Stability of differential equations under nonlinear pertubations

Question

I am trying to understand the proof of the following theorem :

Let $A$ be an $n\times n$ matrix such that it has only negative eigenvalues and let $f: \mathbb{R}\times \mathbb{R}^n$ continuous and locally Lipschits in $x$. If $f(t,0)=0$ for every $t\in \mathbb{R}$ and $\lim_{x\rightarrow 0}\frac{||f(t,x)||}{||x||}=0$ for any $t\in \mathbb{R}$ then the zero solution of the equation $x'=Ax+f(t,x)$ is asymptotically stable .

Now in the proof of this, the authors claim that given $x_0\in \mathbb{R}^n$ with $||x_0||<\delta$ the solution satisfies $||x(t)||<\delta $ for any $t\in [t_0,t_1]$ sufficiently small. And then they use that fact to prove that if $||x_0||<\frac{\delta}{M}$ then $||x(t)||<\delta$ using the following \begin{align} x(t)&=e^{A(t-t_0)}x_0+\int_{t_0}^t e^{A(t-s)}f(s,x(s))ds \\ \|x(t)\|&\leq Me^{-\alpha(t-t_0)}\|x_0\|+\int_{t_0}^tMe^{-\alpha(t-s)}\epsilon \|x(s)\|ds \\ e^{\alpha t}\|x(t)\|&\leq Me^{\alpha t_0}\|x_0\|+\int_{t_0}^t M\epsilon e^{\alpha s}\|x(s)\|ds \\ e^{\alpha t}\|x(t)\|&\leq Me^{\alpha t_0}e^{M\epsilon (t-t_0)} \\[.5em] \|x(t)\|&\leq Me^{(-\alpha +M\epsilon) (t-t_0)}\|x_0\| \end{align} I am just a bit confused why we are doing this for $||x_0||<\frac{\delta}{M}$ since we know that if $||x_0||<\delta$ then $||x(t)||<\delta$. I thought this might be an attempt to enlarge the interval where we have that $||x(t)||<\delta$, to use the fact that $|f(s,x(s)|<\epsilon||x(t)||$ we need the fact that $||x(t)||<\delta$ so we stay in the same interval. Can anyone help me figure out why we care that $||x_0||>\frac{\delta}{M}$? Thanks in advance.

Answer 1

The logic of this kind of continuation argument goes like this. What has been shown is that supposing $\|x_0\|<\frac\delta M$, if $\|x(t)\|<\delta$ for all $t$ in any interval $[t_0,T]$ with $T>t_0$, then $\|x(t)\|\le M\|x_0\|$ for all such $t$.

The number $M\|x_0\|$ is strictly less than $\delta$, so whatever $T$ you have, by continuity it can be made a bit larger. To be precise, let $T_*$ be the supremum of the set $S$ consisting of those $T\ge t_0$ such that $\|x(t)\|<\delta$ for all $t\in[t_0,T]$. Then it must be that $T_*=\infty$, since otherwise we infer $\|x(T_*)\|\le M\|x_0\|<\delta$ by continuity, so $T_*\in S$, and (again by continuity) there is some $T>T_*$ with $T\in S$.

Answer 2

The constant $M$ is there as guard for the behavior of $e^{(A+αI)t}$ in the case of eigenvalues with "proper" multiplicities. Then the matrix exponential may also have polynomial terms, and as one can see in examples like $(1+2t) e^{-t}$ for $t\ge 0$, while the general behavior is monotone convergence to zero, initially this function is growing above its initial value of $1$. So with $$ M=\sup_{t>0}\|e^{(A+αI)t}\| $$ one gets $$ \|e^{At}x\|\le Me^{-αt}\|x\|. $$ One might get a more accurate upper bound $\|e^{At}x\|\le \min(1+Nt,M)e^{-αt}\|x\|$ which would be tight at $t=0$, but that complicates the ensuing computations unnecessarily.

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For $u(t)=e^{αt}\|x(t)\|$ the inequality $$ u(t)\le Mu(t_0)+\int_{t_0}^t Mϵ u(s)\,ds $$ has the solution $$ u(t_0)+\int_{t_0}^t ϵ u(s)\,ds\le u(t_0)e^{Mϵ(t-t_0)} \\~\\ \implies \|x(t)\|\le Me^{-(α-Mϵ)(t-t_0)}\|x(t_0)\| $$

So the flow of the proof is that you select some $α$ so that the spectrum of $A$ has all real parts smaller $α$. Then with this $α$ the bound $M$ is determined. Now select some $0<ϵ<\frac{α}{M}$. This then determines via the properties of $f$ a $δ$ so that $\|f(s,x)\|\le ϵ\|x\|$ for all $\|x\|\le δ$. Then for $\|x_0\|\le\frac{δ}{M}$ the given estimates guarantee (in a slightly circular way) that the solution remains in the admissible region where all assumptions and constants above are valid. Additionally that gives the defining property of stability and asymptotic stability.