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

# Second-order non-linear kinematic wave; FDM principle

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

- Asked in: 2079 Shrawan, 5+1 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-nonlinear-kinematic-principle
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5

## Question

Derive the space-time discretization of the second order accurate non-linear kinematic wave model. Write the principle of the finite difference method.

## Short answer

The nonlinear kinematic wave equation ∂(αQ^β)/∂t+∂Q/∂x=q is discretized with a second-order four-point scheme centred at (i+½, n+½), leading to (Δt/Δx)Q+αQ^β−K=0 solved by Newton-Raphson. The finite difference principle: derivatives are replaced by differences on a grid to convert a PDE into algebraic equations.

## Answer

#### Non-linear kinematic wave model
Continuity: $\dfrac{\partial A}{\partial t}+\dfrac{\partial Q}{\partial x}=q$. Kinematic approximation ($S_f=S_0$) with Manning's equation gives a unique $Q$-$A$ relation, $Q=\dfrac{S_0^{1/2}}{nP^{2/3}}A^{5/3}$, or
$$
A=\alpha Q^{\beta},\qquad \alpha=\left(\frac{nP^{2/3}}{S_0^{1/2}}\right)^{0.6},\quad \beta=0.6
$$
Therefore
$$
\frac{\partial(\alpha Q^\beta)}{\partial t}+\frac{\partial Q}{\partial x}=q
$$

#### Space-time discretisation (second order, four-point/box)
Centre the derivatives at the middle of the cell, $(i+\tfrac12,\ n+\tfrac12)$:
```text
 t
 n+1 :  (i,n+1) o------o (i+1,n+1)   <- Q(i,n+1) known,
                |  *   |                  Q(i+1,n+1) unknown
 n   :  (i,n)   o------o (i+1,n)    <- both known
              i        i+1       --> x
```
$$
\frac{\partial A}{\partial t}\approx\frac{(A_i^{n+1}-A_i^n)+(A_{i+1}^{n+1}-A_{i+1}^n)}{2\Delta t},\qquad
\frac{\partial Q}{\partial x}\approx\frac{(Q_{i+1}^{n+1}-Q_i^{n+1})+(Q_{i+1}^n-Q_i^n)}{2\Delta x},\qquad
q\approx\bar q
$$
Taylor expansion about the cell centre shows that both errors are $O(\Delta t^2)$ and $O(\Delta x^2)$, so the scheme is second-order accurate. Substituting $A=\alpha Q^\beta$:
$$
\frac{\alpha\left[(Q_i^{n+1})^\beta+(Q_{i+1}^{n+1})^\beta-(Q_i^n)^\beta-(Q_{i+1}^n)^\beta\right]}{2\Delta t}+\frac{Q_{i+1}^{n+1}+Q_{i+1}^n-Q_i^{n+1}-Q_i^n}{2\Delta x}=\bar q
$$

#### Solution for the one unknown $Q_{i+1}^{n+1}$
Multiply by $2\Delta t$ and collect the unknown $x=Q_{i+1}^{n+1}$ on the left:
$$
f(x)=\frac{\Delta t}{\Delta x}\,x+\alpha x^{\beta}-K=0
$$
$$
K=\alpha\left[(Q_i^n)^\beta+(Q_{i+1}^n)^\beta-(Q_i^{n+1})^\beta\right]-\frac{\Delta t}{\Delta x}\left(Q_{i+1}^n-Q_i^{n+1}-Q_i^n\right)+2\Delta t\,\bar q
$$
Because $x$ appears non-linearly, solve by Newton-Raphson:
$$
x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)},\qquad f'(x)=\frac{\Delta t}{\Delta x}+\alpha\beta x^{\beta-1}
$$
starting from $x_0=Q_{i+1}^n$ (or $Q_i^{n+1}$) and repeating until $|x_{k+1}-x_k|$ is small. Marching along $i$ from the upstream boundary and then in time gives the whole hydrograph.

#### Principle of the finite difference method
The solution domain is covered by a grid; every derivative in the differential equation is replaced by a difference of grid-point values (from the Taylor series), converting the PDE into algebraic equations that are solved for the unknown values at the grid points.

## Check yourself

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

1. What does the kinematic assumption S_f = S_0 give?
   A. Q depends on time only
   B. A unique Q-A relation, A = alpha Q^beta (beta = 0.6 for Manning)
   C. No relation between Q and A
   D. A is constant
   Answer: B. With S_f = S_0 the flow is at normal depth, so Manning gives Q proportional to A^(5/3), hence A = alpha Q^0.6. (Key point: A-Q relation)

2. Where are the derivatives centred in the second-order four-point scheme?
   A. At (i, n)
   B. At (i+1, n+1)
   C. At (i-1, n)
   D. At the cell centre (i+1/2, n+1/2)
   Answer: D. Centring at the middle of the cell makes both time and space errors second order. (Key point: Box scheme centring)

3. Why is Newton-Raphson used for Q_{i+1}^{n+1}?
   A. Because the unknown appears non-linearly through x^beta
   B. Because the grid is irregular
   C. Because dt is small
   D. Because the equation is linear
   Answer: A. The term alpha x^0.6 makes the equation non-linear, so iterate x = x - f/f'. (Key point: Solving the unknown)

4. What is the principle of the finite difference method?
   A. Integrate the PDE over the whole domain analytically
   B. Replace the grid by a continuous function
   C. Use trial functions over elements
   D. Cover the domain with a grid and replace derivatives by differences from Taylor series
   Answer: D. The PDE becomes algebraic equations in the grid-point values. (Key point: Principle of FDM)

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