Geometric interpretation of Q in Lyapunov's equation

Question

Lyapunov's equation says: given any $Q > 0$ ($Q$ positive definite) there is $P > 0$ such that $A^T P + P A + Q = 0$ if and only if for $\frac{dx(t)}{dt}=A x(t)$ it is the case that the real part of each eigenvalue of $A$ is negative. Then, the ellipsoid $x^T P x \leq 1$ is an invariant of $\frac{dx(t)}{dt}=A x(t)$.

The geometric interpretation of $P$ is such that the eigenvectors of $P$ form the principal axes of the ellipsoid and each eigenvalue is related to the length of the ellipsoid along the axis represented by the corresponding eigenvector.

What is the geometric interpretation of $Q$ resp. how does the choice of $Q$ affect $P$?

Answer 1

It turns out that $P$ is dependent on the properties $Q$. If you define $P$ as

$$P:=\int_{0}^{+\infty} e^{A^Tt}Qe^{At}dt$$

Since the eigenvalues of $A$ have negative real part, this integral exists and the results are clearly dependent on the eigenvalues and eigenvectors of $Q$. One can see that $P$ is a solution to $A^TP+PA=-Q$, that is,

$$A^TP+PA=\int_{0}^{+\infty} (A^Te^{A^Tt}Qe^{At}+e^{A^Tt}Qe^{At}A)dt =\int_{0}^{+\infty}\frac{d}{dt}(e^{A^Tt}Qe^{At})dt = -Q$$

Further, the matrix $Q$ can be understood as the energy dissipated in a specific point $x$, and $P$ as the energy stored in $x$.

Answer 2

Here's another point of view: Let's say you have an autonomous linear dynamical system $\dot x = Ax$ with a cost function $\int_0^\infty x^T Q x \; dt$. (This is like linear-quadratic control, but without the control!)

Now let's say we happen to have a cost-to-go function $V(x)$ such that $\dot V(x) + x^TQ x = 0$. Intuitively, this equation is simply accounting: if we accumulate a lot of cost at time $t$ (i.e., $x^TQx$ is large at time $t$), then our function is decreasing by the same amount, so that's an amount we won't have to pay in the future, so it gets subtracted from the cost-to-go function.

Mathematically, we can integrate from $0$ to $T$, we have $\int_0^T \dot V(x) dt = V(x(T)) - V(x(0)) = -\int_0^T x^T Q x \; dt$. So long as $V(x(T)) \to 0$ as $T \to \infty$, then we have $V(x(0)) = \int_0^\infty x^T Q x \; dt$. This means that $V(x)$ accurately represents the amount of cost left to pay when you start at $x$.

Returning to your question, we assume $V(x) = x^T P x$, for some symmetric matrix $P$. (Note that the condition $V(x) \to 0$ holds if the system is asymptotically stable, because then $x\to 0$.) Then our accounting equation is $\dot V(x) + x^T Q x = x^T (A^TP + PA + Q) x = 0$. (Here we used the chain rule and $\dot x = Ax$.) This must hold for all $x$, so we have $A^TP + PA + Q = 0$. This is your Lyapunov equation.

To summarize, if your stable system accumulates cost at rate $x^T Qx$, then $x^TPx$ is the cost-to-go (or Lyapunov) function.