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Is the fundamental solution (Green's function) of the 1D advection-diffusion equation

$$\frac{\partial{\phi}}{\partial{t}} = D(t)\frac{\partial{^{2}\phi}}{\partial{x^{2}}} - c(t)\frac{\partial{\phi}}{\partial{x}}-k(t)\phi$$

(where the diffusivity $D(t)$, drift speed $c(t)$ and decay coefficient $k(t)$ are functions of $t$ only and not $x$) given by

$$ \phi(x,t) = \frac{1}{\sqrt{4\pi\int_{t_0}^{t}D\left(t'\right)dt'}}\, \exp\left({-\frac{\left\lbrack x-\int_{t_{0}}^{t}c(t')\,{\rm d}t'\right\rbrack^{2}}{4\int_{t_0}^{t}D\left(t'\right){\rm d}t'}-\int_{t_0}^{t}k\left(t'\right)\,{\rm d}t'}\right) $$

provided that $\int{D dt}$, $\int{c dt}$ and $\int{k dt}$ exist?

I believe that this solution is correct, I can't find an error in my derivation. I'll post more details if this result is incorrect, in order to identify the error. But is it correct?

Thanks!

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  • $\begingroup$ Have you tried substituting it into the equation and checking yourself? $\endgroup$ – Kirill Oct 24 '13 at 22:14
  • $\begingroup$ Yes, but as I do so I get a huge mess, which then I find difficult to reduce reduce with pen and paper... But I am also surprised that all textbooks contain the constant-coefficient solution but not this most general one (if it is true of course!) since it would be quite useful, given the applications. $\endgroup$ – JMK Oct 24 '13 at 22:24
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    $\begingroup$ Checking if it actually solves the equation is the only way of checking if it solves the equation. $\endgroup$ – Kirill Oct 24 '13 at 22:31
  • $\begingroup$ Yes, it is indeed! But I was also wondering if happens to be a well-established result that people instantly recongise (if so, it would be encouraging!) $\endgroup$ – JMK Oct 24 '13 at 22:37
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Yes, it is correct assuming your initial condition is $\phi(x,t_0)=\delta(x)$ and the domain is $x\in\mathbb{R}$, $t>t_0$.

First verify that it solves the equation for $t>t_0$.

input: $$\frac{1}{\sqrt{4\pi \int _0^td[s] ds}}\text{Exp}\left[\frac{-\left(x-\text{x0}-\int _0^tc[s] ds\right){}^2}{4\int _0^td[s] ds}-\int _0^tk[s] ds\right]\text{//}\{-1,d[t],-c[t],-k[t]\}.\{D[\#,t],D[\#,x,x],D[\#,x],\#\}\&\text{//}\text{Simplify}$$ output: $0$

To verify that it satisfies that initial condition, compare it with the solution for constant coefficients, and note that $\int_0^t f(s)\,ds = t f(0) + O(t)$ for continuous functions $f$.

I can't resist making a general note that you prove an expression to be correct by proving it to be correct, rather than comparing with an expression (in a textbook maybe) that you suspect is correct. Otherwise, you seem to have got it right yourself.

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  • $\begingroup$ Thanks! I did check the constant coefficient (which of course was correct) and was checking this one as well, but I kept getting inconsistent results (this just meant that I was making some mistakes whilst checking), which prompted me to ask this. BTW, the solution was rigorously derived, I didn't conjure it out of thin air (although the latter may sound more impressive) $\endgroup$ – JMK Oct 24 '13 at 23:04

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