An ODE with maximum

Question

The equation has unic solution? :

$$y^{´}(t) = \left\{ \begin{array}{@{} l c @{}} \text{max}\{t,y(t)\}\\ y(0)=0 \end{array}\right.$$

where $t\in \mathbb{R}$.

My idea was to separate in cases, when :

$$ y(t)\leq t \implies y^{´}(t)=t \implies dy = t dt $$ or

$$ y(t)>t \implies y^{´}(t)=y(t) \implies \dfrac{dy}{y} = dt$$

But I do not know how to guarantee the uniqueness. Thanks for read!

Could you apply the Picard–Lindelöf theorem? The only non-trivial part is to check that the max is Lipschitz in $y$, however, it is doable. — A.Γ., May 05 '19 at 16:21
Yes, that was another of my ideas. To apply the aforementioned theorem, we need to see that it is lipchitz with respect to the second variable. Now, I check your answer! — , May 05 '19 at 16:53

zwim · Answer 1 · 2019-05-05T16:00:13.890

1

You get solutions $y=Ae^t$ and $y=B+\dfrac{t^2}2$

Note that the initial condition $y(0)=0$ kills the exponential in the neighbourhood of zero, so we are left with $y=\dfrac{t^2}2$.

This solution in only valid for $t\in[0,2]$ (we need $y(t)\le t$), outside this interval, it should be the exponential solution.

So you will have a piecewise solution $\begin{cases} y=0 & y\le 0\\ y=\frac {t^2}2 & y\in[0,2]\\ y=Ae^t & y\ge 2\end{cases}$

But for the solution to be $C^1$ we need $y'(2)=t\bigg|_{t=2}=Ae^t\bigg|_{t=2}\iff A=\frac 2{e^2}\approx 0.27$

edited May 05 '19 at 16:00

answered May 05 '19 at 15:52

zwim

Thanks :) and it´s unic? – May 05 '19 at 16:05
Well do you have other choice for gluing the pieces in terms of $A_{<0}$, $A_{>2}$ and $B$ ? It seems to me they are uniquely determined. What do you think ? The solution is unique if there is no other triplet that gives a valid solution. – zwim May 05 '19 at 16:08

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