Given a wave that follows these two conditions, is there some valid and non-trivial physical interpretation to it?

Consider the vectorial wave equation:

$$\frac{\partial^2 \vec{u}}{\partial t^2} = v^2\nabla^2 \vec{u}$$

A way you can define transverse and longitudinal waves is, respectively: $$\nabla \cdot \vec{u}=0\quad\text{ and }\quad\nabla\times\vec{u}=0 .$$ Now say we have some solution $$\vec{u}(\vec{r},t)$$ that follows both conditions. Through the definition of the vector laplacian, this implies:

$$\nabla^2 \vec{u}=0 \ ; \ \frac{\partial^2 \vec{u}}{\partial t^2} =0$$

The first equation is just three Laplace equations, one for each coordinate of the vector. Assume we have a separable solution of the form: $$u_i(x,y,z,t) = X_i(x)Y_i(y)Z_i(z)T_i(t)$$. Then, through the second equation:

$$\frac{d^2T_i}{dt^2}=0 \to T_i(t) = A_i +B_it$$

This means that the perturbation that the wave equation models will have a linear increase in amplitude with time. Ignoring the trivial solution $$B_i=0$$ which just results in a wave that does not evolve with time, do you know of any systems or situations this can model; or of any physical interpretation for this?

By Helmholtz's decomposition theorem if $$\nabla \cdot \vec{u}=0$$ and $$\nabla\times\vec{u}=0$$ in all space then $$\vec{u}=0$$.
• I assume this means you need some "sources" for this to be of any interest. In the electrostatic analogy, if $\nabla^2 \phi = 0$ everywhere then $\phi=0$ everywhere. Commented May 18, 2023 at 14:12