Derive wave equations from stationary action

Try

An equation can be derived by requiring a functional to be stationary under small changes of the field. For a real scalar u(x,t), consider action density L=(uₜ²−c²uₓ²−m²u²)/2, with constants c positive and m nonnegative. The action is the space-time integral of L over a chosen region.

The minus signs matter. This action is not the positive mechanical energy, and stationarity does not automatically mean minimization.

Predict

Replace u by u+εh, differentiate with respect to ε at zero and predict the sign of each term after integration by parts. Initially choose h smooth and compactly supported in space-time so that no boundary term survives. Later, relax that condition to see what boundary data arise.

Experiment

The first variation is ∫(uₜhₜ−c²uₓhₓ−m²uh). After integration by parts it becomes ∫(−uₜₜ+c²uₓₓ−m²u)h. Arbitrary h therefore gives uₜₜ−c²uₓₓ+m²u=0.

Contribution to the variation	After integration by parts
uₜhₜ	−uₜₜh
−c²uₓhₓ	+c²uₓₓh
−m²uh	Unchanged

Substitute a plane wave to obtain ω²=c²k²+m². For c=1,m=2,k=3, this gives ω²=13. The mass term changes the frequency relation without changing the sign of the spatial propagation term.

Understand

If the field is fixed on the spatial boundary, allowed variations vanish there. If the spatial endpoint values are free, the boundary term −c²uₓh must vanish for all endpoint variations, producing a natural zero-normal-derivative condition. Time endpoint variations are typically fixed when deriving the equation. These variational conditions should not be confused with an arbitrary initial-boundary problem chosen afterward.

Replace m²u²/2 by a differentiable potential V(u). The equation becomes uₜₜ−c²uₓₓ+V′(u)=0. Multiplying by uₜ and using the chain rule yields

∂ₜ[(uₜ²+c²uₓ²)/2+V(u)]−∂ₓ(c²uₜuₓ)=0.

When the integrals are finite and the boundary flux vanishes, total energy is conserved. If V is nonnegative, this energy controls derivative sizes; a potential unbounded below does not supply the same positivity. Conservation and coercive control are distinct statements.

This is an example of the relation between time-translation symmetry and energy conservation. A full Noether theorem treats more general fields and symmetries; here the conservation identity has been derived directly, so its signs and hypotheses are inspectable. Stationary action can have an indefinite second variation because time and spatial derivative terms enter with opposite signs. The positive conserved energy is a different functional from the action.

The course has proved concrete characteristic, modal, energy, maximum and weak-form mechanisms, while explicitly relying on selected spectral, density and existence theorems. It does not establish a general theory of nonlinear PDE existence, shocks in systems, arbitrary-domain regularity or graduate dispersive analysis.

Apply

Choose V(u)=u⁴/4 and derive both the PDE and energy density. Identify the regularity needed to use the chain rule, then state a boundary condition that removes energy flux. Explain why the formal conservation identity alone does not prove a global smooth solution exists for every datum.

Continue

As a synthesis, formulate one boundary problem, justify uniqueness, construct an exact or modal reference and compare a stable numerical approximation against it. Report the norm, data class, discretization error and theorem dependencies. Successful calculation on a finite example is evidence for that task, not a claim of complete PDE proficiency.