11. (20 points) For all non-negative integers $m, n$ $(m \geqslant n)$, the sequence $a_{0}, a_{1}, \cdots$ satisfies
$$
a_{m+n}+a_{m-n}+2 m-2 n-1=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right) \text {. }
$$

If $a_{1}=0, a_{p}=2019^{2}$, find the value of $p$.
11. When $m=0, n=0$, $a_{0}=1$.

When $n=0$, $a_{2 m}=4 a_{m}+4 m-3$.
Let $m=1$, we get $a_{2}=4 a_{1}+4-3=1$;
Let $m=2$, we get $a_{4}=4 a_{2}+8-3=9$.
When $m=2, n=1$,
$$
\begin{array}{l}
a_{3}+a_{1}+4-2-1=\frac{1}{2}\left(a_{4}+a_{2}\right) \\
\Rightarrow a_{3}=4 .
\end{array}
$$

We will prove by mathematical induction that: $a_{n}=(n-1)^{2}$.
When $k=0$ and 1, we have $a_{k}=(k-1)^{2}$, the conclusion holds.
Assume that when $k \leqslant m$, the conclusion holds.
$$
\begin{array}{l}
\text { By } a_{2 m}=4 a_{m}+4 m-3 \\
=4(m-1)^{2}+4 m-3 \\
=4 m^{2}-4 m+1=(2 m-1)^{2},
\end{array}
$$

Let $n=1$, we get
$$
\begin{array}{l}
a_{m+1}+a_{m-1}+2 m-3=\frac{1}{2}\left(a_{2 m}+a_{2}\right) \\
\Rightarrow a_{m+1}=\frac{1}{2}\left(a_{2 m}+a_{2}\right)-a_{m-1}-2 m+3 \\
\quad=\frac{1}{2}\left((2 m-1)^{2}+1\right)-(m-1-1)^{2}-2 m-3 \\
\quad=m^{2}=((m+1)-1)^{2} .
\end{array}
$$

Thus, when $k=m+1$, the conclusion also holds.
In summary, $a_{n}=(n-1)^{2}$ holds for all positive integers.
Since $a_{p}=2019^{2}$, we have,
$$
(p-1)^{2}=2019^{2} \Rightarrow p=2020
$$