 Solution

For the usual scalar field theory with a nonnegative potential $U$, shifted to zero in the vacuum, write its static energy in $d$ dimensions asApply Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$. The gradient acquires a factor $\lambda$ and the volume a factor ${\lambda }^{-d}$, givingA smooth finite-energy static solution is a stationary point of this energy, so its Derrick virial identity isFor $d\ge 3$, both terms are nonnegative, forcing $T=P=0$. The field is a constant vacuum: there is no nontrivial finite-energy static soliton in this scalar-gradient-plus-nonnegative-potential model. The assumptions matter: extra derivative terms, gauge fields, time dependence or a potential not bounded below by its vacuum value can change the argument.

For $d=2$, the relation becomes $P=0$ and says nothing further about $T$. A gradient-only energy is scale invariant, so Derrick theorem alone does not rule out nonconstant solutions; sigma-model lumps are examples. For a scalar field with isolated potential minima, $P=0$ constrains the continuous field to one vacuum and hence makes it constant. If the vacuum values form a nontrivial manifold, the scaling relation alone leaves open a nonconstant map into it. In neither case does the two-dimensional scaling argument itself prove existence.