# On asymptotically stable systems and bounded solutions

• psie

#### psie

Homework Statement
Assume that the homogenous system ##x'=Ax## is asymptotically stable. Show that if ##b(t)## is bounded for ##t\geq t_0## then every solution of the system ##x'=Ax+b(t)## is bounded for ##t\geq t_0##.
Relevant Equations
A system ##x'=Ax## is said to be asymptotically stable iff all eigenvalues of ##A## have a negative real part. Moreover, the general solution to ##x'=Ax+b(t)## and ##x(t_0)=x_0## is given by ##x(t)=e^{tA}x_0+\int_{t_0}^te^{(t-\tau)A}b(\tau)d\tau##.
We need to show ##\lVert x(t)\rVert## is bounded. It is given that ##\lVert b(t)\rVert\leq c_1## for ##t\geq t_0##. A TA has claimed that ##\lVert e^{tA}\rVert\leq ce^{-\epsilon t}## holds for some ##\epsilon>0## and a constant ##c##, when ##t\geq0##. I have a hard time confirming this claim and I'd be grateful if anyone could comment on this. If this bound is true, then the statement in the exercise follows from the following estimates

First, for ##t\geq t_0##, \begin{align}\left\lVert \int_{t_0}^te^{(t-\tau)A}b(\tau)d\tau\right\rVert&\leq \int_{t_0}^t \left\lVert e^{(t-\tau)A}\right\rVert \left\lVert b(\tau) \right\rVert d\tau \nonumber \\
&\leq c\cdot c_1 \int_{t_0}^te^{-\epsilon(t-\tau)}d\tau \nonumber \\
& =c\cdot c_1\left(\frac1{\epsilon}-\frac{e^{-\epsilon(t-t_0)}}{\epsilon}\right) \nonumber \\
&\leq \frac{c\cdot c_1}{\epsilon} \nonumber\\
& =\frac{C}{\epsilon}
\nonumber\end{align}

Second, for ##t\geq t_0##, $$\lVert e^{tA}x_0\rVert\leq ce^{-\epsilon t}\lVert x_0\rVert \leq ce^{-\epsilon t_0} \lVert x_0\rVert=D$$

Finally then, ##\lVert x(t)\rVert## must be bounded by ##\frac{C}{\epsilon}+D## when ##t\geq t_0##.

But why does ##\lVert e^{tA}\rVert\leq ce^{-\epsilon t}## hold? I know that every element in ##e^{tA}## is a linear combination of terms of the form ##t^je^{\lambda t}##, where ##\lambda## is an eigenvalue of ##A## and ##j## is less than the multiplicity of that eigenvalue. Moreover, I know of ##\lVert e^{A}\rVert\leq e^{\lVert A\rVert}##, but I don't know if this is helpful.

Also, I have assumed in my computations that ##t_0\geq 0##. I guess it makes no sense for ##t_0<0##, right?

nuuskur said:
If ##A## is a Jordan block, then ##\|\exp(At)\|\to 0## as ##t\to\infty##. More specifically, if we write ##A=\lambda I_n + N##, where ##N## is nilpotent, then ##\exp (At) = \exp (\lambda I_nt)\exp (Nt)##. In particular ##\exp(Nt) = \sum _{k=0}^M \frac{(Nt)^k}{k!}##, where ##N^{M+1}=0## due to nilpotency. Putting ##\lambda = a+bi##, where by assumption ##a<0## it follows that
$$\left\|\exp(At)\right\| = \left\|e^{at}\exp(Nt)\right\| =: Ce^{at}.$$
In general, assume ##A## is in its Jordan normal form and since there are finitely many blocks, you can take maximums.
Interesting. I am not too familiar with this decomposition, and Wikipedia is not so helpful, but we have as many Jordan blocks as there are distinct eigenvalues, right?

My understanding is also that when we take the matrix exponential of a matrix in Jordan normal form, then this is simply the matrix exponential applied to each block. However, how do I compute the norm for a matrix in Jordan normal form?

Lastly, why does ##\lVert\exp(Nt)\rVert## evaluate to a constant? Shouldn't it depend on ##t##?

• nuuskur
Correct, it does depend on ##t##. I was getting ahead of myself. It holds that
##\|e^{-\delta t}\exp(Nt)\| \to 0## as ##t\to\infty## for any ##\delta >0##. Indeed we see that
\begin{align*}
\|e^{-\delta t}\exp(Nt)\| = \left\|e^{-\delta t}\sum _{k=0}^M\frac{N^k}{k!}t^k\right\| \leqslant e^{-\delta t}\sum _{k=0}^M \frac{\|N\|^k}{k!}t^k \xrightarrow[t\to\infty]{}0
\end{align*}
because polynomial grows slower than exponential goes to zero. So we have boundedness ##\|e^{-\delta t}\exp(Nt)\| \leqslant C##.

Let ##A = \lambda I_n + N## be a Jordan block, where ##\lambda = a+bi## (and ##a<0##).
\begin{align*}
\left\|\exp(At)\right\| = \left\|e^{a t}\exp(Nt)\right\| = \left\|e^{(a+\delta - \delta)t}\exp(Nt)\right\| = e^{(a+\delta)t}\left\|e^{-\delta t}\exp(Nt)\right\| \leqslant C e^{(a+\delta)t }.
\end{align*}

In general, assume ##A## is in its Jordan normal form. There are finitely many blocks, each corresponding to respective eigenvalue ##\lambda _k := a_k+ b_ki##. Pick ##\delta>0## sufficiently small such that ## \max a_k + \delta \leqslant -\varepsilon## for some ##\varepsilon >0##.

---

Regarding exponential of Jordan normal form. Yes, if ##J## is Jordan matrix, then ##J= \bigoplus J_k##, where the ##J_k## are Jordan blocks. So ##\exp(Jt) = \bigoplus \exp(J_kt)##. I confess I haven't tried computing (2-)norm of a Jordan matrix. I know that spectral radius is a lower bound, but that's about all I can say.

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• psie
nuuskur said:
So we have boundedness ##\|e^{-\delta t}\exp(Nt)\| \leqslant C##.
Thanks. Do we have boundedness for all ##t## or only for some ##t##?

This bound applies for all ##t\geqslant 0##. If ##t_0## is sufficiently large, then ##\|e^{-\delta t}\exp(Nt)\|\leqslant \varepsilon ## for ##t>t_0## and ##\|e^{-\delta t}\exp(Nt)\|## is continuous (because it's a composition of continuous maps) on any closed interval ##[0,t_0]##.

edit: I should mention the ##C## is a bound for this particular Jordan block. But again, you can take the maximum of these bounds to obtain a bound for the entire matrix.

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• psie