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Greetings to members here. The question is how to calculate the solution $S(k)$ of the following recursive equation $$J(k)S(k+1)J^{T}(k)=A(k)S(k)A^{T}(k)+R(k)$$ where $J$ and $A$ are rectangular not square. $R$ is positive-definite. Furthermore, $J$ and $A$ are with full-row rank.

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  • $\begingroup$ What is the source of this problem? Looks a lot like a discrete-time difference Riccati equation. This, and what Robert Israel said. $\endgroup$
    – Federico Poloni
    Commented Apr 21, 2013 at 11:13
  • $\begingroup$ This is a Lyapunov equation. We met with the problem for treating desciptor system with white noise. $$J(k)x(k+1)=A(k)x(k)+w(k)$$ $\endgroup$
    – eolithr
    Commented Apr 22, 2013 at 12:51

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If $J$ has more columns than rows, the map $S \to J S J^T$ is not one-to-one, so your equation does not determine $S(k+1)$.

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