Physical Interpretation of d'Alembert Operator $$\mathop{{}\Box}\nolimits=\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\mathop{{}\bigtriangleup}\nolimits$$ is the d'Alembert-operator. It seems to consist of an oscillation and a diffusion. Is there a simple physical interpretation of it?
 A: This is the wave operator. For example, a solution is the plane wave $A e^{i(\vec{k}\cdot\vec{x}\pm\omega t)}$ where $\omega=|\vec{k}|c$.
Diffusion involves a first-order time derivative, not a second-order time dervative.
Physically, the meaning is that the “acceleration” of the field at a point is negatively proportional to the difference between the field value at that point and the average of the field values on an infinitesimal sphere around that point. If the value is higher than nearby, it will decrease, and vice versa.
This sounds similar to diffusion, in that the field tries to “even itself out”, but the second-order time derivative means that it overshoots and oscillates rather than relaxing to an evened-out value.
A: Re.
$$\mathop{{}\Box}\nolimits=\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\mathop{{}\bigtriangleup}\nolimits$$
This is the operator for the classical wave equation which in one dimension  is  
$$\frac{\partial^2f}{\partial x^2}-\frac{1}{c^2}\frac{\partial^2f}{\partial t^2}=0$$
It may be easiest to get a physical interpretation (or intuition) in the one dimensional time space domain where the solution is of the form $f(x-ct)$ or $f(x+ct)$. 
see
https://physics.stackexchange.com/a/395212/45664
"From the geometry alone, it was only needed to note that a change in t multiplied by the velocity yields the same results (as measured by the second derivative) as a change in x --that is, a translation of f(x−ct)." 
The details are given in the link above.
Note that $\bigtriangleup x=\bigtriangleup ct$ in the denominators of the wave equation.  So it is about a change in $x$ being the same as a change in $t$ when scaled by $c$, which is apparent by examining $f(x-ct)$ or $f(x+ct)$.
