Past exam of the approximation theory and finite element methods course of the University of Oxford / 2007 / B (Finite Element Methods)

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.

Besides $A,B,C$, set $D=(1,2)$, $E=(2,1)$ and $F=(2,2)$. The four triangles are $ABD$, $BCE$, $BDE$ and $DEF$, each of area $1/2$. The nodal function is one at $B$, zero at every other node, and affine on each triangle. The vertex conditions giveThe formulas agree on every shared edge, proving continuity.

Let $V=\{v\in H^1(\Omega ):v=0\text{ on }{\Gamma }_D\}$, with boundary values interpreted as traces. The outward unit normal on ${\Gamma }_N$ is $(-1,-1)/\sqrt{2}$, so the stated derivative condition is precisely ${\partial }_{n}u=0$. integration by parts givesThe Dirichlet boundary has positive length, so the Poincaré inequality makes the form coercive. Let $V_h\subset V$ be the continuous piecewise affine space on the four triangles. The finite element approximation satisfies the same identity for every $v_h\in V_h$.

Nodes $A,C,D,E,F$ have zero values, leaving $V_h=\mathrm{span}\{{\varphi }_B\}$. Although isolated endpoints have zero boundary measure, continuity along the Dirichlet edges forces the nodal values at $A$ and $C$ to vanish.

Write $u_h=U_B{\varphi }_B$. On $ABD,BCE,BDE$, the gradients are $(0,-1)$, $(-1,0)$ and $(-1,-1)$, respectively. ThusThe integral of an affine nodal function over a triangle equals its area times the mean of its three vertex values. Each supported triangle therefore contributes $1/6$ to $\int {\varphi }_B$, giving total $1/2$. The reduced system is $[2][U_B]=[1/2]$, and the complete nodal answer isOnly one algebraic unknown remains despite the two-dimensional domain.

Taking $w=z$ in the dual problem, and using $e^x\ge 1$, givesHence $\Vert z{\Vert }_{H^1}\le \Vert \psi {\Vert }_2$. Testing with smooth compactly supported functions identifies $-z^{\prime \prime }+e^{x}z=\psi$ in distributions. Therefore $z^{\prime \prime }\in L^2$ and $z\in H^2(0,1)$, withintegration by parts with arbitrary endpoint test values also gives the natural Neumann boundary conditions $z^{\prime }(0)=z^{\prime }(1)=0$.

Put $e_h=u-u_h$ and $w=z-I_{h}z$. Galerkin orthogonality gives $a(e_h,I_{h}z)=0$, so the dual identity becomesOn $I_i=(x_{i-1},x_i)$, $u_h^{\prime }$ is constant and $w$ is zero at both endpoints. Consequently ${\int }_{I_i}{u}_h^{\prime }{w}^{\prime }=u_h^{\prime }[w{]}_{x_{i-1}}^{x_i}=0$. Defining the element residual bygivesNo derivative-jump terms survive, because this interpolation error vanishes at every mesh node. This is the mechanism of the explicit one-dimensional dual residual bound in L2.

Two interval Poincaré inequalities give $\Vert z-I_{h}z{\Vert }_{2,I_i}\le h_i^2\Vert z^{\prime \prime }{\Vert }_{2,I_i}/{\pi }^2$: the error has zero endpoint values and its derivative has zero mean. Apply Cauchy-Schwarz inequality on each interval and then to the sum:Here ${\eta }_h^2={\sum }_i{h}_i^4\Vert R(u_h){\Vert }_{2,I_i}^2$. Taking $\psi =u-u_h$ and cancelling its norm, unless it is zero already, yieldsFor adaptive refinement, solve the discrete problem, compute ${\eta }_i^2=h_i^4\Vert R(u_h){\Vert }_{2,I_i}^2$, and stop when $(e+1){\eta }_h/{\pi }^2\le \mathrm{TOL}$. Otherwise mark intervals carrying a fixed positive fraction of ${\eta }_h^2$, bisect them, and repeat. Residual integration must be exact or have a certified upper bound if the stopping test is to certify the requested tolerance.

Use residual adaptation with a uniform-refinement safeguard, for example global bisection every second refinement, to ensure eventual termination. Discrete energy stability gives $\Vert u_h{\Vert }_{H^1}\le \Vert f{\Vert }_2$, so $\Vert R(u_h){\Vert }_2\le (1+e)\Vert f{\Vert }_2$ independently of the mesh. Thus ${\eta }_h\le h_{\max }^2(1+e)\Vert f{\Vert }_2$ tends to zero under the safeguard. Other iterations concentrate work where the residual is large.

Use $V=\{v\in H^1(0,1):v(0)=0\}$ and its continuous piecewise linear subspace $V_h=\mathrm{span}\{{\varphi }_1,\dots ,{\varphi }_N\}$. The condition at $x=1$ is natural, so the last hat function remains free. Set $u_h^0$ to the $L^2$ orthogonal projection of $u_0$ onto $V_h$: the stated $L^2$ initial data need not have nodal values.

For each $m$, find $u_h^{m+1}\in V_h$ such thatThis applies Backward Euler method to diffusion and reaction together. The step bilinear form is coercive, so each step has a unique solution.

Test with $v_h=u_h^{m+1}$ and apply $2(a-b,a)=\Vert a{\Vert }^2-\Vert b{\Vert }^2+\Vert a-b{\Vert }^2$. The exact energy identity isThe printed identity omits the last reaction contribution while explicitly describing its last term as the $H^1$ seminorm. It requires the full $H^1$ norm instead of that seminorm. With the printed seminorm its left side equals $-\Vert u_h^{m+1}{\Vert }_2^2$, rather than zero. The source reaction term forces this correction.

All terms other than the difference of successive squared norms are nonnegative. ConsequentlyThe second inequality uses contractivity of the initial orthogonal projection. No restriction relates $\Delta t$ to $h$, giving unconditional stability. The retained reaction term strengthens dissipation.

Define the mass matrix $M_{ij}={\int }_0^1{\varphi }_j{\varphi }_i$ and stiffness matrix $K_{ij}={\int }_0^1{\varphi }_j^{\prime }{\varphi }_i^{\prime }$ for $1\le i,j\le N$. With coefficient vectors $U^m$, the step system isThe matrices are symmetric, and $V^T\mathsf{A}V=\Vert v_h{\Vert }_2^2+\Delta t(\Vert v_h^{\prime }{\Vert }_2^2+\Vert v_h{\Vert }_2^2)>0$ for nonzero $V$, proving positive definiteness.

For $h=1/2$, the first basis function is the full tent at $1/2$, and the last is the half tent ending at $1$. Integration givesSince $\Delta t=2/2=1$, the requested matrix in this normalization isHere the matrix symbol $M$ is distinct from the printed integer number of time steps.