Explicit Finite Differences: Too Many Time Steps or Too Many Space Steps?

Worked Solution

How to Think About It: Both errors add work, but only one can destroy the answer. Stability of the explicit scheme depends on the *ratio* $a = \delta\tau/\delta x^2$. More time steps shrink $\delta\tau$ and hence $a$: safe. More space steps shrink $\delta x$ and grow $a$ like $1/\delta x^2$: eventually unstable.

Quick Estimate: Suppose you use $N = 400$ time steps on $[0, 0.1]$ ($\delta\tau = 2.5 \times 10^{-4}$) and start with $J = 20$ space intervals on $[0, 1]$: $a = 2.5\times10^{-4}/(0.05)^2 = 0.1$, fine. Refine to $J = 50$: $a = 2.5\times10^{-4}/(0.02)^2 = 0.625 > 0.5$, and the computed solution reaches $10^{53}$ instead of $O(1)$. Conversely, with $J = 20$ fixed, going from $N = 100$ ($a = 0.4$, error $10^{-3}$) to $N = 10{,}000$ ($a = 0.004$, error $7\times10^{-4}$) just costs a hundred times more work.

Formal Solution:

*Step 1 -- The scheme and its weights.* The explicit scheme is $$u_j^{n+1} = a\,u_{j-1}^n + (1 - 2a)\,u_j^n + a\,u_{j+1}^n, \qquad a = \frac{\delta\tau}{\delta x^2}.$$ If $a \le 1/2$ all three weights are nonnegative and sum to $1$, so each new value is a convex combination of old values and $\max_j |u_j^{n}|$ can never grow (a discrete maximum principle). If $a > 1/2$ the middle weight is negative and this argument fails.

*Step 2 -- Von Neumann stability analysis.* Insert a Fourier mode $u_j^n = g^n e^{i\theta j}$: $$g = 1 + a\left(e^{i\theta} - 2 + e^{-i\theta}\right) = 1 - 2a(1 - \cos\theta) = 1 - 4a\sin^2(\theta/2).$$ The mode is amplified unless $|g| \le 1$ for every $\theta$. The worst case is $\theta = \pi$ (the sawtooth mode), where $g = 1 - 4a$; requiring $g \ge -1$ gives $$a = \frac{\delta\tau}{\delta x^2} \le \frac12.$$ For $a > 1/2$ the sawtooth mode grows by $|1 - 4a| > 1$ per step: at $a = 0.625$ that is $1.5^{400} \approx 10^{70}$ over $400$ steps, which matches the blow-up seen numerically.

*Step 3 -- Time steps versus space steps.* Holding $\delta x$ fixed and adding time steps reduces $\delta\tau$, so $a$ falls and the condition is satisfied more comfortably; the only cost is run time (and slightly more accumulated rounding). Holding $\delta\tau$ fixed and adding space steps raises $a \propto 1/\delta x^2$; halving $\delta x$ quadruples $a$, so a modest refinement crosses the threshold and the scheme diverges. To refine the space grid safely you must refine time quadratically, $\delta\tau \propto \delta x^2$, which makes fine explicit grids expensive. Hence too many space steps is the worse mistake.

*Step 4 -- Implicit and Crank-Nicolson.* For the implicit scheme the amplification factor is $g = 1/(1 + 4a\sin^2(\theta/2))$, which satisfies $0 < g \le 1$ for every $a > 0$; for Crank-Nicolson $g = (1 - 2a\sin^2(\theta/2))/(1 + 2a\sin^2(\theta/2))$, with $|g| \le 1$ for every $a$. Both are unconditionally stable, so the time and space steps can be chosen independently for accuracy alone. Numerically, at $a = 0.64$ the implicit and Crank-Nicolson errors were $2\times10^{-4}$ and $5\times10^{-5}$ while the explicit scheme produced $10^{176}$.

Answer: Too many space steps is worse. The explicit scheme is stable only if $a = \delta\tau/\delta x^2 \le 1/2$ (all weights $a$, $1 - 2a$, $a$ nonnegative; von Neumann factor $1 - 4a\sin^2(\theta/2)$). Extra time steps shrink $a$ and only waste computation, whereas shrinking $\delta x$ with $\delta\tau$ fixed raises $a$ quadratically and makes the solution blow up. Implicit and Crank-Nicolson schemes are unconditionally stable, so the concern does not apply to them.