Past exam of the approximation theory and finite element methods course of the University of Oxford / 2008 / B / 5

The Sobolev space $H^1(0,1)$ consists of square-integrable functions whose weak first derivative is square-integrable, with norm $\Vert v{\Vert }_{H^1}^2={\int }_0^1(v^2+v^{\prime 2})$. Its one-dimensional representatives have well-defined endpoint traces.

Multiplying by $v\in H^1$ and integrating by parts givesThe two Robin conditions give $u^{\prime }(0)=u(0)$ and $u^{\prime }(1)=-u(1)$, so defineA weak solution is $u\in H^1(0,1)$ satisfying $a(u,v)=\ell (v)$ for every $v\in H^1(0,1)$. The boundary conditions are natural; no endpoint derivative is imposed in the trial space.

Since $\cos x\ge \cos 1>0$ on this interval,This proves coercivity on the full space. Trace continuity and Cauchy-Schwarz give continuity of the form, so the Lax-Milgram theorem also ensures a unique weak solution.

Let $V_h$ be the space spanned by all $N+1$ hats, including the endpoint hats. The solution in a conforming finite element space satisfies $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in V_h$. The continuous form remains coercive and continuous on this subspace, so the discrete solution is unique.

With $u_h={\sum }_j{U}_j{\phi }_j$, its equations areThe formula is symmetric in $i,j$. For a nonzero vector $V$, the corresponding spline $v_h={\sum }_j{V}_j{\phi }_j$ is nonzero by basis independence, andThus $A$ is symmetric positive definite, proving algebraic solvability as well.

Testing against compactly supported smooth functions gives the distributional identity $u^{\prime \prime }=\cos xu-f$. The right side belongs to $L^2$, since $u\in H^1$ and $\cos x$ is bounded. Therefore $u^{\prime }\in H^1$ and $u\in H^2(0,1)$. This also makes the endpoint derivative traces meaningful, and integrating the weak form by parts recovers the Robin conditions.

The standard piecewise linear finite element interpolation estimate iswith constants independent of $h$. Hence $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u^{\prime \prime }{\Vert }_{L^2}$ for $h\le 1$. By Céa's lemma, with the fixed continuity and coercivity constants of $a$,The constant depends on the continuous problem but not on mesh refinement.