# Determining the largest interval of definition of a Cauchy problem

I have some problems trying to solve this exercise:
Consider the following Cauchy problem: $$\left\{\begin{matrix}x'=t+\frac{y}{1+x^2} \\ y'=txe^{-ty^2} \\ x(0)=y(0)=1 \end{matrix}\right.$$
Discuss the existence of a solution and its uniqeness. Then I'm asked to prove or to disprove the fact that the supremum of the max interval on which the solution is definted is $$[0,+\propto)$$.

I've defined a function as following $$F:\mathbb{R}^3 \rightarrow \mathbb{R}^2$$, $$F(t,**z**=(x,y))=\begin{bmatrix}t+\frac{y}{1+x^2} \\ txe^{-ty^2} \end{bmatrix}$$
Since $$F$$ is infinitely differentiable (which implies it's continuous and locally lipschitz) on $$\mathbb{R}^3$$, we can conclude that $$\exists \delta>0, \ \exists \ \phi:[-\delta,\delta] \rightarrow \mathbb{R}^2$$ which is a unique solution of the Cauchy problem on that interval. I have some troubles while trying to compute the max interval of definition of the solution. Can someone suggest me how to procede?
Thanks in andvance for the help.

If the function is sub-linear in state-space direction, $$\|F(t,u)\|\le a(t)\|u\|+b(t)$$ then the solution is bounded per Grönwall lemma by the solution of the linear ODE $$v'(t)=a(t)v(t)+b(t),~~ v(0)=\|u(0)\|.$$ As solutions of linear DE exist wherever the coefficients are continuous, the bound and thus the solution exist wherever the sub-linearity can be established.
Sublinearity is easy to see for $$t>0$$, for $$t<0$$ the second equation is not sub-linear due to the exponential term.
Lutz Lehmann gives a good solution. Here is another one which uses argument by contradiction. Suppose there is $$t_0\in(0,\infty)$$ such that $$\lim_{t\to t_0^-}\|(x(t),y(t))\|=\infty.$$ Let $$u=\|(x,y)\|^2$$ and then (1) becomes $$\lim_{t\to t_0^-}\|u(t)\|=\infty. \tag1$$ Multiplying both sides of the firs, second equations by $$x$$ and $$y$$, respectively, one has, for $$t\in[0,t_0)$$, $$\begin{eqnarray} \frac12(x^2+y^2)'&=&tx+\frac{xy}{1+x^2}+txye^{-ty^2}\\ &\le& t_0|x|+\frac12|y|+t_0|x||y|\\ &\le&\frac12(t_0^2+x^2)+\frac14(1+y^2)+\frac{t_0}2(x^2+y^2) \tag2 \end{eqnarray}$$ Let $$a=t_0^2+\frac12, b=\frac12+t_0$$ and then (2) becomes $$u'\le a+bu.$$ It is easy to see $$u(t)\le u(0)e^{bt}+\frac{a}{b}(e^{-bt}-1).$$ Letting $$t\to t_0^-$$ gives $$\limsup_{t\to t_0^-}u(t)<\infty$$ which is against (1). So $$t_0=\infty$$.