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Consider the following second order linear ODE with mixed boundary condition: $$\frac{d^2f}{dt^2}+a(t)f(t)=0,~\frac{df}{dt}(0)=u,~f(1)=0,$$ where $u\in R$ and $a\in C[0,1]$ are fixed.

Is the solution to this equation unique? If so, how to prove it? Thanks!

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Before someone tells you some answers, it is rather instructive to work out examples yourself. If you let $a$ be a constant (the sign matters!), you can find the general solution to the ODE (without the boundary conditions) and figure out when the boundary conditions can be met or not. After you've done that, look up "Sturm-Liouville theory" for self-adjoint second-order ODE's. –  Deane Yang May 6 '11 at 21:15
I don't think this is quite MO-level, though I'm sure there could be a very interesting discussion to be had around your question. My advice would be to assume that the solution is not unique in general, because boundary conditions seldom lead to uniqueness. So I would look for a counterexample. A much trickier question will be: are there easy conditions to check for which we do have uniqueness. This might be what Deane has in mind, but I can't be sure. –  Thierry Zell May 7 '11 at 1:19
Generically, both existence and uniqueness will be satisfied (solution space is 2-dimensional, there are two constraints on it). However, if $u\ne 0$, a solution may fail to exist at all if $f(1)\implies df/dt(0)=0$. On the other hand, if $u=0$, existence is trivial ($f(t)=0$), but uniqueness is not guaranteed. –  Igor Khavkine May 7 '11 at 8:56
Each of the above cases can be explored with $a$ constant, as per Deane's suggestion. For arbitrary $u$ and $a(t)$, it is a non-trivial problem to figure out which of the cases you are in, and has to be tackled separately for each case (using for instance exact solutions, qualitative estimates, or numerics). BTW, a nice modern reference is Zettl, Sturm-Liouville Theory. amazon.com/dp/0821839055 –  Igor Khavkine May 7 '11 at 8:59
I would like to thank all of you for your thinking on my some what stupid question. For the case when $a(t)$ is a constant function, it is easy to disscus and, as Deane said to me, for the general case the theory of Sturm-Liouville can help. Well, if you do think this question is not suitable for MO, we could delete it. Thanks! –  ProbLe May 8 '11 at 9:39

1 Answer 1

up vote 1 down vote accepted

I'm not sure whether this is really appropriate for MathOverflow or not. Still, let me say a little more: As I've mentioned above, you can work out completely the case where $a$ is constant using explicit solutions. If $a$ is not constant, I'm not aware of a definitive answer but you can get separate necessary conditions and sufficient conditions, involving upper or lower bounds on $a$ using the Sturm comparison theorem. Last, I believe that it is possible to find an integral condition on $a$ that is sufficient for there to be a unique solution to the boundary value problem.

The vector-valued version of this problem is used to analyze how geodesics on a Riemannian manifold behave given assumptions on the curvature.

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Deane, thank you very much for all your thinking on my question, your answer is quite instructive! If you find this is not an MO question, I will try to delete it. Before turning to help to MO, I have already considerd the case when $a(t)$ is a constant, which corresponds to my original conceived application to locally symmetric Riemannian manifold and it seems that every thing goes well thanks to the invariance of the sectional curvature by parallel transport along a geodesic. But for the general Riemannian manifold, $a(t)$ is not a constant and I was confused... –  ProbLe May 8 '11 at 9:57
The use of Sturm-Liouville theory to study geodesics in a Riemannian manifold is discussed in almost every Riemannian geometry book I know. I learned a lot from the classic book by Cheeger and Ebin. Another book I like is Gallot-Hulin-Lafontaine. –  Deane Yang May 8 '11 at 14:32
Thanks for the two references. It seems that, generally, the geodesics are studied by considering the Jacobi fields with initial conditions on $J(0),J^{\prime}(0)$ or boundary conditions on $J(0), J(1)$. But the mixed boundary conditions on $J(0),J^{\prime}(1)$ are much less used. It is easily seen that $J(0)=J^{\prime}(1)=0$ does not imply that $J$ vanishes identically even if there is no cut points on the geodesic. I think that the mixed boundary condition is related to the degeneracy of the hessian (restricted to the orthogonal subspace of geodesic direction) of the distance function. –  ProbLe May 8 '11 at 16:07
Yes, you're right. You want to use the same ideas and techniques for your specific problem. I'm pretty sure they can be adapted to get what you want. –  Deane Yang May 8 '11 at 17:14
Thanks! I will try to generalize my previous method to the general case along this direction! –  ProbLe May 8 '11 at 17:44

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