---
title: "Explicit scheme for transient conduction"
subject: "Applied Computational Fluid Dynamics"
chapter: 3
key: "semester_7/applied_computational_fluid_dynamics:explicit-ftcs-stability-numerical"
source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/explicit-ftcs-stability-numerical
site: BCT Engineering
---

# Explicit scheme for transient conduction

Applied Computational Fluid Dynamics · Chapter 3: Discretization Methods

- Asked in: year not given, 8 marks
- Video: [Explicit Methods for Solving the Diffusion Equation | Lecture 69 | Numerical Methods for Engineers](https://www.youtube.com/watch?v=MEqYYjGcnww) — Jeffrey Chasnov
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/explicit-ftcs-stability-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-3

## Question

A metal bar has thermal diffusivity $\alpha = 9.7\times10^{-5}$ m²/s. It is modelled with nodes 0 to 5 at spacing $\Delta x = 0.02$ m. Initially the temperatures of nodes 0 to 5 are 100, 20, 20, 20, 20, 0 °C, and the end nodes are held at 100 °C and 0 °C. (a) Write the explicit FTCS scheme for $\partial T/\partial t=\alpha\,\partial^2T/\partial x^2$ and its stability limit. (b) Find the largest stable time step. (c) With $\Delta t = 1$ s, compute the interior temperatures after two time steps.

## Short answer

FTCS scheme: T_i^{n+1}=T_i^n + r(T_{i+1}^n-2T_i^n+T_{i-1}^n) with r=αΔt/Δx^2. Stability requires r≤0.5. For α=9.7e-5 m²/s, Δx=0.02 m, max Δt=2.06 s. With Δt=1 s (r=0.2425), after two steps interior temperatures: 49.39, 24.70, 18.82, 12.65 °C.

## Answer

#### (a) FTCS scheme
Use a forward difference in time and a central difference in space:
$$
\frac{T_i^{n+1}-T_i^n}{\Delta t}=\alpha\frac{T_{i+1}^n-2T_i^n+T_{i-1}^n}{\Delta x^2}
$$
$$
T_i^{n+1}=T_i^n+r\,(T_{i+1}^n-2T_i^n+T_{i-1}^n),\qquad r=\frac{\alpha\Delta t}{\Delta x^2}
$$
The scheme is first order in time and second order in space. By von Neumann analysis the amplification factor is $G=1-4r\sin^2(\theta/2)$; $|G|\le1$ requires
$$
r\le\frac12
$$
(equivalently the coefficient of $T_i^n$, $1-2r$, must not be negative.)

#### (b) Maximum time step
$$
\Delta t_{max}=\frac{0.5\,\Delta x^2}{\alpha}=\frac{0.5\times(0.02)^2}{9.7\times10^{-5}}=2.06\ \text{s}
$$

#### (c) Two steps with $\Delta t=1$ s
$$
r=\frac{9.7\times10^{-5}\times1}{(0.02)^2}=0.2425\quad(<0.5,\ \text{stable})
$$
**Step 1** (from 100, 20, 20, 20, 20, 0):
- $T_1=20+0.2425(20-40+100)=20+19.40=39.40$
- $T_2=20+0.2425(20-40+20)=20.00$
- $T_3=20+0.2425(20-40+20)=20.00$
- $T_4=20+0.2425(0-40+20)=20-4.85=15.15$

**Step 2** (from 100, 39.40, 20, 20, 15.15, 0):
- $T_1=39.40+0.2425(20-78.8+100)=39.40+9.99=49.39$
- $T_2=20+0.2425(20-40+39.40)=20+4.70=24.70$
- $T_3=20+0.2425(15.15-40+20)=20-1.18=18.82$
- $T_4=15.15+0.2425(0-30.3+20)=15.15-2.50=12.65$

| Node | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| t = 0 | 100 | 20 | 20 | 20 | 20 | 0 |
| t = 1 s | 100 | 39.40 | 20.00 | 20.00 | 15.15 | 0 |
| t = 2 s | 100 | 49.39 | 24.70 | 18.82 | 12.65 | 0 |

Heat diffuses in from the hot end, and node 4 cools towards the 0 °C end.

**Answer:** $r\le0.5$, $\Delta t_{max}=2.06$ s; after 2 s the interior nodes are 49.39, 24.70, 18.82, 12.65 °C.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In FTCS, T_i^{n+1} is found from:
   A. Forward time difference and central space difference at the old time level
   B. Backward time difference at new level
   C. Average of old and new levels
   D. A matrix solve at each step
   Answer: A. FTCS is explicit: T_i^{n+1} = T_i^n + r (T_{i+1}^n - 2T_i^n + T_{i-1}^n). (Key point: FTCS scheme)

2. The stability limit for FTCS diffusion is:
   A. r >= 0.5
   B. r <= 1
   C. r <= 0.5, where r = alpha dt/dx^2
   D. r = 2
   Answer: C. Von Neumann analysis requires |G| <= 1, giving r <= 1/2. (Key point: Stability limit)

3. For alpha = 9.7e-5 and dx = 0.02 m, dt_max is:
   A. 1.03 s
   B. 4.12 s
   C. 0.5 s
   D. 2.06 s
   Answer: D. dt = 0.5 x 0.0004/9.7e-5 = 2.06 s. (Key point: Maximum time step)

4. With dt = 1 s, r equals:
   A. 0.485
   B. 0.2425
   C. 0.97
   D. 0.0485
   Answer: B. r = 9.7e-5 x 1/0.0004 = 0.2425, which is below 0.5, so stable. (Key point: Value of r)

5. After one step node 1 is:
   A. 20.00
   B. 49.39
   C. 39.40
   D. 100
   Answer: C. T1 = 20 + 0.2425 (20 - 40 + 100) = 39.40. (Key point: Node 1 after step 1)

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