11. (20 points) Given positive real numbers $p, q$. It is known that the sequence of positive real numbers $\left\{a_{n}\right\}$ satisfies:
$$
a_{0}=1, a_{n+2}=p a_{n}-q a_{n+1}(n \in \mathbf{N}) .
$$

Find all possible values of $a_{1}$ (expressed in terms of $p, q$).
11. The characteristic equation corresponding to the sequence $\left\{a_{n}\right\}$ is
$$
\lambda^{2}+q \lambda-p=0 \text {. }
$$

Obviously, the two roots of this equation are one positive and one negative. Let the positive root be $\lambda_{1}$, and the negative root be $\lambda_{2}$. Then $\left|\lambda_{2}\right|>\lambda_{1}$.
For $n \in \mathbf{N}$, there exist constants $c_{1}$ and $c_{2}$ such that
$$
a_{n}=c_{1} \lambda_{1}^{n}+c_{2} \lambda_{2}^{n} \text {. }
$$

If $c_{2}>0$, then
$$
a_{2 n+1}=\lambda_{2}^{2 n+1}\left(c_{2}+c_{1}\left(\frac{\lambda_{1}}{\lambda_{2}}\right)^{2 n+1}\right) \text {. }
$$

By $\lim _{n \rightarrow \infty}\left(\frac{\lambda_{1}}{\lambda_{2}}\right)^{2 n+1}=0$ and $\lambda_{2}^{2 n+1}<0$, we know that when $n$ is sufficiently large, $a_{2 n}<0$, which is a contradiction.
Therefore, $c_{2}=0$.
Hence, $a_{n}=c_{1} \lambda_{1}^{n}(n \in \mathbf{N})$.
By $a_{0}=1$, we know $c_{1}=1$.
Thus, $a_{1}=\lambda_{1}=\frac{-q+\sqrt{q^{2}+4 p}}{2}$.