Symmetric cancellation and the nonsymmetric L2 boundary defect

The triangle has edges $y=0$, $x=0$, and $x+y=1$. On the first edge ${\mathrm{\partial }}_{n}v=0$; on the second $v=0$; on the diagonal $n=(1,1)\mathrm{/}\sqrt{2}$, ${\mathrm{\partial }}_{n}v=1\mathrm{/}\sqrt{2}$, and the arclength parametrization $(t,1-t)$ has $ds=\sqrt{2}dt$. Therefore ${\sum }_F{\int }_{F}v{\mathrm{\partial }}_{n}v={\int }_0^{1}tdt=1\mathrm{/}2$. Substitution in [F1] gives the asserted finite difference $-1\mathrm{/}2$.

For the adjoint Poisson equation, [F3] gives $\mathrm{\parallel }e{\mathrm{\parallel }}_2^2={\int }_{\mathrm{\Omega }}\mathrm{\nabla }e\cdot \mathrm{\nabla }{\zeta }_e-{\sum }_F{\int }_{F}e{\mathrm{\partial }}_n{\zeta }_e$. The symmetric form contains the transposed term $-{\sum }_F{\int }_{F}e{\mathrm{\partial }}_n{\zeta }_e$; after subtracting the zero-trace discrete test, the remaining flux is paired with the adjoint interpolation error. The nonsymmetric form has no transposed consistency term; its zero-trace discrete tests satisfy $a_h^0(e,z_h)={\int }_{\mathrm{\Omega }}\mathrm{\nabla }e\cdot \mathrm{\nabla }{z}_h$, leaving the full pairing $-{\sum }_F{\int }_{F}e{\mathrm{\partial }}_n{\zeta }_e$ in the error identity. Thus the symmetric method reduces the adjoint boundary defect to an approximation residual, while the nonsymmetric method retains it. The zero boundary traces used here follow from [F2], and the two $L^2$ theorem items retain their distinct hypotheses and rates.