I know $$\begin{cases} y’=|y|\\ y(0)=0 \end{cases} $$ Has the solution $y\equiv 0$... does it have another?

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



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

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 $x\ge 0$ and $y'(x) = y(x)$ for $x\ge 0$. 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 $x\ge 0$.

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

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


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