Dear all
I want to have two regions and then combine them into 1 region (th=th1+th2) because some integrals are solved on th and some on th1. so I made the mesh as follow:
real xr=6., lsponge=60., Xmax1=100., yr=3., yside=50.;
border in3(tt=1.,0.){real t=tt^1.7; x=0.; y=(yside-yr)t+yr;label=2;}
border interior1(t=0.,1.){x=xrt; y=yr;}
border interior2(t=1.,0.){x=xr; y=yrt;}
border axis2(tt=0.,1.){real t=tt^1.7; x=(lsponge-xr)t+xr; y=0.; label=5;}
border axis3(t=0.,1.){x=(Xmax1-lsponge)t+lsponge; y=0.; label=5;}
border out(t=0.,1.){x=Xmax1; y=ysidet; label=3;}
border side(t=1.,0.){x=Xmax1t; y=yside; label=4;}
border interior3(t=1.,0.){x=xrt; y=yr;}
border interior4(t=0.,1.){x=xr; y=yrt;}
border in2(t=1.,0.){x=0.; y=(yr-1.)t+1.; label=2;}
border in1(t=1.,0.){x=0.; y=t; label=1;}
border axis1(t=0.,1.){x=xrt; y=0.; label=5;}
mesh th1=buildmesh(in3(60)+interior1(10xr)+interior2(10yr)+axis2(80)+axis3(30)+out(20)+side(30));
mesh th2=buildmesh(in2(20)+in1(18)+interior3(10xr)+interior4(10*yr)+axis1(50));
mesh th=th1+th2;
but when I solve the navier stokes equation and put the boundary conditions , it seems that interiors (1,2,3,4) cant be considered as interfaces between th1 and th2. in the picture you can see what happens. it seems that interiors cant connect two domains.
