Riccati differential equation of optimal control

Remarks

Under these hypotheses the solution exists on all of $(-\infty, T]$, and $P(t)$ is symmetric positive semidefinite for every $t$. Example: $n = m = 1$, $A = 0$, $B = Q = R = 1$ and $F = 0$ give $-P' = 1 - P^2$, $P(T) = 0$. The solution is $P(t) = \tanh(T - t)$: indeed $P'(t) = -(1 - \tanh^2(T - t)) = -(1 - P(t)^2)$ and $P(T) = 0$. As $t \to -\infty$, $P(t) \to 1$, the positive solution of the algebraic equation $1 - P^2 = 0$. Non-example: the linear equation $-P' = A^{T} P + P A + Q$ is a Lyapunov differential equation, not a Riccati equation, since it lacks the quadratic term. It gives the cost of the zero control; with $A - BK$ in place of $A$ and $Q + K^{T} R K$ in place of $Q$ it gives the cost of a fixed linear feedback $u = -Kx$. The quadratic term of the Riccati equation accounts for optimizing the control. For $n = 1$ the equation is a scalar Riccati equation $P' = q_0 + q_1 P + q_2 P^2$ with constant coefficients.