Solve the reccati differential equation.

dy/dx=1-(1/x)y+(1/x²)y². Take a particular solution of this differential equation and use it to find its general solution..

# Tag: Riccati

## 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 $mge 3$

## Implicit Riccati equation solver (to find the anti-stabilizing solution)

I have found that the latest Matlab iCARE solver offers the ability to also find the anti-stabilizing solution to the Riccati equation.

I haven't found a similar option on Mathematica. Is there something undocumented or an alternative way?

## 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.

## Proof of the general solution of a Riccati equation

To let $ y_1 $ and $ y_2 $ two special solutions of a Riccati equation be in the form:

$ y = = p (x) y ^ 2 + q (x) y + r (x) $

Prove that every other solution has the form:

$ frac {y-y_1} {y-y_2} = Ce ^ { int (p (x)[y_1(x) – y_2(x)]dx} $