A simulation model of the Triple Helix of university–industry–government relations and the decomposition of the redundancy

Abstract

A Triple Helix (TH) network of bi- and trilateral relations among universities, industries, and governments can be considered as an ecosystem in which uncertainty can be reduced when functions become synergetic. The functions are based on correlations among distributions of relations, and therefore latent. The correlations span a vector space in which two vectors (P and Q) can be used to represent forward “sending” and reflexive “receiving,” respectively. These two vectors can also be understood in terms of the generation versus reduction of uncertainty in the communication field that results from interactions among the three bi-lateral channels of communication. We specify a system of Lotka–Volterra equations between the vectors that can be solved. Redundancy generation can then be simulated and the results can be decomposed in terms of the TH components. Furthermore, we show that the strength and frequency of the relations are independent parameters in the model. Redundancy generation in TH arrangements can be decomposed using Fourier analysis of the time-series of empirical studies. As an example, the case of co-authorship relations in Japan is re-analyzed. The model allows us to interpret the sinusoidal functions of the Fourier analysis as representing redundancies.

Appendices

Appendix 1

Communication of information and meaning in a TH dynamic symmetry model

In a TH dynamic symmetry model, communication field W is described by the following equation:

$$ \partial^{j} W_{ij} = - gW^{j} \times W_{ij} $$

(A.1)

Here i, j = 0,1; $\partial_{0} = \partial_{t}$; $\partial_{1} = - \partial_{x}$; $\partial^{0} = \partial_{t}$; $\partial^{1} = \partial_{x}$; summing on repeating indexes j is implied. Eq. (A.1) can be rewritten in a more explicit form:

$$ \begin{aligned} \partial^{1} W_{01} & = - gW^{1} \times W_{01} \\ \partial^{0} W_{10} & = - gW^{0} \times W_{10} \\ \end{aligned} $$

(A.2)

W _ij is defined by the formula:

$$ W_{ij} = \partial_{i} W_{j} - \partial_{j} W_{i} + gW_{i} \times W_{j} $$

(A.3)

W _i , W _j are three-component vectors in TH internal symmetry space, and two component vectors in (x, t) space. We can define these as: $W_{i} = \left( {P,Q} \right)$; $W^{i} = \left( { - P,Q} \right)$; P, Q are three-component vectors in the case of university–industry–government (TH) relations. Eq. (A.3) can be expressed in components:

$$ \left\{ {\begin{array}{l} W_{01} = Q_{t} + P_{x} + g\left( {P \times Q} \right) \\ W_{10} = - Q_{t} - P_{x} - g\left( {P \times Q} \right) \\ \end{array} } \right. $$

(A.4)

and substituting this into (A.2) produces:

$$ \left\{ {\begin{array}{*{20}c} {P_{xx} + Q_{tx} = - g\left( {P_{x} \times Q} \right) - g\left( {P \times Q_{x} } \right) - gQ \times \left[ {P_{x} + Q_{t} + g\left( {P \times Q} \right)} \right]} \\ { - P_{xt} - Q_{tt} = g\left( {P_{t} \times Q} \right) + g\left( {P \times Q_{t} } \right) - gP \times \left[ {P_{x} + Q_{t} + g\left( {P \times Q} \right)} \right]} \\ \end{array} } \right. $$

(A.5)

Setting: $P = P\left( \xi \right)$; $Q = Q\left( \xi \right)$; $\xi = x - t$, so that $P_{x} = - P_{t} = P_{\xi }$; $Q_{x} = - Q_{t} = Q_{\xi }$, summing first and second equations in (A.5), and assuming that: $Q \times Q_{t} = gQ \times \left( {P \times Q} \right)$; $P \times P_{x} = gP \times \left( {P \times Q} \right)$, we can rewrite Eq. (A.5) in the form:

$$ 2P_{\xi \xi } - 2Q_{\xi \xi } = 2\left( {P \times Q} \right) \times \left( {g^{2} Q + g^{2} P} \right) - g\left( {P \times Q_{\xi } } \right) - g\left( {Q \times P_{\xi } } \right) $$

(A.6)

This can be interpreted as describing two fields: P, Q propagating in a positive direction of the x axes.

We can show that Eq. (A.6) can be coupled to the equation describing these two fields P, Q propagating in a negative direction of the x axes (or propagating in a positive direction but reversed in time). Setting:

$$ \left\{ {\begin{array}{*{20}l} {{P_{t}} = - 2\beta \left( {P \times Q} \right)} \\ {{Q_{t}} = + 2\delta \left( {Q \times P} \right)} \\ \end{array} } \right. $$

(A.7)

where P, Q are vectors: $P = \left( {\begin{array}{*{20}c} {P_{1} } \\ {P_{2} } \\ {P_{3} } \\ \end{array} } \right)$; $Q = \left( {\begin{array}{*{20}c} {Q_{1} } \\ {Q_{2} } \\ {Q_{3} } \\ \end{array} } \right)$, and the right sides of (A.7) are cross-products. Alongside the “time” version of Eq. (A.7) can be introduced the “space” version of this equation:

$$ \left\{ {\begin{array}{*{20}c} {2P_{x} = - \beta \left( {P \times Q} \right)} \\ {2Q_{x} = + \delta \left( {Q \times P} \right)} \\ \end{array} } \right. $$

(A.8)

Each pair of Eqs. (A.7) and (A.8) describes the evolution of functions P and Q in time and space. We can obtain a combined space–time equation. Differentiating the first and second equations of the system (A.8) by t and taking into consideration (A.7) produces:

$$ \left\{ {\begin{array}{*{20}c} {2P_{xt} = + 2\beta^{2} \left( {P \times Q} \right) \times Q - \beta \left( {P \times Q_{t} } \right)} \\ {2Q_{xt} = - 2\delta^{2} \left( {P \times Q} \right) \times P + \delta \left( {Q \times P_{t} } \right)} \\ \end{array} } \right. $$

(A.9)

Setting further: $P = P\left( \xi \right)$; $Q = Q\left( \xi \right)$; $\xi = x + t$, so that $P_{x} = P_{t} = P_{\xi }$; $Q_{x} = Q_{t} = Q_{\xi }$, and subtracting in (A.9) the second equation from the first, we obtain:

$$ 2P_{\xi \xi } - 2Q_{\xi \xi } = 2\left( {P \times Q} \right) \times \left( {\beta^{2} Q + \delta^{2} P} \right) - \beta \left( {P \times Q_{\xi } } \right) - \delta \left( {Q \times P_{\xi } } \right) $$

(A.10)

Equation (A.10) coincides in full with Eq. (A.6) if we set $\beta = \delta = g$, but refers to solutions which evolve reversed in time.

Appendix 2

Analytical solution for a TH communication field

Defining the functions

$$ \begin{aligned} f_{1} & = 2g\left( {P_{2} Q_{3} - P_{3} Q_{2} } \right) \\ f_{2} & = 2g\left( {P_{3} Q_{1} - P_{1} Q_{3} } \right) \\ f_{3} & = 2g\left( {P_{1} Q_{2} - P_{2} Q_{1} } \right) \\ \end{aligned} $$

(B.1)

We can write a system of equations for a TH communication field (coupling constant 2g in the right-hand side of systems (B.2), (B.3) are not mentioned)

$$ \begin{array}{*{20}c} {P_{1t} = - f_{1} } & {Q_{1t} = f_{1} } \\ {P_{2t} = - f_{2} } & {Q_{2t} = f_{2} } \\ {P_{3t} = - f_{3} } & {Q_{3t} = f_{3} } \\ \end{array} $$

(B.2)

from system (2) we obtain:

$$ \begin{array}{*{20}l} {P_{1t} + Q_{1t} = 0} \\ {P_{2t} + Q_{2t} = 0} \\ {P_{3t} + Q_{3t} = 0} \\ \end{array} \quad {\text{or}}\;{\text{otherwise}}:\quad \begin{array}{*{20}l} {P_{1} + Q_{1} = \alpha } \\ {P_{2} + Q_{2} = \beta } \\ {P_{3} + Q_{3} = \gamma } \\ \end{array} $$

(B.3)

Expressing Q in terms of P from system (B.3) and substituting the result into system (B.2), we get:

$$ \begin{aligned} P_{1t} & = cP_{2} - bP_{3} \\ P_{2t} & = aP_{3} - cP_{1} \\ P_{3t} & = bP_{1} - aP_{2} \\ \end{aligned} $$

(B.4)

Here: $a = - 2g\alpha$; $b = - 2g\beta$; $c = - 2g\gamma$. System (B.4) can be solved analytically and has a well-known solution (Kamke 1971)

$$ \begin{aligned} P_{1} & = aC_{0} + rC_{1} \cos \left( {rt} \right) + \left( {cC_{2} - bC_{3} } \right)\sin \left( {rt} \right) \\ P_{2} & = bC_{0} + rC_{2} \cos \left( {rt} \right) + \left( {aC_{3} - cC_{1} } \right)\sin \left( {rt} \right) \\ P_{3} & = cC_{0} + rC_{3} \cos \left( {rt} \right) + \left( {bC_{1} - aC_{2} } \right)\sin \left( {rt} \right) \\ \end{aligned} $$

(B.5)

Here: $r^{2} = a^{2} + b^{2} + c^{2}$; $aC_{1} + bC_{2} + cC_{3} = 0.$ From the equation:

$$ aP_{1t} + bP_{2t} + cP_{3t} = 0 $$

(B.6)

we get after integration:

$$ aP_{1} + bP_{2} + cP_{3} = C_{4} $$

(B.7)

and from the equation:

$$ P_{1} P_{1t} + P_{2} P_{2t} + P_{3} P_{3t} = 0 $$

(B.8)

we also get:

$$ P_{1}^{2} + P_{1}^{2} + P_{1}^{2} = C_{5} $$

(B.9)

This means that integral curves are situated simultaneously on the sphere surface. The same equation holds for $Q$:

$$ Q_{1}^{2} + Q_{1}^{2} + Q_{1}^{2} = C_{6} $$

(B.10)

From (B.5) we get:

$$ a^{2} C_{0} + b^{2} C_{0} + c^{2} C_{0} = aP_{1} \left( 0 \right) + bP_{2} \left( 0 \right) + cP_{3} \left( 0 \right) + r\left( {aC_{1} + bC_{2} + cC_{3} } \right) $$

So that:

$$ C_{0} = \left( {aP_{1} \left( 0 \right) + bP_{2} \left( 0 \right) + cP_{3} \left( 0 \right)} \right)/r^{2} $$

(B.11)

Constants C ₁, C ₂, C ₃ can be calculated from system (B.5):

$$ \begin{aligned} C_{1} & = \left( {P_{1} \left( 0 \right) - aC_{0} } \right)/r \\ C_{2} & = \left( {P_{2} \left( 0 \right) - bC_{0} } \right)/r \\ C_{3} & = \left( {P_{3} \left( 0 \right) - cC_{0} } \right)/r. \\ \end{aligned} $$

(B.12)