Past exam of the approximation theory and finite element methods course of the University of Oxford / 2008 / B / 7

Put $e=u-u_h$ and $V=\{v\in H^1(0,1):v(1)=0\}$. The primal solution and its finite element approximation lie in the same affine space with value one at the right endpoint, so $e\in V$. UseDefine the dual solution byEquivalently $z\in V$ satisfies $a(v,z)=(e,v)$ for every $v\in V$. Positivity of $p,q$ gives a unique solution, and one-dimensional regularity gives $z\in H^2$.

Then $\Vert e{\Vert }_{L^2}^2=a(e,z)$. The nodal interpolant $I_{h}z$ belongs to the homogeneous finite element test space, so Galerkin orthogonality gives $a(e,I_{h}z)=0$. Set $w=z-I_{h}z$. Using the primal weak equation,Integrate the derivative term separately on each element. Since $w$ vanishes at every mesh node, all endpoint flux and flux-jump terms vanish. ThereforeThe absence of jump terms here follows specifically from one-dimensional nodal interpolation; it is not an assumption that the finite element flux itself is continuous.

Let $p_{\ast }=\min p>0$, $q_{\ast }=\min q>0$, $q^{\ast }=\max q$, and $P_1=\Vert p^{\prime }{\Vert }_{\infty }$. Testing the dual equation with $z$ givesConsequently $\Vert z{\Vert }_2\le \Vert e{\Vert }_2/q_{\ast }$ and $\Vert z^{\prime }{\Vert }_2\le \Vert e{\Vert }_2/\sqrt{p_{\ast }{q}_{\ast }}$. From $p{z}^{\prime \prime }=qz-e-p^{\prime }{z}^{\prime }$,On an element of length $h_i$, the interpolation error vanishes at both endpoints and its derivative has zero mean. Two Poincaré inequalities giveApply Cauchy-Schwarz to the residual identity, with factors $h_i^2$ and $h_i^{-2}$:If the error is nonzero, cancel its norm after the regularity estimate; if it is zero the conclusion is immediate. One explicit reliable constant isThis proves the requested dual residual bound entirely in terms of the coefficient functions.

Compute $C$ from part (b), solve the finite element equations, and calculate the element indicators ${\eta }_i^2=h_i^4\Vert R_h{\Vert }_{L^2(I_i)}^2$. Let $\eta =({\sum }_i{\eta }_i^2{)}^{1/2}$. Stop when $C\eta <\mathrm{TOL}$; the proved reliability bound then certifies the requested true-error tolerance.

Otherwise mark elements carrying a chosen fraction of the total indicator, bisect them, and repeat. To ensure termination as well as a valid stopping test, include a uniform-refinement safeguard, for example bisect every element on every second unsuccessful pass. This preserves adaptive refinement on the other passes while forcing $h_{\max }\to 0$.

Here termination can be justified directly. On each element $R_h=f+p^{\prime }{u}_h^{\prime }-q{u}_h$, so its global $L^2$ norm stays bounded if $\Vert u_h{\Vert }_{H^1}$ stays bounded. That stability follows from coercivity after subtracting the fixed lift one at the right endpoint. HenceThe safeguarded algorithm therefore eventually satisfies the strict tolerance. Reliable numerical evaluation of the residual integrals, or an upper bound for their quadrature errors, is part of using the certificate in an implementation.