As I was reading the following theorem (from Nonlinear Systems by H. K. Khalil) concerning a closeness of solution from the nominal solution in the case when the initial states and parameter vector are perturbed from their nominal values, I encountered a bit of difficulty in grasping some parts of the proof. I present only the part of proof along with its necessary details where I am stuck.

Theorem Let $f(t,x,\lambda)$ be a continuous function in $(t,x,\lambda)$ and locally Lipschitz in $x$ (uniformly in $t$ and $\lambda$) on $[t_0,t_1]\times{D}\times\{\|\lambda-\lambda_0\|\leq{c}\}$, where $D\subset\mathbb{R}^{n}$ is an open connected set. Let $y(t,\lambda_0)$ be a solution of $\dot{x}=f(t,x,\lambda_{0})$ with $y(t_0,\lambda_0)=y_0\in{D}$. Suppose $y(t,\lambda_0)$ is defined and belongs to $D$ for all $t\in[t_0,t_1]$. Then, given $\epsilon>0$, there exists a $\delta=\delta(\epsilon)$ such that if $$\|z_0-y_0\|<\delta,~\|\lambda-\lambda_0\|<\delta$$ then there is a unique solution $z(t,\lambda)$ of $\dot{x}=f(t,x,\lambda)$ defined on $[t_0,t_1]$, with $z(t_0,\lambda)=z_0$ and $z(t,\lambda)$ satisfies


Proof: By continuity of $y(t,\lambda_0)$ in $t$ and compactness of $[t_0,t_1]$, we know that $y(t,\lambda_0)$ is bounded on $[t_0,t_1]$. Define a tube ''U'' around the solution $y(t,\lambda_0)$ by

$$U=\{(t,x)\in[t_0,t_1]\times\mathbb{R}^{n}| \|x-y(t,\lambda_0)\|\leq\epsilon\}$$

Suppose that $U\subset{[t_0,t_1]\times{D}}$; if not replace $\epsilon$ by $\epsilon_1<\epsilon$ that is small enough to ensure that $U\subset{[t_0,t_1]\times{D}}$ and continue the proof with $\epsilon_1$. The set $U$ is compact; hence, $f(t,x,\lambda)$ is Lipschitz in $x$ on $U$ with a Lipschitz constant $L$.

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My question is as follows:

Why is $f(t,x,\lambda)$ Lipschitz in $x$ on $U$? I am asking this because in the theorem statement it says that $f$ is locally Lipschitz in $x$ on $[t_0,t_1]\times{D}\times\{\|\lambda-\lambda_0\|\leq{c}\}$ but in the proof $U$ does not include any information about $\lambda$ or $\|\lambda-\lambda_0\|$. Therefore, how does the compactness of $U$ imply that $f$ is Lipschitz on $U$.

It's been quite some time that I have been looking for an answer to this question. Thus, I greatly appreciate your thoughts and inputs. Thanks in advance!!


1 Answer 1


By the assumptions, $f(t,x,λ)$ is still uniformly $x$-Lipschitz on $U$ for all fixed $λ\in \bar B(λ_0,c)$. Apparently (?) the stronger statement of uniform $x$-Lipschitz continuity on $U\times \bar B(λ_0,c)$ is not required.

The proof probably goes over the Grönwall lemma for $\|z(t;λ)-y(t;λ_0)\|\le\|z(t;λ)-z(t;λ_0)\|+\|z(t;λ_0)-y(t;λ_0)\|$, where the terms on the right satisfy integral inequalities which has the integral inequality $$ \|z(t;λ_0)-y(t;λ_0)\|\le \|z_0-y_0\|+\int_{t_0}^tL\|z(s;λ_0)-y(s;λ_0)\|\,ds $$ and $$ \|z(t;λ)-z(t;λ_0)\|\le\int_{t_0}^tL\|z(s;λ)-z(s;λ_0)\|\,ds+ \int_{t_0}^t\|f(t,z(t;λ_0),λ)-f(t,z(t;λ_0),λ_0)\|\,ds $$ In the last inequality the uniformity in $λ$ of the Lipschitz condition is used, so it was wrong to not include it in the statement.

  • $\begingroup$ Thank you very much. $\endgroup$
    – jbgujgu
    Oct 5, 2020 at 20:50

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