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Consider the equation, $$ x'' + ax^2x' - x + x^3 = 0,\ x ∈ \Bbb R $$ where $0 ≤ a$ is a constant.

It asked me to find all equilibria and determine their stability/instability.

When $a = 0$, I found the conserved quantity $$E(x,y) = 2y^2 - 2x^2 + x^4$$ with $y$ being equal to $y = x'$. However when $0 ≤ a$, I don't know how to find equilibria and their stability.

Also, it asked me to prove there is no periodic orbits or homoclinic orbits and instead prove there exists a heteroclinic orbits. How to prove them?

Thanks in advance for help.

Lutz Lehmann
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James333
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2 Answers2

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With $v=x'+\frac a3 x^3$ you get the first order system $$ x' = v-\frac a3 x^3\\ v' = x-x^3 $$ This gives you the stationary points at $x_k=-1,0,1$ and $v_k=\frac13x_k^3$. For the stability the first thing to try is to compute the eigenvalues of the Jacobian, if that is not sufficient other methods need to be employed. $$ J=\pmatrix{-ax_k^2&1\\1-3x_k^2&0} $$ with characteristic equation $$0=λ(λ+ax_k^2)-(1-3x_k^2)=(λ+\tfrac12ax_k^2)^2-(1-3x_k^2+\tfrac14a^2x_k^4)$$ which allows to compute the eigenvalues and examine their real parts.

As observed, $E(x,y)=2y^2−2x^2+x^4$ is a conserved quantity of the equation without the friction term. Computing the time derivative of $E(x,x')$ for the equation with friction results in $$ \frac{d}{dt}E(x,x')=4x'(x''-x+x^3)=-4ax^2x'^2 $$ This means that any solution curve constantly intersects the level curves of $E$ in direction of its minimal values. As $E(x,y)=2y^2+(x^2-1)^2-1$, these minima can be found at $x'=0$, $x=\pm 1$.

A consequence is that there can be no homoclinic or periodic orbits, as no solution curve can return to the starting level curve of $E$. Second all solution curves outgoing from $(0,0)$ remain inside the region $E<0$, which is contained in $[-\sqrt2,\sqrt2]\times [-\sqrt{\frac12},\sqrt{\frac12}]$, so can only go to one of the stationary points at the minima of $E$ Which means they are heteroclinic orbits..

Lutz Lehmann
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  • Thank you for your answer. Then, how would I prove if there are no periodic orbits or homoclinic orbits? – James333 Dec 03 '17 at 20:27
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    @James333 Most of my answer applies to your problem as well. You only need maybe few additional tricks. In your system divergence can be zero somewhere, but has same sign everywhere else. I think that you still can use Dulac-Bendixson criterion here, they say something about zero measure set where divergence can be zero. As for homoclinics, show that one separatrix stays in some invariant domain and another one doesn't enter this domain at all for any positive moment of time. – Evgeny Dec 04 '17 at 04:33
  • In the special case $a=3$, it's actually easy to find a heteroclinic orbit explicitly: $(x,v)=(f(t),f(t))$ will do, where $f'=f-f^3$. But for $a \neq 3$ it will be more complicated, of course... – Hans Lundmark Dec 04 '17 at 08:06
  • @HansLundmark "Heteroclinic trajectory" might be a slight abuse of notation in this problem. As far as I see at least for small $a$ we have only one saddle and two foci. To me this means that they call "heteroclinic trajectory" any trajectory between two different equilibria, they don't require both of them to be saddle (and maybe definition of "heteroclinic" doesn't require that, but I'm too used to heteroclinic structures between saddles). The existence of saddle-sink or saddle-source connection is much easier to prove. – Evgeny Dec 04 '17 at 10:05
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Also, it asked me to prove there is no periodic orbits or homoclinic orbits and instead prove there exists a heteroclinic orbits. How to prove them?

Can you expand on this in more detail please?

Amit Alok
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