1
$\begingroup$

Suppose we have two functions $f=f(x,a):\mathbb{R}^n\times A\to \mathbb{R}^n$, where $A\subset \mathbb{R}^m$ is compact, and also $\alpha:[0,+\infty)\to A$. Under the hypothesis that

A')$f$ and $\alpha$ are continuous

B')$f$ is lipschitz continuous in $x$ uniformly with respect to $a$, ie $$\exists \,L>0: |f(x_1,a)-f(x_2,a)|\leq L|x_1 -x_2|\,\,\forall \, (x_1,a),(x_2,xa)\in \mathbb{R}^n\times A$$

C') $f$ is bounded

I have to prove that for any choice of $x_0\in \mathbb{R}^n$, $$\begin{cases}y'(t)=f(y(t),\alpha(t))\qquad t>0\\y(0)=x_0\end{cases}$$ has a unique global solution $y:[0,+\infty)\to \mathbb{R}^n$. Moreover $y\in C^1([0,+\infty))$ and $$y(t)=x_0+\int_{0}^{t}f(y(s),\alpha(s))ds\qquad t\geq 0.$$

I know that from Picard- Lindelöf theorem we have

Theorem Let $-\infty\leq a\leq b \leq+\infty$ and $f=f(t,x):(a, b)\times \mathbb{R}^n\to \mathbb{R}^n$ be a function. Under the following hypothesis:

A) $f$ is continuous

B) $f$ is lipschitz continuous in $x$ uniformly with respect to $t$, ie $$\exists \,L>0: |f(t,x_1)-f(t,x_2)|\leq L|x_1 -x_2|\,\,\forall \, (t,x_1),(t,x_2)\in (a, b)\times \mathbb{R}^n$$

C) $f$ is bounded

Then, for every choice of $(t_0,x_o)\in (a, b)\times \mathbb{R}^n$ the Cauchy problem $$\begin{cases}y'(t)=f(t,y(t))\\y(t_0)=y_0\end{cases}$$ has a unique global solution $y:(a, b)\to \mathbb{R}^n$. Moreover $y\in C^1((a, b))$ and $$y(t)=x_0+\int_{t_0}^{t}f(s,y(s))ds\qquad t\in(a, b).$$

I think that I can use the previous theorem in order to prove my question by defining $g:[0,+\infty)\times\mathbb{R}^n\to \mathbb{R}^n$ such that $g(t, x)=f(x,\alpha(t))$. In fact $g$ satisfies A) B) and C) in $[0,+\infty)\times\mathbb{R}^n$. By the way $[0,+\infty)$ is not open, and for this reason I don't know how to proceed.

Can anyone give me a hint? Thanks a lot in advance.

$\endgroup$
  • $\begingroup$ Solve it for $(0,\infty)$, then consider the behavior of the solution as $t \to 9$, $\endgroup$ – Paul Sinclair Jan 19 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.