[proofplan]
We apply the Taylor expansion of the second central difference (Theorem 1363) separately in the $x$-direction and the $y$-direction, then add the two results. The second-order terms combine to give $h^2(\partial_{xx} u + \partial_{yy} u) = h^2 \Delta u$, and the fourth-order remainders contribute $\mathcal{O}(h^4)$.
[/proofplan]
[step:Apply the second central difference expansion in the $x$-direction]
Fix $(x, y) \in \Omega$ with $\operatorname{dist}((x, y), \partial\Omega) > h$. Define the function
\begin{align*}
g_1: [a, b] &\to \mathbb{R} \\
s &\mapsto u(s, y),
\end{align*}
where $[a, b]$ is an interval containing $x - h$ and $x + h$ inside $\Omega$. Since $u \in C^4(\Omega)$, the map $g_1$ inherits $C^4$ regularity. By the [Taylor Expansion of the Second Central Difference](/theorems/1363) applied to $g_1$ at $s = x$:
\begin{align*}
\Delta_{h,x}^* u(x, y) &= u(x - h, y) - 2u(x, y) + u(x + h, y) \\
&= h^2 \frac{\partial^2 u}{\partial x^2}(x, y) + \frac{h^4}{12} \frac{\partial^4 u}{\partial x^4}(x, y) + \mathcal{O}(h^6).
\end{align*}
[/step]
[step:Apply the second central difference expansion in the $y$-direction]
Similarly, define the function
\begin{align*}
g_2: [c, d] &\to \mathbb{R} \\
t &\mapsto u(x, t),
\end{align*}
where $[c, d]$ is an interval containing $y - h$ and $y + h$ inside $\Omega$. Again $g_2 \in C^4$ since $u \in C^4(\Omega)$. By the [Taylor Expansion of the Second Central Difference](/theorems/1363) applied to $g_2$ at $t = y$:
\begin{align*}
\Delta_{h,y}^* u(x, y) &= u(x, y - h) - 2u(x, y) + u(x, y + h) \\
&= h^2 \frac{\partial^2 u}{\partial y^2}(x, y) + \frac{h^4}{12} \frac{\partial^4 u}{\partial y^4}(x, y) + \mathcal{O}(h^6).
\end{align*}
[/step]
[step:Sum the two expansions to obtain the five-point approximation to the Laplacian]
Adding the expansions from the two previous steps:
\begin{align*}
\bigl(\Delta_{h,x}^* + \Delta_{h,y}^*\bigr) u(x, y) &= u(x - h, y) + u(x + h, y) + u(x, y - h) + u(x, y + h) - 4u(x, y) \\
&= h^2 \left(\frac{\partial^2 u}{\partial x^2}(x, y) + \frac{\partial^2 u}{\partial y^2}(x, y)\right) + \frac{h^4}{12}\left(\frac{\partial^4 u}{\partial x^4}(x, y) + \frac{\partial^4 u}{\partial y^4}(x, y)\right) + \mathcal{O}(h^6).
\end{align*}
Since the Laplacian in two dimensions is $\Delta u = \partial_{xx} u + \partial_{yy} u$, the leading term is $h^2 \Delta u(x, y)$. The fourth-order terms $\frac{h^4}{12}(\partial_{xxxx} u + \partial_{yyyy} u)$ are $\mathcal{O}(h^4)$, and the sixth-order remainders are absorbed into $\mathcal{O}(h^4)$ as well. Therefore:
\begin{align*}
\bigl(\Delta_{h,x}^* + \Delta_{h,y}^*\bigr) u(x, y) = h^2 \Delta u(x, y) + \mathcal{O}(h^4).
\end{align*}
[guided]
The five-point stencil $u(x - h, y) + u(x + h, y) + u(x, y - h) + u(x, y + h) - 4u(x, y)$ is the sum of two independent one-dimensional second central differences: one in $x$ (holding $y$ fixed) and one in $y$ (holding $x$ fixed).
The [Taylor Expansion of the Second Central Difference](/theorems/1363) tells us that each one-dimensional second central difference applied to a $C^4$ function satisfies
\begin{align*}
g(s - h) - 2g(s) + g(s + h) = h^2 g''(s) + \frac{h^4}{12} g^{(4)}(s) + \mathcal{O}(h^6).
\end{align*}
Applying this to $g_1(s) = u(s, y)$ (with $y$ fixed) recovers the $x$-direction contribution $h^2 \partial_{xx} u + \frac{h^4}{12}\partial_{xxxx} u + \mathcal{O}(h^6)$. Applying it to $g_2(t) = u(x, t)$ (with $x$ fixed) recovers $h^2 \partial_{yy} u + \frac{h^4}{12}\partial_{yyyy} u + \mathcal{O}(h^6)$. Summation gives:
\begin{align*}
&u(x - h, y) + u(x + h, y) + u(x, y - h) + u(x, y + h) - 4u(x, y) \\
&\quad = h^2 \underbrace{(\partial_{xx} u + \partial_{yy} u)}_{\Delta u} + \frac{h^4}{12}(\partial_{xxxx} u + \partial_{yyyy} u) + \mathcal{O}(h^6) \\
&\quad = h^2 \Delta u(x, y) + \mathcal{O}(h^4).
\end{align*}
Note that the $\mathcal{O}(h^4)$ error involves only the unmixed fourth-order partial derivatives $\partial_{xxxx} u$ and $\partial_{yyyy} u$, not the mixed derivative $\partial_{xxyy} u$. This is because each one-dimensional central difference operates on a single variable. The absence of the mixed derivative is a structural feature of the five-point stencil, not a coincidence.
[/guided]
[/step]
