 Solution

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.