Parabolic finite-element gradient errors include the time-dependent Ritz residual

Assume Countable Choice and the differentiable continuous state/adjoint setting of A parabolic adjoint gives the gradient when the state map is differentiable. Let $u,p$ and $u_h,p_h$ denote the continuous and semidiscrete state/adjoint pairs. Write $r[\eta ]=D_{\theta }\varphi (u(T),\theta )[\eta ]+{\int }_0^T{D}_{\theta }L(t,u(t),\theta )[\eta ]dt,r_h[\eta ]=D_{\theta }{\varphi }_h(u_h(T),\theta )[\eta ]+{\int }_0^T{D}_{\theta }{L}_h(t,u_h(t),\theta )[\eta ]dt\mathrm{.}$ Assume their gradients $g,g_h\in ({\mathbb{R}}^m{)}^{\ast }$ have the representations, for every $\eta \in {\mathbb{R}}^m$, $g[\eta ]=r[\eta ]+(p(0),D_{\theta }{u}_0[\eta ]{)}_H+{\int }_0^T(⟨D_{\theta }f[\eta ],p⟩-D_{\theta }a[\eta ](u,p))dt,$ $g_h[\eta ]=r_h[\eta ]+(p_h(0),D_{\theta }{u}_{0,h}[\eta ]{)}_H+{\int }_0^T(⟨D_{\theta }{f}_h[\eta ],p_h⟩-D_{\theta }{a}_h[\eta ](u_h,p_h))dt\mathrm{.}$ The continuous representation is supplied by the cited adjoint theorem; the semidiscrete representation is an explicit hypothesis here. Assume $u,p\in H^1(0,T;V)$, the parabolic form and its parameter derivative are uniformly bounded, and the form is coercive with constant $\alpha >0$. Suppose the forward state and time-reversed adjoint both satisfy the hypotheses of Semidiscrete Ritz error splitting and energy estimate, with operator-norm $C^1$ comparison maps $P_{u,h}(t),P_{p,h}(t):V\to V_h$. For the state comparison map, set ${\rho }_u=P_{u,h}u-u$, ${\vartheta }_u=u_h-P_{u,h}u$, and ${\mathcal{R}}_u(t)(v_h)=({\rho }_u^{\mathrm{\prime }}(t),v_h{)}_H+a_{\theta }(t;{\rho }_u(t),v_h),{\rho }_u^{\mathrm{\prime }}=P_{u,h}^{\mathrm{\prime }}(t)u+P_{u,h}(t)u^{\mathrm{\prime }}-u^{\mathrm{\prime }}\mathrm{.}$ In reverse time $s=T-t$, put $q(s)=p(T-s)$ and ${\tilde{P}}_{p,h}(s)=P_{p,h}(T-s)$; define ${\rho }_p(s)={\tilde{P}}_{p,h}(s)q(s)-q(s)$, ${\vartheta }_p(s)=p_h(T-s)-{\tilde{P}}_{p,h}(s)q(s)$, and ${\mathcal{R}}_p(s)(v_h)=({\rho }_p^{\mathrm{\prime }}(s),v_h{)}_H+a_{\theta }(T-s;v_h,{\rho }_p(s))$. The projection product rule gives ${\rho }_p^{\mathrm{\prime }}={\tilde{P}}_{p,h}^{\mathrm{\prime }}q+{\tilde{P}}_{p,h}{q}^{\mathrm{\prime }}-q^{\mathrm{\prime }}$, with ${\tilde{P}}_{p,h}^{\mathrm{\prime }}(s)=-P_{p,h}^{\mathrm{\prime }}(T-s)$. The Ritz supplier gives, for $z=u,p$, $\mathrm{\parallel }z-z_h{\mathrm{\parallel }}_{L^2(0,T;V)}\le E_z:=\mathrm{\parallel }{\rho }_z{\mathrm{\parallel }}_{L^2(0,T;V)}+{\alpha }^{-1\mathrm{/}2}{(\mathrm{\parallel }{\vartheta }_z(0){\mathrm{\parallel }}_H^2+{\alpha }^{-1}\mathrm{\parallel }{\mathcal{R}}_z{\mathrm{\parallel }}_{L^2(0,T;V_h^{\ast })}^2)}^{1\mathrm{/}2},$ where the adjoint's initial trace means its terminal trace in original time. In particular, ${\mathcal{R}}_z$ contains the derivative of the time-dependent projection; it cannot be dropped from this hypothesis.

For $\mathrm{\parallel }\eta {\mathrm{\parallel }}_2\le 1$, assume $\mathrm{\parallel }{D}_{\theta }f[\eta ]{\mathrm{\parallel }}_{L^2(V^{\ast })}\le M_f$, $\mathrm{\parallel }{D}_{\theta }a(t)[\eta ]{\mathrm{\parallel }}_{\mathcal{L}(V,V^{\ast })}\le M_a$ a.e., and define $U_{\ast }:=\max (\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(V)},\mathrm{\parallel }{u}_h{\mathrm{\parallel }}_{L^2(V)})$, $P_{\ast }:=\max (\mathrm{\parallel }p{\mathrm{\parallel }}_{L^2(V)},\mathrm{\parallel }{p}_h{\mathrm{\parallel }}_{L^2(V)})$, $P_{0,\ast }:=\max (\mathrm{\parallel }p(0){\mathrm{\parallel }}_H,\mathrm{\parallel }{p}_h(0){\mathrm{\parallel }}_H)$, and $U_0^{\mathrm{\prime }}:=\mathrm{\parallel }{D}_{\theta }{u}_0{\mathrm{\parallel }}_{\mathcal{L}({\mathbb{R}}^m,H)}$. Suppose $\mathrm{\parallel }{r}_h-r{\mathrm{\parallel }}_{({\mathbb{R}}^m{)}^{\ast }}\le {\epsilon }_J$, $\mathrm{\parallel }{D}_{\theta }{u}_{0,h}-D_{\theta }{u}_0{\mathrm{\parallel }}_{\mathcal{L}({\mathbb{R}}^m,H)}\le {\epsilon }_0$, $\mathrm{\parallel }{D}_{\theta }{f}_h-D_{\theta }f{\mathrm{\parallel }}_{\mathcal{L}({\mathbb{R}}^m,L^2(V^{\ast }))}\le {\epsilon }_f$, and $\mathrm{\parallel }{D}_{\theta }{a}_h-D_{\theta }a{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(0,T;\mathcal{L}({\mathbb{R}}^m,\mathcal{L}(V,V^{\ast })))}\le {\epsilon }_a$. Put $E_{p,0}=\mathrm{\parallel }{p}_h(0)-p(0){\mathrm{\parallel }}_H$. Then $\mathrm{\parallel }{g}_h-g{\mathrm{\parallel }}_{({\mathbb{R}}^m{)}^{\ast }}\le {\epsilon }_J+U_0^{\mathrm{\prime }}{E}_{p,0}+P_{0,\ast }{\epsilon }_0+M_f{E}_p+{\epsilon }_f{P}_{\ast }+M_a(E_u{P}_{\ast }+U_{\ast }{E}_p)+{\epsilon }_a{U}_{\ast }{P}_{\ast }\mathrm{.}$ The derivative-data and trace defects are additional assumptions; an energy estimate for the state alone does not imply this gradient bound.