[proofplan] We prove both directions. For the forward direction (stability + consistency $\implies$ convergence), we write the error at time level $n$ as a telescoping sum of amplified local truncation errors, bound each term using stability ($\|A_h^n\| \leq C_S$ for all $n$), and use consistency ($\|\eta^n\| \to 0$) to show the total error vanishes as $h \to 0$. For the converse (convergence $\implies$ stability, given consistency), we argue by contradiction: if the scheme were unstable, there would exist initial data for which the iterated amplification matrix $A_h^n$ produces unbounded growth, and this growth cannot be masked by the vanishing truncation error, contradicting convergence. [/proofplan] [step:Express the global error as a telescoping sum of amplified truncation errors] Let $u^n$ denote the numerical solution at time level $n$ and let $\widehat{u}^n$ denote the restriction of the exact solution to the grid at time $t_n$. The scheme gives $u^{n+1} = A_h u^n$, while the exact solution satisfies \begin{align*} \widehat{u}^{n+1} = A_h \widehat{u}^n + \eta^n, \end{align*} where $\eta^n$ is the local truncation error at time level $n$. Consistency means $\|\eta^n\| \to 0$ as $h \to 0$ (with $k/h^2$ fixed). Define the error vector $e^n := u^n - \widehat{u}^n$. Subtracting: \begin{align*} e^{n+1} = A_h e^n - \eta^n. \end{align*} Unrolling the recurrence from $e^0 = 0$ (exact initial data): \begin{align*} e^n = -\sum_{j=0}^{n-1} A_h^{n-1-j} \eta^j. \end{align*} [guided] The error recurrence $e^{n+1} = A_h e^n - \eta^n$ is a discrete analogue of Duhamel's principle: the error at time $n$ is the superposition of the truncation errors from all previous time steps, each propagated forward by the amplification matrix $A_h$. Starting from $e^0 = 0$: - $e^1 = A_h e^0 - \eta^0 = -\eta^0$, - $e^2 = A_h e^1 - \eta^1 = -A_h \eta^0 - \eta^1$, - $e^n = -\sum_{j=0}^{n-1} A_h^{n-1-j} \eta^j$. Each truncation error $\eta^j$ is amplified by $A_h^{n-1-j}$ — the number of remaining time steps. The question of convergence reduces to: does this sum stay bounded and tend to zero? [/guided] [/step] [step:Prove the forward direction: stability and consistency imply convergence] Stability means there exists a constant $C_S > 0$ such that $\|A_h^n\| \leq C_S$ for all $n \geq 0$ and all sufficiently small $h > 0$. Consistency means that for every $\varepsilon > 0$, there exists $h_0 > 0$ such that $\|\eta^j\| \leq \varepsilon$ for all $j$ whenever $h < h_0$. Taking norms in the telescoping representation and applying the triangle inequality: \begin{align*} \|e^n\| = \left\|\sum_{j=0}^{n-1} A_h^{n-1-j} \eta^j\right\| \leq \sum_{j=0}^{n-1} \|A_h^{n-1-j}\| \, \|\eta^j\| \leq \sum_{j=0}^{n-1} C_S \, \|\eta^j\|. \end{align*} Let $\tau_h := \max_{0 \leq j \leq n-1} \|\eta^j\|$ denote the maximum local truncation error. Then \begin{align*} \|e^n\| \leq n \, C_S \, \tau_h. \end{align*} For a consistent scheme of order $p$ in time, $\tau_h = O(k^p + h^q)$ for appropriate powers $p, q > 0$. Since $n = T/k$: \begin{align*} \|e^n\| \leq \frac{T}{k} \cdot C_S \cdot \tau_h. \end{align*} For a first-order method ($\tau_h = O(k)$), this gives $\|e^n\| \leq T \cdot C_S \cdot O(1) \cdot k \cdot (1/k) = C_S T \cdot O(k + h^q/k \cdot k) \to 0$ as $h \to 0$ with $k/h^2$ fixed. More generally, for any consistent scheme $\tau_h \to 0$ as $h \to 0$, and the product $n \cdot \tau_h = (T/k) \cdot \tau_h \to 0$ because the consistency order ensures $\tau_h$ decays faster than $k$ grows. Therefore $\|e^n\| \to 0$ as $h \to 0$. [guided] The forward direction is the workhorse of numerical PDE theory. The argument proceeds by a simple norm estimate on the telescoping sum. Stability provides the uniform bound $\|A_h^m\| \leq C_S$ for all $m \geq 0$. This is the key hypothesis — it says the amplification matrix does not blow up no matter how many times it is applied. Without this bound, the sum $\sum_{j=0}^{n-1} A_h^{n-1-j} \eta^j$ could grow without control even if each $\eta^j$ is small. Taking norms: \begin{align*} \|e^n\| \leq \sum_{j=0}^{n-1} \|A_h^{n-1-j}\| \, \|\eta^j\| \leq C_S \sum_{j=0}^{n-1} \|\eta^j\| \leq n \, C_S \, \tau_h, \end{align*} where $\tau_h = \max_j \|\eta^j\|$. Substituting $n = T/k$: \begin{align*} \|e^n\| \leq \frac{C_S \, T}{k} \cdot \tau_h. \end{align*} For convergence, we need $\tau_h / k \to 0$ as $h \to 0$. This is precisely what consistency guarantees: for a method of temporal order $p \geq 1$, $\tau_h = O(k^p + h^q)$, so $\tau_h / k = O(k^{p-1} + h^q / k) \to 0$ under a suitable mesh ratio. The mechanism is the same as in the [Convergence of Explicit Euler for the Heat Equation](/theorems/1373): each time step introduces a small truncation error, stability prevents amplification, and the total accumulated error is (number of steps) $\times$ (error per step), which vanishes when the per-step error decays faster than the step count grows. [/guided] [/step] [step:Prove the converse: convergence implies stability (given consistency)] Assume the scheme is consistent and convergent, and suppose for contradiction that it is not stable. Then there exist sequences $h_\nu \to 0$ and $n_\nu \in \mathbb{N}$ such that $\|A_{h_\nu}^{n_\nu}\| \to \infty$. By the definition of operator norm, for each $\nu$ there exists a unit vector $w_\nu$ with $\|w_\nu\| = 1$ and $\|A_{h_\nu}^{n_\nu} w_\nu\| \geq \frac{1}{2}\|A_{h_\nu}^{n_\nu}\|$. Choose the initial data $u^0 = \widehat{u}^0 + \delta_\nu w_\nu$ where $\delta_\nu > 0$ is a small perturbation parameter to be chosen. Then \begin{align*} e^0 = \delta_\nu w_\nu, \qquad e^{n_\nu} = A_{h_\nu}^{n_\nu}(\delta_\nu w_\nu) - \sum_{j=0}^{n_\nu - 1} A_{h_\nu}^{n_\nu - 1 - j} \eta^j. \end{align*} By the triangle inequality: \begin{align*} \|e^{n_\nu}\| \geq \delta_\nu \|A_{h_\nu}^{n_\nu} w_\nu\| - \sum_{j=0}^{n_\nu - 1} \|A_{h_\nu}^{n_\nu - 1 - j}\| \, \|\eta^j\|. \end{align*} Convergence requires $\|e^{n_\nu}\| \to 0$ for any choice of initial data converging to the exact initial data. But the first term grows as $\delta_\nu \|A_{h_\nu}^{n_\nu}\| / 2 \to \infty$ (since $\|A_{h_\nu}^{n_\nu}\| \to \infty$) for any fixed $\delta_\nu > 0$, while the truncation error sum remains bounded by consistency. Choosing $\delta_\nu$ small enough that $u^0 \to \widehat{u}^0$ but large enough that $\delta_\nu \|A_{h_\nu}^{n_\nu}\| \to \infty$ (for instance, $\delta_\nu = 1/\|A_{h_\nu}^{n_\nu}\|^{1/2}$), we obtain $\|e^{n_\nu}\| \to \infty$, contradicting convergence. Therefore the scheme must be stable. [guided] The converse direction shows that stability is not merely sufficient for convergence — it is necessary (given consistency). The argument is by contradiction. Suppose stability fails: there exist $h_\nu \to 0$ and integers $n_\nu$ with $\|A_{h_\nu}^{n_\nu}\| \to \infty$. We will construct initial data that converges to the exact initial data but for which the numerical error blows up. Since $\|A_{h_\nu}^{n_\nu}\| \to \infty$, there exist unit vectors $w_\nu$ with $\|A_{h_\nu}^{n_\nu} w_\nu\| \geq \frac{1}{2}\|A_{h_\nu}^{n_\nu}\|$. Perturb the initial data by $\delta_\nu w_\nu$ where $\delta_\nu = \|A_{h_\nu}^{n_\nu}\|^{-1/2}$. Then $\delta_\nu \to 0$, so the perturbed initial data converges to the exact initial data. The error at time $n_\nu$ satisfies \begin{align*} e^{n_\nu} = A_{h_\nu}^{n_\nu}(\delta_\nu w_\nu) - \sum_{j=0}^{n_\nu - 1} A_{h_\nu}^{n_\nu - 1 - j} \eta^j. \end{align*} The first term has norm at least $\delta_\nu \cdot \frac{1}{2}\|A_{h_\nu}^{n_\nu}\| = \frac{1}{2}\|A_{h_\nu}^{n_\nu}\|^{1/2} \to \infty$. The second term (truncation error accumulation) is bounded by $n_\nu \cdot \max_j \|A_{h_\nu}^{n_\nu-1-j}\| \cdot \tau_{h_\nu}$, but regardless of how this term behaves, the first term dominates for large $\nu$ because $\|A_{h_\nu}^{n_\nu}\|^{1/2} \to \infty$. Therefore $\|e^{n_\nu}\| \to \infty$, contradicting convergence. The essential point is that instability means certain directions in the solution space are amplified without bound. By placing initial data in those directions (even with vanishingly small amplitude), we force the error to grow, regardless of how small the truncation error is. Consistency cannot compensate for instability. [/guided] [/step]