Natural/Volumn coordinates in element matrix

by Yi Zhang

When evaluating element matrix, one needs perform integration over local element domain, during which coordinate transform usually happens. For example, for shape function N_i where subscript is for local node index in a heat equation with constant conductivity, one does the integral during applying variational principle:

\int_{\Omega^e}{\nabla N_i\cdot\nabla N_j}dxdydz

In a tetrahedron the natural/volumn coordinate can be described as:

L_1=\zeta(x,y,z)

L_2=\eta(x,y,z)

L_3=\xi(x,y,z)

L_4=1-\zeta-\eta-\xi

where (\zeta,\eta,\xi) is the “standard” tetrahedron coordinate between which and the element a transform is constructed:

(\zeta,\eta,\xi)\Longleftrightarrow (x,y,z)

By this partial derivatives can be performed like:

\nabla_{x,y,z} N_i=J^{-1} \nabla_{\zeta,\eta,\xi} N_i

where J is the Jacobian matrix of the transformation J=\frac{\partial(x,y,z)}{\partial(\zeta,\eta,\xi)}, etc. In turn we have

\frac{\partial N_i}{\partial \zeta}=\frac{\partial N_i}{\partial L_1}\frac{\partial L_1}{\partial \zeta}+\frac{\partial N_i}{\partial L_2}\frac{\partial L_2}{\partial \zeta}+\frac{\partial N_i}{\partial L_3}\frac{\partial L_3}{\partial \zeta}+\frac{\partial N_i}{\partial L_4}\frac{\partial L_4}{\partial \zeta}

and similar expressions for \zeta,\eta. This is equal to:

\frac{\partial N_i}{\partial \zeta}=\frac{\partial N_i}{\partial L_1}-\frac{\partial N_i}{\partial L_4}

In summary, the integral in calculating element matrix can be performed always in the standard element \Omega_0^e:

\int_{\Omega^e}{\nabla_{x,y,z} N_i\cdot\nabla_{x,y,z} N_j}dxdydz=\int_{\Omega_0^e}{(J^{-1}\nabla_{\zeta,\eta,\xi} N_i)\cdot(J^{-1}\nabla_{\zeta,\eta,\xi} N_j) \det J}d\zeta d\eta d\xi

For example, in linear element, the shape functions are exactly the same as natural coordinates: N_i=L_i, i=1,2,3,4. Then we have:

\frac{\partial (N_1,N_2,N_3,N_4)}{\partial \zeta}=(1,0,0,-1)

\frac{\partial (N_1,N_2,N_3,N_4)}{\partial \eta}=(0,1,0,-1)

\frac{\partial (N_1,N_2,N_3,N_4)}{\partial \xi}=(0,0,1,-1)

while in quadratic element, as the shape function N_1\sim N_{10} shown in last post, the gradient of \mathbf{N}=(N_1,N_2,N_3,N_4,N_5,N_6,N_7,N_8,N_9,N_{10}) against natural coordinates would be:

\frac{\partial \mathbf{N}}{\partial \zeta}=(4L_1-1,0,0,-4L_4+1,4L_2,4L_3,4L_4-4L_1,0,-4L_3,-4L_2)

\frac{\partial \mathbf{N}}{\partial \eta}=(0,4L_2-1,0,-4L_4+1,4L_1,0,-4L_1,4L_3,-4L_3,4L_4-4L_2)

\frac{\partial \mathbf{N}}{\partial \xi}=(0,0,4L_3-1,-4L_4+1,0,4L_1,-4L_1,4L_2,4L_4-4L_3,-4L_2)

Advertisements