Past exam of the approximation theory and finite element methods course of the University of Oxford / 2009 / B / 5

The Sobolev space $H^1(b,c)$ consists of $L^2$ functions with a weak first derivative in $L^2$, with normIn one dimension these functions have absolutely continuous representatives and well-defined endpoint Sobolev traces. PutThis is a closed subspace, hence a Hilbert space. Both trial and test functions belong to $V$; one does not impose a derivative boundary condition in this space.

Integrate the differential equation against $v\in V$:The printed condition uses the ordinary derivative, $u^{\prime }(c)=-u(c)$, so the weak form isSeek $u\in V$ with $a(u,v)=\ell (v)$ for all $v\in V$. The coefficient $e^c$ in the endpoint term must be retained: replacing the ordinary derivative with the conormal flux would change the boundary problem.

Let $L=c-b$. The Poincaré inequality on $V$ can be stated as $\Vert v{\Vert }_{L^2}\le L\Vert v^{\prime }{\Vert }_{L^2}$. Also $|v(c)|\le \sqrt{L}\Vert v^{\prime }{\Vert }_{L^2}$. Consequentlyproving coercivity. By Cauchy-Schwarz inequality and the trace bound,Thus continuity holds with $M=e^c(1+L)$. The load is continuous if $f\in L^2$, and the Lax-Milgram theorem gives a unique weak solution.

Choose a finite-dimensional conforming subspace $V_h\subset V$, such as continuous piecewise linear functions vanishing at $b$, and solve $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$. Subtracting the weak equations gives Galerkin orthogonality $a(u-u_h,v_h)=0$.

Céa's lemma states, for a continuous coercive bilinear form with constants $M,\alpha$ on a Hilbert space,To prove it, put $e=u-u_h$ and use orthogonality to write $a(e,e)=a(e,u-v_h)$ for any $v_h\in V_h$. Coercivity and continuity give $\alpha \Vert e{\Vert }^2\le M\Vert e\Vert \Vert u-v_h\Vert$. If $e=0$ the claim is immediate; otherwise cancel its norm and take the infimum. Any approximation theorem making the infimum tend to zero under mesh refinement therefore gives strong convergence $u_h\to u$ in $H^1$.

Finally, $a$ is symmetric and positive definite on $V$. It is an inner product, so $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ is an energy norm on all of $V$: homogeneity and definiteness are immediate, and Cauchy-Schwarz inequality for this inner product gives the triangle inequality. No smaller space is required. In this norm the continuity and coercivity constants are one. More precisely, Galerkin orthogonality givessoThe finite element solution is the exact best energy-norm approximation.