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Economic growth and the transition from non-renewable to renewable energy

Published online by Cambridge University Press:  22 November 2013

Alfred Greiner
Affiliation:
Department of Business Administration and Economics, Bielefeld University, P.O. Box 100131, Bielefeld 33613, Germany. E-mail: agreiner@wiwi.uni-bielefeld.de
Lars Gruene
Affiliation:
Department of Mathematics, University of Bayreuth, Germany. E-mail: lars.gruene@uni-bayreuth.de
Willi Semmler
Affiliation:
Department of Economics, New School for Social Research and Department of Business Administration and Economics, Bielefeld University, Germany. E-mail: semmlerw@newschool.edu
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Abstract

The paper considers the transition of an economy from non-renewable to renewable energy. We set up a canonical growth model with damages in the household's welfare function and two energy sources – non-renewable and renewable energy. To produce renewable energy a capital stock must be built up. A socially optimal solution is considered that takes into account the negative externality from the non-renewable energy. We also study how the optimal solution can be mimicked in a market economy by policies using subsidies and tax rates. To solve the model numerically, we use Nonlinear Model Predictive Control. We study when a transition to renewable energy takes place and whether it occurs before the non-renewable resource is exhausted. In addition, we analyze the impact of the initial values of the non-renewable resource and of the capital stock on the time of paths of the variables.

Information

Type
Research Article
Copyright
Copyright © Cambridge University Press 2013 
Figure 0

Figure 1. NMPC results with R(0)=1, K(0)=1, M(0)=1, N=8

Figure 1

Figure 2. NMPC results for R(0)=0.5, K(0)=1, M(0)=1, N=8

Figure 2

Figure 3. NMPC results for R(0)=5, K(0)=1, M(0)=1, N=8

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Figure 4. NMPC results for R(0)=0.5, and R(0)=2.5, and in both cases K(0)=3, M(0)=1, N=8

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Figure 5. NMPC results for R(0)=5, K(0)=8, M(0)=1, N=20

Figure 5

Figure 6. NMPC results for R(0)=2.5, and R(0)=0.5, and in both cases K(0)=3, M(0)=1, Ap=1, N=8