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I have a Lipschitz function $X=X(t)$ with the property that, at all points $t$, the right derivative $\lim_{\epsilon \downarrow 0} \epsilon^{-1}(X(t+\epsilon)-X(t))$ exists and is given by $f(X)$ for some (discontinuous) function $f$. (Of course, at all regulat points, i.e. almost everywhere, this is then the ordinary derivative.) Is it true that such $X$ is necessarily unique?

Of course, if I just specify that a Lipschitz function satisfies $X'(t) = f(X(t))$ at all regular points, the solution doesn't have to be unique; but I'm requiring that this hold for the right derivative at all times, which intuitively seems like it ought to work.

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up vote 2 down vote accepted

I suppose you want uniqueness assuming some initial condition.

Even when $F$ is continuous and $X$ is $C^1$ you don't have uniqueness. For $t\geq 0$ $X(t)=0$ and $X(t)=t^2$ are both solution of $X'=2\sqrt{X}$.

Maybe with stronger assumptions on $F$ you can do something assuming only that $X$ is Lipschitz but I'm not sure. Without the Lipschitz condition you don't have uniqueness assuming that the differential equation is satisfied a.e., Cantor stair is a counterexample.

Edit : The uniqueness statement you want is true when $F$ is Lipschitz. In this case, with your assumptions, $X$ has a right derivative which is continuous. From that it's not too complicated to show that $X$ has a derivative everywhere which is continuous. And then the classical Cauchy-Lipschitz applies.

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What I actually wanted is "$f$ piecewise linear, but discontinuous, i.e. with jumps between the pieces". (E.g., $f(X)=I[X>0]$.) This particular example gives multiple solutions from $X(0) = 0$ if I only require $X'(t) = f(X)$ whenever the two-sided derivative exists, but forcing the right derivative to be $f(X)$ everywhere squashes them out. – Elena Yudovina Sep 7 '12 at 17:29

You have a logarithmic extension of the classical Cauchy-Lipschitz theorem: the equation $$\dot x=f(x),\quad x(0)=x_0\tag{ODE}$$ has a unique solution if $f$ has a modulus of continuity $\omega$ ($ \vert f(x+h)-f(x)\vert\le \omega(\vert h\vert) $) such that $$ \int_0^a\frac{dr}{\omega(r)}=+\infty,\quad \text{for some positive $a$}. $$ On the top of this $\omega$ should be positive increasing, $\omega(0_+)=0$. This is of course satisfied by $\omega(r)=Cr$ (Lipschitz case) but also by $$ \omega(r) =Cr\ln(1/r),\quad\text{the so-called Log-Lipschitz case.} $$ You have as well $r\ln(1/r)\ln(\ln(1/r))$ and so on.

Let me add a remark on the 1D case. When $x(t)\in \mathbb R^1$, $f$ continuous and $f(x_0)\not=0$, then (ODE) has the uniqueness property (just separate the variables). This is of course compatible with the counterexample in the previous answer where $f(x_0)=0$.

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Thank you. I was specifically fishing for results where $f$ has jump discontinuities -- in my case, the functions $f$ tend to be piecewise-nice but may, for example, depend on which of the coordinates of $X$ is positive (with, potentially, a jump at the boundary). – Elena Yudovina Sep 17 '12 at 17:43
@Elena You have plenty of Eulerian results (I mean for the vector field PDE) for that type of vector fields. In particular a MR search on Ambrosio as author and keyword "nearly incompressible" or "$BV$ vector field" will provide plenty of references. – Bazin Sep 20 '12 at 12:18

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