FEM: periodic boundary conditions (1D)

[Niles]

I am trying to set up the mass matrix for a 1D system which I want to solve using finite elements. So the mass matrix is defined as

$$M = \int{NN^T}dL,$$
where N is the finite element linear basis functions. I use hat functions.

Say I have 10 elements, corresponding to 11 nodes running from -5 to 5 so the spacing is 1. Node 1 is equal to node 11 since I want to employ periodic boundary conditions.

My issue is that I am not sure how to construct the mass matrix for the 10th node. As shown here, the elements for the 10th node will be (I use periodic boundary conditions, so $x_{N+1}=x_1$)

$$M_{10,10} = \frac{x_{1}-x_{10}}{3} = -10/3\\ M_{10,1} = \frac{x_{1}-x_{10}}{6} = -10/6$$
All other elements have positive values given by 1/3 and 1/6, respectively.

Are my values for $M_{10,10}$ and $M_{10,1}$ correct? I find it odd that their values are so much different than the values in the "bulk".

 

[RUber]

This is 1D, so is it a line?
Is x_11 actually x_1 or is it x_1 + P where P is your period in X?
Thus f(x_11) = f(x_1) but x_11 is not actually x_1.
If this is the case, use x_11 for x_11 and in the matrix assign it the position of x_1 since it will be multiplied by that node.

 

[Niles]

Thanks, that is also what I thought. But then my matrix has dimensions 10x10, but my field will have 11 values since x_11 and x_1 are not the same. But that won't work when I multiply them together (?)

Am I missing something?

 

[RUber]

x_11 and x_1 are not the same in reality, but in the matrix they have the same role.
The main thing is that you are keeping with the periodic nature in your x, if all other entries are 1/3 and 1/6, then so should the ones at the edge.
You aren't actually putting the 11th value in, but when you plot it, you can manually put the 11th point in.