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Second-order non-linear kinematic wave; FDM principle

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.

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

Non-linear kinematic wave model

Continuity: ∂A∂t+∂Q∂x=q\dfrac{\partial A}{\partial t}+\dfrac{\partial Q}{\partial x}=q. Kinematic approximation (Sf=S0S_f=S_0) with Manning's equation gives a unique QQ-AA relation, Q=S01/2nP2/3A5/3Q=\dfrac{S_0^{1/2}}{nP^{2/3}}A^{5/3}, or

A=αQβ,α=(nP2/3S01/2)0.6,β=0.6A=\alpha Q^{\beta},\qquad \alpha=\left(\frac{nP^{2/3}}{S_0^{1/2}}\right)^{0.6},\quad \beta=0.6

Therefore

∂(αQβ)∂t+∂Q∂x=q\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+12, n+12)(i+\tfrac12,\ n+\tfrac12):

 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
∂A∂t≈(Ain+1−Ain)+(Ai+1n+1−Ai+1n)2Δt,∂Q∂x≈(Qi+1n+1−Qin+1)+(Qi+1n−Qin)2Δx,q≈qˉ\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(Δt2)O(\Delta t^2) and O(Δx2)O(\Delta x^2), so the scheme is second-order accurate. Substituting A=αQβA=\alpha Q^\beta:

α[(Qin+1)β+(Qi+1n+1)β−(Qin)β−(Qi+1n)β]2Δt+Qi+1n+1+Qi+1n−Qin+1−Qin2Δx=qˉ\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 Qi+1n+1Q_{i+1}^{n+1}

Multiply by 2Δt2\Delta t and collect the unknown x=Qi+1n+1x=Q_{i+1}^{n+1} on the left:

f(x)=ΔtΔx x+αxβ−K=0f(x)=\frac{\Delta t}{\Delta x}\,x+\alpha x^{\beta}-K=0 K=α[(Qin)β+(Qi+1n)β−(Qin+1)β]−ΔtΔx(Qi+1n−Qin+1−Qin)+2Δt qˉK=\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 xx appears non-linearly, solve by Newton-Raphson:

xk+1=xk−f(xk)f′(xk),f′(x)=ΔtΔx+αβxβ−1x_{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 x0=Qi+1nx_0=Q_{i+1}^n (or Qin+1Q_i^{n+1}) and repeating until ∣xk+1−xk∣|x_{k+1}-x_k| is small. Marching along ii 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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