The interpolation functions φi(x,y) for a triangular element are no good for the following reasons:

a. They are not defined for all reasonable values of the nodal coordinates (xi,yi)
b. They may violate compatibility in the sense that the resulting functions one is approximating the solution with are not necessarily continuous at inter-element boundaries.	(The compatibility requirement states that the functions and derivatives must be continuous up to one order less than the highest derivative that appears in the variational formulation of the problem. E.g. if the highest derivative is a 1st derivative, as is the case for the wave equation of the homework if put in variational form, then derivatives up to the 0th order must be continuous; i.e. just the function itself.)
c. They may also violate the completeness requirement.

Further details and an example in which compatibility & completeness is violated follow. (Note that in some cases incompatible elements can still be convergent, if compatibility is achieved approximately in the limit as the mesh is much refined.)
Undefined Interpolation Functions
Consider the case when the nodal coordinates are given by the table below:
i	xi	yi
0	0	0
1	0	1
2	1	0

These are perfectly reasonable values for the nodal coordinates yet they make some of the factors in the numerator zero. E.g. (x1 – x0) is zero.	This renders the interpolation functions undefined.

Incompatibility
Consider a mesh shown in the figure:
The arrows show the x (horizontal) and y (vertical) axes. The mesh consists of 2 elements with the following nodal coordinates:

Element Number	Global Node Label	Element Node Number	Coordinates
					i			xi		yi
1		B			0			1		0
		C			1			0		1
		A			2			α		α

2		B			0			1		0
		C			1			0		1
		D			2			1		1


In the above α can take any reasonable value, e.g. α=0.1.	(A zero value does not work because it makes certain interpolation functions undefined, and the point here is to show that compatibility is violated, even if the interpolation functions are defined.)


Next suppose that the function is value is 1 at node B, but zero at all other nodes. Then the value of the function is φ0,1 in element 1 and φ0,2 in element 2, where φi,j denotes the value of φi in element number j. The inter-element boundary is described by
y=1–x
and there the value of the function in one element must be equal to that in the adjacent one. I.e. for compatibility we need:
φ0,1(x,y) = φ0,2(x,y) for all (x,y) along the element boundary. This is true if φ0,1(x,1-x) = φ0,2(x,1-x)	
for all x in the interval (0,1). 


Using the formula
φ0 =((x1-x)(y2-y))/((x1-x0)(y2-y0))

to calculate these interpolation functions gives, for element 1:
φ0,1(x,y)
=(x1 –x)(y2 –y)/[(x1 –x0)(y2 –y0)] 
= (0–x)(α–y)/[(0–1)(α–0)] 
= x(α–y)/α=x(1–y/α)

Substituting y = 1 – x to get the value at the element boundary yields: 
φ0,1(x, 1-x) = x (1 – (1-x)/ α) = (1-1/α) x + x2/α 
Further for α = 0.1, this becomes: 
φ0,1(x,1-x)= -9x+10x2


Similarly, for element 2:
φ0,2(x,y)
=(x1 – x) (y2 – y)/[ (x1 – x0) (y2 – y0)] 
= (0–x)(1–y)/[(0–1)(1–0)]
= x (1-y) 

So again substituting y = 1 – x to get the value of φ0,2 at the inter-element boundary gives: 
φ0,2(1,1-x)	
= x (1-(1-x)) = 
x2

As can be seen this is very different from the value of φ0,1! Thus compatibility is not satisfied, and for this particular case it is rather badly off.

Completeness
If these elements are used for a problem in which the highest derivative in the variational formulation is of 1st order, then it is necessary to be able to represent any linear function, like u = 1 – y within the element. Yet u = 1 – y at the nodes gives u = x ( 1 – y) within element 2 of the above example, which is a violation of completeness.