Past exam of the finite element methods for partial differential equations course of the University of Oxford / 2023 / 1

For a sufficiently regular solution, put $w=-\Delta u$, with $\Delta$ the Laplacian. Then $-\Delta u-w=0$ and $u-\Delta w=u+{\Delta }^{2}u=f$. The zero Dirichlet boundary conditions on $u$ and $\Delta u$ become zero Dirichlet boundary conditions on $u$ and $w$. Conversely, eliminating $w$ from these two equations recovers the fourth-order equation with the biharmonic operator. For the weak solution we will construct, these interior identities hold as distributional identities; zero Sobolev traces are imposed on $u$ and $w$, without assuming classical fourth derivatives of $u$.

Let $V=H_0^1(\Omega )\times H_0^1(\Omega )$, the Cartesian product of two zero-boundary Sobolev spaces, with normThis is a Hilbert space. Applying integration by parts separately to the two equations gives the weak formulationEquivalently, for $U=(u,w)$ and $T=(v,z)$, define the bilinear form and linear functionalWe seek $U\in V$ with $b(U,T)=F(T)$ for every $T\in V$.

The skew-coupled elliptic system has a useful bounded bilinear form estimate with constant one. Indeed, write the integrand as the dot product of $(\nabla p,\nabla q,-q,p)$ and $(\nabla v,\nabla z,v,z)$. The Cauchy-Schwarz inequality givesFor equal arguments, the two zeroth-order coupling terms cancel. If $K$ is the constant in the supplied Poincaré inequality, thenThus $b$ is a coercive bilinear form, although it is not a symmetric bilinear form. The Lax-Milgram theorem applies and yieldsIn particular, $\Vert (u,w){\Vert }_V\le (1+K)\Vert f{\Vert }_0$. Testing with compactly supported smooth test functions identifies $-\Delta u=w$ and $-\Delta w=f-u$ as distributional identities. This gives the asserted relation to the original boundary value problem at the available regularity.

Take $V_h={\mathcal{U}}_h\times {\mathcal{W}}_h\subset V$. The Galerkin method asks for $U_h=(u_h,w_h)\in V_h$ such thatSince a finite-dimensional subspace is closed, $V_h$ is itself a Hilbert space. The bounded bilinear form and coercive bilinear form estimates in part (a) restrict to $V_h$ with the same constants. The Lax-Milgram theorem therefore gives exactly one $U_h$ for every $h$.

Put $E=U-U_h=(e_u,e_w)$. Subtracting the continuous and discrete weak formulations gives Galerkin orthogonality:For any $Y_h=(v_h,z_h)\in V_h$, the difference $Y_h-U_h$ also belongs to $V_h$, soThe last step uses the constant-one Cauchy-Schwarz inequality estimate from part (a). Bounding the four terms separately by a sum would obscure the precise constant requested here.

The two minimization problems over ${\mathcal{U}}_h$ and ${\mathcal{W}}_h$ are independent. Each has a best approximation because its vector subspace is finite-dimensional. Consequently, ifthen the preceding inequality gives exactlyBoth errors belong to the zero-boundary Sobolev space. Applying the Poincaré inequality componentwise now yieldsWhen $U=U_h$ the desired estimate is automatic; otherwise divide by the nonzero norm. ThusThe constant $C=1+K$ is independent of $h$. This is the Céa lemma estimate for this particular skew-coupled elliptic system.

Use the continuous piecewise-linear nodal interpolants $I_{h}u$ and $I_{h}w$ on the shape-regular mesh. Since $u,w\in H^2(\Omega )$ in two dimensions, the Sobolev embedding theorem supplies continuous representatives, so the nodal values are well-defined. Their zero Sobolev traces imply zero boundary nodal values. Hence $I_{h}u\in {\mathcal{U}}_h$ and $I_{h}w\in {\mathcal{W}}_h$ for the usual full conforming finite element spaces.

The permitted finite element interpolation estimates give constants $C_u,C_w$ determined by the shape regularity and $\Omega$, independent of $h$, such thatSubstitute these choices in the best approximation bound from part (b):For fixed $u,w$, this constant is independent of $h$. A shape-regular mesh suffices; no uniformity of the individual element sizes is needed, since $h$ is their maximum diameter.

The inclusion of these interpolants is an essential assumption. If the wording “consist of” were read as permitting arbitrary smaller vector subspaces, taking ${\mathcal{U}}_h={\mathcal{W}}_h=\{0\}$ for every $h$ would prevent the claimed convergence for a nonzero solution. The full conforming finite element space interpretation is also needed for the dimension assertion in part (d).

Let $x_1,\dots ,x_N$ be the interior nodes, and let ${\varphi }_1,\dots ,{\varphi }_N$ be the nodal basis of the common conforming finite element space, with ${\varphi }_j(x_i)={\delta }_{ij}$ and zero boundary values. These piecewise-linear hat functions give the expansionsTesting the first equation with ${\varphi }_i$ and then the second equation with ${\varphi }_i$ gives the linear systemHere $S$ is the diffusion stiffness matrix and $M$ the mass matrix. Their entries are symmetric in $i,j$, so both are symmetric matrices.

For $c\in {\mathbb{R}}^N$, put $v_c={\sum }_j{c}_j{\varphi }_j$. The two quadratic forms areIf $c\ne 0$, the linear independence of the basis ensures $v_c\ne 0$, so $c^{\mathsf{T}}Mc>0$. If $c^{\mathsf{T}}Sc=0$, the Poincaré inequality and zero Dirichlet boundary condition force $v_c=0$, contradicting $c\ne 0$. Thus both $S$ and $M$ are positive-definite matrices.

For $X=(p,q)\in {\mathbb{R}}^N\times {\mathbb{R}}^N$, symmetry of $M$ cancels the cross terms:ThereforeThis is the positivity of a quadratic form intended by the last sentence of the question. For $N>0$, $A$ is a nonsymmetric matrix, since $M\ne 0$. Under the convention that “positive-definite matrix” includes symmetry, that last sentence needs this qualification: the symmetric part of a matrix is positive definite. The quadratic lower bound for a nonsymmetric matrix still implies that $A$ is an invertible matrix: $AX=0$ would give $X^{\mathsf{T}}AX=0$, hence $X=0$.