What SBP operators are doing at a boundary

Interior rows are the ordinary central stencil, and they are the easy part. The first 4 rows are not: they are dense, they are not the stencil, and they exist so that one identity survives. Write the operator as Q = HD, where H is the diagonal norm. The identity is Q + Qᵀ = E, with E picking out only the two endpoints. Verified on this operator: it holds exactly at N = 12, in rationals, with no floating point anywhere in the check.

What the closure costs

The norm is where it shows. Away from the ends every weight is 1, which is the trapezoid rule with the endpoints already absorbed. The first 4 weights are not:

Row	1	2	3	4	interior
H	17/48	59/48	43/48	49/48	1

Those four numbers are the whole closure, and they are not free parameters that happened to work: they are what the identity forces once the interior stencil is fixed. Read as a quadrature rule, H integrates exactly to degree 2p − 1 and not to 2p, which is the accuracy the closure buys, checked term by term.

Where the order goes, and where it is forfeited

On the interior rows the operator differentiates polynomials exactly up to fourth degree; the closure rows do not, and that is not a defect but the price of the identity. When the scheme is assembled properly the norm is also the energy norm of the problem, and a SAT penalty at the boundary then gives discrete energy stability. When it is not, the same operator quietly forfeits order at the boundary and the convergence rate drops without anything failing.

The MEng project this grew out of closed in 2025. The question did not: it runs on in SBPLab, a private C++ bench for constructing operators and checking them against problems whose answer is known in closed form. The entry case is two-dimensional diffusion of a single Fourier mode, whose discrete decay rate has an exact eigenvalue to answer to.