Past exam of the approximation theory and finite element methods course of the University of Oxford / 2007 / B / 5

The Sobolev space and norm areIn one dimension these functions have continuous representatives and well-defined endpoint traces. integration by parts gives the weak problem: find $u\in H^1(a,b)$ such that $a(u,v)=\ell (v)$ for all $v\in H^1(a,b)$, whereIndeed the integration boundary term is $-u^{\prime }(b)v(b)+u^{\prime }(a)v(a)$; substituting $u^{\prime }(a)=u(a)-A$ and $u^{\prime }(b)=B-u(b)$ gives the two positive boundary terms on the left.

If $0\notin [a,b]$, put $m={\min }_{[a,b]}{x}^2>0$. Then $a(v,v)\ge \Vert v^{\prime }{\Vert }_2^2+m\Vert v{\Vert }_2^2\ge \min (1,m)\Vert v{\Vert }_{H^1}^2$. If $0\in [a,b]$, the supplied trace inequality givesThus the form is coercive in both cases. The trace theorem and boundedness of $x^2$ give continuity of $a$ and $\ell$, and the Lax-Milgram theorem gives existence and uniqueness.

Let $V_h$ be the continuous piecewise linear space including both endpoint hat functions. Robin conditions are natural conditions and do not restrict the endpoint values of this space. The finite element problem is $u_h\in V_h$, $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in V_h$. coercivity restricted to $V_h$ gives existence and uniqueness.

Expanding $u_h={\sum }_{j=0}^N{U}_j{\varphi }_j$ gives $\mathsf{A}U=F$, whereThe stiffness contribution has diagonal $2/h$ at interior nodes, $1/h$ at endpoints, and adjacent entries $-1/h$. For completeness, the weighted mass contribution has interior diagonal $2h{x}_i^2/3+h^3/15$, left endpoint diagonal $h{a}^2/3+a{h}^2/6+h^3/30$, right endpoint diagonal $h{b}^2/3-b{h}^2/6+h^3/30$, and adjacent entry $h{x}_i^2/6+h^2{x}_i/6+h^3/20$. All more distant entries vanish. The Robin contribution adds one to each endpoint diagonal.

Symmetry follows from symmetry of the bilinear form. If $V\ne 0$, nodal independence makes $v_h={\sum }_i{V}_i{\varphi }_i$ nonzero, and $V^T\mathsf{A}V=a(v_h,v_h)\ge \gamma \Vert v_h{\Vert }_{H^1}^2>0$. Thus the matrix is symmetric positive definite.

The weak equation implies $u^{\prime \prime }=x^{2}u-f\in L^2(a,b)$ in distributions, hence $u\in H^2(a,b)$. Its nodal interpolant satisfiesThese follow interval by interval: the interpolation error vanishes at both endpoints, and its derivative has mean zero, so two Poincaré inequalities apply. If $M$ and $\gamma$ are the continuity and coercivity constants, Céa's lemma givesAll constants are independent of the mesh size.