Green's function for fourth order equation

I know the D'Alembert operator ${\frac {1}{c^{2}}}\partial _{t}^{2}-\Delta _{\text{3D}}$ has a well-known Green's function $\frac{\delta(t-\frac{r}{c})}{4 \pi r}$. This is very useful for studying 3D wave equation / fluids.

How about the Green's function of the following operator? $${\frac {1}{\mu^{4}}}\partial _{t}^{2}+\Delta _{\text{3D}}\Delta _{\text{3D}}$$ Is it known? How do I find it? This would be useful for understanding superfluids.

• Does this second operator have a name? Also, are you sure that the second operator has a $-$ and not a $+$ in it? The straight forward way to approach this is to use the Fourier transform, then invert the operator, and then transform back. If you use retarded boundary conditions, the inverse Fourier transform becomes difficult to evaluate because of large (growing exponentially in time) contributions from imaginary frequencies that are present for any wave vector. With the different sign, for each wave vector, there are only real frequencies. Oct 16 '17 at 22:45
• Good catch! The sign is supposed to be a plus. I fixed the question. I don't know if the operator has a name. I could try to go through the standard approach Oct 17 '17 at 5:28

The time-independent equation is the biharmonic equation, $$-\Delta^2f({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}'),$$ with 3D solution $$f({\bf r},{\bf r}')=\frac{1}{8\pi}|{\bf r}-{\bf r}'|,$$ see Fourier expansions for a logarithmic fundamental solution of the polyharmonic equation (2012). More generally, the Green function of the $k$-the power of the Laplacian, $(-\Delta)^k$ is $\propto |{\bf r}-{\bf r}'|^{2k-d}$ in odd dimensions $d$.