If $\|u\|^2 = c^2$, then $\|\dot x\|^2 \to a $ for $\ddot x = -\dot x + u(t)$?

Question

I am dealing with the next simple equation

$$ \ddot x = -\dot x + u(t), $$ where $u, x\in\mathbb{R}^m$, with $m \geq 1$, and I am wondering if for $\|u\|^2 = c^2 > 0$ then $\|\dot x\|^2\to a$, where $a\in\mathbb R^+$.

For the trivial case of having $m = 1$ it is clear, however I do not know whether is true for $m > 1$. Any hints or references?

Edit

The next computations show me a bound for $||\dot x(t)||$, but I do not know how to show that it converges to a constant (if this is indeed the case).

Take $y = \dot x$, then

$$ y(t) = e^{-t}y(0)+\int_0^\tau e^{-I(t-\tau)}u(\tau)d\tau \\ |y(t)| \leq e^{-t}|y(0)|+\int_0^\tau \left|e^{-I(t-\tau)}\right| \, \left|u(\tau)\right|d\tau \\ |y(t)| \leq e^{-t}|y(0)|+ \sup (|u(\sigma)|)_{0\leq\sigma\leq\tau} \int_0^\tau \left|e^{-I(t-\tau)}\right| d\tau, $$ and knowing that $|u(t)| = c, \forall t$, then

$$ |y(t)| \leq e^{-t}|y(0)| + c \\ $$ Then if I take the limit $t\to\infty$ in the before expression. I can conclude that $|y(t)| \leq c$, so $|\dot x(t)| \leq c$ when $t\to\infty$.

How to continue from here? I suspect that it is related with energy. If we are pumping constant energy and the system can only disipate certain energy per unit of time, they will reach an equilibrium in the sense of energy, i.e. $\dot x(t)$ might not be constant but its norm yes.

Answer 1

If $u$ represents an external force that is periodic, we can expect that forced oscillation occurs. In other words, we expect an asymptotically periodic solution of $\dot{x}(t)$ in this case. So there are plenty of rooms to create asymptotic periodicity of $|\dot{x}(t)|$ as well.

Indeed, consider $m = 2$ and identify $(\Bbb{R}^2, \| \cdot \|) \simeq (\Bbb{C}, |\cdot|)$. Let $k > 0$ and consider

$$ u(t) = \exp(ik\sin t). $$

Then we have

\begin{align*} \dot{x}(t) &= e^{-t}\dot{x}(0) + \int_{0}^{t} e^{-\tau}u(t-\tau) \, d\tau \\ &= e^{-t}\dot{x}(0) + \int_{0}^{t} \exp(-\tau + ik\sin(t-\tau)) \, d\tau \\ &= \int_{0}^{\infty} \exp(-\tau + ik\sin(t-\tau)) \, d\tau + \mathcal{O}(e^{-t}). \end{align*}

Let $f : [0, \infty) \to \Bbb{C}$ be defined by

$$f(t) = \int_{0}^{\infty} \exp(-\tau + ik\sin(t-\tau)) \, d\tau. $$

Then $f$ is $2\pi$-periodic. So it suffices to find two values $t_1$ and $t_2$ such that $|f(t_1)| \neq |f(t_2)|$. This can be achieved by taking some appropriate choices of $t_1$, $t_2$ and performing numerical integration.

The trajectory of $f(t)$ for $k = 6$ is given below:

Answer 2

What you are describing is BIBO (Bounded Input Bounded Output) stability if you select $y = \dot{x}$. It is well-known that if a LTI system is asymptotically stable, then it is also BIBO stable. Since your system is clearly asymptotically stable ($A = -I$), your statement is true.

Edit: It is asymptotically stable for all $m$. The solution is

$$\dot{x}(t) = \int_0^t e^{-I(t - \tau)} u(\tau) d \tau$$

for $x(0) = 0$. Since $u$ is bounded and $\lim_{t \to \infty} e^{-It} = 0$, this integration is finite.

Answer 3

The important property of the system under study is that the transition matrix is given by $e^{At}=e^{-t}\mathbb{I}_n$. So the ODE solution is $\dot{x}(t)=e^{-t}\dot{x}(t_0)+\int_0^t{e^{-(t-s)}u(s)ds}$. By direct calculation we then have $\|\dot{x}(t)\|^2=x^T(t)x(t)=e^{-2t}\|\dot{x}(t_0)\|^2+2e^{-t}\int_0^t{e^{-(t-s)}u^T(s)ds}\dot{x}(t_0)+{\int_0^t\int_0^t{e^{-[2t-(s_1+s_2)]}u^T(s_1)u(s_2)ds_1ds_2}}.$

The first two terms converge exponentially to zero and therefore $\lim_{t\rightarrow\infty}\bigg[\|\dot{x}(t)\|^2-{\int_0^t\int_0^t{e^{-[2t-(s_1+s_2)]}u^T(s_1)u(s_2)ds_1ds_2}}\bigg]=0 $

but I cannot see an obvious reason for $\lim_{t\rightarrow\infty}{\int_0^t\int_0^t{e^{-[2t-(s_1+s_2)]}u^T(s_1)u(s_2)ds_1ds_2}}=a$.