Facing a paradox: Earnshaw's theorem in one dimension

Question

Consider a one-dimensional situation on a straight line (say, $x$-axis). Let a charge of magnitude $q$ be located at $x=x_0$, the potential satisfies the Poisson's equation $$\frac{d^2V}{dx^2}=-\frac{\rho(x)}{\epsilon_0}=-\frac{q\delta(x-x_0)}{\epsilon_0}.$$ If $q>0$, $V^{\prime\prime}(x_0)<0$, and if $q<0$, $V^{\prime\prime}(x_0)>0$. Therefore, it appears that the potential $V$ does have a minimum at $x=x_0$, for $q<0$. Does this imply that $x=x_0$ is a point of stable equilibrium? I must be missing something because this appears to violate Earnshaw's theorem (or it doesn't)?

Answer 1

Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.

Answer 2

So technically $V''(x_0)$ doesn't have an actual value, since $\delta(x-x_0)\to\infty$ as $x\to x_0$. However, if you understand the Dirac delta distribution to be a limit of a function whose peak "gets narrower" with its integral remaining constant, then this is fine and you could say there is a minimum at $x_0$ for $q<0$

This can be more easily understood by just thinking about the motion of a positive charge in this potential. It will move towards the negative charge, i.e. towards the minimum of the potential.