Existence and uniqueness of solution for first order Cauchy problems

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.

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