Let $\mathbf{C}$ be such that $u, v \in \mathbf{R}^{+}$, find the necessary and sufficient conditions that $u, v$ must satisfy, so that for a given $n$, there exist real numbers $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant 0$, satisfying
$$
a_{1}+a_{2}+\cdots+a_{n}=u, a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=v .
$$

Find the maximum and minimum values of $a_{1}$.
If there exist $a_{1}, a_{2}, \cdots, a_{n}$ that satisfy the conditions, then by the Cauchy inequality,
$$
n\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right) \geqslant\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}
$$

It is also clear that
$$
\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} \geqslant a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} .
$$

From (1) and (2), we get
$$
n v \geqslant u^{2} \geqslant v \text {. }
$$
(3) is a necessary and sufficient condition. In fact, when (3) holds, we can take
$$
\begin{array}{l}
a_{1}=\frac{u+\sqrt{(n-1)\left(n v-u^{2}\right)}}{n}, \\
a_{2}=a_{3}=\cdots=a_{n}=\frac{u-a_{1}}{n-1} .
\end{array}
$$
The maximum value of $a_{1}$ is $\frac{u+\sqrt{(n-1)\left(n v-u^{2}\right)}}{n}$. If $a_{1}$ is greater than this value, then
$$
\begin{aligned}
& n a_{1}^{2}-2 u a_{1}+u^{2}-(n-1) v>0 \\
\Rightarrow & (n-1)\left(v-a_{1}^{2}\right) < 0,
\end{aligned}
$$
which contradicts the condition that $v \geqslant a_{1}^{2}$. Therefore, the maximum value of $a_{1}$ is indeed $\frac{u+\sqrt{(n-1)\left(n v-u^{2}\right)}}{n}$.

Because if $a_{1}$ is larger than this value, then
$$
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant k a_{1}^{2}+\left(u-k a_{1}\right)^{2},
$$

which means
$$
v \leqslant k a_{1}^{2}+\left(u-k a_{1}\right)^{2} .
$$

This can be written as a quadratic inequality:
$$
k(k+1) a_{1}^{2}-2 k u a_{1}+u^{2}-v \geqslant 0,
$$

We discuss the following cases:
(i) If $v \leqslant \frac{u^{2}}{k+1}$, then (10) always holds, in which case $\min a_{1}=\frac{u}{k+1}$.
(ii) If $v>\frac{u^{2}}{k+1}$, then (10) holds if and only if
$$
a_{1} \geqslant \frac{k u+\sqrt{k(k+1) v-u^{2}}}{k(k+1)}
$$

Since $a_{1} \leqslant \frac{u}{k}$, we have
$$
\frac{k u+\sqrt{k(k+1) v-u^{2}}}{k(k+1)} \leqslant \frac{u}{k} .
$$

Thus, $v \leqslant \frac{u^{2}}{k}$. Therefore, when $\frac{u^{2}}{k} \geqslant v>\frac{u^{2}}{k+1}$,
$$
\min a_{1}=\frac{k u+\sqrt{k(k+1) v-u^{2}}}{k(k+1)} \quad(k=1,2, \cdots, n-1) .
$$