linear algebra – A matrix Riccati differential equation with constant coefficients? Is there a solution for this in closed form?

The following is a matrix Riccati differential equation with constant coefficient matrices.

$$Dfrac{partial{C(t)}}{partial{t}}S + frac{1}{n}C(t)QDC(t)S – EC(t)Q = 0$$ or
$$Ddot{C}(t)S + frac{1}{n}C(t)QDC(t)S – EC(t)Q = 0$$
given initial condition $C(0) = C_0$.

I stumbled upon this from some other problem and I don’t have any background in matrix differential equations and I’d like to know if there is any way to solve this equation. I read it can be reduced to an algebraic Riccati equation. Is there any closed form expression for solution of this equation? Or anything that is closest to solving this equation?

Matrix dimensions

$C(t)$———-> $(m+1)times n$

$S$————–>$ntimes 1$

$Q$————–>$ntimes(m+1)$

$D$————–>$(m+1)times(m+1)$ diagonal matrix. (it is also singular, as there is a diagonal entry that is 0).

$E$————–>$1times (m+1)$

If its useful to know, $n>>m$ and $mge 3$

Solve a two-dimensional problem of optimal control with the Riccati nonlinear equation

Consider the two-dimensional problem of optimal control of the LQR type
$$
min_u int_0 ^ infty (x ^ TQ x + u ^ TRu) , dt quad text {so that} quad begin {cases} Punkt x (t) = Ax (t) + Bu (t ) \ x (0) = begin {pmatrix} 1 \ – 1 end {pmatrix} end {case}
$$

With $ x = begin {pmatrix} x_1 \ x_2 end {pmatrix} $, $ u = begin {pmatrix} u_1 \ u_2 end {pmatrix} $, $ Q = B = I $ (Identity), $ R = gamma I $, $ A = begin {pmatrix} – alpha & alpha \ beta & – beta end {pmatrix} $, $ alpha> beta> 0 $ and $ gamma> 0 $.

Solve it with the stationary Riccati equation
$$
0 = Q + A ^ TS + SA-SBR ^ {- 1} BS, quad text {with} quad S = begin {pmatrix} s_1 & s_2 \ s_2 & s_3 end {pmatrix}.
$$

To find $ x $ and $ u $ we have to combine the equation for $ dot x $ with the equation for optimal control $ u = -R ^ {- 1} B ^ TSx = – frac1 gamma Sx $. So we have to find the expression for first $ S $.

The Riccati equation is simplified as
$$
0 = I + A ^ TS + SA- frac1 gamma SS
$$

This corresponds to the following non-linear system
$$
begin {cases}
– dfrac {s_1 ^ 2} { gamma} -2 alpha s_1 – dfrac {s_2 ^ 2} { gamma} +2 beta s_2 + 1 = 0 \
alpha s_1 – alpha s_2 – beta s_2 + beta s_3 – dfrac {s_1s_2} { gamma} – dfrac {s_2s_3} { gamma} = 0 \
– dfrac {s_2 ^ 2} { gamma} +2 alpha s_2 – dfrac {s_3 ^ 2} { gamma} – 2 beta s_3 + 1 = 0
end {cases}
$$

How to solve such a nonlinear system to find expressions for $ s_1, s_2 $ and $ s_3 $? I think there is a quick way to either solve the system or directly solve the Riccati equation in matrix form, but I don't know how.

The solutions provided by Matlab using the following code are so long that I think this is not the right way to solve the problem

syms x y z a b g
eqn1 = 0 == -x^2/g-2*a*x-y^2/g+2*b*y+1;
eqn2 = 0 == a*x-a*y-b*y+b*z-x*y/g-y*z/g;
eqn3 = 0 == -y^2/g+2*a*y-z^2/g-2*b*z+1;
(x,y,z) = solve((eqn1, eqn2, eqn3), (x, y, z))

Evidence Supplement – The Need for a Condition for the Riccati Comparison Argument

I want to understand why the need of $ lim_ {t to 0} f (t) = + infty $ in the following result: (Corollary 1.2 of https://www.math.upenn.edu/~wziller/math660/TopogonovTheorem-Myer.pdf)

To let $ k $ be a constant. If $ g: (0, a) to mathbb {R} $ (to take $ a leq pi / sqrt {k} $ if $ k> 0 $)
is such a feature that $ g & # 39; + g ^ 2 leq -k $ and
$ lim_ {r to 0 ^ +} g (r) = + infty $, then $$ g (r) leq ct_k (r) quad text {on
} , , (0, a), $$

Where $ ct_k (r) = frac {sn & # 39; k (r)} {sn_k (r)} $ and $ sn_k (r) $
is the solution of the Jacobi equation $ f & # 39; & gt; + kf = 0 $. $ f (0) = 0, f & # 39; (0) = 1 $,

The proof is as follows: Suppose it exists $ r_0 in (0, a) $ so that $ g (r_0)> ct_k (r_0) $, then for some $ epsilon> 0 $ we have $ g (r_0) geq ct_k (r_0- epsilon) $, To take $ G (r) = ct_k (r- epsilon) $it satisfies $ G & # 39; + G ^ 2 = k $ on $ ( epsilon, r_0) $, Then with the fact that $ left ((G-g) e ^ { int (G + g)} right) & # 39; sq 0 $ and $ g (r_0) geq G (r_0) $ we understand that $ g (r) geq G (r) $ on $ ( epsilon, r_0) $, Finally take $ r to epsilon ^ + $ we get a contradiction ($ g ( epsilon) geq + infty $).

Classical analysis and Odes – Algebraic Riccati and WKB

It is a one-liner to show that the algebraic Riccati equation (ARE) and the lowest-order form of WKB are the same for a linear ode. But I've looked everywhere on the Internet and it seems that this connection, despite its triviality is not recognized. It seems that there is a total separation between physicists who make many WKB and control people who do a lot of Riccati.

Another trivial point is that a linear ode in psi has the Lie symmetry psi -> lambda psi, so the Riccati variable is the Lie invariant, and this leads to a reduction in order (or more precisely, a reduced one Order) followed by a quadrature.) Is this so obvious that it can never be mentioned? This finding means that not only linear equations are used, but every equation that is homogeneous in degree 1 and whose order is reduced by the Riccati transformation.