1
$\begingroup$

According to Picard–Lindelöf theorem, IVP $$\begin{cases}y^\prime(t) = f(t,y(t))\\y(x_0)=y_0 \end{cases}$$ has a unique solution if $f$ is lipschitz continuous. What if my ODE contain piecewise defined function $g:\mathbb{R}^2\rightarrow\mathbb{R}$ \begin{equation} g(x,y) = \begin{cases} x^2\tanh(y)\text{ for } x>0\\ 0 \text{ for } x\leq 0 \end{cases} \end{equation} Is $g(x,y)$ lipschitz continuous in $\mathbb{R}^2$ (what is the $K$ then?) and if not, does it automatically mean, that ODE with such function has no solution?

Edit. I am curious about the situation like $$ \mathbf{x}^\prime(t) = \mathbf A^{n\times n}\mathbf{x}(t) + \begin{bmatrix} x_1(t)\\ x_2(t)\\ \vdots\\ g(x_1,x_2)\\ x_n(t)\\ \end{bmatrix} $$

$\endgroup$
2
  • $\begingroup$ You don't need Lipschitz continuity in $\Bbb R^2$ but Lipschitz continuity in the second variable ($y$). $\endgroup$
    – Martin R
    Aug 27, 2021 at 20:29
  • $\begingroup$ Maybe I messed up my original question. Please see edit $\endgroup$
    – struct
    Aug 27, 2021 at 20:52

1 Answer 1

0
$\begingroup$

No, $g$ is not Lipschitz, as $x^2$ is not Lipschitz. But $g$ is Lipschitz on $(-\infty, r)\times\mathbb R$ for all $r$.

Thus the ODE has a solution on each such set, so it also has a solution on $\mathbb R$. Picard Lindelöf only needs local lipschitz, and that only in the second variable.

$\endgroup$
1
  • $\begingroup$ Maybe I messed up my original question. Please see edit $\endgroup$
    – struct
    Aug 27, 2021 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.