Solution uniqueness for ODE - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:07:41Z http://mathoverflow.net/feeds/question/91098 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91098/solution-uniqueness-for-ode Solution uniqueness for ODE Benoît Kloeckner 2012-03-13T17:30:13Z 2012-03-14T11:39:19Z <p>I have a vectorial, non-linear second order ordinary differential equation $$Z''=f(Z)$$ for which I have a solution $Z^0$ on $[0,1]$ with $Z^0_i(0)=0$ and $Z^0_i(1)=1$. I would like to know under which kind of conditions on $f$ it is true that no other solution with same endpoint values can exist. The dimension $1$ case would already be interesting for me to understand, even if I am in fact truly interested in systems.</p> <p>I am pretty sure this kind of question is very classical, but I have some trouble finding relevant keywords to make my way in the literature. Any pointer would be appreciated.</p> <p>By the way, my initial motivation is a Riemannian geometry problem, but I do not think it is especially relevant to this question.</p> http://mathoverflow.net/questions/91098/solution-uniqueness-for-ode/91124#91124 Answer by Pietro Majer for Solution uniqueness for ODE Pietro Majer 2012-03-13T21:34:03Z 2012-03-14T11:39:19Z <p>There is of course a whole theory behind, and the right pointer is the Sturm-Liouville problem as indicated by Deane Yang. However, just the matter of proving the uniqueness of solutions to your equation, can be established quickly under suitable hypotheses. </p> <p>To start with, assume $f:\mathbb{R}^n\to \mathbb{R}^n$ is a continuous and monotone map, that is</p> <p>$$\big(f(x)-f(y)\big)\cdot(x-y)\ge0\ ,$$ for all $x$ and $y$ in $\mathbb{R}^n$. Then, if $u$ and $v$ solve your equation on some interval $[a,b]$ with the same boundary conditions we have, integrating by parts</p> <p>$$\int_a^b|\dot u-\dot v|^2 dt = - \int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\ dt\le0\ ,$$ implying $u-v$ is constant, hence $u=v\ .$ </p> <p>Also, we may gain something exploiting the fact that the interval is given. Assume that $f$ is continuous and $f+cI$ is monotone, for some $c &lt; \pi^2$. So now we just have $$\int_0^1 |\dot u-\dot v|^2 dt= -\int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\ dt\le c \int_0^1 |u-v|^2dt\ .$$</p> <p>By the Poincaré inequality, since each component of $u-v$ is in $H^1_0([0,1])$ we also have</p> <p>$$\pi^2 \int_0^1 |u-v|^2dt \le \int_0^1 |\dot u-\dot v|^2 dt\ ,$$ and we conclude $u=v$ as before.</p> http://mathoverflow.net/questions/91098/solution-uniqueness-for-ode/91154#91154 Answer by Benoît Kloeckner for Solution uniqueness for ODE Benoît Kloeckner 2012-03-14T09:11:33Z 2012-03-14T09:11:33Z <p>The answers and comments so far are perfectly relevant, but let me add one I came across overnight and that suits my problem (not something anyone could guess since I gave little specifics).</p> <p>If $n=1$, and we can prove that $z'>0$ on $[0,1]$ for any solution (for example if $f(z)$ has the sign of $z$), then uniqueness holds. Indeed, look at the integral curves of two solutions with initial value $0$ and different initial velocities. Both curves lie in the first quadrant, one above the other. But the height is exactly the horizontal speed, so the lower one takes more time to reach any given value than the upper one.</p> <p>I said I was interested in the higher-dimensional case, but it appears that the coupling between coordinates in my equation is not that bad.</p>