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?


1 Answer 1


By Helmholtz's decomposition theorem if $\nabla \cdot \vec{u}=0$ and $\nabla\times\vec{u}=0$ in all space then $\vec{u}=0$.

  • $\begingroup$ 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. $\endgroup$
    – agaminon
    May 18 at 14:12
  • $\begingroup$ yes, if in all space. If not then you have to provide boundary conditions which themselves represent surface charges or surface currents. I find the Helmholtz decomposition theorem the easiest way to understand where the concept of surface distributions comes from. $\endgroup$
    – hyportnex
    May 18 at 14:16

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