# Continuity of solutions to the nonlinear time-varying differential equations in terms of initial states and parameters

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

$$\|z(t,\lambda)-y(t,\lambda_0)\|<\epsilon,~\forall{t}\in[t_0,t_1].$$

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 $$L$$.

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!!

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.