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Given $\dot{x} = y, \dot{y} = -f(x)$ where $f\in C^{1}$. A point $(x_0, y_0)$ is an equilibrium iff $f(x_0) = y_0 = 0$.

(a) Find the necessary and sufficient conditions on $f'(x_0)$ for the equilibrium $(x_0 , 0)$ to be hyperbolic?

(b) Assume $f$ has only simple zeros, and there exists an equilibrium $(x_1 , 0)$ with $f'(x_1) < 0$. Also, assume that for $x_2 > x_1$, $\int_{x_1}^{x_2} f(x)dx = 0$, and $\int_{x_1}^{y} f(x)dx\neq 0$ for any $y\in (x_1, x_2)$.

Prove that there is either a homoclinic or a heteroclinic orbit which has $x_1$ as one of its endpoints. Find an additional assumption to guarantee a heteroclinic orbit?

(c) Find an example for eac of the two possibilities in part (b).

My attempt: I'm currently only able to make some minor progresses on part (a). Namely, the sufficient condition is that if none of the eigenvalues of $f'(x_0)$ lie on the imaginary axis. For the necessary condition, I think it is $f'(x_0)$ has all eigenvalues of positive real parts, but I'm not quite certain. For the latter parts, I'm completely stuck on them so far. Any help on either of these 3 parts, especially (b) and (c), would greatly be appreciated. Hope somebody can help me with this problem.

ghjk
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  • On Part (a). $f'(x_0)$ is just a number and this "linear transformation" has only one eigenvalue :) so, write down Jacobi matrix for this system and study its eigenvalues as usual. – Evgeny Nov 16 '15 at 10:37
  • On Part (b). This system has first integral: $y^2 + \int f(x) , dx = {\rm const}$. This should help you a lot. Do you see the way it helps you? – Evgeny Nov 16 '15 at 10:40
  • On Part (c). If you figured out the way in which first integral helps you to prove existence of homoclinic or heteroclinic curves, you just have to invent function $f(x)$ such that level sets of $y^2 + \int f(x) , dx$ are quite specific. – Evgeny Nov 16 '15 at 10:41
  • And you can refer here http://math.stackexchange.com/questions/1382060/trajectories-that-connect-equilibrium-points/1382284#1382284 if you can't figure out why first integral rules in such problems :) – Evgeny Nov 16 '15 at 10:44
  • @Evgeny: thanks a lot for your great hint! On part (a), I was able to derive the sufficient and necessary condition is that $f'(x_0)<0$. For art (a), I think the level set $H(x,y)$ should be $\frac{y^2}{2} + F(x) = constant$, where $F$ is an antiderivative of $f$. But from here, it means every orbits can be described as $\frac{y^2}{2} = c-F(x)$ for every $c\in R$. From the given assumption in part (b), we get 2 things: $x_1$ is another hyperbolic equilibrium point, and $F(x_1) = F(x_2)$. It means that every orbits, which must be on the level set $H(x,y)$, would move from $x_1$ to $x_2$. – ghjk Nov 16 '15 at 20:22
  • @Evgeny: at the same point on the y-axis (note that between $x_1$ and $x_2$, there doesn't exist any point $y$ for which $F(x_1) = F(y)$, so no "coming back" during those time period) for any values of $c$ and $t\rightarrow \infty$ or $-\infty$. Those orbits are either heteroclinic or homoclinic and have $x_1$ as one of its endpoints. Is this a correct argument? But for the additional assumption to guarantee a heteroclinic part, I wasn't able to figure out:p Can you help with this later part? – ghjk Nov 16 '15 at 20:34
  • For part (c), I think the example: $\dot{x} = y, \dot{y} = -2y-sin(x)$ would only have heteroclinic orbit. For the one which only has homoclinic orbit, I haven't been able to figure it out unfortunately:p – ghjk Nov 16 '15 at 20:37
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    On homoclinic part: you are right that we are moving from $(x_1, 0)$ to $(x_2, 0)$ along the level set that contains them both. But you are quietly assuming that $(x_2, 0)$ is an equilibrium too (which is not necessarily true). If you note that level set curves are symmetric w.r.t. $Ox$ axis, you can predict where we should move after we pass point $(x_2, 0)$. And I think that I've accidentally answered the question about "additional assumption" :) – Evgeny Nov 16 '15 at 23:04
  • @Evgeny: Thank you for your great hints. But I'm still quite stuck on inferring why we would have either heteroclinic or homoclinic part when moving from $(x_1, 0)$ to $(x_2, 0)$. I understand your hint, and it seems to me the solution, after passing the point $(x_2,0)$, would either come back to $(x_1,0)$ due to symmetric w.r.t. O$x$ axis, or move forward at a lower value of $y_axis$ . But how does the later guarantee the heteroclinic part? For the "additional assumption,"I think we need $f'(x_2) > 0$? Can you also help with part (c), as I'm stucked on it for a while:( – ghjk Nov 17 '15 at 03:44
  • It seems to me you haven't used ANY assumptions in part (b). Can you please present your argument so that I can see whether it's true/wrong? – ghjk Nov 17 '15 at 03:55
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    My reasoning is the following. We know that the level set of $H(x, y)$ which contains point $(x_1, 0)$ also contains point $(x_2, 0)$.There are two options here: $(x_2, 0)$ is an equilibrium too and $(x_2, 0)$ is a regular point.For point $(x_2, 0)$ being an equilibrium guarantees that we have a heteroclinic connection between $(x_1, 0)$ and $(x_2, 0)$. Because of the symmetry, one incoming and one outgoing separatrices of $(x_1, 0)$ join it with $(x_2, 0)$.If there is no equilibrium at $(x_2, 0)$ the part of level set which contains this point forms a loop and it is homoclinic connection. – Evgeny Nov 17 '15 at 07:19
  • As for part (c): non-linearized pendulum gives an example of heteroclinic connection and $H(x, y) = \frac{y^2}{2} + \frac{x^3}{3} - \frac{x^2}{2}$ gives a nice example with homoclinic connection. – Evgeny Nov 17 '15 at 09:25
  • @Evgeny: Many thanks for your great reasoning on part (b). I'm so stupid not to realize the simple condition on $(x_2,0)$ (I was close):P For part (c), I got another example for homoclinic is $H(x,y) = \frac{y^2}{4} -2q^2+q^4$ where $\dot{q} = \frac{p}{2}$, $\dot{p} = 4(q-q^3)$ – ghjk Nov 17 '15 at 15:20
  • @Evgeny: My apology, but do you know why do we need the condition $f'(x_1) < 0$ for part (b)? Since your argument doesn't require the equilibrium point $(x_1,0)$ to be HYPERBOLIC, this condition seems to be "extraneous"? – ghjk Nov 17 '15 at 17:02
  • I am not really sure, but I think that this kinda fixes possible degeneracies of level set. For hyperbolic saddle (and generic saddle-point of fucntion $H(x,y)$) this level set is "$X$-shaped". But in more degenerate case it might look much more complicated and might have more branches. I don't really know. – Evgeny Nov 17 '15 at 17:18
  • Can you give an example of a degeneracy case for the level set? – ghjk Nov 17 '15 at 18:53
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    Hm, I had another thought and I think that in this case such degeneracy is not possible. For functions of form $\frac{y^2}{2} + F(x)$ near the saddle point we always have four branches of level set. I had in mind an example when you have level set which looks like figure eight inside another figure eight. – Evgeny Nov 17 '15 at 19:07
  • Interesting. I guess the 4 branches you talked about are: 2 branches of hyperbolas with vertices on the x-axis, and 2 branches of a graph of a 4-degree polynomials, plus some eclipses centered at the origin? Do you mind giving an example of the one with figure eight inside another figure eight? – ghjk Nov 17 '15 at 19:17
  • wait a minute... I have another question: for my example of a homoclinic orbit, which is $\dot{x} = \frac{y}{2}$, and $\dot{y} = 4(x-x^3)$, we see that $(1,0)$ and $(-1,0)$ are 2 equilibrium points. But its phase portrait doesn't have ANY heteroclinic orbit between these 2 points, which contradicts to your required assumption in part (b) to guarantee a heteroclinic orbit. Something is missing here:p – ghjk Nov 17 '15 at 19:25
  • Check other condtions ;) you are trying to connect two center equilibria with heteroclinic orbits. These two centers lie in the same level set, but you forgot about other conditions. Also this level set contains only these two points and nothing more. – Evgeny Nov 17 '15 at 19:38
  • So you meant the condition $f(x_2) = 0$ is enough to guarantee the existence of a heteroclinic orbit? I did check the other conditions, but couldn't see where it violates if I choose $x_1 = -1$ (then $x_2=1$ is the only point that satisfies, and $f(x_2) = 0$ in this case as well). I don't see which condition it violates in this case. Also, in your example, you choose $x_1=1$, not $0$, right? – ghjk Nov 17 '15 at 19:59
  • $x_2 = 1$ is the only point satisfying $\int_{x_1}^{x_2} f(x) dx = 0$, and no $y\in (-1, 1)$ satisfies $\int_{x_1}^{y} f(x) dx = 0$ is what I meant. – ghjk Nov 17 '15 at 20:14
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    The key fact here is level set structure. Level set that contains points $(-1, 0)$ and $(1, 0)$ is disconnected, nothing connects these two points => no heteroclinics. – Evgeny Nov 17 '15 at 20:44
  • Oh I see now. What kind of assumption we need to impose to prevent this case from happening then? I meant, this case shows us that requiring $f(x_2) = 0$ is not quite sufficient to guarantee heteroclinic orbit, as my example shows that's not the case? – ghjk Nov 17 '15 at 20:54
  • Well, equilibria of this system are crtical points of corresponding first integral. If critical point has non-degenerate Hessian matrix, there are two opportunities: level set is one point (when Hessian matrix is positive definite) or level set is locally X-shaped (when Hessian matrix has positive and negative eigenvalue). When we are asking $f"(x_1) < 0$ we guarantee that we have a saddle at $(x_1, 0)$ and this becomes a saddle point for first integral. Hence level set "connects" points $(x_1, 0)$ and $(x_2, 0)$. – Evgeny Nov 17 '15 at 21:04
  • Sorry, I meant $f'(x_1)$ – Evgeny Nov 17 '15 at 21:54
  • We still need $f(x_2)=0$ then? But my example shows that that $f'(x_1) < 0$ is not enough to guarantee there are some orbits connecting $x_1$ and $x_2$. That's why I think some additional conditions need to be added to prevent that case of my example from happening. – ghjk Nov 17 '15 at 23:35
  • Pose your question in a clearer way. Because right now you have every fact that allows you to finish the answer by yourself. Just stick to original formulation of question, if you haven't grasp level set stuff. Also, check your calculations for my example, they are just wrong. In example with $\frac{y^2}{2} + \frac{x^3}{3} - \frac{x^2}{2}$ derivative $f'(x)$ is negative at $x = 0$ and positive at $x = 1$. – Evgeny Nov 18 '15 at 09:00
  • @Evgeny: my apology:( I kept messing up between $-f(x)$ and $f(x)$. You're definitely correct.My question is: Find a condition to guarantee we have a HETEROCLINIC orbit. Based on your answers above, I have thought through and really think $f(x_2) = 0$ is the right condition to have HETEROCLINIC orbit. Can you confirm if this is a necessary and sufficient condition to have a HETEROCLINIC orbit between equilibrium points $(x_1,0)$ and $(x_2,0)$? An example is a system $\dot{x} = y$, $\dot{y} = -(x-x^3)$, which has a HETEROCLINIC orbit between $(1,0)$ and $(-1,0)$, and $f(1) = f(-1) = 0$. – ghjk Nov 18 '15 at 15:37
  • Let's fix that $f'(x_1) <0$. Then the only additional assumption that is needed is that $f(x_2) = 0$. – Evgeny Nov 18 '15 at 17:05

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