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Given $\ddot{x} = F(x)$, $x(0) = x_0$ and $\dot{x}(0) = y_0$. Assume $c$ is a simple zero of $F$. Let $V(x) = -\int_{x_0}^{x} F(s)ds$.

(a) If $V$ has a local min at $c$, find the stability and type of the equilibrium (i.e, saddle, focus, center, center-focus, other) at the point $(x_0, y_0) = (c, 0)$. Use $E(x) = \frac{\dot{x}^{2}}{2} + V(x)$ to explain your answer. What is the stability and type if $V$ has a local max. Explain?

(b) Find the type and stability of the equilibria of $\ \ddot{x} = -4x(1-x^2)$.

(c) Find the type and stability of $x = \dot{x} = 0$ for the equation: $\ \ddot{x} = -4x(1-x^2) - x^2\dot{x}$

My attempt: For part (a), let $\dot{x} = y$. It's easy to see that we can rewrite the 2nd-order ODE as the system of 1st-order ODE: $\dot{x} = y, \dot{y} = F(x)$, and $E(x,y) = \frac{y^2}{2} + V(x)$.

Now, since $\frac{\partial E}{\partial y} = \dot{x}$, and $-\frac{\partial E}{\partial x} = \dot{y}$, our system is Hamiltonian system. Thus, all the solutions $(x(t), y(t))$ belongs to the level sets $E(x,y)= C$ for any constant $C > 0$. Now, since $V(x)$ has a local min at $c$, by Fundamental Theorem of Calculus, $F(c)=0$, and $F'(c)< 0$ (since $c$ is a simple zero of $F$, so $c$ is a strict local min). So the point $(c,0)$ is an equilibrium of our system. By principle of linearization, since $Df((c,0)) = (0 1, F'(c) 0)^T$, which has 2 real eigenvalues $\pm \sqrt{F'(c)}$, we conclude that $(c,0)$ is a hyperbolic equilibrium, and it is an unstable saddle.

For the case when $V$ has local max at $c$, we have $F'(c) < 0$, so $Df((c,0))$ would have 2 imaginary eigenvalues with real parts $= 0$, so $(c,0)$ is not hyperbolic equilibrium, thus Principle of Linearized Stability cannot apply in this case. However, as we know that $(c,0)$ would be on some level sets $E(x,y) = C_1$ for some $C_1$, $(c,0)$ is stable. In addition, due to the 2 eigenvalues are all imaginary, $(c,0)$ can't be saddle, focus, center or center-focus, so it must be OTHER.

Is my solution correct for this part?

Part (b) is quite simple by using part (a)'s result (only need to check whether $0$, $1$ or $-1$ is a local min or max of F(x), and choose $y_0=0$), so I omitted the proof here. For part (c), it's quite hard, as we have the term $x^2\dot{x}$, so I cannot use the result above as the RHS is not purely in terms of $x$. Principle of Linearization also does not work in this case, as $Df((0,0))$ has $2$ eigenvalues $\pm 2i$.

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    $\begingroup$ And?$\quad\quad$ $\endgroup$
    – Artem
    Commented Dec 30, 2015 at 19:29
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    $\begingroup$ The definition of $V$ is inconsistent with its use in the supposedly first integral $E$. $\endgroup$ Commented Dec 30, 2015 at 20:03
  • $\begingroup$ @LutzL: I fixed my typo! Can you help with part (c)? $\endgroup$
    – ghjk
    Commented Dec 31, 2015 at 2:30
  • $\begingroup$ @Artem: I added my attempted solution. Please help review my solution for part (a), and help with part (c) if you can. $\endgroup$
    – ghjk
    Commented Dec 31, 2015 at 2:31
  • $\begingroup$ @Evgeny: can you try doing part (c) to help me out:) $\endgroup$
    – ghjk
    Commented Dec 31, 2015 at 6:56

1 Answer 1

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For part c), you can still multiply with $\dot x$ and integrate to obtain $$ E(x,\dot x)=\frac12\dot x^2-(1-x^2)^2=E(x_0,\dot x_0)-\int_{t_0}^t x^2\dot x^2\,dt $$ which tells you that the energy will be reduced with time.

Thus the solution will traverse the level sets of $E$ towards a minimum of $E$.

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  • $\begingroup$ what a solution!! Thanks a lot for your help, Lutzl:) So $(x_0, \dot{x_0}) = (0,0)$ is a stable saddle? Btw, can you please help let me know if my solution to part (a) above is indeed correct? I'm not really sure why we need the "simple condition" on a root $c$ of $F$. $\endgroup$
    – ghjk
    Commented Dec 31, 2015 at 7:37
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    $\begingroup$ with non-simple roots you get eigenvalues $0$ making the case undecided. -- A stable point is not a saddle, a saddle has stable and unstable directions. One might call it a sink (as opposed to a source). -- a) looks good, thus no comment on it. $\endgroup$ Commented Dec 31, 2015 at 8:21
  • $\begingroup$ thank you for your answer:) But even in this case, we still get eigenvalues $0$ for $Df(0,0)$. Do you mean that with the simple root, we would get a STRICT local min/max? So for part (c), is the reason you call $(0,0)$ a sink (in other words, a stable FOCUS) due to the fact that all solutions would converge to $(0,0)$? I personally think the equilibrium $(0,0)$ is a stable OTHER (i.e, it's NOT saddle, center, center focus, etc) Sorry but I have a hard time classify the type of equilibrium for points on the level sets:( $\endgroup$
    – ghjk
    Commented Dec 31, 2015 at 20:07

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