Shows how to compute interpolation functions for linear triangle elements. Steps involve defining the interpolation function on the reference triangle and map taking point in the given triangle to the reference triangle.

1. Interpolation function for linear triangle element
Let T0 ⊂ R2 is a reference triangle element with vertices v1 = (0, 0), v2 = (1, 0), v3 = (0, 1)
and let T ⊂ R2 is any arbitrary triangle with vertices w1, w2, w3, see figure.
v1 v2
v3
Figure: Reference triangle
T0
w1
w2
w3
Figure: Arbitrary triangle
T
For reference triangle, the shape functions, at point (ξ, η) ∈ T0,
associated to each vertices are:
N0
1 (ξ, η) = 1 − ξ − η, N0
2 = ξ, N0
3 = η. (1)
Verify that N0
1 = 1 at v1 and 0 at v2, v3. Similar properties hold for
N0
2 and N0
3 .
Consider a map Φ = (Φ, Φ2) that takes point (ξ, η) ∈ T0 to point
(x, y) ∈ T defined as
Φ1(ξ, η) =
3
X
i=1
N0
i (ξ, η)wx
i , Φ2(ξ, η) =
3
X
i=1
N0
i (ξ, η)wy
i , (2)
where wi = (wx
i , wy
i ), and wx
i , wy
i denote the x and y coordinate of ith
vertex of triangle T.
Then, Φ−1 is the inverse map that takes point (x, y) ∈ T to
(ξ, η) ∈ T0. We will compute Φ−1 and then use it to find the shape
functions for arbitrary triangle T.
PK Jha April 27, 2021 1 / 3

2. Interpolation function for linear triangle element ...
To compute Φ−1, we substitute the formula of N0
i from (1) into (2) and get the following
system of equations:
x − wx
1
y − wy
1
=
wx
2 − wx
1 wx
3 − wx
1
wy
2 − wy
1 wy
3 − wy
1
| {z }
=:B
ξ
η
. (3)
Let C denote the inverse matrix B. C is expressed as
C = B−1
=
1
det B
wy
3 − wy
1 −(wx
3 − wx
1 )
−(wy
2 − wy
1) wx
2 − wx
1
. (4)
Then the inverse map Φ−1 = (Φ−1
1 , Φ−1
2 ) that takes point on T to point on T0 can be written
as
Φ−1
1 (x, y) = C11(x − wx
1 ) + C12(y − wy
1), Φ−1
2 (x, y) = C21(x − wx
1 ) + C22(y − wy
1) . (5)
PK Jha April 27, 2021 2 / 3

3. Interpolation function for linear triangle element ...
Now that we know how to map given triangle T to reference triangle T0,
we can compute the shape functions of vertices of triangle T. These are
give by
N1(x, y) := N0
1 (Φ−1
(x, y)) = 1 − Φ−1
1 (x, y) − Φ−1
2 (x, y),
N2(x, y) := N0
2 (Φ−1
(x, y)) = Φ−1
1 (x, y),
N3(x, y) := N0
3 (Φ−1
(x, y)) = Φ−1
2 (x, y) . (6)
Note that Φ−1
1 , Φ−1
2 can be explicitly computed using (4) and (5) given
the coordinates w1, w2, w3.
PK Jha April 27, 2021 3 / 3