---
title: "Second-order scheme, linear kinematic wave"
subject: "Computational Techniques in Civil Engineering"
chapter: 5
key: "semester_8/computational_techniques_in_civil_engineering:second-order-linear-kinematic-scheme"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/second-order-linear-kinematic-scheme
site: BCT Engineering
---

# Second-order scheme, linear kinematic wave

Computational Techniques in Civil Engineering · Chapter 5: Finite difference method

- Asked in: 2070 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/second-order-linear-kinematic-scheme
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5

## Question

Derive a second order accurate finite difference scheme of the linear kinematic wave equation which computes discharge for unknown time and location.

## Short answer

Central differences in time and space about the cell centre give a second-order accurate scheme. Solving yields the unknown discharge at (i+1,n+1): Q_{i+1}^{n+1} = Q_i^n + [(1-C)/(1+C)](Q_{i+1}^n – Q_i^{n+1}) + [2C Δx/(1+C)] q̄, where C = c Δt/Δx and q̄ is the average lateral inflow over the cell.

## Answer

#### Linear kinematic wave equation
$$
\frac{\partial Q}{\partial t}+c\,\frac{\partial Q}{\partial x}=c\,q
$$
where $c=dQ/dA$ is taken as constant over the step and $q$ is the lateral inflow per unit length (zero if none). To get second-order accuracy, the derivatives are centred at the middle of the cell $(i+\tfrac12,\ n+\tfrac12)$ using all four corner points.

```text
 t
 n+1 :  (i,n+1) o-------o (i+1,n+1)  <- unknown
                |   *   |
 n   :  (i,n)   o-------o (i+1,n)
              i         i+1       --> x
```

#### Central differences about the cell centre
$$
\left.\frac{\partial Q}{\partial t}\right|_{i+\frac12}^{n+\frac12}\approx\frac{(Q_i^{n+1}-Q_i^n)+(Q_{i+1}^{n+1}-Q_{i+1}^n)}{2\Delta t},\qquad
\left.\frac{\partial Q}{\partial x}\right|_{i+\frac12}^{n+\frac12}\approx\frac{(Q_{i+1}^{n+1}-Q_i^{n+1})+(Q_{i+1}^n-Q_i^n)}{2\Delta x}
$$
Expanding each of the four values in a Taylor series about the centre, the first-derivative terms are retained and all second-derivative terms cancel, so the truncation errors are $O(\Delta t^2)$ and $O(\Delta x^2)$. The lateral inflow is also averaged:
$$
q\approx\bar q=\tfrac14\left(q_i^n+q_{i+1}^n+q_i^{n+1}+q_{i+1}^{n+1}\right)
$$

#### Finite difference equation
$$
\frac{Q_i^{n+1}+Q_{i+1}^{n+1}-Q_i^n-Q_{i+1}^n}{2\Delta t}+c\,\frac{Q_{i+1}^{n+1}+Q_{i+1}^n-Q_i^{n+1}-Q_i^n}{2\Delta x}=c\,\bar q
$$
Multiply by $2\Delta t$ and let $C=c\Delta t/\Delta x$:
$$
(1+C)\,Q_{i+1}^{n+1}=(1+C)\,Q_i^n+(1-C)\left(Q_{i+1}^n-Q_i^{n+1}\right)+2c\,\Delta t\,\bar q
$$

#### Result: discharge at the unknown time and location
$$
\boxed{\,Q_{i+1}^{n+1}=Q_i^n+\frac{1-C}{1+C}\left(Q_{i+1}^n-Q_i^{n+1}\right)+\frac{2C\,\Delta x}{1+C}\,\bar q\,}
$$
where the last term is for lateral inflow. Check of the coefficient: $2c\Delta t\bar q/(1+C)=2C\Delta x\bar q/(1+C)$.

#### Remarks
- $Q_i^n$, $Q_{i+1}^n$ are known from the previous time level and $Q_i^{n+1}$ from the upstream boundary (or the previous point), so $Q_{i+1}^{n+1}$ follows directly: march downstream along $i$ for each time step.
- Stability: the scheme is unconditionally stable. Accuracy is best for $C\approx1$; for $C\gg1$ or $C\ll1$ it can show numerical oscillations (dispersion).
- If $C=1$, the formula gives $Q_{i+1}^{n+1}=Q_i^n$ (exact translation of the wave).

## Check yourself

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

1. Second-order accuracy comes from centring derivatives at:
   A. point (i, n)
   B. the cell centre (i+1/2, n+1/2) using all four corners
   C. point (i+1, n+1)
   D. point (i, n+1)
   Answer: B. Central differences make second-derivative terms cancel, giving O(dt^2), O(dx^2). (Key point: Centring)

2. With zero lateral inflow, Q(i+1, n+1) equals:
   A. Q_i^n + (1-C)/(1+C) (Q_{i+1}^n - Q_i^{n+1})
   B. Q_i^n + C (Q_{i+1}^n - Q_i^{n+1})
   C. Q_{i+1}^n + (1+C)/(1-C)(Q_i^{n+1}-Q_i^n)
   D. (Q_i^n + Q_{i+1}^n)/2
   Answer: A. Multiplying by 2dt and solving for Q_{i+1}^{n+1}. (Key point: Final formula)

3. If C = 1 the formula gives:
   A. Q_{i+1}^{n+1} = Q_{i+1}^n
   B. Q_{i+1}^{n+1} = Q_i^n (exact translation)
   C. Q_{i+1}^{n+1} = 0
   D. it is undefined
   Answer: B. The coefficient (1-C)/(1+C) is zero. (Key point: C = 1)

4. The scheme's stability is:
   A. conditional, C <= 1
   B. unconditional, though large or small C may give oscillations
   C. unstable for any C
   D. stable only if C > 1
   Answer: B. It is unconditionally stable; accuracy is best near C = 1. (Key point: Stability)

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