1
$\begingroup$

I have some data: number of nodes $N$ and error in energy norm corresponing to it.

I have seen in some references that the rate of convergence is reported by

$$\| u-u_h\| _E=CN^{\alpha} $$

How can I find $C$ and $\alpha$ by MATLAB, I tried by :

polyfit $(N,\| u-u_h\| _E,1)$

But the answer was not true.

How can I find them.

$\endgroup$
6
$\begingroup$

First of all, rates of convergence are usually given in the form $$ \|u-u_h\| \leq C N^\alpha,$$ rather than equality. Furthermore, rates are asymptotic, i.e., only have to hold for $N\to \infty$. This means that you're unlikely to find a single $C$ and $\alpha$ such that your equation holds.

Another reason why your approach doesn't work is that what you're trying to fit is not a polynomial, since $\alpha$ is a) unknown and b) not an integer (for one thing, it must be negative since the error goes down as $N$ goes up).

What people usually do is look at a double-logarithmic plot: If you take the logarithm of your equation (or my inequality), you get $$ log(\|u-u_h\|) \leq \log(CN^\alpha) = \log(C) + \alpha\log(N).$$ This is a linear polynomial $ax+b$ in $\log (N)$ with coefficients $a=\alpha$ and $b=\log(C)$ (from which you can find $C=\exp(b)$).

So if you apply polyfit to $\log(N)$, $\log(\|u-u_h\|)$, you should get an array $[a,b]$ with $\alpha=a$ and $C=\exp(b)$.

$\endgroup$
1
$\begingroup$

I think that you should be looking for a rate of convergence in the form $$\|u-u_h\|_E \leq Ch^{\alpha}$$ where $h$ is characteristic mesh size. In your case, a rough estimate would be $$h = \frac{1}{\sqrt{N}}$$ in two dimensions.

You need to evaluate the discretization error on several grids with decreasing $h$ and then use the polynomial fit as [alpha, C] = polyfit(lc, error, 1) where lc is a vector of logarithms of characteristic lengths lc = [log(h1), log(h2), log(h3), ..., log(hn)] and error is a vector of logarithms of corresponding error norms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.