Suppose $p(t)>2$ and is continuous for all $t\in\Bbb R$, $$x'=2y,\\ y'=-2x-p(t)y^3,$$ prove that for each solution $(x(t),y(t))$ there exists a point $(x^*,0)$ to which it converges.

I guess the most probable approach is Lyapunov stability theory, which is within the range of my ODE course. But this is a non-autonomous system (with time-varying input $p(t)$), and all I have learned from class is criteria for stability for autonomous systems.

It seems kinda obvious that $x^*$ is relevant to $p(t)$, but I have trouble even proving the existence of a singular point $(x^*,0)$ for this system, let alone dealing with it analytically.

The case $p(t)\equiv\text{const}$ is easy, with $(0,0)$ apparently being the only singular point which is also apparently globally Lyapunov asymptotically stable on $\Bbb R^2$, and hence globally attracting (meaning all solutions converge to this singular point.)

But for a time-varying input it is entirely different. Maybe I'll need a non-autonomous version of criterion for Lyapunov stability? But since this is beyond my course's level I think it there must be some more elementary alternatives. I'd be very grateful if anybody can provide me with one. (Of course if using non-autonomous versions can't be helped I would also be glad to learn about such tricks as long as they are effective.)

EDIT Terribly sorry. I made a mistake. $(x^*,0)$ may be dependent upon the solution (or the initial condition).

  • $\begingroup$ Note that $x(t) \equiv y(t) \equiv 0$ is still a solution in the non-autonomous case. $\endgroup$ – Evgeny Jan 4 '16 at 16:29
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    $\begingroup$ As far as I see, linearization has pair or purely imaginary eigenvalues, so Lyapunov First Thm isn't applicable... Function $V(x, y) = x^2+y^2$ seems to be a nice candidate, but it requires few workarounds. $\endgroup$ – Evgeny Jan 4 '16 at 17:01
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    $\begingroup$ Well, to some extent you are right :) but if you were told about extended phase space, then you can draw new picture: $(0, 0, t)$ is a line in 3dimensional extended phase space which corresponds to equilibrium point of autonomous system. Level sets of $V(x, y) = x^2+y^2$ are cylinders that surround $(0, 0, t)$. If you take a look at vector field at any cylinder it is either tangent to it or "goes inside". And thus all trajectories come closer and closer to this line. $\endgroup$ – Evgeny Jan 4 '16 at 17:39
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    $\begingroup$ This condition $p(t) > 2$ was bugging me too. I thought that maybe if we find more clever Lyapunov function for constant case, we'll get condition this condition for $p$. But no, we are still using $V(x, y) = x^2+y^2$ and I have no better idea. Also I understand that it's kinda strange that we still see only line $(0, 0, t)$ as a candidate for limit set, but our "Lyapunov" functions suggests it quite convincing. $\endgroup$ – Evgeny Jan 5 '16 at 6:53
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    $\begingroup$ @Evgeny not yet. I'm not with my computer now. (Yeah it does go inside any cylinder that it can reach, but there may be one that it will never reach, like a limit cylinder. ) $\endgroup$ – Vim Jan 6 '16 at 10:26

Multiplying the second equation by $y$ and using the first equation we have $$y\dot{y}=-2xy-p(t)y^4=-x\dot{x}-p(t)y^4$$ i.e. as correctly pointed in the comments by @Evgeny $$\frac{d}{dt}[x^2+y^2]=-2p(t)y^4\leq -4 y^4$$

Thus $x,y$ are bounded and the Lyapunov-like function $V=x^2+y^2$ is decreasing and lower bounded (from zero). Hence, $V$ converges to some constant $V_{\infty}\geq 0$. If we integrate the above inequality over $[0,\infty)$ we obtain $$V_{\infty}-V(0)\leq -4\int_0^{\infty}{y^4(s)ds}$$ which yields $$\int_0^{\infty}{y^4(s)ds}\leq \frac{1}{4}V(0)$$

If we assume a bounded $p(t)$ then the boundedness of $x,y$ and the state equations result in the boundedness of $\dot{x},\dot{y}$.

A continuous differentiable function with bounded derivative is uniformly continuous.

You can use now Barbalat's lemma to prove convergence. Barbalat's lemma states that if

i) $\int_0^{\infty}{\phi(t)dt}$ exists and is finite and

ii) $\phi(\cdot)$ is uniformly continuous

then $\lim_{t\rightarrow\infty}\phi(t)= 0$.

From Barbalat lemma we then have that $\lim_{t\rightarrow\infty}y(t)=0$ and since $V$ converges there also exists some $x^*$ (with $V_{\infty}={x^*}^2$) such that $\lim_{t\rightarrow\infty}x(t)=x^*$.

Edit to remove the upper bounded $p(t)$ restriction: I will prove now a variation of Barbalat lemma that does not need uniform continuity of $\phi(\cdot)$ but only an upper bounded derivative.

Barbalat lemma variation: Let $\phi:\mathbb{R}\rightarrow\mathbb{R}_+$ continuous differentiable nonnegative function. If

i) $\int_0^{\infty}{\phi(t)dt}$ exists and is finite and

ii) $\dot{\phi}(\cdot)$ is upper bounded

then $\lim_{t\rightarrow\infty}\phi(t)= 0$.

Proof: The proof uses a contradiction argument. Assume the opposite, then there exists a constant $k_1>0$ and a sequence of times $\{T_i\}$ with $\lim_{i\rightarrow\infty}T_i=\infty$ such that $$\phi(T_i)\geq k_1$$ For $t\leq T_i$ we have from the mean value theorem that $$\phi(t)=\phi(T_i)-\dot{\phi}(\theta t +(1-\theta)T_i)(T_i-t)$$ for some $\theta\in(0,1)$. Since $\dot{\phi}$ is upper bounded there exists some $c$ such that $\dot{\phi}(t)\leq c$ for all $t\geq 0$ and $$\phi(t)\geq \phi(T_i)-c(T_i-t)\qquad \forall t\leq T_i$$ From the above relationship we have that $$\phi(t)\geq \frac{k_1}{2}\qquad \forall t\in\left[T_i-\frac{k_1}{2|c|},T_i\right]$$ Therefore $$\int_{T_i-\frac{k_1}{2|c|}}^{T_i}{\phi(t)dt}\geq \frac{k_1^2}{4|c|} $$ Thus $\int_0^{t}{\phi(\tau)d\tau}$ cannot converge to a finite limit as $t\rightarrow \infty$ which is the desired contradiction.

In our example $\phi(t):=y^4$ and its derivative is upper bounded since $$\dot{\phi}(t)=4y^3\dot{y}=-8xy^3-4p(t)y^6\leq -8xy^3$$ and $x,y$ are bounded. The proposed variation of Barbalat lemma yields now the desired $\lim_{t\rightarrow\infty}y(t)=0$. This completes the proof and indeed there is no need for an upper bound of $p(t)$.

  • $\begingroup$ Point is, we don't know that $p$ is bounded. $\endgroup$ – Vim Jan 6 '16 at 1:26
  • $\begingroup$ @Vim I have edited my answer to consider the general case. The proof is completed by a modification of Barbalat lemma. $\endgroup$ – RTJ Jan 6 '16 at 10:41

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