9.65 Let $u, v$ be positive real numbers, and for a given positive integer $n$, find the necessary and sufficient conditions that $u, v$ must satisfy so that there exist real numbers $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant 0$ satisfying
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=u, \\
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=v .
\end{array}
$$

When these numbers exist, find the maximum and minimum values of $a_{1}$.
[Solution] If $a_{1}, a_{2}, \cdots, a_{n}$ satisfying the conditions exist, by the Cauchy-Schwarz inequality, we have
$$
\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} \leqslant n\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right) .
$$

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

Thus, the necessary conditions for $u, v$ are:
$$
v \leqslant u^{2} \leqslant n v .
$$

It can be proven that (1) is also a sufficient condition. In fact, if (1) holds, then take a positive number
$$
a_{1}=\frac{u+\sqrt{(n-1)\left(n v-u^{2}\right)}}{n} \text {. }
$$

Since $n v-u^{2} \leqslant(n-1) u^{2}$, we have $a_{1} \leqslant u$. If $n>1$, then take
$$
a_{2}=a_{3}=\cdots=a_{n}=\frac{u-a_{1}}{n-1},
$$

then it is easy to verify that $a_{1}+a_{2}+\cdots+a_{n}=u$,
$$
\begin{array}{l}
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=v^{2}, \\
a_{1} \geqslant \frac{u-a_{1}}{n-1} .
\end{array}
$$

It can be proven that the maximum value of $a_{1}$ is $\frac{u+\sqrt{(n-1)\left(n v-u^{2}\right)}}{n}$. In fact, if $a_{1}>\frac{u+\sqrt{(n-1)\left(n v-u^{2}\right)}}{n}$, then $n>1$ and
$$
n a_{1}^{2}-2 u a_{1}+u^{2}-(n-1) v>0,
$$

i.e., $(n-1)\left(v-a_{1}^{2}\right)a_{1}$ when using (4), repeating the above steps a finite number of times, we get
$$
v=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant k a_{1}^{2}+\left(u-k a_{1}\right)^{2} .
$$

Furthermore, if $\frac{u}{k+1} \leqslant a_{1} \leqslant \frac{u}{k}$, by (5) and the properties of quadratic functions, we have
$$
v=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant k a_{1}^{2}+\left(u-k a_{1}\right)^{2} \leqslant \frac{u^{2}}{k} .
$$

By (1), we know $\frac{u^{2}}{n} \leqslant v \leqslant u^{2}$, clearly $v=\frac{u^{2}}{n}$ if and only if $a_{1}=a_{2}=\cdots=a_{n}=\frac{u}{n}$. If $v>\frac{u^{2}}{n}$, then there exists $k \in\{1,2, \cdots, n-1\}$ such that $\frac{u^{2}}{k+1}<v \leqslant \frac{u^{2}}{k}$. If $\frac{u^{2}}{k+1}<v \leqslant \frac{u^{2}}{k}$, then by (8) and (7), we have
$$
a_{1} \leqslant \frac{k u-\sqrt{k\left[(k+1) v-u^{2}\right]}}{k(k+1)}<\frac{u}{k+1},
$$

which contradicts $\frac{u^{2}}{k+1}<v \leqslant \frac{u^{2}}{k}$. Therefore, (7) holds.
On the other hand, under the condition $\frac{u^{2}}{k+1}<v \leqslant \frac{u^{2}}{k}$, if
$$
a_{1}=\frac{k u+\sqrt{k\left[(k+1) v-u^{2}\right]}}{k(k+1)},
$$

then $\frac{u}{k+1}<a_{1} \leqslant \frac{u}{k}$, and $k(k+1) a_{1}^{2}-2 k u a_{1}+u^{2}-v=0$ i.e.,
$$
v=k a_{1}^{2}+\left(u-k a_{1}\right)^{2} \text {. }
$$

Let $a_{1}=\cdots=a_{k}=\frac{u}{k}, a_{k+1}=u-k a_{1}, a_{k+2}=\cdots=a_{n}=0$, then $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the required conditions.
Combining all the above cases, we get: When $\frac{u^{2}}{k+1}<v \leqslant \frac{u^{2}}{k}$, where $k \in\{1,2, \cdots, n-1\}$, the minimum value of $a_{1}$ is
$$
\frac{k u+\sqrt{k\left[(k+1) v-u^{2}\right]}}{k(k+1)}
$$