Introduction

With the development of natural science, people have encountered various new problems. According to the complexity of the problem, solutions emerge in endlessly. Researches show that a host of problems can be solved by establishing resistor network model1,2,3,4,5,6,7,8,9,10 and neural network model11,12,13,14,15,16,17. In the past few years, many results have been achieved in the study of resistor networks. For example, the establishment of graph theory, the researches on Laplacian matrix (LM) method of resistor network, infinite network, finite network, and corner-to-corner resistance and so on7,8,9,10,18,19,20,21,22,23,24,25,26,27,28,29,30,31. Shi11,12 et al. studied a novel discrete-time recurrent neural network. Liu13 et al. and Sun14 et al. proposed different types of the zeroing neural network. And Jin15,16,17 et al. studied an innovative control theory stimulated gradient neural network algorithm.

In recent years, Tan32,33,34,35,36,37,38,39,40,41,42,43 proposed the Recursion-Transform (RT) method which is different from the Laplacian matrix method. It only depends on one matrix and one direction, and the calculation is simpler. In 2013, Tan32 studied the problem of two-point resistance on cobweb with a 2r boundary, which has never been solved before. Under finite and infinite conditions, the \(m \times n\) cobweb general formula of resistance between any two nodes is proposed. In 2014, Tan36,38 made a new breakthrough and introduced the equivalent resistance of the globe network and fan network model by RT method. Shortly after that many researches on resistor networks are based on RT method. RT method needs eigenvalues of a tridiagonal matrix to represent the potential formula. At present, there have been many results on tridiagonal matrices45,46,47,48,49,50,51, which are also widely used. It can be said that it is a powerful tool to solve the resistor network33,34,35,36,37,38,39,40,41,42,43.

Figure 1
figure 1

An m \(\times\) n globe network which has n longitude and \(m-1\) latitude. The resistances in the longitude and latitude directions are \(r_{0}\) and r, respectively.

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Figure 2
figure 2

Segment of the globe network with current directions and parameters.

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In 2018, Tan44 proposed an \(m \times n\) globe network, as shown in Fig. 1. The resistances in the latitude and longitude directions are r and \(r_{0}\), respectively, where m and n are the resistance numbers along the latitude and longitude directions. The nodes of the network are represented by coordinates \(\{x, y\}\). Considering \(O_{0}=0\) as the origin of the coordinate system, the potential of node d(x, y) is \(U_{m\times n}(x , y )\) as shown in Fig. 2. The potential of any node in the \(m\times n\) globe network is as follows

$$\begin{aligned} \frac{U_{m\times n}(x,y)}{J} = \frac{y(y_{1}-y_{2})}{mn}r_{0} + \frac{2r}{m}\sum ^{m}_{i=2}\frac{\beta ^{(i)}_{x_{1},x}S_{y_{1},i}-\beta ^{(i)}_{x_{2},x}S_{y_{2},i}}{\lambda ^{n}_{i}+\bar{\lambda }^{n}_{i}-2}S_{y,i}, \end{aligned}$$
(1)

where

$$\begin{aligned} S_{k,i}= & \sin (y_{k}\theta _{i}),\ \theta _{i} = (i-1)\pi /m, \end{aligned}$$
(2)
$$\begin{aligned} \beta ^{(i)}_{x_{s},x_{k}}= & F_{n-|x_{s}-x_{k}|}+F_{|x_{s}-x_{k}|},\ F^{(i)}_{k} = (\lambda ^{k}_{i}-\bar{\lambda }^{k}_{i})/(\lambda _{i}-\bar{\lambda }_{i}), \end{aligned}$$
(3)
$$\begin{aligned} \lambda _{i}= & 1+b-b\cos \theta _{i}+\sqrt{(1+b-b\cos \theta _{i})^{2} - 1},\nonumber \\ \bar{\lambda }_{i}= & 1+b-b\cos \theta _{i}-\sqrt{(1+b-b\cos \theta _{i})^{2} - 1}, ~~b=r/r_{0}. \end{aligned}$$
(4)

New formula of potential represented by Chebyshev polynomials

Since the potential formula (1) contains exponential function operation, it is less efficient to calculate potential of globe resistor networks by computer. The purpose of this section is to give an improved potential formula of the globe network44, i.e. a new formula of potential expressed by Chebyshev polynomial of the second class is given53. it saves time to calculate the potential by computer.

Let the current J from \(d_{1}(x_{1}, y_{1})\) to \(d_{2}(x_{2}, y_{2})\), the potential formula of two nodes in an \(m \times n\) resistor network can be written as

$$\begin{aligned} \frac{U_{m\times n}(x,y)}{J} = \frac{y(y_{1}-y_{2})}{mn}r_{0} + \frac{2r}{m}\sum ^{m}_{j=2}\frac{\gamma ^{(j)}_{x_{1},x}C_{y_{1},j}-\gamma ^{(j)}_{x_{2},x}C_{y_{2},j}}{t_{j}U^{(j)}_{n-1}-2U^{(j)}_{n-2}-2}C_{y,j}, \end{aligned}$$
(5)

where

$$\begin{aligned} & \gamma ^{(j)}_{x_{s},x_{k}} = U^{(j)}_{n-|x_{s}-x_{k}|-1}+U^{(j)}_{|x_{s}-x_{k}|-1}, \ s = 1,2, \end{aligned}$$
(6)
$$\begin{aligned} & \quad C_{p,j} = \sin (\frac{p(j-1)\pi }{m}),\ p=y_{1},\ y_{2},\ y, \end{aligned}$$
(7)
$$\begin{aligned} & \quad t_{j}=2+\frac{2r}{r_{0}}-\frac{2r}{r_{0}}\cos \frac{(j-1)\pi }{m}, \end{aligned}$$
(8)
$$\begin{aligned} & \quad U^{(j)}_{k} = U^{(j)}_{k}(\cos \psi _{j}) = \frac{\sin (k+1)\psi _{j}}{\sin (\psi _{j})},\ \psi _{j}=\arccos (1+\frac{r}{r_0}-\frac{r}{r_0}\cos \frac{(j-1)\pi }{m}),\nonumber \\ & \quad k=n-|x_1-x|,~|x_1-x|,~n-|x_2-x|,~|x_2-x|,~n-1,~n-2,~~j=1,2,\ldots , m. \end{aligned}$$
(9)

In particular, the input and output points can be one or more, so formula (5) applies to all coordinate points \((x_{k}, y_{k})\) \((0\le k \le n)\). Therefore, p can also be \(y_3,\ y_4,\ldots ,\ y_k\).

Suppose that \(O_{0} = 0\), then calculating the node potential between any two points by Ohm’s law can be described as

$$\begin{aligned} U_{m\times n}(x,y) = U(0,0) - r_{0}\sum ^{y}_{j=1}I^{(j)}_{x} = -r_{0}\sum ^{y}_{j=1}I^{(j)}_{x}, \end{aligned}$$
(10)

where \(I_{x}^{(j)}\) is denoted by the current in the vertical direction.

Base on formula (5), using the visualization of data to realize the dynamic visualization of the potential change. With the change of the current input and output point change, the potential change of any two nodes is shown in Fig. 3.

Figure 3
figure 3

A 3D dynamic view for the changing graph of \(U_{90\times 90}(x,y)/J\) with the current input and output point change.

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Three terms recurrence sequence and discrete cosine transform

In this section, in order to improve the actual performance and realize the fast algorithm, the three terms recurrence sequence represented by Chebyshev polynomial of the second class and two types of discrete cosine transform are introduced.

The three terms recurrence sequence is defined by the following conditions:

$$\begin{aligned} W_k=dW_{k-1}-qW_{k-2},~~W_0=A,~~W_1=B, \end{aligned}$$
(11)

where \(k\in \textbf{N},~~k\ge 2,~~A,B,d,q\in \textbf{C}\), \(\textbf{N}\) is the set of all natural numbers and \(\textbf{C}\) is the set of all complex numbers.

The three terms recurrence sequence52 represented by Chebyshev polynomial of the second class is

$$\begin{aligned} W_{k} =(\sqrt{q})^k\left( \frac{B}{\sqrt{q}}U_{k-1}\left( \frac{d}{2\sqrt{q}}\right) - AU_{k-2}\left( \frac{d}{2\sqrt{q}}\right) \right) , \end{aligned}$$
(12)

where

$$\begin{aligned} U_{k} = \frac{\sin (k+1)\psi }{\sin \psi },~~\psi =\arccos (\frac{d}{2\sqrt{q}})\in \textbf{C}, \end{aligned}$$
(13)

is the Chebyshev polynomial of the second class53.

Remark

Based on the above formula, how to replace Eqs. (1), (2), (3) and (4) with Eqs. (5), (6), (7) and (9) will be showed, respectively. According to the equation (4), \(\lambda _{j}+\bar{\lambda }_{j}=t_{j}\) and \(\lambda _{j}\cdot \bar{\lambda }_{j}=1\) are obtained . Take it into the equation (11), and a three term recursive formula is obtained

$$\begin{aligned} F^{(j)}_k=t_{j}F^{(j)}_{k-1}-F^{(j)}_{k-2},~~F^{(j)}_{0}=0,~~F^{(j)}_{1}=1, \end{aligned}$$
(14)

where \(d=t_{j},\ q=1\), \(t_{j}\) and \(F^{(j)}_{k}\) are the same as Eqs. (8) and (3), respectively. By Eq. (14) and formula (12), the replacement result can be obtained

$$\begin{aligned} {F^{(j)}_{k} = (\lambda ^{k}_{j}-\bar{\lambda }^{k}_{j})/(\lambda _{j}-\bar{\lambda }_{j})=U^{(j)}_{k-1}(\frac{t_{j}}{2}).} \end{aligned}$$
(15)

Similarly, Let

$${Q^{(j)}_{n}=\lambda ^{n}_{j}+\bar{\lambda }^{n}_{j},}$$
(16)

where \(Q^{(j)}_{0}=2,\ Q^{(j)}_{1}=t_{j}\). Then the recursive relation of \(Q^{(j)}_{n}\) is obtained as follows

$$\begin{aligned} {Q^{(j)}_{n}=t_{j}Q^{(j)}_{n-1}-Q^{(j)}_{n-2},\ Q^{(j)}_{0}=2,\ Q^{(j)}_{1}=t_{j},} \end{aligned}$$
(17)

where \(d=t_{j},\ q=1\), \(t_{j}\) and \(Q^{(j)}_{n}\) are the same as Eqs. (8) and (16), respectively. By Eq. (17) and formula (12), the replacement result can be obtained as follows

$$\begin{aligned} {Q^{(j)}_{n}=\lambda ^{n}_{j}+\bar{\lambda }^{n}_{j}=t_{j}U_{n-1}(\frac{t_{j}}{2})-U_{n-2}(\frac{t_{j}}{2}).} \end{aligned}$$
(18)

According to Eqs. (1), (3),(15) and (18), a new potential formula (5) is got.

In order to obtain a fast numerical algorithm for computing the potential, the orthogonal diagonalization of the matrix \(\textbf{B}_m\) is given.

$$\begin{aligned} \textbf{B}_{m}= \left( \begin{array}{cccccc} 2+b & -b & 0& \cdots & 0\\ -b & 2(1+b) & -b & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \ddots & -b & 2(1+b) & -b\\ 0 & \cdots & 0 & -b & 2+b \end{array} \right) _{m\times m}, \end{aligned}$$
(19)

where \(b = r/r_0\).

Then the eigenvalues \(t_1,\ldots ,t_m\) of \(\textbf{B}_{m}\) are given by

$$\begin{aligned} t_j=2+2b-2b\cos \frac{(j-1)\pi }{m},~~j=1,2,\ldots ,m, \end{aligned}$$
(20)

and the corresponding eigenvectors \(\zeta ^{(j)}=(\zeta ^{(j)}_1,\ldots ,\zeta ^{(j)}_m)^\dag\) are given by

$$\begin{aligned} \zeta ^{(j)}_k=\sqrt{\frac{2}{m}}d_j\cos \frac{(2k-1)(j-1)\pi }{2m},~~k=1,2,\ldots ,m,j=1,2,\ldots , m, \end{aligned}$$
(21)

where \(d_j={\left\{ \begin{array}{ll}1,&j\ne 1,\\ \frac{\sqrt{2}}{2},&j=1.\end{array}\right. }\)

Let

$$\begin{aligned} \mathbb {C}_m^{III}=\left( \sqrt{\frac{2}{m}}d_j\cos \frac{(2k-1)(j-1)\pi }{2m} \right) _{k,j=1}^m, \end{aligned}$$
(22)

where \(d_j={\left\{ \begin{array}{ll}1,&j\ne 1,\\ \frac{\sqrt{2}}{2},&j=1.\end{array}\right. }\)

Obviously, the matrix \(\mathbb {C}_m^{III}\) is the famous third type of discrete cosine transform (DCT-III)54,55,56,57. \(\mathbb {C}_m^{III}\) is an orthogonal matrix and the inverse of \(\mathbb {C}_m^{III}\) is actually \(\mathbb {C}_m^{II}\), i.e.

$$\begin{aligned} (\mathbb {C}_m^{III})^{-1}=(\mathbb {C}_m^{III})^T=\mathbb {C}_m^{II}, \end{aligned}$$
(23)

where the orthogonal matrix \(\mathbb {C}_m^{II}\) is the famous second type of discrete cosine transform (DCT-II)54,55,56,57.

By calculation we obtain the orthogonal diagonalization of the matrix \(\textbf{B}_{m}\) as follows

$$\begin{aligned} (\mathbb {C}_m^{III})^{-1}\textbf{B}_{m}(\mathbb {C}_m^{III})=\textrm{diag}(t_1,t_2,\ldots ,t_m), \end{aligned}$$
(24)

i.e.

$$\begin{aligned} \textbf{B}_{m}=(\mathbb {C}_m^{III})\textrm{diag}(t_1,t_2,\ldots ,t_m)(\mathbb {C}_m^{III})^{-1}, \end{aligned}$$
(25)

where \(t_j, j=1,2,\ldots , m\) are given by Eq. (20).

Tan44 used Kirchhoff’s law to analyze the resistor network and to establish the resistor network model. The following general matrix equation is

$$\begin{aligned} \textbf{I}_{k+1} = \textbf{B}_{m}\textbf{I}_{k} - \textbf{I}_{k-1} - b \textbf{H}_{y}\delta _{k,x}, \end{aligned}$$
(26)

where \(\textbf{B}_{m}\) is given by Eq. (19), the function \(\delta _{k,x}\) is defined as: \(\delta _{k,x}(x =k) = 1\), \(\delta _{k,x}(x\ne k) = 0\), \(\textbf{I}_{k}\) and \(\textbf{H}_{x}\) are the \(m \times 1\) column matrices, and can be written as

$$\begin{aligned} & \textbf{I}_{k} = [I^{(1)}_{k},I^{(2)}_{k},...I^{(m)}_{k}] ^{T}\ (0 \le k \le n), \end{aligned}$$
(27)
$$\begin{aligned} & \quad (\textbf{H}_{1})_{k} = J(-\delta _{k,y_{1}}+\delta _{k,y_{1}+1}),~~ (\textbf{H}_{2})_{k} = J(\delta _{k,y_{2}}-\delta _{k,y_{2}+1}). \end{aligned}$$
(28)

To solve the Eq. (26), using the matrix transformation method to multiply \(\mathbb {C}_m^{III}\) at the same time. Through Eq. (24) and the transformed matrix equation we appoint

$$\begin{aligned} \sqrt{\frac{m}{2}}\mathbb {C}_m^{III}\textbf{I}_{k} = \textbf{W}_{k}\ and\ \textbf{I}_{k} = \sqrt{\frac{2}{m}}(\mathbb {C}_m^{III})^T\textbf{W}_{k} =\sqrt{\frac{2}{m}}\mathbb {C}_m^{II}\textbf{W}_{k}, \end{aligned}$$
(29)

where the \(m\times 1\) column matrix \(\textbf{W}_{k}\) is

$$\begin{aligned} \textbf{W}_{k} = [W^{(1)}_{k},W^{(2)}_{k},...,W^{(m)}_{k}]^{T} (0 \le k \le n). \end{aligned}$$
(30)

Based on the Chebyshev polynomial of the second class, the general solution formula of current is rewrited . When \(j = 1\), the following formula is obtained

$$\begin{aligned} W^{(1)}_{k} = \frac{J(y_{2}-y_{1})}{n\sqrt{2}},(0\le k\le n). \end{aligned}$$
(31)

When \(j\ge 2\), the following formula is obtained

$$\begin{aligned} W^{(j)}_{x} = bJ\frac{\gamma ^{(j)}_{x_{1},x}\varsigma _{1,j}+\gamma ^{(j)}_{x_{2},x}\varsigma _{2,j}}{t_{j}U^{(j)}_{n-1}-2U^{(j)}_{n-2}-2},(0\le x\le n), \end{aligned}$$
(32)

where \(\gamma ^{(j)}_{x_{s},x_{k}}\) is equal to Eq. (6), \(\varsigma _{1,j}= -2\sin (\frac{(j-1)\pi }{2m})\sin (\frac{y_{1}(j-1)\pi }{m})\), and \(\varsigma _{2,j}= 2\sin (\frac{(j-1)\pi }{2m})\sin (\frac{y_{2}(j-1)\pi }{m})\).

Displaying of some special and interesting potential formulae

Since formula (5) is a general potential conclusion of a globe network including all cases, some special conditions in formula (5) and a series of fascinating results under various parameters will be displayed. In the following, assuming that the potential reference at point O(0, 0) is \(U(0, 0) = 0\).

Special 1

Suppose the current J flows in from the origin \(d_{1}(x_{1}, y_{1})\) and out from \(d_{2}(x_{2}, y_{2})=O(0, 0)\), the potential of any two points can be written as

$$\begin{aligned} \frac{U_{m\times n}(x,y)}{J} = \frac{yy_{1}}{mn}r_{0} + \frac{2r}{m}\sum ^{m}_{j=2}\left( \frac{\sin (\frac{y_{1}(j-1)\pi }{m}))\sin (\frac{y(j-1)\pi }{m})}{t_{j}U^{(j)}_{n-1}-2U^{(j)}_{n-2}-2}\right) \gamma ^{(j)}_{x_{1},x}, \end{aligned}$$
(33)

where \(\gamma ^{(j)}_{x_{s},x_{k}} = U^{(j)}_{n-|x_{s}-x_{k}|-1}+U^{(j)}_{|x_{s}-x_{k}|-1}\) is defined in Eq. (6), and \(U^{(j)}_{k}\) is the same as Eq. (9).

When \(m = n = 90,\ J = 10,\ x_{1} = y_{1} = 50,\ x_{2} = y_{2} = 0,\ and \ r_{0}=r=1,\) the following formula is obtained

$$\begin{aligned} \frac{U_{90\times 90}(x,y)}{J} = \frac{y}{162} + \frac{1}{45}\sum ^{90}_{j=2}\left( \frac{\sin (\frac{5(j-1)\pi }{9})\sin (\frac{(j-1)\pi }{90})}{t_{j}U^{(j)}_{89}-2U^{(j)}_{88}-2}\right) \gamma ^{(j)}_{50,x}, \end{aligned}$$
(34)

where

$$\begin{aligned} t_{j}&= 4 - 2\cos \frac{(j-1)\pi }{90}, \end{aligned}$$
(35)
$$\begin{aligned} \gamma ^{(j)}_{50,x}&= U^{(j)}_{89-|50-x|}+U^{(j)}_{|50-x|-1}, \end{aligned}$$
(36)
$$\begin{aligned} U^{(j)}_{k}&= U^{(j)}_{k}(\cos \psi _{j}) = \frac{\sin (k+1)\psi _{j}}{\sin (\psi _{j})},\ \psi _{j}=\arccos (2 - \cos \frac{(j-1)\pi }{90}),\nonumber \\ k&= 89-|50-x|,~|50-x|-1,~89,~88,~~j=1,2,\ldots ,90. \end{aligned}$$
(37)

And a 3D dynamic view is shown in Fig. 4 by Matlab.

Figure 4
figure 4

The process of forming a 3D dynamic view for \(U_{90\times 90}(x,y)/J\) in Eq. (34).

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Figure 5
figure 5

The process of forming a 3D dynamic view for \(U_{90\times 90}(x,y)/J\) in Eq. (39).

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Special 2

Consider an arbitrary \(m \times n\) globe network as shown in Fig. (1). Assume the electric current J outflow the network from the pole \(d_{2}(x_{2}, y_{2})\), and the current J input at the node \(d_{1}(x_{1}, y_{1})=O(0, 0)\), the potential of an arbitrary node d(x, y) can be written as

$$\begin{aligned} \frac{U_{m\times n}(x,y)}{J} = -\frac{yy_{2}}{mn}r_{0} + \frac{2r}{m}\sum ^{m}_{j=2}\left( -\frac{\sin (\frac{y_{2}(j-1)\pi }{m})\sin (\frac{y(j-1)\pi }{m})}{t_{j}U^{(j)}_{n-1}-2U^{(j)}_{n-2}-2}\right) \gamma ^{(j)}_{x_{2},x}, \end{aligned}$$
(38)

where \(\gamma ^{(j)}_{x_{s},x_{k}}\) is defined in Eq. (6), and \(U^{(j)}_{k}\) is the same as Eq. (9).

In the network with \(m = 90\) and \(n = 90\), the current flows in from \((x_{1},y_{1})(x_{1} =0,y_{1} = 0)\) and out from \((x_{2},y_{2})(x_{2} =30,y_{2} = 30)\). Let \(r_{0} = r = 1\) and \(J = 10\), the formula is deduced.

$$\begin{aligned} \frac{U_{90\times 90}(x,y)}{J} = -\frac{y}{270} + \frac{1}{45}\sum ^{90}_{j=2}\left( \frac{-\sin (\frac{(j-1)\pi }{3})\sin (\frac{(j-1)\pi }{90})}{t_{j}U^{(j)}_{89}-2U^{(j)}_{88}-2}\right) \gamma ^{(j)}_{30,x}, \end{aligned}$$
(39)

where

$$\begin{aligned} \gamma ^{(j)}_{30,x} = U^{(j)}_{89-|30-x|}+U^{(j)}_{|30-x|-1}, \end{aligned}$$
(40)

\(t_{j}\) and \(U^{(j)}_{k}\) are the same as Eqs. (35) and (37), respectively, \(k=89-|30-x|,~|30-x|-1,~89,~88,~~j=1,2,\ldots ,90.\) And a 3D dynamic view is shown in Fig. 5 by Matlab.

Special 3

Presume that when the current J is injected at node \(d_{1}(x_{1}, y_{1})\) and withdrawn at node \(d_{2}(x_{2}, y_{2})(x_{2}=y_{1},y_{2}=x_{1})\), the potential of any node d(x, y) is

$$\begin{aligned} \frac{U_{m\times n}(x,y)}{J} = \frac{yy_{1}}{mn}r_{0} + \frac{2r}{m}\sum ^{m}_{j=2}\frac{(U^{(j)}_{n-x}+U^{(j)}_{x})\sin (\frac{y_{1}(j-1)\pi }{m})\sin (\frac{y(j-1)\pi }{m})}{t_{j}U^{(j)}_{n-1}-2U^{(j)}_{n-2}-2}, \end{aligned}$$
(41)

where \(t_{j}\) is defined in Eq. (8), and \(U^{(j)}_{k}\) is the same as Eq. (9).

In the scale of \(m \times n (90\times 90)\), we get \(J = 10,\ x_{1} = y_{2}= 0,\ x_{2}=y_{1} = 30, \ and \ r_{0}=r=1,\) the following formula is

$$\begin{aligned} \frac{U_{90\times 90}(x,y)}{J} = \frac{y}{270} + \frac{1}{45}\sum ^{90}_{j=2}\frac{(U^{(j)}_{90-x}+U^{(j)}_{x})\sin (\frac{(j-1)\pi }{3})\sin (\frac{y(j-1)\pi }{90})}{t_{j}U^{(j)}_{89}-2U^{(j)}_{88}-2}, \end{aligned}$$
(42)

where \(t_{j}\) and \(U^{(j)}_{k}\) are the same as Eq. (35) and Eq. (37), respectively, and \(k=90-x,~x,~89,~88,~~j=1,2,\ldots ,90.\)

And a 3D dynamic view is shown in Fig. 6 by Matlab.

Figure 6
figure 6

The process of forming a 3D dynamic view for \(U_{90\times 90}(x,y)/J\) in Eq. (42).

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Figure 7
figure 7

The process of forming a 3D dynamic view for \(U_{90\times 90}(x,y)/J\) in Eq. (44).

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Special 4

When we inject current J at node \(d_{1}(x_{1}, y_{1})\) and exit the current J at node \(d_{2}(x_{1}, y_{2})(x_{2} = x_{1})\), the potential of an arbitrary node d(x, y) is

$$\begin{aligned} \frac{U_{m\times n}(x,y)}{J} = \frac{y(y_{1}-y_{2})}{mn}r_{0} + \frac{2r}{m}\sum ^{m}_{j=2}\left( \frac{(C_{y_{1},j} - C_{y_{2},j})C_{y,j}}{t_{j}U^{(j)}_{n-1}-2U^{(j)}_{n-2}-2}\right) \gamma ^{(j)}_{x_{1},x}, \end{aligned}$$
(43)

where \(C_{k,j}\), \(U^{(j)}_{k}\) and \(\gamma ^{(j)}_{x_{1},x}\) are equal to Eqs. (7), (9) and Eq. (6), respectively.

When \(m = n = 90,\ J = 10,\ x_{1} = x_{2} = y_{1} = 50,\ y_{2} = 30,\ and \ r_{0}=r=1,\) the following formula is obtained

$$\begin{aligned} \frac{U_{90\times 90}(x,y)}{J} = \frac{y}{405} + \frac{1}{45}\sum ^{90}_{j=2}\left( \frac{(C_{50,j} - C_{30,j})C_{y,j}}{t_{j}U^{(j)}_{89}-2U^{(j)}_{88}-2}\right) \gamma ^{(j)}_{50,x}, \end{aligned}$$
(44)

where

$$\begin{aligned} C_{50,j}= & \sin (\frac{5(j-1)\pi }{9}),\nonumber \\ C_{30,j}= & \sin (\frac{(j-1)\pi }{3}), \end{aligned}$$
(45)

\(t_{j}\), \(U^{(j)}_{k}\) and \(\gamma ^{(j)}_{50,x}\) are the same as Eq. (35), Eq. (37)(\(k=~89,~88,~~j=1,2,\ldots ,90\)) and Eq. (36), respectively.

And a 3D dynamic view is shown in Fig. 7 by Matlab.

Special 5

When the current J enters from node \(d_{1}(x_{1}, y_{1})\) to node \(d_{2}(x_{2}, y_{1})\), in other words, \(y_{1} = y_{2}\). From Eq. (5), the potential of any node d(x, y) can be written as

$$\begin{aligned} \frac{U_{m\times n}(x,y)}{J} = \frac{2r}{m}\sum ^{m}_{j=2}\left( \frac{\gamma ^{(j)}_{x_{1},x} - \gamma ^{(j)}_{x_{2},x}}{t_{j}U^{(j)}_{n-1}-2U^{(j)}_{n-2}-2}\right) C_{y_{1},j}C_{y,j}, \end{aligned}$$
(46)

where \(\gamma ^{(j)}_{x_{s},x_{k}}\), \(U^{(j)}_{k}\) and \(C_{k,j}\) are equal to Eq. (6), Eq. (9) and Eq. (7), respectively.

When \(m = n = 90,\ J = 10,\ y_{1} = y_{2} = x_{1} = 50,\ x_{2} = 30,\ and \ r_{0}=r=1,\) the following formula is obtained

$$\begin{aligned} \frac{U_{90\times 90}(x,y)}{J} = \frac{1}{45}\sum ^{90}_{j=2}\left( \frac{\gamma ^{(j)}_{50,x} - \gamma ^{(j)}_{30,x}}{t_{j}U^{(j)}_{89}-2U^{(j)}_{88}-2}\right) C_{50,j}C_{y,j}, \end{aligned}$$
(47)

where \(\gamma ^{(j)}_{50,x}\), \(\gamma ^{(j)}_{30,x}\), \(C_{50,j}\), \(t_{j}\) and \(U^{(j)}_{k}\) are the same as Eq. (36), Eq. (40), Eq. (45), Eq. (35) and Eq. (37), respectively, \(k=89-|50-x|,~|50-x|-1,89-|30-x|,~|30-x|-1,~89,~88,~~j=1,2,\ldots ,90.\) And a 3D dynamic view is shown in Fig. 8 by Matlab.

Figure 8
figure 8

The process of forming a 3D dynamic view for \(U_{90\times 90}(x,y)/J\) in Eq. (47).

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Figure 9
figure 9

The process of forming a 3D dynamic view for \(U_{90\times 90}(x,y)/J\) in Eq. (49).

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Special 6

Presuming that \(d_{k}(x_{k}, y_{1}) (k = 1, 2, \ldots k)\) entering the node at the same latitude is J/k and the current flowing out from O(0, 0) is J, the potential equation is obtained.

$$\begin{aligned} \frac{U_{m\times n}(x,y)}{J} = r_{0}\frac{yy_{1}}{mn} + \frac{2r}{km}\sum ^{m}_{j=2}\frac{C_{y_{1},j}C_{y,j}}{t_{j}U^{(j)}_{n-1}-2U^{(j)}_{n-2}-2}\sum ^{k}_{s=1}\gamma ^{(j)}_{x_{s},x}, \end{aligned}$$
(48)

where \(C_{k,j}\), \(\gamma ^{(j)}_{x_{s},x_{k}}\) and \(U^{(j)}_{k}\) are equal to Eq. (7), Eq. (6) and Eq. (9), respectively.

When \(m = n = 90,\ J = 10,\ y_{1} = y_{2} = x_{1} = 50,\ x_{2} = 30,\ x_{3} = 40,\) \(r_{0}=r=1,\ and \ k=3\) the following formula is obtained

$$\begin{aligned} & \quad \frac{U_{90\times 90}(x,y)}{J} = \frac{y}{162}+\frac{1}{45}\sum ^{90}_{j=2}\frac{C_{50,j}C_{y,j}}{t_{j}U^{(j)}_{89}-2U^{(j)}_{88}-2}(\gamma ^{(j)}_{50,x}+\gamma ^{(j)}_{40,x}+\gamma ^{(j)}_{30,x}),\nonumber \\ & \quad \gamma ^{(j)}_{40,x} = U^{(j)}_{89-|40-x|}+U^{(j)}_{|40-x|-1}, \end{aligned}$$
(49)

where \(C_{50,j}\), \(t_{j}\), \(\gamma ^{(j)}_{50,x}\), \(\gamma ^{(j)}_{30,x}\) and \(U^{(j)}_{k}\) is defined in Eq. (45), Eq. (35), Eq. (36), Eq. (40) and Eq. (37)\((k=89-|50-x|,~|50-x|-1,89-|40-x|,~|40-x|-1,89-|30-x|,~|30-x|-1,~89,~88,~~j=1,2,\ldots ,90)\), respectively.

And a 3D dynamic view is shown in Fig. 9 by Matlab.

Fast numerical algorithm for computing potential

In order to realize fast calculation of potential for large-scale resistor networks. In this section, by summarizing the previous discussion and analysis, a fast numerical algorithm of computing potential by the Eqs. (8), (9), (10), (26), (27), (28), (29), (30), (31), (32) and the famous second type of discrete cosine transform (DCT-II) are given.

figure a
figure b

As is well known, the complexity of tridiagonal matrix-vector multiplication is O(n), which is the same as Algorithm 1. Moreover, one DCT-II needs \(2n\log _2 n+ O(n)\) real arithmetic operations57,58. So the complexity of Algorithm 2 is \(4n\log _2 n+ O(n)\) consist of two DCT-II and Algorithm 1.

According to the above two algorithms, two examples are used to vividly show the computational efficiency for the large scale globe resistor networks.

Example 1

In the network with \(m = 1000\) and \(n = 10\), the current flows in from \((x_{1},y_{1})(x_{1} =3,y_{1} = 200)\) and out from \((x_{2},y_{2})(x_{2} =5,y_{2} = 300)\), \(r = 1,~~r_{0} = 100\), and \(J = 10\). The results calculated using Algorithm 2 are shown in Fig. 10.

Example 2

In the scale of \(m \times n (300\times 10)\), when the current \(x_{1} =3,\ x_{2} = 5,\ y_{1} =100,\ y_{2} = 200\), and \(r=1, r_{0}=100\). The results calculated using Algorithm 2 are shown in Fig. 11.

Figure 10
figure 10

A 3D graph for the changes of \(U_{1000\times 10}(x,y)/J\) with x and y.

Full size image
Figure 11
figure 11

A 3D graph for the changes of \(U_{300\times 10}(x,y)/J\) with x and y.

Full size image

Conclusions

This paper achieved a series of improved exact potential formulae in an \(m \times n\) globe network by the RT-I method. Chebyshev polynomial of the second kind is introduced to improve the potential formula of the globe network44. Some applications of the new potential formula of the globe network are presented, such as some special and interesting potential formulae are given in Eqs. (33), (38), (41), (43), (46) and (48), respectively. The image numerical simulation using matlab has produced many interesting 3D dynamic views. Finally, we also put forward a fast numerical algorithm by the famous second type of discrete cosine transform, which can realize fast calculation of potential for large-scale resistor networks. Furthermore, by using our research ideas of resistor network, we can also explore neural networks11,12,13,14,15,16,17. That will be our next step.