= Solution
For the <real scalar field>, the two derivatives entering the <Euler-Lagrange field equation> are
$$
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}=\partial^\mu\phi,
\qquad
\frac{\partial\mathcal L}{\partial\phi}=-m^2\phi-\frac{\lambda}{6}\phi^3.
$$
Consequently the <equation of motion> is
$$
\boxed{(\Box+m^2)\phi+\frac{\lambda}{6}\phi^3=0.}
$$
The <quartic interaction> therefore adds a local cubic term to the free <Klein-Gordon equation>. Below, write $F=(\Box+m^2)\phi+\lambda\phi^3/6$; thus $F=0$ means that the <field> is <on shell>.
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