Resistor network research involves a wide range of fields, not only of electrical problems but also of non-electric problems, such as chaotic quantum billiards that are simulated by a circuit network,^[1] waveguides in photonic crystals,^[2] a simulation of a non-Abelian Aharonov–Bohm effect,^[3] the electrical properties of conducting meshes,^[4] field theory for scale-free random networks,^[5] lattice Green’s functions,^[6–8] finite difference time-domain method for electromagnetic waves,^[9]etc. In particular, researchers can study the Laplace equation and Poisson equation^[10] by the resistor network model. In addition, the resistor network model has been published in journals of various disciplines, including chemical, physical chemistry, discrete mathematics, applied mathematics, engineering technology, physics, and so on. Specifically, the muhigrid method for three-dimensional (3D) modeling of Poisson equation published in Ref. [11]; resistance distance published in Refs. [12,13]; resistance distance and Laplacian spectrum published in Ref. [14]; a recursion formula for resistance distances and its applications and resistance distance in complete n-partite graphs published in Refs. [15,16]; resistances between two nodes of a path network published in Ref. [17]; resistance distances in corona and neighborhood corona networks based on Laplacian generalized inverse approach published in Ref. [18]; resistance distances in composite graphs and some rules on resistance distance with applications published in Refs. [19,20]; resistance between two nodes of a ring network published in Ref. [21]; universal relation for transport in non-sparse complex networks published in Ref. [22]; two-point resistance on the centered-triangular lattice published in Ref. [23]; exact evaluation of the resistance in an infinite face-centered cubic network published in Ref. [24]; resistance calculation of infinite three-dimensional triangular and hexagonal prism lattices published in Ref. [25], and so on. The above researches show that the research of resistor network model has important theoretical value and potential application value in many fields.

As is well known, computing the effective resistance between any two nodes in a resistor network is a difficult problem because it is required to solve the complex circuits and complex matrix equations. For example, it may be difficult to obtain the explicit expression of potential and resistance of the complex networks with arbitrary boundaries when the boundary resistor is complex. In fact, the boundary conditions are very binding and will affect the calculation method and process of the problem. Therefore, the solution of each complex resistor network problem needs to create innovative ideas and methods.

The resistor network research has been done for a long time. In 1845 Kirchhoff established the basic circuit theory. 150 years later, Cserti^[6] studied the infinite resistor network by Green’s function technique, which is not suitable for computing finite lattices. After some applications,^[23–25] some new issues were investigated by the Green’s function technique. In order to solve the problem of finite resistor network, in 2004 Wu^[26] presented a Laplacian matrix method, the method is suitable for the lattice in definite and canonical boundary conditions. The main weakness of this method is that it needs to find the eigenvalues and eigenvectors of the matrices with two directions, which makes it impossible to solve the resistor network with arbitrary boundaries. After 2004, several new problems of resistor network were studied by the Laplacian matrix approach.^[27–32] From the above analysis, the Green function method and the Laplace method cannot solve the resistor network problem with arbitrary boundaries, but the resistor networks with arbitrary boundaries come from reality, and they need to be solved by researchers. Fortunately, in 2011 Tan created a new theory for studying arbitrary resistor networks,^[33] which now is called recursion–transform (RT) theory of Tan.^[29] The advantage of the RT method is that it depends on a matrix in only one direction and the result is expressed by a single sum. With the development of the RT technique, a series of new resistor networks with zero resistor edges was solved.^[34–44] Recently, the Recursion–Transform method was subdivided into two ways: one way is to use current parameters to set up matrix equations,^[36–42] which is simply called the RT-I method; another way is to use potential parameters to set up matrix equations,^[43,44] which is simply called the RT-V method.

Investigations showed that many previous applications of the RT (including RT-I and RT-V) theory focus on resistor networks with zero resistor boundaries or special cases, such as the globe network^[34,42] belongs to cylindrical network with two zero resistor boundaries, the cobweb network^[32,36,43] belongs to cylindrical network with one zero resistor boundary, et al. Obviously, the complex resistor network without zero resistance boundary condition also needs to be studied. Very recently, new progress has been made: in Ref. [45] the n-step network with Δ structure was studied, in Ref. [46] the electrical characteristics of rectangular network was investigated by using the RT-V method, In Ref. [47] the electrical characteristics of arbitrary rectangular network with an arbitrary right boundary was studied by the RT-I approach. In Ref. [48] a new resistor network theory was developed by unifying the rectangular network and cylindrical network. However, because of the multifunctional nature of a cylindrical network with two arbitrary boundaries, the authors in Ref. [48] have not completely studied the conventional m × n cylindrical network (it sees cylindrical networks as just one example of the basic theory), and it is difficult for readers understand the results it gives. Therefore, in this paper we will systematically introduce the complete research process of cylindrical network, and take □ × n and Δ × n for example to help readers understand the physical implications of the results.

Consider a complex and anisotropic m × n cylindrical resistor network as shown in Fig. 1, in which the grid layout is continuous and the resistors are distributed anisotropy. In this figure n and m denote the numbers of resistors along the horizontal and cycle directions respectively, and the resistors r and r₀ in the respective horizontal (longitude) and loop (cycle or latitude) directions except for two arbitrary boundary resistors of r₁ and r₂. The difference between r and r₀ in horizontal and cycle directions implies the anisotropy of the network. This paper focused on studying the electrical characteristics (resistance and potential) of a cylindrical m × n resistor network with two arbitrary boundaries by using the advanced RT-V method, and we build three new theoretical formulae, thus lead large problems to be resolved. Studies have shown that the complex cylindrical networks with arbitrary boundaries are the multifunctional network model because it can deduce various geometrical structures (Figs. 3–6 and 12). Thus a large number of problems of resistor networks will be resolved in this paper. We emphasize that what was studied in Ref. [40] is only a special cylindrical network (with a zero resistance on the bottom), but our research on a cylindrical network is a general case of the network, which is completely different from the scenario in Ref. [40] and to our knowledge, it has not been studied before.

For the sake of comparative study, here we introduce a main result of cylindrical network. In 2004 Wu^[26] gave the accurate equivalent resistance of the regular cylindrical network by the Laplacian matrix approach for the first time. The so-called regular network refers to the boundary resistors r₁ = r₂ = r₀ in Fig. 1.

Consider a normative m × n cylindrical resistor network (r₁ = r₂ = r₀), where n and m are the numbers of resistors along the horizontal and cycle directions respectively, and r and r₀ are, respectively, the resistors along the horizontal and loop directions, Wu^[26] gave the resistance between two nodes d₁ (x₁,y₁) and d₂ (x₂,y₂) as follows:

$$ \begin{eqnarray}\begin{array}{lll} & & {R}_{m\times n}({d}_{1},{d}_{2})\\ & = & \displaystyle \frac{{r}_{0}}{n+1}\left(|{y}_{1}-{y}_{2}|-\displaystyle \frac{{({y}_{1}-{y}_{2})}^{2}}{m}\right)\\ & & +\displaystyle \frac{r}{m}|{x}_{1}-{x}_{2}|+\displaystyle \frac{1}{m(n+1)}\displaystyle \sum _{i=1}^{m-1}\\ & & \times \displaystyle \sum _{j=1}^{n}\displaystyle \frac{{C}_{{x}_{1},j}^{2}+{C}_{{x}_{2},j}^{2}-2{C}_{{x}_{1},j}{C}_{{x}_{2},j}\cos ({y}_{2}-{y}_{1}){\theta }_{i}}{{r}_{0}^{-1}(1-\cos {\theta }_{i})+{r}^{-1}(1-\cos {\phi }_{j})},\end{array}\end{eqnarray}$$ (1)

Formula (1) is found for the first time by Wu. However, when the boundary resistor r₁ and r₂ are the arbitrary elements, the problem becomes an unresolved difficulty. In addition, the equivalent resistance of Eq. (1) is in the double summations not in a single sum.

The innovation and contribution of this paper is reflected in the four aspects as follows. The first aspect is to generalize the RT theory, for example, previous RT theory relies on the boundary condition of the zero resistance, but it no longer depends on this condition in this paper. The second aspect is the innovation of matrix calculation, for example, the previous matrix transformation is all real numbers, but in this paper, the plural matrix transformation is established (see Eqs. (23) and (44) below). The third aspect is the analytical expressions of the potential function and the equivalent resistance of a complex cylindrical network with two arbitrary boundaries are given for the first time in this paper, and a series of applications are given. The fourth aspect is the discovery of a new mathematical identity from the point of view of physics, which promotes the research and development of mathematical identity.

This article involves a more complex network problem, and the expression of the result is more complex. Some parameters are specifically defined here to simplify the expressions of results

$$ \begin{eqnarray}\begin{array}{lll} & & {C}_{{y}_{k}-y}^{(i)}=\cos ({y}_{k}-y){\theta }_{i},{\theta }_{i}=2i\pi /m,\end{array}\end{eqnarray}$$ (2)

$$ \begin{eqnarray}\begin{array}{lll} & & {\lambda }_{i}=h+1-h\cos {\theta }_{i}+\sqrt{{(h+1-h\cos {\theta }_{i})}^{2}-1},\\ & & {\bar{\lambda }}_{i}=h+1-h\cos {\theta }_{i}-\sqrt{{(h+1-h\cos {\theta }_{i})}^{2}-1},\end{array}\end{eqnarray}$$ (3)

$$ \begin{eqnarray}\begin{array}{lll} & & {F}_{k}^{(i)}=({\lambda }_{i}^{k}-{\bar{\lambda }}_{i}^{k})/({\lambda }_{i}-{\bar{\lambda }}_{i}),{\rm{\Delta }}{F}_{k}^{(i)}={F}_{k+1}^{(i)}-{F}_{k}^{(i)},\end{array}\end{eqnarray}$$ (4)

$$ \begin{eqnarray}\begin{array}{lll} & & {\alpha }_{s,x}^{(i)}={\rm{\Delta }}{F}_{x}^{(i)}+({h}_{s}-1){\rm{\Delta }}{F}_{x-1}^{(i)},{h}_{s}={r}_{s}/{r}_{0},\end{array}\end{eqnarray}$$ (5)

$$ \begin{eqnarray}\begin{array}{lll} & & {\beta }_{x\vee {x}_{s}}^{(i)}=\left\{\begin{array}{ll}{\beta }_{x,{x}_{s}}^{(i)}={\alpha }_{1,x}^{(i)}{\alpha }_{2,n-{x}_{s}}^{(i)}, & \,\,{\rm{if}}\,x\leqslant {x}_{s},\\ {\beta }_{{x}_{s},x}^{(i)}={\alpha }_{1,{x}_{s}}^{(i)}{\alpha }_{2,n-x}^{(i)}, & \,\,{\rm{if}}\,x\geqslant {x}_{s},\end{array}\right.\end{array}\end{eqnarray}$$ (6)

$$ \begin{eqnarray}\begin{array}{lll} & & {G}_{n}^{(i)}={F}_{n+1}^{(i)}+({h}_{1}+{h}_{2}-2){F}_{n}^{(i)}+({h}_{1}-1)({h}_{2}-1){F}_{n-1}^{(i)}.\end{array}\end{eqnarray}$$ (7)

In a nutshell, the above definitions of Eqs. (2)–(7) are used throughout the paper. When you look at the calculation below you will see that it is necessary to define a series of functions because the electrical properties of the resistor network are more complex, and these definitions are innovative.

Consider a complex m × n cylindrical network shown in Fig. 1, two arbitrary resistors are placed on the left- and right-hand side of the network, where the resistor parameters r_i and voltage parameters Vx(y) are shown in Figs. 1 and 2, and the origin of the rectangular coordinate system is specified at point A₀ (0,0). Assume that the electric current J goes from the d₁ (x₁,y₁) to the d₂ (x₂,y₂). Expressing the nodal potential at d(x,y) by Um×n(x,y)=Vx(y) and choosing the reference potential such that ∑i=0m−1V0(i)=0, which means hat the sum of the voltages of all the nodes on the left boundary is zero, the analytic expression of the potential function of d(x,y) in the cylindrical m × n resistor network can be written as

$$ \begin{eqnarray}\begin{array}{ll}\displaystyle \frac{{U}_{m\times n}(x,y)}{J}= & \displaystyle \frac{{x}_{1}-{x}_{\tau }}{m}r\\ & +\displaystyle \frac{{r}_{0}}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{{x}_{1}\vee x}^{(i)}{C}_{{y}_{1}-y}^{(i)}-{\beta }_{{x}_{2}\vee x}^{(i)}{C}_{{y}_{2}-y}^{(i)}}{(1-\cos {\theta }_{i}){G}_{n}^{(i)}},\,\end{array}\end{eqnarray}$$ (8)

$$ \begin{eqnarray}\begin{array}{lll}{x}_{\tau } & = & \{{x}_{1},0\leqslant x\leqslant {x}_{1}\}\cup \{x,{x}_{1}\leqslant x\leqslant {x}_{2}\}\\ & & \cup \{{x}_{2},{x}_{2}\leqslant x\leqslant n\},\end{array}\end{eqnarray}$$ (9)

When taking the reference voltage by ∑i=0m−1Vn(i)=0, which means that the sum of the voltages of all the nodes on the right boundary is zero, the potential function of d(x,y) is

$$ \begin{eqnarray}\begin{array}{lll}\displaystyle \frac{{U}_{m\times n}(x,y)}{J}= & & \displaystyle \frac{{x}_{2}-{x}_{\tau }}{m}r\\ & & +\displaystyle \frac{{r}_{0}}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{{x}_{1}\vee x}^{(i)}{C}_{{y}_{1}-y}^{(i)}-{\beta }_{{x}_{2}\vee x}^{(i)}{C}_{{y}_{2}-y}^{(i)}}{(1-\cos {\theta }_{i}){G}_{n}^{(i)}},\,\,\,\,\,\,\end{array}\end{eqnarray}$$ (10)

Why do we define Eq. (9)? This is because equations (8) and (10) have to be expressed by three piecewise equations if there is no definition of Eq. (9). To understand Eqs. (8) and (9), given here is the explanation: when 0 ⩽ x ⩽ x₁, there is x₁ – x_ρ = 0; when x₁ ⩽ x ⩽ x₂, there is x₁ – x_τ = x₁ – x; when x₂ ⩽ x ⩽ n, there is x₁ – x_τ = x₁ – x₂.

Consider an arbitrary cylindrical m × n resistor network shown in Fig. 1, where two arbitrary resistors r₁ and r₂ are placed on the left- and right-hand side of the network. The analytic expression of effective resistance between two arbitrary nodes d₁ (x₁,y₁) and d₂ (x₂,y₂) in the network is given by

$$ \begin{eqnarray}\begin{array}{lll}{R}_{m\times n}({d}_{1},{d}_{2})= & & \displaystyle \frac{|{x}_{2}-{x}_{1}|}{m}r\\ & & +\displaystyle \frac{{r}_{0}}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{1,1}^{(i)}-2{\beta }_{1,2}^{(i)}\cos (y{\theta }_{i})+{\beta }_{2,2}^{(i)}}{(1-\cos {\theta }_{i}){G}_{n}^{(i)}},\,\,\,\,\,\,\,\end{array}\end{eqnarray}$$ (11)

The above three main results of Eqs. (8), (10), and (11) are the first finding of this paper, which is a theoretical and technical innovation. Their proofs and applications are given below.

RT-V method is pioneered by Tan^[43] in 2017. We are going to derive analytic formulae (8), (10), and (11) by using the RT-V method. The derivation of the potential function is a systematic work, including a series of steps, we will derive them through the following five stages.

Building a discrete Poisson equation by the sub-network of Fig. 2. Using Kirchhoff law (∑ri−1Vk=0) to establish the voltage equations along the latitude (vertical) direction, we obtain two sets of difference equations for the network of Fig. 1

$$ \begin{eqnarray}\begin{array}{lll} & & {V}_{k+1}^{(0)}=(2+2h){V}_{k}^{(0)}-{V}_{k-1}^{(0)}-h{V}_{k}^{(m-1)}-h{V}_{k}^{(1)},\\ & & {V}_{k+1}^{(i)}=(2+2h){V}_{k}^{(i)}-{V}_{k-1}^{(i)}-h{V}_{k}^{(i-1)}-h{V}_{k}^{(i+1)},\,\,\,\,\,\,\,\,\end{array}\end{eqnarray}$$ (12)

$$ \begin{eqnarray}{{\boldsymbol{V}}}_{k+1}={{\boldsymbol{B}}}_{m}{{\boldsymbol{V}}}_{k}-{{\boldsymbol{V}}}_{k-1}-r{{\boldsymbol{I}}}_{k}{\delta }_{k,x}({\delta }_{y,{y}_{1}}-{\delta }_{y,{y}_{2}}),\end{eqnarray}$$ (13)

$$ \begin{eqnarray}{{\boldsymbol{V}}}_{k}={\left[{V}_{k}^{(0)},{V}_{k}^{(1)},{V}_{k}^{(2)},\ldots,{V}_{k}^{(m-1)}\right]}^{{\rm{T}}},\end{eqnarray}$$ (14)

$$ \begin{eqnarray}{{\boldsymbol{I}}}_{k}={[J,J,J,\ldots,J]}^{{\rm{T}}},\end{eqnarray}$$ (15)

$$ \begin{eqnarray}{{\boldsymbol{B}}}_{m}=\left(\begin{array}{ccccc}2+2h & -h & 0 & 0 & -h\\ -h & 2(1+h) & -h & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & -h & 2(1+h) & -h\\ -h & 0 & 0 & -h & 2+2h\end{array}\right).\end{eqnarray}$$ (16)

Setting up the boundary condition equations on the left and right edges in the network of Fig. 1. Applying the Kirchhoff’s law (∑ri−1Vk=0) to each boundary of left and right edges, we obtain two matrix equations that relate the left and right boundary as follows:

$$ \begin{eqnarray}\begin{array}{lll} & & {h}_{1}{{\boldsymbol{V}}}_{1}=[{{\boldsymbol{B}}}_{m}-(2-{h}_{1}){\boldsymbol{E}}]{{\boldsymbol{V}}}_{0},\end{array}\end{eqnarray}$$ (17)

$$ \begin{eqnarray}\begin{array}{lll} & & {h}_{2}{{\boldsymbol{V}}}_{n-1}=[{{\boldsymbol{B}}}_{m}-(2-{h}_{2}){\boldsymbol{E}}]{{\boldsymbol{V}}}_{n},\end{array}\end{eqnarray}$$ (18)

Equations (13)–(18) are all the equations we need in order to derive the potential function. However, we cannot directly solve this equation. In order to solve this difficulty we create new techniques by using the RT theory.^[36–38,43] In the following we first give the transform technique, and then give their solution.

Building matrix transform. Firstly, we work out the eigenvalue t_i of matrix B_m, solving det | B_m – tE | = 0 to yield

$$ \begin{eqnarray}{t}_{i}=2(1+h)-2h\cos {\theta }_{i},\end{eqnarray}$$ (19)

Next, transform Eqs. (13), (17), and (18) by the following methods

$$ \begin{eqnarray}\begin{array}{lll} & & {{\boldsymbol{P}}}_{m}{{\boldsymbol{B}}}_{m}={\rm{diag}}\{{t}_{0},{t}_{1},\ldots,{t}_{m-1}\}{{\boldsymbol{P}}}_{m},\end{array}\end{eqnarray}$$ (20)

$$ \begin{eqnarray}\begin{array}{lll} & & {{\boldsymbol{X}}}_{k}={{\boldsymbol{P}}}_{m}{{\boldsymbol{V}}}_{k}\,\,\,{\rm{or}}\,\,\,{{\boldsymbol{V}}}_{k}={({{\boldsymbol{P}}}_{m})}^{-1}{{\boldsymbol{X}}}_{k},\end{array}\end{eqnarray}$$ (21)

Using a transformation technique similar to the one above, applying P_m to Eqs. (17) and (18), we obtain

$$ \begin{eqnarray}\begin{array}{lll} & & {h}_{1}{X}_{1}^{(i)}=({t}_{i}+{h}_{1}-2){X}_{0}^{(i)},\end{array}\end{eqnarray}$$ (25)

$$ \begin{eqnarray}\begin{array}{lll} & & {h}_{2}{X}_{n-1}^{(i)}=({t}_{i}+{h}_{2}-2){X}_{n}^{(i)}.\end{array}\end{eqnarray}$$ (26)

Solving Eqs. (24)–(26). For solving Xk(i) we should unfold Eq. (24) since it contains several piecewise functions. Thus, we obtain the piecewise solution after solving Eq. (24) as follows:

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{k}^{(i)}={X}_{1}^{(i)}{F}_{k}-{X}_{0}^{(i)}{F}_{k-1},\,\,\,0\leqslant k\leqslant {x}_{1},\end{array}\end{eqnarray}$$ (27)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{{x}_{1}+1}^{(i)}={t}_{i}{X}_{{x}_{1}}^{(i)}-{X}_{{x}_{1}-1}^{(i)}-rJ\exp ({\rm{i}}{y}_{1}{\theta }_{i}),\end{array}\end{eqnarray}$$ (28)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{k}^{(i)}={X}_{{x}_{1}+1}^{(i)}{F}_{k-{x}_{1}}-{X}_{{x}_{1}}^{(i)}{F}_{k-{x}_{1}-1},\,\,\,{x}_{1}\leqslant k\leqslant {x}_{2},\,\,\,\,\,\,\end{array}\end{eqnarray}$$ (29)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{{x}_{2}+1}^{(i)}={t}_{i}{X}_{{x}_{2}}^{(i)}-{X}_{{x}_{2}-1}^{(i)}+rJ\exp ({\rm{i}}{y}_{2}{\theta }_{i}),\end{array}\end{eqnarray}$$ (30)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{k}^{(i)}={X}_{{x}_{2}+1}^{(i)}{F}_{k-{x}_{2}}-{X}_{{x}_{2}}^{(i)}{F}_{k-{x}_{2}-1},\,\,\,{x}_{2}\leqslant k\leqslant n,\,\,\end{array}\end{eqnarray}$$ (31)

Please note that we must consider two cases: i = 0 and i ⩾ 1, because there is θ₀ = 0 (θ_i = 2iπ / m) if i = 0. By Eq. (19) we have the eigenvalue t₀ = 2, but 1 – cos θ_i appears in the denominator of Eqs. (8), (10), and (11). So we need to consider the additional solution of equations when θ₀ = 0.

First we consider the case of i ⩾ 1, by Eqs. (25)–(31), we obtain

$$ \begin{eqnarray}{X}_{k}^{(i)}=\displaystyle \frac{{\beta }_{k\vee {x}_{1}}^{(i)}\exp ({\rm{i}}{y}_{1}{\theta }_{i})-{\beta }_{k\vee {x}_{2}}^{(i)}\exp ({\rm{i}}{y}_{2}{\theta }_{i})}{({t}_{i}-2){G}_{n}^{(i)}}rJ,\end{eqnarray}$$ (32)

Next, we consider the case of i = 0⇒ θ₀ = 0, by Eqs. (3) and (23), we have

$$ \begin{eqnarray}{\lambda }_{0}={\bar{\lambda }}_{0}=1\,\,\,{\rm{and}}\,\,\,{g}_{k,0}=\exp ({\rm{i}}k{\theta }_{0})=1.\end{eqnarray}$$ (33)

Applying limit λ₀ → 1 to Eqs. (4) and (5), we have

$$ \begin{eqnarray}\begin{array}{lll} & & {F}_{k}^{(0)}=k,\,\,\,{\rm{\Delta }}{F}_{k}^{(0)}=1,\\ & & {\alpha }_{s,x}^{(0)}={\rm{\Delta }}{F}_{x}^{(0)}+({h}_{s}-1){\rm{\Delta }}{F}_{x-1}^{(0)}={h}_{s}.\end{array}\end{eqnarray}$$ (34)

$$ \begin{eqnarray}{X}_{1}^{(0)}={X}_{0}^{(0)},{X}_{n-1}^{(0)}={X}_{n}^{(0)}.\end{eqnarray}$$ (35)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{k}^{(0)}={X}_{0}^{(0)},\,\,(0\leqslant k\leqslant {x}_{1}),\end{array}\end{eqnarray}$$ (36)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{{x}_{1}+1}^{(0)}=2{X}_{{x}_{1}}^{(0)}-{X}_{{x}_{1}-1}^{(0)}-rJ,\end{array}\end{eqnarray}$$ (37)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{k}^{(0)}=(k-{x}_{1}){X}_{{x}_{1}+1}^{(0)}-(k-{x}_{1}-1){X}_{{x}_{1}}^{(0)},\,\,({x}_{1}\leqslant k\leqslant {x}_{2}),\end{array}\end{eqnarray}$$ (38)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{{x}_{2}+1}^{(0)}=2{X}_{{x}_{2}}^{(0)}-{X}_{{x}_{2}-1}^{(0)}+rJ,\,\,\,\,\,\,\,\,\,\,\end{array}\end{eqnarray}$$ (39)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{k}^{(0)}=(k-{x}_{2}){X}_{{x}_{2}+1}^{(0)}-(k-{x}_{2}-1){X}_{{x}_{2}}^{(0)},\,\,({x}_{2}\leqslant k\leqslant n).\end{array}\end{eqnarray}$$ (40)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{k}^{(0)}={X}_{0}^{(0)},\,\,\,0\leqslant k\leqslant {x}_{1},\end{array}\end{eqnarray}$$ (41)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{k}^{(0)}={X}_{0}^{(0)}+({x}_{1}-k)rJ,\,\,\,{x}_{1}\leqslant k\leqslant {x}_{2},\end{array}\end{eqnarray}$$ (42)

$$ \begin{eqnarray}\begin{array}{lll} & & {X}_{k}^{(0)}={X}_{0}^{(0)}+({x}_{1}-{x}_{2})rJ,\,\,\,{x}_{2}\leqslant k\leqslant n.\end{array}\end{eqnarray}$$ (43)

From Eqs. (21)–(23), we obtain

$$ \begin{eqnarray}\begin{array}{lll}\left(\begin{array}{c}{X}_{k}^{(0)}\\ {X}_{k}^{(1)}\\ \vdots \\ {X}_{k}^{(s)}\end{array}\right) & = & \left(\begin{array}{ccccc}1 & 1 & 1 & \ldots & 1\\ 1 & \exp ({\rm{i}}{\theta }_{1}) & \exp ({\rm{i}}2{\theta }_{1}) & \ldots & \exp ({\rm{i}}s{\theta }_{1})\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & \exp ({\rm{i}}{\theta }_{s}) & \exp ({\rm{i}}2{\theta }_{s}) & \vdots & \exp ({\rm{i}}s{\theta }_{s})\end{array}\right)\\ & & \times \left(\begin{array}{c}{V}_{k}^{(0)}\\ {V}_{k}^{(1)}\\ \vdots \\ {V}_{k}^{(s)}\end{array}\right),\end{array}\end{eqnarray}$$ (44)

$$ \begin{eqnarray}{X}_{k}^{(0)}=\displaystyle \sum _{i=0}^{m-1}{V}_{k}^{(i)}.\end{eqnarray}$$ (45)

$$ \begin{eqnarray}\displaystyle \sum _{i=0}^{m-1}{V}_{0}^{(i)}=0\iff {X}_{0}^{(0)}=0.\end{eqnarray}$$ (46)

$$ \begin{eqnarray}{X}_{k}^{(0)}=({x}_{1}-{x}_{\tau })rJ,\,\,\,\,0\leqslant k\leqslant n,\end{eqnarray}$$ (47)

In Eq. (46), the sum of the voltages on the left boundary is chosen to be zero, we can also choose the sum of the voltages on the right boundary to be zero, so, equation (45) is used to obtain (taking k = n)

$$ \begin{eqnarray}\displaystyle \sum _{i=0}^{m-1}{V}_{n}^{(i)}=0\iff {X}_{n}^{(0)}=0,\,\,{X}_{0}^{(0)}=({x}_{2}-{x}_{1})rJ.\end{eqnarray}$$ (48)

$$ \begin{eqnarray}{X}_{k}^{(0)}=({x}_{2}-{x}_{\tau })rJ,\,\,\,0\leqslant k\leqslant n,\end{eqnarray}$$ (49)

Using inverse transformation to derive the general formulae (8) and (10). According to the exact calculation, by Eq. (44) we can obtain the inverse transformation equation

$$ \begin{eqnarray}\begin{array}{lll}\left(\begin{array}{c}{V}_{k}^{(0)}\\ {V}_{k}^{(1)}\\ \vdots \\ {V}_{k}^{(s)}\end{array}\right)= & & \displaystyle \frac{1}{m}\left(\begin{array}{cccc}1 & 1 & \cdots & 1\\ 1 & \exp (-{\rm{i}}{\theta }_{1}) & \cdots & \exp (-{\rm{i}}s{\theta }_{s})\\ \vdots & \vdots & \ddots & \vdots \\ 1 & \exp (-{\rm{i}}s{\theta }_{1}) & \cdots & \exp (-{\rm{i}}s{\theta }_{s})\end{array}\right)\\ & & \times \left(\begin{array}{c}{X}_{k}^{(0)}\\ {X}_{k}^{(1)}\\ \vdots \\ {X}_{k}^{(s)}\end{array}\right),\end{array}\end{eqnarray}$$ (50)

Thus, by Eq. (50), we obtain (y ⩾ 1)

$$ \begin{eqnarray}{V}_{k}^{(y)}=\displaystyle \frac{1}{m}\left({X}_{k}^{(0)}+\displaystyle \sum _{i=1}^{m-1}{X}_{k}^{(i)}\exp (-{\rm{i}}y{\theta }_{i})\right).\end{eqnarray}$$ (51)

Again, when taking ∑i=0m−1Vn(i)=0, there is Eq. (49), the substitution of Eqs. (32) and (49) into Eq. (51), then equation (10) is proved.

The above five stages are the specific elaboration of RT-V theory, and can be used to calculate the electrical characteristics of cylindrical networks. Such as stage-1 setting up the main matrix equation, stage-2, setting up the matrix equation with boundary conditions of the left and right edges, stage-3, creating matrix transform, stage-4, solving the matrix equations, and stage-5, deriving the potential by the inverse transform.

Next, we derive Eq. (11). By using Ohm’s law, we obtain

$$ \begin{eqnarray}{R}_{m\times n}({d}_{1},{d}_{2})=[U({x}_{1},{y}_{1})-U({x}_{2},{y}_{2})]\displaystyle \frac{1}{J}.\end{eqnarray}$$ (53)

$$ \begin{eqnarray}\displaystyle \frac{{U}_{m\times n}({x}_{1},{y}_{1})}{J}=\displaystyle \frac{{r}_{0}}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{{x}_{1},{x}_{1}}^{(i)}-{\beta }_{{x}_{1},{x}_{2}}^{(i)}\cos ({y}_{2}-{y}_{1}){\theta }_{i}}{(1-\cos {\theta }_{i}){G}_{n}^{(i)}},\end{eqnarray}$$ (54)

$$ \begin{eqnarray}\displaystyle \frac{{U}_{m\times n}({x}_{2},{y}_{2})}{J}=\displaystyle \frac{{x}_{1}-{x}_{2}}{m}r+\displaystyle \frac{{r}_{0}}{2m}\times \displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{{x}_{1},{x}_{2}}^{(i)}\cos ({y}_{2}-{y}_{1}){\theta }_{i}-{\beta }_{{x}_{2},{x}_{2}}^{(i)}}{(1-\cos {\theta }_{i}){G}_{n}^{(i)}}.\end{eqnarray}$$ (55)

$$ \begin{eqnarray}\displaystyle \frac{{U}_{m\times n}({x}_{1},{y}_{1})}{J}-\displaystyle \frac{{U}_{m\times n}({x}_{2},{y}_{2})}{J}=\displaystyle \frac{{x}_{2}-{x}_{1}}{m}r+\displaystyle \frac{{r}_{0}}{m}\displaystyle \sum _{i=1}^{m-1}\left(\displaystyle \frac{{\beta }_{{x}_{1},{x}_{1}}^{(i)}-{\beta }_{{x}_{1},{x}_{2}}^{(i)}\cos ({y}_{2}-{y}_{1}){\theta }_{i}}{2(1-\cos {\theta }_{i}){G}_{n}^{(i)}}-\displaystyle \frac{{\beta }_{{x}_{1},{x}_{2}}^{(i)}\cos ({y}_{2}-{y}_{1}){\theta }_{i}-{\beta }_{{x}_{2},{x}_{2}}^{(i)}}{2(1-\cos {\theta }_{i}){G}_{n}^{(i)}}\right).\end{eqnarray}$$ (56)

$$ \begin{eqnarray}{R}_{m\times n}({d}_{1},{d}_{2})=\displaystyle \frac{{U}_{m\times n}({x}_{1},{y}_{1})}{J}-\displaystyle \frac{{U}_{m\times n}({x}_{2},{y}_{2})}{J}=\displaystyle \frac{{x}_{2}-{x}_{1}}{m}r+\displaystyle \frac{{r}_{0}}{2m}\displaystyle \sum _{i=1}^{m}\displaystyle \frac{{\beta }_{{x}_{1},{x}_{1}}^{(i)}-2{\beta }_{{x}_{1},{x}_{2}}^{(i)}\cos ({y}_{2}-{y}_{1}){\theta }_{i}+{\beta }_{{x}_{2},{x}_{2}}^{(i)}}{(1-\cos {\theta }_{i}){G}_{n}^{(i)}}.\end{eqnarray}$$ (57)

In particular, since the RT-V theory is matured, and all results can be strictly calculated by the RT theory. All the calculation processes and conclusions are self-consistent, without any guessing factors, so our results are necessarily correct, and the following special cases verify their correctness again.

In subsequent sections we consider the applications of formulae to arbitrary lattices. In all applications, we stipulate that all parameters in Eqs. (2)–(7) are applied to all resistor networks, and denote the resistors along the two principal directions by r and r₀ except for resistors on the left-right boundaries, and the input and output node of current are respectively at d₁ (x₁,y₁) and d₂ (x₂,y₂).

Formula (8) is a general result because the network of Fig. 1 is very complex and has not been resolved before, which includes a lot of resistor network models, where each of the different boundary resistors represents a different network structure. So formula (8) can create many interesting results.

Consider an regular m × n cylindrical network of Fig. 1 with r₁ = r₂ = r₀, selecting ∑V0(i)=0, by Eq. (8) we have the nodal potential as follows:

$$ \begin{eqnarray}\displaystyle \frac{U(x,y)}{J}=\displaystyle \frac{{x}_{1}-{x}_{\tau }}{m}r+\displaystyle \frac{{r}_{0}}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{{x}_{1}\vee x}^{(i)}{C}_{{y}_{1}-y}^{(i)}-{\beta }_{{x}_{2}\vee x}^{(i)}{C}_{{y}_{2}-y}^{(i)}}{(1-\cos {\theta }_{i}){F}_{n+1}^{(i)}},\,\,\,\,\end{eqnarray}$$ (58)

Consider a non-regular m × n cylindrical network as shown in Fig. 1. Defining Cyk−y(i)=cos(yk−y)θi with θ_i = 2iπ / m, and selecting ∑V0(i)=0, when x₂ = x₁, the potential of any node d(x,y) in the finite and sem-infinite networks can be written respectively as

$$ \begin{eqnarray}\begin{array}{lll} & & \displaystyle \frac{{U}_{m\times n}(x,y)}{J}=\displaystyle \frac{{r}_{0}}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{C}_{{y}_{1}-y}^{(i)}-{C}_{{y}_{2}-y}^{(i)}}{(1-\cos {\theta }_{i}){G}_{n}^{(i)}}{\beta }_{{x}_{1}\vee x}^{(i)},\end{array}\end{eqnarray}$$ (59)

$$ \begin{eqnarray}\begin{array}{lll} & & \displaystyle \frac{{U}_{m\times \infty }(x,y)}{J}=\displaystyle \frac{r}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{C}_{{y}_{1}-y}^{(i)}-{C}_{{y}_{2}-y}^{(i)}}{\sqrt{{(1+h-h\cos {\theta }_{i})}^{2}-1}}{\bar{\lambda }}_{i}^{|{x}_{1}-x|}.\,\,\,\,\,\,\,\,\end{array}\end{eqnarray}$$ (60)

Consider an m × n cylindrical network of Fig. 1. When r₂ = 0, figure 1 degrades into a cobweb network as shown in Fig. 3. Selecting ∑V0(i)=0 as the reference potential, by Eq. (8) we have the nodal potential

$$ \begin{eqnarray}\displaystyle \frac{U(x,y)}{J}=\displaystyle \frac{{x}_{1}-{x}_{\tau }}{m}r+\displaystyle \frac{r}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{{x}_{1}\vee x}^{(i)}{C}_{{y}_{1}-y}^{(i)}-{\beta }_{{x}_{2}\vee x}^{(i)}{C}_{{y}_{2}-y}^{(i)}}{{\rm{\Delta }}{F}_{n}^{(i)}+({h}_{1}-1){\rm{\Delta }}{F}_{n-1}^{(i)}},\end{eqnarray}$$ (61)

In particular, when d₁ (0,y₁) is on the left edge, and d₂ (n,y₂) is on the right edge, equation (61) reduces to

$$ \begin{eqnarray}\displaystyle \frac{U(x,y)}{J}=-\displaystyle \frac{x}{m}r+\displaystyle \frac{{r}_{1}h}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{F}_{n-x}^{(i)}\cos ({y}_{1}-y){\theta }_{i}}{{\rm{\Delta }}{F}_{n}^{(i)}+({h}_{1}-1){\rm{\Delta }}{F}_{n-1}^{(i)}}.\end{eqnarray}$$ (62)

Consider an m × n resistor network of Fig. 1. When r₂ = 0, figure 1 degrades into a cobweb network as shown in Fig. 3. Selecting ∑Vn(i)=0 as the reference potential, by Eq. (10) we have the nodal potential

$$ \begin{eqnarray}\displaystyle \frac{U(x,y)}{J}=\displaystyle \frac{{x}_{2}-{x}_{\tau }}{m}r+\displaystyle \frac{r}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{{x}_{1}\vee x}^{(i)}{C}_{{y}_{1}-y}^{(i)}-{\beta }_{{x}_{2}\vee x}^{(i)}{C}_{{y}_{2}-y}^{(i)}}{{\rm{\Delta }}{F}_{n}^{(i)}+({h}_{1}-1){\rm{\Delta }}{F}_{n-1}^{(i)}},\end{eqnarray}$$ (63)

In particular, when d₁ (0,y₁) is on the left edge, and d₂ (n,y₂) is on the right edge, equation (63) reduces to

$$ \begin{eqnarray}\displaystyle \frac{U(x,y)}{J}=\displaystyle \frac{n-x}{m}r+\displaystyle \frac{{r}_{1}h}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{F}_{n-x}^{(i)}\cos ({y}_{1}-y){\theta }_{i}}{{\rm{\Delta }}{F}_{n}^{(i)}+({h}_{1}-1){\rm{\Delta }}{F}_{n-1}^{(i)}}.\,\,\,\,\,\end{eqnarray}$$ (64)

Consider an arbitrary m × n globe network as shown in Fig. 4. That is to say, figure 1 degrades into a globe network when r₁ = r₂ = 0. Selecting ∑V0(i)=0, from Eq. (8) we have the nodal potential

$$ \begin{eqnarray}\displaystyle \frac{{U}_{m\times n}(x,y)}{J}=\displaystyle \frac{{x}_{1}-{x}_{\tau }}{m}r+\displaystyle \frac{r}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{{x}_{1}\vee x}^{(i)}{C}_{{y}_{1}-y}^{(i)}-{\beta }_{{x}_{2}\vee x}^{(i)}{C}_{{y}_{2}-y}^{(i)}}{{F}_{n}^{(i)}},\,\,\,\,\end{eqnarray}$$ (65)

In particular, when d₁ (0,y₁) is at the left pole, and d₂ (n,y₂) is at the right pole, equation (65) reduces to

$$ \begin{eqnarray}\displaystyle \frac{U(x,y)}{J}=-\displaystyle \frac{x}{m}r.\end{eqnarray}$$ (66)

In addition, when d₁ (0,y₁) is at the left pole, and d₂ (n,y₂) is at the right pole, and r₁ = r₂ = 0, selecting ∑Vn(i)=0, by Eq. (10) we have the nodal potential as follows:

$$ \begin{eqnarray}\displaystyle \frac{U(x,y)}{J}=\displaystyle \frac{x}{m}r.\end{eqnarray}$$ (67)

When m = 4, figure 2 degrades into a 3D □ × n resistor network as shown in Fig. 5. By Eq. (2) we have θ_i = iπ / 2, and substituting it into Eq. (3) yields

$$ \begin{eqnarray}\begin{array}{lll} & & {\lambda }_{1}={\lambda }_{3}=1+h+\sqrt{{(1+h)}^{2}-1},\\ & & {\lambda }_{2}=1+2h+\sqrt{{(1+2h)}^{2}-1}.\end{array}\end{eqnarray}$$ (68)

It is not hard to see that Eq. (69) is still quite profound in terms of understanding its physical meaning. Since the node potential value is related to the input position of the power source, different input positions of the power supply will generate different potential values. To understand the nature of the problem easily, we list the following five cases.

Assume that the input current J is at A₀ (0,0) and the output current J is at B₀ (0,1), and select V_A0 + V_B0 = 0, then by Eq. (69) we will have the nodal potential

$$ \begin{eqnarray}\begin{array}{lll}\displaystyle \frac{{U}_{\square \times n}(x,y)}{J}= & & {r}_{1}\displaystyle \frac{{C}_{y}^{(1)}-{C}_{1-y}^{(1)}}{4{G}_{n}^{(1)}}{\alpha }_{2,n-x}^{(1)}\\ & & +\,{r}_{1}\displaystyle \frac{{C}_{y}^{(2)}-{C}_{1-y}^{(2)}}{16{G}_{n}^{(2)}}{\alpha }_{2,n-x}^{(2)},\end{array}\end{eqnarray}$$ (70)

Assume that the input current J is at A₀ (0,0), and the output current J is at C₀ (0,2), and select V_A0 + V_C0 = 0 as the reference potential, then by Eq. (69) we will have the nodal potential

$$ \begin{eqnarray}\begin{array}{ll}\displaystyle \frac{{U}_{\square \times n}(x,y)}{J}= & {r}_{1}\displaystyle \frac{{C}_{y}^{(1)}-{C}_{2-y}^{(1)}}{4{G}_{n}^{(1)}}{\alpha }_{2,n-x}^{(1)}\\ & +{r}_{1}\displaystyle \frac{{C}_{y}^{(2)}-{C}_{2-y}^{(2)}}{16{G}_{n}^{(2)}}{\alpha }_{2,n-x}^{(2)},\end{array}\end{eqnarray}$$ (75)

Assume that the input current is at A₀ (0,0), and the output current is at A_n (n,0), and select V_A0 + V_B0 + V_C0 + V_D0 = 0 as the reference potential, then by Eq. (69) we will have the nodal potential

$$ \begin{eqnarray}\begin{array}{lll}\displaystyle \frac{{U}_{\square \times n}(x,y)}{J} & = & -\displaystyle \frac{x}{4}r+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n-x}^{(1)}-{h}_{2}{\alpha }_{1,x}^{(1)}}{4{G}_{n}^{(1)}}{C}_{y}^{(1)}\\ & & +{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n-x}^{(2)}-{h}_{2}{\alpha }_{1,x}^{(2)}}{16{G}_{n}^{(2)}}{C}_{y}^{(2)},\end{array}\end{eqnarray}$$ (79)

Assume that the input current is at A₀ (0,0), and the output current is at B_n (n,1), and select V_A0 + V_B0 + V_C0 + V_D0 = 0 as the reference potential, then by Eq. (69) we will have the nodal potential

$$ \begin{eqnarray}\begin{array}{lll}\displaystyle \frac{{U}_{\square \times n}(x,y)}{J} & = & -\displaystyle \frac{x}{4}r+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n-x}^{(1)}{C}_{y}^{(1)}-{h}_{2}{\alpha }_{1,x}^{(1)}{C}_{1-y}^{(1)}}{4{G}_{n}^{(1)}}\\ & & +{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n-x}^{(2)}{C}_{y}^{(2)}-{h}_{2}{\alpha }_{1,x}^{(2)}{C}_{1-y}^{(2)}}{16{G}_{n}^{(2)}},\end{array}\end{eqnarray}$$ (83)

Assume that the input current is at A₀ (0,0), and the output current is at C_n (n,3), and select V_A0 + V_B0 + V_C0 + V_D0 = 0 as the reference potential, then by Eq. (69) we will have the nodal potential

$$ \begin{eqnarray}\begin{array}{lll}\displaystyle \frac{{U}_{\square \times n}(x,y)}{J} & = & -\displaystyle \frac{x}{4}r+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n-x}^{(1)}{C}_{y}^{(1)}-{h}_{2}{\alpha }_{1,x}^{(1)}{C}_{2-y}^{(1)}}{4{G}_{n}^{(1)}}\\ & & +{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n-x}^{(2)}{C}_{y}^{(2)}-{h}_{2}{\alpha }_{1,x}^{(2)}{C}_{2-y}^{(2)}}{16{G}_{n}^{(2)}},\end{array}\end{eqnarray}$$ (88)

The above series of results (69)–(91) is the first to be found in this paper, which basically explain the basic meaning of formula (8) and can effectively help readers understand and use formula (8).

Formula (11) is a precise and profound result since the network of Fig. 1 is very complex and has not been resolved before and includes a lot of resistor network models, where each of the different boundary resistor r_i represents a different network structure. In particular, taking some specific values of r₁ and r₂, and assuming x₁ ⩽ x₂, equation (11) gives rise to a series of special cases below.

When r₁ = r₂ = r₀, figure 1 degrades into a normal m × n cylindrical network, then the resistance of Eq. (11) reduces to

$$ \begin{eqnarray}\begin{array}{ll}{R}_{m\times n}({d}_{1},{d}_{2})= & \displaystyle \frac{|{x}_{2}-{x}_{1}|}{m}r\\ & +\displaystyle \frac{{r}_{0}}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{1,1}^{(i)}-2{\beta }_{1,2}^{(i)}\cos (y{\theta }_{i})+{\beta }_{2,2}^{(i)}}{(1-\cos {\theta }_{i}){F}_{n+1}^{(i)}},\end{array}\end{eqnarray}$$ (92)

Consider a non-regular m × n cylindrical network of Fig. 1, when r₁ = r₀ and r₂ is an arbitrary resistor, the resistance of Eq. (11) reduces to

$$ \begin{eqnarray}{R}_{m\times n}({d}_{1},{d}_{2})=\displaystyle \frac{|{x}_{2}-{x}_{1}|}{m}r+\displaystyle \frac{{r}_{0}}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{1,1}^{(i)}-2{\beta }_{1,2}^{(i)}\cos (y{\theta }_{i})+{\beta }_{2,2}^{(i)}}{(1-\cos {\theta }_{i})[{F}_{n+1}^{(i)}+({h}_{2}-1){F}_{n}^{(i)}]},\,\,\,\,\,\,\,\,\end{eqnarray}$$ (93)

When r₂ = 0, the right boundary of Fig. 1 collapses to a pole, so the non-regular cylindrical network degrades into a cobweb network with an arbitrary boundary resistor r₁ as shown in Fig. 3. By Eq. (11) we obtain the resistance formula

$$ \begin{eqnarray}{R}_{m\times n}({d}_{1},{d}_{2})=\displaystyle \frac{|{x}_{2}-{x}_{1}|}{m}r+\displaystyle \frac{r}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{1,1}^{(i)}-2{\beta }_{1,2}^{(i)}\cos (y{\theta }_{i})+{\beta }_{2,2}^{(i)}}{{\rm{\Delta }}{F}_{n}^{(i)}+({h}_{1}-1){\rm{\Delta }}{F}_{n-1}^{(i)}},\end{eqnarray}$$ (94)

In particular, selecting h₁ = 1 and h₂ = 0, the resistor network of Fig. 3 degrades into a regular cobweb network, the resistance of Eq. (94) reduces to

$$ \begin{eqnarray}{R}_{m\times n}({d}_{1},{d}_{2})=\displaystyle \frac{|{x}_{2}-{x}_{1}|}{m}r+\displaystyle \frac{r}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{1,1}^{(i)}-2{\beta }_{1,2}^{(i)}\cos (y{\theta }_{i})+{\beta }_{2,2}^{(i)}}{{\rm{\Delta }}{F}_{n}^{(i)}},\end{eqnarray}$$ (95)

Please note that the regular cobweb network was studied in Refs. [28] and [35] but the results in Refs. [28] and [35] are different from those from Eq. (95). That is because reference [35] chooses the coordinates in different directions to set up the matrix equations, where reference [35] set up matrix along the longitude, but this paper sets up matrix along the latitude, and reference [28] studied the problem by the WU’s method which gives the result in double sum.

When h₁ = h₂ = 0 (r₁ = r₂ = 0), the left and right boundaries collapse respectively to two poles, the network of Fig. 1 degrades into an m × n globe network as shown in Fig. 4, by Eq. (11) we have

$$ \begin{eqnarray}{R}_{m\times n}({d}_{1},{d}_{2})=\displaystyle \frac{|{x}_{2}-{x}_{1}|}{m}r+\displaystyle \frac{r}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{F}_{{x}_{1}}^{(i)}{F}_{n-{x}_{1}}^{(i)}-2{F}_{{x}_{1}}^{(i)}{F}_{n-{x}_{2}}^{(i)}\cos (y{\theta }_{i})+{F}_{{x}_{2}}^{(i)}{F}_{n-{x}_{2}}^{(i)}}{{F}_{n}^{(i)}},\,\,\,\,\,\,\,\,\end{eqnarray}$$ (96)

Please note that the globe network was studied in Ref. [34], but the result in Ref. [34] is different from that from Eq. (96). That is because they chose coordinates in different directions to set up the matrix equations, which implies that the effective resistance can be expressed in different forms for the same resistor network.

Consider a complex m × n cylindrical network of Fig. 1, when d₁ (0,0) is on the left edge and d₂ (n,y) is on the right edge, the resistance between two edges is

$$ \begin{eqnarray}{R}_{m\times n}(\{0,0\},\{n,y\})=\displaystyle \frac{n}{m}r+\displaystyle \frac{{r}_{0}}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{h}_{1}{\alpha }_{2,n}^{(i)}+{h}_{2}{\alpha }_{1,n}^{(i)}-2{h}_{1}{h}_{2}\cos (y{\theta }_{i})}{(1-\cos {\theta }_{i}){G}_{n}^{(i)}}.\end{eqnarray}$$ (97)

$$ \begin{eqnarray}{R}_{m\times n}(\{0,0\},\{n,y\})=\displaystyle \frac{n}{m}r+\displaystyle \frac{{r}_{0}}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\rm{\Delta }}{F}_{n}^{(i)}-\cos (y{\theta }_{i})}{(1-\cos {\theta }_{i}){F}_{n+1}^{(i)}}.\end{eqnarray}$$ (98)

$$ \begin{eqnarray}{R}_{m\times n}(\{0,0\},\{n,y\})=\displaystyle \frac{n}{m}r+\displaystyle \frac{r}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{F}_{n}^{(i)}}{{\rm{\Delta }}{F}_{n}^{(i)}}.\end{eqnarray}$$ (99)

Consider a complex m × n cylindrical network of Fig. 1, when both d₁ (x₁,0) and d₂ (x₂,0) are on the same horizontal axis, we have

$$ \begin{eqnarray}\begin{array}{lll}{R}_{m\times n}(\{{x}_{1},0\},\{{x}_{2},0\})= \displaystyle \frac{|{x}_{2}-{x}_{1}|}{m}r\\ + \displaystyle \frac{{r}_{0}}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{1,1}^{(i)}-2{\beta }_{1,2}^{(i)}+{\beta }_{2,2}^{(i)}}{(1-\cos {\theta }_{i}){G}_{n}^{(i)}}.\,\,\,\,\,\,\,\,\end{array}\end{eqnarray}$$ (100)

When m = 3, figure 1 degrades into a 3D Δ × n resistor network as shown in Fig. 6. By θ_i = 2iπ / m we have θ_i = 2iπ / 3 (i = 1, 2), substituting them into Eq. (3) yields

$$ \begin{eqnarray}{\lambda }_{1}={\lambda }_{2}=1+\displaystyle \frac{3}{2}h+\sqrt{{\left(1+\displaystyle \frac{3}{2}h\right)}^{2}-1}.\end{eqnarray}$$ (101)

Please note that figure 6 is a new model since the boundary resistors of r₁ and r₂ are two arbitrary values which can lead to different geometries. In Ref. [33] studied is only the network of canonical structure of r₁ = r₂ = r₀. So, equation (102) is a general formula, which can produce several specific results below.

Considering the resistance between two nodes A_x1 and A_x2, there be y = 0 and equation (102) reduces to

$$ \begin{eqnarray}{R}_{{\rm{\Delta }}\times n}({A}_{{x}_{1}},{A}_{{x}_{2}})=\displaystyle \frac{|{x}_{2}-{x}_{1}|}{3}r+\displaystyle \frac{2{r}_{0}}{9}\left(\displaystyle \frac{{\beta }_{{x}_{1},{x}_{1}}^{(1)}-2{\beta }_{{x}_{1},{x}_{2}}^{(1)}+{\beta }_{{x}_{2},{x}_{2}}^{(1)}}{{G}_{n}^{(1)}}\right),\end{eqnarray}$$ (103)

In particular, when x₁ = 0 and r₁ = r₂ = r₀, equation (103) reduces to

$$ \begin{eqnarray}\begin{array}{lll} & & {R}_{{\rm{\Delta }}\times n}({A}_{0},{A}_{x})=\displaystyle \frac{x}{3}r+\displaystyle \frac{2{r}_{0}}{9}\left(\displaystyle \frac{{\rm{\Delta }}{F}_{n}^{(1)}-2{\rm{\Delta }}{F}_{n-x}^{(1)}+{\rm{\Delta }}{F}_{x}^{(1)}{\rm{\Delta }}{F}_{n-x}^{(1)}}{{F}_{n+1}^{(1)}}\right),\end{array}\end{eqnarray}$$ (104)

$$ \begin{eqnarray}\begin{array}{lll} & & {R}_{{\rm{\Delta }}\times n}({A}_{0},{A}_{n})=\displaystyle \frac{n}{3}r+\displaystyle \frac{4{r}_{0}}{9}\left(\displaystyle \frac{{\rm{\Delta }}{F}_{n}^{(1)}-1}{{F}_{n+1}^{(1)}}\right).\end{array}\end{eqnarray}$$ (105)

$$ \begin{eqnarray}{R}_{{\rm{\Delta }}\times n}({A}_{{x}_{1}},{B}_{k})={R}_{{\rm{\Delta }}\times n}({A}_{{x}_{1}},{C}_{k})=\displaystyle \frac{|k-{x}_{1}|}{3}r+\displaystyle \frac{2{r}_{0}}{9}\left(\displaystyle \frac{{\beta }_{1,1}^{(1)}+{\beta }_{1,2}^{(1)}+{\beta }_{2,2}^{(1)}}{{G}_{n}^{(1)}}\right).\end{eqnarray}$$ (106)

$$ \begin{eqnarray}\begin{array}{lll} & & {R}_{{\rm{\Delta }}\times n}({A}_{0},{B}_{k})={R}_{{\rm{\Delta }}\times n}({A}_{0},{C}_{k})=\displaystyle \frac{k}{3}r+\displaystyle \frac{2{r}_{0}}{9}\left(\displaystyle \frac{{\rm{\Delta }}{F}_{n}^{(1)}+{\rm{\Delta }}{F}_{n-k}^{(1)}+{\rm{\Delta }}{F}_{k}^{(1)}{\rm{\Delta }}{F}_{n-k}^{(1)}}{{F}_{n+1}^{(1)}}\right),\end{array}\end{eqnarray}$$ (107)

$$ \begin{eqnarray}\begin{array}{lll} & & {R}_{{\rm{\Delta }}\times n}({A}_{0},{B}_{n})={R}_{{\rm{\Delta }}\times n}({A}_{0},{C}_{n})=\displaystyle \frac{n}{3}r+2{r}_{0}\displaystyle \frac{2{\rm{\Delta }}{F}_{n}^{(1)}+1}{9{F}_{n+1}^{(1)}}.\end{array}\end{eqnarray}$$ (108)

$$ \begin{eqnarray}{R}_{{\rm{\Delta }}\times n}({A}_{k},{B}_{k})={R}_{{\rm{\Delta }}\times n}({A}_{k},{C}_{k})=2{r}_{0}\displaystyle \frac{{\rm{\Delta }}{F}_{k}^{(1)}{\rm{\Delta }}{F}_{n-k}^{(1)}}{3{F}_{n+1}^{(1)}}.\end{eqnarray}$$ (109)

In order to clearly understand the change rule of equivalent resistances R_{Δ × n} (A₀,A_k) and R_{Δ × n} (A₀,B_k), we use Matlab to draw their change curve as shown in Figs. 7–9.

Comparing Fig. 7 with Fig. 8, we find that their variation patterns are basically similar, but there are slight differences between them, that is, R_{Δ × n} (A₀,A₀) = 0 and R_{Δ × n} (A₀,B₀) > 0 which are exactly the same as the results of the actual circuit. The equivalent resistance R(A₀, B_k) increases with n and x increasing. Figure 9 shows that R(A₀, B₀) decreases with the increase of n, but R(A₀, B₀) increases with h increasing except for n = 0. Since our formulae are all accurate and not estimates, all the computer simulation results are exactly the same as those in the three figures, i.e., Figs. 7–9, which indirectly indicates the correctness of our results.

When m = 4, figure 2 degrades into a 3D □ × n resistor network as shown in Fig. 5. We can obtain the equivalent resistance between any two nodes by Eq. (11). Firstly, by Eq. (2) we have θ_i = iπ / 2, substituting it into Eq. (3) yields

$$ \begin{eqnarray}\begin{array}{lll} & & {\lambda }_{1}={\lambda }_{3}=1+h+\sqrt{{h}^{2}+2h},\\ & & {\lambda }_{2}=1+2h+2\sqrt{{h}^{2}+h}.\end{array}\end{eqnarray}$$ (110)

When x₁ = 0 and x₂ = k, equation (111) reduces to

$$ \begin{eqnarray}{R}_{\square \times n}({A}_{0},{P}_{k})=\displaystyle \frac{k}{4}r+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n}^{}-2{h}_{1}{\alpha }_{2,n-k}^{(1)}\cos (y\pi /2)+{\alpha }_{1,k}^{(1)}{\alpha }_{2,n-k}^{(1)}}{4{G}_{n}^{(1)}}+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n}^{(2)}-2{h}_{1}{\alpha }_{2,n-k}^{(2)}\cos (y\pi )+{\alpha }_{1,k}^{(2)}{\alpha }_{2,n-k}^{(2)}}{16{G}_{n}^{(2)}},\end{eqnarray}$$ (112)

When x₁ = 0 and x₂ = n, equation (112) reduces to

$$ \begin{eqnarray}{R}_{\square \times n}({A}_{0},{P}_{n})=\displaystyle \frac{n}{4}r+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n}^{}-2{h}_{1}{h}_{2}\cos (y\pi /2)+{h}_{2}{\alpha }_{1,n}^{(1)}}{4{G}_{n}^{(1)}}+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n}^{(2)}-2{h}_{1}{h}_{2}\cos (y\pi )+{h}_{2}{\alpha }_{1,n}^{(2)}}{16{G}_{n}^{(2)}}.\end{eqnarray}$$ (113)

$$ \begin{eqnarray}{R}_{\square \times n}({A}_{{x}_{1}},{P}_{{x}_{1}})={r}_{0}\displaystyle \frac{{\beta }_{1,1}^{(1)}[1-\cos (y\pi /2)]}{2{G}_{n}^{(1)}}+{r}_{0}\displaystyle \frac{{\beta }_{1,1}^{(2)}[1-\cos (y\pi )]}{8{G}_{n}^{(2)}},\end{eqnarray}$$ (114)

In particular, when r₁ = r₂ = r₀, equation (114) reduces to

$$ \begin{eqnarray}{R}_{\square \times n}({A}_{{x}_{1}},{P}_{{x}_{1}})={r}_{0}\displaystyle \frac{{\rm{\Delta }}{F}_{{x}_{1}}^{(1)}{\rm{\Delta }}{F}_{n-{x}_{1}}^{(1)}[1-\cos (y\pi /2)]}{2{F}_{n+1}^{(1)}}+{r}_{0}\displaystyle \frac{{\rm{\Delta }}{F}_{{x}_{1}}^{(2)}{\rm{\Delta }}{F}_{n-{x}_{1}}^{(2)}[1-\cos (y\pi )]}{8{F}_{n+1}^{(2)}}.\end{eqnarray}$$ (115)

Considering the resistance between two nodes A₀ and A_k, there is y = 0, so equation (112) reduces to

$$ \begin{eqnarray}{R}_{\square \times n}({A}_{0},{A}_{k})=\displaystyle \frac{k}{4}r+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n}^{(1)}-2{h}_{1}{\alpha }_{2,n-k}^{(1)}+{\alpha }_{1,k}^{(1)}{\alpha }_{2,n-k}^{(1)}}{4{G}_{n}^{(1)}}+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n}^{(2)}-2{h}_{1}{\alpha }_{2,n-k}^{(2)}+{\alpha }_{1,k}^{(2)}{\alpha }_{2,n-k}^{(2)}}{16{G}_{n}^{(2)}}.\end{eqnarray}$$ (116)

$$ \begin{eqnarray}{R}_{\square \times n}({A}_{0},{A}_{n})=\displaystyle \frac{n}{4}r+{r}_{0}\displaystyle \frac{{\rm{\Delta }}{F}_{n}^{(1)}-1}{2{F}_{n+1}^{(1)}}+{r}_{0}\displaystyle \frac{{\rm{\Delta }}{F}_{n}^{(2)}-1}{8{F}_{n+1}^{(2)}},\end{eqnarray}$$ (117)

Considering the resistance between two nodes A₀ and B_k, there is y = 1, so equation (112) reduces to

$$ \begin{eqnarray}{R}_{\square \times n}({A}_{0},{B}_{k})=\displaystyle \frac{k}{4}r+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n}^{(1)}+{\alpha }_{1,k}^{(1)}{\alpha }_{2,n-k}^{(1)}}{4{G}_{n}^{(1)}}+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n}^{(2)}+2{h}_{1}{\alpha }_{2,n-k}^{(2)}+{\alpha }_{1,k}^{(2)}{\alpha }_{2,n-k}^{(2)}}{16{G}_{n}^{(2)}},\end{eqnarray}$$ (118)

In particular, when r₁ = r₂ = r₀ and k = n, from Eq. (118) we have

$$ \begin{eqnarray}{R}_{\square \times n}({A}_{0},{B}_{n})=\displaystyle \frac{n}{4}r+{r}_{0}\displaystyle \frac{{\rm{\Delta }}{F}_{n}^{(1)}}{2{F}_{n+1}^{(1)}}+{r}_{0}\displaystyle \frac{{\rm{\Delta }}{F}_{n}^{(2)}+1}{8{F}_{n+1}^{(2)}}.\end{eqnarray}$$ (119)

$$ \begin{eqnarray}\begin{array}{lll}{R}_{\square \times n}({A}_{0},{C}_{k}) & = & \displaystyle \frac{k}{4}r+{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n}^{(1)}+2{h}_{1}{\alpha }_{2,n-k}^{(1)}+{\alpha }_{1,k}^{(1)}{\alpha }_{2,n-k}^{(1)}}{4{G}_{n}^{(1)}}\\ & & +{r}_{0}\displaystyle \frac{{h}_{1}{\alpha }_{2,n}^{(2)}-2{h}_{1}{\alpha }_{2,n-k}^{(2)}+{\alpha }_{1,k}^{(2)}{\alpha }_{2,n-k}^{(2)}}{16{G}_{n}^{(2)}}.\,\,\,\,\,\,\,\,\,\end{array}\end{eqnarray}$$ (120)

$$ \begin{eqnarray}{R}_{\square \times n}({A}_{0},{C}_{n})=\displaystyle \frac{n}{4}r+{r}_{0}\displaystyle \frac{{\rm{\Delta }}{F}_{n}^{(1)}+1}{2{F}_{n+1}^{(1)}}+{r}_{0}\displaystyle \frac{{\rm{\Delta }}{F}_{n}^{(2)}-1}{8{F}_{n+1}^{(2)}}.\end{eqnarray}$$ (121)

From Figs. 10 and 11, the equivalent resistances R(A₀, A_k) and R(A₀, C_k) increase with t n and x increasing. Comparing Fig. 10 with Fig. 11, we find that their variation patterns are basically similar, but there are slight differences between them, which are R_{□ × n} (A₀,A₀) = 0 and R_{□ × n1} (A₀,C₀) > 0, these are exactly the same as the results of the actual circuit.

In Refs. [33] and [49] the □ × n circuit network with r₁ = r₂ = r₀ was studied specifically and the results of equivalent resistance between two special nodes, such as the points of A₀,A_n and A₀,C_n and A₀,B_n were obtained (Notice that it is only under the special condition of r₁ = r₂ = r₀). Comparing these results shows that they are exactly the same as the results from Eqs. (115), (117), (119), and (121) in this paper. The comparison among these results verifies their correctness.

The Case H indicates again that general formula (11) is a meaningful and multipurpose result since just a 3D □ × n resistor network has rich contents and many functions such as Eqs. (111)–(121).

When m = 4, n = 3, and r₁ = r₂ = 0, figure 1 degrades into an anisotropic crystal as shown in Fig. 12. We will give the equivalent resistance between any two nodes by Eq. (96), which is to further understand the physical meaning of Eq. (96).

We define the coordinates of four nodes: B_k (1,k), C_k (2,k). As θ_i = iπ / 2, (i = 1,2,3), then λ_k and λ¯k satisfies Eq. (110), and cos (yθ₁) = cos (yθ₃). By Eq. (96) we have a general formula

$$ \begin{eqnarray}R({d}_{1},{d}_{2})=\displaystyle \frac{|{x}_{2}-{x}_{1}|}{4}r+\displaystyle \frac{{\beta }_{1,1}^{(1)}-2{\beta }_{1,2}^{(1)}\cos (y\pi /2)+{\beta }_{2,2}^{(1)}}{2{F}_{3}^{(1)}}r+\displaystyle \frac{{\beta }_{1,1}^{(2)}-2{\beta }_{1,2}^{(2)}\cos (y\pi )+{\beta }_{2,2}^{(2)}}{4{F}_{3}^{(2)}}r,\end{eqnarray}$$ (122)

The above results (123)–(128) are a series of equivalent resistances derived from formula (122). The actual calculation is conducted based on Fig. 13, and the obtained results are found to be exactly the same as the above results. This indirectly verifies the correctness of the conclusion (11) given in this paper.

According to the above discussion, the readers should be able to understand the essential meaning of Eq. (11) and understand that equation (11) has broad application value for a variety of lattice structures, which will provide a new theoretical basis for relevant scientific research. As explained in Section 1 and the numerous special cases given above, the theoretical results of this paper will have potential application value in relevant fields.

Consider a regular m × n cylindrical network with r₁ = r₂ = r₀ as shown in Fig. 1. In Ref. [26] there was a resistance formula (1) given by the Laplacian matrix method, which is in the form of double sum. However, in this paper equation (92) is given by the RT-V method, where the condition and network structure are consistent with those in Ref. [26]. Obviously, the two results in the different form, which appear in two independent articles, are necessarily equivalent because they are from the same network with the same coordinates. Comparing formula (92) with formula (1), and taking y, m, n and x₁,x₂ as natural numbers, when 0 ⩽ x₁ ⩽ x₂ ⩽ n, 0 ⩽ y ⩽ m – 1, we obtain the following trigonometric identity:

$$ \begin{eqnarray}\displaystyle \frac{1}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \sum _{j=1}^{n}\displaystyle \frac{{C}_{{x}_{1},j}^{2}+{C}_{{x}_{2},j}^{2}-2{C}_{{x}_{1},j}{C}_{{x}_{2},j}\cos (y{\theta }_{i})}{(1-\cos {\theta }_{i})+{h}^{-1}(1-\cos {\phi }_{j})}=\left(\displaystyle \frac{{y}^{2}}{m}-y\right)+\displaystyle \frac{n+1}{2m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{1,1}^{(i)}-2{\beta }_{1,2}^{(i)}\cos (y{\theta }_{i})+{\beta }_{2,2}^{(i)}}{(1-\cos {\theta }_{i}){F}_{n+1}^{(i)}},\end{eqnarray}$$ (129)

In particular, when setting y,m,n and x₁,x₂ to be special number values, we have the following interesting identities.

When y = 0, from Eq. (129) we have

$$ \begin{eqnarray}\displaystyle \frac{2}{n+1}\displaystyle \sum _{i=1}^{m-1}\displaystyle \sum _{j=1}^{n}\displaystyle \frac{{[\cos ({x}_{1}+1/2){\phi }_{j}-\cos ({x}_{2}+1/2){\phi }_{j}]}^{2}}{(1-\cos {\theta }_{i})+{h}^{-1}(1-\cos {\phi }_{j})}=\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\beta }_{1,1}^{(i)}-2{\beta }_{1,2}^{(i)}+{\beta }_{2,2}^{(i)}}{(1-\cos {\theta }_{i}){F}_{n+1}^{(i)}}.\end{eqnarray}$$ (131)

$$ \begin{eqnarray}\displaystyle \frac{1}{n+1}\displaystyle \sum _{j=1}^{n}\displaystyle \frac{{[\cos ({x}_{1}+1/2){\phi }_{j}-\cos ({x}_{2}+1/2){\phi }_{j}]}^{2}}{2+{h}^{-1}(1-\cos {\phi }_{j})}=\displaystyle \frac{{\beta }_{1,1}^{(1)}-2{\beta }_{1,2}^{(1)}+{\beta }_{2,2}^{(1)}}{4{F}_{n+1}^{(1)}},\end{eqnarray}$$ (132)

When x₁ = x₂ = x, equation (129) reduces to

$$ \begin{eqnarray}\displaystyle \frac{2}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \sum _{j=1}^{n}\displaystyle \frac{{\cos }^{2}[(x+1/2){\phi }_{j}](1-\cos y{\theta }_{i})}{(1-\cos {\theta }_{i})+{h}^{-1}(1-\cos {\phi }_{j})}=\left(\displaystyle \frac{{y}^{2}}{m}-y\right)+\displaystyle \frac{n+1}{m}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{{\rm{\Delta }}{F}_{x}^{(i)}{\rm{\Delta }}{F}_{n-x}^{(i)}}{{F}_{n+1}^{(i)}}\left(\displaystyle \frac{1-\cos (y{\theta }_{i})}{1-\cos {\theta }_{i}}\right),\end{eqnarray}$$ (134)

When m = 2, y = 1, we have cos θ₁ = –1, ϕ_j = jπ / (n + 1), from Eq. (129), we have

$$ \begin{eqnarray}\displaystyle \frac{1}{n+1}\displaystyle \sum _{j=1}^{n}\displaystyle \frac{{({C}_{{x}_{1},j}+{C}_{{x}_{2},j})}^{2}}{2+{h}^{-1}(1-\cos {\phi }_{j})}=\displaystyle \frac{{\beta }_{1,1}^{(1)}+2{\beta }_{1,2}^{(1)}+{\beta }_{2,2}^{(1)}}{4{F}_{n+1}^{(1)}}-\displaystyle \frac{1}{n+1},\end{eqnarray}$$ (135)

When x = n = 0, θ_i = 2iπ / m, 0 ⩽ y ⩽ m – 1, and y ∈ N, by Eq. (134) we have

$$ \begin{eqnarray}\displaystyle \sum _{i=1}^{m-1}\displaystyle \frac{1-\cos (yi\pi /m)}{1-\cos (i\pi /m)}=y(m-y).\end{eqnarray}$$ (136)

This paper shows a new progress of studying the electrical characteristics (resistance and potential) of an arbitrary cylindrical m × n resistor network with complex boundaries by the advanced RT-V method, which reveals the electrical characteristics of complex cylindrical network for the first time, such as three general formulae (8), (10), and (11). As the general applications of these three formulas, many interesting results are produced for various types of resistor networks. When the boundary resistor r₂ = 0, the non-regular cylindrical network degrades into a cobweb network as shown in Fig. 3; when the boundary resistor r₁ = r₂ = 0, the non-regular cylindrical network degrades into a globe network as shown in Fig. 4; when considering m = 3 and m = 4 figure 1 degrades into the Δ × n and □ × n networks. These show that the cylindrical network in Fig. 1 is a kind of complex network with multi-function and multi-structure, which is a network model and scientific problem with great research value. In particular, a new mathematical identity is discovered in the comparative study, which provides a new proposition for mathematical researchers. These problems discussed above give rise to a series of new results presented in this paper, which fill in the deficiencies of previous literature research,and provide a new theoretical basis for multidisciplinary application. This shows that the RT method has a wide range of applications and can solve a variety of complex resistance network problems, and it has become the basic theory of relevant scientific research in the future.

As is well known, the research of resistor network is mainly a model study, which can explain more problems that have been solved or not solved previously. The present research focuses on how to study and establish the complex network model. Therefore, the discussion emphasizes the research methodology and calculation results of resistor network model. This paper presents the theoretical results of the overall electrical properties of cylindrical networks and discuss a series of special cases for illustrating that the cylindrical network with complex boundary has many special structures, so it has more potential application value. The RT method is mainly used to accurately study the resistor network model with complex boundary conditions, and then the analytical expressions of electrical properties obtained from the network model can be applied to other relevant scientific problems. The limitation of this paper is that it does not give concrete applications in practical problems. However, the relevant problems in the future new technology can be abstracted into the network model at first, and then solved by the theory of electrical properties given in this paper, and the examples can be found from neural networks, artificial intelligence, discrete mathematics, etc.