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We know that

Lyapunov equation: $A^TP + PA + Q = 0$
Algebraic Riccati equation: $A^TP + PA + Q + PBR^{-1}B^TP= 0$

It seems that the difference between the two lies in $B = 0$ (zero input) in Lyapunov Eq and both are infinite horizon in the case above.

Is there any other engineering-sense difference (not mathematics) between the two Eqs?

Thanks!

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    $\begingroup$ Riccati equation is nonlinear (if $B\ne 0$), Lyapunov equation is linear. $\endgroup$
    – daw
    Commented Jul 7, 2014 at 9:02

4 Answers 4

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Lyapunov equation is used for the stability analysis of a relaxed system, I.e. no input signal. There exists a unique positive definite $P$ for any given positive definite $Q$ if and only if the system $\dot{x}=Ax$ is globally asymptotically stable. This means we can make $\dot{V}$ arbitrarily small where $V$ is the quadratic Lyapunov function $V(x)=x^T Px$.

On the other hand, Riccati equation appears in LQR and LQE problems. The first one is used to find the optimal regulator that minimizes quadratic cost function. The second one is used to estimate the state variables from the output when there is measurement noise. First problem deals with input and second deals with output. This is why the extra term appears.

These two problems are dual and convertible to each other. Together these problems makes LQG problem.

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The Lyapunov equation is used to show stability. Solution to the Riccati shows more: optimality. Roughly speaking, if you find an optimal controller it is stabilizing; and if you solve the Riccati you can find a solution to a related Lyapunov.

From the engineering point of view it is thus not surprising that the Riccati is more difficult to solve.

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Replace $A$ by $A-BK$ with $K=R^{-1}B^\top P$ in the Lyapunov equation yields the algebraic Riccati equation. Hope this helps.

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As mentioned in @bersou's answer, here is how to go from the continuous-time Algebraic Riccati equation $$A^T P + P A - P B R^{-1} B^T P + Q = 0$$ to the continuous-time Lyapunov equation $$(A - BK)^T P + P(A - BK) + Q = 0$$ where $A - BK$ is stable (all of its eigenvalues have negative real part): \begin{align} A^T P + P A - P B R^{-1} B^T P + Q &= 0 \\ A^T P + P A - \frac{1}{2}P B R^{-1} B^T P - \frac{1}{2}P B R^{-1} B^T P + Q &= 0 \\ A^T P - \frac{1}{2}P B R^{-1} B^T P + PA - \frac{1}{2}P B R^{-1} B^T P + Q &= 0 \\ \left(A^T - \frac{1}{2}PBR^{-1}B^T\right)P + P\left(A - \frac{1}{2} B R^{-1} B^T P\right) + Q &= 0 \end{align} Let $K = \frac{1}{2}R^{-1}B^TP$. Then, \begin{align} \left(A^T - \frac{1}{2}PBR^{-1}B^T\right)P + P\left(A - \frac{1}{2} B R^{-1} B^T P\right) + Q &= 0 \\ \left(A^T - K^TB^T\right)P + P\left(A - B K\right) + Q &= 0 \\ \left(A - BK\right)^TP + P\left(A - B K\right) + Q &= 0 \end{align}

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