Hello,
Your approach is correct. Using buildmesh you can put a border that is inside the domain.
Then it has a label (1 in your case) that can be used to compute int1d(Th,1)().
This “internal border” is considered as a boundary. In particular if you write int1d(Th) without mentioning a label, it will integrate on all boundaries, including the “internal boundary”.
When you have a finite element function u that is discontinuous through the internal boundary (for example if u is P0 on Th), your question is “what computes int1d(Th,1)(u)?”.
The answer is that it uses the value of u only from one side: the side opposite to the direction of the normal (like if the normal were pointing outside the domain, and the value would be taken from inside the domain).
The orientation of the normal to the internal boundary [N.x,N.y] is determined by its orientation.
If you write for example
border b1(t=0.,1.){x=t ; y=1. ;label=1;}
then the orientation is from left to right, and the normal points down (N.y=-1), the value of u is taken from the up side.
If you write
border b1(t=0.,1.){x=1.-t ; y=1. ;label=1;} then the orientation is from right to left, the normal points upwards (N.y=1), the value of u is taken from the down side.
However the operators dx() and dy() are applied globally to a finite element function v.
For example if v is P1 on Th, then dx(v) is computed globally, it is P0 on Th.
If you write int1d(Th,1)(dx(v)) it will take the value of dx(v) only on the side determined by the orientation (as above).