Multidimensional P1 SUPG consistency, coercivity, and error bound

For $v_h\in V_h$, apply [F4] on each cell to $G=b{v}_h^2\mathrm{/}2$. Interior fluxes cancel because $b$ and $v_h$ are continuous and the outward normals oppose; boundary fluxes vanish because $v_h=0$. Hence $(b\cdot \mathrm{\nabla }{v}_h,v_h{)}_{\mathrm{\Omega }}=-\frac{1}{2}(\mathrm{\nabla }\cdot b,v_h^2{)}_{\mathrm{\Omega }}$.

For the exact $u$, apply [F4] cellwise to $G=ϵv_h\mathrm{\nabla }u$. The interior fluxes cancel and the boundary flux vanishes, so $ϵ(\mathrm{\nabla }u,\mathrm{\nabla }{v}_h)+(b\cdot \mathrm{\nabla }u+cu,v_h)=(f,v_h)$. Since $R_K(u)=f$, adding ${\sum }_K{\delta }_K(f,b\cdot \mathrm{\nabla }{v}_h{)}_K$ gives $a_h(u,v_h)=F_h(v_h)$.

Let $\eta =u-I_{h}u$. Enlarge the interpolation constant if necessary so $C_I\ge 1$; this preserves [F3]. By [F3], $I_{h}u\in V_h$, $\mathrm{\parallel }\mathrm{\nabla }\eta {\mathrm{\parallel }}_2\le C_{I}h{M}_2$, and $\mathrm{\parallel }\eta {\mathrm{\parallel }}_2\le C_I{h}^2{M}_2$. Cellwise, ${\mathrm{\Delta }}_K{I}_{h}u=0$, so $R_K(\eta )=-ϵ\mathrm{\Delta }u+b\cdot \mathrm{\nabla }\eta +c\eta$. The Euclidean matrix norm and Cauchy–Schwarz on the diagonal give $\mathrm{\mid }\mathrm{\Delta }u(x)\mathrm{\mid }=\mathrm{\mid }\mathrm{tr}{D}^{2}u(x)\mathrm{\mid }\le \sqrt{d}\mathrm{\mid }{D}^{2}u(x)\mathrm{\mid }$; integrating yields $\mathrm{\parallel }\mathrm{\Delta }u{\mathrm{\parallel }}_2\le \sqrt{d}{M}_2$. The triangle inequality in the finite product of cellwise $L^2$ spaces then gives ${({\sum }_K{\delta }_K\mathrm{\parallel }{R}_K(\eta ){\mathrm{\parallel }}_{0,K}^2)}^{1\mathrm{/}2}\le \sqrt{{\delta }_{\ast }}(\sqrt{d}ϵM_2+B{C}_{I}h{M}_2+C{C}_I{h}^2{M}_2)$. Also ${\sum }_K{\delta }_K\mathrm{\parallel }b\cdot \mathrm{\nabla }\eta {\mathrm{\parallel }}_{0,K}^2\le {\delta }_{\ast }{B}^2\mathrm{\parallel }\mathrm{\nabla }\eta {\mathrm{\parallel }}_2^2$, so the product norm satisfies $\mathrm{\parallel }\eta {\mathrm{\parallel }}_{\mathrm{S}\mathrm{D},h}\le C_I{M}_2(\sqrt{ϵ}h+\sqrt{\omega }{h}^2+B\sqrt{{\delta }_{\ast }}h)$.

Since $v_h$ is affine on each cell, ${\mathrm{\Delta }}_K{v}_h=0$. Substituting step 1.1 into [F2] gives $a_h(v_h,v_h)=ϵ\mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_2^2+(c-\frac{1}{2}\mathrm{\nabla }\cdot b,v_h^2{)}_{\mathrm{\Omega }}+{\sum }_K{\delta }_K\mathrm{\parallel }b\cdot \mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{0,K}^2+{\sum }_K{\delta }_K(c{v}_h,b\cdot \mathrm{\nabla }{v}_h{)}_K$. On each cell, Cauchy–Schwarz and [F2] give $\mathrm{\mid }{\delta }_K(c{v}_h,b\cdot \mathrm{\nabla }{v}_h{)}_K\mathrm{\mid }\le \frac{1}{4}(\omega \mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{0,K}^2+{\delta }_K\mathrm{\parallel }b\cdot \mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{0,K}^2)$: write the left bound as $(\sqrt{{\delta }_K}\mathrm{\parallel }c{\mathrm{\parallel }}_{\mathrm{\infty },K}\mathrm{/}\sqrt{\omega })(\sqrt{\omega }\mathrm{\parallel }{v}_h\mathrm{\parallel })(\sqrt{{\delta }_K}\mathrm{\parallel }b\cdot \mathrm{\nabla }{v}_h\mathrm{\parallel })$, use the first factor $\le 1\mathrm{/}2$, then $ab\mathrm{/}2\le (a^2+b^2)\mathrm{/}4$. Since $c-\frac{1}{2}\mathrm{\nabla }\cdot b\ge \omega$, this proves $a_h(v_h,v_h)\ge \frac{3}{4}\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{\mathrm{S}\mathrm{D},h}^2$.

The linear map $V_h\to V_h^{\ast }$ induced by $a_h$ has zero kernel: if $a_h(w_h,v_h)=0$ for all $v_h$, test with $v_h=w_h$ and use step 2.1 to get $w_h=0$. The spaces have equal finite dimension, so this map is bijective and the discrete problem has a unique solution $u_h$.

Subtract the discrete equation from step 1.2 to obtain $a_h(u-u_h,v_h)=0$ for every $v_h\in V_h$. Put $e_h:=u_h-I_{h}u\in V_h$. Taking $v_h=e_h$ and using $u-u_h=\eta -e_h$ gives $a_h(e_h,e_h)=a_h(\eta ,e_h)$.

Cauchy–Schwarz bounds the diffusion, convection, reaction, and stabilization terms in $a_h(\eta ,e_h)$ by $\sqrt{ϵ}\mathrm{\parallel }\mathrm{\nabla }\eta {\mathrm{\parallel }}_2\mathrm{\parallel }{e}_h{\mathrm{\parallel }}_{\mathrm{S}\mathrm{D},h}$, $B\mathrm{\parallel }\mathrm{\nabla }\eta {\mathrm{\parallel }}_2\mathrm{\parallel }{e}_h{\mathrm{\parallel }}_{\mathrm{S}\mathrm{D},h}\mathrm{/}\sqrt{\omega }$, $C\mathrm{\parallel }\eta {\mathrm{\parallel }}_2\mathrm{\parallel }{e}_h{\mathrm{\parallel }}_{\mathrm{S}\mathrm{D},h}\mathrm{/}\sqrt{\omega }$, and $({\sum }_K{\delta }_K\mathrm{\parallel }{R}_K(\eta ){\mathrm{\parallel }}_{0,K}^2{)}^{1\mathrm{/}2}\mathrm{\parallel }{e}_h{\mathrm{\parallel }}_{\mathrm{S}\mathrm{D},h}$, respectively. If $e_h=0$ the bound below is immediate; otherwise divide step 4.1 by its norm and use coercivity in step 2.1 to get $\mathrm{\parallel }{e}_h{\mathrm{\parallel }}_{\mathrm{S}\mathrm{D},h}\le \frac{4}{3}(\sqrt{ϵ}\mathrm{\parallel }\mathrm{\nabla }\eta {\mathrm{\parallel }}_2+\frac{B}{\sqrt{\omega }}\mathrm{\parallel }\mathrm{\nabla }\eta {\mathrm{\parallel }}_2+\frac{C}{\sqrt{\omega }}\mathrm{\parallel }\eta {\mathrm{\parallel }}_2+({\sum }_K{\delta }_K\mathrm{\parallel }{R}_K(\eta ){\mathrm{\parallel }}_{0,K}^2{)}^{1\mathrm{/}2})$.

The norm in the statement is the product $L^2$ norm of $(\sqrt{ϵ}\mathrm{\nabla }w,\sqrt{\omega }w,(\sqrt{{\delta }_K}b\cdot \mathrm{\nabla }w{)}_K)$, so $\mathrm{\parallel }u-u_h{\mathrm{\parallel }}_{\mathrm{S}\mathrm{D},h}\le \mathrm{\parallel }\eta {\mathrm{\parallel }}_{\mathrm{S}\mathrm{D},h}+\mathrm{\parallel }{e}_h{\mathrm{\parallel }}_{\mathrm{S}\mathrm{D},h}$. Using step 1.3 in step 5.1 bounds the sum by $C_I{M}_2(\sqrt{ϵ}h+\sqrt{\omega }{h}^2+B\sqrt{{\delta }_{\ast }}h)+\frac{4}{3}{C}_I{M}_2((\sqrt{ϵ}+B\mathrm{/}\sqrt{\omega })h+(C\mathrm{/}\sqrt{\omega })h^2+\sqrt{{\delta }_{\ast }}(\sqrt{d}ϵ+Bh+C{h}^2))$. Since $C_I+\frac{4}{3}{C}_I=\frac{7}{3}{C}_I$, this is at most the displayed right-hand side with $C_{\ast }:=\frac{7}{3}{C}_I$, which has the claimed dependence.