Vtyjkyuk

I know
$$begincases y’=|y|\ y(0)=0 endcases $$
Has the solution $yequiv 0$... does it have another?

The question is about the theorem of existence and uniqueness because $partial_y f(x,y)$ is not continuous

With

$f(y) = vert y vert, tag 1$

we have

$vert f(y_1) - f(y_2) vert = vert vert y_1 vert - vert y_2 vert vert le vert y_1 - y_2 vert, tag 2$

which shows that $f(y) = vert y vert$ is Lipschitz continuous with Lipschitz constant $1$; therefore by the Picard-Lindelof theorem, the equation

$dot y = vert y vert tag 3$

has a unique solution in some interval about any initial point $y_0 = y(t_0)$. So the only solution with $y(0) = 0$ is identically zero, $y(t) equiv 0$.

This shouldn't really come as too much of a surprise since $vert 0 vert = 0$, so $0$ is an equilibrium of (3).

Finally, we don't need $f(y)$ differentiable, merely Lipschitz continuous, for existence and uniqueness to apply.

Note Added in Edit, Friday 17 August 2018 11:14 PM PST: To really flesh this answer out, we should probably also show that the unique solution $y(t) equiv 0$ on an interval about $0$ extends uniquely to $(-infty, infty)$, but I'm leaving this discussion for a later date. End of Note.

Direct proof: any solution $y$ will be nondecreasing as $y'(x) = |y(x)|ge 0$. Then $y(x)ge 0$ for $xge 0$ and $y'(x) = y(x)$ for $xge 0$. Integrating,
$$y(x) = y(x) - y(0) = int_0^x y = 0 + int_0^x y.$$
Now, applying the Gronwall lemma, $y(x)le 0$ for $xge 0$.

EDIT: even more direct (Gronwall-less) proof.

Take $x_0in(0,1)$. Integrating as before and using that $y$ is nondecreasing,
$$y(x_0) = int_0^x_0 yle x_0,y(x_0)implies y(x_0) = 0,$$
i.e., $yequiv 0$ in $[0,1)$, so in $[0,1]$ by continuity. The same argument works in $[1,2]$, $[2,3]$,...