fem_check21_tria.c running 
Given 3 ux + 2 uy + u = 2 x + 3 y + 13 
Analytic solution u(x,y) =  1 + 2 x + 3 y 
about to read triangles from 22_t.tri 
triangles read from 22_t.tri 
tri  0 has vertices  1   2   0 
tri  1 has vertices  1   0   3 
tri  2 has vertices  2   3   0 
subtracting minvert=0 from all vertices, using base zero
finding unique of nold=9 
found unique of n=4 
about to read boundary from 22_t.bound 
Dirichlet boundaries from 22_t.bound 
boundary segment  0 has vertices  1   2 
boundary segment  1 has vertices  2   3 
boundary segment  2 has vertices  3   1 
subtracting minvert=0 from all vertices, using base zero
finding unique of nold=6 
found unique of n=3 
freevert na=4, nb=3 
freevert nc free = 1 
number of free vertices is 1 
free vertex 0 
about to read coordinates from 22_t.coord 
coordinates read from 22_t.coord 
coordinate  0 at   0.00 ,  0.00 
coordinate  1 at  -1.00 , -1.00 
coordinate  2 at   0.00 ,  1.00 
coordinate  3 at   1.00 , -1.00 
xmin=-1, xmax=1, ymin=-1, ymax=1 
nvert=4, nbound=3, nfree=1, ntri=3 
compute global stiffness matrix 
computing local stiffness matrix for triangle 0 
nodes i1=1, i2=2, i3=0, area=0.500000 
coord (-1.000000,-1.000000) (0.000000,1.000000) (0.000000,0.000000) 
cm matrix  of  am cm = ym   solve for cm 
     0.00    -1.00    -0.00
cm matrix  of  am cm = ym   solve for cm 
     0.00    -1.00     1.00
cm matrix  of  am cm = ym   solve for cm 
     1.00     2.00    -1.00
tri_int1 running with np=30, nv=496 
integral=0.749376, at i=0, j=0 
intg fg =2.082709, at i=0 
integral=-0.458021, at i=1, j=0 
integral=-0.124688, at i=2, j=0 
finished k=0 
computing local stiffness matrix for triangle 1 
nodes i1=1, i2=0, i3=3, area=1.000000 
coord (-1.000000,-1.000000) (0.000000,0.000000) (1.000000,-1.000000) 
cm matrix  of  am cm = ym   solve for cm 
     0.00    -0.50    -0.50
cm matrix  of  am cm = ym   solve for cm 
     1.00     0.00     1.00
cm matrix  of  am cm = ym   solve for cm 
     0.00     0.50    -0.50
tri_int1 running with np=30, nv=496 
integral=0.832085, at i=0, j=0 
intg fg =3.829589, at i=0 
integral=-0.749376, at i=1, j=0 
integral=0.250624, at i=3, j=0 
finished k=1 
computing local stiffness matrix for triangle 2 
nodes i1=2, i2=3, i3=0, area=0.500000 
coord (0.000000,1.000000) (1.000000,-1.000000) (0.000000,0.000000) 
cm matrix  of  am cm = ym   solve for cm 
     0.00     1.00     1.00
cm matrix  of  am cm = ym   solve for cm 
     0.00     1.00     0.00
cm matrix  of  am cm = ym   solve for cm 
     1.00    -2.00    -1.00
tri_int1 running with np=30, nv=496 
integral=-1.250624, at i=0, j=0 
intg fg =2.250624, at i=0 
integral=0.875312, at i=2, j=0 
integral=0.541979, at i=3, j=0 
finished k=2 
tri  0 has vertices  1   2   0 
tri  1 has vertices  1   0   3 
tri  2 has vertices  2   3   0 
vertices, coordinates and analytic values 
coordinate  0 at   0.00 ,  0.00, uana=1.000000 
coordinate  1 at  -1.00 , -1.00, uana=-4.000000 
coordinate  2 at   0.00 ,  1.00, uana=4.000000 
coordinate  3 at   1.00 , -1.00, uana=0.000000 
k computed stiffness matrix, if debug 
K(0,0)=0.330837 
K(0,1)=-1.207397 
K(0,2)=0.750624 
K(0,3)=0.792603 
K(1,0)=0.000000 
K(1,1)=1.000000 
K(1,2)=0.000000 
K(1,3)=0.000000 
K(2,0)=0.000000 
K(2,1)=0.000000 
K(2,2)=1.000000 
K(2,3)=0.000000 
K(3,0)=0.000000 
K(3,1)=0.000000 
K(3,2)=0.000000 
K(3,3)=1.000000 
f computed forcing function and boundary, if debug 
F(0)=8.162922, ug[0]=0.000000 
F(1)=-4.000000, ug[1]=0.000000 
F(2)=4.000000, ug[2]=0.000000 
F(3)=0.000000, ug[3]=0.000000 
ug computed Galerkin, Ua analytic, error 
ug[0]= 1.000, Ua= 1.000, err=1.06581e-14 
ug[1]=-4.000, Ua=-4.000, err=0 
ug[2]= 4.000, Ua= 4.000, err=0 
ug[3]= 0.000, Ua= 0.000, err=0 
                                        maxerr=1.06581e-14, avgerr=2.66454e-15