1
$\begingroup$

I stumbled upon this question which says;

The following partial differential equation is called the diffusion equation:

enter image description here

The function ϕ(x,t) represents (for example) the number density of a gas species as a function of position x and time t. The “source function” f(x,t) describes the rate at which atoms of the gas are emitted at each point of space and time. The “diffusion constant” D describes how quickly the atoms diffuse through space.

a) The Green’s function is a solution to the partial differential equation

enter image description here

Derive the relationship between ϕ(x,t) and G(x − x',t − t').

(b) Let us Fourier transform G in the space coordinate (but not in time): enter image description here . (3)

Derive the ordinary differential equation (ODE) in t satisfied by G(k, t − t').

I have no clue how should I start with, I really need help and guidance. Thanks...

$\endgroup$
2
  • $\begingroup$ is part a) ϕ(x,t) = G(x − x',t − t'), if I compare the 1st eqn with the 2nd eqn??? $\endgroup$
    – kevin
    Commented Nov 6, 2017 at 14:12
  • $\begingroup$ if I do this, is this considered deriving their relationship?? $\endgroup$
    – kevin
    Commented Nov 6, 2017 at 14:13

1 Answer 1

1
$\begingroup$

$\phi(x,t) = G(x-x', t-t')$ only if $f(x,t) = \delta(x-x')\delta(t-t')$.

But you are indeed in the right neighborhood. Hint:

$$f(x,t) = \int\int f(x',t') \delta(x- x')\delta(t-t')dx'dt'.$$

$\endgroup$
6
  • $\begingroup$ hi , @Piotr Benedysiuk thanks for helping but how did you get the equation?? Sorry this is my first time doing green's function $\endgroup$
    – kevin
    Commented Nov 6, 2017 at 14:49
  • $\begingroup$ can I also ask to integrate green's function, do I have to just multiply the differential operator for both sides of equation of f(x,t)= delta(x-x')delta(t-t') to make my right side to have delta function ??? $\endgroup$
    – kevin
    Commented Nov 6, 2017 at 14:51
  • $\begingroup$ The equation for $f(x,t)$ is simply the definition of a delta function. What can you exchange the product of delta's with? $\endgroup$ Commented Nov 6, 2017 at 14:56
  • $\begingroup$ hi, @Piotr Benedysiuk, to exchange for f(x)?? But If I got f(x,t) =double integral f(x',t') f(x,t) dx'dt' what does it means?? $\endgroup$
    – kevin
    Commented Nov 6, 2017 at 15:45
  • $\begingroup$ No, not for $f(x)$. Look at the equation in the question a). $\endgroup$ Commented Nov 6, 2017 at 16:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .