Nitsche methods for weak boundary enforcement — Examples

The function $u=x$ is affine, hence belongs to $V_h$ for every $k\ge 1$, and it has the prescribed trace $g$. Apply [F2] to $G=w_h\mathrm{\nabla }u$: since $\mathrm{\Delta }u=0$, its divergence is $\mathrm{\nabla }{w}_h\cdot \mathrm{\nabla }u$, and its boundary flux is $w_h{\mathrm{\partial }}_{n}u$. Thus the volume term in $a_h^0(u,w_h)$ cancels its first boundary-flux term. The remaining term is ${\gamma }_0{\sum }_F{h}_F^{-1}{\int }_{F}g{w}_h$, exactly ${\mathrm{\ell }}_h^0(w_h)$ because $f=0$.

For $\theta =1$, the extra form term is $-{\sum }_F{\int }_{F}u{\mathrm{\partial }}_n{w}_h$, and the extra load term is $-{\sum }_F{\int }_{F}g{\mathrm{\partial }}_n{w}_h$; these agree since $u{\mathrm{\mid }}_F=g$. Thus $u$ satisfies both weak equations. Coercivity [F3] implies uniqueness: the difference $d_h$ of two solutions obeys $a_h^{\theta }(d_h,w_h)=0$ for every $w_h$, so testing $w_h=d_h$ gives ${\lambda }_{\theta }\mathrm{\parallel }{d}_h{\mathrm{\parallel }}_{E,h}^2\le 0$ and hence $d_h=0$. Therefore $u_h^0=u_h^1=u$. For strong imposition, $u\in V_h$ has the required boundary trace and satisfies the Poisson weak equation on each test in $Y_h$: the simplex Green identity [F2] and the zero boundary restriction [F4] make its boundary flux term vanish. Thus it is the strong Galerkin solution as well.

For $v_h=1$, each volume-gradient and normal-flux term in the definition vanishes. At ${\gamma }_0=0$ the penalty term vanishes too, so $a_h^{\theta }(1,1)=0$. Since each facet has positive measure and $h_F>0$, the finite sum ${\sum }_F{h}_F^{-1}\mathrm{\mid }F\mathrm{\mid }$ is positive; hence no positive coercivity constant can bound this form below by $\mathrm{\parallel }1{\mathrm{\parallel }}_{E,h}^2$.

For any ${\gamma }_0$ strictly above the displayed threshold, [F2] gives the positive constant ${\lambda }_{\theta }$ and proves coercivity on the entire finite-element space, including the constant function. This establishes the stated below/above comparison without making a necessity claim for intermediate positive penalties.

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.