6
$\begingroup$

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.

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.

By the way, my initial motivation is a Riemannian geometry problem, but I do not think it is especially relevant to this question.

Improve this question
$\endgroup$
2
  • $\begingroup$ A relevant keyword might be "two point boundary value problem". $\endgroup$
    – BS.
    Mar 13, 2012 at 17:45
  • 2
    $\begingroup$ It seems to me that without knowing more about $f$ it is difficult to say much about this. It is essentially an eigenvalue problem. If the ODE is linear, then it's called Sturm-Liouville theory. I suppose you could try searching for "nonlinear Sturm-Liouville theory". $\endgroup$
    – Deane Yang
    Mar 13, 2012 at 18:44

2 Answers 2

Reset to default
7
$\begingroup$

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.

To start with, assume $f:\mathbb{R}^n\to \mathbb{R}^n$ is a continuous and monotone map, that is

$$\big(f(x)-f(y)\big)\cdot(x-y)\ge0\ , $$ 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

$$\int_a^b|\dot u-\dot v|^2 dt = - \int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\ dt\le0\ ,$$ implying $u-v$ is constant, hence $u=v\, .$

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 $c < \pi^2$. So now we just have $$\int_0^1 |\dot u-\dot v|^2 dt= -\int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\ dt\le c \int_0^1 |u-v|^2dt\ . $$

By the Poincaré inequality, since each component of $u-v$ is in $H^1_0([0,1])$ we also have

$$\pi^2 \int_0^1 |u-v|^2dt \le \int_0^1 |\dot u-\dot v|^2 dt\ ,$$ and we conclude $u=v$ as before.

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ Notice that if $f=\nabla F$, then $f$ is monotone in the sense above if and only if $F$ is convex. In this case, the solution is the unique minimizer of $u\mapsto\int_a^b(\frac12|\dot u-\dot v|^2+F(u))dt$ in the subspace of $H^1(a,b)$ defined by the boundary conditions. $\endgroup$
    – Denis Serre
    Mar 14, 2012 at 6:22
0
$\begingroup$

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).

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.

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ More generally, you can try to study the map from $Z'(0)$ to $Z(1)$ and try to show that it is 1-1. Using standard Sturm-Liouville theory, you should be able to understand when this map is at least a local diffeomorphism. $\endgroup$
    – Deane Yang
    Mar 14, 2012 at 9:15
  • $\begingroup$ The other suggestion, also motivated by Sturm-Liouville theory, is to use a comparison argument and compare solutions of your system to a linear constant coefficient system or some other special case that you can solve in closed form. $\endgroup$
    – Deane Yang
    Mar 14, 2012 at 9:17
  • $\begingroup$ One more comment: Since you mentioned Riemannian geometry, I presume that you're familiar with the use of Sturm-Liouville theory to study Jacobi fields along a geodesic using curvature bounds. You might want to see if you can adapt what is done in that setting to yours. $\endgroup$
    – Deane Yang
    Mar 14, 2012 at 9:18
  • $\begingroup$ @Deane Yang: indeed, the problem is to study Jacobi fields under curvature bounds --but the curvature condition is unusual, and the equation we get has different features than the usual ones. $\endgroup$
    – Benoît Kloeckner
    Mar 14, 2012 at 11:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.