tg

tg
tgt

Wednesday, 27 March 2013

Riccati Equations


Before we give the formal definition of Riccati equations, a little introduction may be helpful. Indeed, consider the first order differential equation

\begin{displaymath}\frac{dy}{dx} = f(x,y).\end{displaymath}


If we approximate f(x,y), while x is kept constant, we will get

\begin{displaymath}f(x,y) = P(x) + Q(x) y + R(x) y^2 + \cdots\end{displaymath}


If we stop at y, we will get a linear equation. Riccati looked at the approximation to the second degree: he considered equations of the type

\begin{displaymath}\frac{dy}{dx} = P(x) + Q(x) y + R(x) y^2.\end{displaymath}


These equations bear his name, Riccati equations. They are nonlinear and do not fall under the category of any of the classical equations. In order to solve a Riccati equation, one will need a particular solution. Without knowing at least one solution, there is absolutely no chance to find any solutions to such an equation. Indeed, let y1 be a particular solution of

\begin{displaymath}\frac{dy}{dx} = P(x) + Q(x) y + R(x) y^2.\end{displaymath}


Consider the new function z defined by

\begin{displaymath}z = \frac{1}{y - y_1}.\end{displaymath}


Then easy calculations give

\begin{displaymath}\frac{dz}{dx} = -\Big(Q(x) + 2y_1 R(x)\Big) z - R(x)\end{displaymath}


which is a linear equation satisfied by the new function z. Once it is solved, we go back to y via the relation

\begin{displaymath}y = y_1 + \frac{1}{z}.\end{displaymath}


Keep in mind that it may be harder to remember the above equation satisfied by z. Instead, try to do the calculations whenever you can.
Example. Solve the equation

\begin{displaymath}\frac{dy}{dx} = -2 -y + y^2.\end{displaymath}


knowing that y1 = 2 is a particular solution.Answer. We recognize a Riccati equation. First of all we need to make sure that y1 is indeed a solution. Otherwise, our calculations will be fruitless. In this particular case, it is quite easy to check that y1 = 2 is a solution. Set

\begin{displaymath}z = \frac{1}{y - 2}.\end{displaymath}


Then we have

\begin{displaymath}y = 2 + \frac{1}{z}\end{displaymath}


which implies

\begin{displaymath}y ' = - \frac{z'}{z^2}.\end{displaymath}


Hence, from the equation satisfied by y, we get

\begin{displaymath}- \frac{z'}{z^2} = -2 -\left(2 + \frac{1}{z}\right)+ \left(2 + \frac{1}{z}\right)^2.\end{displaymath}


Easy algebraic manipulations give

\begin{displaymath}- \frac{z'}{z^2} = \frac{3}{z} + \frac{1}{z^2}.\end{displaymath}


Hence

z' = -3z -1.


This is a linear equation. The general solution is given by

\begin{displaymath}z = \frac{-1/3 e^{3x} + C}{e^{3x}} = -\frac{1}{3} + C e^{-3x}.\end{displaymath}


Therefore, we have

\begin{displaymath}y = 2 + \frac{1}{\displaystyle-\frac{1}{3} + C e^{-3x}}.\end{displaymath}


Note: If one remembers the equation satisfied by z, then the solutions may be found a bit faster. Indeed in this example, we have P(x) = -2, Q(x) = -1, and R(x) = 1. Hence the linear equation satisfied by the new function z, is

\begin{displaymath}\frac{dz}{dx} = -\Big(Q(x) + 2y_1 R(x)\Big) z - R(x) = -\Big(-1 + 4\Big) z - 1 = - 3 z - 1.\end{displaymath}



Example. Check that $y_1 = \sin(x)$ is a solution to

\begin{displaymath}\frac{dy}{dx} = \frac{2 \cos^2(x) - \sin^2(x) + y^2}{2 \cos(x)}.\end{displaymath}


Then solve the IVP

\begin{displaymath}\left\{\begin{array}{lll}
\displaystyle \frac{dy}{dx} = \frac...
... - \sin^2(x) + y^2}{2 \cos(x)}\\
y(0) = -1
\end{array} \right.\end{displaymath}


We will let the reader check that $\sin(x)$ is indeed a particular solution of the given differential equations. We also recognize that the equation is of Riccati type. Set

\begin{displaymath}z = \frac{1}{y - \sin(x)}\end{displaymath}


which gives

\begin{displaymath}y = \sin(x) + \frac{1}{z}.\end{displaymath}


Hence

\begin{displaymath}y' = \cos(x) - \frac{z'}{z^2}.\end{displaymath}


Substituting into the equation gives

\begin{displaymath}\cos(x) - \frac{z'}{z^2} = \frac{2 \cos^2(x) - \sin^2(x) + \left(\displaystyle \sin(x) + \frac{1}{z} \right)^2}{2 \cos(x)}.\end{displaymath}


Easy algebraic manipulations give

\begin{displaymath}- \frac{z'}{z^2} = \frac{\left(\displaystyle 2\sin(x) \frac{1...
...sin(x)}{\cos(x)} \frac{1}{z} + \frac{1}{2\cos(x)}\frac{1}{z^2}.\end{displaymath}


Hence

\begin{displaymath}z' = -\frac{\sin(x)}{\cos(x)} z - \frac{1}{2\cos(x)}.\end{displaymath}


This is the linear equation satisfied by z. The integrating factor is

\begin{displaymath}u(x) = e^{\displaystyle \int \frac{\sin(x)}{\cos(x)}} = e^{-\ln(\cos(x))} = \frac{1}{\cos(x)} = \sec(x).\end{displaymath}


The general solution is

\begin{displaymath}z = \frac{-1/2 \int \sec^2(x) dx + C}{u(x)} = \cos(x) \left(-\frac{1}{2} \tan(x) + C\right) = -\frac{1}{2} \sin(x) + C \cos(x).\end{displaymath}


Now it is time to go back to the original function y. We have

\begin{displaymath}y = \sin(x) + \frac{1}{ \displaystyle -\frac{1}{2} \sin(x) + C \cos(x)}.\end{displaymath}


The initial condition y(0) = -1 implies 1/C = -1, or C = -1. Therefore the solution to the IVP is

\begin{displaymath}y = \sin(x) + \frac{1}{ \displaystyle -\frac{1}{2} \sin(x) - \cos(x)}.\end{displaymath}
Post a Comment