Past exam of the approximation theory and finite element methods course of the University of Oxford / 2007 / B / 7

Taking $w=z$ in the dual problem, and using $e^x\ge 1$, givesHence $\Vert z{\Vert }_{H^1}\le \Vert \psi {\Vert }_2$. Testing with smooth compactly supported functions identifies $-z^{\prime \prime }+e^{x}z=\psi$ in distributions. Therefore $z^{\prime \prime }\in L^2$ and $z\in H^2(0,1)$, withintegration by parts with arbitrary endpoint test values also gives the natural Neumann boundary conditions $z^{\prime }(0)=z^{\prime }(1)=0$.

Put $e_h=u-u_h$ and $w=z-I_{h}z$. Galerkin orthogonality gives $a(e_h,I_{h}z)=0$, so the dual identity becomesOn $I_i=(x_{i-1},x_i)$, $u_h^{\prime }$ is constant and $w$ is zero at both endpoints. Consequently ${\int }_{I_i}{u}_h^{\prime }{w}^{\prime }=u_h^{\prime }[w{]}_{x_{i-1}}^{x_i}=0$. Defining the element residual bygivesNo derivative-jump terms survive, because this interpolation error vanishes at every mesh node. This is the mechanism of the explicit one-dimensional dual residual bound in L2.

Two interval Poincaré inequalities give $\Vert z-I_{h}z{\Vert }_{2,I_i}\le h_i^2\Vert z^{\prime \prime }{\Vert }_{2,I_i}/{\pi }^2$: the error has zero endpoint values and its derivative has zero mean. Apply Cauchy-Schwarz inequality on each interval and then to the sum:Here ${\eta }_h^2={\sum }_i{h}_i^4\Vert R(u_h){\Vert }_{2,I_i}^2$. Taking $\psi =u-u_h$ and cancelling its norm, unless it is zero already, yieldsFor adaptive refinement, solve the discrete problem, compute ${\eta }_i^2=h_i^4\Vert R(u_h){\Vert }_{2,I_i}^2$, and stop when $(e+1){\eta }_h/{\pi }^2\le \mathrm{TOL}$. Otherwise mark intervals carrying a fixed positive fraction of ${\eta }_h^2$, bisect them, and repeat. Residual integration must be exact or have a certified upper bound if the stopping test is to certify the requested tolerance.

Use residual adaptation with a uniform-refinement safeguard, for example global bisection every second refinement, to ensure eventual termination. Discrete energy stability gives $\Vert u_h{\Vert }_{H^1}\le \Vert f{\Vert }_2$, so $\Vert R(u_h){\Vert }_2\le (1+e)\Vert f{\Vert }_2$ independently of the mesh. Thus ${\eta }_h\le h_{\max }^2(1+e)\Vert f{\Vert }_2$ tends to zero under the safeguard. Other iterations concentrate work where the residual is large.