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In the first-order upwind scheme, numerical diffusion can be quantified as:

$$\frac{dT}{dt} = -w\left(\frac{dT}{dz}\right) + \left(\frac{wdz}{2}\right)\left(\frac{d^2T}{dz^2}\right)$$

For Lax-Wendroff, it is:

$$\frac{dT}{dt} = -w\left(\frac{dT}{dz}\right) + \left(\frac{w^2dt}{2}\right)\left(\frac{d^2T}{dz^2}\right)$$

It is said that the second order upwind scheme produces less numerical diffusion than the first order counterpart. How can it be quantified for a second-order upwind scheme?

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  • $\begingroup$ Can you provide the exact second-order stencil you have in mind? And the precise equation you want to solve? Presumably linear transport? $\endgroup$
    – Daniel
    Mar 23 at 16:06
  • $\begingroup$ Hi @Daniel, thanks for answering! I edited the post. I want to calculate the advection of temperature T with ice flow that has a velocity w and grid resolution dz. It is an advection equation (see figures). $\endgroup$ Mar 23 at 16:18
  • $\begingroup$ Out of curiosity, did you test your scheme? Did it turn out to actually be unstable? $\endgroup$
    – Daniel
    Mar 25 at 10:27
  • $\begingroup$ @Daniel Hi, yes I did test it. It is indeed unconditionally unstable. Just like the first order upwind the numerical stability seems to be determined by the CFL criterion. $\endgroup$ Mar 25 at 16:06

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The approach I would try here is that of a modified equation. I am not sure if it goes through for you full problem as your transport speed appears to change with $k$ and $t$. But I will outline how to analyse diffusion of your scheme when $w$ is constant.

First, assuming $w < 0$, I rewrite your scheme as

$$ T_k^{t+1} = T_k^t - \frac{w \Delta t}{2\Delta z} \left( -T^t_{k+2} + 4 T^t_{k+1} - 3 T^t_{k} \right) $$

Next, to determine the local error, I plug the analytic solution $T(x,t)$ into your numerical method. I will play fast and loose with notation but from now on, $T^t_k$ stands for $T(x_k, t)$. This allows me to do Taylor expansions

$$ T^{t+1}_k = T^t_k + \Delta t \partial_t T^t_k + \frac{\Delta t^2}{2} \partial_{tt} T^t_k + \mathcal{O}(\Delta t^3) $$

and

$$ T^t_{k+2} = T^t_k + 2 \Delta z \partial_z T^t_k + 2 \Delta z^2 \partial_{zz} T^t_k + \frac{8 \Delta z^3}{6} \partial_{zzz} T^t_k + \mathcal{O}(\Delta z^4) $$

and

$$ T^t_{k+1} = T^t_k + \Delta z \partial_z T^t_k + \frac{\Delta z^2}{2} \partial_{zz} T^t_k + \frac{\Delta z^3}{6} \partial_{zzz} T^t_k + \mathcal{O}(\Delta z^4) $$

If I plug these three expansions into the method, a lot of stuff cancels out and I am left with

$$ \Delta t \partial_t T^t_k + \frac{\Delta t^2}{2} \partial_{tt} T^t_k + \mathcal{O}(\Delta t^3) = -\frac{w \Delta t}{2 \Delta z} \left( 2 \Delta z \partial_z T^t_k + \frac{2 \Delta z^3}{3} \partial_{zzz} T^t_k + \mathcal{O}(\Delta z^4) \right) $$

Simplifying this gives me

$$ \partial_t T^t_k + w \partial_z T^t_k = - \frac{\Delta t}{2} \partial_{tt} T^t_k + \frac{w \Delta z^2}{3} \partial_{zzz} T^t_k + \mathcal{O}(\Delta t^2) + \mathcal{O}(\Delta z^3) $$

Lastly, because $\partial_{tt} T^t_k = w^2 \partial_{zz} T^t_k$, we get

$$ \partial_t T^t_k + w \partial_z T^t_k = \underbrace{- \frac{\Delta t w^2}{2} \partial_{zz} T^t_k}_{\text{numerical diffusion}} + \frac{w \Delta z^2}{3} \partial_{zzz} T^t_k + \mathcal{O}(\Delta t^2) + \mathcal{O}(\Delta z^3) $$

First, you can see that if your derivatives are all bounded, the error is of order $\mathcal{O}(\Delta t) + \mathcal{O}(\Delta z^2)$ which fits a first order explicit Euler time stepper combined with second order finite differences in space.

However, the problem is that the diffusion has the wrong sign: because $w^2 \Delta t > 0$, the modified equation for the local error is of the form $\partial_t T + w \partial_z T = -\alpha \partial_{zz} T$ with $\alpha > 0$ which is unstable.

This indicates that your scheme is unconditionally unstable. Have you tried to implement it to see what it does? I would expect it to blow up after some time, no matter how small you make $\Delta t$.

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