2 typo; added 10 characters in body

There is of course a whole theory behind, and the right pointer is the Sturm-Liouville problem as indicated by Deane YoungYang. However, just the matter of proving the uniqueness of solutions of 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

$$\int_a^b|\dot u-\dot v|^2 dt = - \int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\le0\ big(f(u)-f(v)\big)\cdot(u-v)\ dt\le0\ ,$$ 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 $$\int_0^1 |\dot u-\dot v|^2 dt= -\int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\le big(f(u)-f(v)\big)\cdot(u-v)\ dt\le c \int_0^1 |u-v|^2dt\ .$$

By the PoincarĂ© 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.

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There is of course a whole theory behind, and the right pointer is the Sturm-Liouville problem as indicated by Deane Young. However, just the matter of proving the uniqueness of solutions of 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

$$\int_a^b|\dot u-\dot v|^2 dt = - \int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\le0\ ,$$ 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 $$\int_0^1 |\dot u-\dot v|^2 dt= -\int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\le c \int_0^1 |u-v|^2dt\ .$$

By the PoincarĂ© 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.