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I am working in extensions of General Relativity Theory and at the moment of taking the Newtonian limit of this extension theory (essentialy, mathematically speaking, this is just linearizing the field equations obtained via the variational principle, but this is not important) I arrive to the following partial differential equation: \begin{equation} \nabla^2 h+b\nabla^4h=\alpha \delta^3(\vec r)+\beta \dfrac{1}{r}e^{-r/\gamma}. \end{equation} Here, $\nabla^2$ is the Laplacian operator, $\nabla^4=\nabla^2\nabla^2$ is the ''squared'' of Laplacian operator, $\alpha,\beta$ and $\gamma$ are just real constants, $\delta^3(\vec r)$ is the three dimensional Dirac delta, $r$ is the variable and $h(r)$ is the function we're solving for (physically, it is basically the Newtonian potential).

I am having a lot of trouble to solve this differential equation, I tried to solve it by Fourier transforms but I'm not capable to find the analytic expression for $h(r)$. Even though, I know the solution must be what in physics we call Yukawa-like potentials, i.e, the solution must be of the form $\dfrac{k_1}{r}$ plus terms like $\dfrac{k_2}{r}e^{-mr}$.

Can someone help me out to find the solution?

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  • $\begingroup$ Have you written it as an ordinary differential equation involving only derivatives with respect to $r$? $\endgroup$
    – G. Smith
    Sep 28, 2020 at 20:35
  • $\begingroup$ The equation is linear in $h$ with two source terms. Have you tried solving it considering each source separately? (BTW, I don’t know the solution.) $\endgroup$
    – G. Smith
    Sep 28, 2020 at 20:37
  • $\begingroup$ @G.Smith Yes, this was directly the first method i considered, but I haven't been able to solve it this way. The problem I had is that the second integral (Yukawa-like potential term) contributes with a divergent integral, which means I did something wrong. $\endgroup$
    – ALPs
    Sep 28, 2020 at 20:42

1 Answer 1

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Take the Fourier transform of each side with $$ h(x) = \int \tilde h(k) e^{-ikx} \frac{d^3k}{(2\pi)^3} $$ so that $$ \nabla^2 h(x)= \int \left\{-|k^2|\tilde h(k)\right\} e^{-ikx} \frac{d^3k}{(2\pi)^3}, \quad etc. $$

Here $|k^2| = k_x^2+k_y^2+k_z^2$.

As (I think!) $$ \frac{e^{-m|x|}}{4\pi r}= \int e^{ikx}\frac 1 {|k^2|+m^2}\frac{d^3k}{(2\pi)^3}, $$

we get $$ (-|k^2|+ b |k^2|^2) \tilde h(k) = \left(\alpha +\beta \frac{1}{k^2+m^2}\right) . $$ This leads to $$ \tilde h(k)= \left(\alpha +\beta \frac{1}{|k^2|+m^2}\right) \left(\frac {1}{|k^2|-b |k^2|^2}\right)\\ = \left(\alpha +\beta \frac{1}{|k^2|+m^2}\right) \left(\frac {1}{|k^2|(1-b |k^2|)}\right). $$ Now invert the FT to get $$ h(x)= -\int \left(\alpha +\frac{\beta}{|k|^2+m^2}\right)\left(\frac{e^{-ikx}}{|k|^2(1-b |k|^2)}\right) \frac{d^3k}{(2\pi)^3}. $$ Resolve the various factors using partial fractions to get a sum of Yukawas.

I have not tried doing the latter, so I have not yet seen your divergence problem.

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  • $\begingroup$ I think this expression you've written doesn't have any divergencies and with the partial fraction method the $1/r$ and Yukawa term should appear without problem. The question is, can you please show me how do you obtain this expression for $h(x)$? In my Fourier approach I didn't had the same result. $\endgroup$
    – ALPs
    Sep 28, 2020 at 22:24
  • $\begingroup$ I just evaluated the Fourier transform of both sides using $e^{-m|x|}/4\pi r= \int {d^3k}/{(2\pi)^3} e^{ikx}/{(k^2+m^2)}$. $\endgroup$
    – mike stone
    Sep 28, 2020 at 23:31
  • $\begingroup$ Ah -- maybe a sign error. Ft of $\nabla^2$ is $-k^2$ so it should be $1-k^2b$ and this problematic if $b$ is positive. $\endgroup$
    – mike stone
    Sep 29, 2020 at 0:14
  • $\begingroup$ I can't really see how you obtain the expression from t'he Fourier transform of my equation. Can you please write the steps with detalle? I would really apreciate it. About the sign of $b$ don't worry, it's up to my choice since it's the coupling constant of the higher derivative term in the action. $\endgroup$
    – ALPs
    Sep 29, 2020 at 12:13
  • $\begingroup$ I'll edit my answer to fill in the details. $\endgroup$
    – mike stone
    Sep 29, 2020 at 12:17

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