Green's functions: Logic behind this step

[WWCY]

Homework Statement

Hi all,

I came across these steps in my notes, relating to a step whereby,
$$\hat{G} (k, t - t') = \int_{-\infty}^{\infty} e^{-ik(x - x')}G(x-x' , t-t')dx$$
and performing the following operation on $\hat{G}$ gives the following expression,
$$[\frac{\partial}{\partial t} - D\frac{\partial ^2}{\partial x^2}] \hat{G} = \frac{\partial \hat{G}}{\partial t} + Dk^2 \hat{G}$$

These steps are in the context of a more complex problem of solving the Heat Equation using the Green's function.

Homework Equations

The Attempt at a Solution

I can partially understand the writing of $\frac{\partial \hat{G}}{\partial t}$ at the LHS as the same thing, $\frac{\partial \hat{G}}{\partial t}$ at the RHS; we do not know how $G$ (found in the first integral) depends on time exactly.

But the other terms that imply $- D\frac{\partial ^2 \hat{G}}{\partial x^2} = Dk^2 \hat{G}$ I can't follow. First of all, we don't know how $G$ varies with $x$. Add to that that the fact that the integral isn't independent of $x$, I don't see how we can apply differentiation under the integral sign to bring down the factor of $(-ik)^2$ down from $e^{-ik(x-x')}$

Could someone explain the logic behind the steps? Help is greatly appreciated!

 

[Orodruin]

$\hat G$ does not depend on $x$ (it is an internal integration variable). The equation you quoted is not true. What is true is that
$$
\partial_t \hat G + k^2 D \hat G
$$
is the Fourier transform of the left-hand side of the heat equation.

 

[WWCY]

$\hat G$ does not depend on $x$ (it is an internal integration variable). The equation you quoted is not true. What is true is that
$$
\partial_t \hat G + k^2 D \hat G
$$
is the Fourier transform of the left-hand side of the heat equation.

Hi @Orodruin , am I right to say that $\hat{G}$ has no explicit dependence of $x$ as it is the Fourier transform of $G$?

Do you also mind illustrating how the factor of $(-ik)^2$ was brought down?

Thanks for your assistance.

 

[Orodruin]

Hi @Orodruin , am I right to say that $\hat{G}$ has no explicit dependence of $x$ as it is the Fourier transform of $G$?

Yes.

Do you also mind illustrating how the factor of $(-ik)^2$ was brought down?

This is a general property of the Fourier transform. In general, for any function $f(x)$, it holds that
$$
F[f'] = \int f'(x) e^{-ikx} dx = - \int f(x) \frac{de^{-ikx}}{dx} dx = ik \int f(x) e^{-ikx} dx = ik F[f],
$$
where $F[f]$ is the Fourier transform of $f$.

 

[WWCY]

Yes.This is a general property of the Fourier transform. In general, for any function $f(x)$, it holds that
$$
F[f'] = \int f'(x) e^{-ikx} dx = - \int f(x) \frac{de^{-ikx}}{dx} dx = ik \int f(x) e^{-ikx} dx = ik F[f],
$$
where $F[f]$ is the Fourier transform of $f$.

Thank you!