Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 13C (Variational Principles)

Take a variation $y+\epsilon \eta$, where the differentiable function $\eta$ obeysbecause both $y$ and its first derivative have fixed endpoint values. The first variation of the functional isApplying integration by parts once to the second term and twice to the third givesThe endpoint conditions on $\eta$ and ${\eta }^{\prime }$ make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation

HereSubstitution into the higher-order Euler-Lagrange equation gives the fourth-order ordinary differential equation

Integrating the ordinary differential equation four times gives a quartic polynomial. It is useful to use the linearity of the equation and split the solution into a gravity part and a force part:wheresatisfies the clamped and torque-free boundary conditions with $F=0$, whilesatisfies the homogeneous equation and contributes the endpoint force. Direct differentiation verifiesThusthe clamped-free beam under uniform load and endpoint force.

Evaluating the function at the endpoint gives the displacementThereforeThe force-induced displacement $h$ is a linear map of $F$.

Substitute $y=y_0+y_F$ into the energy functional and integrate the resulting polynomial:The term linear in $F$ cancels. HenceAs required, $E_0$ is independent of the force and $\mathcal{E}$ is a quadratic function of it.

Taking the derivative of the force-dependent energy givesThus the derivative of the additional minimized internal energy with respect to the applied force equals the resulting endpoint displacement. This is the endpoint force derivative of clamped-free beam energy.