Past exam of the finite element methods for partial differential equations course of the University of Oxford / 2023 / 3

The infinitesimal strain tensor $\epsilon (v)=\frac{1}{2}(\nabla v+(\nabla v{)}^{\mathsf{T}})$ and the divergence are linear in $v$. Consequently the given $a$ is a symmetric bilinear form, andFor a matrix-valued function, $\Vert \cdot {\Vert }_0$ here uses the Frobenius inner product pointwise and then the L2 norm. All expressions are finite for $v\in H^1(\Omega ;{\mathbb{R}}^3)$.

Expand the quadratic term at $\theta v+(1-\theta )w$. The linearity of $\ell$ cancels its contribution to the difference, givingFor $0\le \theta \le 1$, this is nonnegative. Thereforeso $J$ is a convex functional on the whole Sobolev space. No boundary condition is needed for this conclusion. The argument uses nonnegativity of the quadratic term; uniqueness on the zero-boundary Sobolev space will instead use the Korn inequality in part (b).

Write $V=H_0^1(\Omega ;{\mathbb{R}}^3)$. For any $v\in V$ and any real $t$, $u+tv$ is admissible. Expanding the functional givesSince the left side is nonnegative for positive and negative $t$, its derivative at zero vanishes. Thus the first variation gives the weak formulation

To identify its differential equation, define the stress tensorThis is a symmetric matrix field. The Frobenius inner product of a symmetric matrix with a skew-symmetric matrix is zero, so $\sigma (u):\nabla v=\sigma (u):\epsilon (v)$. Since $\mathrm{tr}\epsilon (v)=\mathrm{div}v$, we obtainFor smooth compactly supported test functions, integration by parts in each row givesHence $-\mathrm{div}\sigma (u)=f$ as a distributional identity, that is,Membership in $V$ supplies the zero Sobolev trace. This identifies precisely the required weak solution of the linear elasticity problem, with row-wise divergence.

For uniqueness, use the exact form of the Korn inequality supplied in the paper. Put $E=\epsilon (v)$ and $d=\mathrm{tr}E=\mathrm{div}v$. The deviatoric strain is $D=E-\frac{1}{3}dI$. Because $\mathrm{tr}D=0$ and $I:I=3$, orthogonality for the Frobenius inner product givesIf $C_K$ is the supplied Korn inequality constant and $K$ the permitted Poincaré inequality constant, thenConsequently,If $u_1,u_2$ are two weak solutions, their difference $d_u\in V$ satisfies $a(d_u,v)=0$ for every $v\in V$. Choosing $v=d_u$ and using this coercive bilinear form estimate forces $d_u=0$. The minimizer therefore gives the unique weak solution.

For later use, $|\epsilon (v){|}_F\le |\nabla v{|}_F$ and $|\mathrm{div}v|\le \sqrt{3}|\nabla v{|}_F$ also give the bounded bilinear form and linear functional estimatesTogether with the coercive bilinear form estimate and the Lax-Milgram theorem, these would establish existence independently of the assumed minimizer.

The finite-dimensional vector subspace $V_h\subset V$ is closed, so it is a Hilbert space with the inherited Sobolev norm. The bounded bilinear form, coercive bilinear form, and bounded linear functional estimates from part (b) restrict to $V_h$. Applying the Lax-Milgram theorem proves

For any $v_h\in V_h$, put $d_h=v_h-u_h\in V_h$. Expansion using symmetry of the bilinear form givesThus $u_h$ minimizes $J$ on $V_h$, uniquely by the coercive bilinear form estimate. Similarly, the continuous weak formulation gives the exact identityCombining the two identities provesMore precisely, $J(v_h)-J(u_h)=\frac{1}{2}\Vert v_h-u_h{\Vert }_a^2$ and $J(u_h)-J(u)=\frac{1}{2}\Vert u_h-u{\Vert }_a^2$, with the energy norm justified next. These identities connect the energy excess directly to the squared error.

On $V$, the bilinear form $a$ is symmetric, and part (b) proves $a(v,v)\ge \alpha \Vert v{\Vert }_1^2$. Thus $a$ is an inner product: $a(v,v)\ge 0$, and equality forces $v=0$. Its induced energy norm is $\Vert v{\Vert }_a=\sqrt{a(v,v)}$. For any real scalar $t$,The Cauchy-Schwarz inequality for this inner product follows directly by expanding $a(p-tq,p-tq)\ge 0$ and minimizing over $t$ when $q\ne 0$. Hencewhich proves the triangle inequality. This verifies every norm axiom. In fact, the estimates from part (b) give the equivalent norms

Subtract the discrete weak formulation from the continuous one to get Galerkin orthogonality,For any $v_h\in V_h$, write $u-v_h=(u-u_h)+(u_h-v_h)$. The second summand is in $V_h$, so expansion and Galerkin orthogonality give the Pythagorean theorem in the energy norm:The last term is nonnegative and vanishes exactly when $v_h=u_h$. Choosing $v_h=u_h$ attains the lower bound, provingThus $u_h$ is the unique best approximation; the map $u\mapsto u_h$ is the orthogonal projector onto $V_h$ for the linear elasticity energy norm.