Riccati equations

Riccati equations DEFAULT

Preface

Jacopo Francesco Riccati (1676--1754) was an Venetian mathematician and jurist from Venice. He is best known for having studied the differential equation which bears his name:

\begin{equation} y' + p(x)\,y = g(x)\,y^2 + h(x) , \label{Eq.riccati.1} \end{equation}

where p, g, and hare some real-valued given functions. Riccati himself was concerned with solutions to so called special Riccati equation

\begin{equation} y' = a\,y^2 + x^{\alpha} . \label{Eq.riccati.2} \end{equation}

Riccati was educated first at the Jesuit school for the nobility in Brescia, and in 1693 he entered the University of Paduato study law. He received a doctorate in law in 1696. Encouraged by Stefano degli Angeli to pursue mathematics, he studied mathematical analysis. Riccati received various academic offers, but declined them in order to devote his full attention to the study of mathematical analysis on his own. Peter the Greatinvited him to Russiaas president of the St. Petersburg Academy of Sciences. He was also invited to Viennaas an imperial councilor and was offered a professorship at the University of Padua. He declined all these offers. He was often consulted by the Senate of Venice on the construction of canals and dikes along rivers.

When h(x) = 0, we get a Bernoulli equation. The Riccati equation has much in common with linear equations; for example, it has no singular solution. Except special cases, the Riccati equation cannot be solved analytically using elementary functions or quadratures, and the most common way to obtain its solution is to represent it in series. Moreover, the Riccati equation can be reduced to the second order linear differential equation by substitution

\begin{equation} y(x) = - \frac{u'}{g(x)\,u(x)} . \label{Eq.riccati.3} \end{equation}

Substitution of the above expression into the Riccati equation yields the second order linear differential equation for u(x):

\begin{equation} u'' + a(x)\, u' (x) + b(x)\, u =0, \qquad \mbox{where} \quad a(x) = p(x) - \frac{g' (x)}{g(x)} , \quad b(x) = g(x)\,h(x) . \label{Eq.riccati.4} \end{equation}

Conversely, every linear homogeneous differential equation with variable coefficients \( u'' + a\,u' + b\,u = 0 \) can be reduced to the Riccati equation

\[ y' + y^2 + a(x)\,y + b(x) =0 \]

by the substitution

\[ u(x) = \exp \left\{ y(x)\,{\text d} x \right\} . \]

It is sometimes possible to find a solution of a Riccati equation by guessing. Without knowing a solution to the Riccati equation, there is no chance of finding its general solution explicitly. If one solution ϕ is known, then substitution w = y - ϕ reduces the Riccati equation to a Bernoulli equation. Another substitution y = ϕ + 1/v also reduces the Riccati equation to a Bernoulli type.

Theorem(Liouville, 1841). The special Riccati equation \( y' = a\,y^2 + b\, x^{\alpha} \) can be integrated in closed form via elementary functions if and only if

\[ \frac{\alpha}{2\alpha +4} \qquad\mbox{is an integer}. \qquad ▣ \]

The special Riccati equation can be represented as \( y' = -u' /(au) , \) where

\begin{equation} u(x) = \sqrt{x}\,\begin{cases} C_1 J_{1/2q} \left( \frac{\sqrt{ab}}{q} \, x^q \right) + C_2 Y_{1/2q} \left( \frac{\sqrt{ab}}{q} \, x^q \right) , & \quad \mbox{if } ab> 0, \\ C_1 I_{1/2q} \left( \frac{\sqrt{-ab}}{q} \, x^q \right) + C_2 K_{1/2q} \left( \frac{\sqrt{-ab}}{q} \, x^q \right) , & \quad \mbox{if } ab< 0, \end{cases} \label{Eq.riccati.5} \end{equation}

where \( q= 1+ \alpha /2 \) and J(t), Y(t)are Bessel functions, , while I(t), K(t)are modified Bessel functions. Note that the general solution depends on the ratio \( C_1 / C_2 \) of two arbitrary constants.
We make transformation y= urvs, which we substitute into the special Riccati equation:

\[ r\,u^{r-1} u' \,v^s + s\,u^r v^{s-1} v' = a\,u^{2r} v^{2s} + b\, x^2 . \]

Choosing

\[ \begin{split} r\,u^{r-1} u' \,v^s &= b\,x^2 , \\ s\,u^r v^{s-1} v' &= a\,u^{2r} v^{2s} \qquad \Longleftrightarrow \qquad s\,v' = a\, u^r v^{s+1} . \end{split} \]

Differentiation of the latter, we obtain

\[ s\, v'' = ar\, u^{r-1} u' \, v^{s+1} + a \left( s+1 \right) u^r v^s v' . \]

The nonlinearity can now be removed by choosing s= -1. Then using the latter, we get

\[ s\, v'' = ab\, u^{r-1} u' \, v^{s+1} = ab\, x^2 v \qquad\mbox{or} \qquad v'' = -ab\, x^2 v . \]

Since the equation \( v'' + ab\, x^2 v = 0 \) has the general solution expressed through Bessel functions (see section)

\[ v(x) = \sqrt{x} \left[ C_1 J_{1/4} \left( \frac{\sqrt{ab}\,x^2}{2} \right) + C_2 Y_{1/4} \left( \frac{\sqrt{ab}\,x^2}{2} \right) \right] , \]

where C1and C2are arbitrary constants, we arrive at the required formula. Using Mathematica

v[x_] = Sqrt[ x]*(k*BesselJ[1/4, alpha*x^2 /2] + BesselY[1/4, alpha*x^2 /2])
y[x_] = Simplify[v'[x]/v[x]]
Assuming[alpha > 0 && x > 0, Series[%, {x, 0, 0}]]
Normal[%]
Simplify[%]

we obtain the general solution of the special Riccati equation:

\[ y(x) = \frac{1}{2x} \,\frac{\sqrt{ab}\,k x^2 J_{3/4} \left( \frac{\sqrt{ab}\, x^2}{2} \right) + k\, J_{1/4} \left( \frac{\sqrt{ab}\, x^2}{2} \right) - \sqrt{ab} \, k x^2 J_{5/4} \left( \frac{\sqrt{ab}\, x^2}{2} \right) + \sqrt{ab}\, x^2 Y_{-3/4} \left( \frac{\sqrt{ab}\, x^2}{2} \right) + Y_{1/4} \left( \frac{\sqrt{ab}\, x^2}{2} \right) - \sqrt{ab} \,x^2 Y_{5/4} \left( \frac{\sqrt{ab}\, x^2}{2} \right) }{k\, J_{1/4} \left( \frac{\sqrt{ab}\, x^2}{2} \right) + Y_{1/4} \left( \frac{\sqrt{ab}\, x^2}{2} \right)} , \]

where kis an arbitrary constant, and abis positive.    ▣

 

We can find the limit of the solution to the special Riccati equation when x→ 0:

\begin{align} \lim_{x\to 0} y(x) &= \frac{\sqrt{ab}}{8\,\Gamma^2 \left( \frac{1}{4} \right) \Gamma \left( \frac{5}{4} \right)} \left[ \sqrt{2}\, \Gamma \left( \frac{1}{4} \right) \left( \Gamma \left( -\frac{1}{4} \right) - 4\,\Gamma \left( \frac{3}{4} \right) \right) \Gamma \left( \frac{5}{4} \right) -2k\pi \left( \Gamma \left( \frac{1}{4} \right) + 4\, \Gamma \left( \frac{5}{4} \right)\right) \right] \notag \\ &= -\frac{\sqrt{ab}}{\pi} \,\Gamma^2 \left( \frac{3}{4} \right) \left( 1+ k\right) \approx -0.477989 \left( 1 + k \right) \sqrt{ab} . \label{Eq.riccati.6} \end{align}

Example 1: The Riccati equation

\[ y' = x^2 + y^2 \tag{1.2} \]

has the general solution \( y= -u' /u , \) where

\[ u(x) = \sqrt{x} \left[ C_1 J_{1/4} \left( x^2 /2 \right) + C_2 Y_{1/4} \left( x^2 /2 \right) \right] , \]

where C1and C2are arbitrary constants. The solution of the Riccati equation subject to the homogeneous initial condition y(0) = 0 is

\begin{equation} y(x) = x\,\frac{-Y_{-3/4} \left( \frac{x^2}{2} \right) + J_{-3/4} \left( \frac{x^2}{2} \right)}{Y_{1/4} \left( \frac{x^2}{2} \right) - J_{1/4} \left( \frac{x^2}{2} \right)} . \label{Eq.riccati.7} \end{equation}

This solution blows up at x≈ 2.00315 where the denominator is zero.

FindRoot[BesselY[1/4, x^2/2] - BesselJ[1/4, x^2/2] == 0, {x, 2}]

2.003147359426885

When you try to find the solution of the above initial value problem with Mathematicaby entering

DSolve[{y'[x] == x^2 + (y[x])^2, y[0] == 0}, y[x], x]

DSolve::bvnul: For some branches of the general solution, the given boundary conditions lead to an empty solution.

So Mathematicafails to provide you the solution mostly because the function in the denominator is zero. However, there is another way to go around it by changing the initial condition:

soln = DSolve[{y'[x] == y[x]^2 + x^2, y[0] == C}, y[x], x]
hold = Part[soln, 1, 1, 2]
dansoln = FullSimplify[ hold /. {C -> 0} ]
danseries = FullSimplify[Series[dansoln, {x, 0, 15}]]
dansoln /. {x-> 0.5}
Plot[dansoln, {x,-5,5}, PlotStyle->{Thick, Blue}]

0.0417911

Looking at the output, we find another solution of the given initial value problem \( y' = y^2 + x^2 , \quad y(0) =0 \) to be expressed through Bessel functionsof the first kind only:

\begin{equation} y(x) = x\,\frac{J_{3/4} \left( \frac{x^2}{2} \right)}{J_{-1/4} \left( \frac{x^2}{2} \right)} . \label{Eq.riccati.8} \end{equation}

To convince you that this is the same solution presented in another form, we do two things. First, we expand the above solution into Maclarin series and compare it with true expansion. Second, we verify that our function satisfies the given Riccati equation and the initial condition because the solution of such initial value problem is unique.

AsymptoticDSolveValue[{y'[x] == x^2 + (y[x])^2, y[0] == 0}, y[x], {x, 0, 16}]

\[ y(x) = \frac{1}{3}\, x^3 + \frac{1}{63}\, x^7 + \frac{2}{2079}\, x^{11} + \frac{13}{218295}\,x^{15} + O\left( x^{16} \right) \] First, we define two solution-functions in Mathematica:

f[x_] = x*BesselJ[3/4, x^2 /2]/BesselJ[-1/4, x^2 /2];
g[x_] = x*(BesselJ[-3/4, x^2 /2] - BesselY[-3/4, x^2 /2])/(BesselY[1/4, x^2 /2] - BesselJ[1/4, x^2 /2]);

We check their power series:

N[Series[f[x], {x, 0, 16}]]
N[Series[g[x], {x, 0, 16}]]

Then we check whether these two functions are solutions of the Riccati equation (1.2):

FullSimplify[D[f[x], x] - x^2 - (f[x])^2]
FullSimplify[D[g[x], x] - x^2 - (g[x])^2]

Finally, we check the initial conditions

f[0]
Limit[g[x], x -> 0]

In order to determine where the solution blows up, we find the zero of the denominator:

FindRoot[BesselY[1/4, x^2 /2] - BesselJ[1/4, x^2 /2], {x, 2.0}]

{x -> 2.003147359426885}

FindRoot[BesselJ[-1/4, x^2 /2], {x, 2.0}]

{x -> 2.003147359426885}

Then we evaluate some numerical values:

f[x_] = x*BesselJ[3/4, x^2 /2]/BesselJ[-1/4, x^2 /2];
g[x_] = x*(BesselJ[-3/4, x^2 /2] - BesselY[-3/4, x^2 /2])/(BesselY[1/4, x^2 /2] - BesselJ[1/4, x^2 /2]);
N[f[0.75]-g[0.75]]

-2.77556*10^-17

which is numerically zero.

The solution of the Riccati equation \( y' = y^2 + x^2 \) subject to the initial condition y(0) = 1 is

\[ y(x) = x\,\frac{\left( \pi - \Gamma^2 \left( \frac{3}{4} \right) \right) J_{-3/4} \left( \frac{x^2}{2} \right) - \Gamma^2 \left( \frac{3}{4} \right) Y_{-3/4} \left( \frac{x^2}{2} \right)}{\left( \pi - \Gamma^2 \left( \frac{3}{4} \right) \right) J_{1/4} \left( \frac{x^2}{2} \right) + \Gamma^2 \left( \frac{3}{4} \right) Y_{1/4} \left( \frac{x^2}{2} \right)} . \tag{1.2} \]

This solution blows up at x≈ 0.969811 where the denominator is zero.

DSolve[{u'[x] == x^2 + (u[x])^2 , u[0] == 1}, u, x]
FindRoot[BesselJ[1/4, x^2/2] Gamma[1/4] - 2 BesselJ[-(1/4), x^2/2] Gamma[3/4] == 0, {x, 1}]

{x -> 0.969811}

The solution of the initial value problem

\[ y' = x^2 + y^2 , \qquad y(0) = -1 , \]

is

\[ y(x) = x\,\frac{\left( \pi + \Gamma^2 \left( \frac{3}{4} \right) \right) J_{-3/4} \left( \frac{x^2}{2} \right) - \Gamma^2 \left( \frac{3}{4} \right) Y_{-3/4} \left( \frac{x^2}{2} \right)}{\left( \pi + \Gamma^2 \left( \frac{3}{4} \right) \right) J_{1/4} \left( \frac{x^2}{2} \right) + \Gamma^2 \left( \frac{3}{4} \right) Y_{1/4} \left( \frac{x^2}{2} \right)} . \tag{1.3} \]

This solution blows up at x≈ 4.1785 where the denominator is zero.

DSolve[{u'[x] == x^2 + (u[x])^2 , u[0] == -1}, u, x]
FindRoot[ BesselJ[1/4, x^2/2] Gamma[1/4] + 2 BesselJ[-(1/4), x^2/2] Gamma[3/4] == 0, {x, 4}]

{x -> 4.17851}

   ■

Example 2: We consider a problem similar to the previous example that involves another standard Riccati equation

\[ y' = x^2 - y^2 . \tag{2.1} \]

Its solution is expressed as \( y= u' /u , \) where

\[ u(x) = \sqrt{x} \left[ C_1 I_{1/4} \left( x^2 /2 \right) + C_2 K_{1/4} \left( x^2 /2 \right) \right] , \]

When we impose the homogeneous conditions y(0) = 0, then its solution becomes

\[ y(x) = x\,\frac{I_{-3/4} \left( \frac{x^2}{2} \right) \pi \sqrt{2} - 2\,K_{3/4} \left( \frac{x^2}{2} \right)}{\pi \sqrt{2}\,I_{1/4} \left( \frac{x^2}{2} \right) + 2\, K_{1/4} \left( \frac{x^2}{2} \right)} . \tag{2.2} \]

The initial value problem

\[ y' = x^2 - y^2 , \qquad y(0) =1 , \tag{2.3} \]

has the solution

\[ y(x) = x\,\frac{\left( \pi + \Gamma^2 \left( \frac{3}{4} \right) \sqrt{2} \right) I_{-3/4} \left( \frac{x^2}{2} \right) \pi - 2\, \Gamma^2 \left( \frac{3}{4} \right) K_{3/4} \left( \frac{x^2}{2} \right)}{\left( \pi + \Gamma^2 \left( \frac{3}{4} \right) \sqrt{2} \right) I_{1/4} \left( \frac{x^2}{2} \right) \pi + 2\,\Gamma^2 \left( \frac{3}{4} \right) K_{1/4} \left( \frac{x^2}{2} \right)} . \tag{2.4} \]

      We plot the separatrix (in red) for the Riccati equation \( y' = x^2 - y^2 \) that separates solutions approaching +∞ from solutions that go to -∞.

sp = StreamPlot[{1, x^2 - y^2}, {x, -2, 2}, {y, -2, 1}, StreamScale -> Medium, LabelStyle -> {FontSize -> 18, Black}];
curve = NDSolve[{y'[x] == x^2 - (y[x])^2, y[0] == -0.7134}, y, {x, -2, 2}];
cp = Plot[Evaluate[y[x] /. curve], {x, -2, 2}, PlotStyle -> {Thickness[0.015], Red}];
Show[sp, cp]

       Separatrix for the Riccati equation (2.1).            Mathematica code

If we consider a similar Riccati equation

\[ y' = y^2 - x^2 \tag{2.5} \]

its direction field is very similar. Th4e solution of Eq.(2.5) subject to the homogeneous initial condition is

\[ y = x\, \frac{I_{-3/4} \left( \frac{x^2}{2} \right) \pi\sqrt{2} - 2\, K_{3/4} \left( \frac{x^2}{2} \right)}{\pi\sqrt{2}\,I_{1/4} \left( \frac{x^2}{2} \right) + 2\, K_{1/4} \left( \frac{x^2}{2} \right)} , \qquad y(0) = 0. \tag{2.6} \]

However, for another initial condition, its solution becomes

\[ y = x\, \frac{I_{-3/4} \left( \frac{x^2}{2} \right) \pi \left( \sqrt{2}\,\Gamma^2 \left( \frac{3}{4} \right) + \pi \right) - 2\, K_{3/4} \left( \frac{x^2}{2} \right) \Gamma \left( \frac{3}{4} \right)}{\pi \left( \sqrt{2}\,\Gamma^2 \left( \frac{3}{4} \right) + \pi \right) I_{1/4} \left( \frac{x^2}{2} \right) + 2\, K_{1/4} \left( \frac{x^2}{2} \right) \Gamma \left( \frac{3}{4} \right)} , \qquad y(0) = 1. \tag{2.7} \]

      We plot two nullclines (in black) for the Riccati equation \( y' = y^2 - x^2 \) and some typical solutions.

dfield = VectorPlot[{1, y^2 - x^2}, {x, -2, 2}, {y, -2, 2}, Axes -> True, VectorScale -> {Small, Automatic, None}, AxesLabel -> {"x", "dydx=y^2 - x^2"}]
l1 = Plot[x, {x, -2, 2}, PlotStyle -> {Thickness[0.01], Black}];
l2 = Plot[-x, {x, -2, 2}, PlotStyle -> {Thickness[0.01], Black}];
sol1 = NDSolve[{y'[x] == (y[x])^2 - x^2, y[-1.5] == -1.5}, y, {x, -2, 2}]
s1 = Plot[y[x] /. sol1, {x, -2, 2}, PlotStyle -> {Thickness[0.008], Blue}, PlotRange -> {-2, 2}]
sol2 = NDSolve[{y'[x] == (y[x])^2 - x^2, y[-1.0] == 1.6}, y, {x, -2, 2}]
s2 = Plot[y[x] /. sol2, {x, -2, 2}, PlotStyle -> {Thickness[0.008], Blue}, PlotRange -> {-2, 2}]
sol3 = NDSolve[{y'[x] == (y[x])^2 - x^2, y[0] == 0.6}, y, {x, -2, 2}]
s3 = Plot[y[x] /. sol3, {x, -2, 2}, PlotStyle -> {Thickness[0.008], Blue}, PlotRange -> {-2, 2}]
sol4 = NDSolve[{y'[x] == (y[x])^2 - x^2, y[-1.0] == -1.0}, y, {x, -2, 2}]
s4 = Plot[y[x] /. sol4, {x, -2, 2}, PlotStyle -> {Thickness[0.008], Blue}, PlotRange -> {-2, 2}]
Show[s1, s2, l1, l2, dfield, s3, s4, AspectRatio -> 1]

       Two nullclines for the Riccati equation (2.5).            Mathematica code

   ■

Example 3: : Consider the Riccati equation

\[ y' = y^2 -x . \tag{3.1} \]

      We plot the nullcline (in black) y² = x and phase portrait) for the Riccati equation \( y' = y^2 - x. \)

dfield = VectorPlot[{1, y^2 - x}, {x, -2, 2}, {y, -2, 2}, Axes -> True, VectorScale -> {Small, Automatic, None}, AxesLabel -> {"x", "dydx=y^2 - x^2"}];
l1 = Plot[Sqrt[x], {x, -2, 2}, PlotStyle -> {Thickness[0.01], Black}];
l2 = Plot[-Sqrt[x], {x, -2, 2}, PlotStyle -> {Thickness[0.01], Black}];
sol1 = NDSolve[{y'[x] == (y[x])^2 - x, y[-1.5] == -1.5}, y, {x, -2, 2}];
s1 = Plot[y[x] /. sol1, {x, -2, 2}, PlotStyle -> {Thickness[0.008], Blue}, PlotRange -> {-2, 2}]
sol2 = NDSolve[{y'[x] == (y[x])^2 - x, y[-2.0] == -1.6}, y, {x, -2, 2}];
s2 = Plot[y[x] /. sol2, {x, -2, 1.5}, PlotStyle -> {Thickness[0.008], Blue}, PlotRange -> {-2, 2}]
sol3 = NDSolve[{y'[x] == (y[x])^2 - x, y[0] == 0.6}, y, {x, -2, 2}];
s3 = Plot[y[x] /. sol3, {x, -2, 2}, PlotStyle -> {Thickness[0.008], Blue}, PlotRange -> {-2, 2}]
sol4 = NDSolve[{y'[x] == (y[x])^2 - x, y[-1.0] == -1.5}, y, {x, -2, 2}];
s4 = Plot[y[x] /. sol4, {x, -2, 2}, PlotStyle -> {Thickness[0.008], Blue}, PlotRange -> {-2, 2}]
Show[s1, s2, l1, l2, dfield, s3, s4, AspectRatio -> 1]

       The nullcline and phase portrait for the Riccati equation (3.1).            Mathematica code

   ■

Example 4: : Consider the Riccati equation

\[ y' = 2y/x+y^2 -x^4 . \tag{4.1} \]

We plot the separatrix and the direction field

field = StreamPlot[{1, 2*y/x + y^2 - x^4}, {x, -2, 2}, {y, -2, 2}, StreamColorFunction -> "Rainbow", StreamPoints -> 42, StreamScale -> {Full, All, 0.04}];
sol = NDSolve[{y'[x] == 2*y[x]/x + (y[x])^2 - x^4 , y[0.1] == 0.01}, y[x], {x, 0, 2}];
curve = Plot[Evaluate[y[x] /. sol], {x, 0, 2}, PlotRange -> {{0, 2}, {-2, 2}}, PlotStyle -> {Red, Thickness[0.01]}];
Show[field, curve]

         Direction field and separatrix (in red).       Mathematica code.

The given Riccati equation can be solved by substitution \( y =x^2 +1/v(x) , \) where y1= x² is a particular solution of the given Riccati equation.

R[x_, y_] = (y'[x] - 2 y[x]/x - y[x]^2 + x^4 )
y1[x_] = x^2
R[x, y1]
Simplify[Expand[v[x]^2 R[x, Function[t, t + t/v[t]]]]]
DSolve[% == 0, v[x], x] (* solve linear equation for v *)
y2[x_] = Simplify[(y1[x] + 1/v[x]) /. %[[1]]]

Out[11]= x^4 - (2 y[x])/x - y[x]^2 + Derivative[1][y][x]
Out[12]= x^2
Out[13]= 0
Out[14]= -(1 + 2 x^2) v[x] + (-1 - x^2 + x^4) v[x]^2 - x (x + Derivative[1][v][x])
Out[15]= {{v[x] -> -(x (-(E^((2 x^3)/3 - 1/6 x^2 (3 + 2 x))/(2 x)) + (
E^((2 x^3)/3 - 1/6 x^2 (3 + 2 x)) (1 + x))/(2 x^2) - (
E^((2 x^3)/3 - 1/6 x^2 (3 + 2 x)) (1 + x) ((5 x^2)/3 - 1/3 x (3 + 2 x)))/(
2 x) + (-((E^(-(1/6) x^2 (3 + 2 x)) (-1 + x))/x^2) +
E^(-(1/6) x^2 (3 + 2 x))/x + ( E^(-(1/6) x^2 (3 + 2 x)) (-1 + x) (-(x^2/3) -
1/3 x (3 + 2 x)))/x) C[1]))/((-1 - x^2 + x^4) (-((E^((2 x^3)/3 - 1/6 x^2 (3 + 2 x)) (1 + x))/(
2 x)) + (E^(-(1/6) x^2 (3 + 2 x)) (-1 + x) C[1])/x))}}
Out[16]= (E^((2 x^3)/3) (-1 - x + x^2) + 2 (-1 + x + x^2) C[1])/(E^((
2 x^3)/3) + 2 C[1])

   ■

 

  1. Haaheim, D.R. and Stein, F.M., Methods of Solution of the Riccati Differential Equation, Mathematics Magazine, 1969, Vol. 42, No. 5, pp. 233--240; https://doi.org/10.1080/0025570X.1969.11975969
  2. Robin, W., A new Riccati equation arising in the theory of classical orthogonal polynomials, International Journal of Mathematical Education in Science and Technology, 2003, Volume 34, 2003 - Issue 1, pp. 31--42. https://doi.org/10.1080/0020739021000018746

 

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Riccati equation

In mathematics, a Riccati equation in the narrowest sense is any first-order ordinary differential equation that is quadratic in the unknown function. In other words, it is an equation of the form

y'(x)=q_{0}(x)+q_{1}(x)\,y(x)+q_{2}(x)\,y^{2}(x)

where q_{0}(x)\neq 0 and q_{2}(x)\neq 0. If q_{0}(x)=0 the equation reduces to a Bernoulli equation, while if q_{2}(x)=0 the equation becomes a first order linear ordinary differential equation.

The equation is named after Jacopo Riccati (1676–1754).[1]

More generally, the term Riccati equation is used to refer to matrix equations with an analogous quadratic term, which occur in both continuous-time and discrete-timelinear-quadratic-Gaussian control. The steady-state (non-dynamic) version of these is referred to as the algebraic Riccati equation.

Conversion to a second order linear equation[edit]

The non-linear Riccati equation can always be converted to a second order linear ordinary differential equation (ODE):[2] If

y'=q_{0}(x)+q_{1}(x)y+q_{2}(x)y^{2}\!

then, wherever q_{2} is non-zero and differentiable, v=yq_{2} satisfies a Riccati equation of the form

v'=v^{2}+R(x)v+S(x),\!

where S=q_{2}q_{0} and {\displaystyle R=q_{1}+{\frac {q_{2}'}{q_{2}}}}, because

v'=(yq_{2})'=y'q_{2}+yq_{2}'=(q_{0}+q_{1}y+q_{2}y^{2})q_{2}+v{\frac {q_{2}'}{q_{2}}}=q_{0}q_{2}+\left(q_{1}+{\frac {q_{2}'}{q_{2}}}\right)v+v^{2}.\!

Substituting v=-u'/u, it follows that u satisfies the linear 2nd order ODE

u''-R(x)u'+S(x)u=0\!

since

v'=-(u'/u)'=-(u''/u)+(u'/u)^{2}=-(u''/u)+v^{2}\!

so that

u''/u=v^{2}-v'=-S-Rv=-S+Ru'/u\!

and hence

u''-Ru'+Su=0.\!

A solution of this equation will lead to a solution y=-u'/(q_{2}u) of the original Riccati equation.

Application to the Schwarzian equation[edit]

An important application of the Riccati equation is to the 3rd order Schwarzian differential equation

S(w):=(w''/w')'-(w''/w')^{2}/2=f

which occurs in the theory of conformal mapping and univalent functions. In this case the ODEs are in the complex domain and differentiation is with respect to a complex variable. (The Schwarzian derivativeS(w) has the remarkable property that it is invariant under Möbius transformations, i.e. S((aw+b)/(cw+d))=S(w) whenever ad-bc is non-zero.) The function y=w''/w' satisfies the Riccati equation

y'=y^{2}/2+f.

By the above y=-2u'/u where u is a solution of the linear ODE

u''+(1/2)fu=0.

Since w''/w'=-2u'/u, integration gives w'=C/u^{2} for some constant C. On the other hand any other independent solution U of the linear ODE has constant non-zero Wronskian U'u-Uu' which can be taken to be C after scaling. Thus

w'=(U'u-Uu')/u^{2}=(U/u)'

so that the Schwarzian equation has solution w=U/u.

Obtaining solutions by quadrature[edit]

The correspondence between Riccati equations and second-order linear ODEs has other consequences. For example, if one solution of a 2nd order ODE is known, then it is known that another solution can be obtained by quadrature, i.e., a simple integration. The same holds true for the Riccati equation. In fact, if one particular solution y_{1} can be found, the general solution is obtained as

y=y_{1}+u

Substituting

y_{1}+u

in the Riccati equation yields

y_{1}'+u'=q_{0}+q_{1}\cdot (y_{1}+u)+q_{2}\cdot (y_{1}+u)^{2},

and since

{\displaystyle y_{1}'=q_{0}+q_{1}\,y_{1}+q_{2}\,y_{1}^{2},}

it follows that

u'=q_{1}\,u+2\,q_{2}\,y_{1}\,u+q_{2}\,u^{2}

or

u'-(q_{1}+2\,q_{2}\,y_{1})\,u=q_{2}\,u^{2},

which is a Bernoulli equation. The substitution that is needed to solve this Bernoulli equation is

z={\frac {1}{u}}

Substituting

y=y_{1}+{\frac {1}{z}}

directly into the Riccati equation yields the linear equation

z'+(q_{1}+2\,q_{2}\,y_{1})\,z=-q_{2}

A set of solutions to the Riccati equation is then given by

y=y_{1}+{\frac {1}{z}}

where z is the general solution to the aforementioned linear equation.

See also[edit]

References[edit]

  1. ^Riccati, Jacopo (1724) "Animadversiones in aequationes differentiales secundi gradus" (Observations regarding differential equations of the second order), Actorum Eruditorum, quae Lipsiae publicantur, Supplementa, 8 : 66-73. Translation of the original Latin into English by Ian Bruce.
  2. ^Ince, E. L. (1956) [1926], Ordinary Differential Equations, New York: Dover Publications, pp. 23–25

Further reading[edit]

  • Hille, Einar (1997) [1976], Ordinary Differential Equations in the Complex Domain, New York: Dover Publications, ISBN 
  • Nehari, Zeev (1975) [1952], Conformal Mapping, New York: Dover Publications, ISBN 
  • Polyanin, Andrei D.; Zaitsev, Valentin F. (2003), Handbook of Exact Solutions for Ordinary Differential Equations (2nd ed.), Boca Raton, Fla.: Chapman & Hall/CRC, ISBN 
  • Zelikin, Mikhail I. (2000), Homogeneous Spaces and the Riccati Equation in the Calculus of Variations, Berlin: Springer-Verlag
  • Reid, William T. (1972), Riccati Differential Equations, London: Academic Press

External links[edit]

Sours: https://en.wikipedia.org/wiki/Riccati_equation
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Matrix Riccati Equations in Control and Systems Theory

The aim of the book is to present the state of the art of the theory of symmetric (Hermitian) matrix Riccati equations and to contribute to the development of the theory of non-symmetric Riccati equations as well as to certain classes of coupled and generalized Riccati equations occurring in differential games and stochastic control. The volume offers a complete treatment of generalized and coupled Riccati equations. It deals with differential, discrete-time, algebraic or periodic symmetric and non-symmetric equations, with special emphasis on those equations appearing in control and systems theory. Extensions to Riccati theory allow to tackle robust control problems in a unified approach.

The book is intended to make available classical and recent results to engineers and mathematicians alike. It is accessible to graduate students in mathematics, applied mathematics, control engineering, physics or economics. Researchers working in any of the fields where Riccati equations are used can find the main results with the proper mathematical background.

Sours: https://www.springer.com/gp/book/9783764300852
Riccati Equation

American Institute of Mathematical Sciences

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Equations riccati

Riccati Equation

General Riccati Equation

The Riccati equation is one of the most interesting nonlinear differential equations of first order. It's written in the form:

\[y' = a\left( x \right)y + b\left( x \right){y^2} + c\left( x \right),\]

where \(a\left( x \right),\) \(b\left( x \right),\) \(c\left( x \right)\) are continuous functions of \(x.\)

The Riccati equation is used in different areas of mathematics (for example, in algebraic geometry and the theory of conformal mapping), and physics. It also appears in many applied problems.

The differential equation given above is called the general Riccati equation. It can be solved with help of the following theorem:

Theorem.

If a particular solution \({y_1}\) of a Riccati equation is known, the general solution of the equation is given by

\[y = {y_1} + u.\]

Indeed, substituting the solution \(y = {y_1} + u\) into Riccati equation, we have

\[\left( {{y_1} + u} \right)^\prime = a\left( x \right)\left( {{y_1} + u} \right) + b\left( x \right){\left( {{y_1} + u} \right)^2} + c\left( x \right),\]

\[\underline {{y_1}^\prime } + u' = \underline {a\left( x \right){y_1}} + a\left( x \right)u + \underline {b\left( x \right)y_1^2} + 2b\left( x \right){y_1}u + b\left( x \right){u^2} + \underline {c\left( x \right)} .\]

The underlined terms in the left and in the right side can be canceled because \({y_1}\) is a particular solution satisfying the equation. As a result we obtain the differential equation for the function \(u\left( x \right):\)

\[u' = b\left( x \right){u^2} + \left[ {2b\left( x \right){y_1} + a\left( x \right)} \right]u,\]

which is a Bernoulli equation.

Substitution of \(z = \frac{1}{u}\) converts the given Bernoulli equation into a linear differential equation that allows integration.

Besides the general Riccati equation, there is an infinite number of particular cases of Riccati equation at certain coefficients of \(a\left( x \right),\) \(b\left( x \right),\) and \(c\left( x \right).\) Many of these particular cases have integrable solutions.

Returning to the general Riccati equation, we see that we can construct the general solution if a particular solution is known. Unfortunately, there is no strict algorithm to find the particular solution, which depends on the types of the functions \(a\left( x \right),\) \(b\left( x \right),\) and \(c\left( x \right).\)

Below we consider some well known particular cases of the Riccati equation.

Special Case \(1:\) Coefficients \(a, b, c\) are constants

If the coefficients in the Riccati equation are constants, this equation can be reduced to a separable differential equation. The solution is described by the integral of a rational function with a quadratic function in the denominator:

\[y' = ay + b{y^2} + c,\;\; \Rightarrow \frac{{dy}}{{dx}} = ay + b{y^2} + c,\;\; \Rightarrow \int {\frac{{dy}}{{ay + b{y^2} + c}}} = \int {dx} .\]

This integral can be easily calculated at any values of \(a,\) \(b\) and \(c\) (For more information see Integration of Rational Functions).

Special Case \(2:\) Equation of type \(y' = b{y^2} + c{x^n}\)

Consider a Riccati equation of type \(y' = b{y^2} + c{x^n},\) where the function \(a\left( x \right)\) at the linear term is zero, the coefficient \(b\) at \({y^2}\) is a constant, and \(c\left( x \right)\) is a power function:

\[a\left( x \right) \equiv 0,\;\; b\left( x \right) = b,\;\; c\left( x \right) = c{x^n}.\]

This case of Riccati equation has nice solutions!

First of all, if \(n = 0,\) we get the Case \(1\) where the variables are separated and the differential equation can be integrated.

If \(n = -2,\) the Riccati equation is converted into a homogeneous equation with help of the substitution \(y = \frac{1}{z}\) and then also can be integrated.

This differential equation can be also solved at

\[n = \frac{{4k}}{{1 - 2k}},\;\; \text{where}\;\; k = \pm 1, \pm 2, \pm 3, \ldots \]

Here the general solution is expressed through cylinder functions.

At all other values of the power \(n,\) the solution of the Riccati equation can be expressed through integrals of elementary functions. This fact was discovered by the French mathematician Joseph Liouville \(\left( {1809 - 1882} \right)\) in \(1841.\)

Solved Problems

Click or tap a problem to see the solution.

Example 1

Solve the differential equation \[y' = y + {y^2} + 1.\]

Example 2

Solve the Riccati equation \[y' + {y^2} = \frac{2}{{{x^2}}}.\]

Example 1.

Solve the differential equation \[y' = y + {y^2} + 1.\]

Solution.

The given equation is a simple Riccati equation with constant coefficients. Here the variables \(x, y\) can be easily separated, so the general solution of the equation is given by

\[\frac{{dy}}{{dx}} = y + {y^2} + 1,\;\; \Rightarrow \frac{{dy}}{{y + {y^2} + 1}} = dx,\;\; \Rightarrow \int {\frac{{dy}}{{y + {y^2} + 1}}} = \int {dx} ,\;\; \Rightarrow \int {\frac{{dy}}{{{y^2} + y + \frac{1}{4} + \frac{3}{4}}}} = \int {dx} ,\;\; \Rightarrow \int {\frac{{dy}}{{{{\left( {y + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} = \int {dx} ,\;\; \Rightarrow \frac{1}{{\frac{{\sqrt 3 }}{2}}}\arctan \frac{{y + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} = x + C,\;\; \Rightarrow \frac{2}{{\sqrt 3 }}\arctan \frac{{2y + 1}}{{\sqrt 3 }} = x + C.\]

Example 2.

Solve the Riccati equation \[y' + {y^2} = \frac{2}{{{x^2}}}.\]

Solution.

We will seek for a particular solution in the form:

\[y = \frac{c}{x},\;\; \Rightarrow y' = - \frac{c}{{{x^2}}}.\]

Substituting this into the equation, we obtain:

\[- \frac{c}{{{x^2}}} + {\left( {\frac{c}{x}} \right)^2} = \frac{2}{{{x^2}}}\;\; \text{or}\;\; - \frac{c}{{{x^2}}} + \frac{{{c^2}}}{{{x^2}}} = \frac{2}{{{x^2}}}.\]

We get a quadratic equation for \(c:\)

\[{c^2} - c - 2 = 0,\;\; \Rightarrow D = 1 - 4 \cdot \left( { - 2} \right) = 9,\;\; \Rightarrow {c_{1,2}} = \frac{{1 \pm 3}}{2} = - 1,2.\]

We can take any value of\(c.\) For example, let \(c = 2.\) Now, when the particular solution is known, we make the replacement:

\[y = z + \frac{2}{x},\;\; \Rightarrow y' = z' - \frac{2}{{{x^2}}}.\]

Now substitute this into the original Riccati equation:

\[z' - \frac{2}{{{x^2}}} + {\left( {z + \frac{2}{x}} \right)^2} = \frac{2}{{{x^2}}},\;\; \Rightarrow z' - \cancel{\frac{2}{{{x^2}}}} + {z^2} + \frac{4}{x}z + \cancel{\frac{4}{{{x^2}}}} = \cancel{\frac{2}{{{x^2}}}},\;\; \Rightarrow z' + \frac{4}{x}z = - {z^2}.\]

As it can be seen, we have a Bernoulli equation with the parameter \(m = 2.\) Make one more substitution:

\[v = {z^{1 - m}} = \frac{1}{z},\;\; \Rightarrow v' = - \frac{{z'}}{{{z^2}}}.\]

Divide the Bernoulli equation by \({z^2}\) (assuming that \(z \ne 0\)) and rewrite it in terms of \(v:\)

\[\frac{{z'}}{{{z^2}}} + \frac{{4z}}{{x{z^2}}} = - 1,\;\; \Rightarrow - \frac{{z'}}{{{z^2}}} - \frac{4}{{xz}} = 1,\;\; \Rightarrow v' - \frac{4}{x}v = 1.\]

The last equation is linear and can be easily solved using the integrating factor:

\[u = {e^{\int {\left( { - \frac{4}{x}} \right)dx} }} = {e^{ - 4\int {\frac{{dx}}{x}} }} = {e^{ - 4\ln \left| x \right|}} = {e^{\ln \frac{1}{{{{\left| x \right|}^4}}}}} = \frac{1}{{{{\left| x \right|}^4}}} = \frac{1}{{{x^4}}}.\]

The general solution of the linear equation is given by

\[v = \frac{{\int {u\left( x \right)f\left( x \right)dx} + C}}{{u\left( x \right)}} = \frac{{\int {\frac{1}{{{x^4}}} \cdot 1dx} + C}}{{\frac{1}{{{x^4}}}}} = \frac{{\int {{x^{ - 4}}dx} + C}}{{\frac{1}{{{x^4}}}}} = \left( { - \frac{1}{3}{x^{ - 3}} + C} \right){x^4} = - \frac{x}{3} + C{x^4}.\]

From now on we will subsequently return back to the previous variables. Since \(z = \frac{1}{v},\) the general solution for \(z\) is written as follows:

\[\frac{1}{z} = - \frac{x}{3} + C{x^4},\;\; \Rightarrow z = \frac{1}{{ - \frac{x}{3} + C{x^4}}} = - \frac{3}{{x + 3C{x^4}}} = - \frac{3}{{x\left( {1 + 3C{x^3}} \right)}}.\]

Hence,

\[y = z + \frac{2}{x} = - \frac{3}{{x\left( {1 + 3C{x^3}} \right)}} + \frac{2}{x} = \frac{{ - 3 + 2\left( {1 + 3C{x^3}} \right)}}{{x\left( {1 + 3C{x^3}} \right)}} = \frac{{ - 3 + 2 + 6C{x^3}}}{{x\left( {1 + 3C{x^3}} \right)}} = \frac{{6C{x^3} - 1}}{{x\left( {1 + 3C{x^3}} \right)}}.\]

We can rename the constant: \(3C = {C_1}\) and write the answer in the form:

\[y = \frac{{2{C_1}{x^3} - 1}}{{x\left( {1 + {C_1}{x^3}} \right)}},\]

where \({C_1}\) is an arbitrary real number.

See more problems on Page 2.

Sours: https://math24.net/riccati-equation.html
Riccati ode

Numerical Solution of Riccati Equations by the Adomian and Asymptotic Decomposition Methods over Extended Domains

We combine the Adomian decomposition method (ADM) and Adomian’s asymptotic decomposition method (AADM) for solving Riccati equations. We investigate the approximate global solution by matching the near-field approximation derived from the Adomian decomposition method with the far-field approximation derived from Adomian’s asymptotic decomposition method for Riccati equations and in such cases when we do not find any region of overlap between the obtained approximate solutions by the two proposed methods, we connect the two approximations by the Padé approximant of the near-field approximation. We illustrate the efficiency of the technique for several specific examples of the Riccati equation for which the exact solution is known in advance.

1. Introduction

It is well known that the Riccati equation asfinds surprisingly many applications in physics and mathematics such as random processes, optimal control, and diffusion problems [1]. In fact, the Riccati equation naturally arises in many fields of quantum mechanics, such as in quantum chemistry [2], the Wentzel-Kramers-Brillouin approximation [3], and super symmetry theories [4]. In addition, the Riccati equation plays a prominent role in variational calculus [5], nonlinear physics [6], renormalization group equations for coupling constants in quantum field theories [7, 8], and thermodynamics [9]. It is well known that one-dimensional static Schrödinger equation is closely related to the Riccati equation. Solitary wave solutions of a nonlinear partial differential equation can be expressed as a polynomial in two elementary functions satisfying a projective Riccati equation [10]. Beside important engineering and scientific applications that are well known, the newer applications include areas such as mathematical finance [11, 12].

Adomian and his coauthors have presented a systematic methodology for practical solution of linear or nonlinear and deterministic or stochastic operator equations, including algebraic equations, ordinary differential equations, partial differential equations, and integral and integrodifferential equations [13–18]. Adomian decomposition method is a powerful technique, which provides efficient algorithms for analytic approximate solutions and numerical simulations for real-world applications in the applied sciences and engineering. Using the ADM, we calculate a series solution, but, in practice, we approximate the solution by a truncated series. The series sometimes coincides with the Taylor expansion of the exact solution in the neighborhood of the point . Although the series can be rapidly convergent in a small region, it has a slower convergence rate in the wider region.

Several investigators have proposed a variety of approaches to solve the Riccati equation, approximately [19–24]. In order to obtain the global approximate solution of the Riccati equation, we combine the Padé approximant of the near-field approximation as derived from the ADM with the far-field approximation as derived from the AADM [25–28] to overcome the difficulty of a finite domain of convergence. Adomian introduced a variation of his decomposition method in [25] that can be used to obtain the asymptotic value of solutions. In this method, the recursion is the same as that in the ADM, but it uses a different canonical form of the differential equation such that it yields to a steady state solution of the equation. In fact, rather than nested integrations as in decomposition, we have nested differentiations, which will be expounded later. Haldar and Datta [29] applied the AADM to calculate integrals neither expressible in terms of elementary functions nor adequately tabulated.

This paper is arranged as follows. In the next section, we present a brief review of the ADM for nonlinear IVPs. In Section 3, we present a description of the AADM for solving the Riccati equation. In Section 4, we investigate several numerical examples. In Section 5, we present our conclusions and summarize our findings.

2. Review of the Adomian Decomposition Method

We review the salient features of the Adomian decomposition method in solving IVPs for first-order nonlinear ordinary differential equations aswhere the functions , , and are analytic.

We rewrite (2) in Adomian’s usual operator-theoretic formwhere and then , , and . Next we rewrite (3) asand we apply the integral operator to both sides of (4):where since . In the case of a first-order ordinary differential equation, we have . ThereforeFor the sake of simplicity, we define the function as , and then, upon substitution, we obtainIn the ADM, the solution is represented by a series; sayand the nonlinearity comprises the Adomian polynomialswhereis called an Adomian polynomial, which were first defined by Adomian [13] asFor convenient reference, we list the first five Adomian polynomials

Several algorithms for the Adomian polynomials have been developed by Rach [30, 31], Adomian and Rach [32], Wazwaz [33], Biazar et al. [34], and several others. New, efficient algorithms with their subroutines written in Mathematica for rapid computer-generation of the Adomian polynomials have been provided by Duan in [35–37].

From (7)–(9) the solution components are determined by the classic Adomian recursion scheme:Thus the -term approximation as obtained from the ADM can serve as the near-field approximation of the solution , where is in the neighborhood of the initial point .

We remark that the convergence of the Adomian decomposition series has been previously proven by several researchers [30, 38–41]. For example, Abdelrazec and Pelinovsky [41] have recently published a rigorous proof of convergence for the ADM in accordance with the Cauchy-Kovalevskaya theorem.

3. Description of Adomian’s Asymptotic Decomposition Method

In this section, we advocate Adomian’s asymptotic decomposition method for solving the Riccati equation. We remark that Adomian’s asymptotic decomposition method does not need use of the initial condition to obtain the asymptotic solution or the solution in the large, which is another, convenient advantage in computations using this technique. Rather than nested integrations as required by decomposition, we now have nested differentiations. In effect our aim is to solve for the solution by not inverting the linear differential operator , but instead by decomposing the nonlinear operator and hence determining the asymptotic solution . Toward this end, we rewrite (1) asFor the case when the coefficient , we can divide both sides of (14) by , and we haveSubstituting the respective decomposition series (i.e., and ), we obtainfrom which we design the asymptotic recursion schemeWe note that the Adomian polynomials for the quadratic nonlinearity areUsing the form of the Adomian polynomials in (19), we rewrite the recursion schemes (17) and (18) asIn view of (21) and after appropriate manipulations, we obtainConsequently, with this result, the solution components are given by the following recursion scheme:Thus the obtained -terms asymptotic approximation for the Riccati equation can serve as the far-field approximation, where is far from the initial point .

4. Numerical Examples

In this section, several numerical examples are given to illustrate the efficiency of our technique as presented in this paper. We remark that all calculations are performed by Mathematica package 8.

Example 1. Consider the following Riccati equation:subject to initial condition .
The exact solution is known in advance to beIn Adomian’s operator notation, we havewhere , , and .
To apply Adomian decomposition method, equation (26) should be written as the following,

Sours: https://www.hindawi.com/journals/ijde/2015/580741/

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Algebraic Riccati equation

Nonlinear equation which arises on linear optimal control problems

An algebraic Riccati equation is a type of nonlinear equation that arises in the context of infinite-horizon optimal control problems in continuous time or discrete time.

A typical algebraic Riccati equation is similar to one of the following:

the continuous time algebraic Riccati equation (CARE):

{\displaystyle A^{T}P+PA-PBR^{-1}B^{T}P+Q=0\,}

or the discrete time algebraic Riccati equation (DARE):

{\displaystyle P=A^{T}PA-(A^{T}PB)(R+B^{T}PB)^{-1}(B^{T}PA)+Q.\,}

P is the unknown n by n symmetric matrix and A, B, Q, R are known real coefficient matrices.

Though generally this equation can have many solutions, it is usually specified that we want to obtain the unique stabilizing solution, if such a solution exists.

Origin of the name[edit]

The name Riccati is given to these equations because of their relation to the Riccati differential equation. Indeed, the CARE is verified by the time invariant solutions of the associated matrix valued Riccati differential equation. As for the DARE, it is verified by the time invariant solutions of the matrix valued Riccati difference equation (which is the analogue of the Riccati differential equation in the context of discrete time LQR).

Context of the discrete-time algebraic Riccati equation[edit]

In infinite-horizon optimal control problems, one cares about the value of some variable of interest arbitrarily far into the future, and one must optimally choose a value of a controlled variable right now, knowing that one will also behave optimally at all times in the future. The optimal current values of the problem's control variables at any time can be found using the solution of the Riccati equation and the current observations on evolving state variables. With multiple state variables and multiple control variables, the Riccati equation will be a matrix equation.

The algebraic Riccati equation determines the solution of the infinite-horizon time-invariant Linear-Quadratic Regulator problem (LQR) as well as that of the infinite horizon time-invariant Linear-Quadratic-Gaussian control problem (LQG). These are two of the most fundamental problems in control theory.

A typical specification of the discrete-time linear quadratic control problem is to minimize

\sum_{t=1}^T (y_t^T Qy_t + u_t^T Ru_t)

subject to the state equation

y_t = Ay_{t-1} + Bu_t,

where y is an n × 1 vector of state variables, u is a k × 1 vector of control variables, A is the n × n state transition matrix, B is the n × k matrix of control multipliers, Q (n × n) is a symmetric positive semi-definite state cost matrix, and R (k × k) is a symmetric positive definite control cost matrix.

Induction backwards in time can be used to obtain the optimal control solution at each time,[1]

{\displaystyle u_{t}^{*}=-(B^{T}P_{t}B+R)^{-1}(B^{T}P_{t}A)y_{t-1},}

with the symmetric positive definite cost-to-go matrix P evolving backwards in time from {\displaystyle P_{T}=Q} according to

{\displaystyle P_{t-1}=Q+A^{T}P_{t}A-A^{T}P_{t}B(B^{T}P_{t}B+R)^{-1}B^{T}P_{t}A,\,}

which is known as the discrete-time dynamic Riccati equation of this problem. The steady-state characterization of P, relevant for the infinite-horizon problem in which T goes to infinity, can be found by iterating the dynamic equation repeatedly until it converges; then P is characterized by removing the time subscripts from the dynamic equation.

Solution[edit]

Usually solvers try to find the unique stabilizing solution, if such a solution exists. A solution is stabilizing if using it for controlling the associated LQR system makes the closed loop system stable.

For the CARE, the control is

{\displaystyle K=R^{-1}B^{T}P}

and the closed loop state transfer matrix is

{\displaystyle A-BK=A-BR^{-1}B^{T}P}

which is stable if and only if all of its eigenvalues have strictly negative real part.

For the DARE, the control is

{\displaystyle K=(R+B^{T}PB)^{-1}B^{T}PA}

and the closed loop state transfer matrix is

{\displaystyle A-BK=A-B(R+B^{T}PB)^{-1}B^{T}PA}

which is stable if and only if all of its eigenvalues are strictly inside the unit circle of the complex plane.

A solution to the algebraic Riccati equation can be obtained by matrix factorizations or by iterating on the Riccati equation. One type of iteration can be obtained in the discrete time case by using the dynamic Riccati equation that arises in the finite-horizon problem: in the latter type of problem each iteration of the value of the matrix is relevant for optimal choice at each period that is a finite distance in time from a final time period, and if it is iterated infinitely far back in time it converges to the specific matrix that is relevant for optimal choice an infinite length of time prior to a final period—that is, for when there is an infinite horizon.

It is also possible to find the solution by finding the eigendecomposition of a larger system. For the CARE, we define the Hamiltonian matrix

 Z = \begin{pmatrix} A & -B R^{-1} B^T \\ -Q & -A^T \end{pmatrix}

Since {\displaystyle Z} is Hamiltonian, if it does not have any eigenvalues on the imaginary axis, then exactly half of its eigenvalues have a negative real part. If we denote the {\displaystyle 2n\times n} matrix whose columns form a basis of the corresponding subspace, in block-matrix notation, as

{\displaystyle {\begin{pmatrix}U_{1,1}\\U_{2,1}\end{pmatrix}}}

then

{\displaystyle P=U_{2,1}U_{1,1}^{-1}}

is a solution of the Riccati equation; furthermore, the eigenvalues of {\displaystyle A-BR^{-1}B^{T}P} are the eigenvalues of Z with negative real part.

For the DARE, when A is invertible, we define the symplectic matrix

 Z = \begin{pmatrix} A + B R^{-1} B^T (A^{-1})^T  Q & -B R^{-1} B^T (A^{-1})^T \\ -(A^{-1})^T Q & (A^{-1})^T \end{pmatrix}

Since Z is symplectic, if it does not have any eigenvalues on the unit circle, then exactly half of its eigenvalues are inside the unit circle. If we denote the {\displaystyle 2n\times n} matrix whose columns form a basis of the corresponding subspace, in block-matrix notation, as

{\displaystyle {\begin{pmatrix}U_{1,1}\\U_{2,1}\end{pmatrix}}}

where {\displaystyle U_{1,1}} and {\displaystyle U_{2,1}} result from the decomposition[2]

{\displaystyle Z={\begin{pmatrix}U_{1,1}&U_{1,2}\\U_{2,1}&U_{2,2}\end{pmatrix}}{\begin{pmatrix}\Lambda _{1,1}&\Lambda _{1,2}\\0&\Lambda _{2,2}\end{pmatrix}}{\begin{pmatrix}U_{1,1}^{T}&U_{2,1}^{T}\\U_{1,2}^{T}&U_{2,2}^{T}\end{pmatrix}}}

then

{\displaystyle P=U_{2,1}U_{1,1}^{-1}}

is a solution of the Riccati equation; furthermore, the eigenvalues of {\displaystyle A-B(R+B^{T}PB)^{-1}B^{T}PA} are the eigenvalues of  Z which are inside the unit circle.

See also[edit]

References[edit]

External links[edit]

Sours: https://en.wikipedia.org/wiki/Algebraic_Riccati_equation


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