 Solution

Interpret $\varphi$ as the small transverse displacement of the elastic beam. The first term is the positive kinetic energy density $\rho {\varphi }_t^2/2$, with $\rho$ the mass per unit length. The Lagrangian density is kinetic minus potential energy, so the potential density encoded by the other two terms isFor a compressed rod, transverse deflection releases compression energy: the small-slope geometric change in length is proportional to ${\varphi }_x^2/2$, so the applied compressive load lowers the potential by $\alpha {\varphi }_x^2/2$. This explains the positive sign of the $\alpha$ term in the Lagrangian. Tension would give the opposite sign and a restoring gradient term.

Bending costs elastic energy. At small slopes the curvature is ${\varphi }_{xx}$, so positive bending stiffness $\beta$ gives potential $\beta {\varphi }_{xx}^2/2$, and hence a negative term in the Lagrangian. For a uniform slender beam $\beta$ is its flexural rigidity. The competition between destabilizing compression and stabilizing bending is responsible for Euler buckling of an elastic filament.