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For $x\in\mathbb{R}^{n}$ and $t\in[0,\infty)$, consider the linear PDE initial value problem $$\dfrac{\partial u}{\partial t} = \left(a \Delta - \dfrac{b}{|x|}\right)u, \quad u(x,0) = u_0(x)\quad\text{(given)},$$ where $\Delta$ is the standard Laplacian, $\vert\cdot\vert$ denotes the Euclidean 2-norm, and the constants $a,b>0$. I want to write its solution as $$u(x,t) = \int_{\mathbb{R}^n} K(x,y;t)u_0(y)\:{\rm{d}}y.$$

Is there a known explicit (in terms of $x,y,t,n$) expression for this kernel/Green's function $K$?

It feels like something well-known but I am having difficulty finding a reference stating/deriving such $K$. The solution may be related to time-dependent Schrödinger equation with Coulomb potential but I do not know enough background to be sure.

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    $\begingroup$ The eq. is called heat equation with Coulomb potentials arxiv.org/pdf/1707.07744.pdf $\endgroup$
    – Nemo
    Commented Jan 21 at 7:27
  • $\begingroup$ @Nemo: Not really, the potential is minus the Coulomb potential here. This means that there is no spectrum below zero. $\endgroup$ Commented Jan 21 at 22:55
  • $\begingroup$ @Nemo: Unless there is a typo in the basic equation... (the OP should perhaps clarify) $\endgroup$ Commented Jan 21 at 22:56
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    $\begingroup$ Yes, $b>0$ here. The problem came from gravitational potential. $\endgroup$ Commented Jan 21 at 23:32

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The time-dependent Schrödinger equation with Coulomb potential (hydrogen atom) has the form $$ i\frac{\partial u}{\partial t} = \left( -a^{\prime } \Delta -\frac{b^{\prime } }{|x|} \right) u $$ with $a^{\prime } ,b^{\prime} >0$, so that problem is related to yours by continuing to imaginary $t$. That problem has been treated for $n=2$ and $n=3$ in terms of the Fourier-transformed Green's function $$ K(x,y;E)=\int_{0}^{\infty } dt\, e^{iEt} K(x,y;t) $$ in I.H.Duru and H.Kleinert, Fortschritte der Physik 30 (1982) 401, available on the webpage of one of the authors. It becomes messy - for $n=2$, an integral representation of $K(x,y;E)$ is given in Eq. (35), and for $n=3$, it is given in Eq. (109). Not sure how useful this is for your purposes - they do present the extraction of (well-known) central properties of the hydrogen atom, such as wave functions, from these integral representations.

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  • $\begingroup$ The operator is not the hydrogen operator in the OP, though; the potential has the opposite sign. $\endgroup$ Commented Jan 21 at 22:52
  • $\begingroup$ @ChristianRemling - I think most of the calculation of $K(x,y;E)$ will still go through with a repulsive potential, just towards the end one will have to be careful with the contours (which one will have to be anyway to connect the imaginary time to the real time case). But it needs to be checked carefully. $\endgroup$ Commented Jan 21 at 23:18
  • $\begingroup$ I don't think this is similar to the potential with the opposite sign. For starters, there are no eigenvalues now. $\endgroup$ Commented Jan 21 at 23:35
  • $\begingroup$ @ChristianRemling - sure, but what will typically happen when you evaluate $K(x,y;E)$ is that the formal expressions will be very similar, just that the poles corresponding to the bound states disappear because certain denominators cannot become zero anymore. At energies $E$ corresponding to scattering states, one cannot tell from the scattering amplitude whether the potential is attractive or repulsive! $\endgroup$ Commented Jan 21 at 23:59

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