The attached worksheet develops a procedure for extrapolating boundary data from a square to its interior.  Specifically, let's consider the square [−1,1]×[−1,1] and the continuous functions u1, u2, u3, u4 defined on its edges. The procedure constructs a continuous function u(x,y) in the interior of the square which matches the boundary data.

The function u(x,y) is necessarily discontinuous at a corner if the boundary data of the edges meeting at that corner are inconsistent.  However, if the boundary data are consistent at all corners, then u(x,y) is continuous everywhere including the boundary.

Here is what the extrapolating function looks like:

proc (x, y) options operator, arrow; (1/2)*(1-y)*(u__1(x)-(1/2)*(1-x)*u__1(-1))+(1/2)*(x+1)*(u__2(y)-(1/2)*(1-y)*u__2(-1))+(1/2)*(y+1)*(u__3(x)-(1/2)*(x+1)*u__3(1))+(1/2)*(1-x)*(u__4(y)-(1/2)*(y+1)*u__4(1))+(1/4)*(u__1(-1)-u__4(-1))*(-x^2+1)*(1-y)/((x+1)^2+(y+1)^2)^(1/2)+(1/4)*(u__2(-1)-u__1(1))*(-y^2+1)*(x+1)/((y+1)^2+(1-x)^2)^(1/2)+(1/4)*(u__3(1)-u__2(1))*(-x^2+1)*(y+1)/((1-x)^2+(1-y)^2)^(1/2)+(1/4)*(u__4(1)-u__3(-1))*(-y^2+1)*(1-x)/((1-y)^2+(x+1)^2)^(1/2) end proc


The worksheet contains the details of the derivation, and an example.

Edit: The worksheet presents a slightly different version of the previous result. Here the square's corners are treated symmetrically, leading to a more pleasing interpolating function.

PS: A challenge.  Extend the result to 3D, that is, construct a function u(x,y,z) on the cube [−1,1]×[−1,1]×[−1,1] which matches prescribed functions on the cube's six faces.

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