Past exam of the approximation theory and finite element methods course of the University of Oxford / 2010 / B

The first-order Sobolev space is $H^1(\Omega )=\{v\in L^2(\Omega ):{\partial }_{1}v,{\partial }_{2}v\in L^2(\Omega )\}$, with derivatives understood as weak derivatives and norm squared ${\int }_{\Omega }(v^2+|\nabla v{|}^2)$.

The printed normal boundary condition must be distinguished from a conormal condition. Take $K$ symmetric positive definite, as required for the symmetric matrix requested later. Integration by parts givesThe natural boundary term contains the conormal derivative $\mathbf{n}\cdot K\nabla u$, not the ordinary normal derivative ${\partial }_{n}u$. On a smooth boundary, writing $\mathbf{t}$ for its unit tangent, the literal condition ${\partial }_{n}u=0$ leaves the term $b{\partial }_{\mathbf{t}}u$, where $b=\mathbf{n}\cdot K\mathbf{t}$. Its weak form therefore retains a boundary pairing:Here the tangential derivative of the Sobolev trace can be interpreted in $H^{-1/2}$ and paired against $H^{1/2}$ on a smooth boundary. This form is generally nonsymmetric; the requested usual symmetric energy construction does not follow from the literal boundary condition.

For an explicit failure of the claimed mean-zero compatibility, take the unit disk, $K=\mathrm{diag}(2,1)$ and $u=(x^2-y^2)[1-(x^2+y^2)/2]$. It has ${\partial }_{n}u=0$ on the circle, butThus replacing the printed normal condition by a zero conormal flux is not harmless.

For the intended zero-conormal derivative problem, the requested variational and finite element conclusions are as follows. Seek $u\in H^1(\Omega )$ withPutting $v=1$ gives ${\int }_{\Omega }f=0$. On a connected domain this condition permits a solution unique modulo constants; the Poincare-Wirtinger inequality gives coercivity after fixing the mean.

The energy is $J(v)=\frac{1}{2}a(v,v)-\ell (v)$. A minimizer satisfies $\frac{d}{dt}J(u+tv){|}_{t=0}=a(u,v)-\ell (v)=0$. Conversely, a weak solution satisfiesso it is a minimizer. Testing by $u$ gives the minimum valueThis value is independent of the arbitrary additive constant when $\int f=0$. If the compatibility fails, constant shifts make $J$ unbounded below.

Write $u_h={\sum }_j{c}_j{\varphi }_j$. The Galerkin method givesSymmetry of $K$ gives $A^T=A$. For $v_h={\sum }_j{z}_j{\varphi }_j$, $z^{T}Az={\int }_{\Omega }\nabla v_h\cdot K\nabla v_h\ge 0$. Equality forces $v_h$ constant by positive definiteness and connectedness. Since the basis sums to one, the coefficient vector $1=(1,\dots ,1{)}^T$ represents that constant, givingThus the stiffness matrix is symmetric positive semidefinite and singular; the displayed compatibility condition is necessary and sufficient for the finite system. Positive definiteness of $K$ controls every gradient direction, preventing additional zero-energy directions and supplying ellipticity. A degenerate coefficient can introduce further kernel modes, while an indefinite one destroys the minimizing-energy argument.

Take the affine trial space $V_1=\{v\in H^1(a,b):v(a)=1\}$ and the test space $V_0=\{v\in H^1(a,b):v(a)=0\}$. Multiplication by $v\in V_0$ and integration by parts incorporate the right Neumann boundary condition as a load:For the continuous piecewise-quadratic trial space, impose the left value exactly and use the same equation for every piecewise-quadratic test function vanishing at $a$. With a basis $({\psi }_j)$ of that test space and $u_h=1+{\sum }_j{c}_j{\psi }_j$, the finite element equations areThe Endpoint Poincaré inequality, reflected about the interval midpoint, makes the derivative norm a norm on $V_0$. Galerkin orthogonality then gives the exact best-approximation identity: for every admissible $v_h$,Therefore $u_h$ minimizes the derivative error over all admissible piecewise-quadratic functions.

To prove the requested error bound, assume $f\in H^1(a,b)$. Then $u\in H^3(a,b)$ and $u^{\prime \prime \prime }=-f^{\prime }$. We need an elementary Poincare-Wirtinger inequality on each element $I$ of length $h_I$: if $w\in H^1(I)$ has mean zero, its continuous representative has a zero somewhere, and the fundamental theorem of calculus and Cauchy-Schwarz inequality give $\Vert w{\Vert }_{L^2(I)}\le h_I\Vert w^{\prime }{\Vert }_{L^2(I)}$.

Choose an affine function $q_I$ whose slope is the average of $u^{\prime \prime }$ on $I$ and whose average equals the average of $u^{\prime }$. Then $w=u^{\prime }-q_I$ and $w^{\prime }=u^{\prime \prime }-q_I^{\prime }$ both have mean zero. Applying the just-proved bound twice yieldsDefine a quadratic interpolant on $I$ by integrating $q_I$ from the true left-endpoint value. The equal-average condition ensures that it reaches the true right-endpoint value, so these pieces join to an admissible continuous piecewise-quadratic function $I_{h}u$. Sum the squared estimates and use the best-approximation property:Only the maximum element length is used; equal spacing is unnecessary.

For the Aubin–Nitsche duality argument, let $e=u-u_h$ and solve $-z^{\prime \prime }=e$, $z(a)=0$, $z^{\prime }(b)=0$. Explicitly $z^{\prime }(x)={\int }_x^{b}e(t)dt$, so $\Vert z^{\prime \prime }{\Vert }_2=\Vert e{\Vert }_2$. Integration by parts and Galerkin orthogonality givewhere $J_{h}z$ is its nodal piecewise-linear interpolant, which belongs to the quadratic test space. On each element $(J_{h}z{)}^{\prime }$ is the average of $z^{\prime }$, so the same mean-zero inequality gives $\Vert z^{\prime }-(J_{h}z{)}^{\prime }{\Vert }_2\le h\Vert z^{\prime \prime }{\Vert }_2$. Canceling $\Vert e{\Vert }_2$ when nonzero proves the additional order:

The bilinear square finite element contains the terms $1,x,y,xy$. In anticlockwise corner order beginning at $(0,0)$, its shape functions areThe mass matrix has entries $M_{ij}={\int }_0^h{\int }_0^h{N}_i{N}_{j}dxdy$. Each product separates into one-dimensional factors, with ${\int }_0^1(1-s{)}^{2}ds={\int }_0^1{s}^{2}ds=1/3$ and ${\int }_0^{1}s(1-s)ds=1/6$. Diagonal, adjacent-corner and opposite-corner entries are respectively $h^2/9$, $h^2/18$, $h^2/36$. ThusNumber global grid nodes by $k=i+(n+1)j$, with $x_i=ih$, $y_j=jh$ and $0\le i,j\le n$. For an interior node, four elements contribute to its diagonal, two to each axial neighbour and one to each diagonal neighbour. Its global row therefore has the nine-point stencilIn the stated ordering the axial entries occur at columns $k\pm 1$ and $k\pm (n+1)$; the four diagonal ones occur at $k\pm (n+1)\pm 1$. No boundary degrees of freedom have been removed.

For general valid quadrilaterals, use an isoparametric finite element. Map the unit square to an element by $F(\xi ,\eta )={\sum }_{i=1}^4{\mathbf{x}}_i{N}_i(\xi ,\eta )$ and define physical shape functions $N_i\circ F^{-1}$. Require a one-to-one nondegenerate map and compatible common edges. On a shared edge the map and the trace of each shape function are linear between the same endpoints; the neighbouring traces therefore agree. The assembled functions are continuous and lie in $H^1$, giving a conforming finite element space. On a non-parallelogram the pulled-back functions are generally not bilinear polynomials in the physical coordinates themselves.