Show that every equation of the form $$ax'' + b(x^2 - 1) x' + cx = 0$$ where $a, b, c > 0$ can be transformed into a van der Pol equation by a change in the independent variable.

I am unable to find this replacement. If anyone could help me or give a hint I would be grateful.

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    $\begingroup$ Maybe you could include Van der Pol's equation in the post. $\endgroup$ – coffeemath Jul 5 '14 at 23:21
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    $\begingroup$ Van Der Pol's equation is $x''-\mu(1-x^2)x'+x=0$ @coffeemath $\endgroup$ – ClassicStyle Jul 5 '14 at 23:51

By dividing through by $c$ and pulling a negative out of the $x^2$ term we get the equation:

$$\frac acx''-\frac bc(1-x^2)x'+x=0 $$

Now, let the dependent variable (usually $t$) be: $t=\sqrt{\frac ca} u$.

So now the derivatives via the chain rule will be:

$$ \sqrt{\frac ca}x' $$ And $$ \frac cax'' $$

So plug this in and we see that the $a/c$ term cancels and we are left with:

$$ x''-\frac{b\sqrt c}{c\sqrt a}(1-x^2)x'+x=0 $$

Then let $\mu=\frac{b\sqrt c}{c\sqrt a} $.

$$ x''-\mu(1-x^2)x'+x=0 $$


From the variable change our functions are now $x\left(\sqrt{\frac ca} u\right)$.

The primed notation can be a little confusing sometimes, so what we really have is:

$$ \frac{d}{du}x\left(\sqrt{\frac ca} u\right)=x'\left(\sqrt{\frac ca} u\right)\frac{d}{du}\left(\sqrt{\frac ca} u\right)=\sqrt{\frac ca}x' $$

Then just repeat this process for the second derivative to obtain the $c/a$.

  • $\begingroup$ Do not quite understand how you used the chain rule there. $\endgroup$ – Croos Jul 6 '14 at 1:08
  • $\begingroup$ I'll update it shortly. Let me know if you still don't get it. $\endgroup$ – ClassicStyle Jul 6 '14 at 1:25
  • $\begingroup$ I understand, thanks for the help $\endgroup$ – Croos Jul 6 '14 at 1:53

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