= Solution
Take the free <squared mass> $m^2\geq0$, as for the ordinary massive free theory. The quartic part of the <scalar potential> can be written
$$
V_4=\lambda(\phi_1^2-\phi_2^2)^2+2(\lambda+\kappa)\phi_1^2\phi_2^2.
$$
It is nonnegative for every real field value exactly when
$$
\boxed{\lambda\geq0,\qquad \lambda+\kappa\geq0.}
$$
Necessity follows from the axis $\phi_2=0$ and the diagonal $\phi_1=\phi_2$: a negative coefficient on either ray makes $V$ tend to minus infinity under rescaling. The displayed sum proves sufficiency, since $m^2(\phi_1^2+\phi_2^2)/2$ is also nonnegative. This is a <copositive matrix> condition on the quadratic form in the nonnegative variables $(\phi_1^2,\phi_2^2)$, rather than positive definiteness on all independent signed variables.
The strict inequalities $\lambda>0$, $\kappa>-\lambda$ make the quartic part grow in every nonzero direction. They are not necessary for a stable <vacuum> when $m^2>0$. For example, $\lambda=\kappa=0$ gives two stable free massive fields; $\lambda>0$, $\kappa=-\lambda$ has a flat quartic diagonal but a confining quadratic term. If $m^2=0$, boundary cases have flat vacuum directions and need not have a unique isolated minimum. If one instead allows negative <squared mass>, a flat quartic ray makes the full potential unbounded below; the stability conditions must then be strengthened. This distinguishes boundedness from strict quartic coercivity.
Under the generator of <1/a>, the variation is
$$
\delta V_4=4(\lambda-\kappa)\phi_1\phi_2(\phi_1^2-\phi_2^2).
$$
It vanishes identically if and only if $\kappa=\lambda$. In that case $V_4=\lambda(\phi_1^2+\phi_2^2)^2$ preserves the entire <global U(1) symmetry>. More explicitly, the interacting <Euler-Lagrange equations> give
$$
\partial_\mu j^\mu=-\delta V_4
=4(\kappa-\lambda)\phi_1\phi_2(\phi_1^2-\phi_2^2).
$$
Hence
$$
\boxed{Q\text{ is a conserved operator of the interacting theory exactly when }\kappa=\lambda.}
$$
A special solution lying on an axis does not establish an operator conservation law. Among the stable couplings, the symmetry-preserving line is $\kappa=\lambda\geq0$.
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