On a grid $x_i = i\Delta x$, $t_n = n\Delta t$, write $u_i^{n} \approx u(x_i, t_n)$. The explicit scheme is $$u_i^{n+1} = u_i^{n} + r\left(u_{i+1}^{n} - 2u_i^{n} + u_{i-1}^{n}\right),\qquad r = \frac{k\Delta t}{\Delta x^{2}},$$ with the end values given by the boundary conditions.
🔑 Key idea
The explicit scheme is stable only when $r \le \frac{1}{2}$, that is, $\Delta t \le \frac{\Delta x^{2}}{2k}$. Refining the space grid by a factor of 2 forces 4 times as many time steps.
💡 Why it works
Two views. First, rewrite the update as $u_i^{n+1} = ru_{i-1}^{n} + (1 - 2r)u_i^{n} + ru_{i+1}^{n}$. When $r \le \frac{1}{2}$ all three weights are nonnegative and add to 1, so each new value is an average of old ones and nothing can grow. Second, try the sawtooth $u_i = (-1)^{i}$: then $u_{i+1} - 2u_i + u_{i-1} = -4u_i$, so the sawtooth is multiplied by $1 - 4r$ each step. $|1 - 4r| \le 1$ exactly when $r \le \frac{1}{2}$; otherwise it grows and flips sign every step. (In method-of-lines terms, the fastest mode is $y' = \lambda y$ with $\lambda \approx -\frac{4k}{\Delta x^{2}}$, and Euler needs $\Delta t|\lambda| \le 2$.)
The explicit heat scheme on a rod with its ends held at 0. Below r = 1/2 the bump spreads smoothly; above it, a tiny sawtooth (like round-off) grows until it swamps the solution.
✏️ Worked example: One step
With $r = \frac{1}{4}$, the current row (ends held at 0) is $u = (0, 1, 2, 1, 0)$. Find the next row.
$u_1 = 1 + \frac{1}{4}(2 - 2 + 0) = 1$.
$u_2 = 2 + \frac{1}{4}(1 - 4 + 1) = 1.5$.
$u_3 = 1$ by symmetry. The peak drops and the heat spreads.
Answer: $(0, 1, 1.5, 1, 0)$
✏️ Worked example: The cost of refining
For $k = 1$ and $\Delta x = 0.1$, find the largest stable $\Delta t$. What happens if $\Delta x$ is halved?
$\Delta t \le \frac{(0.1)^{2}}{2} = 0.005$.
With $\Delta x = 0.05$: $\Delta t \le \frac{0.0025}{2} = 0.00125$, 4 times smaller.
Twice the grid points and 4 times the steps: 8 times the work to reach the same final time. In two space dimensions the same refinement costs 16 times as much.
Answer: $0.005$, then $0.00125$
✏️ Worked example: How fast instability wins
With $r = 0.6$, a sawtooth of size $10^{-16}$ (round-off) is present. How big is it after 100 steps?
Each step multiplies it by $|1 - 4(0.6)| = 1.4$.
$1.4^{100} \approx 4 \times 10^{14}$, so it reaches about $4 \times 10^{14} \times 10^{-16} = 0.04$, and a few dozen steps later it dominates the solution.
Answer: about $0.04$
⚠️ Common mistake
Refining the space grid without shrinking the time step. Halving $\Delta x$ quadruples $r$, which can push a stable computation past $r = \frac{1}{2}$. Always recompute $r = \frac{k\Delta t}{\Delta x^{2}}$ after changing either step. (Implicit schemes, such as backward Euler or Crank–Nicolson in time, remove this limit.)
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.
This is one of the top rungs of Numerical Analysis.
Sources and further reading
Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation (University of Delaware, free online edition): chapter "Diffusion equations", sections "The method of lines" and "Method of lines for parabolic PDEs"
MIT OpenCourseWare 18.330, Introduction to Numerical Analysis: lecture notes on finite differences for the heat equation