# If $\vert F(t,x)\vert \leq \alpha (t)\vert x\vert + \beta(t)$ the maximal solutions are global for an ODE

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.

• (by considering $x=0$, $\beta$ maps into $[0,\infty)$, and by considering $x$ sufficiently large, the same is true for $\alpha$) – Calvin Khor Jan 3 at 17:16

## 2 Answers

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.

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 $$t gives $$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).