# Does length unit in FEM affect numerical condition?

I try to solve a system of coupled PDEs using FEM. Unfortunately, the originating matrix has very poor condition. After days of double checking and thinking, I suspect the following reason:

Given a PDE

a * Laplacian(u) + b * u = 0


For FEM, one has to calculate two operator matrices, one for Laplacian(u) and one for u.

When L is the element size, as far as I understand, the matrix entries for Laplacian(u) scale with 1 / L while the ones for u scale with L.

My system has a size of several micrometers, but I use meters as unit for lengths. So the elements for Laplacian(u) are very big, while the ones for u are very small.

So, could I improve the matrix condition by using a shorter length unit for the elements, maybe millimeters or micrometers? (I mean not the unit of u, but the unit of the geometry.)

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Really, it makes no difference... –  Jan May 9 at 13:43
Btw., if your system gets poor conditioned for finer resolutions this indicates that your actual system may be ill-posed. –  Jan May 9 at 13:47

No, a different unit will not alter the condition of the system. Say your FEM system is $$Au + \beta Mu = 0. \quad (*)$$ Then the parameter $\beta$, here something like $b/a$ from your example, will depend on the units in a way that only allows you to add "$A$" and "$M$" in terms of units. A rescaling will then mean a multiplication of $(*)$ by a constant which will not change the condition number of your system matrix.

The entries of $A$ and $M$ are intregals involving the shape functions. Their value is independent of the units in which you express them and they scale like $L$ or $1/L$. Say $L=1mm$ and $\beta = \frac{1}{mm^2}$. Expressing $L$ as $L=10^{-3}m$ will result in $$A \left [\frac{1}{mm}\right] + \beta\left[\frac{1}{mm^2}\right]M\left[mm\right] = 10^3A\left[\frac{1}{m}\right]+10^6\beta\left[\frac{1}{m^2}\right]10^{-3}M\left[m\right]$$ which is a simple scaling of the equations (*): $$10^3\left \{A\left[\frac{1}{m}\right] + \beta\left[\frac{1}{m^2}\right]M\left[m\right]\right\}$$

However, if your matrix entries drop below machine precision, then a rescaling will improve accuracy.

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I see... Roughly speaking, the constants (in my case diffusion coefficient) would scale in a contrary way as the size of the elements, compensating each other. –  Michael May 9 at 16:01