Direct derivation of the semidiscrete Ritz residual

The four triangles meet along whole edges and vertices and cover the closed square. On each triangle the center barycentric coordinate defines $\varphi$; these traces match, and its boundary trace is zero. An affine P1 function is determined by its three vertex values on each triangle, so the global zero-boundary P1 space has only the center value free and equals $\mathrm{span}\{\varphi \}$. Thus [F2] puts this trial space in $H_0^1(\mathrm{\Omega })$; [F3] supplies the Hilbert and form hypotheses. The function $u=q{\varphi }^2$ is also a zero-boundary conforming piecewise polynomial and is in the energy space. Since $q$ is smooth, $u^{\mathrm{\prime }}=q^{\mathrm{\prime }}{\varphi }^2\in V$; therefore $v\mapsto (u^{\mathrm{\prime }},v{)}_{L^2}$ and $v\mapsto a(u,v)$ are continuous functionals on $V$, and the defined load is square-integrable. By [F5] the initial Ritz value is defined.

On each triangle $K$, $\mathrm{\mid }K\mathrm{\mid }=1\mathrm{/}4$, the center barycentric coordinate $\lambda$ satisfies ${\int }_K{\lambda }^m=2\mathrm{\mid }K\mathrm{\mid }{\int }_0^1{\int }_0^{1-x}(1-x-y{)}^{m}dydx=2\mathrm{\mid }K\mathrm{\mid }\mathrm{/}((m+1)(m+2))$ for $m=1,2,3$ by affine change of variables and iterated integration. Hence ${\int }_{\mathrm{\Omega }}\varphi =1\mathrm{/}3$, ${\int }_{\mathrm{\Omega }}{\varphi }^2=1\mathrm{/}6$, and ${\int }_{\mathrm{\Omega }}{\varphi }^3=1\mathrm{/}10$. The altitude from the center to every boundary edge is $1\mathrm{/}2$, so $\mathrm{\mid }\mathrm{\nabla }\varphi {\mathrm{\mid }}^2=4$ on each triangle. Thus $a(\varphi ,\varphi )=4$ and $a({\varphi }^2,\varphi )=2{\sum }_K{\int }_K\varphi \mathrm{\mid }\mathrm{\nabla }\varphi {\mathrm{\mid }}^2=8\mathrm{/}3$. Also $({\varphi }^2,\varphi {)}_{L^2}=1\mathrm{/}10$.

By [F5], $R_{h}u(t)=(2\mathrm{/}3)q(t)\varphi$, since $a((2\mathrm{/}3)q\varphi ,\varphi )=q(8\mathrm{/}3)=a(q{\varphi }^2,\varphi )$. Therefore $\rho =q((2\mathrm{/}3)\varphi -{\varphi }^2)$ and $({\rho }^{\mathrm{\prime }},\varphi {)}_{L^2}=q^{\mathrm{\prime }}((2\mathrm{/}3)(1\mathrm{/}6)-1\mathrm{/}10)=q^{\mathrm{\prime }}\mathrm{/}90$.

Testing the semidiscrete equation with $\varphi$ and using step 1.2 gives $c^{\mathrm{\prime }}\mathrm{/}6+4c=q^{\mathrm{\prime }}\mathrm{/}10+8q\mathrm{/}3.$ For $\theta =(c-2q\mathrm{/}3)\varphi$, direct substitution yields $({\theta }^{\mathrm{\prime }},\varphi {)}_{L^2}+a(\theta ,\varphi )=c^{\mathrm{\prime }}\mathrm{/}6-q^{\mathrm{\prime }}\mathrm{/}9+4c-8q\mathrm{/}3=-q^{\mathrm{\prime }}\mathrm{/}90=-({\rho }^{\mathrm{\prime }},\varphi {)}_{L^2}\mathrm{.}$ This is [F6] for the sole basis vector, hence for every $v_h\in V_h$ by linearity. The calculation uses the initial Ritz projection, so $\theta (0)=0$.