# Determining the surface with given polynomial borders

Let's say we'd like to guess the shape (I'm not sure the word 'approximate' is appropriate here) of some surface when we are given its borders via third order polynomials (i.e. we are given their coefficients). To make things a bit simpler, let's assume we consider only a rectangle $[a,b]\times[c,d]$ - so if $f(x,y)$ is a function representing the sought surface, we have polynomials $f_1(x), \ f_2(x), \ f_3(y), \ f_4(y)$ of degree $3$ such that $f(x,c) = f_1(x), \ f(x,d) = f_2(x), \ f(a,y) = f_3(y), \ f(b,y) = f_4(y)$.

Now, we'd like to have nice and easily calculable formula for $f$ that meets the criteria and is somehow smooth.

I thought that looking for $f$ in the bivariat polynomial form $$\sum_{0 \le i, j \le 3} \alpha_{i,j}x^iy^j$$ would be a good idea. Unfortunately, as I performed a few tests, it turns out that quite often it cannot be done - system of linear equations that emerge from constraints doesn't have a solution.

And my question is - what would be other 'human-friendly' form in which $f$ will always be expressible?

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What about $f(x,y)=\frac{\frac1{y-c}f_1(x)+\frac1{d-y}f_2(x)+\frac1{x-a}f_3(y)+\frac1{b-x}f‌​_4(y)}{\frac1{y-c}+\frac1{d-y}+\frac1{x-a}+\frac1{b-x}}$ for interior points? – Hagen von Eitzen Dec 28 '13 at 16:10
@HagenvonEitzen - wow, it looks good to me! And just out of curiosity - is it evident that $f$ will look 'just as we want' on borders by a careful perusal? (I checked it manually with some use of de l'Hospital rule). – socumbersome Dec 28 '13 at 16:56

A surface that interpolates its boundary curves is called a Coons patch. The Wikipedia page has a good description. There are many different variants, but the simplest is the one that uses bilinear blending, as described in the first section of the Wikipedia page. The discussion assumes $a=c=0$ and $b=d=1$, but you can easily rescale later, if you need to.
We have to assume some compatibility conditions, otherwise the interpolation problem is unsolvable. Specifically, we need to assume that $f_1(a)=f_3(c)$, $f_1(b)=f_4(c)$, $f_2(a)=f_3(d)$, $f_2(b)=f_4(d)$.