---
title: "Upwind versus central scheme for convection"
subject: "Applied Computational Fluid Dynamics"
chapter: 3
key: "semester_7/applied_computational_fluid_dynamics:upwind-central-convection-numerical"
source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/upwind-central-convection-numerical
site: BCT Engineering
---

# Upwind versus central scheme for convection

Applied Computational Fluid Dynamics · Chapter 3: Discretization Methods

- Asked in: year not given, 6 marks
- Video: [Discretization of the Convective Term – Upwind Schemes](https://www.youtube.com/watch?v=zFwtF-WkBoU) — Applied Computational Fluid Dynamics
- 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/upwind-central-convection-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-3

## Question

For the linear convection equation $\partial\phi/\partial t+u\,\partial\phi/\partial x=0$ with $u = 2$ m/s, grid spacing $\Delta x = 0.1$ m and time step $\Delta t = 0.04$ s, find the Courant number. The values of $\phi$ at nodes 0 to 5 are 300, 300, 320, 360, 400, 400. Compute the new values at nodes 1 to 4 using (a) the first-order upwind scheme and (b) the central difference scheme in space with forward Euler in time. Comment on the results and on numerical diffusion.

## Short answer

Courant number C=uΔt/Δx=0.8. Upwind scheme (bounded, stable): φ₁=300, φ₂=304, φ₃=328, φ₄=368. Central difference (FTCS) gives unphysical undershoots: φ₁=292, φ₂=296, φ₃=328, φ₄=384. Upwind introduces numerical diffusion Γ_num=uΔx(1-C)/2=0.02 m²/s smearing gradients; central scheme is unconditionally unstable for pure convection.

## Answer

#### Courant number
$$
C=\frac{u\Delta t}{\Delta x}=\frac{2\times0.04}{0.1}=0.8
$$
The physical shift in one step is $u\Delta t=0.08$ m, i.e. 0.8 of a cell, so the exact solution moves the profile to the right by 0.8 cell.

#### (a) First-order upwind (flow in +x, use the upstream node)
$$
\phi_i^{n+1}=\phi_i^n-C(\phi_i^n-\phi_{i-1}^n)
$$
- $\phi_1=300-0.8(300-300)=300.0$
- $\phi_2=320-0.8(320-300)=304.0$
- $\phi_3=360-0.8(360-320)=328.0$
- $\phi_4=400-0.8(400-360)=368.0$

#### (b) Central difference (FTCS)
$$
\phi_i^{n+1}=\phi_i^n-\frac C2(\phi_{i+1}^n-\phi_{i-1}^n)
$$
- $\phi_1=300-0.4(320-300)=292.0$
- $\phi_2=320-0.4(360-300)=296.0$
- $\phi_3=360-0.4(400-320)=328.0$
- $\phi_4=400-0.4(400-360)=384.0$

| Node | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Old | 300 | 320 | 360 | 400 |
| Upwind | 300.0 | 304.0 | 328.0 | 368.0 |
| Central | 292.0 | 296.0 | 328.0 | 384.0 |

#### Comments
- Upwind results stay between the old minimum (300) and maximum (400): the scheme is **bounded** and stable for $C\le1$.
- The central scheme gives 292 and 296 at nodes 1 and 2, below the lowest old value of 300 (an unphysical undershoot). Forward Euler with central space differencing is **unconditionally unstable** for pure convection, so these oscillations grow with time.
- Upwind is first order. Its truncation error contains $\dfrac{u\Delta x}{2}(1-C)\dfrac{\partial^2\phi}{\partial x^2}$, which acts like an extra diffusion (**numerical or false diffusion**) with coefficient
$$
\Gamma_{num}=\frac{u\Delta x}{2}(1-C)=\frac{2\times0.1}{2}(1-0.8)=0.02\ \text{m}^2/\text{s}
$$
that smears sharp gradients. For $C=1$ it vanishes and the scheme is exact in 1D.
- Remedies: finer grids, higher-order bounded schemes (QUICK, TVD, second-order upwind) with limiters.

**Answer:** $C=0.8$; upwind 300, 304, 328, 368; central 292, 296, 328, 384; numerical diffusion 0.02 m²/s.

## Check yourself

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

1. For u = 2 m/s, dt = 0.04 s and dx = 0.1 m, what is the Courant number and what does it mean?
   A. C = 0.8; the profile should move 0.8 of a cell per time step
   B. C = 8; the profile should move 8 cells per time step
   C. C = 0.08; the profile should move 0.08 of a cell per time step
   D. C = 1.25; the profile should move 1.25 cells per time step
   Answer: A. C = u dt/dx = 2 x 0.04/0.1 = 0.8, so the exact solution shifts the profile by 0.8 of a cell each step. (Key point: Courant number)

2. For flow in the +x direction, which expression gives the first-order upwind update?
   A. phi_i - (C/2)(phi_(i+1) - phi_(i-1))
   B. phi_i - C(phi_i - phi_(i-1))
   C. phi_i - C(phi_(i+1) - phi_i)
   D. phi_i + C(phi_i - phi_(i-1))
   Answer: B. Upwind uses the upstream node i-1 when u > 0: phi_i^(n+1) = phi_i - C(phi_i - phi_(i-1)). (Key point: Upwind update)

3. With the central (FTCS) scheme, node 1 becomes 292, below the old minimum of 300. What does this show?
   A. The scheme is bounded and the result is physically correct
   B. The time step is too small
   C. The scheme is first order, so it adds numerical diffusion
   D. An unphysical undershoot; forward Euler with central differencing is unconditionally unstable for convection
   Answer: D. Values outside the old range indicate oscillations, which grow with time because FTCS is unconditionally unstable for pure convection. (Key point: Central scheme result and stability)

4. What is the numerical diffusion coefficient of the upwind scheme here, and when does it vanish?
   A. 0.02 m^2/s; it vanishes when C = 1
   B. 0.1 m^2/s; it vanishes when C = 0
   C. 0.02 m^2/s; it vanishes when C = 0.5
   D. 0.2 m^2/s; it vanishes when C = 1
   Answer: A. Gamma_num = (u dx/2)(1 - C) = 0.1 x 0.2 = 0.02 m^2/s, which is zero at C = 1 where upwind is exact in 1D. (Key point: Numerical diffusion)

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