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.



Solution uniqueness for ODE
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.


