 Solution

Vary the four-potential with compactly supported variation. Antisymmetry of the electromagnetic field tensor gives $\delta (F_{\mu \nu }{F}^{\mu \nu })=4F^{\mu \nu }{\partial }_{\mu }\delta A_{\nu }$. Integration by parts therefore givesThus the Euler-Lagrange field equation is $\boxed{{\partial }_{\mu }{F}^{\mu \nu }-{\lambda }^2{A}^{\nu }=-{\mu }_0{J}^{\nu }}$.

A gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. Commuting derivatives leave $F_{\mu \nu }$ unchanged. At $\lambda =0$, the source term changes by $c^{-1}\int {\partial }_{\mu }\chi J^{\mu }$, a boundary term because the current is conserved. With suitable boundary conditions the action is gauge invariant. For $\lambda >0$, the mass term changes by terms involving $2A^{\mu }{\partial }_{\mu }\chi +(\partial \chi {)}^2$, so the same arbitrary gauge transformations are not a symmetry.

Take the divergence of the field equation. Antisymmetry gives ${\partial }_{\nu }{\partial }_{\mu }{F}^{\mu \nu }=0$, and conservation gives ${\partial }_{\nu }{J}^{\nu }=0$. Thus ${\lambda }^2{\partial }_{\nu }{A}^{\nu }=0$. For $\lambda >0$ this is the Lorenz constraint in Proca theory, and expanding $F$ givesFor $\lambda =0$ the same reduced equation holds when the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$ is imposed. More generally it is enough that the divergence be spacetime constant, but the conventional vanishing boundary conditions set that constant to zero.