2
$\begingroup$

I have the following simultaneous ODE before me: $x'=(x^2+y^2)y$ and $y'=-(x^2+y^2)x$ where $x$ and $y$ are functions of $t$.

This system can be rewritten as : $xx'+yy'=0$ On integration, I get $x^2+y^2=c^2$.

But now if I express $x$ and $y$ in terms of $t$ as below: $x=c \cos(t)$ and $y=c \sin(t)$,

I find that these values of $x$ and $y$ do not satisfy the original ODE. I am not able to understand why is this the case. Please help.

$\endgroup$
7
  • 1
    $\begingroup$ Note that the original equations become $x'=c^2y, y'=-c^2x$. $\endgroup$
    – copper.hat
    Dec 5 '21 at 5:46
  • 1
    $\begingroup$ Consider time scaling. $\endgroup$
    – copper.hat
    Dec 5 '21 at 5:53
  • $\begingroup$ @copper.hat If I substitute $x=c \cos(t)$ and $y=c \sin(t)$ in $x'=c^2y$, I get $c^2=-1$ which means c is no longer an arbitrary constant. $\endgroup$ Dec 5 '21 at 5:56
  • 1
    $\begingroup$ What if the solution you proposed was incorrect but $x=c\sin c^2t$, $y=c\cos c^2t$ worked out fine? (spoiler: it does). The system of ODEs had an orientation that was not respected in your choice of parameterization. $\endgroup$ Dec 5 '21 at 6:01
  • 2
    $\begingroup$ @HARVEERRAWAT That is one solution. $\endgroup$
    – copper.hat
    Dec 5 '21 at 6:07
1
$\begingroup$

You can assemble the two equations into a complex-scalar equation as $$ \dot z=-i|z|^2z,~~~z=x+iy. $$ As observed, the radius has then the equation $$ r\dot r=\frac12\frac{d}{dt}|z|^2=Re(\bar z\dot z)=Re(-i)|z|^4=0, $$ giving $r=c$ constant. Then the original equation has a simple linear form $$ \dot z=\lambda z,~~~\lambda =-ic^2\\ \implies z(t)=z_0e^{-ic^2t},~~~|z_0|=c\text{ or }z_0=ce^{i\phi}\\ \implies x(t)+iy(t)=c\,\Bigl(\cos(c^2t-\phi)-i\sin(c^2t-\phi)\Bigr) $$

$\endgroup$
1
  • $\begingroup$ This is a nice answer Lutz +1 $\endgroup$ Dec 5 '21 at 20:47

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.