[proofplan] We establish convergence by combining a local truncation error estimate with a stability bound. First, we derive the error recurrence by subtracting the Taylor expansion of the exact solution from the discrete scheme, obtaining $e_m^{n+1} = \mu e_{m-1}^n + (1 - 2\mu) e_m^n + \mu e_{m+1}^n + \tau_m^n$ where $\tau_m^n = O(h^4)$ is the local truncation error. Second, we take the $\ell^\infty$ norm and show that when $\mu \leq \frac{1}{2}$, the amplification factor is exactly $1$, yielding the recurrence $\|\mathbf{e}^{n+1}\|_\infty \leq \|\mathbf{e}^n\|_\infty + ch^4$. Induction over $n$ time steps, combined with the relation $n \leq T/k = T/(\mu h^2)$, converts the accumulated truncation error into an $O(h^2)$ bound that vanishes as $h \to 0$. [/proofplan]

[step:Derive the error recurrence from the scheme and Taylor expansion]Let $u: [0,1] \times [0,T] \to \mathbb{R}$ denote the exact solution of the heat equation $\partial_t u = \partial_{xx} u$ and let $U_m^n$ denote the numerical approximation at grid point $(x_m, t_n) = (mh, nk)$. Define the pointwise error \begin{align*} e_m^n := U_m^n - u(x_m, t_n), \qquad m = 0, \ldots, M+1, \quad n = 0, 1, 2, \ldots \end{align*} The explicit Euler scheme reads \begin{align*} U_m^{n+1} = \mu U_{m-1}^n + (1 - 2\mu) U_m^n + \mu U_{m+1}^n, \qquad \mu = \frac{k}{h^2}. \end{align*} For the exact solution, Taylor expansion in $t$ at $(x_m, t_n)$ gives $u(x_m, t_{n+1}) = u(x_m, t_n) + k \, \partial_t u(x_m, t_n) + O(k^2)$. Since $\partial_t u = \partial_{xx} u$, and the [Taylor Expansion of the Second Central Difference](/theorems/1363) applied to the smooth function $x \mapsto u(x, t_n)$ gives \begin{align*} u(x_{m-1}, t_n) - 2u(x_m, t_n) + u(x_{m+1}, t_n) = h^2 \, \partial_{xx} u(x_m, t_n) + O(h^4), \end{align*} we obtain \begin{align*} u(x_m, t_{n+1}) = \mu \, u(x_{m-1}, t_n) + (1 - 2\mu) \, u(x_m, t_n) + \mu \, u(x_{m+1}, t_n) + \tau_m^n, \end{align*} where the local truncation error $\tau_m^n = O(k^2) + O(k h^2) = O(h^4)$ (the last equality uses $k = \mu h^2$ with $\mu$ fixed). Subtracting this from the scheme equation: \begin{align*} e_m^{n+1} = \mu \, e_{m-1}^n + (1 - 2\mu) \, e_m^n + \mu \, e_{m+1}^n - \tau_m^n. \end{align*}[/step]

[guided]The idea is to subtract two equations — one satisfied by the numerical solution (exactly, by definition of the scheme) and one satisfied by the exact solution (approximately, up to truncation error). The difference isolates the error. The explicit Euler scheme defines the numerical solution: \begin{align*} U_m^{n+1} = \mu U_{m-1}^n + (1 - 2\mu) U_m^n + \mu U_{m+1}^n, \qquad \mu = \frac{k}{h^2}. \end{align*} For the exact solution $u$, we need to see how well it satisfies the same recurrence. Expanding $u(x_m, t_{n+1})$ in a Taylor series about $t_n$: \begin{align*} u(x_m, t_{n+1}) = u(x_m, t_n) + k \, \partial_t u(x_m, t_n) + \frac{k^2}{2} \, \partial_{tt} u(x_m, t_n^*), \end{align*} for some $t_n^* \in (t_n, t_{n+1})$. Since $u$ solves the heat equation, $\partial_t u = \partial_{xx} u$, so \begin{align*} u(x_m, t_{n+1}) = u(x_m, t_n) + k \, \partial_{xx} u(x_m, t_n) + O(k^2). \end{align*} By the [Taylor Expansion of the Second Central Difference](/theorems/1363), the spatial second difference satisfies \begin{align*} u(x_{m-1}, t_n) - 2u(x_m, t_n) + u(x_{m+1}, t_n) = h^2 \, \partial_{xx} u(x_m, t_n) + O(h^4). \end{align*} Substituting $k \, \partial_{xx} u = \mu \bigl[u(x_{m-1}, t_n) - 2u(x_m, t_n) + u(x_{m+1}, t_n)\bigr] + O(\mu h^4/1)$: \begin{align*} u(x_m, t_{n+1}) = \mu \, u(x_{m-1}, t_n) + (1 - 2\mu) \, u(x_m, t_n) + \mu \, u(x_{m+1}, t_n) + \tau_m^n, \end{align*} where $\tau_m^n = O(k^2) + O(kh^2)$. Since $k = \mu h^2$, both terms are $O(h^4)$, so there exists a constant $c_1 > 0$ (depending on bounds for $\partial_{tt} u$ and $\partial_{xxxx} u$) such that $|\tau_m^n| \leq c_1 h^4$. Subtracting the exact-solution relation from the scheme equation, and writing $e_m^n = U_m^n - u(x_m, t_n)$: \begin{align*} e_m^{n+1} = \mu \, e_{m-1}^n + (1 - 2\mu) \, e_m^n + \mu \, e_{m+1}^n - \tau_m^n. \end{align*} This is the error recurrence: the error at the next time step is a weighted combination of errors at the current time step, plus a local truncation error forcing term.[/guided]

[step:Show the amplification factor equals $1$ when $\mu \leq \frac{1}{2}$ and bound $\|\mathbf{e}^{n+1}\|_\infty$]Define $\|\mathbf{e}^n\|_\infty := \max_{1 \leq m \leq M} |e_m^n|$. Taking the absolute value of both sides of the error recurrence and applying the triangle inequality: \begin{align*} |e_m^{n+1}| \leq |\mu| \, |e_{m-1}^n| + |1 - 2\mu| \, |e_m^n| + |\mu| \, |e_{m+1}^n| + |\tau_m^n|. \end{align*} Since $\mu = k/h^2 > 0$, we have $|\mu| = \mu$. Under the hypothesis $0 < \mu \leq \frac{1}{2}$, the coefficient $1 - 2\mu \geq 0$, so $|1 - 2\mu| = 1 - 2\mu$. Bounding each $|e_{m \pm 1}^n|$ and $|e_m^n|$ by $\|\mathbf{e}^n\|_\infty$: \begin{align*} |e_m^{n+1}| \leq \bigl(\mu + (1 - 2\mu) + \mu\bigr) \|\mathbf{e}^n\|_\infty + c_1 h^4 = 1 \cdot \|\mathbf{e}^n\|_\infty + c_1 h^4. \end{align*} Taking the maximum over $m$: \begin{align*} \|\mathbf{e}^{n+1}\|_\infty \leq \|\mathbf{e}^n\|_\infty + c_1 h^4. \end{align*}[/step]

[guided]This is the stability step — we need the amplification factor (the coefficient of $\|\mathbf{e}^n\|_\infty$) to be at most $1$ so that errors do not grow from one time step to the next. From the error recurrence, taking absolute values and applying the triangle inequality: \begin{align*} |e_m^{n+1}| \leq \mu \, |e_{m-1}^n| + |1 - 2\mu| \, |e_m^n| + \mu \, |e_{m+1}^n| + |\tau_m^n|. \end{align*} The three coefficients are $\mu$, $|1 - 2\mu|$, and $\mu$. Their sum is $2\mu + |1 - 2\mu|$. When $\mu \leq \frac{1}{2}$, we have $1 - 2\mu \geq 0$, so $|1 - 2\mu| = 1 - 2\mu$, and the sum equals $2\mu + 1 - 2\mu = 1$. This is exactly the CFL condition for the explicit Euler method: the scheme is stable precisely when $\mu \leq \frac{1}{2}$. What goes wrong when $\mu > \frac{1}{2}$? Then $1 - 2\mu < 0$, so $|1 - 2\mu| = 2\mu - 1$, and the coefficient sum becomes $2\mu + 2\mu - 1 = 4\mu - 1 > 1$. The amplification factor exceeds $1$, and errors grow exponentially in $n$ — the scheme becomes unstable. With the amplification factor equal to $1$ and the truncation error bounded by $c_1 h^4$: \begin{align*} \|\mathbf{e}^{n+1}\|_\infty \leq \|\mathbf{e}^n\|_\infty + c_1 h^4. \end{align*}[/guided]

[step:Induct over time steps and use $n \leq T/(\mu h^2)$ to obtain convergence]At $t = 0$, the initial data is exact: $U_m^0 = u(x_m, 0)$, so $\|\mathbf{e}^0\|_\infty = 0$. Applying the recurrence $\|\mathbf{e}^{n+1}\|_\infty \leq \|\mathbf{e}^n\|_\infty + c_1 h^4$ inductively over $n$ steps: \begin{align*} \|\mathbf{e}^n\|_\infty \leq n \, c_1 h^4. \end{align*} The number of time steps to reach time $T$ is $n = T/k = T/(\mu h^2)$. Substituting: \begin{align*} \|\mathbf{e}^n\|_\infty \leq \frac{T}{\mu h^2} \cdot c_1 h^4 = \frac{c_1 T}{\mu} \, h^2. \end{align*} Since $\mu$ and $T$ are fixed positive constants, $\|\mathbf{e}^n\|_\infty \leq C h^2 \to 0$ as $h \to 0$, where $C = c_1 T / \mu$. This bound holds uniformly over all time levels $n \leq T/k$, establishing convergence of order $O(h^2)$ in the $\ell^\infty$ norm.[/step]

[guided]Starting from $\|\mathbf{e}^0\|_\infty = 0$ (exact initial data), the recurrence $\|\mathbf{e}^{n+1}\|_\infty \leq \|\mathbf{e}^n\|_\infty + c_1 h^4$ telescopes: \begin{align*} \|\mathbf{e}^n\|_\infty \leq \|\mathbf{e}^0\|_\infty + n \, c_1 h^4 = n \, c_1 h^4. \end{align*} The truncation error accumulates linearly in the number of time steps. The crucial question is: how many time steps are there? Since $k = \mu h^2$ and we integrate to time $T$: \begin{align*} n = \frac{T}{k} = \frac{T}{\mu h^2}. \end{align*} Substituting: \begin{align*} \|\mathbf{e}^n\|_\infty \leq \frac{T}{\mu h^2} \cdot c_1 h^4 = \frac{c_1 T}{\mu} \, h^2. \end{align*} This reveals the mechanism of convergence: each time step introduces an $O(h^4)$ error, but there are $O(h^{-2})$ time steps, so the total accumulated error is $O(h^4) \times O(h^{-2}) = O(h^2)$. The constraint $\mu \leq \frac{1}{2}$ ensures that the amplification factor is exactly $1$, so errors neither grow nor shrink between steps — they merely accumulate additively. This separation of local accuracy (consistency, $O(h^4)$ per step) from global error control (stability, amplification $\leq 1$) is the pattern formalised by the [Lax Equivalence Theorem](/theorems/1374): consistency plus stability implies convergence.[/guided]