Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 341 / 4

The clamped energy space isTwice applying integration by parts, with the boundary terms killed by the clamped conditions, gives the symmetric bilinear formConsequentlyfor every nonzero $u\in \mathcal{H}$: equality forces $u^{\prime \prime }=0$ almost everywhere, and the clamped boundary values then force $u=0$. Hence $L$ is symmetric and positive definite.

Strictly under the stated assumption $p\in L^2$ rather than $L^{\infty }$, the first integral need not be finite for every $u\in H_0^2$. The literal energy domain is therefore $\{u\in H_0^2:\sqrt{p}{u}^{\prime \prime }\in L^2\}$; under the usual coefficient assumption $0<p_0\le p\le p_1<\infty$, it is exactly $H_0^2$ and the form is coercive there.

Define the energyIts first variation in direction $w\in \mathcal{H}$ isThus its minimizer $u$ satisfies the weak formulationwhich is the weak equation $Lu=f$. Positive definiteness makes $J$ strictly convex, so this stationary point is the unique minimizer; under uniform positivity of $p$, existence follows from the Lax-Milgram theorem.

Choose a mesh of $[-1,1]$ and the conforming space of Cubic Hermite finite elements: piecewise cubic functions that are globally $C^1$ and satisfy $v(\pm 1)=v^{\prime }(\pm 1)=0$. Let ${\varphi }_1,\dots ,{\varphi }_n$ be its nodal value-and-slope basis. For $u_n={\sum }_k{a}_k{\varphi }_k$, the Ritz method imposesHence the coefficient vector solvesThe stiffness matrix is symmetric positive definite by part a.