Timeline for How to solve a differential equation in the form $\frac{\partial}{\partial t}g(x,t)=g(x-\Delta,t)+\frac{\partial^2}{\partial x^2} g(x,t)$?
Current License: CC BY-SA 4.0
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| Jul 22, 2020 at 12:07 | comment | added | Carlo Beenakker | the inverse Fourier transform does not have a closed form expression for the delta function initial condition; you would need to calculate integrals of the form $\int_0^{\infty } e^{\cos k-k^2} \cos( k x) \cos (\sin k) \, dk$, which can only be done numerically. | |
| Jul 22, 2020 at 11:39 | comment | added | user1611107 | The thing is, the idea of Fourier I am familiar with, but the backward transformation of the exp[t exp(i k Delta)] is something that I can't work out. Is there a simple form of g(x,t) that you can demonstrate the inverse transform on? E.g., if g(x,t)=delta(x) or a narrow Gaussian? Thanks again | |
| Jul 22, 2020 at 11:34 | vote | accept | user1611107 | ||
| Jul 22, 2020 at 11:34 | comment | added | user1611107 | Ok, thank you. About the normalization: I was thinking that it is simpply a diffusion equation, with a drift term (even if the drift dependson x-\Delta). But you are right, it's a source. My mistake. | |
| Jul 22, 2020 at 10:41 | comment | added | Carlo Beenakker | to transform back to $x$-space you calculate $g(x,t)=(2\pi)^{-1}\int_{-\infty}^\infty e^{-ikx}G(k,t)dk$. This is the general solution, it will be real if $g(x,0)$ is real; it cannot be worked out further without further information on $g(x,0)$. | |
| Jul 22, 2020 at 9:58 | comment | added | user1611107 | Thanks. Can you please continue the solution for few more steps? Is it real, or have an imaginary part? How do you than transform back from k to x-space? | |
| Jul 22, 2020 at 9:52 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |