In this section, we develop a certain basis for the PBH functions on the product of two Sierpinski Gaskets. As in the case of the unit interval and the Sierpinski Gasket, there is a very natural or ``easy" basis of the biharmonic elements. The basis for the harmonic elements, H₀, on the unit interval are simply the functions annihilated by the D, 1 and x. The biharmonic basis consists of the functions that are annihilated by D²: 1, x, x², and x³. The situation for the Sierpinski gasket is similarly simple. We first define a class of functions, f_nk^l such that

The harmonic functions,H₀ will be linear combinations of f_nk⁰. The n-harmonic functions, H_n will be a linear combination of f_nk^l for 0 Ј l Ј n. The pluri-n-harmonic functions will consist simply of H_n ДH_n. The easy basis is very easily determined, but not very easy to work with using splines. In the case of the biharmonic spline, we want to be able to control the normal derivative, not the Laplacian. By matching the normal derivatives where the piecewise PBH function meet up, we are able to guarantee an additional level of smoothness. In the Splines paper, the ``better" basis is derived for the Sierpinski Gasket. They are uniquely determined by the following (refer to the paper for an explicit formula for them):

For a product, there basis of H₁ ДH₁ has N² elements, yet for each boundary value, we only want to describe f(x), ¶_n^ўf(x), and ¶_n^ўўf(x) - calling for only 3N basis functions. To be able to compute the normal derivative in one factor, we only need a harmonic function in the other, so we can use H₁ ДH₀ and H₀ ДH₁. The resulting basis functions will be a product of two harmonic functions, or a harmonic function in one factor and a biharmonic in the other. For the unit square, this would amount to the product of two linear functions, or a cubic function and a linear one. The 3N = 27 functions that form the better basis of the PBH functions on the Sierpinski gasket are as follows:

These functions must be a linear combination of elements of H₀ ДH₀, H₁ ДH₀, and H₀ДH₁. We must have (note k = kўkўў)

Notice that we do not use the products of the better basis for f in SG²- f in the better basis of SG is in H₁, so f^ў f^ўў would be an element of H₁ ДH₁, which we do not want. This set of equations give the correct values for the function on the boundary, and to verify the correct normal derivatives. Taking the normal derivative in both directions, we get

Notice that ¶_n f_0k(v_m) and ¶_n g_k(v_m) are given explicitly in the Spline's paper as -H_mk and d_m,k, respectively. This is a system of 2N² equations in 2N² variables. We think that it is reasonable to believe that, since both H and I are invertible, that this set of equations will be solvable. However, no proof is attempted here. The equations are solvable in the case of the SG², and are given below.