False $\Sigma_1$-sentences consistent with PA I'm preparing for an exam and encounter the following exercise in the notes I use.

In the next chapter we shall see that there are $\Sigma_1$-sentences
  which are false in $\mathcal{N}$ but consistent with PA. Use this to show that the
  following implication does not hold: for a $\Sigma_1$-formula $\phi(w)$ with only free
  variable $w$, if $\exists!w\phi(w)$ is true in $\mathcal{N}$, then $\mathrm{PA} \vdash \exists!w\phi(w)$.

So we want to find a $\Sigma_1$-formula $\phi(w)$ such that $\mathcal{N} \models \exists!w\phi(w)$, but $\mathrm{PA} \not\vdash \exists!w\phi(w)$.
We are given the fact that there exists a $\Sigma_1$-sentence $\psi$, such that $\mathcal{N} \not\models \psi$ and $\mathrm{PA} \not\vdash \psi \to \bot$. But then also $\mathcal{N} \models \psi \to \bot$, hence, the sentence $\psi \to \bot$ is already what we are looking for. 
But this $\psi$ doesn't have a free variable. Now we could tack on a $\wedge w = 0$ or something like that, but I suspect I've made an error somewhere.

The answer you give works, but I think here's what they're looking for: replace $\varphi(x)$ with $\varphi'(x)\equiv\varphi(x)\wedge\forall y(y<x\implies \neg\varphi(y)$). By induction, $\varphi$ has a solution iff $\varphi'$ has a unique solution. This is a trick which is frequently useful and nontrivial.
