Standing wave operator?

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Everyone knows the d'Alembert wave operator acting on a scalar function $\Psi(\boldsymbol{r},t)$ as:

$$ ( \partial_t^2/c^2 - \Delta ) \Psi(\boldsymbol{r},t) = 0 $$

The homogeneous solution in 3 dimensions ($\boldsymbol{r}$ is the position vector) is:

$$ \Psi = \frac{1}{r} F(r \pm c t) $$

with $F$ an arbitrary function. We can write the wave equation using the d'Alembert-box operator:

$$ \Box = \partial_t^2/c^2 - \Delta $$

Now let us consider this equation:

$$ ( \partial_t^2/c^2 + \Delta ) \Psi_s(\boldsymbol{r},t) = 0 $$

In this case, the 3d homogeneous solution should become:

$$ \Psi_d = \frac{1}{r} F(r \pm i c t) $$

An example of function $F$ could be $F(x) = e^{-k x}$, $k$ constant (wave vector like) and so we could have:

$$ \Psi_d \propto \frac{1}{r} e^{- k (r + i c t) } $$

Substracting the complex conjugate solution we obtain:

$$ \Psi_d \propto \frac{1}{r} e^{- k r } \sin(k c t) $$

That is a kind of standing wave (it oscillates but at the same location, no propagation). So if we define $\widehat{L}$ as:

$$ \widehat{L} = \partial_t^2/c^2 + \Delta $$

Can we talk about a standing wave operator ? Do the mathematicians already encounter this form of operator (related to physics or other fields)?

Thank you for reading.

Hint

Replacing $t$ by $i t$ transforms also the $\Box$ operator into $\widehat{L}$