PDE by Fourier Transform

1. The problem statement, all variables and given/known data
(a) Solve [tex]\frac{\partial u}{\partial t}=k\frac{\partial ^{2} u}{\partial x^{2}} - Gu[/tex]

where -inf < x < inf
and u(x,0) = f(x)

(b) Does your solution suggest a simplifying transformation?

2. Relevant equations

I used the fourier transform as:
F[f(x)] = F(w) = [tex] \frac{1}{2*pi} \int_{-inf}^{inf} f(x) e^{iwx} dx [/tex]

3. The attempt at a solution

I solved part a using fourier transform. Although I'm not 100% certain, I think my answer is pretty plausible. I'm happy to elaborate on how I solved this, but I didn't want to type it all out for naught, because that's not really my question. Anyway, I got:

[tex] u(x,t) = \int_{-inf}^{inf} [ \frac{1}{2*pi} \int _{-inf}^{inf} f(x) e^{iwx} dx ] e^{(-w^{2}k-G)t} e^{-iwx} dw [/tex]

I'm not sure how to answer part b. Any ideas?
 
(b) Does your solution suggest a simplifying transformation?

I'm not sure how to answer part b. Any ideas?
So I think maybe the problem statement is asking me to change the order of integration and take the middle integral "offline" by substituting for x (as x - xbar, for example). Or maybe this is expected in part A...
 
So I think maybe the problem statement is asking me to change the order of integration and take the middle integral "offline" by substituting for x (as x - xbar, for example). Or maybe this is expected in part A...
following my own logic, I find:

[tex]
u(x,t) = \frac{1}{2 \pi} \int_{-inf}^{inf} f(X) ( \int_{-inf}^{inf} e^{-iw(x-X)} e^{-(w^{2}k-G)t}dw)dX
[/tex]

and I need to find a function g(x-X) such that the Fourier transform is:

[tex] G(w) = e^{(-w^{2}k-G)t} [/tex]


So the substitution must simplify G(w) such that i can get an analytical form of the inverse fourier transform... Any idears?
 
So it turns out that the substitution is a = t(K-G). Thus you can take the integral offline by evaluating the resultant Gaussian.

Thanks! At least PF lets me talk to myself better :P
 
So it turns out that the substitution is a = t(K-G). Thus you can take the integral offline by evaluating the resultant Gaussian.

Thanks! At least PF lets me talk to myself better :P
Well, I dunno if this qualifies as a "transformation" -- again, welcoming comments.
 

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