Past exam of the approximation theory and finite element methods course of the University of Oxford / 2008 / B (Finite Element Methods)

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.

For any $v\in H^1(\Omega )$, expand $J(u+ϵv)$ in $ϵ$. Its linear term must vanish at a minimizer, givingThus the required functionals areThe form is continuous, symmetric, and coercive, with $a(v,v)=\Vert v{\Vert }_{H^1}^2$. Conversely, this weak equation implies $J(u+w)=J(u)+\frac{1}{2}a(w,w)$, so its unique solution is indeed the global minimizer.

For smooth $u$, Green's first identity rewrites the weak equation asCompactly supported test functions yield $-\Delta u+u=f$. Since the test space is the full $H^1(\Omega )$ and has unrestricted boundary trace, the remaining identity gives the homogeneous Neumann condition.

For a merely weak solution on a polygon, the normal derivative is understood as its weak flux trace. Here $\nabla u\in L^2$ and $\mathrm{div}\nabla u=u-f\in L^2$, so this trace is defined through Green's formula. It vanishes by the same identity. Thus the natural boundary condition is ${\partial }_{n}u=0$, without assuming global $H^2$ regularity on an arbitrary polygon.

Take $V_h$ to be all continuous functions affine on each triangle of the stated negative-diagonal mesh. No boundary value is fixed, because the Neumann condition is natural. The discrete problem isIts matrix is symmetric positive definite: the reaction contribution $\int u_h{v}_h$ removes the constant null mode that would occur for pure Neumann Poisson equations. Thus the solution exists uniquely.

For every $v_h\in V_h$, expand the quadratic functional about $u_h$ and use the discrete weak equation to remove its linear term:The continuous solution minimizes $J$ over the larger space $H^1$, so

Put $e=u-u_h$ and $V=\{v\in H^1(0,1):v(1)=0\}$. The primal solution and its finite element approximation lie in the same affine space with value one at the right endpoint, so $e\in V$. UseDefine the dual solution byEquivalently $z\in V$ satisfies $a(v,z)=(e,v)$ for every $v\in V$. Positivity of $p,q$ gives a unique solution, and one-dimensional regularity gives $z\in H^2$.

Then $\Vert e{\Vert }_{L^2}^2=a(e,z)$. The nodal interpolant $I_{h}z$ belongs to the homogeneous finite element test space, so Galerkin orthogonality gives $a(e,I_{h}z)=0$. Set $w=z-I_{h}z$. Using the primal weak equation,Integrate the derivative term separately on each element. Since $w$ vanishes at every mesh node, all endpoint flux and flux-jump terms vanish. ThereforeThe absence of jump terms here follows specifically from one-dimensional nodal interpolation; it is not an assumption that the finite element flux itself is continuous.

Let $p_{\ast }=\min p>0$, $q_{\ast }=\min q>0$, $q^{\ast }=\max q$, and $P_1=\Vert p^{\prime }{\Vert }_{\infty }$. Testing the dual equation with $z$ givesConsequently $\Vert z{\Vert }_2\le \Vert e{\Vert }_2/q_{\ast }$ and $\Vert z^{\prime }{\Vert }_2\le \Vert e{\Vert }_2/\sqrt{p_{\ast }{q}_{\ast }}$. From $p{z}^{\prime \prime }=qz-e-p^{\prime }{z}^{\prime }$,On an element of length $h_i$, the interpolation error vanishes at both endpoints and its derivative has zero mean. Two Poincaré inequalities giveApply Cauchy-Schwarz to the residual identity, with factors $h_i^2$ and $h_i^{-2}$:If the error is nonzero, cancel its norm after the regularity estimate; if it is zero the conclusion is immediate. One explicit reliable constant isThis proves the requested dual residual bound entirely in terms of the coefficient functions.

Compute $C$ from part (b), solve the finite element equations, and calculate the element indicators ${\eta }_i^2=h_i^4\Vert R_h{\Vert }_{L^2(I_i)}^2$. Let $\eta =({\sum }_i{\eta }_i^2{)}^{1/2}$. Stop when $C\eta <\mathrm{TOL}$; the proved reliability bound then certifies the requested true-error tolerance.

Otherwise mark elements carrying a chosen fraction of the total indicator, bisect them, and repeat. To ensure termination as well as a valid stopping test, include a uniform-refinement safeguard, for example bisect every element on every second unsuccessful pass. This preserves adaptive refinement on the other passes while forcing $h_{\max }\to 0$.

Here termination can be justified directly. On each element $R_h=f+p^{\prime }{u}_h^{\prime }-q{u}_h$, so its global $L^2$ norm stays bounded if $\Vert u_h{\Vert }_{H^1}$ stays bounded. That stability follows from coercivity after subtracting the fixed lift one at the right endpoint. HenceThe safeguarded algorithm therefore eventually satisfies the strict tolerance. Reliable numerical evaluation of the residual integrals, or an upper bound for their quadrature errors, is part of using the certificate in an implementation.

Use the full continuous piecewise linear space $V_h\subset H^1(0,1)$, including endpoint basis functions, because the boundary conditions are Neumann. DefineChoose $u_h^0$ as the $L^2$ projection of $u_0$ onto $V_h$. For each time step seek $u_h^{m+1}\in V_h$ withThis is the conforming Crank-Nicolson method. The reaction coefficient and the diffusion are both treated at the midpoint; integration by parts has removed the boundary flux through the homogeneous Neumann condition.

Test the discrete equation with $v_h=u_h^{m+1/2}$. The identity $(a-b,a+b)=\Vert a{\Vert }^2-\Vert b{\Vert }^2$ yieldsSince $e^x\ge 1$ on the interval, $a(v,v)\ge \Vert v{\Vert }_{H^1}^2$. Therefore the requested inequality holds:Discarding the nonnegative second term and iterating proves $\Vert u_h^m{\Vert }_2\le \Vert u_h^0{\Vert }_2$ for every positive $\Delta t$, with no restriction relating it to $h$. This is unconditional $L^2$ stability; the initial projection also satisfies $\Vert u_h^0{\Vert }_2\le \Vert u_0{\Vert }_2$.

Write $u_h^m={\sum }_i{U}_i^m{\phi }_i$ and define the mass matrix and reaction-diffusion stiffness matrixThen each step isBoth component matrices are symmetric, so $A$ is symmetric. For a nonzero coefficient vector $v$, its finite element function $v_h$ is nonzero andThus $A$ is also positive definite and every update has a unique solution. Because nonneighboring hats have disjoint interior supports, the matrix is tridiagonal in nodal order.