Error Estimates. L1 or L2 norm?

Question

I simulate random walk on a divide difference grid to solve heat equation 1D. I want to prove numerically that this method has $Ν^{-1/2}$ error accuracy. My problem is that I don't know which norm must choose. I use $L_{2}$ norm to simulate error for Laplace(2D) equation and I got these results: \

The formula is $$\epsilon=\frac{100}{\psi_{max}}\sqrt{\frac{1}{M}\sum_{i=1}^{M}(\psi_{i}^{n}-\psi_{i}^{e})^2}$$ where $\psi_{i}^{e}$ is the analytical solution and $\psi_{i}^{n}$ is the random walk solution and M is the number of randomly chosen points in the domain where the unknown $\psi$ is evaluated.page 14

I use this formula for heat 1D and I don't take the same results.

Can anybody help me?

Answer 1

There is no reason to have $N^{-\frac{1}{2}}$ accuracy because it's Monte carlo method. Maybe it's more slow from this , if you plot the log log plot you'll see it .

Answer 2

If your error is (eventually, asymptotically) proportional to $N^{-1/2}$: $$\epsilon=\alpha N^{-1/2}$$ You should plot this on a log-log scale: $$\log(\epsilon)=\log(\alpha)-\frac{1}{2}\log(N)$$ and verify that, for sufficiently large $N$, your curve is (approximately) a straight line with slope $-\frac{1}{2}$.