= Solution
Interpret $\phi$ as the small transverse displacement of the <elastic beam>. The first term is the positive <kinetic energy> density $\rho\phi_t^2/2$, 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 is
$$
\boxed{\mathcal V=-\frac\alpha2\phi_x^2+\frac\beta2\phi_{xx}^2.}
$$
For a compressed rod, transverse deflection releases compression energy: the small-slope geometric change in length is proportional to $\phi_x^2/2$, so the applied compressive load lowers the potential by $\alpha\phi_x^2/2$. 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 $\phi_{xx}$, so positive <bending stiffness> $\beta$ gives potential $\beta\phi_{xx}^2/2$, 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>.
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