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Let $F: \mathbb{R} \times\mathbb{R}^2 \rightarrow \mathbb{R}$ localy lipschitz in its second variable. Let the Cauchy problem be:

$$x'=F(t,x),\\ x(0) = x_0 $$ Let $\alpha : \mathbb{R} \rightarrow \mathbb{R}$ and $\beta : \mathbb{R} \rightarrow \mathbb{R}$. Show that if $\forall x \in \mathbb{R}^2, \forall t \in \mathbb{R}$ $$|F(t,x)| \leq \alpha(t)|x| + \beta (t) $$ then any maximal solution is global.

I am unable to do this. I suppose I have to either show that $F$ is uniformly bounded or that it is uniformly Lipschitz, but I have no clue how to do either. Any help will be appreciated.

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  • $\begingroup$ (by considering $x=0$, $\beta$ maps into $[0,\infty)$, and by considering $x$ sufficiently large, the same is true for $\alpha$) $\endgroup$ – Calvin Khor Jan 3 at 17:16
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See Grönwall lemma on a bound for solutions, using the solution of $$u'(t)=α(t)u+(α(t)|x_0|+β(t)), ~~ u(0)=0.$$ Then after establishing that $$|x(t)-x_0|\le |u(t)|$$ wherever the solution exists, apply theorems on the maximal solution, and why the bound prevents finite boundaries of the maximal domain.

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Suppose $x(t)$ is not global, namely $\exists t_0>$ such that $$ \lim_{t\to\to t_0^-}|x(t)|=\infty. \tag{1}$$ Integrating from $0$ to $t<t_0$ gives $$ x(t)=x(0)+\int_0^tF(s,x)ds $$ which implies $$ |x(t)|\le |x(0)|+\int_0^t|F(s,x)|ds. $$ Using $F(t,x)\le\alpha(t)|x|+\beta(t)$, one has $$ |x(t)|\le |x(0)|+\int_0^t(\alpha(s)|x(s)|+\beta(s))ds.\tag{2}$$ Define $$ g(t)=\int_0^t(\alpha(s)|x(s)|+\beta(s))ds \tag{3} $$ and then $$ g'(t)=\alpha(t)|x(t)|+\beta(t). \tag{4} $$ By (1), it is not hard to see $$ \lim_{t\to\to t_0^-}g(t)=\infty. \tag{5}$$ Using (3) and (4) in (2), one has $$ g'(t)-\beta(t)\le|x_0|\alpha(t)+\alpha(t)g(t) $$ or $$ g'(t)-\alpha(t)g(t)\le\beta(t)+|x_0|\alpha(t). $$ It is easy to see $$ \left(e^{-\int_0^t\alpha(s)ds}g(t)\right)'\le (\beta(t)+|x_0|\alpha(t))e^{-\int_0^t\alpha(s)ds}. $$ Integrating from $0$ to $t$, one has $$ g(t)\le g(0)e^{-\int_0^t\alpha(s)ds}+ e^{-\int_0^t\alpha(s)ds}\int_0^t(\beta(r)+|x_0|\alpha(r))e^{-\int_0^r\alpha(s)ds}dr. $$ Using this, one has $$ g(t)\le C, t\in[0,t_0]$$ for some constant $C$, which is against (5).

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