Electric displacement up to a constant?

Question

The electric displacement field $\vec D$ satisfies two equations, $$\vec\nabla \cdot \vec D=\rho_f,$$ and $$\vec\nabla \times \vec D=\vec\nabla \times \vec p,$$ where $\rho_f$ and $\vec p$ are the free charge density and the polarization respectively.

My question is quite simple. Suppose there is no free charge so $\rho_f=0$ and the polarization has no curl, so $\vec\nabla \times \vec p=0$. Still, the electric displacement field does not need to vanish, it could be constant. Actually, we have $\vec D=\vec p+\vec\nabla f$ for any harmonic function, $\nabla^2 f=0$.

The reason this is bugging me is because in Problem 4.15 of the book by Griffiths, he asks the following question: in a system without free charges but with a given irrotational polarization $\vec p$, find $\vec D$. In the solution manual, he states that in the absence of free charges (and $\vec\nabla \times \vec p=0$) it follows that $\vec D=0$ (see figure). This seems incorrect to me.

Answer 1

If $\vec\nabla \times \vec D=0$ then $\vec D$ must be a gradient, $\vec D=\vec\nabla f$.

If $\vec\nabla \cdot\vec D=0$ then $f$ must be harmonic, $\nabla^2 f=0$.

A very common physical boundary condition is that the field should vanish at infinity, so as not to have infinite energy.

If $f$ is harmonic, all its partial derivatives are harmonic. For the gradient to vanish at infinity in all directions, all partial derivatives must vanish at infinity in all directions.

The only harmonic function that vanishes at infinity in all directions is the zero function. So all partial derivatives of $f$ are identically zero, so $\vec D=0$.