Uniqueness of Lipschitz function satisfying differential equation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T23:26:44Z http://mathoverflow.net/feeds/question/106609 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106609/uniqueness-of-lipschitz-function-satisfying-differential-equation Uniqueness of Lipschitz function satisfying differential equation Elena Yudovina 2012-09-07T15:00:38Z 2012-09-08T07:58:49Z <p>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?</p> <p>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 <em>all</em> times, which intuitively seems like it ought to work.</p> http://mathoverflow.net/questions/106609/uniqueness-of-lipschitz-function-satisfying-differential-equation/106610#106610 Answer by Thomas Richard for Uniqueness of Lipschitz function satisfying differential equation Thomas Richard 2012-09-07T15:18:14Z 2012-09-07T15:37:48Z <p>I suppose you want uniqueness assuming some initial condition. </p> <p>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 $X'=2\sqrt{X}$.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/106609/uniqueness-of-lipschitz-function-satisfying-differential-equation/106656#106656 Answer by Bazin for Uniqueness of Lipschitz function satisfying differential equation Bazin 2012-09-08T07:58:49Z 2012-09-08T07:58:49Z <p>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. </p> <p>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 $f(x_0)=0$.</p>