electromagnetism - About the uniqueness of the displacement current

In the Maxwell-Ampère equation, i.e.: \begin{equation} \nabla\times\vec{B} = \mu_0 \vec{J} + \mu_0\epsilon_0 \frac{\partial \vec{E}}{\partial t} \end{equation} the $\vec{J}_d$ term: $$ \vec{J}_d := \epsilon_0 \frac{\partial \vec{E}}{\partial t} $$ was derived by taking the divergence of the left hand side of the equation. Explicitly, before Maxwell's addition of the $\vec{J}_d$ term Ampère's law was: $\nabla\times\vec{B} = \mu_0 \vec{J}$, but when acting with $\nabla \cdot$ we had: $$ 0 \equiv \nabla \cdot \left(\nabla\times\vec{B}\right) = \mu_0 \nabla\cdot\vec{J} = -\mu_0 \frac{\partial \rho}{\partial t} $$ from the $\text{div}(\text{curl}\ \bullet)$ identity and the continuity equation. But $\frac{\partial \rho}{\partial t} $ is not necessarily zero so we need to add a new term let's call it $\vec{J}_d$. And now comes my question. We need $\vec{J}_d : \nabla\cdot\left( \vec{J} + \vec{J}_d \right) = 0 \Rightarrow \nabla \cdot \vec{J}_d = \frac{\partial \rho}{\partial t} $. And indeed $\vec{J}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t}$ is a solution, but for this "test of the divergence" $$\vec{J'}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t} + \vec{k}$$ where $\vec{k}$ is a constant vector, or even $$\vec{J''}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t} + \nabla \times \vec{T}$$ where $\vec{T}$ is any vector, satisfy $\nabla\cdot\left(\nabla\times\vec{J}_d\right) = 0$. Why, then does $\vec{J}_d$ has the form it has and not any of the other possible solutions presented above?

Thanks in advance.