How to find the general solution of a differential equation with a shift, in the following form?

$$\frac{\partial}{\partial t}g(x,t)=g(x-\Delta,t)+\frac{\partial^2}{\partial x^2} g(x,t)$$

where $\Delta > 0$. And what about the following?

$$\frac{\partial}{\partial t}g(x,t)=g(x,t-\Delta)+\frac{\partial^2}{\partial x^2} g(x,t)$$

Edit1: Here are few follow-up details about my question. Is there a ``nice" way to represent the solution in $x$-space, as opposed to e.g., Fourier? Is the solution real + positive + normalizable? Does it have the correct properties of a probability density function?

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    $\begingroup$ as you can see from the general solution I wrote down, the normalization $N(t)=\int_{-\infty}^\infty g(x,t)dt=G(0,t)=e^{t}G(0,0)=e^{t}N(0)$ increases exponentially with time; so this is not a probability density function (why did you expect that?) $\endgroup$ – Carlo Beenakker Jul 22 '20 at 10:45

Fourier transform $G(k,t)=\int_{-\infty}^\infty e^{ikx} g(x,t)dx$ with respect to $x$, then $$\frac{\partial}{\partial t}G(k,t)=e^{ik\Delta}G(k,t)-k^2 G(k,t),$$ hence $$G(k,t)=\exp\left(te^{ik\Delta}-tk^2\right)G(k,0).$$ For the second differential equation you would similarly Fourier transform with respect to $t$.

  • $\begingroup$ 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? $\endgroup$ – user1611107 Jul 22 '20 at 9:58
  • $\begingroup$ 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 $g(x,0)$. $\endgroup$ – Carlo Beenakker Jul 22 '20 at 10:41
  • $\begingroup$ 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. $\endgroup$ – user1611107 Jul 22 '20 at 11:34
  • $\begingroup$ 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 $\endgroup$ – user1611107 Jul 22 '20 at 11:39
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    $\begingroup$ 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. $\endgroup$ – Carlo Beenakker Jul 22 '20 at 12:07

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