---
title: "Explicit, implicit and Crank-Nicolson schemes"
subject: "Applied Computational Fluid Dynamics"
chapter: 3
key: "semester_7/applied_computational_fluid_dynamics:explicit-implicit-crank-nicolson"
source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/explicit-implicit-crank-nicolson
site: BCT Engineering
---

# Explicit, implicit and Crank-Nicolson schemes

Applied Computational Fluid Dynamics · Chapter 3: Discretization Methods

- Asked in: year not given, 5 marks
- Video: [Implicit vs Explicit Scheme](https://www.youtube.com/watch?v=1HM8JkkiVIQ) — Knowledge Engineering
- 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-implicit-crank-nicolson
- Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-3

## Question

Differentiate between explicit, implicit and Crank-Nicolson time-marching schemes for the one-dimensional transient diffusion equation. Write the discretised form of each and compare accuracy, stability and cost.

## Short answer

Explicit (FTCS): T_i^{n+1}=T_i^n+rδ²T_i^n, O(Δt,Δx²), conditionally stable (r≤0.5). Implicit (BTCS): -rT_{i-1}^{n+1}+(1+2r)T_i^{n+1}-rT_{i+1}^{n+1}=T_i^n, O(Δt,Δx²), unconditionally stable. Crank-Nicolson: average of explicit/implicit, O(Δt²,Δx²), unconditionally stable. Explicit cheap per step but tiny Δt; implicit/CN require tridiagonal solves but allow larger Δt.

## Answer

For $\partial T/\partial t=\alpha\,\partial^2T/\partial x^2$, let $r=\alpha\Delta t/\Delta x^2$ and $\delta^2T_i=T_{i+1}-2T_i+T_{i-1}$.

#### Discretised forms
- **Explicit (FTCS):** $T_i^{n+1}=T_i^n+r\,\delta^2T_i^{\,n}$. The right side uses only old values, so each new value is found directly.
- **Fully implicit (BTCS):** $T_i^{n+1}=T_i^n+r\,\delta^2T_i^{\,n+1}$, or
$-rT_{i-1}^{n+1}+(1+2r)T_i^{n+1}-rT_{i+1}^{n+1}=T_i^n$. This needs a tridiagonal system solved each step (Thomas algorithm).
- **Crank-Nicolson:** average of the two, $T_i^{n+1}=T_i^n+\dfrac r2\left(\delta^2T_i^{\,n}+\delta^2T_i^{\,n+1}\right)$. Also tridiagonal.

#### Comparison

| Property | Explicit | Implicit | Crank-Nicolson |
|---|---|---|---|
| Time level of spatial terms | $n$ | $n+1$ | average of $n$ and $n+1$ |
| Accuracy | $O(\Delta t,\Delta x^2)$ | $O(\Delta t,\Delta x^2)$ | $O(\Delta t^2,\Delta x^2)$ |
| Stability | Conditional: $r\le0.5$ | Unconditional | Unconditional |
| Work per step | Very small | Matrix solve | Matrix solve |
| Time step | Restricted by stability | Chosen by accuracy | Chosen by accuracy |
| Weakness | Tiny $\Delta t$ on fine mesh | Only first-order in time; damps | May oscillate for large $r$ |

#### Remarks
- With the explicit scheme, halving $\Delta x$ forces $\Delta t$ to be one quarter, so cost rises about 8 times in 1D.
- Implicit methods are used in most commercial CFD solvers because stable large time steps are possible, especially for steady-state solutions and stiff problems.
- Unconditional stability does not mean accuracy: too large a $\Delta t$ still gives a poor transient.

## Check yourself

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

1. In the explicit scheme the spatial terms are evaluated at:
   A. The new time level n+1
   B. The average of n and n+1
   C. The old time level n, so each new value is found directly
   D. A fixed boundary
   Answer: C. FTCS uses only old values on the right-hand side. (Key point: Explicit scheme)

2. Which statement about stability is correct?
   A. Explicit is unconditionally stable
   B. Implicit and Crank-Nicolson are unconditionally stable; explicit needs r <= 0.5
   C. Crank-Nicolson needs r <= 0.5
   D. All three are conditionally stable
   Answer: B. Only the explicit scheme has a stability restriction on the time step. (Key point: Stability comparison)

3. Crank-Nicolson accuracy is:
   A. O(dt, dx^2)
   B. O(dt, dx)
   C. O(dt^2, dx^2)
   D. O(dt^2, dx^4)
   Answer: C. Averaging the time levels makes it second order in time. (Key point: Crank-Nicolson accuracy)

4. If dx is halved in the explicit scheme, the work increases about:
   A. 2 times
   B. 4 times
   C. 16 times
   D. 8 times
   Answer: D. dt falls to one quarter, and there are twice as many nodes, so about 8 times more work in 1D. (Key point: Cost of explicit scheme)

## Related in Applied Computational Fluid Dynamics

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