Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 13C (Variational Principles)

Let $u\mapsto u+\epsilon \eta$ and $v\mapsto v+\epsilon \xi$, where the variations vanish on the boundary. The first variation isApplying integration by parts to the four derivative terms and discarding the boundary contributions givesThe variations $\eta$ and $\xi$ are independent and arbitrary in the interior. The fundamental lemma of the calculus of variations therefore gives the two Euler-Lagrange equations for two fields

For the displacement gradient tensor specified in the question,while the divergence is $\nabla \cdot \mathbf{u}=u_x+v_y$. Hence the isotropic linear-elastic energy density isExpanding the square and collecting terms yields

Let the displayed integrand be $F$. Since it has no explicit dependence on $u$ or $v$, the Euler-Lagrange equations for two fields areIts derivatives areThusUsing the Laplacian, gradient, and divergence, these combine into the static Navier-Cauchy equation

In the one-dimensional limit, $v=0$ and $u=u(x)$. The first component of the Navier-Cauchy equation reduces toFor a nondegenerate one-dimensional elastic material, $\lambda +2\mu \ne 0$, so $u^{\prime \prime }=0$ and $u(x)=Ax+B$. The boundary conditions $u(0)=0$ and $u(L)=\Delta$ give $B=0$ and $A=\Delta /L$. Therefore the uniform extension of a one-dimensional elastic body is