y2 are finite. As θk = (2k − 1)π/(2m + 1), we have   m+p+q (2i − 1)π lim (y1 + y2 )θi = lim m→∞ m→∞ 2m + 1   1 = i− π. 2 Thus,

(65)

Substituting (64) and (65) into (1), we obtain   h1 h2 (y2 − y1 )2 r0 Rm×n (d1 ,d2 ) = m (n − 1)h1 h2 + h1 + h2  m  r0  (i) sin y2 θi (i) sin y1 θi + Xx2 − Xx1 . J m i=2 sin(θi /2) sin(θi /2)

m→∞

1 cos(y1 − y2 )θi . 2 Applying (68) and (69) to (33), we obtain lim S1,i S2,i =

m→∞

(69)

m |x −x | 2r  1 − λ¯ i 1 2 cos(y1 − y2 )θi  m→∞ 2m + 1 (1 + h − h cos θi )2 − 1 i=1

R∞×∞ (d1 ,d2 ) = lim =

(66) When x1 = 0 or n, we replace (61) by (62) so that we obtain the same result (66). Finally, we obtain the main result (53) after further substituting Xx(i)1 and Xx(i)2 from (43) and (44) into (66). In particular, when n → ∞, x1 ,x2 → ∞ with x1 − x2 finite, we substitute (48) into (53); thus formula (54) in the semi-infinite case is proved.

2  2 + S2,i = 1, lim S1,i

r π



0

π

|x −x | 1 − λ¯ θ 1 2 cos(y1 − y2 )θ  dθ. (1 + h − h cos θ )2 − 1

(70)

Thus, Eq. (67) is proved when h3 = h. Similarly, when h3 = 0 and 2h, we have the same result (67). That is to say the resistance of the infinite resistor network has nothing to do with the boundary conditions. Thus, formula (67) is a general formula of the infinite two-dimensional resistor network with arbitrary boundaries. IV. APPLICATIONS OF RESISTANCE FORMULA (32)

E. A general formula of the infinite resistor network

When m,n → ∞, but x1 − x2 and y1 − y2 are finite, the equivalent resistance between any two nodes d1 (x1 ,y1 ) and d2 (x2 ,y2 ) in an arbitrary m × n resistor network with arbitrary boundaries can be written as  |x −x | r π 1 − λ¯ θ 2 1 cos(y1 − y2 )θ  R∞×∞ (d1 ,d2 ) = dθ, (67) π 0 (1 + h − h cos θ )2 − 1  where λ¯ θ = 1 + h − h cos θ − (1 + h − h cos θ )2 − 1. The derivation process of the formula (67) is as follows. In order to prove Eq. (67), we need to refer an integral theory as follows. If θk = (2k − 1)π /(2m + 1), we have θk = θk+1 − θk = 2π/(2m + 1). In the limit of m → ∞, we have  π m  1 1 g(θk ) = g(θ )dθ, (68) lim m→∞ 2m + 1 2π 0 k=1 which is an identity valid for any function g(θk ).

The above formulas obtained in this paper are concise and significant but complicated to understand; we will give several special cases of (32) to understand its application and meaning. We consider a nonregular m × n resistor network with a null resistor boundary and two other boundaries made by resistors with arbitrary resistances r1 and r2 as shown in Fig. 1, where r3 = r, and r and r0 , respectively, are the resistances in the horizontal and vertical directions (except for the left, right, and bottom boundaries). When h1 and h2 , m or n are special number values, from (32) we have the following special cases. Case 1. When h1 = 1 and h2 is an arbitrary constant, from (32) we have Rm×n (d1 ,d2 ) (i) 2 (i) (i) 2 m − 2β1,2 S1,i S2,i + β2,2 S2,i 2r0  β1,1 S1,i =  , (71)  (i) 2m + 1 i=1 (1 − cos θi ) Fn+1 + (h2 − 1)Fn(i)

where θi = (2i − 1)π/(2m + 1) (which is used in all (i) (i) cases of the following), and βk,s = Fx(i) [Fn−x + (h2 − s k (i) 1)Fn−x ] is reduced from (31). Please note that formula s −1

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RECURSION-TRANSFORM METHOD FOR COMPUTING . . .

(71) is fully consistent with the result in Ref. [26], which shows that the result in [26] is our special case. Meanwhile, this comparison verifies the correctness of formula (32) primitively. Case 2. When d1 = (x1 ,0) is at the bottom edge, and d2 = (x2 ,y2 ) is an arbitrary node, from (32) we have m  (i)  2r0  β2,2 sin2 (y2 θi ) Rm×n ({x1 ,0},{x2 ,y2 }) = , 2m + 1 i=1 G(i) 1 − cos θi n (72) (i) β2,2

and G(i) where n are defined in (31). Case 3. When d1 = (0,y1 ) and d2 = (0,y2 ) are both at the left edge, from (32) we have  (i)  m 2r1  α2,n (S1,i − S2,i )2 , Rm×n ({0,y1 },{0,y2 }) = 2m + 1 i=1 G(i) 1 − cos θi n (73) (i) α2,n

where = Case 4. When have

(i) + (h2 − 1)Fn−1 is defined in (31). n → ∞, with x1 = x2 = 0, from (73)

Fn(i)

Rm×∞ ({0,y1 },{0,y2 }) =

m  

Note that the fan network was resolved by [24] in 2014; formula (78) is completely equivalent to the result of [24]. Obviously the result of [24] is a special case of formula (32). Case 9. When h2 = 0, from (32) we have Rm×n (d1 ,d2 ) (i) 2 (i) (i) 2 m − 2β1,2 S1,i S2,i + β2,2 S2,i 4r  β1,1 S1,i = , (i) (i) 2m + 1 i=1 Fn + (h1 − 1)Fn−1

(79)

(i) (i) where we re-define βk,s = [Fx(i) + (h1 − 1)Fx(i) ]Fn−x . s k k −1 Case 10. When h1 = h2 = 0, from (32) we have

Rm×n (d1 ,d2 ) =

(i) 2 (i) (i) 2 m − 2β1,2 S1,i S2,i + β2,2 S2,i 4r  β1,1 S1,i , (i) 2m + 1 i=1 Fn

(80)

(i) (i) where we re-define βk,s = Fx(i) Fn−x . s k Case 11. When h1 = 2,h2 = 0, from (32) we have

we Rm×n (d1 ,d2 )



λi − 1 2r1 2m + 1 i=1 λi + h1 − 1 2 S1,i − S2,i × . (74) 1 − cos θi

Case 5. When m,n → ∞, with x1 = x2 = 0, applying (68) to (74), we have R∞×∞ ({0,y1 },{0,y2 })  r1 π (λθ − 1)[1 − cos(y1 − y2 )θ ] = dθ. π 0 (λθ + h1 − 1)(1 − cos θ )

PHYSICAL REVIEW E 91, 052122 (2015)

(75)

Case 6. When d1 = (0,y1 ) is at the left edge, and d2 = (n,y2 ) is at the right edge, from (32) we have Rm×n ({0,y1 },{n,y2 }) (i) 2 (i) 2 m + h2 α1,n S2,i − 2h1 h2 S1,i S2,i 2r0  h1 α2,n S1,i = , 2m + 1 i=1 (1 − cos θi )G(i) n

(76) (i) (i) = Fn(i) + (hk − 1)Fn−1 . where αk,n Case 7. When d1 = (x1 ,y) and d2 = (x2 ,y) are both at the same horizontal axis, from (32) we have

Rm×n ({x1 ,y},{x2 ,y})  m  2r0  β1,1 − 2β1,2 + β2,2 sin2 (yθi ). (77) = 2m + 1 i=1 (1 − cos θi )G(i) n

=

(i) 2 (i) (i) 2 m − 2β1,2 S1,i S2,i + β2,2 S2,i 4r  β1,1 S1,i , n n 2m + 1 i=1 λi + λ¯ i

(81)

(i) (i) = (λxi k + λ¯ xi k )Fn−x . where we re-define βk,s s The above 11 cases show our main formulas have good applications and can deduce a lot of new results. In the following we are going to continue giving several results of a simple example to test the reliability of the formula. Case 12. When m = 1, Fig. 1 degrades into an 1 × n resistor network with the left and right arbitrary boundaries as shown in Fig. 3. When we compute the resistance between nodes Ak (k,0) and Bk (k,1), we have y1 = 0,y2 = 1 and i = 1. Thus, from θi = (2i − 1)π /(2m + 1) we have θ1 =√π/3, and S1,i = sin(y1 θ1 ) = 0, S2,i = sin(y2 θ1 ) = sin(θ1 ) = 3/2, and    1 2 1 1 + h − 1. λ1 = 1 + h + (82) 2 2

From (32) we obtain the resistance, RAk ,Bk (n) r0   (1)  (1) (1) Fn−k +(h2 −1)Fn−k−1 Fk(1) + (h1 −1)Fk−1 = . (1) (1) Fn+1 +(h2 −1)(h1 − 1)Fn−1 +(h1 + h2 − 2)Fn(1) (83)

Case 8. When h1 = h2 = 1, the network degrades into a fan network [24]; from (32) we have Rm×n (d1 ,d2 ) =

(i) 2 (i) (i) 2 m − 2β1,2 S1,i S2,i + β2,2 S2,i 2r0  β1,1 S1,i , (i) 2m + 1 i=1 (1 − cos θi )Fn+1

(i) (i) (i) reduces to βk,s = Fx(i) Fn−x . where βk,s s k

(78) FIG. 3. The 1 × n resistor network with two arbitrary boundaries. 052122-9

ZHI-ZHONG TAN

PHYSICAL REVIEW E 91, 052122 (2015)

In particular, when h1 = h2 = 1, from (83) we have (1) Fk(1) Fn−k + λ¯ n+1 + λn−2k + λ¯ n−2k λn+1 RAk ,Bk (n) 1 1 1 1 = . = √ (1) n+1 2 + 4h ¯ r λn+1 − λ h hFn+1 1 1

(84) If we define λ1 = eL1 , λ¯ 1 = e−L1 , then (84) can be written as RAk ,Bk (n) cosh[(n + 1)L1 ] + cosh[(n − 2k)L1 ] . = √ r h2 + 4h sinh[(n + 1)L1 ]

(85)

When set k = 0 in (83), we have RA0 ,B0 (n) =

 (1)  r1 Fn(1) + (h2 − 1)Fn−1

(1) (1) Fn+1 + (h2 − 1)(h1 − 1)Fn−1 + (h1 + h2 − 2)Fn(1)

.

(86) When h1 = 1, from (86) we have RA0 ,B0 (n) = r0

(1) Fn(1) + (h2 − 1)Fn−1 (1) Fn+1 + (h2 − 1)Fn(1)

λn − λ¯ n + (h2 − 1)(λn−1 − λ¯ n−1 ) λn+1 − λ¯ n+1 + (h2 − 1)(λn − λ¯ n ) sinh(nL1 ) + (h2 − 1) sinh[(n − 1)L1 ] = 1− , (87) sinh[(n + 1)L1 ] + (h2 − 1) sinh(nL1 )

= 1−

where λ1 = eL1 and λ¯ 1 = e−L1 are used. Formula (87) fully meets with the result of (59) in [26]. When m = 1 and n = 1, there is F0 = 0; from (86) we have   F1(1) + (h2 − 1)F0(1) RA0 ,B0 (1) = r1 F2(1) + (h1 + h2 − 2)F1(1) =

r1 (r + r2 ) r1 (h + h2 ) = . h + h1 + h2 r + r1 + r2

(88)

Obviously, formula (88) is in complete accord with the result of the actual circuit. In fact, the results we obtained are inevitably perfect since all of the processes of algebra and reduction are strict and self-consistent. V. SUMMARY AND DISCUSSION

We considered a profound problem of two-point resistance in the two-dimensional resistor network with a null resistor edge and three arbitrary boundaries by the improved R-T method. Actually many real-life problems involve resistor networks with complex geometry but disordered structure. Our study shows the boundary condition of the resistor network is a very important problem which affects the behavior of the finite network; for example, from case 1 to case 12 all of the different resistance formulas are from the same formula (32), but the resistance formulas are changed with the change of the boundary conditions. In spite of this, the resistance formulas (32), (33), (49), (50), (53), and (54) are still concise and symmetrical because of excellent characteristics of the R-T method and we find a good parameter definition of Eq. (31). These exact analytical solutions provide the significant theoretical formulas for the interdisciplinary

research and applications in physics, mathematics, biology, chemistry, numerical studies of model systems, and so forth. We verified that our conclusions are correct by comparing two results (71) and (78) with the results of Refs. [26] and [24], respectively, and further used two special results (87) and (88) to test the correctness of our conclusion. The reason why the Green’s function technique and the Laplacian matrix approach are invalid to resolve the nonregular network in Fig. 1 is that they depend on the two matrices along two directions. Besides matrix (6), there also is another matrix with two arbitrary elements of r1 and r2 when we set up a matrix along the horizontal directions, in which it is impossible for us to obtain the explicit eigenvalues and eigenvectors of a matrix with arbitrary elements. In particular, the resistance formula (67) of the infinite resistor network has nothing to do with the boundary conditions. The previous resistance formula of the infinite network expressed by a double integral, such as Ref. [17], gives  2π  2π 1 R∞×∞ (d1 ,d2 ) = (2π )2 0 0 1 − cos[(x1 − x2 )φ] cos[(y1 − y2 )θ ] × dθ dφ. r −1 (1 − cos φ) + r0−1 (1 − cos θ ) (89) It is obvious that (89) and (67) are equivalent because they come from the same resistor network, thus  0

2π



2π

0



1 − cos[(x1 − x2 )φ] cos[(y1 − y2 )θ ] dθ dφ (1 − cos φ) + h(1 − cos θ ) π

= 4π 0

|x

|

1 − λ¯ θ 2 −x1 cos(y1 − y2 )θ  dθ. (1 + h − h cos θ )2 − 1

(90)

This discovery is interesting as it is found not in mathematics but in physics. According to the identity (90) we can derive a series of special trigonometric equalities when we make different coordinates (xi ,yi ). In addition, an important usefulness is that the R-T method can be extended to impedance networks, since the Ohm’s law based on which method is formulated is applicable to impedances. For example, the grid elements rk can be either a resistor or an impedance, rk = iωLk +

Rk . (k = 0,1,2,3). 1 + iωCk Rk

Thus we can study the m × n complex impedance networks. Based on the plural analysis (see [29,30]), the formula of the equivalent impedances of the m × n RLC network with a multivariate boundary can be obtained. ACKNOWLEDGMENTS

This work is supported by Jiangsu Province Education Science 12-5 Plan Project (Grant No. D/2013/01/048). We thank Professor F. Y. Wu and Professor J. W. Essam for giving some useful guiding suggestions.

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