Is a Lyapunov equation always solvable when matrix $A$ is negative definite?

Question

Given a negative definite matrix $A$, is the following Lyapunov equation in $P$

$$P A + A^T P = -I$$

always solvable? What kind of form does the solution have?

I would appreciate if examples could be given.

Answer 1

$A$ needs to be Hurwitz (i.e. All real parts of the eigenvalues must be negative) in general, not necessarily negative definite. Then the solution is

$\displaystyle P = \int_0^\infty e^{A^T \tau} Q e^{A \tau} d \tau$

See that

$ \begin{align} \displaystyle A^T P + PA &= \int_0^\infty \left[ A^T e^{A^T \tau} Q e^{A \tau} + e^{A^T \tau} Q e^{A \tau} A \right] d \tau \\ &= \int_0^\infty \left( e^{A^T \tau} Q e^{A \tau} \right)' d \tau \\ &= e^{A^T \tau} Q e^{A \tau} \vert_0^\infty \\ &= -Q \end{align}$

where prime notation denotes derivative with respect to $\tau$. Note that this integral only converges when $A$ is Hurwitz.

The inverse is also true. Hence, Lyapunov equation has a unique positive definite solution $P$ for any given positive definite $Q$, if and only if $A$ is Hurwitz.