 Solution

A constant shift $\varphi \mapsto \varphi +ϵ$ leaves every derivative unchanged, and the density has no undifferentiated $\varphi$. Thus it is exactly invariant under this scalar-field shift symmetry. Localize the shift parameter temporarily to $ϵ(x,t)$:Integrating the last term by parts once identifies the Noether current for a constant shift with second derivatives:The Euler-Lagrange field equation is $\rho {\varphi }_{tt}+\alpha {\varphi }_{xx}+\beta {\varphi }_{xxxx}=0$, precisely the continuity equation ${\partial }_t{j}^0+{\partial }_x{j}^x=0$. Therefore the conserved quantity, when the endpoint current vanishes or the field has suitable periodic/decaying conditions, isthe total transverse momentum. More explicitly $dQ/dt=-[\alpha {\varphi }_x+\beta {\varphi }_{xxx}{]}_{\mathrm{ends}}$. The fixed endpoints introduced later can exert transverse forces and break the global shift symmetry of the boundary problem, so vanishing endpoint current must not be inferred merely from fixed displacement.