We will show by means of an example that, stated this way, this assertion is in general not correct. Therefore, we reconsider in this paper first the conditions under which the finite-planning- horizon linear-quadratic differential game has an open-loop Nash equilibrium solution. We present both necessary and sufficient conditions for the existence of a unique solution in terms of an invertibility condition on a matrix. Moreover, we present sufficient conditions, which can be computationally easily verified, that guarantee the existence of the unique solution. Next, we consider the infinite-planning-horizon case. We show that even in case the corresponding game has for an arbitrary finite planning horizon a unique solution, a solution for the infinite-planning- horizon game may either fail to exist or it may have multiple equilibria. A sufficient condition for existence of a unique equilibrium is presented. Moreover, we show how the(se) solution(s) can be easily calculated from the eigenstructure of a related matrix. Furthermore, we present a sufficient condition under which the finite open-loop Nash equilibrium solution converges to a solution of the infinite planning horizon case. In particular we show that if in the above cost criteria, a discount factor is included that is large enough in a certain sense, this convergence will always take place. The scalar case is studied in a worked example. We show for this case that solvability of the associated Riccati equations is both necessary and sufficient for existence of a unique solution for the finite- planning-horizon case, and that the infinite-planning-horizon has a unique solution.
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