1
$\begingroup$

I'm trying to use the Local existence and uniqueness theorem for this ODE: $$\begin{cases} y'(x)=xye^{-y^{2}} \\ y(0)=y_0\\ \end{cases}$$ Putting $f(x,y)=xye^{-y^{2}}$ I see that $f(x,y) \in C^{\infty}(\mathbb{R}^{2})$ and this fact should provide local and uniqueness existence for the solution $y(x)$ (due to $f(x,y)$ is locally Lipschitz in $y$ with respect to $x$).

I have more troubles when I have to discuss global existence and uniqueness for the solution. I know that $f(x,y)$ must be globally Lipschitz in $y$ respect to $x$.

Now I see that $f(y)=e^{-y}$ is limited when $y\rightarrow \infty$ and $f(x,y)=xy$ is Lipshitz in $y$ with respect to $x$ in all $\mathbb{R^{2}}$. But I think that this is not enough to prove the claim.

Could someone help me to finish proving global and uniqueness existance? Thank you.

$\endgroup$

1 Answer 1

2
$\begingroup$

$f(x,y)$ is in fact globally Lipschitz with respect to $y$ on any domain of the form $[a,b]\times(-\infty,+\infty)$. To see this, observe that $$ \frac{\partial f}{\partial y}(x,y)=y\,e^{-y^2}+-2\,x\,y^2\,e^{-y^2} $$ is bounded.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .