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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.

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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.

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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 $t_0=\infty$.

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