Backward Euler stability, first-order error for exact initial data, and reconstruction bounds

Fix a finite basis of $V_h$ and write the $j$th variational step as a square linear system for the coefficients of $U^j$. If its homogeneous system has solution $x\in V_h$, testing by $x$ gives ${\tau }_j^{-1}\mathrm{\parallel }x{\mathrm{\parallel }}_H^2+{\bar{a}}_j(x,x)=0$. Coercivity [F4] implies $x=0$. Thus the square matrix has zero kernel and is invertible, proving a unique solution at each step without a choice axiom.

With the instantiation in [F1], ${\bar{a}}_j(v_h,w_h)={\tau }_j^{-1}{\int }_{I_j}a(t;v_h,w_h)dt$ for all $v_h,w_h\in V_h$; changing the form on its common null set does not change this average. Integrate the semidiscrete equation [F2] over $I_j$ and subtract it from the backward Euler step. With $e^j=U^j-u_h(t_j)$, this gives ${\tau }_j^{-1}(e^j-e^{j-1},v_h{)}_H+{\bar{a}}_j(e^j,v_h)=r_j(v_h),$ where $r_j(v_h)={\tau }_j^{-1}{\int }_{I_j}a(t;u_h(t)-u_h(t_j),v_h)dt\mathrm{.}$ The fundamental theorem of calculus in finite-dimensional $V_h$ yields $\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_{V_h^{\ast }}\le M{\int }_{I_j}\mathrm{\parallel }{u}_h^{\mathrm{\prime }}(s){\mathrm{\parallel }}_{V_h}ds$. Cauchy–Schwarz gives ${\sum }_j{\tau }_j\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_{V_h^{\ast }}^2\le M^2{\tau }_{\max }^{}\mathrm{\parallel }{u}_h^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(V_h)}^2$.

Test the step equation with $U^j$, multiply by $2{\tau }_j$, use [F3], and bound $2{\bar{f}}_j(U^j)$ by $\alpha \mathrm{\parallel }{U}^j{\mathrm{\parallel }}_V^2+{\alpha }^{-1}\mathrm{\parallel }{\bar{f}}_j{\mathrm{\parallel }}_{V_h^{\ast }}^2$. Summing and applying Jensen's inequality gives, for every $m$, $\mathrm{\parallel }{U}^m{\mathrm{\parallel }}_H^2+{\sum }_{j=1}^m\mathrm{\parallel }{U}^j-U^{j-1}{\mathrm{\parallel }}_H^2+\alpha {\sum }_{j=1}^m{\tau }_j\mathrm{\parallel }{U}^j{\mathrm{\parallel }}_V^2\le C_U\mathrm{.}$ Taking the maximum of the endpoint term over $m$ proves the maximum bound; setting $m=N$ proves the cumulative bound.

On $(t_{j-1},t_j)$, ${\widehat{U}}^{\mathrm{\prime }}=(U^j-U^{j-1})\mathrm{/}{\tau }_j$ and the step equation gives ${\widehat{U}}^{\mathrm{\prime }}={\bar{f}}_j-{\bar{A}}_j{U}^j$ in $V_h^{\ast }$, where ${\bar{A}}_j$ is the averaged form operator. Its norm is at most $M\mathrm{\parallel }{U}^j{\mathrm{\parallel }}_V$, proving the derivative bound by squaring and integrating. Also $\widehat{U}(t)-U^j=(1-s)(U^{j-1}-U^j)$ with $s=(t-t_{j-1})\mathrm{/}{\tau }_j$; integrating $(1-s{)}^2$ gives ${\tau }_j\mathrm{/}3$. Since $H\to V_h^{\ast }$ has norm at most $c_i$, this proves the reconstruction difference bound.

Test the error equation from step 1.2 with $e^j$, multiply by $2{\tau }_j$, use [F3], and absorb $2{\tau }_j\mathrm{\parallel }{r}_j\mathrm{\parallel }\mathrm{\parallel }{e}^j{\mathrm{\parallel }}_{V_h}$ by $\alpha {\tau }_j\mathrm{\parallel }{e}^j{\mathrm{\parallel }}_V^2+{\alpha }^{-1}{\tau }_j\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_{V_h^{\ast }}^2$. Summing to any $m$ and using the residual estimate in step 1.2 gives, for every $m$, $\mathrm{\parallel }{e}^m{\mathrm{\parallel }}_H^2+\alpha {\sum }_{j=1}^m{\tau }_j\mathrm{\parallel }{e}^j{\mathrm{\parallel }}_V^2\le \mathrm{\parallel }{e}^0{\mathrm{\parallel }}_H^2+{\alpha }^{-1}{\sum }_{j=1}^m{\tau }_j\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_{V_h^{\ast }}^2\le C_e\mathrm{.}$ Taking the maximum of the endpoint term and setting $m=N$ for the cumulative term prove the two asserted bounds.