2
$\begingroup$

I am trying to solve a simple Ricatti equation and am not sure i'm correct. Since this is probably the simplest common case I thought I'd post it here:

Write the Riccati equation as \begin{equation*} \frac{d E (\tau) }{d \tau} = e_2 E(\tau)^2 + e_1 E(\tau) + e_0 \end{equation*} where we can reduce to a second order linear equation via the substitution $E(\tau) = - \frac{1}{e_2} \frac{w^\prime (\tau)}{w(\tau)}$ which yields \begin{equation*} w^{\prime \prime} - e_1 w^\prime + e_2 e_0 w = 0. \end{equation*} The general solution to this ODE is $w(\tau) = C_{1} e^{r_{1} \tau} + C_{2} e^{r_{2} \tau}$ where $r_1$ and $r_2$ are the positive and negative roots \begin{align*} r_{1,2} = \frac{e_1 \pm \sqrt{e_1^2 - 4 e_2 e_0}}{2} = \frac{e_1 \pm q}{2} \end{align*} and $C_{1,2}$ are constants determined by the boundary condition. A general solution for $E(\tau)$ is then \begin{align*} E(\tau) &= - \frac{1}{e_2} \frac{C_1 r_{1} e^{r_{1} \tau} + C_2 r_{2} e^{r_{2} \tau}}{C_1 e^{r_{1} \tau} + C_2 e^{r_{2} \tau}} = - \frac{1}{e_2} \frac{C r_{1} e^{r_{1} \tau} + r_{2} e^{r_{2} \tau}}{C e^{r_{1} \tau} + e^{r_{2} \tau}} \end{align*}

Does this look correct? How else can these ODEs be solved?

$\endgroup$
  • 1
    $\begingroup$ Am I missing the question? $\endgroup$ – Amzoti May 27 '13 at 21:24
  • $\begingroup$ yes - sorry! edited. $\endgroup$ – Luap Nalehw May 27 '13 at 21:57
  • $\begingroup$ If your $e_0,\,e_1,\,e_2$ are constants, then you equation is separable and can be integrated directly. $\endgroup$ – Artem May 28 '13 at 8:17
1
$\begingroup$

I suppose that your non linear OE $$\frac{d E (\tau) }{d \tau} = e_2 E(\tau)^2 + e_1 E(\tau) + e_0 $$ is rewritten as $$\frac{d E }{d x} = R(x) E^2 + Q(x)E + P(x),~~~(*)$$ just for the simplicity in indexes and the variable $\tau$ is our usual variable $x$. Another method in which you can overcome the OE would be to use a known particular solution of $(*)$, say $y_1$. Then set $y=y_1+u$ wherein $u$ is a solution of $$\frac{du}{dx}-(Q+2y_1R)u=Ru^2$$ Since the latter OE is a Bernoulli equation with $n=2$, it can be turned into a linear OE: $$\frac{dw}{dx}+(Q+2y_1R)w=-R$$ where $w=u^{1-2}=u^{-1}$.

$\endgroup$
  • $\begingroup$ @flashdesign2550: Thanks. :) but let us to be seen by the OP. $\endgroup$ – mrs May 28 '13 at 8:38
  • $\begingroup$ Thanks. So three methods: (1) Direct integration (2) Reduction to second order linear ODE (3) Reduction to Bernoulli equation $\endgroup$ – Luap Nalehw May 28 '13 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.