Example 5 Find a function $f(m, n)$ that satisfies the following conditions: for every pair of non-negative integers $m, n$,
$$
\text { (1) } 2 f(m, n)=2+f(m+1, n-1)+f(m-1, n+1)(m \geqslant 1, n \geqslant 1) \text {; }
$$
(2) $f(m, 0)=f(0, n)=0$.
(1993 Korean Olympiad Problem)
From condition $(1)$, we have $f(m, n)-f(m-1, n-1)=f(m+1, n-1)-f(m, n)+2$.
In the above equation, substitute $m-k, n+k$ (non-negative integer $k \leqslant m$) for $m, n$, respectively, to get
$$
\begin{array}{l}
f(m-k, n+k)-f(m-k-1, n+k+1) \\
=f(m-k+1, n+k-1)-f(m-k, n+k)+2 .
\end{array}
$$

Again, in the above equation, let $k=0,1,2, \cdots, m-1$ successively, to get $m$ equations. Adding these $m$ equations, we get
$$
\begin{array}{l}
\sum_{k=0}^{m-1}[f(m-k, n+k)-f(m-k-1, n+k+1)] \\
=\sum_{k=0}^{m-1}[f(m-k+1, n+k-1)-f(m-k, n+k)+2] .
\end{array}
$$

Using condition (2), simplify to get
$$
\begin{array}{l}
f(m, n)=f(m+1, n-1)-f(1, n+m-1)+2 m \text {, i.e., } \\
f(m+1, n-1)-f(m, n)=f(1, n+m-1)-2 m .
\end{array}
$$

In the above equation, substitute $m+k, n-k$ (non-negative integer $k \leqslant n$) for $m, n$, respectively, to get
$$
f(m+k+1, n-k+1)-f(m+k, n-k)=f(1, n+m-1)-2(m+k) \text {. }
$$

Again, in the above equation, let $k=0,1,2, \cdots, n-1$ successively, to get $n$ equations. Adding these $n$ equations, we get
$$
\begin{array}{l}
\sum_{k=0}^{n-1}[f(m+k+1, n-k-1)-f(m+k, n-k)] \\
=n f(1, n+m-1)-2 \sum_{k=1}^{n-1}(m+k) .
\end{array}
$$

Using condition (2) again, simplify to get
$$
-f(m, n)=n f(1, n+m-1)-2 m n-n(n-1),
$$

i.e.,
$$
f(m, n)=(2 m+n-1) n-n f(1, n+m-1) \text {. }
$$

In equation (1), substitute $m-k, n+k$ (non-negative integer $k \leqslant n$) for $m, n$, respectively, to get
$$
f(m-k+1, n+k-1)-f(m-k, n+k)=f(1, n+m-1)-2(m-k) \text {. }
$$

In the above equation, let $k=1,2, \cdots, m$ successively, to get $m$ equations. Adding these $m$ equations, we get
$$
\begin{array}{l}
\sum_{k=1}^{m}[f(m-k+1, n+k-1)-f(m-k, n+k)] \\
=m f(1, n+m-1)-2 \sum_{k=1}^{m}(m-k) .
\end{array}
$$

Using condition (2) and simplifying, we get
$$
f(m, n)=m f(1, n+m-1)-m(m-1) \text {. }
$$

By multiplying equation (2) by $m$ and equation (3) by $n$, we get
$$
(m+n) \cdot f(m, n)=m n(m+n),
$$

Therefore, when $m, n$ are not both zero, we have $f(m, n)=m n$.
From condition (2), we know that the above equation also holds for $m=0$ and $n=0$.
Upon verification, the function $f(m, n)=m n$ satisfies all the conditions given in the problem. Therefore, the required function is
$$
f(m, n)=m n .
$$