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Consider a Riccati differential equation $$ \dot P + A(t)^{T}P + PA(t) -PB(t)R(t)B(t)^{T}P + Q(t) = 0,\;\;\; P(t_0) = P_0 = P_0^{T} \geqslant 0 $$ where $Q(t) = Q(t)^{T} \geqslant 0$, $R(t) = R(t)^{T} > 0$, all matrices are real and continuous. How to show that for any $t_0$ there exists (and unique) a solution defined on $(-\infty,t_0]$?

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The uniqueness of solution, when $P$ remains continuous, as well as the existence of a local solution, follows from Picard-Lindelof. Hence it remains to establish that $P$ doesn't blow-up in finite time. It is clear from the form of the evolution equation that $P$ will remain symmetric. Hence it suffices to show that $P$ has eigenvalues bounded above and below. A sketch of the idea of the proof:

  1. $P \geq 0$ in $(-\infty,t_0]$. This follows from the sign on $Q$. Let $\tau\in (-\infty,t_0]$ be the smallest time such that $P$ is positive semidefinite on $[\tau,t_0]$. Necessarily, there exists some vector $v\in \ker P(\tau)$ and some $\epsilon > 0$ such that $v^TP(t)v < 0$ for all $t\in (\tau-\epsilon,\tau)$. However, for $v\in \ker P$, we have that $$ v^T\dot{P}v + (Av)^TPv + v^TPAv - v^TPBRB^TPv + v^TQv = v^T\dot{P}v + v^TQv = 0 $$ Using that $Q$ is positive semidefinite, we see that it is impossible for increase to 0 from somewhere negative.

  2. $P$ cannot blow-up in finite backwards time. Since $PBRB^TP$ is by definition a positive semi-definite form, we have that $$ -\dot{P} \leq A^TP + PA + Q $$ which implies that in backwards time, the growth of $P$ cannot exceed that of the linear equation $-\dot{P} = A^TP + PA + Q$. Since the latter cannot blow-up in finite time, the same is true for the original equation.


Let me work out the scalar case in more detail (mainly for convenience of notation, since integrating factors are a bit easier to write when the products commute). From this you can see how to write it up for the matrix case.

The analogous scalar equation is $$ \dot{p} + 2ap - rp^2 + q = 0 $$ where $q,r$ are non-negative functions, and $p$ is given initial data $p(t_0) \geq 0$. Let $\alpha(t) = \int_{t_0}^t 2a(s)\mathrm{d}s$. And write $\tilde{p} = e^{\alpha}p$. We have that $$ \dot{\tilde{p}} = e^\alpha\left(\dot{p} + 2a p\right) = e^\alpha (rp^2 - q) $$ So that $$ \dot{\tilde{p}} = e^{-\alpha}r \tilde{p}^2 - e^\alpha q $$

Now since $e^\alpha$ is positive, we have that $\tilde{r} = e^{-\alpha}r$ is still non-negative, as is $\tilde{q} = e^\alpha q$.

For the equation $$ \dot{\tilde{p}} = \tilde{r} \tilde{p}^2 - \tilde{q} $$ upper boundedness in backwards time is immediate. Since we have that $$ \dot{\tilde{p}} \geq -\tilde{q} \implies \tilde{p}(t) - \tilde{p}(t_0) \leq \int_{t}^{t_0} \tilde{q}(s) \mathrm{d}s$$

For lower boundedness we see that on an interval $[t_1,t_2]$ on which $p$ is non-vanishing, we must have that $$ \dot{\tilde{p}} \leq \tilde{r} \tilde{p}^2 \implies \frac{1}{\tilde{p}(t_1)} - \frac{1}{\tilde{p}(t_2)} \leq \int_{t_1}^{t_2} \tilde{r}(s)\mathrm{d}s $$ hence if $\tilde{p}(t_1) < 0$ at some point, $\tilde{p}$ cannot approach 0 in finite time to the future. The contrapositive implies that since $\tilde{p}(t_0) \geq 0$, for every $\tau < t_0$ we cannot have $\tilde{p}(\tau) < 0$, establishing the lower-bound. This last argument, however, is not so easy to translate to the matrix case. So instead, let me give another one which is easier to use in the matrix case.

As before, suppose $\tilde{p} = 0$ at some point $\tau < t_0$ and $\tilde{p} < 0$ in $[\tau - \epsilon,\tau)$ (the existence of the solution, and its continuity on this interval, we assume by appealing to the local existence part of Picard-Lindelof). We can further assume that $0\leq \int_{\tau-\epsilon}^\tau \tilde{r}(\sigma) \tilde{p}(\sigma)\mathrm{d}\sigma < \frac12$, by choosing $\epsilon$ sufficiently small. We have the expression $$ \dot{\tilde{p}} \leq \tilde{r} \tilde{p}^2 $$ Now let $\tau' \in [\tau-\epsilon,\tau)$ be a point where $\tilde{p}$ achieves its minimum (in that interval). Integrate from $\tau'$ to $\tau$ we get $$ - \tilde{p}(\tau') = |\tilde{p}(\tau')| \leq \int_{\tau'}^\tau \tilde{r}|\tilde{p}|^2 \mathrm{d}\sigma $$ Using the integral bound we assumed we have $$ |\tilde{p}(\tau')| \leq \frac12 \sup_{\sigma\in[\tau',\tau]} |\tilde{p}(\sigma)| = \frac12 |\tilde{p}(\tau')|$$ which can only occur if $\tilde{p}(\tau') = 0$. This shows that it is impossible to have $\tilde{p}(\tau') < 0$ and $\tilde{p}(\tau) = 0$ with $\tau' < \tau$.

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Thank you for a great answer! –  Nimza Jun 11 '12 at 12:43
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