---
title: "First-order implicit non-linear kinematic wave"
subject: "Computational Techniques in Civil Engineering"
chapter: 5
key: "semester_8/computational_techniques_in_civil_engineering:first-order-implicit-nonlinear-kinematic"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/first-order-implicit-nonlinear-kinematic
site: BCT Engineering
---

# First-order implicit non-linear kinematic wave

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

- Asked in: 2071 Bhadra, 6 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/first-order-implicit-nonlinear-kinematic
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5

## Question

Derive the first order accurate implicit finite difference equation for the kinematic wave model in the non-linear form.

## Short answer

Backward time and backward space at new time level yield: (Δt/Δx)Q_{i+1}^{n+1} + α(Q_{i+1}^{n+1})^0.6 = (Δt/Δx)Q_i^{n+1} + α(Q_{i+1}^n)^0.6 + Δt q̄, where α=(nP^{2/3}/√S₀)^{0.6}. Solved iteratively by Newton-Raphson.

## Answer

#### Starting equations
Continuity with lateral inflow $q$: $\dfrac{\partial A}{\partial t}+\dfrac{\partial Q}{\partial x}=q$. The kinematic approximation $S_f=S_0$ with Manning's equation gives the unique relation
$$
A=\alpha Q^{\beta},\qquad \alpha=\left(\frac{nP^{2/3}}{S_0^{1/2}}\right)^{0.6},\quad \beta=0.6
$$
so that the non-linear kinematic wave equation is
$$
\frac{\partial(\alpha Q^\beta)}{\partial t}+\frac{\partial Q}{\partial x}=q
$$

#### Grid and differences
```text
 t
 n+1 :  (i,n+1) o-----------o (i+1,n+1)
                |           * unknown
 n   :          .           o (i+1,n)
              i          i+1        --> x
```
The equation is written at the point $(i+1,\ n+1)$, using
- a backward difference in time: $\dfrac{\partial A}{\partial t}\approx\dfrac{A_{i+1}^{n+1}-A_{i+1}^n}{\Delta t}$
- a backward difference in space, evaluated at the **new** time level $n+1$: $\dfrac{\partial Q}{\partial x}\approx\dfrac{Q_{i+1}^{n+1}-Q_i^{n+1}}{\Delta x}$
- $q\approx\bar q$ (average lateral inflow per unit length over the step).

Both differences have errors $O(\Delta t)$ and $O(\Delta x)$, so the scheme is **first-order accurate**. Because the spatial difference uses new-time values it is implicit and unconditionally stable.

#### Finite difference equation
$$
\frac{\alpha\left(Q_{i+1}^{n+1}\right)^{\beta}-\alpha\left(Q_{i+1}^{n}\right)^{\beta}}{\Delta t}+\frac{Q_{i+1}^{n+1}-Q_i^{n+1}}{\Delta x}=\bar q
$$
Multiplying by $\Delta t$ and putting the unknown $Q_{i+1}^{n+1}$ on the left:
$$
\frac{\Delta t}{\Delta x}\,Q_{i+1}^{n+1}+\alpha\left(Q_{i+1}^{n+1}\right)^{\beta}=\frac{\Delta t}{\Delta x}\,Q_i^{n+1}+\alpha\left(Q_{i+1}^{n}\right)^{\beta}+\Delta t\,\bar q
$$
The right side contains only known values ($Q_i^{n+1}$ from the upstream point at the new time, $Q_{i+1}^n$ from the previous time).

#### Solution (Newton-Raphson)
Let $x=Q_{i+1}^{n+1}$ and define
$$
f(x)=\frac{\Delta t}{\Delta x}x+\alpha x^\beta-\left[\frac{\Delta t}{\Delta x}Q_i^{n+1}+\alpha\left(Q_{i+1}^n\right)^\beta+\Delta t\,\bar q\right]=0
$$
$$
x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)},\qquad f'(x)=\frac{\Delta t}{\Delta x}+\alpha\beta x^{\beta-1}
$$
Start with $x_0=Q_{i+1}^n$ and iterate until the change is negligible. The calculation proceeds from the upstream boundary ($i=1$) down the channel for each time step.

## Check yourself

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

1. The non-linear kinematic relation used is:
   A. A = alpha Q^beta with beta = 0.6
   B. Q = alpha A^0.6
   C. A = alpha Q^(5/3)
   D. A = alpha Q
   Answer: A. From Manning with Sf = S0, A = alpha Q^0.6. (Key point: Area-discharge law)

2. The scheme is implicit and first-order because:
   A. dQ/dx is a backward difference at n+1 and dA/dt a backward difference in time, with O(dt), O(dx) errors
   B. central differences are used at n+1/2
   C. dQ/dx is taken at level n
   D. it uses four points
   Answer: A. Using new-level values in the space difference makes it implicit; one-sided differences are first-order. (Key point: Why implicit)

3. The non-linear equation for Q(i+1,n+1) is solved by:
   A. direct substitution only
   B. Newton-Raphson starting from Q(i+1, n)
   C. Thomas algorithm
   D. Taylor series
   Answer: B. f(x)=0 is iterated with x_{k+1} = x_k - f/f'. (Key point: Solving)

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