11 Let $a_{1}, a_{2}, \cdots, a_{2001}$ be non-negative real numbers, satisfying:
(1) $a_{1}+a_{2}+\cdots+a_{2001}=2$;
(2) $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2000} a_{2001}+a_{2001} a_{1}=1$.

Find the extremum of $S=a_{1}^{2}+a_{2}^{2}+\cdots+a_{2001}^{2}$.
11. First, calculate the maximum value of $f\left(a_{1}, a_{2}, \cdots, a_{2001}\right)=a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{2000} a_{2001}+a_{2001} a_{1}$.

Lemma: There exists an $i \in\{1,2, \cdots, 2001\}$ such that $a_{i}>a_{i+4}$ (with $a_{2001+i}=a_{i}$).
Proof: If not, then $a_{1} \leqslant a_{5} \leqslant a_{9} \leqslant \cdots \leqslant a_{2001} \leqslant a_{4} \leqslant a_{8} \leqslant \cdots \leqslant a_{2000} \leqslant a_{3} \leqslant a_{7} \leqslant \cdots \leqslant a_{1999} \leqslant a_{2} \leqslant a_{6} \leqslant \cdots \leqslant a_{1998} \leqslant a_{1}$.

Thus, $a_{1}=a_{2}=\cdots=a_{2001}=\frac{2}{2001}$, which contradicts the problem's conditions. The lemma is proved.
Assume $a_{1998}>a_{1}$. For given $a_{1}, a_{2}, \cdots, a_{1998}, a_{2000}$, let $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}, \cdots, a_{1998}^{\prime}=a_{1998}, a_{2000}^{\prime}=a_{2000}, a_{2001}^{\prime}=0, a_{1999}^{\prime}=a_{1999}+a_{2001}$, then $\sum_{i} a_{i}^{\prime}=2$.
$f\left(a_{1}^{\prime}, a_{2}^{\prime}, \cdots, a_{2001}^{\prime}\right)=a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{1997} a_{1998}+a_{1998}\left(a_{1999}+a_{2001}\right)+\left(a_{1999}+a_{2001}\right) \cdot a_{2000}>f\left(a_{1}, a_{2}, \cdots, a_{2001}\right)$.

Next, let $a_{1}^{\prime \prime}=a_{1}^{\prime}, \cdots, a_{1998}^{\prime \prime}=a_{1998}^{\prime}+a_{2000}^{\prime}, a_{1999}^{\prime \prime}=a_{1999}^{\prime}, a_{2000}^{\prime}=0$, then it is easy to prove $f\left(a_{1}^{\prime \prime}, a_{2}^{\prime \prime}, \cdots, a_{2000}^{\prime \prime}, 0\right)>f\left(a_{1}^{\prime}, a_{2}^{\prime}, \cdots, a_{2000}^{\prime}, 0\right)$.

That is, each time the last non-zero $a_{i}$ is set to 0, and the sum of $a_{i}$ and $a_{i-2}$ is set as the new $a_{i-2}$, the value of $f$ increases, until only 3 numbers remain, $a_{1}+a_{2}+a_{3}=2, f\left(a_{1}, a_{2}, a_{3}, 0, \cdots, 0\right)=a_{1} a_{2}+a_{2} a_{3}=a_{2}\left(2-a_{2}\right) \leqslant 1$, i.e., the maximum value of $f\left(a_{1}, a_{2}, \cdots, a_{2001}\right)$ is 1, which is achieved when $a_{4}=a_{5}=\cdots=a_{2001}=0, a_{2}=1$.

Therefore, $S=a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=1+a_{1}^{2}+a_{3}^{2} \geqslant 1+\frac{\left(a_{1}+a_{3}\right)^{2}}{2}=\frac{3}{2}$.
And $S=2\left(a_{1}-1\right)^{2}+\frac{3}{2} \leqslant 2\left(0 \leqslant a_{1} \leqslant 1\right)$. Hence, $S_{\max }=2, S_{\min }=\frac{3}{2}$.