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I am stuck on this question:

$$x'(t) = x(x-p(t))(x-1),$$ where $p$ is $1$-periodic and $0< p(t) <1$ for all $t$. We want to show the existence of a STABLE periodic solution $\mu$ such that $0< \mu(t) <1$ for all $t$.

If we require $0< x_0 <1$, where we impose the initial condition that $x(0) = x_0$, then it is clear that there exists a solution $\mu$ that satisfies the ODE with this initial condition. Standard result also shows the existence of such periodic solutions. My question is how to show the stability of such solutions by the definition of stable periodic solutions.

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    $\begingroup$ Looks to me like: (i) If $x(t) \in [p(t), 1)$ then $x'(t) \leq 0$. (ii) If $x(t) \in (0, p(t)]$ then $x'(t) \geq 0$. $\endgroup$ – Michael Nov 6 '15 at 4:18
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    $\begingroup$ So in particular of there are constants $p_{min}, p_{max}$ such that $0 < p_{min} \leq p(t) \leq p_{max} < 1$ for all $t$, then $p_{min} \leq x(t) \leq p_{max}$ for all $t$, provided this holds for the initial condition $x(0)$. That seems pretty stable to me. $\endgroup$ – Michael Nov 6 '15 at 4:42
  • $\begingroup$ @Michael Usually it's meant to be (asymptotic) Lyapunov stability, but your variant is not close to any of these two variants. $\endgroup$ – Evgeny Nov 6 '15 at 14:34
  • $\begingroup$ @Evgeny : Well now, I solved what I could. Would be interested in seeing a more detailed analysis that says solutions that start within $\epsilon$ of the 1-periodic solution stay within that distance, for all sufficiently small $\epsilon>0$. This seems true in the special case $p(t)=constant$ for all $t$. Also, I think if $y(t)=x^*(t)+g(t)$, where $x^*(t)$ is the 1-periodic solution, you get an ODE $g'(t) = -R(t)g(t)$, where $R(t)$ is nonnegative if $g(0)$ is sufficiently small and if $1/2<x^*(t)< 2/3$ for all $t$, though there is no reason for $x(t)$ to satisfy that. $\endgroup$ – Michael Nov 6 '15 at 20:27
  • $\begingroup$ I suppose another observation is that $|x'(t)| \leq 1/4$ for all $t$, so $|x^*(t)-x^*(0)| \leq 1/8$ for all $t$ (where $x^*(t)$ is any 1-periodic solution). $\endgroup$ – Michael Nov 6 '15 at 20:28
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Okay, so here is a proof that $f$ is continuous (which takes care of a detail in Evgeny’s proof of existence of a fixed point of $f$).

Setup:

Recall the ODE is: $$ x’ = x(x-p)(x-1) $$ Assume $p$ is integrable and there are values $p_{min}, p_{max}$ such that for all $t$ we have $$ 0<p_{min}\leq p(t) \leq p_{max} < 1 $$
Further impose any additional mild assumptions so that solutions with a given initial condition exist, are unique, and do not cross (perhaps we need $p$ continuous for that?)

Let $a,b$ be values such that $0<a<p_{min}<p_{max}<b<1$. Define Evgeny’s function $f:[a,b]\rightarrow\mathbb{R}$ such that for each $z \in [a,b]$, the value $f(z)$ is defined as $x(1)$ for a solution $x(t)$ to the ODE with initial condition $x(0)=z$. We want to show $f$ is continuous over $[a,b]$.

Proof:

Fix $z \in [a,b]$. Fix a small value $\delta \neq 0$ such that $z+\delta \in [a,b]$. We want to show that $|f(z+\delta)-f(z)| \leq |\delta|D$ for some positive constant $D$. Define $x(t)$ as the ODE solution with $x(0)=z$. Define $y(t)$ as the ODE solution with $y(0)=z+\delta$. Define $g(t) = y(t) - x(t)$, so that $g(0) = \delta$. Since $y=x+g$ we get: $$ y’ = x’ + g’ = x(x-p)(x-1) + g’ $$ On the other hand, since $y$ satisfies the ODE we get: $$ y’ = y(y-p)(y-1) = (x+g)(x+g-p)(x+g-1) $$ Equating the above two expressions gives: $$ x(x-p)(x-1) + g’ = (x+g)(x+g-p)(x+g-1) $$ and so: $$ g’ = y’ - x’ = g\underbrace{[g^2 - gp + 3gx - g - 2px + p + 3x^2 -2x]}_{R(t)} $$ That is, for all $t$ we have $ g’(t) = g(t)R(t) $. Now, we already know (from above comments) that any solution to the ODE with initial condition in $[a,b]$ must stay in $[a,b]$. Thus, $x,y, g$ are bounded for all time, and there is some positive constant $C$ such that $|R(t)| \leq C$ for all $t$. For all $t$ we have $g(t) \neq 0$ (since $\delta\neq 0$ and solutions cannot cross). Thus: $$ \frac{g’(t)}{g(t)} = R(t)$$ So for all $t \geq 0$: $$ \log[|g(t)|/|g(0)|] = \int_0^t R(\tau)d\tau $$ So $$ |g(t)| = |g(0)|e^{\int_0^t R(\tau)d\tau} = |\delta| e^{\int_0^tR(\tau)d\tau} \leq |\delta| e^{Ct} $$ Thus: $$ |f(z+\delta) - f(z)| = |y(1) - x(1)| = |g(1)| \leq |\delta| e^{C} $$ $\Box$

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  • $\begingroup$ Aha, an unexplained downvote several weeks after the answer. Not a very helpful or mathematical thing to do. $\endgroup$ – Michael Nov 21 '15 at 18:58
  • $\begingroup$ This downvote is mine. Although everything is correct here and I like the idea that you've used, in my opinion your answer doesn't serve the purpose. The fact that ODE solutions continuously depend on time and initial conditions is fundamental and it is presented in all more or less serious ODE courses. The continuity of mapping $f$ simply follows from this theorem. Again, in my opinion, I don't see why your proof of this fact can be counted as an answer to original question of stability. Hence the downvote. $\endgroup$ – Evgeny Dec 12 '15 at 16:34
  • $\begingroup$ @Evgeny : You have an interesting way to show your appreciation for my help. $\endgroup$ – Michael Dec 12 '15 at 17:41
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Well, let me explain my idea in full extent. First of all, instead of considering first order ODE, I consider an equivalent system of ODEs:

$$ \dot{x} = x(x - p(\tau))(x - 1), \;\; \dot{\tau} = 1 . $$

It's easy to verify that integral curves of original ODE are the same as integral curves of this system. Also, from this form we get that vector field that is defined by this system is 1-periodic in $\tau$: if $(\tilde{x}(t), t)$ is a solution then $(\tilde{x}(t), t + k)$ is also a solution for any $k \in \mathbb{Z}$. This symmetry naturally allows us to consider behaviour of trajectories only in strip $(t, x) \in \lbrack 0, 1 \rbrack \times (-\infty, +\infty)$ because behaviour of any full trajectory can be "glued" using the "data" from the strip.

From here I'll assume that $p(\tau)$ is a continuous function. This allows us to use two theorems immediately:
1. Existence and uniqueness theorem. The uniqueness part simply means that integral curves either coincide or have empty intersection.
2. Denote by $w(t; w_0) = (\tau(t), x(t))$ the solution that satisfies IVP $w(0; w_0) = w_0 = (\tau_0, x_0)$. Then function $F(t, u) = w(t; u)$ is continuous w.r.t. $t$ and $u$.

The purpose of the following part is two-fold: it proves that there is a 1-periodic solution and it shows that it is asymptotically Lyapunov stable. I don't know what kind of theorem you used for showing that there is a periodic motion, but I believe that my explanation will be quite simple. Here's an illustration:

Cross-section

As I've said earlier, choose $a$ and $b$ in such way that $0 < a < p_{min} < p_{max} < b < 1$ holds. You can simply check that vector field points along boundary in the same way as on the figure. Such configuration allows us to tell that all trajectories that start at $(\tau, x) \in \lbrace 0 \rbrace \times \lbrack a, b \rbrack$ can leave $(\tau, x) \in \lbrack 0, 1 \rbrack \times \lbrack a, b \rbrack$ only through $(\tau, x) \in \lbrace 1 \rbrace \times \lbrack a, b \rbrack$. To be absolutely precious, you might check vector field at points $(1, a)$ and $(1, b)$, see that trajectories that start at them leave $(\tau, x) \in \lbrack 0, 1 \rbrack \times \lbrack a, b \rbrack$ in backward time and conclude that all trajectories escape strip only through $(\tau, x) \in \lbrace 1 \rbrace \times ( a, b )$.

It's time to define "time-1 mapping". We define it by formula

$$ f(x) = F(1, (0, x)).$$

This is a continuous mapping of segment $\lbrack a, b \rbrack$ into itself. If we were interested only in proving existence of periodic solution (i.e., which has $f(x) = x$), we could use a Brouwer fixed-point theorem (which states that such fixed point exists). But we need to investigate the stability and we can use simpler arguments. Note that $f(a) > a$ and $f(b) < b$, so $f(a) - a > 0$ and $f(b) - b < 0$; because $f$ is continuous there exists a point $\hat{x}$ where $f(\hat{x}) - \hat{x} = 0$. Therefore, we proved that $f(x)$ has fixed point and our system of ODEs has a 1-periodic solution.

But we can prove more! Recall that I've said that in this case $f(x)$ is monotonically increasing function. It's very simple to prove. Pick any point $(0, x^\ast)$ and consider the integral curve that is passing through this point. As we already know, it will intersect line $\tau = 1$ at point $(1, f(x^\ast)$. Now consider the closed domain $G$ which is bounded by lines $\gamma$, $\lbrace 0 \rbrace \times \lbrack a, x^\ast \rbrack$, $\lbrack 0, 1 \rbrack \times \lbrace a \rbrace$ and $\lbrace 1 \rbrace \times \lbrack a, f(x^\ast) \rbrack$. By construction there is only one way to escape $G$ -- only through the $\lbrace 1 \rbrace \times \lbrack a, f(x^\ast) \rbrack$ (because escaping through $\gamma$ will violate uniqueness theorem). But this simply means that for any $x \in (a, x^\ast)$ follows $f(x) < f(x^\ast)$. And it holds for any $x^\ast \in (a, b \rbrack$, which simply means that $f(x)$ is increasing function.

But what does that all mean for stability of periodic motion? Here is another illustration:

Cobweb diagram

It's a cobweb diagram for our mapping $f$. It's easy to prove that iterations of any point from segment $\lbrack a, b \rbrack$ converge to fixed point. It means that after each period any other solution "starts" at cross-section closer and closer to periodic solution. Using some $\epsilon$-$\delta$ machinery you can prove that periodic solution is asymptotic Lyapunov stable, but we already have the key fact that fixed point on cross-section is attractive. After that the proof should be quite simple.

EDIT

As @Michael pointed out, my illustration is not the only possibility for cobweb diagram. There could be other fixed points as well: at some of them $y = f(x)$ might touch $y=x$ and at some of them they might be intersecting transversely. The key observation is that there should exist a fixed point $\hat{x}$ such that $f(x) > x$ for $x \in \lbrack \hat{x}-\varepsilon, \hat{x} )$ and $f(x) < x$ for $x \in ( \hat{x}, \hat{x} +\varepsilon \rbrack$. Why such fixed point exists? It's due to these facts:
1) $y = f(x)$ starts in the $y > x$ part of square $\lbrack a, b \rbrack \times \lbrack a, b \rbrack$ and ends in the $y < x$ part;
2) we definitely have fixed points for $y = f(x)$.

By reductio ad absurdum we could show that if we have no such fixed points, then curve $y = f(x)$ stays only in $y \geqslant x$ part of square which contradicts the first fact.

The fixed point with such properties would be attractive and will correspond to the stable periodic solution.

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  • $\begingroup$ Your comments and above answer suggest this is true at every point $x^*$ such that $x^*=f(x^*)$, but your cobweb picture need not apply to all such cases. For example, the red curve might be under the black on the left-hand-side, in which case the point $x^*$ is unstable. If there are many crossings and many points $x_i$ such that $f(x_i)=x_i$ you might argue that there is one with the above picture, but there are also stranger possibilities of infinite crossings with "infinitessimal" distances between crossings. $\endgroup$ – Michael Nov 8 '15 at 22:06
  • $\begingroup$ @Michael why are you pointing to the general case when this cobweb diagram is made specially for this time-1 mapping? It represents $f(x)$ that I've defined and nothing more. $\endgroup$ – Evgeny Nov 9 '15 at 7:07
  • $\begingroup$ Evgeny: I do not see the purpose of your comment above. Overall, it seems you are unintentionally (or intentionally?) claiming that for all points $x^*$ that satisfy $f(x^*)=x^*$ in your $f$ function, we must also have $f(x^*-\epsilon)\geq f(x^*) - \epsilon$ and $f(x^*+\epsilon) \leq f(x^*)+\epsilon$ for all sufficiently small $\epsilon>0$. $\endgroup$ – Michael Nov 9 '15 at 19:02
  • $\begingroup$ I don't understand what are you trying to show that I claim. I've listed all properties of function $f(x)$ that I use and gave an explanation why they appear here. What am I unintentionally claiming? $\endgroup$ – Evgeny Nov 9 '15 at 19:48
  • $\begingroup$ Why does the (second) red curve look that way? Might it not have many such points $f(x_i)=x_i$? Might it not "kiss" the 45 degree line several times before crossing? All of these are possible and consistent with $f$ increasing. $\endgroup$ – Michael Nov 9 '15 at 19:59

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